TITLE: Is there a notion of "space" such that vector bundles can be understood in this way? QUESTION [9 upvotes]: Is there a notion of "space" satisfying the following requirements? Spaces form (at least) a category; morphisms between spaces are called "continuous maps." Every topological space is a space, and the inclusion of $\mathbf{Top}$ in the category of spaces is fully faithful and injective on objects. The category $\mathbb{R}\mathbf{Mod}$ is a space. If $X$ is a topological space, a vector bundle over $X$ can be described as a continuous map $X \rightarrow \mathbb{R}\mathbf{Mod}.$ (I don't mind if some of these requirements are partially violated, they're just meant to be guidelines.) REPLY [7 votes]: As Qiaochu said, probably what you want are topological stacks. Let $T$ be a small full subcategory of $\mathrm{Top}$, with the Grothendieck topology of open covers, and consider the 2-category of stacks of groupoids on $T$. Call its objects "spaces" and its morphisms "continuous maps". You could call its 2-cells "continuous transformations". Any topological space $X$ determines a sheaf $\mathrm{Top}(-,X)$ on $T$ and hence a stack. This functor is fully faithful on $T$, and often on a much larger subcategory of $\mathrm{Top}$. (For instance, if $T=\{\mathbb{R}^n\}$ then the functor is fully faithful on at least all topological manifolds.) The functor $T^{op} \to \mathrm{Gpd}$ defined by sending $X\in T$ to the groupoid of real vector bundles over $T$ is a stack, because vector bundles can be glued together over open covers. Call it your $\mathbb{R}\mathbf{Mod}$. If $X\in T$ (and often for many more $X\in\mathrm{Top}$), then by the Yoneda lemma, the groupoid of continuous maps and continuous transformations $\mathrm{Top}(-,X) \to \mathbb{R}\mathbf{Mod}$ is equivalent to $\mathbb{R}\mathbf{Mod}(X)$, i.e. the groupoid of real vector bundles on $X$.<|endoftext|> TITLE: Is there any definition of $H^1$ in one dimensional Arakelov geometry QUESTION [5 upvotes]: Consider a number field $K$ with ring of integers $O_K$. On the affine scheme $\overline X=\operatorname{Spec}(O_K)$ we have the well known one dimensional Arakelov geometry. Let $\overline D=\sum_{\mathfrak p\neq 0} r_\mathfrak p\mathfrak p+\sum_{\sigma} \lambda_{\sigma}\sigma$ be an Arakelov divisor on $\overline X$ (of course $r_{\mathfrak p}\in\mathbb Z$ and $\lambda_{\sigma}\in\mathbb R$), then we have a well defined notion of "$0$-cohomology group associated to $\overline D$". The definition is the following: $$H^0(\overline D):=\{f\in K^\times\colon v_{\mathfrak p}(f)\ge -r_{\mathfrak p}, v_\sigma(f)\ge -\lambda_\sigma \}\,.$$ Clearly $v_{\mathfrak p}$ is the discrete valuation associated to $\mathfrak p$, on the other hand $ v_\sigma=-\log |\cdot|_\sigma$ is the valuation associated to the complex norm induced by the embedding $\sigma:K\to\mathbb C$. Does exist any reasonable version of $H^1(\overline D)$? I expect a relation with $H^0(\kappa-\overline D)$, where $\kappa$ is the "canonical divisor" in Arakelov geometry, i.e. the divisor with zero component at infinity associated to the inverse of the different of $K$. In this question the OP mentions a modified version of $H^0$, but I'd prefer to remain stick to the "classical one". REPLY [10 votes]: There is a definition of $h^{1}(\overline{D})$ by many people, and a definition of $H^{1}(\overline{D})$ by Alexrander Borisov using the notion of ghost spaces of the second kind. But neither is the same as the "original ones" you are talking about. As Neukirch pointed out in his book (page 210), in the classical setting an analgous definition of $H^{1}(\overline{D})$ is completely missing. The classical way of getting around this issue is to define $\chi(D)$ instead, and to generalize either you sacrifice the exactness of Riemann-Roch or you re-write the theory using $K$- groups. Both have been extensively discussed in Neukirch. As far as I know, apparently the same situation happens for two dimensional Arakelov theory over surfaces as well: we do not have a reasonable definition of $H^{i}$, instead we have a "Faltings volume" that can be used in some ways to form the Euler characteristic. There has been works done by many people to generalize this to higher dimensional situations in 1980s using index theorem, including efforts replacing Faltings volume by Quillen metric and substantial use of characteristic classes. If we step out of the comfort zone and willing to compromise, then there are "regularization procedure" that allows you to define $h^{1}(\overline{D})$ which essentially boils down to Poisson summation formula, or Pontrajin duality if you prefer. And it has been invented and re-invented by many people. But the essential obstacle is still the same. In particular it seems rather unlikely that we will have an Arakelov cohomology theory built upon derived functors or Cech cohomology alone in the case of arithemetic surface. But we do have something analgous to Serre-duality proved by Borisov in this case, and generalized to the case of vector bundles using a construction similar to adeles by Ichiro Miyada. However the higher dimensional analog is missing. The difficulty seems to be concentrated on the particular case of the vertical divisor at infinity, for which the "ghost space approach" collapses and classical approach thrives (see Lang, page 114). In all other cases (horizontal divisor, vertical divisor at finite places) the ghost space gives what we wanted. In particular, it is not clear how to define a united theory that can recover Faltings-Riemann-Roch in two dimensional case. There are quite a few Arakelov theory experts in the forum (Minhyong Kim, for example) who wrote their theses around this topic. I am sure they can provide much more professional feedback. Update: I learned it from Professor Soule, but I think this is well known among experts in the research circle, that the Faltings' volume has an explicit relationship with Quillen's metric. The exact formula may be found from Professor Soule's Bourbaki lecture paper on arithemetic surfaces. If I am not mistaken there is no higher dimensional analog of Falting's volume for general arithemetic varieties. As far as I know, current work attempting further generalizing Gillet-Soule's work (in the case of manifold with singularity, for example) uses analytic torsion instead. So far even very simple technical questions like "Can one derive analytic torsion using Faltings volume directly without using Faltings-Riemann-Roch?" can be difficult to answer.<|endoftext|> TITLE: Why is the Euler-Mascheroni constant not a Liouville number? QUESTION [8 upvotes]: Let $\gamma$ be the Euler-Mascheroni constant. Why is $\gamma$ not a Liouville number? Are there any upper bounds for the irrationality measure of $\gamma$ known? Any pointers to the literature are welcome. I don't find much on this topic online. Thanks. REPLY [18 votes]: At the present time, we do not even know how to prove that the Euler-Mascheroni constant $\gamma=\lim_{n\to\infty} \sum_{k=1}^n\frac{1}{k} - \log n$ is irrational, much less transcendental; although it is conjectured to be transcendental. The reason you won't find a lot about this topic online (or in the research literature) is because so little is known about the algebraic properties of $\gamma$. There is an analgoue of the Euler-Mascheroni constant for Carlitz modules, and I know that there are many transcendence results in the Carlitz (and Drin'feld and T)-module universe that aren't currently provable over number fields. See for example Tensor Powers of the Carlitz Module and Zeta Values, Greg W. Anderson; Dinesh S. Thakur, The Annals of Mathematics, 2nd Ser., Vol. 132, No. 1. (Jul., 1990), pp. 159-191. But I don't know if irrationality (or transcendence) has even been proven in that setting.<|endoftext|> TITLE: Is a 4-dimensional submanifold of a spin manifold always spin? QUESTION [9 upvotes]: Let $M^d$ be a $d$-dimensional orientable spin manifold, and $N^4$ is a closed $4$-dimensional orientable submanifold of $M^d$. Is $N^4$ always spin? If $d=5$, is $N^4$ always spin? If $N^4$ is a boundary in $M^d$, is $N^4$ always spin? REPLY [17 votes]: Let $i$ denote an immersion $N \to M$. There is an exact sequence of vector bundles on $N$ given by $$0 \to TN \to i^*TM \to \nu \to 0$$ where $\nu$ is the normal bundle. As total Stiefel-Whitney classes are multiplicative in short exact sequences (alternatively, $i^*TM \cong TN\oplus\nu$ smoothly), it follows that \begin{align*} i^*w_1(M) &= w_1(N) + w_1(\nu)\\ i^*w_2(M) &= w_2(N) + w_1(N)w_1(\nu) + w_2(\nu). \end{align*} If $M$ and $N$ are orientable, then we see that $w_1(\nu) = 0$ (i.e. $\nu$ is an orientable bundle) and hence $i^*w_2(M) = w_2(N) + w_2(\nu)$. If $M$ is spin, then we see that $N$ is spin if and only if $w_2(\nu) = 0$. More generally, if two of $M$, $N$, $\nu$ are spin, then so is the third. No. There are examples of $N$ non-spin which embed in a spin manifold $M$ with $w_2(\nu) \neq 0$. In fact, we don't need any of the above to see that the answer is no. Note that any manifold $N$ embeds in $M = \mathbb{R}^d$ (or $S^d$ if you want something closed) for $d$ large enough by the Whitney embedding theorem, regardless of whether or not $N$ is spin. Yes. In general, if $\dim M = \dim N + 1$, then $N$ has codimension one so $w_2(\nu) = 0$ and hence $w_2(N) = 0$. If $N$ is also orientable, we see that $N$ is spin. No. The non-spin four-manifold $N = \mathbb{CP}^2\#\overline{\mathbb{CP}^2}$ is orientedly null-cobordant, so there is a compact five-dimensional manifold with boundary $X$ with $\partial X = \mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. Again by the Whitney embedding theorem, $X$ embeds into $M = S^d$ for $d$ large enough. Even though $N$ is a boundary in $M$, $N$ is not spin.<|endoftext|> TITLE: How hard is it to tell when a finite set tiles the integers? QUESTION [19 upvotes]: Given a nonempty set $B$ of integers between 1 and $n$, we wish to determine whether or not $\mathbb{Z}$ can be tiled with translates of $B$ (that is, covered by disjoint translates of $B$). I know an algorithm for answering this question whose running time can be bounded by a doubly-exponential function of $n$; is there a better algorithm? On the one hand, it seems plausible that some sort of greedy approach to fitting together translates of $B$ might be guaranteed to yield a tiling if one exists (and to terminate quickly if none exists), in which case the decision problem might be efficiently solvable. On the other hand, it seems plausible (and to me more likely) that the one-dimensional tiling problem is computationally complete for a seemingly much broader class of one-dimensional existence problems, much as is the case for two-dimensional tiling. REPLY [12 votes]: If I understand what you are asking, there is a directed graph on $2^{n-2}$ nodes with in and out degrees bounded by $1$. The set tiles the integers exactly if this graph has any directed cycles (one of which might be a loop). There is an easy sufficient condition which is necessary if $|B|$ is of the form $p^i q^j$. The proof which establishes that won't work for $|B|=30$ but Ethan and I conjectured that it is always necessary. Izabella Laba and associates may have some advances. LATER Here is the basic idea. I'll have the prototile $B$ be a set of integers with $\min(B)=0$ and $\max(B)=n-1.$ Define a state to be any subset $S$ of $\{{1,\cdots, n-2 \}}.$ A tiling of $\mathbb{Z}$ by translates of $B$ is a disjoint union $$\mathbb{Z}=\bigcup_{t \in T}\left( B+t\right)$$ Where $T$ is the set of translations. A state describes the equivalence class of (potential) partial tilings $(- \infty,k) \bigcup (S+k)$ which might be $$\bigcup_{t \in T \cap (-\infty,m]}\left( B+t\right)$$ for some $m$ (the initial position of the most recently placed translate of $B.$) The rest of the tiling is uniquely determined any one partial tiling, the next translate of $B$ must be $B+q$ where $q$ is the first integer not in the partial tiling. We can also determine the evolution of the (partial) tiling. The last tile placed must have been $B+v$ where $v+n-1$ is the largest member of the partial tiling. THE GRAPH: The nodes are the states. If $S$ is a state disjoint from $B$ then add a directed arc from $S$ to $S'$ where $S'$ is obtained from $S^*=S \cup B$ by taking the first positive integer $j \notin S^*$ and setting $S'=\left( S^*-j\right) \cap [0,\infty)$. If $S$ is not disjoint from $B$ then no arc leaves $S.$ A directed cycle in this digraph gives a tiling and vice-versa. With obvious adjustments one could have a collection of several different prototiles and essentially the same analysis. In that extended situation what I will call the algebraic approach won't work. The algebraic approach hardly pays attention to $n$ being more interested in $|B|.$ For details I will modestly recommend the paper Gerry links. It uses cyclotomic polynomials to give a strong necessary condition (T1) for a prototile to tile $\mathbb{Z}$. The addition of a second condition (T2) is sufficient. There are no known prototiles which tile $\mathbb{Z}$ and fail (T2.) EVEN LATER The shortest cycle can be as large as the width of $B.$ That happens for $B=\{{0,2^j\}}.$ My conjectures imply that that is as large as it can get and that that is the only way it happens. Given a tiling of the integers by $B$, there is a minimal $N$ so that the tiling is $B \oplus \left(A \oplus N\mathbb{Z}\right)=\mathbb{Z}.$ I call $N$ the period. In terms of the graph I described, the cycle length would be $\frac{N}{|B|}$ (which is an integer.) Just to spell out the conjecture (in part), let $B(x)=\sum_Bx^b$ and $S_B$ be the set of prime powers such that the cyclotomic polynomial $\Phi_{p^j}(x)$ divides $B(x).$ FACTS: if $B$ tiles the integers then it must be the case that $$|B|=\prod_{p^j \in S_B}p. \tag{T1}$$ The period of any tiling will be a multiple of $\operatorname{lcm}(S_B).$ Consider the further condition: (T2): $\Phi_k$ divides $B(x)$ for every $k=\prod p_i^{e_i}$ with the $p_i$ distinct primes and each $p_i^{e_i} \in S_B.$ (T1) and (T2) together are sufficient for $B$ to tile and imply that the minimal period of a tiling is $\operatorname{lcm}(S_B).$ CONJECTURE: (T2) is necessary for a tiling so T1 & T2 is necessary and sufficient If the conjecture is true then it is quite easy indeed to decide if $B$ tiles. But if the conjecture is wrong then maybe longer periods(cycles) are possible. However (T1) alone is still a quick necessary condition likely to quickly reject a (sparse) random set.<|endoftext|> TITLE: Is every sequence that looks like an AP really an AP? QUESTION [18 upvotes]: Caveat: I am not at all a number theorist, and I randomly came up with the following question while I was hiking. But I already asked two serious number theorists, and since they did not know the answer to my question, I decided to pose it here. Let $c > 0$ be given. Suppose $a_1,a_2,a_3,\dots$ is a sequence of positive integers such that for all primes $p > c$, we have $a_i \equiv a_j \ (\text{mod}. p)$ if and only if $i \equiv j \ (\text{mod}. p)$. Does it follow that $(a_i)$ is an arithmetic progression, i.e. there are integers $a$ and $b$ such that $a_k = a + kb$? REPLY [22 votes]: For each $n$, the differences $a_{n+1}-a_n$, $a_{n+2}-a_{n+1}$, and $a_{n+2}-a_n$ can only be divisible by powers of $2$ and primes less than or equal to $c$. Since $$ \frac{a_{n+2}-a_{n+1}}{a_{n+2}-a_n}+\frac{a_{n+1}-a_n}{a_{n+2}-a_n}=1, $$ this is a solution to the S-unit equation, where $S=\{2\}\cup\{p:p\leq c\}$. This means the sequence $x_n:=(a_{n+1}-a_n)/(a_{n+2}-a_n)$ can only take on finitely many values as $n$ varies. Let $p$, $p'>c$ be distinct primes each not dividing the numerator of any of the finitely many distinct non-zero values of $x_n - x_{n'}$. Then the sequence $x_n$ has period dividing $p$ and $p'$, hence is constant. Say $x_n=k$ for all $n$. This implies $$a_{n+2}-a_{n+1}=\frac{1-k}{k}(a_{n+1}-a_n),$$ i.e. the differences $a_{n+1}-a_n$ form a geometric progression. Hence either $a_n$ is an arithmetic progression (if the common ratio $(1-k)/k$ is $1$), or $a_n$ has the form $$ a_n = bc^n+d, $$ with $b$, $c$, $d\in\mathbb{Z}$. The latter case cannot happen, as Fermat's Little Theorem would imply $a_{n+p-1}\equiv a_n\mod p$ for $p$ sufficiently large.<|endoftext|> TITLE: Can extensions of $Q$ contradict Löb with recursive reflection? QUESTION [5 upvotes]: It is an odd and arguably unacceptable situation that $PA$ does not have $\vdash_{PA}(Pr_{PA}\ulcorner A\urcorner\to A)$ for false recursive sentences $A$. However, it is not clear to me that Löb's theorem is already derivable in Robinson arithmetic $Q$, for one cannot assume that the provability predicate of $Q$ obeys all the Löb derivability conditions. (Compare to these matter question A question on the provability predicate of Q). Are there natural omega consistent extensions $Q*$ of $Q$ such $\vdash_{Q*}(Pr_{Q*}\ulcorner A\urcorner\to A)$ for all $\Delta_1$ sentences? REPLY [6 votes]: No consistent recursively axiomatized extension $T$ of $Q$ can prove $\mathrm{Pr}_T\ulcorner\bot\urcorner\to\bot$, that is, $\mathrm{Con}_T$. In fact, no consistent r.e. theory $T$ can interpret $Q+\mathrm{Con}_T$. In fact, no consistent r.e. theory $T$ can interpret $Q+\{\mathrm{RCon}_T(\overline n):n\in\mathbb N\}$, where $\mathrm{RCon}_T(x)$ denotes the consistency of $T$ with respect to proofs using only formulas of “complexity” $x$. See Pudlák’s Cuts, Consistency Statements and Interpretations. (He proves it for interpretations on a cut, but this does not really make a difference.)<|endoftext|> TITLE: Behavior of the sum $\sum_p e^{-p^2 \pi x}$ around 0 and $+\infty$ QUESTION [5 upvotes]: A formula found by Jacobi for $\psi(x)=\sum_{n=1}^{\infty} e^{-n^2\pi x}$ says $$\psi(\frac{1}{x})=\sqrt{x} \psi(x)+\frac{\sqrt{x}-1}{2}$$ If we define $$\phi(x)=\sum_p e^{-p^2\pi x}$$ where $p$ runs through all prime numbers. Now we want to know the behavior of this function near 0 and $+\infty$, then what can be told? Does there exist a similar relation like this $$\phi(\frac{1}{x})=\sqrt{x} \phi(x)+f(x)$$ where $f(x)$ is easy to handle? REPLY [6 votes]: There is no Jacobi-like functional equation for your $\phi(x)$. Nevertheless, with a bit of analysis, we can determine its asymptotic behavior as $x\to\infty$ and as $x\to 0+$. When $x>0$ is large, the leading term $p=2$ dominates, so we have the asymptotics $$\phi(x)\sim e^{-4\pi x},\qquad x\to\infty.$$ When $x>0$ is small, we use the prime number theorem in the form $$\pi(x)\sim\mathrm{Li}(x):=\int_2^x\frac{dt}{\log t}.$$ Using this result and integration by parts, we get, as $x\to 0+$, \begin{align*} \phi(x)&=\int_{2-}^\infty e^{-t^2\pi x}\,d\pi(t) =\int_2^\infty \pi(t)\,d(-e^{-t^2\pi x})\\ &\sim \int_2^\infty \mathrm{Li}(t)\,d(-e^{-t^2\pi x}) =\int_2^\infty e^{-t^2\pi x}\,d\,\mathrm{Li}(t) =\int_2^\infty e^{-t^2\pi x}\,\frac{dt}{\log t}. \end{align*} Making the change of variable $t=u/\sqrt{x}$ in the last integral, we get $$\phi(x)\sim\frac{1}{\sqrt{x}}\int_{2\sqrt{x}}^\infty e^{-u^2\pi}\frac{du}{\log(u/\sqrt{x})},\qquad x\to 0+.$$ Now the denominator $\log(u/\sqrt{x})$ always exceeds $2$, but for the bulk of the integral it is $\sim\log(1/\sqrt{x})$. More precisely, with the notation $r(x)=\sqrt{\log(1/x)}$, we can estimate the integral as \begin{align*} \int_{2\sqrt{x}}^\infty e^{-u^2\pi}\frac{du}{\log(u/\sqrt{x})}&=\int_{2\sqrt{x}}^{\exp(-r(x))}+\int_{\exp(-r(x))}^{\exp(r(x))}+\int_{\exp(r(x))}^\infty\\[8pt] &=O(e^{-r(x)})+\frac{1+o(1)}{\log(1/\sqrt{x})}\int_{\exp(-r(x))}^{\exp(r(x))}e^{-u^2\pi}\,du+O(e^{-r(x)})\\[8pt] &=\frac{1/2+o(1)}{\log(1/\sqrt{x})}=\frac{1+o(1)}{\log(1/x)}. \end{align*} Hence we have proved that $$ \phi(x)\sim\frac{1}{\sqrt{x}\log(1/x)},\qquad x\to 0+.$$<|endoftext|> TITLE: Do there exist nonzero identically vanishing polynomials over infinite (or characteristic zero) reduced indecomposable commutative rings? QUESTION [11 upvotes]: Let $R$ be an infinite, characteristic zero, commutative ring. I can furthermore suppose it is reduced and indecomposable (no nontrivial nilpotents or idempotents). My question is whether there is a nonzero polynomial $f\in R[x]$ which is identically zero on $R$. Note: it is easy to show that there are polynomials with infinitely many roots: let $R=\mathbb Z[s]/(2s)$ and consider $f\in R[x]$ given by $f(x)=\overline{s}x^2+\overline{s}x$. All integers are roots of $f$. But my question is whether we can have $f$ vanishing on all of $R$, not just on an infinite subset. On the other hand, if we further mod out by $s^2$, turning $f$ (I believe) identically vanishing, we create a nilpotent element. A technique that I tried is trying and produce a Vandermonde matrix $V$ associated to the elements $a_1,...,a_k$ of $R$ that be a nonzero element, so that $V$ multiplied by the matrix of coefficients of the canonical basis $e_i$ of polynomials of degree up to $k$, with its $i$-th element replaced by the coefficients of $f$, $f_i$'s, would have two proportional columns and yield $\det(V)\,f_i=0$ and therefore, if I can manage to make $\det(V)$ regular, I will get $f_i=0$. But I believe we may have reduced rings where all elements are zerodivisors, so I am currently trying to modify this Vandermonde argument, by using the very coefficients of $f$ as $a_i$'s and create a nice Vandermonde lattice. REPLY [12 votes]: This is impossible. We deal with the Noetherian case first, and then deduce the general case. Lemma. Let $R$ be a reduced infinite Noetherian ring without idempotents, and let $\mathfrak p \subseteq R$ be a prime. If $R/\mathfrak p$ is finite, then there exists a prime $\mathfrak q \subsetneq \mathfrak p$. For every such $\mathfrak q$, the ring $R/\mathfrak q$ is infinite. Proof. Since $R/\mathfrak p$ is a finite domain, it is a finite field, hence $\mathfrak p$ is maximal. If there does not exist such $\mathfrak q$, then $\mathfrak p$ is minimal as well. This means that the constructible set $V(\mathfrak p) = \{\mathfrak p\}$ is stable under generisation and specialisation, hence it is a clopen subset of $\operatorname{Spec}(R)$; see e.g. [Tag 0542]. Since $R$ has no nontrivial idempotents, $\operatorname{Spec} R$ is connected, so this forces $\mathfrak p$ to be the unique prime ideal. But since $R$ is reduced, we conclude that $\mathfrak p = (0)$, so $R$ itself is finite, contradiction. Hence, $\mathfrak p$ cannot be minimal. Thus, there exists $\mathfrak q \subsetneq \mathfrak p$. But $R/\mathfrak q$ cannot be finite, because a finite domain is a field, which would say that $\mathfrak q$ is maximal. $\square$ Remark. We used the Noetherian hypothesis in the statement that $V(\mathfrak p)$ is constructible. In a general ring, there is a retrocompactness assumption on the complement $D(\mathfrak p)$, which is satisfied for example if $\mathfrak p$ is (the radical of) a finitely generated ideal. See [Tag 04ZC] for details. There is a more elementary argument if you don't want to use constructible sets. Indeed, assume $R$ has minimal primes $\mathfrak p_i$ different from $\mathfrak p$. Since $R$ is Noetherian, there are finitely many such, and since $\operatorname{Spec}(R)$ is connected, there must be an $i$ such that $V(\mathfrak p) \cap V(\mathfrak p_i) \neq \varnothing$. But that means that $\mathfrak p \in V(\mathfrak p_i)$, i.e. $\mathfrak p_i \subseteq \mathfrak p$, contradicting minimality of $\mathfrak p$. Lemma. Let $R$ be a reduced infinite Noetherian ring without idempotents, and let $f \in R[x]$. If $f(r) = 0$ for all $r \in R$, then $f = 0$. Proof. Let $\mathfrak p$ be a prime for which $R/\mathfrak p$ is infinite. Denote by $\bar{r}$ the reduction of $r$ in $R/\mathfrak p$. We clearly have $\bar f(\bar r) = \overline{f(r)}$ for all $r \in R$, so we conclude that $\bar f \in (R/\mathfrak p)[x]$ vanishes at all elements of $R/\mathfrak p$. Since $R/\mathfrak p$ is an infinite domain, this forces $\bar f = 0$, i.e. $f \in \mathfrak pR[x]$. Hence, $$f \in \bigcap_{\substack{\mathfrak p\\ |R/\mathfrak p| = \infty}} \mathfrak p R[x].\label{Eq 1}\tag{1}$$ But by the lemma above, every prime for which $R/\mathfrak p$ is finite contains a prime $\mathfrak q$ for which $R/\mathfrak q$ is infinite, so the intersection in (\ref{Eq 1}) is equal to $$\bigcap_{\mathfrak p} \mathfrak p R[x].$$ That is, every coefficient of $f$ is in $\bigcap_{\mathfrak p} \mathfrak p = \mathfrak{nil}(R)$, which is $0$ by assumption. $\square$ Corollary. Let $R$ be any reduced infinite ring without idempotents, and let $f \in R[x]$. If $f(r) = 0$ for all $r \in R$, then $f = 0$. Proof. If $R$ is a field, then it is an infinite field, so we are done. Otherwise, let $r_0 \in R\setminus\{0\}$ be an element that does not have an inverse, and let $R'\subseteq R$ be a finitely generated subring containing $r_0$ such that $f \in R'[x]$. For example, we can take the subring generated by the coefficients of $f$ and $r_0$. Then $R'$ is reduced and has no nontrivial idempotents, and $R'$ is Noetherian since it is finitely generated. If $R'$ is finite, then it is Artinian, hence local since it has no idempotents, hence a field since it has no nilpotents. This contradicts the assumption that $r_0$ does not have an inverse. Thus, $R'$ is infinite, and $f(r') = 0$ for all $r' \in R'$. Therefore, $f = 0$ by the lemma above applied to the ring $R'$. $\square$<|endoftext|> TITLE: Do compact complex manifolds fall into countably many families? QUESTION [39 upvotes]: Do there exist countably many proper holomorphic submersions of complex manifolds $\mathcal{X}_n \to B_n$ such that every compact complex manifold appears as a fiber in at least one of the families? (Conventions: Assume that each $B_n$ is Hausdorff, connected, and paracompact; these conditions imply that $B_n$ is also second countable. For the question at hand, one could equivalently require each $B_n$ to be an open subset of $\mathbf{C}^m$ for some $m$ depending on $n$.) REPLY [3 votes]: This is an attempt to express a variant of Jason's answer in slightly more down-to-earth terms, so that maybe less has to be referred to his experts. (This answer is too long for a comment, so I made it a community wiki.) It seems to me that there is really only one difficult point in the argument beyond the existence of the Kuranishi deformation space, namely the "Theorem" below. Step 1: Nash's article "Real algebraic manifolds" in Annals of Math. (1952) proved that every connected compact $C^\infty$ real manifold is diffeomorphic to a component of a real algebraic variety. There are only countably many families of real algebraic varieties, and Ehresmann's fibration theorem (1950) shows that the diffeomorphism type is locally constant within each family. Thus there are only countably many diffeomorphism types. Therefore we now restrict attention to one. Step 2: Let $M$ be a compact $C^\infty$ real manifold, let $T$ be its tangent bundle, and let $E$ be the bundle with fibers $E_x := \operatorname{End} T_x$. Fix $r$ not too small, and let $C^r(M,E)$ be $\{\text{$C^r$ functions $M \to E$}\}$ with the $C^r$ compact-open topology (Hirsch, Differential topology, 2.1), which is second countable. The space $\operatorname{Comp}$ of complex structures (i.e., integrable almost complex structures) on $M$ is a subset of $C^r(M,E)$; give it the subspace topology, so it too is second countable. Explicitly, complex structures $I$ and $I'$ are close if uniformly on $M$ they are within $\epsilon$ (with respect to some metric) and the same holds for their derivatives up to order $r$. Step 3: On the other hand, let $\mathcal{M}$ be the set of isomorphism classes of compact complex manifolds whose underlying real manifold is diffeomorphic to $M$. Kuranishi proved that each compact complex manifold $X$ admits a versal deformation over a pointed connected complex analytic space $(B,0)$. Each such versal deformation gives rise to a subset of $\mathcal{M}$, namely the set of isomorphism classes of the fibers; define a topology on $\mathcal{M}$ in which these subsets form a basis of neighborhoods of the point $[X] \in \mathcal{M}$ determined by $X$. Step 4: Let $\Phi \colon \operatorname{Comp} \to \mathcal{M}$ be the map sending $(M,I)$ to its isomorphism class as a complex manifold. Then $\Phi$ is open because for any open subset $U \subset \operatorname{Comp}$ and $[X] \in \Phi(U)$, a choice of $C^\infty$ trivialization of a versal deformation over $B$ defines a continuous lifting $B \to \operatorname{Comp}$ of $B \to \mathcal{M}$. The key point is now the following (which seems to be a consequence of what Jason's experts claimed): Theorem: $\Phi$ is continuous. (In other words, given a versal deformation of $X = (M,I)$, any complex structure $I'$ sufficiently close to $I$ in the $C^r$ sense defines a complex manifold $(M,I')$ isomorphic to a fiber of the deformation.) Given this, $\mathcal{M}$ is second countable because it is the image of a second countable space under an open continuous map. This means that $\mathcal{M}$ is covered by countably many subsets each arising from a family over a complex analytic subspace in $\mathbf{C}^n$ for some $n$. Each such base can be stratified into complex manifolds, so we are done.<|endoftext|> TITLE: Doesn't completion of a representation ring preserve its indecomposables? QUESTION [8 upvotes]: For $G = PSU(3)$, it is known that $\dim I(G;\mathbb Q) / I(G;\mathbb Q)^2 = 3$, while $H^{**}(BG;\mathbb Q)$ is obviously a power series ring in two indeterminates since $G$ has rank 2. This would would be fine except that there is an elementary argument the natural map should induce a bijection on spaces of indecomposable elements and $3\neq2$. What goes wrong? Here's the argument. Let $T$ be compact torus, $R(T)$ its representation ring, and $\widehat R(T)$ the completion with respect to the augmentation ideal $I(T)$. One has maps $$R(T) \longrightarrow \widehat R(T) \overset\sim\longrightarrow K^*(BT) \overset{\textrm{ch}}\longrightarrow H^{**}(BT;\mathbb Q),$$ where the double-star insists we view the cohomology ring as the direct product of the $H^n$. If we tensor $R(T)$ with $\mathbb Q$ beforehand, we get $$R(T;\mathbb Q) \longrightarrow \widehat R(T;\mathbb Q) \longrightarrow H^{**}(BT;\mathbb Q).$$ The last map sends a one-dimensional representation $t \in R(T)$ to $e^u$ for $u = c_1(ET \times_T \mathbb C_t)$, and hence is an isomorphism, as both rings are power series rings on $\dim T$ indeterminates and $\log t$ is a power series in $t-1$. Since $t-1 \mapsto e^u - 1 = u + u^2/2 + \cdots$, the map also sends $\hat I(T;\mathbb Q)$ to the augmentation ideal $H^{\geq 1}(BT;\mathbb Q)$ of $H^{**}(BG;\mathbb Q)$. Now consider a compact, connected Lie group $G$ with maximal torus $T$. If we make the identification $BT = EG / T$, then $BT$ admits a right action of the Weyl group $W$ of $G$. The maps are equivariant with respect to the action of $w \in W$ since $t w$ is sent to $\exp c_1(ET \times_T \mathbb C_{t w})$ and $ET \times_T \mathbb C_{t w}$ is the pullback of $ET \times_T \mathbb C_{t}$ under $w\colon BT \to BT$. This equivariance and the standard isomorphisms $$R(T)^W = R(G),$$ $$\widehat R(T)^W = \widehat R(G),$$ $$H^{**}(BT;\mathbb Q)^W = H^{**}(BG;\mathbb Q)$$ then show $\widehat R(G;\mathbb Q) \to H^{**}(BG;\mathbb Q)$ is also an isomorphism preserving the augmentation ideal. Then, since $R(G)$ is Noetherian, the maps $$R(G;\mathbb Q) \longrightarrow \widehat R(G;\mathbb Q) \overset\sim\longrightarrow H^{**}(BG;\mathbb Q)$$ induces isomorphisms $$\frac{I(G;\mathbb Q)\phantom{2}}{I(G;\mathbb Q)^2} \overset\sim\longrightarrow \frac{\hat I(G;\mathbb Q)\phantom{2}}{\hat I(G;\mathbb Q)^2} \overset\sim\longrightarrow \frac{H^{\geq 1}(BG;\mathbb Q)\phantom{2}}{H^{\geq 1}(BG;\mathbb Q)^2}$$ of modules of indecomposables. REPLY [2 votes]: In fact $\text{dim }I(G)/I(G)^2=2$. Let $x=[V_{3L_1}]$, $y=[V_{2L_1+L_2}]$ and $z=[V_{3L_1+3L_2}]$. Then \begin{eqnarray}R(PSU(3))\cong\mathbb{Z}[x, y, z]/(y^3-y^2-xz-2y(x+z)-x-y-z).\end{eqnarray} Note that $\text{dim }V_{3L_1}=\text{dim }V_{3L_1+3L_2}=10$ and $\text{dim }V_{2L_1+L_2}=8$. If $\overline{x}=x-10$, $\overline{y}=y-8$, $\overline{z}=z-10$, then \begin{eqnarray}R(PSU(3))\cong\mathbb{Z}[\overline{x}, \overline{y}, \overline{z}]/(\overline{y}^3+23\overline{y}^2-2(\overline{x}+\overline{z})\overline{y}-\overline{x}\overline{z}-27(\overline{x}+\overline{z})+135\overline{y})).\end{eqnarray} Note that $I(G)=(\overline{x}, \overline{y}, \overline{z})$. It follows that \begin{eqnarray}I(G; \mathbb{Q})/I(G; \mathbb{Q})^2=\text{span}_\mathbb{Q}\{\overline{x}, \overline{y}, \overline{z}\}/\text{span}_\mathbb{Q}\{-27(\overline{x}+\overline{z})+135\overline{y}\}\end{eqnarray} which is of dimension 2. Remark: If we let \begin{cases}x&=X+1-2Y\\ y&=Y-1\\ z&=Z+1-2Y\end{cases} we will get the more compact description \begin{eqnarray}R(PSU(3))\cong\mathbb{Z}[X, Y, Z]/(Y^3-XZ)\end{eqnarray} (see Lemma 7.1 of this paper). Added: The following is the SAGE code which verifies the relation $y^3-y^2-xz-2y(x+z)-x-y-z=0$. A2(a, b) means the irreducible representation $V_{(a+b)L_1+bL_2}$. sage: A2=WeylCharacterRing("A2", style="coroots") sage: x=A2(3, 0) sage: y=A2(1, 1) sage: z=A2(0, 3) sage: y^3-y^2-x*z-2*y*(x+z)-x-y-z 0<|endoftext|> TITLE: volume over a hypercube, over simplex: twist by Euler numbers QUESTION [7 upvotes]: Let $\square_n=\{(x_1,\dots,x_n): 0\leq x_i\leq1,\, \forall i\}$ be an $n$-dimensional unit hypercube, and let $\Delta_n=\{(u_1,\dots,u_n):u_1+\cdots+u_n\leq\frac{\pi}2,\, u_i\geq0,\, \forall i\}$ be $n$-simplex. Also, let $E_{2m}$ be secant numbers (even-index Euler numbers) given by $$\sum_{m\geq0}E_{2m}\frac{y^{2m}}{(2m)!}.$$ The following has been experimentally tested. Question. What is the transformation that makes this equality possible? $$\int_{\square_{2m+1}}\frac{d\mathbf{x}}{1+x_1^2\cdots x_{2m+1}^2}= \frac{(2m+1)E_{2m}}2\int_{\Delta_{2m+1}}d\mathbf{u}$$ REPLY [10 votes]: This is only a partial answer. The Beukers-Kolk-Calabi change of variables $$x_1=\frac{\sin{u_1}}{\cos{u_2}},\;\;x_2=\frac{\sin{u_2}}{\cos{u_3}},\ldots, \;x_{n-1}=\frac{\sin{u_{n-1}}}{\cos{u_n}},\;\;x_n=\frac{\sin{u_n}}{\cos{u_1}}$$ has the Jacobian $$\frac{\partial(x_1,\ldots,x_n)}{\partial(u_1,\ldots,u_n)}= 1-(-1)^n\,x^2_1x^2_2\cdots x^2_n.$$ Therefore you integral is a volume of the polytope $\delta_{2m+1}=\left \{(u_1,\ldots,u_{2m+1}): u_i\ge 0,\; u_i+u_{i+1}\le \pi/2 \right \}$. Here $i=1,\ldots 2m+1$ and it is assumed that $u_i$ are indexed cyclically so that $u_{2m+2}=u_1$. It remains to relate the volumes of the polytope $\delta_{2m+1}$ and $(2m+1)$-simplex $\Delta_{2m+1}$. See https://arxiv.org/abs/math/0101168 P.S. The volume of $\delta_{2m+1}$ can be calculated by using (37) and (40) from https://arxiv.org/abs/1003.3602 and the result is $$Vol(\delta_{2m+1})=(-1)^m\frac{2^{2m-1}}{(2m)!}E_{2m}(1/2)\left(\frac{\pi}{2}\right)^{2m+1}=(-1)^m\frac{E_{2m}}{2(2m)!}\left(\frac{\pi}{2}\right)^{2m+1},$$ while (see, for example, https://eudml.org/doc/141172) $$Vol(\Delta_{2m+1})=\frac{1}{(2m+1)!}\left(\frac{\pi}{2}\right)^{2m+1}.$$ Therefore we indeed get the desired identity provided $(-1)^m$ is incorporated in the definition of Euler numbers through Euler polynomials (we used definitions from https://eudml.org/doc/49338 which doesn't incorporate $(-1)^m$).<|endoftext|> TITLE: Boundary of the image of a compact manifold in the complex plane QUESTION [5 upvotes]: The Question Consider the trace of an $n \times n$ unitary matrix with determinant 1 \begin{align} f: SU(n) &\rightarrow \mathbb{C}\\ U \mapsto \text{tr}\, U &= \sum\limits_{i=1}^{n-1} z_i + \frac{1}{z_1 \cdots z_{n-1}} \end{align} where the $z_i$ are the eigenvalues of $U$ and we have used $\det U =1$ to write $z_n$ in terms of the other eigenvalues, without loss of generality. In section 3 of the paper "Mean eigenvalues for simple, simply connected, compact Lie groups," the author argues that the image of $f$ is the $n$-cusp hypocycloid. A critical step in the argument relies on the statement that on the boundary of the image, we can set $n-2$ of the partial derivatives of $f$ equal to zero, that is \begin{align} \frac{\partial f}{\partial z_1} = \cdots = \frac{\partial f}{\partial z_{n-2}} = 0 \end{align} Why is it true that imposing this condition gives the boundary of the image of $f$? I'm currently trying to use this argument for a generalization of $f$ (determining the image of sums and products of traces of $SU(n)$ matrices by first finding the boundary of the image). Attempt at a solution 1 Confusion over this argument in the paper was mentioned in the comment section of this blog post. Greg Egan writes: "I guess the idea is that we have a compact manifold without boundary of real dimension $n-1$ being projected onto the complex plane, and where the manifold projects to the boundary of its shadow the linearised map has to change from having an $(n-3)$-dimensional kernel to an $(n-2)$-dimensional kernel, so you can choose coordinates there such that $n-2$ of the coordinate vectors lie in the kernel." "Generically there will be some choice of coordinates where the derivatives on the boundary vanish for all but one coordinate, but for a more general function than the trace that coordinate system need not line up with the phases. So he’s exploiting a lot of nice symmetries of the problem, but I wish he’d given a more careful account of the things he’s relying on to obtain the result." Is what Greg writes true? I wasn't able to make it rigorous myself, thinking that the tangent space on the boundary of $f(SU(n)) \subset \mathbb{C}$ is still $2$ dimensional. Maybe someone can recommend some resources on the topic of the tangent space at the boundary of the continuous image of a compact connected manifold. Attempt at solution 2 Let $n = 3$ for simplicity. If we instead think of $f$ in this case as \begin{align} \widetilde{f}: U(1) \times U(1) &\rightarrow \mathbb{C}\\ (\theta_1, \theta_2) &\rightarrow e^{i \theta_1} + e^{i \theta_2} + e^{-i( \theta_1 + \theta_2)} \end{align} then with respect to charts $(V_1, \theta_1, \theta_2)$ at some $p \in U(1) \times U(1)$ and the obvious charts (projecting real and imaginary parts) on $\mathbb{C}$, the pushforward/differential/Jacobian is given by \begin{align} J(p) = \left( \begin{array}{cc} -\sin (\text{$\theta $1})-\sin (\text{$\theta $1}+\text{$\theta $2}) & -\sin (\text{$\theta $2})-\sin (\text{$\theta $1}+\text{$\theta $2}) \\ \cos (\text{$\theta $1})-\cos (\text{$\theta $1}+\text{$\theta $2}) & \cos (\text{$\theta $2})-\cos (\text{$\theta $1}+\text{$\theta $2}) \\ \end{array} \right) \end{align} Then we can see that the pushforward/differential/Jacobian is not of maximal rank at $p$ if $\theta_1 = \theta_2$, which maps out the hypocycloid. edited below to reflect Igor Rivin's comment It is not true in general that for a compact connected manifold, if the pushforward fails to be of maximal rank, this must occur on the boundary of the continuous image of $f$. Then what additional assumption is needed? I know that in this situation the regular values of $f$ must lie in the interior of the image of $f$, but I have not been able to prove that critical values cannot lie in the interior. REPLY [2 votes]: Here is one way to prove the claims about the image of the map $\mathrm{tr}:\mathrm{SU}(n)\to\mathbb{C}$. I'll just outline the steps. For simpicity, I'll always assume $n\ge 3$. (For completeness, observe that $\mathrm{tr}\bigl(\mathrm{SU}(2)\bigr)$ consists of the interval $[-2,2]\subset\mathbb{R}\subset\mathbb{C}$.) First, some notation: Let $\mathbb{T}\subset\mathrm{SU}(n)$ denote the subgroup consisting of of diagonal $n$-by-$n$ special unitary matrices and let $D(z_1,\ldots,z_n)\in\mathbb{T}$ denote the diagonal matrix whose $(j,j)$-entry is $z_j\in S^1\subset\mathbb{C}$. Of course, $z_1\cdots z_n = 1$. Since every matrix in $\mathrm{SU}(n)$ is conjugate to a diagonal matrix, $$ X_n = \mathrm{tr}\bigl(\mathrm{SU}(n)\bigr) = \mathrm{tr}\bigl(\mathbb{T}\bigr),$$ and $X_n\subset\mathbb{C}$ is compact and connected. Second, it's easy to show that the critical points of $\mathrm{tr}:\mathbb{T}\to\mathbb{C}$ are the diagonal matrices $D(z_1,\ldots,z_n)$ for which at most two of the $z_i$ are distinct. In particular, the interior of $X_n$ is nonempty and connected (obvious when $n>3$ and true for $n=3$ by inspection) and its 'boundary' (i.e., $X_n$ minus its interior) is a finite union of analytic arcs consisting entirely of critical values of $\mathrm{tr}$ on $\mathbb{T}$. Since the noncritical points of $\mathrm{tr}$ are open and dense in $\mathbb{T}$, $X_n$ is the closure of its interior. I am going to show that a critical point $D(z_1,\ldots,z_n)$ is mapped into the interior of $X_n$ whenever it has two distinct eigenvalues, each of which has multiplicity at least $2$. Now, the locus of critical values of $\mathrm{tr}:\mathbb{T}\to\mathbb{C}$, is the union of the $\mathrm{tr}$-images of the curves $$ \gamma_{p,k}(t) = \omega^k\,D(\,\underbrace{e^{i(n-p)t},\ldots,e^{i(n-p)t}}_{\text{p times}}, \underbrace{e^{-ipt},\ldots,e^{-ipt}}_{\text{$n-p$ times}}\,) $$ where $\omega = e^{2\pi i/n}$ is the primitive root of unity, $1\le p TITLE: A characterization of flat metrics via global vector fields QUESTION [7 upvotes]: Let $(M,g)$ be a Riemannian manifold with $LC$ conncection $\nabla$. Assume that for every three global vector fields $X,Y,Z \in \chi^{\infty}(M)$ with $[X,Y]=0$ we have $\nabla_{X} \nabla_{Y} Z=\nabla_{Y} \nabla_{X} Z$ Is the metric necessarily a flat metric? REPLY [2 votes]: To see that $g$ is flat, we need enough pairs of commuting global vector fields. To construct these, let $f(r)\colon[0,\infty)\to[0,1)$ be a monotone function with $f(r)=1-\frac1{\log r}$ for $r\gg 1$. Consider the diffeomorphism $\mathbb R^n\to B^n$ given by $\Phi(x)=f(|x|)\,\frac x{|x|}$. We can map constant (and therefore commuting) vector fields $V$, $W$ on $\mathbb R^n$ to vector fields on $B^n$ that decay sufficiently fast near the boundary of $B^n\subset\mathbb R^n$ so that we can extend them by $0$ to compactly supported vector fields $\bar V$, $\bar W$ on $\mathbb R^n$. By naturality of the Lie bracket, these vector fields still commute. The same construction in local coordinates gives sufficiently many pairs of commuting vector field of manifolds to conclude that the connection $\nabla$ is flat. In the comments you ask for global vector fields that are independent almost everywhere. It seems to me that the construction above can be modified to give pairs of commuting vector fields that vanish along the $(n-1)$-skeleton of a smooth triangulation of a manifold $M$, and are linearly independent otherwise.<|endoftext|> TITLE: Are the unipotent and nilpotent varieties isomorphic in bad characteristics? QUESTION [17 upvotes]: In characteristic 0 or good prime characteristic, there are standard ways to relate the unipotent variety $\mathcal{U}$ of a simple algebraic group $G$ and the nilpotent variety $\mathcal{N}$ of its Lie algebra $\mathfrak{g}$. Recall that $p$ is good just when it fails to divide any coefficient of the highest root (the root system being irreducible), bad otherwise. The only possible bad primes are $2,3,5$. In characteristic 0, algebraic versions of the exponential and logarithm maps provide Ad $G$-equivariant isomorphisms in both directions, whereas in good characteristic $p>0$, the less direct arguments of Springer in The unipotent variety of a semi-simple group yield similar results. [The isogeny type of $G$ adds some complications here.] There is scattered literature on the varieties $\mathcal{U}$ and $\mathcal{N}$ when $p$ is bad, often treated case-by-case, e.g., four papers by Lusztig posted on arXiv starting with Unipotent elements in small characteristic, along with papers by his student T. Xue. A serious challenge when $p$ is bad is to find a uniform explanation for the failure of the numbers of unipotent classes and nilpotent orbits to agree in some cases: the details were worked out by Holt–Spaltenstein and others. In spite of this breakdown in $G$-equivariance, a natural question can be raised: Are the two varieties $\mathcal{U}$ and $\mathcal{N}$ isomorphic in all characteristics, for example when $G$ is simply connected? The answer does not seem to be written down explicitly (?), but for example one can see indirectly that both varieties have the same dimension in all characteristics: the number of roots. Existence of regular nilpotent elements in the Lie algebra of a simple algebraic group in bad characteristics by S. Keny, a former student of Steinberg, showed case-by-case that regular nilpotent elements always exist and form a dense orbit in $\mathcal{N}$. By definition, the isotropy group in $G$ of such an element has dimension equal to the rank of $G$. REPLY [5 votes]: Update: Paul Levy points out in the comments that a reasonable way of defining the nullcone in $\operatorname{Lie}(G) = \mathfrak{g}$ is as the zero set of the homogeneous invariants of positive degree — i.e. of $(k[\mathfrak{g}]_+)^G = (S^+\mathfrak{g}^*)^G$. But with this definition, my original comment isn't valid. Indeed, if $G = \operatorname{PGL}_2$ then the co-adjoint representation of $G$ on $\mathfrak{g}^*$ has a fixed vector, which is a linear invariant in $(k[\mathfrak{g}]_+)^G$ whose zero locus is — as in Dave Stewart's original answer — the span of the root vectors. I suppose all my original objection (below) really amounted to was that if $X$ is the affine scheme defined by the $\mathbf{Z}$-algebra $R=\mathbf{Z}[A,B,C]/\langle A^2 + 4BC\rangle$, then for any field $k$ of char. not 2, $X_k$ identifies with the nilpotent variety of $\mathfrak{pgl}_{2,k}$, but if $k$ has char. 2, $X_k$ is not reduced. Original post: I'm going to write this as an "answer", though I think it mostly amounts to a comment on Dave Stewart's answer. It is not completely clear to me that the statement "the nilpotent variety of $\mathfrak{pgl}_2$ is reduced" is correct when $p=2$. Well, I suppose that more precisely I mean: it isn't clear that the scheme of nilpotent elements should be viewed as reduced. Taking a basis $x,y,h$ of $\mathfrak{pgl}_2$ (say, in its 3-dimensional representation), one finds that $ah + bx + cy$ is nilpotent just in case $a^2+4bc=0$. Of course, in char. 2 this amounts to $a^2=0$, which suggests that the scheme of nilpotent elements shouldn't be viewed as a reduced subscheme. (If you don't want to write down the matrices, see e.g. Jantzen "Nilpotent Orbits in Representation Theory" §2.7.) I do doubt (?) that this nilpotent scheme is isomorphic to the scheme of unipotent elements of $\operatorname{PGL}_2$, but (assuming that doubt is correct — I didn't think too carefully about it) I think the reason is more complicated than the statement "one is reduced and the other isn't".<|endoftext|> TITLE: Citing exercises in an article QUESTION [45 upvotes]: I'm writing a paper in which I cite a lot of results that appear in Schikhof's Ultrametric Calculus. Some of these results are exercises in Schikhof's book. These exercises are not difficult, but are laborious. Thus, if I write the proofs, the article may extend by about two or three pages. Should I write the proofs or simply cite them? Schikhof is a very well respected mathematician, and I have never found any errors in his book. Obviously, I have checked that the exercises are correct. (If it were one exercise, I would write the proof in my article, as I have seen in other articles, but in my case there are about five exercises.) REPLY [6 votes]: Here is one additional data point. The standard reference for symmetric function is Macdonald's "Symmetric functions and Hall polynomials". Most of the content of this book is in the exercises; each section of the book contains many pages of useful results and formulas stated without proof. According to Google Scholar, this book has been cited 7735 times, and it seems likely that many (most?) of these citations are references to exercises in the book.<|endoftext|> TITLE: Abel-Jacobi map for Mumford curves analytically QUESTION [6 upvotes]: Let $K$ be a field equipped with a non-Archimedean absolute value, let $\Gamma$ be a Schottky group in $PGL_2(K)$, and let $X_\Gamma$ be the associated Mumford curve, which is a proper smooth rigid analytic space over $K$, but also a proper algebraic curve (due to Mumford's theorem). Let $J_\Gamma$ be the Jacobian variety of degree 0 divisors on $X_\Gamma$, and let $a: X_\Gamma \to J_\Gamma$ be the Abel-Jacobi map which sends a point $x$ to the class of the divisor $[x-x_0]$ for some $x_0 \in X_\Gamma$ fixed in advance. It is known that $J_\Gamma$ admits a uniformisation as the quotient $\mathbb{G}_m^g/\Gamma'$ where $\Gamma'$ is the abelianisation $\Gamma/[\Gamma,\Gamma]$ of the group $\Gamma$, and $g$ is the genus of $X_\Gamma$. Is there a way to describe the Abel-Jacobi map $a$ in analytic terms, similarly to the description of this map over $\mathbb{C}$ in terms of Abelian integrals? REPLY [3 votes]: You can also find an interpretation using integrals, this time multiplicative, and mesures, in the preprint "The Abel Jacobi map for Mumford Curves via Integration": https://arxiv.org/abs/1609.09285 which generalizes the construction given by Darmon and Longhi in the local case.<|endoftext|> TITLE: Two mixed Hodge structures on equivariant cohomology for actions by finite groups QUESTION [7 upvotes]: The answer to the following question might be obvious but I haven’t found a full proof yet (neither by myself nor in the literature). So my apologies if it is trivial. Let $X$ be a (for simplicity quasi-projective and non-singular) complex variety $X$ on which a finite group $G$ acts. Deligne has shown in Hodge III that the equivariant cohomology group $H_G^k(X,\mathbb{Q})$ for any $k$ carries a natural mixed Hodge structure (MHS). On the other hand, the Leray-Serre spectral sequence for the Serre fibration \begin{equation} X \to X\times_G EG \to BG \end{equation} degenerates (complex topology) over $\mathbb{Q}$ and hence yields an isomorphism \begin{equation} H_G^k(X,\mathbb{Q})\cong H^k(X,\mathbb{Q})^G \end{equation} (cf. this MO question: Equivariant cohomology of finite group actions and invariant cohomology classes). Clearly, $H^k(X,\mathbb{Q})^G$ inherits a MHS from the MHS on $H^k(X,\mathbb{Q})$. Now my question is: Are these two MHS on $H_G^k(X,\mathbb{Q})$ naturally isomorphic? More precisely, is there a simplicial version of the above Serre fibration yielding an isomorphism $H_G^k(X,\mathbb{Q})\cong H^k(X,\mathbb{Q})^G$ of MHS? A natural candidate for such a simplicial version is \begin{equation} [X/G]_\bullet \to B_\bullet G, \end{equation} where I (essentially) use Deligne's notation from Hodge III. Since I'm still learning simplicial methods, I was not sure if this is maybe too naive. Any thoughts/references/comments are very welcome! REPLY [6 votes]: As I said in my comment, the mixed Hodge structures are the same. Here is the outline. From [Hodge III, 6.1.2.1], $$[X/G]_n = (G^{n+1}\times X)/G$$ One has a descent spectral sequence $$E_1= H^q([X/G]_p,\mathbb{Q})\cong (G^{p+1}\times H^q(X,\mathbb{Q}))/G$$ abutting to $H^{p+q}([X/G]_\bullet, \mathbb{Q})=H^{p+q}_G(X,\mathbb{Q})$ [Hodge III, (5.2.1.1)], and this is compatible with MHS [Hodge III, (8.1.15)]. Now use the fact that the complex $E_1$ is the bar complex, which computes group cohomology; in this case it is trivial except in degree zero. So in conclusion $$H_G^*(X,\mathbb{Q})\cong H^*(X,\mathbb{Q})^G$$ as MHS.<|endoftext|> TITLE: Given four conditionally convergent series, is there a single sequence of naturals such that each corresponding subseries sums to $\pm\infty$? QUESTION [8 upvotes]: Let us say that a set $A \subseteq \mathbb N$ sends a series $\sum_{n \in \mathbb N}a_n$ of real numbers to infinity if the subseries $\sum_{n \in A}a_n$ sums either to $\infty$ or to $-\infty$. Given four or more conditionally convergent series, is there a single $A \subseteq \mathbb N$ that sends them all to infinity? Notice that I do not require that all of the subseries sum to $\infty$, or that all of them sum to $-\infty$. This may well be impossible, even for just two series $\sum_{n \in \mathbb N}a_n$ and $\sum_{n \in \mathbb N}b_n$, since we may have $a_n = -b_n$ for all $n$. However, we do require that each subseries is made to diverge to either $\infty$ or $-\infty$, and not merely to diverge by oscillation. (If divergence by oscillation is allowed, then the answer to my question becomes a relatively easy yes.) For one series this is trivial: let $A$ be the set of indices of the positive terms. For two series, the question requires thought, but it is doable. [The idea is that two conditionally convergent series naturally partition $\mathbb N$ into four sets, the partition being determined by where each of the series is positive or non-positive. With a little work, you can prove that either one of these four sets works, or else the union of two of them will.] For three series, I was able to prove that the answer is yes, but my proof is long, complicated, and frankly . . . ugly. I can't help but think that I'm going about it the wrong way, and that there must be a better approach leading to an easier proof. For four or more series, I'm stumped. REPLY [11 votes]: There is a counterexample with 4 series. Notice that the problem is equivalent to asking if for every sequence of vectors $X_j$ in $\mathbb R^4$ with lengths tending to $0$ and the infinite sum of absolute values of the projections to each coordinate axis, you can find a sequence of signs $\varepsilon_j=\pm 1$ such that $\sum_j \varepsilon_j X_j$ diverges to plus or minus infinity in each coordinate (in one way, it is trivial: just view taking or not taking a vector as deviating from one half of it up or down). Now build the following blocks of fast increasing lengths $N_k$: all vectors are $(\frac 1{k},\frac 1{k},1/N_k,0)$ if $k$ is odd and $(\frac 1{k},-\frac 1{k},0,1/N_k)$ if $k$ is even. Suppose we try to diverge to $(+\infty,+\infty)$ or $(-\infty,-\infty)$ in the first two coordinates. Then every even block of length $N_k$ will create a terrible setback unless it is balanced up to $(N_1+\dots+N_{k-1})k\approx kN_{k-1}$. But then all we can gain in the fourth position is $\sum_{k\text{ even}}\frac{kN_{k-1}}{N_k}$ and that series hopelessly converges. If we try to get infinities of different signs, the odd blocks and the third position will create a problem. Now, to avoid firing any big cannons to show that the equivalence of the problems holds in both directions, just take the blocks of even length and get the 4 series requested as alternating sums of the specified vectors.<|endoftext|> TITLE: Convention of Address in math journals? QUESTION [9 upvotes]: The paper was written and submitted when I was in Institution A. After (many) years, the paper is accepted when I am in Institution B. Which address shall I put on the paper? The current address (Institution B) will soon expire since I will have to move somewhere else. So when the paper will be finally published online / on paper, neither A or B is valid address. REPLY [4 votes]: The primary purpose of the address, in "days of yore," was so that people could write to you to ask for a reprint and/or start a correspondence with you about your paper. And most institutions were willing to forward snail-mail if you kept them informed of your current address. All of that is pretty much irrelevant now. So I think the primary answer to your question is to make sure that your address includes an email address that will not expire, regardless of your moves. I know many mathematicians who use a gmail address for that purpose. If instead you use the email address at your current institution, make sure that they will forward email to whatever new address you give them. Beyond that, the other advice on how to list the institution(s) seems reasonable, since institution A deserves some credit for supporting your work, as possibly does institution B if, say, you revised the paper while there.<|endoftext|> TITLE: Countable products of total measures QUESTION [8 upvotes]: Suppose $\kappa = \mathfrak{c} =2^{\aleph_0}$ is a real valued measurable cardinal with a witnessing measure $m:\mathcal{P}(\kappa) \to [0, 1]$ - So $m$ is a diffused (points have zero measure) $\kappa$-additive measure on $\mathcal{P}(\kappa)$. Let $\mu = \otimes_{n < \omega} m$ be the countble product measure - So $\mu$ is defined on the sigma algebra generated by the family $\{\Pi_{n < \omega} X_n : (\forall n)(X_n \subseteq \kappa)\}$. Let $\mathcal{N}$ be the sigma ideal of $\mu$-null subsets of $\kappa^{\omega}$. What can be say about the cardinal invariants associated with the ideal $\mathcal{N}$? For example, can either one of $add(\mathcal{N}), non(\mathcal{N}), cov(\mathcal{N})$ be $\aleph_1$? Definitions: $non(\mathcal{N})$ is the least size of a set not in $\mathcal{N}$, $add(\mathcal{N})$ is the least size of a family $F \subseteq \mathcal{N}$ whose union is not in $\mathcal{N}$ and $cov(\mathcal{N})$ is the least size of a family of members of $\mathcal{N}$ whose union is $\kappa^{\omega}$. REPLY [4 votes]: $non(\mathcal{N})=\kappa$. If $S \subseteq \kappa^\omega$ is of size $<\kappa$, then all projections $\pi_i(S)\subseteq \kappa$ are also of size $<\kappa$, hence have measure $0$. But then $S\subseteq \prod_{i<\omega} \pi_i(S)$ also has measure $0$. (It seems to me that it would be enough that one of the projections is null.) $cov(\mathcal{N})$ is bounded by the covering number for the (Borel) null ideal on $2^\omega$, which I will call $\nu$. [EDITED. I do not know if $cov(\mathcal{N})$ can be less than $2^{\aleph_0}$. The "standard" model for real valued measurable is obtained by adding $\kappa$ random reals to a model where $\kappa$ is measurable, so in that model we have $\nu=2^{\aleph_0}$, in which case my claim ``$cov(\mathcal{N})\le \nu$'' is irrelevant.] Let $f:\kappa \to 2$ be such that $f^{-1}(\{i\})$ has measure $\frac12$ for $i=0,1$. Define $F:\kappa^\omega\to 2^\omega$ by $F(x)(n) = f(x(n))$ for all $x\in \kappa^\omega$. Then $F^{-1}(U) \subseteq \kappa^\omega$ has the same measure as $U$ for all open sets $U\subseteq 2^\omega$; hence $F^{-1}(N)$ is a null set whenever $N$ is. Let $(N_j:j<\nu)$ be a family of null sets covering $2^\omega$. Define $M_j = F^{-1}(N_j) \subseteq \kappa^\omega$. Then $(M_j:j<\nu)$ is a cover of $\kappa^\omega$ by null sets.<|endoftext|> TITLE: Is the theory of vector bundles just linear algebra done in a suitable topos? QUESTION [11 upvotes]: Sheaves of sets on a space are somehow "parametrized sets". This is the philosophy by which one can do mathematics internal to a sheaf topos (of which theory I admit I know essentially nothing), with the possibility of defining all the usual mathematical concepts, except that in this case -if I understand correctly- the underlying logic will not be classical, e.g. the excluded middle may not hold. What about linear algebra inside a sheaf topos? Which is the relationship between the category of (say finite dimensional) vector spaces as seen within a sheaf topos, and the category of vector bundles? I imagine that "internal" vector spaces will not be exactly the same thing as "external" vector bundles; rather, they will be something like sheaves of vector spaces. Is it correct? To which "internal" notion -if any- does the category of vector bundles correspond? REPLY [11 votes]: I'll gather all the snippets from the various comments here (and mark this post as community wiki). The answer to the question "Is the theory of vector bundles just linear algebra done in a suitable topos?" is a resounding yes. One needs to fix a ring object $\mathcal{O}_X$ in the topos $\mathrm{Sh}(X)$ of set-valued sheaves on the underlying topological space. Then vector bundles over $X$ are, from the internal point of view of $\mathrm{Sh}(X)$, simply finite free $\mathcal{O}_X$-modules, and any intuitionistic theorem about modules applies to them. This holds for arbitrary ringed spaces (or even ringed locales or ringed toposes). If $\mathcal{O}_X$ is the sheaf of continuous real-valued functions, then this notion will yield topological vector bundles. This example is slightly unique in that the sheaf of continuous real-valued functions can be described internally, as the set of Dedekind real numbers, and doesn't have to be put in as an external ingredient. If $X$ is a smooth manifold and $\mathcal{O}_X$ is the sheaf of smooth real-valued functions, then this notion will yield smooth vector bundles. If $X$ is a complex manifold and $\mathcal{O}_X$ is the sheaf of holomorphic functions, then this notion will yield holomorphic vector bundles. If $X$ is a scheme and $\mathcal{O}_X$ is the structure sheaf of $X$, then this notion will yield scheme-theoretical vector bundles. (In this case, one often relaxes the condition "finite free" to "coherent".) In all these cases, the ring object $\mathcal{O}_X$ will look like a local ring from the internal point of view. Therefore all intuitionistic theorems about modules over local rings apply, for instance that the kernel of a surjection between finite free modules is finite free. In particular, this justifies Qiaochu's comment: A module over any ring is dualizable if and only if it is finitely generated and projective; a module over a local ring is finitely generated and projective if and only if it is finite free. Additionally, in all the cases above except the case of unreduced schemes, the ring object $\mathcal{O}_X$ will satisfy the following field condition from the internal point of view: Any element which is not invertible is zero. This field condition is strong enough to imply that any finitely generated module is not not finite free; this translates to the following external statement: Any sheaf of finite type is finite locally free on a dense open subset. (This field condition is intuitionistically not equivalent to the related condition that any nonzero element is invertible. If one wants a ring object with this property, one has to look at the structure sheaf of an appropriate "big topos".) To conclude, I want to mention just two of many applications of this school of thought. Many constructions which work for vector spaces generalize to vector bundles: for instance tensor products, exterior powers, symmetric powers, and so on. These generalizations "behave like their counterparts for vector spaces". One way to make this observation precise, and at the same time to prove it in full generality, is to employ the internal language: Any intuitionistic theorem about modules carries over to the internal setting. It's easy to show that finite free $R$-modules of rank $n$ are the same thing as $\mathrm{GL}_n(R)$-torsors: map a finite free module $V$ to the torsor of bases of $V$. Conversely, map a $\mathrm{GL}_n$-torsor $T$ to $T \otimes_R R^n$. By the internal language, this automatically implies that vector bundles of rank $n$ are the same thing as sheaves of $\mathrm{GL}_n(\mathcal{O}_X)$-torsors. These are classified by the first Čech cohomology group of $\mathrm{GL}_n(\mathcal{O}_X)$. In this way we painlessly rederive the result that vector bundles of rank $n$ are classified by $\check H^1(X, \mathrm{GL}_n(\mathcal{O}_X))$. More details and examples are in these notes of mine.<|endoftext|> TITLE: Local homology of a space of unitary matrices QUESTION [18 upvotes]: Let $U(n)$ denote the unitary group (this is a manifold of dimension $n^2$). Let $$ {\cal D} \subset U(n) $$ denote the subspace of those matrices having a non-trivial $(+1)$-eigenspace. Background: It is known that $\cal D$ has vanishing homology in dimension $n^2$. It is also not difficult to show that $H_{n^2-1}({\cal D}) \cong \Bbb Z$. Furthermore, it is known that ${\cal D}$ has the structure of a finite CW complex of dimension $n^2-1$. For $g\in {\cal D}$, define the local homology by $$ H_{n^2-1}({\cal D}\, |\, g;\Bbb Q) := H_{n^2-1}({\cal D},{\cal D} \setminus g;\Bbb Q)\, . $$ Question: For arbitrary $g\in {\cal D}$ is the rank of this group known? If yes, can anyone provide a reference? Remark: Based on a direct computation when $n=2$, it seems reasonable to conjecture that the rank of $H_{n^2-1}({\cal D}\, |\, g;\Bbb Q)$ is equal to the dimension of the $(+1)$-eigenspace of $g$ (i.e., the multiplicity of the eigenvalue +1). REPLY [10 votes]: Let us first consider the case when $g=e$ is the identity matrix. Let $U$ be an open neighbourhood of the identity in $\mathcal D$. We want to calculate the local homology of $U$ at $e$. We may assume that $U$ is mapped homeomorphically by the (inverse of) the exponential map onto its image in the tangent space of $U(n)$. The tangent space can be identified with the space of skew-Hermitian $n\times n$ matrices. Elements of $\mathcal D$ that are close to the identity correspond under the exponential map to non-invertible skew-Hermitian matrices. So you are asking about the local homology in degree ${n^2-1}$ of the space of non-invertible skew-Hermitian matrices. By Alexander duality, this is isomorphic to the reduced homology in degree $0$ of the space of invertible skew-Hermitian matrices. So we need to count the path components of this space. Such a matrix will have non-zero purely imaginary eigenvalues, of the form $ir$. The path component of a matrix is determined by the number of eigenvalues for which $r>0$. It follows that there are $n+1$ components, so the reduced homology has rank $n$, which confirms your conjecture. Added later: For the general case, suppose $g$ is a unitary matrix that fixes a subspace ${\mathbb C}^k\subset {\mathbb C}^n$. Let ${\mathcal D}_k\subset U(k)$ be the subspace of matrices that fix a non-zero subspace of ${\mathbb C}^k$. I claim that $g$ has an open neighborhood $U\subset\mathcal D$ that is homeomorphic to $U_k\times {\mathbb R}^{n^2-k^2}$ where $U_k$ is an open neigborhood of the identity in ${\mathcal D}_k$, by a homeomorphism that takes $g$ to $e\times 0$. It follows easily that $H_*(U, U\setminus\{g\})\cong H_{*-(n^2-k^2)}(U_k,U_k\setminus\{e\})$, so the general case follows from the special case $g=e$. It remains to prove the claim. Let $l=n-k$ and let ${\mathbb C}^l$ be the orthogonal complement of ${\mathbb C}^k$ in $\mathbb C^n$. Consider the eigenspace decomposition of $g$. One eigenspace is ${\mathbb C}^k$, with associated eigenvalue $1$. The remaining eigenspaces form an orthogonal decomposition of ${\mathbb C}^l$, and their eigenvalues are unit complex numbers different from $1$. We may identify $g$ with the element $(e_k,g_l)\in U(k)\times \{g_l\}$ where $e_k$ is the identity of $U(k)$ and $g_l$ is the restriction of $g$ to ${\mathbb C}^l$. Since it is a submanifold, $g$ has a product neighborhood of the form $V_1\times V_2$ where $V_1$ is a sufficiently small open neighborhood of the identity in $U(k)$ and $V_2\cong{\mathbb R}^{n^2-k^2}$ is a small tubular neighborhood of $V_1$ in $U(n)$. We want to understand the intersection of $V_1\times V_2$ with $\mathcal D$. Consider the eigenspace decomposition of an element of $V_1\times V_2$. It will have eigenspaces of two types: some are very close to ${\mathbb C}^k$ and some are very close to $\mathbb C^l$. The eigevalues of first type are unit complex numbers close to $1$, and eigenvalues of second type are unit complex numbers distinct from $1$. The element belongs to $\mathcal D$ if an only if at least one of the eigenvalues of first type equals $1$. I think it is easy to see from here that an element of $V_1\times V_2$ belongs to $\mathcal D$ if an only if its $V_1$ complonent belongs to $\mathcal D_k$. It follows that $(V_1\times V_2)\cap {\mathcal D}= (V_1\cap {\mathcal D}_k)\times V_2\cong U_k\times {\mathbb R}^{n^2-k^2}$, which is what we wanted to know.<|endoftext|> TITLE: Independence of embedding for higher sheaf cohomology of local cohomology on projective space QUESTION [7 upvotes]: Suppose $Y$ is a projective variety over a field $k$. Fix an embedding $\iota: Y \hookrightarrow \mathbb{P}^n_k$ for some $n$, and consider the local cohomology sheaves $\mathcal{H}^j_Y(\omega_{\mathbb{P}^n})$ of the canonical sheaf on $\mathbb{P}^n$ supported on $Y$. I believe that the sheaf cohomology groups $H^i(\mathbb{P}^n, \mathcal{H}^{n+1-j}_Y(\omega_{\mathbb{P}^n}))$ should be independent of $n$ and of the embedding, for all $i > 0$. (By considering embeddings $Y = \mathbb{P}^0 \hookrightarrow \mathbb{P}^0$ and $Y = \mathbb{P}^0 \hookrightarrow \mathbb{P}^1$, is is not hard to see that this claim fails for $i = 0$.) As in Hartshorne's paper On the de Rham cohomology of algebraic varieties, one strategy for proving such a statement is to consider two embeddings $\iota: Y \hookrightarrow \mathbb{P}^n$ and $\iota': Y \hookrightarrow \mathbb{P}^m$, together with the diagonal embedding $\Delta: Y \hookrightarrow X = \mathbb{P}^n \times \mathbb{P}^m$. If $\pi: X \rightarrow \mathbb{P}^n$ is the projection, we have $\iota = \pi \circ \Delta$. By symmetry, it would be enough to show that, given such data, we must have $H^i(\mathbb{P}^n, \mathcal{H}^{n+1-j}_Y(\omega_{\mathbb{P}^n})) \simeq H^i(X, \mathcal{H}^{n+m+1-j}_Y(\omega_X))$ for all $i > 0$. The sheaves $\mathcal{H}^{n+m+1-j}_Y(\omega_X)$ (indeed, any quasi-coherent $\mathcal{O}_X$-modules supported on $\Delta(Y)$) are $\pi_*$-acyclic: given any $P \in \mathbb{P}^n$, $\pi^{-1}(\{P\}) \cap \Delta(Y)$ contains at most one point, so we can use, e.g., the theorem on formal functions to show the higher direct images have zero stalks. Therefore we have $H^i(X, \mathcal{H}^{n+m+1-j}_Y(\omega_X)) \simeq H^i(\mathbb{P}^n, \pi_*\mathcal{H}^{n+m+1-j}_Y(\omega_X))$ for all $i \geq 0$. To compute the local cohomology sheaves of $\omega_X$ and $\omega_{\mathbb{P}^n}$, we can use their Cousin resolutions (since the sheaves are line bundles, these are injective resolutions). In fact, if $E^{\bullet}(\omega_X)$ (resp. $E^{\bullet}(\omega_{\mathbb{P}^n})$) are these resolutions, we have a quasi-isomorphism $\pi_*E^{\bullet}(\omega_X)[m] \xrightarrow{\sim} E^{\bullet}(\omega_{\mathbb{P}^n})$ of complexes of sheaves on $\mathbb{P}^n$. The reason for this is that the map $\pi$ can be viewed as a "projective space" $\mathbb{P}^m_{\mathbb{P}^n} \rightarrow \mathbb{P}^n$, and so the corresponding trace map is a derived category isomorphism (Proposition III.4.3 of Residues and duality) This trace map takes the form $\mathbf{R}\pi_*\pi^!E^{\bullet}(\omega_{\mathbb{P}^n}) \rightarrow E^{\bullet}(\omega_{\mathbb{P}^n})$, and that the left-hand side coincides with $\pi_*E^{\bullet}(\omega_X)[m]$ is essentially worked out on pp. 31-32 of Hartshorne's de Rham paper. It follows that we can use $\pi_*E^{\bullet}(\omega_X)[m]$ as an injective resolution of $\omega_{\mathbb{P}^n}$. Then we get $\mathcal{H}^{n+1-j}_Y(\omega_{\mathbb{P}^n}) = h^{n+m+1-j}(\underline{\Gamma}_Y(\pi_*E^{\bullet}(\omega_X)))$, where $\underline{\Gamma}_Y$ is the sheafified "sections with support on $Y$" functor. On the other hand, since $E^{\bullet}(\omega_X)$ is an injective resolution of $\omega_X$, we can compute $\pi_*\mathcal{H}^{n+m+1-j}_Y(\omega_X)$ as $h^{n+m+1-j}(\pi_*\underline{\Gamma}_YE^{\bullet}(\omega_X))$. Here I've used the fact that $\pi_*$ and cohomology objects commute when applied to a complex of quasi-coherent sheaves on $X$ supported on $\Delta(Y)$, again by the acyclicity argument. So finally, we are comparing the sheaves $h^{n+m+1-j}(\pi_*\underline{\Gamma}_YE^{\bullet}(\omega_X))$ and $h^{n+m+1-j}(\underline{\Gamma}_Y(\pi_*E^{\bullet}(\omega_X)))$ on $\mathbb{P}^n$. The functors $\underline{\Gamma}_Y$ and $\pi_*$ do not commute, and in fact the global sections of these sheaves need not be the same. Is there a way to see that their higher sheaf cohomologies on $\mathbb{P}^n$ are the same? In general, is there a homological-algebraic gadget to handle the situation of complexes of sheaves which are not quasi-isomorphic, but whose cohomology objects should have identical higher sheaf cohomology? Edited to add (5/4) If $k$ has characteristic $p > 0$, or if $Y$ is smooth and $k$ has characteristic zero, I believe the independence holds. This is because the $k$-dimension of $H^i(\mathbb{P}^n, \mathcal{H}^{n+1-j}_Y(\omega_{\mathbb{P}^n}))$ is the Lyubeznik number $\lambda_{i+1,j}(Y)$ of $Y$, known already to be independent of the embedding in these cases. The embedding-independence of Lyubeznik numbers in characteristic zero for singular $Y$ is still open, and the above question comes from an attempt to attack this in a characteristic-free way (of course, it says nothing about $\lambda_{0,j}(Y)$ or $\lambda_{1,j}(Y)$, which are sure to be harder). REPLY [2 votes]: It is false! As mentioned in the edit, a positive answer to this question would imply the Lyubeznik numbers $\lambda_{i,j}$ of a projective scheme are independent of the defining projective embedding for $i \geq 2$. The recent preprint https://arxiv.org/abs/1803.07448 by T. Reichelt, M. Saito, and U. Walther, gives counterexamples (even equidimensional ones) in characteristic zero.<|endoftext|> TITLE: "Almost Hankelized" numerical Vandermonde QUESTION [8 upvotes]: One of the more utilized determinant is that of Vandermonde's $$\begin{vmatrix} 1&x_1&x_1^2&\dots&x_1^{n-1}\\ 1&x_2&x_2^2&\dots&x_2^{n-1}\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ 1&x_n&x_n^2&\dots&x_n^{n-1}\\ \end{vmatrix}=\prod\limits_{1\leq i TITLE: A combinatorial expression of Hall-Littlewood polynomials QUESTION [8 upvotes]: This is related to the question Hall-Littlewood functions and functions on the nilpotent cone, and arises in the construction of Coulomb branches of gauge theories. The motivation is explained at the bottom. Let us prepare some notation. Let $\lambda$ be a dominant coweight of $GL(N)$, i.e., tuples of (not necessarily positve) integers $\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_N$. We set $|\lambda| = \lambda_1+\dots+\lambda_N$, and define $b_\lambda(t)$ (denoted by $P_{U(N)}(t,\lambda)$ in 1403.0585) by $$ \prod_{i\in\mathbb Z} \varphi_{m_i(\lambda)}(t), \qquad \varphi_r(t) = (1-t)(1-t^2)\cdots (1-t^r), $$ where $m_i(\lambda)$ denotes the number of times $i$ occurs as a part of $\lambda$. This is the inverse of the Poincare polynomial of the classifying space of the stabilizer of $\lambda$ in $GL(N)$ in view of 1601.03586. Let $\lambda^1$, $\lambda^2$, $\dots$, $\lambda^{N-1}$ be dominant coweights of $GL(N-1)$, $GL(N-2)$, $\dots$, $GL(1)$. We define $ \Delta(\lambda,\lambda^1,\dots,\lambda^{N-1}) $ by $$ \frac12\left( \sum_{j=1}^{N-1} \sum_{i,i'} |\lambda^{j-1}_i - \lambda^{j}_{i'}| \right) - \sum_{j=1}^{N-1} \sum_{i < i'} |\lambda^j_i - \lambda^j_{i'}|, $$ where $\lambda^0 = \lambda$. We fix $\lambda$ and consider $$ H[T_{(1^N)}(SU(N)](t,x_1,\dots,x_N,\lambda) := x_1^{|\lambda|} \sum_{\lambda^1,\lambda^2,\dots,\lambda^{N-1}} t^{\Delta(\lambda,\lambda^1,\dots,\lambda^{N-1})} \times \prod_{j=1}^{N-1} \left(\frac{x_{j+1}}{x_j}\right)^{|\lambda^j|} \frac1{b_{\lambda^j}(t)}. $$ This is the monopole formula in 1403.0585 for the special case $\rho=(1^N)$. Now 1403.0585 claims that $$ H[T_{(1^N)}(SU(N)](t,x_1,\dots,x_N,\lambda) = t^{\frac{(N-1)|\lambda|}2 - n(\lambda)} R_\lambda(x_1,\dots,x_N;t) \prod_{i\neq i'} \frac1{1 - x_i^{-1}x_{i'} t} $$ where $$ n(\lambda) = \sum (i-1)\lambda_i, $$ $$ R_\lambda(x_1,\dots,x_N;t) = \sum_{w\in S_N} w\left( x_1^{\lambda_1}\cdots x_N^{\lambda_N} \prod_{i TITLE: Short proof a monoid is a group iff every splitting is right homogeneous QUESTION [5 upvotes]: In the paper "Schreier split epimorphisms between monoids" by Bourn, Nelson, Martins-Ferreira, Montoli and Sobral, Semigroup Forum June 2014, the authors prove a characterization of groups among monoids. I am hoping for a simpler proof of one direction. Definitions. A splitting is a monoid homomorphism $f:A\to B$ with a specified section $s:B\to A$. A splitting is left homogeneous if for every $b\in B$, multiplication on the left by $s(b)$ is bijective $\mathrm{Ker}f\to f^{-1}(b)$. Analogously for right homogeneous. A splitting is homogeneous if it's both left and right homogeneous. Remark. For the characterization, I think it suffices to assume the multiplication maps are surjective, not necessarily bijective. Theorem. For a monoid $B$, TFAE $B$ is a group; Every splitting $A\overset{s}{\leftrightarrows}B$ over $B$ is homogeneous. That 1$\implies$2 is straightforward and appears in proposition 3.4. The converse implication 2$\implies$1 is part of corollary 5.7 and involves an ordeal with internal relations. Is there a simple(r) proof that 2$\implies$1? REPLY [8 votes]: Frieder Ladisch noticed, that even less of 2. is needed than I thought. By his suggestion I make my comment into an answer. Given 2. (in fact even less than one-sided homogeneity of one particular splitting suffices), let $A=B\times B$, let $f:B\times B\to B$ be the projection $(b',b)\mapsto b$, and let $s$ be the diagonal $s(b)=(b,b)$. Given any $b\in B$, multiplication on the left by $s(b)$ on $\operatorname{Ker}(f)\to f^{-1}(b)$ is the map $B\times\{1\}\to B\times\{b\}$ with $(x,1)\mapsto(b,b)(x,1)=(bx,b)$. From left homogeneity of $s$ we only need that $(1,b)\in f^{-1}(b)$ is in the image of this map, i. e. there is an $(x,1)\in B\times\{1\}$ with $(bx,b)=(1,b)$; that is, there is an $x\in B$ with $bx=1$. Thus every element has a right inverse, so $B$ is a group.<|endoftext|> TITLE: Imbedding of a representation of a compact subgroup QUESTION [6 upvotes]: Let $G$ be a compact subgroup of $O(n)$. Let $\rho$ be a continuous finite dimensional representation of $G$. Question Is it true that there exists a continuous finite dimensional representation $\pi$ of $O(n)$ such that $\rho$ is a direct summand of the restriction of $\pi$ to $G$? I am almost sure that this is true. A reference would be very helpful. REPLY [13 votes]: I don't have a reference, but here is a short argument. Assume $G TITLE: Analogy of Gerstenhaber algebra QUESTION [5 upvotes]: We know that the (endomorphism) operad structure on the cochain complex of an associative algebra induces a Gerstenhaber algebra structure on the cohomology. My query is: if a cochain complex has a (endomorphism) dioperad structure, what structure should we expect in the cohomology? Is there an analogy of Gerstenhaber algebra in this case? REPLY [2 votes]: One possible answer is contained in the paper of Victor Ginzburg and Travis Schedler, "Free products, cyclic homology, and Gauss-Manin connection", https://arxiv.org/abs/0803.3655. You will be in particular interested in Section 7.2 there, where the structure that naturally arises on the chain level is discussed. This structure also appears in my recent work with Sergey Shadrin and Bruno Vallette https://arxiv.org/abs/1510.03261, see in particular Section 3.1 of that paper.<|endoftext|> TITLE: Arguments of exponential sums QUESTION [9 upvotes]: Let $p$ be a prime, let $\zeta_p=e^{2\pi i/p}$, let $g\in{\bf F}_p$ be a non-square and let $\chi:{\bf F}_p^*\rightarrow{\bf C}^*$ be a non-trivial character. Then the complex numbers $$ \chi(n)\sum_{r\in{\bf F}_p}\chi(r^2-g)\zeta_p^{nr},\quad(\hbox{$n\in{\bf F}_p^*$}) $$ have the same arguments modulo $\pi$. This is a corollary of some computations I did with modular forms. I wonder whether this result is known or fits into some known context. I would also be interested in a proof that avoids modular forms. REPLY [13 votes]: There are at least 3 ways of seeing this - an elementary way, via the determinants of $\ell$-adic sheaves, and via Katz's calculus of finite field hypergeometric functions. Elementary: Subtraction inside a multipliciative character is too difficult. Let's add a new variable $$=\chi(n) \sum_{r,t \in \mathbb F_p} \chi(t) \delta_{r^2-g,t} \zeta_p^{nr}$$ and detect using additive characters $$=\chi(n) \frac{1}{p} \sum_{r,t,s \in \mathbb F_p} \chi(t) \zeta_p^{nr+s (r^2-g-t)}$$ and separate variables $$ = \chi(n) \frac{1}{p} \sum_{s \in \mathbb F_p}\zeta_p^{-sg} \left( \sum_{t \in \mathbb F_p} \chi(t) \zeta_p^{-st} \right) \left(\sum_{r \in \mathbb F_p} \zeta_p^{n r + sr^2} \right)$$ Both sums vanish for $s=0$ and for other $s$ are standard Gauss sums $$ = \chi(n) \frac{1}{p} \sum_{s \in \mathbb F_p} \zeta_p^{-sg} \left( \chi^{-1}(s) G(\chi) \right) \left(\chi_2(s) \zeta_p^{- \frac{n^2}{4s} } G(\chi_2) \right) $$ $$ = \frac{ G(\chi) G(\chi_2) }{p} \chi(n) \sum_{s \in \mathbb F_p} \chi_2(s) \chi^{-1}(s) \zeta_p^{ -sg - \frac{n^2}{4s}}$$ We can now calculate the argument with the substitution $s \mapsto - \frac{n^2}{4gs}$, which sends $\zeta_p^{ -sg - \frac{n^2}{4s}}$ to its complex conjugate, $\chi^{-1}(s)$ to its complex conjugate times $\chi^{-1}(\frac{n^2}{4g})$, and $\chi_2(s)$ to its complex conjugate times $\chi_2(g)$, so it sends the whole sum to its complex conjugate times $\chi^{-1}(n^2) \chi(4g)/\chi_2(g)$, and hence the argument of the inner sum is half the argument of $\chi^{-1}(n^2) \chi(4g)/\chi_2(g)$, which when multiplied by $\chi(n)$ is half the argument of $\chi(4g)/\chi_2(g)$ and hence is independent of $n$. Determinant: This family of exponential sum is associated to a rank 2 sheaf (a slight variant of a hypergeometric sheaf). This means the exponential sum corresponds to the trace of Frobenius, a unitary matrix, on a rank two representation. The argument of the trace of a unitary matrix is half the argument of the determinant. So your statement is equivalent to the statement that the determinant of the family of exponential sums is constant. The determinant of an individual exponential sum can be calculated using local epsilon factors. For a one-parameter exponential sum like this, if we only want to calculate the ratios between the different determinants, it can be easier. Because your family of sums can be expressed as a Fourier transform, it is easy to apply Laumon's theory of the $\ell$-adic Fourier transform to calculate the local monodromy of the sheaf, take its determinant, and verify that it is trivial. Because the determinant sheaf has trivial local monodromy it has trivial global monodromy and must be constant. So there are many different determinant formulas like this that can be proven, especially for simple Fourier transforms. Only for rank $2$ sheaves will they give argument formulas. For example, the graph of $\chi(n) \sum_r \chi(r^d-g) \zeta_p^{nr}$ in the complex plane should form a $d$-pointed star (where a $2$-pointed star is a line segment, i.e. a set of bounded numbers with fixed argument modulo $\pi$.) Hypergeometric calculus - TBC<|endoftext|> TITLE: Which topological spaces contain dense simply connected subspace? QUESTION [7 upvotes]: And when can this subspace be chosen to be open? As the answer to this question indicates, any manifold contains an open dense subset, which is homeomorphic to $\mathbb{R}^{n}$, and so for manifolds the stronger of the two properties holds. I wonder how widespread this phenomenon is. Note that simply connected spaces are assumed to be connected, and so we only consider connected topological spaces. Also feel free to narrow the class of spaces further, e.g. add local path connectedness, local compactness, etc. REPLY [3 votes]: Any real semialgebraic set $X \subset \mathbb{R}^N$ has a dense, open subset that is a submanifold: just take the complement of its singular set. In fact, the singular set is Zariski closed in $X$, hence nowhere dense in the euclidean topology of $X$. Using your remark about manifolds, it follows that any semialgebraic set contains an open, dense subspace homeomorphic to $\mathbb{R}^n$, with $n = \dim X$.<|endoftext|> TITLE: Solve the functional equation $f(4x(1-x))=\sin(\pi f(x))$ to find an invariant measure of a dynamical system $x_{n+1}=\sin(\pi x_{n})$ QUESTION [7 upvotes]: During the investigation of my thesis I found the following problem: I need find a injective function $f$ such that $$f(4x(1-x))=\sin(\pi f(x))\tag{1}$$ and $f(0)=0$ and $f(1)=1$. Remark: I have tried to solve this equation but I have only reached a reformulation: $$ \frac{4(1-2y)f'(4y(1-y))}{\sqrt{1-f^{2}(4y(y-1))}}=\pi f'(y) .$$ You might think that this problem is not suitable for this site but for me it is important because it arises in trying to address the following situation. The situation: This question is motivated by a problem of Ergodic Theory in my thesis, the problem is find invariant measure of dynamical system $$x_{n+1}=\sin(\pi x_{n}).\tag{2}$$ For this purpose it is sufficient to find its invariant density. In this sense, the idea is find a function $f$ injective such that satisfies (1) with $f(0)=0$ and $f(1)=1$. If that function existed then we have a change of coordinates given by $$x=f(y).$$ Therefore, substituting in (2) we have $f(y_{n+1})=\sin(\pi f(y_{n}))$, but by (1) we have $$ f(y_{n+1})=f(4 y_{n}(1-y_{n})) .$$ Since we assume that $f$ injective then $$y_{n+1}=4y_{n}(1-y_{n}) \tag{3}.$$ But we know the invariant probability density function of dynamical system (3) is the function $$\rho(x)=\frac{1}{4\sqrt{x(1-x)}}.$$ Therefore, the invariant probability density function $g$ of dynamical system (1) is the function $$g(x)=\left|\frac{df(x)}{dx}\right|\rho(x).$$ Therefore, the whole problem is reduced to finding $f$. REPLY [6 votes]: Following up on Anthony Quas's post, let $g(x) = 4x(1-x)$ and $h(x) = \sin(\pi x)$. The equations $g^{(n)}(x) = 1/2$ and $h^{(n)}(x) = 1/2$ each have $2^n$ roots in $[0,1]$. If we sort those roots in order as $y^n_1 < y^n_2 < \cdots < y^n_{2^n}$ and $z^n_1 < z^n_2 < \cdots < z^n_{2^n}$, we should have $f(y_i) = z_i$. To see that there is unlikely to be a closed form solution, note that $y^n_1$ behaves like $c_1 4^{-n}$ for some constant $c_1$ and $z^n_1$ behaves like $c_2 \pi^{-n}$. So we must have $f(y) \sim c_3 y^{\log(\pi)/\log(4)}$ as $y \to 0$. Similar arguments should show that we have $f(y^n_i+\delta) - z^n_i \sim c \delta^{\log(\pi)/\log(4)}$ as $\delta \to 0$ for every $y^n_i$ (with the constant $c$ depending on $(n,i)$). This is not how a nice function behaves! Here is a plot, and list of values, for $n=6$: 0.0002 0.0005 0.0014 0.0033 0.0038 0.0079 0.0074 0.0132 0.0121 0.0211 0.0181 0.0282 0.0252 0.0369 0.0335 0.0465 0.0429 0.0604 0.0534 0.0716 0.0650 0.0832 0.0776 0.0942 0.0912 0.1092 0.1058 0.1220 0.1214 0.1375 0.1379 0.1546 0.1552 0.1790 0.1734 0.1974 0.1924 0.2153 0.2121 0.2310 0.2325 0.2505 0.2536 0.2659 0.2752 0.2833 0.2974 0.3012 0.3201 0.3262 0.3432 0.3454 0.3666 0.3651 0.3904 0.3835 0.4145 0.4082 0.4388 0.4292 0.4632 0.4543 0.4877 0.4813 0.5123 0.5187 0.5368 0.5457 0.5612 0.5708 0.5855 0.5918 0.6096 0.6165 0.6334 0.6349 0.6568 0.6546 0.6799 0.6738 0.7026 0.6988 0.7248 0.7167 0.7464 0.7341 0.7675 0.7495 0.7879 0.769 0.8076 0.7847 0.8266 0.8026 0.8448 0.821 0.8621 0.8454 0.8786 0.8625 0.8942 0.878 0.9088 0.8908 0.9224 0.9058 0.9350 0.9168 0.9466 0.9284 0.9571 0.9396 0.9665 0.9535 0.9748 0.9631 0.9819 0.9718 0.9879 0.9789 0.9926 0.9868 0.9962 0.9921 0.9986 0.9967 0.9998 0.9995<|endoftext|> TITLE: Characterization of Krasner analytic functions on the complement of $p$-adic integers QUESTION [7 upvotes]: Let $\mathbb{Z}_p$ be the ring of $p$-adic integers, $\mathbb{Q}_p$ the field of fractions of $\mathbb{Z}_p$, and $\mathbb{C}_p$ the completion of the algebraic closure of $\mathbb{Q}_p$. Let $v_p$ be the $p$-adic valuation on $\mathbb{C}_p$ with $v_p(p)=1$. For each $m\ge1$ define the set $$ A_m=\mathbb{C}_p\setminus\left( \bigcup_{a=0}^{p^m-1} \left\{x\in\mathbb{C}_p:v_p(x-a)\ge m\right\} \right). $$ Then we can write $\mathbb{C}_p\setminus\mathbb{Z}_p=\cup_{m\ge1}A_m$. A Krasner analytic function $f$ on $\mathbb{C}_p\setminus\mathbb{Z}_p$ is a function $f:\mathbb{C}_p\setminus\mathbb{Z}_p\to\mathbb{C}_p$ such that, for each $m\ge1$, $f$ restricted to $A_m$ is a uniform limit of a sequence of rational functions with poles outside $A_m$. In other words, for all $m\ge1$, $f$ is an analytic element on $A_m$. This makes sense because each $A_m$ is a quasi-connected subset of $\mathbb{C}_p$. Question: Is there any known characterization of Krasner analytic functions on $\mathbb{C}_p\setminus\mathbb{Z}_p$? By a characterization I mean something like the Amice-Fresnel theorem (see Alain Robert's book on $p$-adic analysis, page 348). My main motivation is that the second derivative of Diamond's $p$-adic log-gamma function is an analytic element on $\mathbb{C}_p\setminus\mathbb{Z}_p$ (see Jack Diamond's 1977 paper, Theorem 12), and I was wondering if this is a general phenomenon for some "nice" functions defined by means of the Volkenborn integral (Kubota-Leopoldt sums), as is the case of Diamond's $p$-adic log-gamma function. REPLY [3 votes]: If you impose some growth conditions on the boundary of ${\mathbb C}_p\setminus {\mathbb Z_p}$ to the Krasner-analytic function $F$, plus some invariance under the absolute Galois group of ${\mathbb Q}$, you have a caracterization in the spirit of Amice-Fresnel theorem. I use $p$-adic absolute value instead of $p$-adic valuation. For example if you impose the conditions Suppose that $\sup_{x\in A_m}\vert F(x)\vert_p\leq p^{m}$ for all $m\in {\mathbb N}$ Suppose that the Krasner-analytic function $F$ is ${\rm Gal_{\text{cont}}(\mathbb C}_p/{\mathbb Q}_p)$-equivariant, that is for all $\sigma\in {\rm Gal_{\text{cont}}(\mathbb C}_p/{\mathbb Q}_p)$ and for all $x\in {\mathbb C}_p\setminus {\mathbb Z_p}$ one has $F(\sigma(x))=\sigma(F(x))$. Suppose that $\lim_{\vert x \vert\to \infty}\vert F(x) \vert=0$ Then there exists a ${\mathbb Z}_p$-valued measure, $\mu$, on ${\mathbb Z}_p$ such that: \begin{equation*} F(x)= \int_{t\in{\mathbb Z}_p}\frac{d\mu(t)}{t-x} \end{equation*} (see the references below). Let $e_n^*$ be the measure on ${\mathbb Z}_p$ such that $\int_t\binom{t}{m}de_n^*=\delta_m^n$ ${\scr C}({\mathbb Z}_p,{\mathbb Z}_p)$. It is known that any ${\mathbb Z}_p$-valued measure $\mu$ on ${\mathbb Z}_p$ can be written \begin{equation*} \mu=\sum_{n\geq 0} b_ne_n^*, \ b_n \in {\mathbb Z}_p\, . \end{equation*} From this one get easily, using Mahler's expansion of the $p$-adic continuous function $t\mapsto \frac{1}{t-x}$, that the Krasner-analytic function $F$ subject to conditions 1,2,3 has the following expansion on ${\mathbb C}_p\setminus {\mathbb Z_p}$ \begin{equation*} F(x)=\sum_{n\geq 0} b_n\sum_{k=0}^n(-1)^{k}\binom{n}{k}\frac{1}{k-x}= \sum_{n\geq 0} \frac{b_nn!}{x(x-1)\dots(x-n)} \end{equation*} This result can be extended in many directions. (references: D. Barsky, Transformation de Cauchy $p$-adique et algèbre d'Iwasawa, Math. Ann. 232 (1978), 255-266, or V. Alexandru, C.C. Niţu, M. Vâjâitu, A. Zaharescu, On the norm of Krasner analytic functions with applications to transcendence results, Journal of Pure and Applied Algebra 219 (2015) 4607–4618)<|endoftext|> TITLE: Does each discrete solvable group admit an injective homomorphism to a compact topological group? QUESTION [8 upvotes]: It is well-known that each abelian group admits an injective homomorphism to some compact topological group (for example to its Bohr compactification). Is the same fact true for solvable groups? Question 1. Does every solvable group admit an injective homomorphism to a compact Hausdorff topological group? Equivalently: Question 2. Is the Bohr topology of every discrete solvable group Hausdorff? The Bohr topology is the largest totally bounded group topology on the group. REPLY [2 votes]: According to Proposition 3.3 from Dikranjan and Toller [Topology and its Applications 159 (2012) 2951-2972], the Heisenberg group H_K over an infinite field K of characteritic 0 is not maximally almost periodic. This provides a negative answer to the question even for nilpotent groups of class 2.<|endoftext|> TITLE: Ramsey type theorems and Magidor's forcing QUESTION [6 upvotes]: Consider Prikry's forcing for changing the cofinality of a measurable cardinal into $\omega.$ The forcing has the Prikry property and one can prove this either directly or using Rowbottom's theorem which is a Ramsey type theorem. Now consider Magidor's forcing for changing the cofinality of a large cardinal into $\omega_1$ (or any other uncountable regular cardinal). The proof of Prikry property for it (that I have seen) is direct. My question is: Question. Is there a suitable Ramsey type theorem such that one can use it to present a simplified proof of the Prikry property for Magidor's forcing? Any references, if there are, are appreciated. REPLY [4 votes]: There is an analog of Rowbottom's theorem which can be extracted from the proof of the Magidor forcing: Suppose that $u$ is a sequence of ultrafilters on $\kappa$ of length $\lambda<\kappa$ and $f\colon [\kappa]^{<\omega}\to 2$. Then there are sets $A_\nu\in u(\nu)$ for $\nu<\lambda$ such that whenever $\langle \nu_0,\dots,\nu_{k-1}\rangle$ is a finite sequence of ordinals less than $\lambda$, the function $f$ is constant on the set of increasing sequences $\langle \alpha_0,\dots,\alpha_{k-1}\rangle\in A_{\nu_0}\times\dots\times A_{\nu_{k-1}}$. This can be extended to Radin forcing on sequences of ultrafilters of length less than $\kappa^+$; I'm not sure what could be done for longer sequences.<|endoftext|> TITLE: An ordinary differential equation QUESTION [5 upvotes]: While I was working on a variational problem, I met this equation as its Euler-Lagrange equation, but I cannot solve it: $ x= \frac{af'(x)}{\sqrt{1+af'(x)^{2}}} + \frac{bf'(x)}{\sqrt{1+bf'(x)^{2}}} \ (a\neq b) $. REPLY [7 votes]: You can solve it parametrically as follows: Write $$ x(t) = \frac{at}{\sqrt{1+at^2}}+\frac{bt}{\sqrt{1+bt^2}} $$ and $$ y(t) = c - \frac{\sqrt{1+at^2}+\sqrt{1+bt^2}}{\sqrt{1+at^2}\sqrt{1+bt^2}} $$ where $c$ is a constant. Then this gives the general solution as $c$ varies. If you want an explicit relation between $x$ and $y$, you can eliminate $t$, but it won't be pretty. REPLY [5 votes]: Set $p:=f'(x)$ and $$ \Phi(p)=\Phi_{a,b}:=\frac{ap}{\sqrt{1+ap^2}}+\frac{bp}{\sqrt{1+bp^2}}. $$ The differential equation you wrote can be rewritten as $$ x=\Phi(p). $$ If we could invert $\Phi$, then we could write $$ f'(x)= p=\Phi^{-1}(x). $$ For $a, b>0$ the function $\Phi$ seems to be increasing. The animation below depicts the graphs of $\Phi_{1,t}$ for $t\in [0,6]$ It already shows that the solution blows up in finite time. (Here I think of $x$ as time.) The next animation depicts $\phi_{1,t}$ for $t=-1..0$ and you can see that the injectivity of $\Phi$ is lost for some values of $t$. Remark. Here is an animation of the curve described by Robert Bryant for $a=1$ and $b\in [-0.1,0.2]$, $t\in[-3,3]$<|endoftext|> TITLE: $\omega_2$-sequence of Suslin trees QUESTION [6 upvotes]: Is it possible to have an $\omega_2$-length sequence of ($\omega_1$-)Suslin trees such that if one builds the product of finitely many trees in that sequence, one ends up with a Suslin tree again? The existence of such a sequence of length $\omega$ follows from $\diamondsuit$, as was shown by Jensen. By Shelah and independently Todorcevic, already a Cohen real gives rise to a Suslin tree, so it could be possible that a adding $\aleph_2$ many Cohen reals produces such a sequence. REPLY [5 votes]: The answer is yes, and indeed, one can even have that any countable number of the Suslin trees join to a Suslin tree. To see this, simply force with countable support to add $\omega_2$ many Suslin trees. The forcing to add one Suslin tree has conditions consisting of countable normal $\alpha$-tree, for $\alpha<\omega_1$, and this tree will become an initial segment of the desired generic tree. This forcing is countably closed and isomorphic to $\text{Add}(\omega_1,1)$, adding a Cohen subset of $\omega_1$. This forcing also adds a $\diamondsuit$-sequence, and there is a tight connection between the argument from diamond that there is a Suslin tree and the proof that this forcing adds a Suslin tree. Namely, given any name for a maximal antichain, one undertakes a bootstrapping argument to decide more and more of the antichain, until one has a condition $t$ that decides $A\cap t$ and such that $A\cap t$ is maximal in $t$. Then, one extends $t$ to $\bar t$ with one more level in a way that seals that antichain in any further extension of $\bar t$. So $\bar t$ forces that the antichain is bounded and hence countable. Consider now adding $\omega_2$ many such Suslin trees, with countable support. An essentially similar sealing argument shows that these trees are Suslin, and furthermore, that they are mutually Suslin over each other. If you force with all but one of these trees, with countable support, then the remaining tree will still be Suslin. The reason is that if $\dot A$ is a name for antichain in $T$, the unforced tree, where $\dot A$ is a name in the product of the other trees, then it will be dense in the product of the other tree-forcing conditions that forces that $\dot A$ is bounded. The argument is fundamentally similar to the methods used in my paper Fuchs, Gunter; Hamkins, Joel David, Degrees of rigidity for Souslin trees, J. Symb. Log. 74, No. 2, 423-454 (2009). ZBL1179.03043. (blog post) The point is that to add a generic Suslin tree $T$ and then force with it, is the same as forcing with conditions $(t,b)$ where $t$ is a normal $(\alpha+1)$-tree and $b$ is an element on the top level (determining the generic path). This forcing is very nice, and can be used to show that a generic Suslin tree is Suslin off the generic branch.<|endoftext|> TITLE: How to compute with the Stark conjectures? QUESTION [8 upvotes]: I would like a convenient basis for the elements of a fixed abelian extension $E$ of a real quadratic field $\mathbb{Q}(\sqrt{d})$. The accepted answer to this MO question suggests that the Stark conjectures give explicit generators for $E$ which can then be verified using the computer algebra system PARI/GP. Question: Given $d$, how do I use PARI/GP to find and verify the desired generators? REPLY [14 votes]: The main reference here is the very useful User's Guide, C. Batut, K. Belabas, D. Bernardi, H. Cohen, M. Olivier, "User's Guide to PARI / GP" (2003) Particularly the sections about bnrstark (pp. 108), quadhilbert (pp. 87) and quadray (pp. 88). For more information and examples you may want to look at Roblot's work, for instance Xavier-François Roblot, Brett A. Tangedal, "Numerical Verification of the Brumer-Stark Conjecture" (2000) Xavier-François Roblot, "Checking the Brumer-Stark conjecture using PARI/GP" (2004)<|endoftext|> TITLE: Virtual mixed Tate motives QUESTION [6 upvotes]: Let $\mathbf{Sch}_k$ be the category of $k$-schemes of finite type, and let $K_0(\mathbf{Sch}_k)$ be the Grothendieck ring of $k$-schemes. Let $\mathbb{Z}[\mathbb{L}]$ the subring generated by the virtual Lefschetz motive $\mathbb{L} := [\mathbb{A}^1_k]$. Is every element of $\mathbb{Z}[\mathbb{L}]$ the class of a $k$-scheme (or at least a $k$-constructible set) ? It might be convenient to recall that $K_0(\mathbf{Sch}_k)$ is generated by isomorphism classes of $k$-schemes $[X]$, and that $$[X] = [X \setminus C] + [C]$$ whenever $C$ is a closed subscheme in $X$; the product is given by $$[X] \cdot [Y] = [X \times Y].$$ REPLY [6 votes]: Here is a proof of the fact that over an infinite field $f(\mathbb{L})$ is a class of a scheme if and only if the leading term of $f$ is positive, as suggested by Will Sawin. Example. For $f(t)=a_nt^n+a_{n-1}t^{n-1}+\dots+a_0$ with $a_n>0$ throw away from affine space of dimension $n$ planes $\mathbb{A}^i$ in amount $-a_i$ for each negative $a_i$ such that these planes are pairwise disojoint(this is possible due to the infinitness of the base field). Then take the disjoint union of the resulting complement with $a_n-1$ copies $\mathbb{A}^n$ and with $a_j$ copies of $\mathbb{A}^j$ for each positive $a_j$. The variety which we get clearly equals to $f(\mathbb{L})$ in $K_0$(one can construct an irreducible example by doing blow-ups instead of disjoint unions). All other polynomials do not give a class of a scheme. 1) First, consider the case $\mathrm{char}\, k=0$ and reduce by a Lefschetz principle type argument to the case $k=\mathbb{C}$. Assume that $[f(\mathbb{L})]=[X]$ and $a_n<0$. This identity is a linear combination of a finite number of standard identities. There exists a subfield $F$ of $k$, finitely generated over $\mathbb{Q}$ such that all schemes involved in these identities are defined over $F$. Thus, we have $[p(\mathbb{L})]=[X_F]$ already in $K_0(Sch_F)$. Then embedding $F$ into $\mathbb{C}$ and base changing to $\mathbb{C}$ we arrive at the case $k=\mathbb{C}$. Over $\mathbb{C}$ we can use the following motivic measure(strictly speaking, this formula defines it only on $K_0(Var_{\mathbb{C}})$ but if I got the definition of $K_0(Sch)$ right, the relations imply $[X]=[X_{red}]$ -- there is still something to be said about not-separatedness, but let's omit that): $$\mu([X])=\sum_{k,p,q}(-1)^k\dim H^k_c(X,\mathbb{C})^{p,q}u^pv^q\in \mathbb{Z}[u,v]$$ where by $H^k_c(X,\mathbb{C})^{p,q}$ I mean the piece of weight $(p,q)$ in the MHS on $H^k_c(X)$. For an irreducible variety of dimension $n$ the leading term of $\mu([X])$ is $u^nv^n$ so it is at least positive for arbitrary varirty. But for $f(\mathbb{L})$ this motivic measure gives $f(uv)$, qed. 2)The case $\mathrm{char}\, k=p>0$. Exactly as in the first case, we may assume that we have a counterexample defined over a ring which is a finite extension of $\mathbb{F}_p[t_1,\dots,t_n]$(say, Noether normalization). Specializing at some values of $t_i$ we arrive at the case $k=\mathbb{F_q}$ where we can use the motivic measure suggested by THC: $$\mu_n([X])=\#X(\mathbb{F}_{q^n})$$ and for $f(\mathbb{L})$ with negative leading coefficient there exists a big enough $n$ such that $\mu_n(f(\mathbb{L}))<0$. For finite fields everything seems more delicate -- a necessary condition is $f(q^{nm})\geq f(q^n)\geq 0$ for any $n,m$.<|endoftext|> TITLE: Coordinate free isomorphism between $d+1$-dimensional antisymmetric rank $2$ tensors and $d$-dimensional symmetric rank $2$ tensors QUESTION [8 upvotes]: The space of $(d+1)$-dimensional antisymmetric matrices has the same dimension as the space of $d$-dimensional symmetric matrices, $\frac12d(d+1)$. There are isomorphisms between the two spaces, e.g. for $d=2$ $$ \begin{pmatrix} a & b\\ b & c \end{pmatrix} \leftrightarrow \begin{pmatrix} 0 & a & b\\ -a & 0 & c\\ -b & -c & 0 \end{pmatrix}. $$ This isomorphism is obviously only defined after we have chosen basis vectors. My question: given a space of $(d+1)$-dimensional antisymmetric rank $2$ tensors and a space of $d$-dimensional symmetric rank $2$ tensors, what is the least that we can get away with specifying in order to provide an isomorphism between them? In the above example we needed bases. Ideally it would be desirable to have something coordinate free, but my intuition tells me that this may not be possible, in particular because they transform differently. REPLY [7 votes]: Here is a revised partial answer and some comments: It seems that you are asking for some kind of isomorphism between $S^2(\mathbb{F}^d)$ and $\Lambda^2(\mathbb{F}^{d+1})$ that would 'have the greatest symmetry', in the sense that it would commute with some group action on $\mathbb{F}^d$ and $\mathbb{F}^{d+1}$, the bigger the better, so that it would be as independent of a choice of basis as possible. Ryan's suggestion of a basis-dependent identification has a kind of symmetry based on the symmetric group $S_d$, but it isn't induced by a corresponding symmetry of the underlying vector spaces $\mathbb{F}^d$ and $\mathbb{F}^{d+1}$ because, if you permute the bases of these spaces, you'll see that the $x_ix_j$ permute among themselves, but the $\mathrm{d}x_i\wedge\mathrm{d}x_j$ permute and change signs at the same time, so it's not really based on the geometry of the underlying vector spaces. It seems that what you want to find a subgroup $H\subset\mathrm{GL}(d,\mathbb{F})\times \mathrm{GL}(d{+}1,\mathbb{F})$ such that $S^2(\mathbb{F}^d)$ and $\Lambda^2(\mathbb{F}^{d+1})$ would be isomorphic as $H$-modules, and you'd want $H$ to be maximal with this property. It is an interesting questions as to what the maximal such subgroups $H$ could look like. Here is an example of the kind of thing that you might consider as an answer to your question: For a $2$-dimensional vector space $V$, there is a natural isomorphism $$ \Lambda^2\bigl(S^2(V)\bigr) = S^2(V)\otimes\Lambda^2(V), $$ i.e., this isomorphism is $\mathrm{GL}(V)$-equivariant, and this provides an isomorphism between the skew-symmetric $2$-forms on $S^2(V^*)\simeq \mathbb{F}^3$ and the quadratic forms on $V^*\simeq \mathbb{F}^2$, twisted by the determinant $\Lambda^2(V)$, a $1$-dimensional vector space. You can get rid of this twisting if you fix a volume form on $V$ and consider only volume perserving automorphisms, i.e., $\mathrm{SL}(V)$. Alternatively, you can consider $H = \mathrm{SL}(V)\times \mathbb{F}^\times$ and let the $\mathrm{SL}(V)$ factor act on $S^2(V)$ and $V$ as above, but let an invertible element $\lambda\in F^\times$ act on $V\times S^2(V)$ as $(\lambda,\lambda)$ (i.e., with equal weights, instead of as $(\lambda,\lambda^2)$), and this will give an $H$ acting on $V\times S^2(V)$ for which $S^2(V)$ and $\Lambda^2\bigl(S^2(V)\bigr)$ are isomorphic as $H$-modules on the nose (with no $\Lambda^2(V)$-twisting). This generalizes to the case of all $d$, because, by the Clebsch-Gordan formulae, there is an isomorphism of $\mathrm{GL}(V)$-modules $$ S^2\bigl(S^{d-1}(V)\bigr)\otimes\Lambda^2(V) = \Lambda^2\bigl(S^{d}(V)\bigr). $$ Since $S^{d-1}(V)\simeq\mathbb{F}^d$ and $S^{d}(V)\simeq\mathbb{F}^{d+1}$, this seems to correspond to the kind of thing you are thinking about. (As before, you can get rid of the $\Lambda^2(V)$-factor by restricting to $\mathrm{SL}(V)$ and instead making the scalars act with equal weights on $S^{d-1}(V)$ and $S^{d}(V)$, instead of with their 'natural' weights $d{-}1$ and $d$. The vector spaces $S^{d-1}(V)$ and $S^{d}(V)$ remain irreducible under this action.) Note, however, that when $k\ge2$, the module $S^2\bigl(S^{k}(V)\bigr)$ is no longer an irreducible $\mathrm{SL}(V)$ module, instead, we have $$ S^2\bigl(S^{k}(V)\bigr)\simeq \Lambda^2\bigl(S^{k+1}(V)\bigr) \simeq \bigoplus_{0\le j\le k/2} S^{2k-4j}(V). $$ A good test case would be $d=3$. I suspect that, in this case, the scalar-modified embedding as above of $H=\mathrm{GL}(V)$ into $$ \mathrm{GL}\bigl(S^2(V)\bigr)\times \mathrm{GL}\bigl(S^3(V)\bigr) \simeq \mathrm{GL}(3,\mathbb{F})\times \mathrm{GL}(4,\mathbb{F}) $$ makes $\mathrm{GL}(V)$ into a maximal subgroup of $\mathrm{GL}(3,\mathbb{F})\times \mathrm{GL}(4,\mathbb{F})$ for which $S^2(\mathbb{F}^3)$ and $\Lambda^2(\mathbb{F}^4)$ are isomorphic as $H$-modules, and I can show that any such $H$ that is a connected Lie group that contains a simple Lie subgroup must be this one, up to isomorphism. What happens in higher dimensions, I don't know.<|endoftext|> TITLE: History of deletion-contraction formula QUESTION [10 upvotes]: The following is known as deletion-contraction formula: Assume $\Gamma$ is a connectted graph with edge $\rho$ then $$t(\Gamma)=t(\Gamma\backslash\rho)+t(\Gamma/\rho),$$ where $\Gamma\backslash\rho$ is $\Gamma$ with the edge $\rho$ removed and $\Gamma/\rho$ obtained from $\Gamma$ by contracting $\rho$ and $t(\Gamma)$ is the number of spanning trees in $\Gamma$. The statement appear as Proposition 4.9 in Graph Theory by Bondy Murty with no reference. (Haghighi and Bibak claim that it is due to Kirchhoff, but I was not able to find the statement in his paper on the subject) I wonder who made this observation first, or at least what is the earlies reference to this. REPLY [8 votes]: Not exactly a reference, but I found some discussion of this formula in Bill Tutte's graph-theoretic memoirs, Graph Theory as I Have Known it (Oxford, 1998). In Chapter 5 ("Algebra in Graph Theory"), he writes: In their study of complexities or tree-numbers the Four made much use of the recursion formula $$C(G) = C(G'_A) + C(G''_A)$$ When I was doing my PhD research I began to collect other functions of graphs that satisfied similar recursions. (p.53) The "Four" that Tutte is referring to here are: R. L. Brooks, C. A. B. Smith, A. H. Stone, and W. T. Tutte. (He mentions that they all met as students at Cambridge in the Trinity Mathematical Society.) So perhaps this deletion-contraction formula was folklore at the time, or at least "Fourlore". I think that the connection to Kirchhoff's theorem is indirect, corresponding to the fact that the "complexity" and the "tree-number" of a graph coincide. Tutte explains in Chapter 1 ("Squaring the square"): We learned that Kirchhoff's Laws for a network $N$ were associated with a special matrix $K(N)$ called, by us at least, the Kirchhoff matrix of $N$. [...] We learned that the determinant of the matrix $K_i(N)$ obtained from $K(N)$ by striking out the $i$th row and column was independent of $i$. We decided to call its value the "complexity" $C(N)$ of $N$. (p.7) [...] By expanding a determinant it can be shown that the number of spanning trees of $G$ is the complexity $C(G)$ of $G$. For a general electrical network $N$ the complexity $C(N)$ proves to be the sum over all the spanning trees of $N$ of the products of the conductances of their edges. This result is known as the Matrix-Tree Theorem, and it goes back to Kirchhoff. (p.11)<|endoftext|> TITLE: Can the full shift be embedded in a flow? QUESTION [9 upvotes]: Write $I=[0,1]$, and let $S$ be the shift on $X=\{ (x_n)_{n\in\mathbb Z} : x_n\in I^k \}$. Is there a flow $\phi_t$ on $X$ with $\phi_1=S$? Here I require that $\phi_t$, for fixed $t$, is at least a homeomorphism on $X$, with respect to the product topology. I'm really mainly interested in $k=2$ (though I'd also be interested in the variant where we replace the cube by a torus of that dimensions, which should make it easier to flow around). For $k=1$, the answer is no, for the trivial reason that $p$-periodic sequences must be invariant under $\phi_t$, so we won't be able to evolve $(\ldots 010101\ldots)$ towards its shift without crossing through a constant sequence, but these are invariant. For $k\ge 2$, this specific problem disappears, but of course there could be other obstructions, even of this nature, since $\phi_t$ has to preserve any property that can be described dynamically in terms of $S$ (such as being an almost periodic sequence). This sounds like the kind of question someone must have thought about already, so will perhaps be easy for the experts. REPLY [8 votes]: For odd $k$, the shift in $X^k$ has no square root (and hence does not lie in a 1-parameter subgroup). Indeed, the the set of 2-periodic points can be identified with $(I^k)^2$, where the shift acts as $(x,y)\mapsto (y,x)$. It would then preserve the set of fixed points, namely the diagonal $(x,x)$, so any root should preserve the subset of points of period 2, the set $M^k_2$ or pairs of distinct points of $\mathbf{R}^n$. There is an obvious homotopy retract of the latter to the set of pairs $(x,-x)$ where $x$ ranges over the unit sphere $\mathbf{S}^{k-1}$, commuting with the flip (first homotope the center to 0 and then renormalize...). Here the flip can be identified with the antipode map. Since $k$ is odd, the antipode of the $(n-1)$-sphere is orientation-reversing and hence its action of the antipode on the real cohomology $H^{k-1}(\mathbf{S}^{k-1},\mathbf{R})\simeq H^{k-1}(M^k_2,\mathbf{R})\simeq\mathbf{R}$ is given by sign change and hence has no square root. So the flip of $M^k_2$ has no square root for odd $k$. For even $k$, $I^k$ is a square and hence the shift map has an obvious square root (and we can even find a 1-parameter subgroup interpolating the flip in $M^k_2$), so another argument would be needed.<|endoftext|> TITLE: Is the configuration space of ordered triples of distinct points in the four-edge banana graph homotopy equivalent to a surface of genus 13? QUESTION [10 upvotes]: If $X$ is a topological space, write $C_n(X)$ for the configuration space of distinct ordered tuples of points in $X$: $$C_n(X) = \{(x_1, \ldots, x_n) \in X^n \mbox{ so that $i \neq j \implies x_i \neq x_j$ } \}.$$ Define the banana graph $\beta_k$ to be the suspension of the discrete space $\{1, \ldots, k\}$. Finally, write $\Sigma_g$ for the closed orientable surface of genus $g$. A) If $X = \beta_4$ is the four-edge banana graph, is there a homotopy equivalence $C_3(X) \simeq \Sigma_{13}$? As background, the thesis of Aaron Abrams (available on his website http://home.wlu.edu/~abramsa/publications/index.html) gives homotopy equivalences $$ \begin{align*} C_2(K_5) & \simeq \Sigma_6 \\ C_2(K_{3,3}) & \simeq \Sigma_4. \end{align*} $$ Abrams also proves that a related combinatorial approximation of configuration space $C_n^{approx}(X)$ satisfies $$ \begin{align*} C_3^{approx}(K_5) & \simeq \Sigma_{16} \\ C_4^{approx}(K_{3,3}) & \simeq \Sigma_{37}, \end{align*} $$ where $K_5$ is a complete graph and $K_{3,3}$ is complete bipartite. (Thank you to user j.c. for pointing out the difference between $C_n^{approx}$ and $C_n$, which I had not understood.) So this sort of thing has happened before! Also, using an explicit simplicial model of $C_3(\beta_4)$ that Sage tells me has 336 vertices and 840 facets, I am able to compute that $$ H^*(C_3(\beta_4) \, ; \mathbb{Z}) \simeq H^*(\Sigma_{13} \, ; \mathbb{Z}), $$ and that the cup product in rational cohomology gives a non-degenerate pairing on $H^1$. B) How might I check if a finite simplicial complex has the homotopy type of some $\Sigma_g$? I say "might" because it's probably not computable in general. Finally, I'll ask what might be a tricky question: C) For what graphs $G$ and $n \in \mathbb{N}$ does $C_n(G)$ have the homotopy type of a surface? At Ryan Budney's suggestion, I have collapsed many free faces. (The algorithm I used comes from the paper https://arxiv.org/pdf/1303.6422.pdf by Benedetti and Lutz). The result is a complex with 120 vertices and 288 facets. It is a pseudomanifold! Sage computes a presentation for $\pi_1$ $$ \langle a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z \rangle / \langle \omega \rangle $$ where $\omega$ is the impressive-looking word $$ y^{-1}a^{-1}bf^{-1}en^{-1}ml^{-1}x^{-1}wo^{-1}vkg^{-1}hpu^{-1}tdh^{-1}ic^{-1}e^{-1}car^{-1}qlod^{-1}fq^{-1}i^{-1}s^{-1}b^{-1}xt^{-1}m^{-1}p^{-1}rsk^{-1}uznjv^{-1}w^{-1}yz^{-1}j^{-1}g $$ in which every variable appears exactly twice, once with an inverse. So $\pi_1$ is also correct! REPLY [4 votes]: It has come to my attention that this example has been computed independently (and more directly) by Safia Chettih and Daniel Lütgehetmann. The result is given as Proposition 4.3 in v2 of their preprint https://arxiv.org/abs/1612.08290v2 which appeared in July 2017. They use a combinatorial model due to Swiatkowski to show that $C_3(B_3)$ is a surface of genus 13 glued from 144 2-cells.<|endoftext|> TITLE: Inconsistencies in MathSciNet citation database QUESTION [14 upvotes]: Starting this January, I observe more and more inconsistencies in the MathSciNet citation database. A typical situation is that the number of citations shown in the description of a paper differs from the number of matches that one actually gets by following the link. For example, take papers by Gerd Faltings (sorted by the number of citations). MR0718935 is listed with 347 citations, but has only 340 matches; MR1083353 is listed with 304 citations, but has only 303 matches; MR1463696 is listed with 133 citations, but has only 129 matches, and so on. Moreover, the number of matches depends on the mirror. The above numbers are provided by the Providence cite. The Bielefeld mirror gives, respectively, 340 (the same number), 302 (one less) and 131 (two more). Is there any good explanation for this phenomenon? REPLY [12 votes]: As @ZachTeitler pointed out by quoting my blog post, one of our new features acted a bit like a bug: related papers. We have made adjustments to avoid the "counting with multiplicity" issue, but some unexpected cases still slipped in. One of the examples involving Faltings given by @AlekVainshtein had related papers involved. Some of Vainshtein's other examples uncovered a different issue, which had to do with timing of updates. Updates directly related to individual items in the database are done (almost) continuously. Some other tables are updated continuously at Math Reviews, but pushed out to the web less frequently. We have now synced them up. As @AlexSuciu pointed out, there were some growing pains when we released the updates to MathSciNet. We completely changed the search software. We had tested it at both the Math Reviews offices and the AMS headquarters in Providence, but nothing provides a stress test like releasing it publicly. People gave us great feedback, especially about citations, that allowed us to make improvements. If you find anything odd, be it citations or something else, you can write to us at either msn-support@ams.org or mathrev@ams.org. The first address doubles for problems related to access, including subscription problems. The second address doubles as the address reviewers use for communicating with the editorial staff. Both addresses have some traffic direction that will get messages to the right person.<|endoftext|> TITLE: Can the "real" Peano Arithmetic be inconsistent? QUESTION [7 upvotes]: Assuming $\text{PA}$ is consistent. Then $\text{PA} + \neg\text{Con}(\text{PA})$ is consistent and have a model, say $M$. We know $M$ must be nonstandard, in which case, there is a nonstandard proof of $0=1$ from ($M$'s version of) $\text{PA}$ in $M$. The fact that $M$'s version of $\text{PA}$ is different from the "real" $\text{PA}$ make the assertion $$ M\vDash\text{PA} + \neg\text{Con}(\text{PA}) $$ kind of less stunning than we might expect it to be. My question is if we can compose some other expressions of $\text{PA}$ so that if $M\vDash\neg\text{Con}(\text{PA})$ then there is a (nonstandard) proof of absurdity making use of only the standard part of $\text{PA}$ and logic axioms. If it is impossible, how to prove it? REPLY [7 votes]: It seems that the Feferman-style description of PA will exhibit your requirements. Specifically, consider the theory $P$ defined as follows. Begin to enumerate the usual PA axioms, but include the next axiom in $P$ only if doing so keeps $P$ consistent. Never add an axiom to $P$ that would make it inconsistent. Since PA proves of every finite subset of PA that it is consistent, it follows that PA proves of any particular axiom of PA, that it is in $P$. In this sense, this theory $P$ is the same as PA, just described in a different way. But because of how it is described, it follows that PA proves $\text{Con}(P)$. We never include an axiom in $P$ that would enable it to prove a contradiction. Therefore, there is no model $M\models PA+\neg\text{Con}(P)$, and so the theory vacuously has your desired property. Of course, notice that all instances of your requested property must be vacuous like this, because if the proof of a contradiction inside $M$ were somehow forced to be standard, then it would be actual proof of a contradiction, and so $M$ couldn't exist in the first place, if it is a model of this theory. So it seems to me that the only way to get your situation is in the vacuous way that the Feferman theory achieves it.<|endoftext|> TITLE: Elliptic curves with the same mod $p$ representation QUESTION [8 upvotes]: What is the largest prime number $p$ for which one knows examples of nonisogenous elliptic curves $E_1$ and $E_2$ over $\mathbb{Q}$ with isomorphic mod $p$ Galois representations: $E_1[p] \cong E_2[p]$? I do not require the isomorphism to preserve the Weil pairing. REPLY [15 votes]: The largest known is for $p = 17$. This can be found by searching Cremona's tables. In particular, in this paper, Tom Fisher mentions the examples of curves 3675b1 and 47775b1, which are $E_{1} : y^{2} + xy + y = x^{3} + x^{2} - 393x - 9654$ and $E_{2} : y^{2} + xy + y = x^{3} + x^{2} + 398095697x + 4077685984826$. The conjecture that if $p > 17$, there are no such curves is known as the Frey-Mazur conjecture. (Frey and Mazur originally asked whether this can occur for $p > 5$, because in that situation the curve $X_{E}(p)$, the moduli space of elliptic curves with the same mod $p$ Galois representation as $E$, has genus $> 1$.)<|endoftext|> TITLE: Is every ordinal the nimber of a ring? QUESTION [9 upvotes]: This question is about the game of Noetherian rings, see MO/93276. Here I will include the zero ring in order to get better formulas. The nimber of a Noetherian ring is an ordinal number. It is defined recursively by $$\alpha(R) = \mathrm{mex} \bigl\{\alpha\bigl(R/\langle x \rangle\bigr) : 0 \neq x \in R\bigr\}.$$ where $\mathrm{mex} \, S$ denotes the smallest ordinal not contained in $S$. I am only interested in commutative rings here. You can find some basic computations in Section 5.4 here. But I do not know yet the nimber of $K[X,Y]$ for instance (this is work in progress). What I would like to know: Does every ordinal number arise as the nimber of a Noetherian commutative ring? For example, every ordinal number $<\omega^2$ is realizable: For all $k,n<\omega$ we have $$\alpha(R_1 \times \dotsc \times R_k \times S/p^n) = \omega \cdot k + n,$$ where $R_1,\dotsc,R_k,S$ are principal ideal domains which are no fields, and $p \in S$ is a prime element. Examples and partial results as answers are very much appreciated! REPLY [3 votes]: I know how to construct all ordinals less than $\omega^3$. Let $k$ be an infinite field. Lemma 1: For any Noetherian ring $R$, $$\alpha( R \times k ) \geq \alpha(R)+1.$$ Proof: $R \times k$ can move to $R$ and it can also move to every element $R$ can move to, so $\alpha(R \times k) \geq \alpha(R)$. Lemma 2: For any Noetherian ring $R$, $$\alpha\left(R \times k[x]\right) \geq \alpha(R) + \omega.$$ Proof: By inductively applying Lemma 1, every ordinal $< \alpha(R) + \omega$ is less than the ordinal of some ring of the form $R$ times a finite product of copies of $k$. Hence every ordinal $< \alpha(R) + \omega$ is equal to the ordinal of a quotient of such a ring by a single element, which is necessarily a quotient of $R$ by zero or one elements times a finite product of copies of $k$, which is manifestly a quotient of $R \times k[x] $ by one element. Lemma 3: For any Noetherian ring $R$, $$\alpha\left( R \times k[x,y] \times k[x]\right) \geq \alpha(R) + \omega^2$$ Proof: By inductively applying Lemma 2, every ordinal $< \alpha(R) + \omega^2$ is less than the ordinal of some ring of the form $R$ times a finite product of copies of $k[x]$. So it is equal to the ordinal of a quotient of $R$ by $0$ or $1$ elements, times a finite product of copies of $k[x]$, times a finite product of quotients of $k[x]$ by one element. The first is a quotient of $R$ by $0$ or $1$ elements, the second is a quotient of $k[x,y]$ by one element, and the third, crucially, is a quotient of $k[x]$ by one element, because we can fit any finite number of zero-dimensional schemes that individually embed into the line together in the same line so that they do not intersect. Hence every ordinal less than $\alpha(R) + \omega^2$ is the ordinal of a quotient of $R \times k[x,y] \times k[x]$ by one element. This last step prevents us from extending the argument one dimension higher, because there is no Noetherian affine scheme such that a finite union of hypersurfaces in $\mathbb A^2$ is itself a hypersurface in that scheme - at least I don't think there is. However, by iterating Lemma 3, we can construct rings with nimbers at least $ n \omega^2$ for all natural numbers $n$, and hence verify the existence of rings with any given nimber $< \omega^3$. Note that we have done this inductive argument without at any point upper bounding the nimber of any ring. I am guessing that at some point upper bounds for certain rings will be needed to lower bound the nimbers of other rings.<|endoftext|> TITLE: Kunneth Theorem and localisation QUESTION [13 upvotes]: I have two, somehow related questions, and I would be very grateful if you point out at some references if the answer is known. If these are too elementary or not research level, then please feel free to move it mathexchange. I am not an expert in this, so I thought it is worth asking here. According to Ravenel's orange book, the Nilpotence Theorem implies that $H\mathbb{F}$ and $K(n)$ are ``essentially'' the only homology theories for which the Kunneth isomorphism holds; I suppose this latter means that $$E_*(X\times Y)\simeq E_*(X)\otimes_{E_*(pt)}E_*(Y).$$ So, my first question is whether this conclusion is immediate or whether it is not that trivial and a proof is recorded somewhere? I have not tried to prove it. Second, I wonder if there is a spectrum $E$ so that $$\pi_*L_E(S^0\wedge S^0)\simeq\pi_*L_ES^0\otimes\pi_*L_ES^0$$ where the tensor product is over some suitable coefficient ring, $L_E$ is the Bousfield localisation with respect to $E$, and $S^0$ is the sphere spectrum. Or, equivalently on the category of spaces $$\pi^s_*L_E(X\times Y)\simeq\pi^s_*L_EX\otimes\pi^s_*L_EY$$ again with tensor product over say $\pi_*L_ES^0$. A first guess is $E=H\mathbb{F}$ and $K(n)$, but these do not seem that immediate to me either. Edit. Perhaps it is better to replace $\times$ with $\wedge$, and ask for an isomorphism $$\pi^s_*L_E(X\wedge Y)\simeq\pi^s_*L_EX\otimes_{\pi_*L_ES^0}\pi^s_*L_EY$$ as also in Neil's update to his answer. The motivating example for the second question is the following. Consider a compact Lie group and an embedding $1^{\times n} TITLE: Derivatives of norm of vector-valued holomorphic functions QUESTION [6 upvotes]: Let $G$ be a connected domain in $\mathbb{C}^{n}$, let $H$ be a Hilbert space and let $f,g:G\to H\backslash \{0\}$ be holomorphic (in my particular situation they are also injective, but I don't think it helps). Is it true that if $\frac{\partial^2}{\partial z_i\partial \overline{z_j}}\log \|f(z)\|=\frac{\partial^2}{\partial z_i\partial \overline{z_j}}\log \|g(z)\|$, for all $i,j\le n$, then there is a holomorphic function $h:G\to\mathbb{C}$ and an isometry $U:H\to H$, such that $g(z)=h(z)Uf(z)$? Motivation. Consider the canonical quotient $p:H\to PH$, where the latter is the projective space over $H$. Suppose $\varphi:G\to G$ is a biholomorphism, such that $f$ and $g=f\circ\varphi$ satisfy that condition. Then $pf\varphi(pf)^{-1}$ is an isometry of $pf(G)$ with respect to the Fubini-Study metric on $PH$. I wonder, if this isometry can be lifted to an isometry $H$, i.e. if there is an isometry $U:H\to H$, such that $pUf=pf\varphi$. Finally the last condition means that there is a holomorphic function $h:G\to\mathbb{C}$, such that $h(z)Uf(z)=f(\varphi(z))$. REPLY [4 votes]: Yes, this is true, and this is called the Calabi rigidity, though it was proved (for finite dimensional Hilbert space) long before Calabi, see, for example, Polya-Szego, revised edition, part IV, problem 207, where there is a reference on the original paper. Calabi's paper is: Isometric imbedding of complex manifolds, Ann. Math. 58 (1953), 1–23. Another source is Theorem 2 in https://arxiv.org/pdf/math/0007030.pdf.<|endoftext|> TITLE: 3-adic valuation of a sum involving binomial coefficients QUESTION [16 upvotes]: Let $$a(n) = \sum_{0 \leq k \leq n} {n \choose k}{{n+k} \choose k},$$ and define $b(n) = \nu_3 \bigl(a(n)\bigr)$, where $\nu_3$ is the $3$-adic valuation. About twenty years ago or so, I discovered (empirically) the following conjectured expression for $b(n)$: $$b(n) = \begin{cases} b\bigl(\lfloor n/3 \rfloor\bigr) + \bigl(\lfloor n/3 \rfloor \bmod 2\bigr), & \text{if $n \equiv 0,2$ (mod 3); } \\ b\bigl(\lfloor n/9 \rfloor\bigr) + 1, & \text{if $n \equiv 1$ (mod 3).} \end{cases} \tag{$*$}$$ But I have not been able to prove it. For some background, the problem has some similarity to the following theorem, a weaker version of which was originally suggested by N. Strauss: $$\text{If}\quad r(n) = \sum_{0 \leq i < n} {{2i} \choose i},\quad\text{then}\quad \nu_3 \bigl(r(n)\bigr) = \nu_3\left ( n^2 {{2n} \choose n}\right),$$ which I proved by a kind of tedious argument, with Jean-Paul Allouche. Later, another more elegant proof was given by Don Zagier. See here. Can anybody prove $(*)$? REPLY [11 votes]: The $3$-adic evaluation you seek is compactly given by $$\nu_3(a_{2n})=\nu_3\left(\binom{2n}n\right) \qquad \text{and} \qquad \nu_3(a_{2n+1})=\nu_3\left(3(2n+1)\binom{2n}n\right),$$ which can be proved inductively using the well-known recurrence $$na_n=3(2n-1)a_{n-1}-(n-1)a_{n-2}$$ according to the parity of $n$. To address Cigler's request, here is an illustration for the case even: $(2n)a_{2n}=3(4n-1)a_{2n-1}-(2n-1)a_{2n-2}$ and by induction assumption $$\nu_3(3(4n-1)a_{2n-1})=\nu_3\left(9(4n-1)(2n-1)\binom{2n-2}{n-1}\right) \qquad\text{and}\qquad \nu_3((2n-1)a_{2n-2})=\nu_3\left((2n-1)\binom{2n-2}{n-1}\right).$$ So, $\nu_3(3(4n-1)a_{2n-1})>\nu_3((2n-1)a_{2n-2})$ and $\nu_3((2n)a_{2n})=\nu_3((2n-1)\binom{2n-2}{n-1})$; or, \begin{align} \nu_3(a_{2n})&=\nu_3\left(\frac{(2n-1)}{2n}\binom{2n-2}{n-1}\right) =\nu_3\left(\frac14\binom{2n}n\right)=\nu_3\left(\binom{2n}n\right) \end{align} as desired.<|endoftext|> TITLE: Bi-Lipschitz version of Kirszbraun's extension theorem QUESTION [8 upvotes]: Kirszbraun's theorem for $\mathbb{R}^2$ states the following: Given any set $S\subset \mathbb{R}^2$ and any Lipschitz function $f:S\rightarrow \mathbb{R}^2$ with Lipschitz constant $k$, $0< k< \infty$, for any set $F$ which contains $S$ there exists a function $\tilde f:F\rightarrow C$ such that $Lip(\tilde{f})=k$, $\tilde{f}|_{S}=f$ and $C$ is any closed convex set containing $f(S)$. I'm wondering if something similar could hold true for bi-Lipschitz functions, for example is the following claim true? Given any two discrete sets of $n\ge 3$ linearly independent points $S_1,S_2\subset \mathbb{R}^2$, and any bi-Lipschitz function $f:S_1\rightarrow S_2$ with bi-Lipschitz constant $k$, $00$ so that $\ell=0$; in this case the triangle $[x_1'x_2'x_3']$ is degenerate).<|endoftext|> TITLE: Brauer groups and field extensions QUESTION [6 upvotes]: Let $k$ be a field and $\mathrm{Br}(k)$ the Brauer group of $k$. Let $k \subset L$ be a field extension. Let $b \in \mathrm{Br}(k)$ and denote by $b \otimes L \in \mathrm{Br}(L)$ the base-change of $b$ to $L$. If $b \otimes L = 0$, then does this exist a subextension $k \subset K \subset L$ such that $K/k$ has finite degree and such that $b \otimes K = 0$? i.e. if $b$ is killed by some field extension $L$, then must $b$ be killed by some finite field extension of $k$ which is contained in $L$? REPLY [12 votes]: No: the conic $C:X^2+Y^2+1=0$ splits over the field $L=\mathbb{Q}(x)[y]/(x^2+y^2+1)$, since $(X,Y)=(x,y)$ is an $L$-point of $C$. However $L$ has no subfields algebraic over $\mathbb{Q}$ other than $\mathbb{Q}$ itself, since it is the function field of a geometrically irreducible variety.<|endoftext|> TITLE: Representability of the sum of homology classes QUESTION [10 upvotes]: This is probably a very simple question, but I have not found it addressed in the references that I know. Let $M$ be a closed and connected orientable $d$-dimensional manifold ($d\leq 8$) and let $[\alpha_{i}]\in H_{k}(M,\mathbb{Z})$, $k\leq 6$, $i=1,2,$ be $k$-homology classes of $M$ that admit representations $\iota_{i}\colon Z_{i}\hookrightarrow M$ in terms of closed oriented submanifolds $Z_{i}$, that is: $\iota_{i \ast}[Z_{i}] = [\alpha_{i}]$ Since $k\leq 6$, every element in $H_{k}(M,\mathbb{Z})$ is representable by a closed oriented submanifold, so in particular $[\alpha] := [\alpha_{1}] + [\alpha_{2}]\in H_{k}(M,\mathbb{Z})$ is also representable by a closed oriented submanifold $\iota\colon Z\hookrightarrow M$. Is there any relation between $Z$ and $Z_{i}$? Can I "construct" a representative of $[Z]$ once I know $Z_{1}$ and $Z_{2}$? Thanks. REPLY [10 votes]: There are a few well-known (at least to low-dimensional topologists) cases of this. First, the obvious observation that if $k TITLE: Subcategories of the Verdier quotient? QUESTION [5 upvotes]: Let $\mathcal T$ be a triangulated category and $\mathcal C$ a thick triangulated subcategory. We consider the Verdier quotient $\mathcal T/\mathcal C$. Is there a bijective correspondence between thick triangulated subcategories of the quotient $\mathcal T/\mathcal C$ and thick triangulated subcategories of $\mathcal T$ containing $\mathcal C$? It "feels" true, but I have learned never to take chances with technicalities of triangulated categories. REPLY [3 votes]: Yes, this is true and works as you expect. See Proposition 2.3.1 (pages 125-127) of Verdier's thesis: Jean-Louis Verdier -- Des catégories dérivées des catégories abéliennes (Astérisque No. 239, 1996) [MR1453167]. Since it is in French, let me just remark that "sous-catégorie triangulée pleine" means "full triangulated subcategory", "sous-catégorie triangulée strictement pleine" means "strictly full triangulated subcategory" a.k.a "full triangulated subcategory closed under isomorphism", and Verdier uses "saturée" to mean "thick" (closed under direct summmands). Let $\mathcal D$ be a triangulated category, $\mathcal B\subset \mathcal D$ a thick triangulated subcategory, and let $Q:\mathcal D \rightarrow \mathcal D / \mathcal B$ denote the Verdier quotient. That proposition establishes a bijection between the (strictly full) triangulated subcategories of $\mathcal D$ which contain $\mathcal B$ and the (strictly full) triangulated subcategories of $\mathcal D/\mathcal B$. It restricts to a bijection between the (strictly full) thick triangulated subcategories of $\mathcal D$ which contain $\mathcal B$ and the (strictly full) thick triangulated subcategories of $\mathcal D/\mathcal B$.<|endoftext|> TITLE: Is determinacy on an infinite Dedekind finite set consistent? QUESTION [11 upvotes]: Consider $\mathrm{AD}_X$, determinacy for games where players pick moves from $X$. We know that it is consistent for $X = \omega$ or $\mathbb{R}$ (under large cardinal assumptions), but inconsistent for $X = \omega_1$. Since this implies determinacy is inconsistent for any set with $\omega_1 \leq X$, this answers the question of consistency for most cardinalities that we might consider (e.g. $\omega_1 \cup \mathbb{R}$, $\mathcal{P}(\mathbb{R})$, all uncountable ordinals). Of course, in the absence of choice other cardinalities might exist. This leads me to the following two questions: Q1: Is the existence of infinite Dedekind finite sets consistent with $\mathrm{AD}$? Q2: If so, is 'there exists an infinite Dedekind finite set $X$ such that $\mathrm{AD}_X$ holds' consistent? (Let's call this statement $\mathrm{AD}_\mathrm{DF}$.) (Note that assuming the consistency of $\mathrm{AD_\mathbb{R}}+\mathrm{DC}$, we have that $\mathrm{AD}_\mathrm{DF}$ can't be a consequence of either $\mathrm{AD}$ or $\mathrm{AD_\mathbb{R}}$.) Games on infinite Dedekind finite sets seem very strange. For example, consider the game where the first player to play a move that's already been played loses. By Dedekind finiteness somebody has to win this game (and in finite time!), but both players always have not-losing moves at every point. I initially thought this game would be nondetermined, but someone pointed out the following to me: Let $X$ be infinite Dedekind finite. Then $2 \times X$ is as well. Player 2 has a winning strategy for this game on $2 \times X$: given that an odd number of moves have been played, if nobody has lost yet then there must be some $x \in X$ such that one but not both of $(0, x)$ and $(1, x)$ have been played. Thus we play the other one. (This means that Player 1 has a winning strategy for this game on $2 \times X \setminus (0, x)$, by pretending they are Player 2 in the game on $2 \times X$ and that the first move was $(0, x)$.) REPLY [18 votes]: The answer is no, you cannot have determinacy for all games on an infinite Dedekind-finite set. Indeed, one cannot even have clopen determinacy for games on such a set. So the answer to question 2 is negative. Theorem. If $X$ is an infinite Dedekind-finite set, then there is a non-determined clopen game on $X$. In particular, $\text{AD}_X$ fails, and hence also $\text{AD}_{\text{DF}}$ is false. Proof. Suppose that $X$ is an infinite Dedekind-finite set. Fix some particular element $a\in X$, and consider the following game. Player I plays distinct elements of $X$, while player II responds with the element $a$, until Player I plays $a$, at which time player II must play an element not yet played. If he can do so, he wins, and otherwise player I wins. This game must terminate in finitely many moves, because player I cannot play distinct elements forever, as $X$ has no countably infinite subset. So this is a clopen game. It is clear that player I cannot have a winning strategy, since that strategy would call for her to play some elements of $X$ and eventually the element $a$, during which time all the while player II does something very easy, just playing the element $a$, until player I plays $a$, at which time player II can defeat the strategy by simply playing some new element. But I claim that also player II cannot have a winning strategy. If there were a winning strategy for player II, then we could use that strategy to generate a countable subset of $X$, since it provides a systematic way to extend any particular finite sequence of distinct elements of $X$. Just have player I play that finite sequence, and then $a$, and player II will respond with a fresh element of $X$. Since $X$ has no countably infinite subsets, there can be no such strategy. So the game is not determined. QED This game is related to the classical fact that clopen determinacy implies the axiom of choice, since in the comparatively trivial game, where player I plays a nonempty set from a fixed collection, and player II playing an element of that set (and winning upon doing so), it is clear that the first player cannot have a winning strategy, but a winning strategy for player II provides a choice function.<|endoftext|> TITLE: Why Grothendieck's Homotopy Hypothesis is so difficult? QUESTION [12 upvotes]: Recall that Grothendieck's Homotopy Hypothesis states that the homotopy category of weak globular $n$-groupoids (as in https://arxiv.org/pdf/1009.2331.pdf) is equivalent to the category of homotopy $n$-types. It's also well known that strict globular groupoids are not enough since $\Pi_3 (S^2)$ cannot be rectified to a strict $3$-groupoid due to the non-vanishing of the Whitehead bracket. On the other side, it's well known that every quasi-category can be rectified into a strict $(\infty, 1)$-category. More generally, there's Berger-Moerdjik result that says that a bunch of algebras over operads can be rectified. In particular, $A_{\infty}$-algebras can be rectified and also homotopy coherent diagrams (Vogt's theorem). In view of these observations, I have the following questions: 1) By considering $\Pi_3 (S^2)$ as a quasi-category (i.e., extend it by degenerated simplices), we can rectify it. Why doesn't it contradict my first observation? (EDIT : As Yonatan mentioned in the answer below, only the level $0$ and $1$ can be rectified. Therefore there's no contradiction.) 2) Why can't we rectify weak globular groupoids? By Vogt's theorem, homotopy coherent diagrams can be rectified. So, what fails if one views a weak globular groupoid as a homotopy coherent diagram and, then, rectify it? Further comments about the intuition of why one cannot rectify weak globular groupoids would be also of great utility. REPLY [14 votes]: First, let me remark that your question does not seem to be about the homotopy hypothesis, but about rectification. More specifically, the homotopy hypothesis concerns the question of whether (some version of) weak globular groupoids is equivalent to, say, Kan complexes, while the rectification problem you mention asks whether weak globular groupoids are the same as strict globular groupoids. The confusion then seems to arise from the fact that on the side of Kan complexes, there are some aspects which can be rectified. For example, we can rectify a Kan complex into a strict simplicial groupoid. One can then maybe rephrase the question as: given that we believe the homotopy hypothesis, how can it be that Kan complexes can be rectified while weak globular groupoids cannot? The answer to this question is simply that Kan complexes cannot be rectified either. Indeed, if one replaces a Kan complex by a simplicial groupoid, the mapping spaces of this simplicial groupoid will still be Kan complexes. This can be informally described as saying that we have rectified the levels of objects and morphisms, but we have left unrecitifed the level of homotopies between morphisms, homotopies between homotopies etc. all the way up. Note that by only rectifying the 0 and 1 dimensional levels you can still avoid the counterexample $S^2$. On the other hand, that's the maximum you can do. For example, $S^2$ cannot be modeled by a strict 2-category enriched in simplicial sets: indeed, the group of self equivalences of an identify morphism in any simplicially enriched strict 2-category is a simplicial abelian group, but the corresponding group of automorphism for $S^2$ is an $\mathbb{E}_2$-group whose $\mathbb{E}_2$-structure does not refine to an $\mathbb{E}_{\infty}$-structure.<|endoftext|> TITLE: Hooks in a staircase partition: Part I QUESTION [11 upvotes]: This quest has its impetus in a paper by Stanley and Zanello. I became curious about What is the sum of all hooks lengths of all partitions that fit inside the $n$-th staircase partition? On the basis of experimental data, I'm prompted to ask: Question. Let $\lambda=(n,n-1,\dots,1)$ be the staircase partition, $h_{\square}$ the hook-length of a cell $\square$ in the Young diagram of a partition. Then, $$\sum_{\mu\subset\lambda}\sum_{\square\in\mu}h_{\square}=\frac{(n+4)(2n+3)}6\binom{2n+2}{n+1} - (n+2) 2^{2n+1}.$$ Is this true? If so, any proof? Update. I'm still hoping for a complete solution for this evaluation; it's not common to get such a nice closed from in general. Part II of this problem has received a fine answer. Update. To help Stanley's mention of a recurrence, denote the RHS quantity by $w_n:=\frac{(n+4)(2n+3)}6\binom{2n+2}{n+1} - (n+2) 2^{2n+1}$ then we have ("shaloshable") $$n(n-2)w_n-2(4n^2-5n-3)w_{n-1}+8n(2n-1)w_{n-2}=0.$$ REPLY [4 votes]: The following method should give a proof, but I haven't done the computation. Let $f_n$ denote the desired sum. I use the result of Douglas Zare's answer that $$ f_n = \sum_{\mu\subseteq \delta_n} \sum_i \mu_i^2, $$ where $\delta_n=(n,n-1,\dots,1)$. Let $$ g_n = \sum_{\mu\subseteq \delta_n} \sum_i \mu_i. $$ By OEIS A029760 we have $$ g_n = \frac 12 (n+2)^2C_n-2^{2n+1}, $$ where $C_n$ is a Catalan number. We now break up the sum for $f_n$ into the following parts: (1) No $\mu_i=n-i+1$. In other words, $\mu\subseteq\delta_{n-1}$, so the contribution from these terms is $f_{n-1}$. (2) $\mu_1=n$. Below the first row is any partition $\nu\subseteq \delta_{n-1}$, so the contribution from these terms is $f_{n-1}+n^2C_n$ since there are $C_n$ choices for $\nu$. (3) $\mu_k=n-k+1$ for some $k\geq 2$, and $\mu_j TITLE: Some questions on division algebras QUESTION [5 upvotes]: Given a field $K$, is there a finite dimensional quiver algebra, such that any finite dimensional division algebra is isomorphic to End(M)/rad(End(M)) for some indecomposable finite dimensional module M? Can fields with finite Brauer group be somehow characterised? What are the field with Brauer group equal to $\mathbb{Z}/\mathbb{Z}3$? I do not even know one example. ($\mathbb{Z}/\mathbb{Z}3$ might be replaced by any finite abelian group with at least 3 elements) REPLY [10 votes]: Questions 2 and 3 are addressed in section 11 of Auel, Asher; Brussel, Eric; Garibaldi, Skip; Vishne, Uzi, Open problems on central simple algebras., Transform. Groups 16, No. 1, 219-264 (2011). ZBL1230.16016. In particular, the Brumer-Rosen conjecture implies that there are no fields with Brauer group $\mathbb{Z}/3\mathbb{Z}$ (and this is known), and only elementary abelian 2-groups appear. But this is still a conjecture, and $\mathbb{Z}/7\mathbb{Z}$ isn't ruled out yet. What I is very funny is that Ford has shown that every finite abelian group is the Brauer group of some commutative ring. Edit: At the end of section 11 there is a reference to the very conveniently titled Efrat, Ido, On fields with finite Brauer groups, Pac. J. Math. 177, No.1, 33-46 (1997). ZBL0868.12005.<|endoftext|> TITLE: Comparing Frobenius weights with Mixed Hodge theory QUESTION [5 upvotes]: For a variety over a finite field Deligne define weights by looking at the eigenvalues of the Frobenius. On the other hand, if we take a variety over $\mathbb{Q}$, at least for its constant sheaf we have the weight filtration which gives the mixed Hodge structure. How are these two comparable? Can someone point out some references? Say we have a variety $X$ defined over $\mathbb{Z}[\frac{1}{p}]$. We can consider $p$-etale sheaves on $X$, or rather, the bounded derived category of constructible sheaves. If $F$ belongs to this category, for each prime $q\neq p$ we get a complex in the derived category of the fiber, which is a variety over a finite field, for which we have a weight filtration. Could we patch together these weights theories to get a weight filtration on $F$? Then I would like this weight filtration to induce the filtration of Mixed Hodge theory after restriction to generic fiber and extension of scalars. Thank you very much. EDIT: I should specify I want $F$ to induce a mixed sheaf on the fiber over each $q\neq p$. REPLY [4 votes]: For constant coefficients, the comparison statement, along with a sketch, appears in Deligne's ICM talk, Poids dans la cohomologie.... For things to work the way you seem to want in your second paragraph, you're going to need a lot more structure for $F$ than what you've given. At the very least, when $F$ is a local system (lisse), you need the weights with respect to the various Frobenius's, at different primes, to be compatible, and you would also need to specify an admissible variation of mixed Hodge structures on $X(\mathbb{C})$ plus appropriate compatibilities with the Hodge theoretic weights. Take a look at Saito's paper Arithmetic mixed sheaves in Inventiones for a general framework.<|endoftext|> TITLE: Solving the equation $a^a+b^b=c^c$ in positive integers QUESTION [8 upvotes]: Under what all conditions on $(a,b,c)$ where $a,b,c$ are positive integers can we say about the non-existence or existence of any solution to the equation $$a^a+b^b=c^c$$ (Or in other words : Solve $a^a+b^b=c^c$ ) $ $ $ $ About this note that : For a solution we must have $gcd(a,b,c) = 1$ (follows from Fermat's Last Theorem) and all the elements of the set $\{a,b,c\}$ cannot be prime numbers (proof shown below). We will prove there are no positive prime numbers $p,q,r$ such that the following equation is satisfied : $$p^p+q^q=r^r$$ This is easy to prove. Note that in fact we have something stronger : The equation $4+m^m=n^n$ has no solutions in positive integers. Assume the contrary, then we get $m>1$ (by checking the case $m=1$ separately) and then note that $n \ge m+2$ and thus $n^n > (m+2)^m > m^m+2^m \ge n^n$ and thus a contradiction. REPLY [18 votes]: The sequence $(n^n)_{n \ge 1}$ grows too fast for there to exist a solution: if $b \ge 2$, then $(b+1)^{b+1} > b^{b+1} \ge 2 b^b$ (and also $2^2 > 2 \cdot 1^1$), so if $1 \le a \le b$, then $b^b < a^a + b^b \le 2 b^b < (b+1)^{b+1}$, and no $c$ can exist such that $a^a + b^b = c^c$.<|endoftext|> TITLE: How to use Hilbert series to count combinatorial objects? QUESTION [8 upvotes]: In THE SLOPES DETERMINED BY n POINTS IN THE PLANE by JEREMY L. MARTIN, Page 2, Theorem 1.1, a Hilbert series is used to compute some combinatorial objects: Let $R_n=k[m_{1,2}, \ldots, m_{n-1,n}]$, and let $I_n$ be the ideal generated by the tree polynomials of all rigidity circuits in the complete graph $K_n$. Then the coefficient $h(n,k)$ of the numerator of the Hilbert series of $R_n/I_n$ counts the number of perfect matchings on $[1, 2n − 4]$ with exactly $k$ long pairs, that is, pairs not of the form $\{i,i + 1\}$. In general, if we want to compute some combinatorial objects using Hilbert series, how to define the ring $R_n$ and the ideal $I_n$? Are there some other examples which count some combinatorial objects using Hilbert series? Thank you very much. REPLY [2 votes]: See: Mordechai Katzman, Counting monomials This paper presents two enumeration techniques based on Hilbert functions. The paper illustrates these techniques by solving two chessboard problems.<|endoftext|> TITLE: Is there a monoidal category that coclassifies enriched category structures for a given set? QUESTION [5 upvotes]: Let $S$ be a set. Is there a monoidal category $TS$ that we can construct from $S$ such that monoidal functors $F: TS \to M$ (up to monoidal natural isomorphism) correspond to $M$-enriched categories with underlying object set $S$ (up to $M$-enriched equivalence)? The construction I had in mind was to take $S^2$, add elements $(a,b)\otimes (b,c)$, add maps $\circ_{a,b,c} : (a,b)\otimes (b,c)\to (a,c)$, add a unit element $\mathbf{1}$ and maps $\mathbf{1}_{x}:\mathbf{1}\to (x,x)$, freely add the rest of the structure needed for it to be monoidal, and then impose commutativity of diagrams corresponding to associativity of the composition maps and unit axioms. My hope is that a monoidal functor out of this category into some monoidal category $M$ designates all of the structure necessary to specify an $M$-enriched category with objects $S$. Is such a construction known? I am not specifying any particular level of strictness on the monoidal functors involved here, but I suspect that has an impact on the answer to my question. REPLY [4 votes]: I don't think you can construct $T(S)$ as a monoidal category. It's more common to construct $T(S)$ as a bicategory with object set $S$. (see the end for comments on regarding it as a monoidal category) Let $I(S)$ denote the indiscrete category on $S$. Then a lax 2-functor $I(S) \to M$ (where $M$ is considered as a 1-object bicategory -- more generally $M$ could be an arbitrary bicategory) is the same as an $M$-enriched category with object set $S$ (this goes back to Benabou's original paper on bicategories). For any bicategory $B$ there is a lax morphism classifier $B'$ such that a weak 2-functor $B' \to C$ is the same as a lax 2-functor $B \to C$ (where $C$ is a bicategory) [actually, I'm not quite certain of this -- if we were talking about strict 2-categories and strict 2-functors rather than weak ones I'd be more certain. You can fit into this framework by replacing $M$ with a monoidal category that is strictly associative]. So $I(S)'$ has the property you're looking for. But it's not exactly a monoidal category -- it's bicategory with many objects. I think this construction is "well-known". For example, a construction with a similar flavor comes in Gepner and Haugseng's definition of an enriched $\infty$-category: it $S$ is a Kan complex, then let $I(S)$ be the "indiscrete simplicial Kan complex on $S$", with $I(S)_n = S^{n+1}$. A monoidal $\infty$-category $M$ is also sort of simplicial quasicategory, and an $M$-enriched category with object space $S$ is a map of bisimplicial sets $I(S) \to M$ (well, I think they speak in terms of cartesian fibrations over $\Delta$ rather simplicial objects, but these are equivalent by straightening/unstraightening). If you want $TS$ to be a monoidal category rather than a bicategory, then I think you're in luck because the inclusion from monoidal categories into bicategories should have a left adjoint, just like the inclusion of monoids into category. So just apply this left adjoint to $I(S)'$ above. This adjoint is kind of weird -- you freely add composites for morphisms with non-composable domain and codomain. But in the case of such a simple bicategory as this, I suppose it's not so bad -- and should look basically like you describe.<|endoftext|> TITLE: Connected compact manifolds with unique Lie group structure QUESTION [11 upvotes]: I am sorry if my question is stupid (or very hard) or common knowledge, or should be placed at math.stackexchage.com. As long as a math student read the definition of Lie group, several natural questions appear instantly: Is it true, that any compact manifold admits Lie group structure? (NO) Is it true, that there exists compact manifold that admits different Lie group structures? (YES) Answers to these questions can be found here - Lie Groups and Manifolds But I was not able to find the answer for third most natural question: Which connected compact manifolds admits unique Lie group structure? (As pointed out by YCor, without the connectedness assumption, there are trivial non-uniqueness examples.) Thanks a lot for your answers! REPLY [4 votes]: Your question seems to be of a tall order! There are many,vmany necessary conditions involving the entire spectrum of algebraic topology such as homotopy and cohomology. This 3rd edition book by Kakl H. Hofmann and Sidney A. Morris on compact groups is full of these. For instance if you asked: which compact connected n-dimensional manifold can support an abelian Lie group, one would tell you "only $\mathbb{T}^n =(\mathbb{R}/\mathbb{Z})^n$" but without the qualification "unique". There are tons of discontinuous automorphisms of $\mathbb{T}$, and so you have millions of group topologies (all isomorphic!). By contrast, the $3$-dimensional real projective space supports only one compact Lie group topology; namely, that of $SO(3)$. This has to do with the automatic continuity of algebraic automorphisms of certain compact connected Lie groups (see loc.cit., Corollary 5.66 and Exercise E5.21 following it, page 168). For more stuff see (just as a for instance) 8.59, 8.83, 9.59, A3.90, A3.92. Life is not so simple that we can hope for a positive answer to your question as it stands. However, information in the area is around so that as soon as you pose a question of a more specific nature on the topic of compact connected manifolds and Lie groups, there is hope for an answer.<|endoftext|> TITLE: Squareful values of polynomials QUESTION [8 upvotes]: Recall that an integer $n$ is called squareful if for every prime $p$ with $p \mid n$, we also have $p^2 \mid n$. Any squareful number can be written uniquely as $n= x^2 y^3$ where $y$ is squarefree. From this, it is easy to see that $$\#\{ n \in \mathbb{Z}: |n| \leq X, \, n \text{ is squareful} \} \ll X^{1/2}.$$ I would like a version of this for polynomials. Let $f \in \mathbb{Z}[x]$ be non-constant and separable. Then does there exist $\delta > 0$ such that $$\#\{ n \in \mathbb{Z}: |n| \leq X, \, f(n) \text{ is squareful} \} \ll X^{1 - \delta} \quad ?$$ Hopefully there are not some necessary local conditions here that I overlooked. In my application I am happy to change $f$ as required so that one can assume that $f$ is sufficiently "general". Moreover, I can even assume that $f$ is of very large degree if necessary to simplify things. I would normally try to prove something like this using the large sieve, however the large sieve gives the poor upper bound $X/(\log X)$, whereas I would like a power saving. If necessary, I'm happy to assume some standard conjectures (e.g. the abc conjecture). REPLY [4 votes]: In the sieve book of Cojocaru and Murty they give a simple application of the square sieve of Heath-Brown, namely in Theorem 2.3.5 of their book they prove that $$\#\{1\leq n \leq x:f(n)=\square\}\ll_{f,\epsilon} x^{1-\epsilon}$$ for all $0<\epsilon<\frac{1}{3}$. Squares and squarefulls have the same asymptotic density and there is definitely a chance to make the square-sieve approach work for $$\#\{1\leq n \leq x:f(n)=\text{ squarefull}\}$$ without abc. Note that if $f(n)$ is square-full then $f(n)=a^2b^3$ hence there exists $b \in \mathbb{N}$ such that $f(n)b$ is a square. The case $b=1$ is what they do however their proof can be easily adopted to work for $bf(n)$. It gives a bound independent of $b$ by changing $\mathcal{P}$ in their proof to be the set of all primes in $(x^{1/3},2x^{1/3}]\setminus \{p \text{ prime } : p|b\}$. The prime divisors of $b$ that are excluded are at most $\log b\ll \log x$, while the set of primes is $\asymp x^{1/3} (\log x)^{-1}$, therefore all estimates go through unaltered. This does not solve your problem since one has to do something extra to get more saving when $b$ gets larger, but there is some hope.<|endoftext|> TITLE: lowest degree of polynomial that removes the first digit of an integer in base p QUESTION [11 upvotes]: Let $p$ be an odd prime and $n \geq 2$. (1) Does there exist an integer-coefficient polynomial $f$ such that $f(x) = x - (x \bmod p)$ for all $x \in \mathbb{Z}/p^n \mathbb{Z}$? The polynomial effectively removes the first base-$p$ digit of $x$. For example, if $n = 2$, then the defining property is $f(a+bp) = bp$. (2) If so, then how can we find such a polynomial with lowest possible degree? How does this lowest degree change with $n$? UPDATE: I am super amazed. Because the reason I asked the question is because by trial and error I have an algorithm to construct polynomials of degree $(n-1)(p-1) + 1$ that works for small $n$, but can't quite finish the proof by induction to all $n$. This has been bugging me for past few days. Seeing the answers confirm with my heuristic is just thrilling! SECOND UPDATE: Can we generalize this to functions that removes the lower $k$-digits, i.e., $f(x) = x - (x \mod p^k)$ for all $x \in \mathbb{Z}/p^n \mathbb{Z}$? Thank you all for your nice answers! REPLY [9 votes]: Yes, this can be done, and the bound for the degree given by Neil Strickland (that is, $(n-1)\cdot(p-1)+1$ holds. Let $F_A(x)$ be the formal sum in binomial coefficients $$ F_A(x):=\sum_{j=0}^\infty (-1)^j \begin{pmatrix}A-1+j\\j\end{pmatrix}\begin{pmatrix}x\\A+j\end{pmatrix}.$$ Then $F_A(x)$ converges on every integer, and for $M\in \mathbb Z$ we have $$ F_A(M)=\begin{cases}1 & M\geq A\\ 0 & \text{otherwise.}\end{cases} $$ The 'otherwise' part of the claim is easy. The rest follows by noticing that $F_A(M)$ is the same as coefficient of $X^{M-A}$ found by taking the product of the Taylor series of $(1+X)^{-A}$ and the polynomial $(1+X)^M$. The result is of course $1$. Let $$G(x):= p\cdot \sum_{j=1}^\infty F_{j\cdot p}(x)=\sum_{m=0}^\infty a(m)\begin{pmatrix}x\\m\end{pmatrix},$$ where $$a(m):= p\sum_{k=0}^\infty (-1)^{m-kp} \begin{pmatrix}m-1\\m-kp\end{pmatrix}.$$ Then for $M$ a positive integer, we have $0\leq M-G(M)(n-1)(p-1)+1$. Thus, truncating the sum as described above does not change the valuer, mod $p^n$. We now just need to show that $\tilde G$ can be expanded in a polynomial with $p$-integral coefficients. Each binomial coefficient $\begin{pmatrix}x\\m\end{pmatrix}$ introduces a denominator of $m!$, which has $p$-adic valuation bounded by $$v_p(m!)<\frac{m}{p-1}$$. But as we already saw, $v_p(a(m))>\frac{m}{p-1}$ so the coefficient of $\tilde G(x)$ are indeed $p$-integral.<|endoftext|> TITLE: Is there a natural inner model of AD$_\mathbb{R}$? QUESTION [10 upvotes]: The question is as in the title, but let me explain a bit. Assuming a proper class of Woodin cardinals, $L(\mathbb{R})$ satisfies AD (and DC). And $L(\mathbb{R})$ is a very natural inner model. I'm curious if there is a similarly natural inner model for AD$_\mathbb{R}$. Now, ZF + $V=L(\mathcal{P}(\mathbb{R}))$ + large cardinals proves AD$_\mathbb{R}$ if I recall correctly; however, this is somewhat misleading as $L(\mathcal{P}(\mathbb{R}))$ is never a model of AD, let alone AD$_\mathbb{R}$, since in it the reals are well-ordered. So I'm curious: assuming large cardinals, is there a reasonably canonical inner model of AD$_\mathbb{R}$ which is "easy to describe"? REPLY [6 votes]: A Wadge initial segment (of $\mathcal P(\mathbb R)$) is a subset $\Gamma$ of $\mathcal P(\mathbb R)$ such that whenever $A\in\Gamma$ and $B\le_W A$, where $\le_W$ denotes Wadge reducibility, then $B\in\Gamma$. Note that if $\Gamma\subseteq\mathcal P(\mathbb R)$ and $L(\Gamma,\mathbb R)\models \Gamma=\mathcal P(\mathbb R)$, then $\Gamma$ is a Wadge initial segment. The relevance of this notion is that if $M$ is an inner model containing all the reals and satisfying $\mathsf{AD}_{\mathbb R}$, then $\Gamma=\mathcal P(\mathbb R)^M$ is a Wadge initial segment and $L(\Gamma,\mathbb R)\models\mathsf{AD}_{\mathbb R}$. Under appropriate large cardinal assumptions, there is a Wadge initial segment $\Gamma=\Gamma_{min}$ such that $L(\Gamma,\mathbb R)\models\mathsf{AD^+}+\mathsf{AD_{\mathbb R}}+\Gamma=\mathcal P(\mathbb R)$. Moreover, given any inner model $M$ containing all the reals and satisfying $\mathsf{AD}^++\mathsf{AD}_{\mathbb R}$, we have $\Gamma_{min}\subset M$. The mention of $\mathsf{AD}^+$ may well be superfluous here (or, perhaps, we should redefine $\mathsf{AD}_{\mathbb R}$ as $\mathsf{AD^+}+\mathsf{AD_{\mathbb R}}$); the situation does not seem entirely understood otherwise. Surely $\Gamma_{min}$ admits a purely descriptive set-theoretic description as well (in terms of the complexity of the iteration strategies of the hybrid or hod mice that it captures), but I do not know how to specify it. I suspect all of this is written up in reasonable detail nowadays. I suggest to read first MR3362806 Reviewed. Sargsyan, Grigor. Hod mice and the mouse set conjecture (English summary), Mem. Amer. Math. Soc. 236 (2015), no. 1111, viii+172 pp. ISBN: 978-1-4704-1692-8 (with all the technical details of the underlying theory) and MR3087400 Reviewed. Sargsyan, Grigor(1-RTG). Descriptive inner model theory (English summary), Bull. Symbolic Logic 19 (2013), no. 1, 1–55 (for a more leisurely introduction). The result can of course be generalized to other strengthenings of $\mathsf{AD}^+$, but you will eventually run into difficulties, as it is possible that there are incompatible (or ``divergent'') $\mathsf{AD}^+$ models, that is, it is consistent to have sets of reals $A,B$ such that $L(A,\mathbb R)$ and $L(B,\mathbb R)$ are both models of $\mathsf{AD}^+$, but $A$ and $B$ are Wadge-incomparable so $A\notin L(B,\mathbb R)$, $B\notin L(A,\mathbb R)$, and $L(A,B,\mathbb R)$ is not a model of $\mathsf{AD}^+$. In such a setting, it may well be that there is no minimal pointclass $\Gamma$ playing the role of $\Gamma_{min}$ for your strengthening of determinacy. What saves us for $\mathsf{AD}^++\mathsf{AD}_{\mathbb R}$ is that it is a theorem of Woodin that if $A,B$ are as above, and $\Gamma=\mathcal P(\mathbb R)^{L(A,\mathbb R)}\cap\mathcal P(\mathbb R)^{L(B,\mathbb R)}$, then ($\Gamma$ is again a Wadge initial segment, and) $L(\Gamma,\mathbb R)\models\mathsf{AD}^++\mathsf{AD}_{\mathbb R}$.<|endoftext|> TITLE: Example of Banach spaces with non-unique uniform structures QUESTION [10 upvotes]: While it is known that compact Hausdorff spaces admit unique uniform structures, it is further shown by Johson and Lindenstrauss's result that Banach spaces are characterized by their uniform structures [Johson&Lindenstrauss&Schechtman]. [Ribe] pointed out that two non-isomorphic spaces may have homeomorphic uniform structures (See [Benyamini&Lindenstrauss, Chap 10, Sec 10.4]). The classical nonunique example in [Benyamini&Lindenstrauss, Chap 10, Prop 10.32-10.34] shown that p-convexified Tsirelson spaces $\mathcal{T}^{(p)}$ for $1 TITLE: Connectedness in the plane QUESTION [11 upvotes]: There are several open problems in topology which concern connectedness and subsets of the plane. The biggest of these is undoubtedly: Question. Does every non-separating plane continuum have the fixed point property? This is a very beautiful problem. In essence, it is asking whether every non-separating plane continuum is like a closed disk, or conversely, if every separating plane continuum is like a circle. Many other open problems are not explicitly about subsets of the plane, but it helps to think about whether they are true for spaces that can be embedded in the plane. The plane is, in many cases, the nicest embedding space that does not make a problem trivial (the real line lacks sufficient complexity). In this post I would like to compile a list of topological properties of the plane. They should be fairly easy to state, and it is perfectly okay to state a result in (non-trivially) different ways. I know that this is a general request, but it would be extremely helpful to me. Jordan Curve Theorem. Every simple closed curve separates the plane into two components. Denjoy–Riesz Theorem. Every compact totally disconnected subset of the plane is contained in an arc. (Kuratowski). If a plane continuum $X$ is the common boundary of three of its complementary components, then $X$ is either indecomposable or the union of two proper indecomposable subcontinua. (Rudin). If there are three arcs in the plane which do not contain the point $x$ and which have only their end-points in common, then one of these arcs is separated from $x$ by the union of the other two. Nice theorems of the geometric variety are also welcome. Sylvester–Gallai Theorem. Given a finite number of points in the plane, either all the points are collinear, or there is a line which contains exactly two of the points. Erdős–Anning Theorem An infinite number of points in the plane can have mutual integer distances only if all the points lie on a straight line. REPLY [2 votes]: The Annulus Theorem: If $J_1$ is a Jordan curve in the interior region of another Jordan curve $J_2$, then the complement of the exterior region of $J_2$ and the interior region of $J_1$ is the closed annulus. The Torhorst Theorem: If $X$ is a Peano continuum in the plane and $B$ is a complementary region, then $\partial(B)$ is a Peano continuum. The Moore Triod Theorem: If $X$ is a triodic continuum then there is no uncountable collection of mutually disjoint copies of $X$ in the plane. The Roberts Embedding Theorem: If $X$ is an arc-like or circle-like continuum in the plane, then there is an uncountable collection of mutually disjoint copies of $X$ in the plane. Classification of Homogeneous Planar Continua: A point, a circle, a pseudo-arc, and a circle of pseudo-arcs. The Isotopy Extension Theorem: Any isotopy between compact $A$ and $B$ extends to an isotopy on the plane. Tarski Circle-Squaring Theorem: A closed disc can be decomposed into finitely many pieces, then reassembled into a closed square by translating the pieces. As for open problems, how about the Mandelbrot Conjecture, the Siegel Disc Conjecture, the Toeplitz Square problem . . . for some 'smaller' ones how about a characterization of planar dendroids, of continua which admit uncountably many copies in the plane, or the classification of isotopy classes of arc-like continua based on their inverse limit representations?<|endoftext|> TITLE: Applications of Kirchhoff's circuit laws to graph theory QUESTION [15 upvotes]: Is there a good survey on applications of Kirchhoff's circuit laws to graph theory or/and discrete geometry? Examples: Matrix tree theorem, Squaring the square, Electrician’s proof of Euler’s formula. REPLY [3 votes]: I really liked the discussion of electrical circuits in the recent book "Probability on Trees and Networks" by Lyons and Peres. Chapters 2, 4 and 9 seem the most relevant to what you want.<|endoftext|> TITLE: Is $L^2(\mathbb R)$ homeomorphic to $L^1(\mathbb R)$? QUESTION [32 upvotes]: Is $L^2(\mathbb R)$ homeomorphic to $L^1(\mathbb R)$? More generally, are there instances of surprising homeomorphisms between non-isomorphic Banach spaces? REPLY [40 votes]: According to Are $L^\infty(\Bbb R)$ and $L^2(\Bbb R)$ homeomorphic?, the map $f \mapsto \operatorname{sgn}(f) |f|^2$ is a homeomorphism from $L^2$ to $L^1$. It's clearly a bijection. Suppose $f_n \to f$ in $L^2$; set $g_n = \operatorname{sgn}(f_n) |f_n|^2$ and $g$ likewise. Passing to a subsequence we can assume $f_n \to f$ a.e., so that $g_n \to g$ a.e. also. And we have $\|g_n\|_1 = \|f_n\|_2^2 \to \|f\|_2^2 = \|g\|_1$. By this lemma due to Riesz (a clever use of Fatou's lemma) we have $g_n \to g$ in $L^1$. So the map is continuous. The same argument shows the inverse is also continuous. This works over any measure space. Thanks to Yemon Choi for pointing out that this observation is apparently due to S. Mazur, Studia Math., 1929 (EuDML). Much more powerfully, all separable infinite-dimensional Banach spaces are homeomorphic. MR0209804 Kadec, M. I. A proof of the topological equivalence of all separable infinite-dimensional Banach spaces. (Russian) Funkcional. Anal. i Priložen. 1 1967 61–70. (Reviewer: G. L. Krabbe) English translation at: M.I. Kadets, Functional Analysis and Its Applications 1 (1967) p 53. (https://doi.org/10.1007/BF01075865) The MathSciNet review points out that, combined with other contemporaneous work, all separable infinite-dimensional Fréchet spaces are homeomorphic. I found this reference in a Math.SE post by Tomek Kania, where Tomek says that two infinite-dimensional Banach spaces are homeomorphic iff they have dense subsets of the same minimal cardinality.<|endoftext|> TITLE: Definitional complexity of truth in $L$ without $0^{\#}$ QUESTION [9 upvotes]: Cross posted from MSE at commenter's suggestion: https://math.stackexchange.com/questions/2269319/definability-of-truth-in-l-without-0 I'm interested in the relationship between the existence of $0^{\#}$ and the definability of truth in $L.$ My advisor showed me an argument that you don't need $0^{\#}$ for truth in $L$ to be definable (from a Mahlo cardinal you can make a model where regular cardinals $\kappa$ such that $L_{\kappa} \prec L$ are precisely those where GCH fails). Of course, this definition requires quantification over the set-theoretic universe, which is much less impressive than the $\Delta^1_3$ definability of $0^{\#},$ and I don't see how to "mark" countable ordinals in the way forcing failures of GCH can mark regular cardinals. So my question is, how low complexity can a definition of truth (without parameters) in $L$ be if $0^{\#}$ doesn't exist? Can there still be a definition with only real quantification? Could it even be $\Delta^1_3?$ REPLY [10 votes]: The answer is yes, the theory of $L$ can be definable by a low-complexity definition quantifying over reals, even when $0^\sharp$ does not exist. Here is one way to achieve this. Let me assume that the theory of $L$ is an element of $L$. This happens, for example, if $L_\kappa\prec L$ for some ordinal $\kappa$, because in this case the theory of $L$ is the same as the theory of $L_\kappa$, which is an element of $L$. So let $t$ be the theory of $L$, which I have assumed is (coded by) a real in $L$. This real is therefore the $\alpha^{th}$ real in the $L$-order, and in order to define the theory $t$, it will suffice to define the ordinal $\alpha$. Let $L[G]$ be a forcing extension of $L$ forcing to collapse $\aleph_{\alpha}^L$ to $\omega$. So in $L[G]$, the true $\omega_1$ is the same as $\omega_{\alpha+1}^L$, and we can determine this inside $H_{\omega_1}$. In that structure, we can define the class of ordinals that are cardinals in $L$, and there will be exactly $\alpha$ of them. This makes the theory of $L$ definable inside $V=L[G]$ by a formula quantifying only over reals, and it will be $\Delta^1_n$ for some smallish $n$. Let's try to find out how complex the definition is. My proposed definition is that in $V=L[G]$, the theory of $L$ is the theory coded by the real $t$ which is the $\alpha^{th}$ real in the $L$ order, where $\alpha$ is the number of infinite $L$-cardinals that are countable in $V$. So, $t$ is as desired if there is a countable transitive model $L_\beta$ that thinks $t$ is the $\alpha^{th}$ real and which has exactly $\alpha$-many infinite cardinals, which do not get collapsed in any larger countable transitive $L_\gamma$, whereas all larger countable ordinals above those ordinals do get collapsed in some larger $L_\gamma$. What is the complexity? It seems to be $\Sigma^1_4$. My initial thought that it might be $\Sigma^1_3$ are not right, as explained in the comments, since in that case it would be upward absolute by further forcing, but it clearly is not, since we could collapse more cardinals and thereby change the meaning of $\alpha$.<|endoftext|> TITLE: Endomorphism ring spectrum of the Eilenberg-MacLane spectrum QUESTION [24 upvotes]: Consider the endomorphism ring spectrum $R = \mathrm{End}_S(H\mathbb{F}_p)$ of the mod $p$ Eilenberg-MacLane spectrum $H\mathbb{F}_p$. The homotopy groups of $R$ are the Steenrod algebra $A^*$ with reversed grading: $$\pi_n R = [\Sigma^n H\mathbb{F}_p, H\mathbb{F}_p] = A^{-n}.$$ This spectrum $R$ is an associative $S$-algebra (or $A_{\infty}$ ring spectrum). Moreover, $R$ is an $H\mathbb{F}_p$-module spectrum, say, using the $H\mathbb{F}_p$-module structure on the target $H\mathbb{F}_p$. In particular, $R$ is an $H\mathbb{Z}$-module spectrum. However, $R$ is known not to be an $H\mathbb{F}_p$-algebra spectrum. Question. Is $R = \mathrm{End}_S(H\mathbb{F}_p)$ an $H\mathbb{Z}$-algebra spectrum? My hunch is that the answer is no, but I couldn't find that statement in the literature. Perhaps it can be shown using an invariant of structured ring spectra, some flavor of $THH$. Or perhaps a dg-algebra over $\mathbb{Z}$ doesn't have enough room to encode the homotopical structure of the Steenrod algebra [1]. Remark. For the sake of definiteness, feel free to pick a model of spectra such as $S$-modules or symmetric spectra. The question is meant to be about the underlying symmetric monoidal $\infty$-category. In light of [2], working with your favorite model of spectra should be fine. [1] Shipley, Brooke, $H\Bbb Z$-algebra spectra are differential graded algebras, Am. J. Math. 129, No. 2, 351-379 (2007). ZBL1120.55007. [2] Mandell, M.A.; May, J.P.; Schwede, S.; Shipley, B., Model categories of diagram spectra, Proc. Lond. Math. Soc., III. Ser. 82, No.2, 441-512 (2001). ZBL1017.55004. REPLY [30 votes]: No, $A$ is not an $H\Bbb Z$-algebra. Suppose $R$ is an $H\Bbb Z$-algebra. Then the category of left $R$-modules is $H\Bbb Z$-linear: for any $R$-modules $M$ and $N$, the function spectrum $F_R(M,N)$ naturally has the structure of an $H\Bbb Z$-module. One reason for this is that $R$ is now an algebra object in the symmetric monoidal closed category of $H\Bbb Z$-modules, and the internal $R$-module function objects can be given weakly equivalent constructions there instead of in the category of $\Bbb S$-modules. In particular, the unit $\Bbb S \to F_R(M,M)$ of the endomorphism ring factors through $\Bbb S \to H\Bbb Z \to F_R(M,M)$ because the endomorphism ring is now an algebra in $H\Bbb Z$-modules. However, if we take $R$ to be the Steenrod algebra spectrum and $M = H\Bbb F_p$ with the action it has by definition, there is an equivalence of the $R$-linear endomorphisms of $M$ with the $p$-completed sphere: $$ F_A(H\Bbb F_p, H\Bbb F_p) \simeq \Bbb S^\wedge_p $$ (This, for example, is what gives rise to the Adams spectral sequence.) In particular, the unit $\Bbb S \to \Bbb S^\wedge_p$ doesn't factor through $H\Bbb Z$ for any prime. I think that many people find this pretty surprising when they first encounter it; I certainly did.<|endoftext|> TITLE: Derived noncommutative geometry includes derived, or spectral algebraic geometry? QUESTION [8 upvotes]: Let $k$ be a commutative ring. In derived noncommutative (algebraic) geometry a "noncommutative space over $k$" is a $k$-linear $\mathrm{DG}$-category. This is motivated by the fact that homological invariants of an "ordinary" scheme and of its $\mathrm{DG}$-category of quasicoherent sheaves coincide, so we can study a scheme by studying it's $\mathrm{DG}$-category of quasicoherent sheaves. Then we generalize to general $\mathrm{DG}$-categories over $k$ and get a generalized, noncommutative space (over $k$). Note, that another, but quite similar framework for derived noncommutative geometry is defining a space to be a linear $A_{\infty}$-category. And both approaches ($\mathrm{DG}$ and $A_{\infty}$) are really the models for linear stable $\infty$-categories. But this noncommutative geometry was motivated by the desire to study noncommutative versions of schemes, wasn't it? However, in Lurie's "Structured Spaces" (http://www.math.harvard.edu/~lurie/papers/DAG-V.pdf) there is an abstract definition of "generalized" schemes which seems to include derived and spectral schemes, derived and spectral stacks. It really seems that (at least according to ncatlab: https://ncatlab.org/nlab/show/derived+algebraic+geometry#RelationToDerivedNoncommutativeGeometry) that such "generalized" schemes in the sense of Lurie are also can be represented by stable $\infty$-categories of quasicoherent sheaves, that is, a noncommutative "generalized" scheme over $k$ would also be $k$-linear $\mathrm{DG}$-category, thus a special case of Kontsevich's noncommutative algebraic geometry? That said, is it true that in derived noncommutative geometry (such as studied by Kontsevich, Katzarkov, Kaledin, Orlov, Tabuada) one studies "noncommutative" versions not only of ordinary schemes but also of such geometric objects as: algebraic stacks (Artin or Deligne-Mumford) derived schemes (in the "simplicial commutative rings" sense) spectral schemes (in the "$E_{\infty}$-rings" sense, see Lurie "Spectral Algebraic Geometry) derived algebraic stacks (Artin or Deligne-Mumford) spectral Deligne-Mumford stacks If not, which objects are included in "derived noncommutative geometry" framework, and which are not? REPLY [4 votes]: There are several functors (for instance, $QCoh$ or $Perf$) from derived or spectral stacks (geometric or not) to stable $\infty$-categories. It is thus possible to study objects of derived/spectral algebraic geometry à la Toën-Vezzosi-Lurie from the point of view of derived noncommutative geometry à la Kontsevich. Though it is not clear to me at all how much information is loss by applying these functors (some information is, for sure). For instance, I don't think one can detect geometricity from the non-commutative view-point.<|endoftext|> TITLE: Hyperbolic planes inside hyperbolic 3-space quotients QUESTION [8 upvotes]: Let $\mathcal{H}_2 = \{(x,t) \in \mathbf{R}^2: t > 0\}$ be the upper half-plane, and let $\mathcal{H}_3$ be the hyperbolic 3-space $\{(x,t) \in \mathbf{C} \times \mathbf{R}: t > 0\}$. Clearly $\mathcal{H}_2$ embeds in $\mathcal{H}_3$. There is an action of $PSL(2, \mathbf{C})$ on $\mathcal{H}_3$, extending the familiar action of $PSL(2, \mathbf{R})$ on $\mathcal{H}_2$. If $\gamma \in PSL(2, \mathcal{O}_K)$, where $K$ is an imaginary quadratic field which is not $\mathbf{Q}(\zeta_3)$, is it the case that we have must have either $\gamma \cdot \mathcal{H}_2 = \mathcal{H}_2$ or $\mathcal{H}_2 \cap \gamma \cdot\mathcal{H}_2 = \varnothing$? For a general $PSL(2, \mathbf{C})$, or for non-integral $\gamma \in PGL(2, K)$, the intersection $\mathcal{H}_2 \cap \gamma \cdot\mathcal{H}_2$ can certainly be a nonempty proper subspace of $\mathcal{H}_2$ (in which case it's a geodesic arc). This case also seems to occur for some $\gamma \in PSL(2, \mathbf{Z}[\zeta_3])$, but I have found no other examples for any other quadratic fields. Is there a reason for this, or have I just not looked hard enough? REPLY [10 votes]: Let $\Gamma = PSL_2(\mathcal{O}_K)$. Note that complex conjugation acts as an orientation-reversing isometry on $\mathcal{H}_3$, the action being compatible with that on $\Gamma$. Let $\gamma\in \Gamma$ be such that $a = \mathcal{H_2}\cap \gamma^{-1}\mathcal{H}_2$ is a nontrivial geodesic. For all $x\in a$ we have $\bar{x}=x$ and $\gamma x = \overline{\gamma x} = \bar{\gamma}\bar{x} = \bar{\gamma}x$, so $\sigma = \gamma^{-1}\bar{\gamma}\in \Gamma$ fixes $a$ pointwise, so $\sigma$ is elliptic. Elementary considerations on the entries of the matrix $\sigma$, using $\sigma \in \Gamma$ $\bar{\sigma} = \sigma^{-1}$ $tr(\sigma)\in\mathcal{O}_K\cap (-2,2) = \{-1,0,1\}$ quickly rule out all cases where $K \neq \mathbb{Q}(\zeta_3)$.<|endoftext|> TITLE: Is there a closed-form solution for $\frac{dy}{dx} = 1 + \frac{a}{y} + \frac{b}{x}$? QUESTION [8 upvotes]: I am looking for an exact solution for the following special case of Chini Equation with $2\geq a > 1 > b > 0, x, y \in \mathbb{R}^+$, $$\frac{dy}{dx} = 1 + \frac{a}{y} + \frac{b}{x}$$ I have tried to approach this using multiple methods and substitutions, but none has gotten me far. I know that Chini equation tends not to have closed form solution, but this looks so simple to not have a closed form solution! Could someone please suggest a way to try obtaining the solution? REPLY [9 votes]: Maple does not find a closed-form solution. Note that the substitution $y = 1/u$ produces the Abel DE $$u' = - a u^3 - \frac{b+x}{x} u^2$$ but Maple can't find a closed-form solution for that either: it doesn't seem to fall into one of the known exactly solvable classes. EDIT: You might also note that scaling the dependent and independent variables by $y = cY$, $x = cX$ gives you a similar equation for $Y(X)$ with $a$ and $b$ replaced by $a/c$ and $b/c$ respectively. So effectively there is just one parameter here.<|endoftext|> TITLE: Which motivic cohomology groups of complex numbers are non-torsion? QUESTION [17 upvotes]: I would like to know which motivic cohomology groups of complex numbers are non-zero and ("better") non-torsion, i.e., for which $(i,j)$ the $i$th cohomology of the complex ${\mathbb{Q}}(j)$ over $\mathbb{C}$ is not zero. I would like to know both which of these groups are known to be non-zero (and also known to be "large") unconditionally and also which of them shouldn't vanish according to some conjectures. Certainly, my question can be easily translated into the language of $K$-theory. REPLY [12 votes]: This is going to be a slightly extended explanation, I apologize. The short version is basically that little is actually known (and even that is hard to prove), but conjecturally everything permitted by Beilinson-Soulé vanishing should be infinite-dimensional. What is known unconditionally (I bet everybody knows the first three items and I make no claim that this is an exhaustive list): the groups ${\rm H}^i({\rm Spec} \mathbb{C},\mathbb{Q}(n))$ vanish for $i>n$ (follows from Nisnevich cohomology dimension reasons). the groups ${\rm H}^n({\rm Spec} \mathbb{C},\mathbb{Q}(n))$ are uniquely divisible groups, uncountably-dimensional as $\mathbb{Q}$-vector spaces. (Milnor K-theory, related to elements in $\mathbb{C}^\times$) the groups ${\rm H}^{1}({\rm Spec} \mathbb{C},\mathbb{Q}(n))$ (corresponding to weight $n$ part of $K_{2n-1}$) are non-trivial infinite-dimensional $\mathbb{Q}$-vector spaces. The K-theory of $\mathbb{C}$ contains the K-theory of $\overline{\mathbb{Q}}$ as direct summand, and the groups ${\rm H}^{1}({\rm Spec}\overline{\mathbb{Q}},\mathbb{Q}(n))$ are vector spaces of countable dimension, detected by Borel's regulator (this computation is due to Borel). The rigidity conjecture voices the expectation that the inclusion $\overline{\mathbb{Q}}\subset \overline{\mathbb{C}}$ induces an isomorphism on ${\rm H}^{1}(-,\mathbb{Q}(n))$ for $n\geq 2$. C. Soulé. $p$-adic K-theory of elliptic curves. Duke Math. J. 54 (1987), 249--269. His Theorem 2.1 provides unconditional lower bounds for high-dimensional K-groups (related to the argument given below). From curves over number fields we would get contributions in ${\rm H}^2({\rm Spec}\mathbb{C},\mathbb{Q}(n))$, which is just above the Borel case above. For example, Soule's unconditional result tells us that weight 82 in $K_{162}(\mathbb{C})$ is unbounded (as well as all the bigger weights in the appropriate K-groups...). What is expected: (I think) Of course, the first thing to mention is that the Beilinson-Soulé vanishing conjecture predicts the vanishing of ${\rm H}^i({\rm Spec}\mathbb{C},\mathbb{Q}(n))$ for any $i\leq 0$ except for $i=n=0$. In particular, it would be conjectured that the motivic cohomology of $\mathbb{C}$ can only be nontrivial for $1\leq i\leq n$. Now I would like to claim that whatever nontrivial motivic cohomology groups are allowed by Beilinson-Soulé vanishing also contain nontrivial elements and are in fact infinite-dimensional vector spaces, depending on conjectures of Beilinson and Serre. First step: if $X/F$ is a smooth projective variety over a number field $F$, then the function field $F(X)$ is finitely generated and therefore embeds into $\mathbb{C}$. Second, the rational motivic cohomology of a function field embeds into the rational cohomology of $\mathbb{C}$. This follows from the existence of transfers on motivic cohomology; in a finite field extension, composition of the transfer with the natural presheaf restriction is multiplication by the degree and therefore an isomorphism for rational coefficients. So we cannot kill any motivic cohomology by finite extensions. On the other hand, finite transcendental extensions of algebraically closed fields also induce split injections (by evaluating on a rational point of the respective affine space). So, no rational motivic cohomology is lost by passing from the function field to $\mathbb{C}$. Third, (in the range we are interested in) the motivic cohomology of a smooth projective variety $X/F$ embeds into the motivic cohomology of its function field. This is essentially a form of the Gersten resolution. For ease of citation, I would replace motivic cohomology by Nisnevich cohomology of the Milnor K-sheaf (see my answer to this MO-question: Motivic cohomology and cohomology of Milnor K-theory sheaf) and use the Gersten resolution of the latter. Fourth, we now employ the conjectures of Beilinson and Serre (but note that all reduction steps so far were unconditional). For more information, see Schneider's survey "Introduction to the Beilinson conjectures" (see here) or this great MO-answer: Beilinson conjectures. If we have a smooth projective variety $X/F$, Beilinson's conjecture predicts an isomorphism between "the integral part" of motivic cohomology of $X$ with Deligne-Beilinson cohomology of $X(\mathbb{R})$ (under suitable condition on the indices). Put differently, the rank of the integral part of motivic cohomology (a subspace in ${\rm H}^{i+1}(X,\mathbb{Q}(n))$) should be equal to the order of vanishing of the L-function at $s=i+1-n$, assuming that $i/2+1 TITLE: Is there some relation between cluster algebras and crystal graphs? QUESTION [14 upvotes]: Cluster algebras are closely related to totally positivity in algebraic groups and canonical bases in quantum groups. Is there some relation between cluster algebras and crystal graphs? Can the crystal graph on page 19 of the book be described using cluster variables (or cluster monomials)? Thank you very much. REPLY [8 votes]: Yes, there are many relations between cluster algebras and crystal graphs. I am by no means an expert on these things, but let me mention one connection. Cluster algebras were originally discovered in the study of totally positive matrices and total positivity. This theory is ultimately interested in Lusztig's canonical basis $\mathcal{B}$ of the quantum group $U_q(\mathfrak{u})$, so crystal graphs are already not too far away. In fact, Berenstein-Zelevinsky relate Lusztig's parametrization of the canonical basis with Kashiwara's string parametrization of the dual canonical basis by studying totally positive varieties. The key technique is to use generalized minors $\Delta_{\lambda,\gamma}$, which turn out to be cluster variables for the cluster algebra structure on the coordinate algebra of the unipotent radical $U$. This cluster algebra structure gives rise to many rational, subtraction-free expressions among functions of representation theoretic interest, enabling tropicalization. In fact, Berenstein-Kahzdan have developed the theory of geometric crystals, which are highly-structured varieties that may be tropicalized to given normal Kashiwara crystal. Generalized minors play an important role here as well. More generally, double Bruhat cells $G^{u,v}=BuB\cap B_-vB_-$ are also cluster varieties: their coordinate rings $\mathbb{C}[G^{u,v}]$ each have the structure of a cluster algebra with cluster variables given by generalized minors. These also have connections to crystal graphs, but I know next to nothing about this more general set up. I think these examples served as a primary motivation development of cluster algebras by Fomin-Zelevinsky. The crystal graph from the book you link to is a highest weight crystal graph of type $A_2$, and its basis vectors are being parametrized by string data. String data may be related to generalized minors (hence cluster variables) via the results of Berenstein-Zelevinsky and tropicalization.<|endoftext|> TITLE: Cauchy identity in three sets of variables? QUESTION [10 upvotes]: The Cauchy identity states that $$ \prod_{i,j} \frac{1}{1-x_i y_j} = \sum_\lambda s_\lambda(x) s_\lambda(y), $$ where $s_\lambda(x)$ is the Schur function. Is there a known decomposition of the product $$ \prod_{i,j,k} \frac{1}{1-x_i y_j z_k} $$ as a sum of Schur functions? Essentially equivalently, let $U$, $V$ and $W$ be finite dimensional complex vector spaces. Is the irreducible decomposition of $Sym(U \otimes V \otimes W)$ known as a $GL(U) \times GL(V) \times GL(W)$-module? REPLY [13 votes]: Yes, up to the hard problem of determining Kronecker coefficients. Let $\Delta^\lambda$ be the Schur functor for the partition $\lambda$ of $r$ and let $S^\lambda$ be the corresponding irreducible representation of $S_r$ with character $\chi^\lambda$. Schur–Weyl duality states that $$ U^{\otimes r} \cong \bigoplus_{\lambda \in \mathrm{Par}(r)} \Delta^\lambda(U) \otimes S^\lambda $$ as a representation of $\mathrm{GL}(U) \times S_r$. Take $\dim U, \dim V, \dim W \ge r$. It follows by taking $S_r$ invariants in $U^{\otimes r} \otimes V^{\otimes r} \otimes W^{\otimes r}$ that the multiplicity of the irreducible $\mathrm{GL}(U) \times \mathrm{GL}(V) \times \mathrm{GL}(W)$-module $\Delta^\lambda(U) \otimes \Delta^\mu(V) \otimes \Delta^\nu(W)$ in $\mathrm{Sym}^r (U \otimes V \otimes W)$ is the Kronecker coefficient $g_{\lambda\mu\nu} = \langle \chi^\lambda\chi^\mu\chi^\nu, 1_{S_n} \rangle$. Restated in symmetric polynomials, this says $$ \prod_{ijk}\frac{1}{1-x_iy_jz_k} = \sum_{\lambda\mu\nu}g_{\lambda\mu\nu} s_\lambda(x)s_\mu(y)s_\nu(z)$$<|endoftext|> TITLE: Kindda-Perfect number: Is there a sequence of numbers which are equal to the sum of its proper divisors excluding itself as well as 1? QUESTION [6 upvotes]: Perfect number is a positive integer that is equal to the sum of its proper divisors. The smallest perfect number is 6, which is the sum of 1, 2, and 3. Is there a sequence of numbers which are equal to the sum of its proper divisors excluding itself as well as 1? If yes, what are they called? Name first few numbers in the sequence. Is there a general formula for them? Please provide with some link to read more. REPLY [20 votes]: Such an abundant number with abundance 1 is called a quasiperfect number (which is a more professional way to say "kindda-perfect"). None have been found, according to Wikipedia. This 1982 article says that if a quasiperfect number exists, it must be an odd square number greater than $10^{35}$ and have at least seven distinct prime factors. A recent article on this topic is here.<|endoftext|> TITLE: Other Arabic translations of the Arithmetica QUESTION [10 upvotes]: From what I've read, in 1975, Jacques Sesiano in his Ph.D. thesis managed to translate four 'new' books from Diophantus' Arithmetica. This brings the total number of books of Arithmetica salvaged to 10 distinct books, if we include the 6 from Heath's Greek translation. Does anyone know if any of the 6 Greek version of the books were found in Arabic? Has there been other major progress in finding the last 3 books? REPLY [4 votes]: As Sesiano writes in his book, the first three books that once existed in Arabic translation (by Qusta ibn Luqa) are lost. But al-Karaji quoted extensively from Diophantus Book III (and gives almost all of Book IV) in his work Fahri, which proves that the Books I-III must have been known in Arabic times. The situation has not changed since the publication of the Arabic books by Rashed and Sesiano: Rashed and Houzel do not mention any new discoveries in their book Les Arithmétiques de Diophante from 2013.<|endoftext|> TITLE: $\log \log p / \log \log n$, where $p|n$, gets equidistributed in [0,1] (for almost all $n$) QUESTION [7 upvotes]: According to Hardy-Ramanujan/Erdős-Kac we know that usually there are $\sim\log\log n$ prime numbers in a factorization. But if you pick up a natural number at random, and you factor it, what is the expected distribution of its factors? By the Turán–Kubilius inquality, we see that this is the expectation: $$ \left\{\frac{\log \log p}{\log \log n}:\ p\mid n\right\}$$ tend to equidistribute on $[0,1]$. Where can I find a more precise statement of the above? [Edit: a previous version contained comments irrelevant to the question] REPLY [12 votes]: For precise information on many questions of this type you should consult the Cambridge Tract, Divisors by Hall and Tenenbaum. Here is Hildebrand's review in the Bulletin of this book. A much more precise version of your question is a result of Erdős. Suppose $n =p_1^{\alpha_1} \cdots p_k^{\alpha_k}$ is the prime factorization of $n$ with $p_1 < p_2 < \ldots TITLE: A not quite theta not quite basic hypergeometric function QUESTION [8 upvotes]: The study of matrix quantum group coactions on the noncommutative disk algebra turns up the following series, which is a $q$-deformation of the negative binomial series, for integer $t\ge 0$, complex $z$ and $q\in[0,1]$: $$\sum_{n\ge 0}(-1)^n q^{n(n-1)/2}\frac{[n+t]_q!}{[t]_q![n]_q!}\,z^n = \sum_{n\ge 0} (-1)^n q^{n(n-1)/2}\frac{(q^{1+t},q)_n}{(q,q)_n}\,z^n$$ written in terms of $q$-factorials or alternatively $q$-Pochhammer symbols. Now this is not a theta function (in the $t=0$ case), as the sum is over positive $n$ only. It is not a basic hypergeometric function, as the standard formula would not have the $q^{n(n-1)/2}$ factor. If anybody could give me a reference for such a function, or has other comments, I would be very grateful. REPLY [4 votes]: Let's observe that $[n]_q!=1(1+q)(1+q+q^2)\cdots(1+q+\cdots+q^{n-1})=\frac{(q)_n}{(1-q)^n}$ so that \begin{align} \sum_{n\geq0}(-1)^nq^{\binom{n}2}z^n\frac{[n+t]_q!}{[n]_q![t]_q!} &=\sum_{n\geq0}(-1)^nq^{\binom{n}2}z^n\frac{(q)_{n+t}}{(q)_n(q)_t} \\ &=\sum_{n\geq0}(-1)^nq^{\binom{n}2}z^n\frac{(q^{t+1})_n}{(q)_n} \\ &=\lim_{\tau\rightarrow0}\,\,\, {}_2 \phi_1 \left({{\frac1{\tau},q^{t+1}}\atop{\tau}}\middle|\,\,q,z\tau\right) \\ &=\lim_{\tau\rightarrow0}\,\,\frac{(q^{t+1})_{\infty}(z)_{\infty}}{(\tau)_{\infty}(z\tau)_{\infty}}\,\,\cdot\,\, {}_2 \phi_1 \left({{\frac{\tau}{q^{t+1}},z\tau}\atop{z}}\middle|\,\,q,q^{t+1}\right) \\ &=(q^{t+1})_{\infty}(z)_{\infty}\,\sum_{n\geq0}\frac{q^{n(t+1)}}{(q)_n(z)_n}; \end{align} where we have applied a Heine transformation in the penultimate step. Perhaps other Heine transformations would lead to something more interesting.<|endoftext|> TITLE: "sinc'n determinant" QUESTION [14 upvotes]: The function $\text{sinc}(x)=\frac{\sin x}x$ permeates mathematics and physics in several aspects, and it carries multiple presentations/formulations. My interest is to inject yet another one of such. Let's understand the determinant $\det(M_{ij})_1^{\infty}$ to mean $\lim_{n\rightarrow\infty}\det(M_{ij})_1^n$. The following has experimental basis, therefore I like to ask: Question. Is there a proof for the determinantal representation $$\det\left[\frac{(i-1)!}{(2j-1)!}\binom{i^2-\theta^2}j\right]_{i,j=1}^{\infty}=\text{sinc}(\pi\theta) \,\,\,?$$ REPLY [4 votes]: This is just another way to compute $\det\left[\binom{i^2-\theta^2}j\right]$, which will spin out of a more generalized determinant evaluation. Define the determinants with indeterminates $\mathbf{X}=(X_1,X_2,\dots)$, $$M_n(a,b;\mathbf{X})=\det\left[\binom{X_{i+a}}{j+b}\right]_{i,j}^{1,n}.$$ Lemma. For $a, b\in\Bbb{N}_0$, we have \begin{align} \det\left[\binom{X_{i+a}}{j+b}\right]_{i,j}^{1,n} &=\prod_{1\leq i TITLE: Kunen (2011) exercise IV.8.17: $M[G]\vDash HOD^\mathbb{R}\subseteq L[\mathbb{R}]$ QUESTION [11 upvotes]: Assume that $V = L$ is true in the ground model $M$, and let $G$ be generic for $\mathbb{P} = Fn(I, 2)$, which is just the set of finite functions from $I$ to $2$. Then, ${\rm HOD}^{\mathbb{R}} = L[\mathbb{R}]$ is true in $M[G]$. I assume that the proof is just a modification of the proof that ${\rm HOD}$ in $M[G]$ is contained in $L[\mathbb{P}]$ in $M$, but I can't figure out how to do it. As in Kunen, ${\rm HOD}^\mathbb{R}$ are the sets hereditarily definable from ordinal and real parameters, and $L[\mathbb{R}]$ is the constructible hierarchy over the reals, i.e., $L[\mathbb{R}]_0 = \{\mathbb{R}\} \cup {\rm tcl}(\mathbb{R})$, $L[\mathbb{R}]_{\alpha +1} = {\rm Def}(L[\mathbb{R}]_\alpha)$, etc., and where ${\rm tcl}(x)$ is the transitive closure of $x$. Hopefully the exercise isn't too trivial for mathoverflow. REPLY [8 votes]: This question seems to be a little more subtle than the usual homogeneity arguments. Here is one way that I found to do it.$\newcommand\R{{\mathbb{R}}}\newcommand\HOD{\text{HOD}}\newcommand\Ord{\text{Ord}}$ Allow me to describe the problem like this. We start in $L$ and force with $\newcommand\P{\mathbb{P}}\P=\text{Add}(\omega,\kappa)$ to add $\kappa$ many Cohen reals. In the forcing extension $L[G]$, we form the class $\HOD_\R^{L[G]}$ of hereditarily ordinal-definable sets. Meanwhile, $L(\R)$ is the class of sets constructible from reals in $L[G]$, defined by starting with the real numbers of $L[G]$ and iteratively taking the definable power set. Theorem. $\HOD_\R^{L[G]}=L(\R)$. Proof. The inclusion $L(\R)\subseteq\HOD^\R$ is easy to see, as alluded to in the question, because every real is definable using itself as a parameter and, as can be seen by induction, every object in $L(\R)$ is definable in $L(\R)$ using finitely many ordinal and real parameters, and so every object there is definable in $L[G]$ using real and ordinal parameters. Indeed, in $L(\R)$ there is a definable surjection $s:\Ord\times\R\to L(\R)$, and this surjection is definable in $L[G]$. It remains to see, conversely, that $\HOD^\R\subseteq L(\R)$. Suppose that $A$ is hereditarily definable in $L[G]$ using ordinal and real parameters. By $\in$-induction, we may assume that $A\subseteq L(\R)$. Let $B=\{(\alpha,z)\mid s(\alpha,z)\in A\}$, which is also definable in $L[G]$, using the definable surjection. It suffices to show that $B\in L(\R)$. Since $B\in\HOD_\R^{L[G]}$, there is a formula $\varphi$ for which $$B=\{(\alpha,z)\mid L[G]\models\varphi(\alpha,z,\beta,a)\},$$ for some ordinal parameter $\beta$ and real parameter $a$. Fix names $\dot B$ and $\dot a$ for which $\dot B_G=B$ and $\dot a_G=a$, and fix a condition $p_0\in G$ forcing that $\dot B$ is defined by $\varphi(\cdot,\cdot,\check\beta,\dot a)$. Claim. $(\alpha,z)\in B$ if and only if there is $\dot z$ and $L$-generic $g$ for a countable complete suborder of $\P$ in $L$, with $p_0\in g$ and $\dot a_g=a$ and $\dot z_g=z$ and $\exists p\in g$ with $L\models p\Vdash \varphi(\check\alpha,\dot z,\check\beta,\dot a)$. Proof. The forward implication is easy, since we can take $g$ to be a fragment of $G$. Conversely, suppose that we have $g$ and $\dot z$ as stated. Since $g$ is an $L$-generic real in $L[G]$, we may extend $g$ to a full $L$-generic filter $H\subset\P$ with $L[G]=L[H]$. It follows that there is an automorphism of the forcing $\pi:\P\cong\P$ in $L$ with $\pi[H]=G$. Since $p\Vdash\varphi(\check\alpha,\dot z,\check\beta,\dot a)$, it follows that $\pi(p)$ forces $\varphi(\check\alpha,\dot z^\pi,\check\beta,\dot a^\pi)$, where the superscript $\pi$ means that we have applied the induced automorphism on names. Since $\dot a_G=a=\dot a_g=\dot a_H=\dot a^\pi_{\pi[H]}=\dot a^\pi_G$, it follows that there is some condition $q\in G$ forcing that $\dot a=\dot a^\pi$. Without loss, $q\leq \pi(p)$. So $q$ forces $\varphi(\check\alpha,\dot z^\pi,\check\beta,\dot a^\pi)$. Since $z=\dot z_g=\dot z_H=\dot z^\pi_{\pi[H]}=\dot z^\pi_G$, this implies $(\alpha,z)\in B$, as desired. QED (claim) The claim implies the theorem, since that definition can be carried out in $L(\R)$, using $a$ and $\beta$ as parameters and the fact that the forcing relation is definable in $L$. QED This argument is a little more complicated than I expected, and I'm not sure if there is a more direct argument.<|endoftext|> TITLE: From Gassmann-Sunada triples to isospectral manifolds QUESTION [10 upvotes]: A Gassmann-Sunada triple is a triple $(U,V,W)$ of groups, with $V, W$ subgroups of $U$, such that $U$ and $V$ meet every conjugacy class in $U$ in the same number of elements, and such that $V$ and $W$ are not conjugate. Such triples have been widely used for constructing isospectral manifolds. However, usually when I encounter these triples in the manifold context, I read something like "... and if one now can find a suitable manifold such that $U$ acts in the right way, then ...." My first question is: (a) Starting from a triple $(U,V,W)$ as above, how does one naturally construct isospectral manifolds, which are not isometric ? (So I want to see the manifolds arise only from the data $(U,V,W)$, and I also want a guarantee that the manifolds are not isometric.) Next, I wonder if someone can give me a reference on the second question: (b) Construct isospectral manifolds $\mathcal{M}_U$ and $\mathcal{M}_V$ from $(U,V,W)$ in the "natural way" (see the answers on (a)). Now let $T$ be a group properly containing $U$, and assume it "induces the same Gassmann-Sunada triple" (for instance, $T$ also acts on the same sets $X$ and $Y$ which give the permutation groups $(U,U/V)$ and $(U,U/W)$). My guess is that this "new" triple gives the same manifolds as those coming from $(U,V,W)$. Can anyone fill in the details here ? Bye ... REPLY [5 votes]: In my opinion, given a Gassmann-Sunada triple $(U,V,W)$, there is no any "natural" way to associate a pair of non-isometric isospectral manifolds. In general, one looks for a manifold $M$ such that $U$ can be embbeded (or something similar) in the isometry group of $M$. Now, I will give you a way to associate a pair of isospectral manifolds to any triple $(U,V,W)$, but I wouldn't call it "natural". It is just a very particular (and simple in my opinion) way to do it. We assume that $U$ is a finite group. Then, $U$ can be embbeded in some symmetric group $S_n$ of $n$ letters (is there a natural way?). Of course, $S_n$ is a subgroup of $SO(n)$. Then, $SO(n)/V$ and $SO(n)/W$ are isospectral manifolds! As Ben mentioned in his answer, these manifolds are going to be non-isometric if $V$ and $W$ are not conjugate in $SO(n)$, which I think it doesn't necessarily hold by assuming that $V$ and $W$ are not conjugate in $U$. I hope this helps you to your first question (a). I'm sorry, but I couldn't understand your question (b), particularly when you wrote "induces the same Gassmann-Sunada triple". REPLY [5 votes]: The essential computation isn't that hard. Let $G$, $H_1$, $H_2$, $M$ and $f:\pi_1(M) \to G$ be as in Ben Linowitz's answer. Let $X \to M$ be the Deck cover with Galois group $G$ corresponding to $f$ and let $Y_i = X/H_i$. Fix a real number $\lambda$. Let $V$ be the $\lambda$-eigenspace of $\nabla^2$ on $C^{\infty}(X)$; we note that $V$ is finite dimensional. Write $\rho: G \to GL(V)$ for the action. A function on $Y_i$ is a $\lambda$-eigenfunction of $\nabla^2$ if and only if it pulls back to a $\lambda$-eigenfunction on $X$, and a function on $X$ is pulled back from $Y_i$ if and only if it is $H_i$-invariant. So the multiplicity of $\lambda$ as an eigenvalue on $Y_i$ is the dimension of the space of invariants $V^{H_i}$. Note that $\pi_i := \tfrac{1}{H_i} \sum_{h \in H_i} \rho(h_i)$ is an idempotent with image $V^{H_i}$. So $$\dim V^{H_i} = \mathrm{Tr}(\pi_i) = \frac{1}{|H_i|} \sum_{h \in H_i} \mathrm{Tr} \ \rho(h).$$ Using conjugacy invariance of trace, and the obvious fact that $|H_1| = |H_2|$ concludes the result. Note that we could replace $C^{\infty}(X)$ with sections of some other vector bundle $\mathcal{E}$ on $X$, and $\nabla^2$ with some other differential operator $D$, so long as the $G$-action extended to an action on $\mathcal{E}$ and this action commutes with $D$.<|endoftext|> TITLE: computing the Weyl character formula QUESTION [20 upvotes]: Suppose $G$ is a connected compact group and $\pi$ is an irreducible representation of $G$. The Weyl character formula gives a formula for $\text{trace}(\pi(t))$ for $t\in G$. If $t$ is of order $n$ then $\text{trace}(\pi(t))$ is an integral linear combintation of $n^{th}$ roots of unity. Question: Is there software which will compute $\text{trace}(\pi(t))$ for arbitrary $t$ of finite order? This should include singular elements, in which case there is a modification of the Weyl character formula, which essentially uses L'Hopital's to evaluate the quotient (similar to the Weyl dimension formula). The reason I ask is that we are implementing such a formula in the Atlas of Lie groups and Representations software www.liegroups.org, (using Mark Reeder's The Compleat Weyl Character Formula for singular elements). On the one hand we don't want to reinvent the wheel. On the other, if we do reinvent it, we'd like to compare it to other wheels. REPLY [8 votes]: The Lie software from the 1990's (still available here) can do this easily. One of its fundamental functions is to compute weight multiplicities for characters of irreducible modules, as specified by their highest weight. The function domchar combines these multiplicities at all dominant weights; by the $W$-symmetry of the character this is really all information one needs, but provided you've got the memory to store the result, the software can compute the full formal character as well (a usually huge polynomial with natural number coefficients and weights as exponents). Note that while this is the same formal character you would get from the Weyl character formula (doing the division as an exact division of polynomials), LiE uses a different method (Freudenthal's recursion) to compute it. Now a torus element of finite order$~n$ can be represented as a rational co-weight, modulo the lattice of integral co-weights. Evaluating this co-weight on each of the exponents of the formal character, and reducing the results modulo$~1$, gives a formal linear combination of elements in $\Bbb Q/\Bbb Z$ that through the exponential mapping $\exp_1: t\mapsto\exp(2\pi\mathbf i t)$ correspond do roots of unity; their sum is the trace you asked for. The LiE function spectrum gives you the linear combination of elements in $\Bbb Q/\Bbb Z$, but scaled by the order$~n$ of the element to values in $\Bbb Z/n\Bbb Z$ so that the combination can be represented as a polynomial with integer exponents. One can represent this trace as an element of the cyclotomic field for$~n$, for which one takes the remainder of the output from spectrum by the $n$-th cyclotomic polynomial. Doing that in the LiE user programming language is not a hard exercise; for completeness I'll give below the code that does this (badly, but improving the efficiency of cyclotomic polynomials was not necessary to get satisfactory results). # first some univariate polynomial division stuff monic_quotient (pol dividend, divisor) = { loc quot = 0X[0]; loc d=degree(divisor) ; if divisor|[d] != 1 then error("Division by non monic polynomial") fi ; while degree(dividend)>=d do loc dd=degree(dividend); loc c = dividend|[dd] ; loc term = c*X[dd-d] ; quot += term; dividend += -term*divisor od ; quot } modulo (pol dividend, divisor) = { loc quot = 0X[0]; loc d=degree(divisor) ; if divisor|[d] != 1 then error("Division by non monic polynomial") fi ; while degree(dividend)>=d do loc dd=degree(dividend); loc c = dividend|[dd] ; loc term = c*X[dd-d] ; quot += term; dividend += -term*divisor od ; dividend } # the set of divisors of a number, needed for cyclotomic polynomials Cartesian (vec v, w) = { loc sv=size(v); loc result = null(sv*size(w)) ; for i=1 to size(v) do for j=1 to size(w) do result[i+sv*(j-1)] = v[i]*w[j] od od ; result } divisors (int b,n) = # divisors of n, assuming it has only 1 less than b { if b*b>n then [1,n] ; else loc r=n%b ; if r!=0 then divisors(b+1,n) else loc l=[1] ; while n%b==0 do l = 1 + b*l; n=n/b od ; if n>b then Cartesian(l,divisors(b+1,n)) else l fi fi fi } divisors(int n) = if n>=2 then divisors(2,n) else all_one(n) fi Phi(int n) = # cyclotomic polynomial, very inefficient version { loc divs=divisors(n); loc result=X[n]-X[0] ; for i=1 to size(divs)-1 do result=monic_quotient(result,Phi(divs[i])) od ; result } trace (vec lambda, t; grp g) = { loc denom=t[size(t)]; modulo(spectrum(lambda,t,g),Phi(denom)) }<|endoftext|> TITLE: Seeking for a meaning: a curious symmetry QUESTION [6 upvotes]: Suppose $\Phi(m,n)=(2m)!^n\prod_{k=1}^n\binom{2m+2k+x}{2k+x}$. Then, algebraically, it is trivial to see that $$(2m)!^n\prod_{k=1}^n\binom{2m+2k+x}{2k+x}=(2n)!^m\prod_{k=1}^m\binom{2n+2k+x}{2k+x}.$$ Question. Is there a combinatorial (or anything but algebraic) reason why $\Phi(m,n)=\Phi(n,m)$? UPDATE. I still await for any suggestion. REPLY [2 votes]: The question is a bit strange, because it would also be true if one removes all the $2$ from the statement. Namely: $$\Phi'(m, n):=m!^n \prod_{k = 1} ^ n \binom{m+k+x}{m}$$ is also symmetric. Let me explain the symmetry of this modified function $\Phi'$, and you can certainly adapt it to your $\Phi$. But the explanation is somewhat artificial. Assume harmlessly that $x$ is a non-negative integer. Consider a grid of size $m \times n$. The rows and columns are indexed $0, \cdots, m - 1$ and $0, \cdots, n - 1$, respectively, and a cell is indexed $(i, j)$ if it is in row $i$ and column $j$. Now one wants to fill in the grid with integers, one in a cell, with the following restriction: every integer is at most $m + n + x$, and the integer in the cell $(i, j)$ should be strictly larger than $i + j$. Here is the claim: For every $k \in \{0, \cdots, n - 1\}$, the number of different ways to fill in the column $k$ is equal to $$m!\binom{m + n - k + x}{m}.$$ This explains everything. To prove the claim, it is (again) algebraically obvious, but if you really want a combinatorial proof, then there is the following bijection: $$\{(a_1, \cdots, a_m): a_i \in \{1, \cdots, M\}, a_i \neq a_j \} \longleftrightarrow \{(c_1, \cdots, c_m): c_i \in \{1, \cdots, M - i\}\},$$ which is described as follows: given such $(a_i)_i$, define $c_i$ to be the integer such that $a_i$ is the $c_i$-th largest number in the set $\{1, \cdots, M\} \backslash \{a_1, \cdots, a_{i - 1}\}$. The procedure is obviously reversible, hence gives a bijection.<|endoftext|> TITLE: Linearizing a power series by conjugation QUESTION [8 upvotes]: Let $\mathfrak{I}:=\big\{ \, f:=\sum_{k=0}^\infty f_k z^k \in\mathbb{C}[[z]]\; : \text{s.t. }\; f_0=0 \;\text{ and }\; f_1=1\big\}$. A most basic result about linearization states that, for any $f\in\mathfrak{I}$ and for any $\lambda\in\mathbb{C}$ not a root of unity, there exists a unique $h\in\mathfrak{I}$ that linearizes $f(\lambda z)$: $$h(\lambda z)=f(\lambda h(z)).$$ Existence and uniqueness of this conjugation in the setting of formal power series is indeed easily established, as the coefficients of $h$ can be determined inductively from the data $(f_1,f_2,\dots) $ expanding the composition: one finds $$h_1=1$$ $$h_{n+1}={1\over \lambda^{n}-1 } \sum_{k\ge2,\,j} f_k \lambda^{k-1} h_{j_1}\dots h_{j_k},$$ the latter sum being extended over all integers $k\ge2$ and multi-indices $j\in\mathbb{Z}_+^k$ of length $|j|:=j_1+\dots+j_k=n+1$, thus with all components $j_i\le n$. The above recursion implies that the coefficients $h_n$ should be rational functions of $\lambda$, with denominators in the form $\prod_{k=1}^{n-1}(\lambda^k-1)^{m_{n,k}}$, where the exponents $m_{k,n}$ may be larger than $1$, at least at a first glance. However, after experiments with various $f$ and up to $n=50$ I observed that the form of $h_n$ always shows the simpler denominator, with all $m_{k,n}=1$: $$h_n={P_n(\lambda)\over (\lambda-1)(\lambda^2-1)\dots(\lambda^{n-1}-1)},$$ for some polynomials $P_n(\lambda)$ of degree at most ${n\choose 2}$. Is this really a general phenomenon? If so, why? Is there a simpler formula (maybe still recursive) for $h_n$ than the above recursion? REPLY [6 votes]: You may find useful information in a recent article by D. Sauzin and al. "Explicit linearization of one-dimensional germs through tree-expansions" here, where they use "mould calculus" (introduced by J. Écalle 40 years ago) to write down the coefficients $h_n$ and explore the combinaotrial structure of their family.<|endoftext|> TITLE: Incorporating Divisors (D4-branes) into Donaldson-Thomas Theory? QUESTION [6 upvotes]: Let $X$ be a Calabi-Yau threefold. Ordinary Donaldson-Thomas theory is formulated as a virtual count of ideal sheaves $\mathcal{I}$ with discrete invariants $\text{ch}(\mathcal{I}) = (1,0, -\beta, -n)$, which is equivalent to counting one-dimensional subschemes $Y \subseteq X$ with $[Y] = \beta \in H_{2}(X, \mathbb{Z})$ and $\chi(Y)=n$. In the derived category, we have the equivalence $\mathcal{I} \cong [\mathcal{O}_{X} \to \mathcal{O}_{Y}]$. For this reason, in the physical literature, they describe the DT theory as enumerating "bound states of D0-D2 branes with a single D6-brane. This is because $\text{ch}(\mathcal{O}_{X})=(1,0,0,0)$ represents a single D6-brane while $\mathcal{O}_{Y}$ represents the D0-D2 branes. I was thinking it would be nice if divisors (or what a physicist would call a D4-brane) could be placed on the same footing and incorporated into the DT partition function as well. Mathematically, I feel like it would be nice to sort of have all the holomorphic sub-geometry of $X$ in one generating function and physically, it would be preferable to have all the possible D-branes in the Topological B-model (D0,D2,D4,D6) branes sort of on the same footing. Of course divisors are special as they're given by vanishing of sections of line bundles, so I'm wondering if it's possible to write something like $$\text{Hilb}_{D, \beta, n}(X) \cong \text{Pic}_{D}(X) \times \text{Hilb}_{\beta' n'}(X)$$ where $\text{Hilb}_{D, \beta, n}(X)$ would be the Hilbert scheme of points, curves, and divisors. Note that $\beta$ and $n$ will be different from $\beta'$ and $n'$. We have a deformation/obstruction theory for $\text{Hilb}_{\beta' n'}(X)$ with a virtual class. Since $\text{Pic}_{D}(X)$ is smooth, can we compute $[\text{Hilb}_{D, \beta, n}(X)]^{\text{vir}}$ to define enumerative invariants? I was thinking that since the Picard factor is a torus, maybe by some localization argument it wouldn't contribute to an integral over the moduli space? Perhaps what I'm asking for is nonsense, but even if it's possible, I guess it would probably destroy modularity properties of the partition function? And would almost certainly lose connection to the Gromov-Witten theory via the MNOP conjecture. REPLY [5 votes]: The original definition of DT invariants ( https://arxiv.org/abs/math/9806111 ) works for any ch such that there is no strictly semistable objects. Later, this was generalized by Joyce and Song to arbitrary ch (giving rational invariants satisfying a conjectural multicovering formula). In general these invariants depend on a choice of Kähler class defining the notion of stability. The case of D4-D2-D0 branes has been studied by Gholampour and Sheshmani : see https://arxiv.org/abs/1309.0050 . The most recent work along this line is https://arxiv.org/abs/1601.04030 where you can find further references. The product definition given is the question is too naive because it neglects "interactions" between D4 and D2-D0 branes. I don't understand the remark on modularity properties: there is no modularity properties in the usual 1D6-D2-D0 case. But there are modularity properties in the D4-D2-D0 cases, which are partially predicted by S-duality. Concerning the relation with Gromov-Witten theory, there is a very surprising proposal connecting count of D4-D2-D0 branes with Gromov-Witten theory (or via MNOP with 1D6-D2-D0): the OSV conjecture https://arxiv.org/abs/hep-th/0405146 . The OSV conjecture has generated a lot of activity in physics but is still poorly understand mathematically: even its precise mathematical statement is not clear, but it is an interesting direction to pursue. In the D4-D2-D0 cases, a lot of the geometry can be reduced to some surface geometry which is much easier than the 3-fold geometry. The general case D6-D4-D2-D0 should obviously be the most interesting but I think that almost nothing is known for cases with multiple D6 (except very simple universal cases like those with only D6-D0).<|endoftext|> TITLE: Does the optimal strategy converge in poker if the SPR tends to infinity? QUESTION [9 upvotes]: This a a theoretical question about poker type games. I'm sure I don't have to explain the rules - you can consider No Limit Texas Hold'em or some simple theoretical model, where each player holds a number from the $(0,1)$ interval. We only consider heads-up, i.e., when two players play. Suppose that at the beginning of the game the size of the pot is $1$, and both players have $n$ chips (their stack). The stack-to-pot-ratio of the game, SPR, is $n$. Does the optimal strategy of the players converge as $n$ tends to $\infty$? To me this seems like a fundamental question, but I could find practically no information about it. I'm interested in any sort of related results, which is why I haven't given precise definitions. REPLY [3 votes]: The Clairvoyant Game Here is a well-known toy problem (the Clairvoyant Game) that doesn't converge: Suppose your hand is face-up. You have no hidden information. You don't know whether your opponent's hand is stronger. There is no drawing. The optimal strategy is for your opponent to bet all-in for value with all strong hands, and bluff with some weaker hands. If the chance for your opponent to have a stronger hand $p$ is too large, then you have to fold to all bets, even smaller bets. If $p$ is not too large, then you can call to neutralize bluffs while your opponent bluffs to neutralize your folds. Calling risks $n$ to gain $n+1$ so your opponent should bluff $n$ times for every $n+1$ value bets, with absolute probability $\frac{n}{n+1}p$, if that is possible. If $\frac{2n+1}{n+1}p \gt 1$ then you have to give up. Bluffing risks $n$ to gain $1$ so if $p \lt \frac{n+1}{2n+1}$ then you should call with probability $\frac{1}{n}$. As $n\to\infty$, the bet size does not converge. Further, your calling probability goes to $0$. However, the conditional probability that your opponent bluffs, given a bet, converges to $1/2$. The probability that your opponent bluffs converges to $p$. If you have a strong hand less than $1/2$ of the time, betting $n$ into a pot of size $1$ is not optional. It's not that you make an uncallable overbet, but a smaller one would do. If you don't bet all-in, you lose equity because you can't protect as many bluffs, or if your opponent calls with the right frequency, you don't get paid off enough when you do get called. If there are $r$ betting rounds, the optimal strategy is for your opponent to bet so the pot grows geometrically, $\frac{1}{2}(\sqrt[r]{2n+1}-1)$ times the pot. Your opponent sometimes bluffs $i$ times for each $i \in \{1,...,r\}$. As $n\to\infty$, the chance to bet on round $i$ goes to $2^{r+1-i}p$. Big Pots for Big Hands Will Sawin mentioned a version of this result in his answer. Suppose each player is dealt a hidden hand in $[0,1]$, with the lower number winning in case of a showdown. Claim: For any SPR $n$, in the optimal strategy, the probability that you put more than $b$ into the pot against any strategy is $O(1/b)$. Proof idea: If you put in more than $b$ more frequently, then you will often do so with hands that are not in the strongest $1/b$. So, your strategy will lose more than the pot to the suboptimal strategy of folding all hands except for the top $1/b$, and playing to provoke you to put at least $b$ into the pot, then reveal the hand and play the Clairvoyant Game above. If this bound were $o(1/b)$ then it would imply convergence of the optimal strategy as $n\to \infty$ since we could play the optimal strategy for SPR $b$ and forfeit if the pot reached $b$, costing $o_b(1)$. I believe the solutions to some restricted games indicate that there is some power law with a power strictly less than $-1$. As Will Sawin mentioned, this lets you prove some sorts of convergence of subsequences for $[0,1]$ games as long as you ensure that the ways for the pot to get up to $b$ are compact. Nut-Blockers So, what about poker games people play? Should we see people use the full stack, so that we might get convergence of equities and probabilities of betting but not amounts, or does poker resemble a $[0,1]$ game where the pot can't get very big because players are afraid of paying off extremely strong hands? The consensus among poker players may be that it almost never makes sense to bet more than the pot. However, the consensus of poker players is unreliable and it's wrong here. I've posted examples from my own play (in poker strategy forums) where I argued that particular overbets were much better than pot-sized bets. In some situations, this is to exploit suboptimal play, such as that people rarely want to fold a hand that has just improved, or to fold a full house or better, even if it is quite possible that someone has a stronger hand. However, there are also times when people have a limited range and the game resembles the Clairvoyant Game. You have to make large overbets to maximize the power of your range. Will Sawin mentioned the example of a nut-blocker hand, a hand that might not be strong, but which blocks your opponent from having the strongest hand. For example, if you are playing Texas Hold'em and the community cards are KQQ, the nut hand is QQ for four-of-a-kind. The second-best hand is KK for a high full house. The third best hand is KQ for the low full house. It also blocks your opponent from having the nuts. So, if a significant portion of your opponent's range is KK, and you have just QQ and KQ in your range, then you should overbet with QQ, and for every time you do that, you can overbet almost as often with KQ, playing the Clairvoyant Game.<|endoftext|> TITLE: Should you bet in poker against Darth Vader? QUESTION [12 upvotes]: This is a theoretical question about poker-type games. I'm not going to specify the rules. You can consider No Limit Texas Hold'em or some simple theoretical model, where each player holds a number from the $(0,1)$ interval. We only consider playing heads-up, i.e., when two players play. Playing against Darth Vader is quite tough because he can read your mind, and thus he knows your hand. You don't know Darth Vader's hand, but you have no hidden information. Should you ever bet against the Dark Lord? In games where cards are revealed at later stages, like Texas Hold'em, the answer can be yes, but even there it seems that you should bet/raise at most once after each time a new card is revealed. I'm interested in any sort of related results, which is why I haven't given precise definitions. Update. After Douglas's examples, the following interesting question is left open. If you bet/raise, should you always go all-in in a No Limit game? REPLY [3 votes]: Here is an answer to the updated question: Suppose that there are two betting rounds. Darth Vader has three types of hands. Type 1 wins with probability 1. Type 2 is a draw that hits (becomes a winning hand) with probability $1/10$ between the betting rounds. You can't see whether type 2 hands hit. Vader has Type 1 $1/100$ of the time, and type 2 $99/100$ of the time. You act first. The pot is $1$. The effective stack depth is $4$. Option 1: Push all-in (bet $4$). Then Vader calls with type-1 hands and folds type-2 hands. You gain $1$ unit $99\%$ of the time but lose not just the pot but an extra $4$ units $1\%$ of the time, for a net gain of $95\%$ of the pot. Option 2: Check to Vader. Vader can choose to check behind. Then Vader has a winning hand with probability $1/100 + 99/1000 = 109/1000$ on the river with an effective stack depth of $4$. The Nash equilibrium strategy is for you to check to Vader, who bets $4$ for value with all winning hands, and neutralizes your folds by bluffing $4$ times for every $5$ value bets. So, Vader bets $981/5000$ of the time with $10.9\%$ value bets and $8.72\%$ bluffs. This means you might as well fold all of the time, although you call $1/5$ of the time to neutralize Vader's bluffs. If you fold every time Vader bets, you win $80.38\%$ of the pot. We have not considered all of Vader's possible actions. (Vader can get more by raising some of the time.) However, this is enough to say that checking is worse than pushing all-in. Option 3: Bet $0.5$, then play to neutralize Vader's raises and later bets with type-2 hands. Before we do the calculations, as long as it is possible to neutralize drawing (type-2) hands, this is better hand-by-hand than pushing all in (option 1) because Vader might as well fold type-2 hands, and we might fold against type-1 hands, saving some of the remaining $3.5$ units. Suppose when Vader raises to $x$ total, then bets $4-x$ all-in on the next street, you call the raise with probability $p(x)$ and you call the push with probability $q(x)$. We can set the river calling frequency $q(x)$ so that these have the same equity, so that bluffing on the river has the same expected value as giving up. Vader risks $4-x$ to try to gain $2x+1$, so we call with probability $q(x) = \frac{2x+1}{x+5}$. That means if Vader raises with a draw and gets called, this costs Vader $x$ to get a situation worth $\frac{1}{10}(2x+1 + \frac{2x+1}{x+5}(4-x))$, a net cost of $\frac{10x^2+32x-9}{10x+50}$. On the first round, Vader risks $\frac{10x^2+32x-9}{10x+50}$ to get a reward of $\frac{3}{2}$. We can neutralize his decision by calling with probability $p(x) =\frac{\text{reward}}{\text{risk+reward}} = \frac{15x+75}{10x^2+47x+66}$. This means Vader's payoff comes only from type 1 hands. Suppose we call with probability $p(x) = \frac{15x+75}{10x^2+47x+66}$ on the first betting round and $q(x) = \frac{2x+1}{x+5}$ on the second round. Given a type-1 hand, Vader gets $\frac{3}{2}$ with probability $\frac{10x^2+32x-9}{10x^4+47x+66}$, $x+1$ with probability $\frac{60-15x}{10x^2+47x+66}$, and $5$ with probability $\frac{30x+15}{10x^2+47x+66}$. The average is $ \frac{486x+243}{20x^2+94x+132}$. If Vader knows your strategy, the best he can do is to set $x=\frac{3\sqrt{2}-1}{2}=1.62132$ and get $3.05944$ every time he has a type-1 hand, or $3.06\%$ of the pot. This is lower than the $5\%$ Vader gets when you push, and lower than the $18+\%$ Vader gets if you check, so betting less than all-in into Vader is better than checking or pushing, though another amount may be better than $0.5$.<|endoftext|> TITLE: The rank of a perturbed triangular matrix QUESTION [19 upvotes]: $\DeclareMathOperator{\rk}{rk}$ The question below is implicit in this MO post, but I believe it deserves to be asked explicitly, particularly now that I have some more numerical evidence. Suppose that $A$ is a real, square matrix of order $n$, such that all main-diagonal elements of $A$ are distinct from $0$, and of any two elements symmetric about the main diagonal, at least one is equal to $0$. That is, writing $A=(a_{ij})_{1\le i,j\le n}$, we have $a_{ii}\ne 0$ and $a_{ij}a_{ji}=0$ whenever $i,j\in[1,n]$, $i\ne j$. In other words, $A$ can be obtained from a non-singular triangular matrix by switching some pairs of symmetric elements. How small can the rank of such a matrix be, in terms of $n$? Since the pointwise product $A\circ A^t$ is full-rank, as an immediate corollary of the inequality $\rk(B\circ C)\le\rk(B)\,\rk(C)$ we have $\rk(A)\ge\sqrt n$. How sharp this estimate is? Is it true that $\liminf_{n\to\infty} \log(\rk(A))/\log(n)=\frac12$? Computations show that there are matrices of order $6$ and rank $3$ satisfying the assumptions above; taking $A$ to be a tensor power of such a matrix, we will have $\rk(A)=n^c$ with $=\log_6(3)\approx0.6131$. I do not know whether there exist matrices of order $7$ and rank $3$ with the property in question. REPLY [3 votes]: $\DeclareMathOperator{\rk}{rk}$It is possible to construct a matrix with $\rk(A)\leq 2\sqrt{n}$. Assuming that $n=r^2$ with an integer $r$, introduce two matrices $B$ and $C$, whose rows and columns are indexed by elements of $\{1,2,\dotsc,r\}^2$, and whose entries are defined by $$ B_{(x,y),(x',y')}=\begin{cases}1&\text{if }x=x',\\0&\text{otherwise},\end{cases} $$ and $$ C_{(x,y),(x',y')}=\begin{cases}-1&\text{if }y TITLE: The group of $k$-automorphisms of $k[x_1,\ldots,x_n,x_1^{-1}]$ QUESTION [7 upvotes]: Let $k$ be a field (of characteristic zero). For $k[x_1,\dotsc,x_n]$ it is known that the affine and triangular automorphisms generate $G_n$, the group of automorphisms of $k[x_1,\dotsc,x_n]$, see, for example, van den Essen's book "Polynomial automorphisms and the Jacobian conjecture". It is also known that $G_2$ is a free amalgamated group, see, for example, Dicks's paper "Automorphisms of the polynomial ring in two variables". This question asks what is $\hat G_n$, the group of automorphisms of $k[x_1,\dotsc,x_n,x_1^{-1},\dotsc,x_n^{-1}]$. Now, my question is: What is $\tilde G_n$, the group of automorphisms of $k[x_1,\dotsc,x_n,x_1^{-1}]$? Namely, we only invert one variable, say $x_1$. Thus far, what I was able to obtain is as follows: (1) $k[x,x^{-1}]$: This case was already done in the answer to the above mentioned question. (2) $k[x,x^{-1},y]$: If $f$ is an automorphism, then it is necessary that $f(x)$ is invertible in $k[x,x^{-1},y]$, so (if I am not wrong), $f(x)=\lambda x^d$, for some non-zero scalar $\lambda$ and some integer $d$. In order to be surjective, $d \in \{\pm1\}$. Then, unless I am missing something, $f(y)$ must be of the following form: $\varphi(y)= \mu y + \sum_{i=s}^{t}c_i x^{i}$, where $\mu$ is a non-zero scalar, $c_i$ are scalars, and $s \leq t$ are integers. So there are only ‘a few’ automorphisms, which are similar to the usual triangulars (the difference is the existence of negative exponents for $x$). (3) Perhaps the $n \neq 3$ is similar to the $n=2$ case. REPLY [7 votes]: I repeat my comment as an answer as suggested. Your argument for $n=2$ generalises and shows that $x_1$ is always mapped to some $\lambda x_1^m$ for $\lambda\in k^\times$ and $m\in\{\pm1\}$. Therefore if you consider the subring $S:=k[x_1^{\pm1}]$, a general $k$-linear automorphism of $R:=k[x_1^{\pm 1},x_2,\ldots,x_n]$ is a composition of a $S$-linear automorphism of $S[x_2,...,x_n]$ and a $k$-linear automorphism of $S$ (acting on $S[x_2,...,x_n$] by acting on coefficients). Phrased slightly differently $$Aut_k(R) = Aut_S(S[x_2,\ldots,x_n])\rtimes Aut_k(S) = Aut_S(S[x_2,\ldots,x_n]) \rtimes (k^\times \rtimes \{\pm1\})$$ But note that $Aut_S(S[x_2,\ldots,x_n])$ is not the group $G_{n-1}$, only a subgroup of it (and for a different field of coefficients: $k(x_1)$ instead of $k$), because $S$ is not a field. Nevertheless, this gives you a lot more information than you had before.<|endoftext|> TITLE: What are some kinds of models where DC holds? QUESTION [10 upvotes]: There are a lot of ways to build a model where DC fails. However, all of them that I'm aware of involve adding at least a messy set of reals (or rather, taking a forcing extension and then passing to an intermediate structure where some prescribed set becomes messy). It occurred to me recently that I don't know how to kill DC by only changing the reals; and after playing around with it for a couple days, I still don't see how to do it. I suspect I'm missing some very obvious ideas, but I haven't been able to find it on my own. And it nags at me a bit that I don't even know these basic facts about breaking DC, hence my asking here. Specifically, there are a few ways to phrase the question. The simplest one is: Q1. Does ZF+$V=L(\mathbb{R})$ imply DC? EDIT: I didn't quite ask the question I intended here. Although I'm very pleased to have a (negative) answer to the above question, the one I had in mind when I wrote this was: Q1'. If $M\models$ ZFC, does $L(\mathbb{R})^M\models$ DC? (I am very very tired right now, and I managed to convince myself that every model of ZF+$V=L(\mathbb{R})$ is the $L(\mathbb{R})$ of some ZFC-model, which is of course completely bonkers.) The negative answer Jing Zhang gives is the $L(\mathbb{R})$ of a very non-choicey model. At the more ambitious end, we have: Q2. Suppose $M\models$ ZFC and $G$ is a set of reals which is generic over $M$. Is DC necessarily true in $HOD^{M[G]}(M\cup G)$? I do mean "$M\cup G$" instead of "$M\cup G\cup \{G\}$" - I want to include only the individual reals in $G$. Also, note that $HOD(X)$ makes sense even when $X$ is a transitive class. Even Q1 seems implausible; Q2 seems downright ridiculous. But I can't cook up a counterexample at the moment. In general, I'm interested in learning when a substructure of a forcing extension can be easily seen to have DC: Q3. What are some properties of symmetric submodel constructions which "quickly" imply DC? REPLY [6 votes]: I believe the answer to Q1' is yes; here's a sketch of a proof. First notice that we can code elements of $L(\mathbb R)$ by ordinals and reals. That's because (except for the reals and their elements, which are thrown in to start the construction of $L(\mathbb R)$ and are easily coded), an element of $L(\mathbb R)$, say in $L_{\alpha+1}(\mathbb R)$, looks like $\{x\in L_\alpha:(L_\alpha,\in)\models\phi(x,\vec p)\}$ for some formula $\phi$ (codable by a natural number) and some parameters $\vec p$ from $L_\alpha$ (codable thanks to induction hypothesis). Since finitely many ordinals can (without AC) be coded as one ordinal and since finitely many reals can (without AC) be coded as one real, we have a set-theoretically definable surjection $F:\text{ORD}\times\mathbb R\to L(\mathbb R)$. Note that this coding scheme is absolute between $V$ and $L(\mathbb R)$. So DC in $L(\mathbb R)$ reduces to the special case where the choices are to be made, not from an arbitrary set, but from $\lambda\times\mathbb R$ for some ordinal $\lambda$. Working in $V$ where AC holds, we can get a sequence $\langle(\alpha_n,r_n):n\in\omega\rangle$ as required by DC, and we can do this in such a way that each of the ordinals $\alpha_n$ is as small as possible given all the earlier choices --- at each step, you just pick the smallest possible ordinal to serve as $\alpha_n$ and then, for this $\alpha_n$, pick some arbitrary appropriate $r_n$. This sequence, though produced in $V$, is in fact in $L(\mathbb R)$. The point is that the $\omega$-sequence of reals $\langle r_n\rangle$ can be coded as a single real, so we can give it to an inhabitant of $\langle(\alpha_n,r_n):n\in\omega\rangle$. Using that real, the inhabitant of $L(\mathbb R)$ can produce the sequence $\langle(\alpha_n,r_n):n\in\omega\rangle$, by always using the smallest possible $\alpha_n$ along with the real $r_n$ that we supplied, and this sequence is as required by DC. Note that I didn't need AC in $V$; DC in $V$ suffices. So (unless I've made a mistake) $\text{DC}^{L(\mathbb R)}$ is a theorem of ZF+DC. REPLY [4 votes]: Here's a negative answer to Q1 (it looks like the comment addressing this was flawed): It is consistent with ZF that there is an infinite Borel (in fact, $F_{\sigma\delta}$) set $B$ with no greatest element which is Dedekind-finite. Now every Borel set is coded by a real, so it is consistent with ZF + $V=L(\mathbb{R})$ that there is such a Borel set. But now consider the relation $<$ restricted to $B$: DC fails for this relation, since otherwise we could embed $\omega$ into $B$, contradicting the Dedekind-finiteness of $B$. Why we can assume $B$ has no greatest element: let $B_0=B$, and $B_{i+1}=B_i$ minus $B_i$'s greatest element, if it exists, and $B_i$ otherwise. This sequence has to stabilize at a finite stage, since otherwise we get an embedding of $\omega$ into $B$; but the resulting $B_i$ is still Borel (in fact, $F_{\sigma\delta}$), infinite, and Dedekind-finite, so we can WLOG assume that this was the original $B$ we picked.<|endoftext|> TITLE: "strange" diophantine and parity of the partition function QUESTION [11 upvotes]: Let $\{x_i\}:=\{x_1=5, x_2=13, x_3=29, x_4=37, x_5=45, \dots \}$ be the sequence of those positive integers of the form $$ p^{4\alpha+1}n^2$$ in increasing order where $p\equiv 5\pmod 8$ is prime and $\gcd(n,2p)=1$. Next, define the sequence $\{y_i\}$ by letting $y_i:=\frac{x_i-5}{8}$. The first few terms of the $q$-series $\sum_{i=1}^{\infty}q^{y_i}$ are $$\sum_{i=1}^{\infty}q^{y_i}=1+q+q^3+q^4+q^5+q^6+q^7+q^{12}+q^{13}+q^{14}+q^{18}+q^{19}+\cdots.$$ If $p(N)$ denotes the number of ordinary integer partitions of $N$, then $$ \sum_{N=0}^{\infty}p(N)q^N\equiv 1+q+q^3+q^4+q^5+q^6+q^7+q^{12}+q^{13}+q^{14}+q^{16}+q^{17} +q^{18}+\cdots\mod 2. $$ By comparing these two $q$-series one finds for $N\leq 15$ that $p(N)$ is odd precisely for those $N$ that are elements in the sequence $\{y_i\}$. If one searches for long strings of consecutive even values of the partition function in arithmetic progressions, one immediately finds in the progression $10\pmod{16}$ that $$p(10)\equiv p(26)\equiv p(42)\equiv p(58)\equiv p(74)\equiv 0\pmod 2,$$ but $p(90)\equiv 1\pmod 2$. It is curious to note that $10, 26, 42, 58, 74\not \in \{y_i\}$ but $90=y_{45}$. This motivates me to inquire: Question. If $N< 16^k$, is the following true? $$p(N)\equiv \# \left \{ (z_0,z_1,\dots z_{k-1})\ | \ z_j\in \{y_i\}\ \text {and}\ \sum_{j=0}^{k-1}16^jz_j=N \right \} \mod 2. $$ REPLY [12 votes]: Your conjecture is true! Here is one way to get it using some mod 4 generatingfunctionology. The answer got a bit long, so I divided it into two parts, as an attempt to improve readability. Part1: Let's denote by $Y(q)$ your generating function $\sum_{i\geq 1}q^{y_i}$. We will prove that $$Y(q)\equiv \prod_{k\geq 1}(1+q^{2k-1})(1+q^{16k})\pmod{2}.$$ Proof: One very classical theorem that dates back to Legendre states that the number of ways of writing $n\in \mathbb N$ as a sum of four triangular numbers is equal to $\sigma(2n+1)$, the sum of divisors of $2n+1$. Since the generating function of triangular numbers admits an infinite product expression from Jacobi's triple product formula, we can rewrite this theorem in the language of generating functions: $$\sum_{n\geq 0}\sigma(2n+1)q^n=\prod_{k\geq 1}\frac{(1-q^{2k})^8}{(1-q^k)^4}.$$ I made use of this in an older answer, where we also pointed out the following simple corollary: $$\sum_{n\geq 0}\sigma(4n+1)q^n\equiv\prod_{k\geq 1}\left(\frac{1-q^{4k}}{1-q^k}\right)^2=\prod_{k\geq 1}(1+q^{2k-1})^2(1+q^{2k})^4\pmod{4}.$$ Let us denote $$A(q)=\sum_{n\geq 0}\sigma((2n+1)^2)q^{\binom{n+1}{2}}.$$ The next lemma will allow us to split $\sum_{n\geq 0}\sigma(2n+1)q^n \pmod{4}$ into its even and odd parts. More precisely it will show $$\sum_{n\geq 0}\sigma(2n+1)q^n\equiv A(q^2)+2qY(q^2) \pmod{4}$$ Lemma: The value of $\sigma(4n+1)$ is $1,3\pmod{4}$ iff 4n+1 is a perfect square, and it is $2\pmod{4}$ iff $n=2y_i+1$ for some $i$. Proof of lemma: An odd number $m$ has a factorization $\prod p_i^{\alpha_i}q_j^{\beta_j}$ where $p_i$'s are primes $1\pmod 4$ and $q_j$'s are primes $3\pmod{4}$. The function $\sigma$ is multiplicative, so we just need to check the prime powers individually. $$\sigma(p_i^{\alpha_i})=\frac{p_i^{\alpha_i+1}-1}{p_i-1}\equiv \alpha_i+1\pmod{4}$$ $$\sigma(q_j^{\beta_j})=\frac{q^{\beta_j+1}-1}{q_j-1}\equiv\left\{ \begin{array}{ll} 1\pmod{4} & \beta_j\text{ is even} \\ 0\pmod{4} & \beta_j\text{ is odd} \\ \end{array} \right.$$ Using the fact that $\sigma(m)=\prod \sigma(p_i^{\alpha_i})\sigma(q_j^{\beta_j})$ we see that $\sigma(m)$ is odd iff each $\alpha_i$ and $\beta_j$ are even, iff $m$ is a perfect square. We also see that $\sigma(m)=2\pmod{4}$ iff all $\beta_j$ are even, and exactly one $\alpha_i$ is $1\pmod{4}$, which is equivalent to saying that $m$ is one of the elements of your $x_i$ sequence. Now, $4n+1=x_i \implies n=2y_i+1$, so this concludes the proof of the lemma. Some algebraic manipulations give us $$\prod_{k\geq 1}(1+q^{2k-1})^2(1+q^{2k})^4=\prod_{k\geq 1}(1+q^{4k-2})(1+q^{2k})^4\prod_{k\geq 1}(1+2\frac{q^{2k-1}}{1+q^{4k-2}})$$ $$\equiv \prod_{k\geq 1}(1+q^{4k-2})(1+q^{2k})^4\left(1+2\sum_{k\geq 1} \frac{q^{2k-1}}{1+q^{4k-2}}\right)\pmod{4}.$$ Therefore we have an expression for $Y(q)\pmod{2}$ $$Y(q)\equiv \prod_{k\geq 1}(1+q^{2k-1})(1+q^{k})^4\left(\sum_{k\geq 1} \frac{q^{k-1}}{1+q^{2k-1}}\right)\pmod{2}$$ $$\equiv \prod_{k\geq 1}(1+q^{2k-1})(1+q^{k})^4\left(\sum_{k\geq 1} \frac{q^{k-1}}{1-q^{2k-1}}\right)\pmod{2}.$$ Next we notice that the coefficient of $q^n$ in $\sum_{k\geq 1} \frac{q^{k-1}}{1-q^{2k-1}}$ counts the number of integers $r\geq 0,k\geq 1$ such that $n=k-1+r(2k-1)$, which can be rewritten as $2n+1=(2k-1)(2r+1)$. The number of solutions to this equation is precisely the number of divisors of $2n+1$, and since the divisors of an odd number are odd, the number of divisors of $2n+1$ has the same parity as the sum of divisors of $2n+1$. Therefore $\sum_{k\geq 1} \frac{q^{k-1}}{1-q^{2k-1}}$ is also equal to $\sum_{n\geq 0}\sigma(2n+1)q^n\pmod{2}$. Plugging this in our equation gives $$Y(q)\equiv \prod_{k\geq 1}(1+q^{2k-1})(1+q^{k})^4\frac{(1-q^{2k})^8}{(1-q^k)^4}\equiv \prod_{k\geq 1}(1+q^{2k-1})(1+q^{16k})\pmod{2}$$ and this finishes our proof. $\blacksquare$ Part2: Your conjecture would follow easily from the equality $$\prod_{k\geq 1}\frac{1}{1-q^k}\equiv \prod_{i\geq 0}Y(q^{16^i})\pmod{2}.$$ Indeed the coefficient of $q^n$ on the left represents the parity of $p(n)$, the number of partitions of $n$, whereas the right hand side represents the number of ways of writing $n=z_0+16z_1+16^2z_2+\cdots$ where each $z_j$ is equal to some $y_i$. In order to prove this we will use the mod 2 identity $$\prod_{k\geq 1}(1+q^{2k-1})=\prod_{k\geq 1}\frac{1+q^k}{1+q^{2k}}=\prod_{k\geq 1}\frac{1+q^k}{(1+q^k)^2}=\prod_{k\geq 1}\frac{1}{1+q^k}\pmod{2}$$ which implies that $$\prod_{k\geq 1}(1+q^{16k})=\prod_{k\geq 1}\frac{1}{1+q^{16(2k-1)}}\pmod{2}.$$ We can use this in our expression for $Y(q)$ to get $$Y(q)=\prod_{k\geq 1}\frac{1+q^{2k-1}}{1+q^{16(2k-1)}}$$ which shows that the product $Y(q)Y(q^{16})Y(q^{16^2})\cdots$ telescopes to $\prod_{k\geq 1} (1+q^{2k-1})$ which is equal to the partition function $\pmod{2}$ thanks to the same identity above.<|endoftext|> TITLE: Why a random variable is better described by its cumulants than by its characteristic function? QUESTION [5 upvotes]: It is a classical and well known problem that a random variable $X$ is not uniquely determined by its moments $\mathbb{E}(X_n)$. The moment problem is the problem of determining the probability density of a random variable in terms of its moments, as well as the uniqueness of such a density given by the moments. I have read this from the Algebraic Combinatorics and Computer Science of H. Crapo and D. Senato: "In the first half of this century, the method of moments was replaced by Paul Levy by a more pliable method relying upon the characteristic function $\mathbb{E}(\mathrm{e}^{\mathrm{i}tX })$, which is used to this day to derive the limit theorems of probability in their sharpest form. There is however one drawback to the characteristic function: it has no obvious probabilistic significance. My teacher William Feller was aware that the religious invocation of characteristic functions is extraneous to probabilistic reasoning. He managed to avoid characteristic functions in his treatise on probability. To be sure, characteristic functions made an occasional appearance in the second volume, when he just could not do without them. However, it was his intention to write a third volume, dealing with Brownian motion and diffusion processes, in which characteristic functions would be relegated to the dustbin of history. Unfortunately, he died before he could accomplish this task." Other important objects that can better represent many properties of random variables than moments are cumulants. For $n\geq 1$, we consider a vector of real-valued random variables $X_{[n]}= (X_1,\ldots, X_n)$ such that $\mathbb{E}(|X_j|^{n}) <\infty, \forall\ j = 1,\ldots,n$. For every subset $b = \{j_1,\ldots, j_k\} \subset [n]=\{1,\ldots,n\}$ , one writes $X_b=(X_{j_1},\ldots, X_{j_k})$ and $X^{b}=X_{j_1}\times \ldots\times X_{j_k}$. From the $k$-dimensional vector $X_b$, one cane define the multivariate characteristicfunction of this vector as $$\phi_{X_b}(z_1,\ldots,z_k)=\mathbb{E}\Bigg[\exp\Big(\mathrm{i}\sum_{l=1}^{k}z_l X_{j_l}\Big)\Bigg].$$ The joint cumulant of the components of the vector $X_b$ is defined as \begin{align*} k(X_b)=(-\mathrm{i})^{k} \frac{\partial^{k} }{\partial z_1\ldots\partial z_k}\log\phi_{X_b}(z_1,\ldots,z_k)|_{z_1=\ldots=z_k=0}. \end{align*} I know that cumulants are polynomials in moments, invariants by the translation and additive if the the components of the vector $X_b$ are partly independents. They have also common properties with characteristic and generating function in the sense they characterize uniquely the distribution of a random variable as also they can characterize the independence of random variables. My question: why is a random variable is better described by its cumulants, which are combinatoric nature, than by its generating function or its characteristic function, which are analytical nature. REPLY [3 votes]: I think it is a bit misleading to contrast cumulants as combinatorial quantities with the characteristic function as analytic object. The characteristic function is an analytical device containing information about the moments. In the same way there is an analytic object (namely the logarithm of the characteristic function) which contains information about the cumulants. So I would say that there are moments and there are cumulants, dealing with them has often a combinatorial flavor, and there are analytic reformulations of moments and of cumulants which allow the use of more analytic tools (and, in particular, allow to deal with situations where no moments/cumulants exist). In some cases moments are better suited for the problem at hand, in some cases cumulants are. In the classical case, the closeness between the analytic avatars of moments and of cumulants (the first is the characteristic function, the second is the logarithm of the characteristic function) might be a reason that one usually does not talk so much about the analytic version of cumulants. In free probability theory the difference between the Cauchy transform (the analytic function for moments) and the $R$-transform (the analytic function for free cumulants) is much bigger and the parallelism between the combinatorial and the analytical side of moments/cumulants is more visible.<|endoftext|> TITLE: Homeomorphisms vs Borel automorphisms QUESTION [10 upvotes]: Let $\mathrm{Homeo}(M)$ and $\mathrm{Borel}(M)$ be the groups of homeomorphic and Borel automorphisms of a space $M$, respectively. Question: Are $\mathrm{Homeo}(M)$ and $\mathrm{Borel}(M)$ isomorphic as abstract groups for any reasonable non-discrete space $M$? Say, $M=\mathbb{R}$ or $M=\mathbb{S}^1$. Obviously $\mathrm{Homeo}(M)\subsetneq\mathrm{Borel}(M)$ if $M$ is non-discrete, but with infinite cardinality this does not prohibit an algebraic isomorphism. The literature on $\mathrm{Homeo}(\mathbb{R})$ and $\mathrm{Borel}(\mathbb{R})$ that I have seen so far has not been explicit enough for me to figure out the answer in this case. Thank you. REPLY [2 votes]: For every uncountable Polish space $M$ (hence every non-discrete metrizable manifold with countably many components), the groups $\mathrm{Homeo}(M)$ and $\mathrm{Borel}(M)$ are non-isomorphic. Furthermore, $\mathrm{Borel}(M)$ is not isomorphic to any subgroup of $\mathrm{Homeo}(M)$. Indeed, $\mathrm{Homeo}(M)$ is a separable metrizable group, and I claim that the subgroup $\mathfrak{S}_{\mathrm{fin}}(M)$ of finitely supported permutations of $M$ (which is a subgroup of $\mathrm{Borel}(M)$) does not embed into any separable metrizable topological group $G$. Indeed, let $(M_i)_{i\in I}$ be pairwise disjoint finite subsets of $M$, each of cardinal $\ge 3$, with $I$ uncountable, and let $\Gamma_i$ be a non-abelian subgroup of permutations of $M_i$, extended to the identity outside $M_i$. By contradiction, let $f$ be an embedding of $\mathfrak{S}_{\mathrm{fin}}(M)$ into $G$. Since every subset of a separable metrizable space is separable, there exists a countable subset $J$ of $I$ such that $f\Big(\bigcup_{j\in J}f(M_j)\Big)$ is dense in $f\Big(\bigcup_{i\in I}f(M_i)\Big)$. Choose $i\in I-J$. Then $[M_i,M_j]=\{1\}$ for all $j\in J$, so $[f(M_i),f(M_j)]=\{1\}$ . By density, we deduce $[f(M_i),f(M_i)]=\{1\}$. But the latter equals $f([M_i,M_i])$, which is not trivial since $M_i$ is non-abelian and $f$ is injective. Contradiction. For countable Polish spaces, one can give a full answer too: basically Goldstern's example is the only one along with discrete ones. Let $M$ be a countable Polish space. Then $\mathrm{Homeo}(M)$ is non-isomorphic to $\mathrm{Borel}(M)$ (which equals $\mathrm{S}(M)$, the whole permutation group), with the only exceptions when $X$ is finite, infinite discrete or the 1-point compactification of an infinite countable set. Namely, beyond these exceptions, $\mathrm{Homeo}(M)$ is infinite and has at least 5 normal subgroups. For $M$ infinite, by the Onofri-Schreier-Ulam theorem, $\mathrm{S}(M)$ has exactly 4 normal subgroups (trivial, whole, finite support, even finite support). In $\mathrm{Homeo}(M)$ we already have such a chain, where the intermediate subgroups are the group $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$ of finitely supported permutations of the subset $M_{\mathrm{iso}}$ of isolated points of $M$ ($M_{\mathrm{iso}}$ is dense in $M$, hence infinite) and its index 2 subgroup. Hence, if $\mathrm{Homeo}(M)$ is homeomorphic to $\mathrm{S}(M)$, then every normal subgroup is one of these. Let $M_{\mathrm{acc}}$ be the set of accumulation points. If $M_{\mathrm{acc}}$ is empty then $M$ is discrete and this case is clear. Next we assume $M_{\mathrm{acc}}$ non-empty. Let $H_M$ be the group of those self-homeomorphisms of $M$ that are identity on $M_{\mathrm{acc}}$, so $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})\subset H_M$. Hence we either have (a) $H_M=\mathrm{Homeo}(M)$ or (b) $H_M=\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$. Also denote $G_x$ as the subgroup of $\mathrm{Homeo}(M)$ consisting of those $g$ that are identity at the neighborhood of $x\in M$; if $x$ is $\mathrm{Homeo}(M)$-invariant then $G_x$ is a normal subgroup. Lemma: for every topological metrizable space and point $x_0$ which is isolated among accumulation point and neighborhood $V$ of $x_0$, there exists a self-homeomorphism fixing $x_0$, not with finite support, and identity outside $V$. Indeed, one can suppose that $V\smallsetminus\{x_0\}$ consists of isolated points of $X$; choose an injective sequence $(y_n)$ tending to $x_0$. Then permuting $y_{2n-1}$ and $y_{2n}$ and fixing all the remainder yields the desired permutation.$\Box$ Choose $x_0$ an isolated point in $M_{\mathrm{acc}}$. Applying the lemma to $x_0$, we see that $H_M$ is not reduced to $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$, which excludes (b). So assume (a): $\mathrm{Homeo}(M)$ acts trivially on $M_{\mathrm{acc}}$. In particular, $G_{x_0}$ is a normal subgroup; by the lemma it is a proper subgroup of $\mathrm{Homeo}(M)$. If we assume that $X$ is not the 1-point compactification of a discrete set, we either (a') have another point in $M_{\mathrm{acc}}$, and hence another isolated point in $M_{\mathrm{acc}}$ or (b') there exists an open infinite discrete subset in $M$. In both cases, we deduce (using the lemma in Case (a')) that $G_{x_0}$ is not reduced to $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$.<|endoftext|> TITLE: Curious fact about number of roots of $\mathfrak{sl}_n$ QUESTION [20 upvotes]: The Lie algebra $\mathfrak{sl}_n $ has many special features which are not shared by other simple Lie algebras, for example all of its fundamental representations are minuscule. I recently discovered another curious fact. The number of positive roots of $ \mathfrak{sl}_n $ is the same as the dimension of $ Sym^2 \mathfrak h$ --- both are $ \binom{n}{2} $. All other simple Lie algebras have more positive roots. Is this fact is connected to the special properties enjoyed by $\mathfrak{sl}_n$? Is there some a priori reason to expect this equality? I came upon this fact because I was thinking about the following linear map (here $\mathfrak g $ is any semisimple Lie algebra and $ \mathfrak h $ is its Cartan subalgebra): \begin{align*} Sym^2 \mathfrak h &\rightarrow U \mathfrak g \\ xy &\mapsto \sum_{\alpha \in \Delta_+} \langle \alpha, x \rangle \langle \alpha, y \rangle E_\alpha F_\alpha \end{align*} The set $ \{ E_\alpha F_\alpha \}_{\alpha \in \Delta_+} $ in $ U \mathfrak g $ is linearly independent and I was hoping that this map would be an isomorphism onto its span, but this can only be true for $ \mathfrak{sl}_n $. Has anyone seen this map before? REPLY [5 votes]: I will mostly answer spin's approach and question here, but it is rather long to write as a comment only. I am not sure if it will answer the original OP's question, but it does seem like a good step. Given a positive root $\alpha$, we associate to it, we associate to it (following spin's comment) a symmetric bilinear form $\varphi_\alpha$ on $\mathfrak{h}$, defined by: $\varphi_\alpha(h,h') = \alpha(h)\alpha(h')$ for all $h,h'\in \mathfrak{h}$. We claim that the set $\{\varphi_\alpha; \alpha \text{ is a positive root}\}$ spans $\operatorname{Sym}^2(\mathfrak{h})$ for simple algebras. This is equivalent to showing that any real homogeneous quadratic form $B(-,-)$ on $\mathfrak{h}^*$ which vanishes identically on the set of positive roots must vanish identically. Lemma: if $V$ is a real vector space, with a symmetric bilinear form $B(-,-)$ on $V$, and if $v_1, v_2\in V$ are linearly independent, and suppose further that $B(v_i,v_i) = 0$ for $i=1,2$ and that $B(v_1+cv_2, v_1+cv_2)=0$ for some real nonzero constant $c$, then $B(-,-)$ must vanish identically on the span of $v_1$ and $v_2$. The proof of the lemma is easy. Indeed it implies in particular that the $v_i$ are null, and that $B(v_1,v_2) = 0$. We now apply this lemma inductively. We start on one end of the Dynkin diagram, and apply the lemma to $v_1$ being the first simple root, and $v_2$ being the second simple root, which is connected to the first simple root by either a simple, a double or triple edge. Applying reflection to $v_2$ about the hyperplane orthogonal to $v_1$, we get the required third vector of the form either $v_1+cv_2$ or $v_2+cv_1$ for some $c>0$, which is also a positive root. We thus get that $B(-,-)$ vanishes identically on the span of the first 2 simple roots, and then apply the lemma again to the second simple root, and a new simple root connected to the second simple root by a simple, double or triple edge, and so on. This would show that $B(-,-) = 0$ identically, thus proving the claim. Hence this would show that the number of positive roots is always greater or equal to the dimension of $\operatorname{Sym}^2(\mathfrak{h})$, answering spin's question at least. The proof is conceptual and does not rely on the classification result. I am not sure if it really answers the OP's original question though. Edit 1: building up on my previous argument, we would like to show, in order to answer the OP's question, that if the $\varphi_\alpha$'s are linearly independent over $\mathbb{R}$ in $\operatorname{Sym}^2(\mathfrak{h})$, then the root system is of the A type. Thus we have to rule out the existence of multiple edges, and rule out the existence of Dynkin subdiagrams which are isomorphic to that of $SO(8)$ (here, note that I am unfortunately relying on the classification result). So let us assume that the $\varphi_\alpha$ are linearly independent. Then for any positive root $\alpha$, there exists a symmetric bilinear form $B(-,-)$ on $\mathfrak{h}^*$, such that $B(\beta,\beta) = 0$ for any positive root $\beta \neq \alpha$ and $B(\alpha,\alpha) \neq 0$. By restricting to a rank 2 Dynkin subdiagram, corresponding to 2 simple roots connected by a multiple edge, one can remove one positive root spanned by these 2 simple roots, and still get at least 3 positive roots satisfying the lemma above, so that $B$ must vanish identically on the span of these 2 simple roots, thus leading to a contradiction. It remains to rule out the existence of rank 4 Dynkin subdiagrams isomorphic to that of $SO(8)$ under the linear independence assumption on the $\varphi_\alpha$. This can be done explicitly by removing a specific positive root, and showing that if $B$ vanishes on all other positive roots, then it must be 0 on this rank 4 subspace of $\mathfrak{h}^*$, thus leading to a contradiction too. Alternatively, one can simply count the dimensions in this case: $SO(8)$ has 12 positive roots, while $\operatorname{Sym}^2(\mathfrak{h})$ has dimension 10, thus ruling out the existence of such Dynkin subdiagrams under the linear independence hypothesis on the $\varphi_\alpha$. Remark: the first part of my post does not rely on the classification result, while edit 1 does unfortunately rely on at least part of the classification result. Edit 1 consists in ruling out multiple edges and Dynkin subdiagrams isomorphic to that of $SO(8)$. I hope that over all, my post answers the OP's question (it could use some editing though).<|endoftext|> TITLE: Distribution of the area statistic for Catalan paths QUESTION [10 upvotes]: A Catalan path of semilength $n$ is a path from $(0,0)$ to $(2n,0)$ that proceeds by taking northeast (1,1) or southeast (1,-1) steps, and never goes below the $x$-axis. The area of a path $P$ is the area beneath the path and above the $x$-axis. So, for example, the path of semilength 3 that goes $(0,0)-(1,1)-(2,2)-(3,1)-(4,0)-(5,1)-(6,0)$ has area $5$. The sum of the area of all Catalan paths of semilength $n$ is known to be $4^n - \binom{2n+1}{n}$, so that the average area of a path is on the order of $n^{3/2}$. In an application I'm working on, I've found that it would be useful to have upper bounds on how many paths have area significantly greater than $n^{3/2}$. The largest possible area is $n^2$, and from Markov's inequality I can say that a vanishing proportion of paths have area greater than $n^2/1000$, say; specifically, there are no more than $O\left(4^n/n^2\right)$ such paths (whereas there are $\Theta\left(4^n/n^{3/2}\right)$ paths in total). I suspect that there are far fewer paths with large area, and for my application I need a much lower upper bound. Are there quantitative results for this problem? I can translate the problem into that of estimating the number of partitions of the number $N$ into at most $c\sqrt{N}$ parts each of size at most $c\sqrt{N}$, which in turn becomes a problem of estimating a coefficient of a Gaussian polynomial, but I'm not aware of quantatitive results here, either. REPLY [6 votes]: An attempt to provide the whole distribution using analytic combinatorics (though one answer is already accepted). In particular, we can obtain the first and the second moment using this technique (and maybe some more by induction if you want a limiting distribution and not only first two moments). The generating function for Catalan paths $$ Cat(z) = \sum_{n \geq 0} c_n z^n $$ satisfies an equation $$ Cat(z) = z^2 Cat(z) + 1 \enspace , \quad Cat(z) = \dfrac{1 - \sqrt{1 - 4z^2}}{2z^2} $$ where 1 corresponds to an empty path, then $z Cat(z) \cdot z$ corresponds to $ \nearrow$ times Catalan path times $\searrow$, then another Catalan path $Cat(z)$. The idea is to mark the area inside Catalan path, i.e. consider a bivariate generating function $$ C(z,u) = \sum_{n, k \geq 0} c_{n,k}z^n u^k \enspace , $$ where $ c_{n,k}$ is equal to the number of Catalan paths of semilength $n$ and area $k$. An equation here is the following one: $$ C(z,u) = z^2 u C(zu^2, u)C(z,u) + 1 $$ Explanation: the first part which starts from $\nearrow \ldots \searrow$ is a Catalan path of area $k + 2n + 1$, where $k$ is the area of upper part, $n$ is the semilength and $1$ is because you glue two parts of a triangle. For that reason, $$ uC(zu^2,u) = u\sum_{n, k \geq 0} c_{n,k}(zu^2)^n u^k = \sum_{n,k \geq 0} c_{n,k} z^n u^{k + 2n + 1} \enspace . $$ The second Catalan part is glued without chantings, i.e. we count already existing area as an additional summand. Next step is to obtain the moments from the functional equation. Though the equation cannot be solved explicitly, we can still obtain some information. Recall that k-th factorial moment is given by $$ \dfrac{[z^n] \partial_u^k C(z,u) |_{u=1}}{[z^n] Cat(z)} $$ I denote $ C^\square(z) = \partial_u C(z,u)|_{u=1} $, $ C^{\square\square}(z) = \partial^2_u C(z,u)|_{u=1} $ for brevity. Let's take a derivative then: $$ C^\square(z) = z^2 C^2(z) + z^2 C^\square(z) C(z) + z^2(2C_z + C^\square)C(z) \enspace , $$ where $ C_z(z) := \partial_z C(z) $. Thus, $C^\square(z)$ can be explicitly expressed through $C(z) $ and $\partial_z C(z)$, and that's how you can obtain (already present) result on expectation of the area. Then you can repeat and obtain explicit expression for the second derivative, thus having a bound for the variance. In principle, you can try to derive some general pattern on the moments of this distribution (or more concretely, on the singularity $\rho(u)$ and nature of singularity of complex function $C^{k\square}(z)$), and apply Theorem IX.8 from Analytic Combinatorics of Flajolet and Sedewick. I believe that some of the summands in the functional equation will be negligible which will simplify the analysis. UPD: Actually, it happens that corresponding distribution is of area Airy type. See the article "Analytic Variations on the Airy Distribution" by Flajolet and Louchard: http://algo.inria.fr/flajolet/Publications/FlLo01.pdf<|endoftext|> TITLE: Generalizing Big O notation to arbitrary vector spaces QUESTION [5 upvotes]: I'm constructing a Coq library for Big-O notation. Naturally, I'd like it to be as general as possible. The Wikipedia page on Big-O notation says The generalization to functions taking values in any normed vector space is straightforward (replacing absolute values by norms) Of course, there's no inline citation. My attempt at doing this yielded the following: Definition big_O (f g : V -> V) : Prop := ∃ k : K, 0 < k ∧ exists n0 : K, 0 < n0 ∧ ∀ n : V, n0 ≤ ∥n∥ -> ∥f n∥ ≤ k * ∥g n∥. Which translates to the following in informal language: Definition 1. Given a vector space $V$ with (semi)norm $\lVert-\rVert$ over a totally ordered field $(K,\leq)$ and functions $f,g:V\to V$, we say that $f\in O(g)$ iff there exists some positive $k$ and $n_0$ in $K$ such that for all vectors $n\in V$ with $n_0 \leq \lVert n\rVert$, $\rVert f(n)\rVert\leq k\cdot\lVert g(n)\rVert$. However, I feel like the following definition is somewhat more elegant and general: Definition big_O (f g : V -> V) : Prop := ∃ k : K, k ≠ 0 ∧ ∃ n0 : V, ∀ n : V, ∥n0∥ ≤ ∥n∥ -> ∥f n∥ ≤ ∥k · g n∥. Which translates to the following: Definition 2. Given a vector space $V$ with (semi)norm $\lVert-\rVert$ over a totally ordered field $(K,\leq)$ and functions $f,g:V\to V$, we say that $f\in O(g)$ iff there exists some nonzero $k$ and some $n_0$ in $V$ such that for all vectors $n\in V$ with $\lVert n_0\rVert \leq \lVert n\rVert$, $\lVert f(n)\rVert\leq \lVert k\cdot g(n)\rVert$. So I guess I have a few questions: Does anyone have a good reference for the generalization to vector spaces? Do you think these definitions are equivalent? Any proofs or counterexamples? This is somewhat opinion-based, but any arguments for using one or the other? REPLY [4 votes]: Here is a straightforward generalization of the asymptotic notation to functions taking values in normed vector spaces: Let $X$ be a topological space, $\overline X\supseteq X$ a superspace in which $X$ is dense, and $a\in\overline X$. Let $Y$ be a normed vector space, and $f,g\colon X\to Y$. Then we say $f(x)=O(g(x))$ as $x\to a$ if there is a neighbourhood $U\ni a$ and $q>0$ such that $\|f(x)\|\le q\left\|g(x)\right\|$ for all $x\in U\cap X$; $f(x)=o(g(x))$ as $x\to a$ if for every $q>0$, there is a neighbourhood $U\ni a$ such that $\|f(x)\|\le q\left\|g(x)\right\|$ for all $x\in U\cap X$. This is for standard, real-valued norms. You seem to be interested also in norms valued in nonarchimedean ordered fields $K$. In this case, there are at least two useful possibilities how to cast the definition, namely (1) by taking $q\in\mathbb Q$, or (2) by taking $q\in K$. Which one is more suitable depends on the intended use case. Further generalizations are possible, seeing as we didn’t actually use most of the structure of $Y$. You are apparently trying to define only the special case with $a=\infty$. Thus, you need to decide what are appropriate neighbourhoods of infinity in your spaces (complements of bounded sets? complements of compact sets?).<|endoftext|> TITLE: Can a simple group be equivalent to a non-simple group? QUESTION [12 upvotes]: Two abstract groups $G$ and $H$ are called equivalent, $G\sim H$, if each of them is isomorphic to a subgroup of another. Question: Can a simple group $G$ be equivalent to a non-simple group $H$? Of course, we are talking about infinite groups here. Thanks. REPLY [14 votes]: Yes. (This is corrected and expanded since the first version.) There are easy examples of simple groups $G$ such that $G\times G$ is isomorphic to a subgroup of $G$. One example is the group of finitely supported even permutations of a countable infinite set. Another is the quotient of all permutations of a countably infinite set by the finitely supported ones. Another is the infinite special linear group (direct limit of $SL_n(k)$) of a field.<|endoftext|> TITLE: Maps to the universal punctured elliptic curve QUESTION [5 upvotes]: I have just started reading Hain's paper On the Universal Elliptic KZB Connection. I am a bit confused about a comment made there about base points on orbifolds. I am still very new to the idea of orbifolds so I apologise in advance for this likely being very trivial. Let $\mathcal{E}\to\mathcal{M}_{1,1}$ denote the universal elliptic curve over the moduli space of genus 1 curves with 1 marked point (over the complex numbers). Following Hain's Lectures on moduli spaces of elliptic curves, I am thinking about these as Riemann surfaces in the category of orbifolds, although they are also treated as stacks in the present paper. Let $\mathcal{E}'$ denote $\mathcal{E}$ with its identity section removed (i.e. the "universal punctured elliptic curve"). On page 5 Hain writes: A non-zero point $x$ on an elliptic curve $E$ determines (and is determined by) an orbifold map $[E,x]: \mathbb{C}\to\mathcal{E}'$. I don't really understand why this is. The picture in my head is that a point of $\mathcal{E}'$ is (roughly) a nonzero point $x$ on some elliptic curve $E$. Therefore constant maps $\mathbb{C}\to \mathcal{E}'$ correspond to pairs $[E,x]$ as above. But I don't understand why other possibly nonconstant maps $\mathbb{C}\to\mathcal{E}'$ should also correspond to such a pair $[E,x]$. Or are all such maps constant? REPLY [6 votes]: It is indeed true that a holomorphic map $\mathbb C \to \mathcal E'$ is constant. This is because it must factor through the universal cover of $\mathcal E'$, since $\mathbb C$ is simply connected. But the universal cover is a product of two copies of the complex unit disk, so the result follows by the maximum principle. Nevertheless I don't think that this is what he meant. My guess is that he accidentally left out the word "Spec", so that he meant to say that maps $\mathrm{Spec}\,\mathbb C \to \mathcal E'$ are canonically in bijection with pairs $(E,x)$ of an elliptic curve and a non-identity point.<|endoftext|> TITLE: General questions on the eigenfunctions of Laplacian and Dirac operators QUESTION [7 upvotes]: We know that the eigenvalues of the Laplacian contains a lot of information of a Riemannian manifold, but they do not determine the full information ( Hearing the shape of a drum). And the eigenfunctions of the Laplacian seem to have much more information (see the reference). Now my question is that whether the eigenfunctions of the Dirac operator would contain more information than that of Laplacian, since it seems to me that the Dirac operator is a more refined version of the Laplacian ( a Riemmanian manifold could have several spin structure). Here the Laplacian means the Laplace-Beltrami operator and the Dirac operator means the Dirac operator on the spinor bundle. Thank you. REPLY [2 votes]: Most of the questions raised above are answered in the article "The Dirac Operator on Nilmanifolds and Collapsing Circle Bundles" by Christian Bär and myself published in Annals of Global Analysis and Geometry June 1998, Volume 16, Issue 3, pp 221–253. http://link.springer.com/article/10.1023/A:1006553302362. A preprint version is available on the arxive as https://arxiv.org/abs/math/9801091. All examples of Laplace-isospectral tori are also Dirac-isospectral for the trivial spin structure. In the reference above we constructed Dirac-isospectral families of 3-step nilmanfolds, inspired by a construction by Ruth Gornet. There are also families which are Dirac-isospectral for some spin structures and not Dirac-isospectral for other spin structures.<|endoftext|> TITLE: Absoluteness, reflection to ctms, and choice in outer models QUESTION [7 upvotes]: Last night I was thinking about some related statements which follow from ZF+DC, but it actually seems they only need DC to hold in some outer model of the universe. In particular, let $M \models ZF.$ Consider the following claims (all relativized to $M$): For any sentence $\sigma$ such that $M \models \sigma,$ there is a countable transitive model $M_0 \in M$ such that $M_0 \models \sigma.$ $L^M$ and $M$ agree on $\Sigma_1$ sentences (Levy's version of Shoenfield absoluteness). $HC^M \prec_1 M.$ I believe these follow from there being some outer model $N \supset M$ where $N \models ZF + DC,$ or even $M$ existing in some ambient universe $V$ such that $V \models ZF+DC \wedge ``M \text{ is transitive}" \wedge \text{ } \omega_1 \subset M.$ E.g., to prove (1), use DC in $V$ to construct a countable $S \subset M$ which collapses to $M_0'$ such that $M_0' \models \sigma.$ The claim that such a ctm exists is $\Sigma_2^1$ if I'm not mistaken, so Shoenfield absoluteness implies $M$ also has such a ctm. So I'm wondering if these claims can be proven directly in ZF; I've heard (2) can be, but I've never seen claim (1) proven without DC. Is there a way to formalize "using choice in an ambient universe" within a model? I know there's a theorem of Woodin that says collapsing a supercompact cardinal in a model of ZF forces DC to hold, but that seems overkill. REPLY [8 votes]: Yes, all three of these statements can be proved in ZF, without any DC assumption. For statement 1, assume $M\models\newcommand\ZF{\text{ZF}}\ZF+\sigma$. By the reflection theorem, there is some ordinal $\theta$ with $(V_\theta)^M\models\sigma$. One doesn't need DC to prove the reflection theorem, since the argument is about finding an ordinal that is closed under the ranks of witnesses, rather than being able to pick out particular witnesses. So in $M$, we have a transitive set, $(V_\theta)^M$, which is a model of $\sigma$. Let $M[G]$ be a forcing extension of $M$ in which $|V_\theta|^M$ is countable. In the forcing extension $M[G]$, there is a countable transitive model of $\sigma$, namely, the set $(V_\theta)^M$, which is countable in $M[G]$. But the assertion "there is a countable transitive model of $\sigma$" is a $\Sigma^1_2$ assertion, and so it is absolute to $M$. So $M$ has a countable transitive model of $\sigma$, as desired. A similar argument works for statement 2. If $M\models \sigma$ and $\sigma$ is $\Sigma_1$, then by the above, there is a countable transitive model of $\sigma$ in $M$. By Shoenfield absoluteness again, there is a countable transitive model of $\sigma$ in $L$. But if $\sigma$ is $\Sigma_1$, then $\sigma$ is upward absolute to $L$, and so $L\models\sigma$. An essentially similar argument works for statement $3$. Namely, if $M\models\sigma$ and $\sigma$ is $\Sigma_1$, then there is countable transitive model of $\sigma$, and this countable transitive model is contained in $H_{\omega_1}$. So by upward absoluteness of $\Sigma_1$ assertions, it is true in $H_{\omega_1}$, as desired.<|endoftext|> TITLE: Closed symmetric monoidal structure on the derived category of modules whose unit is a dualizing complex? QUESTION [5 upvotes]: Let $A$ be non-positively graded commutative DG-algebra almost of finite type over a field $k$ of characteristic $0$. Most of these assumptions (affine, commutative, characteristic, bound) are only to ensure that the question isn't answered negatively by exhibiting a not so convincing counter-example. Let $Coh(A)$ be the stable infinity category of DG-modules over $A$ with coherent cohomologies in finitely many degrees. Let $Perf(A)$ be the stable infinity category of perfect $DG$ modules over $A$. Does there exist a closed symmetric monoidal structure (in the $\infty-$sense) on $Perf(A)$ which induces the familiar symmetric monoidal structure on the homotopy category $Ho(Perf(A))$ corresponding to $\{\mathcal{Hom}_A(-,-),-\otimes-\}$ (both understood as derived bi-functors between triangulated categories)? If so, is this easy? If not where can I find a proof? Same question for $Coh(A)$ the stable $\infty$-category of perfect complexes. Should we expect that the stable category of all modules $QCoh(A)$ to admit a closed symmetric monoidal structure? In the classical non-derived setting $QCoh(A)$ is a symmetric monoidal abelian category however it doesn't have an internal hom as the sheaf hom is not quasi-coherent in general. Despite this I've seen in several places claims that the derived $\infty$-category $QCoh(A)$ has a closed symmetric monoidal structure which is odd for me even from the classical perspective. What am I missing? Assume $A$ is perfect as a bi-module over itself and suppose there's monoidal structure on $Perf(A)$ as above. Let $A\otimes_k A \to A$ be the diagonal morphism and $\pi_1, \pi_2$ the 2 projections $A \to A \otimes_k A$. and consider the following bi-functor: $$Perf(A)\times Perf(A) \to Perf(A)$$ $$(M,N) \mapsto M \overset{!}{\otimes}_A N := \mathcal{Hom}_{A \otimes_k A}(A, \pi_1^* M \otimes_{A \otimes_k A} \pi_2^* N)$$ I've seen in several places cryptic remarks suggesting that this functor gives a symmetric monoidal structure on the derived category for which the unit is a dualizing complex. Does the shriek tensor product $\overset{!}{\otimes}$ defined above extend to a symmetric monoidal structure on $Perf(A)$? If so is it closed? What is its unit? What will be the correct generalization for non-smooth $A$? Would one must enlarge to the category $IndCoh$ or $QCoh$? The main point of this question is trying to understand if there's something behind said cryptic remarks about $\overset{!}{\otimes}$. REPLY [5 votes]: Let me answer the question in the title and explain the notation $- \otimes^{!}-$, forgive me for ignoring some finiteness conditions. Let's start with something somewhat simpler - take any field $k$, and for any finitely generated $k$-algebra $A$ with structure map $f:k \to A$, let $R_A = f^{!}(k)$, the canonical (or rigid) dualizing complex of $A$. Let $D_A(-) = RHom_A(-,R_A)$ be the dualizing functor. Given any functor $F:D_f(A) \to D_f(B)$ between such $k$-algebras, we can define its twist as follows: $F^{!}(-) = D_B(F(D_A(-))$. For example, if $g:A \to B$ is a $k$-algebra map and you take $G(-) = B\otimes^L_A -$ then (with suitable boundedness conditions), you will get that $G^{!}$ is just $g^{!}$, so this construction can be thought of a generalization of the construction of the twisted inverse image from the usual inverse image functor. Now, we can co the same construction for functors in several variables. For instance, twisting $-\otimes^L_A -$, we would get the following functor: $D_A(D_A(-)\otimes^L_A D_A(-))$, and so it makes sense to denote this by $-\otimes^{!}-$. Observe now that since $D_A(D_A(-)) \cong 1$, at least when restricted to things with f.g cohomologies, the operation $-\otimes^{!}-$ will remain associative and commutative up to natural isomorphisms. What is the unit of this operation? it is $D_A(A) = R_A$, the dualizing complex. You have asked about another operation, which I will write as $(M,N) \mapsto RHom_{A\otimes^L_k A} (A, M\otimes^L_k N)$. How are they related? turns out, they are the same! This was essentially shown for $k$-algebras in Theorem 4.1 of Avramov, L. L., Iyengar, S. B., Lipman, J., & Nayak, S. (2010). Reduction of derived Hochschild functors over commutative algebras and schemes. Advances in Mathematics, 223(2), 735-772. (you don't need $k$ to be a field). It is also true for commutative DG-algebras (again, under some assumptions), see Theorem 5.5 of my preprint https://arxiv.org/abs/1510.05583<|endoftext|> TITLE: Abel hidden by Cauchy? QUESTION [24 upvotes]: When I was a student, I learned from some of my teachers that Abel submitted an important part of his work to Cauchy, as a member of the "Académie des Sciences de Paris". But Cauchy hid it in a drawer of his office. Abel's work was discovered longtime after. How much of this is true? What is exactly Abel's work (hidden by Cauchy)? REPLY [29 votes]: The story is described on page 320 of E.T Bell's Men of Mathematics. The work is Abel's Memoir on a general property of a very extensive class of transcendental functions, presented to the Paris Academy of Sciences in 1826. The memoir contains what has come to be known as Abel's theorem. Legendre and Cauchy were appointed as referees. Legendre complained about the poor quality of the manuscript: "ink almost white, letters badly formed" [however, see my comment below]. Cauchy would ask Abel for a better copy, but apparently this did not happen. Abel died in 1829 and in 1830 a revolution broke out in France. When Cauchy followed the king into exile, the manuscript remained behind and was forgotten. It was not published until 1841, after the Norwegian consul in Paris raised an inquiry. There is an intriguing follow-up to this story: Abel’s manuscript disappeared again a short time after it was printed. Most of it was found in recent years in Florence, as described by Andrea Del Centina in Abel’s manuscripts in the Libri collection: Their history and their fate (2002). microfiche image of the first page of Abel's 1826 memoir. It does seem quite legible actually, perhaps Legendre's explanation that he had rejected the manuscript because it was illegible (in a letter to Jacobi in 1829) was just an excuse three years later for having neglected his duties as referee.<|endoftext|> TITLE: Inequality for functions on [0,1] QUESTION [10 upvotes]: Let $a\in (0,1), \;\;\psi_a(x):=\prod_{j=0}^\infty (1-a^{2j+1}x).$ Question. Is it true that, for all $x\in [0,1]$ and all $k\in\mathbb{N},$ the following inequality holds: $$\frac{x^k}{(1-a)(1-a^3)\dots (1-a^{2k-1})}\leq\frac{1}{\psi_a(x)}\quad ?$$ For $a\leq 0.5,$ this is obvious. When $a$ approaches 1, it becomes difficult (for me). The inequality arises in the study of the dependence of the limit $q$-Bernstein operator on parameter $q.$ REPLY [2 votes]: This was meant to cover the "calculus part" in fedja's initial answer (now completed independently), $0\le a\le a_0:=(1/9+1/25)^{1/3}$, and may possibly be completed to an independent proof. The functions $\phi(x):=\prod_{j=0}^\infty {1\over 1-a^{2j+1}x}$ and $x^{-k}$ are positive and logarithmically convex on $[0,1]$, therefore such is their product $x^{-k}\phi(x)$ too, which is therefore convex. Moreover $\big(x^{-k}\phi(x)\big)'_{x=1}=-k\phi(1)+\phi'(1)\le0$ iff $k\ge \phi'(1)/\phi(1)=\sum_{j=0}^\infty{a^{2j+1}\over 1-a^{2j+1}},$ in which case $x^{-k}\phi(x)$ has a minimum at $x=1$. We conclude that the improved inequality $$\prod_{j=0}^\infty {1\over 1-a^{2j+1}x}\ge \Big( \prod_{j=0}^\infty {1\over 1-a^{2j+1}}\Big)x^k,\qquad 0\le x\le 1$$ holds whenever $k\ge \sum_{j=0}^\infty{a^{2j+1}\over 1-a^{2j+1}};$ in particular, (by a simple numeric estimates of the sum) for all $k\ge 2$ and $0\le a\le a_0$. (For $k=1$ the inequality can be proven directly as shown in T. Amdeberhan's answer)<|endoftext|> TITLE: Schrödinger eigenfunctions are bounded QUESTION [6 upvotes]: Let $V:\mathbb{R}\rightarrow \mathbb{R}^{+ *}$ a real positive function such that $\displaystyle \lim_{ x \to \pm\infty} V(x)= +\infty $. Then the Schrödinger operator $H=-\frac{d^2}{dx^2}+V(x)$ has compact resolvant, in particular it has a pure discrete spectrum $(\lambda_i)_{i\geq 0}$ such that $\displaystyle \lim_{i\to +\infty} \lambda_i = + \infty$. Associated to each eigenvalue, there is an eigenfunction $\phi_i\in L^2(\mathbb{R})$, ie $$-\phi_i''(x)+V(x)\phi_i(x)=\lambda_i \phi_i(x), \quad \forall x\in\mathbb{R} $$ satisfying $||\phi_i||_{L^2(\mathbb{R})}=1$, My question is: Are eigenfunctions uniformly bounded? i.e. Does exist $M>0$ such that for all $n\geq 0$, $$|| \phi_i ||_\infty TITLE: How did ancient greek geometers represent solids? QUESTION [17 upvotes]: I've seen plenty of ancient diagrams representing plane figures.* But I'd like to know how ancient geometers, especially around the time of Euclid, might have represented solids. Did they use diagrams with perspective? Three dimensional models? *cf. David Fowler, "The Mathematics of Plato's Academy", see the plates between pages 6 and 7 REPLY [4 votes]: Below is a dodecahedron, dated to the 1st century, with faces labeled by the Greek alphabet. It's a figure from "A Dodecahedron of Rock Crystal from the Idaean Cave and Evidence for Divination in the Sacred Cave of Zeus," by Angelos Chaniotis. Here's a place to download the full article: https://archiv.ub.uni-heidelberg.de/propylaeumdok/1865/1/Chanotis_Dodecahedron_2006.pdf. A cursory search shows that we have found at least a few handfuls of carved dodecahedra and icosahedra in Graeco-Roman Egypt. See https://www.metmuseum.org/art/collection/search/551072 for a nice icosahedron. So the Platonic solids were tangible objects a few centuries after Euclid. Scholarship on earlier dodecahedra and icosahedra gets a bit sketchy. As for diagrams, I'd stick with Fowler's uncertainty for now.<|endoftext|> TITLE: Can the homological dimension of a coherent sheaf explode along a formal deformation? (is the resolution property hereditary for formal deformations?) QUESTION [6 upvotes]: Let $X_0$ be a locally noetherian scheme and $\mathcal{F}_0$ a coherent $\mathcal{O}_{X_0}$-module. Let $C$ be an artin ring with residue field $k$ and let $X \to Spec C$ be a (flat) deformation of $X_0$ over $C$ (meaning the closed fiber isomorphic to $X_0$). Definition 1: A deformation of $\mathcal{F}_0$ over $X$ consists of the following data: A coherent $\mathcal{O}_X$-module $\mathcal{F}$ flat over $C$ An epimorphism $q:\mathcal{F} \to \mathcal{F}_0$ inducing isomorphism $\mathcal{F}\otimes_{\mathcal{O}_X } \mathcal{O}_{X_0} \cong \mathcal{F}_0$ Definition 2: The homological dimension of a coherent sheaf $\mathcal{F}$ is the minimal length of coherent locally free resolutions of $\mathcal{F}$. If $\mathcal{F}$ doesn't have a coherent locally free resolution or all it's resolutions are infinite then define the $hd(\mathcal{F})=\infty$. The basic question is: Question: Suppose in the above situation $q:\mathcal{F} \to \mathcal{F}_0$ is a deformation of $\mathcal{F}_0$ which is of finite homological dimension $n < \infty$. Could $hd(\mathcal{F}) > hd(\mathcal{F}_0)$ ? In other words can homological dimension jump in a formal deformation? In the affine case this is not possible: Proposition: If $X_0$ is affine homological dimension can't jump. Proof: Let $\mathcal{F} \to \mathcal{F}_0$ be a deformation of $\mathcal{F}_0$ over $X$. The kernel of $\mathcal{O}_X \to \mathcal{O}_{X_0}$ is nilpotent. By factoring $Speck \to SpecC$ into small extensions we may assume that the kernel $J$ of $\mathcal{O}_X \to \mathcal{O}_{X_0}$ is square zero where $X_0 \to SpecC_0$ is over an artin ring now and s.t. $J$ is annihilated by the maximal ideal of $C_0$. Then by the local criteria for flatness over noetherian rings we conclude that $\mathcal{F}$ sits in an exact sequence: $$ 0 \to \mathcal{F}_0 \otimes_k J \to \mathcal{F} \to \mathcal{F}_0 \to 0$$ By the long exact sequence for the functor $Ext^{j}(-,Q)$ (with $Q$ arbitrary) we know that the $pd(\mathcal{F})$ is at most $pd(\mathcal{F}_0)$ and in fact they are equal since a resolution of $\mathcal{F}$ will give a resolution of $\mathcal{F}_0$. I will now use a somewhat strengthened version of the characterization of projective dimension which is a special case of what's proved here: $(*)$ Over a noetherian ring projective dimension (defined by the vanishing of Ext groups) equals the minimal possible length of a resolution by finitely generated proejctives. In other words $hd=pd$. We have therefore $hd(\mathcal{F})=hd(\mathcal{F}_0)$. Q.E.D. In the non-affine case one can use the above proof to show that all deformations $\mathcal{F}$ that admit coherent locally free resolutions have the same homological dimension as $\mathcal{F}_0$. So in fact if everything I said until now is correct then either homological dimension is constant along a formal deformation or it explodes. Therefore the main problem that arises in the non-affine case is the following: Not enough vector bundles: There might not exist any resolution of $\mathcal{F}$ by finitely generated locally free sheaves (even an infinite one). If the scheme $X$ has the resolution property (meaning every coherent sheaf is a quotient of a coherent locally free sheaf) then this problem disappears and we are left with the following questions: Is the resolution property inherited by formal deformations? Suppose $X_0$ has the resolution property and $X$ is a deformation of $X_0$ over an artin ring $C$, does $X$ have the resolution property? If $X$ doesn't have the resolution property everything seems to be stuck. Therefore I must ask the following imprecise question: Is there a deformation theoretic way to detect the resolution property? REPLY [8 votes]: You ask many questions. I will answer the question that you labelled "question". The way that you phrase the question is ambiguous. When you write "finite homological dimension $n<\infty$", you do not specify whether you are assuming that $\mathcal{F}$ has finite homological dimension, or whether you are assuming that $\mathcal{F}_0$ has finite homological dimension. I will answer both formulations. If $hd(\mathcal{F})$ is finite, then it equals $hd(\mathcal{F}_0)$. As you explain, if $\mathcal{F}$ has finite homological dimension, then the homological dimension of $\mathcal{F}$ equals the finite homological dimension of $\mathcal{F}_0$. As you point out, the homological dimension of $\mathcal{F}$ is at least as large as the homological dimension of $\mathcal{F}_0$. If the homological dimension of $\mathcal{F}_0$ equals $n$, then for every exact complex of locally free $\mathcal{O}_X$-modules, $$ \mathcal{E}_{n-1}\xrightarrow{d_{n-1}} \mathcal{E}_{n-2} \xrightarrow{d_{n-2}} \dots \xrightarrow{d_2} \mathcal{E}_1\xrightarrow{d_1} \mathcal{E}_0\xrightarrow{\eta} \mathcal{F}\to 0,$$ the kernel of $d_{n-1}$ is locally free. This can be checked locally on open affines. Then you can use the argument that you outline. Thus, if the homological dimension of $\mathcal{F}$ is finite, then the homological dimension of $\mathcal{F}$ equals the homological dimension of $\mathcal{F}_0$. On a nonseparated scheme, even if $hd(\mathcal{F}_0)$ is finite, $hd(\mathcal{F})$ may be infinite. There do exist examples where $\mathcal{F}_0$ has finite homological dimension, yet $\mathcal{F}$ has infinite homological dimension. Let $C$ be $k[\epsilon]/\langle \epsilon^2 \rangle$. Consider polynomial rings $A=C[a,b]$ and $R=C[r,s]$. Denote by $\phi$ the isomorphism of $C$-algebras, $$ C[a,b]\leftrightarrow C[r,s], \ \ a\leftrightarrow r, \ \ b\leftrightarrow s.$$ This gives rise to a $C$-isomorphism $f$ between $\text{Spec}(A)$ and $\text{Spec}(R)$ that identifies the closed points $\langle \epsilon, a,b\rangle$ and $\langle \epsilon, r,s\rangle$. Thus, $f$ identifies the open complements of these closed points, say $U=\text{Spec}(A)\setminus \langle \epsilon,a,b\rangle$ and $V=\text{Spec}(R)\setminus\langle \epsilon,r,s\rangle$. Denote by $X$ the locally Noetherian (non-separated) flat $C$-scheme containing $\text{Spec}(A)$ and $\text{Spec}(R)$ as open subschemes obtained by glueing $U$ and $V$ via $f$. Denote this common open by $O$. Let $\mathcal{F}_0$ be the ideal sheaf whose restriction to $\text{Spec}(A/\epsilon A)$ equals $\widetilde{\langle a,b\rangle}$ and whose restriction to $\text{Spec}(R/\epsilon R)$ equals $\widetilde{\langle r,s\rangle}.$ This $\mathcal{O}_{X_0}$-module has homological dimension $1$. Indeed, let $\eta:\mathcal{O}_{X_0}^{\oplus 2}\to \mathcal{F}_0$ be the homomorphism of coherent sheaves that on $\text{Spec}(A)$ equals $$\eta_A:(A/\epsilon A)^{\oplus 2} \to \langle a,b\rangle, \ \ (g,h) \mapsto g \cdot a + h \cdot b$$ and that on $\text{Spec}(R)$ equals $$\eta_R:(R/\epsilon R)^{\oplus 2} \to \langle r,s\rangle, \ \ (g,h) \mapsto g \cdot r + h \cdot s.$$ Then $\eta$ is surjective, and the kernel is a locally free sheaf of rank $1$. Via the usual Koszul complex, this kernel is actually isomorphic to $\bigwedge^2_{\mathcal{O}_{X_0}}(\mathcal{O}_{X_0}^{\oplus 2}) \cong \mathcal{O}_{X_0}$. At any rate, $\mathcal{F}_0$ has homological dimension $1$. Now let $\mathcal{F}$ be the ideal sheaf in $\mathcal{O}_X$ whose restriction to $\text{Spec}(A)$ equals $\widetilde{\langle a,b\rangle}$ and whose restriction to $\text{Spec}(R)$ equals $\widetilde{\langle r,s-\epsilon \rangle}$. Observe that the restriction of each of these ideal sheaves to $U$, resp. to $V$, is the full structure sheaf. Thus, there is a glueing of these sheaves compatible with $f$. So $\mathcal{F}$ is an $\mathcal{O}_X$-module. Checking locally, $\mathcal{F}$ is a coherent $\mathcal{O}_X$-module that is $C$-flat. I claim that there is no surjection from a locally free sheaf to $\mathcal{F}$. First, there is an observation about trivializations of locally free sheaves on small opens that contain both $\langle \epsilon,a,b\rangle$ and $\langle \epsilon,r,s\rangle$. For every open affine neighborhood $\text{Spec}(A[1/\alpha])$ of $\langle \epsilon,a,b\rangle$, then $\text{Spec}(R[1/\phi(\alpha)])$ is an open affine neighborhood of $\langle \epsilon,r,s\rangle$. Thus, for every locally free $\mathcal{O}_X$-module, there exists $\alpha \in A\setminus \langle \epsilon,a,b\rangle$ such that the locally free sheaf is trivialized on $\text{Spec}(A[1/\alpha])$ and on $\text{Spec}(R[1/\phi(\alpha)])$. Denote by $W$ the union of these two open affines. By the $S2$ property, every section of $\mathcal{O}_X$ on $U\cap D(\alpha)$ extends to a section on $\text{Spec}(A[1/\alpha])$, and similarly for sections on $V\cap D(\phi(\alpha))$. It follows that every automorphism of $\mathcal{O}_X^{\oplus m}$ on $U\cap D(\alpha)$ extends to an automorphism on $\text{Spec}(A[1/\alpha])$, and similarly for $V\cap D(\phi(\alpha))$. Thus, the locally free sheaf on $W$ is isomorphic to $\mathcal{O}_W^{\oplus m}$ for some integer $m$. Finally, let $\theta:\mathcal{O}_W^{\oplus m}\to \mathcal{F}|_W$ be a homomorphism of coherent sheaves. The composition with the inclusion $\mathcal{F}_W\subset \mathcal{O}_W$ defines a homomorphism $\theta':\mathcal{O}_W^{\oplus m} \to \mathcal{O}_W$. Again using the $S2$ property, it follows that $\theta'$ is uniquely determined by the restriction of $\theta'$ to the open $W\cap O$. Thus, on the open $\text{Spec}(A)$, it follows that $\theta'_A:A^{\oplus m}\to A$ factors through both $\langle a,b\rangle$ and through $\langle \phi^{-1}(r),\phi^{-1}(s+\epsilon)\rangle = \langle a,b+\epsilon \rangle$. Thus, $\theta'$ factors through $$\langle a,b\rangle \cap \langle a,b+\epsilon \rangle = \langle a, b^2,\epsilon b\rangle.$$ In particular, $\theta'$ is not surjective to $\langle a,b\rangle$. This contradiction proves that $\mathcal{F}$ is not a quotient of a locally free $\mathcal{O}_X$-module.<|endoftext|> TITLE: The Ultimate L in a Nutshell: On Descriptive Articles QUESTION [18 upvotes]: Everybody who catches a fleeting glimpse of Woodin's central papers on Ultimate $L$ (i.e. Suitable Extender Models I & II), admits that they aren't so tempting for lazy readers who don't like to deal with a lot of technicalities in their very first approach towards such an important topic. On the other hand there are a lot of lazy but curious readers (like me) out there who seek royal roads leading to the main ideas of such long technical papers; the descriptive articles explaining the history, philosophy and general ideas behind the proofs and related conjectures (e.g. $HOD$ conjecture). In this direction a simple Googling brings up several related articles and blog posts as well as uncountablly many lecture slide shows (mainly by Woodin himself) including the followings: Woodin's "The realm of the infinite" in the Infinity: New Research Frontiers. Caicedo's "Luminy – Hugh Woodin: Ultimate L" on his website. This question is intended to provide a list of descriptive articles on Woodin's Ultimate $L$ project. Please let me know if you are aware of any related under preparation article too. REPLY [12 votes]: You may also look at the following: 1) How Woodin changed his mind: new thoughts on the continuum hypothesis. 2) Tutorial outline: suitable extender sequences. Computational prospects of infinity. Part I. Tutorials, 195–253. 3) The weak ultimate L conjecture. Infinity, computability, and metamathematics, 309–329, Tributes, 23, Coll. Publ., London, 2014. Regarding reference 3, the following is the review of it in Mathscinet by Kanamori: The author provides an accessible, very helpful account of pivotal issues surrounding his current work toward the inner model "Ultimate L''. Most of the work appeared in his first long paper on suitable extender sequences [J. Math. Log. 10 (2010), no. 1-2, 101–339; MR2802084] in a general, very technical context, and here, in 20 pages, the author provides a thematically clear line through the heart of the main issues. Regarding works in preparation, I may refer to the references given in Woodin's paper mentioned by Todd Eisworth (see also http://nrs.harvard.edu/urn-3:HUL.InstRepos:34649600 for an open access to the paper). You can see: W. Hugh Woodin. The Axiom V = Ultimate-L. In preparation, 2016. W. Hugh Woodin. Fine Structure at the Finite Levels of Supercompactness. 710 pages, 2016. W. Hugh Woodin. The Ultimate-L Conjecture. 419 pages, 2016.<|endoftext|> TITLE: The sinc function strikes again QUESTION [7 upvotes]: Recall $\text{sinc}(x)=\frac{\sin x}x$. It's a familiar exercise that $\int_0^{\infty}\text{sinc}(x)\,dx=\frac{\pi}2$. But, at present, I wish to ask about the following claim on a "sinc-ing" product which is supported by extensive numerical computations. Question. Is it true that $$\int_0^{\infty}dx\prod_{n=1}^{\infty}\text{sinc}\left(\frac{x}{2n-1}\right) =2\int_0^{\infty}dx\prod_{n=1}^{\infty}\cos\left(\frac{x}n\right)\,\,?$$ REPLY [20 votes]: We can start with a substitution $x=2y$ $$\int_0^{\infty}dx\prod_{n=1}^{\infty}\text{sinc}\left(\frac{x}{2n-1}\right)=2\int_0^{\infty}dy\prod_{n=1}^{\infty}\text{sinc}\left(\frac{2y}{2n-1}\right).$$ Now the double angle formula $$\text{sinc}(2a)= \text{sinc}\left(a\right)\cos(a)$$ can be iterated to arrive to the familiar formula for $\text{sinc}$ $$\text{sinc}(2a)=\prod_{i=0}^{\infty}\cos\left(\frac{a}{2^i}\right).$$ This means that $$\prod_{n=1}^{\infty}\text{sinc}\left(\frac{2y}{2n-1}\right)=\prod_{m= 1}^{\infty}\cos\left(\frac{y}{m}\right)$$ because every $m\in \mathbb N$ can uniquely be written as $2^i(2n-1)$ for some $i\geq 0$ and $n\geq 1$. And this, in turn, proves the integral identity.<|endoftext|> TITLE: "Nearly" Fermat triples: case cubic QUESTION [6 upvotes]: Suppose $a^2+b^2-c^2=0$ are formed by a (integral) Pythagorean triple. Then, there are $3\times3$ integer matrices to generate infinitely many more triples. For example, take $$\begin{bmatrix}-1&2&2 \\ -2&1&2 \\ -2&2&3\end{bmatrix}\cdot \begin{bmatrix}a\\ b\\ c\end{bmatrix}=\begin{bmatrix}u\\ v\\ w\end{bmatrix}.$$ One may wish to do the same with $a^3+b^3-c^3=0$, but Fermat's Last Theorem forbids it! Alas! one settles for less $a^3+b^3-c^3=\pm1$. Here, we're in good company: $9^3+10^3-12^3=1$, coming from Ramanujan's taxicab number $1729=9^3+10^3=12^3+1^3$. There are plenty more. Question. Does there exist a concrete $3\times3$ integer matrix $M=[m_{ij}]$ such that whenever $a^3+b^3-c^3\in\{-1,1\}$ (integer tuple) then $u^3+v^3-w^3\in\{-1,1\}$ provided $$\begin{bmatrix}m_{11}&m_{12}&m_{13} \\ m_{21}&m_{22}&m_{23} \\ m_{31}&m_{32}&m_{33}\end{bmatrix}\cdot \begin{bmatrix}a\\ b\\ c\end{bmatrix}=\begin{bmatrix}u\\ v\\ w\end{bmatrix}.$$ REPLY [6 votes]: This writeup addresses the orginal question, as stated, just marginally. However, I hope to answer in more detail Amdeberhan's "other thoughts when he did ask the question, as he wrote. Actually, for every rational integer solution $(a,b,c)$ to $a^3+b^3-c^3=1$ there is a matrix $M$ of infinite order, depending on $(a,b,c)$, such that for all integer $n$ we have that $(a',b',c')=M^n\cdot (a,b,c)^T$ is another triple of integers satisfying the equation. To see this, we can observe that there are parametrizations $x_i=f_i(p,q)$ of solutions to $x_1^3+x_2^3+x_3^3+x_4^3=0$, such that: $f_1$, $f_2$, $f_3$, $f_4$ are quadratic polynomials in $p,q$; $f_4(p,q)=p^2-Dq^2$ for some squarefree $D$; $f_1(1,0)=-a$, $f_2(1,0)=-b$, $f_3(1,0)=c$. Then we solve the quadratic Pell equation $f_4(p,q)=1$. In other words, we find a primitive solution $(p_1,q_1)$ and we generate the sequence $(p_n,q_n)$ by $p_n+\sqrt D q_n = (p_1+\sqrt D q_1)^n$. We plug in the parametrization and we find $(a_n,b_n,c_n)$. Now, the key observation is that the transition from $n$ to $n+1$ is a linear transformation of the triple, so we get our matrix that does the job. Below, I give a numerical example. I use the parametrization (5) given in this MSE post, with $a=-10, b=12, c=-9, d=1$, after the linear change of coordinates $p=x+22y$, $q=2y$. Unfortunately a solution to the Pell equation with $D=85$ is quite big: $p_1=285769$, $q_1=30996$. But it does give a matrix of integers $$M=\begin{pmatrix}-331671644135 & -512714878704& -314190334080\\ 266762362608 & 412374813481 & 252702205056\\ 259637126112 & 401360260896 & 245952516097\end{pmatrix} $$ that predictably works: $M\cdot (-10,12,-9)= (-8149096378, 6554290188, 6379224759)$, and $(-8149096378)^3+6554290188^3=(-6379224759)^3+1$. To help comparing with Stadnicki's analysis, I report that $M$ has characteristic polynomial $(x - 1) \cdot (x^2 - 326655685442 x + 1)$ and eigenvectors $v_{max}=(-\frac 1 8 (1+ \sqrt{85}),-\frac 1 8 (1- \sqrt{85}), 1)$, $v_1=(-16,-16,43)$, $v_{min}=(-\frac 1 8 (1- \sqrt{85},-\frac 1 8 (1+ \sqrt{85}),1)$.<|endoftext|> TITLE: Decompostition of a Lipschitz domain QUESTION [5 upvotes]: We say that $\Omega$ is a strongly star shaped domain (with respect to $0$ for example) in $\mathbb R ^n$ if: $$\Omega = \{x\in \mathbb R ^n : \left \| x \right \| < g(\frac{x}{\left \| x\right \|})\} $$ and $$\partial \Omega = \{x\in \mathbb R ^n : \left \| x \right \| = g(\frac{x}{\left \| x \right \|})\} $$ with $g$ is a continuous, positive function on the unit sphere. In this paper, Bramble uses the fact that : Any Lipschitz domain can be written as the union of strongly star shaped Lipschitz domains: $\Omega=\cup_{i=1}^{M}\Omega_i$ Can you help me to find why do we have this result? Do you have any references in which I can find this proposition? PS: Sorry to ask this question again, but this did not get answered (as I had desired) on M.SE. REPLY [4 votes]: A Lipschitz domain $\Omega$ is an open set and any open set is a union of balls, which are strongly star-shaped. So I assume you meant $\overline\Omega$. By definition, any point of $\partial\Omega$ has a nbd in $\overline\Omega$ which is isometric to a sub-graph of a positive $k$-Lipschitz function $f:B(0,r)\to(0,+\infty)$, $$\{(x,t)\, :\, |x| TITLE: On the coincidence (or non-coincidence) of two norms defined on the quotient of a given Hilbert $ C^{\ast} $-module by a certain linear subspace QUESTION [5 upvotes]: Let $ A $ be a $ C^{\ast} $-algebra, $ I $ a closed two-sided ideal of $ A $, and $ \mathcal{E} $ a Hilbert $ A $-module. Let $$ \mathcal{E}_{I} \stackrel{\text{df}}{=} \{ x \in \mathcal{E} \mid \langle x,x \rangle_{\mathcal{E}} \in I \}. $$ Using the Cauchy-Schwarz Inequality for Hilbert $ C^{\ast} $-modules, it is not difficult to show that $$ \mathcal{E}_{I} = \{ x \in \mathcal{E} \mid (\forall y \in \mathcal{E})(\langle x,y \rangle_{\mathcal{E}} \in I) \}. $$ This implies that $ \mathcal{E}_{I} $ is a linear subspace of $ \mathcal{E} $. Furthermore, as the $ A $-valued inner product on $ \mathcal{E} $ is continuous, $ \mathcal{E}_{I} $ is a $ \| \cdot \|_{\mathcal{E}} $-closed subset of $ \mathcal{E} $. All of this implies that $ \mathcal{E}_{I} $ is a Hilbert $ I $-module. Now, the quotient space $ \mathcal{E} / \mathcal{E}_{I} $ is a Banach space w.r.t. the quotient norm $ \| \cdot \|_{\text{q}} $ defined by $$ \forall x \in \mathcal{E}: \qquad \| x + \mathcal{E}_{I} \|_{\text{q}} \stackrel{\text{df}}{=} \inf_{y \in \mathcal{E}_{I}} \| x + y \|_{\mathcal{E}}. $$ It is also a right $ (A / I) $-module equipped with an $ (A / I) $-valued pre-inner product $ [\cdot,\cdot] $ defined by $$ \forall x_{1},x_{2} \in \mathcal{E}: \qquad [x_{1} + \mathcal{E}_{I},x_{2} + \mathcal{E}_{I}] \stackrel{\text{df}}{=} \langle x_{1},x_{2} \rangle_{\mathcal{E}} + I. $$ Question. Is the norm on $ \mathcal{E} / \mathcal{E}_{I} $ that is induced by $ [\cdot,\cdot] $ the same as $ \| \cdot \|_{\text{q}} $, i.e., $$ \forall x \in \mathcal{E}: \qquad \| x + \mathcal{E}_{I} \|_{\text{q}} = \sqrt{\| \langle x,x \rangle_{\mathcal{E}} + I \|_{A / I}}? $$ Thank you for your help! REPLY [4 votes]: Here are a couple of supplements to Omar's answer. The first thing to note is that the module $\mathcal{E}_I$ is equal to the module $\mathcal{E}I=\{ei\ |\ e\in \mathcal{E},\ i\in I\}$. The inclusion $\mathcal{E}I\subset \mathcal{E}_I$ follows easily from the linearity of the inner product and from the fact that $I$ is an ideal. For the reverse inclusion, let $i_\lambda$ be an approximate unit for $I$, and then show (by expressing the norm in terms of the inner product) that for each $e\in \mathcal{E}_I$ one has $\| e-ei_\lambda\|\to 0$ as $\lambda\to\infty$. This gives $\mathcal{E}_I\subset \overline{\mathcal{E}I}$, and the latter is equal to $\mathcal{E}I$ by the Cohen factorisation theorem. For full details see Lemma 3.23 in: Raeburn, Iain; Williams, Dana P., Morita equivalence and continuous-trace $C^*$-algebras, Mathematical Surveys and Monographs. 60. Providence, RI: American Mathematical Society (AMS). xiv, 327 p. (1998). ZBL0922.46050. Now, turning to the inner product on $\mathcal{E}/\mathcal{E}_I$: here are three arguments showing that the norm defined by the inner product is equal to the quotient norm. (1) A short direct computation is given in Lemma 3.1 in: Zettl, Heinrich H., Ideals in Hilbert modules and invariants under strong Morita equivalence of C*-algebras, Arch. Math. 39, 69-77 (1982). ZBL0498.46034. (2) An argument based on the uniqueness of the $C^*$-norm on the linking algebra is given in Proposition 3.25 in the book of Raeburn-Williams cited above. (The authors attribute the argument to Siegfried Echterhoff.) (3) Here is a third proof, using ideas around operator modules and the Haagerup tensor product. All of the necessary background can be found in Blecher, David P.; Le Merdy, Christian, Operator algebras and their modules -- an operator space approach, London Mathematical Society Monographs. New Series 30; Oxford Science Publications. Oxford: Oxford University Press (ISBN 0-19-852659-8/hbk). x, 387~p. (2004). ZBL1061.47002. Consider $I\to A\to A/I$ as an exact sequence of operator $A$-bimodules. Taking the Haagerup tensor product with $\mathcal E$ (viewed as a right operator module over $A$) gives $$ \mathcal E\otimes_A I \to \mathcal E\otimes_A A \to \mathcal E\otimes_A (A/I). $$ By the exactness property of the Haagerup tensor product over a $C^*$-algebra (a theorem of Anantharaman-Delaroche and Pop), the first map in the display is a (completely) isometric embedding, and the second map induces a (completely) isometric isomorphism $$ (\mathcal E\otimes_A A)/ (\mathcal E\otimes_A I) \cong \mathcal E\otimes_A (A/I)\qquad (*) $$ where the left-hand side carries its canonical quotient operator space structure (and in particular, the usual Banach-space quotient norm). The module action gives a (completely) isometric isomorphism $\mathcal E\otimes_A A \to \mathcal E$, which restricts to an isomorphism $\mathcal E\otimes_A I \to \mathcal E I=\mathcal E_I$. Making these identifications, the isomorphism $(*)$ is given by the formula $$ \mathcal E/\mathcal E_I\to \mathcal E\otimes_A (A/I),\qquad (ea+\mathcal E_I)\mapsto e\otimes(a+I).\qquad (**) $$ Now, $A/I$ is a (right) Hilbert $C^*$-module over itself, and the quotient map $A\to A/I$ gives a (left) action of $A$ on this $C^*$-module by adjointable operators. We can thus form the Hilbert $C^*$-module tensor product $\mathcal E\otimes^{C^*}_A (A/I)$, which will be a Hilbert $C^*$-module over $A/I$. A theorem of Blecher asserts that the identity map on the algebraic tensor products extends to a (completely) isometric isomorphism between $\mathcal E\otimes^{C^*}_A (A/I)$ and the Haagerup tensor product $\mathcal E\otimes_A (A/I)$. A straightforward computation shows that the map $(**)$ is isometric with respect to the inner product $[\cdot,\cdot]$ on $\mathcal E/\mathcal E_I$, and the canonical inner product on $\mathcal E\otimes^{C^*}_A (A/I)$. Since $(**)$ is also isometric for the quotient norm, the latter must coincide with the norm induced by $[\cdot,\cdot]$.<|endoftext|> TITLE: Who first proved ergodicity of irrational rotations of the circle? QUESTION [26 upvotes]: It is a classic result that the irrational rotations of the circle are ergodic. Formally, let $T:\mathbb{T}\to \mathbb{T}$ be defined by $Tz=ze^{2\pi i\alpha}$. If $\alpha$ is irrational, then $T$ is ergodic. This result appears in many textbooks (e.g., Walters, An Introduction to Ergodic Theory), and even in Wikipedia. However, none of these refer to the original. A Google Scholar search didn't help either. My question is simply, who was the first to prove or to notice this result, and is there a reference to the original paper? REPLY [2 votes]: Weyl had only one "big" paper on uniform distribution theory, which is the 1916 paper. However, the fact that $(n \alpha)$ is equidistributed for irrational $\alpha$ is usually attributed to Bohl, Spierpinski and Weyl (independently), who proved it in 1909-1910. However, they did not prove anything about ergodicity. The ergodicity question does not ad-hoc have any close relation to the equidistribution problem.<|endoftext|> TITLE: What is the precise relationship between forcing on a poset and the topos of double-negation sheaves on this poset? QUESTION [9 upvotes]: I've seen various statements that the Boolean-valued models of ZFC occurring in model-theoretic forcing are "really" the topos of sheaves on an appropriate site, but never a fully precise statement. What exactly is the relationship, and where can I find this written down in a clear, simple way? More precisely, let $\mathbb{P}$ be a poset and $B$ its completion to a Boolean algebra. Then, the machinery of forcing gives a Boolean-valued model of ZFC (let's ignore issues related to the universe, countable transitive models, etc. unless they really become crucial) $V^{(B)}$ such that the truth values of the axioms of ZFC are $1$, and the forcing relation can be defined as $p \Vdash \phi$ for $p \in B$ iff the truth value of $\phi$ is at least $p$. On the other hand, we may consider $\mathbb{P}$ as a site with the double negation topology (for some $p \in \mathbb{P}$, a collection $S$ of elements $s \in \mathbb{P}$, $s \leq p$ covers $p$ iff for all $q \leq p$ there is some $s \in S$ with $s \leq q$; i.e. $S$ is "dense" below $p$) and take the Boolean topos $\mathscr{V}_\mathbb{P}$ of sheaves on this site. The subsheaves of the terminal object in this topos form a Boolean algebra $\Omega$, and we can assign elements of $\Omega$ as "truth values" of sentences $\phi$ via the Kripke-Joyal semantics. This lets us define a "forcing relation" for $p \in \Omega$ as $p \Vdash' \phi$ iff the terminal map from the Yoneda image of $p$ factors through the truth value of $\phi$. What is the exact relationship between $V^{(B)}$ and $\mathscr{V}_\mathbb{P}$? In what sense is $\mathscr{V}_\mathbb{P}$ a model of ZFC? (i.e. can it be used directly to prove consistency results in the same way that $V^{(B)}$ can?) I would like to say something like "$p \Vdash \phi$ iff $p \Vdash' \phi$", but strictly speaking this doesn't quite make sense since the formulas are expressed in two slightly different languages (i.e. the language for Kripke-Joyal semantics is that of type theory). I would love to see this written down in detail somewhere. (As far as I can tell, a generic filter is not strictly relevant to this question; however, I would be curious if it has an interesting interpretation on the sheaf side). REPLY [9 votes]: One good way of seeing the connection is given in Section 4 of Michael Fourman’s paper Sheaf models for set theory (JPAA, Vol 19, 1980). For any Grothendieck topos $\newcommand{\E}{\mathcal{E}}\E$, there’s a canonical ‘cumulative hierarchy’ in $\E$, i.e. a sequence of objects $V^\E_\alpha$, for $\alpha \in \mathrm{On}$, given by taking power-objects at successor stages and colimits at limit stages, so in particular $V^\E_0 = 0$. Moreover, these carry membership relations $(\in_\alpha) \rightarrowtail V^\E_\alpha \times V^\E_\alpha$; and the total sequence can be seen as a model of IZF, and when the topos is Boolean, a model of ZF. (There are several ways to deal with the fact that it’s a sequence of objects not a single object of $\E$ — I won’t go into that here. Edit: OK, I will — see below!) Now, when the topos is $\newcommand{\P}{\mathbb{P}} \newcommand{\Sh}{\mathrm{Sh}}\Sh(\P)$, this model $V^{\Sh(\P)}$ ‘is’ the standard Boolean model over $\P$ in several senses. The simplest is Theorem 4.1 in Fourman’s paper. For any sentence $\varphi$ in the language of set theory, we can interpret it either in $V^{\Sh(\P)}$ or in the set-theorist’s Boolean-valued model $V\P$, and either way we get an element of $B(\P)$; then Theorem 4.1 says that these interpretations are the same. A slightly stronger statement, fairly straightforward to prove, is as follows. (Indeed, I can’t see how to prove Fourman’s Thm 4.1 without going via something like this statement.) There’s a (class) bijection between (the colimit over $\alpha$ of) global sections of the objects $V^{\Sh(\P)}_\alpha$, and $\P$-names in the set-theorist’s sense modulo forcing-equality. Then for any formula $\varphi(x_1, \ldots, x_n)$, and $\P$-names $a_1,\ldots,a_n$, and $p \in \P$, we have $p \Vdash \varphi(a_1,\ldots,a_n)$ in the set-theorist’s sense exactly if $(a_1,\ldots,a_n)\mathord{\upharpoonright_p} \in [\![\, \vec x \,|\, \varphi(\vec x)\, ]\!](p) \subseteq V^{\Sh(\P)}(p)$. A crucial point in proving this is that the sheaves $V^{\Sh(\P)}$ are flabby, so any local section can be extended to a global section. This is why the set-theoretic presentation of the model can get away with only using $\P$-names, i.e. global sections, even though in general the sheaf-theoretic interpretation of quantification ranges over all elements of a sheaf. Regarding the interpretation of logic in the sequence $V_\alpha$: The simplest approach is to assume an inaccessible cardinal $\kappa$ larger than $\P$, and cut off at $\kappa$ on both the topos-theoretic and set-theoretic sides. So the model on the topos-theoretic side is just $(V_\kappa,\in_\kappa)$, and the interpretation of (I)ZF in it is the standard interpretation of first-order logic in a structure in a topos using the Kripke–Joyal semantics. On the set-theoretic side, one then correspondingly restricts the Boolean-valued model to “hereditarily $\kappa$-small $\P$-names”, i.e. $V\P \cap H_\kappa$. Then there’s Fourman’s approach, which is to work with the full sequence $V_\alpha$ for $\alpha \in \mathrm{On}$, and define the interpretation of logic in it by hand, in a way which clearly follows the normal Kripke-Joyal semantics except that its quantifications quantify over the whole sequence. Finally, the modern viewpoint on Fourman’s interpretation is the categories of classes setting, developed for Algebraic Set Theory. One can embed the topos $\E$ into a super-large category of classes $\E'$, whose objects are to the objects of $\E$ what classes are to sets in (I)ZF. In general, $\E'$ can be taken to be a category of ideals on $\E$; when $\E$ is a topos of sheaves on a small site, $\E$ can be taken to be a topos of “large sheaves” on the same site. Either way, $\E'$ is a Heyting category, so one can interpret first-order logic in it, and under the “large sheaves” presentation of $\E'$, this interpretation looks exactly the usual Kripke–Joyal semantics of a topos. And then in $\E'$, we can take the colimit of the sequence $(V_\alpha,\in_\alpha)$, and use that as our model.<|endoftext|> TITLE: "Universal" differential identities QUESTION [6 upvotes]: (This is a cross-post from MSE). Let $f:\mathbb{R}^d \to \mathbb{R}$ be smooth. The mixed derivatives commute: $f_{xy}=f_{yx}$. This identity is "universal" in the sense that it holds for any smooth map. Question: Are there any universal identities which are not consequences of the commutation of the mixed derivatives? More explicitly, let $D_i$ be the differential operator which takes the partial derivative with respect to $x_i$. The symmetry can be written as an algebraic statement $$ D_i \circ D_j = D_j \circ D_i \tag{1}.$$ So, the ring of differential operators with constant coefficients, generated by the $D_i$, is commutative. (When we choose the domain for these operators to be the space of smooth maps $\mathbb{R}^d \to \mathbb{R}^d$, so we can compose operators). Are there relations in this ring which are not consequences of the fundamental relation $(1)$?. Edit: Actually, this is not the "right" algebraic formulation which interests me: I want to be able to talk about relations of the form of $f_x f_y=f_yf_x$ (trivially true) or $f_{xx}=f_yf_{xy}$ (clearly not universal). Thus I need to add multiplication. (Without multiplication, there are no additional relations as observed in this answer). In particular, I am interested to know whether the "Cofactor Lemma" (divergence-free rows, see below) is a "consequence" of the commutation of mixed derivatives (for dimension $d>2$ this involves multiplication, as well as addition and composition). So, we need to consider some algebraic structure $A$ which is a subset of differential operators (which map $C^{\infty}(\mathbb{R}^d) \to C^{\infty}(\mathbb{R}^d)$), that is "generated" by the $D_i$ via the $3$ operations - addition, composition, and multiplication*. (I am not sure if there is a term for such an "algebraic creature", $A$ is a ring w.r.t both operations $(+,\cdot)$ and $(+,\circ)$, but these two "multiplicative" operations have relations, namely $$(f \cdot g) \circ h = (f \circ h) \cdot (g \circ h).$$ Does such a structure have a name? *By multiplication (as opposed to composition) of operators I mean the following: $$D_x \times D_y(f)=f_x \cdot f_y \, , \, D_x∘D_y(f)=f_{xy} \, , \, (D_x \circ D_x) \times D_y(f)=f_{xx}f_y$$ etc. (Here $f$ is a scalar function, to extend the operations to $\mathbb{R}^d$-valued maps, jut act on each component separately. I am also allowing for the $i$-th component of output to depend on partial derivatives of all components of $f:\mathbb{R}^d \to \mathbb{R}^d$). The Cofactor Lemma: Let $f:\mathbb{R}^d \to \mathbb{R}^d$ be smooth. Then the Cofactor of $df$ has divergence-free rows: $$\sum_{j=1}^n \frac{\partial(Cof(Du))_{kj}}{\partial x_j} = 0, k=1,...,d.$$ In dimension $d=2$, it reduces to relation $(1)$: Given $A= \begin{pmatrix} a & b \\\ c & d \end{pmatrix}$, $\operatorname{Cof}A= \begin{pmatrix} d & -c \\\ -b & a \end{pmatrix}$, so $$ df= \begin{pmatrix} (f_1)_x & (f_1)_y \\\ (f_2)_x & (f_2)_y \end{pmatrix}, \operatorname{Cof}df= \begin{pmatrix} (f_2)_y & -(f_2)_x \\\ -(f_1)_y & (f_1)_x \end{pmatrix} .$$ We see that $\operatorname{div} (\operatorname{Cof}df)=0$ is equivalent to $(f_1)_{xy}=(f_1)_{yx},(f_2)_{xy}=(f_2)_{yx}$. As stated above, for dimension $d>2$ we need multiplication to even phrase the question properly. REPLY [7 votes]: Let's work in two dimensions for notational simplicity. We claim that there is no non-trivial polynomial identity of the form $$ P( f, f_x, f_y, f_{xx}, f_{xy}, \dots ) = 0$$ relating some finite number of derivatives of $f$, for any smooth $f$. Suppose for contradiction that this is the case. Taking a functional derivative (i.e. replacing $f$ by $f+ \varepsilon g$ for arbitrary smooth $g$, and then differentiating at $\varepsilon=0$, one obtains a linearised (in $g$) identity of the form $$ P_f(f,f_x,\dots) g + P_{f_x}(f,f_x,\dots) g_x + \dots = 0.$$ Now, at any point in space, the values of finitely many of the various derivatives $g, g_x, g_y, \dots$ are completely unconstrained (as can be seen by taking $g$ to be a finite Taylor series around that point). So the only way the above identity can hold for all $f$ and $g$ is if one has $$ P_f(f,f_x,\dots) = 0$$ $$ P_{f_x}(f,f_x,\dots) = 0$$ etc.. By induction on the degree of $P$, this can only happen if $P_f, P_{f_x}, \dots$ vanish identically, so that $P$ is constant, hence zero, so the identity is trivial.<|endoftext|> TITLE: Cobordisms and Euler characteristics QUESTION [8 upvotes]: I am trying to understand exactly which role the Euler characteristic plays in (smooth) cobordism theory, and especially why the answer seems to depend on the dimensions of the manifolds in question. Suppose $M$ and $N$ are two closed smooth $n$-manifolds and that we have a smooth cobordism $(W;M,N)$ (i.e. $W$ is a smooth compact $(n+1)$-manifold with boundary $\partial W=M\sqcup N$). Poincare-Lefschetz duality then predicts that $H_k(W,N;\mathbb{Z}_2)\cong H_{n+1-k}(W,M;\mathbb{Z}_2)$ for every $k\in \mathbb{Z}$. In particular, if we denote by $\chi(W,M)$ the Euler characteristic of the homology $H_{*}(W,M;\mathbb{Z}_2)$ we see that $\chi(W,N)=(-1)^{n+1}\chi(W,M)$. By additivity of the Euler characteristic we also have $$ \chi(W,N)+\chi(N)=\chi(W)=\chi(W,M)+\chi(M). $$ If $n$ is odd we therefore conclude that $\chi(M)=\chi(N)$, so in this case $\chi$ is a cobordism invariant. My question is: What happens when $n$ is even? Is the Euler useful in any way for understanding cobordisms of even dimensional manifolds? Any interesting statement would be much appreciated, even if additional assumptions such as orientability, spin etc. are needed. (Note that the sphere $S^2$ is null-cobordant, the cobordism given by the 3-ball, so clearly $\chi$ is not a cobordism invariant of even dimensional manifolds...) Thanks in advance :). REPLY [15 votes]: René Thom proved that two closed $n$-manifolds $M$ and $N$ are (unoriented) cobordant iff their Stiefel-Whitney numbers agree: for any partition $i_1 + \dotsb + i_k = n$, $$[M]\frown w_{i-1}(M)w_{i_2}(M)\dotsm w_{i_k}(M) = [N]\frown w_{i-1}(N)w_{i_2}(N)\dotsm w_{i_k}(N).$$ The mod 2 Euler characteristic is a Stiefel-Whitney number: $\chi(M)\bmod 2 = [M]\frown w_n(M)$. Thus, if $M$ and $N$ are unoriented cobordant, then their Euler characteristics differ by a multiple of 2. (There might be a more direct proof of this using handles.) You can get rid of the mod 2 assumption if you work with (stably almost) complex cobordism, equipping all manifolds and cobordisms with a complex structure on the stable normal bundle. (In particular, this determines an orientation of your manifold.) Milnor and Novikov showed that $M$ and $N$ are complex cobordant iff their Chern numbers agree: these are defined in the same way as Stiefel-Whitney numbers, but for Chern classes. The Euler characteristic is a Chern number: $\chi(M) = [M]\frown c_n(M)$ if $M$ is $2n$-dimensional, and $\chi(M) = [M]\frown 0$ if $M$ is odd-dimensional. Thus in any dimension, the Euler characteristic is a cobordism invariant for stably almost complex cobordism.<|endoftext|> TITLE: On some sense of representing an endofunctor of the category of modules over polynomial rings QUESTION [5 upvotes]: If $R$ is commutative ring, $n\in\mathbb{N}$, $\mathsf{M}$ the category of $R[x_1,\dotsc,x_n]$-modules,and $F\colon\mathsf{M}\to\mathsf{M}$ an endofunctor of $\mathsf{M}$ which preserves all finite limits, does it follow in general that there exists an $M\in\mathsf{M}$ and a natural transformation $\tau\colon F\rightarrow \mathrm{Hom}_{\mathsf{M}}(M,-)$ such that for each injective $N\in\mathsf{M}$ the induced morphism $F(N)\rightarrow \mathrm{Hom}_{\textsf{M}}(M,N)$ is an isomorphism of $\mathsf{M}$? REPLY [5 votes]: Replacing $R$ by $R[x_1,\ldots,x_n]$ if necessary, we may assume $n = 0$. Note that preserving all finite limits is equivalent to being (additive and) left exact (see e.g. Tag 010N). Lemma. Let $R$ be a commutative ring, let $F \colon \operatorname{Mod}_R \to \operatorname{Mod}_R$ be a left exact functor, and let $\tau \colon F \to \operatorname{Hom}_R(M,-)$ be a natural transformation that is an isomorphism on injective $R$-modules $I$. Then $\tau$ is a natural isomorphism. Proof. Let $N$ be an arbitrary $R$-module, and choose a resolution $0 \to N \to I \to J$ with $I$ and $J$ injective. Then we get a commutative diagram $$\begin{array}{ccccccc}0 & \to & F(N) & \to & F(I) & \to & F(J) \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & \operatorname{Hom}_R(M,N) & \to & \operatorname{Hom}_R(M,I) & \to & \operatorname{Hom}_R(M,J) \end{array}$$ with exact rows. Since the right two vertical maps are isomorphisms, so is the left vertical map by the five lemma (three lemma?). $\square$ Thus, we get that $F$ is representable. This is not always the case under your assumptions: Example. Let $R = \mathbb Z$, and let $F \colon \operatorname{\underline{Ab}} \to \operatorname{\underline{Ab}}$ be the functor $A \mapsto A_{\operatorname{tors}}$, the subgroup of torsion elements of $A$. This is clearly additive and left exact. To show that $F$ is not representable, note that representable functors preserve all limits (not just finite ones). This is not the case for $(-)_{\operatorname{tors}}$. For example, the natural (injective) map $$\left(\prod_{n \in \mathbb Z_{>0}} \mathbb Z/n\mathbb Z\right)_{\operatorname{tors}} \to \prod_{n \in \mathbb Z_{>0}} (\mathbb Z/n\mathbb Z)_{\operatorname{tors}}$$ is not an isomorphism, since the left hand side does not contain the element $(1,1,\ldots)$. $\square$ Remark. If one really insists on using polynomial rings, one can replace $\mathbb Z$ by $k[x]$ where $k$ is a field, and use a similar argument. Remark. For $F$ to be representable, it is necessary and sufficient that $F$ preserves all limits and satisfies some solution set condition. See for example Thm. V.6.3 of Mac Lane.<|endoftext|> TITLE: Vector field with holomorphic flow QUESTION [11 upvotes]: Let $(M,J)$ be a complex manifold. Suppose that $X$ is a real vector field such that the flow of $X$ is by biholomorphisms.Question Show the flow of $JX$ is by biholomorphisms. I know one reference to show the answer is yes (where it is stated that: flow of $X$ is by biholomorphisms $\iff$ $X-iJX$ is a holomorphic vector field) but I am interested to find a simple explanation (if indeed one exists). I would also be interested if anyone knows about the corresponding question for an almost complex manifold $(M,J)$ with a vector field with flow by J-isomorphism's? - for this question I do not have a reference (or clue if it is true). REPLY [2 votes]: Without lose of generality we may work locally in $\mathbb{R}^{2n}$ with its standard complex structure $J$. Let $\phi$ be the flow of the vector field $x'= f(x)$. So the statement in the question is a consequence of the variational equation $$ \partial_t \partial_x \phi_t = Df(\phi). \partial_x \phi_t$$ If the flow is holomorphic then $\partial_x \phi_t$ and its time derivative $\partial_t \partial _x \phi_t $ commute with $J$. This implies that $Df.J= J.Df$. Q.E.D In the same manner one can shows the following: Fact: If the flow of a vector field $x'=f(x)$ on $\mathbb{R}^n$ is harmonic then $f$ is harmonic, too. Proof: The Laplacian, the gradient and the Hessian of a vector valued function is defined naturally by componentwise corresponding operators. Hence the Hessian of a vector valued function is a $3$ dimensional matrix whose trace is a "vector". With such notation we have the following formula: $$\partial_t \Delta \phi= trace( {(\partial _x \phi_t)}^{tr}Hess(f)\partial_x \phi_t)+ \Delta \phi . \nabla f$$ This shows (using the time independence of $f$ to evaluate at $t = 0$) that if the flow of a vector field is harmonic then the vector field is a harmonic vector valued function. But what about the converse? Is the flow of a harmonic vector field, a harmonic function? Moreover, how can we rephrase the above Euclidean fact in an abstract Riemannian manifold?<|endoftext|> TITLE: Is it true that every subspace contain a primitive element? QUESTION [5 upvotes]: Let $R = GF(q), q = p^r$, be a field with identity $e$, where $p$ is a prime number. Let $S=GF(q^n)$ be an extension of $R, n\geq 2$ and $K = GF(q^{mn})$ be an extension of $S$, where $m$ is prime. Let $_RW$ be a subspace of $_RK$ such that $$ \operatorname{dim}_RW =n, e\in W, W\neq S $$ I want to prove that there exist a primitive element of $K$ (that is the element of multiplicative order $q^{mn}-1$) in $W$. But so far my attempts have not been successful. Experiments on the computer confirm this hypothesis. REPLY [4 votes]: This does not work in general: the problem is you have no control over $m$. Let $r = 1$. There's a sequence of embeddings of fields $$\mathit{GF}(p)\subset\mathit{GF}(p^2)\subset\mathit{GF}(p^4)\subset\mathit{GF}(p^8)\subset\dotsb.$$ Let $S = \mathit{GF}(p^2)$, so $n = 2$. Let $W$ be any $2$-dimensional subspace of $\mathit{GF}(p^4)$ containing the identity but not equal to $S$. Let $K = \mathit{GF}(p^8)$. $W$ is still a $2$-dimensional subspace of $K$, but since $\mathit{GF}(p^8)\hookrightarrow K$ as a subfield, any element of $W$ can generate at most $\mathit{GF}(p^4)$, which is not all of $K$. For $r\ne 1$, you can do the same thing with $$\mathit{GF}(p^r)\subset\mathit{GF}(p^{2r})\subset\mathit{GF}(p^{4r})\subset\mathit{GF}(p^{8r})\subset\dotsb.$$<|endoftext|> TITLE: Two abelian groups, each being direct factor of the other QUESTION [6 upvotes]: Let $M$ and $N$ be two abelian groups. Suppose that $M$ is a direct factor of $N$ (i.e. there are homomorphisms $i:M\rightarrow N$ and $p:N\rightarrow M$ such that $p\circ i=id_M$) and $N$ is also a direct factor of $M$. Is $M$ isomorphic to $N$? REPLY [6 votes]: Arturo Magidin's answer is absolutely correct, but there's an earlier counterexample than Corner's. This question is Kaplansky's first "test problem" in his 1954 book on Infinite Abelian Groups, and was solved by Sąsiada in 1961: Sąsiada, E., Negative solution of I. Kaplansky's first test problem for Abelian groups and a problem of K. Borsuk concerning cohomology groups, Bull. Acad. Pol. Sci., Ser. Sci. Math. Astron. Phys. 9, 331-334 (1961). ZBL0105.25902.<|endoftext|> TITLE: About the abelian category of endofunctors of $\mathsf{Vect}$ QUESTION [23 upvotes]: Let $k$ be a field, $\mathsf{Vect}$ the category of finite dimensional vector spaces, and $\mathsf{C} = Fun(\mathsf{Vect},\mathsf{Vect})$ the abelian category of pointed endofunctors (sending $0$ to $0$). The question has 2 parts, one for characteristic $0$ fields and the other for characteristic $p>0$. Assume $char(k)=0$ Definition: An endofunctor $F \in \mathsf{C}$ is called polynomial if it can be expressed as a sum $$F(V) = \bigoplus_{n \in \mathbb{N}}P(n) \otimes_{k[S_n]}V^{\otimes n}$$ Where $P(n)$ are $k$-linear representations of the symmetric group s.t. $P(n)=0$ for $n \gg 0$. Questions: 1. Is there a characterization of the polynomial functors among all endofunctors? 2. What's an example of a non-polynomial functor in $\mathsf{C}$? 3. Is $\mathsf{C}$ a semi-simple category? If so is there an explicit description of it? (hopefully identifying it with something built out of categories of representations). Assume $char(k) = p \gt 0$ I think that in this case $\mathsf{C}$ will probably never be semi-simple (my intuition being that the exact sequence $0 \to \bigwedge^2 \to \otimes^{2} \to S^2 \to 0$ doesn't seem to be split in general) and so there looks to be some interesting (in my opinion) homological algebra going on here. Question: Is $\mathsf{C}$ well understood in this case? By which I mean that there's a full list of all isomorphism classes and all Ext groups can be calculated in principle. And finally, where can I read more about this topic? REPLY [26 votes]: There are a couple of equivalent ways to characterise polynomial functors. One is to say that $F$ is a polynomial functor of degree $n$ if the function $$\hom(U, V)\to \hom(F(U), F(V))$$ is polynomial of degree n. A second, equivalent formulation goes via cross-effects: A functor is polynomial of degree $n$ if its $n+1$-th cross-effect vanishes. As far as I know, these definitions were first introduced in the classic paper Eilenberg, Samuel; MacLane, Saunders, On the groups $H(\Pi,n)$. II, Ann. Math. (2) 60, 49-139 (1954). ZBL0055.41704. In this paper the concept of cross-effect was introduced, and the equivalence of the two definitions proved. In characteristic $0$, the category of polynomial functors is equivalent to the direct sum of representation categories of symmetric groups. In particular, it is semi-simple. This seems to be due to MacDonald Macdonald, I.G., Polynomial functors and wreath products, J. Pure Appl. Algebra 18, 173-204 (1980). ZBL0455.18002. In positive characteristic, the category is not semi-simple. The following papers introduced the systematic study of functor categories between vector spaces over $\mathbb F_p$, and in particular related them to modules over the Steenrod algebra Henn, Hans-Werner; Lannes, Jean; Schwartz, Lionel, The categories of unstable modules and unstable algebras over the Steenrod algebra modulo nilpotent objects, Am. J. Math. 115, No.5, 1053-1106 (1993). ZBL0805.55011. Kuhn, Nicholas J., Generic representations of the finite general linear groups and the Steenrod algebra. I, II, and III, Am. J. Math. 116, No.2, 327-360 (1994). ZBL0813.20049. There has been a lot of work on the homological algebra in this category, with striking applications in particular by Friedlander and Suslin to the homology of general linear group (Edit: for this application one needs the category of strict polynomial functors - a variant of the "naive" notion of polynomiality). However I don't think there is an effective procedure known for calculating the ext groups between general polynomial functors. For an up to date survey of the subject you may want to consult, for example, the following book Franjou, Vincent, Touze, Antoine Lectures on functor homology., , Proceedings of the conference on functor homology, Nantes, France, April 2012. Cham: Birkh\"auser/Springer (ISBN 978-3-319-21304-0/hbk; 978-3-319-21305-7/ebook). Progress in Mathematics 311, 7-39 (2015). ZBL1338.18011. Lastly let me mention that Kuhn also proved that the some results about the category of polynomial functors between vector spaces over different characteristics is in fact semi-simple. See his comment below. Kuhn, Nicholas J., Generic representation theory of finite fields in nondescribing characteristic, Adv. Math. 272, 598-610 (2015). ZBL1354.18001.<|endoftext|> TITLE: Why is $\mathbb{Z}$ not a Kähler group? QUESTION [11 upvotes]: Is there some simple proof that $\mathbb{Z}$ is not isomorphic to the fundamental group of any compact Kähler manifold? This follows from the main result of https://arxiv.org/abs/0709.4350 which states that any $3$-manifold group which is Kähler must be finite. The simplest non-finite 3-manifold group is $\mathbb{Z} = \pi_{1}(S^{2} \times S^{1})$. So I was wondering, is there an argument to rule out this particularly simple group, without applying to a lot of machinery? (also historically when were we first able to rule this group out?) REPLY [14 votes]: Every free group is not Kähler (fundamental group of compact Kähler manifold). In particular, a free group of rank 1 is not Kähler. This follows since the rank of the abelianization of a Kähler group is even, even rank free groups contain finite index subgroups of odd rank (see here for example), and finite index subgroups correspond to covers of Kähler manifolds which are also Kähler. This and other interesting examples can be found in Fundamental Groups of Compact Kähler Manifold by Amorós, Burger, Corlette, Kotschick, and Toledo.<|endoftext|> TITLE: Direct sum of injective modules is injective QUESTION [6 upvotes]: By the Bass-Papp Theorem, for a unital ring $R$, any direct sum of injective left $R$-modules is injective if and only if $R$ is left Noetherian. I would like to restrict my consideration to an arbitrary abelian subcategory $\mathcal{C}$ of the category $R\text{-mod}$ of unitary left $R$-modules. We say that an abelian subcategory $\mathcal{C}$ of $R\text{-mod}$ is injectively closed if it satisfies the property that, for an arbitrary family $\left(I_\alpha\right)_{\alpha\in A}$ of injective objects in $\mathcal{C}$ such that $I:=\bigoplus\limits_{\alpha \in A}\,I_\alpha$ is an object in $\mathcal{C}$, $I$ is an injective object in $\mathcal{C}$. Is it true that if $R$ is left Noetherian, then any abelian subcategory of $R\text{-mod}$ is injectively closed? If not, can you please provide a counterexample? Is there a sufficient condition for $\mathcal{C}$ to be injectively closed? References are greatly appreciated. For a nontrivial example, let $\mathfrak{g}$ be a finite-dimensional semisimple Lie algebra over an algebraically closed field of characteristic $0$ with a triangular decomposition $$\mathfrak{g}=\mathfrak{n}^-\oplus \mathfrak{h}\oplus \mathfrak{n}^+\,.$$ Denote by $\bar{\mathcal{O}}$ the full subcategory of the category of $\mathfrak{U}(\mathfrak{g})$-modules (where $\mathfrak{U}(\mathfrak{g})$ is the enveloping algebra of $\mathfrak{g}$) consisting of $\mathfrak{U}(\mathfrak{g})$-modules $M$ with the following properties: $M$ is a weight module with respect to the Cartan subalgebra $\mathfrak{h}$, each weight space of $M$ is finite dimensional, and $M$ is locally $\mathfrak{n}^+$-finite (that is, $\mathfrak{U}\left(\mathfrak{n}^+\right)\cdot v$ is a finite-dimensional vector subspace of $M$ for any $v\in M$). Then, $\bar{\mathcal{O}}$ is injectively closed. (In this example, note that $\mathfrak{U}(\mathfrak{g})$ is both left and right Noetherian.) P.S. I: The Bass-Papp Theorem can be found, for example, in Theorem 3.39 on Page 123 of An Introduction to Homological Algebra by Joseph Rotman. P.S. II: I posted this question here as well, but didn't receive any answer. I figured that MathOverflow users may be able to help. I don't know how to move my Math.StackExchange post to MathOverflow. REPLY [3 votes]: EDIT Thanks to Jeremy Rickard for several corrective insights in the comments! In the direction of positive conditions, I'm not sure whether the following conditions are reasonable for your purposes: Proposition: Let $\mathcal{C}$ be an abelian category which is locally finitely presentable (i.e. it has a generator of finitely presentable objects and is complete, or equivalently cocomplete) and "strongly Noetherian" in the sense that any subobject of a finitely-presentable object is finitely-presentable, and similarly for higher presentability degrees. Then $\mathcal{C}$ is injectively closed. Proof sketch: One shows that in such a category, an object is injective if and only if it lifts against monomorphisms between finitely-presentable objects. This uses the "strong Noetherian" property. This implies, in conjunction with the Noetherianity property again, that injective objects are closed under filtered colimits; since they are also closed under finite direct sums, they are thus closed under arbitrary direct sums. Corollary: Let $R$ be a "strongly Noetherian ring" in the sense that $R$-$\mathrm{Mod}$ is "strongly Noetherian", and suppose that $\mathcal{C}$ is a full abelian subcategory of $R$-$\mathrm{Mod}$ which is closed under kernels and colimits in $R$-$\mathrm{Mod}$, and generated under colimits by finitely-presentable $R$-modules. Then $\mathcal{C}$ is injectively closed. Proof sketch: Check that $\mathcal{C}$ satisfies the hypotheses of the proposition. Notes: As Jeremy Rickard points out, for countable $R$, $R$-$\mathrm{Mod}$ is "strongly Noetherian" iff it is Noetherian. So it seems that it is not that much stronger a condition than simply being Noetherian. In Jeremy Rickard's example, where $\mathcal{C}$ is the subcategory of $R$-$\mathrm{Mod}$ given by the image of $S$-$\mathrm{Mod}$, it is the case that $\mathcal{C}$ is closed under kernels and colimits, and is itself locally finitely presentable. But it doesn't satisfy the hypotheses of the corollary because $\mathcal{C}$ is not generated under colimits by modules which are finitely-presentable as $R$-modules. Another way to say this is that the embedding $S$-$\mathrm{Mod} \to R$-$\mathrm{Mod}$ doesn't preserve finite presentability: the module $S$ itself is finitely-presentable as an $S$-module but not as an $R$-module.<|endoftext|> TITLE: How to construct the mirror partner of a blowup? QUESTION [8 upvotes]: Question: Let's assume we have a pair $(X,\check{X})$ that are mirror dual to each other in the sense of Homological mirror symmetry (EDIT: this does not have to be CY n-folds, but can also be a Fano variety and a Landau-Ginzburg partner etc). Now take a subvariety $Y \subset X$ and consider $Z = Bl_Y X$. Are there any examples where such mirror pairs is known? Ideally: is there a general recipe for what should be the mirror $\check{Z}$? Motivation: Derived categories behave nicely under blowup; In fact by Bondal-Orlov, we can construct the derived category of sheaves on $Z$ from that of $X$ and $Y$. Thus one should expect the same for Fukaya categories (e.g., Ivan Smith's Floer cohomology and pencils of quadrics provides some evidence in that direction). Now, such an equivalence should probably have some geometric meaning via SYZ perhaps and be explained in terms of a nice mirror recipe. REPLY [5 votes]: You can find a lot of such examples in the paper of Abouzaid-Auroux-Katzarkov: https://link.springer.com/article/10.1007/s10240-016-0081-9. Basically, they studied the case when $X$ is $(\mathbb{C}^\times)^{n-1}\times\mathbb{C}$, and $Y\subset X$ is codimension 2 and is a hypersurface in $(\mathbb{C}^\times)^{n-1}$. For Fukaya categories, first one should expect a fully faithful embedding $\Phi_\mathbb{L}:\mathcal{F}(Y)\hookrightarrow\mathcal{F}(Z).$ Such an embedding has already been proved in the case when $X$ is a smooth even-dimensional quadric, and $Y\subset X$ is the base locus of a pencil which includes $X$ as one of its members. The functor $\Phi_\mathbb{L}$ should arise from a Lagrangian correspondence $\mathbb{L}\subset Y^-\times Z$. In the simplest case when $X=\mathbb{C}^2$ and $Y$ is a point, $\mathbb{L}$ is simply the Clifford torus $T\subset\mathbb{C}^2$, which arises as the composition of two correspondences $\mathbb{L}_1\subset\mathit{pt}\times E$ and $\mathbb{L}_2\subset E^-\times Z$. In this case, $\mathbb{L}_1$ is the large circle in the exceptional divisor $E\cong\mathbb{P}^1$, and $\mathbb{L}_2$ is the standard $S^3$ in $Z$, namely the boundary of the unit disc bundle associated to $\mathcal{O}(-1)$. The construction in the pencil of quadric case is just a family version of this. In general, the Lagrangian correspondence $\mathbb{L}_2\rightarrow E$ arises as a bundle over the exceptional divisor $E\subset Z$, whose fibers are Clifford tori in the projective space $\mathbb{P}^k$. You can find constructions of similar flavor in the paper of Abouzaid-Auroux-Katzarkov, in which case you will be forced to deal with non-compact Lagrangian correspondences. It is expected by many people that the Fukaya category is well-behaved under birational maps, see the introduction of https://arxiv.org/abs/1508.01573. The main result of the paper referred above deals with the case when the center is trivial, namely $Y$ is a point. For related works in the case of Landau-Ginzburg models, see the paper of Kerr: https://arxiv.org/abs/1603.08074. There are many difficulties in this direction. For example, when the center is not trivial, then you don't get a toric neighborhood in $Z$ after blowing up $Y$, and it's not known in general how to show that the blow-up creates new non-displaceable Lagrangian tori. In the toric case, of course, the existence problem reduces to the analysis of broken holomorphic discs, and this has been done in the paper of Charest-Woodward. In this particular case when the center is trivial, the understanding of $\mathcal{F}(Z)$ reduces essentially to the understanding of $\mathcal{F}(X)$ and $\mathcal{F}(\mathcal{O}(-1))$. The latter one is well-understood by the work of Ritter-Smith.<|endoftext|> TITLE: $L^{p}$ isoperimetric inequalities on the Hamming cube QUESTION [5 upvotes]: Let $A \subset \{-1,1\}^{n}$ be a subset of the Hamming cube with cardinality $|A|=2^{n-1}$. Define $w_{A} : \{-1,1\}^{n} \to \mathbb{N}\cup \{0\}$ so that $w_{A}(x)$ to be number of boundary edges to $A$ containing $x$, i.e., $w_{A}(x)$ counts number of edges of $\{-1,1\}^{n}$ with endpoints in $A$ and in the complement of $A$ so that one of the endpoint is $x$. Clearly $w_{A}(x)=0$ if $x$ is in the "strict interior" of $A$, or in the "strict complement" of $A$, and it is nonzero if and only if $x$ is on the "boundary" of $A$. Notice that $w_{A}(x)$ can be nonzero for some $x \notin A$. The classical edge--isoperimetric inequality on the Hamming cube of Harper implies that $$ \frac{1}{2^{n}}\sum_{x \in \{-1,1\}^{n}}w_{A}(x) \geq 1 $$ and the constant $1$ on the right hand side is sharp. My question is what is known about the best possible $C(p)\geq 0$ such that $$ \frac{1}{2^{n}}\sum_{x \in \{-1,1\}^{n}}w_{A}^{p/2}(x) \geq C(p), \quad 1\leq p\leq 2 $$ for all sets $A \subset \{-1,1\}^{n}$ with cardinality $2^{n-1}$. REPLY [2 votes]: The bound $C(1) \geq \frac{1}{\sqrt{2 \pi}}$ follows from "Bobkov's Inequality" [Bob97], which says that for any $f : \{-1,1\}^n \to [0,1]$ it holds that $\mathcal{U}(\mathbb{E}[f]) \leq \mathbb{E}\sqrt{\mathcal{U}(f)^2 + |\nabla f|^2}$, where $\nabla f$ is the discrete gradient $\nabla f(x) = \left(\frac{f(x_1, \dots, x_{i-1}, 1, x_{i+1}, \dots, x_n) - f(x_1, \dots, x_{i-1}, -1, x_{i+1}, \dots, x_n)}{2}\right)_{i=1\dots n}$ and $\mathcal{U}$ denotes the "Gaussian isoperimetric function" $\phi \circ \overline{\Phi}^{-1}$ (with $\phi$ the Gaussian pdf and $\overline{\Phi}$ the Gaussian complementary cdf). In particular, if $f$ is the $0$-$1$ indicator of a set $A \subseteq \{-1,1\}^n$, then this reduces to $\mathbb{E}_x[\sqrt{w_A(x)}] \geq \mathcal{U}(\text{vol}(A))$, where $\text{vol}(A) = |A|/2^{n}$. Thus in your case where $\text{vol}(A) = 1/2$ we get a lower bound of $\mathcal{U}(1/2) = \frac{1}{\sqrt{2 \pi}}$. This is sharp up to a factor of $\sqrt{2}$ by virtue of the "Majority" function -- i.e., $A$ being a Hamming ball of radius $n/2$ -- which has $\mathbb{E}_x[\sqrt{w_A(x)}] \sim \frac{1}{\sqrt{\pi}}$. Bobkov remarks on this missing factor of $\sqrt{2}$; presumably $\frac{1}{\sqrt{\pi}}$ is the correct answer but it's not known how to improve $\frac{1}{\sqrt{2\pi}}$ as far as I'm aware. Given this result, one can get $C(p) \geq \left(\frac{1}{2\pi}\right)^{p/2}$ for any $1 < p < 2$ by monotonicity of $\ell_p$-norms, which is a decent bound but is surely not optimal. Getting the optimal answer in this range is presumably harder than getting the optimal answer for $p = 1$. By the way, it was Talagrand who first studied the $p = 1$ case; this notion of looking at the average square-root of the number of sensitive coordinates seems to be a pretty good notion of "discrete surface area". Talagrand's original paper [Tal93] proved Bobkov's Inequality (for sets $A$) up to a universal constant factor on the LHS; in particular, he established $C(1)$ is bounded away from $0$ by a univeral constant. [Bob97] Bobkov, S.G., An isoperimetric inequality on the discrete cube and an elementary proof of the isoperimetric inequality in Gauss space, Ann. Probab. 25, No.1, 206-214 (1997). ZBL0883.60031. [Tal93] Talagrand, M., Isoperimetry, logarithmic Sobolev inequalities on the discrete cube, and Margulis' graph connectivity theorem, Geom. Funct. Anal. 3, No.3, 295-314 (1993). ZBL0806.46035.<|endoftext|> TITLE: Sufficient conditions for the finite model property QUESTION [6 upvotes]: Let $L$ be a finite relational language. Let $T$ be a complete theory with infinite models. If $T$ is an almost sure theory then it has the finite model property (in the sense that any model $M$ of $T$ has the property that if $M\models \varphi$, where $\varphi$ is an $L$ sentence, then there is some finite $L$ structure $A$ such that $A\models\varphi$). Also, I believe, if $T$ is $\forall\exists$-axiomatizable where the sentences use at most two variables then you again have the finite model property (This result can be found in chapter 4.1 of "finite model theory" by Ebbinghaus and Flum: http://www.springer.com/us/book/9783540287872). What other conditions guarantee the finite model property? What would be a good reference to these results? REPLY [2 votes]: In my paper Disjoint $n$-amalgamation and pseudofinite countably categorical theories, I did some work on this question for the restricted class of countably categorical theories. (The question is already very hard for these theories! For example, the finite model property for the theory of the generic triangle-free graph is a famous open problem.) A theory $T$ has disjoint $n$-amalgamation if, for any system of complete types $\{p_X((x_i)_{i\in X})\mid X\subsetneq [n]\}$ which agree on their intersections, there is a type $p_{[n]}((x_i)_{i\in [n]})$ extending all of them. For example, disjoint $3$-amalgamation means that whenever three $2$-types $p_{\{0,1\}}(x_0,x_1)$, $p_{\{0,2\}}(x_0,x_2)$, and $p_{\{1,2\}}(x_1,x_2)$ agree on their intersection $1$-types $p_0(x_0)$, $p_1(x_1)$, and $p_2(x_2)$, then there is a $3$-type $p_{\{0,1,2\}}(x_0,x_1,x_2)$ extending all three. This condition fails for the generic triangle-free graph, since you can't amalgamate three edges into a triangle. I observed that if $T$ is countably categorical and has disjoint $n$-amalgamation for all $n$, then $T$ has the finite model property. The proof is by a probabilistic argument very similar to the usual proof of the finite model property for the theory of the random graph. This sufficient condition generalizes other sufficient conditions identified previously (see p. 9 of the linked paper), such as the parametric universal theories (Oberschelp), and local Fraïssé classes (Brooke-Taylor & Testa). In my view, the really interesting thing is that all of the countably categorical theories which I know to have the finite model property have either a "structured/algebraic" flavor (think infinite-dimensional vector spaces over a finite field) or a "random/combinatorial" flavor (think the random graph). And every "combinatorial" example that I'm aware of can be explained by disjoint $n$-amalgamation, either directly (as in the random graph, for example) or indirectly by approximating the theory and taking finite expansions of the language (as in the theories $T^*_{feq}$ and $T_{CPZ}$ analyzed in the linked paper). It would be very interesting to somehow make these ideas precise, but I haven't yet seen how to do it (again, it seems very hard: a successful realization of this goal would show that the theory of the generic triangle-free graph does not have the finite model property, since it is definitely "combinatorial", but it has a very strong obstruction to $3$-amalgamation). So instead I'll just leave you with two quotes: “In all those homogeneous structures which I know to have the finite model property, [it] arises either from probabilistic arguments as above [i.e. 0-1 laws], or from stability, or conceivably from a mixture of these.” - Macpherson, A survey of homogeneous structures “When does a homogeneous structure for a finite relational language have the finite model property? More broadly, is there anything of interest in graph theory besides randomness and algebra?” - Cherlin, Exercises for logicians (worth a read, if you haven't seen them before!)<|endoftext|> TITLE: Generalized limits on $\ell^\infty(\mathbb{N})$ QUESTION [9 upvotes]: Let $\ell^\infty(\mathbb{N})$ denote the set of bounded real sequences $(a_n)_{n\in\mathbb{N}}$. The $\lim$ operator is a partial linear operator from $\ell^\infty(\mathbb{N})$ to $\mathbb{R}$. With the Hahn-Banach theorem, $\lim$ can be extended to a generalized limit functional. The Hahn-Banach theorem uses the Axiom of Choice. Are there models of $\mathsf{ZF}$ in which no generalized limit functionals exist? REPLY [18 votes]: Yes. In fact, if you work in $\mathsf{ZF}+\mathsf{DC}+$ "all sets of reals have the property of Baire" ($\mathsf{BP}$), say the Solovay or Shelah models, you can prove that $(\ell^\infty)^* = \ell^1$, so that the only continuous linear functionals on $\ell^\infty$ are those coming from $\ell^1$. The generalized limits are certainly not of this form, since they vanish on $c_0$ but not identically. Under these axioms, $\ell^1$ is reflexive. You can find a proof in Schechter, Handbook of Analysis and its Foundations, sections 29.37–38. Interestingly, under these axioms all linear functionals on any Banach space are continuous, so in fact the algebraic dual of $\ell^\infty$ also equals $\ell^1$. And the limit operator has no linear extension to $\ell^\infty$. Digression: This is a fun example because there are several different ways to prove in $\mathsf{ZFC}$ that $(\ell^\infty)^* \ne \ell^1$. It is interesting to see how all of them fail in this model, and where they used choice. For instance: Use Hahn-Banach as in your example. Hahn-Banach doesn't work without choice. Use Alaoglu's theorem to produce a weak-* limit point of the usual basis $\{e_i\}$ of $\ell^1 \subset (\ell^\infty)^*$. This is another sort of generalized limit. So Alaoglu also must fail without choice; of course, the usual proof uses Tychonoff. Recall the following exercise: "If $X^*$ is separable then so is $X$." Note that $\ell^1$ is separable and $\ell^\infty$ is not. But the "exercise" used Hahn-Banach, and isn't necessarily true without choice.<|endoftext|> TITLE: Fundamental groups of compact Kähler manifolds QUESTION [10 upvotes]: This is a sort of a follow-up to this question, and especially to Sean Lawton's answer: The book Fundamental Groups of compact Kähler manifolds (which, in my opinion, is one of the best mathematics books on any subject) is fantastic, but is over twenty years old now. Can someone summarize how the state of the art in that field has advanced in the intervening period? REPLY [5 votes]: As Donu mentioned, it is an open problem whether every Kähler group is projective, i.e. the fundamental group of a complex projective manifold (equivalently, a complex projective surface). This question is closely related to the Kodaira problem, which asks whether every compact Kähler manifold can be deformed to a projective variety (at least after passing to a minimal model). Using this relationship, the problem on Kähler groups has recently been answered positively in dimension three, in a series of papers. So let $X$ be a compact Kähler threefold. We make a case distinction according to the Kodaira dimension $\kappa(X)$. $\kappa = -\infty$: $X$ is uniruled, and the MRC quotient $\varphi \colon X \dashrightarrow Z$ induces an isomorphism $\pi_1(X) \cong \pi_1(Z)$. Since $\dim Z \le 2$, $\pi_1(Z)$ is projective. $\kappa = 0$: A minimal model $X'$ of $X$ admits a locally trivial deformation to a projective variety, so $\pi_1(X) \cong \pi_1(X')$ is projective (Graf). $\kappa = 1$: Same line of argument as above (Lin). $\kappa = 2$: Claudon and Höring proved that $\pi_1(X)$ is projective using the theory of elliptic fibrations. $\kappa = 3$: A Kähler manifold which is Moishezon is already projective, so there is nothing to show.<|endoftext|> TITLE: Explicit zero density estimate for Dirichlet $L$-functions QUESTION [6 upvotes]: Let's define $N(\alpha,T,\chi)=\sharp\lbrace \rho=\sigma+i\gamma: L(\rho,\chi)=0, \alpha\leq \sigma<1, |\gamma|\leq T\rbrace$ , where $\chi$ is a primitive Dirichlet character. We know, from Gallagher's paper, that $$\sum_{q\leq T}\sum_{\chi\pmod{q}}N(\alpha,T,\chi)\ll T^{c(1-\alpha)}$$ for every $0\leq \alpha<1$ with $c>0$ an absolute costant. But what about an explicit value of such costant $c$? I need some good and explicit estimate of the sum above. Can anyone help me? REPLY [5 votes]: Matti Jutila (On Linnik's Constant, Math. Scand. 41 (1977), 45-62) proved that if $Q\geq 2$, $T\geq 1$, and $4/5\leq\alpha\leq 1$, and $\epsilon>0$, then $\sum_{q\leq Q}\sideset{}{'}\sum_{\chi\bmod{q}}N(\alpha,\chi,T)\ll_{\epsilon}(Q^2 T)^{(2+\epsilon)(1-\alpha)},$ where the decoration ' indicates a sum over primitive characters. (The restriction to primitive characters is necessary.) The density hypothesis (which follows from the generalized Riemann hypothesis for Dirichlet $L$-functions) predicts that the above estimate holds for $1/2\leq\alpha\leq 1$, and so we have the density hypothesis in a limited range of $\alpha$. Montgomery's bound $\sum_{q\leq Q}\sideset{}{'}\sum_{\chi\bmod q}N(\alpha,T,\chi)\ll(Q^2 T)^{\frac{3(1-\alpha)}{2-\alpha}}(\log QT)^{13},\qquad 1/2\leq\alpha\leq 4/5$ should suffice for most applications when $\alpha$ is far from 1.<|endoftext|> TITLE: Deligne Mumford Compactification of Moduli Space Of Annuli QUESTION [9 upvotes]: I am reading Abouzaid's paper "A geometric criterion for generating the Fukaya category" (https://arxiv.org/abs/1001.4593), and it is claimed there, without proof, in section C.4 in the appendix (pp.34) that the compactificaiton of the moduli space of holomorphic annuli is given by two discs meeting in an interior node at $R\to\infty$ and by two discs meeting at two boundary nodes at $R\to0$, where $e^R$ is the conformal parameter (i.e. annuli is given by $\{1\le|z|\le e^R\}$. I tried searching the literature for a proof, but other than finding the same claim in other papers I have found none. 1) Could someone please suggest a proof, or refer me to one? 2) Is there an explicit way to "calculate" the DM-compactification of a space of Riemann surfaces with boundary (and maybe marked points).? 3) While I can imagine the limit as $R\to\infty$ as a thin neck being developed, I am especially curious about the limit as $R\to 0$ as it seems somewhat arbitrary, why do only 2 chords shrink to a point? why not 1 or 3? Thank you! REPLY [4 votes]: The study of the Deligne-Mumford compactification of Riemann surfaces with boundary can be reduced to the study of the Deligne-Mumford compactification of closed Riemann surfaces together with the study of separating antiholomorphic involutions on them. More precisely, the data of a compact Riemann surface A with (nonempty) boundary can be recovered from its double S=double(A) (a closed Riemann surface), the natural antiholomorphic involution $\iota$ on $S$ and the choice of one connected component C of $S\setminus Fix(\iota)$. Thus Isomorphism classes of compact Riemann surfaces A of a given type are naturally identified with Isomorphism classes of triples $(S,\iota,C)$ (for a separating antiholomorphic involution $\iota$) such that C is of the same topological type as A. In your specific case, the doubles are tori. As Riemann surfaces these doubles can only degenerate in one way, but there are two distinct isomorphism types of involutions on the limit: In one case the fixed point set of the involution has a component consisting of the node and a circle disjoint from the node; in the other case the fixed point set consists of two circles, both passing through the node, but otherwise disjoint. The first case gives rise to a disk with an interior puncture, the second to a disk, meeting with two punctures on the boundary, which form together a boundary node. If the conformal modulus of the annulus A goes to zero (i.e. $R\rightarrow 0$, the latter case), then the geodesic contracting is simply obtained by doubling a segment joining the two boundary components of $A$.<|endoftext|> TITLE: Existence of generalized inverse-like operator QUESTION [5 upvotes]: Does there exist an operator, $\star$, such that for all full rank matrices $B$ and all $A$ of appropriate dimensions: $$ B(B^\intercal AB)^\star B^\intercal = A^\star, $$ and such that $A^\star=0$ if and only if $A=0$? Edit: Also, $\star : \operatorname{M}(m,n,\mathbb R) \to \operatorname{M}(n,m,\mathbb R)$. Edit: If possible, we would also like $\operatorname{rank}(A^\star)=\operatorname{rank}(A)$. REPLY [7 votes]: No. We suppose it is defined for $2\times 2$ matrices and we get a contradiction. Unless you drop the requirement on $\operatorname{rank} A^\star$ in which case $A^\star=0$ trivially works. Let $A=\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$ and let $B=\begin{pmatrix} a & b\\ c & d\end{pmatrix}$ with arbitrary $a,b,c,d\in\mathbb R$ satisfying $ad-bc\neq 0$. The key observation is that $B^T A B = \begin{pmatrix} a^2 & ab \\ ab & b^2\end{pmatrix}$ does not depend on $c,d$. Therefore $(B^T A B)^\star=B^{-1}A^\star (B^T)^{-1}$ should be independent of $c,d$ as well. Let $A^\star =\begin{pmatrix} x & y\\ z & w\end{pmatrix}$ for some $x,y,z,w\in\mathbb R$. Then we compute $$B^{-1}A^\star (B^T)^{-1}=\frac 1 {(\operatorname{det} B)^2}\begin{pmatrix} x d^2 - (y+z) cd + w c^2 & -x bd +(yad + zbc) -wac\\ -x bd +(ybc + zad) -wac & x b^2 - (y+z) ab + w a^2 \end{pmatrix}$$. In case $B=\begin{pmatrix} 1 & 0\\ t & 1\end{pmatrix}$ we get that $B^{-1}A^\star (B^T)^{-1}=\begin{pmatrix} x - (y+z)t + w t^2& \dots\\ \dots & \dots\end{pmatrix}$ is independent of $t$. So $y+z=0$ and $w=0$. Now, in case $B=\begin{pmatrix} 0 & -1\\ 1 & t\end{pmatrix}$ we get $B^{-1}A^\star (B^T)^{-1}=\begin{pmatrix} x t^2 + 0 & \dots\\ \dots & \dots\end{pmatrix}$ from which $x=0$. This is already a contradiction, because $A^\star=\begin{pmatrix} 0 & y \\ -y & 0\end{pmatrix}$ has either rank 0 or 2. But in case $B=\begin{pmatrix} 1 & 0\\ 0 & t\end{pmatrix}$ we get that $B^{-1}A^\star (B^T)^{-1}=\frac 1 {t^2}\begin{pmatrix} 0 & yt \\ -yt & 0\end{pmatrix}$ is independent of $t$, so $y=0$.<|endoftext|> TITLE: References for existence of solutions to overdetermined system of partial differential inequalities QUESTION [7 upvotes]: I want to show the existence of solutions to an overdetermined system of second-order partial differential inequalities in a given region $\Omega\subset\mathbb{R}^2$, \begin{equation*} \begin{cases} P_1(u)\geq 0,&\\ P_2(u)\geq 0,&\\ P_3(u)\geq 0.& \end{cases} \end{equation*} Updated: $P_i$ are linear/quasilinear/nonlinear second-order differential operators, \begin{equation*} \begin{cases} P_1(u)=&\hspace{-0.3cm}xu_{xx}+yu_{xy}+x^2u_{yy}+4u_x.\\ P_2(u)=&\hspace{-0.3cm}u_{xx}u_{yy}-u_{xy}^2+(xu_x-yu_y)u_{xx}+(yu_x-x^2u_y)u_{xy}+x(xu_x-yu_y)u_{yy}.\\ P_3(u)=&\hspace{-0.3cm}(u_{xx}u_{yy}-u_{xy}^2)u_x+(xu_x^2-yu_xu_y-x^2u_y^2)u_{xx}+(yu_x^2-x^2u_xu_y-xyu_y^2)u_{xy}+(x^2u_x^2-xyu_xu_y-y^2u_y^2)u_{yy}. \end{cases} \end{equation*} To do this, I am trying to solve for $(u,f_1,f_2,f_3)$ of the following system of PDEs, \begin{equation*} \begin{cases} P_1(u)=f_1^2,&\\ P_2(u)=f_2^2,&\\ P_3(u)=f_3^2.& \end{cases} \end{equation*} Could anyone suggest any references to deal with either system? Any suggestion is appreciated. Updated: Is it true that if I find all compatible conditions for $f_1,f_2,f_3$ and they are satisfied, then this system has a solution? REPLY [3 votes]: The Cartan--Kaehler theorem gives sufficient conditions for the existence of local solutions of any real analytic system of partial differential equations. However, the solutions are only local. See Bryant, Chern, Gardiner, Goldschmidt and Griffiths, Exterior Differential Systems, MSRI.<|endoftext|> TITLE: Lagrangian Grassmannian from an Involution QUESTION [10 upvotes]: I don't know if this is already answered somewhere in MO. The dynkin involution of $SL_{2n}$ that is $\alpha_i \mapsto \alpha_{2n-i}$ gives an outer automorphism of $SL_{2n}$ and then the maximal parabolic $P_n$ corresponding to the simple root $\alpha_n$ is invariant under this automorphism. So it induces an automorphism of $Gr(n,2n)=SL_{2n}/P_n$. Do you get the Lagrangian Grassmannian from here say as a fixed points of this automorphism or you need to twist it by an inner automorphism of $SL_{2n}$ to get the Lagrangian Grassmannian $LG(n,2n)$ ? How is this automorphism related to the involution $V \mapsto V^{\perp}$ on $Gr(n,2n)$ ? Of course here one needs to fix a symplectic form on $\mathbb C^{2n}$ in order to define the map. REPLY [3 votes]: I think the confusion comes from the term "gives an outer automorphism" which has to be made precise. The automorphism, say $\phi$, is only well-defined up to multipliciation by conjugation with an element of the torus. To make it unique one usually requires that a "pinning" $e_\alpha$ is fixed. For the $SL(n)$ this means that, on the Lie algebra level, the elementary matrix $E_{i,i+1}$ is mapped to $E_{2n-i,2n-i+1}$. It is easily verfied that $\phi(A)=J^{-1}(A^t)^{-1}J$ does the job where $J=\text{antidiag}(1,-1,1,\ldots,-1)$. Observe that $J$ is swewsymmetric so the fixed point set of $\phi$ is a symplectic group. To get the symplectic form $\omega$ one wants $A\mapsto\phi(A)^{-1}$ to be the adjoint $A^*$ with respect to $\omega$: $$ \omega(Ax,y)=\omega(x,\phi(A)^{-1}y)\text{ or, equivalently, }\omega(Ax,\phi(A)y)=\omega(x,y). $$ Another easy calculation shows that $\omega(x,y)=x^tJy$ fits. Any other solution is a non-zero multiple. At this stage, the perpendicular space $U^\perp$ with respect to $\omega$ is defined. It is intertwined with $\phi$ via the formual $$ (AU)^\perp=\phi(A)U^\perp. $$ Thus if you let $\phi$ act on $Gr(n,2n)$ as $\phi(U)=U^\perp$ then one gets $\phi(AU)=\phi(A)\phi(U)$ and then $LG(n,2n)$ is precisely the fixed point set of $\phi$.<|endoftext|> TITLE: Why is conformal invariance only possible for massless theories? QUESTION [34 upvotes]: I'm conscious that this isn't necessarily a research level question, but I've asked this question on mathstackexchange, and received no answer. So I'm trying it here. A usual mantra in field theories is the assertion that only massless theories can be conformally invariant. By a theory I mean an action $$ S = \int \mathcal{L} \, \mathrm{dVol}, $$ where $\mathcal{L}$ is the Lagrangian density, and the integral is taken over a 4-dimensional Lorentzian manifold with metric $g$. By conformal invariance I mean the statement that under the conformal rescaling of the metric $$ \hat{g} = \Omega^2g, $$ the Lagrangian transforms as $\hat{\mathcal{L}} = \Omega^{-4} \mathcal{L}$. Then, as the volume form transforms as $\widehat{\mathrm{dVol}} = \Omega^{4} \mathrm{dVol}$, the action $S$ is invariant, and the theory is said to be conformally invariant. The usual physics explanation given is that "if a theory is supposed to be conformally invariant, then there cannot exist an intrinsic scale to it, such as mass or a Compton wavelength". Of course, this is a load of hand waving. I guess I don't strictly know what I mean by a massless theory. Maxwell's equations, for example, are a massless conformally invariant theory. My guess would have been that the mass of a theory is its ADM mass, but as has been pointed out in the comments, this is a property of a solution to a theory, not the theory itself. So, if $m$ is the mass of a theory, whatever it stands for exactly, and $m \neq0$, why must conformal invariance fail? REPLY [8 votes]: Why is "conformal invariance only possible for massless theories?" Actually, the answer is NO. Not only the massless, but also the infinite large massive theory can have "conformal invariance." As long as the theory sets no scale in order to be scale invariance. Some massless theories (or gapless, which means the energy gap $\Delta_E$ for excitations away from the vacuum ground state is zero, $$\Delta_E\equiv E_{exc}- E_0 \to 0,$$ where $E_{exc}$ and $E_0$ are the energies of excited states and ground state respectively) theory can be "conformal invariance" or be a "conformal field theory (CFT)." The $\Delta_E=0$ sets no scale in the theory, thus it can be scale invariant. The infinite large massive theory, or the infinite large gap theory can still be "conformal invariance" or be a "conformal field theory." The infinite large massive theory means there is an infinitely large mass $M\to \infty$, or say the energy gap $\Delta_E$ for excitations away from the vacuum ground state has $$\Delta_E\equiv E_{exc}- E_0 \to \infty.$$ So the length/time scale of the system (based on the dimensional analysis $\Delta_p \Delta_x \sim \Delta_E \Delta_t \sim \hbar$) $$\Delta_l \to 0$$ The well-known examples of such a massive $\Delta_E \to \infty$ is Topological Quantum Field Theory (TQFT). Indeed, Topological Quantum Field Theory at its ground state has no scales (thus scale invariance) and it is also Poincaré (translations + Lorentz) invariance, and also have a special conformal transformation (if needed). Thus TQFT at ground state is also conformal invariant. The $\Delta_E=\infty$ also sets no scale in the theory, thus it can be scale invariant. TQFTs may be among the most simplest/trivial examples of CFT. But TQFTs are still complicated enough to have many different distinct classes. The "conformal invariance" requires the scale invariance/dilations and the special conformal transformation, and Poincaré group (the semi direct product of translations group and Lorentz group).<|endoftext|> TITLE: Schematic locus of algebraic spaces in fibers QUESTION [10 upvotes]: Let $S$ be a scheme and let $X$ be a quasi-separated algebraic space over $S$. Does there exist an open subspace of $X$ which is a scheme and which is dense in each fiber $X_s$, $s \in S$? I am happy to make the following additional assumptions: (1) $S$ is the spectrum of a complete dvr with algebraically closed residue field and $X$ is flat and of finite type over $S$. (2) All fibers of $X$ are geometrically irreducible. (3) The generic fiber of $X$ is geometrically normal and the special fiber of $X$ is geometrically integral (then $X$ is normal). REPLY [6 votes]: A counter example to the question in the presence of (1), (2), and (3) is to take $\mathbf{G}_{m, S}$ and divide out by the etale equivalence relation given by $(x, s) \sim (x, s)$ and for $s \not = 0$ also $(x, s) \sim (-x, s)$ (assume the residue characteristic of your dvr is not $2$). Then the resulting morphisms $$ \mathbf{G}_{m, S} \longrightarrow X \longrightarrow S $$ have the following properties: the first is surjective \'etale and the second is smooth of relative dimension 1 and both $\mathbf{G}_{m, S}$ and $X$ have geometrically integral fibres over $S$. The schematic locus of X is exactly the complement of the special fibre. It seems to me the "problem" is that the degree of the function field extensions induced between the fibres of $\mathbf{G}_{m, S}$ and $X$ drops from $2$ to $1$ from the generic to the special fibre. PS: As Jason points out on locally Noetherian separated algebraic spaces the schematic locus contains all codimension $1$ points, see Tag 0ADD<|endoftext|> TITLE: Do Riemann-Weil formulas exist for functions other than the Mangoldt function $ \Lambda (n) $ QUESTION [5 upvotes]: Are there formulas similar to the Riemann-Weil formula for other arithmetical functions like $ \mu (n) $ or $ \lambda (n) $, for example a sum of the form $ \sum_{n=1}^{\infty}a(n) f(n) $ with this sum being related to the sum over the imaginary part of the Riemann zeros involving the Fourier transform of $ f(x)$? To summarize: can the Riemann-Weil explicit formulas be generalized to other arithmetical functions? For example: $$ \sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}g(\log n)=\sum_{\gamma}\frac{h( \gamma)}{\zeta '( \rho )}+\sum_{n=1}^{\infty} \frac{1}{\zeta ' (-2n)} \int_{-\infty}^{\infty}dxg(x)e^{-(2n+1/2)x} $$ REPLY [8 votes]: The reason that $\mu(n)$ and $\lambda(n)$ have such expressions is that the corresponding Dirichlet generating functions can be expressed in terms of the Riemann zeta function: $$ \sum_n \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)} $$ and $$ \sum_n \frac{\lambda(n)}{n^s}=\frac{\zeta(2s)}{\zeta(s)}. $$ In general, if the Dirichlet series coefficients are multiplicative: $a(mn)=a(m)a(n)$ for $(m,n)=1$, then there will be an Explicit Formula in terms of the zeros (and poles) of the corresponding $L$-function. For example, when $a(n)=\chi(n)$ is a Dirichlet character modulo $d$, there is a formula involving the zeros of $L(s,\chi)$. The classical version (with a cutoff for the sums rather than a test function and its transform) is in Montgomery and Vaughan's Multiplicative Number Theory, chapter 12.1. You can imitate the treatment of the zeta function in chapter 12.2 to get a version with a test function. See also the notes in chapter 12.3, which point to Weil's treatment of the case of an $L$-function attached to a Grossencharaktere $\chi$.<|endoftext|> TITLE: An interesting identity: in search of a proof -Part I QUESTION [12 upvotes]: I like the following binomial identity in that the RHS extracts the indeterminate $w$ from the LHS. Question. Can you show that $$\sum_{k=0}^n\binom{x+kw}k\binom{y-kw}{n-k}=\sum_{k=0}^n\binom{x+y-k}{n-k}w^k\,\,\,?$$ It would be great if we can see alternative proofs? I've a bias for combinatorial arguments. :-) REPLY [13 votes]: Oh no, I was too slow... sorry for the double reference. This is Jensen's identity. It first appeared (in a slightly modified form) in: Jensen, Sur une identité d'Abel et sur d'autres formules analogues, Acta Mathematica, vol 26 (1902) pp.307-318. There is the following paper that gives elementary proofs to Jensen's identity and some generalizations or related formulas: Guo, On Jensen's and related combinatorial identities. Applicable Analysis and Discrete Mathematics Vol. 5, No. 2 (2011), pp. 201-211. The elementary proof is very short, and consists just in a clever use of the Chu-Vandermonde convolution formula $$\sum_{k=0}^n \binom A k \binom B {n-k}=\binom {A+B} n\quad (1)$$ on the term $\binom {y-k\omega} k $ with $A=x+y+1$ and $A+B=y-k\omega $, followed by a straightforward change in the order of summation.  Then you get an equivalent identity  (just change of the names of the variables). The proof of Chu's and Mohanty-Handa's multinomial generalization of Jensen's identity is similar, and involves a multinomial Chu-Vandermonde used iteratively. Since (1) has combinatorial significance, it makes sense to rewrite Guo's computations avoiding changes of variables and negative binomials. You get an equivalent way of presenting the proof, which is slightly more amenable to combinatorial proofs. A key summation here is: $$ \sum_{0\leq k\leq i\leq n} \binom {x+k\omega} k (-1)^{i-k} \binom {x+k\omega-k} {i-k} \binom {x+y-i} {n-i}. $$ Now it is possible to show that if you sum over $i$ first, you get the LHS of Jensen's identity, while you get the RHS summing over $k$. Moreover, both directions can possibly be proved combinatorially (assuming at least that $x$,$y$, and $\omega$ are integers).<|endoftext|> TITLE: How can Wall's theorem be generalised to non-simply connected manifolds? QUESTION [6 upvotes]: In a sense, this is a follow-up to this question. By work of Freedman and Wall, it is known that if two simply-connected 4-manifolds $M$ and $N$ are homeomorphic, then there is $k \in \mathbb{N}$ such that $M \times \#^k (S^2 \times S^2)$ is diffeomorphic to $N \times \#^k (S^2 \times S^2)$. (By $\#^k$, we denote $k$-fold connected sum.) Can this statement be generalised to non-simply-connected manifolds, possibly including Whitehead torsion? REPLY [5 votes]: For orientable 4-manifolds, it was shown by Gompf as Jeff says. For compact nonorientable 4-manifolds with universal cover not spin, homeomorphic also implies stably diffeomorphic. On the other hand, Kreck showed that for every finitely presented group G with a surjective homomorphism $w \colon G \to C_2$, there is a nonorientable 4-manifold $M$ with 1-type $(G,w)$, with the following property. The 4-manifold ($M$ plus the K3 surface) is homeomorphic to ($M$ plus eleven copies of $S^2 \times S^2$), but these two manifolds do not become diffeomorphic after adding copies of $S^2 \times S^2$. https://link.springer.com/content/pdf/10.1007/BFb0075570.pdf Thus there is a sense in which Wall's theorem fails quite badly to generalise to nonorientable 4-manifolds. Here 1-type means (fundamental group, orientation character).<|endoftext|> TITLE: Analogue of j-invariant for CM fields QUESTION [6 upvotes]: For any imaginary quadratic field $F$, the Hilbert class field $H$ is generated by the $j$-invariant of any elliptic curve with complex multiplication (CM) by $\mathcal O$, the ring of algebraic integers of $F$. What is the simplest generalization of this well known and useful fact? Since an imaginary quadratic field is the simplest example of a CM field, one is led to ask: Is there an (albeit conjectural) invariant of abelian surfaces which generates the Hilbert class field of a CM field $K$ (of degree $4$ over $\mathbb Q$)? Note that $K$ is a totally imaginary quadratic extension of a real quadratic field. REPLY [6 votes]: The simplest generalisation to abelian surfaces is, I believe, the statement (which is a theorem, not a conjecture) that the Igusa invariants of an abelian surface with CM by $K$ generate an abelian unramified extension of the reflex field of $K$. Note that the reflex field is not, in general, equal to $K$, and that there is no claim that the full Hilbert class field is obtained in this way. There is also a generalisation of this to higher dimensional principally polarised abelian varieties.<|endoftext|> TITLE: A "quantum" identity: in search of a proof -Part II QUESTION [12 upvotes]: As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write $$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$ As a follow up on this MO question, I propose a $q$-analogue identity. Question. Can you show that $$\sum_{k=0}^nq^{(y-n+1)k}\binom{x+k}k_q\binom{y-k}{n-k}_q =\sum_{k=0}^nq^{n-k}\binom{x+y-k}{n-k}_q\,\,\,?$$ It would be great if we can see alternative proofs? I've a bias for combinatorial arguments. :-) REPLY [7 votes]: Actually this is a partial case of $q$-Vandermonde identity. In the notation $$[a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n,$$ as in this terminology question, we have $$ [a;c]_q^n=\sum {\binom{n}{k}}_q[a;b]_q^{n-k}\,[b;c]_q^{k}, $$ which is an equivalent form of $q$-Vandermonde identity. Substituting $a=q^{n-y-1}$, $b=1$, $c=q^{x+1}$ we get $$\sum_{k=0}^nq^{(y-n+1)k}\binom{x+k}k_q\binom{y-k}{n-k}_q=\binom{x+y+1}n_q,$$ replacing $x,y$ to $x+y-n,n$ (and $k$ to $n-k$ in the sum) we get that RHS also equals $\binom{x+y+1}n_q$.<|endoftext|> TITLE: Is there anything significant about GAP's SmallGroup(512,2045)? QUESTION [15 upvotes]: Here's the output of the GAP command "SmallGroupsInformation(512)" There are 10494213 groups of order 512. 1 is cyclic. 2 - 10 have rank 2 and p-class 3. 11 - 386 have rank 2 and p-class 4. 387 - 1698 have rank 2 and p-class 5. 1699 - 2008 have rank 2 and p-class 6. 2009 - 2039 have rank 2 and p-class 7. 2040 - 2044 have rank 2 and p-class 8. 2045 has rank 3 and p-class 2. 2046 - 29398 have rank 3 and p-class 3. 29399 - 30617 have rank 3 and p-class 4. 30618 - 31239 have rank 3 and p-class 3. 31240 - 56685 have rank 3 and p-class 4. 56686 - 60615 have rank 3 and p-class 5. 60616 - 60894 have rank 3 and p-class 6. 60895 - 60903 have rank 3 and p-class 7. 60904 - 67612 have rank 4 and p-class 2. 67613 - 387088 have rank 4 and p-class 3. 387089 - 419734 have rank 4 and p-class 4. 419735 - 420500 have rank 4 and p-class 5. 420501 - 420514 have rank 4 and p-class 6. 420515 - 6249623 have rank 5 and p-class 2. 6249624 - 7529606 have rank 5 and p-class 3. 7529607 - 7532374 have rank 5 and p-class 4. 7532375 - 7532392 have rank 5 and p-class 5. 7532393 - 10481221 have rank 6 and p-class 2. 10481222 - 10493038 have rank 6 and p-class 3. 10493039 - 10493061 have rank 6 and p-class 4. 10493062 - 10494173 have rank 7 and p-class 2. 10494174 - 10494200 have rank 7 and p-class 3. 10494201 - 10494212 have rank 8 and p-class 2. 10494213 is elementary abelian. This size belongs to layer 7 of the SmallGroups library. IdSmallGroup is not available for this size. Even if I only barely know what "rank" and "p-class" are, 2045 on that list stands out. Is there any other way to describe it? For example, is it the Sylow subgroup of any simple group? REPLY [19 votes]: It is the "free" group of nilpotency class at most $2$, exponent dividing $4$, on three generators. (In other words, every group in that class is a quotient of it.) According to https://etd.ohiolink.edu/rws_etd/document/get/osu1086112148/inline (ON SYLOW $2$-SUBGROUPS OF FINITE SIMPLE GROUPS OF ORDER UP TO $2^{10}$, Sergey Malyushitsky, M.S. Thesis, The Ohio State University) it is not the Sylow $2$-subgroup of a finite simple group. In general, the "free" group of nilpotency class at most $2$ and exponent at most $4$ on $d$ generators has order $4^d\cdot 2^{\binom{d}{2}}$. So, for example, you should see a similar phenomena at order 32, where there is a unique group of rank $2$ and $p$-class $2$.<|endoftext|> TITLE: A labelling of the vertices of the Petersen graph with integers QUESTION [16 upvotes]: The vertices of the Petersen graph (or any other simple graph) can be labelled in infinitely many ways with positive integers so that two vertices are joined by an edge if, and only if, the corresponding labels have a common divisor greater than 1. (One such labelling is with the numbers 645, 658, 902, 1085, 1221, 13243, 13949, 14053, 16813, and 17081, whose sum is 79650). Of all such ways of labelling the Petersen graph, what is the minimum the sum of the 10 corresponding integers can be? The reason to single out the Petersen graph over all other graphs of order ten or less is that it is one particularly difficult to label if one is trying to achieve the minimum sum. Is there some systematic why of finding that minimum other than brute force? REPLY [11 votes]: The minimum value is $37294$ as described by F. Barrera. I broke the symmetry a little by identifying $9$ inequivalent triples of edges to which the primes $\{41,43,47\}$ can be assigned, wrote a constraint satisfaction program for the problem, and then used Minion to solve it. (I am sure there are more efficient ways to do this.)<|endoftext|> TITLE: Maximizing a ratio of determinants QUESTION [9 upvotes]: Let $D\in\mathbb{R}^{n\times n}$ be a diagonal positive definite matrix s.t. $D\leq I$ ($I$ denotes the $n$-dim. identity matrix) and let $\alpha$ be a strictly positive real number. Consider the optimization problem over the set of positive semidefinite matrices with trace less or equal than one $$ \max_{A\in\mathbb{R}^{n\times n}\,:\,A\geq 0,\, \mathrm{tr}(A)\leq 1}\frac{\det(A+\alpha I)}{\det(AD + \alpha I)}. $$ My question: Is the optimal matrix $$ A^\star:=\arg\max_{A\in\mathbb{R}^{n\times n}\,:\, A\geq 0,\,\mathrm{tr}(A)\leq 1}\frac{\det(A+\alpha I)}{\det(AD + \alpha I)} $$ diagonal? REPLY [6 votes]: Yes, of course (provided that you mean that the maximum is attained on a diagonal matrix, not that every matrix on which it is attained is diagonal). First notice that it is a bit more convenient to have the denominator in the form $\det(A+Q)$ where $Q$ is $\alpha D^{-1}$. Now $A=R^*BR$ where $B$ is diagonal and $R$ is orthogonal. Notice that the numerator does not depend on $R$ or on the order of the diagonal elements in $B$. Using just all permutations $R$, we see that WLOG we may assume that the diagonal elements of $B$ and $Q$ are arranged in the same order. So, WLOG, $B_{11}\ge B_{22}\ge\dots$ and the same for $Q$. We want to show that in this case $\det(R^*BR+Q)$ over all orthogonal $R$ is minimized at $R=I$. Recall that for a positive definite $X$, $(\det X)^{\frac 1n}=\frac 1n\inf \operatorname{Tr}YX$ where $Y$ runs over all positive definite matrices of determinant $1$. Now fix the eigenvalues of $Y$ (say $y_1\le y_2\le\dots\le y_n$) (note that I arranged them in the opposite order comparing to the order of $B_{ii}$ and $Q_{ii}$. Note also that $\operatorname{Tr}YR^*BR=\operatorname{Tr}RYR^*B$ and $RYR^*$ has the same eigenvalues as $Y$. At last, note that if $Y=\operatorname{diag }(y_1,\dots,y_n)$ and $R=I$, then $\operatorname{Tr}RYR^*B=\sum_i y_iB_{ii}$ and similarly for $Q$. Thus it will suffice to show that if $Y$ and $X$ are positive semi-definite with the eigenvalues $y_1\le y_2\le\dots\le y_n$ and $x_1\ge x_2\ge\dots\ge x_n$, then $\operatorname{Tr}YX\ge\sum_i x_iy_i$. That, without any doubt, can be found in textbooks, but let me provide a proof just in case. Using the spectral decomposition $X=x_nP_n+(x_{n-1}-x_{n})P_{n-1}+\dots+(x_1-x_2)P_1$, we see that it is enough to prove the desired inequality for the case when $X$ is an orthogonal projection to a $k$-dimensional space. Let $e_i$ be the eigenvector of $Y$ corresponding to the eigenvalue $y_i$. There is a unit vector $u_k$ in $H$ orthogonal to $e_1,\dots,e_{k-1}$. Then $\langle YPu_k,u_k\rangle\ge y_k$. Next, there exists a unit vector $u_{k-1}\in H$ orthogonal to $e_1,\dots,e_{k-2};u_k$. Thus, $\langle YPu_{k-1},u_{k-1}\rangle\ge y_{k-1}$, and so on. This gives an orthonormal basis $u_1,\dots,u_k$ in $H$ with $\sum_{i=1}^k\langle YPu_i,u_i\rangle\ge \sum_{i=1}^k y_i$. Complementing it by an orthonormal basis of $H^\perp$ (which will be killed by $P$ anyway), we see that the LHS equals $\operatorname{Tr}YP$ and we are done.<|endoftext|> TITLE: Reference request: Reduced reflection length in Coxeter groups QUESTION [6 upvotes]: I recently read this paper, where the authors define on page 26 what they call the reduced reflection length. For that we take a Coxeter group $G$ with Coxeter generators $S$ and transpositions $T$. The function $\ell$ gives the classical Coxeter length on $G$, that is $\ell(w)$ is the length of a shortest word in $S$ that yields $w$. The said reduced length is then defined as $$\ell_R(w) := \min \{r \mid w = t_1t_2\ldots t_r, \,\, t_i \in T, \,\, \ell(w) = \sum_{i=1}^r \ell(t_i)\}$$ and the authors show that the depth statistic discussed in the given paper satisfies $$dp = \frac{\ell + \ell_R}{2}$$ in many cases of interest. I tried to find more information on $\ell_R$, but couldn't find the term in any other work. Has anyone already heard of the term and knows where to look? What I am most interested in is the joint distribution of length and reduced length or length and depth (by the above formula, both problems are equivalent), that is the polynomials $$\sum_{g \in G} x^{\ell(g)}y^{\ell_R(g)} \text{ and } \sum_{g \in G} x^{\ell(g)}y^{dp(g)}$$ where we assume $G$ to be finite. I also couldn't find any paper discussing the length with either one of these functions together. While I would love to find answers for all finite Coxeter groups, it would already be great to get information on the special case $G = S_n$. REPLY [6 votes]: After contacting the authors of the paper mentioned above, I want to share the information they gave me with anyone who might ever stumble upon this question. For that, define an ordering on $G$ by saying $g \leq h$ if and only if there are reflections $t_1,t_2,\ldots, t_k$ such that $g = h t_1t_2\ldots, t_k.$ $\ell(g) = \ell(h) + \sum_{i=1}^k \ell(t_i).$ Then the reduced reflection length is the rank function for this order. Now this order appears in some works under different names (some only defined for symmetric groups): The Grassmannian order in Schubert polynomials, the Bruhat order, and the geometry of flag manifolds, Nantel Bergeron, Frank Sottile, 1998. The $T_G$ order in Partial orders generalizing the weak order on Coxeter groups, Curtis D Bennetta, Rieuwert J Blok, 2003. The BeSo order in A simple definition for the universal Grassmannian order, Curtis D. Bennetta, Lakshmi Evanib, David Grabinerc, 2003 Note that I don't claim this list to be complete. If anyone happens to find this question and is interested in discussing the problems I mentioned above or related topics, feel free to leave me a message.<|endoftext|> TITLE: Equivariant cohomology defined by restrictions? QUESTION [5 upvotes]: Suppose that $G=S^1$ acts on a smooth, connected, compact manifold with discrete fixed points, additionally assume that there is at least one fixed point. Let $\alpha \in H^{2}_{S^1}(M)$ be such that $\alpha|_{p} = 0$ for any fixed point $p$. The space of such $\alpha$ forms an additive sub-group $A \subset H^{2}_{S^{1}}(M)$. Question: is there an example where $A \neq \{0\}$? How about if we assume that $M$ is a compact symplectic manifold with a Hamiltonian $S^1$-action with discrete fixed points? REPLY [2 votes]: This subgroup $A$ is precisely the torsion subgroup in $H^2_{S^1}(M)$ (since $H^2_{S^1}(M^{S^1})$ is obviously free, and the relative $H^2_{S^1}(M,M^{S^1})$ is torsion). So, it will be trivial if and only if the equivariant cohomology in degree 2 is free. This is true for compact $M$ with Hamiltonian $S^1$-action by Kirwan (see, for example, Theorem 14.1 of http://www.math.ias.edu/~goresky/pdf/equivariant.jour.pdf), but I think this is too much to hope for in general (I don't know a connected counterxample off-hand; obviously, you can take a free action on a compact manifold, and take disjoint union with a fixed point).<|endoftext|> TITLE: Surgery along an embedded surface in a 4-manifold QUESTION [7 upvotes]: Let $X$ be a 4-manifold and $\Sigma_g\subset X$ be an embedded closed orientable genus = $g$ surface. Suppose $\Sigma_g\subset X$ has a trivial closed normal bundle $N(\Sigma_g) = \Sigma_g\times D^2$. For a given self-diffeomorphism $h$ on $\Sigma_g\times S^1$, we can define a new 4-manifold: $$X(\Sigma_g,h):=(X-\text{Int}N(\Sigma_g))\cup_{h}(\Sigma_g\times D^2)$$ For example, in the case of $g=0$ this operation is called Gluck construction, and in the case of $g=1$ it is well known as (generalized) logarithmic transformation. My question is: Is there any research in the case of $g\geq2$? Thank you for your help. REPLY [8 votes]: For $g\geq 2$, this construction doesn't yield anything new. In this situation, any self-diffeomorphism of $\Sigma_g \times S^1$ extends to a self-diffeomorphism of $\Sigma_g \times D^2$, and hence your $X$ is unchanged.<|endoftext|> TITLE: Centralizers of subtori in reductive groups, derived subgroups QUESTION [7 upvotes]: Let $G$ be a split, almost-simple connected reductive group over a field $F$ with split maximal torus $T$. I am trying to understand precisely the groups $[G_{\alpha}, G_{\alpha}]$, where $\alpha$ is a root, $G_{\alpha} = \mathcal{Z}_G(T_{\alpha})$, and $T_{\alpha} = (\mathrm{ker} \ \alpha)^{\circ}$. Sometimes this group is isomorphic to $SL_2$, and sometimes to $PGL_2$. Given a root $\alpha$, let's say ``in coordinates'' (for example, $G = PSO(2n)$ and $\alpha = e_{n-1} + e_n$, or $G = SL(10) / \mu_2$, and $\alpha = e_2 - e_3$), how can I tell which group $[G_{\alpha}, G_{\alpha}]$ is? I am especially interested in any isogeny of type $A_n$, although really I will be interested in all types $A$ through $G$, and all isogenies. Perhaps you can give me a reference for how to do this computation in general, or an explanation for how to do this? REPLY [3 votes]: Since I erred in my initial comment, to compensate here is a characteristic-free argument over general fields $k$ (that also adapts to work over general commutative rings). As in Paul Levy's answer, we will show that beyond the trivial rank-1 case, PGL$_2$ is obtained precisely for the short root of adjoint type B$_n$ with $n \ge 2$ (and one could make the statement more uniform including rank 1 via type B$_n$ for $n \ge 1$ if we declare that the roots for type ${\rm{B}}_1 = {\rm{A}}_1$ are short, or perhaps both long and short). The simply connected case always gives ${\rm{SL}}_2$ for the general reason that the derived group of any torus centralizer in a simply connected semisimple group is always simply connected. A proof of this general fact, based on the characterization of "simply connected" in terms of simple positive coroots being a basis of the cocharacter lattice, is sketched in Corollary 9.5.11 of https://www.ams.org/open-math-notes/omn-view-listing?listingId=110663 (where the argument is given over fields, and a reference is provided to make the argument work over rings). Now consider a general split connected semisimple $k$-group $G$ that is (absolutely) simple with split maximal $k$-torus $S$, and let $f:\widetilde{G} \rightarrow G$ be the split simply connected central cover. The preimage $\widetilde{S} := f^{-1}(S)$ is a split maximal $k$-torus of $\widetilde{G}$, and $\Phi(G,S)= \Phi(\widetilde{G},\widetilde{S})$ via the finite-index inclusion ${\rm{X}}(S) \subset {\rm{X}}(\widetilde{S})$ due to the centrality of the subgroup scheme $\ker f$ (even though ${\rm{Lie}}(f)$ may not be surjective). Here is the main trick to avoid too much case-work: by the settled simply connected case we know $[\widetilde{G}_a, \widetilde{G}_a] = {\rm{SL}}_2$ for all $a \in \Phi$, and this maps onto $[G_a,G_a]$ via a central isogeny. In particular, if the group scheme $\ker f$ has odd degree then we're done (as SL$_2$ has scheme-theoretic center $\mu_2$ of order 2). Thus, the case of type A$_2$ (i.e., PGL$_3$ and SL$_3$) always gives SL$_2$ since 3 is odd. This will dispose of nearly all remaining cases below because most vertices in Dynkin diagrams are adjacent to one with the same length and such a pair of vertices along with the edge joining them is the A$_2$ diagram. Let $n>1$ be the rank of the reduced and irreducible root system $\Phi$. We may assume that our root $a$ belongs to a chosen basis $\Delta$ of $\Phi$. We may also assume $G$ is not simply connected or else there's nothing to do, so we're not in types G$_2$ and F$_4$. We shall now show that we get SL$_2$ away from the short roots of B$_n$ and the long roots of type C$_n$ (and then we will analyze these remaining cases). Being away from those two classes of cases, note that we're not in type ${\rm{B}}_2 = {\rm{C}}_2$. Thus, $n \ge 3$ and in the Dynkin diagram the root $a \in \Delta$ is adjacent to another root $b \in \Delta$ with the same length. Thus, the codimension-2 subtorus $S_{a,b} = (\ker a \cap \ker b)^0_{\rm{red}} \subset S$ killed by $a$ and $b$ has centralizer whose derived group $G_{a,b}$ is of type A$_2$ (with split maximal $k$-torus $S' = S \cap G_{a,b} = a^{\vee}(\mathbf{G}_m)b^{\vee}(\mathbf{G}_m)$ and Dynkin diagram having nodes $a|_{S'}, b|_{S'}$). The original commutator subgroup of interest for $(G, S, a)$ agrees with the one for $(G_{a,b}, S', a|_{S'})$. This reduces our task to the settled case of type A$_2$! It remains to consider long roots of type C$_n$ with $n \ge 2$ and short roots of type B$_n$ with $n \ge 2$. We will reduce these to consideration of adjoint type ${\rm{B}}_2 = {\rm{C}}_2$ (concretely, SO$_5$). Using the action of the Weyl group, we may arrange that our root $a$ corresponds to the unique node that is long for type C and short for type B. Let $b \in \Delta$ be the unique node adjacent to $a$ in the diagram. The center of simply connected type B$_n$ is $a^{\vee}(\mu_2)$ (this is the key fact, as also noted in Paul Levy's comment and answer), so by open cell considerations we see that the group $G_{a,b}$ made as above but now for the nodes $a$ and $b$ as just defined is also of adjoint type when we are in type B. Keeping in mind what we are aiming to show (that we get SL$_2$ for the long roots of adjoint type C$_n$ with $n \ge 2$ and PGL$_2$ for the short roots of adjoint type B$_n$ with $n \ge 2$), since the simply connected case gives only SL$_2$ in general and in types B and C the only options are simply connected and adjoint type (as the fundamental group of the root system has order 2) we can pass to $G_{a,b}$ to reduce to the case of adjoint type ${\rm{B}}_2={\rm{C}}_2$. (It doesn't matter that for type C this passage to rank 2 might leak back to the simply connected case, since that case always gives SL$_2$ anyway.) Letting $\{a, b\}$ be a basis for type ${\rm{B}}_2 = {\rm{C}}_2$ with $a$ short and $b$ long, we want to show that $[G_a, G_a] = {\rm{PGL}}_2$ and $[G_b, G_b] = {\rm{SL}}_2$. The center of $\widetilde{G}$ is $a^{\vee}(\mu_2)$ as noted already (contained in the split maximal torus $a^{\vee}(\mathbf{G}_m)$ of $[\widetilde{G}_a, \widetilde{G}_a] = {\rm{SL}}_2$), so for $G = \widetilde{G}/Z_{\widetilde{G}} = \widetilde{G}/a^{\vee}(\mu_2)$ we have $[G_a,G_a] = [\widetilde{G}_a,\widetilde{G}_a]/a^{\vee}(\mu_2) = {\rm{SL}}_2/\mu_2 = {\rm{PGL}}_2$. Likewise, $[\widetilde{G}_b,\widetilde{G}_b]$ meets the rank-2 split maximal torus $\widetilde{S} = \mathbf{G}_m^{\Delta^{\vee}}$ in precisely its own split maximal torus $b^{\vee}(\mathbf{G}_m)$, so its intersection with $a^{\vee}(\mathbf{G}_m)$ is trivial. In particular, its intersection with $a^{\vee}(\mu_2)$ is trivial, so the central isogeny ${\rm{SL}}_2 = [\widetilde{G}_b, \widetilde{G}_b] \twoheadrightarrow [G_b,G_b]$ has trivial (scheme-theoretic) kernel and thus is an isomorphism.<|endoftext|> TITLE: Can a "stable" $G$-torsor have automorphisms which do not lie in the center $Z(G)$? QUESTION [5 upvotes]: Let $X$ be a connected reduced stable curve of genus $g$ over an algebraically closed field $k$ of characteristic not dividing $|G|$. Let $G$ be a finite group acting faithfully on $X$. Let $A := Aut(X\rightarrow X/G)$ be the group of automorphisms $\alpha : X\rightarrow X$ which induce the identity on $X/G$ and commutes with the action of $G$. Clearly the center $Z(G)$ can be embedded in $A$. A paper (second to last paragraph on p136, $\S$8.2) I'm reading implies that $A$ might be strictly larger than $Z(G)$. I'd like to understand how this might be possible, and if it is, if it is still possible under the assumptions: $G$ acts freely on the smooth locus of $X$. $G$ acts with determinant 1 on the cotangent spaces of the nodes if it fixes the branches, and with determinant -1 if it swaps the branches. REPLY [3 votes]: Assume $k$ has characterstic not $2$. Take two curves $Y$ and $Z$ of genus $> 1$ over $k$ each with an automorphism of order $2$ having the same number of fixed points. Say the fixed points are $y_1, \ldots, y_n$ on $Y$ and $z_1, \ldots, z_n$ on $Z$. Let $X$ be obtained by glueing $y_i$ to $z_i$ in $Y \amalg Z$ and let $G = \mathbf{Z}/2\mathbf{Z}$ with $1$ acting via our order $2$ automorphism on both $Y$ and $Z$. Then $A$ will contain at least $4$ elements. For example, you can use the order $2$ automorphism of $Y$ and the identity on $Z$ and this will commute with $G$.<|endoftext|> TITLE: Subtle point in definition of BNS invariant QUESTION [11 upvotes]: Let $G$ be a finitely generated group. Let $S(G)$ be the quotient of $\text{Hom}(G,\mathbb{R}) \setminus \{0\}$ by the equivalence relation that identifies two homomorphisms if they differ by scaling by a positive constant. For a nonzero $\phi \in \text{Hom}(G,\mathbb{R})$, write $[\phi] \in S(G)$ for the associated equivalence class. The Bieri-Neumann-Strebel (BNS) invariant of $G$ is a certain subset of $S(G)$. The definition in the original paper is hard to parse, but in several places I have seen the following definition of it: Fix a generating set $S$ of $G$. Let $\text{Cay}(G,S)$ be the Cayley graph of $G$ with respect to $S$. Consider a nonzero $\phi \in \text{Hom}(G,\mathbb{R})$. Define $X_{\phi} = \{\text{$g \in G$ $|$ $\phi(g) \geq 0$}\}$. Then $[\phi] \in S(G)$ is in the BNS invariant if and only if the full subgraph of $\text{Cay}(G,S)$ spanned by $X_{\phi}$ is connected. I have seen it asserted without proof or reference that the above property does not depend on the choice of generating set $S$. Question 1: Can someone give me either a proof or reference for this? Question 2: Can someone explain how to relate this to the definition in the original paper defining the BNS invariant? A reference here would also be fine. REPLY [2 votes]: This is a reference of question 2. Various definitions (including the original one and the one you have given) of BNS invariant are discussed in Chapter C of this lecture notes by Ralph Strebel who call the BNS invariant as "Sigma invariants".<|endoftext|> TITLE: Are the Fourier coefficients of $\eta(q^m)^m / \eta(q)$ non-negative? QUESTION [6 upvotes]: In this paper, the following result is proved. For any prime $p$, all the Fourier coefficients of $$\eta(q^p)^p / \eta(q) = q^{\frac{p^2-1}{12}} \prod_{n=1}^\infty (1 - q^{pn})^p (1 - q^{n})^{-1}$$ are non-negative. As seeing this sequence, I conjectured all the Fourier coefficients of $$\eta(q^m)^m / \eta(q) = q^{\frac{m^2-1}{12}} \prod_{n=1}^\infty (1 - q^{mn})^m (1 - q^{n})^{-1}$$ are non-negative for any positive integer $m$, . Is this conjecture true? REPLY [15 votes]: Yes this is true and these Fourier coefficients actually enumerate some combinatorial objects called $m$-cores. These are partitions with no hooklength divisible by $m$. In fact for $m\geq 4$ these coefficients are positive. This last result was proved in "Defect zero $p-$blocks for finite simple groups", by Granville and Ono. I want to also point out that Saito's conjecture on nonnegativity of particular eta quotients (which was the subject of the paper you link to) has been fully proven in "K. Saito's Conjecture for Nonnegative Eta Products and Analogous Results for Other Infinite Products" by Berkovich and Garvan.<|endoftext|> TITLE: The $\ell$- part of the class groups of the $p$-cyclotomic fields QUESTION [5 upvotes]: Let $K_n = \Bbb Q(\mu_{p^{n+1}})$ and let $A_n$ be it's class group. Iwasawa theory tells us a lot about the $p$-part of $A_n$. For instance, we know quite a lot about how it varies with $n$. I am interested in the $\ell$-part of $A_n$ where $\ell \neq p$ is a prime. What is known about these groups? For instance, do we know which primes can occur in the class numbers of these extensions? Do we know the order of growth of the prime powers that do occur? Since none of the standard sources seem to cover these questions (despite being quite natural), I suspect this is quite a difficult question. What is the main difficulty in extending the methods of classical Iwasawa theory to this case? For instance, going through the proof of the p-part of the class number in a $\Bbb Z_p$ extension (Chapter 13, Sections 1-3 of Washington's Cyclotomic fields), we have a problem while classifying modules over the corresponding Iwasawa algebra $\Bbb Z_\ell[[\Bbb Z_p]]$. If $\ell = p$, then this ring would be isomorphic to $\Bbb Z_p[[t]]$ and one can classify modules over this field (upto finite kernel and cokernel). What other problems arise like this? REPLY [2 votes]: About the question which primes can occur in the class numbers you may find some basic facts in my paper in Expo. Math. 25 (2007) 325--340.<|endoftext|> TITLE: A closed form for an integral expressed as a finite series of $\zeta(2k+1)$, $\pi^m$ and a rational? QUESTION [22 upvotes]: In this paper the following beautiful integral expression for $\zeta(3)$ is derived: $$\zeta(3)=\frac{1}{7}\,\int_0^{\pi} x\,(\pi-x)\csc(x)\, dx$$ In a comment at the end of this question, I mentioned the following tweak: $$f(n):= \int_0^{\pi} x^n\,(\pi-x)\csc(x)\,dx$$ Maple and Mathematica both return closed forms for $f(n)$ as a finite series of $\zeta(2k+1)$, weighted by $π^m$ and a rational ($n,k,m \in \mathbb{N}$). This is what I get for $n=1..12$ : I then tried to find a closed form for the above and got as far as: For $n=$ odd: $$f(n)=\scriptsize \left(\Gamma(n+1)\,\sum_{k=1}^{n}(-1)^{k+1}\,\frac{(4-2^{1-2k})\,k\,\pi^{n+1-2k}}{\Gamma(2+n-2k)}\,\zeta(2k+1)\right)-\left(\frac{i}{2}\right)^{n+1}\,(2^{n+2}-1)\,\Gamma(n+2)\,\zeta(n+2)$$ For $n=$ even: $$f(n)=\scriptsize \left(\Gamma(n+1)\,\sum_{k=1}^{n}(-1)^{k+1}\,\frac{(4-2^{1-2k})\,k\,\pi^{n+1-2k}}{\Gamma(2+n-2k)}\,\zeta(2k+1)\right)+\left(\frac{i}{2}\right)^{n}\,(2^{n+1}-1)\,\pi\,\Gamma(n+1)\,\zeta(n+1)$$ Note that only the last part differs, which encourages me to believe that a simpler formula might exist. Could this be simplified? Since the CAS-tools evaluate the integral very quickly into the series, I am also keen to better understand which mathematical steps they take. Thanks. REPLY [11 votes]: I think the following articles can give a clue: http://www.tandfonline.com/doi/abs/10.1080/10652460701688125?journalCode=gitr20 (Closed-form evaluation of some families of cotangent and cosecant integrals, by D. Cvijović) and http://www.sciencedirect.com/science/article/pii/S0377042702003588 (Integral representations of the Riemann zeta function for odd-integer arguments, by D. Cvijović and J. Klinowski). Update. In particular, using $2x^n=E_n(x)+E_n(1+x)$ and $E_n(1+x)=\sum\limits_{k=0}^n\binom{n}{k}E_k(x)$, we get $$f(n)=\frac{\pi^{n+2}}{2}\left[I_n+\sum\limits_{k=0}^n\binom{n}{k}I_k-I_{n+1}-\sum\limits_{k=0}^{n+1}\binom{n+1}{k}I_k\right]=\frac{\pi^{n+2}}{2}\left[I_n-2I_{n+1}-\sum\limits_{k=1}^n\binom{n}{k-1}I_k\right], \tag{1}$$ where $$I_n=\int\limits_0^1E_n(x)\csc{(\pi x)}\,dx.$$ Odd indices in (1) doesn't contribute because $E_n(1-x)=(-1)^nE_n(x)$ leads to the integrand which is antisymmetric with regard to $x\to 1-x$. For even indices, $I_n$ was calculated in the second paper indicated above as $$I_{2n}=(-1)^n\frac{(4-2^{1-2n})(2n)!}{\pi^{2n+1}}\zeta(2n+1),$$ and we obtain, for even $n=2m$: $$f(2m)=(-1)^m\frac{\pi}{2}(4-2^{1-2m})(2m)!\,\zeta(2m+1)- \sum\limits_{p=1}^m(-1)^p\;\frac{\pi^{2(m-p)+1}}{2}\;\binom{2m}{2p-1}(4-2^{1-2p})(2p)!\,\zeta(2p+1),$$ and for odd $n=2m-1, m>1$: $$f(2m-1)=(-1)^{m+1}(4-2^{1-2m})(2m)!\,\zeta(2m+1)-\sum\limits_{p=1}^{m-1}(-1)^p\;\frac{\pi^{2(m-p)}}{2}\;\binom{2m-1}{2p-1}(4-2^{1-2p})(2p)!\,\zeta(2p+1).$$<|endoftext|> TITLE: Adem relations of Steenrod square without modding out the coboundaries QUESTION [8 upvotes]: In the paper Products of Cocycles and Extensions of Mappings, Steenrod introduced the cup-$i$ product and Steenrod square $Sq^k$: $$ Sq^k(x_n) \equiv x_n \smile_{n-k} x_n,\ \ \ x_n \in C^n(M^d;\mathbb{Z}_2) $$ Note that the above definition works for any $\mathbb{Z}_2$-valued cochain $x_n$. My question is that if the Adem relations $$ Sq^i Sq^j = \sum_{k=0}^{[i/2]} {j-k-1 \choose i-2k} Sq^{i+j-k} Sq^k $$ are still valid when acting on cochains (or cocycles) without modding out the coboundaries? (All the references that I can find state Adem relations up to the coboundaries.) REPLY [5 votes]: Theorem 3.1 in Peter May's "A General Algebraic Approach to Steenrod Operations" gives a better behaved definition for $Sq^k(x)$ for (co)chains $x$ which are not assumed to be cocycles. It probably makes sense to think about these questions (Adem relations or Cartan formula on the cocycle level) using them rather than just the cup-(n-k) formula. I find it highly implausible that either of these (Adem relations or Cartan formula) should hold 'on the nose', no matter what definition extending the usual one on (co)cycles one gives.<|endoftext|> TITLE: Is the triple product in a Freudenthal triple system fully symmetric? QUESTION [6 upvotes]: I'm trying to learn about Freudenthal triple systems. Here is the definition given by Helenius [1], start of Section 5: A Freudenthal triple system is a finite-dimensional vector space $V$ over a field $F$ (with characteristic not 2 or 3) such that • There is a nonzero quartic form $q$ defined on $V$. A corresponding 4- linear form, also called $q$, is given by linearization, with $q(x, x, x, x) = q(x)$ for all $x \in V$. • There is a nondegenerate skew-symmetric bilinear form $\langle , \rangle$ defined on $V$. Thus for given $x, y, z \in V$ we may define the triple product $xyz$ to be the unique vector in $V$ such that $q(w, x, y, z) = \langle w, xyz \rangle$ for all $w \in V$. • The triple product satisfies the following identity: $$2(xxx)xy = \langle y, x \rangle xxx + \langle y, xxx \rangle x.$$ So far so good. Helenius then remarks that other sources give slightly different definitions, that are equivalent to his definition by rescaling the bilinear and/or the quartic form by a scalar. In particular, Helenius notes, the definition of Ferrar in [2] omits the 2 in the last equality. I looked up [2] on JSTOR and indeed, the definition (beginning of section 1) is the same as in Helenius. EXCEPT that Ferrar list the additional axiom (A1) that "xyz is symmetric in all arguments". This sounds like a big deal. I find it very weird that Helenius, who explicitly states that his definition is equivalent of that of Ferrar doesn't mention this property at all. Moreover, I am a bit distrustful of the symmetry claim. If the definitions are indeed equivalent then the symmetry should follow from the three bullet points in the definition above but I can't see how. Also, there is a third article, by Faulkner [3], in which he considers a similar situation (a vector space $V$ with a triple product, a 4-linear form $q$ and a skew symmetric bilinear form $\langle, \rangle$ satisfying $q(w, x, y, z) = \langle w, xyz \rangle$) satisfying some different axioms that make it even more explicit that the triple product in general is not totally symmetric. Now Faulkner's construction need not be equivalent to the other two, but I am pretty convinced that in one special case: Faulkner's 'Example 2' starting at page 399, Ferrar's 'Prototype FTS' starting at page 314 and Helenius' '$G = E_8$'-case discussed in detail in chapter 10, they are all really describing the same thing: the case where the vector space $V$ is 56-dimensional and the subalgebra of $\mathfrak{gl}(V)$ preserving $q$ and $\langle, \rangle$ being a Lie algebra of type $E_7$. So at least in this case the symmetry has to come from somewhere. However, I have a hard time explicitly localizing it in [1] (Proposition 5 comes close, though.) In summary: is Ferrar's claim that $xyz$ is totally symmetric correct and if so, how do we see it from the definitions given by Helenius and Faulkner? [1] Fred Helenius, 2010: Freudenthal Triple Systems from Root System Data: https://arxiv.org/pdf/1005.1275.pdf [2] J. C. Ferrar, 1972: Strictly Regular Elements in Freudenthal Triple Systems: http://www.jstor.org/stable/1996111?seq=1#page_scan_tab_contents [3] John R. Faulkner, 1971: Construction of Lie algebras from a Class of Ternary Algebras: http://www.ams.org/journals/tran/1971-155-02/S0002-9947-1971-0294424-X/S0002-9947-1971-0294424-X.pdf REPLY [4 votes]: I figured out the precise relationship between the 'Class of Ternary Algebras' of Faulkner and the 'Freudenthal Triple Systems' of Ferrar/Helenius and I will write it down here for the benefit of future readers (mostly my future self). Following Helenius (as in the OP) I write $\langle . , . \rangle$ for the skew-symmetric biliear form. Following John Baez above I write $q$ for the quartic form and $Q$ for the 4-linear form. Following Faulkner I write $\langle . , . , . \rangle$ for the triple product. In all cases I use the subscript $f$ for Faulkners definitions and $h$ for Ferrar's definition (as I mostly learned his definition from Helenius). With these notational conventions we have: $$\langle x, y \rangle_f = \langle y, x \rangle_h = - \langle x, y \rangle_h$$ $$Q_h(x_1, x_2, x_3, x_4) = \frac{1}{4!} \sum_{\pi \in S_4} Q_f(x_{\pi(1)}, x_{\pi(2)}, x_{\pi(3)}, x_{\pi(4)})$$ from which it follows that $$q_h(x) = Q_h(x, x, x, x) = Q_f(x, x, x, x) = q_f(x).$$ The relation between the triple products can be deduced from $$Q_f(x, y, z, w) = \langle \langle x, y ,z \rangle_f, w \rangle_f$$ $$Q_h(x, y, z, w) = \langle w, \langle x, y ,z \rangle_h \rangle_h = \langle \langle x, y ,z \rangle_h, w \rangle_f$$ This is not perfect. It allows us to deduce $Q_h$ from $Q_f$ but not the other way around and obtaining an explicit description of $\langle . , . , . \rangle_h$ in terms of $\langle . , . , . \rangle_f$ is messy. However, the article of Helenius provides a great way of clearing this up. Faulkner starts with his version of the triple system and uses it to create a Lie algebra out of it; similar constructions of (the same) Lie algebra from Ferrar's version of the triple system have been given elsewhere. Helenius turns this viewpoint upside down: he starts with the Lie algebra ($\mathfrak{g}$ in his notation) identifies a special subspace $\mathfrak{g}_1$ on which he can define $\langle . , . \rangle_h, \langle . , . , . \rangle_h$ and $Q_h$ in terms of the Lie algebra structure and then shows they satisfy the Freudenthal triple system axioms. I will give here the analogous presentation of $\langle . , . \rangle_f, \langle . , . , . \rangle_f$ and $Q_f$. We may assume that the Lie-algebra $\mathfrak{g}$ is simple, apparently this is equivalent to the non-degeneracy of $\langle . , . \rangle$. Helenius starts with picking a long root $\rho$ and defines the space $\mathfrak{g}_1$ as the sum of all the root spaces $\mathfrak{g}_\alpha$ for roots $\alpha$ satisfying $2 \frac{(\alpha, \rho)}{(\rho, \rho)} = 1$. (I try to avoid introducing another use of $\langle . , . \rangle$ here.) More generally he defines a grading of $\mathfrak{g}$ where $\mathfrak{g}_k$ is the span of the rootspaces $\mathfrak{g}_\beta$ where $2 \frac{(\beta, \rho)}{(\rho, \rho)} = k$. We fix a generator $x_{\rho}$ of $\mathfrak{g}_\rho$ and define $x_{-\rho} \in \mathfrak{g}_{-\rho}$ in such a way that $\{x_\rho, x_{-\rho}, [x_\rho, x_{-\rho}]\}$ is a standard $\mathfrak{sl}_2$-triple. Now we have for all $x, y, z, w \in \mathfrak{g}_1$ that $$[x, y] = \langle y, x \rangle_f x_\rho$$ $$\langle x, y, z \rangle_f = [x, [y, [z, x_{-\rho}]]] = (ad(x)\circ ad(y) \circ ad(z)) (x_{-\rho})$$ $$Q_f(x, y, z, w) = [w, [x, [y, [z, x_{-\rho}]]]] = (ad(w) \circ ad(x)\circ ad(y) \circ ad(z)) (x_{-\rho})$$ Comparing to the equalities $$[x, y] = \langle x, y \rangle_h x_\rho$$ $$Q_h(x_1, x_2, x_3, x_4) = \frac{1}{4!} \sum_{\pi \in S_4} (ad(x_{\pi(1)}) \circ ad(x_{\pi(2)}) \circ ad(x_{\pi(3)}) \circ ad(x_{\pi(4)})) (x_{-\rho})$$ from Helenius, chapter 3, yields the above relations. The equation $\langle x, y, z \rangle_f = [x, [y, [z, x_{-\rho}]]]$ is given explicitely in Faulkner as part of the proof of Lemma 3. Verifying that the above descriptions of $\langle . , . \rangle_f, \langle . , . , . \rangle_f$ and $Q_f$ when taken as the definitions satisfy Faulkner's axioms $T1 - T4$ is mostly a long list of application of the Jacobi-idenity together with an occasional use of the fact that the space $\mathfrak{g}_k$ is zero for $k < - 2$ and $k > 2$.<|endoftext|> TITLE: Torsion points on twists of elliptic curves and products of fine modular curves over $\mathcal{M}_{1,1}$ vs over the $j$-line QUESTION [6 upvotes]: Let $Y_1(n)$ (for $n\ge 4$) be the fine moduli scheme over $\mathbb{Q}$ parametrizing elliptic curves with a rational point of order $n$. Let $\mathbb{A}^1_j$ be the $j$-line over $\mathbb{Q}$, the coarse moduli scheme of the moduli stack of elliptic curves $\mathcal{M}_{1,1}$ (over $\mathbb{Q}$). Let $\mathcal{M}_{1,1}^\circ$ be the open substack classifying elliptic curves with $j$-invariant $\notin\{0,1728\}$. Let $Y_1(n)^\circ$ be the open subscheme obtained by removing the fibers above $j = 0,1728$, and let $\mathbb{A}_j^{1\circ}$ be the $j$-line minus $0,1728$. One may consider the fiber products $$A := Y_1(5)^\circ\times_{\mathcal{M}_{1,1}^\circ}Y_1(7)^\circ,\qquad B := Y_1(5)^\circ\times_{\mathbb{A}_j^{1\circ}}Y_1(7)^\circ$$ Both $A,B$ are schemes. $Y_1(7)^\circ$ is finite etale over $\mathcal{M}_{1,1}^\circ$ of degree $[SL_2(\mathbb{Z}):\Gamma_1(7)] = 48$, in the sense that the fiber category above a geometric point $\text{Spec }k\rightarrow\mathcal{M}_{1,1}^\circ$ is a groupoid with 48 objects, but with isomorphisms between them (48 should be the degree according to the definition of the Galois category of an algebraic stack). Thus, since we're staying away from $j = 0,1728$, the resulting scheme theoretic fiber should have degree 24 over $\text{Spec }k$. For the same reason, $A$ should be finite etale over $Y_1(5)^\circ$ of degree 24. Similarly, $Y_1(7)^\circ$ is finite etale over $\mathbb{A}_j^{1\circ}$, and so the pullback $B$ is finite etale over $Y_1(5)^\circ$, also of degree 24. The universal property of fiber products provides a map $A\rightarrow B$ (commuting with the finite etale maps to $Y_1(5)^\circ$). This map must then also be finite etale, and by comparing degrees, must be an isomorphism. Is this correct? (That $A\cong B$?) In particular, this would seem to say that if we have an elliptic curve $E_1/K$ with a point $P_1$ of order 5, and an elliptic curve $E_2/K$ with a point $P_2$ of order 7, such that $j(E_1) = j(E_2) \ne 0,1728$, then this pair $((E_1,P_1),(E_2,P_2))$ would correspond to a $K$-point of $B$, and hence a $K$-point of $A$, but by definition of the fiber product of stacks, giving such a $K$-point of $A$ would imply that in fact $E_1\cong E_2$ over $K$ - which says more than just $j(E_1) = j(E_2)$ - ie, $E_1$ is not a twist of $E_2$. If so this seems quite striking. In particular, this would seem to imply that if an elliptic curve $E/K$ with automorphism group $\{\text{id},[-1]\}$ has a $K$-point of order $n\ge 4$ (so that $Y_1(n)$ is a fine moduli scheme), then none of the twists of $E/K$ can have $K$-points of order $\ge 4$. Ie, amongst the set of all twists of $E$, at most one can admit a $K$-rational point of order $\ge 4$. REPLY [3 votes]: I think your degree calculation is bogus. You write: $Y_1(7)^\circ$ is finite etale over $\mathcal{M}_{1,1}^\circ$ of degree $[SL_2(\mathbb{Z}):\Gamma_1(7)] = 48$, in the sense that the fiber category above a geometric point $\text{Spec }k\rightarrow\mathcal{M}_{1,1}^\circ$ is a groupoid with 48 objects, but with isomorphisms between them [...] The resulting scheme theoretic fiber should have degree 24 over $\text{Spec }k$. But the map $Y_1(7)^\circ \to \mathcal{M}_{1,1}^\circ$ is representable, and this means in particular that the fiber category over $\text{Spec }k$ (or over any scheme for that matter) is an algebraic space, not a stack. So the only isomorphisms in your 48-object category are identities, and the "scheme theoretic fiber" should have degree 48 over $\text{Spec }k$.<|endoftext|> TITLE: Determinants (and traces) of linear maps of matrices QUESTION [8 upvotes]: Let $k$ be a field or a commutative ring with unit and let $F:M_n(k)\to M_n(k)$ be a $k$-linear map. Suppose that $F$ is given in the form $F(X) = A_1XB_1 + \cdots + A_m X B_m$ for some $A_i,B_i\in M_n(k)$ (note that any $k$-linear $F$ can be written in this form, though not uniquely). Is there a formula that allows one to determine $\mathrm{det}(F)$ (no pun intended) or $\mathrm{tr}(F)$ directly from the matrices $A_i, B_i$ ($1\leq i\leq m$), i.e. without having to compute a representation matrix of $F$? REPLY [5 votes]: We have $F(X)=\sum_i A_i X B_i = \sum_i (B_i^T \otimes A_i) vec(X)$ (see here), i.e. $F \sim \sum_i (B_i^T \otimes A_i)$. Because of some formulas here we have $tr(F)=\sum_i tr(A_i)tr(B_i)$. For the determinant I don't think there is a nice formula.<|endoftext|> TITLE: A question about the quivers with potentials QUESTION [8 upvotes]: Let $Q=(Q_0,Q_1,h,t)$ be a quiver consisted of a pair of finite sets $Q_0$(vectors),and $Q_1$ (arrows) supplied with two maps $h : Q_1 → Q_0$ (head) and $t : Q_1 → Q_0$ (tail ). This definition allows the underlying graph to have multiple edges and (multiple) loops. Then the path algebra of $Q$ is the graded tensor algebra $R\langle Q \rangle=\bigoplus_{d=0}^{\infty}A^d$, where $R=k^{Q_0},A=k^{Q_1}$, $k$ is a field of characteristic zero. Any path algebra has a maximal ideal generated by elements of positive degree, we define the completed path algebra $R\langle\langle Q \rangle\rangle=\prod_{d=0}^{\infty}A^d$ as the completion of $R\langle Q \rangle$ with respect to powers of the maximal ideal. Let $R\langle\langle Q \rangle\rangle_{cyc}$ be the linear subspace of the completed path algebra generated by cyclic paths, i.e. products of the form $\prod_{i=1}^{n}a_i$ such that $t(a_1)=s(a_n)$. A potential on $Q$ is an element of $R\langle\langle Q \rangle\rangle_{cyc}$ considered up to cyclic shift: $a_1a_2...a_n\leftrightarrow a_na_1...a_{n-1}$. A quiver with potential is a pair $(Q,W)$ where $Q$ is a quiver and $W$ is an element of $R\langle\langle Q \rangle\rangle_{cyc}$ considered up to cyclic shifts. You can see https://arxiv.org/pdf/0704.0649.pdf and https://arxiv.org/pdf/1701.00672.pdf for more details. I want to know whether the definition of potentials for quivers with weighted arrows or weighted vertices coincide with the above definition? For example, Let $B = (b_{ij})$ is a sign-skew-symmetric real square matrix. $B$ is 2-finite if it has integer entries, and any matrix $B'=\mu_{k}(B)=(b_{ij}^{'})$ mutation equivalent to $B$ is sign-skew-symmetric and satisfies $\mid b_{ij}^{'}b_{ji}^{'}\mid \leq 3$ for all $i$ and $j$. In the paper https://arxiv.org/pdf/math/0208229.pdf. On page 26, Definition 7.3. S. Fomin and A. Zelevinsky define a 2-finite diagrams. The diagram of a sign-skew-symmetric matrix $B = (b_{ij}),i,j\in I$ is the weighted directed graph $\Gamma (B)$ with the vertex set $I$ such that there is a directed edge from $i$ to $j$ if and only if $b_{ij} > 0$, and this edge is assigned the weight $|b_{ij}b_{ji}|$. On page 28, Proposition 8.1, gives a mutation rule for diagrams. Whether the definition of potentials for these diagrams with weighted edges have to be done? Any help will be appreciated. REPLY [6 votes]: Let $B$ be an $n\times n$ skew-symmetrizable matrix with integer coefficients. The first question seems to be "how to 'realize' $B$ through a 'path algebra'?". There are at least three different approaches to this question (Demonet,Dlab-Ringel,Geiss-Leclerc-Schröer). The three approaches seem to fit in the following general idea: A 'modulation' or 'species realization' of $B$ is a pair $(\mathbf{F},\mathbf{A})$ consisting of an $n$-tuple $\mathbf{F}=(F_k)_{1\leq k\leq n}$ of rings and a tuple $\mathbf{A}=(A_{jk})_{b_{jk}\geq 0}$ of bimodules (A_{jk} being an $F_j$-$F_j$-bimodule) satisfying certain properties (one of which is that each $A_{jk}$ is free when separately considered as left and as right module, with its right and left dimensions being prescribed by the entries of the matrix; I can list the conditions explicitly if you want, but at the moment I'd rather keep it short). With the pair $(\mathbf{F},\mathbf{A})$ at hand, you can define a ring $R=\times_{1\leq k\leq n}F_k$ and an $R$-$R$-bimodule $A=\oplus_{b_{jk}\geq 0}A_{jk}$. The tensor algebra $T_R(A)$ can be thought of as a 'path algebra' associated to $B$ (when $B$ is skew-symmetric and $Q$ is the corresponding quiver, $T_R(A)$ is the usual path algebra of $Q$ if you take $F_k$ to be $\mathbb{C}$ for all $k\in\{1,\ldots,n\}$ and $A_{jk}$ to be $\mathbb{C}$-vector space with basis the set of all arrows that go from $k$ to $j$). Then you can define a 'potential' to be an element of $T_R(A)/[T_R(A),T_R(A)]$. Depending on which $\mathbf{F}$ you take, a tuple $\mathbf{A}$ of bimodules giving a species realization of $B$ may or may not exist. Wether one can define cyclic derivatives (and hence Jacobian algebras), perform mutations in this setting, or show that non-degenerate potentials exist, is a different matter, and things can become rather technical. To finish, the difference between the approaches of Demonet, Dlab-Ringel and Geiss-Leclerc-Schröer lies mainly in which rings $F_k$ they take: Demonet takes each $F_k$ to be a group algebra of certain cyclic group; Dlab-Ringel take $F_k$ to be a division ring; Geiss-Leclerc-Schröer take $F_k$ to be a truncated polynomial ring $\mathbb{C}[X]/X^{d_k}$. Edit (this is the first time I post in MO, I hope to be doing it the right way...): Just as a follow-up, it is probably a good idea to say a few words on how you can actually construct an explicit 'modulation' or 'species realization' for a given skew-symmetrizable matrix $B\in\mathbb{Z}^{n\times n}$. Fix a matrix $D=\operatorname{diag}(d_1,\ldots,d_n)$, with $d_1,\ldots,d_n\in\mathbb{Z}_{>0}$, such that $DB$ is skew-symmetric. The matrix $C=(c_{ij})_{1\leq i,j\leq n}$ defined by $$ c_{ij}=\frac{b_{ij}\operatorname{gcd}(d_i,d_j)}{d_j} $$ has integer entries and is skew-symmetric. As such, it has an associated quiver $Q_C$ with $c_{ij}$ arrows from $j$ to $i$ whenever $c_{ij}\geq 0$. Fix a degree-$d$ cyclic Galois extension $E/F$, where $d=\operatorname{lcm}(d_1,\ldots,d_n)$. For each $k\in\{1,\ldots,n\}$, set $F_k$ to be the unique subfield of $E$ such that $[F_k:F]=d_k$, and for every pair $(j,k)$ such that $b_{jk}\geq 0$, set $A_{jk}:=\oplus_{\ell=1}^{c_{jk}}F_j\otimes_{F_j\cap F_k}F_k$. The pair $((F_k)_{1\leq k\leq n},(A_{jk})_{b_{jk}\geq 0})$ then constitutes a species realization of $B$. If you are interested in mutating the above species realization (with some potential), then you are forced to consider some 'twisted' version of the bimodules $F_j\otimes_{F_j\cap F_k}F_k$.<|endoftext|> TITLE: Possible behaviors of integer sequences that arise from powering nonnegative integer matrices QUESTION [22 upvotes]: Let's call a sequence of nonnegative integers $x_1,x_2,\ldots$ matrix-realizable, if there exists a $k\times k$ nonnegative integer matrix $A$ (for some $k$), as well as nonnegative integer vectors $u,v$, such that $x_n = u^T A^n v$ for all $n$. (So in particular, all matrix-realizable sequences are linearly recurrent sequences; because of the nonnegative integer constraints, I don't know whether the converse holds.) I'm interested in the possible patterns of decreases in matrix-realizable sequences. For example: Is there any matrix-realizable sequence $(x_n)$ such that $x_{2n} > x_{3n+1}$ for all $n$? Is there any matrix-realizable sequence $(x_n)$ such that $x_n > x_{n+1}$ whenever $n$ is composite? In case you care about the motivation: these questions arose from a project I've been working on with Marijn Heule and Luke Schaeffer about computer-generated proofs. If the answers to the questions are "no," then it follows that there can be no computer-generated proofs of a very particular format (which we call "unary matrix proofs") for various interesting statements. In the above two examples, those statements are the Collatz Conjecture and the infinitude of primes, respectively. REPLY [16 votes]: The answer is "no" for both questions. The rational functions $a_0+a_1x+a_2x^2+\cdots$ with non-negative integer entries which can be obtained by $a_n=u^{T}A^nv$ for some nonnegative vectors $u,v$ and matrix $A$ are called $\mathbb N$-rational in the literature. Here are some slides by I. Gessel that provide a quick introduction. For a deeper treatment there is the book "Noncommutative Rational Series with Applications" by Jean Berstel and Christophe Reutenauer, especially the first 8 chapters (the rest of the book deals with noncommutative generalizations). First, as it was mentioned by others, not only is there no $\mathbb N$ rational series satisfying the second bullet point, there is in fact no such linear recurrence relation. If we take the power series $$h(t)=(t-1)(x_0+x_1t+x_2t^2+\cdots)$$ we get a rational function with positive coefficients at every composite integer. The condition that the $x_i$ are nonnegative implies that $h$ must have infinitely many negative coefficients at a sparse set. However a theorem of Bell and Gerhold says that this is impossible (reference): the set of indices where the coefficients are negative must have positive density! For the first bullet point I'm not sure what happens for general linear sequences, but we can restrict to $\mathbb N$-rational since these are a bit more special, and are the ones relevant to your question. A theorem of Berstel (see page 318 here, or chapter 8 of the book above), says that for such sequences there exist some $m,p\in \mathbb N$ for which the subsequences $\{x_{m+pk+i}\}_{k=1}^{\infty}$, for $i\in\{0,2,\dots,p-1\}$ are all of the form $P_i(k)\alpha_i^k +o(P_i(k)\alpha_i^k)$. Therefore each such subsequence has terms that are constant or strictly increasing. However if we take $n=qp-1$ for large enough $q$ we have $(3n+1)-(2n)=qp$, so $x_{2n},x_{3n+1}$ are in the same subsequence and therefore we can't have $x_{2n}>x_{3n+1}$.<|endoftext|> TITLE: Discriminant of a composition of binary forms QUESTION [5 upvotes]: Let $F(x,y), A(x,y), B(x,y)$ be homogeneous polynomials with integer coefficients (a.k.a. binary forms) such that the degrees of $A(x,y)$ and $B(x,y)$ match. Define $$R(x,y) := F\left(A(x,y), B(x,y)\right).$$ Then $R(x,y)$ is a homogeneous polynomial with integer coefficients of degree $dr$, where $d := \deg(F)$ and $r := \deg(A) = \deg(B)$. We identify the discriminant $D(R)$ of $R(x,y)$ with the discriminant of the polynomial $R(x,1)$. Similarly, we define the discriminants of $F, A, B$. I would like to obtain the formula for $D(R)$. Alternatively, I need to bound $D(R)$ from below non-trivially in terms of the invariants of $F, A, B$, such as their discriminants. Has this question been explored at all? It seems to me that John Cullinan in his short article The discriminant of a composition of two polynomials considers a somewhat similar problem, though restricting himself to the composition of univariate polynomials, not homogeneous polynomials in two variables. REPLY [4 votes]: When $R = h(f,g)$ with $h,f,g$ quadratic forms, with coefficients $h_i, f_i, g_i$ respectively for $i = 0,1,2$, the $K$ polynomial is given explicitly as $$\displaystyle K = g_1^4h_0^2 - 8 g_0 g_1^2 g_2 h_0^2 + 16 g_0^2 g_2^2 h_0^2 - 4 f_2 g_0 g_1^2 h_0 h_1 + 2 f_1 g_1^3 h_0 h_1 + 16 f_2 g_0^2 g_2 h_0 h_1 - 8 f_1 g_0 g_1 g_2 h_0 h_1 - 4 f_0 g_1^2 g_2 h_0 h_1 + 16 f_0 g_0 g_2^2 h_0 h_1 + f_1^2 g_1^2 h_1^2 - 4 f_0 f_2 g_1^2 h_1^2 - 4 f_1^2 g_0 g_2 h_1^2 + 16 f_0 f_2 g_0 g_2 h_1^2 + 16 f_2^2 g_0^2 h_0 h_2 - 16 f_1 f_2 g_0 g_1 h_0 h_2 + 2 f_1^2 g_1^2 h_0 h_2 + 8 f_0 f_2 g_1^2 h_0 h_2 + 8 f_1^2 g_0 g_2 h_0 h_2 - 16 f_0 f_1 g_1 g_2 h_0 h_2 + 16 f_0^2 g_2^2 h_0 h_2 - 4 f_1^2 f_2 g_0 h_1 h_2 + 16 f_0 f_2^2 g_0 h_1 h_2 + 2 f_1^3 g_1 h_1 h_2 - 8 f_0 f_1 f_2 g_1 h_1 h_2 - 4 f_0 f_1^2 g_2 h_1 h_2 + 16 f_0^2 f_2 g_2 h_1 h_2 + f_1^4 h_2^2 - 8 f_0 f_1^2 f_2 h_2^2 + 16 f_0^2 f_2^2 h_2^2.$$ Not only is this homogeneous in the variables here, it is bi-homogeneous in $(h_2,h_1,h_0)$ and $(f_2, f_1, f_0, g_2, g_1, g_0)$. With $h_2, h_1, h_0$ fixed, $K$ is the product of two quadratic forms in $f_2, f_1, f_0, g_2, g_1, g_0$. I suspect in general it might be very hard to study the nature of $K$.<|endoftext|> TITLE: categorification of q-series QUESTION [9 upvotes]: In his talk, S. Gukov asked two questions: What is the categorification of a $q$-serie ? How to associate to a 3-manifold a $q$-serie ? As far as I understand, he was looking for a bigarded cohomology theory $\mathcal{H}^{a,b}$ associated to 3-manifold $X$ which should be given by $$\sum_{a,b \in \mathbb{Z}}(-1)^{a} q^{b}dim \mathcal{H}^{a,b}(X) $$ Question Could someone explain the real motivation of S. Gukov ? Is it realistic to ask such questions? Is there any progress on that direction ? REPLY [4 votes]: As far as I know, question 1 is open. Let me say a few words about the first part of question 2. One way to associate a q-series to a 3-manifold is via the 3-d index developed by Dimofte, Gaiotto and Gukov: Dimofte, Tudor; Gaiotto, Davide; Gukov, Sergei, 3-manifolds and 3d indices, Adv. Theor. Math. Phys. 17, No. 5, 975-1076 (2013). ZBL1297.81149. Dimofte, Tudor; Gaiotto, Davide; Gukov, Sergei, Gauge theories labelled by three-manifolds, Commun. Math. Phys. 325, No. 2, 367-419 (2014). ZBL1292.57012. To concretely associate this q-series to an invariant of a 3-manifold, Garoufalidis, Hodgson, Rubinstein, and Segerman were able to show that for any cusped orientable hyperbolic 3-manifold there is a standard 3-d index which can be computed from its canonical cell decomposition. The proof of their main theorem is worth flushing out here. They show that for any triangulation of cusped 3-manifold which is 1-efficient (a property involving normal surfaces) the 3-d index makes sense to compute. Furthermore, if two 1-efficient triangulations are connected by a single 2-3 or 3-2 move (plus 0-2 and 2-0 moves) the 3-d index is the same for both triangulations and if a manifold does not admit a canonical triangulation then all simple refinements of its canonical cell decomposition are 1-efficient and related by simple moves as above. Finally, their paper also includes some computations and shows that the 3-d index of a triangulation converges as a formal q-series (that is it can be written down) if and only if the triangulation is 1-efficient. Garoufalidis, Stavros; Hodgson, Craig D.; Rubinstein, J.Hyam; Segerman, Henry, 1-efficient triangulations and the index of a cusped hyperbolic 3-manifold, Geom. Topol. 19, No. 5, 2619-2689 (2015). ZBL1330.57029. Garoufalidis, Hodgson, Rubinstein, and I did a follow up to this paper, which tries to better understand the connection between the 1-efficiency and convergence of the 3-d index: https://arxiv.org/pdf/1604.02688.pdf An interesting question along these lines, is what is the 3-d index counting? There is some hope that trying to 'categorify' this invariant might shed light on that question as well. In my opinion, this is a relevant interesting question which has already motivated novel approaches to better understanding 3-manifolds.<|endoftext|> TITLE: Do bubbles between plates approximate Voronoi diagrams? QUESTION [33 upvotes]: For example, soap bubbles:                   Image from UPenn: "A 2-dimensional foam of wet soap bubbles squashed between glass plates, after 10 hours of coarsening by the diffusion of gas from smaller to larger bubbles." I suspect the answer is No, even though, for example, the average number of sides of a cell is $6$ just as it is for Voronoi diagrams (which is just a consequence of Euler's formula). Is there a characterization of the graphs (or dual graphs) realized by such planar soap bubbles? REPLY [16 votes]: There are two nice connections to generalizations of Voronoi diagrams that I'm aware of. Moukarzel showed that 2D soap bubbles are sectional multiplicative Voronoi partition (SMVP), i.e. 2D slices (sections) of a 3D generalized (multiplicative) Voronoi diagram where each source $f_i$ has a weight $a_i$ so that a cell $\Omega_i$ in the diagram consists of all points $x$ such that $d(x,i)/a_i TITLE: Codes with a twisted cyclic action QUESTION [7 upvotes]: Are there interesting examples of linear binary codes that are closed under the operation of removing the first bit of the codeword and appending the complement of that bit to the end of the codeword? REPLY [7 votes]: This is an extended comment that may be too long for the comment section. If I understand your question correctly, the linear codes you described are automatically trivial. Define $\pi_{c}$ to be the map that takes any $n$-dimensional vector $\boldsymbol{v} = (v_0, v_1, \dots, v_{n-1}) \in \mathbb{F}_2^n$ over the finite field $\mathbb{F}_2$ of order $2$ to its complement shift $\pi_c(\boldsymbol{v}) = (v_{n-1} + 1, v_0, v_1, \dots, v_{n-2})$. If I understand your question right, the codes you are thinking of are those that are closed under $\pi_c$ (or the equivalent thing under the shift in the opposite direction). To see why they are all trivial, let $\mathcal{B} \subseteq \mathbb{F}_2^n$ be a binary linear code of length $n$ closed under $\pi_c$. Because $\mathcal{B}$ is a linear space, it contains the zero vector $\boldsymbol{0} = (0,\dots,0)$, which implies that $\pi_c(\boldsymbol{0}) = (1,0,0,\dots,0) \in \mathcal{B}$. Similarly, because $\mathcal{B}$ is a linear space closed under $\pi_c$, we have \begin{align*}\pi_c(\pi_c(\boldsymbol{0})) + \pi_c(\boldsymbol{0}) &= (1,1,0,\dots,0) + (1,0,0,\dots,0)\\ &= (0,1,0,0,\dots,0)\\ &\in \mathcal{B}.\end{align*} By the same token, applying $\pi_c$ and taking the sum with $\pi_c(\boldsymbol{0})$ recursively shows that $\mathcal{B}$ contains all binary vectors of weight $1$, which form the basis of $\mathbb{F}_2^n$. I am not sure if this is of any help, but there are nontirival, nonbinary codes that are closed under a map that is very similar to what you described. Let $n$ be a positive integer and $\lambda$ a nonzero element of the finite field $\mathbb{F}_q$ of order $q$. A linear code $\mathcal{C} \in \mathbb{F}_q^n$ of length $n$ over $\mathbb{F}_q$ is $\lambda$-constacyclic if for any codeword $\boldsymbol{c} = (c_0, c_1, \dots, c_{n-1}) \in \mathcal{C}$, its $\lambda$-cyclic shift $(\lambda c_{n-1}, c_0, c_1, \dots, c_{n-2})$ is also a codeword. When $\lambda = -1$, the codes are called negacyclic. Just like the standard cyclic codes, it is known that the constacyclic codes can be identified by certain ideals of polynomial rings. There are many papers on constacyclic codes and the like (e.g., https://arxiv.org/pdf/1301.0369.pdf), so if this kind of code is of interest to you, looking up constacyclic or negacyclic codes in the literature may help.<|endoftext|> TITLE: Relationship between the Radon transform and Twistor spaces QUESTION [12 upvotes]: I have often heard that the theory of Twistor spaces is ``a complex analogue" of the Radon transform. What is the precise connection ? REPLY [5 votes]: The correspondence between the Radon transform (from a space of real-valued functions on $\mathbb{R}^2$ to the space of functions on the manifold of straight lines in $\mathbb{R}^2$) and the Penrose transform (from $\mathbb{CP}_2$ to the dual projective space $\mathbb{CP}^\ast_2$) is easiest to work out when lines in the Radon transform are replaced by great circles on a sphere. This special spherical case of the Radon transform is called the Funk transform, and it corresponds to the Penrose transform when $\mathbb{CP}_2$ is restricted to $\mathbb{RP}_2$, see The Funk transform as a Penrose transform (1999). REPLY [4 votes]: To understand the relationship between the Penrose and Radon transforms, it's hard to do better than the outline given by Atiyah in [1]. See chapter VI, section 5 (pages 78--81). (It even looks like there's a PDF available online if you search.) [1] Atiyah, M. "Geometry of Yang-Mills Fields", Annali della Scuola Normale Superiore di Pisa (1979).<|endoftext|> TITLE: $C^k$ version of Hadamard's Lemma. Differentiability of the remainder QUESTION [5 upvotes]: Let $f:\Omega\to\mathbb{R}$ be of $C^k$ class, let $0\in\Omega\subset\mathbb{R}^n$ and let $\Omega$ be star shaped at $0.$ From Hadamard's Lemma we know that we can write function $f$ as $$f(x)=f(0)+\sum_{i=1}^n \overbrace{x_i\int_0^1\frac{\partial f}{\partial x_i}(tx)dt}^{R_i(x)}.$$ This is in fact multivariate case of Taylor's Theorem with the remainder of first order. Question. Is $R_i$ of $C^k$ class for $i=1,\dots,n$? For $n=1$ the answer is yes, cause then $R(x)=f(x)-f(0).$ Problem is with $n>1.$ Remarks. Few days ago I posted similar question on Math.SE, but it didn't draw much attention. After reading Whitney's article I thought that the the answer may be nontrivial so this is why I post here as well. I just can add that from Whitney's paper we immediately know that $$ x\int_0^1\frac{\partial f}{\partial x}(tx,y)dt$$ is of $C^k$ class, but in our case we have $$ x\int_0^1\frac{\partial f}{\partial x}(tx,ty)dt.$$ I can't find a way to transform Whitney's argument to my case. Whitney, Hassler, Differentiability of the remainder term in Taylor's formula, Duke Math. J. 10, 153-158 (1943). ZBL0063.08234. REPLY [9 votes]: No, not in general. Consider $f(x,y)=(x+y)|x+y|$. This is $C^1$ with partial derivatives $f_x=f_y=2|x+y|$, but you lose one derivative when you form $$ R= x \int_0^1 2|tx+ty|\, dt = x|x+y| . $$<|endoftext|> TITLE: Can the limit set of an infinitely generated Schottky group have positive area? QUESTION [9 upvotes]: Dear Mathoverflow Community, Suppose that $\Omega$ is a domain in the Riemann Sphere $\widehat{\mathbb{C}}$ with $\infty \in \Omega$, and assume that every connected component of $\partial \Omega$ is a round circle. Let $\Gamma(\Omega)$ denote the Schottky group of $\Omega$, that is, the free group of Möbius and anti-Möbius transformations generated by the family of reflections across the boundary circles. Question : Can the limit set $\widehat{\mathbb{C}} \setminus \bigcup_{T \in \Gamma(\Omega)} T(\Omega)$ have positive area? Remark : If there are only finitely many boundary circles, then the limit set is a Cantor set and it is not difficult to prove that it must have zero area. We can therefore assume that there are infinitely many boundary circles. Thank you REPLY [5 votes]: Two relevant references: W. Abikoff, Some remarks on Kleinian groups. 1971 Advances in the Theory of Riemann Surfaces (Proc. Conf., Stony Brook, N.Y., 1969) pp. 1–5. Ann. of Math. Studies, No. 66. Princeton Univ. Press, Princeton, N.J. Among other things, he constructs an infinitely generated free Kleinian subgroup $\Gamma< PSL(2,C)$ whose limit set is a Jordan curve of positive planar measure. It is worth looking more closely at his construction to see if it can be made using fundamental domains with a single accumulation point of boundary faces. Remark 1. I looked: Abikoff's argument is a variation on the construction in Remark 2 and, hence, is useless for your purposes. K. Matsuzaki, The Hausdorff dimension of the limit sets of infinitely generated Kleinian groups. Math. Proc. Cambridge Philos. Soc. 128 (2000), no. 1, 123–139. In Theorem 2' he proves that if $\Gamma< PSL(2,R)$ is a discrete subgroup such that the convex core $C$ of $H^2/\Gamma$ consists of infinitely many boundary loops $c_i$ which satisfy $$ \sum_{i} \ell(c_i)^{1/2}<\infty $$ then the limit set of $\Gamma$ (in $S^1$) has positive linear measure. (Here $\ell$ denotes the length of a curve) Using this it is easy to construct an example of an infinite rank Schottky subgroup of $PSL(2,R)$ whose fundamental domain has unique accumulation point (of pairwise disjoint boundary arcs) and such that the limit set has positive linear measure. This is a 1-dimensional version of an example you are asking for. In order to construct an example, note that there exists a convex infinitely-sided right-angles polygon $P\subset H^2$ whose boundary is connected, whose ideal boundary is a single point, such that $\partial P$ is a concatenation of alternating odd and even edges $e_i$, $I\in {\mathbb N}$, such that the sequences of hyperbolic lengths $$ (\ell(e_2i))_{i>0}, (\ell(e_2i))_{i<0} $$ converge to zero as fast as you wish. Thus, you can assume that $$ \sum_{i\in {\mathbb Z}} \ell(e_{2i})^{1/2}<\infty. $$ Now, take the subgroup of isometries of $H^2$ generated by reflections in the odd-numbered edges of $P$. Let $\Gamma$ denote the orientation preserving index 2 subgroup of this reflection group. The convex core $C$ of $H^2/\Gamma$ is isometric to the surface obtained by doubling $P$ across its odd-numbered edges. Hence, $\Gamma$ satisfies Matsuzaki's condition. Remark 2. There is a variety of inequivalent definitions of Schottky groups in the literature. An easy and well-known construction (which some people call "Schottky" but you do not!) is the following: Start with a compact nowhere dense subset $K\subset R^2$ of positive measure (say the product of two thick Cantor sets in $R$). Let $D_i\subset R^2$ be a collection of pairwise disjoint closed disks disjoint from $K$, such that the closure of $$ \bigcup_{i} D_i $$ contains $K$. Now, take the group $\Gamma_K< Mob(S^2)$ generated by inversions in the boundary circles of the disks $D_i$. Its limit set will contain $K$ and, hence, will have positive measure. By taking $K$ totally disconnected one obtains examples with totally disconnected limit sets. I think this is what Rich Schwartz had/has in mind. Furthermore, the examples of Stratmann and Urbanski mentioned by Igor are variations on this construction (there is a minor and inessential difference that instead of reflections they use pairwise matchups wings of "isometric spheres"): Their $K$, while it may have zero measure, is never a singleton, it always has positive Hausdorff dimension. (You have to dig through the proof of Theorem 5.3 of their paper in order to understand this, their $W$ is my $K$.) This is all fine and well, but you want the union of disks to accumulate at a single point in $S^2$. Here is the essential difference between the two classes of groups: The above example $\Gamma_K$ will have dissipative action on the limit set (i.e. there is a measurable wondering domain of positive measure, namely, $K$). On the other hand, you are asking for a Schottky group such that the action on the limit set is conservative, i.e. admits no measurable wondering domains (this is not immediate, but follows from Sullivan's work; the key thing is that the closure of the $H^3$-fundamental domain in your case intersects the limit set in a subset of measure zero, since it is a singleton). Edit. Here is a conjectural construction (in arbitrary dimension), but making it work requires doing some computations and I do not have time for this. Let $\{D_i\}_{i\in {\mathbb N }}$ be a collection of pairwise disjoint closed round disks in the $n$-dimensional sphere which accumulate to a single point $p\in S^n$. Suppose that this collection has the following property: Given $R>0$ let $B(p,R)\subset S^n$ denote the $R$-ball centered at $p$ and set $$ F_R:= B(p,R) - \bigcup_{i\in {\mathbb N }} D_i. $$ Assume that $$ \lim_{R\to 0} \frac{Vol(F_R)}{R^n}= 0. $$ Conjecture. Then the discrete group $\Gamma$ of Moebius transformations generated by inversions in the spheres $\partial D_i$ has limit set $\Lambda$ of positive $n$-dimensional measure and, moreover, the point $p$ is a density point of the limit set: $$ \lim_{R\to 0} \frac{mes_n(\Lambda \cap B(p,R))}{Vol(B(p,R))} = 1 $$<|endoftext|> TITLE: On Wilson's claim that Lyapunov function level sets are not exotic spheres QUESTION [6 upvotes]: In Wilson's paper "The structure of the level surfaces of a Lyapunov function," he states in Corollary 1.3 that the level sets of a smooth Lyapunov function are diffeomorphic to a standard sphere. (The Lyapunov function is for a globally asymptotically stable equilibrium point for a flow on $\mathbb{R}^n$.) To prove this for a Lyapunov function $V:\mathbb{R}^n\to \mathbb{R}$ and $c > 0 $, he uses the flow to show that there are diffeomorphisms $V^{-1}(c) \approx \mathbb{R}^n\setminus \{0\} \approx S^{n-1}\times \mathbb{R}$. Thus $V^{-1}(c)$ is a homotopy sphere, and since $c$ is a regular value of $V$ it follows from the generalized Poincaré conjecture in Top that $V^{-1}(c)$ is homeomorphic to a sphere. Wilson claims that $V^{-1}(c)$ is diffeomorphic to a sphere. Why is this true? He also makes this comment on the third page: "Our spheres will always have the standard differentiable structure, since it is induced by the embedding in $\mathbb{R}^n$." But all exotic spheres also embed into some $\mathbb{R}^N$ by Whitney's theorem -- so I am not sure what Wilson means. Could this be due to the fact that $V^{-1}(c) \subset \mathbb{R}^n$ is codimension-1? REPLY [6 votes]: Matthew. I had a look at Wilson's paper; he is of course rigourous; he says that $V^{-1}(c)$ is homotopy-equivalent to $S^{n-1}$ for every $n$; $V^{-1}(c)$ is diffeomorphic to $S^{n-1}$ for every $n\neq 4, 5$ which corresponds to what was known in 1967. Nowadays, one can say a little more: $V^{-1}(c)$ is also diffeomorphic to $S^{n-1}$ for $n=4$ (thanks to Perelman's proof of the Poincare conjecture); $V^{-1}(c)$ is homeomorphic to $S^{n-1}$ for every $n$ (thanks to Friedman's topological h-cobordism theorem). Good reading!<|endoftext|> TITLE: Understanding a quip from Gian-Carlo Rota QUESTION [34 upvotes]: In the chapter "A Mathematician's Gossip" of his renowned Indiscrete Thoughts, Rota launches into a diatribe concerning the "replete injustice" of misplaced credit and "forgetful hero-worshiping" of the mathematical community. He argues that a particularly egregious symptom of this tendency is the cyclical rediscovery of forgotten mathematics by young mathematicians who are unlikely to realize that their work is fundamentally unoriginal. My question is about his example of this phenomenon. In all mathematics, it would be hard to find a more blatant instance of this regrettable state of affairs than the theory of symmetric functions. Each generation rediscovers them and presents them in the latest jargon. Today it is K-theory yesterday it was categories and functors, and the day before, group representations. Behind these and several other attractive theories stands one immutable source: the ordinary, crude definition of the symmetric functions and the identities they satisfy. I don't see how K-theory, category theory, and representation theory all fundamentally have at their core "the ordinary, crude definition of the symmetric functions and the identities they satisfy." I would appreciate if anyone could give me some insight into these alleged connections and, if possible, how they exemplify Rota's broader point. REPLY [45 votes]: I think Abdelmalek Abdesselam and William Stagner are completely correct in their interpretation of the words "Behind" and "one immutable source" as describing one theory, the theory of symmetric functions, being the central core of another. The issue that led to this question instead comes from misunderstanding this sentence: Today it is K-theory yesterday it was categories and functors, and the day before, group representations. The listed objects are not a list of theories. If they were, he would say "category theory" and "representation theory". Instead, it is a list of different languages, or as Rota calls them, jargons. The function of this sentence is to explain what jargons he is referring to in the previous sentence. If we delete it, the paragraph still makes perfect sense, but lacks detail: In all mathematics, it would be hard to find a more blatant instance of this regrettable state of affairs than the theory of symmetric functions. Each generation rediscovers them and presents them in the latest jargon. [...] Behind these and several other attractive theories stands one immutable source: the ordinary, crude definition of the symmetric functions and the identities they satisfy. The "theories" in question are not K-theory, category theory, and representation theory but rather the theory of symmetric functions expressed in the languages of K-theory, category theory, and representation theory. For instance presumably one of them is the character theory of $GL_n$, expressed in the language of group representations. The reason I am confident in this interpretation is nothing to do with grammar but rather the meaning and flow of the text. The claim that symmetric function theory is the source of three major branches of mathematics seems wrong, but if correct, it would be very bizarre to introduce it in this way, slipped in the end of a paragraph making a seemingly less shocking point, and then immediately dropped (unless the quote was truncated?). One would either lead with it, or build up to it, and in either case then provide at least some amount of explanation. Thus instead I (and Joel, and Vladimir) interpret it as making a less dramatic claim. REPLY [12 votes]: Le teorie vanno e vengono ma le formule restano.--G.C. Rota. (The theories may come and go but the formulas remain.) Perhaps the Wiki on the Adams operation and "Formal groups, Witt vectors, and free probability" by Friedrich and McKay provide a quick intro to the connections the OP is questioning. "Today the jargon is that of K-theory, yesterday it was that of categories and functors, and, the day before, group representations." All three jargons are used in the refs above (and those in my comments), serving to present different perspectives on, or even generalizations of, the basic, originally discovered relations among the symmetric functions. Territorial instincts may compel some camps to claim the superiority (and even priority) of their insights, or approach, which is probably what Rota decries even though he was certainly guilty of this same behavior. (Read the introductory paragraph of "Alphabet Splitting" by Lascoux: ... meals were followed by long discussions about the comparative merits of algebraic structures, Gian Carlo for his part tirelessly asking me to repeat the definition of λ-rings that he copied each time in his black notebook with a new illustrative example.) Added Nov. 8, 2019 I'm fairly convinced that Rota expressed exactly what he meant to express--that the identities/properties of the symmetric polynomials lie at the foundations of "these and several other attractive theories." See the refs and comments in the MO-Q "Canonical reference for Chern characteristic classes," in particular, "Characteristic classes and K-theory" by Randal-Williams, the linked Wikipedia article on Chern classes, the relevant sections in "Manifolds and Modular Functions" by Hirzebruch et al., and the Wikipedia article on the Splitting Principle. Added 5/27/21: Donald Knutson, in "$\lambda$-Rings and the Representation Theory of the Symmetric Group," states, "the notion of $\lambda$-ring is built upon the classical Fundamental Theorem of Symmetric Functions," that much of classical algebra is based on this theorem, and "the general definition (of a $\lambda$-ring) is somewhat complicated ... and will be best understood by first analyzing one manifestation of the ring Z, its appearance in the simplest example of K-theory." In addition, "the main technical tool (In proving the Fundamental Theorem) is the notion of $\lambda$-ring, first introduced by Grothendieck in 1956 ... in an algebraic-geometric context, and later used in group theory by Atiyah and Tall ... ." Edit 8/23/2021: From "Ten lessons I wish I had learned before I started teaching differential equations" by Rota: I have always felt excited when telling the students that even though there is no formula for the general solution of a second order linear differential equation, there is nevertheless an explicit formula for the Wronskian of two solutions. The Wronskian allows one to find a second solution if one solution is known (by the way, this is a point on which you will find several beautiful examples in Boole’s text). ... every differential polynomial in the two solutions of a second order linear differential equation which is independent of the choice of a basis of solutions equals a polynomial in the Wronskian and in the coefficients of the differential equation (this is the differential equations analogue of the fundamental theorem on symmetric functions, but keep it quiet).<|endoftext|> TITLE: (Efficient) computation of symmetric powers of square matrices QUESTION [6 upvotes]: I'm looking for software that can compute symmetric powers of medium-size square (say rational, 100 by 100) matrices, and ideally can do so efficiently if the matrix is sparse enough. I haven't found any function for symmetric power in Sage or sympy, and a google search hasn't turned up anything. (An aside: Do people studying matrix algorithms have a different name for symmetric power, or they just aren't interested?) Any suggestions? Edit: Macaulay2 only does symmetric powers of one-rowed matrices, it seems. Singular seems hopelessly inefficient on medium-sized sparse matrices (20 by 20). REPLY [4 votes]: Sage can compute symmetric powers of matrices via a Singular function, we just discussed this here the other day. Or, indeed, you can use Singular directly. Posted from mobile, sorry for lack of details. PS. GAP has an undocumented function SymmetricPower, which computes symmetric powers of square matrices, but it is unfortunately quite slow.<|endoftext|> TITLE: Kaplansky conjecture (consequences) QUESTION [9 upvotes]: The Kaplansky conjecture says that: for any field $F$ and any torsion free group $G$, the group ring $F[G]$ does not have nontrivial idempotent elements. Questions Do we assume that $F$ has any characteristic ? Does the conjecture imply that any finitely generated projective $F[G]$-module is free ? or eventually stably free? For which field $F$ and which class of groups $G$, the conjecture is known to be true? REPLY [11 votes]: Kaplansky's zero divisor conjecture: Let $\mathbb{F}$ be a field and $G$ be a torsion-free group. Then $\mathbb{F}[G]$ does not contain a zero divisor. The existence of a nontrivial idempotent $a$ in a ring $R$ implies the existence of a zero divisor in $R$, because $a(a-1)=0$. So, the zero divisor conjecture implies the idempotent conjecture. The idempotent conjecture has been confirmed in special cases. For example, Formanek (1973) showed that if $G$ is a torsion-free group satisfying the ascending chain condition on cyclic subgroups and $\mathbb{F}$ is a field of characteristic $0$, then $\mathbb{F}[G]$ has no nontrivial idempotents. Also, Bass (1976) proved that if $G$ is a torsion-free linear group, then $\mathbb{C}[G]$ has no nontrivial idempotents. These conjectures have not been confirmed for any fixed field and it seems that confirming the conjecture even for the finite field $\mathbb{F}_2$ is still out of reach. For the zero divisor conjecture case, our work Zero divisors and units with small supports in group algebras of torsion-free groups may be helpful. Also for more details about these two conjectures you can see Zero-divisors and idempotents in group rings.<|endoftext|> TITLE: What would be the cotangent bundle of a Banach manifold? QUESTION [6 upvotes]: Reference: Lang - Differential manifolds p.123 Quick question: Lang defines the cotangent bundle as the dual vector bundle of the tangent bundle, but shouldn't there be additionally a somewhat canonical differentiable structure on the cotangent bundle? (Or does the dual vector bundle naturally determine the differentiable structure?) How do I define such canonical differentiable structure? Let me illustrate my approach in detail below: Let $X$ be a Banach $C^k$-manifold. Define $C_p(X)$ as the class of $(c,v)$ where $c=(U,\phi,E)$ is a chart at $p$ and $v\in E$. (Note that $U$ denotes a coordinate domain and $\phi$ denotes a coordinate map and $E$ denotes a Banach space where $\phi$ is mapping to) Now, define an equivalence relation on $C_p(X)$ as follows: $((U,\phi,E),v)\sim((V,\psi,F),w)$ iff $D(\psi\circ \phi^{-1})(\phi(p))(v)=(w)$ Define a map $\mathscr{O}_c^p:E \rightarrow C_p(X)/\sim: v\mapsto [(c,v)]$. Then, it can be shown that $\mathscr{O}_c^p$ is a bijection. Using this map, we can isomorphically give a vector space structure on $C_p(X)/\sim$, and this structure can be shown independent of choice of charts $c$. Let $T_p(x)$ denote this set equipped with the natural vector space structure. Moreover, we can homeomorphically give a Banachable topology on $T_p(X)$ using those maps $\mathscr{O}_c^p$. Now, define $T_p^*(X)$ as the collection of continuous linear functionals on $T_p(X)$. So that $T_p^*(X)$ is a Banachable space too. With these terminologies (which are standard as far as I know), let's define the cotangent bundle. Define $T^*X:=\bigcup_{p\in X} (\{p\}\times T_p^*(X))$, and let $\pi^*:T^*X\rightarrow X$ be the map such that $\pi^*(p,y)=p$. Let $c=(U,\phi,E)$ be a chart of $X$. Define $\mathscr{D}_c ^p:T_p^*(X)\rightarrow E^*: A\mapsto A\circ \mathscr{O}_c^p$. Note that this map is linear homeomorphism. Define $U^*:=(\pi^*)^{-1}(U)$, and $\phi^*(p,A)=(\phi(p),\mathscr{D}_c^p(A))$ for each $(p,A)\in U^*$. Now define $c^*:=(U^*,\phi^*,E\times E^*)$. Then, $\phi^*:U^*\rightarrow E\times E^*$ is an injection and $\phi^*(U^*)=\phi(U)\times E^*$ is open and $E\times E^*$ is Banachable. Hence, $c^*$ is a chart on the set $T^*X$. Take the unique differentiable structure generated by those $c^*$'s, and call the set $T^*X$ together with this differentiable structure the cotangent bundle of $X$. Would it be the correct one? REPLY [2 votes]: Yes. That unique differentiable structure is given by Lang in Proposition 2, p. 43 which applies by Theorem 1, p. 54. (A later edition has these and the cotangent bundle example on pp. 45, 59 and 61.)<|endoftext|> TITLE: Smith normal form for specialized matrices QUESTION [6 upvotes]: Consider an $n\times n$ matrix $M_n$ where the sequence $1,2,3,\dots,n^2$ forms a clock-wise spiral, in that given order. For example, $$M_4=\begin{bmatrix} 1&2&3&4\\ 12&13&14&5\\ 11&16&15&6 \\ 10&9&8&7 \end{bmatrix} \qquad \text{and} \qquad M_5=\begin{bmatrix} 1&2&3&4&5\\ 16&17&18&19&6 \\ 15&24&25&20&7 \\ 14&23&22&21&8 \\ 13&12&11&10&9 \end{bmatrix}.$$ Question. What are the diagonal entries in the Smith normal form of the matrix $M_n$, over $\mathbb{Z}$? REPLY [5 votes]: not a complete answer, but too long for a comment: some experimentation$^*$ suggests that the diagonal entries of the diagonal matrix $S_n$ in the integer decomposition $M_n=U_nS_nV_n$, with $U_n,V_n$ unimodular, have the following form for $n> 2$: $${\rm diag}\,S_n=1,1,2,2,2,\cdots 2,2,2,\{Z_n\},c_n$$ where the $\cdots$ indicate padding to length $n$ with a string of $2$'s and $\{Z_n\}$ is a string of integers given by $$\{Z_n\}=\emptyset\;\;{\rm for}\;\;n\leq 5,$$ $$\{Z_n\}=\{6\},\{30\},\{6\},\{6,30\},\{6,30\}\;\;{\rm for}\;\;n=6,7,8,9,10,$$ $$\{Z_n\}=\{6,30\},\{6,30,210\},\{6,6,210\},\{6,30,210\},\{6,6,30,6930\}\;\;{\rm for}\;\;n=11,12,13,14,15,$$ $$\{Z_n\}=\{6,6,30,630\},\{6,30,30,630\},\{6,6,6,30,6930\},\{6,6,30,210,6930\},\{6,6,30,210,6930\}\;\;{\rm for}\;\;n=16,17,18,19,20,$$ $$\{Z_n\}=\{6,6,6,30,210,90090\},\{6,6,30,30,210,90090\},\{6,6,6,30,210,1531530\},\{6,6,6,30,30,630,90090\},\{6,6,6,30,30,630,90090\}\;\;{\rm for}\;\;n=21,22,23,24,25,$$ and so on. Unfortunately, I have been unable to detect a pattern in this sequence. The final diagonal entry of $S_n$ is $$c_n=\frac{2^{3-n}|{\rm det}\,M_n|}{\prod_{i}\tfrac{1}{2}(Z_n)_i}$$ following from the formula OEIS A023999 for the determinant of a spiral matrix: $$|{\rm det}\,M_n|={\rm det}\, S_n=(3n-1) \frac{ (2n-3)!}{(n-2)!}$$ $^*$ if you would like to experiment further with spiral matrices, here are a few lines of relevant Mathematica code; I am intrigued by this $Z_n$ pattern, what is the logic behind it?<|endoftext|> TITLE: Splitting of tangent bundle QUESTION [13 upvotes]: Is it possible to give an example of $n$ dimensional manifold with the property that the tangent bundle $TM$ cannot be expressed as Whitney sum of two subbundles? It is certain true for two sphere; it is certainly not true for three dimensional manifold since every three manifold is parallelizable. You can always split $TM$ as $\gamma \oplus Q$ where $\gamma$ is one dimensional if you can find non vanishing vector field (i.e. where the Euler class is zero which is the case for odd dimension) but what about examples with $n$ even, larger than $2$? REPLY [2 votes]: A condition that prevents a vector bundle from splitting a Whitney sum is that the projection of the associated sphere bundle induces a non-surjective map on some homotopy group. This is proved by L.Guijarro and G.Walschap in Transitive holonomy groups and rigidity in nonnegative curvature who show that the above condition forces the normal holonomy of the bundle to be transitive (which is clearly impossible if the bundle splits as a Whitney sum). For example, they explain in the paper that the tangent bundle to $CP^n$ with $n>1$ splits as a Whitney sum if and only if $n$ is odd.<|endoftext|> TITLE: Tauberian theorem $\sum_{k=1}^{\infty}e^{-\lambda_{k}t}c_{k} \xrightarrow{t\to 0} \sum_{k=1}^{\infty}c_{k} $ QUESTION [14 upvotes]: I am trying to prove or disprove $$\sum_{k=1}^{\infty}e^{-\lambda_{k}t}c_{k} \xrightarrow{t\to 0} \sum_{k=1}^{\infty}c_{k} ,$$ where $\sum c_{k}<\infty, \sum c_{k}^{2}<\infty\text{ and }\frac{\lambda_{k}}{k}\to c$(Weyl's law). The $c_{k},\lambda_{k},t\in \mathbb{R}$ and Update: the $\lambda_{k}$ are eigenvalues of any domain D in $\mathbb{R}^{2}$: $\lambda_{k}<\lambda_{k+1}<....$ $\lambda_{k}=\frac{4\pi}{|D|}k+c_{2}\sqrt{k}+o(\sqrt{k})$ $\sum^{k}_{j\geq 1}\lambda_{j}\geq \frac{2\pi}{|D|}k^{2}\Rightarrow \lambda_{k}\geq \frac{2\pi}{|D|}k$ (Li-Yau) The counterexamples below were before the update, where we only assumed $\frac{\lambda_{k}}{k}\to c$ (but they seem to include monotonicity too). Q:Since the line of convergence is $Re(z)=0$, is there any way to analytic continue this series in a neighbourhood of zero, in order to apply some of the known tauberian theorems. Please, I prefer just hints to help me practice. Attempts 0)The $\lambda_{k}$ correspond to the Dirichlet-Laplacian eigenvalues of a domain in $\mathbb{R}^{2}$, so I was hoping to have them fixed and only modify the $c_{k}$ for a counterexample. Or come up with the $\lambda_{k}$ and the corresponding domain, that gives a counterexample. 1)The proof of Abel's theorem doesn't work because it requires linear error term $|\lambda_{k}/k-c|<\frac{c'}{k}$ (from Weyl's law we have $\frac{1}{\sqrt{k}}$ error): for $|s_{N}|=|\sum_{N}c_{k}|\leq \varepsilon$ we have $|\sum_{N} c_{k} e^{-\lambda_{k}t}|\leq \varepsilon \sum_{N} ( e^{-\lambda_{k}t}- e^{-\lambda_{k+1}t})\approx \varepsilon e^{-N c_{1} t}\frac{(1-e^{-\sqrt{N+1}t c_{2}})}{(1-e^{t c_{2}})}\to \varepsilon c \sqrt{N+1}$ assuming the unproved but plausible $\lambda_{k}\geq \frac{4\pi}{|D|}k$ for large k. 2)Littlewood tauberian theorems requiring $c_{k}k=o(1)$ don't apply because we might have $c_{k}=\frac{(-1)^{k}}{k}$. 3)By having $c_{k}=\frac{(-1)^{k}}{k}$ the line of convergence of abstract Dirichlet series $$\sum e^{-\lambda z}c_{k}$$. So analytic continuation is not clear. is $\{z:Re(z)=0\}$. The Ostrowski–Hadamard gap theorem doesn't apply because $\lambda_{k+1}/\lambda_{k}\to 1$. 4)Borel summation method might apply: Because $\sum c_{k}^{2}<\infty$ and so $$\sqrt{k}c_{k}=o(1).$$ Moreover, $e^{-x}\sum s_{k}\frac{1}{k!}z^{k}$ is weakly Borel summable. So indeed $$\sum_{k=1}^{\infty}e^{-kt}c_{k} \xrightarrow{t\to 0} \sum_{k=1}^{\infty}c_{k}.$$ So maybe the proof can be modified for $\lambda_{k}$ instead. 5)Possible counterexample via letting $c_{k}=\frac{(-1)^{k}}{k}$ to disallow any dominated convergence. 6)Another Tauberian theorem requires showing for $f(z):=\sum_{k=1}^{\infty}e^{-\lambda_{k}z}c_{k}$ and some function F $$\frac{f(z)-f(0)}{z}\to F(iIm(z)),$$ uniformly or in L1; as $Re(z)\to 0$ and $Im(z)\in [-\lambda,\lambda]$, where in our case $\lambda=\infty$. So assuming we can show convergence, the limit is $$\frac{\sum_{k=1}^{\infty}c_{k}(e^{-\lambda_{k}iy}-1)}{iy}$$ and since $\sum a_{k}<\infty$ we ask that $\sum_{k=1}^{\infty}c_{k}e^{-\lambda_{k}iy}$ is well-defined. However, I think we can concoct $c_k$ s.t. $\sum_{k=1}^{\infty}c_{k}e^{-\lambda_{k}iy_{0}}=\infty$ for some fixed $y_{0}$. REPLY [2 votes]: The $\lambda_k$ correspond to the Laplacian eigenvalues of a domain in $\mathbb{R}^2$ implies that \begin{equation} 0<\lambda_1\le \lambda_2\le\cdot\cdot\cdot\le \lambda_n\le\cdot\cdot\cdot \end{equation} and $\lambda_n\rightarrow\infty $ as $n\rightarrow\infty$. Hence for any $t>0$ \begin{align} \sum_{n=1}^{\infty}e^{-\lambda_nt}c_n&=\sum_{n=1}^{\infty}[C(n)-C(n-1)]e^{-\lambda_nt}\\ &=\sum_{n=1}^{\infty}C(n)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\\ &=C(\infty)+\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right], \end{align} by Abel transform, where $$C(n)=\sum_{k=1}^{n}c_k ~\mbox{for}~ n\ge 1 $$ and $$o_n(1)=\sum_{k=1}^{\infty}c_k-C(n)=C(\infty)-C(n)\rightarrow 0~as~ n\rightarrow \infty.$$ Then it is obviously that $$\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\rightarrow 0~ as~t\rightarrow 0+.$$ In fact, for any $\varepsilon>0$, there exist an $N_{\varepsilon}\in\mathbb{N}$ such that for all $n>N_{\varepsilon}$, $|o_n(1)|< \varepsilon$ holds. Furthermore, $$\left|\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\right|\le \left|\sum_{n=1}^{N_{\varepsilon}}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\right|+\epsilon\sum_{n>N_{\varepsilon}}\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right].$$ Namely, $$\left|\sum_{n=1}^{\infty}o_n(1)\left[e^{-t\lambda_n}-e^{-t\lambda_{n+1}}\right]\right|\le \max_{1\le k\le N_{\varepsilon}}|o_k(1)|e^{\lambda_1t}+\epsilon.$$ The next let $t\rightarrow 0$ and then let $\varepsilon\rightarrow 0$, then the results is obvious.<|endoftext|> TITLE: The number of $2$-simplices and the number of $1$-simplices in a $4$-dimensional simplicial complex QUESTION [7 upvotes]: Given a $4d$ simplicial complex (a triangulation of $4$-manifold), is there any relation between the number of $2$-simplices (triangles) and the number of $1$-simplices (edges)? Generically, is the number of $2$-simplices always greater than the number of $1$-simplices? Thanks! REPLY [5 votes]: The answer is yes for pure abstract simplicial complexes. Thus we consider a finite set $X$ and a system of subsets $S$ in $X$, such that $S$ is closed under taking subsets, every $s \in S$ has at most $5$ elements and is contained in at least one $t\in S$ with $5$ elements. Denote by $m$ the number of $s\in S$ with $2$ elements and by $n$ the number of elements $t\in S$ with $3$ elements. (These are the numbers of one-dimensional and two-dimensional simplices in the abstract simplicial complex $(X,S)$, respectively.) Let $Y$ be the set of pairs $(s,t)$, such that $s$,$t\in S$, $\# s=2$, $\# t=3$, and $s$ is contained in $t$. Then the number of elements in $Y$ equals $3n$. In fact every $t\in S$ with three elements, contains exactly three subsets of cardinality two. On the other hand, every $s\in S$ with $\# s=2$, is contained in at least one $v\in S$ with $\# v=5$. Now, there are three different $t$ such that $s\subset t \subset v$ and $\# t =3$. All these $t$ are elements of $S$. Therefore the number of elements in $Y$ is at least $3m$. Hence $3n \ge 3m$, which is equivalent to $n\ge m$.<|endoftext|> TITLE: Riemann surface from Riccati equation QUESTION [6 upvotes]: I have quite a practical question motivated by physics. Consider the Riccati equation whose solution gives a quantum-mechanical (QM) analogue of the classical momentum: $$ (p(x))^2 + \dfrac{\hbar}{i}p'(x)= 2 (E-V(x)) \quad. $$ Clearly, in the $\hbar\to0$ limit one obtains the definition of the classical momentum: $$ p_{c}(x)= \sqrt{ E-V(x)}\quad. $$ It's easy to determine what's the Riemann surface on which $p_c(x)$ is defined $-$ a two-dimensional oriented manifold with, typically, a finite number of punctured points (let's limit ourselves with polynomial potentials). Now, the standard approach in QM is to consider an expansion of $p(x)$ in powers of $\hbar$, which leads to the (Generalised) Bohr-Sommerfeld qantisation: $$ \begin{gathered} p(x) = \sum \limits_{k=0}^\infty \left(\dfrac{\hbar}{i}\right)^k p_k(x)\quad,\\ p_0(x) \equiv p_c(x) \quad. \end{gathered} $$ Here $p_k(x)$ are, of course, assumed to be $\hbar$-independent. It is now a simple exercise to show that all the $p_k(x)$ live on the same Riemann surface as $p_c(x)$: they all are obtained from $p_c(x)$ recursively by means of algebraic operations and taking derivatives $-$ none of these can drag us out of the Riemann surface. This result suggests me to make a way stronger statement: namely, that the solutions of the original Riccati equation have to live on the same Riemann surface as $p_c(x)$. Basically, I'm saying that $\hbar\to0$ limit does not change the Riemann surface (well, that's a tricky limit since it turns a differential equation into a trivial equality). It may be temptingly to say that the last equation (the infinite sum) is exactly what I need. However, those who are familiar with things like asymptotic series know that it's not like that. The reason for this is that $p(x)$ may depend on $\hbar$ in a non-polynomial way, like $\exp(-1/\hbar)$ or $\log (1/\hbar)$ (non-perturbatively, in physical jargon). So my question is: Given the Riccati equation (the top one in the question), is it possible to prove that its solutions live on the same Riemann surface as the function $p_c(x)$ defined by the $\hbar\to 0$ limit of the equation? I would prefer to get the answer which will not rely on employing any (trans-)series expansions of $p(x)$, but would rather be based on a global analysis of the differential equation's Riemann surface (or smth like that). P.S. If I'm asking something trivial, any references to the relevant textbooks are greatly appreciated. UPDATE I'm still very much interested to hear any useful comments/references on the topic, however I've just realised that the answer to my question is negative. Different solutions may live on different Riemann surfaces. Unfortunately, this conclusion relies not on strict mathematical statements, but rather on my knowledge from physics. The quantum momentum $p(x)$ has a first-order pole wherever the wave function $\psi(x)$ has a zero: $$ p(x) =\dfrac{\hbar}{i} \dfrac{1}{\psi(x)} \dfrac{\operatorname{d} \psi(x)}{\operatorname{d} x} $$ These wave function may have arbitrary number of poles on the branch cut of the classical momentum. In the classical limit this sequence of first-order poles coalesces into a branch cut, just like $\int \dfrac{\operatorname{d}x}{x}=\log x$. Which tells us that for non-zero $\hbar$ the Riemann surface is very different from the $\hbar=0$ case. REPLY [9 votes]: The answer to the highlighted question is "no". When $V$ is a polynomial, the general solution of the Riccati equation is single valued, it is a meromorphic function in the complex plane. To prove the statement, make the coefficient at $p'$ equal to $1$ by scaling of $x$, and then reduce your Riccati equation to a linear second order equation by setting $p=w'/w$. You obtain a second-order linear equation with polynomial coefficient, so its general solution is entire. What really happens when $h\to 0$, is that the poles of these meromorphic solutions accumulate to certain ``branch cuts'' of your multivalued classical approximation, and the convergence to this classical approximation happens away from these poles and is not uniform. I recommend Heading, J. An introduction to phase-integral methods. Methuen & Co., Ltd., London; John Wiley & Sons, Inc., New York 1962 for a short and clear exposition of this topic.<|endoftext|> TITLE: Definability of isomorphisms between class well-orderings QUESTION [6 upvotes]: Inspired by this question, I've been trying to figure out for myself the basic properties of definable class well-orderings in transitive models $M$ of ZFC: What is $\omega_1^{CK}(\mathsf{Ord})$? There were a couple things I couldn't figure out. Is it necessarily the case that $M$ is correct about well-orderings? I.e., if $M \models ``\varphi(\cdot, \cdot)$ is a total ordering of $Ord" \wedge \forall x(x=\emptyset \vee \exists y \in x(\forall z \in x(\varphi(y, z)))),$ then $\varphi$ truly (in $V$) defines a well-ordering of $Ord^M?$ If $\varphi_1$ and $\varphi_2$ define in $M$ isomorphic well-orderings of $Ord^M,$ is there necessarily a definable isomorphism $\psi(\cdot, \cdot)$ between them? E.g., if $\alpha$ is the first ordinal to exceed a proper class of ordinals under $\varphi_1$ and $\beta$ is the first such ordinal under $\varphi_2,$ then we would have $\psi(\alpha, \beta).$ (1) is easy to prove in the case $Ord^M$ has uncountable cofinality in $V,$ but it isn't obvious to me whether it can fail in models of countable cofinality height. (2) seems unlikely, since I don't really see how to extend transfinite recursion past the ordinals (especially in light of potential failures of (1)), but I have trouble imagining what a counterexample would like. REPLY [9 votes]: The answer to question 1 is that no, a transitive model $M$ can be wrong about whether a definable class relation is a well-order. To see this, consider a transitive model $M$, and let me assume that there is no worldly cardinal in $M$. For example, perhaps we have cut off the universe at the smallest worldly cardinal. By the reflection theorem, we know that for any given $n$, there are many ordinals $\theta$ in $M$ with $V_\theta^M\prec_{\Sigma_n} M$. In particular, we can make an increasingly elementary chain $$V_{\theta_0}^M\prec_{\Sigma_1} V_{\theta_1}^M\prec_{\Sigma_2}\cdots\prec_{\Sigma_n} V_{\theta_n}^M\prec_{\Sigma_{n+1}}\cdots$$ that unions up to $M$. Let $T$ be the tree of all finite sequences that obey this increasingly elementary substructure relation with one another, and where also the $n^{th}$ element in the sequence also models the $\Sigma_n$ fragment of ZFC. It follows from our observation above that there are a proper class of such instances in $M$, and so $T$ is a proper class. Elements of $T$ amount to finite sequences of ordinals, which can be coded by single ordinals, and so we may view $T$ as a partial order relation on $\text{Ord}$, if you like. We order this tree growing downward, and I claim it is well-founded in $M$. That is, $M$ has no $\omega$-sequence that is a descending sequence in this tree, because the union of that chain would be a $V_\theta$ that models ZFC, and so $\theta$ would be worldly, but we assumed there are none in $M$. So $M$ thinks this tree order is well-founded, and therefore it thinks the Kleene-Brouwer order on the tree is a well-order. But $M$ is wrong about both of these things, since we have already observed that $M$ is the union of an increasingly elementary chain, and this is exactly a descending sequence in the tree order and hence a descending sequence in the Kleene-Brouer order. So $M$ thinks the relation was a well-order, but it was mistaken.<|endoftext|> TITLE: Determinants: periodic entries $0,1,2,3$ QUESTION [12 upvotes]: Consider an $n\times n$ matrix $M_n$ where the sequence $$\{1,2,3,\dots,n^2\} \mod 4=\{1,2,3,0,1,2,3,\dots\}$$ forms a clock-wise spiral, in that given order. For example, $$M_4=\begin{bmatrix} 1&2&3&0\\ 0&1&2&1\\ 3&0&3&2 \\ 2&1&0&3 \end{bmatrix} \qquad \text{and} \qquad M_5=\begin{bmatrix} 1&2&3&0&1\\ 0&1&2&3&2 \\ 3&0&1&0&3 \\ 2&3&2&1&0 \\ 1&0&3&2&1 \end{bmatrix}.$$ Question. Is it true that $$\det(M_{2n})=3(2n-1)4^{n-1} \qquad \text{and} \qquad \det(M_{2n+1})=-(3n^2-1)4^n\,\,\,?$$ Added clarification. To understand the construction of the above matrices, take a look at the matrices from my other MO question. Then, reduce the entries modulo $4$ and follow through by computing the determinants. REPLY [5 votes]: Yes, it is true. More generally, the entries $1,2,3,0$ can be replaced by arbitrary numbers $a,b,c,d$, in which case the determinant of $M_n$ can be computed in terms of the four numbers $u = d-b$, $v = a-c$, $U = d+b$ and $V = a+c$ as follows: If $n=4k$ for some positive integer $k$, then $$ \det\left( M_{n} \right) =\dfrac{1}{4} v^{n-4}\left( v^{4}-u^{2}v^{2}+\left( U^{2}-V^{2}\right) \left( \left( 2k-1\right) ^{2}v^{2}-\left( 2k\right) ^{2}u^{2}\right) \right) . $$ If $n=4k+2$ for some positive integer $k$, then $$ \det\left( M_{n} \right) =-\dfrac{1}{4}% v^{n-4}\left( v^{4}-u^{2}v^{2}+\left( U^{2}-V^{2}\right) \left( \left( 2k+1\right) ^{2}v^{2}-\left( 2k\right) ^{2}u^{2}\right) \right) . $$ If $n=4k+1$ for some positive integer $k$, then $$ \det\left( M_{n} \right) =\dfrac{1}{2} u^{n-3}\left( u^{2}\left( v+V\right) -\left( 2k\right) ^{2}v\left( U^{2}-V^{2}\right) \right) . $$ If $n=4k+3$ for some positive integer $k$, then $$ \det\left( M_{n} \right) =\dfrac{1}{2} vu^{n-3}\left( u^{2}+vV-\left( 2k+1\right) ^{2}\left( U^{2}-V^{2}\right) \right) . $$ As you can imagine, this is not very fun to prove. I have a writeup (The 4-periodic spiral determinant) in which I attempt at making the idea clear without going into all the details; in particular, annoying computations are relegated to SageMath and to the reader (and on some occasions to a combination of both). Even at that level of terseness, it is 24 pages long. I would normally hope that something nicer can be found, but with the complexity of the answer I am not too hopeful. The proof starts out as suggested by @user44191 in one of the comments to the original post; thus the matrix is brought to a form where all entries are zero except for those in northwesternmost $4\times 4$-submatrix and on four sub-antidiagonals (namely, the $1$-st, the $3$-rd, the $5$-th and the $7$-th sub-antidiagonals) below the main antidiagonal. Then, I turn the matrix upside down, so that the sub-antidiagonals become the super-diagonals. I then perform Laplace expansion with respect to the last $4$ rows. All $4 \times 4$-minors from the last $4$ rows can be explicitly computed (only $\dbinom{7}{4}$ of them nonzero, and this can be further reduced by looking at the vanishing of the complementary minors), so it remains to compute the complementary $\left(n-4\right)\times\left(n-4\right)$-minors. For this, Jacobi's complementary minor theorem turns out to be of use, along with an explicit computation of the inverse of a certain power series.<|endoftext|> TITLE: Endoscopic group that is not a subgroup QUESTION [9 upvotes]: The question is a very little more than what's in the title. It is easy (for some values of ‘easy’) to produce examples of endoscopic groups that are not subgroups. When I asked a colleague, he mentioned $\mathrm{PGL}_3$ as an endoscopic group of $\mathrm G_2$. However, in this example, $\mathrm{PGL}_3$ is isogenous to $\mathrm{SL}_3$, which is an actual subgroup of $\mathrm G_2$, and similarly for the other examples that I know. Is this kind of ‘central failure’ as bad as it can get, or is there an example of an endoscopic group that is not even isogenous to a subgroup? REPLY [11 votes]: If I understand the definition correctly, a connected reductive group $H$ is an endoscopic group for a connected reductive group $G$ if its Langlands dual $H^\vee$ is a connected centralizer in $G^\vee$. So $H=SO(2p+1)\times SO(2q+1)$ is endoscopic in $G=SO(2p+2q+1)$ since $H^\vee=Sp(2p)\times Sp(2q)$ is a centralizer in $G^\vee=Sp(2p+2q)$. But clearly $H$ is not isogenous to a subgroup of $G$. Another example is $H=SO(2p)\times Sp(2q)$ for $G=SO(2p+2q)$. Using Borel-de Siebenthal the problem comes from the fact that for finite Dynkin diagrams one may have $$ \text{extended dual}\ne\text{dual extended}. $$<|endoftext|> TITLE: Higgs paper ``A category approach to Boolean valued set theory'' QUESTION [10 upvotes]: As Philip Scott says about Denis Higgs: In category theory, he wrote an influential and beautiful long paper, "A category approach to Boolean valued set theory", which initiated many early students in topos theory in the 1970s to Omega-valued sets. I wonder to know if there is a way to get access to the following paper by Higgs: ``A category approach to Boolean valued set theory'' Remarks. I am aware of some similar works, for example A category-theoretic approach to boolean-valued models of set theory or the book Sheaves in geometry and logic by Mac Lane and Moerdijk. REPLY [5 votes]: Definitely not a answer, but too long for a comment, and just for completeness: As pointed out by მამუკა ჯიბლაძე, Higgs writes in Injectivity in the Topos of Complete Heyting Algebra Valued Sets that this paper is related to the unpublished paper you are looking for (cited as [5] by Higgs): An earlier version ( [5] ) of this paper was accepted for publication in the Canadian Mathematical Bulletin but for various reasons I discontinued the process of seeing it through into print. In addition to the present contents, [5] contained: a brief account of logic in $\mathcal L(\mathcal A)$; the Lawvere Tierney version (for $\mathcal L(\mathcal A)$) of the independence of the continuum hypothesis [14]; a sketch of a proof that $\mathcal L(\mathcal A)$ is equivalent to the category of sets within the universe $V^{\mathcal A}$ of $\mathcal A$-valued set theory; a mention of sheaves on an arbitrary site from the point of view of $\mathcal A$-valued sets; and some elementary remarks on boolean powers and ultrapowers in relation to $\mathcal L(\mathcal A)$. [5] D. Higgs A category approach to boolean valued set theory, preprint, University of Waterloo (1973) [14] M. Tierney, Sheaf theory and the continuum hypothesis, in Toposes, algebraic geometry and logic (F. W. Lawvere, éd.), 13-42, Lecture Notes in Mathematics 274 (SpringerVerlag, Berlin, Heidelberg, New York, 1972).<|endoftext|> TITLE: Are there other dualities on finite vector spaces besides the canonical one? QUESTION [8 upvotes]: Let $\text{FinVec}$ denote the category of finite dimensional vector spaces over some field $k$, and let $F:\text{FinVec}\to \text{FinVec}$ be a contravariant functor such that $F^2$ is naturally isomorphic to the identity. Is $F$ naturally isomorphic to the canonical duality functor $V\mapsto V^*=\text{Hom}(V,k)$? I suspect this question has been asked before on MO, but I couldn't find it. REPLY [10 votes]: $FinVec$ and its opposite are enriched in finite dimensional $k$-vector spaces. Assume that $F$ is an enriched functor. Then consider the covariant functor: $$F(-)^*: FinVec \to FinVec$$ It is a $Vec$-enriched functor. Finite direct sums are absolute limits and so are preserved by this functor (see here). Since every finite dimensional vector space is isomorphic to a finite direct sum of copies of $k$ it follows that this functor is completely determined by the single vector space $L=F(k)^*$, and there is a natural isomorphism: $$F(V)^* \cong V \otimes L$$ Thus $F(V) \cong V^* \otimes L^*$ as functors. Your condition that $F^2 \cong id$ implies that $L$ must be one-dimensional (hence isomorphic to $k$) and so $F(-) \cong (-)^*$ is the standard duality functor.<|endoftext|> TITLE: n-categorical pasting diagram overview QUESTION [12 upvotes]: There has been a lots of approaches to the notion of n-categorical diagram and n-categorical pasting diagram: Street : "Parity complexes" Power : " An n-categorical pasting theorem" Johnson : "The combinatorics of n-categorical pasting" Steiner : "Omega-category and chain complexes" I'm wondering about the relations between these approaches. It seem very clear for each of these how to translate from one approach to the other, but it is not obvious at all, and probably not true in general that these notion are exactly equivalent. In particular the way the "loop free" condition is formulated for each of these is rather subtle and they do not look equivalent. Are there some known equivalences between these notions ? or maybe inclusion of some of them into others ? I would also be interested in examples that can be covered by one notion and not by the others... Another way to put it: each notion of n-categorical diagram in the papers above define a class of polygraph, what are there known inclusion/equality between these classes. REPLY [5 votes]: The answer above by Simon does address Steiner's much newer and immensely more computable structure, the Augmented Directed Complexes, but he doesn't really cover its benefits over all of the previous methods. ADCs have been used to extend the lax/oplax tensor product to strict $\omega$-categories as well as the lax/oplax join and its various slice adjoints by Dimitri Ara and Georges Maltsiniotis because working with chain complexes of free modules is, all told, pretty easy. It reduces a lot of the combinatorics of pasting to doing linear algebra in a chain complex of free abelian groups with a particular basis. The morphisms are just morphisms that preserve the free commutative monoids of 'positive elements' generated by the basis. Steiner does dip back into some of his earlier directed complex work in order to prove that Augmented Directed Complexes with a unital loop-free basis embed fully and faithfully into strict $\omega$-categories and are dense in it. This answer is mostly supplementary to Simon's but if you're looking into working with pasting diagrams, I would highly highly recommend you look into the work of Steiner. Also important is the fact that in Steiner's case, because the embedding is full and faithful, you can find 'relevant' elements as the images of maps from the ADC associated with an $n$-globe. Edit: Avoid Steiner's 1993 version and read the 2004 version. They are (almost) totally unrelated. Basically, the 1993 version sort of describes the image of the embedding directly. It is not really useful for computation.<|endoftext|> TITLE: Monoid of continuous self-maps of (real) surfaces QUESTION [5 upvotes]: Let $S$ be a closed surface of genus $g > 0$ and $[S,S] = Hom(\pi_{1}(S),\pi_{1}(S))$ be the monoid of (homotopy classes of) continuous maps from $S$ to itself. Consider the semi-group $A$ of elements that induce the $0$ map on $H_{2}(S,\mathbb{Z})$. Question. Is there a "nice" set of generators for $A$? (Maybe we quotient by natural conjugacy action of the homeomorphisms of $S$.) The map on $\mathbb{S}^{1} \times \mathbb{S}^{1}$, $(x,y) \rightarrow (x,1)$ would be an example of such a map on the torus, I guess that this generates up to conjugacy? EDIT More generally, we can generate many classes in $A$ by picking any continuous mapping $f: S \rightarrow \mathbb{S}^1$ (of which there are many different classes; precisely $\mathrm{Hom}(\pi_{1}(S),\mathbb{Z})$), and choosing some continuous mapping $\mathbb{S}^{1} \hookrightarrow S$. More specifically, do maps of the above form generate $A$? (This seems quite unlikely at the algebraic level, but if there is a counterexample I would be interested to see a geometric realisation of the map as well) REPLY [2 votes]: Here's some sort of geometric description of maps of degree 0. I don't know how realistically you can find a generators-and-relations description of the semigroup of such maps. It is a theorem due to Kneser that I learned from this answer that any map of degree 0 between closed oriented surfaces is homotopic to a non-surjective map. So $\Sigma_g \to \Sigma_h$ factors through the 1-skeleton up to homotopy. You're reduced to asking for a classification of maps $\Sigma_g \to \vee_1^{2h} S^1$ up to homotopy (which again is equivalent to the algebraic statement about maps on the fundamental group). The largest free group $\pi_1(\Sigma_g)$ surjects onto is $F_g$. To see this, look at the induced map on cohomology of the corresponding Eilenberg MacLane spaces; the fact that the map is a surjection implies that $\Bbb Z^k \cong H^1(F_k) \to H^1(\Sigma_g)$ is injective. The cup product must be zero on the image; because it's a nondegenerate bilinear form on $H^1(\Sigma_g)\cong \Bbb Z^{2g}$, the image can be at most $g$-dimensional. Dualizing, we see that when we have a surjection $\pi_1(\Sigma_g) \to F_g$, the kernel of the map on $H_1$ is $\Bbb Z^g$, which furthermore can be given a basis of pairwise disjoint simple closed curves with connected complement. I'd like to be able to say that this is true further at the level of fundamental groups, so that (after precomposing with a diffeomorphism of the domain) every map $\Sigma_g \to \Sigma_g$ of degree 0 extends over the handlebody $H_g$. (Note that algebraically this semigroup is not at all generated by the maps that factor through $S^1$ - the image of the fundamental group of $gf$, where $f$ is such a map, necessarily has image a quotient of $\Bbb Z$.) But I don't know how to get the statement at the level of fundamental groups, though I think it's true.<|endoftext|> TITLE: Identifying a group without 2-torsion QUESTION [11 upvotes]: Suppose we have a finitely presented group $G$ with solvable word problem. (For instance, the command RWSGroup in Magma terminates giving us a finite [but possibly gigantic] rewrite system.) Is there then an algorithm to determine whether $G$ has an element of order 2? If so, where is the best place to find such an implementation? If not, what more needs to assumed? What standard methods exist to disprove the existence of 2-torsion? Edited to add: Here is a specific example with three generators and four relations. Take $$ G=\langle x,y,z\ :\ xy^{-1}x^{-1}y^{-1}z^{-2}=1,\ x^{-1}z^{-2}xz^{-2}=1,\\ xyx^{-1}yz^{-2}=1,\ y^{2}xzx^{-1}z^{-1}=1\rangle. $$ I can prove that this group is not a unique product group, but I wish to show it is torsion-free. I can show it has no odd torsion. Any advice? REPLY [11 votes]: Using a mixture of computation and thought I believe that I have established that this group is indeed torsion-free. I don't know of any general approach to solving that particular problem. Even if the group is hyperbolic (which this example is not, because it has free abelian subgroups of rank $2$), I am not aware of any practical implementable algorithm for deciding torsion-freeness. Here is some Magma code for this problem. Following Pace Nielsen's suggestion that there might be an easily identifiable subgroup of index $32$, I found such a subgroup $K$ of index $4$, and we can compute a presentation of it on its four defining generators. > G:=Group y^2*x*z=z*x >; > K := sub; > Index(G,K); 4 > Rewrite(G,~K); > K; Finitely presented group K on 4 generators Index in group G is 4 = 2^2 Generators as words in group G a = x^2 b = z^2 c = x * z * y^-1 d = y^2 Relations (c^-1, a) = Id(K) (a^-1, b) = Id(K) (a^-1, d^-1) = Id(K) (d^-1, b^-1) = Id(K) (b, c) = Id(K) d * c * b^-1 * d^-1 * c^-1 * b^-1 = Id(K) b^-1 * a * c^-1 * a^-1 * b * c = Id(K) We see that all pairs of generators of $K$ commute except for $c$ and $d$. The final relation collapses completely, and the preceding one is equivalent to $dcd^{-1}c^{-1} =b^2$, so $K$ is a direct product of $\langle a \rangle$ and a torsion-free nilpotent group of class $2$ generated by $b,c,d$. So $K$ is torsion-free. > Transversal(G,K); {@ Id(G), x, y, z @} So a nontrivial torsion-element would have to have order $2$ and be of the form $xk$, $yk$ or $zk$ for some $k \in K$. I spent some time trying to prove that this does not happen, but eventually found an easy way to do it by calculating in a finite quotient. We let $K_2$ be the kernel of the map of $K$ on the to the largest elementary abelian quotient of $K$, which has order $16$, and then check for corresponding elements of order $2$ in $G/K_2$. We see that there are none, so $G$ is torsion-free. > PK, phi := ElementaryAbelianQuotient(K,2); > Order(PK); 16 > K2 := Kernel(phi); > Index(K,K2); 16 > T2 := Transversal(K,K2); > exists{k : k in T2 | (x*k)^2 in K2 }; false > exists{k : k in T2 | (y*k)^2 in K2 }; false > exists{k : k in T2 | (z*k)^2 in K2 }; false<|endoftext|> TITLE: Rigorous definition of the commutator $[a(k_1), a^\ast(k_2)]$ of creation and annihilation operators in boson quantum field models QUESTION [7 upvotes]: In their lecture notes "Boson Quantum Field Models" (in "Mathematics of Contemporary Physics", R.Streater (ed.)), Glimm and Jaffe define an annihilation operator $a(k), k \in \mathbb{R}$ on a certain domain $\mathcal{D}$ in Fock space by \begin{equation} (a(k)\theta)_n(k_1, \ldots, k_n) = (n+1)^{1/2}\theta_{n+1}(k, k_1, \ldots, k_n) . \end{equation} (This is the equation between equations (4.4) and (4.5) in the paper.) They then make the point that the adjoint, $a^\ast(k)$, is not a well-defined operator on Fock space, although it is a quadratic form. However, in equation (4.9) on the next page, they state that \begin{equation} [a(k_1), a^\ast(k_2)] = \delta(k_1 - k_2) , \end{equation} and claim that this commutator ``can be verified directly''. I have no trouble verifying the commutator directly at the level of formal manipulation, but I am having trouble understanding how to make the calculation rigorous, or even how to make rigorous sense of the term $a(k_1) a^\ast(k_2)$ in the commutator, if $a^\ast(k_2)$ is only a quadratic form. Is there a standard way to represent $a^\ast(k_2)$ as an operator on some larger space in order to make sense of the commutator? Failing that, is there some other rigorous interpretation of the commutator? (Glimm and Jaffe do not give an explicit formula for $a^\ast(k)$, but it is easy to derive, and can be found in many other references: \begin{equation} (a^\ast(k)\theta)_n(k_1, \ldots, k_n) = n^{-1/2} \sum_{l=1}^n \delta(k - k_l) \theta_{n-1} (k_1, \ldots, k_{l-1}, \hat{k_l}, k_{l+1}, \ldots k_n) , \end{equation} where $\hat{k_l}$ indicates that $k_l$ is omitted. I have not yet found a reference that explains a rigorous interpretation of the commutator, however.) REPLY [6 votes]: I have the book right in front of me and from a quick glance at the relevant section it seems that GJ chose a somewhat non-standard presentation. I think the most commonly used approach is the one explained by yuggib or suggested by user1504 which involves smearing and understanding everything in some suitable "sense of distributions". However, as you rightly noticed, GJ are really talking about statements which are pointwise in the momentum variables. Your (formal) formula (I know that sounds bad) for $a^*(k)$ which contains a delta function immediately shows that there is a problem with the interpretation as an unbounded operator on $\mathscr{F}$. GJ are well aware of that and mention on page 96 that the domain of this "operator" would have to be $\{0\}$. Then, they (adroitly!) pirouette this by saying that they do not define $a^*(k)$, for fixed $k$, as an operator but rather as a bilinear form $\mathscr{D}\times\mathscr{D}\rightarrow\mathbb{C}$. For $\theta_1,\theta_2\in \mathscr{D}$, the definition of this bilinear form $$ \langle \theta_1, a^*(k)\theta_2\rangle $$ is: compute à la physicist using the "formal formula" and then declare the outcome (which has no delta function) to be the definition. Similarly, $a^*(k)a(k')$ is not the composition of two linear maps (with suitable domains) but a bilinear form defined by the same recipe. I should add that what is really happening behind the scene here is multilinear algebra (the way for instance representation theorists use it, often with the help of diagrammatic algebra for duality pairings or tensor contractions) in infinite dimension. This is what Laurent Schwartz calls Volterra composition in the ultimate reference on the subject: volume 3 and volume 4 of his book on distributions. He also gave a quick summary in his ICM contribution "Théorie des noyaux". I like to call this body of knowledge the Schwartz-Grothendieck Theory because of important input of Alexander Grothendieck needed to make it work, namely, the concept of nuclear space.<|endoftext|> TITLE: Applications of cut locus structure theorems QUESTION [14 upvotes]: I am preparing a presentation discussing the structure of the cut locus on Riemannian and Finsler manifolds. Since my audience will consist mostly of applied mathematicians, I was hoping to include some nice applications to other (non-geometric) areas of mathematics which might appeal to the majority of the audience. While I'm vaguely aware that there are some applications to regularity theory, I do not understand these results. I am therefore looking for a nice introduction to this topic in addition to any other applications which might appeal to a group of applied mathematicians. REPLY [3 votes]: Here is a possibility: Villani, Cédric. "Regularity of optimal transport and cut locus: From nonsmooth analysis to geometry to smooth analysis." Discrete Contin. Dyn. Sys. 30.2 (2011): 559-571. Abstract. In this survey paper I describe the convoluted links between the regularity theory of optimal transport and the geometry of cut locus. Theorem 4.1 (convex Earth). [...] Informally speaking: assume that the Earth is very smooth, but not exactly round (there are hills etc.) The theorem says that if you draw a map of the Earth from one given point, it may not look round, but at least it will look convex"<|endoftext|> TITLE: Open bilinear maps that are not uniformly open QUESTION [15 upvotes]: A map $f\colon X\to Y$ between metric spaces is uniformly open whenever for each $\varepsilon >0$ there is $\delta >0$ such that for any $x\in X$ one has $$B_Y\big(f(x),\delta\big)\subseteq f\big(B_X(x,\varepsilon)\big), $$ where $B_X, B_Y$ denote open balls in the respective spaces. Uniformly open maps have the property that they are surjective as long as the codomain is connected. The exponential function is an example of an open map that is not uniformly open. A continuous, surjective bilnear map $A\colon X\times X\to Y$ between Banach spaces need not be open, contrary to the linear case. (Already matrix multiplication is an example; another example is multiplication in $C[0,1]$ in the case of real scalars; this is due to Fremlin). Is there an example of a continuous, bilinear and surjective map between Banach spaces that is open but not uniformly open? I have a suspicion that it could be already convolution in $\ell_1(\mathbb Z)$ but I am not sure how to approach this (in the latter case I know that it is not uniformly open as $\ell_1(\mathbb{Z}/n\mathbb{Z})$ do not have equi-uniformly open convolutions). Edit (27.06.2017). This question was also asked in a paper by Balcerzak, Behrends, and Strobin (Banach J. Math. Anal. 10 (2016), no. 3, 482-494). I would also welcome examples of $n$-linear maps with this property, where $n$ is arbitrary. REPLY [2 votes]: Recently it was proved that multiplication in the space of functions of bounded variation is open but not uniformly so it is a counterexample to the above question. This also reflects the above comments, which predicted that. Stanisław Kowalczyk, Małgorzata Turowska, Multiplication in the space of functions of bounded variation, Journal of Mathematical Analysis and Applications 472 (April 2019), 696–704.<|endoftext|> TITLE: Why define étale cohomological dimension as it is defined? QUESTION [11 upvotes]: I am learning étale cohomology, and we defined the étale cohomological dimension of a scheme $X$ as the minimum of $n$ where $H_{ét} ^n (X,F)$ vanishes for all the torsion sheaves $F$, yet I don't get why we only restrict our view to torsion sheaves. This seems to be a very stupid question, but I really have no idea why we only talk about torsion sheaves as while discuss étale cohomological dimensions. Is there some kind of fundamental reason or intuition behind this? This may be too elementary, so if it is, please feel free to tell me. REPLY [11 votes]: I think it's a combination of two things: First, we do etale cohomology with torsion sheaves only, because of pathologies with non-torsion sheaves, and so we care about the cohomological dimension with torsion sheaves only. Second, the cohomological dimension is actually different if we include nontorsion sheaves. Thus we lose information about torsion sheaves if we include nontorsion sheaves. Both of these can be demonstrated already with the multiplicative group $\mathbb G_m$. We expect $H^1$ to be one-dimensional by analogy with classical cohomology, but it is easy to calculate $H^1_{et}(\mathbb G_m, \mathbb Z)=0$ by the relationship with the fundamental group, say. So the nontorsion cohomology is bad for $\mathbb G_m$. Next consider the exact sequence $0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/n\mathbb Z \to 0$, which by the previously mentioned vanishing induces an exact sequence $0 \to H^1(\mathbb G_m, \mathbb Z/n\mathbb Z) \to H^2(\mathbb G_m, \mathbb Z) \to H^2(\mathbb G_m, \mathbb Z)$ which, because $H^1(\mathbb G_m , \mathbb Z/n)$ is non-trivial, implies $H^2(\mathbb G_m, \mathbb Z)$ is nontrivial, and hence the cohomlogical dimension allowing nontorsion sheaves is at least $2$, while the cohomological dimension with torsion sheaves only is $1$. Because there is a distinction, we must pick either torsion or nontorsion, and we pick torsion for the better analogies with singular cohomology.<|endoftext|> TITLE: Proportion of polynomials of a fixed degree with a certain number of real roots QUESTION [13 upvotes]: For a polynomial $f(x) = \sum_{i=0}^dc_ix^i \in \mathbb Z[x]$ of degree $d$, let $$ H(f):=\max\limits_{i=0,1,\ldots, d}\{|c_i|\} $$ denote the naive height. Further, define $$ R(M, r, d) := \#\{f(x) \colon \text{$H(f) \leq M$, $\deg f = d$ and $f(x)$ has extactly $r$ real roots}\}. $$ I wonder if anything is known about the quantity $R(M,r,d)$. More precisely, I am interested how it changes as $r$ and $d$ remain fixed and $M$ varies. Has this question been explored at all? I would be thankful for any references. REPLY [8 votes]: The quadratic case can be dealt with as follows. A quadratic polynomial $f(x) = ax^2 + bx + c \in \mathbb{Z}[x]$ has two distinct real roots if and only if $\Delta(f) = b^2 - 4ac > 0$, and a pair of complex conjugate roots if and only if $\Delta(f) < 0$. We now let $a,b,c$ vary in the box $[-X,X]^3$. We first pick a pair $(a,c) \in \mathbb{Z}^2 \cap [-X,X]^2$. If $ac < 0$, that is, if the pair $(a,c)$ lies in two of four quadrants, then $\Delta(f) > 0$; hence 100% of quadratic polynomials with $a,c$ coming from those two quadrants have two real roots. The remaining two quadrants are symmetric to each other, so we might as well consider only the positive quadrant. We can exploit symmetry once more to assume that $a \leq c$, and from density considerations we can carve out the 0-density sets corresponding to $a = 0$ and $a =c$; whence we assume $0 < a < c$. The count of triples $(a,b,c)$ satisfying $0 < a < c \leq X$ and $\Delta(f) < 0$ can be estimated by the triple integral $$\displaystyle \int_1^X \int_1^c \int_{-2\sqrt{ac}}^{2 \sqrt{ac}} db da dc = \frac{8}{9} X^3 + O(X^2).$$ Multiplying by 4 to account for the assumption that $a \leq c$ and $a,c > 0$ (and using standard geometry of numbers arguments), we see that the total number of negative discriminant quadratic polynomials of height at most $X$ is $$\displaystyle N^+(X) = \frac{32}{9} X^3 + O(X^2).$$ The number of positive discriminant forms is then $$\displaystyle N^{-}(X) = 8X^3 - \frac{32}{9} X^3 + O(X^2) = \frac{40}{9} X^3 + O(X^2).$$ One can do a similar (but much more difficult) argument for cubic polynomials (binary forms), by exploiting the fact that for a cubic binary form $g(x,y) \in \mathbb{Z}[x,y]$, its Hessian covariant $q_g(x,y)$ (which is a quadratic form) has discriminant $-3\Delta(g)$; and hence the problem of counting cubic binary forms with three or one real linear factors is reduced to dealing with the Hessians. However, the inequalities involved are no longer linear in general, and hence the application of geometry of numbers methods will be more complicated. Cremona also worked out the exact conditions for quartic polynomials to have 0, 2, or 4 real roots in https://homepages.warwick.ac.uk/~masgaj/papers/r34jcm.pdf I suspect the methods I used above become intractable very quickly with respect to the degree, so perhaps a different formulation is necessary to make progress. Addendum: I should add that the answer to the question is known for degrees 3 and 4 if instead one counts $\operatorname{GL}_2(\mathbb{Z})$-classes of binary forms (of degrees 3 and 4 respectively) with respect to an appropriate $\operatorname{GL}_2(\mathbb{Z})$-invariant height. In particular, when $d = 3$ and we put the height as the discriminant, we have \begin{align*} N_3(X)& = \# \{F = a_3 x^3 + a_2 x^2 y + a_1 xy^2 + a_0 y^3 \in \mathbb{Z}[x,y]: |\Delta(F)| \leq X\} \\ & = \frac{\pi^2}{18} X + O(X^{5/6}), \end{align*} and $N_3^{\pm}(X)$ (which counts the number of forms of bounded positive/negative discriminant, respectively) is given by $$\displaystyle N_3^+(X) = \frac{\pi^2}{72} X + O(X^{5/6}), N_3^-(X) = \frac{\pi^2}{24} X + O(X^{5/6}).$$ The main term was first obtained by Davenport, and the error term as given was obtained by Shintani. Taniguchi and Thorne and Bhargava-Shankar-Tsimerman independently obtained a secondary main term of order $X^{5/6}$. If one includes this secondary term, then the error term is $O(X^{3/4+\varepsilon})$. For the degree 4 case, if we put the height as $H(F) = \max\{|I(F)|^3, J(F)^2/4\}$, as in Bhargava-Shankar (http://annals.math.princeton.edu/2015/181-1/p03), and put $N_4^{(0)}(X), N_4^{(1)}(X), N_4^{(2)}(X)$ for the number of $\operatorname{GL}_2(\mathbb{Z})$-classes of integral quartic forms of height at most $X$ with 0 pairs of complex conjugate linear factors, 1 pair of complex conjugate linear factors, and 2 pairs of complex conjugate linear factors respectively. They showed that $$\displaystyle N_4^{(0)}(X) = \frac{4 \zeta(2)}{135} X^{5/6} + O_\varepsilon(X^{3/4 + \varepsilon}),$$ $$\displaystyle N_4^{(1)}(X) = \frac{32 \zeta(2)}{135} X^{5/6} + O_\varepsilon(X^{3/4 + \varepsilon}),$$ and $$\displaystyle N_4^{(2)}(X) = \frac{8 \zeta(2)}{135} X^{5/6} + O_\varepsilon(X^{3/4 + \varepsilon}).$$ Sorting by a $\operatorname{GL}_2(\mathbb{Z})$-invariant height is likely a more natural question and possibly easier than the original question.<|endoftext|> TITLE: About Riemann-Roch without denominators QUESTION [9 upvotes]: The Riemann-Roch without denominators can be expressed as follows: Let $f: X\rightarrow Y$ be a closed embedding of quasi-projective smooth $k$-varieties of codimension $d$ for some field $k$. Let $E$ be a vector bundle of rank $r$ on $X$. Then we have in $\mathrm{CH}^{\bullet}(Y)$ that \begin{equation*} c_{j}(f_{*}([E])) = \begin{cases} 0, & \text{if $0 TITLE: Is every maximal ideal in a C*-algebra always closed? QUESTION [27 upvotes]: I wonder if every maximal two-sided (self-adjoint) ideal in a C*-algebra is automatically closed. It is a very basic fact of C*-algebra theory that it holds true for the unital case. In the non-unital case, there certainly exists a non-closed dense ideal but the point is that such an ideal may never be maximal. It is a non-trivial fact that the answer is affirmative for the commutative case [D. Rudd, On isomorphisms between ideals in rings of continuous functions. Trans. Amer. Math. Soc. 159 (1971) 335--353]. One can ask a similar question for maximal *-subalgebras. REPLY [7 votes]: Theorem: Let $J\subseteq A$ be a maximal ideal. Then $J$ is hereditary (if $a\in A_+$ satisfies $a\leq b$ for some $b\in J_+$, then $a\in J_+$), strongly invariant (if $x^*x\in J_+$ then $xx^*\in J_+$) and if $a\in J_+$ then $a^t\in J_+$ for every $t>0$. We need some notation and some lemmas first: We define the relation $\sim$ on $A_+$ by setting $a\sim b$ if there exists $x\in A$ such that $a=xx^*$ and $b=x^*x$. Lemma 1: If $a\leq b\sim b'$, then there exists $a'$ such that $a\sim a'\leq b'$. Proof: Let $x\in A$ such that $b=x^*x$ and $b'=xx^*$. Let $x=v|x|$ be the polar decomposition in $A^{**}$. Set $y:=va^{1/2}$ and $a':=yy^*$. Then $y$ belongs to $A$ and we have $a=y^*y$ and $a'=yy^*=vav^*\leq vbv^*=b'$. Lemma 2: Let $a,b,c\in A_+$ and $t>1$ such that $a\sim b\leq c^t$. Then there exists $y\in A$ such that $a=ycy^*$. Proof: Choose $x\in A$ such that $a=xx^*$ and $b=x^*x$. Set $\alpha:=\tfrac{1}{2t}$. Then $0<\alpha<1/2$. Applying the polar decomposition in C*-algebras (see Proposition~II.3.2.1 in Blackadar's Operator algebras book) we obtain $y\in A$ such that $x = y(c^t)^\alpha$. Then $$ a = xx^* = yc^{2t\alpha}y^* = ycy^*. $$ Proof of Theorem: We consider the ideal as suggested by @Black $$ K := \{ a\in A : a^*a \leq b \text{ for some } b\in J_+ \}. $$ We will show that $J=K$. Indeed, as noted by @Black, we either have $J=K$ or $K=A$. We show that $K=A$ leads to a contradiction. So assume that $K=A$. Let $a\in A_+$. Since $a^{1/4}\in K$, we obtain $b\in J_+$ such that $a^{1/2}\leq b$. Then $$ a = a^{1/4}a^{1/2}a^{1/4} \leq a^{1/4}ba^{1/4} \sim b^{1/2}a^{1/2}b^{1/2} \leq b^2. $$ By Lemma 1, we obtain $a'$ such that $a\sim a'\leq b^2$. It then follows from Lemma 2 that $a\in J_+$. Thus $J=A$, a contradiction. With an argument as in the answer of @Black, it now follows that $J$ is hereditary, strongly invariant, and closed under roots of positive elements.<|endoftext|> TITLE: Analogue of Infinitesimal Schwarzian for holomorphic $(G,X)$-manifolds QUESTION [6 upvotes]: If $\Sigma$ is a closed Riemann surface, a complex projective structure $Z$ on $\Sigma$ is an atlas of charts, refining the holomorphic atlas, with values in $\mathbb{CP}^{1}$ such that the transition functions are locally projective transformations in ${PSL}(2,\mathbb{C})$ acting by Mobius transformations. Associated to such a structure is a short exact sequence of sheaves \begin{align} 0\rightarrow \mathcal{P}_{Z}\rightarrow \Theta_{\Sigma}\rightarrow K_{\Sigma}^{2}\rightarrow 0 \end{align} where $\mathcal{P}_{Z}$ is the sheaf of projective vector fields (i.e. those vector fields which in a local chart for the projective structure are identified with an infinitesimal Mobius transformation), $\Theta_{\Sigma}$ is the holomorphic tangent sheaf of $\Sigma,$ and $K_{\Sigma}^{2}$ is the tensor square of the canonical sheaf. The first map is given by inclusion. In a projective coordinate, the second map takes a local holomorphic vector field $v(z)\frac{\partial}{\partial z}$ to $v'''(z) dz^{2}.$ Since projective vector fields, expressed in a projective coordinate, are exactly the quadratic vector fields, the above sequence is left exact and it's not too hard to see that the latter map is surjective. Of course, we know there is some exact sequence of the kind above without ever identifying what the quotient sheaf and quotient map are explicitly: the key here is that we can explicitly identify the quotient sheaf associated to the injection $0\rightarrow \mathcal{P}_{Z}\rightarrow \Theta_{\Sigma}$ with the square of the canonical bundle, a fact which underlies much of the structure of the space of complex projective structures on $\Sigma.$ For those that are somewhat familiar with this story, one can concisely assert that the above described map is the infinitesimal Schwarzian derivative, which explains the title of the question. I am curious is there is a uniform way to extend this picture to holomorphic $(G,X)$-structures on a given compact complex manifold $M.$ To recall, let $G$ be a complex Lie group acting transitively on a complex manifold $X.$ If $M$ is a complex manifold, a holomorphic $(G,X)$-structure on $M$ is an atlas of charts, refining the holomorphic atlas, with values in $X$ and whose transition functions are locally in $G.$ Now, let $W$ be a holomorphic $(G,X)$-manifold with underlying complex manifold $M.$ If $\mathcal{G}_{W}$ denotes the sheaf of $G$-vector fields, that is, the sheaf whose sections are vector fields which in a local $(G,X)$-chart are identified with the infinitesimal $G$-action on $X,$ then there is a short exact sequence \begin{align} 0\rightarrow \mathcal{G}_{W}\rightarrow \Theta_{M}\rightarrow Q\rightarrow 0 \end{align} where $Q$ is the corresponding quotient sheaf. $\textbf{My question is the following}:$ As with the case of projective structures on a Riemann surface, can one explicitly identify the quotient sheaf and the quotient map? For instance, is it a locally free sheaf? The kind of thing I have in mind is the following: I want there to be a differential operator acting on the Pseudo-group of locally defined biholomorphisms of $X$ whose Kernel is precisely $G,$ maybe there is some issue with $G$ not acting effectively but let's ignore that for the moment. The linearization of this operator should be an operator on holomorphic vector fields on $X$ whose kernel consists of the vector fields given by the infinitesimal $G$-action. Since the transition functions of the $(G,X)$-structure are given by elements of $G,$ this data should port over to the $(G,X)$-manifold $W$ and produce a sheaf map from $\Theta_{W}$ to some other sheaf which seems like it should be some kind of jet bundle. If one is familiar with Schwarzian derivative, this is exactly what is happening in the example of complex projective structures. I've found some things in the literature that are in this spirit, but they mostly seem quite ad-hoc, and I am really trying to understand if there is some uniform way to understand this question using the geometry of $X$ as a $G$-homogeneous space. I could go on more, but this post is already long, so please let me know if more information would be useful. As always, thank you for any help you can provide. REPLY [3 votes]: To add to what Ben said: Typically the resolution of the sheaf mapping ${\frak{g}}\to\Gamma(TX)$ is the Spencer resolution, which is a canonical $G$-equivariant exact sequence of differential operators $D_i:\Gamma(V_i)\to \Gamma(V_{i+1})$ for $0\le i< n = \dim X$, where $V_0 = TX$ and $\mathrm{ker}(D_0) = {\frak{g}}$. When you have a flat $(G,X)$-geometry on a manifold $M$, this sequence pulls back (under the local charts) to define the corresponding sequence on $M$. For example, when $G$ is the group of isometries of a (simply-connected) $3$-dimensional Riemannian manifold $X$ of constant curvature, one finds that the ranks $r_i$ of the bundles $V_i$ over $X$ are $$ (r_0,r_1,r_2,r_3) = (3,6,6,3) $$ and that the order of the differential operators $D_0$ and $D_2$ is $1$ while the order of the differential operator $D_1$ is 2. When $G$ is the group of conformal transformations of compactified flat $3$-space~$X = S^3 = \mathbb{R}^3\cup\{\infty\}$, then the ranks are $$ (r_0,r_1,r_2,r_3) = (3,5,5,3) $$ and the order of the differential operators $D_0$ and $D_2$ is $1$ while the order of the differential operator $D_1$ is 3. As another explicit example, if you take $G$ acting on $X = G$ by left-translation, then the Lie algebra of symmetries is the right-invariant vector fields on $G$, and there is a unique (flat) connection $\nabla$ on $TG$ whose parallel sections are the right-invariant vector fields. In this case, $V_i = TG\otimes \Lambda^i(T^*G)$ and $D_i$ is just the $\nabla$-twisted exterior derivative $\mathrm{d}^\nabla:TG\otimes \Lambda^i(T^*G)\to TG\otimes \Lambda^{i+1}(T^*G)$, so $r_i = n{n\choose i}$ and the order of $D_i$ is $1$ for all $i$. Look in any book that treats the Spencer resolution or the topic of Lie equations for references. (I'm traveling, so I don't have access to the literature.) One general purpose reference that treats this topic from a much more general point of view (and contains the important references to the earlier literature from the 50s, 60s, and 70s) is Chapter X of the book Exterior Differential Systems by Bryant, Chern, Gardner, Goldschmidt, and Griffiths (Mathematical Sciences Research Institute Publications, volume 18, 1991). It has long been out of print, but you can download a copy of the whole book from http://library.msri.org/books/Book18/MSRI-v18-Bryant-Chern-et-al.pdf. [Chapters IX and X can be read independently from the first eight chapters.] (Caution: There are many misprints, but there was never a second edition published, so the reader has to be somewhat careful.)<|endoftext|> TITLE: Fascinating moments: equivalent mathematical discoveries QUESTION [63 upvotes]: One of the delights in mathematical research is that some (mostly deep) results in one area remain unknown to mathematicians in other areas, but later, these discoveries turn out to be equivalent! Therefore, I would appreciate any recollections (with references) to: Question. Provide pairs of theorems from different areas of mathematics and/or physics, each proven with apparently different methods, and later the two results were found to be equivalent? The quest emanates from my firm belief that it is imperative to increase our awareness of such developments, as a matter of collective effort to enjoy the value of all mathematical heritage. There is always this charming story about the encounter between Freeman Dyson and Hugh Montgomery. REPLY [7 votes]: My personal favorite example of this phenomenon occurred in 1931. On one hand, Dirac published the paper Quantized Singularities in the Electromagnetic Field which hypothesized the existence of what we would now call a magnetic monopole, and observed using quantum mechanics that the existence of such an object would necessarily imply that electric charge is quantized, meaning it can only occur in integer multiples of some base unit of charge. On the other hand, Heinz Hopf published the paper Über die Abbildungen der dreidimensionalen Sphäre auf die Kugelfläche which introduced what we now know as the Hopf fibration $S^3 \to S^2$ as a nontrivial element of $\pi_3(S^2)$. This came as a surprise since most topologists at the time, based on their experience with $\pi_1$ and with homology, assumed that $\pi_k(S^n) = 0$ if $k > n$. These were both shocking results in their fields that opened up huge areas of future research. There was no reason to believe at the time that the results were connected in any way, but with the advent of gauge theory much later we now know that they were secretly about the same thing. The idea is that the monopole that Dirac wrote about would correspond to a connection on a prinicipal $S^1$-bundle over $S^2$. Monopoles of different charges would correspond to connections on different choices of principal $S^1$-bundle, and so Dirac's observation that electric charge would necessarily be quantized corresponds to the topological fact that principal $S^1$-bundles over $S^2$ are classified up to isomorphism by the group $\pi_3(S^2)$, and by Hopf's results this group is isomorphic to $\mathbb{Z}$ with the Hopf fibration as the generator. REPLY [2 votes]: Brouwer's fixed point theorem and the "Hex theorem". To quote Wikipedia: "John Nash was the first to prove (c. 1949) that Hex cannot end in a draw, a non-trivial result colloquially called the 'Hex theorem', which we now know is equivalent to the Brouwer fixed-point theorem. Apparently, he didn't publish the proof. The first exposition of it appears in an in-house technical report in 1952, in which he states that 'connection and blocking the opponent are equivalent acts'. The earliest published proof of the determinacy of Hex, by David Gale in 1979, also showed that it can be used to prove the two-dimensional Brouwer fixed-point theorem, and that the determinacy of higher-dimensional variants proves the fixed-point theorem in general."<|endoftext|> TITLE: Flat metrics on $n$-toruses, their systoles, and the "shortest vector problem" QUESTION [6 upvotes]: Apologies if this is too basic, but I haven't been able to find any info about the question. Is there anything known about the moduli space of flat metrics on an $n$-torus (i.e., $(S^1)^n$)? Information about the moduli space of lattices in families of Lie groups other than $\mathbb{R}^n$ would be of interest as well. (But I want to stress I am curious about families of Lie groups with the dimension going to infinity rather than a fixed-dimensional example.) Ultimately, I'm interested in any information about the systoles of flat tori (i.e., the length of the shortest non-contractible loop). I do know the Gromov systolic inequality but was wondering what more is known (esp. since the Gromov inequality is straightforward in the flat case.) I am asking (and the question has the computational-complexity tag) because I've been reading about the "shortest vector problem"*, and it struck me that it is equivalent to finding the systole of a flat torus, and so I was curious about any understanding of the problem from this perspective. (I guess this equivalence should predict that there can't be too much understanding of flat metrics on tori since the shortest vector problem is NP hard.) *The shortest vector problem is: Given the generators of a Euclidean lattice, find the shortest (WRT the Euclidean norm say) element in the lattice. REPLY [6 votes]: The moduli space of $n$-tori (of volume 1) is the locally symmetric space (orbifold) $SO(n)\backslash SL(n,\mathbb{R})/SL(n,\mathbb{Z})$. The symmetric space is also the space of positive-definite symmetric matrices of determinant 1, and $SL(n,\mathbb{Z})$ acts on these as the symmetric square representation (i.e. by reparameterizing quadratic forms). Various things are known about its spine (the well-rounded retract), and the systole is known to be a topological Morse function. But really when you are asking about systoles of tori, you are asking about lattice sphere packings, of which much is known. Given a systole of length $l$, there exists a ball at any point in the torus whose radius is $l/2$, and whose interior is embedded, and conversely. Hence lattices with maximal sphere packings achieve the maximal systole for volume 1. In a given dimension, this is equivalent to determining Hermite's constant. The optimal lattice packings are known in dimensions $\leq 8$ and $24$. Check out the encyclopedic Conway, J.H.; Sloane, N.J.A., Sphere packings, lattices and groups. With additional contributions by E. Bannai, R. E. Borcherds, J. Leech, S. P. Norton, A. M. Odlyzko, R. A. Parker, L. Queen and B. B. Venkov., Grundlehren der Mathematischen Wissenschaften. 290. New York, NY: Springer. lxxiv, 703 p. (1999). ZBL0915.52003.<|endoftext|> TITLE: $(\kappa, \kappa, 2)$-saturated ideals? QUESTION [6 upvotes]: Is it consistent to have a $(\kappa,\kappa,2)$-saturated ideal $I$ on $\kappa$ that is $\kappa$-complete and $\kappa$ is not weakly compact? Here $\kappa$ is inaccessible. An ideal is $(\kappa,\kappa, 2)$-saturated, if for any collection $\{A_i: i<\kappa\}\subset I^+$, there exists a sub collection of size $\kappa$ such that any two elements have $I$-positive intersection. In other words, $P(\kappa)/I$ is $\kappa$-Knaster. The motivation comes from Kunen's result on the consistency of $\kappa$-saturated $\kappa$-complete ideal on inaccessible $\kappa$ but $\kappa$ is not weakly compact. In the model, $P(\kappa)/I$ is equivalent to forcing with a Suslin tree, so it's not $\kappa$-Knaster. Further restriction, if $\mathbb{P}$ is $\kappa$-Knaster and $\Vdash_{\mathbb{P}} \kappa$ is weakly compact, then $\kappa$ is weakly compact in $V$. So if the answer to the question is yes, $\kappa$ must not be weakly compact after forcing with $P(\kappa)/I$. A more general question: forget about weak compactness, is it consistent to have a $(\kappa,\kappa, 2)$-saturated ideal at an inaccessible at all? REPLY [2 votes]: Sorry, I missed the inaccessible there. Assuming you want an atomless forcing here, I think this is impossible. For let $I$ be an atomless $\kappa$-additive $\kappa$-Knaster ideal on $\kappa$. Contruct a tree $T = \langle t_{\sigma} : \sigma \in 2^{< \kappa}\rangle$ of $I$-positive sets as follows. At successor step, if $t_{\sigma}$ is $I$-positive, $t_{\sigma0}, t_{\sigma1}$ split it into $I$-positive sets. At limits take intersection. Some branches could die because of $I$-null intersection but inaccessibility of $\kappa$ lets the construction run for $\kappa$ levels. This gives us a $\kappa$-tree with no branch (otherwise $\kappa$-cc is violated). But this is impossible as we argued that $\kappa$ had the tree property.<|endoftext|> TITLE: Invertibility of element in $K(X)$ QUESTION [5 upvotes]: If $\xi$ is a virtual bundle of virtual dimension $1$ in the ring $K(X)$, where $X$ is a compact Hausdorff topological space and $K$ stays for complex topological $K$-theory, then is $\xi$ invertible in the ring $K(X)$ ? REPLY [7 votes]: Yes. Any rank zero element x in K(X) is nilpotent by https://ncatlab.org/nlab/show/virtual%20vector%20bundle, hence 1+x is invertible.<|endoftext|> TITLE: Finiteness properties of mapping class groups QUESTION [5 upvotes]: Question: Is it known if the mapping class groups (of surfaces of finite type) are similar to Gromov-hyperbolic groups in the following senses: 1) Does every finite generating set give us a finite presentation? 2) Are there finitely many cone types with respect to any (some) finite presentation? Definition: A group is called hyperbolic if its Cayley graph is Gromov-hyperbolic, i.e., triangles are $\delta$-thin for some positive $\delta$. Definition: Let $G$ be a group with a finite generating set $S$ and let $g \in G$. The cone type of $g$ w.r.t. $S$ is the following set: $ \mathcal{C}(g) = \{ h \in G | \hspace{2mm} d(e,gh) = d(e,g)+d(e,h)\}$ ,where $d(.,.)$ shows the distance in the Cayley graph w.r.t. to the generating set $S$. REPLY [2 votes]: It looks like question 1) and 2) has been answered, but in case you're still wondering generally "in what ways are mapping class groups similar to Gromov-hyperbolic groups?", you may be interested in reading about hierarchically hyperbolic spaces, introduced by Jason Behrstock, Mark F. Hagen, Alessandro Sisto in this paper: Hierarchically hyperbolic spaces I: curve complexes for cubical groups. For a less technical overview you can also check out Sisto's blog post. (I heard about everything here from Jacob Russel's talk at GSCAGT.)<|endoftext|> TITLE: Relation between Morse Theory and integration against Euler Characteristic QUESTION [10 upvotes]: I'm studying Robert Ghrist papers on integration against Euler Characteristic. I am particularly interested in the relation with Morse Theory. I am trying to understand the proof of Theorem 25.1 (page 52) of this preprint, which I reproduce in an extended version (the parts I understand): Theorem. Suppose $h\colon M\to \mathbb{R}$ is a function integrable against Euler characteristic and a Morse function on the closed (compact without boundary) manifold $M$. Then: $$\int_M [h] \: d \chi=\sum_{p\in Cr(h)}(-1)^{n-\mu(p)}h(p)$$ where $Cr(h)$ is the set of critical points of $h$. Remark. We follow the usual conventions of Milnor's book in Morse theory. The proof as I understand it. First we use a previous result (Proposition 24.8 of the same paper) which says that: $$\int_M [h] \: d \chi=\int_{s=0}^{\infty}\chi\{h\ge s\}-\chi\{h<-s\}\: ds.$$ Now, using Theorem 3.2 from Milnor's book I know that $\chi\{h\le s\}$ is piecewise constant and only changes at critical values of $h$. Moreover, the change is due to the addition of a cell of the dimension of the Morse index of that critical value. So, the change in the Euler Characteristic at a critical value $s=h(p)$ is: $$\chi\{h\le s + \epsilon\}-\chi\{h\le s - \epsilon\}=(-1)^{\mu(p)}.$$ At this point I get lost. So I insert as a picture the proof given in the paper, and a similar proof of the same result in another preprint(page 8): Any help would be appreciated! By the way, there is a proof which involves Stratified Morse Theory, but I am interested in this one using elementary methods. REPLY [2 votes]: I glanced at the paper briefly. Let me try to explain what I understand. Let us suppose that M is compact and without boundary. Let $f: M \to \mathbb{R}$ be a Morse function. Let us further suppose for simplicity that $f$ is positive on $M$, that is it only takes positive values. We could always translate $f$ to make this the case, since $M$ is compact. In that case $\{h< -s\} = \emptyset$ for all non-negative $s$. So the Euler characteristic of that portion is always zero, so by prop 24.8 we then have $$\int_m h [d \chi] = \int_{s=0}^\infty \chi(h \geq s) ds$$ Let $p_i$ be the critical points of the Morse function. And assume for simplicity that these take distinct critical values $h(p_i)$. What does $\chi(h \geq s)$ look like? It starts at $s=0$ with the Euler characteristic of the manifold $\chi(M)$. Then it is locally constant with jumps at exactly the critical values $s = h(p_i)$. These jumps are by one unit up or one unit down according to the index via $(-1)^{\mu(p_i) + 1}$. The extra factor of $-1$ is because we are taking away the handle as $s$ increases. In other words we can write this as $$ \chi(h \geq s) = \chi(M) + \sum_{i, h(p_i) < s} (-1)^{\mu(p_i + 1)}g_{[h(p_i), \infty)} $$ where $g_{[h(p_i), \infty)}$ is the "jump function" which takes value one on the interval $[h(p_i),\infty)$ and zero otherwise. It is useful to fix a large $T > \max\{h(p_i)\}$ and consider the definite integral, which we can compute: $$\int_{s=0}^T \chi(h \geq s) ds = T \cdot \chi(M) - \sum_i (-1)^{\mu(p_i)} \int_{s=0}^T g_{[h(p_i), \infty)} ds $$ $$= T \cdot \chi(M) -\sum_i (-1)^{\mu(p_i)} (T - h(p_i))$$ $$ =T \cdot \chi(M) - T \cdot (\sum_i (-1)^{\mu(p_i)} ) + \sum_i (-1)^{\mu(p_i)} h(p_i) $$ The first two terms cancel and this gives the second claimed formula 25.2. The first one is the same argument with the function $h$ flipped over.<|endoftext|> TITLE: Constructive Central Limit Theorem QUESTION [5 upvotes]: Background: Let $\{a_i\}_{i=1}^n$ be i.i.d. random variables with zero-mean and unit variance, on a probability space $\Omega$. Define $$s_n=\frac{1}{\sqrt{n}}\sum_{i\leq n} a_i$$ Central limit theorem says that the distribution of $s_n$ converges to the standard normal ${\cal{N}}(0,1)$. Short question: For each $n$, give me a Gaussian random variable $g_n$ that is close to $s_n$. Rigorous question: Fix the number $n$. Can we construct an ${\cal{N}}(0,1)$ distributed random variable $g_n:\Omega\to\mathbb R$ so that $g_n,s_n$ are close in some sense. For instance, for some $\alpha>0$, we want $${\mathbb{E}}[(g_n-s_n)^2]={\cal{O}}(n^{-\alpha})$$ REPLY [2 votes]: Did you try given $g_n$ to let $s_n=F_{s_n}^{-1}F_{g_n}(g_n)$ where $F_X$ is the c.d.f. of $X$?<|endoftext|> TITLE: Vector Bundles over Stunted projective spaces QUESTION [5 upvotes]: Is the stunted complex projective space $\mathbb{C}P^{n}/\mathbb{C}P^{m}$ coreducible for $m= 1,3$ and $n\geq m+2$ ? $\mathbb{C}P^{n}/\mathbb{C}P^{m}$ is said to be coreducible if $\exists$ a map $f:\mathbb{C}P^{n}/\mathbb{C}P^{m}\rightarrow S^{2m+2}$ such that the composition of $f$ with the inclusion $i:S^{2m+2}\hookrightarrow \mathbb{C}P^{n}/\mathbb{C}P^{m}$ is of degree one. Also is it true that the stunted quaternionic projective space $\mathbb{H}P^{n}/\mathbb{H}P^{1}$, $n\geq 3$ coreducible? REPLY [4 votes]: If $n\ge 2m+2$ then the generator of $H^{2m+2}({\mathbb C}P^n/{\mathbb C}P^m)$ has a non-trivial cup square, which easily implies that the space is not co-reducible. This leaves open the cases when $(m,n)$ is one of the following pairs $(1,3), (3,5), (3,6)$, or $(3,7)$. Let us consider them one by one. Consider the action of mod $3$ Steenrod algebra on $H^*({\mathbb C}P^\infty; {\mathbb Z}/3)\cong {\mathbb F}_3[x]$. An easy calculation shows that ${\mathcal P}^1(x^4)=x^6$. It follows that in ${\mathbb C}P^{6}/{\mathbb C}P^3$ and ${\mathbb C}P^7/{\mathbb C}P^3$ the bottom cell (in dimension 8) is connected to the cell in dimension 12 by a Steenrod operation, and so these spaces are not co-reducible. ${\mathbb C}P^3/{\mathbb C}P^1$ and ${\mathbb C}P^5/{\mathbb C}P^3$ are two-cell complexes determined by attaching maps $S^5\to S^4$ and $S^9\to S^8$ respectively. Both sets of homotopy classes $[S^5, S^4]$ and $[S^9, S^8]$ have just one non-trivial element, which is the suspension of the Hopf map $S^3\to S^2$. The cofiber of the Hopf map has a non-trivial action of $Sq^2$. But $Sq^2$ acts trivially on $x^2$ and $x^4$ in the cohomology of ${\mathbb C}P^\infty$, which means that the attaching maps in our cases are nullhomotopic, and so these two spaces are in fact co-reducible. Finally for the quaternionic space you also can use the action of ${\mathcal P}^1$ in the mod $3$ Steenrod algebra to show that the cell in dimension 8 is connected to the cell in dimension 12, and so ${\mathbb H}P^n/{\mathbb H}P^1$ is never co-reducible.<|endoftext|> TITLE: Elements of $L^p$ and nice representatives of equivalence classes QUESTION [8 upvotes]: Considering $L^p$ $( 1 \leq p < \infty)$ as a normed vector space, each element of $L^p$ is actually an Equivalent class. Take $[f] \in L^p $ as an Equivalent class, What is the Nicest possible function $g$ such that $g \in [f]$ (i.e. $g=f$ almost everywhere). By the word Nice you are free to consider any good topological or algebraic property like continuity, differentiability, boundedness, etc... (as many as you can) Second question: Let $N$ denotes the set of such Nice Properties, imposing $N$ as set of conditions on $g$, can one identify $g$ uniquely ? In other words can we rewrite $L^p$ in following way $$L^p = \{ g : R\rightarrow R \cup \{\pm \infty\} \quad | \quad \|g\|_p < \infty ,~g\text{ satisfies }N \}$$ This makes we can think about $L^p$ as a set of nice functions, free of any confusing equivalent classes. Clarification: My intention is to find a set of finite conditions called $N$ , when we imposing them, $g$ is determined uniquely, in order to replace this $g$ by $[f]$ to rewrite $L^p$ as above. Lusin's theorem says one can pick $g \in [f]$ such that $g$ is continuous except on very small set (as small as you want but might have positive measure).This might be helpful but still can't give us the set of condition $N$ ! REPLY [2 votes]: In harmonic analysis one frequently defines a modification of $f$ by taking the new value $f(x)$ as the limiti of averages of $f$ over balls centered at $x$, with radius tending to 0. You do this at every point for which the limit exists. If $f$ is equivalent to a continuous function, this method produces the continuous representative. By Lebesgue differentiation this limit exists at almost every $x$ and coincides with the original value of $f(x)$ for any locally $L^1$ function; of course it remains the question of how to redefine $f$ at points where the limit do not exist; maybe one can set $f=0$ there. I guess you can not do better than this, but I might be wrong since I do not know lifting theory.<|endoftext|> TITLE: Ambiguity in Nicolas' criterion for the Riemann Hypotheis? QUESTION [8 upvotes]: A result of Sole, Planat and Omar's paper, ''Quantum mechanics and the Riemann Hypothesis'', (Theorem 2 with $b=2$), says the RH is equivalent to the statement that for every large enough integer $k$, one has $$\dfrac{\sigma(N_k)}{N_{k}\log\log N_{k}} > \dfrac{6e^{\gamma}}{\pi^2} $$ where $\sigma(u)$ is the sum of divisors function, $\varphi(u)$ the Euler totient function, $\gamma$ the Euler constant and $N_k$ is the product of the first $k$ primes. Multiplying both sides by $ \dfrac{(N_{K})^2}{\sigma(N_k)\varphi(N_k)}$ gives $$\dfrac{N_k}{\varphi(N_k)\log\log N_{k}} > \dfrac{6e^{\gamma}(N_{K})^2}{\varphi(N_k)\sigma(N_k) \pi^2}$$ By Nicolas' criterion, the right hand side should be equal to $e^{\gamma}$, which happens only ''at $k=\infty$''. But for this to make analytic sense, one should consider the left-hand side of the preceding inequality as $k$ approaches infinity, thus the RH is in fact the statement that as $k$ tends to infinity, the limit of the the left hand side of Nicolas' inequality is greater than or equal to $\gamma$ ? REPLY [24 votes]: From the paper you mention and the result of Nicolas, the following three statements are known to be logically equivalent: The Riemann hypothesis. The inequality $\frac{N_k}{\phi(N_k) \log\log N_k} > e^\gamma$ holding for all $k$. (Nicolas's criterion) The inequality $\frac{N_k}{\phi(N_k) \log\log N_k} > \frac{6 e^\gamma (N_k)^2}{\varphi(N_k) \sigma(N_k) \pi^2}$ holding for all sufficiently large $k$. In particular, claim 2 and claim 3 are equivalent. This however does not necessarily imply any relationship between the quantities $e^\gamma$ and $\frac{6 e^\gamma (N_k)^2}{\varphi(N_k) \sigma(N_k) \pi^2}$. This is basically because there is a significant gap between the statistics of arithmetic functions such as $\varphi(N_k)$ and $\sigma(N_k)$ under the assumption of RH, and under the opposite assumption of having a non-trivial zero of zeta off the critical line. This gap allows for the existence of multiple criteria that do not immediately look equivalent to each other, but are nevertheless able to separate the scenarios when RH holds from the scenarios where RH fails. A more familiar example of this phenomenon arises with the well known equivalences of RH with various bounds on the error term in the prime number theorem. Namely, the following three statements are known to be logically equivalent: The Riemann hypothesis. The inequality $|\sum_{n \leq x} \Lambda(n) - x| \leq \frac{1}{8\pi} \sqrt{x} \log^2 x$ holds for all $x \geq 74$. (Schoenfeld's criterion) For every $\varepsilon > 0$, there exists $C_\varepsilon$ such that the inequality $|\sum_{n \leq x} \Lambda(n) - x| \leq C_\varepsilon x^{1/2+\varepsilon}$ holds for all sufficiently large $x$. At first glance, the claim 5 is significantly stronger than claim 6, and if the von Mangoldt function $\Lambda$ was replaced by a completely arbitrary function, one could easily cook up examples in which claim 6 was true but claim 5 was false. But the von Mangoldt function is far from being completely arbitrary, being tied to the zeroes of the zeta function by the explicit formula, and the non-trivial zeroes either all lie on the critical line (in which case one can show claim 5), or there is at least one zero off the critical line (in which case one can contradict claim 6 for suitable choices of $x$ and $\varepsilon$). For an even simpler instance of this sort of gap phenomenon that involves no number theory whatsoever, observe that the following claims are equivalent for any polynomial $P: {\bf R} \to {\bf R}$: $P$ is constant. $P$ is bounded. One has $P(x) = o(|x|)$ as $|x| \to \infty$. Again, claim 8 appears to be stronger than claim 9 (and claim 7 stronger than claim 8), but they are all equivalent, because there is a significant gap between the behaviour of the constant polynomials and the behaviour of non-constant polynomials. (But one should mention, though, that it is not difficult to see that the expression $$ \frac{6 e^\gamma (N_k)^2}{\varphi(N_k) \sigma(N_k) \pi^2} = \frac{e^\gamma}{\zeta(2)} \prod_{p \leq p_k} (1 - \frac{1}{p^2})^{-1} = e^\gamma \prod_{p > p_k} (1 - \frac{1}{p^2}) = e^\gamma + O( \frac{1}{k \log k} )$$ does converge reasonably quickly to $e^\gamma$ in the limit $k \to \infty$. So claim 2 and claim 3 are actually not so different from each other, though still not obviously identical. It's more accurate to say that Nicolas's criterion is flexible, rather than ambiguous - one has a little bit of "wiggle room" in that criterion (or in many of the other known criteria for RH), provided by the above-mentioned gap between the RH and non-RH worlds.)<|endoftext|> TITLE: Bounds on Betti numbers of subvarieties? QUESTION [14 upvotes]: Let's say I have a smooth irreducible subvariety $X$ of $\mathbb{CP}^n$ with some fixed Hilbert polynomial. What are the best bounds known for the sum of the Betti numbers of $X$? That such a bound exists follows from the boundedness of (a component of) the Hilbert scheme. I imagine that one could get a reasonable bound by writing down a suitable Morse function and doing some intersection-theoretic calculation for the number of critical points. Or perhaps more simply, just doing some inductive argument using Lefschetz pencils. I'm unsure what the best way of doing this would be though, and it would be nice if there was already a reference in the literature for this. REPLY [17 votes]: There is a very recent paper by Zak : http://mathecon.cemi.rssi.ru/zak/files/Castelnuovo%20Bounds%20for%20Higher%20Dimensional%20Varieties.pdf which deals with this issue. He has found many new bounds on the total Betti numbers. For instance, he proves that if $X \subset \mathbb{P}^n$ is smooth of degree $d$ and of codimension $a$, then: $$ b(X) \leq \dfrac{d^{\dim X +1}}{a^{\dim X}},$$ provided that the $d \geq 2(a+1)^2$ and that $\dim X \geq 2$. He gives other bounds if the degree is lower. He also proves that these bounds are asymptotically sharp. Note the surprising fact that these bounds only depend on the degree, the dimension and the codimension.<|endoftext|> TITLE: The Mittag-Leffler condition as necessary and sufficient QUESTION [6 upvotes]: Let $A_1\leftarrow A_2\leftarrow A_3\leftarrow\dotsb$ be a projective system of abelian groups with the projection maps $p_{ij}\colon A_j\to A_i$, $j\ge i$. The derived functor of projective limit $\varprojlim_n^1 A_n$ is constructed as the cokernel of the map $$ \mathrm{id}-\mathit{shift}\colon\prod\nolimits_{n=1}^\infty A_n\longrightarrow\prod\nolimits_{n=1}^\infty A_n, $$ where the map $\mathit{shift}\colon\prod_{n=1}^\infty A_n\to\prod_{n=1}^\infty A_n$ takes a sequence $(a_n\in A_n)_{n=1}^\infty$ to the sequence $(p_{n,n+1}(a_{n+1})\in A_n)_{n=1}^\infty$. The kernel of the map $\mathrm{id}-\mathit{shift}$ is the projective limit $\varprojlim_n A_n$. The following condition, called the Mittag-Leffler condition, is sufficient for the vanishing of $\varprojlim_n^1 A_n$. Suppose that for every fixed $m\ge1$ the nonincreasing sequence of subgroups $p_{m,n}(A_n)\subset A_m$ stabilizes for $n$ large enough. Then $\varprojlim_n^1 A_n=0$. The Mittag-Leffler condition is not necessary for the vanishing of $\varprojlim_n^1 A_n$. The simplest counterexample would be $A_n=x^nk[[x]]\subset k[[x]]$, where $k[[x]]$ denotes the right of formal Taylor power series in one variable $x$ over a field $k$ and the maps $p_{i,j}\colon x^jk[[x]]\hookrightarrow x^ik[[x]]$ are the identity inclusions. Then $\varprojlim_n^1 A_n=\varprojlim_n A_n=0$, but the sequence of images of the maps $p_{m,n}$ never stabilizes as a sequence of subgroups in $A_m$. Notice that, replacing the power series with the polynomials in the above example, one obtains the projective system of groups/vector spaces $C_n=x^nk[x]$ and identity inclusions between them, whose derived projective limit does not vanish. In fact, one has $\varprojlim_n^1 C_n=k[[x]]/k[x]$. I believe I can prove the following result, which is of relevance in connection with the weak proregularity condition in the MGM duality theory. Let $(A_n)$ be a projective system of abelian groups. Denote by $(B_n)=\bigoplus_\omega (A_n)$ the direct sum of a countably infinite family of copies of the projective system $(A_n)$. Then the following three conditions are equivalent: the projective system $(A_n)$ satisfies the Mittag-Leffler condition; the projective system $(B_n)$ satisfies the Mittag-Leffler condition; $\varprojlim_n^1 B_n=0$. My question is: this sounds like a too elementary result in a too popular area of algebra to remain unknown till the present time. Is there a reference where a proof of the equivalence of these conditions can be found? REPLY [8 votes]: This is due to Emmanouil as far as I know. See this.<|endoftext|> TITLE: ''Are Hermitian metric pullbacks automatically via biholomorphisms?'' QUESTION [8 upvotes]: The awkward title is an attempt at approximating the following specific question: Let $(M^{2n}, J)$ be a complex manifold, suppose $g_0$ is a Riemannian metric $M$ compatible with $J$, and suppose $\phi_t$ is a one parameter family of diffeomorphisms of $M$ generated by a vector field $X$ such that $\phi_t^* g$ is compatible with $J$ for all $t$. The questions is: must $X$ be a (real part of a) holomorphic vector field in the sense that $L_X J = 0$? Assume that $g_0$ is Kahler and that $X = \nabla f$ for some smooth function $f$. As the family $\phi_t^* g_0$ remains compatible with $J$, it follows that $L_{\frac{1}{2} \nabla f} g_0 = \nabla^2 f$ is of type $(1,1)$. Using the K\"ahler hypothesis one can show that $(\nabla^2 f)^{1,1}(\cdot, J \cdot) = \frac{1}{2} d d^c f$. On the other hand, since the associated K\"ahler form $\omega$ is closed, it follows from the Cartan formula that $L_{\frac{1}{2} \nabla f} \omega = \frac{1}{2} d d^c f$. Thus it follows that $L_{\frac{1}{2} \nabla f} \omega = (L_{\frac{1}{2} \nabla f} g)(\cdot, J \cdot) = L_{\frac{1}{2} \nabla f} \omega + g(L_{\frac{1}{2} \nabla f} J\cdot , \cdot)$, and hence $L_{\nabla f} J \equiv 0$, as required. REPLY [7 votes]: The answer is 'no' in specific cases, and it doesn't help that $g$ is assumed to be Kähler (i.e., you have a mistake in your argument in the second paragraph). For example, let $M^{2n}=\mathbb{C}^n$ and let $g$ be the standard Kähler metric, and assume $n>1$. Then $$ g = \mathrm{d}z_1\circ\mathrm{d}\bar z_1 + \cdots + \mathrm{d}z_n\circ\mathrm{d}\bar z_n\,. $$ Then $g$ is just the standard flat metric on $\mathbb{R}^{2n} = \mathbb{C}^n$, and hence the symmetry group of $g$ that fixes the origin is $\mathrm{O}(2n)$. Let $\phi_t$ be the flow of an orthogonal vector field vanishing at the origin that is not holomorphic. These exist because $\mathrm{SO}(2n)$ is much larger than $\mathrm{U}(n)\subset\mathrm{SO}(2n)$. Then $\phi_t$ actually preserves $g$, so $\phi_t^*g=g$ is $J$-compatible for all $t$, but, in general $\phi_t^*J\not= J$, so the the vector field $X$ generating $\phi_t$ not the real part of a holomorphic vector field.<|endoftext|> TITLE: Reproving a known theorem in an article QUESTION [19 upvotes]: Suppose one wants to use a theorem that was published quite a long time ago (+80 years) in a paper that is using a terminology and notations that are very much out-dated (making the paper very hard to read). Is it okay if we want to reformulate the result as well as the proof in an article using a more modern language (of course giving credit to the original paper)? Or, on the other hand, is it something one shouldn't put in a research paper, since it's not a new result ? What is the general policy for this kind of things ? REPLY [4 votes]: Littlewood's Miscellany includes a quote from somebody's paper saying Part 2 [of the theorem], which is trivial, is due to Hardy and Littlewood which met the requirements of both completeness and acknowledging prior publication of something H&L had earlier also included for completeness, while coming across as vaguely demeaning Your aim should be for usefulness and prior attribution of the result without sounding negative<|endoftext|> TITLE: Phantoms and strongly exceptional collections QUESTION [8 upvotes]: I would like to ask a question about phantom categories and strongly exceptional collections. Let $X$ be a smooth prpjective variety over an algebraically closed field. An admissible subcategory $B \subset D^{b}(X)$ is said to be a phantom category if $K(B) = 0$ but $B \neq 0$ (where $K(B)$ is the K-theory of $B$). I know (for instance from here : https://arxiv.org/pdf/1209.6183.pdf) that there exists smooth projective varieties having an exceptional collection of vector bundles $E_1, \ldots, E_n$ such that $n = \textrm{rank}(K(X))$, but the collection is not full. This implies in particular that the left orthogonal to the category generated by the $E_i$ is a phantom category. I was wondering if there exists examples as above, but with the additional assumption that the collection $E_1,\ldots,E_n$ is strongly exceptional? (that is the higher $\mathrm{Ext}$ between the $E_i$'s vanish in both directions). In all examples I have been reading so far, the exceptional collections of length equal to $\mathrm{rank}(K(X))$ which are not full are never strongly exceptional. Does someone know an example where this could be the case? Thanks in advance for your answers! REPLY [2 votes]: In the case of exceptional collection consisting of line bundles, one can say something about it(at least on surface). Let $X$ be a smooth projective surface with strong exceptional collection of line bundles of maximal length(i.e length $l=rk(K_0(X))$, then $X$ is a rational surface. Then in the case of line bundles, your question becomes whether there is an example of a rational surface admitting a strong exceptional collection of line bundle which is not full. Probably not. L.Hille and M.Perling ("Exceptional sequences of invertible sheaves on rational surfaces", Compositio Math. 147 (2011), 1230-1280) introduced a very natural operations on producing exceptional collection of line bundles on rational surface, called augmentations which is a derived categorical analogue of "blow up". They have the following conjecture: Let $X$ be a rational surface, then any strong exceptional collection of line bundles is coming from augmentations from $\mathbb{P}^2$ and $\mathbb{F}_n,n\neq 1$. Since any exceptional collection of line bundles on $\mathbb{P}^2$ and $\mathbb{F}_n$ is full. Thus if the conjecture of Hille-Perling is true, then any strong exceptional collection of line bundles of maximal length on rational surface is full. The conjecture has been verified for several cases: Toric surface, del Pezzo surface and weak del Pezzo surfaces of degree $K_X^2\geq 3$ and any anticanonical rational surface. Thus, such example in your question can not exist on those varieties. Unfortunately, the conjecture above is NOT true. There is an example of weak del Pezzo surface of degree $2$ admitting a strong exceptional collection of line bundles of maximal length but not coming from augmentation in any sense. However, by a result of Kuleshov, it still full! By a computer algorithm checking, the number of counter examples of above conjecture is really small. Thus one may say, in the most cases, strong exceptional collection of line bundles of maximal length is full. In fact, one is able to show that the so called cyclic strong exceptional collection of line bundles of maximal length is full. I and my collaborators are working to prove that any strong exceptional collection of line bundles on rational surface is full by a completely different method.<|endoftext|> TITLE: Combinations of positive semidefinite matrices QUESTION [8 upvotes]: Let $U$ be some convex subset of $\mathbb{C}^d$, and define the set of square matrices $\mathcal{U} \subset \mathbb{C}^{d \times d}$ to be the set of all convex combinations of $x y^*$ such that $x \in U, y \in U$. (Here $^*$ denotes the conjugate transpose.) Is it always true that if a matrix belongs to $\mathcal{U}$ and is Hermitian and positive semidefinite, then it can be written as a convex combination of terms $x x^*$ with $x \in U$? Edit: The conjecture was shown to be false in the most general case in the answer below, but is it possible to formalise the constraints under which it is true? This is a problem I've encountered when trying to generalize the property that each positive semidefinite matrix with trace 1 can be written as a convex combination of rank-one positive semidefinite terms $x x^*$ with trace 1. I was wondering how general it is and I did not manage to prove this. REPLY [3 votes]: No. If $U$ is a segment between $x$ and $y=ix$, then zero matrix equals $(xy^*+yx^*)/2$ and thus belongs to $\mathcal U$. Well, it is not positive definite, but only non-negative definite. This may be easily fixed by considering a neighborhood of the segment $U$ (instead of $U$) and taking a convex combination of zero and other matrices, which is still too close to zero.<|endoftext|> TITLE: Primary definition of a geodesic QUESTION [12 upvotes]: I am wondering if there is a sense in which one of these definitions for a geodesic on a smooth Riemannian manifold is primary to the other. A geodesic has acceleration zero, i.e., it is self-parallel. A geodesic is locally length-minimizing, i.e., it is stationary relative to variations in arc length. I encountered another characterization—in the Einstein, Infeld, Hoffmann paper, "Gravitational Equations and the Problem of Motion," cited in Shlomo Sternberg's book on Curvature in Mathematics and Physics—which seems far from primary, and made me wonder if there is any natural priority between the two definitions above. Of course, each can be mathematically derived from the other, but I am thinking that, e.g., one might more easily generalize than the other, or somehow be more pure to the spirit of Riemannian geometry. Image below included just for aesthetic appreciation.:-)                     Wikipedia image: "A geodesic on a triaxial ellipsoid." Credit: Charles Karney. REPLY [5 votes]: I think that depending on what is the most fundamental structure you consider in your Riemannian manifold, both answers can be true. Let me explain. In Euclidean geometry, one can consider the affine structure as more fundamental than the metric structure, so in this case the lines are more fundamental than lengths and angles. In general, this would mean that one considers the connection as more fundamental, even if the manifold happens to be Riemannian and the connection happens to be metric. So the answer would be 1. But if the Riemannian manifold comes primarily from a metric space, then one can define curves and lengths of these curves, even if it happens that the topological manifold is also differentiable, and there is a metric tensor giving the same distances as the metric structure. In this case the connection appears only because of the metric, as its associated Levi-Civita connection, so the notion of parallel transport and of self-parallelism are secondary. So in this case the answer is 2. I think there are cases in which the path leading to the Riemannian manifold, the order of the layers of different structures, really matters. One example can be in General Relativity - there are approaches where the causal structure is seen as more fundamental than the metric tensor, which allows the recovery of the metric up to the volume form, but even with only the causal structure much of the geometric and physical properties make sense. Of course we are talking here about the causal structure, which seems more basic than the affine structure, since we only know the null geodesics. I can expect that in general, both in applications of geometry and in abstract mathematical problems, spaces of parameters arise, and they may be endowed with one structure or another at the most basic level, and extra information allows adding the other structures on top of these.<|endoftext|> TITLE: Surjective entire functions QUESTION [7 upvotes]: If I have an entire function give as a power series $f(z)=\sum_{i=0}^{\infty}a_iz^i$, is there a way/technique to check if the function is surjective? Weierstarss factorization theorem gives that $f$ is not surjective if and only if $f=\exp(g)+c$ for some $\require{cancel}\cancel{surjective}$ entire function $g$. However I don't know how to check this. Is there, for example, some known result that gives the surjectivity of $f$ depending on the behavior (rate of convergence?) of the coefficients? This question was asked before, but did not receive an answer at the time. When are entire functions surjective? REPLY [18 votes]: Your statement that if $f$ is not surjective, then $f=e^g+c$ where $g$ is surjective is wrong: $g$ does not have to be surjective. Example: $f(z)=e^{e^z}$. By the way, this example permits an infinite iteration: there is an infinite sequence of entire functions $f_n$ such that all of them are zero-free, and $e^{f_{n+1}}=f_n$. There are various results which imply that $f$ is surjective, you have to be more specific, what kind of criterion you want. In terms of growth of coefficients, if $$\rho:=\limsup_{n\to\infty}\frac{n\log n}{-\log|a_n|}<1$$ then $f$ is surjective. If this is equal to $1$, but $$\sigma:=\limsup_{n\to\infty}n|a_n|^{1/n}=0,$$ then $f$ is surjective. Moreover, if $f$ is not surjective then either $\rho$ is a positive integer and $\sigma$ is non-zero and finite, or $\rho=\infty$. This gives a pretty strong sufficient condition of surjectivity in terms of coefficients. Another types of conditions can be obtained when you know something about $f$ beyond the coefficients. For example, if $f$ has infinitely many zeros, but not too many: $$\limsup_{r\to\infty}\frac{\log n(r)}{\log r}<\rho,$$ where $n(r)$ is the number of zeros in the disk $|z|\leq r$, then $f$ is surjective. This can be very much refined, if needed. One can give very many other sufficient conditions in terms of coefficients, for example if many of the coefficients are $0$ (gap series), and in other terms. For example, one can combine growth conditions with gap conditions. But the only reasonable necessary and sufficient condition is that $f=e^g+c$ with some entire $g$. Reference: B. Ya. Levin books on entire functions (any of the two of them).<|endoftext|> TITLE: Number of zeros of a real analytic function QUESTION [5 upvotes]: Let $f(x,y,t):[-1,1]^3\to \mathbb{R}$ be a real-analytic function. Assume that for any fixed $x,y$, $f(x,y;t)$ is not a constant function $[-1,1]\to \mathbb{R}$. Since the zeros of a non-constant real-analytic function of one variable are isolated, we denote the number of the zeros of $f(x,y;t)$ on $[-1,1]$ by $N(x,y)$. Then do we have the uniform bound $$\sup_{(x,y)\in[-1,1]^2}N(x,y)\le C$$ where $C$ is a constant? I can not find a counterexample such that for some sequence $(x_n,y_n)\in[-1,1]^2$, $N(x_n,y_n)\to \infty$, as $n\to \infty$. REPLY [2 votes]: Let's do this for two variables $x,t$ rather than three for ease of notation. Also, I assume that by real analytic on a compact set, you mean real analytic on some open neighborhood of this set. Then your claim follows because $N(x)$, with the zeros counted according to multiplicity, is upper semicontinuous, so if $N$ were unbounded, then $N(x)=\infty$ somewhere, contrary to your assumption. To see that $N(x)\ge\limsup N(x_n)$ if $x_n\to x$, suppose you have consistently at least $k$ zeros $a_n(1),\ldots, a_n(k)$ for each $x_n$. On a suitable subsequence $a_n(j)\to a(j)$, which already gives us zeros at $x$. The only issue is that some of these limit points could agree, but then just look at the derivative(s) to see that they pick up enough multiplicity to compensate for this. Finally, if the original zeros already had multiplicity $>1$, this won't get lost in the limit. This argument actually doesn't use analyticity; it works the same way for smooth functions.<|endoftext|> TITLE: Closed topological embedding of complex algebraic varieties into a smooth manifold QUESTION [5 upvotes]: In the book "Representation Theory and Complex Geometry” by Ginzburg-Chriss page 93 they claim that (the analytification of) every complex algebraic variety admits a closed embedding in a smooth manifold. They do not really provide a reference for this statement. Why is this true? For quasi-projective varieties, the claim is fairly obvious. But it seems to me that they do not impose this restriction. Moreover, I would say that we need the variety to be separated for this but (I can live with that). REPLY [4 votes]: It took me ages to find this reference (and I actually found it in this post). There is a theorem by Acquistapace–Broglia–Tognoli in An embedding theorem for real analytic spaces stating: Theorem. Let $X$ be a paracompact connected $n$-dimensional analytic space and suppose that $q := \sup_{x \in X} \dim T_x X < \infty$. Then $X$ admits a closed $C^\omega$-immersion into $\mathbb{R}^{n+q}$. This is a generalization of Grauert's embedding theorem for real analytic manifolds. From this we obtain: Corollary. Let $X$ be a separated complex algebraic variety. Then $X^{\mathrm{an}}$ admits a topological closed embedding into $\mathbb{R}^N$ for some $N$. Proof. $X^{\mathrm{an}}$ is a paracompact complex (thus also real) analytic space with finitely many connected components and with bounded tangent space dimensions. Now, apply the theorem. Remark 1. The only thing that surprises me is why the ABT paper has only one citation on MathSciNet since 1979; I don't know if there's another source for the above theorem (I hope it's right actually). Remark 2. I don't know if there isn't a simpler argument.<|endoftext|> TITLE: If an equivariant map is smooth on diagonal matrices, is it smooth everywhere? QUESTION [15 upvotes]: This is a followup from a question I asked on math.SE, which received a helpful answer but unfortunately not a complete one. $\def\Sym{\mathrm{Sym}_{n\times n}}$ $\def\s{\mathrm{Sym}}\def\sp{\s^+}$Let $\sp \subset GL(n,\mathbb R)$ denote the space of symmetric positive-definite $n \times n$ matrices. (It might be more fruitful to think of the full subspace of symmetric matrices, since it turns all the group actions below in to representations.) I am interested in functions $A : \sp \to \sp$ that are equivariant under the natural conjugation action of $O(n)$; i.e. such that$$A(R^T X R) = R^T A(X) R$$ for all $X \in \sp, R \in O(n,\mathbb R)$. Since we can diagonalize any $X$, we know that such an $A$ is determined by its restriction to diagonal matrices, which gives a function $a : (0,\infty)^n \to (0,\infty)^n$ which is equivariant under the natural permutation action of the symmetric group $S_n$. Conversely, any such $a$ can be extended uniquely to an equivariant $A$. Thus we can specify an $A$ by just declaring what it does to eigenvalues. Question: If we know $a$ is smooth, can we conclude $A$ is smooth? In the analogous problem for $O(n)$-invariant scalars $F : \sp \to \mathbb R$ (which reduce to symmetric functions $f : (0,\infty)^n \to \mathbb R$ of the eigenvalues), we can solve this problem using Glaeser's "differentiable Newton's theorem" - we get that a smooth symmetric function of the eigenvalues is a smooth function of the symmetric matrix invariants, which are in turn smooth functions of the matrix itself. The key is that $S_n$-invariant polynomials of the eigenvalues and $O(n)$-invariant polynomials of the matrices are the same thing. Since I couldn't find any similar work on equivariant maps (please relieve me of my ignorance!), my only thought was to use something like this theorem of Schwarz to study the scalar $\tilde A : \sp \times \sp\to \mathbb R$ defined by $$\tilde A(X,Y) = \langle A(X), Y\rangle,$$ which is invariant under the action $\rho_R(X,Y) = (R^T X R, R^T Y R)$. We can define an $S_n$-invariant $\tilde a : (0,\infty)^n \times (0,\infty)^n \to \mathbb R$ similarly; but unfortunately it seems to me that there is no obvious relation between $\tilde a$ and $\tilde A$ - we only have $a(\lambda(X)) \cdot \lambda(Y) = \langle A(X), Y \rangle$ when $X,Y$ have the same eigenvectors. Any pointers would be great - this is a tangent from my usual research, so there is probably a whole body of relevant work I'm unaware of. REPLY [9 votes]: I think, one can argue as follows. Let $D\subseteq\text{Sym}$ be the diagonal matrices. Since $\exp:D\to D^+$ and $\exp:\text{Sym}\to\text{Sym}^+$ are compatible diffeomorphisms it suffices to answer the analogous problem for $D\subseteq\text{Sym}$. For $m=0,\ldots,n-1$ let $c_m:D\to D:(x_i)\mapsto(x_i^m)$. These are certainly $S_n$-covariants. It is well known that they freely generate the space of all polynomial $S_n$-covariants as a module over the ring of all polynomial invariants. I claim that this also holds for smooth functions. In other words, for every smooth covariant $a$ there are smooth invariants $f_0,\ldots,f_{n-1}$ with $$ a=\sum_{m=0}^{n-1}f_mc_m $$ There is probably a general theorem but here is an ad hoc argument which independently shows that the $c_m$ form a basis. Let $a=(a_1,\ldots,a_n)$ be the components. Then we have to solve $$ a_i=\sum_{m=0}^{n-1}f_m x_i^m $$ This is a linear system of equations for the $f_m$ with the Vandermonde matrix as coefficients. So we can uniquely solve it. One gets $f_m=\frac{\tilde f_m}{V}$ where $\tilde f_m$ is smooth and $V$ is the Vandermonde determinant. The equivariance of $a$ implies that $\tilde f_m$ vanishes where $V$ vanishes so $f_m$ is a smooth function. Glaeser's theorem shows that $f_i(x)$ can be extended to a smooth $O(n)$-invariant $F_i(X)$ on $\text{Sym}$. Thus $$ A(X)=\sum_{m=0}^{n-1}F_m(X)X^m $$ is a smooth extension of $a$.<|endoftext|> TITLE: Is $\delta(df \wedge df)=0$ an Euler-Lagrange equation? QUESTION [15 upvotes]: $\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\TM}{\operatorname{TM}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\TM}{\operatorname{T\M}}$ $\newcommand{\TN}{\operatorname{T\N}}$ $\newcommand{\TstarM}{T^*\M}$ Edit: I narrowed the focus of the question. Summary: I suspect $\delta(df \wedge df)=0$ is not an E-L equation because it pus too many constraints on $f$. Can this heuristic be formalized? Let $\M,\N$ be $d$-dimensional oriented Riemannian manifolds, and let $f:\M \to \N$ be smooth; $df \in \Omega^1(\M,f^*{\TN})$, and for $1 < k \le d$ let $\bigwedge^k df\in \Omega^k\big(\M,\Lambda_k(f^*{\TN})\big)$ be the induced map. Notation: $\nabla^{\Lambda_k(f^*{\TN})}$ is the induced connection on $\Lambda_k(f^*{\TN})$, $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} $ is the adjoint of $d_{\nabla^{\Lambda_k(f^*{\TN})}}$. Question: Is $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ an Euler-Lagrange equation of some functional $E(f)$? I suspect the answer is negative. Edit: Can this be proved using known results on inverse problems in calculus of variations? e.g if we can show the equation in the Euclidean setting ($M=N=\mathbb{R}^d$) is not an E-L eq, then we are done. (I am not aware of much work on such inverse problems in general Riemannian settings, but if the Euclidean can be decided, it's enough). Heuristic: $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ puts too many constraints on $f$: $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) \in \Omega^{k-1}(\M,\Lambda_k(f^*{\TN}))$, so locally the "equation" is a system of ${d \choose k-1}{d \choose k}$ scalar equations. However, it seems to me an Euler-Lagrange eq of a functional cannot consist of more than $\dim(f^*\TN)=d$ equations, since (roughly) this is the number of degrees of freedom we have in choosing the variation field $V\in\Gamma(f^*\TN)$. More explicitly, since $dE(V)$ is linear in $V$, it should always be in the form of $dE(V)=\langle A(f),V \rangle$, where $A(f) \in \Gamma(f^*\TN)$ so the E-L eq should be in the form of $A(f)=0$. Of course, this argument gives only an upper bound to the number of "independent" constraints - degeneracies can occur (e.g Null-Lagrangians). So, suppose ${d \choose k-1}{d \choose k}>d$. Is it possible $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ is degenerate and reduces to no more than $d$ independent eqs (locally)? Can we prove this is not happening? (If we can, than this shows our equation is not an E-L eq of any functional). Analysis of the borderline cases: We now turn to see what happens when ${d \choose k-1}{d \choose k}=d$. This happens (for $1 \le k \le d$) iff $k=1$ or $k=d$. The former corresponds to harmonicity. Suppose that $k=d$. We will try to understand if our equation is an E-L equation in this case. For concreteness, let's work with $k=d=2$. I proved in my answer that the E-L equation of the functional $$ E_2(f)=\frac{1}{2}\int_{M} \| \bigwedge^2 df\|^2 \text{Vol}_{M}, \, \text{is}$$ $$h_{f^*\TN} \big( \tr_{\TM}\big( df \otimes \delta_{\nabla^{\Lambda_2(f^*{\TN})}}(\bigwedge^2 df)\big) \bigg)=0.$$ Since $h_{f^*\TN}$ is injective in this case, the E-L eq is equivalent to $$\tr_{\TM}\big( df \otimes \delta_{\nabla^{\Lambda_2(f^*{\TN})}}(\bigwedge^2 df)\big)=0. \tag{1}$$ When restricting the discussion to immersions, this reduces to $ \delta_{\nabla^{\Lambda_2(f^*{\TN})}}(\bigwedge^2 df)=0$. It is not clear what happens in the general case; If we don't assume $f$ is an immersion, are the equations equivalent? (If not, perhaps there is another way to realize $\delta_{\nabla^{\Lambda_2(f^*{\TN})}}(\bigwedge^2 df)=0$ as an E-L eq). The problem is that we cannot conclude (at least not immediately) that at every point where $df$ is not invertible, $\delta_{\nabla^{\Lambda_2(f^*{\TN})}}(\bigwedge^2 df)=0$. (We know $\bigwedge^2 df=0$ at the point, but $\delta$ is a differential operator, it sees beyond the pointwise behaviour). REPLY [4 votes]: Consider the following functional: $$ E_k(f)=\frac{1}{2}\int_{M} \| \bigwedge^k df\|^2 \text{Vol}_{M}.$$ Theorem: The Euler-Lagrange equation of $E_2$, is $A(\phi)=0$, where $A(\phi) \in \Gamma(\phi^*\TN)$ is defined by $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$ $$h_{\phi^*TN}:\phi^*TN \otimes \Lambda_2(\phi^*TN) \to \phi^*TN$$ is a linear map, which depend on the metric on $\N$, and is defined precisely below (see eq $(6)$, and replace $W$ with $\phi^*TN$). Proof: Let $\phi$ be a map $\M \to \N$, and $\phi_t:\M \to \N$ as smooth family, where $\phi_0=\phi$ and $\frac{\partial \phi_t}{\partial t}|_{t=0}:=V \in \Gamma(\phi^*(\TN))$. Then $$ \frac{d}{dt}|_{t=0}E(\phi_t)=\frac{1}{2}\int_{\M}\frac{\partial{}}{dt}|_{t=0} \| d\phi_t \wedge d\phi_t \|^2 \text{Vol}_{\M}= \int_{\M} \langle d\phi \wedge d\phi, \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\rangle \text{Vol}_{\M}. \tag{1}$$ It is well-known that $\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0}=\nabla^{\phi^*(TN)}V \in \Gamma(T^*\M \otimes \phi^*(\TN))$. Now, $$\bigg(\nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}\bigg)(X,Y)=$$ $$(\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(X) \wedge d\phi(Y)+d\phi(X) \wedge (\nabla_{\frac{\partial{}}{dt}} d\phi_t|_{t=0})(Y)=$$ $$ \nabla V(X)\wedge d\phi(Y)+d\phi(X) \wedge \nabla V(Y)= $$ $$\big(\nabla V \wedge d\phi+d\phi \wedge \nabla V\big) (X,Y), \tag{2}$$ where $\nabla V \wedge d\phi+d\phi \wedge \nabla V \in \Omega^2\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$ is defined by the last equality. Thus, we have obtained $$ \nabla_{\frac{\partial{}}{dt}} (d\phi_t \wedge d\phi_t)|_{t=0}= \nabla V \wedge d\phi+d\phi \wedge \nabla V. \tag{3}$$ Define also $\xi=V \wedge d\phi \in \Omega^1\Big(\M,\Lambda_2 \big(\phi^*T\N\big)\Big)$. Lemma: $d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)=\nabla V \wedge d\phi+d\phi \wedge \nabla V$. Assuming the lemma, we combine equations $(1),(3)$ and get $$ \frac{d}{dt}|_{t=0}E(\phi_t)=\int_{\M} \langle d\phi \wedge d\phi, d_{{\nabla}^{\Lambda_2(\phi^*T\N)}}(\xi)\rangle \text{Vol}_{\M}=\int_{\M} \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle \text{Vol}_{\M}. \tag{5}$$ To find the exact $E-L$ equations, one further step needs to be taken: $$V \to \langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),\xi\rangle=\langle \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi),V \wedge d\phi\rangle$$ is a linear functional in $V$, so it can be expressed as $V \to \langle V, A(\phi) \rangle_{\phi^*\TN}$, where $A(\phi) \in \Gamma(\phi^*\TN)$. The E-L equation is $A(\phi)=0$. We now turn to finding an explicit expression for this "representation": The corresponding pointwise linear algebra situation is this: We have two oriented $d$-dimensional inner product spaces $V,W$, together with maps $A \in \Hom(V,W),B \in \Hom(V,\Lambda_2(W))$, and we look for a bilinear map $$\psi: \Hom(V,W) \times \Hom(V,\Lambda_2(W)) \to W,$$ satisfying $$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}=\langle w, \psi(A,B) \rangle_W \, \text{ for every $w \in W$}$$ Proposition: With the notation as above, $\psi(A,B)=h_W\big(\tr_{V} (A \otimes B)\big)$ where $h_W:W \otimes \Lambda_2(W) \to W$ is defined by the linear extension of $$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2. \tag{6}$$ Note $A \otimes B \in V^* \otimes V^* \otimes W \otimes \Lambda_2(W)$, so $\tr_{V} (A \otimes B) \in W \otimes \Lambda_2(W)$. Proof: It suffices to prove this for $A,B$ "pure" tensors, i.e $A=\alpha \otimes \tilde w,B=\beta \otimes (w_1 \wedge w_2)$, where $\alpha,\beta \in V^*,\tilde w,w_1,w_2 \in W$. Now, on the one hand $$ \langle w \wedge A,B \rangle_{\Hom(V,\Lambda_2(W))}= \langle \alpha \otimes (w \wedge \tilde w) ,\beta \otimes (w_1 \wedge w_2) \rangle_{\Hom(V,\Lambda_2(W))}=$$ $$ \langle \alpha , \beta \rangle_{V^*} \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}. $$ On the other hand $$ \tr_{V} (A \otimes B)= \langle \alpha , \beta \rangle_{V^*} \tilde w \otimes (w_1 \wedge w_2).$$ Thus, it's enough to show $$ \langle w \wedge \tilde w ,w_1 \wedge w_2 \rangle_{\Lambda_2(W)}=\langle w , h_W\big(\tilde w \otimes (w_1 \wedge w_2)\big) \rangle_W,$$ but this nows follows directly form the definition of the induced inner product on $\Lambda_2(W)$, and the definition of $h_W$ (see $(6)$). Using the above proposition, we deduce that $$ A(\phi)=h_{\phi^*TN}\bigg(\tr_{\TM}\big(d\phi \otimes \delta_{\nabla^{\Lambda_2(\phi^*{\TN})}}(d\phi \wedge d\phi)\big)\bigg).$$<|endoftext|> TITLE: "Weakly" Woodin cardinals QUESTION [10 upvotes]: Recall that an inaccessible cardinal $\kappa$ is a Woodin cardinal if for every $A\subseteq V_\kappa$ there is an unbounded set in $\kappa$ of $\lambda$ such that $V_\kappa\models\lambda$ is $A$-strong. This raises the question, of whether or not intermediate notions have been defined in the literature, and what sort of results can they be used to obtain? For example, $\Sigma^n_m$-Woodin is an inaccessible cardinal $\kappa$ such that for every $A$ which is $\Sigma^n_m$-definable over $V_\kappa$, there is an unbounded set of $\lambda<\kappa$ such that $V_\kappa\models\lambda$ is $A$-strong. It seems probable that there is such hierarchy of definable Woodin-ness, and that it might be used to obtain partial determinacy (and saturation) results. Do these notions exist in the literature, and if so do they have interesting consequences? REPLY [8 votes]: There is a useful notion of "weak Woodinness" according to which every uncountable regular cardinal is weakly Woodin :) See https://ivv5hpp.uni-muenster.de/u/rds/fabiana_ralf.pdf<|endoftext|> TITLE: Decomposing a finite group as a product of subsets QUESTION [18 upvotes]: My friend Wim van Dam asked me the following question: For every finite group $G$, does there exist a subset $S\subset G$ such that $\left|S\right| = O(\sqrt{\left|G\right|})$ and $S\times S = G$? Also, can we describe such an $S$ explicitly? I believe it's not hard to show that if we choose $S$ uniformly at random, then $\left|S\right| = O(\sqrt{\left|G\right| \log \left|G\right| })$ suffices. But this still leaves the problems of giving an explicit construction and of removing the $\log \left|G\right|$ term. Of course, if $G$ happens to have a subgroup $H$ of size $\approx \sqrt{\left|G\right|}$, then we just take the union of $H$ with a collection of coset representatives and are done. So, this suggests the possibility of a win-win analysis, where we identify the relatively rare finite groups that don't have a subgroup of the right size and handle them in a different way. Work on approximate subgroups might also be relevant. In the likely event that this has already been solved, just a link to the paper would be great (a half hour of googling didn't succeed). REPLY [12 votes]: It was brought to my attention by Noga Alon that my previous answer (which I keep to avoid any confusion) was in fact incorrect: the Rohrbach conjecture got solved completely by Finkelstein, Kleitman, and Leighton in 1988 ("Applying the classification theorem for finite simple groups to minimize pin count in uniform permutation architectures"), and independently by Kozma and Lev in 1992 ("Bases and decomposition numbers of finite groups"). Explicitly, there exists a subset $S\subseteq G$ of size $|S|\le(4/\sqrt 3)|G|^{1/2}$ such that every element of $G$ is representable as a product of two elements from $S$.<|endoftext|> TITLE: Explicit cobordism between Wu manifold and Dold manifold P(1,2)? QUESTION [16 upvotes]: The Wu manifold $SU(3)/SO(3)$ and the Dold manifold $P(1,2)$, the latter being defined as $(S^1\times \mathbb{C}P^2) / (p, x) \sim (-p, \overline{x})$, are cobordant because they are both generators of $Ω_{SO}^5$. As the Wu manifold is equipped with an $SU(3)$-action, and the Dold manifold $P(1,2)$ is a fiber bundle over $S^1$ with fiber equipped with an $SU(3)$-action, I am curious is there an explicit construction of this cobordism? Probably a more well-defined question would be: as the Wu manifold and the Dold manifold are both equipped with $SO(3)$-actions, can there be a $6$-manifold equipped with $SO(3)$ action to give the cobordism? REPLY [4 votes]: There is a cobordism relating the two manifolds of the simplest possible kind: it consists of the attachment of a single 6-dimensional 2-handle. Pick the Dold manifold $\bf D$, thicken it to a 6-manifold ${\bf D}\times [-1,1]$, and attach a 2-handle to ${\bf D} \times\{1\}$ along the unique loop (up to isotopy) that generates the fundamental group $\mathbb Z$. The new boundary that you get is diffeomorphic to the Wu manifold ${\bf W}$. That's all. To prove this, we need to verify that if we replace $S^1 \times D^4 \subset {\bf D}$ with $D^2 \times S^3$ we get ${\bf W}$. (Here $S^1 \times D^4$ is a tubular neighbourhood of the loop generating the fundamental group of $\mathbf D$.) The first thing that we notice is that the result of this surgery is clearly simply connected, and that sounds promising. Exercise: The complement ${\bf D} \setminus (S^1 \times D^4)$ is diffeomorphic to a disc bundle over the non-orientable 3-manifold $M = S^2 \tilde \times S^1$ (this is the non-orientable 3-manifold fibering over $S^1$ with fiber $S^2$). We denote this disc bundle as $D^2 \tilde \times M$, or as $D^2 \tilde \times (S^2 \tilde \times S^1)$. Observation: The non-orientable 3-manifold $M$ is Poincaré dual to $w_2({\bf D})\in H^2({\bf D}, \mathbb Z_2)$. The fact that $M$ is non-orientable (and there is no way to find an orientable representative) is a manifestation of the fact that ${\bf D}$ has no ${\rm Spin}^c$ structure, that is $w_2$ has no integral lift. We pick from the 1965 paper Simply connected five-manifolds of Barden a simple description of ${\bf W}$ as the union of two copies of the (unique) orientable non-trivial $D^3$-bundle over $S^2$, that we may indicate as $D^3 \tilde\times S^2$. The boundary of this bundle is the unique orientable non-trivial $S^2$-bundle over $S^2$, usually indicated as $S^2 \tilde\times S^2$. One of the first exercises one proves in dimension 4 is that $S^2 \tilde \times S^2$ is diffeomorphic to the connected sum $\mathbb{CP}^2 \# \overline{\mathbb{CP}}^2$. The two copies of $D^3 \tilde \times S^2$ are glued to each other along a diffeomorphism of $\mathbb{CP}^2 \# \overline{\mathbb{CP}}^2$ that acts homologically like $(p,q) \mapsto (p,-q)$, that is it reverses the generator of the second factor in the connected sum. Consider the $S^3$ that separates the two factors $\mathbb{CP}^2$ and $\overline{\mathbb{CP}}^2$. Its normal bundle is trivial, so it has a tubular neighbourhood $S^3\times D^2$. Exercise : The complement ${\bf W} \setminus (S^3 \times D^2)$ is again diffeomorphic to $D^2 \tilde \times M$. Therefore ${\bf W}$ is obtained from ${\bf D}$ by surgery as stated above.<|endoftext|> TITLE: (Co)homology theories not satisfying the wedge axiom QUESTION [8 upvotes]: Foundational uniqueness and representability results on (co)homology theories in algebraic topology frequently make a point of assuming additivity, indicating at least some people think it's worthwhile considering theories not satisfying the wedge axiom despite the inapplicability of familiar results. What are some natural examples of such theories (and the failure of additivity)? What are some stupid examples? REPLY [7 votes]: The following example of a non-additive homology theory is due to James and Whitehead, in "Homology with zero Coeffients". It appears as an example in Rudyak's book "On Thom Spectra, Orientability, and Cobordism", pg 65. Define $\tilde{h}(X)=\frac{\prod_{n=0}^{\infty} \tilde{H}_n(X)}{\sum_{n=0}^{\infty} \tilde{H}_n(X)}$ where $\tilde{H_*}$ is (reduced) ordinary homology. In particular $\tilde{h}(S^n)=0$ for each $n$ but $\tilde{h}(\vee_{n=0}^\infty\; S^n)\neq 0$. James and Whitehead's paper also contains other examples which may be of interest.<|endoftext|> TITLE: Summation of bounded sequences QUESTION [6 upvotes]: Let $b_{n,j}\in \mathbb{C}$ for each $n,j\in \mathbb{N}$. I was wondering if there is some characterization of those $b_{n,j}$ such that for all bounded sequences $s_j\in \mathbb{C}, j\in \mathbb{N}$, we have that $$\sigma_n := \sum_{j=1}^{+\infty} b_{n,j} s_j$$ converges. Thus we want that the sequence $(\sigma_n)_n$ converges for all bounded sequences. Note that I do not ask that $s_j\rightarrow s$ implies $\sigma_n\rightarrow s$ as is usual in summability methods. So can we characterize exactly which $b_{n,j}$ satisfy this? REPLY [5 votes]: A comment on Christian Remling's answer. As proved there, we know that $(b_n)$ is a sequence in $\ell_1$, and by assumption we have that $\langle b_n, s\rangle$ converges in $\mathbb{C}$ for any $s\in\ell_\infty$. This defines an element $\tilde b$ of $\ell_\infty^*=\ell_1^{**}$ such that $b_n\to \tilde b$ in the weak$^*$ topology of $\ell_\infty^*$, that is $\sigma(\ell_\infty^*,\ell_\infty)$. The key point is that $\ell_1$ is weakly sequentially complete (essentially proven in C.R. answer), so that $\tilde b$ actually is in $\ell_1$ (not just $\ell_1^{**}$) and the convergence is also in the $\ell_1$ norm as said. (Reference for $L^1$ spaces are weakly sequentially complete: Dunford-Schwarz, Linear Operators, Part I, IV.8.6). rmk. In fact Schur's theorem on the equivalence of strong and weak convergence in $\ell_1$, that is the fact that a weak convergent sequence is also a norm convergent sequence, extends to a more complete statement : a weak Cauchy sequence is also a norm Cauchy sequence (whence the weak sequential completeness). Indeed, if $b_n$ is a weak-Cauchy sequence, for any pair of diverging sequences of integers $n_k$ and $m_k$, one has $b_{n_k}- b_{m_k}\to0$ weakly as $k\to0$, therefore also $\|b_{n_k}- b_{m_k}\|\to0$.<|endoftext|> TITLE: Route to Alain Connes'work about classification of injective factors QUESTION [8 upvotes]: I've read R Kadison, Ringrose "fundamental of operator Vol 1" and R G Douglas "Banach technique in operator algebra".I wanna know how far I'm from the Alain Connes' work. Is there any route (paper and books) leading to well-understanding of that topics? REPLY [5 votes]: I would strongly recommend to have a look (and to always keep available) to the books by Takesaki "The theory of Operator algebras" I,II and III You might not need to know everything that is in these books but they contains most of the tools involved in the classifications, and even some part of the classification (all the basics of Von neuman algebra theory, tomita-Takesaki theory, Connes cocycle, injective = AFD etc...).<|endoftext|> TITLE: Topological universal algebra: what is a variety? QUESTION [17 upvotes]: Very roughly, universal algebra is the study of those classes of algebraic structures which can be defined via a set of equations; such a class is called a variety. Of course there is far more to the subject than this, but this is the essential starting point; and (in my opinion) the essential first result is the HSP theorem, which characterizes varieties in an often more tractable way: HSP theorem: $\mathcal{V}$ is a variety iff $\mathcal{V}$ is closed under homomorphic images, substructures, and arbitrary-arity products. I'm interested in whether there is a similar characterization of varieties over a space. An algebra $A$ with underlying set $X$ is compatible with a topology $\tau$ on $X$ if all the operations of $A$ are continuous in the sense of $\tau$. (We can also look at compatibility up to homotopy, but that doesn't obviously make things cleaner here, so I'll use the more restrictive version here, although I'm interested in each.) For instance, the ring of real numbers $(\mathbb{R}; +, \times, 0, 1)$ is compatible with the usual topology on the reals, and by contrast there is no group compatible with the usual topology on $S^2$. Vastly more generally, Adams' Hopf invariant one theorem states that there is no unital magma structure compatible with the usual topology on $S^n$ unless $n\in\{0, 1, 3, 7\}$. Walter Taylor has studied algebras compatible with given topologies; among other things, he extended Adams' theorem to show that if $n\not\in\{0, 1, 3, 7\}$, then $S^n$ doesn't admit any nontrivial algebraic structure at all, in a precise sense. He also studied the satisfiability problem for algebras over a topological space, from a computability-theoretic perspective. I'm interested in a different aspect of "topological universal algebra": we can also ask about those classes of algebras with underlying set $X$, compatible with $\tau$, which are defined by equations; and we can define a $(X, \tau)$-variety as such a class of algebras. This is a perfectly reasonable notion; however, each of the operations H, S, P are terribly behaved in this context! My main question here is: Q1.1. Is there a "nice" alternate characterization of $(X, \tau)$-varieties, for instance in terms of a fixed family of operations which build new $(X, \tau)$-algebras from old ones? This is hopelessly broad, though, so a reasonable thing to do is look at a more restrictive case: Q1.2. Is there a "nice" alternate characterization of algebras compatible with the reals (with the usual topology)? Note that already on $\mathbb{R}$, we see interesting structure - e.g. Taylor showed that the problem of deciding which finite equational theories are compatible with $\mathbb{R}$ is undecidable. Even this, though, seems intractable to me - there are individual equations whose corresponding variety I don't understand at all! Take our language to consist of a single binary function symbol, "$*$". I don't have a good sense of the class of algebras on $\mathbb{R}$ in which $*$ is commutative. However, at least I know a little about the natural topology on this class - it's path-connected, by taking weighted averages of operations: $$a*_{p}b={a*_1b\over p}+{a*_2b\over 1-p}.$$ An equation that by contrast I know nothing about is associativity: Q2.1. What is a good description of the class of continuous associative binary operations on $\mathbb{R}$? It's easy to check that weighted averages no longer preserve associativity in general. And this brings me to my final, most-concrete question: Q2.2. Is there any interesting way to combine two continuous associative binary operations on $\mathbb{R}$ and get a third? Of course "interesting" is a vague term here - I certainly want to rule out trivialities like constant maps and the projection maps, as well as completely ad-hoc constructions. I could try to make this precise (e.g. asking for continuity with respect to the natural topology on the space of such operations), but rather than do that it feels more natural to leave this subjective. Basically: as natural a notion as it appears to me, I have absolutely no idea what a topological variety is, and would like to. REPLY [9 votes]: This is a long comment rather than a complete answer. But before writing it let me insert that I don't agree that universal algebra is the study of varieties. (In my universe, universal algebra is synonymous with algebra.) Nevertheless, it is clear that this question is about varieties, so I will write about them. Let $Var$ be the category of varieties. The morphisms between objects are the homomorphisms between the clones of the varieties. One object in $Var$ is the variety ${\mathcal V}_{\mathbb R}$ generated by the algebra whose underlying set is $\mathbb R$ and whose operations are all continuous operations on $\mathbb R$. Now let $\mathcal U$ be any other variety. Each morphism $\mathcal U\to {\mathcal V}_{\mathbb R}$ corresponds to a way of equipping $\mathbb R$ with compatible continuous operations defining a $\mathcal U$-structure on $\mathbb R$. If you want to know `what kinds of algebras are compatible with the reals', then you want to know the principal ideal in $Var$ defined by ${\mathcal V}_{\mathbb R}$: i.e. $({\mathcal V}_{\mathbb R}]:=\{{\mathcal U}\;|\;\exists \varphi(\varphi\colon {\mathcal U}\to {\mathcal V}_{\mathbb R}\;\textrm{is a hom})\}$. Moreover, if you want to know all ${\mathcal U}$-structures on $\mathbb R$, you want to know all homomorphisms $\varphi\colon {\mathcal U}\to {\mathcal V}_{\mathbb R}$. (The last sentence of the previous paragraph shows that a better object than $({\mathcal V}_{\mathbb R}]$ for this question is the slice category $Var/{\mathcal V}_{\mathbb R}$, since this identifies not just the types of algebras definable on $\mathbb R$ by continuous operations, but also their realizations.) Taylor showed that there is no algorithm to determine if a finitely presented variety belongs to $({\mathcal V}_{\mathbb R}]$. He did this by interpreting the undecidable satisfiability problem for Diophantine equations into this problem. On the other hand, Taylor shows that there is an easy algorithm to determine whether a variety $\mathcal U$ defined by a finite set $\Sigma$ of simple equations belongs to $({\mathcal V}_{\mathbb R}]$. (An equation is simple if each side has at most one function symbol.) He shows that such ${\mathcal U}$ belongs to $({\mathcal V}_{\mathbb R}]$ iff $\mathcal U$ has a 2-element model. From Taylor's work it is easy to see that it is possible to equip $\mathbb R$ with a continuous commutative binary operation, since $x*y=y*x$ is a simple equation that has a $2$-element model. But Taylor's work does not tell us all ways of equipping $\mathbb R$ with a continuous commutative binary operation. Taylor's work says nothing about the associative law, because it is not simple. Regarding Q2.1: There are continuumly many continuous, associative, binary operations on $\mathbb R$. There can't be more, and it is not hard to construct this many. For a trivial construction of many continuous associative operations, consider constant operations $x*_ry = r$ for different $r\in \mathbb R$. Less trivially, select an interval $[a,b]$ and define $f(x) = \min(\max(x,a),b)$. The operation $x*y = \max(f(x),f(y))$ is continuous, associative, depends on both variables, and has range $[a,b]$. You get a lot of these by varying $a$ and $b$. From some given continuous, associative, binary operations you can often generate more by conjugating by continuous permutations of $\mathbb R$. ($x*_{\pi}y:=\pi^{-1}(\pi(x)*\pi(y))$.) Regarding Q2.2: There are obstacles to constructing a third associative binary operation from two given ones, if you intend to construct the third by some form of composition. For example, the projections $\ell(x,y)=x$ and $r(x,y)=y$ are both associative, but the only binary operations you can get from these by composition are $\ell(x,y)$ and $r(x,y)$ themselves. Similarly, $\max(x,y)$ and $\min(x,y)$ are associative, but the only binary operations in the clone they generate are $\max, \min, \ell, r$. So any general construction would have to trivialize in these situations. Nevertheless, here is something. Suppose you are given continuous, associative, binary operations $x+y$ and $x*y$. Assume in addition that (i) addition is commutative, and (ii) multiplication distributes over addition on both sides. Then $x\circ y:= x+y+x*y$ is a third continuous, associative, binary operation.<|endoftext|> TITLE: Precise relationship between elementary and Grothendieck toposes? QUESTION [23 upvotes]: Elementary toposes form an elementary class in that they are axiomatizable by (finitary) first-order sentences in the "language of categories" (consisting of a sort for objects, a sort for morphisms, function symbols for domain and codomain, and a ternary relation symbol on morphisms for composition). Grothendieck toposes do not form an elementary class: Giraud's axioms are inherently infinitary. However, we can form the "elementary theory of Grothendieck toposes" as the theory axiomatized by every (finitary first-order) sentence true in every Grothendieck topos. (Apply your favorite workaround for size issues. If the size issues are important in this context, though, I'd enjoy reading why.) The elementary theory of Grothendieck toposes (which I'll denote $T_{\operatorname{GrTop}}$) is a proper strengthening of the theory of elementary toposes (which I'll denote $T_{\operatorname{ElTop}}$): it stronger because every Grothendieck topos is an elementary topos, and it is properly stronger because, at a minimum, $T_{\operatorname{GrTop}}$ contains a sentence asserting the existence of a natural number object. Beyond the existence of a natural number object, I can't think of any other concrete ways in which $T_{\operatorname{GrTop}}$ is stronger than $T_{\operatorname{ElTop}}$ (but I don't really know what to look for). My question is, what is known about $T_{\operatorname{GrTop}}$? I'm asking out of general interest, so I'd value any information at all including, Can we give explicit axioms for it? If not, is it at least known to be recursively axiomatizable? If not, might its contents depend on set-theoretic issues? If it does depend on set-theoretic issues, do we get interesting results by redefining $T_{\operatorname{GrTop}}$ to be the theory axiomatized by all sentences true in all Grothendieck toposes "in all models of set theory" (in some suitable sense)? REPLY [7 votes]: Regarding the bullet point list of questions at the end, I believe that every true $\Pi^0_1$-sentence holds in the internal logic of any Grothendieck topos, so in that case $T_{\mathbf{GrTop}}$ is definitely not recursively axiomatizable. If I understand correctly what you mean by "its contents depend on set-theoretic issues," this is already answered by some of the axioms listed in Mike Shulman's answer. In order to show WISC we need to assume the axiom of choice in the background set theory, and if we only work over $\mathbf{ZF}$ then e.g. $\mathbf{Set}$ is a Grothendieck topos where we can't prove that WISC holds.<|endoftext|> TITLE: $SL_n(\mathbb{Z})$ is co-hopfian QUESTION [11 upvotes]: Is it true that $SL_n(\mathbb{Z})$ is co-hopfian for $n \geq 3$? I heard about this result, but I can't find any reference. Would be grateful for any references! REPLY [3 votes]: $\text{SL}_n(\mathbb{Z})$ is indeed cohopfian and evenmore, every non-trivial homomorphism $\text{SL}_n(\mathbb{Z})\to \text{SL}_n(\mathbb{Z})$ is onto, and this could be seen in a fairly elementary fashion. My answer here should be seen as an extended comment to YCor's answer. For simplicity I will deal with the case $n=3$. The following theorem, which elementary proof will be sketched below, is a version of "super-rigidity". Theorem: Given a non trivial homomorphism $\phi:\text{SL}_3(\mathbb{Z})\to \text{SL}_3(\mathbb{C})$ there exists a basis $v_1,v_2,v_3\in\mathbb{C}^3$ such that the image of $\phi$ could be identified with the group of orientation preserving automorphisms of $\mathbb{Z}v_1\oplus \mathbb{Z}v_2\oplus \mathbb{Z}v_3\in \mathbb{C}^3$. The proof of the lemma relies only on the following fact concerning elementary matrices in $\text{SL}_3(\mathbb{Z})$. Denote $$ U_1=\{E_{1,2}(t)\mid t\in\mathbb{Z}\},~ U_2=\{E_{1,3}(t)\mid t\in\mathbb{Z}\},~U_3=\{E_{2,3}(t)\mid t\in\mathbb{Z}\},~ $$ $$ U_4=\{E_{2,1}(t)\mid t\in\mathbb{Z}\},~U_5=\{E_{3,1}(t)\mid t\in\mathbb{Z}\},~U_6=\{E_{3,2}(t)\mid t\in\mathbb{Z}\}. $$ This 6 subgroups $U_i$ (with the standing convention $i\in\mathbb{Z}/(6)$) satisfy: they are all isomorphic to $\mathbb{Z}$, they are all conjugated in $\text{SL}_3(\mathbb{Z})$ (by the Weyl group). they sequentially commute, that is $[U_i,U_{i+1}]=1$. for all $i$, $U_i$ is the commutator group of $U_{i-1}$ and $U_{i+1}$, that is $[U_{i-1},U_{i+1}]=U_i$. together they generate $\text{SL}_3(\mathbb{Z})$. It follows that for every $i$, $H_i=\langle U_{i-1},U_{i+1}\rangle$ is an isomorphic copy of the discrete Heisenberg group $H$ whose center is given by $U_i$. By the facts that the $U_i$'s generate and they are all conjugated, it must be that the non-trivial $\phi$ in the theorem is non-trivial on each of the ${U_i}$'s. Thus for every $i$, $\phi|_{H_i}$ is non-trivial on the center. This is useful because of the following lemma, which proof I leave as an exercise. Lemma: Let $\psi:H\to \text{SL}_3(\mathbb{C})$ be a homomorphism which is non-trivial on the center. Then the image of the center is a rank-1 unipotent group (a group of operators $u$ which eigenvalues are all 1, and such that the image of $u-1$ is 1-dimensional). Advice: when trying to follow my notation in the proof below, have in mind the case where $\phi$ is the standard representation. Sketch of the proof of the theorem: By the lemma, for every $i$, $\phi(U_i)$ is a rank-1 unipotent group. Denote by $P_i<\mathbb{C}^3$ the invariant plane of $\phi(U_i)$. By commutation, $P_i$ is preserved under $U_{i-1}$ and $U_{i+1}$. By the fact that $[U_{i-1},U_{i+1}]$ it is easy to see that either $P_{i-1}=P_i$ or $P_{i+1}=P_i$, but not both. It follows that either $$ (1)~~P_1=P_2,~P_3=P_4,~P_5=P_6 \quad \text{or} \quad (2)~~P_1=P_6,~P_3=P_2,~P_5=P_4. $$ Note that applying inverse-transpose to $\text{SL}_3(\mathbb{Z})$ takes $U_i\to U_{i+3}$, so upon precomposing $\phi$ with inverse-transpose we may and will assume that option (2) occurs (as is the for the standard representation). So we have these three planes $P_1,P_3,P_5$ and their three lines of intersection $L'_i=P_{i-1}\cap P_{i+1}$, $i=2,4,6$. For notational convenience it makes sense to denote $L_i=L'_{2i}$, $i=1,2,3$. Reflecting on $\phi(H_2)$ it is easy to see that $L_1$ is the image of the rank one transformations in $\phi(U_1)-1$ and $\phi(U_2)-1$. Similarly, $L_2$ is the image of the rank one transformations in $\phi(U_3)-1$ and $\phi(U_4)-1$ and $L_3$ is the image of the rank one transformations in $\phi(U_5)-1$ and $\phi(U_6)-1$. Fix $0\neq v_1\in L_1$ and set $v_2 = (\phi(E_{2,1}(1))-1)v_1$ and $v_3 = (\phi(E_{3,2}(1))-1)v_1$ it is easy to check that coordinate change $e_1\mapsto v_i$ conjgates $\text{GL}_3(\mathbb{Z})$ to the group of automorphisms of the lattice $\mathbb{Z}v_1\oplus \mathbb{Z}v_2\oplus \mathbb{Z}v_3$. $~~~\square$ The proof we gave above is constructive and actually gives more than demanded: for $i=1,2,3$ the lines $\mathbb{C}v_i$ are identified with the images of the rank-1 operators $\phi(E_{i,i-1}(1))-1$ and an arbitrary choice of $v_1\in \text{Im}(\phi(E_{1,3}(1))-1)$ determines uniquely the choices of $v_2$ and $v_3$ in the corresponding lines. For simplicity of formulation I did not put this into the theorem, but this is useful. For example, if $\phi:\text{SL}_3(\mathbb{Z})\to \text{SL}_3(\mathbb{Z})<\text{SL}_3(\mathbb{C})$, it follows by the construction that the lines $L_i$ are rational, and upon choosing $v_1$ with integer coordinates, the vectors $v_i$ are integer. We get that $\mathbb{Z}v_1\oplus \mathbb{Z}v_2\oplus \mathbb{Z}v_3< \mathbb{Z}^3$. Finally, observe that for a full rank subgroup $\Lambda<\mathbb{Z}^3$, unless $\Lambda$ is charcteristic, there exists an orientation preserving automorphism of $\Lambda$ which does not extend to $\text{SL}_3(\mathbb{Z})$. Comparing with the theorem above, we conclude that $\mathbb{Z}v_1\oplus \mathbb{Z}v_2\oplus \mathbb{Z}v_3< \mathbb{Z}^3$ is characteristic and thus $\phi$ is onto $\text{SL}_3(\mathbb{Z})$.<|endoftext|> TITLE: Example of a non-locally connected continuum QUESTION [7 upvotes]: Continuum $=$ compact connected metric space. Let $X$ be a continuum. $X$ is indecomposable means that every proper subcontinuum of $X$ is nowhere dense in $X$. It is easy to see that if $X$ is indecomposable then every connected open subset of $X$ is dense in $X$. Question. Are these two conditions equivalent? Given the wealth of examples in continuum theory, the answer is likely no. So what is an example of a decomposable continuum all of whose connected open subsets are dense? EDIT: I have constructed an example; the two conditions are NOT equivalent. Before revealing it, I will leave the bounty open in hopes of attracting more examples. REPLY [2 votes]: **Edit 1.a:**This construction tried to provide a counterexample, but is wrong. I give more details below. Glue two solenoids as follows. Let $K$ be a Cantor set (e.g. iterate $[0,1]\to [0,\frac 1 3] \cup [\frac 2 3,1]$). Let $D\subset K$ be a closed nowhere dense Cantor subset (e.g. iterate $[0,1]\to [0,\frac 1 9] \cup [\frac 8 9, 1]$). Let $S_1,S_2$ be two solenoids over circles $C_1,C_2$ respectively. Essentially, a solenoid is a space $S$ that fibres over a circle $S\to C$, with all the fibres homeomorphic to Cantor sets, and with a certain nontrivial twist. A solenoid is an indecomposable continuum, is homogeneous, and the only proper subcontinua are arcs. On each circle $C_i$ consider two points $P_i^0, P_i^\pi$, and let $F_i^\alpha$ be the corresponding fibers on $S_i$, for $i=1,2$ and $\alpha=0,\pi$. Let homeomorphisms $a: F_1^0\to D$, $b:K\to F_2^0$, $c: F_2^\pi\to D$, $d:K\to F_1^\pi$ be given. Glue $F_1^0\cup (d\circ c)(F_2^\pi)\subset S_1$ to $(b\circ a)(F_1^0)\cup F_2^\pi\subset S_2$ via the above identification, to obtain the required space $X$. $X$ is clearly a continuum, and is decomposable, because is the union of two proper subcontinua (copies of $S_1,S_2$). Edit 1.b: The observation motivating the example was that $S_1$ and $S_2$ are separately indecomposable, and an open set $U$ meeting $S_1$ necessarily meets the fiber $F_1^0$ and since $F_1^0$ is nowhere dense in $F_2^0$ it follows that $U$ meets $F_2^\pi$ as well. This property is achieved in a less trivial way than just identifying (an open subset of) a fiber of $S_1$ with a similar subset of $S_2$. Indeed there is no small neighborhood of $F_2^0$ retracting on it. However this does not really help in solving the question.<|endoftext|> TITLE: Schmidt decomposition on infinite-dimensional Hilbert spaces QUESTION [5 upvotes]: The Schmidt decomposition theorem says: If $H_1,H_2$ are Hilbert-spaces (for simplicity: of same dimension) and $x\in H_1\otimes H_2$, then there exist orthonormal bases $\alpha_i,\beta_i$ of $H_1,H_2$, and reals $\lambda_i\geq0$ such that $x=\sum_i \lambda_i\alpha_i\otimes\beta_i$. I know that this holds for separable Hilbert spaces (e.g., Breuer, Petruccione, The theory of open quantum systems). Does this also hold for nonseparable Hilbert spaces? (If possible, please provide a reference to a paper or textbook.) REPLY [5 votes]: Yes. Christian Remling's comment in essence already gave an affirmative answer, and this question is a bit old, but I think it's worthwhile to write a proof that does not require reducing to the separable situation. I encountered the same question myself and have not found any reference---but with the polar decomposition and spectral theorems, the result follows with little fuss; this leads me to be believe it is almost certainly known, if not written down somewhere on some ancient scroll of functional analysis. (But the fact that I haven't seen it written down in general form compels me to put it out into the internet ether.) I take my Hermitian inner products to be antilinear in the first entry and linear in the second. If $\mathcal{H}$ is a Hilbert space, the continuous dual of a Hilbert space $\mathcal{H}$ will be denoted $\mathcal{H}^{\vee}$, the complex conjugate space will be denoted $\overline{\mathcal{H}}$. Every linear map $f: \mathcal{H} \rightarrow \mathcal{K}$ is functorially mapped to a linear map $\overline{f}: \overline{\mathcal{H}} \rightarrow \overline{\mathcal{K}}$. If you don't like conjugate spaces, just decorate the correct terms with ``anti". First note that there is an isometric isomorphism of Hilbert spaces $I: \mathcal{H}_{1} \otimes \mathcal{H}_{2} \rightarrow \mathcal{H}_{1} \otimes \overline{\mathcal{H}_{2}}^{\vee}$ given by linearization and continuous extension of the map $\alpha \otimes \beta \mapsto \alpha \otimes \langle -, \beta \rangle_{2}$. The Hilbert space $\mathcal{H}_{1} \otimes \overline{\mathcal{H}_{2}}^{\vee}$ is canonically identifiable with the space of Hilbert-Schmidt maps from $\overline{\mathcal{H}}_{2}$ to $\mathcal{H}_{1}$; in particular, it is canonically identifiable with a subspace of the space of bounded operators from $\overline{\mathcal{H}}_{2}$ to $\mathcal{H}_{1}$. Define $X := I(x)$. By the polar decomposition theorem \begin{align*} X = U \sqrt{X^{*} X} \end{align*} for a partial isometry $U: \overline{\mathcal{H}}_{2} \rightarrow \mathcal{H}_{1}$. The operator $X^{*} X: \overline{\mathcal{H}}_{2} \rightarrow \overline{\mathcal{H}}_{2}$ is positive semidefinite and trace class (with trace $\|x\|^2$), so, it is in particular self-adjoint and compact. By self-adjoint compact operator spectral theory we have that the non-vanishing spectrum $\sigma_{\neq 0}$ of $X^{*} X$ is countable, consists only of eigenvalues, and moreover the eigenspaces $E_{\lambda}$ associated to each $\lambda \in \sigma_{\neq 0}$ are finite dimensional and orthogonal. Positive semidefiniteness ensures that each eigenvalue is positive. By the spectral theorem \begin{align*} X^{*} X = \sum_{\lambda \in \sigma_{\neq 0}} \lambda \mathrm{Proj}_{\lambda}, \end{align*} where $\mathrm{Proj}_{\lambda}$ is the orthogonal projection onto $E_{\lambda}$. Choosing an orthonormal basis $(\beta^{\lambda}_{i})_{i = 1}^{\dim E_{\lambda}}$ for each finite dimensional $E_{\lambda}$ \begin{align*} X^{*} X &= \sum_{\lambda \in \sigma_{\neq 0}} \lambda \left[\sum_{i_{\lambda} = 1}^{\dim E_{\lambda}} \beta^{\lambda}_{i_{\lambda}} \otimes (\beta^{\lambda}_{i_{\lambda}})^{\vee} \right], \end{align*} (where $\omega^{\vee}$ denotes the linear form $\langle \omega, - \rangle$). Of course I can just reindex things here to get a sum \begin{align*} X^{*} X=\sum_{j = 1}^{r} \lambda_{j} \beta_{j} \otimes \beta_{j}^{\vee}, \end{align*} where $r = \mathrm{rank}(X^{*} X) = \sum_{\lambda \in \sigma_{\neq 0}} \dim E_{\lambda} \leq \aleph_{0}$, the $\lambda_{j}$ are eigenvalues, and the $\beta_{j} \in \overline{\mathcal{H}}_{2}$ are associated eigenvectors with $\langle \beta_{i}, \beta_{j} \rangle = \delta_{ij}$. Using our polar decomposition: \begin{align*} X = \sum_{j = 1}^{r} \sqrt{\lambda_{j}} (U \beta_{j}) \otimes \beta_{j}^{\vee}. \end{align*} Because $U$ is a partial isometry, and one can verify $\ker(U) \leq \ker(X^{*} X)^{\perp}$, the $U \beta_{j} \in \mathcal{H}_{1}$ remain orthonormal. Applying $I^{-1}$, then we simply have \begin{align*} x = \sum_{j = 1}^{r} \sqrt{\lambda_{j}} (U \beta_{j}) \otimes \tilde{\beta}_{j}. \end{align*} where $\tilde{\beta}_{j} \in \mathcal{H}_{2}$ is the image of $\beta^{\vee}_{j}$ under the canonical isomorphism $\overline{\mathcal{H}}^{\vee}_{2} \overset{\sim}{\rightarrow} \mathcal{H}_{2}$. It is an eigenvector of $\overline{X^{*}X}: \mathcal{H}_{2} \rightarrow \mathcal{H}_{2}$ with eigenvalue $\lambda_{j}$. A few side remarks: $U \beta_{i}$ is an eigenvector of $X X^{*}: \mathcal{H}_{1} \rightarrow \mathcal{H}_{1}$ associated to eigenvalue $\lambda_{i}$. This follows by looking at the explicit construction of $U$ as in, e.g. Douglas' Lemma (a generalization of the polar decomposition theorem). I mention this because $\overline{X^{*} X}$ and $X X^{*}$ are the reduced density states associated to the pure state $x \otimes x^{\vee}$. The proof above is just a generalization of the usual finite-dimensional proofs that the Schmidt decomposition is given by tensor products of eigenstates of reduced density states. Because Douglas' lemma constructs a unique $U$, the only choices involved in the procedure above are the choices of bases on each finite dimensional eigenspace $E_{\lambda}$. Any two choices are then related by the action of $\prod_{\lambda \in \sigma_{\neq 0}} U(E_{\lambda})$ (the infinite product of underlying groups). One can check that this group can be realized as a subgroup of $U(\bigoplus_{\lambda \in \sigma_{\neq 0}} E_{\lambda})$, and hence $U(\mathcal{H}_{2})$, or even $U(\mathcal{H}_{1} \otimes \mathcal{H}_{2})$ in a natural way. If one wants to be overly pedantic about the question you asked: you mentioned something about bases. Indeed, every Hilbert space has an orthornormal basis, so there exists a (possibly uncountable) basis $(e_{k})_{k \in K}$ for $\mathrm{span}(\beta_{i})^{\perp} \leq \mathcal{H}_{2}$ (I'm taking all spans to be closed) and a basis $(f_{l})_{l \in L}$ for $\mathrm{span}(U\beta_{i})^{\perp} \leq \mathcal{H}_{1}$; we can combine these with the bases we constructed on the spans to get orthonormal bases $(\tilde{e}_{k})_{k \in K'}$ and $(\tilde{f}_{l})_{l \in L'}$. Then, in a rather stupid way, we have \begin{align*} x = \sum_{m \in M} \omega_{m} \tilde{e}_{m} \otimes \tilde{f}_{m} \end{align*} where $M$ is some (possibly uncountable) indexing set with $|M| = \mathrm{min}(|K'|,|L'|)$. Only countably many of the $\omega_{m}$ are non-zero.<|endoftext|> TITLE: Transfer map of simplicial sets QUESTION [5 upvotes]: Let $ f:X \rightarrow Y$ be finite covering map of simplicial sets of finite degree, say $d$. Let $\varSigma^{\infty}$ denote the functor from the category of simplicial sets to spectra which is an additive category, where $\varSigma^{\infty}Y = \{n \mapsto S^{n} \wedge Y \}$. I want to know the sketch or reference for the proof of the following statement:"$f$ induces a transfer map $ f': $ $\varSigma^{\infty}Y \rightarrow \varSigma^{\infty}X $ such that $f'' \cdot f'$ is multiplication by $d$ in the abelian group $Hom(\varSigma^{\infty}Y, \varSigma^{\infty}Y )$. Note $f''$ is the obvious map $\varSigma^{\infty}X \rightarrow \varSigma^{\infty}Y $ induced by functor $\varSigma^{\infty}$. " REPLY [5 votes]: The statement is not true (in topological spaces or simplicial sets). The composition $f'' \cdot f'$ will only be multiplication by $d$ "up to higher Atiyah--Hirzebruch filtration". For an explicit counterexample one can calculate the effect for the universal cover of $B\mathbb{Z}/2$ in zeroth stable cohomotopy. By the Segal conjecture $\pi^0(\Sigma^\infty B\mathbb{Z}/2_+)$ is the completion of the Burnside ring $A(\mathbb{Z}/2)$ at the augmentation ideal $I$. As an abelian group this is free with basis 1, the singleton $\mathbb{Z}/2$-set, and $T$, the free transitive $\mathbb{Z}/2$-set. Taking bases 1 and $H := T-2$, we have $H^2 = (T-2)^2 = T^2-4T+4 = 4-2T = -2H$, so the $I$-adic filtration agrees with the 2-adic filtration on $\mathbb{Z}\{H\}$. Thus we have $$\pi^0(\Sigma^\infty B\mathbb{Z}/2_+) = \mathbb{Z}\{1\} \oplus \mathbb{Z}_2 \{H\},$$ where $\mathbb{Z}_2$ denotes the 2-adic integers. The Becker--Gottlieb transfer map $\mathrm{trf} : \Sigma^\infty B\mathbb{Z}/2_+ \to \Sigma^\infty E\mathbb{Z}/2_+ = \mathbb{S}^0$ (associated to the covering map $\pi : E\mathbb{Z}/2 \to B\mathbb{Z}/2$) on cohomotopy sends the unit to $$T = H+2 \in \pi^0(\Sigma^\infty B\mathbb{Z}/2_+),$$ so $\pi \circ \mathrm{trf} : \Sigma^\infty B\mathbb{Z}/2_+ \to \Sigma^\infty B\mathbb{Z}/2_+$ on zeroth cohomotopy sends a $\mathbb{Z}/2$-set $X$ to $\vert X \vert \cdot T$. This has twice the cardinality of $X$, but is not equal to $2X$ (for example if $X=1$). (Note that $H$ restricts to zero on the 0-skeleton of $\Sigma^\infty B\mathbb{Z}/2_+$, so has positive Atiyah--Hirzebruch filtration: thus $T=2$ module higher Atiyah--Hirzebruch filtration, as I mentioned above.)<|endoftext|> TITLE: Girth of the symmetric group QUESTION [20 upvotes]: Let $n \in \mathbb N$ and $\{\sigma,\tau\} \subset {\rm Sym}(n)$ be a generating set. Question: What is the maximal possible girth (if one varies $\sigma, \tau$) of the associated Cayley graph? I am interested in asymptotic bounds. In [A. Gamburd, S. Hoory, M. Shahshahani, A. Shalev, B. Virág, On the girth of random Cayley graphs. Random Structures Algorithms 35 (2009), no. 1, 100–117.] it was proved that a random pair will produce girth $\Omega( (n \log(n))^{1/2})$ with probability tending to $1$. However, I am not aware of concrete generators that realize this girth. A natural guess for a lower bound on the maximal possible girth (and maybe even the typical girth) would be $\Omega(n \log(n))$. It is also clear that $O(n \log(n))$ is an upper bound for the maximal possible girth. To the best of my knowledge, it is not known if ${\rm Sym}(n)$ does satisfy a law of length $O(n)$. Hence, any superlinear lower bound on the maximal possible girth would have an immediate application. REPLY [4 votes]: I don't know how to answer your question, but I'll suggest a possible approach, i.e. a source of examples that potentially have large girth. As I'm sure you're aware, Margulis constructed graphs of large girth which were Cayley graphs of $SL_2(\mathbb{Z}/p)$. The point is that reducing a free subgroup $SL_2(\mathbb{Z}) (\mod p)$, one could show that words in the generators must have a certain length before they could be trivial by computing the operator norm of products of matrices. These graphs achieve the optimal growth of girth of $O(\log(n))$, where $n$ is the number of vertices. His approach works for families of simple groups of Lie type, by reducing linear representations $(\mod p)$. By analogy, one could hope to achieve Cayley graphs for the symmetric or alterating groups by considering non-linear actions of free groups on varieties, and reducing $(\mod p)$ to get families of finite actions. Recently Bourgain, Gamburd, and Sarnak have investigated such actions for the Markoff equation: $$ x^2+y^2+z^2=xyz.$$ The automorphism $\theta: (x,y,z)\mapsto (y,z, yz-x)$ (found by exchanging the two solutions of the quadratic formula in the variable $z$ and permuting) together with permutations generates a virtually free group of automorphisms (in fact, it has a finite-index 2-generators subgroup). These act transitively on the fundamental solution $(3,3,3)$, giving the Markoff triples. Subsequently it was shown by Meiri and Puder that the action on the solutions to the Markoff equation $(\mod p)$ is usually symmetric or alternating (the number of solutions is $O(p^2)$, so one gets generators for $Sym(O(p^2))$). Thus, one may ask for the girth of the Cayley graph associated to this action. It seems tricky to implement Margulis' method here, because it is not as easy to identify when a polynomial automorphism is trivial $(\mod p)$, unlike the linear case. Also, the growth of the action on the Markoff numbers is superexponential instead of exponential. One may show that on a ball of radius $c\log\log p$, the fundamental solution $(3,3,3)$ is not sent to itself. In particular, the orbit of $(3,3,3) (\mod p)$ under the element $\theta$ has size at least $\log\log p$. This is far off from optimal of course. But there is a bit of evidence that one can do much better: in this paper, it is shown that $\theta$ has a cycle of size $O(\log p)$. Moreover, Bourgain-Gamburd-Sarnak remark that experiments indicate that the Cayley graphs form an expander family, although they don't explicitly conjecture this or discuss the girth. A heuristic that the girth should be at least $O(\log(p))$: taking the involution generators for the automorphism group $(x,y,z)\mapsto (x,y,xy-z), (x,xz-y,z), (yz-x,y,z)$, one sees that their compositions have degrees which grow at most exponentially (in fact, at most the Fibonacci sequence) with $n$ compositions. Since a non-trivial polynomial in $\mathbb{F}_p[x,y,z]$ must be non-vanishing if the degree of each variable is $ TITLE: Local Galois representation associated to twist of modular form QUESTION [5 upvotes]: Let $f$ be a modular newform of weight $k \geq 2$, level $N$ (square free) and trivial nebentypus. Let $V_{f}$ be the $p$-adic (p odd) Galois representation associated $f$. We denote by $V_{f,l}:= V_{f}|_{G_{l}}$. Let $\chi$ be a quadratic character of conductor $l$. Suppose that $(N,l)=1$. Then $f \otimes \chi$ is a newform of weight $k$, level $Nl^2$ and trivial nebentypus. Case I: $l \neq p$ In this case $V_{f,l} \sim \pmatrix{ \chi_{1} & 0 \\ 0 & \chi_2 }$. But for $f \otimes \chi$, I think the representation $V_{f\otimes \chi, l}$ is irreducible. (Reference: Tilouine - Modular forms and Galois representation, Section 3.2, Bull Greek Math Soc. Vol 46) Can some one explain why it should be irreducible? And how does $V_{f,l} \otimes \chi$ is related to $V_{f \otimes \chi,l}$? Case II: $l=p$ Suppose that $f$ is ordinary at $p$. Is it true that $f \otimes \chi$ is ordinary at $p$? In this case how is $V_{f,p} \otimes \chi$ is related to $V_{f \otimes \chi,p}$? REPLY [4 votes]: I think it helps to put things in a larger perspective. To an eigencuspform $f$ and a prime number $\ell$ is attached on the one hand an irreducible, admissible representation $\pi(f)_{\ell}$ of $\operatorname{GL}_{2}(\mathbb Q_{\ell})$ which is either 1) An irreducible principal series $\pi(\chi,\psi)$, 2) A special representation $\mu\cdot\operatorname{St}$, 3) A (super)cuspidal representation and on the other hand a local $\operatorname{Gal}(\bar{\mathbb Q}_\ell/{\mathbb Q}_\ell)$-representation $\rho_f|G_{\mathbb Q_\ell}$ with coefficients in ${\mathbb Q}_p$ ($p\neq\ell$) which is either 1) Reducible and potentially unramified ($I_\ell$ acts through a finite quotient), 2) Reducible and not potentially unramified ($I_\ell$ acts through an infinite quotient), 3) Irreducible (and $I_\ell$ acts through a finite quotient). If $\chi$ is a finite order character, then $\pi(f\otimes\chi)_\ell=\pi(f)_\ell\otimes\chi$, $\rho_{f\otimes\chi}|G_{\mathbb Q_\ell}=(\rho_f|G_{\mathbb Q_\ell})\otimes\chi$ and the two classifications above are stable by twisting by $\chi$. Finally, the local-global compatibility property of the Langlands Reciprocity Conjecture (for $\operatorname{GL}_2$) (which is a theorem of Deligne and Carayol) states that $\pi(f)_\ell$ belongs to class $i$ if and only if $\rho_f|G_{\mathbb Q_\ell}$ belongs to class $i$. In Tilouine's survey, it seems that section 3.1 is only valid for $f$ with non-zero eigenvalue if not supercuspidal at $\ell$ (an hypothesis which is not satisfied by your $f\otimes\chi$).<|endoftext|> TITLE: What are some of results in low dimensional statistics that do not hold in high dimensions? QUESTION [5 upvotes]: This question is partially inspired by the following MO post: What are some of the surprising results of finite sample statistical estimation? and current heated research front of high dimensional statistics. Instead of asking about surprising results in high dimensions, I will ask what kind of results that holds in low dimensions fails to hold in a higher dimension. And what is its relation with other mathematical branch? (By high dimensional statistics we usually refer to a high dimension covariate space instead of response space. For example, in a regression setting $Y=f(X)$ we tend to say a problem is of high dimension if $\dim\mathcal{X}\gg \dim\mathcal{Y}$) For one simplest example, we know that James-Stein estimator performs better than maximum likelihood estimator in terms of $L^2$ norm when the dimension $\dim\mathcal{X}=d\geq 3$; and it turns out to be an equivalent statement that a symmetric random walk in $\mathcal{X}=\mathbb{R}^d$ is transient when $d\geq 3$ via an infinitely divisible stochastic process. Another example is provided in the answer below. Are there other such examples that can relate high dimensional phenomena in statistics? REPLY [3 votes]: A great example that I have in mind is the concentration phenomena in high dimensions. Consider the simplest multivariate normal distribution $X\sim N_d(0_d,I_d)$, we can compute its $L^2$ norm $\sum_iX^2_i=:\|X\|^2\sim\chi^2(d)$. and $X_i^2\sim \chi^2(1)$ independently. With central limit theorem applied on each component, we have that $$\frac{1}{d}\sum^d_{i=1}X^2_i=\frac{1}{d}\|X\|^2\overset{P}{\rightarrow}N_1(1,\frac{2}{d})$$ as $d\rightarrow\infty$. Use the delta method we can see that $\|X\|\overset{P}{\rightarrow}\sqrt{d}N_1(1,\frac{1}{d})=N_1(\sqrt{d},1)$ and therefore we can actually assert that as dimension $d=dim\mathcal{X}\rightarrow \infty$ the random vectors are concentrated around a sphere. Even more surprising is that if we have another independent $Y\sim N_d(0_d,I_d)$, then we can compute the distribution of $\frac{X\cdot Y}{\|X\|\|Y\|}$ as $d\rightarrow \infty$ is $N_1(0,d)$(multidimensional CLT and delta method) and the distribution of $\|X-Y\|$ as $d\rightarrow \infty$ is $N_1(0,2d)$. These two results claimed that as $d\rightarrow \infty$ two random vectors are most likely to be orthogonal and evenly distributed on the sphere, which is not expected when $d=1,2$. I really hope to know if there are more such examples with a motivation from consideration of the difference between the geometry of high and low dimensional spaces.<|endoftext|> TITLE: Are these two new ways of representing odd zeta values as integrals known? QUESTION [26 upvotes]: This is inspired by the same beautiful integral expression for $\zeta(3)$ as this question, but goes in a slightly different direction. Writing the original integral in the form $$\int_0^1\frac{x(1-x)}{\sin\pi x}dx=7\frac{\zeta(3)}{\pi^3} ,$$ it turns out that for $n\in\mathbb N$ there is a unique monic polynomial $p_n$ of degree $n-1$ such that $$\int_0^1x^np_n(x)\frac{1-x}{\sin\pi x}dx=c_{2n+1}\frac{\zeta(2n+1)}{\pi^{2n+1} } $$ with rational $c_k=4(k-1)! \dfrac{2^k-1}{2^k}= (4-2^{2-k})(k-1)! $. This follows for $\zeta(2n+1)$ from solving the linear system given by the blue lines numbered $n+1,...,2n$ in the other question. As Zurab Silagadze answered it by giving an explicit formula of the coefficients in the blue lines, the $p_n$'s can be calculated. I don't know however if it is possible to give a formula in closed form (meaning here that it should not contain a matrix inversion), but see below. The first polynomials are $$\begin{align} p_1(x)&=1 \\ p_2(x)&=3-x \\ p_3(x)&=25-20x+x^2 \\ p_4(x)&=455-707x+287x^2-x^3 \\ p_5(x)&=14301-34734x+29046x^2-8304x^3+x^4 \\ p_6(x)&=683067-2289309x+2949276x^2-1721434x^3+382547x^4-x^5 \\ \end{align}$$ The constant terms are supposedly the sequence A272482, thus, correcting the oeis typo $1/(2n)!$, $$[x^0]p_n(x)= {(2n)!}[x^{2n}y^n]\frac{\cos\frac{x(1-y)}{2}} {\cos\frac{x(1+y)}{2}} = \frac 1{4^n} {2n\choose n}\sum_{i=0}^n{n\choose i}E_i,$$ where $E_i$ are the Euler numbers. This seems to suggest something similar for the other coefficients, and thus possibly a closed form. Are the $p_n$ known? How to find their closed form or generating function? More generally now, define $$J(m,n,k)=J(n,m,k):= \int_0^1\frac{x^m(1-x)^n}{\sin^k\pi x}dx.$$For this to converge, we need $m,n\geqslant k$. Experimentally, the situation for $k=2$ is quite similar to the $k=1$ case in that $J(m,n,k)$ is a rational combination of values $\dfrac{\zeta(i)}{\pi^{i+1}}$ with $i$ running over all odd numbers between $\min(m,n)$ and $m+n-1$, e.g. $$J(7,4,2)=\dfrac{105}2\left(-\dfrac{\zeta(5)}{\pi^6}+51\dfrac{\zeta(7)}{\pi^8}-405\dfrac{\zeta(9)}{\pi^{10}}\right).$$ This leads to new possibilities of representing odd zeta values as integrals, this time with $\sin^2\pi x$ in the denominator. Writing as a shortcut $h_m:=J(m,m,2)$, we can for example express $\dfrac{\zeta(2n-1)}{\pi^{2n}}$ as a rational combination of $h_2,\dots,h_n$, i.e. as an integral $$\dfrac{\zeta(2n-1)}{\pi^{2n}}=\int_0^1q_n(x-x^2)\frac{x^2(1-x)^2}{\sin^2\pi x}dx,$$ where $q_n$ is a unique polynomial of degree $n-2$. The first of them are: $$\begin{align} \zeta(3)&=\frac{\pi^4}{6}h_2 \\ \zeta(5)&=\frac{\pi^6}{90}(h_2 +2h_3),\qquad \text{ i. e. } q_2(z)=\frac{1}{90}(1+2z) \quad \text{ etc. }\\ \zeta(7)&=\frac{\pi^8}{1890}(2h_2 +4h_3+3h_4)\\ \zeta(9)&=\frac{\pi^{10}}{28350}(3h_2 +6h_3+5h_4+2h_5)\\ \zeta(11)&=\frac{\pi^{12}}{935550}(10h_2 +20h_3+17h_4+8h_5+2h_6)\\ \zeta(13)&=\frac{\pi^{14}}{638512875}(\color{blue}{691}h_2 +1382h_3+1180h_4+574h_5+175h_6+30h_7)\\ \end{align}$$ Experimentally, in $\dfrac{\zeta(2n-1)}{\pi^{2n}}$ the last coefficient (i.e. the one of $h_n$ and the leading term of $q_n$) is $\dfrac{2^{2n-2}}{(2n)!}$ and the one preceding it is $\dfrac{n(n-2)}6\dfrac{2^{2n-2}}{(2n)!}$, while for the first coefficient (equally, the constant term of $q_n$), the occurrence of $\color{blue}{691}$ in the expression for $\zeta(13)$ suggests that it involves the Bernoulli number $B_{2n-2}$. Any ideas about these polynomials? Finally, for $k\geqslant 3$ there does not seem to exist any closed form, at least not in terms of zeta values. What about $J({3,3,3})= \int\limits_0^1\dfrac{x^3(1-x)^3}{\sin^3\pi x}dx$? REPLY [4 votes]: In this recent paper on log tangent integrals, Theorem 1 (2.2, in the published version) expresses that, if $n$ is a positive integer, \begin{equation} \int_0^{\tfrac{\pi}{2}}E_{2n-1}\left( \frac{2}{\pi}x \right)\log(\tan x)\,dx= \frac{(-1)^{n-1}(2n-1)!}{\pi^{2n-1}}\left( 2-2^{-2n} \right)\zeta(2n+1) \end{equation} where $E_n(x)$ are the Euler polynomials. This expression can be written as \begin{equation} \int_0^{1}E_{2n-1}\left( x \right)\log(\tan \frac{\pi}{2}x)\,dx= \frac{(-1)^{n-1}2(2n-1)!}{\pi^{2n}}\left( 2-2^{-2n} \right)\zeta(2n+1) \end{equation} Defining the antiderivatives \begin{equation} F_{2n-1}(x)=\int_0^xE_{2n-1}\left( t \right)\,dt \end{equation} one may notice that $F_{2n-1}(0)=F_{2n-1}(1)=0$, as $E_{2n-1}( 1-x )=-E_{2n-1}( x )$. Then, integrating by parts, it comes \begin{equation} \int_0^1\frac{ F_{2n-1}(x)}{\sin\pi x}\,dx= \frac{(-1)^{n}4(2n-1)!}{\pi^{2n+1}}\left( 1-2^{-2n-1} \right)\zeta(2n+1) \end{equation} As $x=0,1$ are two roots of $F_{2n-1}(x)$, we conclude that the polynomials \begin{equation} f_{2n-1}(x)=\frac{1}{x(1-x)}\int_0^xE_{2n-1}\left( t \right)\,dt \end{equation} verify \begin{equation} \int_0^1 f_{2n-1}(x)\frac{x(1-x) }{\sin\pi x}\,dx= \frac{(-1)^{n}4(2n-1)!}{\pi^{2n+1}}\left( 1-2^{-2n-1} \right)\zeta(2n+1) \end{equation} More generally, for symmetry reasons, any polynomial \begin{equation} g_{2n-1}(x)=f_{2n-1}(x)+P(2x-1) \end{equation} where $P(x)$ is an arbitrary odd polynomial, gives the same result. This result gives an explicit representation of the polynomials derived in my previous answer. Edit: Using the derivative property for the Euler polynomials, $E_{2n-1}(x)=(2n)^{-1}dE_{2n}(x)/dx$, one can express \begin{align} f_{2n-1}(x)&=\frac{1}{2n}\frac{1}{x(1-x)}\left[E_{2n}(x)-E_{2n}(0)\right]\\ &=\frac{1}{x(1-x)}\left[\frac{1}{2n}E_{2n}(x)+\frac{1}{n(2n+1)}\left( 2^{2n+1}-1 \right)B_{2n+1}\right] \end{align} where $B_{2n+1}$ is a Bernoulli number.<|endoftext|> TITLE: the cohomology objects in the t-structure and long exact sequence QUESTION [5 upvotes]: It is known that, given an abelian category $\mathcal B$, the derived category $\mathrm D (\mathcal B)$ is a triangulated category. In particular, given a distinguished triangle $E\to F\to G \to E[1]$ in $\mathrm D ( \mathcal B)$, taking cohomology of complexes gives a long exact sequence $$ \cdots \to H^*(E) \to H^*(F) \to H^*(G) \to H^{*+1}(E) \to \cdots $$ For a general triangulated category, there is no notion of "complexes". Nevertheless, we have something similar provided a bounded t-structure. (Reference: Dirichlet Branes and Mirror Symmetry section 4.4) Indeed, suppose $\mathcal A$ is the heart of a bounded t-structure on a triangulated category $\mathrm D$. Then it is known that for every nonzero object $E\in \mathrm D$ we can define the so-called cohomology objects $H^i(E)\in \mathcal A$ of $E$ in the given t-structure. Question: (1) If $E\to F\to G \to E[1]$ is a distinguished triangle in a triangulated category $\mathrm D$, then can we still have a long exact sequence as above? Can we find a cohomological functor $\mathrm D \to \mathcal A$ to explain this? (2) What is the motivation of the definition of cohomology objects $H^i(E)$ of $E$ given a t-structure? (Especially, do you know how to establish the Lemma 4.56 in the reference mentioned above?) Thanks REPLY [3 votes]: (1) The functor $H^0$ is cohomological [BBD, Théorème 1.3.6]. (2) A t-structure is, basically, a way to determine an abelian subcategory inside a triangulated category together with a cohomological functor with values in this subcategory. This permits a formulation within the derived category of sheaves of perverse sheaves. Perverse sheaves are complexes associated to a stratification and are associated to the theory of Intersection Homology of Goresky and McPherson. This theory was developed to recover duality on stratified spaces. Moreover, this general formalism has allowed Beĭlinson, Bernstein, Deligne and Gabber to transport the construction of intersection cohomology to the context of étale cohomology of algebraic varieties (possibly with singularities). [BBD] Beĭlinson, A. A.; Bernstein, J.; Deligne, P.: Faisceaux pervers. Analysis and topology on singular spaces, I (Luminy, 1981), 5–171, Astérisque, 100, Soc. Math. France, Paris, 1982.<|endoftext|> TITLE: When is the sheaf of isomorphism classes of a reasonable moduli problem an algebraic space? a scheme? QUESTION [8 upvotes]: Let $\mathcal{M}$ be a reasonable moduli problem (ie, at least a separated Deligne-Mumford stack, flat over some base scheme $S$). Let $M$ be the sheafification of the presheaf: $$(U\rightarrow S)\mapsto \{\text{isomorphism classes of objects in }\mathcal{M}(U)\}$$ I'd like to understand when $M$ is an algebraic space, or a scheme, and if not, what sorts of things can go wrong. Here, I'd like to stick to the conventions used in the stacks project, except that I'd like to stick to the etale topology, if it turns out to be relevant. Ie, a DM stack is a stack in groupoids with unramified diagonal representable by algebraic spaces, and admitting an etale surjective cover by a scheme, and an algebraic space is a sheaf with diagonal representable by schemes, also admitting an etale surjective cover by a scheme. In the particular case I'm currently looking at, $\mathcal{M}$ is in addition proper and quasi-finite over $S$. REPLY [4 votes]: As Piotr Achinger observes (but I shall stick to your notation), $\mathcal{M}$ has a coarse moduli space, say $M_0$, and we have morphisms $\mathcal{M}\to M:=\pi_0(\mathcal{M})^\mathrm{sh}\to M_0$. If $M$ is an algebraic space, then $M\to M_0$ has a section, hence $\mathcal{M}\to M_0$ must be an epimorphism for the étale topology (I assume here that $^\mathrm{sh}$ means "associated étale sheaf", but see below). Here is an example where this fails (but at the moment I cannot think of one which is flat and quasifinite over a scheme). Let $k$ be a field of characteristic $\neq2$, and let $\mathcal{M}=[\mathbb{A}^2_k/G]$ where $G=\{\pm1\}$ acts in the obvious way. Then $M_0$ is the usual quotient scheme, and $\mathcal{M}\to M_0$ is an epimorphism iff the quotient map $f:\mathbb{A}^2_k\to M_0$ is. This is clearly not the case: this map has no section locally for the étale topology. In fact, let me prove the stronger result that $\mathbb{A}^2_k\to M_0$ is not an epimorphism of fppf (or even fpqc) sheaves. In other words, we still have a counterexample if $^\mathrm{sh}$ means "associated fppf sheaf". Put $X=\mathbb{A}^2_k$, let $U$ (resp. $V$) be the complement of the origin in $X$ (resp. $M_0$), and $j:V\to M_0$ the inclusion. Then $\mathcal{A}:=f_*(\mathcal{O}_X)$ is a finite $\mathcal{O}_{M_0}$-algebra, étale of rank 2 on $V$ but not locally free on $M_0$ since its rank at the origin is 3. Moreover we have $$j_*(\mathcal{A}_V)=\mathcal{A}\quad\text{ and }\quad j_*(\mathcal{O}_V)=\mathcal{O}_{M_0}$$ because both $X$ and $M_0$ are $\mathrm{S}_2$. Note that these properties are preserved by flat base change. Now assume $p:Z\to M_0$ is faithfully flat and lifts to $p':Z\to X$. Put $W:=p^{-1}(V)\subset Z$, and let $w:W\to Z$ be the inclusion. The étale double cover of $W$ induced by $U\to V$ has a section, so it is trivial and in particular $\mathcal{A'}:=p^*\mathcal{A}$ is free on $W$. Hence, using the above-mentioned base change property, $$\mathcal{A'}\cong w_*\mathcal{A'}_W\cong w_*\mathcal{O}_W^2\cong \mathcal{O}_Z^2$$ which implies by descent that $\mathcal{A}$ is locally free over $M_0$, a contradiction.<|endoftext|> TITLE: Bijection directly from (n,n+1)-core partitions to parking functions? QUESTION [7 upvotes]: It is well-known that the increasing parking functions are counted by the Catalan numbers. The Catalan numbers also count the dominant alcoves in the Shi arrangement of type $A_{n}$. Athanasiadis-Linusson gave a bijection between these two (and in fact between the sets of all Shi alcoves and all parking functions.) In addition, Fishel-Vazirani gave a bijection from dominant Shi alcoves to partitions which are both $n$ and $n+1$-core. This (composed with Athanasiadis-Linusson) gives a bijection from such partitions to increasing parking functions. Has anyone made this direct map explicit? Edit: In conversations in the comments, I've vaguely remembered something that might be helpful in solving this. Somewhere I recall reading about "Inversion-labelled Dyck paths" or something like that. The idea is to turn a Dyck path into a partition in the usual way, but label the partitions in a different way. Specifically, label the diagonal of the partition with a permutation $\phi$, but with the restriction that if $i \phi_{j}$ can only occur if there is not a removable box where $i$ and $j$ meet. (This means the Dyck path has no valley there.) So the partition consisting of a single box could be labelled as 123,213, or 132. Since the box aligns with the first and third entries of the permutation, $\phi_{1} < \phi_{3}$. Similarly, the partition 2 can be labelled as 123,213, or 312. Now, the second box in the first row is removable, forcing $\phi_{2} < \phi_{3}$. There are the same number of these labellings as there are parking functions. Unfortunately, I don't see a reference to something like this in the OEIS entry for parking functions, and I don't immediately see a bijection with parking functions. I'm sure this isn't my original idea, either - I suspect it's due to some subset of Haiman/Haglund/Loehr/Remmel but I'm not finding it right now. I think finding the reference for this idea would be quite helpful. Apologies for the vagueness of my memories of it. REPLY [2 votes]: I believe I have as much of an answer as I'm going to get. An $(n,n+1)$-core turns into an $n$-abacus diagram (this map is at the heart of Anderson's paper Partitions which are simultaneously t1- and t2-core, as pointed out by Christian Stump). Then, we read the abacus diagram as a Dyck path. But now, the bijection from Dyck paths to nonnesting partitions (See Stanley's solution to part uu of Exercise 6.19, where he phrases the bijection essentially in terms of the root poset $\Phi^{+}$) comes into play. Rotate the Dyck path 45 degrees to get the outline of a partition that fits inside the partition $(n,n-1,n-2,\ldots 2,1)$. If this partition has a removable box in row $i$ (counting down from the top starting at $n$) and column $j$ (counting from the left starting at 1), then connect $i$ and $j$ in the resulting set partition. Then, read this set partition as in Athanaisdis-Linusson's paper - $f(x)$ is defined to be the smallest element in the block of the set partition containing $x$. For example, the $(3,4)$-core $(3,1,1)$ becomes the $3$-abacus diagram with beads at 5,2,1 - corresponding to the hook lengths of the first column. This gives the partition $(2,1)$. This partition has two removable boxes, at positions $(1,2)$ and $(2,3)$, which gives us the set partition $\{\{1,2,3\}\}$. By the A-L map, this is the parking function $(1,1,1)$. Conversely, the $(3,4)$-core $\emptyset$ becomse the $3$-abacus diagram with no beads, which gives the empty partition. This in turn corresponds to the finest set partition $\{\{1\},\{2\},\{3\}\}$, which A-L interprets as the parking function $(1,2,3)$. The key insight is that the removable boxes of the Abacus diagram (considered as a partition as we have just done) essentially do same work as the affine symmetric group action in Fishel-Vazirani. Both encode the affine "floors" of the corresponding Shi chamber - the removable boxes label them directly (the connection $(i,j)$ in the set partition means the hyperplane $x_{i} - x_{j} = 1$ is a boundary of the Shi chamber at hand), while F-V connects the floors to roots in the inversion set of the corresponding affine symmetric group element. (I should acknowledge some debt to Armstrong-Rhoades' paper The Shi Arrangement and the Ish Arrangement - a remark there pointed me to the connection between the A-L correspondence and floors of Shi chambers.)<|endoftext|> TITLE: Total positivity of $q$-Pascal matrix? QUESTION [12 upvotes]: A matrix of real numbers is called totally positive if all its minors are non-negative. A well-known example is the Pascal matrix $(\binom{i}{j})$. Is it true that the minors of the $q$-Pascal matrix $({\binom{i}{j}}_q)$ are polynomials with non-negative coefficients? REPLY [12 votes]: Yes. Consider the set $V$ of points with integer coordinates as vertices of a weighted directed acyclic graph. Namely, for any $(i,j)\in V$, the edge from $(i,j)$ to $(i+1,j)$ has weight 1, the edge from $(i,j)$ to $(i,j+1)$ has weight $q^{i}$. Then $\binom{n+k}{k}_q$ is a weighted sum of paths from the origin to $(n,k)$, it follows by induction from the $q$-Pascal identity $\binom{n+k}{k}_q=q^k\binom{n+k-1}{k}_q+\binom{n+k-1}{k-1}_q$. For integers $a,b$ and non-negative integers $n,k$, the weighted sum of paths from $(a,b)$ to $(a+n,b+k)$ equals $q^{an}\binom{n+k}{k}_q$, this is seen from comparing with paths between the origin and $(n,k)$. Now consider the minor $\binom{u_i}{v_j}_q$, $i,j=1,\dots,m$ of your matrix. Denote $A_j=(-v_j,v_j)$, $B_i=(0,u_i)$. Then the weighted sum of paths from $A_j$ to $B_i$ equals $q^{-v_j^2}\binom{u_i}{v_j}_q$. Thus our determinant is a Laurent polynomial with non-negative coefficients by Lindström–Gessel–Viennot lemma. On the other hand, the value of a minor is a genuine polynomial in $q$.<|endoftext|> TITLE: Undecidability of Diophantine equations with disjoint variables? QUESTION [5 upvotes]: Consider a special case of the Hilbert's 10th problem: $f(\vec{x})=g(\vec{y})$, where $\vec{x}$ and $\vec{y}$ are disjoint ( i.e, the LHS and RHS do not have any common variables), moreover, $f$ and $g$ are polynomials with positive coefficients. The question is, with the restrictions, whether the undecidability still holds? For example, the Pell equation $x^2=3y^2+1$ falls into this class and is difficult to solve. I am wondering whether we can obtain undecidability outright. [Important note: the Hilbert 10th question has two versions, ie., whether there is a solution in natural numbers, or in integers. In general, they are equivalent, but in this particular case, they are not. The answer below shows that checking whether there is a solution in integers is undecidable, but it is still open for the case that the solution must be natural numbers.] REPLY [9 votes]: It is undecidable. The only integral point on $x^3+x=y^2$ is $(0,0)$. Let $F(\vec{y})=0$ be undecidable diophantine equation with positive coefficients and not depending on $x$. Take $f(x)=x^3+x$ and $g(\vec{y})=F^2$ leading to $x^3+x=F^2(\vec{y})$. To get $F$ from $F'$ with negative coefficients use sum of squares replacing each negative coefficient $c_i$ with variable $v_i$ and add the square $(v_i + c_i)^2$. Even simpler, the integral solutions of $x^2=1+d^2 y^2$ are $(\pm 1,0)$.<|endoftext|> TITLE: Cohomology of ramified double cover of $\mathbb P^n$ (reference) QUESTION [17 upvotes]: Let $X\rightarrow \mathbb P^n_{\mathbb C}$ be a double cover ramified over a smooth hypersurface $B$ of degre $2d$. In the case of hypersurfaces of $\mathbb P^n$ one can determine the integral cohomology using Lefschetz hyperlane section theorem and universal coefficients theorem. Q. Are there some simple techniques allowing to compute $H^k(X,\mathbb Z)$ for any $k$? REPLY [5 votes]: Expanding a bit Jason Starr's comment: let $R_X$ (or $R_B$) the jacobian ring of $X$ (resp. $B$). Recall that if $X=V(F) \subset w \mathbb{P}(a_0, \ldots, a_n)$, the jacobian ring is defined as $$ R_X := \mathbb{C}[x_0, \ldots, x_n]/(\partial_0(F), \ldots, \partial_n(F)),$$ with the $x_i$ suitably weighted. From Griffiths result (and its extension to weighted projective hypersurfaces) the cohomology of $X$ can be recovered by looking at some homogeneous slices of $R_X$. In your situation, if $B=V(f_{2d}) \subset \mathbb{P}^n$, then $X=V(y^2+f_{2d}) \subset w \mathbb{P}(1^{n+1}, d)$. In particular it is trivial to check that $R_X \cong R_B$ and you can find all the cohomology informations for $X$ by looking only at $B$. (note that in general $H^{dim \ X}(X) \neq H^{dim \ B}(B)$, since you will have to look to different homogeneous components of the Jacobian ring)<|endoftext|> TITLE: The surreal version of $e$ QUESTION [14 upvotes]: For a sequence $(x_{\alpha})$ of surreal numbers indexed by the set of all ordinal numbers, we say that $\lim x_{\alpha}=l$ ($l$ is a surreal number) if for each surreal $\epsilon>0$, there exists an ordinal $\beta$ such that $|x_{\alpha}-l|<\epsilon$ for each ordinal $\alpha>\beta$. Consider the sequence $x_{\alpha}=(1+1/\alpha)^\alpha$, where for each surreal $x,y$, $x^y$ is defined by $\exp(y\log x)$. Q1: What is the normal form of $x_{\omega}$? Is its real part equals to $e$? Q2: What is $\lim x_{\alpha}$? REPLY [7 votes]: $\DeclareMathOperator{\ee}{e}$If $\varepsilon$ is an infinitesimal surreal, the quantity $\log(1+\varepsilon)$ is actually equal to the formal sum à la Hahn series $\sum \limits_{n \in \mathbb{N}} \frac{(-1)^n\varepsilon^{n+1}}{n+1}$, and $\log(1+\varepsilon) - \varepsilon$ is negligeable with respect to $\varepsilon$. So the proof for real numbers can be applied here to show that the sequence converges to $\ee$. As for $x_{\omega}$, this is $\exp(1 - \frac{1}{2\omega} + \frac{1}{3\omega^2} - ...)$ where $\exp(- \frac{1}{2\omega} + \frac{1}{3\omega^2} - ...)$ is infinitesimally close to $1$ because $a:= -\frac{1}{2\omega} + \frac{1}{3\omega^2} - ...$ is an infinitesimal. So the real part of $x_{\omega}$ is $\ee$. For now I don't see what its normal form because some combinatorial cleverness seems to be required in unfolding $\sum \limits_{n \in \mathbb{N}} \frac{a^n}{n!}$. I'll edit this answer if I find something. In the meantime you can try to find it too: we know that exponents of $\omega$ in the $a^n$ and $\exp(a)$ will be negative integers. It might be easier to compute the exponential sum directly if you can find relations between the coeffifients $q_{n,k}$ of $\omega^{-k}$ in $a^n$ for different values of $n$.<|endoftext|> TITLE: Another extension of Catalan: divisibility QUESTION [11 upvotes]: The classical Catalan numbers are given by $C_n=\frac1{n+1}\binom{2n}n$, for which there is a plethora of generalizations of interpretations in the literature. Still, let us consider one more such: $$C_n(k):=\binom{kn}{n,\dots,n}\prod_{j=0}^{k-1}\frac1{jn+1};$$ where the multinomial coefficient is simply $\frac{(kn)!}{n!^k}$. Note. $C_n(2)=C_n$. The following has been rediscovered many times and it is easy to prove: $C_n$ is odd iff $n=2^m-1$ for some $m$ (for example, see this paper by Postnikov and Sagan). Such considerations prompted me to ask: Question. Fix a prime $p$. Is this characterization valid? $$\text{$C_n(p)$ is not divisible by $p$ iff $n=1+p+\cdots+p^m$ for some $m$}.$$ UPDATE. Both solutions are great, in particular, I appreciated the variation of techniques that went into them. REPLY [10 votes]: Here's a slightly different way to look at this. We have $$C_n(k)=\prod_{j=1}^{k}\left(\frac{1}{(j-1)n+1}\binom{jn}{n}\right).$$ And each factor $$\frac{1}{(j-1)n+1}\binom{jn}{n}=\binom{jn}{n}-(j-1)\binom{jn}{n-1}$$ is an integer. For $C_n(p)$ let us look at the last factor $\frac{1}{(p-1)n+1}\binom{pn}{n}$. Since $\binom{pn}{n}$ is always divisible by $p$ we can say that if $C_n(p)$ is not divisible by $p$ then $p$ must divide $(p-1)n+1$, or in other words $n$ must be $1\pmod{p}$. If $n$ is written as $a_r\cdots a_1a_0$ in base $p$ then we have $a_0=1$ and by Lucas's theorem $$\binom{pn}{n-1}\equiv \prod_{i=1}^{r}\binom{a_{i-1}}{a_{i}}\pmod{p}$$ Therefore in order for $\frac{1}{(p-1)n+1}\binom{pn}{n}$ not to be divisible by $p$ we must have $\binom{a_{i-1}}{a_{i}}\neq 0 \pmod{p}$. Therefore $1=a_0\geq a_1\geq \cdots \geq a_r$ which implies that all $a_i=1$. This is equivalent to saying $n=1+p+\cdots+p^r$ for some $r$. This also takes care of the other factors since for such $n$ we have $$\binom{jn}{n}\equiv j^r\neq 0 \pmod{p}.$$<|endoftext|> TITLE: Identity with Pochhammer and harmonic numbers QUESTION [5 upvotes]: This came out of some work on the digamma function. Let $(x)_k=x(x+1)\cdots(x+k-1)$ denote the Pochhammer symbol. Then, Question. Can you prove/disprove this identity? $$\pmb{\frac{(\frac12)_j^2}{j!^2}}\sum_{i=0}^{j-1}\frac4{2i+1} =\sum_{i=0}^{j-1}\pmb{\frac{(\frac12)_i^2}{i!^2}}\frac1{j-i}.$$ I found this fascinating in view of fact that the factors in bold are able "go in and out" of the sum. REPLY [10 votes]: The identity under question can be found in the paper S. Boettner, V.H. Moll The integrals in Gradshteyn and Ryzhik. Part 16: Complete elliptic integrals, pages 11-12 https://arxiv.org/abs/1005.2941 The proof given by OP is miraculously similar to the one in this paper, not only in the method used, but also in the choice of words and formatting of equations. By the way, one of the authors V. Moll, is a frequent collaborator of the OP's, and they have written numerous papers that start in the same way The integrals in Gradshteyn and Ryzhik. Part ..., e.g. https://arxiv.org/abs/1004.2440. Coincidences happen, and with some people they happen more often than with others, this is just law of probabilities. No foul play is suspected here. But it is really intriguing to know, how is the identity under question related to digamma function, as OP claims? Unfortunately his proof does not contain any digamma functions. But in the paper the identity naturally came out of some work on the elliptic integral. Please, @T.Amdeberhan, would you share your insight? It is really intriguing to know, how is digamma function related here? Here is a screenshot of OP's own answer, just in case:<|endoftext|> TITLE: Tate's definition of residues QUESTION [20 upvotes]: In http://www.numdam.org/article/ASENS_1968_4_1_1_149_0.pdf, Tate defines residues on a curve over an arbitrary field as a trace of some commutator. What is the intuition for the definition? If I knew the definition of a residue in complex analysis what might lead me to think about traces of commutators? REPLY [12 votes]: I don't have a conceptual answer, but here is what you get by tracing through Tate's definitions. Let $V$ be the vector space of Laurent series in $t$ and $A$ the subspace of power series. We will work with linear operators $\phi: V \to V$, which we think of as $\infty \times \infty$ matrices with rows and columns labeled by $\mathbb{Z}$. We will often divide these infinte $4$ blocks. For example, the condition that $\phi(A) \subseteq A$ says that $\phi$ has block form $\left( \begin{smallmatrix} \ast & 0 \\ \ast&\ast \end{smallmatrix} \right)$. We can represent Tate's ring $E$ with ideals $E_1$,$E_2$ and $E_0$ visually as follows: $$ E = \begin{pmatrix} \ast & 0' \\ \ast & \ast \end{pmatrix} \qquad E_1 = \begin{pmatrix} 0' & 0' \\ \ast & \ast \end{pmatrix} $$ $$E_2 = \begin{pmatrix} \ast & 0' \\ \ast & 0' \end{pmatrix} \qquad E_0 = E_1 \cap E_2 = \begin{pmatrix} 0' & 0' \\ \ast & 0' \end{pmatrix} $$ where $0'$ means ``confined to finitely many rows". Let $f(t)$ and $g(t)$ be Laurent series and write $f$ and $g$ for the operations of multiplication by $f(t)$ and $g(t)$. Let $f_1$ and $g_1$ be approximations with $f_1$, $g_1 \in E_1$ and $f \equiv f_1$, $g \equiv g_1 \bmod E_2$. Tate's claim is that $\mathrm{Tr}\ [f_1, g_1] = \mathrm{Res} (fdg)$. Let's check for $f(t)=t^a$, $g(t)=t^b$, and one particular choice of $f_1$, $g_1$. Multiplication by $t^c$ is given by the matrix $$M(c)_{ij} = \begin{cases} 1 & i=j+c \\ 0 & \mbox{otherwise} \end{cases}.$$ We choose the approximation $$S(c)_{ij} = \begin{cases} 1 & 0 \leq i=j+c \\ 0 & \mbox{otherwise} \end{cases}.$$ In other words, we take $M(c)$ and change all elements in the top blocks to $0$. We note that $S(a)S(b) - S(b) S(a)$ has finitely many nonzero entries, so it makes sense to take its trace. Those entries lie on the diagonal $i=j+a+b$, so the trace is zero if $a+b \neq 0$. In the case $a+b=0$, I compute $\mathrm{Tr}\ [ S(-b), S(b) ] = b$. So we have $$\mathrm{Tr}\ [S(a),S(b)] = \left\{ \begin{matrix} b & a+b=0 \\ 0 & \mbox{otherwise} \end{matrix} \right\} = \mathrm{Res} {\Big(} x^a d(x^b) {\Big)}$$ as desired.<|endoftext|> TITLE: A Naive Question on Mixed Motives and Mixed Hodge Structures QUESTION [8 upvotes]: As a physicist, I have some naive questions about mixed motives and its mixed Hodge structure (MHS) realization. Any references, comments, answers will be appreciated! The category of mixed motives over $\mathbb{Q}$ has not been constructed, but anyway let us suppose it exists and denote it by $\mathcal{MM}_{\mathbb{Q}}$, and I want to know some expected properties of it. Every mixed motive $M$ then has a Hodge realisation, denoted by $H(M)$, which is a MHS (an obeject in the abelian category of \mathbb{Q}-MHS), i.e. a functor \begin{equation} H:\mathcal{MM}_{\mathbb{Q}} \rightarrow \mathbb{Q}\,MHS \end{equation} First, suppose $H(M)$ is the direct sum of $S_1$ and $S_2$ in the category $\mathbb{Q}$-MHS, do we expect there exist $M_1$ and $M_2$ in $\mathcal{MM}_{\mathbb{Q}}$ such that $H(M_i)=S_i$ and $M=M_1 \oplus M_2$? If yes, does this property have anything to do with Hodge conjecture? Second, suppose there is a sequence in $\mathcal{MM}_{\mathbb{Q}}$, \begin{equation} 0\rightarrow M_1 \rightarrow M \rightarrow M_2 \rightarrow 0 \end{equation} which we do not require to be exact. If its Hodge realization is exact in the category $\mathbb{Q}$-MHS, i.e. the following sequence is exact, \begin{equation} 0\rightarrow H(M_1) \rightarrow H(M) \rightarrow H(M_2) \rightarrow 0 \end{equation} Do we expect the sequence upstairs is exact! Third, Veovodsky has constructed a triangulated category which is candidate for the derived category of the assumed category $\mathcal{MM}_{\mathbb{Q}}$, denote it by $\mathcal{DMM}_{\mathbb{Q}}$, does there exist a functor which looks like a Hodge realization functor? i.e. a functor \begin{equation} \widetilde{H}:\mathcal{DMM}_{\mathbb{Q}} \rightarrow \mathbb{Q}\,MHS \end{equation} Fourth, if third is true, $\widetilde{H}$ exists, is a similar property like (First) true when we replace $\mathcal{MM}_{\mathbb{Q}}$ by $\mathbb{DMM}$? Similarly, is there a similar property like (second) when we replace $\mathcal{MM}_{\mathbb{Q}}$ by $\mathbb{DMM}$? (need to replace SES by a distinguished triangle in the question)? REPLY [3 votes]: I will try to answer. 1) I suspect that the answer is "no", but justifying this should be difficult. The Hodge conjecture gives a positive answer under the assumption that $M$ splits as the direct sum of its weight factors (one may say that $M$ is semi-pure). 2) The answer to this question as well as to its "triangulated version 4.2" should be positive. Indeed, these statements follow from the following (widely believed to be true) conservativity conjecture: the singular realization of a non-zero geometric motif (with rational coefficients) is non-zero. Note here that the exactness of an exact sequence of MHS is equivalent to that of the underlying $\mathbb{Q}$-vector spaces. Yes, there is a Hodge realization. The first construction was described by Huber (see https://pdfs.semanticscholar.org/2b04/2f81bc16df356e7efb35ac2504ef0aadd5ff.pdf and the erratum to this text); there is also a paper by Lecomte and Wach and by Ivorra on this subject. 4.1. This question seems to be easier that question 1; still I don't know how to answer it.:)<|endoftext|> TITLE: Independent families of functions on $\omega$ of size continuum QUESTION [5 upvotes]: In Hausdorff's article "Über zwei Sätze von G. Fichtenholz und L. Kantorovich''(1935) one can find the (simplified) proofs of the following two theorems: 1) There are continuum many essentially different functions (in the original version they are called wesentlich verschieden) from $\omega$ to $\omega$: i.e. there is some $H \subseteq {^\omega \omega}$ such that $|H|=2^{\aleph_0}$ and for every finitely many different $f_0, \dots, f_l \in H$ there is a position $a \in \omega$ such that $f_i(a) \neq f_j(a)$ for $i TITLE: Model existence theorem in topos theory QUESTION [12 upvotes]: One of most classical and somehow striking result in classical model theory states: A consistent first order theory $T$ has a model. Few considerations are needed. This result is not true for infinitary logics, such as geometric or second order. So it tells us something about the logic we are working with. From the perspective of a topos theorist this result tells something about the elementary topos Set. In fact, if we change topos in which we take models, some first order theories have no models. An easy example is that there is no model of Peano arithmetic in FinSet. I want to focus on the topos theorist perspective. A natural question would be the following: Characterize toposes $G$ such that any first order theory has a model in $G$ And strangely it turns out that this is linked to presentability of the topos. A consistent coherent theory (using terminology of Sketches of an Elephant) has a model in any locally presentable elementary topos $G$. Proof: There is a unique geometric morphism $\text{Set} \rightleftharpoons G $. Since a coherent theory is a first order theory, there is a model in Set, i.e. we have a geometric morphism $B(T) \rightleftharpoons \text{Set}$ so one can prolong this geometric morphism $$B(T) \rightleftharpoons \text{Set} \rightleftharpoons G $$ obtaining a model of $T$ in $G$. Same proof shows the following refinement. A first order theory which has a classifying topos has a model in any locally presentable elementary topos $G$. It would be interesting to understand how far one can goes, in any of two direction: Characterize toposes $G$ such that any first order theory $T$ which is classified by a topos has a model in $G$. How big is the class of theories that have models in any Grothendieck topos? And yet, former theorems are a partial answer: Any Grothendick topos is ok. This class contains at least first order theories which have a classifying topos. Locally presentable elementary toposes (i.e. Grothendieck Toposes) have a beautiful notion of cardinality for a model, which is its presentability rank. Find hypotesis on $B(T)$ such that the category $$\text{Mod}(T, G) = \text{Geom}(B(T),G) $$ is locally presentable or at least accessible. Maybe one can hope in the following. If $B(T)$ is locally presentable, than $\text{Geom}(B(T),G)$ is reflective in $G^{B(T)},$ so is locally presentable. $B(T)$ is very often locally presentable. When $\text{Mod}(T, G)$ is locally presentable one can state freely Shealah conjecture for this categories and try to understand what happens. But first, what about Lowenheim-Skolem theorem?! What about Lowenheim-Skolem theorem?! This result has a very partial answer in the book Topos Theory, by Johnstone. Also there are two papers from Zawadowski. How does Shelah conjecture look like in these categories of models? If these questions have not a trivial answer and someone finds them interesting, I would like to discuss them. REPLY [4 votes]: While I won't be able to give full answers to the questions, I would like to point to a few ideas which I believe are relevant. I first would like to respond to your question: ``How big is the class of theories that have models in any Grothendieck topos?''. It was a little unclear whether you intended the question to only be about first order theories or not so I will consider the full case of (set sized) theories of $L_{\infty, \omega}$. First note that if a theory It is worth pointing out that given any geometric theory $T$ which has a model in all Grothendieck toposes must also have a model in SET. Therefore any such $T$ must be classically consistent. However, if $T$ is any classically consistent theory (not just geometric) we can find an equivalent geometric theory by Morleyization. As such if there is some Grothendieck topos $G$ in which $T$ has no models then either: $T$ is classically inconsistent, or $T$ is classically consistent but has no models in SET. It is worth noting that if $T$ is a first order theory or a sentence of $L_{\omega_1,\omega}$ (2) can't happen. This is connected with Barr's Covering Theorem. Next I want to talk about your question on the Lowenheim-Skolem theorem and your observation about in local presentability giving a notion of cardinality in a Grothendieck topos. While the notion of presentability does give a nice notion of size in a Grothendieck topos, once we are in this framework there are several other notions of size which are useful. For a discussion of several of them as well as several model theoretic results that lift to Grothendieck toposes (including the Lowenheim-Skolem theorem) you might look at: On transferring model theoretic theorems of $L_{\infty, \omega}$ in the category of sets to a fixed Grothendieck topos. This paper also discusses Morley's theorem on the number of models in a Grothendieck topos: The Number Of Countable Models In Categories of Sheaves. Also if you are interested in model theory in realizability toposes you can also check out: The Number Of Countable Models In A Realizability Topos. While the specific case of presentability rank isn't discussed in these papers it should be relatively straightforward to prove a Lowenheim-Skolem theorem with respect to presentability rank. Specifically, given a theory $T$ which has a model in a Grothendieck topos $G$ it should be relatively straight forward to find a model with presentability rank at most $|T| + |(C, J)|$ where $G \cong Sh(C, J).$ Roughly the argument should go something like the following: Let $X$ be a transitive set containing $T$ along with $(C, J)$ and with $|X| = |T| + |(C, J)|$. Let $V^*$ be a transitive elementary substructure of the universe $V$ containing $X$ with $|V^*| = |X|$. As $V^*$ is an elementary substructure of $V$ and $V$ thinks there is a model $M$ of $T$ in $Sh(C, J)$ we also have $V^*$ thinks there is a model of $T$ in $Sh(C, J)^{V^*}$. Because the satisfaction relation is absolute $M$ should also model $T$ in $Sh(C, J)^V \cong G$. But $M$ should have presentability rank no more than $|V^*| = |T| + |(C, J)|$. You need to be a little careful here with what we mean by the relativization of a Grothendieck topos, but it is worked out in this paper: Relativized Grothendieck topoi. Finally by the Shelah conjecture, I assume you mean the "Shelah categoricity conjecture"? In general models of theories (even first order theories) inside a Grothendieck topos are like``models of a sentence of $L_{\infty, \omega}$ with a very little second order logic thrown in''. However to the best of my knowledge I don't think his eventual categoricity conjecture is even known for sentences of $L_{\omega_1, \omega}$ (although I could be wrong). If so this seems like a very hard problem in general.<|endoftext|> TITLE: How large is Dcris of certain twists of modular forms? QUESTION [7 upvotes]: I want to determine $\mathrm D_{\mathrm{cris}}$ of certain twists of the Galois representations attached to modular forms. For one particular twist it is not clear to me how $\mathrm D_{\mathrm{cris}}$ looks like. Let $f\in\mathrm S_k(\Gamma_1(N),\psi)$ be a newform of weight $k\ge2$, level $N$, nebentype $\psi$, and assume $p$ is a prime with $p\mid N$ such that $f$ is ordinary at $p$ and the $p$-part of $\psi$ is nontrivial. Let $V_f$ be the representation attached to $f$. Then $$V_f|_{\mathrm G_{\mathbb Q_p}}\cong\begin{pmatrix}\delta&*\\&\varepsilon\end{pmatrix}$$ with characters $\delta$ and $\varepsilon$, and $\delta$ is unramified. To be precise, I want the representation $V_f$ to be characterized by characteristic polynomials of geometric Frobenii, while I normalize class field theory arithmetically. In particular, $V_f$ has determinant $\psi^{-1}\kappa^{1-k}$, where $\kappa$ is the cyclotomic character. If we twist with $\psi\kappa^n$ ($1\le n\le k-1$), tensor with $\mathrm B_{\mathrm{cris}}$ and take cohomology we get a long exact sequence \begin{align*} 0&\rightarrow\mathrm D_{\mathrm{cris}}(\delta\psi\kappa^n)\rightarrow\mathrm D_{\mathrm{cris}}(V_f(\psi)(n))\rightarrow\mathrm D_{\mathrm{cris}}(\varepsilon\psi\kappa^n)\\ &\rightarrow\mathrm H^1(\mathbb Q_p,\mathrm B_{\mathrm{cris}}\otimes\delta\psi\kappa^n)\rightarrow\mathrm H^1(\mathbb Q_p,\mathrm B_{\mathrm{cris}}\otimes V_f(\psi)(n))\rightarrow\dotsm \end{align*} and we know: $\mathrm D_{\mathrm{cris}}(\delta\psi\kappa^n)=0$ since $\delta$ is unramified and $\psi$ is ramified, $\dim\mathrm D_{\mathrm{cris}}(\varepsilon\psi\kappa^n)=1$ since $\varepsilon|_{I_p}=(\delta\varepsilon)|_{I_p}=(\psi^{-1}\kappa^{1-k})|_{I_p}$, so $\varepsilon\psi$ is unramified. I want to determine $\mathrm D_{\mathrm{cris}}(V_f(\psi)(n))$. Clearly, if $f$ is a CM form, then $V_f$ decomposes, and it is easy to see that $\dim\mathrm D_{\mathrm{cris}}(V_f(\psi)(n))=1$ in this case. But I guess there are also cases where the space vanishes. Are there any general results about this? Maybe these cases can be characterized? More specifically, assume that we know in addition that $\mathrm H^i_{\mathrm f}(\mathbb Q,V)=\mathrm H^i_{\mathrm f}(\mathbb Q,V^*(1))=0$ for $V=V_f(\psi)(n)$ and $i=0,1$. Can we say something more under this assumption? My hope would be that $\mathrm D_{\mathrm{cris}}(V_f(\psi)(n))$ vanishes in these cases. Here $\mathrm H^i_{\mathrm f}$ should be as defined by Bloch and Kato, or as in Fukaya-Kato's paper "A formulation of conjectures on $p$-adic zeta functions in non-commutative Iwasawa theory", §2.4.2. In fact, my question arises from calculating certain characteristic polynomials as in 4.2.21 (iii) in this article and the assumption on $\mathrm H^i_{\mathrm f}$ comes from (i) there; the situation described above is the only one in which it is not clear to me how to calculate these. REPLY [4 votes]: The isomorphism class of the $G_{\mathbf{Q}_p}$-representation $V_f$ determines (up to scaling) a class in $H^1(\mathbf{Q}_p, \delta \epsilon^{-1})$. The condition that $\mathbf{D}_{\mathrm{cris}}(V_f(\psi)(n))$ is 1-dimensional is exactly requiring that this extension is in Bloch--Kato's $H^1_{\mathrm{f}}$. Now, you know that $V_f$ is de Rham (because it comes from a modular form) so the extension class automatically lies in $H^1_{\mathrm{g}}$. However, there are formulae for the dimensions of $H^1_{\mathrm{f}}$ and $H^1_{\mathrm{g}}$, and in particular the two are almost always equal. For instance, I'm pretty sure that $H^1_{\mathrm{f}}$ and $H^1_{\mathrm{g}}$ will always coincide for 1-dimensional representations that are not crystalline, since the error terms are all dimensions of subquotients of $D_{\mathrm{cris}}$. So this shows that $\mathbf{D}_{\mathrm{cris}}(V_f(\psi)(n))$ is 1-dimensional in your setting.<|endoftext|> TITLE: Is there a 1-dimensional analogue of the correspondence between the Levin-Wen and Turaev-Viro models? QUESTION [10 upvotes]: Given a spherical fusion category $\mathcal C$, the Levin-Wen model constructs a lattice field theory: to each oriented surface with a triangulation, it assigns a state space $\mathcal H$ and a Hamiltonian $H$, whose space of ground states is independent of the choice of triangulation; and the Turaev-Viro model uses $\mathcal C$ to construct a (2+1)-dimensional oriented TQFT $Z_{\mathcal C}$. Kirrilov (2011) showed that for any spherical fusion category $\mathcal C$ and oriented surface $\Sigma$, the space of ground states of the Levin-Wen model for $\mathcal C$ on $\Sigma$ is canonically isomorphic to the state space $Z_{\mathcal C}(\Sigma)$ in the Turaev-Viro TQFT. Is there an analogous result in one dimension lower? Presumably this construction would, given a semisimple commutative Frobenius algebra $A$, define a lattice model on the circle whose space of ground states can be canonically identified with $A$. Someone's probably worked on this, but I haven't found any references. REPLY [4 votes]: The 1D result is discussed in detail here: 1607.06766. (See also 1607.06504.) The analog of the TVBW invariant is the FHK state sum, which takes as input a separable algebra $A$. As Kevin points out, this algebra has an interpretation as the space of states on the interval with some boundary conditions. The algebra has a canonical special symmetric Frobenius form, which is used to build the invariant. Not all 2d TQFTs, which are classified by commutative Frobenius algebras, have an FHK state sum; but up to deformation (appropriate for gapped phases), they all do. The construction is also redundant, as the TQFT only senses the Morita class of $A$. In particular, the space of states on the circle (the vector space underlying the commutative Frobenius algebra) is the center of $A$. The analog of the LW model is the fixed-point Matrix Product State (MPS) system, which also takes as input a separable algebra. It consists of a parent Hamiltonian, whose ground states are have a simple wavefunction representation called MPS. For each module $V$ over $A$, there is a ground state $\vert\psi_V\rangle$ on the circle; the space of ground states is spanned by the indecomposable modules. Therefore, as was the case for the TQFT, the space of ground states is the center of the separable algebra $A$. The correspondence is made more explicit by extending the state sum to incorporate physical (brane) boundaries. These are associated to modules $V$ over $A$. If one evaluates the state sum for the anulus with boundary condition $V$ on one boundary circle and outgoing cut boundary on the other, one obtains the MPS state $\vert\psi_V\rangle\in A^{\otimes N}$ before projecting to the TQFT state space (the center of $A$). The parent Hamiltonian arises as the projector assigned to the cylinder. I agree with Kevin's opinion that the 1D result only becomes interesting when new structure is added. This is due to the trivial algebra $\mathbb{C}$ being the only indecomposable separable algebra in $\text{Vect}$. See the references above for systems with finite group actions. There, one works with separable algebras in $\text{Rep}(G)$ and has a Morita class of algebras for each pair $(H,\omega)$, an unbroken symmetry group and a cocycle characterizing the SPT order. For the correspondence between 1D fermionic systems and field theories that are sensitive to spin structure, see 1610.10075.<|endoftext|> TITLE: Different definitions of derived functors QUESTION [5 upvotes]: In principle one uses the notion of derived category, and the other doesn't. Suppose $F: \mathcal A \to \mathcal B$ is a left exact (additive) functor between abelian categories, and suppose the category $\mathcal A$ has enough injective objects. Then we have two kinds of terminology of derived functor: (1) The standard one follows Hartshorne's book on algebraic geometry. For an object $A\in \mathcal A$ choose an injective resolution $I^\bullet$ of $A$, i.e. an exact sequence $0 \to A \to I^0 \to I^1 \to \cdots$. Then we define a collection $\{R^iF | i \ge 0 \}$ of the so-called right derived functor by setting $$R^iF(A) := H^i (F(I^\bullet)) = \frac {\mathrm{ker}\big(F(I^i) \to F(I^{i+1})\big) }{ \mathrm{im} \big( F(I^{i-1}) \to F(I^i) \big)}$$ (2) Alternatively, one can resort to the notion of derived category (cf. Dirichlet Branes and Mirror Symmetry, sec. 4.4.5) For instance, suppose for simplicity that $\mathcal A = \mathcal B = \mathbf{Mod}(R)$ is the category of $R$-modules for a given ring $R$. Fix $P \in \mathcal A$, then it is known that the Hom functor $$F\equiv \mathrm{Hom}(P,- ) : \mathbf{Mod}(R)\to \mathbf{Mod}(R)$$ is a left exact functor. It is not hard to check that $F$ trivially induces a functor $\hat F$ on the category $\mathcal C(\mathcal A)$ of complexes of $R$-modules. Now we want to find a way to get a How functor on the derived category as follows. Given a $R$-module $M\in \mathcal A$ we take a injective resolution: $$0 \to M\to L^0 \to L^1 \to \cdots \to L^n \to \cdots$$ Then the $R$-module $M$, considered as a complex, is quasi-isomorphic to the complex $L^\bullet = \{0 \to L^0 \to L^1 \to \cdots \to L^n \to \cdots \}$. Applying the functor $\hat F$ to $L^\bullet$ yields a complex $$ 0 \to F(L^0) \to F(L^1) \to \cdots \to F(L^n) \to \cdots $$ This defines an object of the derived category $D(\mathcal A)$, which we denote $\mathbf RF(M)$. It is easy to see different choices of $L^\bullet$ yields isomorphic object of $D(\mathcal A)$. Hence this defines a derived functor $$ \mathbf RF : D(\mathcal A) \to D(\mathcal A) $$ By definition we immediately have $H^i(\mathbf RF(A)) =R^iF(A)$. My question is how to recover the derived functor $\mathbf RF(-)$ from the collection $\{ R^iF(-): i\ge 0\}$? Indeed, it is generally believed that "an object $E$ of $D( \mathcal A)$ consists of its cohomology objects $H^i(E)\in \mathcal A$ together with some "glue" which holds them together. " which I read from the second reference mentioned above. Hence, it seems that the derived functor $\mathbf RF$ defined in (2) contains more information than $\{ R^iF\}$ defined in (1), right? If so, what should be the additional information here? REPLY [9 votes]: The total right derived functor ${\bf R}F(-)$ contains a bit more information than just its individual cohomologies ${\bf R}^iF(-) = H^i({\bf R}F(-))$. This information can indeed be described as a kind of gluing data, and can be encoded as suitable $k$-invariants. For example, if $C_{\bullet}$ is a cochain complex in ${\cal A}$ of length $2$ then $C_\bullet$ is completely determined by the data of $H^0(C_\bullet), H^1(C_\bullet)$ and a suitable $k$-invariant $\alpha_{C} \in {\rm Ext}^2_{{\cal A}}(H^1(C_\bullet),H^0(C_\bullet))$. An example where this additional gluing information is important is when one is trying to compute ${\bf R}^iF(A)$ for a functor $F = G \circ H$ which is a composition of two left exact functors. In this case one has the Grothendieck spectral sequence $${\bf R}^pG({\bf R}^qH(A)) \Rightarrow {\bf R}^{p+q}F(A) .$$ While the $E_2$-page of this spectral sequence depends only on the individual derived functors $\{{\bf R}^pG(-)\}$ and $\{{\bf R}^qH(-)\}$, the differentials in this spectral sequence depend on the precise gluing data, or $k$-invariants, of the total derived functor ${\bf R}H(A)$.<|endoftext|> TITLE: Is the mirror of a hyperkaehler manifold always a hyperkaehler manifold? QUESTION [9 upvotes]: Is the mirror of a hyperkaehler manifold always a hyperkaehler manifold? What I know so far is as follows: In this paper (https://arxiv.org/pdf/hep-th/9512195.pdf) by Verbitsky, it is claimed that every hyperkaehler manifold is mirror to itself (which would meant the answer to my question is 'yes'). On the other hand, from the answer to this question - Mirror symmetry for hyperkahler manifold , it seems that this is not always the case, i.e., a K3 surface can have a nontrivial mirror. REPLY [3 votes]: The following example from Hausel–Thaddeus say that mirror hyperKahler is HyperKahler. Lets explain it Let $\mathcal M$ be the moduli space of stable $GL_n$-Higgs bundles, non-singular and hyperkahler and $\tilde {\mathcal M}$ moduli space of stable $SL_n$-Higgs bundles, non-singular and hyperkahler and ${\hat{\mathcal{M}}}:= \tilde{\mathcal M} /Γ $ is $PGL_n$-Higgs moduli space which is an orbifold($\Gamma\cong \mathbb Z_n^{2g}$) Hausel–Thaddeus , proved the following theorem about Strominger-Yau-Zaslow conjecture for the hyperKahler mirror pair $(\hat{\mathcal M}, \tilde {\mathcal M})$ \begin{array} ^\tilde{\mathcal M} & \stackrel{}{\longrightarrow} & \hat{\mathcal M}\\ \downarrow{\tilde\chi} & & \downarrow{\hat\chi} \\ \mathcal A & \stackrel{\cong}{\longrightarrow} & \mathcal A \end{array} The generic fibers $\tilde\chi^{-1}(a)$ and $\hat \chi^{-1}(a)$ are dual Abelian varieties. The pair of hyperkahler manifolds $(\tilde{\mathcal M} , J)$ and $(\hat{\mathcal M} , J)$ satisfy SYZ conjecture and mirror to each other See Mirror symmetry, Langlands duality, and the Hitchin system, Inventiones mathematicae, July 2003, Volume 153, Issue 1, pp 197–229 Hausel - Thaddeus interpreted SYZ conjecture in the context of Hitchin system and Langlands duality. Let briefly explain it Let $\pi : E \to Σ$ a complex vector bundle of rank $r$ and degree $d$ equipped with a hermitian metric on Riemann surface $\Sigma$ . Take th moduli space $$\mathcal M(r, d) = \{(A, Φ) \text{ solving }(\star)\}/\mathcal G $$ (which is a finite-dimensional non-compact space carrying a natural hyper-Kähler metric) where $$F^0_A + [Φ ∧ Φ^∗] = 0 ,\; \; \bar ∂AΦ = 0\; \; (\star)$$ Here $A$ is a unitary connection on $E$ and $Φ ∈ Ω^{1,0}(End E)$ is a Higgs field. $F^0$ denotes the trace-free part of the curvature and $\mathcal G$ is the unitary gauge group. $\mathcal M(r, d)$ is the total space of an integrable system(which can be interpreted by the non-abelian Hodge theory due to Corlette), the Hitchin fibration, together with Langlands duality between Lie groups provides a model for mirror symmetry in the Strominger-Yau and Zaslow conjecture.<|endoftext|> TITLE: Decidability and Cluster algebras QUESTION [6 upvotes]: Recall the definition of a cluster algebra, which can be seen as a (possibly infinite) graph, where each vertex is a tuple of a quiver and Laurent expressions at some of the vertices of the quiver. The edges of this graph is given by mutations, and every vertex has the same degree. Whenever the quiver is a Dynkin diagram, the graph is finite, and the Laurent polynomials have nice interpretations. However, for a general seed quiver, the graph is infinite. Given a seed, and a Laurent polynomial, is there an algorithm to determine if the given Laurent polynomial appear as a cluster variable in the cluster algebra? Or might this be undecidable? The reason it might be undecidable, is that a priori, the graph is infinite, so there is no upper bound on how many vertices we need to visit (by successive mutations), before concluding that the expression cannot be constructed, and the problem has "the same feel" as say the Collatz problem, or the problem of deciding if a set of matrices can produce the zero matrix by multiplication (the latter problem is undecidable). REPLY [2 votes]: For rank 2 it is decidable since there is a combinatorial formula for cluster variables. This case is certainly simpler than the general case, but some nice properties on the rank 2 case are conjectured to hold more generally. So, I'll write a bit about the rank 2 case and a bit about how we may hope to generalize. This does not answer the whole question, but hopefully it is still useful to someone. See Theorem 1.11 of Greedy elements in rank 2 cluster algebras by Lee, Li, and Zelevinsky. For the rank 2 cluster algebra $\mathcal{A}(a,b)$ the theorem gives this formula $$x[a_1,a_2] = x_1^{-a_1} x_2^{-a_2} \sum_{(S_1, S_b)}x_1^{b|S_2|}x_2^{a|S_1|} $$ where $S_1$ and $S_2$ are defined in terms of certain Dyck paths (see the paper for details). For each $(a_1,a_2) \in \mathbb{Z} \times \mathbb{Z}$ the theorem describes corresponding the so call greedy element of the cluster algebra. It is known that each cluster variable is a greedy element, and it is known which pairs $(a_1,a_2)$ correspond to cluster variables (see Remark 1.9). As a result we can decide if a given Laurent polynomial is a cluster variable in any rank 2 cluster algebra. The $(a_1,a_2)$ is known as the denominator vector, and such denominator vectors exists in other cluster algebras too. It is conjectured that different cluster variables have different denominator vectors (see this MO question). It seems like a hard problem, but perhaps it's possible to classifying denominator vectors corresponding to cluster variables then give a combinatorial formula for the remaining polynomial like the rank 2 case.<|endoftext|> TITLE: Tensor product of measure spaces QUESTION [6 upvotes]: For a compact topological space $X$, denote by $\mathcal{M}(X)$ the Banach space of finite signed Borel (Radon) measures on $X$ with the total variation norm. This is canonically isometric to the dual space of the space of continuous functions $C(X)$. If $X$ and $Y$ are compact spaces, then we have the isometric isomorphism $$\mathcal{M}(X) \hat{\otimes}_\pi \mathcal{M}(Y) \cong \mathcal{M}(X \times Y),$$ where on the left side, we have the completed projective tensor product (Side question: Does anybody know a reference for this statement? I only found the corresponding statement for the subspaces $L^1$, although the proof is very similar). [EDIT: this claim is incorrect in general -- see below] Now of course, one can also complete with respect to the injective tensor product, where the norm is given by $$\varepsilon(z) = \sup_{\xi, \eta} \bigl\{ (\xi \otimes \eta)(z) \mid \xi \in \mathcal{M}(X)^\prime, \eta \in \mathcal{M}(X)^\prime, \|\xi\| = \|\eta\| = 1\bigr\}$$ for $z \in \mathcal{M}(X) \otimes \mathcal{M}(Y)$ (the algebraic tensor product). However, since the measure space are dual spaces, there is also a third alternative, namely $$\tilde{\varepsilon}(z) = \sup_{f, g} \bigl\{ z(f \otimes g) \mid f \in C(X), g \in C(Y), \|f\| = \|g\| = 1\bigr\}.$$ Notice that (at least formally), the latter a much smaller norm, since die dual space of $\mathcal{M}(X)$ are huge. It seems to me that here we have the isomorphism $$\mathcal{M}(X)\hat{\otimes}_{\tilde{\varepsilon}} \mathcal{M}(Y) \cong B\bigl(C(X) \times C(Y)\bigr),$$ with the space of bounded bilinear maps on $C(X) \times C(Y)$. Is this correct? In any case, we have the inclusions $$\mathcal{M}(X) \hat{\otimes}_\pi \mathcal{M}(Y) \subseteq \mathcal{M}(X)\hat{\otimes}_{{\varepsilon}} \mathcal{M}(Y) \subseteq \mathcal{M}(X)\hat{\otimes}_{\tilde{\varepsilon}} \mathcal{M}(Y)$$ Now the questions: Is there a characterization of the middle space? Are there "good" (i.e. somewhat natural examples) that show that this inclusion is in fact strict? Edit: As Yemon Choi notices, I was possibly wrong by claiming that $\mathcal{M}(X) \hat{\otimes}_\pi \mathcal{M}(Y) \cong \mathcal{M}(X \times Y)$; in fact, it is only clear that it is an isometrically embedded subspace. Furthermore, by the comment of Matthew Daws, we have $\varepsilon = \tilde{\varepsilon}$. Edit2: Thanks to the answers below, we obtain the following picture: First, we have $$ \mathcal{M}(X) \hat{\otimes}_\varepsilon \mathcal{M}(Y) \cong \mathcal{M}(X) \hat{\otimes}_{\tilde{\varepsilon}} \mathcal{M}(Y),$$ using the fact that the inclusion of the closed unit ball of $C(X)$ into the closed unit ball of $\mathcal{M}^\prime(X)$ is dense with respect to the weak-$*$-topology on $\mathcal{M}^\prime(X)$. From this follows that the norms $\varepsilon(z)$ and $\tilde{\varepsilon}(z)$ coincide. Now due to the great answer by Matthew Daws below, the diagonal measure $\mu \in \mathcal{M}([0, 1]^2$ is an example of an element which is not contained in $\mathcal{M}([0, 1]) \hat{\otimes}_\varepsilon \mathcal{M}([0, 1])$, hence neither in $\mathcal{M}([0, 1]) \hat{\otimes}_\pi \mathcal{M}([0, 1])$. In fact studying some literature tells me that $\mathcal{M}(X \times Y)$ can be identified with the subset of integral forms of $\mathrm{Bil}(C(X) \times C(Y))$. However, it is not yet clear to me yet whether every element of $\mathcal{M}(X) \hat{\otimes}_\varepsilon \mathcal{M}(Y)$ is actually integral, i.e. contained in $\mathcal{M}(X, Y)$, and whether the inclusion $$\mathrm{Bil}(C(X) \times C(Y)) \supseteq \mathcal{M}(X \times Y)$$ is really a proper inclusion. REPLY [3 votes]: Let me work with general Banach spaces (good references are Vector Measures by Diestel and Uhl, or Introduction to Tensor Products of Banach Spaces by Ryan). For Banach spaces $E,F$ the bounded bilinear maps on $E\times F$ can be identified with the dual space of $E \hat\otimes_\pi F$ (this is a special case of the universal property of the norm $\pi$) which in turn can be identified with $B(E,F^*)$ in the obvious way. The injective tensor norm agrees with the norm induced by embedding $E^*\otimes F^*$ into $B(E,F^*)$; the closure is the approximable operators. As $M(X)$ has the (metric) approximation property we find that $M(X) \hat\otimes_\epsilon M(Y)$ is the compact operators from $C(X) \rightarrow M(Y)$. An example where $M(X) \hat\otimes_\epsilon M(Y)$ is not the bounded bilinear maps on $C(X) \times C(Y)$. Let $X=Y=[0,1]$ with Lebesgue measure and consider $T:C(X) \rightarrow L^1(X) \subseteq M(X)$. By considering e.g. the functions $f_n(t) = \exp(2\pi itn)$ we see that $T$ is not compact. Furthermore, we do not have that $M(X\times Y) \subseteq M(X) \hat\otimes_\epsilon M(Y) = K(C(X),M(Y))$. Again let $X=Y=[0,1]$ and let $\mu\in M(X\times Y)$ be $$ \langle \mu, f \rangle = \int_{[0,1]} f(t,t) \ dt \qquad (f\in C([0,1]^2) $$ so $\mu(E\times F) = |E\cap F|$ the Lesbegue measure of $E\cap F$, for $E,F\subseteq [0,1]$ measurable. This does induce a bounded linear map $T:C([0,1]) \rightarrow M([0,1])$ by $$ \langle T(f), g \rangle = \langle \mu, f\otimes g \rangle = \int_{[0,1]} f(t) g(t) \ dt \qquad (f,g \in C([0,1])). $$ But this is just the same $T$ as I considered above: the inclusion of $C([0,1])$ into $L^1([0,1]) \subseteq M([0,1])$; this is not compact. This $\mu \in M(X\times Y)$ is also an example of an element not in $M(X) \hat\otimes_\pi M(Y)$. I don't know an elementary way to see this; but the theory is interesting. Again see the references at the start. A vector measure $\lambda$ on a measure space $X$ is a countably additive map $\lambda:X\rightarrow E$, for a Banach space $E$. $\lambda$ is regular if $\varphi\circ\lambda\in M(X)$ is regular for each $\varphi\in E^*$. Every weakly compact operator $T:C(X)\rightarrow E$ has the form $$ T(f) = \int_X f \ d\lambda \qquad (f\in C(X)), $$ for a regular vector measure $\lambda$. Let $\lambda_0$ be a finite positive measure on $X$. Then $\lambda$ is absolutely continuous with respect to $\lambda_0$ exactly when $\lambda_0(E)=0 \implies \lambda(E)=0$. Say that $\lambda$ has the Radon-Nikodym property if, whenever $\lambda$ is absolutely continuous with respect to $\lambda_0$, then $\lambda = \int F \ d\lambda_0$ for some Bochner integrable $F:X\rightarrow E$. (The point is that not all vector measures have this property, in contrast to the scale case). An operator $T:C(X)\rightarrow E$ comes from some member of $M(X) \hat\otimes_\pi E$ (that is, is a nuclear operator) if and only if the representing vector measure has the Radon-Nikodym property. Again with $X=Y=[0,1]$ with $\mu$ and $T$ as before, the associated vector measure $\lambda:[0,1]\rightarrow M([0,1])$ is $$ \lambda(E) : F \mapsto |E\cap F| $$ for measurable $E,F\subseteq [0,1]$. This is absolutely continuous with respect to Lebesgue measure. If $F:[0,1]\rightarrow M([0,1])$ is a Bochner integrable function inducing $\lambda$ then $$ \langle \lambda(E), f \rangle = \int_E f(t) \ dt = \int_E \langle F(s), f \rangle \ ds \qquad (f\in C([0,1])). $$ Thus $F(s) = \delta_s$ the point-mass at $s$, for all $s\in [0,1]$. However, by the Pettis Measurability Theorem, $F$ should be essentially separably valued but clearly this is not the case, giving the required contradiction. My guess would be that you can boost this example up to show that $M(X)\hat\otimes_\pi M(Y)$ is strictly contained in $M(X\times Y)$ unless one of $X$ or $Y$ is discrete (in which case we do have equality).<|endoftext|> TITLE: Quotients of rings with finite free additive group QUESTION [6 upvotes]: Let $R$ be a ring (assumed associative and unital) whose additive group is a finitely generated abelian group. As a reduction step in a paper I'm working on, we need to know that $R$ is a quotient of another ring $S$ whose additive group is a finite rank free abelian group. We believe we have a proof, but it is rather involved. Is this something that has appeared already or at least has a reasonably short proof? (The current proof reduces to showing it for finite rings and then proves it by using the basic ring associated to $R$.) When $R$ is commutative, this has a quick proof: $R$ is generated by $r_1,\dots,r_n$ which are integral over ${\bf Z}$, and hence $R$ is a quotient of ${\bf Z}[x_1,\dots,x_n]/(f_1(x_1),\dots,f_n(x_n))$ where each $f_i$ is a monic univariate polynomial. So I'm wondering if there is something analogous in the general case. REPLY [2 votes]: For a finite ring $R$ you can write it as a quotient of the monoid ring $\mathbb ZR$ (with respect to the multiplicative structure) which has a finitely generated free additive group.<|endoftext|> TITLE: Generalised Hodge Conjecture QUESTION [6 upvotes]: Further to my question, A Naive Question on Mixed Motives and Mixed Hodge Structures that has received very good replies and suggestions, and I really appreciate it. I am going to ask a question on generalised Hodge conjecture which is closely related to last one. From conjecture 3.22 of Marc Levine's Mixed Motives in K-theory hand book https://www.uni-due.de/~bm0032/publ/MixMotKHB.pdf it conjectures a functor from Nori's mixed motives to rational MHS, \begin{equation} H:\mathcal{MM}_{\text{Nori}}(k,\mathbb{Q})\rightarrow \text{MHS}_{\mathbb{Q}} \end{equation} is fully faithful, which could be seen as the generalisation of Hodge conjecture. My first question is why it is Nori's mixed motives that fits into this conjecture? If $H$ is conjectured to be fully-faithful, then (I guess) the derived functor of $H$ (by abuse of notation) \begin{equation} DH:D^b(\mathcal{MM}_{\text{Nori}}(k,\mathbb{Q}))\rightarrow D^b(\text{MHS}_{\mathbb{Q}}) \end{equation} is also fully faithful. In D. Harrer's PhD thesis, Comparison of the Categories of Motives defined by Voevodsky and Nori https://arxiv.org/ftp/arxiv/papers/1609/1609.05516.pdf From main theorem, 7.4.17, there exists a realization functor \begin{equation} R_{\text{Nori}}: DM_{gm}(k,\mathbb{Q})\rightarrow D^b(\mathcal{MM}_{\text{Nori}}(k,\mathbb{Q})) \end{equation} I guess the composition of $R_{\text{Nori}}$ and $DH$ \begin{equation} DH \circ R_{\text{Nori}}:DM_{gm}(k,\mathbb{Q}) \rightarrow D^b(\text{MHS}_{\mathbb{Q}}) \end{equation} is the usual Hodge realisation functor of Voevodsky's motive. My second question is, is there a generalised Hodge conjecture stated using Voevodsky's category instead of Nori's mixed motive? e.g. like $DH \circ R_{\text{Nori}}$ is fully faithful? Any references, comments and answers will be fully faithfully appreciated! REPLY [6 votes]: Here are a series of comments which might help. "Why it is Nori's mixed motives that fits into this conjecture?". The usual Hodge conjecture is known to be equivalent to the full-faithfulness of the realization from pure homological motives to Hodge structures. Suppose that one sought an extension to the mixed setting, where the target is $MHS_\mathbb{Q}$. Then one would like a source which is (expected to be) abelian together with an (exact) realization functor to $MHS$. Nori's category is abelian etc. and it may be the "right" category of mixed motives, so it seems natural to use (at least to me). A side remark: Nori's Hodge conjecture is really more of an analogue rather than a strict generalization of Hodge (it doesn't imply it with unless one also assume's Grothendieck's standard conjectures or something like it). "..then (I guess) the derived functor of H ... is also fully faithful. " I don't think this follows or is even reasonable to expect. There are no higher $Ext$'s beyond $Ext^1$ in $MHS$, whereas there should be on the other side. Similar objections would apply to conjecturing your $DH\circ R_{Nori}$ is fully-faithful.<|endoftext|> TITLE: Geodesic in space of circulant matrices QUESTION [6 upvotes]: I'm trying to find the geodesic that connects the identity with some circulant, symmetric matrix $U\in\mathrm{GL}(N,\mathbb{R})$, meaning we have \begin{align} U=\left(\begin{array}{ccc} u_1 & u_2 & \cdots\\ u_2 & u_1 & \cdots\\ \vdots & \vdots & \ddots \end{array}\right)\quad\text{with}\quad b_{N-i}=b_{1+i}\,. \end{align} Given two matrices $A$ and $B$ in the Lie algebra $\mathrm{gl}(2N,\mathbb{R})$, their inner product is defined by \begin{align} g(A,B)=\mathrm{tr}(\left(\begin{array}{ccccc} a_{11} & \alpha\,a_{12} & \cdots & \alpha^2\,a_{1(N-1)}& \alpha\,a_{1N}\\ \alpha\,a_{21} & a_{22} & \cdots & \alpha^3\,a_{2(N-1)}& \alpha^2\,a_{1N}\\ \vdots & \vdots & \ddots & \vdots & \vdots \end{array}\right)\left(\begin{array}{ccc} b_{11} & \alpha\,b_{12} & \cdots & \alpha^2\,b_{1(N-1)}& \alpha\,b_{1N}\\ \alpha\,b_{21} & b_{22} & \cdots & \alpha^3\,b_{2(N-1)}& \alpha^2\,b_{1N}\\ \vdots & \vdots & \ddots & \vdots & \vdots \end{array}\right)) \end{align} where $\alpha$ is a positive real number. For other tangent vectors, we require that the metric is right-invariant. The metric has the same invariance as circulant & symmetric matrices. I want to argue that the geodesic itself $U(t)$ is itself circulant & symmetric for every $t$. Is there a simple way to prove this? After Robert's very helpful answer, I kept thinking about the following: My $U$ is actually defined by the condition $UU^\intercal=G$ where $G$ is a positive definite, circulant, symmetric matrix. Intuitively, I just thought that $U$ should be itself circulant and symmetry and therefore $U=\sqrt{G}$. However, in general there is full class of matrices $U$ satifying $UU^\intercal=G$, namely $U\to U\cdot O$ where $O\in\mathrm{SO}(N)$ with $UO(UO)^\intercal=UOO^\intercal U^\intercal=UU^\intercal=G$. In this case, there is a geodesic from $1\!\!1$ to $U$ for every $U$ in this set. I still want to argue that the $U$ from above is the one with the shortest distance and I could verify this explicitly for the case $N=2$. Is there a simple argument that generalizes this to larger $N$? REPLY [10 votes]: The answer is as follows: When $U$ is a positive-definite, symmetric, circulant matrix in $\mathrm{GL}(n,\mathbb{R})$, then there is a symmetric circulant matrix $u$ such that $U = e^u$ and the curve $\gamma(t) = e^{tu}$ (which is a positive-definite, symmetric, circulant matrix for all $t$) is a geodesic in the metric $g_\alpha$ on $\mathrm{GL}(n,\mathbb{R})$ described by the OP for each $\alpha>0$. Here is a sketch of a proof. First, fix $\alpha>0$ and define a vector space isomorphism $\phi:{\frak{gl}}(N,\mathbb{R})\to {\frak{gl}}(N,\mathbb{R})$ by $\phi(A_{ij}) = A'_{ij}$ where $A'_{ij} = \alpha^{d(i,j)}A_{ij}$ and where $d(i,j)=d(j,i)\ge0$ is the mininum distance from $i-j$ to an integer multiple of $N$. Let $g_\alpha$ be the right-invariant (pseudo-)Riemannian metric on $\mathrm{GL}(N,\mathbb{R})$ for which the inner product at the identity $I_N\in\mathrm{GL}(N,\mathbb{R})$ is $$ g_\alpha(A,B) = \mathrm{tr}\bigl(\phi(A)\phi(B)\bigr) = \mathrm{tr}\bigl(\phi^2(A)B\bigr)= \mathrm{tr}\bigl(A\phi^2(B)\bigr). $$ (Note that $g_\alpha$ is both left- and right- invariant if and only if $\alpha = 1$.) Let $C_N$ denote the (vector) space of symmetric, circulant $N$-by-$N$ matrices. Then $C_N$ is closed under multiplication and forms a commutative sub-ring of the matrix ring $M_N(\mathbb{R})$ and an abelian subagebra of the Lie algebra ${\frak{gl}}(N,\mathbb{R})$. Let $C_N^+\subset C_N$ denote the set of positive definite elements of $C_N$, so that $C_N^+$ is the connected component of $C_N\cap \mathrm{GL}(N,\mathbb{R})$ that contains $I_n$. The usual matrix exponential mapping $\mathrm{exp}(u) = e^u$ induces a bijective diffeomorphism $\mathrm{exp}:C_N\to C_N^+$. Also, note that $\phi(C_N) = C_N$. Meanwhile, calculation shows that, $\gamma:\mathbb{R}\to \mathrm{GL}(N,\mathbb{R})$ is a geodesic in the metric $g_\alpha$ if and only if $v(t) = \gamma'(t)\gamma(t)^{-1}$ satisfies the Euler equation for $g_\alpha$, namely, $$ v'(t) = - \phi^{-2}\bigl(\bigl[v(t),\phi^2(v(t))\bigr]\bigr). $$ Now, consider the curve $\gamma(t) = e^{tu}$ for $u\in C_n$. It satisfies $v(t) = \gamma'(t)\gamma(t)^{-1} = u$, and, since both $u$ and $\phi^2(u)$ belong to $C_N$, it follows that $$ 0 = v'(t) = - \phi^{-2}\bigl(\bigl[u,\phi^2(u)\bigr]\bigr) = -\phi^{-2}(0) = 0. $$ Thus, $\gamma:\mathbb{R}\to \mathrm{GL}(N,\mathbb{R})$ is a geodesic for all $u\in C_N$, as claimed.<|endoftext|> TITLE: What is the square of the Weyl denominator? QUESTION [8 upvotes]: Let $\Phi$ be a (crystallographic) root system with Weyl group $\mathcal{W}$, and $\Phi^+$ a choice of positive roots, and $$ q := \prod_{\alpha\in\Phi^+} (\exp(\alpha/2) - \exp(-\alpha/2)) = \sum_{w\in\mathcal{W}} \mathrm{sgn}(w)\,\exp(w\rho) $$ be the denominator in the Weyl character formula ("WCF"); here, of course, $\exp$ is a formal exponential, $\rho := \frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha = \sum_i\varpi_i$ is the Weyl vector, $\varpi_i$ are the fundamental weights, and $\mathrm{sgn}$ is the abelian character of $\mathcal{W}$ with value $-1$ on the reflections. Since $q$ is $\mathcal{W}$-anti-invariant (we have $w(q) = \mathrm{sgn}(w)\,q$), it follows that $q^2$ is $\mathcal{W}$-invariant. So $q^2$ can be expressed as a polynomial in the fundamental characters (the fundamental characters being the $x_i := q^{-1} \sum_{w\in\mathcal{W}} \mathrm{sgn}(w)\,\exp(w(\rho+\varpi_i))$ by the WCF); or equivalently, as a polynomial in the averages of the fundamental weights (meaning the $\frac{1}{\#\mathcal{W}} \sum_{w\in\mathcal{W}} \exp(w\varpi_i)$): see the references to Bourbaki and Lorenz in this related question. This invariant quantity $q^2$ can be considered as a kind of discriminant (see PS3 below). My question about this square denominator $q^2$ is: how can we compute it in practice (as a polynomial of the kind I just described)? Is there a convenient expression? Or perhaps, can I get LiE or Sage to compute it? I am also interested in any sort of remarks about it: for example, does it have a standard name (beyond "the square of the Weyl denominator")? Is it the (virtual) character of some naturally defined element in the Grothendieck group of representations? PS 1: I should have made it clear that I am looking for a computational approach that does not involve writing down all the $\#\mathcal{W}$ terms in $q$. (LiE is capable of doing computations on the ring of [multiplicative] $\mathcal{W}$-invariants without fully expanding them; so the question is whether we can do this for a $\mathcal{W}$-anti-invariant like $q$.) PS 2: If we let $R = \mathbb{C}[\exp(\Lambda)]^{\mathcal{W}}$, where $\Lambda := \langle\Phi^\vee\rangle^*$ is the weight lattice, be the ring of multiplicative $\mathcal{W}$-invariants (which is a polynomial ring), then the ring of multiplicative $\mathcal{W}_0$-invariants, where $\mathcal{W}_0 := \ker(\mathrm{sgn})$ is the group of rotations in the Weyl group, is the free quadratic algebra $R \oplus R q$. So asking to describe $q^2$ in $R$ is the missing bit in the presentation of this algebra. Hopefully this helps motivate the question. PS 3: In the case where $\Phi = A_n$, then $q^2$ expressed as a polynomial of $x_1,\ldots,x_n$ is exactly the discriminant of the polynomial (in the $z$ variable) $z^{n+1} - x_1 z^n + x_2 z^{n-1} + \cdots + (-1)^n x_n z + (-1)^{n+1}$ (indeed, consider a diagonal element of $SL_{n+1}$: then $x_1,\ldots,x_n$ give the elementary symmetric functions of the $n+1$ eigenvalues, whose product is $1$, and $q^2$ is their discriminant since $q$ is the Vandermonde determinant). What I'm asking for is a generalization of this to the other root systems. REPLY [3 votes]: WCR = WeylCharacterRing(['A',3], style='coroots') WR = WeightRing(WCR) PR = WR.positive_roots() q = prod([WR(1/2*x) - WR((-1/2*x)) for x in PR]) qs = q*q qs.character() 105*A3(0,0,0) - 6*A3(0,2,0) - 15*A3(0,0,4) + 27*A3(0,1,2) - 45*A3(1,0,1) - 12*A3(1,2,1) + 6*A3(1,1,3) + 9*A3(0,4,0) - 3*A3(0,3,2) + 27*A3(2,1,0) - 9*A3(2,0,2) - 15*A3(4,0,0) - 3*A3(2,3,0) + A3(2,2,2) + 6*A3(3,1,1) - 3*A3(3,0,3) https://cocalc.com/projects/25e624ba-9091-484f-af2e-71deb7120f58/files/Weyl%20denominator%20square.sagews<|endoftext|> TITLE: Isoperimetric dimension for any (metric) measure space? QUESTION [9 upvotes]: $\newcommand{\v}{\operatorname{vol}}$The isoperimetric dimension is the maximum $d$ s.t. $$\v(D)\leq C\cdot \v(\partial D)^{d/d-1}$$ for all open with smooth boundary $D\subset M$, differentiable manifold $M$, and universal constant $C$ for $M$. But it has been defined for graphs as well. Q1: Let's start with metric measure space $(X,\Sigma, \mu, d)$, what are the minimal conditions and objects we need to add to it, to have the notion of isoperimetric dimension? I think having just a measure space $(X,\Sigma, \mu)$ is not enough because we need a metric to define a perimeter length eg. via Minkoswki $\displaystyle \lim_{\varepsilon\to 0} \frac{\mu(A_{\varepsilon})-\mu(A)}{\varepsilon}$. Q2: In other words, what is an example of a measure space $(X,\Sigma, \mu)$, and different metrics $d_1,d_2$ , each $(X,\Sigma, \mu, d_i)$ having different isoperimetric dimensions $\dim_1,\dim_2$ Finally, Q3: Is there any counterexample for metric measure spaces, where we can't define isoperimetric dimension? Probably some counterexample where there is no universal constant $C$ for any $d\in [0,\infty]$. And preferably a continuum example. Because for example in the metric measure space $(\mathbb{Z}^{2},d_{discrete}, \mu_{counting })$ , sets are boundariless, but I think it is natural that the perimeter of say a 2D discrete square $Q$ is the number of edges between $Q$ and $Q^{c}$. Whereas in the continuum I am more comfortable with perimeter of a set being $$\displaystyle \lim_{\varepsilon\to 0} \frac{\mu(A_{\varepsilon})-\mu(A)}{\varepsilon}.$$ REPLY [4 votes]: The main problem in defining isoperimetric inequalities on general measure metric spaces is that in the classical manifold case one actually has to deal not just with the original "volume" measure on state space, but also with a completely different "area" measure on appropriately defined "codimension 1 surfaces", which is not a part of the original definition of a measure metric space. One way to bypass this difficulty consists in trying to mimick the classical differential geometry in the absence of a manifold structure by extending, in a rather involved way, the notion of currents. One disadvantage though is that this approach does not work for the simplest measure metric spaces there are, namely, for countable graphs. An alternative approach is based on the observation that in order to talk about isoperimetric inequalities it might be enough just to agree, for any subset of the state space, on what the size of its boundary is without necessarily specifying the boundary itself. This approach was implemented by Kaimanovich who introduced the isoperimetric inequalities for a general measure space endowed with a Markov operator (i.e., just a collection of transition probabilities associated to points of the state space). No explicit metric is required, although in natural situations the Markov operator can be defined in terms of a metric on the state space (e.g., the simple random walk in the case of graphs, or the Brownian motion in the case of Riemannian manifolds). The classical theorems on the equivalence of isoperimetric inequalities and the embeddings of the associated Sobolev spaces (see the fundamental book of Maz'ya on Sobolev spaces) then carry over to this extended setup as well. PS The Wikipedia entry on isoperimetric dimension contains a number of completely misleading mistakes. To begin with, there is no way to define it for a finite graph.<|endoftext|> TITLE: Proportion of pairs of integer polynomials with a bounded value of the resultant QUESTION [6 upvotes]: Let $f(x), g(x) \in \mathbb Z[x]$ be polynomials of positive degrees $r$ and $s$ respectively, and let $\operatorname{Res}(f,g)$ denote their resultant. Further, let $H(f)$ denote the naive height of a polynomial $f$; that is, $H(f)$ is equal to the maximum of absolute values of its coefficients. Now let us fix positive integers $r, s, H_0, R_0$. Then there are $(2H_0)^2(2H_0+1)^{r}(2H_0+1)^{s}$ pairs of integer polynomials $(f,g)$ such that $\deg f = r$, $\deg g = s$, and $\max\{H(f),H(g)\}\leq H_0$. My question is: what proportion of those pairs of polynomials satisfy the inequality $$|\operatorname{Res}(f,g)|\geq R_0?$$ I would like to derive a non-trivial lower bound on this quantity in terms of $r, s, H_0, R_0$. Was this question studied before? I would be thankful for any references. Also, if we set $g(x)$ to be equal to the derivative of $f(x)$, then this question essentially reduces to the question about the magnitude of the absolute value of a discriminant $D(f)$ of a polynomial $f(x)$. Then the question can be rephrased as follows: what proportion of polynomials $f(x)$ of degree $r$ and height at most $H_0$ satisfy the inequality $|D(f)| \geq R_0$? Perhaps, there are references to this special case? P.S. Note that it is a consequence of Hadamard's inequality that $$|\operatorname{Res}(f,g)| \leq (r+1)^{s/2}(s+1)^{r/2}H(f)^sH(g)^r,$$ so we need to demand $R_0 \leq (r+1)^{s/2}(s+1)^{r/2}H_0^{s+r}$, for otherwise the answer would trivially be zero. REPLY [4 votes]: This is much too complex a question. Firstly, the discriminant. It is a theorem of Mahler that the discriminant of a polynomial $P$ of degree $d$ is bounded above by $$d^d L(P)^{2d-2},$$ where $L(P)$ is the $L^1$ norm of the coefficient vector, which gives a bound of $$d^d (d+1)^{2d-2} H^{2d-2},$$ in terms of naive height. This bound is close to sharp - you can read Mahler's paper (free online). Mahler, K., An inequality for the discriminant of a polynomial, Mich. Math. J. 11, 257-262 (1964). ZBL0135.01702. Secondly, it is an empirical fact (which should be provable, but I don't have a proof at the moment), that for any $H$ (including $H=1$) the distribution of discriminant is lognormal, with mean about $1/2$ of the Mahler bound (this is for height $1$), and very small variance (standard deviation is around $20$ for degree $200$ and around $30$ for degree $400,$ indicating sublinear growth (or, if you prefer, possibly linear growth of variance). This should give you a reasonable estimate of the number of polynomials with discriminant whose log is not too far from the mean. For very large discriminant, we are in the world of large deviations, and I have no idea what to guess. For resultants, I don't know the analogue of Mahler's bound (I am guessing his method can be adapted). The distribution again seems log-normal, but with smaller mean (about half the mean of the discriminant) and larger standard deviation.<|endoftext|> TITLE: The first eigenfunction of Dirac operator for surface QUESTION [5 upvotes]: Let $M$ be a spherical oriented surface with Riemannian metric and with trivial spin structure. We know that the equation $$D \phi = \rho \phi$$, where $\rho: M\rightarrow \mathbb{R}$ is a real scalar function, corresponds to a conformal immersion $f:M\rightarrow \mathbb{R}^3$ (See Friedrich 03). It follows that the eigenvalue problem $D \phi =\lambda_i \phi$, where $\lambda_i\in \mathbb{R}$, gives some special immersion of this surface $f_i:M\rightarrow \mathbb{R}^3$. My questions is, if $f_1$, the immersion corresponding to the first eigenvalue, has some special meaning? Actually I'm highly interested in the Willmore energy of this immersion. I guess the this immersion would have a small Willmore energy from the inequality in (Bär 98): $$\lambda_1^2\leq \frac{\int_M H^2}{\mathrm{area}(M)}$$ However it might be too optimistic to expect that this immersion is a round sphere. Anyone has some insights for this immersion? Thank you very much. REPLY [3 votes]: First some historical comments: The first sufficiently clear reference that I know for the spinorial Weierstrass is a preprint by Rob Kusner and Nick Schmitt link. From the point of view of the results, Friedrich does not add much, the value of Friedrich's article mostly lies in its very explicit presentation in invariant language. Friedrich's article was also available to experts after Baer's article. There is some discussion about the question whether the spinorial Weierstrass representation was known before Kusner-Schmitt. Obviously, by performing some extra computations one can read off the spinorial Weierestrass representationalready in older work, and depending on how much extra work should be done, one gets another inventor. Taimanov points to Eisenhart, but I do not find it visible from Eisenhart. If $M$ is $S^2$, then every solution of $D\phi= \rho \phi$ as above comes from a conformal immersion, potentially with branching points of even order (the "even order" condition is severly neglected in the literature!). This is no longer in the case of higher genus surfaces in contrast to what you indicate above. There is also another unclear point in your question: it is not clear what is meant by a "trivial spin structure". On the sphere there is a unique one. If M is of genus at least 1, then it carries several spin structures that are the boundary of a 3-dimensional compact spin manifold, let us call them "bounding spin structures", and there is at least one spin structure which is not such a boundary. On a torus there is one non-bounding spin structure, which is often called the trivial one (including my early articles), but it is the only spin structure on the torus which defines a non-trivial element in the spin bordism group, so I try to avoid the word "trivial" in this context. On surfaces of genus at least 2 there are several non-bounding and several bounding ones, and there is no particular one which would deserve the name trivial. So you should formulate your question more precisely in order that it can be answered more precisely. Many questions related were considered within my PhD thesis, German, available on link and in my article with C. Bär Dirac eigenvalue estimates on surfaces Math. Z. 240, 423-449 (2002).<|endoftext|> TITLE: Sum of multinomals = sum of binomials: why? QUESTION [16 upvotes]: I stumbled on the following identity, which has been checked numerically. Question. Is this true? If so, any proof? $$\sum_{j=0}^{\lfloor\frac{k}2\rfloor}\binom{n-2k+j}{j,k-2j,n-3k+2j} =\sum_{j=0}^{\lfloor\frac{k}2\rfloor}\binom{n-k-2j-1}{k-2j}.$$ Here, $\binom{m}{a,b,c}$ is understood as $\frac{m!}{a!\,b!\,c!}$. REPLY [13 votes]: For convenience set $m=n-2k$. Then \begin{equation} \begin{split} \binom{n-2k+j}{j,k-2j,n-3k+2j} &= \binom{m+j}{j,k-2j,m-k+2j} \\ &= \binom{m+j}{m} \binom{m}{k-2j} \\ &= [t^j](1-t)^{-(m+1)} \cdot [t^{k-2j}](1+t)^m \\ &= [t^{2j}](1-t^2)^{-(m+1)} \cdot [t^{k-2j}](1+t)^m \end{split} \end{equation} where $[t^a]p$ is the coefficient of $t^a$ in the formal power series $p=p(t)$. Note that $[t^{2j+1}](1-t^2)^{-(m+1)} = 0$. So the left hand side is \begin{equation} \begin{split} [t^k](1-t^2)^{-(m+1)}(1+t)^m &= [t^k](1-t)^{-(m+1)}(1+t)^{-1} \\ &= \sum_{j=0}^k [t^{k-j}](1-t)^{-(m+1)} \cdot [t^j](1+t)^{-1} \\ &= \sum_{j=0}^k \binom{m+k-j}{m} (-1)^j \\ &= \sum_{j=0}^{\lfloor k/2 \rfloor} \binom{m+k-2j}{m} (-1)^{2j} + \binom{m+k-2j-1}{m} (-1)^{2j+1} \\ &= \sum_{j=0}^{\lfloor k/2 \rfloor} \binom{m+k-2j}{m}-\binom{m+k-2j-1}{m} \\ &= \sum_{j=0}^{\lfloor k/2 \rfloor} \binom{m+k-2j-1}{m-1} \end{split} \end{equation} as desired.<|endoftext|> TITLE: Tietze's extension theorem for contractible manifolds QUESTION [7 upvotes]: I've read that the Tietze's extension theorem was still valid for continuous applications from a closed subspace of a normal topological space to a contractible topological manifold (understood as Hausdorf and 2nd countable). But I can't find any clear reference for this result. What I have found is that the theorem generalize to applications from a normal space to an Absolute Retract (but for which family ? Normal spaces ? Metric spaces ? both ?), that manifolds are ANR (Abolute Neighbourhood Retract, once again for which family of spaces ?), and that Contractible ANR implies AR. Is this correct ? Is there a direct proof somewhere that Tietze generalizes to contractible manifolds ? PS: it is not a research-level question, but I can't get any answer on MathStackExchange... sorry. REPLY [5 votes]: Let us say that a topological space $Y$ is a normal absolute extensor if each continuous map $f:Z\to Y$ defined on a closed subspace $Z$ of a normal topological space $X$ extends to a continuous function $\bar f:X\to Y$. By the Tietze Theorem, the real line is a normal absolute extensor and so is the countable product $\mathbb R^\omega$ of real lines and any retract of $\mathbb R^\omega$. Since each Polish (= separable completely metrizable) space admits a closed embedding into $\mathbb R^\omega$, we conclude that Polish AR's are normal absolute extensors. This fact is explicitely written as Theorem 16.1(d) in the old book ``Theory of Rectracts'' of S.-T. Hu. Now it remains to observe that each contractible topological manifold $M$ is a Polish contractible ANR and hence a Polish AR. So, $M$ is a normal absolute extensor.<|endoftext|> TITLE: What is the need for torsion in the definition of lisse sheaves? QUESTION [7 upvotes]: I am studying the basics of constructible and lisse sheaves, and am trying to understand SGA 4, IX. As Grothendieck himself observes at the beginning of the chapter, one is forced to work with torsion sheaves, and there are several "moral" explanations for this: Serre's example of an elliptic curve with quaternionic multiplication which cannot provide a $\mathbb{R}$-valued representation of the quaternions, the fact that constructible/lisse sheaves should correspond to continuous representations of $\pi_1^\mathrm{et}$ and these are of pro-finite image, and so on. But I am trying to understand a more technical detail, basically around Lemma 2.1 of SGA 4, Chap. IX. Grothendieck fixes a topos $T$ and goes on to find sufficient conditions for a subsheaf $\mathcal{F}$ of a constant sheaf $S_T$ to be in turn (locally) constant: this depends a bit on the kind of object $S$ is. If $S$ is an abelian group, and we look at $S_T$ and $\mathcal{F}$ as being $\textbf{Ab}$-valued, then we need $S$ to be finite. If we have fixed a (noetherian) ring $A$, $S$ is an $A$-module and $\mathcal{F}$ to be $\textbf{Mod-A}$-valued, then we need $S$ to be finitely generated. These conditions turn out to be precisely those needed to make the category of locally constant, and hence of constructible, sheaves abelian. As a consequence (see Remark 2.3.1 ibid.) a constructible sheaf of $\mathbb{Z}$-modules is not the same thing as a constructible sheaf of abelian groups. I have tried to come up with examples to see that these finiteness assumptions are really needed, but did not succeed. Is there an example of a subsheaf of a locally constant sheaf with stalks (say) $\mathbb{Z}^n$, which is not locally constant as $\textbf{Ab}$-valued étale sheaf? Being a noetherian object in the category of $\mathbb{Z}$-modules or in that of abelian groups is the same thing: what is the reason for imposing that constructible $\mathbf{Ab}$-sheaves be valued in finite groups (which, in passing, kills injective objects)? Let me add that I'd prefer to see examples in an algebraic setting, so working with either étale or Zariski topology: I was able to cook up an example of an infinitely-generated locally constant sheaf and a non-locally constant subsheaf on the one-point compactification of $\mathbb{Z}$ (with discrete topology) but this is not the kind of topological space which look natural to me. Edit: In a first version, I asked also about $\mathbb{Z}_\ell$-sheaves and Nicolás adressed this in his very correct answer; but I am more interested in the true need for the finiteness conditions mentioned above. Here is the rest of the old question. In the literature, a "constructible $\mathbb{Z}_\ell$-sheaf" is usually defined as a projective system of constructible $\mathbb{Z}/\ell^n$-sheaves (plus some conditions) whereas a definition was a priori already available. What is an example of a $\mathbb{Z}_\ell$-constructible sheaf $\mathcal{F}=(\mathcal{F}_n)$ defined as a projective system which does not come from a "naive" constructible sheaf over the ring $A=\mathbb{Z}_\ell$? Why is the former the "right" definition? REPLY [2 votes]: I think you are misinterpreting things slightly. Lemma 2.1 says nothing about abelian groups. Lemma 2.1(i) is about sets, and Lemma 2.1(ii) is about A-modules. For $A = \mathbb Z$, $\mathbb Z$-modules are equivalent to abelian groups and so Lemma 2.1(ii) applies. It is only 2.1(i) that is used in Remark 2.3(i). That is used to show that a sheaf of groups that is constructible as a sheaf of sets is also constructible as a sheaf of groups - in this case we take $f$ to be the multiplication or inversion map. The last line, that $\mathbb Z_X$ is constructible as a $\mathbb Z$-module, but not as a sheaf of groups, follows immediately from the Definition 2.3, as of course $\mathbb Z$ is constant and of finite presentation, but isn't finite. Nothing about the category of sheaves of $\mathbb Z$-modules or the category of sheaves of groups forces you to do that, because they are equivalent categories. So why did Grothendieck define constructible sheaves of abelian groups to be finite? I don't know, but I believe it's exactly for all the other reasons one must work with torsion coefficients. Remember that if Grothendieck wants to talk about constructible sheaves of groups without the torsion condition, he can just work with constructible sheaves of $\mathbb Z$-modules.<|endoftext|> TITLE: Parabolic cohomology of modular groups and cup-products QUESTION [7 upvotes]: I am stuck with a technical question concerning parabolic cohomology of modular groups and cup-products on them. Basically, I am trying to understand the appendix about cohomology of Hida's book "elementary theory of $L$-functions and Eisenstein Series" and related ideas, and though Hida's exposition is very detailed, I am not illuminated. So, $\Gamma$ is a congruence subgroup of $SL_2(\mathbb Z)$ which acts freely on the Poincaré upper half-plane, for instance $\Gamma_1(\nu)$ for $\nu \geq 3$. We choose for each cusp of $\Gamma$ a generator of the subgroup fixing that cusp, and we define a subset $P$ of $\Gamma$ as the set of all conjugate of those elements. We let $M$ (and later $N$) be left $\Gamma$-modules. So we have the cohomology groups $H^i(\Gamma,M)$ defined as the homology of the usual complex $C^0(\Gamma,M) \rightarrow C^1(\Gamma,M) \rightarrow C^2(\Gamma,M) \rightarrow \dots$ These groups have a geometric interpretation, namely there are canonical isomorphisms $H^i(\Gamma,M) \simeq H^i(Y_\Gamma,\tilde M)$ where $Y_\Gamma$ is the quotient of the upper half-plane by $\Gamma$ and $\tilde M$ the locally free sheaf on it defined by $M$, and the RHS $H^i$ is sheaf cohomology. To define parabolic cohomology we consider a subcomplex $C^\bullet_P(\Gamma,M)$ of $C^\bullet(\Gamma,M)$ as follows: $C^i_P(\Gamma,M) = C^i(\Gamma,M)$ for $i=0,2,3,4,\dots$ and for $i=1$, we set $$C^1_P(\Gamma,M) = \{u \in C^1(\Gamma,M), u(\pi) \in (\pi-1) M \ \ \forall \pi \in P\}.$$ (To check that $C^\bullet_P$ is a subcomplex we just need to check that the differential sends $C^0$ into $C^1_P$ but this is trivial.) We define $H^i_P(\Gamma,M)$ as the homology of that complex. It is trivial that $H^i_P(\Gamma,M)=H^i(\Gamma,M)$ for all $i \neq 1,2$. Hida also gives a geometric interpretation of the new cohomology groups: there are canonical isomorphisms $$H^1_P(\Gamma,M) = H^1_! (Y_\Gamma, \tilde M) := Im \left(H^1_c(Y_\Gamma,\tilde M) \rightarrow H^1(Y_\Gamma,\tilde M)\right),$$ $$H^2_P(\Gamma,M) =H^2_c(Y_\Gamma,\tilde M),$$ where $H^i_c$ is the sheaf cohomology with compact support. (The second isomorphism is given in the last assertion of Prop. 2 of the appendix of Hida's book, and the first is Prop. 1 of his paper Congruences of cusp forms and special values of $L$-functions, invent. math 63 (1981).) So far, so good. Now the cup-product $H^1_c(Y_\Gamma,\tilde M) \otimes H^1_c(Y_\Gamma,\tilde N) \rightarrow H^2_c(Y_\Gamma,\tilde M \otimes \tilde N)$ induces a cup-product $H^1_!(Y_\Gamma,\tilde M) \otimes H^1_!(Y_\Gamma,\tilde N) \rightarrow H^2_c(Y_\Gamma, \tilde M \otimes \tilde N)$ (as explained in Hida's paper, cf (2.2) and the first page of section 2), hence a cup-product $$H^1_P(\Gamma,M) \otimes H^1_P(\Gamma,N) \rightarrow H^2_P(\Gamma,M \otimes N).$$ Now my question: How can this cup-product be described in terms of the definition of $H^\bullet_P$ as the homology of $C^\bullet_P$? That is, how can this cup-product be described in purely group cohomological terms? In theory, one should be able to answer this question by translating the definition of the cup-product in sheaf cohomology through all the isomorphisms involved, but this looks not very engaging. What worries me more is that I can't seem to be able to guess what the result could be, that is I am not able to define a cup-product $H^1_P \otimes H^1_P \rightarrow H^2_P$ in the only natural way I can think of, that is by defining it at the level of cochains. Indeed, the natural map $C^0(\Gamma,M) \otimes C^1_P(\Gamma,N) \rightarrow C^1(\Gamma,M \otimes N)$ does not seem to have image in $C^1_P$. So, I would be happy with any "natural" definition of the cup-product in terms of cochains, and quite ready to believe it corresponds to the one defined using geometric cup-product. REPLY [2 votes]: Since I seem to my stuck on something myself, let me take a crack at your problem. This is just a proposal, I haven't checked if this really works. Let me start with the suggestion in Dan's answer and identify $H^i_P(\Gamma, M)= H^i(X, j_*M)$, here $X=X_\Gamma$ is a smooth compactification of $Y=Y_\Gamma$. I'm conflacting $M$ with $\tilde M$ and dropping various decorations for ease of writing. Then from the triangle $$j_*M\to \mathbb{R}j_*M\to R^1j_*M[-1]\to $$ we get $$H^1_P(\Gamma, M)\to H^1(Y, M)\to \bigoplus_p H^1(C_p,M)\to H_P^2(\Gamma,M)\to0$$ where $p$ (resp. $C_p$) are (resp. circles around) the cusps. This can be translated back to group cohomology $$H^1_P(\Gamma, M)\to H^1(\Gamma, M)\to \bigoplus_p H^1(P_p,M)\stackrel{\partial}{\to}H_P^2(\Gamma,M)\to 0$$ where now $P_p=\pi_1(C_p)$. Given classes $\alpha\in H^1_P(M),\beta\in H_P^1(N)$, their product, whatever it is, would have to be $\partial(?)$. Here is a guess as to what it might be. Let us assume that the above sequence is induced by an exact sequence of suitable complexes $$0\to C_P^\bullet(M)\to C^\bullet(M)\to \bigoplus C^\bullet(P_p, M)\to 0$$ (Perhaps the ones specified in your question will work for this, I haven't checked.) Let $\alpha\in C^1_P(M),\beta\in C_P^1(N)$ denote cocycles representing classes above. Then $\alpha,\beta$ can be mapped to cocycles, and in fact coboundaries, $\alpha',\beta'$ in $\oplus C(P_p,-)$. Write $\alpha'=d \alpha''$. Try $\partial(\alpha''\cup\beta')$.<|endoftext|> TITLE: More on categories of modules over the algebraic cobordism spectrum QUESTION [6 upvotes]: I have the following questions on monoidal model structure(s) for the motivic stable homotopy category $SH(k)$ (where $k$ is a field); certainly, I am also interested in general statements concerning these matters. Which "models" for $SH(k)$ are monoidal model categories (with the extra monoid axiom fulfilled) such that the unit object (sphere spectrum) is cofibrant? For $MGl'$ being a cofibrant ring spectrum in this category that is weakly equivalent to the Voevodsky algebraic cobordism spectrum $MGl$ (over $k$) I would like to have a criterion that ensures that a cohomological functor from $SH(k)$ (into abelian groups) factors through the (triangulated) homotopy category $D^{MGl}$ of highly structured $MGl'$-modules. Any suggestions? 3.[My own idea concerning 2]. The spectrum $MGl$ possesses a universality property (see https://projecteuclid.org/euclid.hha/1251811074) that ensures that various forms of K-theory are represented by spectra that are algebras over $MGl$ in $SH(k)$. To prove that these "K-theories" factor through $D^{MGl}$ I would like to prove that certain "replacements" of these K-theory spectra are highly structured $MGl'$-modules. It seems reasonable to consider the model structure for ring spectra in the model for $SH(k)$ that is provided by http://www.math.uni-bonn.de/~schwede/AlgebrasModules.pdf Yet to use this model structure I probably need a calculation of morphisms in the corresponding homotopy category; is anything known about this? Proposition 1.10 of Hovey's unpublished http://mhovey.web.wesleyan.edu/papers/mon-mod.pdf says that for a left proper cellular monoidal model category $C$ the category of modules over a cofibrant monoid object $R$ of $C$ is (cellular and) left proper also. Does a (published) proof of this fact or something similar exist? Cf. the discussion at Properness of the category of modules over a spectrum (that represents algebraic cobordism or motivic cohomology) I believe that the category of $R$-modules is a $C$-module category in the sense of Definition 4.2.18 of Hovey's book. Does there exist a reference for this statement? REPLY [3 votes]: For (1), I recommend the paper Motivic Functors by Dundas-Rondigs-Ostvaer. They compare many different models for SH(k), with an eye towards monoidal properties (i.e. pushout product axiom, monoid axiom, axiom that cofibrant objects are flat). For (2), you probably know that these categories of modules are Quillen equivalent, going back to Schwede-Shipley's "algebras and modules" paper. Another reference is Fresse's book "Modules over operads and functors" Theorem 12.5.A (which avoids an assumption Schwede-Shipley had to make about $-\otimes_R S$ when you're changing ring from $R$ to $S$), since both categories are algebras over a $\Sigma$-cofibrant operad. Dmitri also has a version of rectification (in his paper #1 from the comments above) that does not require $\Sigma$-cofibrancy, and works for any weak equivalence of colored operads. I've got a version with Donald Yau ("Homotopical Algebraic Lifting Theorem") which is meant to study a more general phenomena than rectification (regarding lifting Quillen equivalences to categories of algebras), and provide easier-to-check conditions for levelwise cofibrant operads, but in your case the Fresse reference already does the job. For (4), a lot of those results ended up in Spitzweck's thesis, and later in Fresse's book. For those that didn't (including the one you cite), I included the version for commutative monoids in my thesis (see Theorem 4.17 in "Model structures on commutative monoids in general model categories", now accepted to JPAA). In your case, you probably don't care that the category of modules is cellular, because it's actually combinatorial, and that's even better. A reference is Beke, "Sheafifiable homotopy model categories", 2.3. For (5) this is Proposition 5.3 in Spitzweck's thesis. The operad in question (whose algebras are $R$-modules) is $\Sigma$-cofibrant. When Spitzweck says that the $J$-semi model category $Alg(O)$ is a $C$-module, this implies that, if there is a full model structure (which there is in this case, by Schwede-Shipley), then it's a $C$-module.<|endoftext|> TITLE: Geometry of Hecke Operators on Jacobi Forms? QUESTION [6 upvotes]: In something I've been thinking about recently, the following object appears: $$\mathcal{F}_{g} = \sum_{n=0}^{\infty} Q^{n} T_{n}\big( \phi_{2g-2}(\tau, z) \big)$$ where $T_{n}$ is the $n$-th Hecke operator and $\phi_{2g-2}(\tau, z)$ is a weak Jacobi form of weight $2g-2$ and index one. For context, this arises from a purely geometric standpoint in enumerative geometry. Therefore, I'm wondering what sort of geometric interpretation exists for these Hecke operators, specifically of this sort of infinite generating function of them which I show above? My interest was peaked by the following. On page 68 of (https://books.google.ca/books?id=Oy2n7wVuREwC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false) Cheng and Duncan's article in String-Math 2011, they consider a Hecke operator acting on a weak Jacobi form of weight zero and index one: $$T_{n} \phi_{0}(\tau, z) = \frac{1}{n}\sum_{ad=n, \,\,b\text{mod} d} \phi \big(\frac{a \tau + b}{d}, az\big).$$ They provide the following geometrical interpretation: "the right-hand side looks like a sum over degree $n$ maps between elliptic curves $E' \to E$. Summing over $n$ gives the instanton contribution to the free energy of the free string theory." Is there a similar geometric interpretation for all weak Jacobi forms of even weight (larger than -2) and index 1? and more simply, about the above example specifically... How exactly does that right-hand side correspond to degree $n$ covers of an elliptic curve? I certainly understand that the range of the sum looks exactly like the degree $n$ covers of a torus, but what about the summand? What role does the Jacobi form play with this interpretation? To be honest, my formula arises in a similar context to Cheng-Duncan (Gromov-Witten theory, string theory) so I'm hoping for a similar geometrical analogy. REPLY [4 votes]: The interpretation of Hecke operators on modular forms and Jacobi forms in terms of a sum over isogenies is classical - this is the worldsheet interpretation. Their "target space" interpretation in terms of genera of symmetric products and second quantization of strings was given by Dijkgraaf-Moore-Verlinde-Verlinde in 1996. The generating function of Hecke operators actually appeared earlier, in the 1980s, in the Koike-Norton-Zagier formula: $$ J(\sigma) - J(\tau) = p^{-1}\prod_{m>0,n \in \mathbb{Z}}(1-p^mq^n)^{c(mn)} $$ where $c(n)$ is the $q^n$ coefficient of $J(\tau)$. To prove this identity, one multiplies by $p$ and takes the logarithm of the right side to get $\sum_{m>0} T_m J(\tau) p^m$. Borcherds used this formula as the denominator formula of the Monster Lie algebra on the way to proving the Monstrous Moonshine conjectures. There are more recent interpretations of this formula in terms of second quantized strings, following the lead of DMVV. If you just want to know how the summand works in the weight zero case, replace the $\tau$ and $z$ variables with pairs given by an elliptic curve $\mathbb{C}/\langle 1, \tau \rangle$ and a point on it, i.e., consider the function as a function on isomorphism classes of marked elliptic curves. Degree $n$ isogenies to the curve are given by index $n$ sublattices of $\langle 1, \tau \rangle$, and if we set $d$ to be a positive generator of the intersection of the sublattice with $\mathbb{Z}$, these are given by $\langle d, a\tau + b \rangle$ as the variables range over non-negative integers satisfying $ad=n$ and $0 \leq b < d$. The point $z$ gets pulled back to $az$. If you consider nonzero weights, then you need to consider a contribution from a tensor power of the canonical bundle, and some powers of $n$ appear.<|endoftext|> TITLE: Is the completion of a CAT(0) open ball a closed ball? QUESTION [6 upvotes]: It is well-known that the completion of a metric space which is homeomorphic to a ball can be very wild; in fact, I think, every compact manifold is the closure of an open ball! But CAT(0) spaces are very different from general manifolds. If I have a CAT(0) with the path metric, that is totally bounded, and that is homeomorphic to an open ball, will its completion be a closed ball? REPLY [11 votes]: The answer is "no". Let $\Sigma$ be the suspension over Poincaré homology sphere. It admits a polyhedral $\mathrm{CAT}[1]$-metric. Let $B$ be the unit ball in the Euclidean cone $\mathrm{Cone}\,\Sigma$. Note that $B$ is a compact $\mathrm{CAT}[0]$-space; its interior is homeomorphic to the ball in $\mathbb{E}^5$ but its boundary is homeomorphic to $\Sigma$, which is not a manifold.<|endoftext|> TITLE: Simplicial mapping spaces, stable $\infty$-categories, and triangles QUESTION [5 upvotes]: Let $C$ be a stable $\infty$-category (presentable, if you like) and let $map(-,-)$ denote the simplicial mapping space. If $X \to Y \to Z$ is a fiber sequence, and $W$ is an object, when is $map(W,X) \to map(W,Y) \to map(W,Z)$ a fiber sequence? I suspect that this does not come for free. I'm more willing to believe that an internal hom object would have this property. I'm sure this is somewhere in Higher Algebra (probably in the first 300 pages), but I can't find it. REPLY [8 votes]: This is always true, even without the hypothesis of stability. In an ∞-category a fiber sequence $X\to Y\to Z$ is a pullback square $$\require{AMScd} \begin{CD} X @>>> Y\\ @VVV @VVV \\ * @>>> Z \end{CD}\,.$$ So you are pretty much asking whether the functor $\mathrm{Map}(W,-)$ preserves pullback squares. In fact it preserves all limits. In fact we can write it as the composition of the functors $$ C\to P(C)\xrightarrow{ev_W} \mathcal{S}$$ where the first arrow is the Yoneda embedding and the second is evaluation at $W$. Proposition 5.1.3.2 of Higher topos theory says that the Yoneda embedding preserves limits, while proposition 5.1.2.3 says that for any functor category (like $P(C)=\mathrm{Fun}(C^{\mathrm{op}},\mathcal{S})$) evaluation preserves (and in fact detects) (co)limits.<|endoftext|> TITLE: Are there any interesting surreal constants? QUESTION [31 upvotes]: In $\mathbf R$, we have all sorts of fascinating constant, like $e$, $\pi$, $\gamma$, ... For ordinal numbers, we have $\omega$, $\epsilon_0$, $\omega_1^{CK}$, $\omega_1$, ... Have we discovered any interesting surreal constants (that are not real or ordinal numbers)? REPLY [15 votes]: $\DeclareMathOperator{\Noo}{\mathbf{No}}$When seen as a big ordered field, $\Noo$ hasn't much to offer in terms of constants besides real numbers; indeed every other surreal can be sent to pretty much every other surreal by an automorphism preserving the reals. They also lack of interesting geometric/analytic constants simply because there are no established "geometric and analytic theories" of surreals. However, because of the profusion of inter-related notions that can be defined on surreals, the study of $\Noo$ is actually laden with encounters of specific surreals. In fact I would say they mainly go unnoticed as "constants" because they usually come in proper classes. Here are a few examples. In each of them, the simplicity relation plays a role as it allows us to chose simplest surreals satisfying given conditions: -First, there are the simplest in any non empty final segment of $\Noo$, but those are just ordinals. -Then there are constants that can be defined using sign sequences with nice properties. For instance simplest surreals $x$ such that given a surreal / strictly positive surreal $a$, $a \cdot x= x$ or $a \star x= x$ where $\cdot $ is the concatenation of sign sequences and $\star$ is the "ordinal-like" product of sign sequences. (though those examples will be real numbers or ordinals if $a$ is) -Then there are singular subclasses of $\Noo$ useful to describe various asymptotic order relations, for instance the $\omega$-map [1], and the corresponding constants, ($\omega^{\frac{1}{\omega}}$ and $\omega^{\frac{\varepsilon_0}{\omega^{\omega}}}$ for instance). -Fixed points for the corresponding maps when they have (for instance generalized $\varepsilon$-numbers, or the fixed point of $x \mapsto \omega^{-x}$[1]) -For well-behaved functions (such that they preserve o-minimality of the structure for instance), one can look at their fixed points if they have some. -Surreals corresponding to given logarithmic-exponential asymptotic classes by the correspondance of Berarducci and Mantova, and their formal derivatives (see this paper). -Infinite irreducible or prime Conway integers (for instance the surreal $1 + \sum \limits_{n \in \mathbb{N}^{>0}} \omega^{\frac{1}{n}}$, see here and here). ... I should add that many of the mentioned constants are mysterious in that it may not be clear what their sign sequence or Conway normal form may be, what their exponential is, what asymptotic growth they may represent, and to what classes they belong... [1] Harry Gonshor, An Introduction to the Theory of Surreal Numbers<|endoftext|> TITLE: The Bochner integral about a semigroup of bounded linear operators on a Banach space QUESTION [5 upvotes]: Let $T(t)$ be a semigroup of bounded linear operators on a Banach space $X$. When does the following hold $$ \int_0^t T(s)x ds = \Big(\int_0^t T(s) ds\Big)x, \quad x \in X \, , $$ where $ t \in (0,1)$? REPLY [7 votes]: Following ideas of John von Neumann, J. von Neumann, Über einen Satz von Herrn M. H. Stone, Ann. Math. (2) 33, 567-573 (1932). ZBL0005.16402, it can be shown that if a function satisfies the semigroup law and is measurable, then it is already continuous. Hence your identity is only true if the generator is bounded. See also Engel, Klaus-Jochen; Nagel, Rainer, One-parameter semigroups for linear evolution equations, Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). ZBL0952.47036., Exercise I.1.7.(2).<|endoftext|> TITLE: On the determinant of a class symmetric matrices QUESTION [15 upvotes]: Consider the matrix $2\times2$ symmetric matrix: $$ A_2=\begin{pmatrix} 1 & a_1 \\ a_1 & 1\end{pmatrix}. $$ It's clear that the restriction $|a_1|<1$ implies that $\det(A_2)>0$. Moreover, this is the best restriction on the modulus of $a_1$ with this property, for $\det\begin{pmatrix} 1 & 1\\ 1 & 1\end{pmatrix}=0$. Now, consider the $3\times3$ symmetric matrix: $$ A_3= \begin{pmatrix} 1 & a_1 & a_2\\ a_1 & 1 & a_3 \\ a_2 & a_3 & 1\end{pmatrix} $$ We have $\det(A_3)=1+2a_1a_2a_3-a_1^2-a_2^2-a_3^2$. Then, the restriction $|a_i|<1/2$ implies that $$ \det(A_3)>1-2(1/8)-3(1/4)=0. $$ Moreover, this is the best restriction on the modulus of the $a_i$'s with this property, for $$ \det\begin{pmatrix} 1 & -1/2 & -1/2\\ -1/2 & 1 & -1/2 \\ -1/2 & -1/2 & 1\end{pmatrix}=0. $$ Now we consider the general case. Let $A_n=[a_{i,j}]_{1\leq 1,j\leq n}$ be a matrix such that $a_{i,j}=a_{j,i}$ for every $1\leq i,j\leq n$ and $a_{i,i}=1$ for every $1\leq i\leq n$. What is the greatest number $\alpha_n$ satisfying $$ \max_{i\neq j}|a_{i,j}| < \alpha_n\ \ \Rightarrow\ \ \det A_n>0? $$ I was able to prove that $\alpha_n\leq 1/(n-1)$, by showing that the determinant of the following $n\times n$ matrix is $0$: $$ M_n=\begin{pmatrix} 1 & -\dfrac{1}{n-1} & -\dfrac{1}{n-1} & \dots & -\dfrac{1}{n-1} \\ -\dfrac{1}{n-1} & 1 & -\dfrac{1}{n-1} & \cdots & -\dfrac{1}{n-1} \\ -\dfrac{1}{n-1} &-\dfrac{1}{n-1}& 1 &\dots & -\dfrac{1}{n-1} \\ \vdots & \vdots & \vdots &\ddots & \vdots \\ -\dfrac{1}{n-1} & -\dfrac{1}{n-1}& -\dfrac{1}{n-1}&\dots & 1\end{pmatrix}. $$ One way to see that $\det(M_n)=0$ is to pick $v_1,...,v_n\in\mathbb R^n$ such that $\|v_i\|=1$ for every $i\in\{1,...,n\}$ and $\{v_1,...,v_n\}$ is the set of vertices of a regular $(n-1)$-symplex. One may se by induction that $\langle v_i, v_j \rangle=-1/(n-1)$ for every $i\neq j$. Therefore, $$ M_n=\left[\langle v_i, v_j \rangle\right]_{1\leq i, j \leq n}. $$ Since $v_1,...,v_n$ are the vertices of an $(n-1)$-symplex, they are linearly dependent, so $$ \det\left(\left[\langle v_i, v_j \rangle\right]_{1\leq i, j \leq n}\right)=0. $$ My intuition says that $\alpha_n=1/(n-1)$. However, I'm not able to prove that $\alpha_n \geq 1/(n-1)$. Any suggestions would be very helpful. Thanks in advice. REPLY [17 votes]: Your guess is correct. If the elements outside the diagonal have absolute values less than $1/(n-1)$, the matrix has 'diagonal dominance', thus it is nonsingular. To make the answer self-contained, I give a proof. If $x=(x_1,\dots,x_n)^t$ satisfies $Ax=0$, take $k$ such that $|x_k|$ is maximal and look at $\sum a_{ki}x_i$. The summand $a_{kk}x_k$ has greater absolute value than all other summands altogether, thus the total sum is non-zero. A contradiction. So, determinant of $A$ is non-zero. And it is non-zero if we multiply all elements outside diagonal by $t\in [0,1]$, call such a new matrix $A(t)$. Since $f(t)=\det A(t)$ is continuous, does not vanish on $[0,1]$ and $f(0)=1$, we get $f(1)>0$ as desired.<|endoftext|> TITLE: Morphism from a surface group to a symmetric group, lifted to the braid group QUESTION [21 upvotes]: Let $\Sigma_g$ be the fundamental group of the closed orientable surface of genus $g\ge 2$; let $B_n$ be the braid group on $n\ge 3$ braids; let $S_n$ be the symmetric group on $n$ letters; let $p:B_n\to S_n$ be the canonical epimorphism. Does every homomorphism $f:\Sigma_g\to S_n$ lift to $B_n$? That is, is there a homomorphism $\bar f:\Sigma_g\to B_n$ such that $f=p\circ\bar f$? This is known for $n=3$, the proof is by ad hoc elementary computations (Hector, Meigniez, Matsumoto, "Ends of leaves of Lie foliations", J. Math. Soc. Japan 57 (2005), no. 3, 753--779.) Is it true for every $n$? The question is crucial for the construction of some Lie foliations. REPLY [6 votes]: In certain very special cases, I think this can be answered. In particular, if $f:\Sigma_g \twoheadrightarrow S_n$ is onto, $n\geq 4$ and $g \gg 0$, then Theorem 6.20 of Dunfield-Thurston implies that the map $f$ is determined up to the action of the mapping class group by the image $f_\ast:H_2(\Sigma_g) \to H_2(S_n) \cong \mathbb{Z}/2$ (when $n\geq 4$). In the case that the image is zero, then $f$ can be chosen to factor through a handlebody of genus $g$, and since the map factors through a free group, this will lift to $B_n$ since there is no obstruction. If $f_\ast$ is non-trivial, then there will be a lift iff $\mathbb{Z}\oplus \mathbb{Z}/2 = H_2(B_n)\twoheadrightarrow H_2(S_n)=\mathbb{Z}/2$ is surjective. I'm pretty sure that this is true, and it should be represented by a torus whose fundamental group is generated by $\sigma_1, \sigma_3$ in the standard braid group generators. One need only check that this torus maps homologically non-trivially into $H_2(S_n)$, which I think follows from the presentation of the double cover of $S_n$. If $f$ is not onto, then one could still attempt to apply Theorem 6.20 to its image. Let $f(\Sigma_g)= H < S_n$. Then Theorem 6.20 implies that for $g$ large enough, $f$ is classified up to the mapping class group by the image of $f_*: H_2(\Sigma_g)\to H_2(H)$, up to the action of $Out(H)$. Let $\tilde{H} = p^{-1}(H)$ be the preimage of $H$ in $B_n$. If $p_{|\ast}: H_2(\tilde{H}) \to H_2(H)$ is not onto (again, up to the action of $Out(H)$), then one could find a counterexample. I think there's a good chance of such a subgroup existing.<|endoftext|> TITLE: Identity involving an improper integral (with geometric application) QUESTION [7 upvotes]: Is it (for some reason) true that $\lim_{c\to 0^+}\int_c^{\pi/2}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt=\frac{\pi}{2}$? Numerical evidence (from Mathematica): when $c=1/5$, the integral is $\approx 1.578$. when $c=1/10$, the integral is $\approx 1.575$. when $c=1/100$, the integral is $\approx 1.571$. Geometric motivation: By Clairaut's relation, if for some $T$, $$\int_c^{T}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt=\frac{\pi}{2},$$ then there are two three distinct geodesics between the points $(T,0, \tfrac{1}{2}T^2)$ and $(-T,0, \tfrac{1}{2}T^2)$ on the paraboloid of revolution $z=\tfrac{1}{2}r^2$: one through the point $(0,0,0)$, $(0, c, \tfrac{1}{2}c^2)$ and one through the point $(0, -c, \tfrac{1}{2}c^2)$. Edit: Deleted the comment about injectivity radius which concluded my original question -- it was false. Note that the answers of Nemo and Christian Remling, below, appear to work for any integration bound $T$, not just for $T=\pi/2$: for any $T>0$, $$\lim_{c\to 0^+}\int_c^{T}\frac{c}{t}\sqrt\frac{1+t^2}{t^2-c^2}dt=\frac{\pi}{2}.$$ REPLY [4 votes]: You can, if you want, calculate the integral. By performing the change of variables $$ t\mapsto \sqrt{\frac{1+t^2}{t^2-c^2}}. $$ you end up with a standard integral. For $0 TITLE: Finding the largest number which cannot be the sum of the labels of the Petersen graph QUESTION [9 upvotes]: The vertices of the Petersen graph (or any other simple graph) can be labelled in infinitely many ways with positive integers so that two vertices are joined by an edge if, and only if, the corresponding labels have a common divisor greater than 1. It has been shown (A labelling of the vertices of the Petersen graph with integers) that the smallest possible sum of these labels is 37294. Is there a largest number which cannot be the sum of the labels of the Petersen graph in such a labelling? If so, what is it? REPLY [10 votes]: I can prove that all sufficiently large integers are representable. Firstly, observe that there is a unique way, up to isomorphism, to choose three edges $p, q, r$ of the Petersen graph such that the subgraph induced by their six endpoints is just the union of $p, q, r$. Now, observe that we can write $2^{16}$ as the sum of semiprimes in three ways such that no prime is repeated: $$ 101 \times 103 + 13 \times 4241 = 65536 $$ $$ 107 \times 109 + 17 \times 3169 = 65536 $$ $$ 3 \times 7 + 5 \times 13103 = 65536 $$ Label the other 12 edges of the Petersen graph with these 12 distinct primes, such that, for each edge $s \in \{p, q, r \}$, the sum (over both endpoints) of the products (over both incident edges excluding $s$) is equal to $65536$. Moreover, do this labelling in such a way that the edges $5, 13, 17$ are incident at a vertex $v$. Now, for each of the $2^{17}$ residue classes $R_i := \{ 2^{17}x + i : x \in \mathbb{N} \}$, we can find some $k_i \in R_i$ which is coprime to each of the aforementioned 12 primes (by Chinese Remainder Theorem). Let $K = \max \{ k_i : 0 \leq i < 2^{17} \}$ be a huge upper bound. We're going to label each vertex with the product of the incident edges, with the exception of $v$ which will be labelled by $(5 \times 13 \times 17)k_i$ for some $i$. This means that we can attain any integer of the form: $$ C + (5 \times 13 \times 17) k_i + 2^{16} (p + q + r) $$ for any $0 \leq i < 2^{17}$, where $C$ is a universal constant (the sum of the labels at the 3 vertices which are neither $v$ nor endpoints of $p,q,r$) and $p,q,r$ are distinct primes larger than $K$. But by Vinogradov's theorem, any sufficiently large odd integer $N$ can be written as $p + q + r$ for three distinct primes larger than $K$ (because the number of representations asymptotically dominates $N^2$, so we can throw away the $O(N^2)$ representations involving repeated primes or primes beneath $K$). So to represent an integer $I$, choose $k_i$ to make the following true: $$ C + (5 \times 13 \times 17) k_i + 2^{16} \equiv I \mod 2^{17} $$ Then $I - C - (5 \times 13 \times 17) k_i = 2^{16} N$, where $N$ is a large odd integer expressible as the sum of three distinct primes larger than $K$. The result follows.<|endoftext|> TITLE: exponential sum over variety QUESTION [11 upvotes]: I am wondering where to find a good reference for bounds of the type $$\sum_{x\in V(\mathbb{F}_p)} \chi(g(x))\psi(f(x))$$ where $V$ is a variety, $\chi$ is a multiplicative character over $\mathbb{F}_p$ and $\psi$ is an additive character over $\mathbb{F}_p$. What conditions for V, $f$, $g$ gives what kind of bounds? When $\chi$ or $\psi$ is trivial, what are known? Similarly, for nontrivial characters, what are known? REPLY [17 votes]: This is a very general sort of problem and you will get very different answers depending on exactly which $V, f,g$ you need. The least savings you could ask for is $\sqrt{p}$ savings over the trivial bound, i.e $O(p^{\dim V- 1/2})$. This essentially always applies, and can be proved without reference to etale cohomology theory, by reducing to the one-dimensonal case - see the paper of Weil. The most savings you could hope for in any nontrivial case is square-root cancellation, i.e. $O(p^{ \dim V/2})$. This is expected to happen for most $V, f,g$, but nothing approaching a classification of cases where it fails is known. Instead we know a number of different results giving specific conditions, usually involving some kind of nonsingularity at infinity, that imply square-root cancellation. Some of the most general work in this area is due to Katz. The $\chi$ trivial case is studied here. The case $V = \mathbb A^n$ is studied here. I think you can find more references by following the links in those. If your case is similar to these but is not exactly covered, then perhaps the methods can be modified, and you can always use slicing to reduce to a lower-dimensional case at the cost of a somewhat weaker bound. Often in practice you end up with a sum of this type where the polynomials do not satisfy any kind of nonsingularity condition but instead have other kinds of structure (e.g. multiplicative or additive symmetries). Then one could try to bring to bear any of the many tools of $\ell$-adic sheaf and etale cohomology theory to the problem. This is hard to summarize as it can get very messy.<|endoftext|> TITLE: Antipode of Hopf algebra in braided monoidal category is an algebra antihomomorphism? QUESTION [10 upvotes]: I realize that the question posed in the title has already been addressed here: Identities that connect antipode with multiplication and comultiplication, where the graphical calculus proof provided by Majid is cited. However, I am wondering if there is a more algebraic proof that can be formulated without the use of graphical calculus. If we have a Hopf algebra $(H, \mu, \eta, \Delta, \varepsilon, S)$ in the category of vector spaces, where $\mu$ is the multiplication map, $\eta$ is the unit map, $\Delta$ is the comultiplication, $\varepsilon$ is the counit, and $S$ is the antipode, then one can show that $S \circ \mu$ is a left convolution inverse of $\mu$ and that $\mu \circ \tau \circ (S \otimes S)$ is a right convolution inverse of $\mu$ (where $\tau$ is the flip map). From here, then $S \circ \mu = \mu \circ \tau \circ (S \otimes S)$. I am aiming to replicate the above proof for an arbitrary braided monoidal category, replacing the flip map $\tau$ with a braiding $\sigma$. Showing that $S \circ \mu$ is a left convolution inverse of $\mu$ is easy, just using that the comultiplication $\Delta$ is an algebra homomorphism. I can not seem to get that $\mu \circ \sigma \circ (S \otimes S)$ is a right convolution inverse of $\mu$, however. We should have \begin{align*} &\mu \ast (\mu \circ \sigma \circ (S \otimes S))\\ &= \mu \circ (\mu \otimes \mu) \circ (\operatorname{id} \otimes \operatorname{id} \otimes \sigma) \circ (\operatorname{id} \otimes \operatorname{id} \otimes S \otimes S) \circ (\operatorname{id} \otimes \sigma \otimes \operatorname{id}) \circ (\Delta \otimes \Delta) \end{align*} but I can not see how to get anywhere with this, particularly trying to figure out what to do with the map $\operatorname{id} \otimes \operatorname{id} \otimes \sigma$. It seems as though the antipode equation for $H \otimes H$ should be used, but I am not quite sure how to implement it. Any ideas would be appreciated. REPLY [4 votes]: This idea should work and Majid's proof uses this. I think your reasoning illustrates how the proof is related to the one in the (non braided) Hopf algebra case. You can verify that $\mu\circ \sigma\circ(S\otimes S)$ indeed is a right convolution inverse of $\mu$ as follows. First, by naturality of $\sigma$ twice you get \begin{align*} &\mu \circ (\mu \otimes \mu) \circ (\operatorname{id} \otimes \operatorname{id} \otimes \sigma) \circ (\operatorname{id} \otimes \operatorname{id} \otimes S \otimes S) \circ (\operatorname{id} \otimes \sigma \otimes \operatorname{id}) \circ (\Delta \otimes \Delta)\\ &=\mu \circ (\mu \otimes \mu(S\otimes\operatorname{id})) \circ (\operatorname{id} \otimes \operatorname{id} \otimes \sigma) \circ (\operatorname{id} \otimes \sigma \otimes \operatorname{id}) \circ ((\operatorname{id}\otimes S)\Delta \otimes \Delta). \end{align*} Then apply naturality of braiding (after using the Yang-Baxter equation of the braiding) to $\Delta$, yielding that the above equals (if you also use associativity of multiplication twice) \begin{align*} \mu\circ(\mu\otimes \operatorname{id})\circ(\operatorname{id}\otimes \mu(\operatorname{id}\otimes S)\Delta\otimes \operatorname{id})\circ(\operatorname{id}\otimes \sigma)\circ((\operatorname{id}\otimes S)\Delta \otimes \Delta). \end{align*} Here, you can find the term $\mu(\operatorname{id}\otimes S)\Delta$ which equals $1\circ\varepsilon$ and simplifies everything (under use of the axioms of the unit), to $$(\mu(\operatorname{id}\otimes S)\Delta)\otimes \varepsilon$$ which now gives $1\circ(\varepsilon\otimes \varepsilon)$. You need to convince yourself that if a map has a right and left convolution inverse, they are equal. The same proof of this fact also works in a braided monoidal category as it only uses (co)associativity. Note that this does not require that $H\otimes H$ is a Hopf algebra (which it is not in general). However, $H\otimes H$ is both a coalgebra and algebra and this is all we need. The above reasoning will again be more clear to me when using graphical calculus. Without it, I would have a hard time coming up with the steps necessary: (Note that the functional identities above do repress $\circ$ when it appears within tensor product legs)<|endoftext|> TITLE: Goldman Lie algebra of a bordered surface vs. a closed surface? QUESTION [5 upvotes]: How are the Goldman Lie algebra of a closed surface $\overline{S}$ and the bordered surface $S$ obtained by taking $\overline{S}$ and removing an open disc (or more generally, $n$ disjoint discs) related? REPLY [3 votes]: I am not really sure what you are asking, but the main theorem in Section 5.17 in Goldman's paper Invariant functions on Lie groups and Hamiltonian flows of surface group representations seems to imply that adding boundary components to $\overline{S}$ introduces central elements to the Lie algebra (but other changes occur too since the fundamental group becomes free). However, quotienting by the ideal generated by the differences of these new central elements and the homotopy class of the identity (or any scalar really) will result in another Lie algebra which should be isomorphic to the original Lie algebra in terms of the closed surface. This should be in parallel with the fact that the Lie algebra on free homotopy classes in $\overline{S}$ induces a symplectic structure on the (smooth locus of the) moduli space of homorphisms from $\pi_1(\overline{S})$ to a Lie group $G$ (admitting an Ad-invariant bilinear form, for example reductive) and introducing boundaries to $\overline{S}$ turns the symplectic structure into Poisson structure. Fixing the invariant functions in terms of the boundary classes (the resulting Casimirs) to be the value at the identity then give a symplectic leaf isomorphic to the original symplectic moduli space of the closed surface.<|endoftext|> TITLE: In the group ring $\mathbb{Z}_p [G]$, what elements satisfy $(\sum a_g g)^p = \sum a_g g^p$? QUESTION [6 upvotes]: Here $\mathbb{Z}_p$ is the ring of integers in $\mathbb{Q}_p$. Preferably I would want to know this for a general group $G$, but I have been concentrating on the case $G = (\mathbb{Z} / p^n \mathbb{Z})^{\times}$ as a starting point. I have also looked at the case when $\sum a_g g$ has finite order and I deduced that this implies $\sum (a_g g)^p = \sum a_g g^p$ (I wrote $\sum a_g g$ as $\sum p^i a_{g,i} g$ so we now have $1 = (\sum p^i a_{g,i} g)^k$. By equating coefficients we see that $\sum a_{g,0} g$ is both a unit and a zero-divisor unless $\sum a_{g,i} g = 0$ for all $i>0$. Finally we just equate coefficients of $(\sum a_g g)^p = \sum a_g g^p$.) Any help on the general case or special case would be greatly appreciated. REPLY [3 votes]: If $G$ has no $p$-torsion, then the Jacobian matrix of this system of equations is invertible modulo $p$ (the derivative of the left side vanishes mod $p$, and the derivative of the right side is a permutation matrix). Hence each mod $p$ solution lifts to a unique solution over $\mathbb Z_p$. If in addition $G$ is abelian, then every function mod $p$ is a solution. We can calculate the solutions in this case using Geoff Robinson's answer. Because in this case composing with the $p$th power is a permutation with the character, the Fourier coefficient associated to each character is a $q-1$ root of unity or zero, where $q$ is the order of the finite field over which that character, modulo $p$, is defined. So we can calculate the lift by taking an arbitrary mod $p$ function on $G$, applying the Fourier transform, obtaining a function on $\hat{G}$, taking Techmuller lifts, and applying the inverse Fourier transform. EDIT : I think my first sentence is actually only true when $G$ is abelian and $p$-torsion free. We can write the derivative as a sum of $p$ terms, which are identical if $g$ is commutative.<|endoftext|> TITLE: Limitations of determinacy hypotheses in ZFC QUESTION [7 upvotes]: When considering (set-theoretic) games, we have three parameters we can adjust: Definability of the payoff set The set of legal moves The length of the game When working in $\textsf{ZFC}$, what are our limitations on the above three parameters? When do we reach inconsistencies? Some more specific subquestions could be: What is the least $\alpha$ such that open (or analytic) determinacy on $\omega$ of length $\alpha$ is inconsistent, if such an $\alpha$ exists? Same question as above but instead consider analytic determinacy on $\alpha$ of length $\omega$. I have found some partial results: If we have no definability requirement then we can't even have determinacy of length $\omega$ games on $\omega$ (as this is just $\textsf{AD}$). If we consider $\textsf{OD}$ games then (according to wiki) it's consistent relative to the sharp of a Woodin limit of Woodins that games on $\omega$ of length $\omega_1$ are determined, but it's inconsistent to consider length $\omega_1+\omega$ games. Is this relative to $\textsf{ZFC}$? (EDIT: Yes it is) Caicedo mentions here that it's consistent that all $\textsf{OD}$ games on ordinals of length $\omega$ are determined - but again, is this relative to $\textsf{ZFC}$? I suppose not. Also, here Noah Schweber asks the same question, but where we fix the length to be $\omega$, the legal moves to be $\mathbb R$ and then ask about the definability. EDIT: I've written up an overview based on Juan's answer below and by taking a closer look at the complexity of the payoff sets in the non-determinacy proofs - this can be found here. Here are a few diagrams to illustrate. REPLY [6 votes]: Here is a non-exhaustive list of limitations. I'm including some from ZF alone, for completeness. As you mentioned, naturally there is a non-determined game of length $\omega$, in ZFC. Without assuming choice, there is a non-determined game of length $\omega_1$. Also without assuming choice, there is a non-determined game of length $\omega$ with moves in $\omega_1$ (this is due to Mycielski, I believe) and a non-determined game of length $\omega$ with moves in $\mathcal{P}(\mathbb{R})$. Back to ZFC, there a definable non-determined game of length $\omega$ with moves in $\mathcal{P}(\mathbb{R})$. One can consider long games on $\omega$ (or on $\mathbb{R}$, it becomes equivalent), but one needs to impose definability. Yes, all definable games of length $\omega$ can be determined, but you can get more - Woodin showed that if there is an iterable model of ZFC with a countable (in V) Woodin cardinal which is a limit of Woodin cardinals, then it is consistent that all ordinal-definable games of length $\omega_1$ are determined. Note that this is a proof of consistency. The assertion "all ordinal-definable games of length $\omega_1$ are determined" is not provable from large cardinals alone (this follows from a theorem of Larson). As for (definable) games of length $\omega$ with moves in $\mathbb{R}$, they are subsumed by long games, so they can be determined (provably from large cardinals). Determinacy for, say, projective games follows from $\omega^3$-many Woodin cardinals (this is due to Neeman). Beyond this, there is a definable game of length $\omega_1+\omega$ that is not determined. Regarding your specific subquestions, it depends on how one defines analyticity for those games. If one naively defines it as in Baire space, then of course there is a non-determined "open" game of length $\omega+1$ (since there is a non-determined game of length $\omega$). If one means "analytic" as in "coded by an analytic set," then those games can be determined for every countable $\alpha$ by the above remarks (for long games, this is due to Neeman; for games on $\alpha$, one simply uses a bijection from $\omega$ to $\alpha$ to code the games).<|endoftext|> TITLE: Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$? QUESTION [29 upvotes]: For $n\geqslant m>1$, the integral $$I_{n,m}:=\int\limits_0^\infty\dfrac{\tanh^n(x)}{x^m}dx$$ converges. If $m$ and $n$ are both even or both odd, we can use the residue theorem to easily evaluate it in terms of odd zeta values, since the integrand then is a nice even function. For example, defining $e_k:=(2^k-1)\dfrac{\zeta(k) }{\pi^{k-1}}$, we have $$ \begin{align} I_{2,2}&= 2e_3 \\ \\ I_{4,2}&= \dfrac83e_3-4e_5 \\ \\ I_{4,4}&= -\dfrac{16}{3}e_5+20e_7 \\ \\ I_{6,2}&=\dfrac{46}{15}e_3-8e_5+6e_7 \\ \\ I_{6,4}&=-\dfrac{92}{15}e_5+40e_7-56e_9 \\ \\ I_{6,6}&=\dfrac{46}{5}e_7-112e_9+252e_{11} \\ \\ I_{3,3}&= -e_3+6e_5 \\ \\ I_{5,3}&= -e_3+10e_5-15e_7 \\ \\ I_{5,5}&= e_5-25e_7 +70e_9 \\ \\ &etc. \end{align}$$ But: Is there a closed form for $I_{3,2}=\int\limits_0^\infty\dfrac{\tanh^3(x)}{x^2}dx$? I am not sure at all whether nospoon's method used here or one of the other ad hoc approaches can be generalized to tackle this. If the answer is positive, there might be chances that $I_{\frac32,\frac32}$ and the like also have closed forms. REPLY [41 votes]: Following the suggestion I made in a comment, the integral can be rewritten as the contour integral $$ I_{3,2} = \frac{1}{2\pi i} \oint \frac{\operatorname{tanh}^3 z}{z^2} \log(-z) \, dz , $$ where the clockwise contour tightly encircles the positive real axis, which coincides with the branch cut of the logarithm. The reason that this integral is equivalent is because the branch jump across the real line of $\frac{1}{2\pi i} \log(-z)$ is precisely $1$. The integrand has poles at all $z=\pm i\pi(k+\frac{1}{2})$, $k=0,1,2,\ldots$. Evaluating the residues we find \begin{align*} I_{3,2} &= \sum_{k=0}^\infty \frac{8\log\pi(k+\frac{1}{2})}{\pi^2 (2k+1)^2} - \frac{96 \log\pi(k+\frac{1}{2})-80}{\pi^4 (2k+1)^4} \\ &= \frac{5}{6} - \gamma - \frac{19 \log 2}{15} + 12 \log A - \log\pi + \frac{90 \zeta'(4)}{\pi^4} \\ &= 1.1547853133231762640590704519415261475352370924508924890\ldots \end{align*} The last two lines can be checked with Wolfram Alpha, where $\gamma$ is the Euler-Mascheroni constant, and $A$ is the Glaisher constant. Edit: Using $\gamma =12\,\log(A)-\log(2\pi)+\frac{6}{\pi^2}\,\zeta'(2)$, this can be simplified to $$I_{3,2}=\frac{5}{6} - \frac{4\log 2}{15} -\frac{6\zeta'(2)}{\pi^2}+ \frac{90 \zeta'(4)}{\pi^4},$$ that is $$\boxed{I_{3,2}=\frac{5}{6} - \frac{4\log 2}{15} -\frac{\zeta'(2)}{\zeta(2)}+ \frac{\zeta'(4)}{\zeta(4)}}. $$ And this form may hint to the existence of similar closed forms for other integrals $I_{n,m}$ with $n+m$ odd. Edit: Indeed... Numerically, it looks like e.g. $$ I(5,2)=\frac1{3}\Bigl(\frac{8}{5}-\frac{44}{63}\log2-3\frac{\zeta'(2)}{\zeta(2)} +5 \frac{\zeta'(4)}{\zeta(4)}-2\frac{\zeta'(6)}{\zeta(6)} \Bigr)$$ $$ I(7,2)=\frac1{45}\Bigl(\frac{7943}{420}-\frac{428}{45}\log2- 45\frac{\zeta'(2)}{\zeta(2)}+ 98\frac{\zeta'(4)}{\zeta(4)}- 70\frac{\zeta'(6)}{\zeta(6)}+ 17\frac{\zeta'(8)}{\zeta(8)}\Bigr)$$ $$ I(9,2)=\frac1{315}\Bigl(\frac{71077}{630}-\frac{10196}{165}\log2- 315\frac{\zeta'(2)}{\zeta(2)}+ 818\frac{\zeta'(4)}{\zeta(4)}- 798\frac{\zeta'(6)}{\zeta(6)}+ 357\frac{\zeta'(8)}{\zeta(8)}- 62\frac{\zeta'(10)}{\zeta(10)}\Bigr)$$ $$I(11,2)=\frac1{14175}\Bigl(\frac{2230796}{495}-\frac{10719068}{4095}\log2- 14175\frac{\zeta'(2)}{\zeta(2)}+ 41877\frac{\zeta'(4)}{\zeta(4)}- 50270\frac{\zeta'(6)}{\zeta(6)}+ 31416\frac{\zeta'(8)}{\zeta(8)}- 10230\frac{\zeta'(10)}{\zeta(10)}+1382\frac{\zeta'(12)}{\zeta(12)}\Bigr)$$ $$I(13,2)=\frac1{467775}\Bigl(\frac{270932553}{2002}-\frac{25865068}{315}\log2- 467775\frac{\zeta'(2)}{\zeta(2)}+ 1528371\frac{\zeta'(4)}{\zeta(4)}- 2137564\frac{\zeta'(6)}{\zeta(6)}+ 1672528\frac{\zeta'(8)}{\zeta(8)}- 771342\frac{\zeta'(10)}{\zeta(10)}+197626\frac{\zeta'(12) }{\zeta(12)}-21844\frac{\zeta'(14) }{\zeta(14)}\Bigr)$$ Here the initial denominators are chosen as the least common denominators of the $\frac{\zeta'(2k)}{\zeta(2k)}$ terms, such that inside the big parentheses, we have integer coefficients here. It turns out that the sequence of those denominators, viz. $1, 3, 45, 315, 14175, 467775$, coincides up to signs with A117972, the numerators of the rational numbers $\frac{\pi^{2n}\;\zeta'(-2n)}{\zeta(2n+1)}$. Moreover, note that the coefficients of the $\frac{\zeta'(2k)}{\zeta(2k)}$ have alternating signs and always sum to 0, meaning that the closed forms can be decomposed into terms $\frac{\zeta'(2k)}{\zeta(2k)}-\frac{\zeta'(2k-2)}{\zeta(2k-2)}$, which are thus "exponential periods". Further, the numerators of the $\frac{\zeta'(2n)}{\zeta(2n)}$ term of $I_{2n-1,2}$, viz. $1, -2, 17, -62, 1382, -21844$, coincide with A002430, the numerators of the Taylor series for $\tanh(x)$, which are also closely related to the rational values $\zeta(1-2n)$. All this seems rather interesting.<|endoftext|> TITLE: Web interface for GAP (or other computer algebra system dealing with finite groups)? QUESTION [10 upvotes]: GAP is computer algebra system which allows to make calculations with finite groups. (See wikipedia link for an example). Is there web interface for it ? (I cannot google it.) Or may be some other computer algebra systems which allows to calculate with finite groups (i.e. obtain information on subgroups, conjugacy classes, irreducible representations etc...) REPLY [8 votes]: There exists a Jupyter kernel for GAP, see https://github.com/gap-packages/jupyter-kernel-gap A simple way to get this actually running is through SageMath: if you have a recent beta(!) version of SageMath installed (or wait until 8.0 gets released), you can run sage -i gap_jupyter to install that kernel. At that point, you start Jupyter with sage -n jupyter and then create a New GAP notebook using the Jupyter menu.<|endoftext|> TITLE: Identify the sphere bundle of a complex line bundle $BD_{2n}\to BU(1)$ QUESTION [10 upvotes]: I'd like to know whether it is possible to identify the sphere bundle arising as follow: Let $\xi \colon BD_{2n}\to BU(1)$ the complex line bundle corresponding to the element $y^2 \in H^2(D_{2n};\Bbb Z) \cong \Bbb Z_2\langle x^2,y^2\rangle$ (we assume $n=0 \pmod{4}$ and $D_{2n}$ is the dihedral group of order $2n$). Let $q\colon S(\xi) \to BD_{2n}$ be the sphere bundle of $\xi$. It's easy to see that it's a Eilenberg MacLane space $K(G,1)$ where $G$ sits in the following s.e.s. $$0 \to \Bbb Z \to G \to D_{2n} \to 0$$ which arises from the following piece of l.e.s. of a fibration $$0 \to \pi_2(\Bbb CP^{\infty}) \to \pi_1(S(\xi)) \xrightarrow{\pi_1(q)} \pi_1(BD_{2n}) \to 0$$ I'd like to identify such $G$. The Serre exact sequence in homology (integer coefficient) gives us the following exact sequence: Since it's know that $H_2(BD_{2n})\cong \Bbb Z_2$, we have that $H_2(BU(1))\hookrightarrow H_1(S(\xi))$ which means that the exact sequence gives us $$0 \to \Bbb Z \to H_1(S(\xi)) \to \Bbb Z_2 \oplus \Bbb Z_2 \to 0$$ which implies that $H_1(S(\xi))$ cannot be a finite abelian group. In particular we already have that that $G \neq D_{\infty}$ since $H_1(D_{\infty})=ab(D_{\infty})=\Bbb Z_2\oplus \Bbb Z_2$. From the exact sequence retrieved by the dual Blakers Massey Theorem (see here, in our case the base space is $1$-connected and the map is $1$-connected) I was able to retrive this informations: $H^1(S(\xi);\Bbb Z)\cong \Bbb Z$, the boundary map $\delta \colon H^1(S(\xi);\Bbb Z) \to H^2(\Bbb C P^{\infty})$ is multiplication by $2$ and $H^2(S(\xi);\Bbb Z)\cong \Bbb Z_2$ since I know explicitly the map $H^2(BU(1))\to H^2(BD_{2n})$. Is there a way to identify such $G$ completely? REPLY [5 votes]: Your cohomology class in $H^2(D_{2n},\mathbb{Z})$ comes from a homomorphism $D_{2n} \to S^1$, which factors as $D_{2n} \to \mathbb{Z}/2 \to S^1$, where the map $D_{2n} \to \mathbb{Z}/2$ sends the rotation to $0$ and the reflection to $1$. So, you have maps $BD_{2n} \to B \mathbb{Z}/2 \to BS^1$, and you want to compute the homotopy fiber of this composition. To do this, first compute the homotopy fiber of the map $B \mathbb{Z}/2 \to BS^1$. This corresponds to the unique nontrivial central extention of $\mathbb{Z}/2$ by $\mathbb{Z}$, i.e. $\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2$. Now, you have a homotopy pullback diagram, where $G$ is the group you are trying to compute: $$ \require{AMScd} \begin{CD} BG @> >> B \mathbb{Z} \\ @VVV @VVV \\ BD_{2n} @> >> B \mathbb{Z}/2 \end{CD} $$ Taking fundamental groups doesn't preserve pullbacks in general, but it does when everything in sight is the classifying space of a discrete group. So, $G$ is the fiber product $D_{2n} \times_{\mathbb{Z}/2} \mathbb{Z}$, where the map $D_{2n} \to \mathbb{Z}/2$ is the one mentioned earlier. Algebraically speaking, you have some central extension of $D_{2n}$ by $\mathbb{Z}$, and it is pulled back from a known central extension of $\mathbb{Z}/2$ by $\mathbb{Z}$, so you can use this to figure out what the extension is.<|endoftext|> TITLE: When is a Fourier coefficient field Galois? QUESTION [11 upvotes]: Let $f$ be a modular form with $q$-coefficients $a_n$, and let $L=\mathbf Q(a_n:n\ge 0)$ be the Fourier coefficient field. Does anyone know of any necessary or sufficient conditions for $L/\mathbf Q$ to be a Galois extension? I know that by Momose, if $f$ is a non-CM newform with weight $\ge 2$ then you can embed the group $\Gamma$ of inner-twists into $\operatorname{Aut}(L/\mathbf Q)$ and then $L/L^\Gamma$ is a Galois extension, and I have seen one paper refer to "$\operatorname{Gal}(L^\Gamma/\mathbf Q)$", which would suggest that $L/\mathbf Q$ is Galois, but since they don't justify referring to the extension as Galois I've decided that either they've made a mistake, or they know something I don't. REPLY [5 votes]: As Will Sawin points out, it is not true in general that the Hecke field of a modular form is a Galois number field. The first example in weight $2$ and level $\Gamma_0(N)$ appears at $N=41$: the space $S_2(\Gamma_0(41))$ has dimension $3$ and is generated by the conjugates of a newform $f$ with Hecke field $L=\mathbf{Q}(\alpha)$ with $\alpha$ being a root of the irreducible polynomial $X^3+X^2-5X-1$. The number field $L$ is not Galois, and its Galois closure has Galois group $\mathfrak{S}_3$. In general, let $f \in S_k(\Gamma_1(N),\varepsilon)$ be a newform without CM, and let $L$ be the Hecke field of $f$. Let $\Gamma$ be the group of inner-twists of $f$, consisting of those automorphisms $\sigma$ of $L$ satisfying $f^\sigma = f \otimes \chi_\sigma$ for some Dirichlet character $\chi_\sigma$. It is known that $L/L^\Gamma$ is a finite abelian extension and that $L^\Gamma=\mathbf{Q}(\{a_p^2/\varepsilon(p)\}_{p \nmid N})$ is a totally real number field. But $L^\Gamma$ need not be Galois over $\mathbf{Q}$, as the preceding example already shows ($\mathrm{Aut}(L)$ is clearly trivial in this case).<|endoftext|> TITLE: Bloch-Kato's dual exponential map in a simple setting QUESTION [9 upvotes]: I have a quick question on Bloch-Kato p-adic Hodge theory. I'm wondering about the following easy case: Note that $H^1(Q_p,Q_p(1))$ is two-dimensional, isomorphic by the Kummer map to the completion of $Q_p^\times$. We have $H^1_e(Q_p,Q_p(1)) = H^1_f(Q_p,Q_p(1)) = Z_p^\times \otimes Q_p$, and the Bloch-Kato logarithm is simply the usual p-adic $\log$ on $Z_p^\times$. -My question is: I thought that the dual exponential on $H^1(Q_p,Q_p(1))$ would simply be the p-adic valuation $\mathrm{ord}_p$ on $Q_p^\times$, but now I see that perhaps not, because $H^1_g = H^1$ in this case, and $\exp^*$ vanishes on $H^1_g$, so the dual exponential is just the zero map on $H^1(Q_p,Q_p(1))$. Is this is correct, or am I wrong and one actually has $\exp^* = \mathrm{ord}_p$? If I am right, can one recover the $\mathrm{ord}_p$ map on $H^1_g/H^1_f = Q_p^\times/Z_p^\times$ as part of the general theory of Bloch-Kato? I'm aware this is a rather special setting, as for many representations one has $H^1_f = H^1_g$. REPLY [9 votes]: As Francois has very clearly explained, the Bloch--Kato dual exponential for $\mathbf{Q}_p(1)$ is indeed zero. Just as a footnote, let me say a few words about how one can "get at" the one-dimensional quotient $H^1 / H^1_{\mathrm{f}}$, given that $\exp^*$ doesn't suffice to do this. The point is that there are two slightly different versions of the Bloch--Kato exact sequence: the first goes $$ 0 \to \mathbf{Q}_p \to \mathbf{B}_{\mathrm{cris}}^{\varphi = 1} \to \tfrac{\mathbf{B}_{\mathrm{dR}}}{Fil^0} \to 0$$ and the second goes $$ 0 \to \mathbf{Q}_p \to \mathbf{B}_{\mathrm{cris}} \to \mathbf{B}_{\mathrm{cris}} \oplus \tfrac{\mathbf{B}_{\mathrm{dR}}}{Fil^0} \to 0,$$ with the last map being $x \mapsto ( (1 - \varphi) x, x \bmod Fil^0)$. If you tensor the first exact sequence with some Galois rep $V$ and take cohomology, you get a boundary homomorphism $$ \exp_{\mathbf{Q}_p, V}: \tfrac{\mathbf{D}_{\mathrm{dR}}(V)}{Fil^0\mathbf{D}_{\mathrm{dR}} + \mathbf{D}_{\mathrm{cris}}^{\varphi = 1}} \cong H^1_{\mathrm{e}}(\mathbf{Q}_p, V).$$ which is the Bloch--Kato exponential. However, if you use the second version, you get a homomorphism $$ \widetilde\exp_{\mathbf{Q}_p, V}: \frac{\mathrm{D}_{\mathrm{cris}}(V) \oplus \tfrac{\mathbf{D}_{\mathrm{dR}}(V)}{Fil^0}}{\{ ((1-\varphi)x, x \bmod Fil^0): x \in \mathrm{D}_{\mathrm{cris}}(V)\}} \cong H^1_{\mathrm{f}}(\mathbf{Q}_p, V).$$ If $\mathbf{D}_{\mathrm{cris}}(V)$ surjects onto $\tfrac{\mathbf{D}_{\mathrm{dR}}(V)}{Fil^0}$, e.g. if $V$ is crystalline, then this simplifies into an isomorphism $$ \widetilde\exp_{\mathbf{Q}_p, V}: \frac{\mathrm{D}_{\mathrm{cris}}(V) }{ (1-\varphi) Fil^0} \cong H^1_{\mathrm{f}}(\mathbf{Q}_p, V).$$ (The quotient $\frac{\mathrm{D}_{\mathrm{cris}}(V) }{ (1-\varphi) Fil^0}$ is a rather natural object: one can see by easy explicit linear algebra that this quotient computes extensions of $\mathbf{Q}_p$ by $\mathrm{D}_{\mathrm{cris}}(V)$ in the category of filtered $\varphi$-modules.) When $V = \mathbf{Q}_p$, the map $\exp$ is zero, but $\widetilde\exp$ is not. If we define $\widetilde\exp^*$ to be the dual of $\widetilde\exp$, then $\widetilde\exp^*$ gives an isomorphism $H^1(\mathbf{Q}_p, \mathbf{Q}_p(1)) / H^1_{\mathrm{f}} \to \mathbf{Q}_p$, which agrees with the $p$-adic order map $\operatorname{ord}: \mathbf{Q}_p^\times / \mathbf{Z}_p^\times \otimes \mathbf{Q}_p \to \mathbf{Q}_p$ up to a sign. See Proposition 2.5.5 of my paper "Local epsilon isomorphisms" with Venjakob and Zerbes (journal website, arxiv).<|endoftext|> TITLE: Generalization of finite-projective-plane with more than one intersection point QUESTION [5 upvotes]: In a finite projective plane, each two points appear together in exactly one line, and each two lines intersect in exactly one point. It is known that, if each line contains $n+1$ points, then the total number of points and of lines is $n^2+n+1$. What is known about a generalization of this concept, in which each two points appear together in exactly $k$ lines, and each two lines intersect in exactly $k$ points? In particular, for what values of $n,k$ such planes exist, and what is the number of lines and points in that case? REPLY [4 votes]: Let me adjust notation slightly -- the $k$ in the original post is more usually a $\lambda$ in the literature. Thus the concept you want is this: Definition. A symmetric $2-(v,k,\lambda)$ design is a pair $(\Omega, \mathcal{B})$ where $\Omega$ is a set of size $v$ and $\mathcal{B}$ is a set of $k$-subsets of $\Omega$ such that: any 2 points of $\Omega$ lie in $\lambda$ elements of $\mathcal{B}$; any 2 elements of $\mathcal{B}$ intersect in $\lambda$ elements of $\Omega$. A simple counting argument asserts that an object has the property that $b=|\mathcal{B}|=v$. If you want to know when these things exist, then the following theorem should be your starting point: The Bruck-Ryser-Chowla Theorem. If a symmetric $2-(v,k,\lambda)$ design exists, then if $v$ is even, then $k-\lambda$ is a square; if $v$ is odd, then the following Diophantine equation has a nontrivial solution: $$x^2-(k-\lambda)y^2 - (-1)^{(v-1)/2}\lambda z^2=0.$$ More is known in special cases. For instance there is a famous result of Lam, using a computer, that asserts that a symmetric $2-(111,11,1)$ design does not exist (there is no projective plane of order $10$).<|endoftext|> TITLE: Local property of split exact sequence QUESTION [6 upvotes]: In the module category of a ring $A$, is a short exact sequence split if and only if the localization of this sequence is split for every prime ideal? Thanks! REPLY [7 votes]: Without extra finiteness assumptions, this is not true in general. Even for $A=\mathbb{Z}$, there are infinitely generated $A$-modules $M$ that are locally free (in the sense that $M_\mathfrak{p}$ is a free $A_\mathfrak{p}$-module for every prime ideal $\mathfrak{p}$) but not projective. Then if $0\to N\to P\to M\to0$ is a (necessarily non-split) short exact sequence with $P$ projective, then for every prime ideal $\mathfrak{p}$, $0\to N_\mathfrak{p}\to P_\mathfrak{p}\to M_\mathfrak{p}\to0$ is a split short exact sequence, since $M_\mathfrak{p}$ is projective. For non-Noetherian rings, there are counterexamples with $M$ finitely generated, since there can be finitely generated flat $A$-modules $M$ that are not projective. Letting $0\to N\to P\to M\to0$ be a (necessarily non-split) short exact sequence where $P$ is projective, then for any prime ideal $\mathfrak{p}$, $M_\mathfrak{p}$ is projective, since every finitely generated flat module for a local ring is projective, and so $0\to N_\mathfrak{p}\to P_\mathfrak{p}\to M_\mathfrak{p}\to0$ is a split short exact sequence. However, as alluded to in comments, one finiteness condition that gives a positive answer is where the short exact sequence $0\to N\to X\to M\to0$ has $M$ finitely presented. I'll expand on the proof sketched in the comments. If $A\subseteq B$ is a flat ring extension (e.g., $B=A_\mathfrak{p}$), then for a finitely generated projective $A$-module $P$, the natural map $$\text{Hom}_A(P,N)\otimes_AB\to\text{Hom}_B(P\otimes_AB,N\otimes_AB)$$ is an isomorphism. Take a projective $A$-module resolution $$\dots\to P_2\to P_1\to P_0\to M\to0$$ with $P_1$ and $P_0$ finitely generated. The natural maps $$\text{Hom}_A(P_i,N)\otimes_AB\to\text{Hom}_B(P_i\otimes_AB,N\otimes_AB)$$ are isomorphisms for $i=0,1$, and taking homology in degree $1$ it follows that the natural map $$\text{Ext}^1_A(M,N)\otimes_AB\to\text{Ext}^1_B(M\otimes_AB,N\otimes_AB)$$ is injective (even an isomorphism if $P_2$ is also finitely generated). Taking the class $\zeta$ of $\text{Ext}^1_A(M,N)$ representing the original short exact sequence, it follows that if every localization of the sequence is split (so the image of $\zeta$ in $\text{Ext}^1_{A_\mathfrak{p}}(M_\mathfrak{p},N_\mathfrak{p})$ is zero for every $\mathfrak{p}$), then $\zeta_\mathfrak{p}$ is zero in $\text{Ext}^1_A(M,N)_\mathfrak{p}$ for every $\mathfrak{p}$. But if $\zeta\neq0$ then the annihilator of $\zeta$ is a proper ideal of $A$, contained in some maximal ideal $\mathfrak{m}$, and so $\zeta_\mathfrak{m}\neq0$.<|endoftext|> TITLE: The Hilbert scheme for 3 points on a surface QUESTION [15 upvotes]: I'm trying to understand the Hilbert scheme of points of a smooth complex algebraic surface, and I think I'll have a much clearer picture of it if I understand the case of $n$ points in the affine plane very concretely. Here's the idea. Let $X$ be the complex affine plane, or $\mathbb{C}^2$ for us dummies. A point in the Hilbert scheme of points $M_{X, n}$ is an ideal $J$ in the polynomial ring $\mathbb{C}[x,y]$ for which the dimension of $\mathbb{C}[x,y]/J$ is $n$. There's a way to make the punctual Hilbert scheme into a smooth complex variety. The easiest way to get a point in $M_{X,n}$ is to take $n$ distinct points $p_1, \dots, p_n$ in $\mathbb{C}^2$ and let $J$ consist of polynomial functions on $\mathbb{C}^2$ that vanish at all these points. This simple recipe works generically: we get an open dense set in the punctual Hilbert scheme this way. All the fun happens in the non-generic situation where some of the points $p_1, \dots, p_n$ collide. When $n = 2$, we get more points in the Hilbert scheme of points this way: take just one point $p$, and let $J$ consist of polynomials that vanish at $p$ and whose derivative at $p$ vanishes in one chosen direction. It's easy to check that $J$ is an ideal, and the dimension of $\mathbb{C}[x,y]/J$ is 2 because we're imposing two equations. We can think of these points in $M_{X,2}$ as arising from 'pairs of infinitesimally nearby points' in the plane. Question 1. Does this trick give all the remaining points in $M_{X,2}$? I feel the answer must be yes. If so we can move on to this: Question 2. What tricks do we need to get all the points in $M_{X,3}$? Here are some tricks I'm guessing that we need: We can take $J$ to consist of all polynomials that vanish at 3 distinct points. We can take $J$ to consist of all polynomials that vanish at one point and vanish along with a chosen directional derivative at a second, distinct, point. More interestingly, we can take $J$ to consist of all polynomials that vanish along with both their first partial derivatives at a single point. Presumably these points in $M_{X,3}$ describe ways for 3 points in the plane to collide. I imagine that in the usual topology of $M_{X,3}$ as a complex manifold they are limits of situations where we have 3 points arranged in a triangle and the triangle shrinks to zero size while remaining the same shape, with one corner not moving. Apparently the shape of the triangle doesn't matter as long as the triangle is nondegenerate, i.e. the 3 points don't lie on a line. More interestingly still, we can take $J$ to consist of all polynomials that vanish at some point along with their first and second directional derivatives in a chosen direction at a single point. Apparently these points in $M_{X,3}$ describe subtler ways for 3 points in the plane to collide: namely, limits of situations where we have 3 points in a line, all moving closer and closer to a chosen point on that line. Perhaps these tricks give all of $M_{X,3}$, but perhaps I'm leaving something out (or making a still worse mistake). Assuming I'm on the right track, the obvious question is: Question 3. What tricks of this sort give all the points in $M_{X,n}$ for arbitrary $n$? We seem to be getting some sort of stratification of $M_{X,n}$. How can we keep track of these strata? There might be some combinatorial way to name them all. It reminds me a bit of the Fulton–MacPherson compactification of a configuration space, but it seems different. Perhaps everything I need to know is lurking in here: José Bertin, The punctual Hilbert scheme: an introduction. but it will take some work to dig it out. REPLY [8 votes]: If $J_n$ is an ideal with $\mathbb C[x,y]/J_n$ of dimension $n$, then there exists an ideal $J_{n-1}$ containing $J_n$, with $J_{n-1}/J_n$ one-dimensional. This lets you understand the possibilities for the points of $M_{X,n}$ inductively if you can calculate for the ideal $J_{n-1}$ all possible subideals with quotient of dimension $1$. You can do this using the fact that $J_{n-1}/J_n$ is necessarily isomorphic to $\mathbb C[x,y]/(x-a,y-b)$ for some $a,b$, and $J_n$ is the kernel of some map $J_{n-1} \to \mathbb C[x,y]/(x-a,y-b)$, which are classified by linear forms on the vector space $J_{n-1} / ( (x-a) J_{n-1}, (y-b) J_{n-1})$. You can calculate this space using generators for $J_{n-1}$. In particular, the dimension is the number in a minimal set of generators over $\mathbb C[[x,y]]$. What does that mean in practice? $J_0$ is just $\mathbb C[x,y]$. It has one generator, so all the spaces $J_0 / ((x-a)J_0,(y-b) J_0)$ are one-dimensional. For each point, we have a unique choice of ideal, which is $(x-a,y-b)$. So $J_1$ has the form $(x-a,y-b)$. For simplicity (and wlog), let's assume $a=0,b=0$. If we pick any other point, the quotient will again be one-dimensional and so we have a unique choice. So let's pick the same point again. Then $J_1 / (x J_1, y J_1) = (x,y)/(x^2,xy,y^2)$ which is two-dimensional. So we have a projective line of possible choices of linear maps. These linear maps can be represented as derivatives in a direction, and so the associated ideal is the space of functions whose derivatives in that direction vanish. Again wlog, the direction is along $x$. Then $J_2$ is $(x^2,y)$. It has two generators, and $J_2/ (xJ_2,yJ_2) = (x^2,y)/ (x^3, xy,y^2)$ is again two-dimensional. You have written down the ideals corresponding to two linear maps - the derivative in $y$ and the second derivative in $x$. However, there are also ideals corresponding to some linear combination of these maps, which are functions $f$ satisfying $f(0,0)=0$, $\frac{\partial f}{\partial x}=0$, $a \frac{\partial^2 f}{\partial x^2} + b \frac{\partial f}{\partial y}=0$. These are exactly what Sasha and Zach Teitler point out - polynomials vanishing to order $3$ along a smooth curve. We can go further, but the number of possibilities multiplies. If $J_3 = (x^2,xy,y^2)$, then we have a three-dimensional space of possibilities for the next linear form. We can view these linear forms as differential operators of order $2$, evaluated at $(0,0)$. These could either be differential operators along a line, or Laplacian-like ones. If $J_3$ is something like $(x^3,y)$, then again you have a two-dimensional quotient, and the linear forms give a projective line. All but one of the points will correspond to the third derivative along a cubic curve, and that one remaining point $\frac{\partial f}{\partial y}$ will force every element of your ideal to vanish to order $2$ at $(0,0)$, making this overlap with the previous case.<|endoftext|> TITLE: Are localization functors always essentially surjective? QUESTION [6 upvotes]: Let $\mathcal{C}$ be a category and $\mathcal{W} \subseteq \text{Arr}(\mathcal{C})$ a set (or class) of arrows. There are (at least) two notions of localization of $\mathcal{C}$ with respect to $\mathcal{W}$, a strict and a weak one. In the strict one, a functor $\lambda: \mathcal{C} \rightarrow \mathcal{D}$ is a localization iff $\lambda$ inverts all arrows in $\mathcal{W}$ and if for every functor $G: \mathcal{C} \rightarrow \mathcal{E}$ with the same property there exists a unique functor $G':\mathcal{D} \rightarrow \mathcal{E}$ with $$ G = G'\circ{}\lambda $$ It is not hard to see, that such a strict localization functor $\lambda$ must actually be surjective on objects. In the weak version of localization, one merely asks for the existence of a functor $G'$ together with an isomorphism $$ \eta: G \stackrel{\sim}{\rightarrow} G'\circ{}\lambda$$ such that for every other such pair $(G'',\eta')$ there exists a unique isomorphism $\kappa:G' \stackrel{\sim}{\rightarrow} G''$ with $$ (\kappa \circ{} \lambda) \eta = \eta' $$ Now, if such a weak localization functor would always be essentially surjective, then I (think I) can prove that the notions of strict and weak localization are essentially equivalent, in the sense that the existence of one implies that of the other. So, are all weak localization functors essentially surjective? REPLY [3 votes]: The question has already been answered by Fernando and Mike, but I think it might be useful to expand on the details and provide another (partial) argument as to why localizations are essentially surjective. As Fernando suggested, a weak localization functor $$ \lambda: \mathcal{C} \rightarrow \mathcal{D}$$ can be factorized (like any other functor) as the composition $\lambda = \lambda_2 \circ{} \lambda_1$ of an essentially surjective followed by a fully faithful one: $$ \begin{array}{ccccc} \mathcal{C} & & \xrightarrow{\lambda} & & \mathcal{D} \\ & \searrow && \nearrow \\ && \mathcal{C}'\end{array} $$ One simply takes $\mathcal{C}'$ to be the category with the same objects as $\mathcal{C}$ and puts $\text{Hom}_{\mathcal{C'}}(X,Y) := \text{Hom}_{\mathcal{D}}(\lambda(X),\lambda(Y))$. The functors $\lambda_1$ and $\lambda_2$ are then defined in the obvious way, with $\lambda_1$ being the identity on objects. From the construction it's clear that $$ \lambda \text{ ess. surj. } \Leftrightarrow \lambda_2 \text{ ess. surj. } \Leftrightarrow \lambda_2 \text{ is an equivalence} $$ This has nothing to do with $\lambda$ being a localization. However since $\lambda$ is a localization (in the following, "localization" refers to the weak notion), it follows by easy arguments that $$ \lambda_2 \text{ is an equivalence } \Leftrightarrow \lambda_1 \text{ is a localization } \Leftrightarrow \lambda_1 \text{ is a strong localization}$$ The first equivalence follows by observing that $\lambda_1$ already inverts all arrows in $\mathcal{W}$, which implies that for all categories $\mathcal{E}$ the diagram $$ \begin{array}{ccccc} [\mathcal{D},\mathcal{E}] & & \xrightarrow{\lambda^\ast} & & \{ F \in [\mathcal{C},\mathcal{E}] : F(\mathcal{W}) \subseteq \text{Isos}(\mathcal{E}) \} \\ & \searrow && \nearrow \\ && [\mathcal{C}', \mathcal{E}] \end{array} $$ of functor categories is well-defined (and commutative). Because $\lambda$ is a localization, the top arrow is an equivalence. If $\lambda_1$ is a localization, it follows that $\lambda_1^\ast$ and therefore $\lambda_2^\ast$ is an equivalence for all categories $\mathcal{E}$, which implies that $\lambda_2$ itself must be an equivalence. This shows the direction $\Leftarrow$ of the first equivalence, and the direction $\Rightarrow$ is obvious. For the second equivalence the direction $\Leftarrow$ is obvious, and the direction $\Rightarrow$ follows easily because given any weak factorization $$\eta:\widehat{F}\circ{}\lambda_1 \stackrel{\simeq}{\rightarrow} F$$ of a functor $F: \mathcal{C} \rightarrow \mathcal{E}$ inverting the arrows in $\mathcal{W}$, we can simply twist $\widehat{F}$ via $\eta$ to obtain a functor $\widehat{F}'$ giving an on-the-nose-factorization $F = \widehat{F}'\circ{}\lambda_1$: $$ \widehat{F}'(X \stackrel{\phi}{\rightarrow} Y) := \eta(Y) \circ{} \widehat{F}(\phi) \circ{} \eta(X)^{-1} $$ Given these preliminary remarks, let us prove that $\lambda$ is essentially surjective, using Mike's argument Fernando's argument My own (partial) argument 1. Essential surjectivity via left orthogonality to fully faithful functors According to this nLab entry, a functor $\lambda: \mathcal{C} \rightarrow \mathcal{D}$ is essentially surjective iff it is left orthogonal to fully faithful functors, meaning that for every fully faithful functor $\mu: \mathcal{A} \rightarrow \mathcal{B}$ the induced diagram $$ \require{AMScd} \begin{CD} [\mathcal{D},\mathcal{A}] @>{\mu_\ast}>> [\mathcal{D},\mathcal{B}] \\ @V{\lambda^\ast}VV @V{\lambda^\ast}VV \\ [\mathcal{C},\mathcal{A}] @>{\mu_\ast}>> [\mathcal{C}, \mathcal{B}] \end{CD} $$ is a 2-categorical pullback. Now whatever this means, it should certainly imply the following: given functors $\phi \in [\mathcal{C},\mathcal{A}]$, $\psi \in [\mathcal{D},\mathcal{B}]$ and an isomorphism $\eta: \mu\circ{} \phi \stackrel{\sim}{\rightarrow} \psi\circ{}\lambda$, there exists a functor $\alpha \in [\mathcal{D},\mathcal{A}]$ together with isomorphism $\varepsilon_1: \alpha\circ{}\lambda \stackrel{\sim}{\rightarrow} \phi$ and $\varepsilon_2: \mu\circ{}\alpha \stackrel{\sim}{\rightarrow} \psi$. This last property implies essential surjectivity of $\lambda$ as follows. Factorize $\lambda = \lambda_2 \circ{} \lambda_1$ as above and consider $\mathcal{A} = \mathcal{C}'$, $\mathcal{B} = \mathcal{C}$, $\mu = \lambda_2$ (fully faithful by construction), $\phi = \lambda_1$, and $\psi = \text{id}$. Because $\psi\circ{}\lambda = \lambda = \lambda_2\circ{}\lambda_1 = \mu \circ{} \phi$ on the nose, we get $\alpha \in [\mathcal{D},\mathcal{C}']$ and isomorphisms $\alpha \circ{} \lambda \simeq \lambda_1$ and $\lambda_2\circ{}\alpha \simeq \text{id}$, showing that $\lambda_2$ is essentially surjective and hence an equivalence, implying that $\lambda = \lambda_2\circ{}\lambda_1$ is essentially surjective (because $\lambda_1$ is by construction). Therefore it only remains to see that a localization functor has the property above. But given $\mu$, $\phi$, $\psi$, and an isomorphism $\eta: \mu\circ \phi \stackrel{\sim}{\rightarrow} \psi\circ \lambda$ as in the statement of the property, because $\lambda$ inverts the arrows in $\mathcal{W}$, so must $\mu \circ{} \phi$; since $\mu$ is fully faitful and therefore detects isomorphisms, $\phi$ itself must already $\mathcal{W}$. The universal property of the localization then gives a functor $\alpha \in [\mathcal{D},\mathcal{A}]$ and an isomorphism $\varepsilon_1: \alpha \circ{} \lambda \stackrel{\sim}{\rightarrow} \phi$ out-of-the-box. Moreover the universal property also states that $\lambda^\ast$ defines a bijection $$ \text{Nat}(\mu\circ \alpha, \psi) \stackrel{\sim}{\rightarrow} \text{Nat}(\mu\circ \alpha \circ \lambda, \psi \circ \lambda)$$ which we can use to lift the isomorphism given by the composition $$ \mu\circ\alpha\circ\lambda \stackrel{\mu\varepsilon_1}{\rightarrow} \mu\circ\phi \stackrel{\eta}{\rightarrow} \psi\circ\lambda$$ This lift is again an isomorphism (because of the fully faithfulness of $\lambda^\ast$), giving the desired second isomorphism $\varepsilon_2$. 2. Constructing a quasi-inverse to $\lambda_2$ Because $\lambda_1$ inverts the arrows in $\mathcal{W}$ as we've seen already, we can apply the universal property of $\lambda$ to get a weak factorization of $\lambda_1$: $$\eta: \mu \circ\lambda \stackrel{\sim}{\rightarrow} \lambda_1$$ From this we deduce an isomorphism $$\lambda_2\eta: \lambda_2\circ\mu\circ\lambda \stackrel{\sim}{\rightarrow} \lambda_2 \circ \lambda_1 = \lambda$$ Applying the second part of the universal property of $\lambda$, the fully faithfulness of $\lambda^\ast$: $$\text{Nat}(\lambda_2\circ\mu , \text{id}) \stackrel{\sim}{\rightarrow} \text{Nat}(\lambda_2\circ\mu\circ\lambda , \lambda)$$ we deduce a natural transformation $\kappa: \lambda_2\circ\mu \rightarrow \text{id}$ that must be an isomorphism for the same argument used in the previous section. 3. Using the description of morphisms in localized categories One can verify directly that $\lambda_1$ is a localization, therefore proving the essential surjectivity of $\lambda_2$, by simply writing down the objects asked for by the universal property. The verification that these objects have the required properties is not obvious, but follows easily if one knows that all morphisms between objects $\lambda(X)$, $\lambda(Y)$ in the localized category are given by images of morphisms in $\mathcal{C}$ and inverses of morphisms in $\mathcal{W}$. Let $F: \mathcal{C} \rightarrow \mathcal{E}$ be a functor inverting $\mathcal{W}$. By the universal property of $\lambda$, we get a weak factorization $$ \eta: \widehat{F}\circ\lambda \stackrel{\sim}{\rightarrow} F$$ which we can also read as an equivalence $\widehat{F}'\circ\lambda_1 \simeq F$ if we put $\widehat{F}' := \widehat{F}\circ\lambda_2$, proving one half of the universal property for $\lambda_1$. To show the other half, we must prove that for any two functors $F,G: \mathcal{C}' \rightarrow \mathcal{E}$ we get a bijection $$\lambda_1^\ast: \text{Nat}(F,G) \stackrel{\sim}{\longrightarrow} \text{Nat}(F\circ\lambda_1 , G\circ\lambda_1 )$$ Since $\lambda_1$ is the identity on objects, this map is obviously injective. To see surjectivity, we would have to know that if for an $\eta:F\circ\lambda_1 \rightarrow G\circ\lambda_1$ the diagrams $$ \require{AMScd} \begin{CD} F(X) @>\eta(X)>> G(X) \\ @VF(\phi)VV @VG(\phi)VV \\ F(Y) @>\eta(Y)>> G(Y)\end{CD} $$ commuted for all $\phi \in \lambda(\text{Hom}_{\mathcal{C}}(X,Y)) \subseteq \text{Hom}_{\mathcal{C}'}(X,Y) = \text{Hom}_{\mathcal{D}}(\lambda(X),\lambda(Y))$, then they commuted for all $\phi \in \text{Hom}_{\mathcal{D}}(\lambda(X),\lambda(Y))$. But given a description of the morphisms in $\mathcal{D}$ as compositions of images of morphisms in $\mathcal{C}$ and inverses of images of morphisms in $\mathcal{W}$, this would be immediate. Such as description would follow for instance from the explicit construction of the localized category via generators and relations (which may not exist due to size issues).<|endoftext|> TITLE: Does a central extension split over some finite index subgroup? QUESTION [7 upvotes]: Consider a central extension of groups $\mathcal{E}$ : $$1\rightarrow A\rightarrow \widetilde{G}\rightarrow G \rightarrow 1$$with $A$ finite cyclic. Does there always exist a finite index subgroup $H$ of $G$ such that the restriction of $\mathcal{E}$ to $H$ becomes trivial? REPLY [7 votes]: No. Deligne gave a famous example of a central extension $$ 1\to\mathbb{Z}/2\to \widetilde{G}\to G\to 1 $$ such that $G$ is a finite-index subgroup of $PSp(2n,\mathbb{Z})$ but $\widetilde{G}$ is not residually finite. (See, for instance, the references in this MO question.) But it's easy to see that if the extension were to virtually split then $\widetilde{G}$ would be residually finite.<|endoftext|> TITLE: A variant of the Stothers-Mason Theorem QUESTION [6 upvotes]: Let $K$ be a function field over $\mathbb{C}$, i.e. a finitely generated extension of $\mathbb{C}$ of transcendence degree 1. Suppose that $x, y \in K^\ast$ are such that $x + y = 1$. Then the Stothers-Mason Theorem states that if $x \not \in \mathbb{C}$ we have $$ H_K(x) \leq |S| + 2g_K - 2, $$ where $g_K$ is the genus of $K$ and $S := \{v \in M_K : v(x) \neq 0 \text{ or } v(y) \neq 0\}$ are the places of $K$ such that $v(x) \neq 0$ or $v(y) \neq 0$. The Stothers-Mason Theorem can be used to count the number of solutions to unit equations in function fields, but it leads to the appearance of $g$ in the resulting upper bounds. Therefore I am wondering if the upper bound in the Stothers-Mason Theorem can be replaced by $$ H_K(x) \leq c \gamma |S|, $$ where $c$ is some absolute constant and $\gamma$ is the gonality, that is $$ \gamma := \min_{t \in K} [K : \mathbb{C}(t)]. $$ Note that this upper bound is weaker than the Stothers-Mason Theorem if $|S|$ is of size at least $2g_K - 2$, but a lot stronger if $|S|$ is very small. So far I have been able to find the following: Brownawell and Masser in their paper ``vanishing sums in function fields'' find examples for every value of $g$ such that equality holds in the Stothers-Mason Theorem for infinitely many values of $|S|$. Unfortunately $|S|$ is of size at least $g$ in their examples, so it does not say anything about the truth of $H_K(x) \leq c \gamma |S|$. Edit: changed Mason's Theorem to Stothers-Mason Theorem. REPLY [8 votes]: No. Take the curve with equation $z^n = x (x-1)$ for $n$ a large odd number. This is a degree $n$ covering of $\mathbb P^1$, totally ramified over $x=0,x=1$, and $x=\infty$. Those $3$ points are the only places where the valuation of $x$ or $y$ is nonzero. Hence $|S|=3$ Furthermore, the map $z$ is a degree $2$ map to $\mathbb P^1$, which shows that $\gamma=2$ The height of $x$ is the degree of the map $x$ (I presume), and so is equal to $n$. This is unbounded, but $c \gamma |S|= 6 c$ is bounded. REPLY [4 votes]: If your intention is to bound the number of solutions to the unit equation $x+y=1, x,y \in G$, Beukers and Schlikewei (Acta Arithmetica 78(1996), 189-199) proved that it is bounded by $2^{8r + 8}$ where $r$ is the rank of the finitely generated group $G$. Now the $S$-units have rank bounded by $|S|-1$ only modulo constants. I am not sure how to deal with the constants.<|endoftext|> TITLE: Existence of a specific mad family QUESTION [7 upvotes]: Preliminaries: Let $[\omega]^{\omega}$ be the set of all infinite subsets of $\omega$, the first countable ordinal (the set of the natural numbers). We say that $\mathcal A\subset [\omega]^{\omega}$ is open if for all $A, B \in [\omega]^{\omega}$ then if $A\subset^*B\in \mathcal A$ then $A\in \mathcal A$. We say that $\mathcal D\subset [\omega]^{\omega}$ is dense if for all $A \in [\omega]^\omega$ there exists $B \in \mathcal D$ such that $B\subset^*A$. It's possible to show that these open sets define a topology on $[\omega]^\omega$ and that in this topology, the dense subsets are what we defined above. Let $\mathfrak h$ be the first cardinal that there exists a colection $\mathscr A$ of dense open subsets of $[\omega]^\omega$ of cardinality $\mathfrak h$ such that $\bigcap \mathscr A=\emptyset$. It's possible to show that $\omega<\mathfrak h\leq \mathfrak c$, that $\mathfrak h$ is regular and it's consistent that $\mathfrak h<\mathfrak c$. We may prove the following theorem, that is called the Base Matrix Tree Lemma: There exists $\mathcal T\subset [\omega]^\omega$ such that: $\supset^*$ is a tree order on $\mathcal T$. The first level of $\mathcal T$ is $\{\omega\}$. The height of $\mathcal T$ is $\mathfrak h$. Every node of $\mathcal T$ has exactly $\mathfrak c$ successors. For every $A\in [\omega]^\omega$ there exists $B \in \mathcal T$ such that $B\subset A$. If $0<\alpha<\mathfrak h$, then the $\alpha$-th level of $\mathcal T$ is a mad family over $\omega$. Reminder: Let $N$ be an infinite countable set. An almost disjoint family over $N$ is an infinite collection $\mathcal A$ of infinite subsets of $N$ such that for every $A, B \in \mathcal A$, $A\cap B$ is finite. A maximal almost disjoint family (mad family) over $N$ is an almost disjoint family that isn't properly contained in any almost disjoint family. Context: I am trying to understand the proof of theorem 4.2 of this article. I have understood every detail but the one I'm gonna ask for help. My Question: Suppose $\mathfrak h<\mathfrak c$ and fix $\mathcal T$ as above. For every $A \subset 2^{<\omega}$, let $\pi_A=\{n \in \omega: A\cap 2^n\neq \emptyset\}\subset \omega$. So there exists $\mathcal A\subset[2^{<\omega}]^\omega$ such that: $\mathcal A$ is a mad family (over $2^{<\omega}$) Every $A \in \mathcal A$ is either a chain or an antichain in $2^{<\omega}$. $\pi_A \in \mathcal T$ for all $A \in \mathcal A$. If $A, B \in \mathcal A$ and $A\neq B$, then $\pi_A\neq \pi_B$. Question: Why does $\mathcal A$ exists? My Progress: The authors claim that the existence of $\mathcal A$ follows from a simple application of the Zorn's Lemma. So I defined: $$\mathcal U=\{\mathcal A \subset [2^{<\omega}]^\omega: |\mathcal A|\geq\omega, \mathcal A \text{ is an adf}, \mathcal A \text{ satisfies (2), (3), (4)}\}.$$ I managed to show that $\mathcal U$ is not empty and it's easy to see that the union of a chain in $\mathcal U$ is again an element of $\mathcal U$, therefore there exists a maximal element $\mathcal M$. It remains to show that $\mathcal M$ is mad, and this is where I'm having trouble. If $\mathcal M$ is not mad, then there exists $X\in [2^{<\omega}]^\omega$ such that for every $A \in \mathcal M$, $A\cap X$ is finite. Then it's not hard to show that there exists an infinite $X' \subset X$ such that $X'$ is either a chain or an antichain in $2^{<\omega}$ and such that $\pi_{X'}\in \mathcal T$. If we can shrink $X'$ so that for every $A \in \mathcal M$, $\pi_A\neq \pi_{X'}$ then we are done but this is where I'm stuck. REPLY [2 votes]: You could define $\mathcal{A}$ recursively: Enumerate all elements of $[2^{<\omega}]^\omega$ as $(X_\alpha)_{\alpha< \mathfrak{c}}$. Then whenever elements $(A_{\alpha})_{\alpha<\gamma}$ have already been constructed, check if $X_\gamma$ is almost disjoint from them. If not, skip step $\gamma$. If yes then find a subset $A_\gamma \subseteq X_\gamma$ so that $A_\gamma$ is a chain or antichain, $\pi_{A_\gamma} \in \mathcal{T}$ and $\pi_{A_\gamma} \neq \pi_{A_\alpha}$ for all $\alpha<\gamma$ (this is easy to achieve because we have only defined $<\mathfrak{c}$ many elements of $\mathcal{A} $ yet and below each node in $\mathcal{T}$ there are enough sets to choose from). Of course there is choice involved and so maybe this is what they meant with Zorn's Lemma. In an application of Zorn's Lemma as you did above, you might have a maximal element which is not a mad family (just pick take a bijection of $2^\omega$ to $\mathcal{T}$ and get an almost disjoint family with the above properties which is not maximal but where each $\pi \in \mathcal{T}$ was already picked by restricting each $f \in 2^\omega$ to its corresponding set in $\mathcal{T}$).<|endoftext|> TITLE: Mapping class group of certain 3-manifolds QUESTION [7 upvotes]: Let $\xi : M^3 \to F$ be an orientable circle bundle over a closed orientable surface $F$ of genus $g \geq 2$. I am mostly interested to the case where the bundle $\xi$ is non-trivial. My question is about the mapping class group $\mathrm{MCG} (M) = \pi_0(\mathrm{Diff}_+(M))$. In particular, is this group generated by liftings of Dehn twists and bundle automorphisms? REPLY [13 votes]: Since you write ${\rm Diff}_+(M)$ you are probably assuming $M$ is orientable and diffeomorphisms of $M$ are orientation-preserving. Every diffeomorphism of $M$ can be isotoped to take fibers to fibers. This is proved using the assumption that the base surface $F$ has genus at least 2, so $M$ contains vertical incompressible tori (unions of fibers) and every incompressible torus can be isotoped to be vertical. If a diffeomorphism that takes fibers to fibers preserves orientations of the fibers then it is isotopic to a product of twists along vertical tori. These are generated by lifts of Dehn twists in $F$ and twists taking each fiber to itself, i.e., bundle automorphisms. The other possibility is a diffeomorphism taking fibers to fibers but reversing their orientations, hence also reversing orientation of $F$ since the orientation of $M$ is preserved. Such diffeomorphisms certainly exist for the product bundle $M=F\times S^1$, and they also exist for any nontrivial bundle whose Euler class is even. Namely, start with an example for $F\times S^1$ and modify this by removing two vertical solid tori $V_1$ and $V_2$ that are interchanged by the diffeomorphism, then glue $V_1$ and $V_2$ back in by diffeomorphisms of their boundary tori preserving fibers and taking a slope $0$ curve to a slope $n$ curve, using the same $n$ in both cases. This gives a circle bundle with Euler class $2n$. The diffeomorphism of $M-(V_1\cup V_2)$ reversing orientations in fiber and base extends over the reglued solid tori since slopes in a torus $S^1\times S^1$ are preserved by a $180$ degree rotation that reflects each $S^1$ factor. Added a few minutes later: It looks like a similar construction can be made more simply using one vertical solid torus instead of two, where this solid torus projects to a disk neighborhood of a fixed point of an orientation-reversing diffeomorphism of $F$. In this case there is no need to assume the Euler class is even.<|endoftext|> TITLE: Constant term in Green's kernel expansion QUESTION [7 upvotes]: Let $(M,g)$ be a closed Riemann surface, and $p\in M$ any point. Let $G_p$ be the Green's kernel of the Laplacian with singularity at $p$, normalized to have zero average. Then locally (in some normal coordinates centered at $p$) $$G_p(x) = \log|x| + A + O(|x|).$$ Is there some geometric significance of the constant $A$? For instance the Green's function of the conformal Laplacian (say in dimension 3) is related to the ADM mass of some asymptotically flat manifold, but I could not find any result for the usual Laplacian. Any reference would be appreciated! REPLY [4 votes]: This turned out to be a highly non-trivial topic. For a detailed analysis, see Jorgenson and Kramer's paper Bounds on canonical Green functions. Their result can be summarized as $$ g_{\textrm{can},X_1}-\sum_{\gamma\in S_{\Gamma_{X_1}}(\delta;z,\omega)}g_{\mathbb{H}}(z,\gamma \omega)=O_{X_0,\delta}\left(1+\frac{1}{\lambda_{X_1,1}}\right). $$ Roughly speaking, their result indicated up to a finite cover, the "canonical Green function" on the cover minus the hyperbolic Green function given by the log function is bounded inversely proportional to the first eigenvalue of the Laplacian on the covering Riemann surface. So up to a certain constant, the growth of $A$ is controlled by the reciprocal of the first eigenvalue of the Riemann Surface.<|endoftext|> TITLE: Alternating power series $\sum_{k\geq 0}(-1)^k z^{(2k+1)^2}$ QUESTION [6 upvotes]: Suppose that $f(x):=\sum_{k\geq 0}(-1)^ke^{-(2k+1)^2x}$ has a holomorphic continuation to a neighborhood of $0$, that is, $f(x)=\sum_{n\geq 0}a_n x^n$ for $x> 0$ small. I want to know the value of $a_1$. Let $z=e^{-x}$, $g(z):=\sum_{k\geq 0}(-1)^k z^{(2k+1)^2}$; then, formally, $g'(1)$ is what we want to compute. Note that $1$ is on the circle of convergence of $g$ so differentiating term-wise is not guaranteed. REPLY [5 votes]: We can indeed prove, as already suspected by მამუკა ჯიბლაძე, that $g(z)=\sum (-1)^k z^{(2k+1)^2}$ cannot be holomorphically continued past $z=1$. In fact, every point on $|z|=1$ is singular. This is a consequence of the following rather general version of the classical results on lacunary power series: Theorem: Suppose that $a_n$ is bounded. Suppose further that there exists a sequence $n_j\to\infty$ such that: (1) $|a_{n_j}|\ge\delta>0$; (2) $a_{n_j-k}\to 0$ as $j\to\infty$ for every fixed $k\ge 1$. Then $\sum a_n z^n$ cannot be holomorphically continued to any open set larger than the unit disk. (Note that $R=1$ under these assumptions.) Of course, this applies to our power series, with $n_j=(2j+1)^2$. A very elegant proof of the Theorem was recently given by Breuer-Simon. See reference 328 here, Theorem 1.6 of the paper.<|endoftext|> TITLE: Reference for map $\operatorname{Hom}^d(C,\mathbb{P}^1) \to \operatorname{Sym}^d(C)$ QUESTION [9 upvotes]: For a curve $C$ over a finite field, I am looking at the map $\phi: \operatorname{Hom}^d(C,\mathbb{P}^1) \to \operatorname{Sym}^d(C)$ where $\operatorname{Hom}^d(C,\mathbb{P}^1)$ are the functions of degree $d$ from $C \to \mathbb{P}^1$ in the $\operatorname{Hom}$-scheme $\operatorname{Hom}(C, \mathbb{P}^1)$, defined by $$\phi: f \mapsto [\sigma]\cdot \Gamma_f$$ i.e. we intersect the graph of $f$ in $C \times \mathbb{P}^1$ with the line $[\sigma]$ in $C \times \mathbb{P}^1$ (which is the inverse image of $\sigma$ under the projection to $\mathbb{P}^1$. For example the $f$ in the picture would map to $\phi(f) = 2\cdot P_1 + 3\cdot P_2 + 2\cdot P_3 + 2\cdot P_4$ I have looked at Kollár's Rational Curves on Algebraic Varieties and Chapter 9 of Nakajima's Lectures on Hilbert Schemes of Points on Surfaces however I can't find the information I need of this map. For example, is $\phi$ flat? What is the dimension of the fiber of $\phi$? Etc. Etc. Does anyone know of a reference for this topic? EDIT: If we take $d > 2g$ we can view any point in $\operatorname{Sym}^d(C)$ as an effective divisor $D$ of degree $d$ of $C$. Then $D$ is base-point free and so we can construct a map $f: C \to \mathbb P^1$ with fiber $D$ above $0$ (or say, $\sigma$). So $\phi$ is surjective for $d > 2g$. In general, I only care for $d$ much greater than 0 as I want to study it's behavior as $d$ grows. REPLY [3 votes]: Remark. Note that $\operatorname{\mathbf{Hom}}(C,\mathbb P^1)$ is an open subscheme of $\operatorname{\mathbf{Div}}_{C \times \mathbb P^1}$, whereas $\operatorname{Sym}^d(C)$ is a component of $\operatorname{\mathbf{Div}}_C$. Specifically, a homomorphism $\phi \colon C \to \mathbb P^1$ corresponds to its graph, which is a section of a line bundle $\mathscr M$ on $C \times \mathbb P^1$. Since \begin{align*} \operatorname{\mathbf{Pic}}_{C \times \mathbb P^1} &= \operatorname{\mathbf{Pic}}_C \times \operatorname{\mathbf{Pic}}_{\mathbb P^1} \times \operatorname{\mathbf{Hom}}(\operatorname{\mathbf{Alb}}_C,\operatorname{\mathbf{Pic}}^0_{\mathbb P^1})\\ &= \operatorname{\mathbf{Pic}}_C \times \operatorname{\mathbf{Pic}}_{\mathbb P^1}, \end{align*} the line bundle $\mathscr M$ is of the form $\mathscr L \boxtimes \mathcal O(n)$ for some line bundle $\mathscr L$ on $C$ and some $n \in \mathbb Z$. Counting intersections with $\{c\} \times \mathbb P^1$, we see that $n = 1$. Restricting to $C \times \{\sigma\}$, we see that $\mathscr L = \phi^* \mathcal O(1)$. Thus, $\phi$ corresponds to a section $s$ of $\mathscr L \boxtimes \mathcal O(1)$. Conversely, a section $s$ of $\mathscr L \boxtimes \mathcal O(1)$ comes from a morphism $\phi$ if and only if $\mathscr L$ is effective and $V(s)$ is irreducible. Indeed, since $\mathscr L \boxtimes \mathcal O(1)$ is primitive (not divisible), $V(s)$ is reduced. Hence, the dominant morphism $V(s) \to C$ is flat, since $V(s)$ is torsion-free over $C$ since it is integral. Since it is of degree $1$, it is an isomorphism. The only other possibility for a section $s$ of $\mathscr L \boxtimes \mathcal O(1)$ is that it contains a vertical component. Indeed, the only decomposition of $\mathscr L \boxtimes \mathcal O(1)$ as a sum of effective divisors must have $\mathscr L' \boxtimes \mathcal O$ as one of its summands, all of whose sections are vertical. The morphism $\phi \colon C \to \mathbb P^1$ and the pair $(\mathscr L, s)$ determine each other uniquely, where two pairs $(\mathscr L, s)$, $(\mathscr L', s')$ correspond if and only if there exists an isomorphism $\alpha \colon \mathscr L \stackrel \sim \to \mathscr L'$ taking $s$ to $s'$. Remark. Thus, the map you give can be extended to the rational map \begin{align*} \sigma^* \colon \operatorname{\mathbf{Div}}_{C \times \mathbb P^1} &\dashrightarrow \operatorname{\mathbf{Div}}_C \\ Z &\mapsto Z \cap (C \times \{\sigma\}), \end{align*} which is defined away from the divisors containing $C \times \{\sigma\}$. We will focus on the locus $U$ of $\operatorname{\mathbf{Div}}_{C \times \mathbb P^1}$ of irreducible divisors in $\mathscr L \boxtimes \mathcal O(1)$ for some $\mathscr L$ of degree $d > 2g$. We get a commutative square $$\begin{array}{ccc} \operatorname{\mathbf{Div}}_{C \times \mathbb P^1} & \stackrel{\sigma^*}\dashrightarrow & \operatorname{\mathbf{Div}}_C \\ \downarrow & & \downarrow \\ \operatorname{\mathbf{Pic}}_{C \times \mathbb P^1} & \stackrel{\sigma^*}\rightarrow &\ \operatorname{\mathbf{Pic}}_C. \end{array}$$ Over the locus of $\operatorname{\mathbf{Pic}}_C$ where $\mathscr L$ is of degree $d > 2g$, the right vertical arrow is a $\mathbb P^r$-bundle, where $r = h^0(\mathscr L) - 1 = d - g$ (which does not depend on $\mathscr L$ but only on $d$). Similarly, over the same locus, the left vertical arrow is a $\mathbb P^{r'}$-bundle, where $r' = h^0(\mathscr L \boxtimes \mathcal O(1)) - 1$. By Künneth we have $r' = 2(r+1) - 1 = 2d - 2g + 1$. Restricting the left hand side to the locus of the form $\mathscr L \boxtimes \mathcal O(1)$, the bottom map becomes an isomorphism. On the locus $U$ of irreducible divisors in $\mathscr L \boxtimes \mathcal O(1)$, the top map on each fibre is a morphism from an open in $\mathbb P^{2d-2g+1}$ to $\mathbb P^{d-g}$. Remark. To describe the map more explicitly, let $[x:y]$ be coordinates on $\mathbb P^1$. For simplicity, assume $\sigma = [0:1]$. Then $H^0(\mathscr L \boxtimes \mathcal O(1))$ is given by $\{\lambda x + \mu y\ |\ \lambda,\mu \in H^0(\mathscr L)\}$, and a section $\lambda x + \mu y$ is mapped to the section $\mu \in H^0(\mathscr L)$. In particular, this is a linear map, so the fibres are linear spaces. The locus of divisors $Z$ not containing $C \times \{0\}$ corresponds to the $\lambda x + \mu y$ with $\mu \neq 0$. This is the maximal set where the map on the projective spaces is defined, and there the fibres are $\mathbb A^{d-g+1}$ corresponding to the affine space of values of $\lambda$. We are further restricting to divisors $Z$ not containing any vertical component; this is probably equivalent to linear independence of $\lambda$ and $\mu$, so this removes an $\mathbb A^1\setminus\{0\}$ from each fibre (corresponding to $\lambda = c \mu$). More canonically, the conditions that $\lambda$ and $\mu$ are linearly independent are cut out in $\mathbb P^{2d-2g+1}$ by the nonvanishing of certain minors. Any further properties you want to deduce should follow from this description.<|endoftext|> TITLE: Sylvester–Gallai theorem with circle version, plane version and curve version? QUESTION [9 upvotes]: The Sylvester–Gallai theorem in geometry states that, given a finite number of points in the Euclidean plane, either All the points are collinear; or There is a line which contains exactly two of the points. I think the theorem is also true for circle, plane version and curve version, can you give a proof of these versions? Conjecture: (circle version): Given a finite number of points in the Euclidean plane, either All the points are concyclic; or There is a circle which contains exactly three points. Conjecture: (Plane version) Given a finite number of points in the three-dimensional space, either All the points in a plane; or There is a plane which contains exactly three points. Conjecture: (Curve version) Let $n, N$ be the natural number with $(\frac{3n+n^2}{2} < N < \infty)$, give $N$ points in the Euclidean plane, either All the points lie on a curve of degree $n$; or There is a curve of degree $n$ which contains exactly $\frac{3n+n^2}{2}$ of the points. REPLY [5 votes]: The plane version is false as stated. Choose two skew lines $\ell_1$ and $\ell_2$ in $\mathbb{R}^3$ and place $\geq 3$ points on each of them. Any three of these points will either all lie on the same $\ell_i$ (and thus be collinear) or two of them will lie on the same $\ell_i$, so the plane they span contains that $\ell_i$ and thus contains three collinear points. A true version is that there is a plane containing precisely $k$ of the points for which $k-1$ of them are collinear. This is due to Motzkin (Trans. AMS, 1951), with a higher dimensional generalization by Hansen (Math. Scand. 1965). I took the above counterexample from Hansen's paper. I think the curve version should be false, but I haven't broken it yet. I have now discovered two papers: Wiseman and Wilson, (Disc. and Comp. Geom 1988) and Czapliński et. al., (Rendiconti del Seminario Matematico della Università di Padova 2016) which pose the curve version as an open question, so it probably isn't easy to resolve.<|endoftext|> TITLE: Thick subcategories QUESTION [6 upvotes]: I hope this question is not too trivial for mathoverfolw. Let $R$ be a commutative ring (with $0\neq 1$) and $D_{Perf}(R)$ the triangulated category of perfect complexes. Let $C$ be a thick subcategory of $D_{Perf}(R)$ is it true that there exists a (commutative ?) ring $A$ such that $C$ is equivalent to $D_{perf}(A)$ as triangulated category? If yes is there a precise description of $A$ ? Thank you. REPLY [6 votes]: The answer is no, in principle. By Thomason's classification of thick subcategories, these correspond to certain stable for specialization subsets, i.e. arbitrary unions of closed subsets (with quasi-compact complement). Even for a closed subset $Z \subset \mathrm{Spec}(R)$ with quasi-compact complement the corresponding thick subcategory has objects the perfect complexes supported at $Z$, something than one may denote $D_{Z}(A)_\mathrm{perf}$. If $I$ is the ideal of $Z$ this is bigger than $D(A/I)_\mathrm{perf}$. In algebraic terms, a complex $M^\cdot \in D_{Z}(A)$ if and only if its homologies are killed by any power of $I$. If you allow DGA then it is possible, see Dwyer, W. G.; Greenlees, J. P. C. Complete modules and torsion modules. Amer. J. Math. 124 (2002), no. 1, 199–220. See, concretely, Theorem 2.1: $\mathbf{A}_\mathrm{tors}$ denotes $D_{Z}(A)_\mathrm{perf}$ and  $\mathcal{E}$ is the DGA you asked for.<|endoftext|> TITLE: Reference for normalization of an algebraic stack? QUESTION [5 upvotes]: Is there a standard reference on stacks which discusses (relative) normalization? This older question seemed to link to someone's notes, but the link is now broken. In any case, it would be nice to have a reference which is a book or a paper. I've tried the stacks project, Martin Olsson's book on stacks, and Laumon/Moret-Bailly's "Champs Algebriques", and none of them seem to have anything on normalization of stacks (at least not in the index). REPLY [4 votes]: Quoting Vistoli, Angelo Intersection theory on algebraic stacks and on their moduli spaces. Invent. Math. 97 (1989), no. 3, 613–670. to be found here http://dx.doi.org/10.1007/BF01388892 : Definition 1.18 p.623 `Definition. The normalization $F$ of a reduced stack $F$ with a presentation $R \rightrightarrows U$ is the stack associated with the groupoid $\overline{R} \rightrightarrows \overline{U}$. The normalization of a general stack is the normalization of its associated reduced stack. If $F$ is a stack, the canonical morphism $\overline{F} \to F$ is representable. If $F$ is of finite type over a universally japanese ring (see EGA IV, 23.1.1) then $\overline{F} \to F$ is also finite.' Here $\overline{R}$ and $\overline{U}$ denote the normalizations of $R$ and $U$, respectively.<|endoftext|> TITLE: When the value of a function in a point is equal to its integral average over the point's neighborhood? QUESTION [7 upvotes]: It is well-known that the harmonic functions have this remarkable Averaging Property: if $f$ is harmonic in a domain $U \subset R^n$, then, for any point $x \in U$, $f(x)$ is equal to the integral average of $f$ over any ball centered at $x$ and lying in $U$. Here a ball is understood with respect to the Euclidean distance -- generated by the norm $\sqrt{\sum_{i=1}^{n} x_i^2}.$ But one can consider the "balls" with respect to other "classical" distances in $R^n$ -- generated by such norms as $\max(|x_1|,|x_2|,...,|x_n| )$ or $\sum_{i=1}^{n} |x_i|$. I have done some research investigating functions satisfying this Averaging Property with respect to the "balls"'corresponding to the distances generated by other norms in $R^n$ (such as the two norms listed above). However I suspect that some studies on this topic have been done already. I would be most grateful if somebody could inform me about publications on this topic. REPLY [2 votes]: You are essentially asking for functions which have the mean value property but for balls with respect to different metrics, and possibly using different measures. This seems amenable (no pun intended) to googling. For example, I googled "harmonic functions on metric measure spaces" and found this paper: https://arxiv.org/pdf/1601.03919.pdf , which contains some regularity and Dirichlet-problem results for such functions in some general metric measure spaces. There is also a very large literature related to (discrete) harmonic functions on graphs, which may be relevant to you, but I know little about this, so whatever you find by googling will be better. Hope it's helpful.<|endoftext|> TITLE: Number of commuting pairs (triples, n-tuples) in GL_n(F_q) (and other groups)? QUESTION [13 upvotes]: Question 1 What is the number of pairs of commuting elements in GL_n(F_q) ? I am aware of many results concerning commuting elements in Mat_n(F_q), but I am interested in GL i.e. non-degenerate matrices. In particular Feit, Fine 1960, results for many kinds of Lie algebras obtained recently Jason Fulman, Robert Guralnick, and even Motivic Donaldson... is related to counting nilpotent matrices - see beautiful answer on math.se by Olivier Schiffmann. [EDIT after answers] Answers say: the number of commuting pairs in ANY group equals to number of conjugacy classes multiply order of group. Number of conjugacy classes in $GL(n,F_q)$ has nice generating function MO8415, MO104457 (or links in answers below): $$ \sum_{n\geq 0} |number~of~conjugacy~classes~in~GL(n,F_q)| x^n = \prod_{j\geq 1} \frac{1-x^j}{1-qx^j} $$ So the question is done. [end EDIT] Question 1b If we take limit q->1 can we get number of commuting elements in S_n ? (Standard analogy S_n = GL_n(F_1) ). [EDIT after answers] There is beautiful formula for $S_n$ which can count m-tuples of commuting elements, number of involutions, number of elements of order k and so on - just one formula can do it all - see MO272045. From the answer there we know that there is certain q-analog of that formula for GL(n,F_q). However it does not immediately cover the case of commuting pairs (m-tuples) and also the limit q->1 is not clear. In particular it is not clear to me how generating function above can in the limit q->1 give generating functions for conjugacy classes in $S_n$ - partition function. So there is place to think more (imho). [end EDIT] Question 2 what about triples, m-tuples ? [EDIT after answers] Is it open problem ? (Since it is not covered by powerful analysis in papers quoted in MO answer. It suggest that should be nice generating function): $$ \sum_{n\geq 0} \frac{|commuting~~(m-tuples)~in~GL(n,F_q)|}{|GL(n,F_q)|} x^n = ??? $$ For commuting pairs we will get generating function above, since order of $GL(n,F_q)$ in numerator cancel one in denominator. [end EDIT] Question 3 what about other finite algebraic groups ? (In view of results Fulman, Guralnick for Lie algebras it seems natural). Question 4 for compact simple groups over R one can put for example Killing Riemnanian metric and consider volumes for such manifolds - are there some results ? PS By the way nice fact on pairs of commuting elements is here: 5/8 bound in group theory PSPS Question 1c By the way can computer algebra systems like GAP or MAGMA get the answer on such question for small "n", but in the form of polynomial in "q", or they can deal only with fixed q ? (See answer in comment by Alexander Konovalov ). REPLY [17 votes]: The probability that two elements of a finite group $G$ commute is $k(G)/|G|$, where $k(G)$ is the number of conjugacy classes of $G$. Hence the number of pairs of commuting elements is $k(G)|G|$. According to group props, the number of conjugacy classes of $\mathrm{GL}_n(\mathbb{F}_q)$ is $q^n + O(q^{n-1})$. The order of $\mathrm{GL}_n(\mathbb{F}_q)$ is $$q^{n(n-1)/2}(q^n-1) \ldots (q-1) = q^{n(n-1)/2}(q-1)^n [n]!_q. $$ So the number of commuting pairs is $q^{n^2+n} + \mathrm{O}(q^{n^2+n-1})$. The number of commuting pairs in $S_n$ is $p(n)n!$, where $p(n)$ is the partition function. (Also $p(n)$ is the number of conjugacy classes of unipotent elements in $\mathrm{GL}_n(\mathbb{F}_q)$, for any $q$.) We have $\lim_{q \rightarrow 1} [n]!_q = n!$, but otherwise the results seem unconnected. The exponential generating function for commuting triples in symmetric groups was found in this paper by John Britnell. The OEIS sequence is A061256. This result was generalized to commuting $r$-tuples by Tad White. REPLY [13 votes]: The number of commuting pairs in any finite group $G$ is well-known to be $k(G)\lvert G\rvert$ where $k(G)$ is the number of conjugacy classes of $G$. There are also well-known generating functions for $k({\rm GL}(n, q))$ (see for example Benson, Feit, and Howe - Finite linear groups, the Commodore 64, Euler and Sylvester (MSN)), where efficient means of computing these numbers are considered). I am unsure what happens to the commuting probability for ${\rm GL}(n, q)$ if we regard it as a rational function of $q$ and let $q \to 1$. Calculating the number of commuting $k$-tuples in ${\rm GL}(n, q)$ is (I believe) much trickier for $k > 2$.<|endoftext|> TITLE: Is there a characterization of the Hausdorff measures? QUESTION [9 upvotes]: It is known that there is a unique measure on the Borel $\sigma$-algebra of $\mathbb{R}^n$ such that the measure of the rectangle $\prod_i [a_i,b_i[$ is $\prod_i (b_i-a_i)$. This is the Lebesgue measure. My question is about the existence of similar result for the Hausdorff measures $H^d$. More precisely, is there a result that says that the Hausdorff measure $H^d$ is the unique measure on the Borel $\sigma$-algebra of $\mathbb{R}^n$ such that it takes a specific value on a subset of the Borel $\sigma$-algebra (as small as possible) ? Edit : to state a more precise question (but directly related) : on $\mathbb{R}$, is the $d$-Hausdorff measure the unique translation-invariant measure that gives the value 1 to the Cantor set of dimension $d$ ? REPLY [8 votes]: The spherical Hausdorff measure coincides with the Hausdorff measure on every rectifiable set. However it is possible to find sets where these two measures differ. See Federer 2.10.6 Hence the class of sets where one should fix the value of the measure must contain (a lot of!) non-rectifiable sets even for integer dimensions.<|endoftext|> TITLE: Bound the eigenvalue of product of matrices? QUESTION [6 upvotes]: Let $H$ be a $n \times n$ real symmetric matrix that has eigenvalues with absolute value less than 1. Define the matrix $M = \prod_{i=1}^n (I - e_ie_i^{\top}H)$ where $e_i$ denotes the $i^\text{th}$ canonical basis vector of $\mathbb{R}^n$. Assuming that $H$ has at least one negative eigenvalue, is it true that $M$ has an eigenvalue with absolute value greater than 1? REPLY [6 votes]: OK, looks like I've got it. Since it is pretty late here now, it would be nice if someone could check that I haven't written some nonsense (I apologize in advance if I have). Let $N(H)$ be the kernel of $H$, let $Q$ be the orthogonal projector to $N(H)$ and let $P$ be the orthogonal projector to $N(H)^\perp$. Note that since $H$ is real symmetric, we have $QH=HQ=0$ and $PHP=H$. Now let $e$ be a unit vector and let $X=x-\langle Hx,e\rangle e$. Then $PX=Px-\langle PHP(Px),Pe\rangle Pe$. Thus, if $x_j\in \mathbb R^n$ satisfy the recursion $x_{j+1}=x_j-\langle Hx_j,e_j\rangle e_j$ where $e_j$ cycle through an orthonormal basis in $\mathbb R^n$, then $y_j=Px_j\in P(\mathbb R^n)$ satisfy the recursion $y_{j+1}=y_j-\langle \widetilde H y_j,v_j\rangle v_j$, where $v_j=Pe_j$ cycle through some complete system of vectors whose norms are bounded by $1$. Moreover, if $H$ has a negative eigenvalue, so does $\widetilde H$, but the advantage is that $\widetilde H$ is invertible in $P(\mathbb R^n)$. Thus, we reduced the problem to the following: Let $E$ be a real Euclidean space, let $v_j\in E$ be a sequence cycling through a fixed complete system of vectors with norm at most $1$, let $H:E\to E$ be any invertible symmetric linear operator of norm at most $1$, let $x_0\in E$ be any vector with $\langle Hx_0,x_0\rangle<0$. Then the sequence of vectors defined by the recursion $x_{j+1}=x_j-\langle Hx_j,v_j\rangle v_j$ grows at least geometrically in norm. The key inequality is $$ \langle Hx_{j+1},x_{j+1}\rangle = \langle Hx_{j},x_{j}\rangle-\langle Hx_{j},v_{j}\rangle^2(2-\langle Hv_{j},v_{j}\rangle) \\ \le \langle Hx_{j},x_{j}\rangle-\langle Hx_{j},v_{j}\rangle^2\,. $$ Since $H$ is invertible, we have $|HX|\ge c|X|$ for all $X\in E$ with some $c>0$. Now, if $|\langle Hx_j,v_j\rangle|<\delta|x_j|$ throughout the whole cycle with very small $\delta$, then all vectors in the cycle differ from the first vector $X$ in the cycle by at most $O(\delta)|X|$ and we derive that $|\langle HX,v_j\rangle|=O(\delta)|X|$, which, due to the completeness of the system of vectors $v_j$ over which we cycle, results in $|HX|=O(\delta)|X|$, contradicting the lower bound if $\delta>0$ is small enough. Thus, in each cycle we never increase $\langle Hx_j,x_j\rangle$ and have the inequality $$ \langle Hx_{j+1},x_{j+1}\rangle \le \langle Hx_j,x_j\rangle-\delta^2|x_j|^2 \\ \le (1+\delta^2)\langle Hx_j,x_j\rangle $$ at least once. That is the desired geometric growth. The end.<|endoftext|> TITLE: Is Somos-8 $\mod 2$ periodic? QUESTION [11 upvotes]: It is known that the Somos-$k$ sequences for $k\ge 8$ do not give integers. But the first terms of Somos-8 sequence $s_n=a_n/b_n$ $$1, 1, 1, 1, 1, 1, 1, 1, 4, 7, 13, 25, 61, 187, 775, 5827, 14815,\frac{420514}{7}, \frac{28670773}{91}$$ defined by $s_1=s_2=s_3=s_4=s_5=s_6=s_7=s_8=1$, $$s_{n+8}s_n=s_{n+7}s_{n+1}+s_{n+6}s_{n+2}+s_{n+5}s_{n+3}+s_{n+4}^2\qquad(n\ge1)$$ have only odd denominators $b_n$. Morever $s_n$ has even numerator $a_n$ only for $n=9k$. It was checked for $n\le 67$. First terms of $s_n\mod 8:=a_n\cdot b_n^{-1}\mod 8$ are $$\begin{array}{l} 1, 1, 1, 1, 1, 1, 1, 1, {\bf 4},\\ 7, 5, 1, 5, 3, 7, 3, 7, {\bf 6},\\ 7, 7, 5, 7, 5, 3, 1, 1, {\bf 6},\\ 5, 1, 5, 5, 3, 3, 7, 1, {\bf 4},\\ 7, 7, 3, 3, 7, 7, 3, 1, {\bf 2},\\ 1, 3, 1, 3, 7, 5, 3, 5, {\bf 2},\\ 5, 7, 3, 7, 7, 3, 7, 3, {\bf 0},\\ 1, 3, 7, 3,\ldots\end{array}$$ Is it possible to prove that the sequence $s_n\mod 2$ is periodic? EDT. It was found by მამუკა ჯიბლაძე that numerator of $s_{71}$ is even while that of $s_{72}$ is odd. The last line of the table above is $$1, 3, 7, 3, 1, 5, 1, {\bf 2}, 5$$ So the conjecture about numerators of $s_n$ is wrong. But the qustion about periodicity is still valid. REPLY [8 votes]: I confirm the observation of @მამუკაჯიბლაძე that $\nu_2(s_{103})=-1$. That is, $s_n\bmod 2$ is not well-defined at first place, which invalidates the question. Moreover, for $n\geq 133$, $\nu_2(s_n)$ seems to form a strictly decreasing function, i.e., $s_n$ accumulates larger and larger powers of $2$ in the denominators.<|endoftext|> TITLE: Presentable small diagrams over a locally presentable category QUESTION [5 upvotes]: It is another question about the category $\mathcal{D}\mathcal{K}$ of all small diagrams over all small categories, its definition is here : About the category of all small diagrams. I suppose $\mathcal{K}$ locally $\lambda$-presentable. Question: Take a diagram $F:I\to \mathcal{K}$ such that $I$ is finitely presentable as a small category and $F$ $\lambda$-presentable in the functor category $\mathcal{K}^I$. Is $F$ $\lambda$-presentable in $\mathcal{D}\mathcal{K}$ ? The motivation of this question is that $\mathcal{D}\mathcal{K}$ has a colimit-dense category of such objects. Therefore, assuming Vopenka's principle, $\mathcal{D}\mathcal{K}$ is locally presentable. I would like to avoid the use of this principle. REPLY [2 votes]: Maybe I have this wrong, but I think that in $\mathcal{DK}$, filtered colimits at least are constructed in the following way. Let $I$ be a $\lambda$-directed poset for simplicity, and let $G: I \to \mathcal{DK}$, $i \mapsto G_i: D_i \to \mathcal{K}$ be a functor, with transition maps $(D_{ii'}: D_i \to D_{i'}, \gamma_{ii'}: G_i \Rightarrow G_{i'} \circ D_{ii'})$ for $i\leq i'$. Let $G_\infty : D_\infty \to \mathcal{K}$ denote the colimit of $G$. Then we should have $D_\infty = \varinjlim_i D_i$, so an object of $\varinjlim_{i \in I} D_i$ is an equivalence class $[(i,d)]$ where $i \in I$ and $d \in D_i$. And $G_\infty$ should be given by $G_\infty([(i,d)]) = \varinjlim_{i \leq i'} G_{i'}(D_{ii'}(d))$ (where the colimit is of a diagram constructed using the $\gamma$ maps; this colimit is $\lambda$-directed). Now suppose that $F: C \to \mathcal{K}$ is such that $C$ is $\lambda$-presentable, and $F$ takes values in the $\lambda$-presentable objects of $\mathcal{K}$. A map $F \to G_\infty$ will consist first of a functor $f: C \to D_\infty$; this data commutes with the colimit because $C$ is $\lambda$-presentable. Secondly, there will be a natural transformation $\mu: F \Rightarrow G_\infty \circ f$. In components this consists of maps $\mu_c : Fc \to G_\infty(fc)$; the data of this component commutes with the colimit because $Fc$ is $\lambda$-presentable. The data of the whole natural transformation then commutes with the colimit because $C$ is $\lambda$-presentable; I'm not sure how to argue this conceptually, but it follows from the idea that a $\lambda$-presentable category is generated by $<\lambda$-many morphisms subject to $<\lambda$-many relations. That is, $F$ is indeed $\lambda$-presentable, with slightly weaker hypotheses than requested. It's interesting that standard results about presentability and fibrations like Makkai-Paré Theorem 5.3.4 don't seem to quite apply here.<|endoftext|> TITLE: Trace of integral trace-class operator QUESTION [8 upvotes]: I have seen many answers to the converse question (which seems to be difficult in general), but I would like to ask the following: Let $T: L^2 \rightarrow L^2$ be a trace-class operator that is also an integral operator $$Tf = \int K(\cdot,y)f(y)dy.$$ Since $T$ is trace-class $\operatorname{tr}(T)$ exists. Now, I would like to ask: Under what conditions is this trace given by $$\operatorname{tr}(T)=\int K(x,x) dx.$$ In a way, continuity would presumably be a sufficient requirement to make sense out of this expression, but I could imagine that much more is known about this. REPLY [4 votes]: As js21 says in the comnent, this is always the case if we understand a restriction $K(x,x)$ of the kernel $K(x,y)$ onto diagonal in appropriate way. In general, a square-summable function of two variables does not have a well-defined restriction to sets of zero measure, like diagonal. But the kernels of nuclear operators belong to a nice class of functions for which integrals against many singular measures, including the diagonal measure, are well-defined. It is defined by Kellerer and recently studied from a different point of view in a series of papers of Vershik, Zatitskiy and myself. We called such functions 'virtually continuous', since an equivalent description is the following: there exist a set of full measure $X'\subset (X,\mu)$ and a certain separable metric $\rho$ on $X'$ (such that measure $\mu$ is Borel) for which $K(x,y)$ is equivalent to a continuous function on $(X',\rho)$. That is, the virtual continuity is defined without referring to any topology, but using only the measure and the direct product structure of the space. Our paper in Russian Mathematical Surveys contains the details on nuclear operators and virtual continuity, and the further paper in St. Petersburg Mathematical Journal explains the connection with another approach to the trace via averaging over the neighborhood of a diagonal.<|endoftext|> TITLE: Does the degree of a finite dominant morphism bound the induced degree on subschemes? QUESTION [5 upvotes]: Suppose $f: \widetilde{X} \to X$ is a finite dominant morphism between connected, normal, Noetherian schemes, and that this morphism induces a dominant morphism $f_W: \widetilde{W} \to W$ between connected normal subschemes of $\widetilde{X}$ and $X$, respectively. My question is: can we bound $\deg(f_W)$ in terms of $\deg(f)$? I know that if $f$ is assumed to be flat, and $\widetilde{W} = \widetilde{X} \times_X W$, then the degrees are equal (see Elencwajg's answer to this question, which cites Q. Liu's "Algebraic Geometry and Arithmetic Curves", ex. 1.25 on pg 176) but I don't see how to obtain a general bound, nor what a likely counterexample would be. REPLY [2 votes]: If you have a degree $d$ finite dominant morphism $f : X \to Y$ between normal integral schemes, then every geometric fibre $X_y$ of $f$ has at most $d$ points. More generally, this is true if we replace the assumption that $X$ is integral by "every irreducible component of $X$ dominates $Y$" (but we keep all the other assumptions). In this form, the statement can be reduced to the case where $Y$ is strictly henselian and $y$ is the closed point. This case is obvious because $X$ splits into one piece for each $x$ lying over $y$. Now apply this to a geometric point of $Y$ lying over the generic point of an integral closed subscheme $W \subset Y$ and see what you get. Enjoy!<|endoftext|> TITLE: Implications of the disproof of the "climb-to-a-prime" conjecture QUESTION [20 upvotes]: Now that James Davis has found a counter example, 13532385396179, to John Conway's climb-to-a-prime conjecture, I would be interested to learn whether this has any implications of interest in number theory. REPLY [44 votes]: Hans said that Conway's point in asking it was that there exist problems easy to state but impossible to prove. The point I took away was that there exist problems that look so hard, nobody has tried anything easy. Read the letter to Conway on Numberphile if you want to see how it was easy. I'll ask Conway if it has any other implications (if I get the chance). I kinda doubt it, but who knows. The idea isn't really limited to base 10. The problem and short search that worked generalizes to other bases. I worked in smaller bases at first. Edit: If anyone does think of a mathematical usefulness to this, please let me know!<|endoftext|> TITLE: Multiplier norm vs cb norm QUESTION [5 upvotes]: Let $f:G\to \mathbb{C}$ be a finitely supported functions and let $m_f$ denote the associated multiplier on $C^*_r(G)$, the reduced group $C^*$-algebra: $$m_f(\alpha)(g)=f(g)\alpha(g)$$ for every $\alpha\in C^*_r(G)$. We can look at two natural norms of $m_f$: the completely bounded norm $\Vert m_f\Vert_{cb}$ of $m_f$ as a multiplier on $C^*_r(G)$ the multiplier norm $\Vert m_f\Vert_m$ of $m_f$ viewed as a bounded linear operator on $C^*_r(G)$. In general, $$\Vert m_f\Vert_m\le \Vert m_f\Vert_{cb}.$$ Is there anything else known about the relation of these two norms? For instance, are there examples of $f$ with $\Vert m_f\Vert_{cb}=1$ and $\Vert m_f \Vert_m\to 0$? Or should they be equal in some cases? I would be grateful for either an answer or direction where to look for this in the literature. REPLY [2 votes]: The norms are equal if $G$ is amenable, and inequivalent if $G$ is e.g. the free group. Losert has stated that the two norms can be equivalent (in fact, equal IIRC) when $G=SL(2,R)$ but to my knowledge this has never been published I can't give an explicit reference off the top of my head, but key words here are "Herz-Schur multiplier" and "multipliers of the Fourier algebra". A good place to find out some of the basic terminology, and to find links to the literature, is this set of minicourse notes by Todorov. See Theorem 3.6 in particular.<|endoftext|> TITLE: When do Gorenstein Stanley-Reisner rings have Du Bois singularities? QUESTION [5 upvotes]: The question is pretty much as in the title. Given a simplicial complex $\Delta$, I can associate a Stanley-Reisner ring. I assume this ring is Gorenstein, when does it have Du Bois singularities? Motivation: I learned from the opening paragraphs of https://arxiv.org/pdf/1106.4977.pdf that the cycle of affine planes (a very simple Stanley Reisner ring) has Du Bois singularities. I am coming from the study of mirror symmetry, but the Du Bois condition seems like an important class of singularity and I would like to understand it better. REPLY [5 votes]: Stanley-Reisner rings have F-injective type (even F-pure type), so they are DuBois by https://arxiv.org/pdf/0806.3298.pdf.<|endoftext|> TITLE: What problem in pure mathematics required solution techniques from the widest range of math sub-disciplines? QUESTION [41 upvotes]: (This is a restatement of a question asked on the Mathematics.SE, where the solutions were a bit disappointing. I'm hoping that professional mathematicians here might have a better solution.) What are some problems in pure mathematics that require(d) solution techniques from the broadest and most disparate range of sub-disciplines of mathematics? The difficulty or importance or real-world application of the problem is not my concern, but instead the breadth of the range of sub-disciplines needed for its solution. The ideal answer would be a problem that required, for instance, number theory, group theory, set theory, formal logic, homotopy theory, graph theory, combinatorics, geometry, and so forth. Of course, most sub-branches of mathematics overlap with other sub-branches, so just to be clear, in this case you should consider two sub-branches as separate if they have separate listings (numbers) in the Mathematics Subject Classification at the time of the result. (Later, and possibly in response to such a result, the Subject Classifications might be modified slightly.) One of the reasons I'm interested in this problem is by analogy to technology. More and more problems in technology require a range of disciplines, e.g., electrical engineering, materials science, perceptual psychology, optics, thermal physics, and so forth. Is this also the case in research mathematics? I'm not asking for an opinion—this question is fact-based, or at minimum a summary of the quantification of the expert views of research mathematicians, mathematics journal editors, mathematics textbook authors, and so forth. The issue can minimize the reliance on opinion by casting it as an objectively verifiable question (at least in principle): What research mathematics paper, theorem or result has been classified at the time of the result with the largest number of Mathematics Subject Classification numbers? Moreover, as pointed out in a comment, the divisions (and hence Subject Classification numbers) are set by experts analyzing the current state of mathematics, especially its foundations. The ideal answer would point to a particular paper, a result, a theorem, where one can identify objectively the range of sub-branches that were brought to bear on the proof or result (as, for instance, might be documented in the Mathematics Subject Classification or appearance in textbooks from disparate fields). Perhaps one can point to particular mathematicians from disparate sub-fields who collaborated on the result. REPLY [4 votes]: The Beilinson regulator is $\frac{1}{2}$ times the Borel regulator. The Beilinson regulator is a map from algebraic K-theory to Deligne cohomology and the above equality generalizes Borel's theorem on the algebraic K-theory of number rings, which in turn generalizes the class number formula from algebraic number theory. A complete proof is contained in this book. To get an impression of the range of involved fields one may just look at its Table of Content, from which I copy the names of chapters: Simplicial and Cosimplicial Objects H-spaces and Hopf Algebras The Cohomology of the General Linear Group Lie Algebra Cohomology and the Weil Algebra Group Cohomology and the van Est Isomorphism Small Cosimplicial Algebras Higher Diagonals and Differential Forms Borel's regulator Beilinson's Regulator<|endoftext|> TITLE: Klee's trick --- more applications QUESTION [15 upvotes]: In his "Some topological properties..." (1955), Klee gave a construction (simple and beautiful) of an isotopy $h_t\colon\mathbb{R}^{2\cdot n}\to \mathbb{R}^{2\cdot n}$ which moves any compact set $K$ in the coordinate $\mathbb{R}^n$-subspace to any other homeomorphic compact $K'$ set in this subspace. The idea is to extend the homeomorphisms $K\to K'$ and $K'\to K$ to continuous functions $f,g\colon\mathbb{R}^n\to\mathbb{R}^n$ and construct the needed isotopy as concatenation of $(x,y)\mapsto (x,y+t\cdot f(x))$ and $(x,y)\mapsto (x-t\cdot g(y),y)$ Later in "Plane separation" (1968), Doyle used this idea to give 5 line proof of the Jordan separation theorem. Do you know any other places where this idea shows up? REPLY [9 votes]: This answer is a correction to my comment above. In these notes from Marshall Cohen's course (from Spring 2001), he uses the Klee trick to give a simple proof that if $f, g: X \to S^n$ are non-surjective embeddings with $X$ compact, then the integral homology of $S^n - f(X)$ is the same as that of $S^n - g(X)$. From this he gave simple deductions of Invariance of Domain (showing that $\mathbb{R^k}$ does not embed into $\mathbb{R^n}$ if $k>n$ and the Jordan-Brouwer separation theorem for embeddings of $S^{n-1}$ into $S^n$.<|endoftext|> TITLE: Using Jordan's theorem to find Galois group for a polynomial QUESTION [9 upvotes]: I'm trying to apply the result of Jordan's theorem (cited below) to find the Galois group for a given polynomial. My goal is to provide an example where Jordan's theorem is useful, so the polynomial I'm using will have to be one whose Galois group is difficult to calculate without using Jordan's theorem. I chose the polynomial $f(x)=x^7-7x-1$, but any other examples that you think will better serve my goal are welcome. First we have to show that the conditions for Jordan's theorem are met: $f$ is irreducible and hence its Galois group $G$ is contained in $S_7$. $G$ is primitive acting on the spilitting field of $f$. $G$ contains a $p$-cycle ($p<7-2=5$). Proving the conditions are met: $f$ is irreducible since $f(x)= x^7+x+1 \pmod 2$ which is irreducible . $G$ contains a permutation of cycle type (2,5) because $f(x)=(x^2+2x+2)(x^5+x^4+2x^3+2x+2) \pmod3$, and hence $G$ contains a 2-cycle ((2,5), whose 5th power is a 2-cycle). $G$ is primitive acting on the roots of $f$ because G is transitive and $f$ has 7 roots. (by the way, is it ok to say "$G$ acts on the roots of $f$" in this context?) Therefore, By Jordan's theorem $G$ is $S_n$ or $A_n$. Since $disc(f)$ is not square we get that $G=S_7$. Is this proof correct? Jordan's theorem: Let $G$ be a primitive permutation subgroup of the symmetric group $S_n$. If $G$ contains a $p$-cycle for some prime number $p TITLE: Triviality of the adjoint and endomorphism bundles QUESTION [5 upvotes]: Let $P$ be be a principal bundle over a manifold $M$ with structure group $G$, where $G$ is a Lie group. Let $E = P\times_{\rho} \mathbb{R}^{k}$ be a vector bundle associated to $P$ through a faithful representation $\rho\colon G\to Gl(k,\mathbb{R})$. Let $\mathrm{Ad}(P) = P\times_{Ad}\mathfrak{g}$ denote the adjoint bundle associated to $P$, where $\mathfrak{g}$ denotes the Lie algebra of $G$. My question is the following: If $\mathrm{Ad}(P)$ is topologically trivial, namely $\mathrm{Ad}(P)\simeq M\times \mathfrak{g}$, does it follow then that the endomorphism bundle $\mathrm{End}(E)\simeq E\otimes E^{\ast}$ of $E$ is also topologically trivial? If I am not mistaken, this is clearly true if $G=Gl(k,\mathbb{R})$. I am interested in the case $G\subset Gl(k,\mathbb{R})$ and $G\neq Gl(k,\mathbb{R})$, with $G$ connected, compact and semi-simple. Thanks. REPLY [5 votes]: This was too long for a comment. There are counterexamples where $G$ is a finite group. The OP correctly notes that the endomorphism bundle of a rank $1$ bundle is trivial. Nonetheless, there are plenty of examples in higher rank with nontrivial endomorphism bundle. For instance, let $G$ be the cyclic group of order $2$. Begin with the real projective plane $\mathbf{RP}^2 = \mathbf{S}^2/G$ with its tautological $G$-bundle, $\mathbf{S}\to \mathbf{RP}^2$. The standard inclusion $G\subset \textbf{GL}_1(\mathbb{R}) = \mathbb{R}^\times$ induces the tautological rank $1$ bundle $\gamma_2$. Let $M$ be a product $X\times Y$ of two real projective planes $X\cong Y\cong \mathbf{RP}^2 = \mathbf{S}^2/G$. The $G\times G$-bundle $$\widetilde{X}\times \widetilde{Y}\to X\times Y$$ and the inclusion $$G\times G\subset \mathbf{GL}_1(\mathbb{R})\times \mathbf{GL}_1(\mathbb{R}) \subset \mathbf{GL}_2(\mathbb{R})$$ induces a rank $2$ bundle $E$ that is simply $\text{pr}_X^*\gamma_2 \oplus \text{pr}_Y^*\gamma_2$. The endomorphism bundle is isomorphic to a rank $4$ bundle, $$\text{End}(E) \cong \mathbb{R} \oplus (\text{pr}_X^*\gamma_2\otimes \text{pr}_Y^*\gamma_2^\vee) \oplus (\text{pr}_X^*\gamma_2^\vee \otimes \text{pr}_Y^*\gamma_2) \oplus \mathbb{R}.$$ By the Whitney sum formula, the total Stiefel-Whitney class is $$(1)(1+\text{pr}_X^*a - \text{pr}_Y^*a)(1-\text{pr}_X^*a+\text{pr}_Y^*a)(1) =$$ $$ 1 - (\text{pr}_X^*a)^2 - (\text{pr}_Y^*a)^2 + 2\text{pr}_X^*a\cup \text{pr}_Y^*a.$$ Of course the last summand is zero since the coefficients are $\mathbb{Z}/2\mathbb{Z}$. However, $a^2$ is nonzero in $H^2(\mathbb{RP}^2;\mathbb{Z}/2\mathbb{Z})$. Using Künneth, the second Stiefel-Whitney class of $\text{End}(E)$ is nonzero. REPLY [5 votes]: There are counterexamples for $G$ connected abelian. Let $X = \mathbb CP_2$, $P$ be the complex line bundle $\mathcal O(1)$, viewed as a $G=SO(2)$-bundle. We take $\rho$ the standard two-dimensional representation of $SO(2)$. Of course the adjoint bundle is trivial in this case. To show that $E \otimes E^*$ is nontrivial, we calculate its first Pontryagin class by tensoring with $\mathbb C$, where we can view it as $(\mathcal O(1) + \mathcal O(-1)) \otimes ( \mathcal O(-1) + \mathcal O(2) ) = \mathcal O(2) + \mathcal O + \mathcal O + \mathcal O(-2)$, which has second chern class $-4 H^2$, which is nontrivial, hence is a nontrivial complex vector bundle (and thus $E \otimes E^*$ is a nontrivial real vector bundle). However, the statement might be possibly be true for $G$ connected, semisimple, and compact. Let $Z$ be the center of $G$. Then the condition that $Ad(P)$ is trivial necessarily implies that the $G/Z$ bundle on $M$ induced by $P$ is pulled back from the obvious $G/Z$-bundle on $Gl_n(\mathfrak g)/ (G/Z)$, where $G/Z$ acts by right multiplication in the adjoint representation. If $\rho$ is a representation on which $Z$ acts by scalars, then $E \otimes E^*$, as a representation of $G$, factors through $G/Z$, and hence is a pullback from $M$. So in this special case, it suffices to check whether the vector bundle $E \otimes E^*$ on $GL_n(\mathfrak g) / (G/Z)$ is nontrivial. However, I don't know how to do this.<|endoftext|> TITLE: Classification of weak 3-groups QUESTION [6 upvotes]: Weak 2-groups can be classified by the data $(\pi_1,\pi_2, t, \omega)$, where $\pi_1$ is a group, $\pi_2$ an Abelian group, $t: \pi_1 \to Aut(\pi_2)$, and $\omega \in H^3(B\pi_1,\pi_2)$. I wonder do we have a similar classification for weak 3-groups? 3-groups contain data $(\pi_1,\pi_2,\pi_3)$ (three groups). I am interested in the simple cases where $\pi_2$ is $Z_2$ or trivial and $\pi_3$ is $U(1)$ or $Z_n$. REPLY [5 votes]: This has been worked out by Conduché in Modules croisés généralisés de longueur 2 (JPAA 34 (1984), 155-178), using simplicial group methods. While it more or less boils down to the stuff described in the answer by Qiaochu, Conduché also describes a nice (2? 3?)-category holding the same information. Its objects are complexes$$G_2\xrightarrow{\partial}G_1\xrightarrow{\partial}G_0$$of $G_0$-groups (composite trivial, $G_0$ acting on itself by conjugation) together with the s. c. Peiffer bracket $\{,\}:G_1\times G_1\to G_2$ satisfying the elaborate but appealing identities $$ \begin{aligned} {}^{x_0}\{x_1,y_1\}&=\{{}^{x_0}x_1,{}^{x_0}y_1\}\\ \{\partial x_2,\partial y_2\}&=[x_2,y_2]\\ \partial\{x_1,y_1\}&=x_1y_1x_1^{-1}\left({}^{\partial x_1}y_1\right)^{-1}\\ \{\partial x_2,x_1\}\{x_1,\partial x_2\}&=x_2\left({}^{\partial x_1}x_2\right)^{-1}\\ \{x_1y_1,z_1\}&=\{x_1,y_1z_1y_1^{-1}\}\ \ {}^{\partial x_1}\{y_1,z_1\}\\ \{x_1,y_1z_1\}&=\{x_1,y_1\}\{x_1,z_1\}\{\partial\{x_1,z_1\}^{-1},{}^{\partial x_1}y_1\}. \end{aligned} $$ Since both $G_2$ and $G_1$ are nonabelian, it is clear that there will be lots of equivalent nonisomorphic objects, but still this description has its advantages. Homotopy groups are, as expected, the homology groups of the complex (i. e. $\pi_1$ is the cokernel of $G_1\to G_0$, $\pi_2$ the quotient of the kernel of $G_1\to G_0$ by the image of $G_2\to G_1$, and $\pi_3$ the kernel of $G_2\to G_1$. Thus given $\pi_1$, $\pi_2$, $\pi_3$, to build a model of the above kind one also needs actions of $\pi_1$ on $\pi_2$ and $\pi_3$ as well as the bracket $\pi_2\times\pi_2\to\pi_3$. One then must build a complex as above such that its homology groups are $\pi_1$, $\pi_2$, $\pi_3$, while the actions and the bracket are induced from the structure on the complex.<|endoftext|> TITLE: Integral cohomology of $SU(n)$ - looking for constants QUESTION [20 upvotes]: I am interested in explicit generators of the cohomology $H^\bullet(SU(n),\mathbb{Z})$. Let $\omega = g^{-1} dg$ be the Maurer-Cartan form on $SU(n)$. The forms $\alpha_3,\alpha_5,\dots,\alpha_{2n-1}$, defined by $$ \alpha_k := \text{Tr}(\omega^{k})$$ are bi-invariant and define classes in de Rham cohomology. It is well-known that the cohomology algebra $H^\bullet(SU(n),\mathbb{R})$ is an exterior algebra in $\alpha_3,\dots,\alpha_{2n-1}$. More precisely, my question is thus the following: find the optimal constants $C_k$ such that $C_k \cdot \alpha_k$ is an integral class. For $n=2$, I have found in the literature that $C_3$ is $(24\pi^2)^{-1}$. But I cannot find a reference for the higher dimensions. I expect that $C_k = q_k \cdot \pi^{-(k+1)/2}$, with $q_k$ some rational number, because of the volume of the sphere $\mathbb{S}^{2k-1}$, but I have no precise proof. It seems from the comments in Cohomology of the unitary group that an interpretation in terms of transgression can solve my problem but I am not very familiar with this. Thanks! REPLY [5 votes]: I got confused by some of the remarks in the other answers and so decided to work it out for myself. Hopefully, the following is helpful to people who stumble upon this question like me. One can indeed use transgression as discussed in Jeremy Daniel's answer in order to settle this problem, since the generators of $H^*(BU; \mathbb{Z})\cong \mathbb{Z}[c_1,c_2,\dots]$ are well known and given by the Chern classes. The following properties of the transgression map $T$ are important for us: The transgression of nontrivial cup products is $0$, and therefore we have that $(-1)^{k-1} (k-1)! T(ch_k) = T(c_k)$ for the Chern character/ Chern classes Transgressing the Chern character $ch_k$ two times in the path loop fibration gives the Chern character $ch_{k-1}$ in one degree lower under the identification of $\Omega^2 BU \sim BU$, since the Chern character is compatible with Bott periodicity Transgression of an integral class must always give an integral class It is clear from this that the transgression $T(ch_k) = (-1)^{k-1}\left(\frac{i}{2 \pi}\right)^k \frac{(k-1)!}{(2k-1)!} \alpha_{2k-1}$ cannot be the image of an integral generator under the coefficient homomorphism $H^{2k-1}(U;\mathbb{Z})\to H^{2k-1}(U;\mathbb{R})$ in general, since transgressing again in the path loop fibration would then yield that $T(T(ch_k)) = ch_{k-1}$ is an integral cohomology class, which is wrong for $k>2$. What is claimed in Matthias Wendt's answer is still true: If we are in the stable range, then on a smooth map $f\colon S^{2k-1}\to U(n)$ which generates the homotopy group $\pi_{2k-1}(U(n)) = \mathbb{Z}$, this form evaluates to $\int_{S^{2k-1}} f^* T(ch_k)=1$. The crucial point here is though that the image of such a generator under the Hurewicz homomorphism $h\colon \pi_{2k-1}(U) \to H_{2k-1}(U;\mathbb{Z})$ is divisible by exactly $(k-1)!$. This can be deduced from the last corollary in chapter 8 in Bott, Raoul. The space of loops on a Lie group. Michigan Math. J. 5 (1958), no. 1, 35--61. doi:10.1307/mmj/1028998010. https://projecteuclid.org/euclid.mmj/1028998010, which gives the corresponding statement for BU. This means that $(k-1)! T(ch_k)$ is an integral generator, which is consistent with the fact that transgression is an integral isomorphism and therefore must map the Chern classes to integral generators. Therefore, the optimal constant that makes the forms $\alpha_{2k-1}$ into integral generators should be $a_k = \left(\frac{i}{2 \pi}\right)^k \frac{((k-1)!)^2}{(2k-1)!}$.<|endoftext|> TITLE: Distance function on a curve on a manifold QUESTION [5 upvotes]: Suppose that we are given a non-negative even function $b\in C^\infty[-1,1]$ satisfying $b(0)=0$, $\sqrt{b(x+y)}\le \sqrt{b(x)}+\sqrt{b(y)}$ for any $x,y\in[-\frac12,\frac12]$. Can we always find a 3-dimensional smooth Riemannian manifold $(M,g)$ and a smooth curve $a:[-\frac12,\frac12]\to M$ parametrized by arc length, such that the square of the distance function $d_g(a(t),a(s))^2=b(t-s)$? For example, if we set $(M,g)=(\mathbb{R}^3,\|\cdot\|)$, can we find a smooth curve $a:[-\frac12,\frac12]\to \mathbb{R}^3$ parametrized by arc length, such that the square of the distance function $\|a(t)-a(s)\|^2=b(t-s)$? REPLY [4 votes]: The answer is "no". Take $b(x)=x^2-x^{1000}$. Assume that there is a manifolds with the required curve $a$. Note that $a$ has vanishing geodesic curvature; in particular $a$ is a geodesic. On the other hand, no arc of $a$ is length-minimizing, a contradiction.<|endoftext|> TITLE: Fefferman's article: Pointwise convergence of Fourier series QUESTION [9 upvotes]: I have some problems reading Pointwise convergence of Fourier series by Fefferman https://www.jstor.org/stable/1970917 When I proceed to Lemma 2, Chapter 6, I could not verify either of the following: "Trivial estimates show that $|T_{p'}T_p^*f(x)|\leq \delta^{10}/|I'^*|\int_{E(p)|f(y)|dy}$, if $\mathrm{distance}(\omega,\omega')>\delta^{-\epsilon/2}|\omega|$, " "$T_{p'}T_p^*f(x)=0$ if $I\not\subseteq I'^*$." For the second, I could only show that this is the case if $I^*\cap I'^*=\varnothing$, but I think it just needs some modification (any suggestions?) to be true. However, for the first one, the only place that involves $\delta$ is the distance condition. However, I simply do not know how to use that in the trivial estimate. I tried using the decay of the Fourier transform, but it turns out I could only obtain a term involving $\delta^{\epsilon/2}$. Thank you for spending time reading my question. If you happened to have read the paper before, could you help me understand what he meant? REPLY [4 votes]: I have solved the problem by reading the following master's thesis: http://diposit.ub.edu/dspace/bitstream/2445/107985/2/memoria.pdf The idea is that we could iterate the process for $\sim \epsilon^{-1}$ times, so as to bootstrap the bounds on the right hand side from the $\delta^{\frac \epsilon 2}$ to $\delta^{10}$. The implicit constant here may become incredibly large, but it does no harm, since in Lemma 2, we tolerate a loss of $\eta$ on the power of $\delta$. However, there seems also a problem in the thesis when I proceed to Chapter 6, Lemma 5. Below is a link to another problem I raised. Fefferman's article: Pointwise convergence of Fourier series, II Also, I would like to correct a typo in the statement: the $\delta^{\frac 1 2-\eta}$ should be replaced by $\delta^{\frac 1 4-\eta}$.<|endoftext|> TITLE: Harmonic congruence QUESTION [11 upvotes]: There are a number of interesting congruences for harmonic sums, not the least of which is Wolstenholme's theorem: $H_{p-1}:=\sum_{j=1}^{p-1}\frac1j\equiv 0\mod p^2$. It appears that $\sum_{j=1}^{p-1}(-1)^{\binom{j}2}\frac1j\equiv 0\mod p$ when $p\equiv1\mod 4$. Anyways, I ask for: For a prime $p\geq5$, what is the value of $$\sum_{j=1}^{p-1}\frac{(-1)^{\binom{j}2}}j \mod p\,\,?$$ Contrast this with $\sum_{j=1}^{p-1}(-1)^{\binom{j}2}j=\frac{(-1)^{\binom{p}2}-1}2$ which equals $0$ when $p\equiv1\mod 4$ and equal $-1$ when $p\equiv3\mod4$. REPLY [7 votes]: Let's use the fact that $$\frac{1}{j}\equiv \frac{(-1)^{j-1}}{p}\binom{p}{j}\pmod{p}$$ to rewrite your sum mod p as $$\sum_{j=1}^{p-1}\frac{(-1)^{\frac{(j+2)(j-1)}{2}}}{p}\binom{p}{j}=\frac{1}{p}\left( \sum_{j=1}^{\frac{p-1}{2}}(-1)^{j-1}\binom{p}{2j-1} \right)+\frac{1}{p}\left( \sum_{j=1}^{\frac{p-1}{2}}(-1)^{j+1}\binom{p}{2j} \right)$$ $$=\frac{1}{2ip}\left((1+i)^p-(1-i)^p-2i^p\right)+\frac{1}{2p}\left(2-(1+i)^p-(1-i)^p\right)$$ $$=\frac{(1+i)^{p-1}-(1-i)^{p-1}+i-i^p}{ip}=\frac{2^{\frac{p-1}{2}}(i^{\frac{p-3}{2}}-(-i)^{\frac{p+1}{2}})+1-(-1)^{\frac{p-1}{2}}}{p}$$ and this can easily be checked to be zero for $p=1\pmod{4}$ and $\frac{2+2^{\frac{p+1}{2}}(-1)^{\frac{p-3}{4}}}{p}$ when $p=3\pmod{4}$. REPLY [6 votes]: For simplicity, let $p'=\lfloor\frac{p-1}2\rfloor, p''=\lfloor\frac{p-1}4\rfloor$ and the so-called Fermat's quotients $q_2=\frac{2^{p-1}-1}p$. Observe that $$\sum_{k=1}^{p-1}\frac{(-1)^{\binom{k}2}}k =\sum_{k=1}^{p'}\frac{(-1)^k}{2k}-\sum_{k=1}^{p'}\frac{(-1)^k}{2k-1}.$$ If we change variables $k\rightarrow p'-k+1$ then $$\sum_{k=1}^{p'}\frac{(-1)^k}{2k-1}=\sum_{k=1}^{p'}\frac{(-1)^{p'-k+1}}{p-2k} \equiv_p (-1)^{p'}\sum_{k=1}^{p'}\frac{(-1)^k}{2k}.$$ The following are known facts: $\sum_{k=1}^{p''}\frac1k\equiv_p-3q_2$ and $\sum_{k=1}^{p'}\frac1k\equiv_p-2q_2$. Therefore, we have \begin{align} \sum_{k=1}^{p'}\frac{(-1)^k}k &=\sum_{k=1}^{p''}\frac1{2k}-\sum_{k=1}^{p''}\frac1{2k-1} =\sum_{k=1}^{p''}\frac1{2k}-\left(\sum_{k=1}^{p'}\frac1k-\sum_{k=1}^{p''}\frac1{2k}\right) \\ &=\sum_{k=1}^{p''}\frac1k - \sum_{k=1}^{p'}\frac1k\equiv_p-q_2. \end{align} Going back to the earlier sums, we conclude that $\sum_{k=1}^{p'}\frac{(-1)^k}{2k-1}\equiv_p-\frac12(-1)^{p'}q_2$ and $$\sum_{j=1}^{p'}\frac{(-1)^{\binom{j}2}}j \equiv_p\frac{(-1)^{p'}-1}2\,q_2 \equiv_p\frac{\left(\frac{-1}p\right)-1}2\,q_2;$$ where $\left(\frac{a}p\right)$ stands for the Legendre symbol.<|endoftext|> TITLE: Rationality of the Tate module of an abelian variety relative to the algebra of its endomorphisms QUESTION [5 upvotes]: Suppose that $K/\mathbb{Q}_p$ is a finite extension and $k_K$ the residue field of $K$. Let $A/K$ be an abelian variety with good reduction. Suppose that $E\to\mathrm{End}^0_K(A)$ is an inclusion of a number field that sends $1$ to the identity. Denote by $T_\ell A$ the Tate module for a prime $\ell\neq p$ and let $V_\ell A=T_\ell A\otimes_{\mathbb{Z}_\ell}\mathbb{Q}_\ell$. The Galois representation $$\rho_\ell:\mathrm{Gal(\bar{K}/K)}\to \mathrm{Aut}(V_\ell A)$$ is then $E_\ell=E\otimes \mathbb{Q}_\ell$-linear. We can decompose $E_\ell=\prod_\lambda E_\lambda$ where $\lambda$ run through the places of $E$ dividing $\ell$ and $E_\lambda$ is the corresponding completion of $E$. Then by $E_\ell$-linearity, $V_\ell A=\prod_\lambda V_\lambda$ as $E_\ell[\mathrm{Gal}(\bar{K}/K)]$-modules. Let $\mathrm{Frob}_K$ be a lift of the Frobenius element. I want to prove that each $E_\lambda$-characteristic polynomial of $\rho_\lambda(\mathrm{Frob}_K)$ is the image of some common polynomial $P_0\in E[X]$ via $E[X]\hookrightarrow E_\lambda[X]$. I believe this follows from a result by Shimura from "Algebraic Number Fields and Symplectic Discontinuous Groups", Prop. 11.09, but I want to understand the argument of his proof. It is well-known that the action of the lift of the Frobenius element is obtained via the Frobenius endomorphism $\pi$ of the reduction $\tilde{A}/k_K$, so the $\mathbb{Q}_\ell$-characteristic polynolmial of $\rho_\ell(\mathrm{Frob}_K)$ has rational coefficients. Then, we need to find an $E[\pi]$-module $U$ so that $V_\ell A\simeq U\otimes \mathbb{Q}_\ell$ as $E[\pi]\otimes \mathbb{Q}_\ell$-modules. But how to get this $U$? REPLY [4 votes]: Let's instead decompose the $\ell$-adic Tate module, tensored up with $\overline{\mathbb Q}_\ell$, as a product over embeddings of $E$ into $\overline{\mathbb Q}_\ell$ of a $ \overline{\mathbb Q}_\ell$-vector space with an action of $\pi$. On each such vector space $\pi$ admits a characteristic polynomial, and you want to know that these characteristic polynomials are all conjugate under the action of $\operatorname{Gal}(\overline{\mathbb Q}_{\ell}|\mathbb Q)$ permuting the embeddings. This follows from two facts: The determinant of every element of $E[\pi]$ acting on this vector space is Galois-invariant, because it equals the degree of the corresponding rational endomorphism and hence is a rational number. The determinant of $\pi$ minus an element of $E$ can be calculated as the product of these different characteristic polynomials applied to the different embeddings of the element of $E$. From the determinant, we can recover the individual factors, because we can write the determinant as a polynomial in the element of $E$ under different embeddings into $\overline{\mathbb Q_\ell}$, which we treat as independent random variables, and then factor this polynomial. Because the determinant is Galois-invariant, and we can use it to calculate the individual polynomials, the polynomials must be appropriately equivariant under the action of Galois. Constructing a module like you suggest might be difficult, because, as Serre's example of a supersingular elliptic curve shows, it is impossible to do so canonically.<|endoftext|> TITLE: Manifold of mappings between $M$ and $N$, with non-compact source $M$ QUESTION [5 upvotes]: EDIT: Let $M$ and $N$ are two smooth manifold and suppose $N$ is compact but $M$ is not necessarily compact. For my purpose, I just need to consider the case $M=\mathbb R \times S^1$ or $\mathbb R \times [0,1]$. Thanks to David's comments that help me a lot, the problem I really concern is just the following. Question: Now that we know if $M$ is non-compact then $C^\infty(M,N)$ is modeled on spaces $C^\infty_c(M,N)$. Since we know $W^{k,p}_0(\Omega)$ is the closure of $C^\infty_c(\Omega) $ in $W^{k,p}(\Omega)$ for $\Omega\subset \mathbb R$, in analogy, can we say $W^{k,p}(M,N)$ is modeled on spaces $W^{k,p}_0(M,f^*TN)$? Any reference? Can we consider $W^{k,p}_0(M,f^*TN)$ as a completion (in some sense?) of $C^\infty_c(M,f^*TN)$, the spaces of smooth sections with compact support of pullback bundles along $f:M\to N$? Previous post: (1) Can we define $W^{1,p}_0(M,N)$ in a similar manner that $W^{1,p}_0(\mathbb R)$ is defined by the completions of $C^\infty_0$ or $C^\infty_c$? For example, Floer in his paper (see Definition 2.1) actually discusses (roughly) $\mathcal W:=W_{loc}^{k,p}(\mathbb R \times [0,1], N)$ with a topology given by open sets as follows: $ \mathcal O_{u,\rho,\epsilon} =\{ v\in \mathcal W \mid v= \exp_u \xi ~\text{on}~ [-\rho,\rho]\times[0,1], ~\text{and}~ ||\xi||_{W^{k,p}} <\epsilon \} $ On the other hand, Audin and Damian in their book (see Definition 8.2.2) consider Banach manifolds $\widetilde {\mathcal W} =W^{k,p}(\mathbb R \times [0,1], N)$ (actually $[0,1]$ should be replaced by $S^1$) in a quite different manner. Here open sets are the space of maps of the form $v=\exp_u \xi$ where $\xi \in W^{k,p}(\mathbb R \times [0,1], N)\equiv W_0^{k,p}(\mathbb R \times [0,1], N)$ and where $u$ is smooth and converges in some decay at the infinity. Heuristically, $\mathcal W$ is like the completion of $C_c^\infty$ while $\widetilde {\mathcal W}$ is like that of $C_0^\infty$. (2) Is $\mathcal W$ the same as $\widetilde{\mathcal W}$? Which one could be a better candidate for the definition of $W^{k,p}_0(M,N)$? Are they both Banach manifolds as in the case $M$ is compact? Recently I notice that when considering (infinite-dimensional) manifolds of mapping, say $C^\infty(M,N)$, we usually require $M$ to be compact. (See Section 4.2 of this paper: The inverse function theorem of Nash and Moser). Notice that as long as the domain $M$ is compact, the questions (1), (2) and (3) become trivial. (3) In general, if $M$ is non-compact, then is $C^\infty(M,N)$ still a Frechet manifold as in the case $M$ is compact? REPLY [6 votes]: Let $M$ and $N$ be Riemannian manifolds. In general, the space of Sobolev mappings $W^{k,p}(M,N)$ should not be defined as a completion of smooth mappings even if $k=1$ and manifolds are compact. The common definition (at least if $N$ is compact) is as follows. Take an isometric embedding of $N$ into a Euclidean space $\mathbb{R^\nu}$ and then define: $$ W^{k,p}(M,N)=\{ f\in W^{m,p}(M,\mathbb{R}^\nu):\, f(x)\in N \text{ a.e.} \}. $$ This space is equipped with the metric inherited from the Sobolev norm and in general smooth mappings are not dense [1]. There are some problems when $N$ is not compact. You can find more papers about higher order Sobolev mappings between manifolds, including the case of non-compact target, at the homepage of Van Schaftingen. [1] P. Bousquet, A. C. Ponce, J. Van Schaftingen, Strong density for higher order Sobolev spaces into compact manifolds. J. Eur. Math. Soc. (JEMS) 17 (2015), no. 4, 763–817.<|endoftext|> TITLE: Formula for the Frobenius-Schur indicator of a finite group? QUESTION [8 upvotes]: Let $G$ be a finite group and let $k$ be an algebraically closed field of characteristic $p \neq 2$. Let $V$ be a finite-dimensional irreducible $kG$-module. If $V \cong V^*$, then $V$ admits a nonzero $G$-invariant bilinear form $(-,-)$, unique up to scalar, such that $(-,-)$ is alternating or symmetric. Is there a formula or a general method to determine whether $(-,-)$ is going to be alternating or symmetric? If $k = \mathbb{C}$, then the answer is yes: for a $\mathbb{C}G$-module $V$ with irreducible character $\chi$ we have the Frobenius-Schur indicator $$\nu_2(\chi) = \frac{1}{|G|} \sum_{g \in G} \chi(g^2)$$ which satisfies $\nu_2(\chi) \in \{-1, 0, 1\}$ and $\nu_2(\chi) \neq 0$ iff $V \cong V^*$. $\nu_2(\chi) = 1$ if there is a nonzero $G$-invariant orthogonal bilinear form on $V$. $\nu_2(\chi) = -1$ if there is a nonzero $G$-invariant alternating bilinear form on $V$. I think similar things should be true when $p \not \mid |G|$. What about in general? Is this an open problem? REPLY [5 votes]: For one answer, here is a theorem due to Thompson and Willems (Bilinear forms in characteristic $p$ and the Frobenius-Schur indicator, Lecture Notes in Mathematics 1185, pg. 221-230). For an irreducible self-dual $kG$-module $V$, set $\varepsilon(V) = 1$ if $G$ preserves a nonzero symmetric bilinear form on $V$, and $\varepsilon(V) = -1$ if $G$ preserves a nonzero alternating form on $V$. Theorem: Let $V$ be an irreducible self-dual $kG$-module with Brauer character $\phi$. There exists an ordinary irreducible character $\chi$ such that (1) $\chi$ is real valued. (2) The decomposition number $d(\chi, \phi)$ is odd. Furthermore, any ordinary irreducible character $\chi$ satisfying (1) and (2) has the property that $\nu_2(\chi) = \varepsilon(V)$. So assuming you know the ordinary character table of $G$, and the Frobenius-Schur indicators and decomposition numbers of all irreducible characters; you can give an answer. I don't know if there are any other general results. REPLY [4 votes]: The answer to the question can be found in Section 14, Frobeius-Schur indicator for Brauer characters, starting on page 320, of the book "Group Representations, Volume 4" by Gregory Karpilovsky. I don't have easy access to this book myself (our library only has Volume 1, and that is in two parts), but fortunately the Google Books preview is letting me see all 14 pages of that section. It says on page 321, "It is natural to ask whether there is a Frobenius-Schur indicator for the Brauer character $\beta$ of $G$ afforded by $V$, which allows us to determine whether $V$ is of symmetric of skew symmetric type. It turns out that the answer is positive and the required indicator may be taken to be the Frobenius-Schur indicator of a certain irreducible ${\mathbb C}$-character of $G$ closely related to $\beta$." The theory of all this follows and takes about 12 pages, so it is not for the faint-hearted. A Google search for "Frobenius Schur indicator Brauer character" comes up with several relevant pointers to technical papers on this topic, so it seems to be known to specialists.<|endoftext|> TITLE: Trying to prove one of C.Taubes' theorems gauge-theory-freely QUESTION [7 upvotes]: One of C.Taubes' theorems says that for a symplectic 4-manifold $X$ with $b^2_+>1$ (where $b^2_+$ denotes the dimension of a maximal positive-definite subspace of $H^2(X;\mathbb R)$ under the intersection form), $\mathrm{Gr}(e)=0$ if $c_1(K)\cdot e-e\cdot e\neq0$. Here $Gr(e)\in\mathbb Z$ is a particular count of $J$-holomorphic curves in $X$ which represent the class $e\in H_2(X;\mathbb Z)$, and $K^{-1}$ denotes the canonical bundle. This theorem is proved using Seiberg-Witten theory. Can one prove this gauge-theory-freely? REPLY [8 votes]: The key use of SW theory was to show that $Gr(e)=Gr(c_1(K)-e)$. For the moment, take this equality as granted. If $Gr(e)\ne0$ then there must exist a $J$-holomorphic curve $C\to X$ such that $[C]=e$, and likewise a $J$-holomorphic curve $C'\to X$ such that $[C']=c_1(K)-e$. To demonstrate the gist of the proof, assume $X$ is not a blow-up, and that $C$ and $C'$ are distinct embedded connected surfaces. By positivity of intersections of holomorphic curves, $e\cdot(c_1(K)-e)=\#(C\cap C')\ge0$, hence $-e\cdot c_1(K)+e\cdot e\le0$. But the dimension of the moduli of $J$-holomorphic curves representing $e$ is $c_1(TX)\cdot e +e\cdot e=-e\cdot c_1(K)+e\cdot e$, which must be nonnegative otherwise the moduli space would be empty. Thus $c_1(K)\cdot e-e\cdot e=0$. So, we need a way to demonstrate the aforementioned equality of Gromov invariants. I don't know yet (though it's part of my research) how to prove it inherent to $J$-holomorphic curve theory. But you just want a SW-free proof, and that is granted by Donaldson-Smith invariants. These are counts of sections of a certain bundle associated with a given Lefschetz fibration of $X$ (giving pseudolomorphic surfaces in some sense), and were shown to recover Taubes' Gromov invariants. Our desired equality (under a small restrictive assumption*) is a consequence of Serre duality between divisors on Riemann surfaces! *The restrictive assumption: $b^2_+(X)>b_1(X)+1$. So let's make $X$ simply connected.<|endoftext|> TITLE: Closed form for $\int_0^T e^{-x}\frac{I_n(\alpha x)}{x}dx$ QUESTION [5 upvotes]: EDIT: Some additional details and corrections, I would appreciate any information about the highlighted expression. I try to solve $\int_0^T e^{-x}\frac{I_n(\alpha x)}{x}dx$ where $I_n(x)$ is the modified Bessel function of the first kind and $0<\alpha<1$. My first approach was to turn this integral into an infinite sum to fit a hypergeometric series: Using the infinite series representation of the Bessel function, I got incomplete gamma functions in the sum, which does not sound promising. The multiplication theorem yields an infinite series of integrals where we get rid of the $\alpha$: $$\int_0^T e^{-x}\frac{I_n(\alpha x)}{x}dx=\alpha^n\sum_{m=0}^{\infty}\frac {\big(\frac {\alpha ^{2}-1}{2}\big)^m}{m!}\int_0^T e^{-x}x^{m-1}I_{n+m}(x)dx$$ According to a table of integrals, the new integrals are: \begin{align} \int_0^T e^{-x}x^{m-1}I_{n+m}(x)dx&=\frac{T^{2m+n}}{2^{m+n}}\frac{\Gamma(2m+n)}{\Gamma(m+n+1)\Gamma(2m+n+1)}\\ &\times{}_2F_2[\{m+n+\frac{1}{2},2m+n\};\{2m+2n+1,2m+n+1\};-2T] \end{align} Expanding ${}_2F_2$ (let's call $k$ the summation index), we get a double infinite series which might fit the definition of a hypergeometric function of 2 variables. However, I get several Pochhammer symbols with coupled summations indices: $$\sum_{m,k=0}^{\infty}\frac{(n+\frac{1}{2})_{m+k}(n)_{2m+k}}{(n+1)_{2m+k}(2n+1)_{2m+k}}\frac{X^mY^k}{m!\,k!}$$ which, apparently, does not fit any hypergeometric function definition (at least, this is not an Appell function). Another approach could be to get inspiration from the limit $T\rightarrow \infty$ which is the Laplace transform of $\frac{I_n(x)}{x}$ (up to a constant) and it has a closed form (according to a table): $$\int_0^\infty e^{-x}\frac{I_n(\alpha x)}{x}dx=\frac{\big(\frac{\alpha}{1+\sqrt{1-\alpha^2}}\big)^n}{n}$$ However, I don't find any reference on the way to compute this. EDIT: This comes from the recurrence identity $I_{n-1}(x)-I_{n+1}(x)=2n\frac{I_n}{x}$ and the calculation of the Laplace transform of $I_n(x)$ is well documented. Do you have any information or suggestion about the above formulae ? REPLY [4 votes]: as requested by the OP in the comment section: $$\int_0^T e^{-x}I_n(x)\frac{1}{x}\,dx=\frac{1}{n}+\frac{1}{n T^{n-1}}e^{-T}\left[a_n(T)I_0(T)+b_n(T)I_1(T)\right]$$ the functions $a_n$ and $b_n$ are polynomials of degree $n-1$, I do not have a closed form expression; the first few are: $$a_1(T)=-1,\;\;a_2(T)=-2T,\;\;a_3(T)=-3 T^2+4 T,$$ $$a_4(T)=-4 T^3+8 T^2-24 T,\;\;a_5(T)=-5 T^4+20 T^3-48 T^2+192 T$$ $$b_1(T)=-1,\;\;b_2(T)=-2T+2,\;\;b_3(T)=-3T^2+4T-8,$$ $$b_4(T)=-4 T^3+12 T^2-16 T+48,\;\;b_5(T)=-5 T^4+20 T^3-88 T^2+96 T-384$$<|endoftext|> TITLE: Concavity of the trace of a matrix power QUESTION [8 upvotes]: Let $B$ be an $n\times n$ matrix, and define $f$ to be the function that maps positive semidefinite (PSD) $n\times n$ matrices $A$ to real numbers by $$ f(A) = \mathrm{trace}( (B^*A^2B)^{1/3}). $$ In other words, $f$ maps $A$ to the sum of $1/3$-powers of the eigenvalues of the PSD matrix $B^*A^2B$. Is the function $f$ concave over the PSD cone? I.e. is it true that for any two PSD matrices $X$ and $Y$, $f((X + Y)/2) \ge f(X)/2 + f(Y)/2$? A more general question is whether the function $f(A) = \mathrm{trace}( (B^*A^2B)^{p})$ is concave over the PSD cone for $0< p < 1/2$. Carlen and Lieb have some closely related results, but I could not find the particular combination of matrix powers in their paper or in other related work on trace inequalities. REPLY [10 votes]: Unfortunately, the conjectured function is not concave. Here is a simple simpler counterexample. \begin{equation*} B = \begin{bmatrix} 1 & 2 \\ 3 & 4\end{bmatrix},\quad A = \begin{bmatrix} 2 & 0 \\ 0 & 3\end{bmatrix},\quad C = \begin{bmatrix} 5 & 0 \\ 0 & 2\end{bmatrix}. \end{equation*} With this choice, and for $f(X)=\text{tr} (B^*X^2B)^{1/3}$ we see that $f((A+C)/2)- (f(A)+f(C))/2 \approx -0.0616$. You may find the following necessary condition interesting: (Prop. 5.1, Hiai, 2013). Assume that $p,s\neq 0$. If $A \in \mathbb{P}_2 \mapsto \text{tr}(B^*A^pB)^s$ is concave for any invertible matrix $B \in \mathbb{C}^{2\times 2}$, then either $0 TITLE: Is primality essential in Varshamov's bound? QUESTION [15 upvotes]: Let $v_q(n,r)=\sum_{i=0}^r \binom{n}i (q-1)^i$ denote a number of points in a ball of radius $r$ in the Hamming metric on the cube $\Sigma^n$, where $|\Sigma|=q$. What is the maximal number of points in $\Sigma^n$ with mutual distances at least $d$ (later: $d$-distant point sets)? Gilbert's bound says that we may find $M$ such points if $(M-1)\cdot v_q(n,d-1)< q^n$: add points one by one, while we have less than $M$ points, the $(d-1)$-balls centered at them do not cover $\Sigma^n$ and we may add another point. If $q$ is a power of prime, Varshamov's bound improves the previous result by claiming that we may find $q^k$ $d$-distant points whenever $q^{k-1}v_q(n,d-1) TITLE: Oesterlé's unpublished bound on Uniform Boundedness QUESTION [11 upvotes]: The bound in Merel's solution to the Uniform Boundedness conjecture is not explicit, as it relies on Falting's work on the Mordell conjecture. I think this still is the case. But there are known explicit bounds for the largest prime divisor. The best one seems to be $(1+3^{d/2})^2$, where $d$ is the degree of the number field, due de Oesterlé (1994!). But as far as I known, the proof of such a bound remains unpublished. Quoting a relatively recent survey on the topic for context ("Torsion subgroups of elliptic curves over number fields" by Andrew Sutherland, 2012): Oesterlé's bound plays a critical role in several of the results discussed here; it is quite unfortunate that no proof has been published. The work of Parent in implies that Oesterlé's bound holds for all suffciently large d, but we are typically interested in particular small values of d (e.g. d = 5; 6; 7). There is current work in progress aimed at addressing this gap in the literature [6]. [6] Maarten Derickx, e-mail regarding innitely many rational points on a modular curve of degree Q-gonality 2, December 2012. My question is, has this gap been filled? Alternatively, has Oesterlé's work been superseded? REPLY [11 votes]: Yes, this is published as appendix A to chapter 3 in Derickx' PhD thesis available here: https://openaccess.leidenuniv.nl/handle/1887/43186 . The thesis contains, of course, many more interesting results.<|endoftext|> TITLE: Steenrod powers of Pontryagin classes QUESTION [12 upvotes]: It is well known that the Stiefel–Whitney classes $w_i$ of a smooth manifold are generated, over the Steenrod algebra, by those of the form $w_{2^{i}}$. I wonder if it the same statement is known/true in the case of odd primes. More precisely Question: Given a smooth manifold $M$, is it true that the Pontrjagin classes $p_i\in H^*(M,\mathbb{Z}/q\mathbb{Z})$ are generated, over the algebra of Steenrod powers, by those of the form $p_{q^i}$? In the case of the Siefel-Whitney classes, that fact can be obtained using the formula $$ Sq^i(w_j)=\sum_{t=0}^i {j+t-i-1 \choose t} w_{i-t}w_{j+t}. $$ As far as I understand, such a nice formula does not exists in the case of odd primes. Nevertheless, my question is much weaker, and I have the feeling it should be known, at least in the case $q=3$ (which I am mainly interested in). EDIT: As Oscar points out, the answer to this question is NO for $p\geq 5$. As far as I see it, though, it is not trivially false for $p=3$. REPLY [8 votes]: Let's first consider Chern classes mod $p$ for an odd prime $p$, or if you like, $H^*(BU; \mathbb{Z}/p)$. Theorem 4 of the paper mod p Wu formulas for the Steenrod algebra and the Dyer-Lashof algebra by Brian Shay gives a formula for the Steenrod powers of Chern classes: $$ P^{j-k}(c_j) = \sum (-1)^{\sum r_\gamma} \prod \frac{(-1)^{l_\alpha}}{l_\alpha} \left\{\frac{(j_\alpha - k_\alpha)p + k_\alpha}{j_\alpha} {j_\alpha \choose k_\alpha} \frac{1}{t_\alpha} {t_\alpha \choose t_{\alpha 1} \cdots t_{\alpha ((j_\alpha - k_\alpha)p + k_\alpha)}} \right\}^{l_\alpha} c_1^{r_1} c_2^{r_2} \cdots c_{(j - k)p + k}^{r_{(j - k)p + k}} $$ for "admissible summands": $\sum_\alpha k_\alpha l_\alpha = k$, $\sum_\alpha j_\alpha l_\alpha = j$, $\sum_\gamma \gamma t_{\alpha \gamma} = (j_\alpha - k_\alpha)p + k_\alpha$, $\sum_\alpha l_\alpha t_{\alpha \gamma} = r_\gamma$, $\sum_\gamma t_{\alpha\gamma} = t_\alpha$, $\sum_\gamma r_\gamma < p$, $j_\alpha > k_\alpha$ with at most one exception, for which $t_\alpha = l_\alpha = 1$, $k_\alpha > 0$ unless $k = 0$. This is indeed not the nicest formula. However, we're only really interested in the coefficient of $c_{(j - k)p + k}$, and the only way to achieve that $r_\gamma = 1$ for $\gamma = (j - k)p + k$ and $r_\gamma = 0$ for all other values of $\gamma$ is to take $l_\alpha = 1$ for a single index $\alpha$, for which $k_\alpha = k$, $j_\alpha = j$, $t_{\alpha\gamma} = 1$ for $\gamma = (j - k)p + k$ and $t_{\alpha\gamma} = 0$ for other $\gamma$. The resulting coefficient is therefore $$ \frac{(j-k)p + k}{k} {j \choose k} . $$ Rewriting this cosmetically by setting $j = n - (p-1)m$ and $k = n - pm$, the coefficient of $c_n$ in the expression for $P^m(c_{n-(p-1)m})$ is $$ \frac{n}{n - (p-1)m} {n - (p-1)m \choose m} = \frac{n}{m} { n - (p-1)m - 1 \choose m-1} . $$ Specialising to $m = 1$, the $c_n$-coefficient in $P^1(c_{n - p + 1})$ is simply $n$. (Sanity check: this is not too hard to check by hand, see Equation 12.3 in Groupes de Lie et Puissances Reduites de Steenrod by Borel and Serre.) Hence, for any $n > p$ that is not divisible by $p$, we can express $c_n \textrm{ mod } p$ in terms of $P^1(c_{n-p+1})$ and products of Chern classes in lower degrees. More generally, for any $s \geq p$ and $r \geq 0$, the coefficient of $c_{p^r s}$ in the expression for $P^{p^r}(c_{p^r(s - p + 1)})$ is $$s \, {p^r(s-p+1) - 1 \choose p^r - 1} . $$ The binomial factor is never divisible by $p$, so this coefficient is non-zero mod $p$ if and only if $s$ is coprime to $p$. Thus $c_{p^r s}$ can be expressed in terms of products and Steenrod powers of Chern classes of lower degree whenever $s \geq p$ is coprime to $p$. The conclusion is that $H^*(BU, \mathbb{Z}/p)$ is generated over the Steenrod algebra by Chern classes in degree $p^r s$ for $r \geq 0$ and $s = 1, 2, \ldots p-1$, and we can also see that the classes in those degrees cannot be removed from the generating set. Combining this with the definition of Pontrjagin classes, we see that the Pontragin classes mod $p$ are generated over the Steenrod algebra by Pontrjagin classes in degree $p^r s$ for $r \geq 0$ and $s = 1, \ldots, \frac{p-1}{2}$.<|endoftext|> TITLE: Motivations for the study of dual connections QUESTION [7 upvotes]: I am intrigued by the notion of dual connections: two affine connections $\nabla$ and $\nabla^*$ are called dual if they satisfy $$X(g(Y,Z))=g(\nabla_XY,Z)+g(Y,\nabla^*_XZ)$$ for a given (pseudo)-riemannian metric $g$. What is the motivation and the deep results behind this notion? What are the main fields of application: information geometry, riemannian foliations, webs ...? I am interested in any nice reference or survey paper (especially any information geometry paper written by a 'true' mathematician). REPLY [4 votes]: One motivation for studying dual (also called conjugate) connections in this sense comes from the study of the equiaffine geometry of a hypersurface. Let $i:\Sigma \to \mathbb{A}$ be a nondegenerate cooriented codimension one immersion in flat $(n+1)$-dimensional affine space $\mathbb{A}$. That the immersion be nondegenerate means that its second fundamental form (a normal bundle valued symmetric two tensor) is nondegenerate. That it be cooriented means that its normal bundle is orientable. The coorientation and the second fundamental form determine a (pseudo-Riemannian) conformal structure $[g]$ on $\Sigma$. A choice of parallel volume form $\Psi$ on $\mathbb{A}$ determines a transverse vector field known as the equiaffine normal distinguished by the following requirements. First, such a transversal $N$ determines two volume densities on $\Sigma$, one, $|\iota(N)\Psi|$ via interior multiplication with the fixed ambient volume form, and the other the volume element of the representative, $g$, of the second fundamental form corresponding to the transversal $N$, viewed as a pseudo-Riemannian metric. These are required to agree. The transversal also determines a torsion-free affine connection $\nabla$ on $\Sigma$, and this connection is required to preserve the volume density. The oriented projectivization of the vector space $\mathbb{A}^{\ast}$ dual to $\mathbb{A}$ carries a flat projective structure. The conormal Gauss map associates to each $p \in \Sigma$ the parallel translate to the origin of the ray in $T_{i(p)}\mathbb{A}$ annihilating $T_{i(p)}i(\Sigma)$. The pullback of the flat projective structure on the oriented projectivization of $\mathbb{A}$ via the conormal Gauss map induces on $\Sigma$ a flat projective structure. The equiaffine normal determines a unique immersion from $\Sigma$ to $\mathbb{A}^{\ast}$ covering the conormal Gauss map; the pullback via this map of the flat affine connection on $\mathbb{A}^{\ast}$ dual to that on $\mathbb{A}$ is a representative $\bar{\nabla}$ of the induced flat projective structure, and it can be checked that $\bar{\nabla}$ is conjugate (dual) to the connection $\nabla$ decribed in the previous paragraph with respect to the metric $g$. In general the connections $\nabla$ and $\bar{\nabla}$ are different. Precisely, they coincide if and only if both equal the Levi-Civita connection of $g$, in which case the image $i(\Sigma)$ is necessarily an open subset of a hyperquadric, by a theorem of Maschke-Pick-Berwald. On the other hand, it is not true that an arbitrary pair of conjugate connections can be obtained in this way. In the pair of conjugate connections arising on a nondegenerate cooriented hypersurface immersion, at least one of the pair is necessarily projectively flat. The preceding can mostly be found in some form in the textbook Affine Differential Geometry by Nomizu and Sasaki. See also section $6$ of the Survey on affine spheres by J. Loftin.<|endoftext|> TITLE: Mathematical induction vis-a-vis primes QUESTION [5 upvotes]: One of the most used proof-techniques is mathematical induction, and one of the oldest subjects is the study of prime numbers. Thanks to Euclid, we can consider the primes as a infinite monotone sequence $2=p_13n$. By the way, Erdös's proof of Bertrand's postulate is not by induction (it depends on some results which can be proven via mathematical induction, but that's a different thing): what Erdös actually does in his proof is compare lower bounds for the central binomial coefficients $\binom{2n}{n}$ with some upper ones which he obtains by means of Legendre's formula, the Erdös-Kalmár inequality, and the assumption that there are no primes in $(n,2n]$.<|endoftext|> TITLE: Are there topological versions of the idea of divisor? QUESTION [34 upvotes]: I am trying to extract a particular, more lightweight and more focussed at the same time, case of my recent question Which of the physics dualities are closest in essence to the Spanier-Whitehead duality (with a subquestion)? From time to time I keep becoming fascinated anew by one of the deepest notions in algebraic geometry - divisors. Specifically, I would like to understand better the interplay of analysis and elementary arithmetic of integers that occurs there. At the first sight, forming the group of divisors is a purely "topological" move: you take formal linear combinations of codimension one subvarieties with integer coefficients. Topologists do such things all the time, and with coefficients in arbitrary abelian groups. However as soon as you add the notion of principal divisor and linear equivalence, two things happen: first, integers become absolutely distinguished among the possible coefficient systems, since the coefficient starts to mean something very analytic and non-topological - order of vanishing/infinity of a function along the subvariety. Second, again from the purely topological point of view, some kind of duality becomes apparent, since involvement of functions on the variety suggests something cohomological, as opposed to homology presumed by considering subvarieties as cycles. If a topologist is given an $n$-dimensional manifold, then something similar can be done relating $(n-1)$-cycles and $1$-cocycles. But the divisor class group is even more interestingly behaved in the nonsmooth case, and here I am not sure what a topologist would relate this to. My question then is whether there exists a purely topological version of this subtle mixture of homology and cohomology. Maybe some version of Spanier-Whitehead-like duality or something like that, but the main point is that there must be a single group incorporating homology and cohomology simultaneously for non-manifolds, looking as basic and fundamental as the divisor class group, and the coefficients (in the generalized version, the spectrum) chosen being distinguished among all other possible choices of coefficients. Turning the same question backwards - is there analog, in algebraic geometry, of taking coefficients other than $\mathbb Z$ for divisors? I am aware of the notion of $\mathbb Q$-divisor but more generally I mean, say, divisors with coefficients in the field of coefficients for varieties defined over that field, or, say, something like $\ell$-adic coefficients, or something similar. Does this have sense in the motivic context, for example? Also, from topological viewpoint, there must be higher versions of that, relating codimension $k$ cycles with zeros/poles of $(k-1)$-forms or some related gadgets (norm residue symbols? higher dimensional local fields? or what?) REPLY [18 votes]: Disclaimer. I am no expert at all in algebraic geometry. Therefore much of the following will be oversimplified or maybe even simply wrong. You are still invited to improve it. EDIT. There is a paper by Totaro where he shows that the cycle class map from the Chow group to singular cohomology factors as $$CH^*(X)\longrightarrow MU^*(X)\otimes_{MU^*}\mathbb Z\longrightarrow H^*(X,\mathbb Z)\;.$$ Here $MU^*$ denotes complex cobordism, as explained below. The fact that Totaro does not construct a map through $MU^*(X)$ seems to indicate that in general, equivalence of cycles can be coarser than complex cobordism. I have not checked the implications for divisors, that is, for $CH^1(X)$. Let $L\to V$ be a line bundle over a complex variety, let $s$ be a meromorphic section of $L$, and let $D$ be the associated divisor. For simplicity, we assume that $s$ is an algebraic section and that $s$ meets $0$ transversely, so all zeros have multiplicity one. Then $D$ `is' the subvariety $s^{-1}(0)$. If $V$ was smooth, then $D$ is a subvariety of complex codimension $1$, which we regard as a cobordism $2$-cocycle. Its normal bundle is a complex line bundle. A linear equivalence between two divisors $D_0$, $D_1$ is given by a meromorphic function $f$ such that $D_a=f^{-1}(a)$ for $a=0$, $1$. Consider the graph $\Gamma$ of $f$ in $V\times\mathbb P^1$. If we are still over $\mathbb C$ and everything is sufficiently regular, for a generic real smooth curve $c\colon[0,1]\to\mathbb P^1$ from $0$ to $1$ in $\mathbb P^1$ (now with analytic topology), the (transverse) intersection $W$ of $\Gamma$ and $V\times\operatorname{im}(c)\cong V\times[0,1]$ is then a cobordism between $D_0\times\{0\}$ and $D_1\times\{1\}$. The normal bundle of $W$ in $V\times[0,1]$ naturally carries the structure of a topological complex line bundle, which restricts to the normal bundles of $D_0$, $D_1$ in $V$. To see that we get a map from the divisor class group to complex cobordism, we recall cocycles and relations. A $k$-cocycle in a manifold $M$ is a submanifold $D\subset M\times\mathbb R^\ell$ of real codimension $(k+\ell)$ together with a complex structure on its normal bundle. If one projects down to $M$, the image may become singular. One can stabilise in $\ell$ by taking $D\times\{c\}\subset M\times\mathbb R^{\ell+m}$. A cobordism between two $K$-cocycles, represented by submanifolds $D_0$, $D_1\subset M\times\mathbb R^\ell$ is a submanifold $W\subset M\times\mathbb R^\ell\times[0,1]$ such that $\partial W=D_0\times\{0\}\sqcup D_1\times\{1\}$, together with a complex structure on the normal bundle that restricts to the given complex structures of the normal bundles of $D_0$, $D_1\subset M\times\mathbb R^\ell$. This shows that sufficiently smooth divisors give rise to cocycles, and linear equivalence implies cobordism. It seems that this construction extends to Chow groups, and I am sure it is described somewhere with greater care. It is not entirely clear to me though how to deal with singular subvarieties. To push the analogy further, to each algebraic line bundle one associated a divisor class $[D]$. This can be regarded as a universal first Chern class, with values in the Chow ring. On the topological side, there is a first Chern class with values in the complex cobordism ring, and it corresponds to $[D]$ under the map above. By Quillen, this class is universal for complex oriented multiplicative cohomology theories. Conversely, in algebraic geometry, a divisor defines a line bundle. If we can choose $\ell=0$ above, then the Pontryagin-Thom construction identifies a complex $2$-cocycle in $M$ with a homotopy class of maps $M\to\mathbb P^\infty$, and $\mathbb P^\infty$ is also the classifying space for (topological) complex line bundles. There is a direct construction: the normal bundle of $D$ comes with a complex structure. Pull it back to a tubular neighbourhood $\pi\colon U\to D$ of $D$, then $\pi^*\nu\to U$ has a tautological section, which can be used to glue it to trivial bundle over $M\setminus D$, giving a complex line bundle $L\to M$. This construction is unstable however, that is, it does not work for higher values of $\ell$.<|endoftext|> TITLE: How is a Stack the generalisation of a sheaf from a 2-category point of view? QUESTION [12 upvotes]: A stack is usually given in terms of: -A category $F$ fibered over another $C$ such that the functor $Hom(x,y), x,y \in F(\alpha), \alpha \in C$ is a sheaf -The descent data are effective. There is an equivalent definition, using the Grothendieck construction, which is a correspondence between fibered categories and pseudofunctors in $Cat$. Given this correspondence a stack becomes a (contravariant) pseudofunctor such that descent is effective. Now, with this definition in mind it seems obvious that a stack is a generalised sheaf. But here is my doubt: When defining a sheaf $T$ over a space $X$, is not unusual to see the following diagram (naively) $ T(X) \rightarrow T(U) \stackrel{\longrightarrow}{\longrightarrow} T(U \cap U)$ While, saying that descent data are effective is like asking a similar diagram, but there is the difference that here descent satisfies cocycle condition, something which is naively in $U \cap U \cap U$. Recalling that (for example) locally constant sheaves automatically satisfy the cocycle condition because of their correspondent covering spaces do. So my question is, at the light of my interpretation, the cocycle condition seems to me the only obstruction to the fact that a stack is a generalised sheaf. Am i wrong? Every sheaf trivially satisfies cocycle condition? REPLY [12 votes]: Let us start with what we know about sheaves, i.e. the "1-level". A sheaf on a (Grothendieck) site $\mathcal{C}$ is a contravariant functor $F : \mathcal{C}^\text{op} \to \textbf{Set}$ such that for any cover $\{ X\ \to Y\}$ , the diagram $$F(Y) \to F(X) \stackrel{\longrightarrow}{\longrightarrow} F (X \times_Y X)$$ is an equalizer in the category of sets. For the sake of exposition, I will only consider the case where the cover consists of a single element. Now suppose we want to move to the "2-level" and talk about stacks. Then we need to throw in the cocycle condition, so we can naively define a stack $F$ to be a contravariant functor $\mathcal{C}^{\text{op}} \to \textbf{Set}$ such that $$F(Y) \to F(X) \stackrel{\longrightarrow}{\longrightarrow} F (X \times_Y X)\stackrel{\stackrel{\longrightarrow}{\longrightarrow}}{\longrightarrow} F(X \times_Y X \times_Y X)$$ is an equalizer in the category of sets. The problem now is the following: If $F$ is some kind of moduli stack, e.g. $F = \mathcal{M}_{1,1}$ then to make $F$ set valued often involves quotiening out isomorphisms. However, this is very bad as the presence of quadratic twists of elliptic curves means $F$ is not injective. So now what do we do? Well, we can try to not quotient out isomorphisms, and think of $F$ as a groupoid valued functor. But now we have a new problem: If $F$ is valued in groupoids, what does it mean to say that $F(Y) \to F(X)$ is "injective"? The solution is the following. Let's go back to situation where $F$ is a plain old sheaf, and let us think of the set $F(X)$ as a category where the only arrow $x \to x'$ is when $x = x'$, otherwise $\operatorname{Hom}(x,x') = \emptyset$. Then now to say that $F(Y) \to F(X)$ is injective is exactly equivalent to the statement that the functor $F(Y) \to F(X)$ is fully faithful. The upshot is that to make the right definition (of a prestack), we now know that: $F(X)$ should be a groupoid. $F(Y) \to F(X)$ should be fully faithful. We're not there yet, and we need one last modification (at least for $F$ to be a prestack. We need to replace $F(X)$ with $F(X \to Y)$, namely the category of covering data. The objects of this category are pairs $(y, \phi)$ where $y \in F(Y)$ and $\phi : \text{pr}_1^\ast y \to \text{pr}_2^\ast y$ is an isomorphism. A morphism of covering data $ (y, \phi) \to (y', \phi')$ is a map $f : y\to y'$ such that an appropriate diagram commutes (see chapter 8 of the book "Neron Models" by BLR for the exact definition). We can now define: A groupoid valued functor $F$ (or pseudofunctor in Vistoli's language) is a prestack if the natural pullback functor $F(Y) \to F(X \to Y)$ is fully faithful. If you unravel what the morphisms are in the category $F(X \to Y)$, you will see this is exactly the condition that the set-valued functor $\underline{\operatorname{Isom}}$ is a plain old sheaf! So we can finally get to your question. In my view, a stack is a generalized sheaf if you replace sets with groupoids, and if you introduce the category of covering data.<|endoftext|> TITLE: Group of order $5p^aq^b$ QUESTION [6 upvotes]: In Lectures by Dan Bump on Modular representation theory, Theorem 13.14 states that whenever $G$ is a non-abelian simple group of order $|G|=p^aq^br$ for distinct primes $p$,$q$, and $r$, every $r$-Sylow $R$ is equal to its own centralizer. He states Corollary 13.15 (with no proof), namely when $r=5$, it follows that $G$ is isomorphic to $A_5$, $A_6$, or $\mathrm{SO}_5(\mathbb{F}_3)$. How can one deduce the corollary? Is there a paper/book that I can refer to? REPLY [9 votes]: Note that Bump has his own notes in his webpage: http://sporadic.stanford.edu/modrep While the contents are in different order with that of Feng's notes, he mentioned in Section~6.3 that this is a result of Brauer. I found that it is in this paper of Brauer: http://www.ams.org/journals/bull/1968-74-05/S0002-9904-1968-12073-7/home.html<|endoftext|> TITLE: Simple identity on Lie algebras in a note of Koszul QUESTION [8 upvotes]: In a 1947 Comptes Rendus note (T224, p. 448), Koszul makes the following claim (paraphrased, hopefully correctly), which seems like it should have a simple proof I am missing. Given a compact, connected Lie group, choose a basis $\alpha, \theta^1,\ldots,\theta^n$ of the dual Lie algebra $\mathfrak g^\vee$ which is orthonormal with respect to an invariant inner product, the closed, invariant Cartan form $$\omega = \frac 1 3 \big( \alpha \wedge d\alpha + \sum \theta^i \wedge d\theta^i\big) \in \bigwedge{}^3 \mathfrak g^\vee.$$ decomposes as $\omega = \omega_2 + \omega_3$, where $$ \omega_2 \in \mathbb R\alpha \otimes \Lambda^2[\vec \theta], \qquad \omega_3 \in \Lambda^3[\vec \theta]. $$ How do we know, as Koszul claims, that $$d(\alpha \wedge d\alpha) = 3 d\omega_2?$$ This would be clear if we knew $d\omega_3$ were $\sum\theta^i \wedge d\theta^i$, but this does not seem immediately obvious as the $d\theta^i$ may nontrivially involve $\alpha$. What am I missing? REPLY [3 votes]: Are you sure about the factor of $3$ in Koszul's formula? The computation below does not give such a factor, and it gives a stronger result. Let $X_i$ be any basis of the left-invariant vector fields (where the indices run from $0$ to $n$), with $[X_i,X_j] = c^k_{ij}X_k$, where, of course, $c^k_{ij}=-c^k_{ji}$ (and where here, as below, we use the summation convention that we sum over repeated indices in any given term if no summation is explicitly given). Let $\langle,\rangle$ be an $\mathrm{ad}$-invariant inner product, with $\langle X_k,X_l\rangle = g_{kl}$. Then $\mathrm{ad}$-invariance is just the equation $$ g_{lk}c_{ij}^k + g_{jk}c_{il}^k = 0. $$ If we assume that the $X_i$ are $\langle,\rangle$-orthonormal, then $g_{kl} = \delta_{kl}$, so $c^k_{ij} = - c^j_{ik}$, i.e., $c^i_{jk}$ is skew-symmetric in all its indices. For simplicity, let's agree to write $c^i_{jk} = c_{ijk}$, when the $X_i$ are orthonormal with respect to an $\mathrm{ad}$-invariant $\langle,\rangle$, which I'll assume from now on. Now consider the dual $1$-forms $\theta_i$ to the vector fields $X_i$. By the formula relating exterior derivative and Lie bracket, they satisfy $$ \mathrm{d}\theta_i = -\tfrac12\,c_{ijk}\,\theta_j\wedge\theta_k = -\sum_{j TITLE: Laplace-Beltrami and the isometry group QUESTION [5 upvotes]: H$\vphantom{a}$i. Consider the Laplacian on $\mathbb R^n$, $$ \Delta=\partial_i^2 $$ It is easy to prove that the most general differential operator that commutes with rotations and translations is of the form $$ \sum_k a_k\Delta^k $$ for some constants $a_k$. Is there a similar result for non-trivial manifolds? In other words, consider the Laplacian on a manifold $\mathcal M$, $$ \Delta=\nabla_i\nabla^i $$ Under what conditions is $\Delta$ the most general differential operator that commutes with all Killing vectors of $\mathcal M$? It is clear that $[\Delta,\mathcal L_\xi]=0$ for all $\xi\in\mathrm{Iso}(\mathcal M)$, but can we rule out the existence of any other quadratic differential operator that commutes with $\xi$? My gut tells me that this should be true at least for maximally symmetric spaces. Is this correct? A good reference will be very appreciated. REPLY [6 votes]: For the maximally symmetric case with the second order operator, the proof is very simple. A second order differential operator acting on scalars can be written in the form $$ b^{ij} \nabla^2_{ij} + c^i \nabla_i + e $$ Consider first the action of all isometries that fixes a point $p\in M$. You must have that $b^{ij}(p)$ is a symmetric two tensor that is fixed by the action of the entire special orthogonal group, and so must be proportional to the inverse metric. Now as the principal part is the Laplacian and is invariant under the isometries, you can then conclude similarly that $c^i$ is a vector that is invariant under the action of the entire special orthogonal group, and hence vanishes. This implies that your differential operator is $$ \lambda \triangle_g + e $$ for scalar functions $\lambda$ and $e$. But then translation symmetry tells you immediately that both $\lambda$ and $e$ are constant. The maximal symmetry is used crucially in the argument: for example, consider the standard torus $\mathbb{T}^2$; it is homogeneous but does not admit any rotation symmetry. In standard coordinates any second order differential operator of the form $a_1 \partial^2_{11} + b \partial^2_{12} + a_2 \partial^2_{22}$ commute with all the Killing vector fields (being spanned by $\partial_1$ and $\partial_2$). Notice that in addition to the "natural" differential operators mentioned in the link in Phillip Andreae's comments, there are also other "less natural" differential operators that can commute with the Killing vector fields. For example, let $(M,g)$ be any Riemannian manifold with only one Killing vector field $X$ (isometry group being $\mathbb{R}$ or $U(1)$; think a generic surface of revolution). Then quite clearly the second order operator $\mathcal{L}_X^2: f \mapsto X(X(f))$ commutes with the (one and only) Killing vector field, yet is not the Laplacian.<|endoftext|> TITLE: Nijmegen 1978 $p$-adic analysis proceedings QUESTION [5 upvotes]: Anyone knows if there is a chance of getting a copy of the following: Proceedings of the Conference on p-adic Analysis. Held in Nijmegen, January 16–20, 1978. Report, 7806. Katholieke Universiteit, Mathematisch Instituut, Nijmegen, 1978. ii+224 pp. The link to MR is http://www.ams.org/mathscinet-getitem?mr=522116 This book has many interesting articles, in particular, "Duals" of Yvette Amice. I have searched on the internet without results, and I have asked some friends in universities outside Chile if they could get a copy, but it seems very hard. There is a link to Google books, but it has only the "limted search" option. Note: A related question was already asked (in 2011) with no satisfactory answer: Paper by Y. Amice REPLY [8 votes]: In the Nijmegen University Repository I only found pages 193-204, Non-archimedean differentiation. Since I presume a copy for private use is OK, I have scanned Amice's and Morita's contributions, you can find them here: Duals, Yvette Amice Krasner's analytic functions and rigid analytic spaces, Yasuo Morita update January 2019: upon request, I have scanned the entire 224 page volume (58 MB pdf file), available at: Proceedings of the Conference on p-adic analysis (1978). Note in particular the list of open problems at the end of the proceedings. How many of these are still open after 40 years?<|endoftext|> TITLE: Hilbert's alleged proof of the Continuum Hypothesis in "On the Infinite" QUESTION [44 upvotes]: As is known, Hilbert attempted a proof sketch of the Continuum Hypothesis in the latter part of his paper, "On the Infinite". It is also known that it is false. Has there ever been a published analysis of this alleged proof showing where the error(s) lie(s)? In particular, does his 'proof' implicitly assume a form of $V=L$? Thanks in advance for any help given. REPLY [45 votes]: The article "Hilbert and Set Theory" by Dreben and Kanamori devotes Section 7 to this argument and an analysis of its flaws. Dreben and Kanamori use the translation provided by van Heijenoort, so that helps things significantly if you have access to that book. I unfortunately don't at the moment; I might add to this answer once I find my copy, if I find anything else to say. I must also make one terminological caveat: Hilbert, and later Godel, used the phrase "recursive function" in a way very different from the modern sense (= computable); they considered a truly huge range of recursions, to the point that plausibly every function of natural numbers could be recursive in this sense. According to Dreben and Kanamori, Hilbert's goal was to use proof theory to show that from any (formal) disproof of the continuum hypothesis, he could produce a proof of the continuum hypothesis. As Dreben and Kanamori observe, even if he could accomplish this he wouldn't have proved CH, merely the consistency of CH, but even this didn't work. They isolate two crucial lemmas in Hilbert's claimed proof. The second lemma was a technical one, comparing forms of definition-by-recursion, and Dreben and Kanamori state that "there is a sense in which Hilbert's Lemma II is correct." The first lemma, however, is the crucial one, where Hilbert claims to be able to transform any disproof of CH into one which only invokes "$\epsilon$-free" definitions of functions - in more detail, "functions defined, without the use of the symbol $\epsilon$, by means merely of ordinary and transfinite recursion, so that the transfinite appears only in the guise of the universal quantifier." From Lemma I Hilbert concludes that "in order to prove the continuum theorem, it is essential to correlate those definitions of number-theoretic functions that are free from the symbol $\epsilon$ one-to-one with [the countable ordinals]." This represented a huge step forwards in recursion theory, at least in spirit: as Dreben and Kanamori write, "Hilbert was the first to consider number-theoretic functions defined through recursions more general than primitive recursion." However, even assuming he could carry out the substitution claimed by Lemma I, his use of the lemma is flawed: why is it enough to consider only those functions which appear in claimed disproofs of CH? This wouldn't even establish consistency! But this was something Hilbert missed; according to Dreben and Kanamori, Hilbert seems to have believed that there can be no number-theoretic functions unless definable in some formal proof. Of course they go on to say more, but I think the above gives a good introduction to their discussion, which you should read for a full answer to the first part of your question. In answer to the second - whether the argument could be salvaged and applied in, say, $L$ - I think the answer is that the proof that CH holds in $L$ could be viewed as a spiritual cousin of Hilbert's attempt, but Hilbert's ideas probably wouldn't get you all the way; and far more fundamentally, the need for an argument that $L$ satisfies ZFC is something that never occurred to Hilbert in any guise. Hilbert's assumption that "there can be no number-theoretic functions unless definable in some formal proof" can be weakened to "there is a model of ZFC in which there can be no number-theoretic functions unless definable in some formal proof," but of course even that would have to be proved. EDIT: Let me elaborate on my last paragraph, specifically the sub-question: To what extent can Hilbert's sketch be viewed as a precursor of Godel's proof of the consistency of ZFC+CH via $L$? A closely related question is: what did Godel think about the two? At least early on, Godel interpreted Hilbert's program quite generously, explicitly connecting it with his own proof(s) in two lectures (see below). Others did, too: for instance, Bernays, in his 1940 review of Godel's paper, wrote "The whole Godel reasoning may also be considered as a way of modifying the Hilbert project." At the same time, though, this analogy - and its implicit interpretation of Hilbert's program - can be seriously criticized. Indeed, Godel's own opinion seems to have changed over time, as can be seen in his 1965 response to a letter from van Heijenoort quoting Bernays (to be found in the Collected Works); I'll quote from this below. In his introduction to the lectures in the Collected Works, Solovay is harsher: he argues that in fact Godel’s generous interpretation was "not only excessively generous but downright misleading." (Personally I find Solovay’s criticisms very on-point, but I might not go that far.) So this is clearly a bit of a subtle question. In what follows I'll try to dig into Godel's original sympathetic interpretation a bit. Keep in mind that I am wildly unqualified to give an historical interpretation of Hilbert and Godel, but hey, this is the internet, that’s what it’s for. (The relevant parts of the Collected Works are pages 114-155 and 175-189 of vol. 3. The lectures in question are a 1939 lecture in Gottingen and a 1940 lecture at Brown University; note that this was right after his proof, and at the same time as Bernays' comparison quoted above.) Godel opened his 1940 lecture at Brown University by explicitly drawing the analogy between his arguments and Hilbert's: "If I want to sketch a proof for the consistency of Cantor's continuum hypothesis, I have a choice between many possibilities. Just recently I have succeeded in giving the proof a new shape which makes it somewhat similar to Hilbert's program presented in his lecture." During the lecture, he then repeatedly compares his approach to Hilbert's program, e.g. "The difference between this notion of recursiveness and the one that Hilbert seems to have had in mind is chiefly that I allow quantifiers to occur in the definiens." The relevant passages in the 1940 lecture are brief. To get a better sense for Godel’s interpretation (at least stated) of Hilbert's sketch at the time, I think it’s valuable to quote a large section of Godel’s introduction to his 1939 lecture, mostly verbatim (pp. 129-130). And, to get at the criticisms of that interpretation, I’ll bold sections of the quote below that I take particular issue with, and say a bit about these at the end: “As you know, the first to outline a program for a consistency proof of the continuum hypothesis was Hilbert … Let me remind you that, in a very broad outline, Hilbert’s program for proving this proposition consisted in the following: first, a certain class of functions of integers was singled out, namely, those that are defined recursively [recall my comment at the top of my answer re: this word in this context]. About these recursive functions, two lemmas were then to be proved; namely: These recursively defined functions can be put in correspondence with the numbers of the second number class. The other definitions that occur in mathematics, namely, those involving quantification, …, do not lead outside the domain of the recursively definable functions, or at any rate, one can consistently assume that they do not lead outside the domain of the recursively definable functions. [At this point, compare with Dreben/Kanamori's summary of Hilbert's two lemmas.] Thereby the proof of the first lemma, about the number of recursive functions, was to rest on the fact that in the recursive definitions of functions of natural numbers one can avoid variables of types higher than those of the second number class. The proof on which I would like to report here is analogous to this program insofar as there is likewise singled out a certain class of functions, or, what comes to the same thing, of sets, which I call constructible sets. … Now likewise two things are shown about these constructible sets, namely: The cardinality of the constructible sets of natural numbers is at most $\aleph_1$ [(and etc.)] … The methods of definition otherwise applied in mathematics (in particular, the impredicative as well) do not lead outside the domain of the constructible sets. The proof of the first lemma, about the number of constructible sets, here too rests on the avoidability in the definition of constructible sets of variables of types that are too high. As to the second lemma, closer examination reveals that this statement means nothing other than that the constructible sets form a model for set theory … As to Lemma 2 therefore it is a matter of proving that the constructible sets form a model for set theory, and on the basis of Lemma 1, one can show that the generalized continuum hypothesis holds in that model, and thus its consistency is proved by the method of models.” (It's worth going back to the 1940 lecture at this point, to note that - having used the analogy with Hilbert's program to situate his proof in his first lecture - he still felt it worthwhile to present a second proof, even closer in spirit (and it is meaningfully closer - it replaces the model-theoretic approach of 1939 with a more proof-theoretic one). To me this indicates that he was serious in his interpretation of Hilbert (as opposed to, say, simply presenting the analogy to quicken acceptance of his proof, which would in my opinion be a reasonable action given the multiple mathematical and philosophical novelties in the argument).) The analogy Godel draws here is clear, but in my opinion faulty, because of the points bolded: Hilbert, unlike Godel, does not seem to understand the subtleties of using a model to provide a consistency proof. First, Hilbert doesn’t recognize (at least, not explicitly) the importance of checking that the class of functions specified actually satisfies the given set of axioms. He avoids this by implicitly claiming something stronger: that all functions have this property. Godel recognizes the problem here, and the first bolded passage above introduces a crucial phrase (“one can consistently assume that”) which was not in Hilbert’s argument. This is important, because it adds a layer of subtlety to the task - namely, the appropriate construction and verification of a model - which was definitely not in Hilbert. Another side of this coin crops up in Godel’s description of his own proof, in the second and third bolded passages (“the constructible sets form a model for set theory” and “its consistency is proved by the method of models”) which are not present in his description of Hilbert’s. Solovay also points out the huge gulf between Godel’s notion of constructible and Hilbert’s (admittedly somewhat vague) notion of recursive, and argues (and I agree) that this gulf isn’t just a technical point but really fundamental; in particular, although Solovay doesn’t explicitly say this, I think that Hilbert’s notion is too restrictive to convincingly cover all “the other definitions that occur in mathematics.” Certainly to the extent that we care about set theory, it isn’t enough: after all, $L$ forms the smallest inner model of ZFC! So the difference in power between Godel’s constructible and Hilbert’s recursive is really reflective of the fact that Godel was aware of the complexities of actually building a model, in a way which Hilbert wasn’t (or at least, wasn’t explicitly in “uber das unendliche.” And according to Solovay, in his later letter to van Heijenoort (which I don’t have) Godel was himself far less generous. Godel, later on (in his 1965 correspondence with van Heijenoort referred to above - pp. 324-325 of Collected Works vol. 5), seems to agree - he writes: "There is a remote analogy ... There is, however, this great difference that Hilbert considers only strictly constructive definitions and, moreover, transfinite iterations of the defining operations only up to constructive ordinals, while I admit, not only quantifiers in the definitions, but also iterations of the defining operations up to any ordinal number, no matter whether or how it can be defined. ... It was exactly by viewing the situation from this highly transfinite, set-theoretical point of view that in my approach the difficulties were overcome and a relative finitary consistency proof was obtained. Of course there is no place in this approach for anything like Hilbert's Lemma 1." (Recall that this lemma was the one singled out by Dreben and Kanamori as the most problematic.) So on the balance, I think that there is an analogy between Hilbert’s ideas and Godel’s proof, but it is fundamentally shaky and arguably even misleading. (Again, though, I have absolutely no expertise as an historian, so I’m not really qualified to do this kind of analysis.)<|endoftext|> TITLE: Harmonic maps are light QUESTION [6 upvotes]: Assume $f\colon \mathbb{D}\to\mathbb{R}^2$ is a harmonic map and $x\notin f(\partial\mathbb{D})$. Is it true that $f^{-1}\{x\}$ is totally disconnected? I hope that the answer is yes. But actually I need a "yes", with a CAT(0) space instead of $\mathbb{R}^2$. So, I need a generalizable proof. P.S. For CAT(0) target the answer is "no" --- if the target is a cone with large total angle then the tip of the cone might have a tree as an inverse image for harmonic map; this example is constructed in "Harmonic maps between flat surfaces with conical singularities" by Ernst Kuwert. REPLY [6 votes]: As your question operates with $f(\partial D)$, I assume that $f$ is continuous in $\overline{D}$, though you do not mention this explicitly. Then the answer is yes, the zero set of $f$ is discrete (not just totally disconnected). Suppose wlog that $0\not\in f(\partial D)$. Suppose by contradiction that zeros of $f$ have an accumulation point $z^*$ in $D$. Then, according to a theorem of Wilmshurst, there is an open analytic arc $\gamma$ which has $z^*$ in its interior, and such that $f(z)=0$ on $\gamma$. Consider now this whole arc $\gamma$ (the maximal arc on which $f(z)=0$). It cannot reach the boundary $\partial D$ because on $\partial D$ we have $|f(z)|>\delta>0$ by assumption. So $\gamma$ must contain a loop, but then $f\equiv 0$ inside this loop and thus everywhere. A. Wilmshurst, The valence of harmonic polynomials, PAMS 126, 7 (1998) 2077-2081, Theorems 3, 4. I don't know whether this argument generalizes to CAT(0) spaces because I do not know what is a harmonic mapping into a CAT(0) space.<|endoftext|> TITLE: Is the solution of this optimization problem always positive semidefinite? QUESTION [7 upvotes]: We are given a set of unit vectors $U \subset \mathbb{C}^n$ which spans the space $\mathbb{C}^n$. Given another unit vector $x$, consider then the following optimization problem: $$ \sup_H \left\{ x^* H x \;:\; H \text{ is Hermitian},\; 0 \leq u^* H u \leq 1 \;\forall u \in U \right\}. $$ This optimization comes up in a problem that I have been solving numerically, and to my surprise, the optimal solution $H$ seems to always be positive semidefinite (when the supremum is actually achieved). I am trying to understand whether this is a general property of the problem. My question is thus: is the optimal $H$ necessarily positive semidefinite? If not, is there some property of the set $U$ that guarantees that the optimal solution is positive semidefinite? One observation is that the problem is actually unbounded when the vectors in $U$ are mutually orthogonal, so we can assume this not to be the case. However, I do not see if this by itself implies the positive semidefiniteness of the optimal $H$ somehow. I would appreciate any help. REPLY [6 votes]: No, it is not always attained at a positive semidefinite matrix. The simplest example I have been able to find to demonstrate this is as follows: \begin{align*} U = \{ (1,0), (0,1), \tfrac{1}{\sqrt{2}}(1,1), \tfrac{1}{\sqrt{2}}(1,-1) \}, \quad \mathbf{x} = (1,2)/\sqrt{5}. \end{align*} For this optimization problem, the optimal value is $6/5$ (easily found by LP solvers), and here is an example of a Hermitian matrix attaining this value: \begin{align*} H = \begin{bmatrix} 0 & 1/2 \\ 1/2 & 1 \end{bmatrix}. \end{align*} However, there is no PSD matrix attaining the same value $6/5$. I don't see a "nice" way to see this, but if you add the constraint that $H$ is PSD, then it is now a semidefinite program (and thus solvable), and now has an optimal value of $(1+\sqrt{2})^2/5 \approx 1.165\ldots \leq 6/5 = 1.2$. This could be proved rigorously by constructing the dual of the SDP if desired.<|endoftext|> TITLE: $q$-(and other)-analogs for counting index-$n$ subgroups in terms of Homs to $S_n$? QUESTION [20 upvotes]: The following formula of astonishing beauty and power (imho): $$ \sum_{n \ge 0} \frac{| \mathrm{Hom}(G,S_n) | }{n! } z^n = \exp\left( \sum_{n \ge 1} \frac{|\text{Index}~n~\text{subgroups of}~ G|}nz^n \right) $$ can be found in e.g. Qiaochu Yuan blog , here $G$ is a finitely generated group, $| \mathrm{Hom}(G,S_n) | $ is the number of homomorphisms from $G$ to the symmetric group $S_n$ (pay attention that we count homomorphisms themselves - not up to conjugation). $|\mathrm{Index}~n~\text{subgroups of}~ G|$ is the number of index $n$ subgroups in $G$. (See examples below). It seems to me that Tao's post is related. Keeping in mind the heuristics $S_n$ = $\mathrm{GL}$ (field with one element), it seems natural to think in the following direction: Question 1a Is there $q$-analog of this formula of the following spirit: let us substitute $S_n$ in the left hand side by $\mathrm{GL}(n,\mathbf{F}_q)$. So we are interested in: $$ \sum_{n \ge 0} \frac{| \mathrm{Hom}(G,\mathrm{GL}(n,\mathbf{F}_q)) | }{ [n]_q! } z^n = ??? $$ or in the closely related: $$ \sum_{n \ge 0} \frac{| \mathrm{Hom}(G,\mathrm{GL}(n,\mathbf{F}_q)) | }{ |\mathrm{GL}(n,\mathbf{F}_q)| } z^n = ??? $$ where $\mathrm{GL}(n,\mathbf{F}_q)$ is the general linear group over the finite field $\mathbf{F}_q$, the number of elements in it - almost $q$-factorial: $ |\mathrm{GL}(n,\mathbf{F}_q)| = [n]_q! (q-1)^n q^{\frac{n(n-1)}{2}}$. Question 1b At least if $G$ is a finite group, can one say something about the series appearing in the left-hand side? In the original formula we get $\exp(\text{Polynomial})$; what is an appropriate $q$-analog of it? Example 1 Consider $G$ to be trivial group $G= \{\mathrm{Id}\}$ (just identity). The original formula reads: $$ \sum_{n \ge 0} \frac{1}{n! } z^n = \exp( z) $$ The $q$-analog-1 reads: $$ \sum_{n \ge 0} \frac{1}{[n]_q ! } z^n = e_q( z) $$ Ура! We got a bit of luck: $q$-exponential appeared in the right-hand side in complete analogy with original formula. Consideration of $q$-analog-2 also gives something closely related to the $q$-exponential expression. Example 2 Consider $G$ to be $\mathbf{Z}$ (integers - free abelian group with one generator). The original formula reads: $$ \sum_{n \ge 0} \frac{n!}{n! } z^n = exp\left( \sum_{n \ge 1} \frac{z^n}n \right) $$ which is indeed true , since $1/(1-z)= \exp(-\log(1-z))$. The $q$-analog-1 reads: $$ \sum_{n \ge 0} \frac{[n]_q ! (q-1)^n q^{\frac{n(n-1)}{2}} }{[n]_q ! } z^n = ??? $$ It is not clear what should we get at the right-hand side, but at least the left-hand side looks not terrible: $[n]_q!$ cancels and we get expression similar to theta function. If we consider $q$-analog-2 we just get $1/(1-z)$ built-in. Example 3 Consider $G$ to be $\mathbf{Z}\oplus \mathbf{Z}$ (pairs of integers - free abelian group with two generators). The left-hand side will involve the number of pairs of commuting elements in $S_n$ - which is equal to number of conjugacy classes in $S_n$ multiplied by the size of $S_n$, so original formula reads: $$\mathrm{LHS} = \sum_{n \ge 0} \frac{p(n) n!}{n! } z^n = \sum_{n \ge 0} p(n) z^n$$ In the $q$-case, we again see that $[n]_q !$ cancels and we will get the generating function for number of conjugacy classes in $\mathrm{GL}(n,\mathbf{F}_q)$, which has a nice expression but is difficult to to recognize as $q$-exponential of what it should be. Example ... Taking $G=\mathbf{Z}^k$, we will get generating function for number of commuting $k$-tuples in $S_n$ in the original formula; what should be the $q$-analog? Taking $G = \mathbf{Z}/2\mathbf{Z}$ we will get number of involutions; taking $G=\mathbf{Z}/n\mathbf{Z}$, the number of elements of order $n$. Question 2 what are the analogs of the original formula if we substitute $S_n$ by the other Weyl groups? And $q$-analogs are other simple algebraic groups over $\mathbf{F}_q$? Question 3 For compact Lie groups the left-hand side also has an analog - take $\mathrm{U}(n)$ instead of $S_n$ - consider volumes of $\mathrm{U}(n)/G$... Is there nice right-hand side? PS: What is the history of the formula: author? references? Lubotzky seems to attribute it to Müller around the nineties... Is there some intuitive explanation of the formula? How subgroups can be be related to homomorphisms to $S_n$? REPLY [14 votes]: Yes, there's a $q$-analogue. See this paper of Yoshida from 1992, where $\sum_{n \ge 0} \frac{| \mathrm{Hom}(G,\mathrm{GL}(n,\mathbf{F}_q)) | }{ |\mathrm{GL}(n,\mathbf{F}_q)| } z^n$ is expressed in terms of some invariants of the group ring $\mathbb{F}_q G$, at least when $G$ is a finite group (page 27). Yoshida only states the result. A proof appears for instance in a follow up paper by Chigira, Takegahara and Yoshida from 2000. From their result it follows that when $(|G|,q)=1$, the generating function is a finite product of power series of the form $P(z^a,q^b)$ for various $a$ and $b$, where $$P(z,q) = \sum_{n \ge 0} \frac{z^n}{|GL(n,\mathbb{F}_{q})|}.$$ The original formula for $S_n$, in terms of generating functions, is due to Wohlfahert, see this paper from 1977. An earlier equivalent result, stated in a different fashion, is due to Dey (1965). Dress and Müller generalized the generating function result to a more general setting their paper from 1997, see Proposition 1.<|endoftext|> TITLE: Geometric realization of the mapping stack QUESTION [5 upvotes]: Some background and notation Let $Sh_{\infty}(Cartsp)$ be the infinity category of smooth simplicial sheaves on the site of cartesian spaces (convex open subsets of $\mathbb{R}^n$ and smooth maps between them), equipped with the topology of good open covers (contractible finite intersections). This infinity topos is cohesive and in particular the functor ${\rm disc}:\infty-\mathscr{G}{\rm rpd}\to Sh_{\infty}(Cartsp)$ admits an $\infty$ left adjoint $\Pi$ which preserves finite products. This is the functor I am calling geometric realization. Question Let me denote the internal mapping simplicial sheaf, coming from the cartesian closed structure on $Sh_{\infty}(Cartsp)$, by $[M,X]$. Given a smooth compact manifold $M$ and a smooth simplicial sheaf $X$, I would like to compare the geometric realization of $[M,X]$ and the mapping space of the geometric realizations of $M$ and $X$ respectively. Ideally, I would hope for an equivalence $$\Pi[M,X]\simeq Map(\Pi(M),\Pi(X))\;.$$ Is this true? I haven't been able to come up with a counter example and I know this is true if one takes sheaves with values in a stable infinity category (there one can use the fact that arbitrary colimits commute with finite limits along with descent with respect to a finite cover of $M$). Any help here would be greatly appreciated! REPLY [4 votes]: Yes. This question was already asked and answered on the nForum: https://nforum.ncatlab.org/discussion/6816/the-shape-of-function-objects/<|endoftext|> TITLE: Question about Wasserstein metric QUESTION [8 upvotes]: Let $\mu$ and $\nu$ be two probability measures on $\mathbb R^n$ with finite first moment. Denote by $d:=W_1(\mu,\nu)$, where $W_1(\cdot,\cdot)$ stands for the Wasserstein distance of order $1$. My question is the following: Let $X$ be a random variable defined on some probability space (rich enough) with law $\mu$, could we find a measurable function $f:\mathbb R^n\times \mathbb R^n\to\mathbb R^n$ and a random variable $G$ independent of $X$ s.t. $$Y:=f(X,G)~\sim~\nu~~~~~~ \mbox{ and }~~~~~~ \mathbb E[|X-Y|]~\le ~2d~?$$ Thought 1: Let $d_0:=\rho(\mu,\nu)$, where $\rho(\cdot,\cdot)$ denotes the Prokhorov distance. Then we have a measurable function $f_0:\mathbb R^n\times \mathbb R^n\to\mathbb R^n$ and a random variable $G_0$ independent of $X$ s.t. $$Y_0:=f_0(X,G)~\sim~\nu~~~~~~ \mbox{ and }~~~~~~ \mathbb P[\{|X-Y_0|\ge2d_0\}]~\le ~2d_0.$$ The above construction is from the paper On a representation of random variables by Skorokhod, but I can't find this paper. Thought 2: Let $\pi(dx,dy)$ be the optimal transport plan, i.e. $\pi(A\times\mathbb R^n)=\mu(A)$ and $\pi(\mathbb R^n\times A)=\nu(A)$ for all measurable $A\subset\mathbb R^n$. Disintegration w.r.t. the first coordinate $x$, one has $\pi(dx,dy)=\mu(dx)\otimes \lambda_x(dy)$, where $(\lambda_x)_{x\in\mathbb R^n}$ denotes the r.c.p.d. (regular conditional probability distribution). But I've no idea how to recover the function $f$ using $\lambda_x$. Any answer, help or comment is highly appreciated. Thanks a lot! REPLY [2 votes]: $\newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}}$ The desired function $f$ and random variable (r.v.) $G$ can be built recursively, by induction, using the increasing rearrangement/inverse transformation method: If $F$ is any cumulative distribution function (cdf), \begin{equation} F^{-1}(u):=\inf\{x\in\mathbb R\colon F(x)\ge u\} \end{equation} for $u\in(0,1)$, and $U\sim\mathcal U(0,1)$ (a r.v. uniformly distributed on the interval $(0,1)$), then the cdf of the r.v. $F^{-1}(U)$ is $F$. Indeed, for $j=0,\dots,n$, let $\pi_j$ be the push-forward image of the probability measure $\pi$ under the projection of $\R^n\times\R^n$ onto $\R^n\times\R^j$, so that $\pi_j(A\times B_j)=\pi(A\times B_j\times\R^{n-j})$ for $A$ in the Borel sigma-algebra $\B(\R^n)$ and $B_j$ in $\B(\R^j)$; naturally, $\R^0=\{0\}$, and we identify $\R^j\times\R^{n-j}$ with $\R^n$. Similarly, let $\nu_j$ be the push-forward image of the probability measure $\nu$ under the projection of $\R^n$ onto $\R^j$, so that $\nu_j(B_j)=\nu( B_j\times\R^{n-j})$ for $B_j$ in $\B(\R^j)$. Then $\pi_0 =\mu$, $\pi_n=\pi$, $\nu_0$ is the only probability measure on $\B(\R^0)=\B(\{0\})$, and $\nu_n=\nu$. Write $Y=(Y_1,\dots,Y_n)$ and let $Y_{1;j}:=(Y_1,\dots,Y_j)$, with $Y_{1;0}:=0$; the r.v.'s $Y_1,\dots,Y_n$ are to be constructed. Accordingly, for $y=(y_1,\dots,y_n)\in\R^n$ let $y_{1;j}:=(y_1,\dots,y_j)$, with $y_{1;0}:=0$. For $j=0,\dots,n$, let $G_j:=(U_1,\dots,U_j)$, where $U_1,\dots,U_n$ are independent $\mathcal U(0,1)$ r.v.'s. In particular, $G_0=0$. For $j=1,\dots,n$, we are going to construct, by induction, a function $f_j\colon\R^n\times(0,1)^j\to\R^j$ such that for $Y_{1;j}:=f_j(X,G_j)$ the distribution of $(X,Y_{1;j})$ is $\pi_j$ and hence the distribution of $Y_{1;j}$ is $\nu_j$. To complete the basis of induction, let $f_0(x,0):=0$ for all $x\in\R^n$, so that $Y_0=f_0(X,G_0)$. Take now any $j=1,\dots,n$. Let $$\R^n\times\R^{j-1}\times\B(\R)\ni(x,y_{1;j-1},C)\longmapsto \la_{x,y_{1;j-1}}(C)$$ be a regular version of the conditional distribution of $Y_j$ given $(X,Y_{1;j-1})$ assuming that the joint distribution of $(X,Y_{1;j})$ is $\pi_j$, so that $$\pi_j(dx\times dy_{1;j-1}\times dy_j)=\pi_{j-1}(dx\times dy_{1;j-1})\la_{x,y_{1;j-1}}(dy_j).$$ For each $(x,y_{1;j-1})\in\R^n\times\R^{j-1}$, let $F_{x,y_{1;j-1}}$ be the cdf of the probability measure $\la_{x,y_{1;j-1}}$ on $\B(\R)$, and define the function $f_j\colon\R^n\times(0,1)^j\to\R^j$ by the formula \begin{equation} f_j(x,u_{1;j}):=\big(y_{1;j-1},F^{-1}_{x,y_{1;j-1}}(u_j)\big)\quad\text{with}\quad y_{1;j-1}=f_{j-1}(x,u_{1;j-1}) \end{equation} for $(x,u_{1;j})\in\R^n\times(0,1)^j$, where the notation $u_{1;j}$ is of course quite similar to $y_{1;j}$. Let now $Y_{1;j}:=f_j(X,U_{1;j})=f_j(X,G_j)$, which is in agreement with the definition $Y_{1;j-1}:=f_{j-1}(X,U_{1;j-1})=f_{j-1}(X,G_{j-1})$ at the previous step of the induction process. Then the distribution of $(X,Y_{1;j})$ is $\pi_j$ and hence the distribution of $Y_{1;j}$ is $\nu_j$. In particular, the distribution of $(X,Y)=(X,Y_{1;n})$ is $\pi_n=\pi$ and hence the distribution of $Y=Y_{1;n}$ is $\nu$. Moreover, $Y=Y_{1;n}=f_n(X,G_n)$, as desired.<|endoftext|> TITLE: Tutte's conjecture on Petersen graphs QUESTION [16 upvotes]: I'm trying to find out whether Tutte's conjecture (that every snark has the Petersen graph as a minor) has been satisfactorily proved. Wikipedia claims that Robertson, Saunders, Seymour and Thomas announced a proof but that it has not been published. Indeed, I cannot find a published paper which confirms the proof of the theorem. Does anyone know what the status of this conjecture and proof are? Thanks! REPLY [5 votes]: I think this is considered proved at this point. A list of relevant papers can be found at http://people.math.gatech.edu/~thomas/FC/generalize.html. In this paper, the conjecture is reduced to proving it for two classes of graphs- apex and doublecross. The doublecross case has been published. I'm not sure if the apex paper has been published yet. Note that the proof does indeed make use of a computer program that is no less complex than the one Robertson et al. used for their proof of the four-colour theorem. To my knowledge there is no accepted proof of 4CT that isn't computer assisted.<|endoftext|> TITLE: Loss of cuspidality by Langlands tranfer QUESTION [11 upvotes]: Given an $L$-homomorphism of Langlands dual groups $${}^LG \to {}^LG'$$ Langlands functoriality contectures predicts the existence of a tranfer map of automorphic representations $$Aut(G) \to Aut(G')$$ However, nothing in the functoriality results or conjectures seems to be concerned with cuspidality. Let us say I am interested in the set $Cusp(G)$ of cuspidal representations of $G$. The above map gives a a tranfer from it to $Aut(G')$. My question is: what do we know about its image? Is it also cuspidal? Is it endowed at least with some extra properties? Since those questions have in general negative answers I believe, I am more precisely interested in unitary groups: what about $G$ be a quasi-split unitary group in 2 or 3 variables, and $G'=GL_4$ or $GL_6$ on the quadratic extension defining the unitary group (from the base change transfer)? REPLY [15 votes]: You are quite correct that the Langlands transfer map does not preserve cuspidality in general. E.g. if you take a modular form of CM type, coming from a Groessencharacter $\psi$ of some imaginary quadratic field K that doesn't factor through the norm map to $\mathbf{Q}$, then this gives you a cuspidal automorphic representation $\pi$ of $GL_2 / \mathbf{Q}$, but the base-change of $\pi$ to $GL_2 / K$ will be Eisenstein. This gives a flavour of what to expect: VERY roughly, the transfer of a cuspidal automorphic rep $\pi$ of $G$ to $G'$ will be cuspidal unless $\pi$ is itself a Langlands transfer from some other group $H$ via a homomorphism ${}^L H \to {}^L G$ whose image in ${}^L G'$ lands inside a Levi subgroup. This naive picture is, of course, far from being the whole story -- for instance, there are "CAP representations" which are cuspidal despite being lifts from Levi subgroups. A lot of these subtleties were first seen in the case of $G = GSp_4$, $G' = GL_4$; this is where CAP forms were first identified, for instance. There is a nice survey article by Arthur from 2005. At the end of the article, he states a classification of discrete-spectrum automorphic reps of $GSp_4$ into six types A-F, of which type A ("general type") has cuspidal transfer to $GL_4$, and the other five types are all functorial liftings from various subgroups of $GSp_4$, whose transfers to $GL_4$ are easily seen to be Eisenstein. A lot of Arthur's work has been generalised to quasisplit unitary groups by Chung Pang Mok (Mem AMS, 2015) and that might go some way towards answering your more specific questions.<|endoftext|> TITLE: Narayana and Fermat's Factorization Method QUESTION [14 upvotes]: In some notes of mine I have found a comment according to which the Indian mathematician Narayana Pandit (14th century) found the prime factors of $1161$ by writing it in the form $1161 = 35^2-8^2$. Unfortunately I can't find the source of this piece of information. Any help is appreciated. REPLY [2 votes]: Unfortunately, sources on the history of Indian mathematics are somewhat hard to find on the internet (and even in libraries). Fortunately, this particular question can be answered. Three scholars of the history of Indian mathematics taught a course called “Mathematics in India - From Vedic Period to Modern Times”, which is available online: see course outline, syllabus, lecture notes, actual course, YouTube playlist. This particular topic (Nārāyaṇa Paṇḍita's treatment of factorization) is covered in Lecture 27 by M. D. Srinivas, starting at slide 12 or from 15:25 to 20:10 in the video. Apparently this is the first (known) book in Indian mathematics that discusses factorization (as late as c. 1356), but the treatment here is already nontrivial and contains what is known as Fermat's method (17th century). After stating the usual method of checking whether the successive acchedyas $2$, $3$, $5$, $7$ etc divide the number (literally, “acchedhya” means “indivisible” or “irreducible”), Nārāyaṇa says: given a non-square number $N$, writing $N = a^2 + r$, if it so happens that $2a + 1 - r$ is a square (say $b^2$), then our work is done: $N = (a + 1 + b)(a + 1 - b)$. [apada-pradasya rāśeḥ padam āsannaṃ dvi-saṅguṇaṃ saikam / mūlāvaśeṣa-hīnaṃ vargaś cet kṣepakaś ca kriti-siddhau //] Else, if it ($2a + 1 - r$) is not a square, add $(2a + 3)$ and so on, and keep doing this [numbers increasing in arithmetic sequence, i.e. successive odd numbers] until you get a square. That is, if $(2a + 1) + (2a + 3) + \dots + (2a + 2k - 1) - r$ is a square (say $b^2$), then $N = (a + k + b)(a + k - b)$. [vargo na bhavet pūrvāsannapadaṃ dvi-guṇitaṃ tri-saṃyuktam / adyād uttara-vṛddhyā tāvad yavad bhaved vargaḥ //] After having stated this rule, he works out two examples: $N = 1161$ and $N = 1001$ — of course both are easy to factorize directly (using the straightforward method he mentioned earlier), but I imagine he chose these examples because they illustrate the two cases: in the case of $1161$, already $2a + 1 - r$ is a square, and in the case of $1001$, we have to add as many as $14$ terms before we get a square. More references: The (Sanskrit) text of Nārāyaṇa's work (Gaṇita-kaumudī, c. 1356) was first edited and printed by Padmakar Dvivedi, son of Sudhakar Dvivedi in two volumes: 1936 (Volume 1) and 1942 (Volume 2). It took me a while to locate it, but this particular section on factorization (part of vyavahāra (Chapter) 11) is on page 246. An English translation of the work, with notes, was published by Paramanand Singh in successive issues of the journal Gaṇita Bhāratī: 20 (1998), pp. 25–82; 21 (1999), pp. 10–73; 22 (2000), pp. 19–85; 23 (2001), pp. 18–82; 24 (2002), pp. 34–98. Some (more easily available) (brief) references to Nārāyaṇa's Gaṇita-kaumudī (focusing more on combinatorics) can be found in The Art of Computer Programming by D. E. Knuth (Volume 4A, section 7.2.1.7, pp. 499–500, earlier published as Fascicle 4B, draft version online), in the book Combinatorics: Ancient and Modern (OUP 2003), etc.<|endoftext|> TITLE: Reduction of integral for geodesic area to elliptic integrals QUESTION [5 upvotes]: In my paper on geodesics on an ellipsoid, I express the area between a geodesic segment and the equator in terms of an indefinite integral $$\int \frac{t(e'^2) - t(k^2\sin^2\sigma)}{e'^2-k^2\sin^2\sigma} \frac{\sin\sigma}2 \,d\sigma,$$ where $$t(x) = x + \sqrt{x^{-1} + 1}\,\sinh^{-1}\!\sqrt x,$$ $e'$ is the second eccentricity, $k = e'\cos\alpha_0$, and $\alpha_0$ is the azimuth of the geodesic when crossing the equator. For oblate ellipsoids, we have $0 < k \le e'$. In the paper, I evaluate this integral by Taylor expanding the integrand in the limit that $e' \rightarrow 0$. I would like to relax this assumption. I have made some unsystematic (and unsuccessful) stabs at expressing the integral in terms of elliptic integrals. I would appreciate help with expressing the integral in terms of elliptic integrals, pointing me to a systematic procedure for doing this, proving that the integral can't be expressed in terms of elliptic integrals, or expressing the integral in terms of other special functions (especially those which can be numerical evaluated easily). REPLY [2 votes]: I have seen your paper. You have worked out a series with $e'\rightarrow 0$. You can do the same on the other side $e'\rightarrow\infty$. Also for this series the integrals can be evaluated. In this way, you will be able to get a satisfactory numerical evaluation of the integral for a wide range of values of $e'$. That this can be done can be easily seen by noting the asymptotic series $$ t(x)=x+\ln 2+\frac{1}{2}\ln x+\left(\frac{1}{2}\ln 2+\ln x+\frac{1}{4}\right)\frac{1}{x}+\ldots $$ that entails both logarithm and power terms. For example, for your integral you will get $$ \int\frac{t(e'^2)-t(k^2\sin^2\sigma)}{e'^2-k^2\sin^2\sigma}\frac{\sin\sigma}{2}d\sigma= -\frac{1}{2}\cos\sigma+O\left(\frac{1}{e'^2}\right) $$ and you can evaluate the higher order terms by any computer based program you prefer.<|endoftext|> TITLE: Constructing $E_8$ from its branching to $A_8$ QUESTION [14 upvotes]: Background/motivation: One of the usual constructions of [the adjoint representation of] the $E_8$ exceptional Lie group (found, e.g., in J. F. Adams's, "Lectures on Exceptional Lie Groups", esp. chap. 6–7) consists of starting from $\mathit{Spin}(16)$ and taking the direct sum of the latter's adjoint representation and one half-spin representation (it remains, of course, to construct a Lie bracket on this direct sum, but at least three of the four cases to consider are clear). The reason for this approach is that this direct sum is the branching of the adjoint representation of $E_8$ (to be constructed) to its maximal subgroup $D_8 = \mathit{Spin}(16)$. The branching $E_8 \to D_8$ is arguably the most manageable one, so it makes sense to use it to construct $D_8$. However, branching to the $A_8$ subgroup is also intelligible, so I ask: Question: Given that the adjoint representation of $E_8$, restricted to its maximal subgroup $A_8$ (meaning $\mathit{SL}(9)$ or $\mathit{SU}(9)$ according as we are working with complex or real compact Lie groups), decomposes as $\bigwedge^3 V \oplus \bigwedge^3 V^* \oplus W$ where $V$ is the ($9$-dimensional) natural representation of $A_8$, $V^*$ is its dual, and $W$ is the ($80$-dimensional) adjoint representation of $A_8$ (the nontrivial factor in $V\otimes V^*$), is there an explicit description of a Lie bracket on $\bigwedge^3 V \oplus \bigwedge^3 V^* \oplus W$ that constructs $E_8$? Is this description somewhere to be found in the literature? (To be honest, what I am really interested in is how to get an intuitive grasp of this branching. But I presume the easiest way to do that is to use it to construct $E_8$, rather than hope to describe it starting from some other construction of $E_8$. However, any such description counts as an answer to this question.) REPLY [8 votes]: Of course, the original description of this branching is due to Élie Cartan, himself. For example, see Chapitre IX of his 1914 paper Les groupes réels simples, finis et continu (Annales scientifiques de l'É.N.S. 3rd serie, 31 (1914), 263–355). His chapter on $\mathrm{E}_8$ starts on page 328, and it is very clear and straightforward. The formulae he gives, both for the complex and real forms, are exactly what you are looking for. I think it's the best reference for this, myself. It's available for free download from Nundam, but I don't have the link in front of me right now.<|endoftext|> TITLE: strong measurability question QUESTION [9 upvotes]: Let $X$ be a separable Banach space and $\mathcal L$ the collection of bounded linear operators on $X$. The strong operator topology has the sub-basis $\{B_{x,y,\epsilon}\colon x,y\in X,\epsilon>0\}$, where $B_{x,y,\epsilon}=\{T\colon \|Tx-y\|<\epsilon\}$. The Borel $\sigma$-algebra generated by this topology is called the strongly measurable $\sigma$-algebra on $\mathcal L$. Now the question (which has arisen in a study of the multiplicative ergodic theorem on Banach spaces): Let $K=\{T\in\mathcal L\colon \text{ker}(T)\ne\{0\}\}$. Is this set strongly measurable? Thanks for any information. REPLY [6 votes]: This is true at least when $X$ is a separable and reflexive. Take a dense sequence $(x_n)_n$ in the unit sphere of $X$. Then, for any $T \in B(X)$, one has $\ker T \neq 0$ iff $\exists m$ $\forall k$ $\exists n$ such that $\| x_m - x_n \| < 1/2$ and $\| Tx_n \| < 1/k$. Indeed, since the closed unit ball of $X$ is weakly compact and any $T \in B(X)$ is weak-weak continuous, if the latter condition holds, then any weak limit point $x$ of the subsequence $(x_{n(k)})_k$ satisfies $\| x \| \geq \| x_m \| - \| x_m - x \| \geq 1/2$ and $Tx = 0$. This proves the "if" direction. The "only if" direction is easy.<|endoftext|> TITLE: Rational eigenforms QUESTION [12 upvotes]: Probably an easy question: let $f$ be an eigenform in $S_k^{\text{new}}(\Gamma_0(N),\chi)$ assumed to be with rational fourier coefficients. Then $\chi$ is necessarily trivial or quadratic. But more precisely, $\chi$ seems to be unique: if $k$ is even, then $\chi$ must be the trivial character, and if $k$ is odd then $N$ must be of the form $N=-Df^2$ with $D$ a negative discriminant, and then $\chi(n)=(D/n)$. Can someone explain why this is true? More generally, in even weight and nontrivial quadratic character the irreducible Galois orbits all have even dimension (equivalently, the irreducible factors of the characteristic polynomial of Hecke operators all have even degree). Is this true, and why? REPLY [2 votes]: I only answer the first part of the question. As it is well-known, $\chi$ and $k$ have the same parity. In particular, if $\chi$ is trivial then $k$ must be even. Conversely, if $\chi$ is non trivial then $f$ has CM by its own Nebentypus $\chi$ because $f$ is assumed to have rational (hence real) coefficients. Therefore $\chi$ is the quadratic character associated with an imaginary quadratic field $K$ of (negative) discriminant $D$ (see Ribet, Galois representations attached to eigenforms with Nebentypus, Modular Functions of One Variable V, LNM Vol. 601). In particular, we have $\chi(-1)=-1$ and $k$ is odd. Let us now assume that we are in that latter case, i.e. $k$ is odd. By a result of Hecke and Shimura (loc. cit. pp. 34-36), $f$ is the newform attached to a Hecke character $\psi$ of $K$ with conductor, say, $\mathfrak{m}$ and infinite type $k-1$. (That is, if we view $\psi$ as a homomorphism from the group of prime-to-$\mathfrak{m}$ fractional ideals of $K$ into $\mathbf{C}^*$, then $\psi(\alpha \mathcal{O}_K)=\alpha^{k-1}$ for all $\alpha\in K^*$ with $\alpha\equiv 1\pmod{\mathfrak{m}}$; here $\mathcal{O}_K$ denotes the integer ring of $K$.) Moreover, in the notation of the question, we have $N=-\mathcal{N}(\mathfrak{m})D$ where $\mathcal{N}$ denotes the norm map from $K$ to $\mathbf{Q}$. If $\mathfrak{p}$ is a prime of $\mathcal{O}_K$, denote by $e_{\mathfrak{p}}$ the exponent of the conductor $\mathfrak{m}$ of $\psi$ at $\mathfrak{p}$. By Proposition 6.1 and Table 1 of Schütt's paper CM newforms with rational coefficients (Ramanujan Journal 19 (2009), 187-205), for all prime ideals $\mathfrak{p}$, we have $e_{\mathfrak{p}}=e_{\overline{\mathfrak{p}}}$ and $e_{\mathfrak{p}}$ even if $\mathfrak{p}$ ramifies in $K$. (For a more detailed proof of these results, see chapter II of the author's dissertation available on his homepage.) This in turn implies that $\mathcal{N}(\mathfrak{m})=N/(-D)$ is a square, as desired.<|endoftext|> TITLE: References: spectral analysis of the Laplacian operator QUESTION [10 upvotes]: I'm looking for several references on the spectral analysis of the Laplacian operator. It is such a well-known topic, but I'm a bit struggling to locate modern systematic expositions in the literature. I'd appreciate multiple suggestions that explore the topic using different approaches too. I'm particularly interested in the variational characterization of the eigenvalues and eigenfunctions. REPLY [2 votes]: A good recent survey is Geometrical Structure of Laplacian Eigenfunctions by Grebenkov and Nguyen, with lots of nice pictures and over 500 references.<|endoftext|> TITLE: Do complex varieties have a dense open subset with residually finite fundamental group? QUESTION [8 upvotes]: Let $S$ be a smooth connected variety over the complex numbers. The fundamental group might not be residually finite (i.e., the homomorphism $\pi_1(S(\mathbb C)) \to \pi_1^{\mathrm{et}}(S)$ might not be injective). Is there a dense Zariski open subset $U\subset S$ such that $\pi_1(U(\mathbb C)) $ is residually finite? REPLY [11 votes]: (I'm converting my comment to an answer.) In SGA4 exp XI, Artin constructs a nonempty Zariski open $U\subset S$ which admits a sequence $U=U_n \to U_{n-1}\to\ldots $ which are topological fibrations with curves as fibres. Without loss of generality, these fibrations, and the base of the fibrations, can be taken to be non proper. It follows that $\pi_1(U)$ admits a finite filtration by normal subgroups such that the quotients are free and finitely generated (cf. YCor's comment). An easy induction shows that $\pi_1(U)$ is residually finite. Note that the base case of the induction, where $\pi_1(U)$ is free, is discussed here: Why are free groups residually finite?<|endoftext|> TITLE: Pull-back of an irreducible ample divisor via an isogeny of abelian varieties QUESTION [9 upvotes]: In the (wonderful) book by C. Birkenhake and H. Lange Complex Abelian Varieties we can find the following result, see Corollary 4.3.4 page 77. It is stated in any dimension $g \geq 2$, but let us consider only the case of abelian surfaces for the sake of simplicity. Proposition. Let $f \colon X \to Y$ be an isogeny of abelian surfaces and let $D$ be a positive definite and irreducible divisor on $Y$. Then $f^*D$ is also irreducible. The proof starts as follows: Assume the contrary, then $f^*D$ is a sum of effective divisors $D_1+ \cdots + D_n$. But necessarily $D_i \cdot D_j=0$ and $D_i$ is numerically equivalent to $D_j$ for all $i \neq j$, the map $f$ being an étale Galois covering $\ldots$ and then a contradiction is reached by using the Nakai-Moishezon theorem. Now, I do not understand the part $D_i \cdot D_j=0$ for all $i \neq j$. If $D$ is smooth then this is immediate: in fact, an étale cover of a smooth curve must be smooth, in particular its irreducible components do not intersect. However, if $D$ is singular (and in the Proposition there is no smoothness assumption) the same argument do not work. Indeed, it is well-known that there are irreducible, $1$-nodal curves with a connected, étale double covering consisting of two copies of the normalization, see for instance example 10.6 in Chapter III of Hartshorne's Algebraic Geometry. The two components intersect at two points, that are the two nodes of the covering. So my question is Q. Is the result stated in the Proposition above also true when $D$ is singular? If not, what is a counterexample? Or maybe am I missing something? REPLY [9 votes]: I think the Proposition is not true if $D$ is singular. Take a smooth curve $C$ of genus 2, and $X=JC$; embed $C$ in $X$ (say, by choosing a point of $C$). Let $\alpha$ be a point of order 2 in $X$; take for $Y$ the quotient of $X$ by the translation $x\mapsto x+\alpha $, and put $D=f(C)$. Then $f^*D=C+C'$, with $C':=C+\alpha $, and $C\cdot C'=C^2=2$. Here $D$ has one singular point, image of the two points of $C\cap C'$.<|endoftext|> TITLE: Do monotone functions on the interval have an "Alexander duality" property? QUESTION [5 upvotes]: Let $X,Y$ be two copies of the unit interval $[0,1]$. Consider functions $X\rightarrow Y$ and $Y\rightarrow X$ both as subsets of the cartesian product $X\times Y$. (More precisely: identify a function $f:X\rightarrow Y$ with its graph $\{(x,f(x)):x\in X\}$, and likewise a function $g:Y\rightarrow X$ with its graph $\{(g(y),y)\subset X\times Y:y\in Y\}$.) One can convince oneself that: (A) Any monotone nondecreasing function $f:X\rightarrow Y$ and any monotone nondecreasing function $g:Y\rightarrow X$ intersect nontrivially in $X\times Y$. (Proof sketch below in the "appendix".) I am curious about the following "converse" statement: Is the following true? (B) If a subset $S\subset X\times Y$ has the property that it intersects nontrivially every monotone nondecreasing function $g:Y\rightarrow X$, then it contains a monotone nondecreasing function $f:X\rightarrow Y$. If so, does this have a name? Do you know a proof or reference? Remarks I'm calling (B) a "converse" to (A) because the latter is equivalent to the statement "if a subset $S\subset X\times Y$ contains a monotone $f:X\rightarrow Y$, then it meets every monotone $g:Y\rightarrow X$." The motivation comes from combinatorial commutative algebra (!): I got curious about this question after hearing a talk yesterday on letterplace and co-letterplace ideals of posets. Let $P$ be a finite poset, and let $[n]$ be the $n$-chain. Fix a field $k$ and let $R$ be a polynomial ring over $k$ with indeterminates indexed by the elements of the cartesian product $[n]\times P$. The squarefree monomials in $R$ are thus in bijection with the subsets of $[n]\times P$. The letterplace ideal of $P$, denoted $L(n,P)$, is the ideal generated by squarefree monomials corresponding with the graphs of poset maps $[n]\rightarrow P$, and the co-letterplace ideal of $P$, written $L(P,n)$, is the ideal generated by squarefree monomials corresponding with graphs of poset maps $P\rightarrow [n]$. In the talk I learned that it is a theorem that $L(n,P)$ and $L(P,n)$ are Alexander dual to each other, in the sense of Stanley-Reisner theory. This translate into the following pair of converse statements: (A') Every poset map $[n]\rightarrow P$, regarded as a subset of $[n]\times P$, meets every poset map $P\rightarrow [n]$. (B') If a subset $S\subset [n]\times P$ meets every poset map $P\rightarrow [n]$, then it contains a poset map $[n]\rightarrow P$. (And vice versa.) Suppose that $P$ is a chain. Then the true statement (A') is a discrete analogue of the true statement (A). This made me want to know if the continuous analogue of (B') was also true. Appendix Proof sketch of statement (A), that $f,g$ have a common point in $X\times Y$: If they don't meet at $(0,0)$ or $(1,1)$, then we have strict inequalities $g(f(0))>0$ and $g(f(1))<1$. Then the function $g(f(x)) - x$ goes from pos. to neg. on $[0,1]$. Let $x^\star$ be the infimum of the $x$'s such that $g(f(x))-x$ is negative. Since $g(f(x))$ is monotone nondecreasing, it is continuous except for an at-most-countable set of (upward) jump discontinuities; hence the same is true of $g(f(x))-x$. The point $x^\star$ cannot be one of these discontinuities because this would force $g(f(x))-x$ to be positive on an open interval to the right of $x^\star$. Thus $g(f(x))-x$ is continuous at $x^\star$. This implies it is not negative at $x^\star$ since this would force it to be negative on an open interval to the left of $x^\star$. Then $g(f(x))-x$ is positive at every point $xx^\star$ that have $x^\star$ as a limit point. Hence it is zero there, so $x^\star = g(f(x^\star))$, and $$(x^\star,f(x^\star)) = (g(f(x^\star)),f(x^\star))\in X\times Y$$ is a point common to $f$ and $g$. REPLY [4 votes]: Let us try to construct $S$ which is a counterexample to (B) by transfinite induction. (In the style of just-do-it1 proofs.) We will use the fact2 that the set $\mathcal M$ of all monotone non-decreasing functions $[0,1]\to[0,1]$ has cardinality $\mathfrak c$. Let $\mathcal M=\{f_\alpha; \alpha<\mathfrak c\}=\{g_\alpha; \alpha<\mathfrak c\}$ be any two enumerations3 of $\mathcal M$. By transfinite recursion we define points $P_\alpha,Q_\alpha \in [0,1]\times[0,1]$ for $\alpha<\mathfrak c$. The points $P_\alpha$ will help us avoid all $f_\alpha$'s (they will be forbidden points). The points $Q_\alpha$ will help us intersect all $g_\alpha$'s (they will be included in $S$.) In the inductive step we simply choose any $P_\alpha\ne Q_\alpha$ such that $P_\alpha,Q_\alpha\notin\{P_\beta,Q_\beta; \beta<\alpha\}$ $P_\alpha \in \{(x,f_\alpha(x)); x\in X\}$ $Q_\alpha \in \{(g_\alpha(y),y); y\in Y\}$ This is always possible since the set $\{P_\beta,Q_\beta; \beta<\alpha\}$ has cardinality less than $\mathfrak c$ and the graphs of $f$ and $g$ have cardinality $\mathfrak c$. Now we simply put $$S=\{Q_\alpha; \alpha<\mathfrak c\}.$$ For any $\alpha$, we get that $S$ intersects $g_\alpha$ (since $Q_\alpha\in S$), but $S$ does not contain $f_\alpha$ (since $P_\alpha\notin S$). 1blog, tricki 2See, for example, here: What is the cardinality of a set of all monotonic functions on a segment $[0,1]$? 3Notice that in this step we are using Axiom of Choice. (In the form of the well-ordering principle.)<|endoftext|> TITLE: Another weak form of the ABC conjecture QUESTION [24 upvotes]: First I will explain why a weaker form is needed. And then I formulate the conjecture (more precisely, the formulation will be clear). It is related to the question https://math.stackexchange.com/questions/40945/triangular-factorials and several Mathoverflow questions from the comments to that question. A number $m$ is called a triangular factorial if $m=\frac{n(n+1)}{2}=k !$ for some $n,k$. It is an open problem whether the set of triangle factorials is finite. Moreover the only known such numbers are $1, 6, 120$. But (somewhat surprisingly for me) it can be shown that the ABC conjecture implies that there are only finitely many triangular factorials. Indeed, suppose that for arbitrary large $k,m$ we have $ \frac{n(n+1)}{2}=k!$. Then $n+1=\frac {2k!}{n}$. Let $a=n, b=1, c= \frac {2k!}{n}$. Then by the ABC conjecture $\frac {2k!}{n}rad(a(a+1))^{d\log\log a}.$$ Note that the exponent in the right hand side may have to be a little different. REPLY [5 votes]: For $a> e$, we have $\log \log a > 0$ and hence $x^{\log \log a}$ is an increasing function, so $$a>\mathrm{rad}(a(a+1))^{\frac {\log a}{\log\log a}}$$ then implies that $a^{\log \log a} > \mathrm{rad}(a(a+1))^{\log a}$. Note that $a^{\log \log a} = (e^{\log a})^{\log \log a} = (e^{\log \log a})^{\log a} = (\log a)^{\log a}$. So we have $(\log a)^{\log a} > \mathrm{rad}(a(a+1))^{\log a}$, so $(\log a) > \mathrm{rad}(a(a+1))$, so with $b=1$ and $c=a+1$ we have $$c > a > \exp(\mathrm{rad}(abc))$$ but for every $\varepsilon > 0$ there is a $K$ such that $c < \exp(K\,\mathrm{rad}(abc)^{\frac13+\varepsilon})$ (Stewart & Yu 2001) , which gives a contradiction for sufficiently large $\mathrm{rad}(abc)$. So we can just show that for every constant $L$, we have that $L > \mathrm{rad}(abc) = \mathrm{rad}(a(a+1))$ for only finitely many $a$. This again follows form this result by Stewart & Yu (2001). If there were an infinitude of $a$ such that $L > \mathrm{rad}(a(a+1))$, since every such triple has a different value of $c$, there are arbitrarily large values of $c$ such that $L > \mathrm{rad}(abc)$, so $c < \exp(K\,\mathrm{rad}(abc)^{\frac13+\varepsilon}) < \exp(K\cdot L^{\frac13+\varepsilon})$, for arbitrarily large values of $c$, but the last is a constant, contradiction.<|endoftext|> TITLE: Bernoulli sum meets golden number QUESTION [18 upvotes]: Let $B_n$ denote the Bernoulli numbers and let $\phi=\frac{1+\sqrt{5}}2$ be the golden ratio. I encountered the following infinite sum and would like to ask: Question. Is this true? If so, any proof? $$\sum_{\pmb{k=0}}^{\infty}\sum_{j=k}^{2k}\binom{k}{j-k}\frac{B_{j+1}}{j+1} =\frac{2\,\log\phi}{1-2\phi}.$$ Caveat. Do not try reversing summations, it diverges! Update. Thanks to Henri Cohen for observing the typo, the sum has been edited to start at $k=0$. Readers are advised that Nemo's answer is given when the sum begins with $k=1$. REPLY [4 votes]: I meant to give a comment (to Henri Cohen's post), but there seemed to be too many characters. The first equation follows from the "reciprocity formula" $$(-1)^{m+1}\sum_{j=0}^k{k\choose j}\frac{B_{m+1+j}}{m+1+j}+(-1)^{k+1} \sum_{j=0}^m{m\choose j}\frac{B_{k+1+j}}{k+1+j}=\frac{k!m!}{(k+m+1)!}$$quoted in Wikipedia just above here by setting $m=k$. The references are [$ $1] M.B. Gelfand, A note on a certain relation among Bernoulli numbers (Russian), Bashkir. Gos. Univ., Uchen. Zap. Ser. Mat. 31 (1968) 215-216. [2] Takashi Agoh and Karl Dilcher, Reciprocity Relations for Bernoulli Numbers, American Mathematical Monthly, Vol. 115, No.3, (2008), p.237-244. [3] L. Saalschütz, Verkürtzte Recursionsformeln für die Bernoullischen Zahlen, Zeit. für Math. und Phys. 37 (1892) 374-378.<|endoftext|> TITLE: For isolated singularity algebra, is every maximal Cohen-Macaulay module locally projective? QUESTION [8 upvotes]: Let $R$ be a Cohen-Macaulay noetherian local ring. Let $\Lambda$ be a noetherian $R$-algebra which is maximal Cohen-Macaulay as an $R$-module, where for every nonmaximal prime $\mathfrak{p}$, $\Lambda_{\mathfrak{p}}$ has finite global dimension. How can I prove that every $\Lambda$-module which is maximal Cohen-macaulay as an $R$-module, is locally projective on the punctured spectrum of $R$? REPLY [5 votes]: The answer is no. Because locally, you can have a module over $\Lambda$ which is mCM over $R$ but not free. There are examples of singular $R$ with a module $M$ such that $\Lambda= End_R(M)$ has finite global dimension. But $M$ itself is a $\Lambda$-module and if it is free, then $R$ is Morita equivalent to $\Lambda$, contradicting the fact that $R$ is not regular. For example, $R=k[x,y,z]/(xy-z^2)$ and $M = R\oplus (x,z)$. To construct a concrete counter example, take $R= \mathbb C[[x,y,z,t]]/(xy-z^2)$ and $M = R\oplus (x,z)$ and $\Lambda = End_R(M)$. $M$ is not free at the point $(x,y,z)$. On the other hand, $\Lambda$ has finite global dimension on the punctured spectrum. (the singular locus is $V(x,y,z)$ so we only need to check at $(x,y,z)$.<|endoftext|> TITLE: Reference for parallel transport around loop and its relation to curvature QUESTION [6 upvotes]: It is a well known fact that the geometric meaning of a linear connection's curvature can be realized as the measure of a change in a fiber element as it is parallel transported along a closed loop. As far as I am aware, the formula goes as $$ F(X,Y)\sigma=-\lim_{t,s\rightarrow 0}\frac{1}{ts}\left(P^X_tP^Y_sP^X_{-t}P^Y_{-s}\sigma-\sigma\right), $$ where $P^X_t$ is parallel transport along $X$'s integral curves for time $t$, and I here assume that $[X,Y]=0$. Unfortunately, none of my references contain a proof of this statement, at least not one that is useful to me now. I can prove this statement using coordinate-based methods that often involve "expanding to first order" and other rather handwave-y methods (Weinberg, Wald etc. though Wald's procedure is actually fairly interesting and rigorous, it is not what I am looking for), but that's not what I am looking for. I have also seen proofs of this statement (Lee's Manifolds and Differential Geometry) which involves starting with vectors $u$ and $v$ at points and extending them to vector fields in special ways, so that they are parallel along some curves, which simplifies things greatly. This is once again not what I am looking for, because it already presupposes that we know that $F$ is tensorial and that $F(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$. In essence, what I am looking for is a derivation that $$ -\lim_{t,s\rightarrow 0}\frac{1}{ts}\left(P^X_tP^Y_sP^X_{-t}P^Y_{-s}\sigma-\sigma\right)=\nabla_X\nabla_Y\sigma-\nabla_Y\nabla_X\sigma $$ for commuting vector fields $X,Y$, without supposing we already know what curvature is (I need this for didactic reasons, for possible use in a general relativity lecture). I have tried to do this on my own, but I am really terrible at formally manipulating these "flow" type of objects like $P^X_t$, and so far my attempts failed. I would appreciate any reference, textbooks, lecture notes, articles, papers, which prove this formula (with the added caveat that I outlined in bold above). If it does so without assuming $X$ and $Y$ commute, it would especially be stellar, but I don't need that. REPLY [4 votes]: A simple way is as follows. Given a point $p$, the vector field $(P_{-t}^X\sigma)(p)$ is obtained from $\sigma(\gamma(s))$ by backward parallel transport along the curve $\gamma(t):=\exp(tX)(p)$, up to time $0$. Denoting this backward parallel transport (from time $t$ to time $0$) by $Q_t^\gamma$, we have $$ \frac{d}{dt}(P_{-t}^X\sigma)(p)=\frac{d}{dt}(Q_t^\gamma \sigma(t))=Q_t^\gamma(D_t\sigma(t))=(P_{-t}^X)(\nabla_X\sigma)(p), $$ where $\sigma(t):=\sigma(\gamma(t))$ is a vector field along $\gamma$ and $D_t$ is the covariant derivative along $\gamma$. So $$P_{-t}^X\sigma=\sigma+\int_0^t P_{-t'}^X(\nabla_X\sigma)\,dt'=\sigma+t\nabla_X\sigma+O(t^2). $$ Hence, $$\begin{align}&P_t^XP_s^YP_{-t}^XP_{-s}^Y\sigma=P_t^XP_s^YP_{-t}^X(\sigma+s\nabla_Y\sigma)+O(s^2)\\ &=P_t^XP_s^Y(\sigma+s\nabla_Y\sigma+t\nabla_X\sigma+ts\nabla_X\nabla_Y\sigma)+O(s^2)+O(t^2)\\ &=P_t^X(\sigma+t\nabla_X\sigma+ts\nabla_X\nabla_Y-ts\nabla_Y\nabla_X\sigma)+O(s^2)+O(t^2)\\ &=\sigma+ts\nabla_X\nabla_Y-ts\nabla_Y\nabla_X\sigma+O(s^2)+O(t^2)\end{align}$$ (of course the parallel transport applied to the error terms has the same order of magnitude, since it is obtained by solving a first-order ODE). Now you just have to observe that the second order expansion of the smooth function $f(t,s):=(P_t^XP_s^YP_{-t}^XP_{-s}^Y\sigma)(p)$ has the form $\sigma(p)+tsV$ for some $V\in T_pM$: the coefficients of $s$, $t$, $s^2$ and $t^2$ all vanish since $f(t,0)\equiv\sigma(p)$ and $f(0,s)\equiv\sigma(p)$.<|endoftext|> TITLE: Picard number of a general fiber of a fiber contraction QUESTION [6 upvotes]: Suppose in the last step of a MMP, we obtain a Mori fiber space $f: X \to Z$, and let $F$ be a general fiber of $f$, then is the Picard number $\rho(F)$ of $F$ equal to $1$? Notice that the relative Picard number $\rho(X/Z)=1$ because I only assume to contract an extremal ray. My feeling is that $\rho(F)$ may not be one, but I don't have a good example. Notice that if $X$ is toric, the fiber always has Picard number $1$. I would also appreciate examples (if any) of fibration (with connect fibers) $f: X \to Y$ such that $\rho(X/Z)=1$ but the general fiber does not have Picard number $1$. REPLY [5 votes]: I do not think so, because of the following result. Proposition. A smooth del Pezzo surface $F$ can be realised as the general fibre of a Mori fibre space if and only if it is not isomorphic to the blow-up of $\mathbb{P}^2$ in one or two points. This shows that all values $\rho(F) \in \{1, \ldots, 9\}$ with $\rho(F) \neq 3$ are realized. See Theorem 1.4 in Fano Varieties in Mori Fibre Spaces, Int. Math. Res. Notices (2016) 2016 (7): 2026-2067.<|endoftext|> TITLE: Regular holonomic D-modules as generalisation of regular singular points QUESTION [5 upvotes]: I'm trying to understand why the definition of a regular holonomic D-module is a good generalisation of the usual definition of a regular singular point for a differential equation. More precisely, here is the definition of regular holonomic D-module I want to use A D-module $M$ on a complex manifold $X$ is holonomic if $\text{char}(M)$ has the same dimension as $X$. Moreover, it is regular if there is a good filtration $\{M_j\}_{j\in \mathbb{Z}}$ such that $$f\cdot\text{gr}(M)=0$$ for all $f\in \text{gr}(D_X)$ vanishing on $\text{char}(M).$ Now I'd like to test this definition on an example. I consider a Bessel equation $$P(f) :=(z^2 \partial_z^2 + z\partial_z + (z^2-4))(f)=0.$$ The point $z=0$ is regular singular thanks to the $z^2$. Of course if the leading coefficient is replaced by $z^3$ it becomes irregular singular. Now I consider the D-module $$M := D_{\mathbb{C}}/D_{\mathbb{C}} P$$ naturally attached to this equation. We should normaly be able to prove that this D-module on $\mathbb{C}$ is regular holonomic. (And not regular if we put $z^3$ as a leading coefficient) First the characteristic variety is given by the null-set of $(z,w)\mapsto z^2w^2$ and so $$\text{char}(M) = \mathbb{C} \cup T_0^{*}\mathbb{C}$$ which has dimension $1$, so the module is holonomic. Now as a candidate for the good filtration, Simon Wadsley proposes to take $$M_n=D_n.(1+D_\mathbb{C}P).$$ Now I'm getting lost, how can I prove that $f\cdot\text{gr}(M)=0$ for all $f\in \text{gr}(D_X)$ vanishing on $\text{char}(M) \,?$ It doesn't make much sense to me. Thank you for any help. REPLY [3 votes]: I'm going to show (eventually) that your proposed filtration does not work. Let $I$ be an ideal of $D_\Bbb{C}$, and let $M=D_\Bbb{C}/I$. Then, letting $u$ be the image in $M$ of $1$ (so $\operatorname{ann}_{D_\Bbb{C}}(u) = I$), the filtration $M_j = D_j u$ of $M$ is a good filtration; the filtration $I_j = D_j \cap I$ of $I$ is a good filtration; the maps in the exact sequence of $D_\Bbb{C}$-modules $$ 0\to I \to D_\Bbb{C} \to M \to 0$$ are strictly filtered (recall that a map $f\colon M\to N$ is strictly filtered if $f(M)\cap N_j = f(M_j)$ for all $j$). By 3., the sequence $$0 \to \operatorname{gr} I \to \operatorname{gr}D_\Bbb{C} \to \operatorname{gr} M \to 0$$ is still exact, so $ \operatorname{gr}M \cong \operatorname{gr}D_\Bbb{C}/\operatorname{gr} I$, and the characteristic variety of $M$ is the zero locus of $\operatorname{gr} I$. Case $I$ is principal: Suppose $I= D_\Bbb{C} P$ is principal. Then (prove this!) $\operatorname{gr} I$ is generated by the principal symbol $\sigma(P)$ of $P$. Hence, $\operatorname{gr} M \cong \operatorname{gr} D_\Bbb{C}/(\sigma(P))$. In particular, if $P = z^2 \partial_z^2 + z\partial_z + z^2-4$, then $\sigma(P)=z^2\partial_z^2$, and therefore $\operatorname{gr} M \cong \operatorname{gr} D_\Bbb{C}/(z^2\partial_z^2)$. However, the ideal of $\operatorname{Ch}(M)$ is $(z\partial_z)$, which means that this filtration that we chose is not the one whose existence is claimed by the proof.<|endoftext|> TITLE: Bimeromorphic equivalence of reduced spaces for Kähler $S^1$-actions QUESTION [7 upvotes]: Let $(X,\omega)$ be a smooth Kähler manifold (not necessarily compact) with an isometric $S^1$-action with a Hamiltonian $H$. It is a well known fact that 1) The reduced spaces $X(c)=H^{-1}(c)/S^1$ have a complex analytic structure. 2) Such spaces are bimeromorphic for $c$ satisfying $\min(H) TITLE: Is this Wikipedia article linking to the wrong notion of coherent space QUESTION [16 upvotes]: I'm reading up on infinite generalizations of the fundamental theorem of distributive lattices. Wikipedia (June 15, 2017) says that there is a duality between distributive lattices and coherent spaces (sometimes called spectral spaces) where I have reproduced the links as they appear. I think that the article on coherent spaces they have linked to is referring to a different, unrelated notion. Could someone in logic check me on this before I correct it? Thanks, and apologies if this is too frivolous a use of MO. REPLY [13 votes]: Yes, the notion of "coherent space" at the page you linked is completely unrelated to spectral spaces. The coherent spaces of that article were introduced by Jean-Yves Girard as a denotational semantics of second-order intuitionistic logic and were instrumental in his discovery of linear logic. People have been aware of this potential confusion since the beginning of linear logic (see footnote 3, p.55 of Girard, Lafont and Taylor's book Proofs and Types), that is why some prefer to call them coherence spaces. I guess that the authors of the Wikipedia page didn't follow this practice. Edit: Just after posting my answer, I realized that this potential confusion is explicitly pointed out in the Wikipedia page, as well as the alternative terminology in Proofs and Types.<|endoftext|> TITLE: Some curious Hankel determinants QUESTION [5 upvotes]: Let $f(n,q)=\prod_{j=1}^na(q^j)$ for a polynomial $a(q)$ and let $d(n)=\det(f(i+j,q))_{i,j=0}^n$ be its Hankel determinant. Computer experiments suggest that $$\lim_{q\to1}\frac{d(n)}{(q-1)^\binom{n+1}{2}}=(a(1)a'(1))^\binom{n+1}{2}\prod_{j=0}^nj!.$$ Has anyone an idea how to prove this? Remark: For $a(q)=a+qb$ it is easy to verify that $$d(n)=((q-1)b)^\binom{n+1}{2}q^\frac{n(n+1)(2n+1)}{6}{\prod_{j=1}^n[j]_{q}!(a+q^jb)^{n+1-j}},$$ if $[n]_{q}=\frac{1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\dots[n]_q.$ Therefore the conjecture is true for linear polynomials and also for $a(q)=q^m.$ REPLY [5 votes]: The identity is clearly invariant to scaling $a$ by a constant. (One has to check the constant appears in the determinant to power $n(n+1)$). So wlog we may assume $a(1)=1$. We can expand $\log a$ as a formal power series in $\log q$, $$ a(q)= e^{ \sum_{i} a_i (\log q)^i}$$ $$a (q^j) = e ^{ \sum_i a_i j^i (\log q)^j}$$ $$\prod_{j=1}^n a(q^j) = e^{ \sum_i a_i (\sum_{j=0}^n j^i) (\log q)^j }$$ the key thing being that $(\sum_{j=0}^n j^i)$ is a polynomial in $n$ of degree $\leq n+1$. Hence if we expand this product $f(n,q)$ out as a power series in $\log q$, the coefficient of the $i$th power of $\log q$ will be a polynomial in $n$ of degree $\leq 2i$, whose leading term is a power of $a_1$. Hence the coefficient of $\log q^i$ in the power series of $f(x+y,q)$ is a sum of monomials in $x$ times monomials in $y$ of total degree $\leq i$. When we take a determinant of a sum of rank one matrices, in this case a sum indexed by powers of $\log q$ and then pairs of monomials, it's the same as the sum of the determinants of all sums of $n+1$ matrices from the set. These determinants are nonzero only if the degreees of all the monomials in $x$ and monomials in $y$ appearing are distinct, which means the total degree in each variable must be at least ${n+1 \choose 2}$, for a total degree of at least $2 {n+1 \choose 2}$, and thus total degree in $\log q$ of at least ${n+1 \choose 2}$. This is only sharp if the monomials we study are the highest degree terms, which we saw come only from $a_1$. So the answer depends only on $a_1$, and the calculations you include in your answer finish the job.<|endoftext|> TITLE: Are functions of bounded variation a.e. differentiable? QUESTION [17 upvotes]: In general, it is well known that, on the real line, say on $[0,1]$, if a function $f$ is of (pointwise) bounded variation, meaning that $$ \sum_{i=1}^n |f(x_i)-f(x_{i-1})| <+\infty $$ for every partition ${x_i}_{0}^n$ of $[0,1]$, then $f$ can be written as the difference of two monotone functions, hence it is differentiable a.e. w.r.t. the Lebesgue measure. I am wondering if the same is true for $BV$ functions in $\mathbb R^d$ for $d \ge 2$. Of course, the right definition of $BV$ in $d$-dimensional domains passes through the theory of distributions: $f \colon \Omega \subset \mathbb R^d \to \mathbb R$ is in $BV(\Omega)$ if it is an $L^1$ function whose distirbutional gradient $Df$ can be represented by a finite Radon measure (see here). Question 1. Let $f \colon \Omega \subset \mathbb R^d \to \mathbb R$ be in $BV(\Omega)$. Is it true that $f$ is differentiable a.e. with respect to the Lebesgue measure? What I know is that they are approximately differentiable a.e. (in view of Calderon-Zygmund Theorem) so an approximate differential exists a.e. but I am not aware of any link between the approximate differentiability and the pointwise a.e. one. Addendum. In view of Mizar's answer, it seems that the answer to Q1 is negative, as it has been exhibited a $BV$ function which does not have even a continuous a.e. representative (in $L^1$). While checking the details of the answer I received, I would like to ask another version of question above (do not know if still meaningful or not). Question 2. Let $f \colon \Omega \subset \mathbb R^d \to \mathbb R$ be in $BV(\Omega)$. Assume further $f \in C^0(\Omega)$ i.e. it is continuous. Is it true that $f$ is differentiable a.e. with respect to the Lebesgue measure? REPLY [22 votes]: No. Take a dense countable set $\{x_1,x_2,\dots\}$ in $\mathbb{R}^d$ and a sequence $(r_i)\subseteq\mathbb{R}^+$ such that $\sum_i r_i^{d-1}<\infty$. Then the function $$f=1_{\bigcup_{i=1}^\infty B_{r_i}(x_i)}$$ is in $BV(\mathbb{R}^d)$ (since $|\bigcup B_{r_i}(x_i)|\le C\sum_i r_i^d$ and $f$ is the limit in $L^1$ of the functions $1_{\bigcup_{i=1}^k B_{r_i}(x_i)}$, whose gradients have total variation bounded by $C\sum_i r_i^{d-1}<\infty$). Now, for any Lebesgue point $x_0$ of $f$ in the closed set $\{f=0\}$, no representative $g$ is continuous at $x_0$ (representative means a function which coincides a.e. with $f$). Indeed, $x_0$ lies in the closure of the open set $\bigcup B_{r_i}(x_i)$, so it belongs to the closure of $\{g=1\}$. On the other hand, since $x_0$ is a Lebesgue point for $f$, it must also belong to the closure of $\{g=0\}$. This shows that $g$ is not even a.e. continuous (since the set $\{f=0\}$ has positive measure). Addendum. The answer is still no even assuming $f$ continuous. Below I construct an example where the differentiability of $f$ fails on a Borel set of positive measure. Choose a countable dense set $\{x_i\}$ in $B_1(0)$ and a sequence $r_i>0$ such that $\sum_i r_i^{d-1}<\infty$ and $\sum_i|B_{r_i}(x_i)|<|B_1(0)|$. In particular, $r_i\to 0$. Using Besicovitch covering theorem, up to a subsequence we can assume that the balls $B_{r_i}(x_i)$ have bounded overlapping (i.e. any point lies in at most $N$ such balls); in doing this, we could lose the density of $\{x_i\}$ but we still have $B_1(0)\subseteq\overline{\cup_i B_{r_i}(x_i)}$. Let $S:=B_1(0)\setminus\cup_i\overline{B_{r_i}(x_i)}$. We remark that $|S|>0$ and that, for any $N\ge 1$, $S\subseteq\overline{\{x_i\mid i>N\}}$. It follows that we can find a sequence of positive radii $R_i\to 0$ such that $S\subseteq\cup_{i\ge j}B_{R_i}(x_i)$ for all $j\ge 1$: by compactness, we can find $n_1>0$ such that $$S\subseteq\overline{\{x_i\mid i>0\}}\subseteq\cup_{i=1}^{n_1}B_1(x_i),$$ then we use the above remark with $N=n_1$ and we find $n_2>n_1$ such that $$S\subseteq\overline{\{x_i\mid i>n_1\}}\subseteq\cup_{i=n_1+1}^{n_2}B_{1/2}(x_i),$$ and so on. Now the function $f(x):=\sum_i R_i(1-r_i^{-1}|x-x_i|)^+$ (a superposition of 'traffic cones' with heights $R_i$ placed on our balls $B_{r_i}(x_i)$) is continuous, as it is a uniform limit of continuous functions, thanks to the bounded overlapping. It also lies in $BV(\mathbb{R}^d)$ thanks to the assumption $\sum_i r_i^{d-1}<\infty$. I claim that $f$ cannot be differentiable at $x$, for any $x\in S$. Indeed, as $x$ is a minimum point for $f$, we would have $\nabla f(x)=0$. But there is a sequence $i_k\to\infty$ with $x\in B_{R_{i_k}}(x_{i_k})$, so $f(x_{i_k})\ge R_{i_k}\ge|x-x_{i_k}|$, which contradicts $\nabla f(x)=0$ (since $x_{i_k}\to x$).<|endoftext|> TITLE: Heegaard splitting of maps between 3-manifolds QUESTION [8 upvotes]: Let $M$ and $M'$ be closed oriented connected 3-manfolds and let $f : M \to M'$ be a continuous map. Do there exist Heegaard splittings $M = H_1 \cup H_2$ and $M' = H_1' \cup H_2'$ and a map $f'$ homotopic to $f$ so that $f'$ preserves the Heegaard splittings in the sense that $f'(H_1) \subset H_1'$ and $f'(H_2) \subset H_2'$? Let $\Sigma = H_1 \cap H_2$ and $\Sigma' = H_1' \cap H_2'$. Note that by naturality of Mayer-Vietoris, the degree of $f'$ will have to equal the degree of $f'|_\Sigma : \Sigma \to \Sigma'$ which gives some restriction on the possible Heegaard splittings. I believe this should not be true for $\deg(f) >1$ but I can not seem to find an obstruction. Thanks! REPLY [4 votes]: Waldhausen, Friedhelm, On mappings of handlebodies and of Heegaard splittings, Topology of Manifolds, Proc. Univ. Georgia 1969, 205-211 (1971). ZBL0282.57003. Bielefeld. See this paper of Waldhausen, which discusses the case of degree 1 maps. However, in the proof of Theorem 2.1 in his paper, he sketches a proof that a map between 3-manifolds may be homotoped to preserve Heegaard splittings (the details are not given, but it seems clear that the hypothesis of degree 1 is not used in this part of the theorem). He seems to think this is "well-known", but doesn't give a reference.<|endoftext|> TITLE: First order formulas for finite groups and invariant theory QUESTION [9 upvotes]: Let $G$ be a finite group, and let $K$ be a field of characteristic zero. Let $\phi(x_1,\ldots,x_n)$ be a first order formula in the language of group theory (so $\phi$ can be for example something of the form $\exists y\in G$ $y^2=x$ or $\exists y_1,y_2\in G $ $ x_1=y_1y_2,x_2=y_2y_1$). For every such formula $\phi$ we consider the element $$t_{\phi}:=\sum_{\{(g_1,\ldots g_n)\in G^n| \phi(g_1,\ldots g_n)\}}g_1\otimes g_2\otimes\cdots\otimes g_n$$ In other words, $t_{\phi}$ is the sum over all tuples who satisfy the formula $\phi$. If now $\mu:G\to G$ is a group automorphism, then $\phi(g_1,\ldots,g_n)$ holds if and only if $\phi(\mu(g_1),\ldots,\mu(g_n))$ holds. It thus follows that the elements $t_{\phi}$ are all $Aut(G)$-invariants inside $(KG)^{\otimes n}$. Is there an elementary proof that the elements $t_{\phi}$ span the invariant subspace $(KG^{\otimes n})^{Aut(G)}$? REPLY [6 votes]: The assumption on the characteristic of $K$ implies that $(KG^{\otimes n})^{Aut(G)}$ is spanned by the indicator functions of the $Aut(G)$-orbits in $G^n$. So it suffices to observe that for every $(g_1,\dots,g_n) \in G^n$ the formula "$(x_1,\dots,x_n)$ belongs to the $Aut(G)$-orbit of $(g_1,\dots,g_n)$" is a first-order formula. But (if I remember correctly what a first-order formula is) this is clear, as it can be written as "$\exists (y_g)_{g \in G} \in G, y_{g h}=y_{g}y_h, y_g \neq y_h$ if $g \neq h$ and $y_{g_i} = x_i$".<|endoftext|> TITLE: Work of plenary speakers at ICM 2018 QUESTION [57 upvotes]: The next International Congress of Mathematicians (ICM) will be next year in Rio de Janeiro, Brazil. The present question is the 2018 version of similar questions from 2014 and 2010. Can you, please, for the benefit of others give a short description of the work of one of the plenary speakers? List of plenary speakers at ICM 2018: Alex Lubotzky (Israel) Andrei Okounkov (Russia/USA) Assaf Naor (USA) Carlos Gustavo Moreira (Brazil) Catherine Goldstein (France) Christian Lubich (Germany) Geordie Williamson (Australia/Germany) Gil Kalai (Israel) Greg Lawler (USA) Lai-Sang Young (USA) Luigi Ambrosio (Italy) Michael Jordan (USA) Nalini Anantharaman (France) Peter Kronheimer (USA) and Tom Mrowka (USA) Peter Scholze (Germany) Rahul Pandharipande (Switzerland) Ronald Coifman (USA) Sanjeev Arora (USA) Simon Donaldson (UK/USA) Sylvia Serfaty (France/USA) Vincent Lafforgue (France) REPLY [15 votes]: Nalini Anantharaman is a french mathematician working in the fields of dynamical systems, partial differential equations and mathematical physics. Her early works deal with the counting of closed geodesics on hyperbolic surfaces in a given homology class. She gave a full asymptotic expansion for the counting function following dynamical methods introduced by D. Dolgopyat, when the homology class lies in the interior of the set of winding cycles of the invariant measures of the geodesic flow. She also gave estimates when the homology class is in the boundary of this set. In that case, the problem is connected to the zero temperature limit in the theory of Markov processes and maximizing measures in the Aubry-Mather theory of Lagrangian systems. She then got interested in semi-classical analysis and used entropy methods to study the weak limits of the sequence of probability measures $$ |\psi_k|^2 d\hbox{vol}$$ where $\psi_k$ are the eigenfunctions of the Laplacian defined on a negatively curved compact manifold. The quantum unique ergodicity conjecture asserts that the sequence should converge to the Liouville measure after a suitable lift of the measures to the unit tangent bundle of the manifold. She showed that any cluster points of the sequence must have positive entropy, thus ruling out a convergence to the Dirac mass on some closed orbit. So this is a remarkable application of ergodic theory to the study of the linear wave equation and Schrodinger equation. See a survey of P. Sarnak for additional details. More recently, she studied related problems on billiards and regular graphs. Interesting pictures of cardioid billiards can be found in her joint work with Arnd Backer. Her webpage contains a few of her lectures in video format.<|endoftext|> TITLE: What techniques are there to prove Schur positivity? QUESTION [12 upvotes]: As the title says, what methods exists for proving that a symmetric polynomial (or function) is Schur positive, perhaps involving extra parameters, in which case coefficients should be polynomials in the parameters with non-negative coefficients. This is what I have seen in the literature: Representation-theoretical proof. Construct a (graded) $S_n$-module, and show that the Frobenius map of the decomposition into irreducibles gives the polynomial. This often involves finding recursions, and does not give formulas for the coefficients. Examples: Modified Macdonald polynomials, and things in diagonal harmonics. This MO-post. RSK-type proof. This essentially gives a way to convert words to semi-standard tableaux, and is therefore a type of bijective proof. Examples: skew-Schur functions (and thus the Littlewoord-Richardson rule). Via Gessel fundamental quasisymmetric expansion. It is usually easy to find the Gessel expansion of a combinatorially defined polynomial. Once this is known, the goal is to gather these pieces into Schur-polynomials. The 'Schur-expansion' is technically known from the Gessel expansion, but it is given in the form of functions $S_{\alpha}$, where one needs to modify the compositions $\alpha$ according to the 'slinky' rule to obtain partitions. This can introduce signs, that needs to be taken care of via a sign-reversing involution or similar. Type-A crystal proof. If the polynomial is given as a sum over combinatorial objects, one can try to define a certain graph structure on these objects, fulfilling some combinatorial (fairly local) axioms (Stembridge did this characterization, if I recall). Each connected component of this graph will each correspond to a Schur polynomial in the Schur expansion. This is related to my old question, before I knew about crystals, and it turns out it is enough to consider three variables at a time (in the Stembridge axioms). Basically, if you can give a crystal graph in three variables, it should generalize to $n$ variables without any issue. The crystal structure is closely related to RSK, and also provides a representation-theoretical connection, as well as a (crystal) bijection to SSYTs. Examples: Stanley symmetric functions. Dual Grothendieck. Dual equivalence graph, introduced by S. Assaf. Similar idea as crystals/RSK/Gessel. From the fundamental quasisymmetric expansion, define a graph structure that gathers these pieces into Schur-positive parts. Example: This article, which has a non-symmetric counterpart as well. REPLY [4 votes]: Some further examples of Schur positivity: Exercises 7.38, 7.46, and 7.91 in EC2, and Problems 116(b,d) and 137(a,b) in http://www-math.mit.edu/~rstan/ec/ch7supp.pdf (version of 14 March 2021).<|endoftext|> TITLE: Dioperads vs polycategories QUESTION [11 upvotes]: As defined by Gan, a dioperad consists of sets of operations $P(n,m)$ with "$n$ inputs and $m$ outputs", which can be composed by joining one output of one operation to one input of another, giving rise to composition operations $$ P(n_1,m_1) \times P(n_2,m_2) \to P(n_1+n_2-1,m_1+m_2-1). $$ Dioperads are apparently motivated by the fact that some PROPs are freely generated by dioperads, because they have a generators-and-relations description in which all the relations only involve composites of the sort that exist in a dioperad. In such a case, the dioperad is evidently a simpler object to study than the entire PROP. As defined by Szabo, a polycategory has a set of objects, and for any $n$-tuple and $m$-tuple of objects $a_1,\dots,a_n$ and $b_1,\dots,b_m$, a set of morphisms $P(a_1,\dots,a_n; b_1,\dots,b_n)$, with composition operations that join one of the targets of one morphism to one of the sources of another. If we write $\Gamma,a,\Delta$ for a list of objects containing $a$ somewhere in the middle, then this composition operation looks like $$P(\Gamma; \Delta,a,\Psi) \times P(\Xi,a,\Upsilon; \Phi) \to P(\Xi,\Gamma,\Upsilon; \Delta,\Phi,\Psi).$$ Polycategories are motivated by providing a semantics for classical linear logic, whose cut rule looks like this composition rule: $$ \frac{\Gamma \vdash \Delta,A,\Psi \qquad \Xi,A,\Upsilon \vdash \Phi}{\Xi,\Gamma,\Upsilon \vdash \Delta,\Phi,\Psi} $$ One generally thinks of the source objects as joined by a "multiplicative conjunction" tensor product $\otimes$ and the target objects as joined instead by a "multiplicative disjunction" tensor product $⅋$, and this can be made precise by showing that "representable" polycategories are precisely linearly distributive categories. It looks to me as though (ignoring questions of enrichment) a dioperad is just a one-object polycategory, in the same way that an ordinary operad is a one-object multicategory — or in other words, a polycategory is a "colored dioperad". Is this true? If so, is it in print anywhere? If not, what is the difference? REPLY [6 votes]: In Martin Markl's article "Operads and PROPs," just after Def. 64, the dioperad-polycategory connection is briefly mentioned. Markl attributed this observation to Leinster. In my book with Mark W. Johnson "A Foundation for PROPs, Algebras, and Modules," in Section 11.5 we wrote down the explicit axioms for a colored dioperad. The relevant graphs and monads were written down earlier in Part I, as explained further down in Section 11.5. If you take out the equivariant structure of a C-colored dioperad (Def. 11.18), the remaining structure is a polycategory with object set C.<|endoftext|> TITLE: Maximal Sylow 2-subgroups of simple groups QUESTION [11 upvotes]: It is provable, using transfer theory and Thompson's J-subgroup, that if a nonabelian finite simple group $G$ has a nilpotent maximal subgroup $M$, that $M$ must be a 2-subgroup. It then follows from Sylow theory that $M$ must be a Sylow 2-subgroup of $G$. (It also follows from Sylow theory that, if this is true, any Sylow 2-subgroup of $G$ will be maximal.) It's not unusual, to my understanding, for group theory textbooks to present $PSL_{2}(\mathbb{F}_{17})$ as an example of a nonabelian finite simple group in which the Sylow 2-subgroups are maximal. But this can be generalized: $PSL_{2}(\mathbb{F}_{p})$ has maximal Sylow 2-subgroups whenever $p$ is prime, $p \geq 17$, and $p$ is 1 away from a power of 2 (which is to say that $p$ is a Fermat prime or a Mersenne prime). (Note that the only composite prime-power that is 1 away from a power of 2 is 9. 9 does not need to be included here because $PSL_{2}(\mathbb{F}_{9}) \cong A_{6}$ does not have maximal Sylow 2-subgroups. So weakening the condition by allowing powers of primes does not help here.) My question is: if a nonabelian finite simple group has maximal Sylow 2-subgroups, it is necessarily isomorphic to one of the $PSL_{2}$ groups described above? REPLY [13 votes]: Yes this follows from work of Baumann and Thompson. See the paper `On finite insoluble groups with nilpotent maximal subgroups' by John Rose https://doi.org/10.1016/0021-8693(77)90301-5 In fact, these are the only examples of simple groups with nilpotent maximal subgroups.<|endoftext|> TITLE: Dyer-Lashof algebra and Steenrod algebra "duality" QUESTION [18 upvotes]: Previously asked on math.stackexchange, but perhaps this is more appropriate for MathOverflow: Let $X$ be a spectrum. I've heard that the action of the Steenrod algebra on $H^*(X; \mathbb{F}_p)$ and the action of the Dyer-Lashof algebra on the homology of the associated infinite loop space, $H_*(\Omega^{\infty}X; \mathbb{F}_p)$, are "dual" in some sense. To my knowledge, this has something to do with Koszul duality, but I've never seen this spelled out in detail. Anyone willing to enlighten me on this? Or point me toward a helpful reference? REPLY [5 votes]: This is not the duality that was asked for in the question, but it does show a way in which the Steenrod algebra structure is related to the Dyer-Lashof algebra structure. The Dyer-Lashof algebra can act both on the homology of infinite loop spaces and $E_{\infty}$ ring spectra. If I consider the latter and look at $H_*(F(X,S^0);\mathbb{F}_p)$ then this has an action of the Dyer-Lashof algebra. It also happens to be isomorphic to the cohomology of $X$ as an algebra. In fact, the action of the Dyer-Lashof algebra "is" the action of the Steenrod algebra (there is an op that needs to be thrown in). I learned this from chapter 3 of http://www.math.uchicago.edu/~may/BOOKS/h_infty.pdf.<|endoftext|> TITLE: Special units in the $11$th cyclotomic field QUESTION [10 upvotes]: In connection with this problem: Do there exist integers $a_0,\dotsc,b_{10}\ge 0$ such that $a_0+\dotsb+a_{10}=36$, $b_0+\dotsb+b_{10}=37$, and $$ (a_0+a_1\zeta+\dotsb+a_{10}\zeta^{10})(b_0+b_1\zeta+\dotsb+b_{10}\zeta^{10})=1, $$ where $\zeta$ a primitive $11$-th root of unity? If they exist, can one explicitly describe / list all of them? Notice that $36\cdot 37=11^3+1$. REPLY [6 votes]: If I did this right there are a total of $1045 = 55 \cdot 19$ solutions, obtained from the following $19$ basic solutions by changing $a_i,b_i$ to $a_{ri+s\bmod 11}$ and $b_{ri+s \bmod 10}$ for all $s\bmod 11$ and $r=1,2,3,4,5$: 33433433433 | 33434343433 33434243433 | 33344344333 43424242434 | 34423532443 43234443234 | 34252525243 34251615243 | 33334543333 32344444323 | 42516161524 32434443423 | 52244344225 43244244234 | 54423132445 65421012456 | 42443334424 55331213355 | 44224542244 35234243253 | 35160706153 35160606153 | 33453135433 23253635232 | 35145154153 44214641244 | 12346564321 32463036423 | 33426162433 14642224641 | 45250505254 50274047205 | 34260706243 41615251614 | 15523532551 31608080613 | 23642324632 The solutions with $\sum_i a_i \zeta_i = 1+\zeta+\zeta^{10}$ and $(2+\zeta+\zeta^{10})^{-1}$ are in the first and fifth orbit respectively. Each orbit is of size $55$, not $11\cdot 10 = 110$, because every solution is fixed by some involution $(a_i,b_i) \leftrightarrow (s-a_i,b-a_i)$. We seek units $\alpha := \sum_{i=0}^{10} a_i \zeta^i \in {\bf Z}[\zeta]$ and $\beta = \alpha^{-1}$ all of whose algebraic conjugates have absolute value at most $37$. (This condition is weaker than the required conditions that $a_i,b_i$ are nonnegative and sum to $36$ and $37$ respectively; we check this stricter condition at the end.) It is well known that every unit in a cyclotomic number field $F_N := {\bf Q}(e^{2\pi i/N})$ is a root of unity times a real unit; in our setting with $N=11$ this is equivalent to the observation that the $a_i$ and $b_i$ are symmetric under some reflection $i \leftrightarrow s-i$. I chose representatives that make the sequences $(a_i)$ and $(b_i)$ visibly symmetric. Now let $v_j$ ($1\leq j\leq 5$) be the five embeddings into $\bf R$ of the real subfield $F_{11}^+ = {\bf Q}(\zeta+\zeta^{-1})$; and let $\lambda: (F_{11}^+)^* \to {\bf R}^5$ be the homomorphism $$ x \mapsto (\log|v_1(x)|, \log|v_2(x)|, \log|v_3(x)|, \log|v_4(x)|, \log|v_5(x)|). $$ The kernel of $\lambda$ is ${\pm 1}$, and the image of any element of norm $1$ is contained in the hyperplane $\{ (c_1,c_2,c_3,c_4,c_5) : \sum_{j=1}^5 c_j = 0 \}$. By the Dirichlet unit theorem, the group $U$ of units maps to a lattice in this hyperplane. For $F_{11}^+$ the units are "well-known"; we could also find generators for $U$, and thus for $\lambda(U)$, by consulting the LMFDB entry for $F_{11}^*$, or using the number-field routines of gp or similar packages. We seek units satisfying the additional condition that $|c_j| \leq \log 37$ for each $j$. Thus they are lattice points in the sphere $$ \sum_{j=1}^5 c_j^2 \leq 5 (\log 37)^2 < 66. $$ I used the qfminim routine in gp to generate a list of all such lattice points; there are $5025$ nonzero $\pm$ pairs. For each one, I checked whether $\alpha = \sum_{i=0}^{10} a_i \zeta^i$ satisfies $A := \sum_i a_i \equiv 3 \bmod 11,$ and thus $B := \sum_{i=0}^{10} b_i \equiv 3^{-1} \equiv 4 \bmod 11.$ If so, then subtracting $\min_i a_i$ from each $a_i$ and $\min_i b_i$ from each $b_i$ makes the coefficients nonnegative. Then if $A \leq 36$ and $B \leq 37$, adding back $(36-A)/11$ to each $a_i$ and $(37-B)/11$ to each $b_i$ makes the sums exactly $36$ and $37$ respectively. Choosing one representative from each orbit yields the list of $19$ displayed at the start of this answer.<|endoftext|> TITLE: The Heegner hypothesis for a mean value result of Murty-Murty/Bump-Friedberg-Hoffstein QUESTION [5 upvotes]: The classical mean value result of Murty and Murty (1991) and Bump, Friedberg, and Hoffstein (1990) on derivatives of modular form L-functions $L(s,f)$ proves (roughly speaking) the existence of infinitely many imaginary quadratic fields $K$ for which the Heegner hypothesis holds, ie., every prime dividing the level of a given modular form $f$ splits in $K$, such that a $L(s,f)$ has a simple zero at $s=k$, with $2k$ being weight of $f$, and such that $L(k,f,\chi)\neq 0$ for the twist $\chi$ associated to $K$ (or vice versa). I am looking to see if this result has been generalized to (any) fields $K$ in which the Heegner hypothesis fails to hold. Searching through the literature has not proved to be a simple task. REPLY [5 votes]: Here is what the Friedberg-Hoffstein result says: In non-technical language, suppose you are given a GL(2) automorphic L-series with the property that there exists some quadratic twist such that the functional equation of the twisted L-series has a negative sign. Then there must exist infinitely many distinct quadratic twists of this L-series with the property that the twisted L-series has a simple zero at the center of the critical strip.<|endoftext|> TITLE: Expected value of the spectral radius of a random nonnegative matrix QUESTION [5 upvotes]: I have two questions, the second of which is related to this question posed by Denis Serre. Let $X$ be a random variable and suppose that $Y=|X|$ (e.g., $Y$ could be the folded normal distribution). If $M$ is an $n$-by-$n$ random matrix with iid entries from $Y$, then: Is the method above an appropriate way to generate a random nonnegative matrix? If the answer to the first question is 'yes', what is the expected value of the spectral radius of $M$? If this is not available, is there an asymptotic description of the spectral radius as $n \longrightarrow \infty$? Thanks in advance for your kind assistance. REPLY [5 votes]: To amplify on Brendan's answer: You are dealing with a matrix of zero mean iid's plus a rank one perturbation: $M=W+ A$ where $A=c11^T$ and $c=EY$. The spectral norm of $W$ is asymptotically $2\sqrt{n} \sigma$ where $\sigma^2=\mbox{ Var}(Y)$, see Geman's 1980 paper (at least when $Y$ has good moment growth). However, the rank one perturbation has norm $cn$, as can be seen by multiplication with the vector $n^{-1/2} (1,\ldots,1)$. So the norm of $M$ is about $cn$ with error at most $2\sqrt{n}\sigma$, with high probability. If $Y$ has heavy tail, everything changes - the norm may be dominated by the maximal entry of $M$. This is the case if the entries are for example $\alpha$-stable with $\alpha<2$; In particular, for one sided Cauchy the norm could be as large as $n^2$. Edit: The answer above deals with the spectral norm, not the spectral radius. However, in the case of good moment bounds of the entries, it also applies to the spectral radius. This is due to separation of eigenvalues of the matrix $A$, and in particular the fact that the top eigenvalue of $A$ (which equals $cn$) is separated from other eigenvalues of $A$ $(=0)$ by more than twice the norm of $W$. See e.g. Theorem 2 of Schonhage's paper http://www.sciencedirect.com/science/article/pii/002437957990154X (with k=m=n) and references there to earlier work.<|endoftext|> TITLE: A curious sin-integral QUESTION [17 upvotes]: While contending with a certain Fourier series, I stumbled on an incredibly simple evaluation (numerically) of a slightly complicated-looking sin-integral. So, I wish ask: Question. Is this really true? If so, any proof? $$I:=\int_0^{\frac{\pi}2}\frac{\sin x}{1+\sqrt{\sin 2x}}\,dx=\frac{\pi}2-1.$$ ADDED. I'm an experimentalist and I find many many results. Some I could find being discovered earlier after checking the literature. For others, either I don't find them easily or I might be tired of looking and hope someone else points them out to me. I'm mostly interested in sharing and having fun, not seeking recognition of any sort. However, one thing is for sure: I don't give oxygen to rude comments. REPLY [11 votes]: Here is another approach: $$\begin{align}2I&=\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{\sin 2t}}\,dt\\&=\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{1-(\sin t-\cos t)^2}}\,dt\\& =\underbrace{ \int_{-\pi/2}^{\pi/2}\frac{\cos u}{1+\cos u}\,du}_{\sin t-\cos t=\sin u }\\&=2\int_0^{\pi/2}\left(1-\dfrac{1}{1+\cos u}\right)\,du\end{align}$$ $$\boxed{I=\frac{\pi}{2}-1}$$<|endoftext|> TITLE: Is $TS^n$ diffeomorphic to an open subset of $\mathbb{R}^{2n}$ QUESTION [5 upvotes]: For what values of $n \neq 1,3,7$ is the tangent bundle $TS^n$ of the $n$-sphere diffeomorphic to an open subset of $\mathbb{R}^{2n}$? REPLY [3 votes]: There are no sphere's with non-trivial normal bundle in that dimension. As far as I know, this is originally a theorem of Massey. See http://www.ams.org/journals/proc/1959-010-06/S0002-9939-1959-0109351-8/S0002-9939-1959-0109351-8.pdf for details. REPLY [3 votes]: Take $n$ odd; this is never possible for $n$ even. Suppose the unit disc bundle of $TS^{n}$ embeds in $S^{2n}$. One may calculate the relative homology of its complement (relative to the boundary) to be supported in degrees $n+1$ and $2n$, and so after appropriate handle cancellation can be obtained by attaching precisely one handle in each of those degrees. But where are we attaching the $(n+1)$-handle on $T^1 S^n$? Its attaching sphere homologous to a section of the bundle; I claim that any sphere in that homology class is isotopic to a section. This is at least true in a sufficiently stable range $(n \geq 6)$ so that homotopy classes of $n$-spheres only contain a single isotopy type. But you can identify the normal bundle of the section with the subbundle of $TS^n$ your section splits off; if this is trivial, then your tangent bundle itself must have been trivial, so $n = 1, 3, 7$ (or I suppose $5$, because I don't know whether there are some extra isotopy classes I don't want for some reason).<|endoftext|> TITLE: Random N-body problem QUESTION [13 upvotes]: Suppose there are $N$ unit-mass particles whose initial positions are uniformly distributed in a unit-radius disk. Each particle is assigned a randomly oriented initial velocity vector $v_i$ of length $1$ (green below). Added: Robert Israel's incisive comment suggests that I should also stipulate that $\sum_i v_i = 0$. Then the particles act upon one another via inverse-square gravity.                 Dots show initial positions inside unit disk. Green vectors: initial velocity; red vectors: final velocity. Q. What is the probability that $k \ge 1$ of the $N$ particles remain within a disk of some radius $R \ge 1$ forever? In the illustration above, $N=8$ and $R=3$. (But I do not trust my crude simulations.) I am wondering if the answer is: zero, independent of $k$ and $R$ and the gravitational constant? Perhaps the answer differs for points in $\mathbb{R}^2$ vs. points in $\mathbb{R}^3$ (or $\mathbb{R}^d$, $d > 3$)? REPLY [6 votes]: What is the probability that $k \ge 1$ of the $N$ particles remain within a disk of some radius $R \ge 1$ forever? If the gravitational constant is small enough, then the probability is definitely less than one, since it's easy for one of the particles to be initially headed away from the others at a high enough speed to escape. For $N\le 3$, the probability appears to be nonzero given a high enough gravitational constant. For $N=1$, the probability is 1. For $N=2$, we have Keplerian orbits. For $N=3$, there are figure-eight configurations known as 3-body choreographies that are believed to be stable in the KAM sense, and there seem to be solutions of this type for a variety of potentials, including $-1/r$. Stability against small perturbations means that you have bound states that occupy a region of phase space with nonvanishing volume, which implies nonzero probability. For $N\ge4$, this appears to be an open problem. In general, we expect such systems to be unbound for thermodynamic reasons. This is because for large $N$ the probability of a given state for a particular particle falls off like $e^{-E/T}$, where $E$ is the kinetic energy and $T$ is the temperature. Since there is no bound on $E$, any system like this is generically expected to evaporate its particles off into the surrounding space. This is really just a phase space argument. There is infinite phase space out there at large distances. Such evaporation is in fact observed in galaxies and globular clusters, where we see stars shooting off with anomalously high velocities. So all such systems should be considered guilty until proven innocent, i.e., we expect them to be unbound by default unless we have some tricky way of constructing a specific example and proving that it's bound. (Numerical simulation doesn't work, because these systems are normally chaotic.) To prove that the probability of being bound is nonzero, we also need to prove that it remains bound under a small perturbation. For $N\ge 4$, I did not come across any mention of any examples that are known to be stable, but I could certainly be missing relevant work. By the way, physically the "right" way to state the problem in two dimensions is probably to use a $1/r$ force, not a $1/r^2$ force, since we expect Gauss's law to hold for fundamental reasons. Gerhard Paseman says: I bet cosmologists thought about this problem in considering origins of the universe. This would be more about the far future than the distant past. The relevant tool is general relativity, not Newtonian mechanics. The thermodynamic equilibrium state of a gravitationally interacting system according to classical GR is basically a black hole, although it can be more complicated than that depending on the large-scale geometry and topology. You don't get stable orbiting states with a purely gravitational interaction, because gravitational radiation sucks energy away.<|endoftext|> TITLE: Reference request: Heyting algebra structure on Catalan numbers QUESTION [12 upvotes]: I've noticed that for every natural number $n\in\mathbb{N}$, there is a finite Heyting algebra with cardinality $C(n)$, where $C(n)$ is the $n$th Catalan number, $$1,1,2,5,14,42,132,\ldots$$ I'm looking for a reference, if this fact is known (to be known). Below I will explain where the Heyting algebra structure comes from, in case it helps. When $n=0$ or $n=1$, we have $C(n)=1$, and there is a unique Heyting algebra structure on a set with one element, so suppose $n\geq 2$. For any $m\in\mathbb{N}$, let $[m]:=\{0,1,\ldots,m\}$ and for any $0\leq a\leq b\leq m$, write $[a,b]$ for the subinterval $\{a,a+1,\ldots,b\}\subseteq[m]$. These subintervals form a poset, which we consider as a topological space with the Alexandrov topology: points are subintervals $[a,b]$ and open sets are down-closed subsets. Write $\Omega[m]$ for the poset of open sets in this space, so it has the structure of a Heyting algebra. It remains to show that the cardinality of $\Omega[m]$ is $C(m+2)$. It is well-known that the Catalan number $C(n)$ counts the Dyck paths of length $2n$. These are paths in a triangle of dots (see below for $n=5$), starting at the southwest point, ending at the northeast point, where each edge in the path moves one unit either northward or eastward. Position the elments of $[m]$ in the $(m+2)$-triangle, as shown here in the case $m=3$: $$ \begin{array}{ccccccccccc} \bullet&&\bullet&&\bullet&&\bullet&&\bullet&&\bullet\\ &&&&&&&3\\ \bullet&&\bullet&&\bullet&&\bullet&&\bullet\\ &&&&&2\\ \bullet&&\bullet&&\bullet&&\bullet\\ &&&1\\ \bullet&&\bullet&&\bullet\\ &0\\ \bullet&&\bullet\\ \\ \bullet\\\\ \end{array} $$ In this setup, a Dyck path $p$ of length $m+2$ can be identified with a downclosed subset, $S(p)\in\Omega[m]$. For example, the Dyck path $p_0$ shown below $$ \begin{array}{ccccccccccc} \bullet&&\bullet&&\bullet&&\bullet&-&\bullet&-&\bullet\\ &&&&&&|&3\\ \bullet&&\bullet&-&\bullet&-&\bullet&&\bullet\\ &&|&&&2\\ \bullet&&\bullet&&\bullet&&\bullet\\ &&|&1\\ \bullet&&\bullet&&\bullet\\ &0&|\\ \bullet&-&\bullet\\ |\\ \bullet\\\\ \end{array} $$ represents the set $S(p_0)=\mathord{\downarrow}[1,2]\cup\mathord{\downarrow}[3]$. In fact, all these Heyting algebras $\Omega[m]$ fit together in a single topos, as we now explain. Consider the additive monoid of natural numbers as a category $BN$ with one object. Let $\mathbf{Int}:=\mathrm{Tw}(BN)$ be the twisted arrow category, and consider the presheaf topos $\mathrm{Psh}(\mathbf{Int})$. The subobject classifier for this topos is a functor $$\Omega'\colon\mathbf{Int}^\mathrm{op}\to\mathbf{Set}.$$ so for each object $n\in\mathbb{N}=\mathrm{Ob}(\mathbf{Int})$, we have a set $\Omega'(n)$. Moreover this set carries the structure of a Heyting algebra. Finally, $\Omega'(n)$ has a well-known description in terms of sieves, i.e. subfunctors of the representable functor $\mathbf{Int}(-,n)$. Unwrapping the definition, these are exactly the open sets of $[n]$. In other words, we have a bijection $\Omega'(n)\cong\Omega[n]$. REPLY [7 votes]: I found a reference: "Dyck algebras, interval temporal logic and posets of intervals", which discusses these Heyting algebras (though not from a topos-theoretic perspective).<|endoftext|> TITLE: Upward Löwenheim–Skolem theorem for well-ordered models with/without measurable cardinals QUESTION [6 upvotes]: Consider a complete first order theory $T$ whose language contains a binary predicate $\leq$. Assume that $T$ has an uncountable model that is well-ordered by $\leq$ so that this question isn't stupid and assume for simplicity that $T$ is countable. If we want to restrict our attention to models $\mathcal{M}$ of $T$ for which $\leq$ actually well-orders $M$ we immediately get a kind of downward Löwenheim–Skolem theorem since any sub-order of a well-order is automatically well-ordered. The naive analog of the upward Löwenheim–Skolem theorem is false because the only well-ordered model of $\text{Th}(\mathbb{N})$ is the standard one, but there is still the possibility of analogous statements requiring the existence of uncountable well-ordered models. Ultraproducts of well-ordered sets with regards to countably complete ultrafilters are well-ordered, so assuming that they exist (i.e. assuming the existence of a measurable cardinal) and assuming $T$ has a well-ordered model whose cardinality can be increased by a countably complete ultrapower, then we get a strictly larger well-ordered model. (I assume that this works for all uncountable cardinals or maybe for all cardinals with uncountable cofinality?) It doesn't necessarily follow that we have arbitrarily large well-ordered models because unions of elementary chains of well-ordered sets aren't necessarily well-ordered, unless countably complete ultrapowers of ordinals are always end-extensions, which I don't know one way or the other. There are a lot of questions: Is there a sufficient condition for the existence of arbitrarily large well-ordered models of $T$ (assuming the existence of a measurable cardinal if necessary)? Is there a way of doing it without measurable cardinals? Is it possible to characterize which order types well-ordered models of $T$ can be? When do models of $T$ have proper elementary end extensions? (And incidentally when do well-ordered models of $T$ have initial segments that are elementary sub-structures?) REPLY [8 votes]: A very nice collection of questions. Here are a few things one can say to get started. If $\kappa$ is a measurable cardinal and $T$ has a well-ordered model of size at least $\kappa$, then it has arbitrarily large well-ordered models. To see this, suppose that $M$ is a model of $T$ in which $\leq$ is a well-order and $M$ has size at least $\kappa$. Since $\kappa$ is measurable, we may by iterating the ultrapower maps find an elementary embedding $j:V\to N$ into a transitive class $N$ with critical point $\kappa$ and $j(\kappa)$ as large as desired. So $j(M)$ is a model of $T$, with the order $j(\leq)$ of size at least $j(\kappa)$ and a well-order in $N$ and hence also still a well-order in $V$. One can do it with less than a measurable. Specifically, if $\kappa$ is merely an unfoldable cardinal (this is consistent with $V=L$, so much smaller than measurable, but above indescribable), and $T$ has a well-ordered model of size at least $\kappa$, then it has arbitrarily large well-ordered models. To see this, note first that a downward Löwenheim-Skolem argument shows that $M$ has well-ordered models of size exactly $\kappa$. Now, we can place this model into a transitive model $N$ of size $\kappa$, and then apply unfoldability to get elementary embeddings $j:N\to \bar N$ with $j(\kappa)$ as large as desired. The same reasoning as before shows that $j(M)$ is a well-ordered model of large size. You can get a kind of converse from the same idea, using the extension property characterization of unfoldability, if one allows the language to become larger. Theorem. A cardinal $\kappa$ is unfoldable if and only if every theory $T$ of size at most $\kappa$ with a model of size at least $\kappa$ in which $\leq$ is a well-order, has well-ordered models of arbitrarily large size. The idea is to write down the diagram of $\langle V_\kappa,\in,A,\leq\rangle$, where $\leq$ is a well-order of $V_\kappa$ and $A\subset V_\kappa$, and then get arbitrarily large extensions $\langle N,\in,A^*,\leq^*\rangle$, which will give you the extension version of the unfoldability property. It seems to me that your questions are very closely related to how one often thinks about unfoldable cardinals. If you are only seeking end-extensions, rather than arbitrarily large extensions, then it is weak compactness, a weaker large cardinal, that you will want. An inaccessible cardinal $\kappa$ is weakly compact if and only if for every set $A\subset V_\kappa$, there is an elementary extension of $\langle V_\kappa,\in,A\rangle$ to a taller well-founded model $\langle M,\in,A^*\rangle$.<|endoftext|> TITLE: Maximal order of elements in SL(n,q) QUESTION [9 upvotes]: The maximal order of an element of $\mathrm{GL}(n,\mathbb{F}_q)$ is $q^n-1$, where the characteristic of $\mathbb{F}_q$ is odd $p$. See here for a nice proof that uses the Cayley-Hamilton Theorem. However, for $\mathrm{SL}(2,\mathbb{F}_q)$ the maximal order is $2p$ if $q=p$ and $q+1$ otherwise. To see this, note that over $\mathbb{F}_{q^2}$ any such matrix $A$ is conjugate to an upper-triangular matrix. If it is not diagonal then its eigenvalues are repeated and $\pm 1$; thus its order is bounded by $2p$. If it is diagonal then the eigenvalues are in $\mathbb{F}_q$ or properly in the degree two extension. If the former then the order divides $|\mathbb{F}_q^*|=q-1$, and if the latter then $x\mapsto x^q$ generates the Galois group of the extension showing the eigenvalue $a$ satisfies $a^q=a^{-1}\implies a^{q+1}=1 \implies |a|{\large \mid} q+1$. I did some searching on MO and other places online but did not find a generalization of this for $n\geq 3$. Is there a nice formula for the maximal order of an element in $\mathrm{SL}(n,\mathbb{F}_q)$? If so, what is the proof or reference? Remark: I would also be interested in the answer to the same question for other finite groups of Lie type, as well as the even characteristic case. REPLY [9 votes]: Yes. It is shown in the paper Darafsheh, M.R., Order of elements in the groups related to the general linear group., Finite Fields Appl. 11, No. 4, 738-747 (2005). ZBL1147.20043. (Theorem 1) that the maximal order is $$\frac{q^n -1}{q-1},$$ except in the case $SL(2, p),$ where the maximum is $2p.$<|endoftext|> TITLE: When minimum of two supporting functionals of convex bodies is convex? QUESTION [5 upvotes]: For a convex compact set $K\subset \mathbb{R}^n$ let us denote by $h_K$ its supporting functional $$h_K(\xi):=\sup_{x\in K}\langle\xi,x\rangle.$$ Thus $h_K\colon \mathbb{R}^n\to \mathbb{R}$ is a convex function. Let $A,B\subset \mathbb{R}^n$ be two convex compact sets. It is well known (and easy to see) that if the union $A\cup B$ is convex then $\min\{h_A,h_B\}$ is convex; in fact in this case $\min\{h_A,h_B\}=h_{A\cap B}$. Is the converse true? Namely assume that $\min\{h_A,h_B\}$ is convex. Does it follow that $A\cup B$ is convex? REPLY [10 votes]: Yes. At first, if $h=\min(h_A,h_B)$ is convex (note that it is also 1-homogeneous), it is a support function of the body $C:=\{x:\forall\xi\in \mathbb{R}^n,\langle \xi,x\rangle\leqslant h(\xi)\}$. Next, $C=A\cap B$, since the inequality $\langle \xi,x\rangle\leqslant h(\xi)$ is equivalent to a system of two inequalities $\langle \xi,x\rangle\leqslant h_A(\xi)$, $\langle \xi,x\rangle\leqslant h_B(\xi)$. Now assume that $A\cup B$ is not convex. It means that there exist $a\in A$, $b\in B$ such that the segment between $a$ and $b$ is not covered by $A\cup B$. Look at a line between $a$ and $b$, it intersects $A$ by a segment $[a_1,a_2]$ and $B$ by a segment $[b_1,b_2]$, we may suppose that $a_1h_C(\xi)$, a contradiction.<|endoftext|> TITLE: Why is Khovanov homology considered a 'categorification'? QUESTION [8 upvotes]: I understand that the Euler characteristic of Khovanov homology is the Jones polynomial. But in what sense does this give category theory structure to the Jones polynomial, i.e., what are the objects and morphisms? REPLY [9 votes]: The idea is that Khovanov homology is a functor out of a category of tangle cobordisms. Bar-Natan's notes are a great reference for this. Khovanov (2002) sets this up in a "tangle 2-category:" the objects are sets of even numbers of oriented points. The 1-morphisms are planar diagrams of cobordisms between them (so that which line in a crossing is on top is important). Hence a 1-morphism is a (diagram of a) tangle, and 2-morphisms are diagrams of oriented tangle cobordisms. Khovanov constructed Khovanov homology as a 2-functor from this category to a category built out of chain complexes of abelian groups. There are subtleties to this setup, however: Jacobsson (2003) showed that it's not functorial when generalized to $\mathbb Z[c]$-modules. However, this is not why people say Khovanov homology is the categorification of the Jones polynomial. Mike Shulman's comment is right: categorification doesn't just mean "turn things into categories," but is a vaguely defined process of adding categorical structure to things: replacing sets with vector spaces or abelian groups; functions with functors or sheaves; rings with tensor categories; and so forth. Replacing a polynomial with a bigraded abelian group fits into this paradigm, even in the absence of an explicit category or functoriality. The many answers to this question are great for explaining how to think about categorification.<|endoftext|> TITLE: Equivariant Riemann-Hurwitz QUESTION [10 upvotes]: The Riemann-Hurwitz formula starts with a genus $g$ algebraic curve $Y$ and a ramified cover $\pi\colon X\to Y$ of degree $N$, with ramification indices $e_P$ and computes invariants of $X$, such as the genus, or more simply the Euler characteristic: $$\chi(X)=N\chi(Y)-\sum_P (e_P-1)$$ This computes the Euler characteristic as a number. Categorifying to the homology suggests generalizations. If the cover is Galois, the quotient by a finite group $G$, then the homology carries an action by that group and I would like to know the representation. An analogue of the Riemann-Hurwitz formula holds in the representation ring $RG$. The number $N=|G|$ is replaced by the regular representation. Thus the first homology is $2(g-1)$ copies of the regular representation, plus a couple more trivial representations, to account for $H_0$ and $H_2$, plus some additional representations from the ramifications. The typical way I have seen this proved is to form an $G$-stable cellular decomposition of $X$ by lifting a cellular decomposition of $Y$ with a vertex for each ramification point. Assuming that there are $r>0$ ramification points, there are $r$ vertices, $(r-1)+2g$ edges, and one face. The edges and faces contribute regular representations, while a vertex with isotropy $I$ contributes $\mathbb C[G/I]$, which is smaller than $\mathbb C[G]$. The chain complex of $X$ is: $\bigoplus_P\mathbb C[G/I_P]\leftarrow \mathbb C[G]^{2g+r-1}\leftarrow \mathbb C[G]$. Together $C_0$ and $C_2$ are generated by $r+1$ elements, so they partially cancel at most $r+1$ regular representations, leaving $2g-2$ regular representations, plus more representations parameterized by the ramifications (specifically $\mathbb C[G]-\mathbb C[G/I_P])$. But what I really want is to understand the combination of the representation and the intersection form, equivalently to understand a Lagrangian subspace as a representation. This depends only on the topology, but it is convenient to exploit the complex structure of the curve, because it provides Lagrangian subspaces $H^0(X;\Omega^1)$ and $H^1(X;\mathcal O)$. I think that there should be a Riemann-Hurwitz formula that says that each has $g-1$ copies of the regular representations, plus a trivial representation, plus additional contributions from the ramification $R$. Is this true? $$H^0(X;\Omega^1)\overset{?}{=}\mathbb C\oplus \mathbb C[G]^{g-1}\oplus ?(R,T_R)$$ In the unramified case, $\pi_*\mathcal O$ is a flat vector bundle, the one corresponding to the regular representation. Note that the isotypic decomposition of the this vector bundle induces the isotypic decomposition of its cohomology. Each of the of the isotypic bundles is flat, so it has degree zero. So each has Euler characteristic equal to $1-g$ times its dimension. So each to the Euler characteristic $1-g$ times its corresponding representation. We started with the regular representation, broke it into pieces, multiplied each piece by $1-g$ and added them together again. Thus $\chi(X)=(1-g)\mathbb C[G]$; or $H^1(X;\mathcal O)=H^1(Y;\pi_*\mathcal O)=\mathbb C[G]^{g-1}\oplus\mathbb C$. But in the ramified case, I don’t see how to exploit flatness. There are two separate questions here. One question is what is the value of $?(R,T_R)$, as a function of the ramification divisor $R$ and representation of the isotropy on the tangent spaces $T_R$. I think that the Woods Hole / Holomorphic Lefschetz / Atiyah-Bott formula says that there is such an equivariant Riemann-Hurwitz formula depending only on those two ingredients; and that $?(R,T_R)$ is linear in the divisor. And it provides a way to check such a formula, if one could write it down. In some sense it even gives a formula. But the other question, the one that really interests me, is whether the correction term is positive, whether there really are regular representations, regardless of anything else. I don’t think that the contributions from individual fixed points are all positive, or even real, so all we can ask is that sum over all ramification points is positive. Is it plausible to prove this without understanding the ramification term? For the earlier theorem, the one identifying the representation, but not the symplectic form, is there any way to prove such a weak version, finding $2g-2$ copies of the regular representation in $H^1(X)$, without going so far as to compute the correction from the ramification? The context for my interest in the presence of the regular representations is that Andy Putman and I made a conjecture about the monodromy of symmetric curves acting on the homology of $X$. We conjectured that if $Y$ has genus at least 2 that the homology of $X$ has no vectors fixed by any finite index subgroup of the monodromy of the space of curves with the same symmetry. An upper bound on the monodromy of symmetric curves is the monodromy of symmetric abelian varieties. Indeed, I think it is a standard conjecture that they have the same Zariski closure. The bigger group is an arithmetic group easily described in terms of the representation and symplectic form. The cotangent space to the space of symmetric abelian varieties is $(S^2\;H^0(X;\Omega^1))^F$. Thus a symmetric (real) representation is never fixed; and a skew (quaternionic) representation is fixed only if it has multiplicity 1. Those cases don’t depend on the symplectic form (which doesn’t vary), only multiplicity, so they can be dealt with by the first equivariant Riemann-Hurwitz formula proved by cellular decomposition. The hard part is a non-self-dual (complex) representation. They have several symplectic forms, depending on how the representation and its dual are distributed between $H^{1,0}$ and $H^{0,1}$. A complex representation is fixed if it appears in $H^{1,0}$ and its dual does not appear there, only in $H^{0,1}$. But if the regular representation appears in $H^{1,0}$, it includes both representations, so it handles this case. REPLY [6 votes]: This is a well known and well understood problem when the base field is $\mathbb C$. It was first studied by Chevalley and Weil (almost a century ago !) who were interested in modular curves (what else ?). For a modern account, see Kani, Ernst : The Galois-module structure of the space of holomorphic differentials of a curve. J. reine angew. Math. 367 (1986), 187-206 . Nakajima, Shoichi : Galois module structure of cohomology groups for tamely ramified coverings of algebraic varieties. J. Number Theory 22, 115-123 (1986). Here is a brief report. Warning : this problem is well understood for a tame action only, over an algebraically closed field $k$, in particular over $\mathbb C$, but it is still open, as far as I know, in the wild action case. So I assume tameness. At each point $P\in X$, define the ramification character $\psi_P$ as the character given by the action of the cyclic stabilizer $G_P$ on the cotangent space $\mathcal m_P/\mathcal m_P^2$. Define the ramification module by $$\Gamma _{G}=\sum_{P\in X}{\rm Ind}_{G_{P}}^{G}\sum_{l=1}^{e_{P}-1}l\psi _{P}^{l}$$ where $e_P$ is the ramification index at $P$. Then the equivariant Hurwitz formula says : there exists a unique $k\left[ G\right]$ -module $\widetilde{\Gamma} _{G}$ such that $$\widetilde{\Gamma}_{G}^{\oplus \#G}=\Gamma _{G}$$ and in the Grothendieck ring $R_{k}(G)$ holds the following equality : $$ \chi(G,\mathcal{O}_{X})=\chi(\mathcal{O}_{Y})\cdot \left[ k \left[ G\right] \right] -[ \widetilde{\Gamma} _{G}] $$ where, as you can assume, $\chi(G,\cdot)$ is the equivariant Euler characteristic, and $\chi(\cdot)$ is the ordinary Euler characteristic. From this, and Serre duality, which is of course equivariant, you get the structure of $H^0(X,\Omega^1_X)$. The proof of the equivariant Hurwitz formula is not hard. Since it is an equality between characters, it is enough to prove it when you evaluate at each $g\in G$. For $g=e$, this is the usual Hurwitz formula. For $g\neq e$, one uses a Lefschetz fix point formula for the corresponding automorphism of $X$.<|endoftext|> TITLE: iterated harmonic numbers vs Riemann zeta QUESTION [13 upvotes]: Define the $m$-th iterated harmonic sums in the manner: $\bar{H}_0(n):=1$ and for $m\geq1$ by $$\bar{H}_m(n):=\sum_{k=1}^n\frac{\bar{H}_{m-1}(k)}k.$$ For example, $\bar{H}_1(n)=\sum_{k=1}^n\frac1k$ are the familiar harmonic numbers. Euler proved that $$\frac12\sum_{n\geq1}\frac{\bar{H}_1(n)}{n^2}=\zeta(3).$$ Hoping for a natural generalization, I ask: Question 1. Is this true? If so, any proof? $$\frac1{m+1}\sum_{n\geq1}\frac{\bar{H}_m(n)}{n^2}=\zeta(m+2).$$ Of course, this works for $m=0$ as well: $\frac1{0+1}\sum_{n\geq1}\frac{\bar{H}_0(n)}{n^2}=\zeta(2)$. Question 2. This might be of auxiliary interest. Any proof? $$\bar{H}_m(n)=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}\binom{n}k.$$ REPLY [3 votes]: Unless I messed up the notation, it seems that Note 5.3 of this paper Luis A. Medina, Victor H. Moll, and Eric S. Rowland, Iterated primitives of logarithmic powers, International Journal of Number Theory 7 (2011) pp 623–634, doi:10.1142/S179304211100423X, arXiv:0911.1325 (pdf) answers Question 2. It seems quite likely that the OP is well aware of the above paper, given this recent paper: Tewodros Amdeberhan, Christoph Koutschan, Victor H. Moll, Eric S. Rowland, The iterated integrals of $\ln(1 + x^n)$, International Journal of Number Theory 8 (2012) pp 71–94 doi:10.1142/S1793042112500042, arXiv:1012.3429 (.ps file).<|endoftext|> TITLE: Analogies supporting heuristic: Weyl groups = algebraic groups over field with one element? QUESTION [32 upvotes]: There is well-known heuristic that Weyl groups are reductive algebraic groups over "field with one element". Probably the best known analogy supporting that heuristic is the limit $q\to1$ for number of elements in $G(F_q)$ - for appropriate "m" it holds: $$ \lim_{q\to1} \frac { |G(F_q) | } { (q-1)^m} = |Weyl~Group~of~G| $$ For example: $|GL(n,F_q)|= [n]_q! (q-1)^{n}q^{n(n-1)/2} $ so divided by $(q-1)^n$ one gets $[n]_q! q^{n(n-1)/2} $ and at the limit $q\to1$, one gets $n!$ which is the size of $S_n$ (Weyl group for GL(n)). (For other groups see Lorscheid 2009 page 2 formula 1). Question What are the other analogies supporting heuristics: Weyl groups = algebraic groups over field with one element ? Subquestion once googling papers on F_1, I have seen quite an interesting analogy from representation theory point of view - it was some fact about induction from diagonal subgroups of symmetric groups $S_{d_1}\times ... \times S_{d_k} \subset S_n$ where $\sum d_i = n$ and similar fact for $GL(n,F_q)$ which was due to Steinberg or Springer or Carter (cannot remember). But I cannot google it again and cannot remember the details :( (Tried quite a lot - I was sure it was on the first or second page of Soule's paper on F_1 - but it is not there, neither many other papers). Knowing that total element count is okay, we may ask about counting elements with certain properties - like: m-tuples of commuting elements (MO271752), involutions, elements of order $m$, whatever ... From answer MO272059 one knows that there are certain analogies for such counting, however it seems the limits $q\to1$ are not quite clear. Question 2 Is there any analogy for counting elements with some reasonable conditions ? Hope to see that count for $G(F_q)$ (properly normalized) in the limit $q\to1$ gives answer for Weyl group. REPLY [5 votes]: [The following comment is too long for the comment box.] On the subquestion: Zelevinsky's Representations of finite classical groups - a Hopf algebra approach (LNM) may be relevant to what you're thinking of. Zelevinsky builds two Hopf algebras: the first coming from induction and restriction of (complex) representations of the symmetric groups along the inclusion $S_n\times S_m\to S_{n+m}$, the second using parabolic induction and restriction (again, of complex representations) for finite general linear groups along the inclusion $GL_n(\mathbb F_q)\times GL_m(\mathbb F_q)\to GL_{n+m}(\mathbb F_q)$. Zelevinsky shows that the second algebra is a tensor product of copies of the first, with one copy for each pair $(n,\pi)$ where $\pi$ is a cuspidal representation of $GL_n(\mathbb F_q)$. Here's an attempt, possibly completely misguided, to extract from this an analogy that would be relevant to your question. (I don't know how this relates to the existing literature on $\mathbb{F}_1$. Sorry if I am repeating something that is well known.) If we identify $S_n=GL_n(\mathbb F_1)$, Zelevinsky's result might be interpreted metaphorically as saying that ''the only cuspidal representation of $GL_n(\mathbb F_1)$ is the trivial representation of the trivial group''. Since cuspidal representations of $GL_n(\mathbb F_q)$ are associated to characters of anisotropic tori, and since (I suppose?) the ''group of $\mathbb F_1$-points of a torus over $\mathbb F_1$'' is always the trivial group, this makes some kind of sense.<|endoftext|> TITLE: How can I endow a "locally product" CW structure on a vector bundle over a CW complex? QUESTION [6 upvotes]: I asked the same question in math stackexchange: https://math.stackexchange.com/questions/2322883/how-can-i-endow-a-locally-product-cw-structure-on-a-vector-bundle-over-a-cw-co but it seems that it's harder than I thought, so I ask here: I'm now learning characteristic classes, and I need a CW structure on the total space of a vector bundle $E\to B$ where $B$ is a CW complex such that the associated sphere bundle and rectriction over any subcomplex of $B$ are both subcomplexes (this is required in "Algebraic Topology from a Homotopical Viewpoint", page 364). I think this should be something like a locally product structure, but I couldn't figure out how to glue them together. I even doubt that this can be done. REPLY [13 votes]: The authors of this book are attempting to use CW structures to justify certain cohomology isomorphisms, but this seems to be the wrong approach since some of their claims about CW structures are just not true. For example, they say a vector bundle over a CW complex base space has a CW structure such that the complement of the zero section is a subcomplex, but this cannot be true since a subcomplex is always a closed subspace. It seems best not to talk about CW structures on vector bundles and instead prove the cohomology isomorphisms using standard tools such as excision for general topological spaces.<|endoftext|> TITLE: Fourier series of $e^{\cos x}$ QUESTION [6 upvotes]: I need to compute the fourier series of $f(t)=e^{\cos(t)}, 0 \leq t < 2\pi$. The fourier series are defined as $f(t) = \sum_{n=-\infty}^\infty c_n e^{2\pi int/T}$ with $c_n = \frac 1 T \int_0^T e^{-2\pi int/T}f(t) \, dt$. I have tried to do this by using the definition of the $c_n$, but i get stuck when with the integration by parts. I also have tried to use the approach used in of this question but i cannot go forward more than expanding $e^{\cos(t)}$. Could any of you help me with a hint? REPLY [22 votes]: $$\int_0^{2\pi} \exp(int) \exp(\cos(t))\; dt = \int_{-\pi}^{\pi} \cos(n t) \exp(\cos(t))\; dt = 2 \pi I_n(1)$$ where $I_n$ is a modified Bessel function of the first kind and thus $$I_n(1)=\frac12\sum_{k\geq0}\frac1{4^kk!(n+k)!}.$$<|endoftext|> TITLE: Can two-point sets be Borel? QUESTION [18 upvotes]: Recall that a two-point set is a subset of the plane which meets every line in exactly two points. Such a set was first constructed by Mazurkiewicz in 1914. I wonder if the following question of Erdos is solved: Question. Is there a two-point set which is Borel? See On sets which meet each line in exactly two points, where Mauldin says that he “believes” he first heard of the problem from Erdos, who in turn said that it had been around since he (Erdos) was a “baby.” Remarks. A two-point set can not be $F_\sigma$ (by a result of Larman). Also it is known that if a two-point set is analytic, then it is Borel. REPLY [15 votes]: A two-point set cannot be $F_\sigma$, as Mohammad mentions in his question. Also, A two-point set cannot contain a dense $G_\delta$ subset of an arc. This was proved by Gareth Davies in his thesis (Oxford, 2011), but I do not think he ever published this result. To my knowledge, no better results are known in the Borel-sets-shouldn't-work direction. In the other direction, the best known results belong to Arnie Miller. He showed that If $V=L$ then there is a co-analytic two-point set. See this paper of his from 1989 where he writes about this question and other such things, like nicely-definable Hamel bases and MAD families. Also see this related paper of Zoltan Vidnyansky's, which appeared more recently and extended/streamlined some of Miller's work. In an unpublished manuscript that he made available here on his website, Arnie Miller also showed that It is consistent for the Axiom of Choice to fail badly, but still to have two-point sets (e.g., two-point sets can exist even when there is no well-ordering of the reals). Ben Chad did quite a bit of work about 5-10 years ago in trying to eliminate the Axiom of Choice from constructions of two-point sets as much as possible. Some of this stuff made it into this paper, joint with Robin Knight and Rolf Suabedissen.<|endoftext|> TITLE: Is there a Yoneda structure on $\bf PDer$? QUESTION [8 upvotes]: The title says it all. A Yoneda structure in a 2-category, as defined in [SW], is given by a coherent choice of a 1-cell $y_A : A \to PA$ such that $\text{Lan}_{y_A}F\dashv \text{Lan}_F {y_A}$ for each $F :A\to B$; $F(-)\cong \text{Lift}_{B(F-,=)}y_A$; $\text{Lan}_yy\cong 1_{PA}$ (read as: ``the Yoneda embedding is dense''); $\text{Lan}_{y_AF}y\circ \text{Lan}_Gy_A \cong \text{Lan}_{GF}y_A \; (\cong \text{Lan}_G\text{Lan}_F y_A)$. Of course, the validity of these statements in $\bf CAT$ is a consequence of your favourite way to compute pointwise Kan extensions. I am wondering if the 2-category of prederivators, i.e. strict functors in $[{\bf Cat}°,{\bf CAT}]$, with pseudonatural transformations and modifications as 1- and 2-cells, has a Yoneda structure. [SW] : Street, Ross, and Robert Walters. "Yoneda structures on 2-categories." Journal of Algebra 50.2 (1978): 350-379, doi: 10.1016/0021-8693(78)90160-6. REPLY [12 votes]: This is contained in $\S$6 (Corollary 6.7) of Ross Street's paper Street, Ross. Conspectus of variable categories. J. Pure Appl. Algebra 21 (1981), no. 3, 307--338, doi: 10.1016/0022-4049(81)90021-9. where it is shown more generally that each 2-category of the form $\mathrm{Hom}(\mathcal{C}^\mathrm{op},\mathbf{CAT})$, where $\mathcal{C}$ is a small 2-category, has a Yoneda structure. Note that $\mathrm{Hom}(\mathcal{C}^\mathrm{op},\mathbf{CAT})$ is the 2-category whose objects are the pseudofunctors $\mathcal{C}^\mathrm{op} \to \mathbf{CAT}$, whereas your 2-category of interest is the full sub-2-category of one of these on the strict 2-functors. But as this full sub-2-category is biequivalent to $\mathrm{Hom}(\mathcal{C}^\mathrm{op},\mathbf{CAT})$, the Yoneda structure should transport to one on the full sub-2-category.<|endoftext|> TITLE: A simple colimit in the derived category? QUESTION [5 upvotes]: I have recently come across the following question : Let $X$ be a (bounded below)chain complex in an arbitrary abelian category, and denote ${\sigma_{\leq n}}$ the stupid truncations functors (i.e. it replaces all $X_k$ by 0 for $k > n $). We have obvious (natural) inclusion morphisms : $$ \sigma_{\leq n}X \hookrightarrow \sigma_{\leq n+1}X$$ and it is very easy to see that this form a diagram in the category of chain complexes such that : $$ \mathrm{colim}\ \sigma_{\leq n}X = X.$$ My question is : does this equality still holds in the derived category? I am not talking about homotopy colimit, but standard colimit in the derived category. More precisely, if $\gamma$ is the localization functor, I am asking if $\gamma$ preserves this colimit. Actually, I wasn't even able to answer this question when you replace the derived category by the category of chain complexes quotiented by homotopies of chain complexes. Thus, I would already be happy with an answer to that simpler question. REPLY [6 votes]: No, not in general. For example, let $R=k[x]/(x^2)$ for a field $k$ and let $X$ be the object $$\dots\stackrel{x}{\to}R\stackrel{x}{\to}R\stackrel{x}{\to}R\to0\to0\to\dots$$ of the derived category of $R$-modules, with the last non-zero term in degree zero. To show that $X$ is not the colimit of its truncations $\sigma_{\leq n}X$ it suffices to find a non-zero map $X\to Y$ such that the restriction $\sigma_{\leq n}X\to X\to Y$ to $\sigma_{\leq n}X$ is zero for every $n$. Take $$Y=\bigoplus_n(\sigma_{\leq n}X)[1],$$ (where $[1]$ denotes a shift to the left), and take the chain map $\alpha:X\to Y$ that in degree $n+1$ is just the map given by multiplication by $x$ from the degree $n+1$ term of $X$ to the first non-zero term of $(\sigma_{\leq n}X)[1]$. It's easy to check that the components $X\to(\sigma_{\leq n}X)[1]$ are all homotopic to zero, but only by a chain homotopy that is non-zero in degree zero. Each restriction to a truncation of $X$ maps into a finite subsum of $Y$, and so is homotopic to zero. But $\alpha$ itself is not homotopic to zero (since a chain homotopy would need a map $R\to\bigoplus_nR$ in degree zero that had non-zero component for every $n$).<|endoftext|> TITLE: Trouble with models of PA and ZFC QUESTION [12 upvotes]: I have a big trouble in my mind, here is my false reasoning: The Goodstein's theorem is undecidable in (first order) Peano Arithmetic. There exist a non standard model N of PA where the Goodstein's theorem is false. The Goodstein's theorem is provable in ZFC, using ordinals below $\epsilon_{0}$. Since a statement provable in a theory T is true in all models of T, the Goodstein's theorem must be true in all models of ZFC. (?) The ordinals of a ZFC model are well-founded from the internal point of view of this model. If there exist a model M of ZFC where the model N is the naturals numbers of M. (?) The Goodstein's theorem is false in M by definition of N, but true by 2), we can prove it in M using the $\epsilon_{0}$ of M, like in our universe. Where I have gone wrong? REPLY [7 votes]: The resolution to your confusion is simply that there is no model $M$ of $ZFC$ for which $N$ is the set of natural numbers. Being the set of natural numbers of a model of $ZFC$ is very rare for models of $PA$, because a set of natural numbers for a model of $ZFC$ has to satisfy all kinds of statements that $ZFC$ can prove about natural numbers and $PA$ cannot, whereas in many models of $PA$ some or all of those statements are false. The contradiction that you demonstrated is actually a proof that there is no model of $ZFC$ which has $N$ as its to set of natural numbers.<|endoftext|> TITLE: dg-categories and fully faithful functor QUESTION [6 upvotes]: dg: is for differential graded Suppose that $F: C\rightarrow D$ is a dg-functor between small dg-categories such that: F: Objects of $C$ $\rightarrow$ Objects of $D$ is injective. $Hom_{C}(a,b)\rightarrow Hom_{D}(F(a),F(b))$ induces an isomorphism in homology for any $a, b \in C$ Let $\hat{C}$ be the category of dg-$C$-modules. Is it true that the induced dg-functor $$\hat{F}:\hat{C}\rightarrow \hat{D}$$ is faithfull in the sense that for any $x,y\in \hat{C} $, $Hom_{\hat{C}}(a,b)\rightarrow Hom_{\hat{D}}(\hat{F}(a),\hat{F}(b))$ induces an isomorphism in homology. Edit: I'm also interested in the particular case when $F:C\rightarrow D$ is an embedding of dg-categories e.g. when $C$ is a full dg-subcategory of $D$. REPLY [2 votes]: Let me start with an answer for your last claim, namely, that a fully faithful dg-functor $F \colon \mathcal C \to \mathcal D$ induces a fully faithful dg-functor $\hat{F} \colon \hat{\mathcal C} \to \hat{\mathcal D}$ between the dg-categories of right dg-modules. This is true, and the reason lies in general facts about Kan extensions. In any case, a sketch of proof would be as follows. First, $\hat{F}$ is usually defined as the "induction" dg-functor $\operatorname{Ind}_F$, and in particular $$ \operatorname{Ind}_F(M)(D) = M \otimes_{\mathcal C} \mathcal D(D,F(-)) = \int^C M(C) \otimes\mathcal D(D,F(C)), $$ if you also like coends. To show that $\operatorname{Ind}_F$ is fully faithful, we first recall that it is by definition the left adjoint of the restriction dg-functor $\operatorname{Res}_F \colon \hat{\mathcal D} \to \hat{\mathcal C}$, which maps $N$ to $N \circ F$. There is a unit morphism $$ M \to \operatorname{Res}_F \operatorname{Ind}_F(M), $$ and if we check that this is an isomorphism, then $\operatorname{Ind}_F$ will be fully faithful. But now we can compute: \begin{align*} \operatorname{Res}_F \operatorname{Ind}_F(M)(C) &= M \otimes_{\mathcal C} \mathcal D(F(C),F(-)) \\ &\cong M \otimes_{\mathcal C} \mathcal C(C,-)\\ & \cong M(C), \end{align*} where I used fully faithfulness and for the last step the "co-Yoneda lemma" or "density theorem". Now, your main claim that a quasi-fully faithful dg-functor induces something which is again quasi-fully faithful between the dg-category of dg-modules is a little more nuanced. Working with this "cohomological" assumptions, you'll want derived stuff all around. So, instead of working with dg-modules, it's best to work with h-projective/cofibrant/semi-free dg-modules, namely your favourite version of the "derived dg-category". Let us work with the dg-category of h-projective dg-modules, just to fix ideas. What is true is that if you have a quasi-equivalence $G \colon \mathcal A \to \mathcal B$ between dg-categories, then $\operatorname{Ind}_G$ will induce a quasi-equivalence $$ \operatorname{Ind}_G \colon \operatorname{h-proj}(\mathcal A) \to \operatorname{h-proj}(\mathcal B). $$ This is quite a well-known fact: if you want to check it yourself, you can work instead with semi-free dg-modules and prove that the unit and counit maps are quasi-isomorphism "by hand", or you can use the fact that you have compact generators: a reference is Drinfeld's famous article (Remark 4.3). Now, that was the hard part. If you have just a quasi-fully faithful dg-functor $F \colon \mathcal C \to \mathcal D$ as in your post (you don't even need injectivity on objects, just that you have quasi-isomorphisms on hom-complexes), then clearly you can factor it as a quasi-equivalence and a (strictly) fully faithful dg-functor. So, in the end, I think you'll end up with a quasi-fully faithful dg-functor $\operatorname{h-proj}(\mathcal C) \to \operatorname{h-proj}(\mathcal D)$.<|endoftext|> TITLE: Loday's characterization and enumeration of faces of associahedra (Stasheff polytopes) QUESTION [8 upvotes]: From "The multiple facets of the associahedra" by Loday: Let us consider the formal power series $$f(x) = x+a_1 x^2 +a_2 x^3 + \cdots+ a_n x^{n+1} + \cdots$$ and let $$ g(x) = x+b_1 x^2 + b_2 x^3 + \cdots + b_n x^{n+1} + \cdots$$ be its inverse for composition, that is, $$f(g(x)) = x.$$ The coefficient $b_n$ is a polynomial in $a_1$ to $a_n$. In low dimension, one gets $$b_1 = a_1$$ $$b_2 = 2a_1^2 - a_2$$ $$b_3 = -5a_1^3+5a_1 a_2-a_3$$ $$b_4 = 14a_1^4 -21 a_1^2 a_2 +6a_1 a_3 +3a_2^2 -a_4$$ and more generally $$b_n = \sum (-1)^{\sum n_i} \lambda(n_1,...,n_k) a_1^{n_1} \cdots a_n^{n_k},$$ where the sum is extended to all the k-tuples of integers $(n_1,...,n_k)$ so that $n_1 +2n_2 + \cdots +k n_k =n$. Here the coefficient $\lambda(n_1,...,n_k)$ is the number of cells of the associahedron $K^{n-1} $ that are isomorphic to the cartesian product $(K^0)^{n_1} \times \cdots \times(K^{k-1})^{n_k}$. In other words, for example, $b_4$ is related to a refined f-vector (face-vector) for the 3-D Stasheff polytope, or 3-D associahedron, with 14 vertices (0-D faces), 21 edges (1-D faces), 6 pentagons (2-D faces), 3 rectangles (2-D faces), 1 3-D polytope (3-D faces). Subtracting one from the index of $a_n$, and ignoring the resulting indeterminates with indices with values less than one, allows one to read off the geometry of the associahedron from cartesian products of the lower dimensional associahedra, e.g., $3\: a^2_2$ becomes $3\: a^2_1$, the cartesian product of the 1-D associahedron with itself, which is a tetragon, or square in some reps. Loday further asserts: There exists a short operadic proof of the above formula which explicitly involves the parenthesizings, but it would be interesting to find one which involves the topological structure of the associahedron. QUESTION: What references contain proofs of the above argument relating the monomials of the inversion formula to distinct faces of an associahedron as cartesian products of the lower dimensional associahedra? Similar relationships hold, with a shift in indices, for the permutahedra (see this MSE-Q) and noncrossing partitions (see OEIS A134264). REPLY [3 votes]: I believe that the proof to which Loday is referring is the one that appears for Proposition 13.11.7 in Algebraic Operads. In this book, the coefficients $\lambda(n_1,\dots,n_k)$ are described as the number of planar rooted trees with certain properties. Such trees are in bijection with the Cartesian products of interest. Also, you might find the example following Corollary 9.2 of this paper helpful.<|endoftext|> TITLE: Do the Lebesgue-null sets cover "all the sets can naturally be regarded as sort-of-null sets"? QUESTION [9 upvotes]: Let $F$ be the set of bijective Borel-measurable functions $f \colon [0,1] \to [0,1]$ that preserve the Lebesgue measure. Is it the case that for every non-Lebesgue-measurable set $A \subset [0,1]$, there exists a countable family $\{f_n\}_{n \in \mathbb{N}} \subset F$ such that $\ \bigcup_{n \in \mathbb{N}} f_n(A)\,$ contains a Lebesgue-measurable set of positive measure? If so, then there is no natural way to extend the Lebesgue measure to include more null sets. (Conversely, if not, then it seems reasonable to regard all the counterexemplary sets as "kind-of-null sets".) REPLY [9 votes]: The answer is no, by a construction using the axiom of choice. We shall build a counterexample set $A$ by a transfinite recursive process of length continuum. At each stage, we shall promise that certain elements are in $A$, in order to ensure that $A$ will be non-null, and that other elements are not in $A$, in such a way so as to prevent $\bigcup_n f_n(A)$ from containing a particular positive-measure Borel set. But at each stage, we will have made fewer than continuum many such promises altogether, and this fact will enable the construction to proceed to later stages. To begin the construction, observe that there are continuum many Borel functions and therefore continuum many countable families $\{f_n\}$ of bijective Borel functions that you consider, and also there are continuum many Borel sets. So let us fix a well-ordered enumeration $\langle \{f_n^\alpha\}_n,B_\alpha\rangle$, for $\alpha<\mathfrak{c}$, of all pairs of such objects. At stage $\alpha$, we consider first the possibility that $B_\alpha$ might be a measure-zero Borel set containing the set $A$ we aim to construct. In order to prevent this, if $B_\alpha$ is measure zero, then there must be some $c_\alpha\notin B_\alpha$ about which we have not yet made any promises, and we promise now that $c_\alpha\in A$. This will ensure that $A$ is not contained in this particular Borel measure-zero set, and therefore, since all such Borel measure-zero sets will eventually be considered, it will ensure that $A$ does not have measure zero. Next, still at stage $\alpha$, we consider the possibility that $B_\alpha$ might be a positive-measure set contained in $\bigcup_n f^\alpha_n(A)$. Since we have made so far fewer than continuum many promises about $A$, it follows that we have promised fewer than continuum many elements altogether to be in this union. But if $B_\alpha$ has positive measure, then it must have size continuum, and so there is a real $b_\alpha\in B_\alpha$ about whose pre-images $a_\alpha=(f^\alpha_n)^{-1}(b_\alpha)$ we have not yet made any promises. Let's now promise that none of these particular $a_\alpha$ are in $A$, which is countably many additional promises at this stage. This will ensure that $\bigcup_nf^\alpha_n(A)$ does not contain this particular positive-measure Borel set. By design, the construction ensures that $A$ is not measure zero, yet there is no positive-measure Borel set contained in $\bigcup_n f_n(A)$ for any Borel bijections $f_n$. And so $A$ has the features necessary to be the desired kind of counterexample.<|endoftext|> TITLE: Resolving $\mathbb Z_n$ action on $\mathbb C^2$ QUESTION [5 upvotes]: Consider a diagonal action of $\mathbb Z_n$ on $\mathbb C^2$ generated by $(z_1,z_2)\to (\mu^pz_1,\mu^qz_2)$, with $\mu^n=1$. Question. Is it always possible to find a smooth blow up $X\to \mathbb C^2$ such that the $\mathbb Z_n$-action lifts to $X$ and such that $X/\mathbb Z_n$ is smooth as well? The same question can be asked for action of any finite group $G$ on any smooth variety (but I am especially interested in the above example). REPLY [7 votes]: I think the answer is no. In your notation, take $n=5$, $p=1$ and $q=2$. If you consider the blowup at the origin $X\to \mathbb C^2$, then $\mathbb Z_5$ acts on $X$ with two isolated fixed points: at one of the points the action has $p=q=1$ and at the other the action has $p=2$, $q=4$, which is the same as $p=1$, $q=2$. So the blow up introduces an isolated fixed point of the same type as the one we started with.<|endoftext|> TITLE: Average size of extreme points of convex hull of $N$ points QUESTION [9 upvotes]: Fix $n$ a (small) integer. Let $N$ be a (big) integer. Consider $N$ random points in the $n$-dimensional unit cube $[0, 1]^n$. The $N$ points are independently uniformly distributed. Define $V(N)$ to be the expectation of the number of extreme points of the convex hull of the $N$ points. Question: when $N$ grows to infinity, how fast does $V$ grow? For $n = 1$, one has $V = 2$; For $n = 2$, my intuition suggests $V \simeq N^{1/2}$, but I'm not quite sure; Similarly, for general $n$, my intuition suggests $V \simeq N^{(n - 1)/n}$. Are there known results on this problem or similar problems? REPLY [8 votes]: The result turns out to be very different depending whether you draw your points from a polytope or a smooth body. This paper (Random Points and Lattice Points in Convex Bodies, Bárány in Bull. AMS 2008) contains the result you desire, and many pointers to the (very abundant) literature no this topic. Turns out your intuition is far from the right answer, which as pointed out by Joseph O'Rourke is $$ V \simeq (\log N)^{n-1}$$ for points drawn in a polytope, but it is quite closer to the answer for points drawn from a smooth convex body: $$ V \simeq N^{\frac{n-1}{n+1}} $$ (see Section 9 of the above paper for both results). The current research is more focused about higher-order results (CLT etc.), but the Poissonian case (where the number $N$ is chosen randomly in precisely the way that gives independence to the events "there is a point in $A$" and "there is a point in $B$" whenever $A$ and $B$ are disjoint) is much better understood, as far as I know. REPLY [3 votes]: Grows as $O(\log^{n−1} N)$. Below, your $n$ is H-P's $d$, and your $N$ is H-P's $n$. So: $O(\log^{d−1} n)$, or, to avoid notational ambiguity: $O((\log n)^{d-1})$ Har-Peled, Sariel. "On the expected complexity of random convex hulls." arXiv:1111.5340 (2011). "We prove that the expected number of points that lie on the boundary of the quadrant hull of $n$ points, chosen uniformly and independently from the axis-parallel unit hypercube in $\mathbb{R}^d$, is $O(\log^{d−1} n)$. This readily imply[sic] $O(\log^{d−1} n)$ bound on the expected number of maxima and the expected number of vertices of the convex hull of such a point set. Those bounds are known [BKST78], but we believe the new proof is simpler and more intuitive." Here is the [BKST78] reference: J. L. Bentley, H. T. Kung, M. Schkolnick, and C. D. Thompson. On the average number of maxima in a set of vectors and applications. Journal of the ACM, 25:536–543, 1978.<|endoftext|> TITLE: Can Alexandrov surfaces of CAT(0) type be approximated by CAT(0) polyhedra? QUESTION [8 upvotes]: The theory of such surfaces goes back to the book by Alexandrov and Zalgaller (1967 English translation) and from a more analytic viewpoint, work by Reshetnyak where everything is translated into Radon measures. It is proved in AZ that any CAT(0) surface can be approximated by a polyhedral surface also with finite total curvature. Can this be done while respecting the CAT(0) condition? REPLY [9 votes]: I am sure it done somewhere, but I do not know a ref. I did something like this in my "Metric minimizing surfaces", but do not want to claim originality. You may fix a finite set of points draw all the geodesics between them. Together these geodesics form a finite graph; they cut finitenumber of discs from your surface. Exchange each disc by the convex plane polygon provided by Reshetnyak's majorization theorem and you get an approximation. Note that the angles of each polygon can not be smaller that the corresponding angle in the surface. Therefore the angle around each point in the approximating surface is at least $2{\cdot}\pi$; it follows that, the approximating surface in CAT[0]. By Reshetnyak's theorem, the constructed polyhedral surface admits a short map to the original one which is isometric on the finite set you started with. From this the convergence follows, assuming you choose right notion of convergence for the surfaces.<|endoftext|> TITLE: "Reversion" of class $J(\theta)$ interpolation property for Besov spaces QUESTION [7 upvotes]: In (function space) interpolation theory, a Banach space $E$ is of class $J(\theta)$ (for $0 < \theta < 1$) if $$X \cap Y \subseteq E \subseteq X+Y,$$ where $(X,Y)$ are Banach spaces and form an interpolation couple, and there exists a constant $C>0$ such that $$\|x\|_E \leq C \|x\|_X^{1-\theta} \|x\|_Y^\theta \quad \text{for all}~x \in X\cap Y.$$ It is known that if $E$ is of class $J(\theta)$, then $(X,Y)_{\theta,1} \hookrightarrow E$, and every real or complex interpolation space of parameter $\theta$ is of class $J(\theta)$. Now I am interested in a sort of reversion of this property in the case of Besov spaces $B^\theta_{p,1}(\Omega)$, where $X = L^p(\Omega)$ and $Y = W^{1,p}(\Omega)$ for $1 < p <\infty$ and $\Omega$ being either $\mathbb{R}^d$ or a bounded extension domain in the $L^p$- and $W^{1,p}$-sense. More specifically: Question: Let $(f_n) \subset W^{1,p}(\Omega)$ be a sequence which converges to zero in $$(L^p(\Omega),W^{1,p}(\Omega))_{\theta,1} = B^\theta_{p,1}(\Omega).$$ Does this imply that $$\|f_n\|_{L^p(\Omega)}^{1-\theta}\|f_n\|_{W^{1,p}(\Omega)}^\theta \to 0 \quad \text{as}~n\to\infty~\text{?}$$ The motivation to study this question is from Corollary 2 in Chapter 1.4.7 in Maz'ja's Sobolev spaces where it is established that $$\|\operatorname{tr} f\|_{L^q(\partial\Omega)} \leq C \|f\|_{L^p(\Omega)}^{1-\theta}\|f\|_{W^{1,p}(\Omega)}^\theta \quad \text{for all}~f \in W^{1,p}(\Omega)$$ for suitable combinations of $p,q$ and $\theta$; $\operatorname{tr}$ being the trace operator. I would like to use this inequality to extend the trace operator from $W^{1,p}(\Omega)$ to $B^\theta_{p,1}(\Omega)$ for which I need to show that the operator is closable there, i.e., if $(f_n) \subset W^{1,p}(\Omega)$ converges to zero in $B^\theta_{p,1}(\Omega)$ and $(\operatorname{tr} f_n)$ converges to some $g \in L^q(\partial \Omega)$, then already $g = 0$. Due to the real and complex interpolation spaces of parameter $\theta$ between $L^p(\Omega)$ and $W^{1,p}(\Omega)$ being of class $J(\theta)$, the property to be shown would also imply that $(f_n) \to 0$ in such spaces; this however is in compliance with the property that $B^\theta_{p,1}(\Omega)$ already continuously embeds into all those. I have tried, amongst other attempts, using the characterization of Besov spaces via an integrated weighted modulus of smoothness (see e.g. this related question) to show that difference quotients of $f_n$ and thus $\|\nabla f_n\|_{L^p(\Omega)}$ stays bounded, but this did not work and I think this claim is too strong (the argument would imply that every convergent sequence in the interpolation space on $\Omega$ also converges weakly in $W^{1,p}(\Omega)$, if I am not mistaken). In the expression to be shown to converge to zero, $\|f_n\|_{W^{1,p}(\Omega)}^\theta$ is allowed to be unbounded in general, because $(f_n)$ goes to 0 in $L^p(\Omega)$ in any case. Please also note that the way I arrived at the question at hand might be sub-optimal in the sense that I know that the closedness-property of $\operatorname{tr}$ is sufficient for what I want to ultimately show, but might not be the weakest condition (it is only the weakest I could come up with..). It might thus be well possible that the question has a negative answer. REPLY [2 votes]: Here is a negative answer for a certain range of $p$ and $\theta$. It shows that one can have convergence to zero in a $\gamma$-Besov norm, where $\gamma$ may be larger than $\theta$. Hope I didn't mess up the parameters. Claim. For any $p\in (1,\infty)$ and $\theta, \gamma \in (0, 1 - 1/p)$ there are $f_\varepsilon\in W^{1,p}(0,1)$ such that, as $\varepsilon \to 0$, $$\|f_\varepsilon\|_{B_{p, 1}^\gamma(0,1)} \to 0, \qquad \|f_\varepsilon\|_{L^p(0,1)}^{1-\theta} \|f_\varepsilon\|_{W^{1, p}(0,1)}^\theta \to \infty.$$ Proof of the claim. Fix $p$ and $\theta$ as above, and let $\varepsilon > 0$ be small. For $\alpha \in (\theta, 1 - 1/p)$ and small $\beta >0$ define $$f_\varepsilon(x) = \varepsilon^\beta (x+\varepsilon)^\alpha, \qquad x\in (0, 1).$$ Since $f_\varepsilon \in C^1[0,1]$, we have $f_\varepsilon \in W^{1,p}(0,1)$ for all $\varepsilon$. We calculate $$\|f_\varepsilon\|_{L^p(0,1)}^p = \varepsilon^{p\beta} \int_0^1(x+\varepsilon)^{p\alpha} dx = \frac{\varepsilon^{p\beta}}{p\alpha + 1} \left ((1+\varepsilon)^{p\alpha + 1} - \varepsilon^{p\alpha + 1}\right),$$ which implies $\|f_\varepsilon\|_{L^p(0,1)} \sim \varepsilon^\beta.$ Moreover, $$\|f_\varepsilon'\|_{L^p(0,1)}^p = \varepsilon^{p\beta} \alpha^p \int_0^1(x+\varepsilon)^{p(\alpha -1)} dx = \frac{ \varepsilon^{p\beta} \alpha^p}{p(\alpha -1) + 1} \left ( (1+\varepsilon)^{p(\alpha -1) + 1} - \varepsilon^{p(\alpha -1) + 1} \right ).$$ The choice $\alpha < 1 - 1/p$ yields $\|f_\varepsilon\|_{W^{1, p}(0,1)} \sim \varepsilon^{\beta + \alpha -1 + 1/p},$ whence $$\|f_\varepsilon\|_{L^p(0,1)}^{1-\theta} \|f_\varepsilon\|_{W^{1, p}(0,1)}^\theta \sim \varepsilon^{(1-\theta)\beta + \theta(\beta + \alpha -1 + 1/p)} = \varepsilon^{\beta + \theta(\alpha - 1 + 1/p)}.$$ This is unbounded for $\varepsilon \to 0$ if $\beta$ is sufficiently small, as $\theta(\alpha - 1 + 1/p) <0$ by the choice of $\alpha$. On the other hand, for $\eta \in (0, \alpha)$ we have $$\|f_\varepsilon\|_{C^\eta[0,1]} = \|f_\varepsilon\|_\infty + \varepsilon^\beta \sup_{x, x+h\in [0, 1]} \frac{|(x+ h + \varepsilon)^\alpha - (x+\varepsilon)^\alpha|}{|h|^\eta} \sim \varepsilon^\beta,$$ as $t \mapsto t^\alpha$ is $\eta$-Hölder continuous on $[0, 1]$ if $\eta < \alpha$. Since $C^\eta \hookrightarrow B_{p,1}^{\gamma}$ for $\gamma \in (0, \eta)$, we conclude that $$\|f_\varepsilon\|_{B_{p, 1}^\gamma(0,1)} \leq C \|f_\varepsilon\|_{C^\eta[0,1]} \sim \varepsilon^\beta \to 0.$$ Because we can choose $\alpha$ as close to $1-1/p$ as we want, the same is true for $\eta$ and thus for $\gamma$ as well.<|endoftext|> TITLE: Reference Request: Derived group of $\mathscr R_u(B)$ QUESTION [5 upvotes]: Let $G$ be a connected, reductive group over an algebraically closed field $k$. Let $B$ be a Borel subgroup with maximal torus $T$ and unipotent radical $U$. Let $\Phi^+ = \Phi(B,T)$ and $\Delta$ the base of $\Phi = \Phi(G,T)$ corresponding to $\Phi^+$. A subset $\Psi$ of $\Phi^+$ is called closed if whenever $\alpha, \beta \in \Psi$, and $\alpha + \beta$ is a root, we have $\alpha + \beta \in \Psi$. If $\Psi$ is closed, then the root subgroups $U_{\alpha} : \alpha \in \Psi$ directly span a closed, connected subgroup $U_{\Psi}$ of $U$ which is normalized by $T$ and whose Lie algebra is $\bigoplus\limits_{\alpha \in \Psi} \mathfrak g_{\alpha}$. Furthermore, if $\alpha \in \Phi^+$, and $\Psi \subseteq \Phi^+$ is closed, and all roots of the form $a \alpha + b \Psi$ for $a, b \in \mathbb{Z}^+$ lie in $\Psi$, then the root subgroup $U_{\alpha}$ normalizes $U_{\Psi}$. These facts are proved in Chapter 14 of Borel, Linear Algebraic Groups. In particular, let $\Psi = \Phi^+ - \Delta$. It is easy to see that $\Psi$ is closed and normalized by every root subgroup $U_{\alpha} : \alpha \in \Phi^+$, and in particular, $U_{\Phi^+ - \Delta}$ is a normal subgroup of $U$. Moreover, it is a consequence of 8.32 in Springer's Linear Algebraic Groups that for any $x \in U_{\alpha}$ and $y \in U_{\beta}$, the commmutator $xyx^{-1}y^{-1}$ lies in $U_{\Phi^+ - \Delta}$. From here one can argue that $U_{\Phi^+ - \Delta}$ contains the derived group of $U$ by producing a homomorphism $U \rightarrow \prod\limits_{\alpha \in \Delta} \mathbf G_a$ with kernel $U_{\Phi^+ - \Delta}$. My question is, is $U_{\Phi^+ - \Delta}$ exactly the derived group of $U$? For $G = \textrm{GL}_n$, this does seem to be the case. REPLY [3 votes]: Let me add a few comments in community-wiki format. There doesn't seem to be a convenient reference, apart from the one in Digne-Michel which Jay Taylor cites. But even here, the authors don't give a full proof of the existence of regular unipotent elements. Such regular elements and their properties are essential to the approach Mikko gives, though it's possible that there is a more elementary method yet to be found. What's clear is that a considerable amount of structure theory for semisimple groups is involved. (Of course, in characteristic 0 one can instead work more straightforwardly in the Lie algebra.) Steinberg's treatment of regular elements (IHES, 1965) is available online through numdam.org, as is Springer's article (IHES, 1966). But note that Springer didn't succeed for bad primes in arriving at a proof of existence for regular unipotents. What he did was more direct than Steinberg's method, relying mainly on Chevalley's basis and commutation formula. Later on, students of Steinberg pushed this technique further for bad primes, but it's unclear how to extract a uniform theoretical approach. (Recently I wrote up some notes attempting to sort out the arguments used for both regular unipotents and regular nilpotents, posted here.) As nsfc23 comments, it's tricky to work directly with Chevalley's commutator formula in some small characteristic cases. On the other hand, Steinberg's approach to regular unipotents requires fairly heavy machinery. As to finiteness of the number of unipotent clases (still conjectural in the mid-1960s), it remains an open problem to use modular representation theory of $G$ or its Lie algebra to get a more self-contained proof.<|endoftext|> TITLE: Conjugation action on the classifying space of circle QUESTION [11 upvotes]: Let $T$ denote the unit sphere in complex plane. $\mathbb{Z}/2$ acts by complex conjugation on $T$. There is an induced action on $BT$. The cohomology of the homotopy orbit space of this action is $$ H^*(BT//(\mathbb{Z}/2),\mathbb{F}_2) = H^*(B\mathbb{Z}/2,\mathbb{F}_2)\otimes H^*(BT,\mathbb{F}_2)=\mathbb{F}_2[t,x] $$ where degrees are given by $|t|=1$ and $|x|=2$. This is because the associated fibration has a section. But I am curious about the structure of the cohomology ring as a module over the Steenrod algebra. REPLY [14 votes]: It is the interesting one, with $Sq^1(x) = tx$ and the remainder determined by the axioms of Steenrod operations. This may be seen as the homotopy orbits is a model for BO(2), where one knows the Steenrod operations. (Here I am assuming the choice of class $x$ which pulls back to zero under the section; under the equivalence to $BO(2)$ it then bevomes the second Stiefel--Whitney class.) Alternatively, you can find the mapping torus $T$ of the complex conjugation map on $\mathbb{CP}^1$ as a natural subspace of $BT /\!\!/ \mathbb{Z}/2$, and the fundamental class of $T$ pairs to 1 with $tx$. By naturality, the question now becomes whether $$\mathrm{Sq}^1 : H^2(T ; \mathbb{F}_2) = \mathbb{F}_2\{x\} \to H^3(T ; \mathbb{F}_2) = \mathbb{F}_2\{tx\}$$ is non-zero or not. But $T$ is a manifold, and Steenrod-squaring into the top degree is always given by cup product with the corresponding Wu class, in this case $v_1$. As the total Stiefel--Whitney class and the total Wu class are related by $w = \mathrm{Sq}(v)$, we have $v_1 = w_1$, so the question becomes to identify the first Stiefel--Whitney class of the 3-manifold $T$. But $T$ is clearly nonorientable, as it is the mapping torus of an orientation-reversing diffeomorphism, so $w_1 \neq 0$; thus $w_1$ must be $t$, so $\mathrm{Sq}^1(x) = t \cdot x$.<|endoftext|> TITLE: curious relation between orders of generators of a finite group QUESTION [5 upvotes]: Let $G$ be a finite group, and $g,h\in G$ be such that $G = \langle g,h\rangle = \langle g,hgh^{-1}\rangle$ - that is, $g,h$ generate $G$, and so does $g,hgh^{-1}$. Let $n := |G|$, $r := |g|$, and $e := |ghg^{-1}h^{-1}|$ ("$|.|$" denotes "order") I was doing some computations with covers of curves, and it seems that my computations give the following relation: $$e \ge \frac{n}{2+n-\frac{2n}{r}}$$ The cleanest special case is if $r = 2$, in which case one gets $e\ge \frac{n}{2}$ (in which case we must have $e = \frac{n}{2}$ barring the trivial case when $G$ is cyclic, so that the commutator subgroup is cyclic of index 2) Firstly, is this true? (I've checked it to be true for the smallest 10 or so finite simple groups, where it seems to hold). If so, this is somewhat curious to me. Is this related to any known results? REPLY [5 votes]: Denote $f=hgh^{-1}$. 1) If $r\geqslant 4$, then $n\geqslant 4$ and either $e=1$, $r=n$ or $e\geqslant 2>\frac{n}{2+n/2}\geqslant \frac{n}{2+n-2n/r}$. 2) If $r=3$, then either $e\geqslant 3\geqslant \frac{n}{2+n/3}=\frac{n}{2+n-2n/3}$ or $e=2$, then we have $f^3=g^3=1$, $(fg)^2=1$. It follows that the group generated by $f,g$ contains at most 12 elements, since it is a factor of the group $\langle f,g|f^3=g^3=1,(fg)^2=1\rangle=A_4$. But for $n\leqslant 12$ we have $e=2\geqslant \frac{n}{2+n/3}$. $r=2$. Then $n$ does not exceed the order of the group generated by relations $f^2=g^2=1$, $(fg)^e=1$. This group contains at most $2e$ elements as desired. Indeed, all elements have the form either $(fg)^k$, or $(fg)^kf$, where $k=0,1,\dots,e-1$.<|endoftext|> TITLE: Creating an additive structure over the set of all finite groups? QUESTION [6 upvotes]: I'm trying to form a ring (or ring-like structure) out of the set of all finite groups. Has anyone created/encountered an operation "+" before with the follow properties. Let $G_1, G_2$ be finite groups and |G| denote the size of G, then $$ |G_1| + |G_2| = |G_1 + G_2|$$ Where the left hand side is addition amongst natural numbers and the right hand side is our "abstract group addition" And let $\times$ denote the direct product. Then: $$ G_1 \times ( G_2 + G_3) = (G_1 \times G_2) + (G_1 \times G_3) $$ The "+" ideally would be commutative and associative, but I'm not sure if such an operator can necessarily exist. My goal is to try to denote a notion of "rational groups" and dedekind cuts to groups to see if I can create a "fractional group theory" so to speak. REPLY [5 votes]: I believe that we can get a commutative, associative operation as follows. Let $G=A\times B$ and $H=A\times C$, where $B$ and $C$ have no common factor (when written as a direct product of indecomposable groups of order greater than one). Define $G+H=A\times D$, where $D$ is a product of cyclic groups of prime order such that $|D|=|B|+|C|$.<|endoftext|> TITLE: A direct proof of a property of symmetric 2x2-determinants QUESTION [8 upvotes]: Let $f(a,b,c)=\det\begin{pmatrix}a &b\\ b& c\end{pmatrix}\in\mathbb{R}[a,b,c]$ be the determinant of a $2 \times 2$ real symmetric matrix. Let $f(x_i,y_i,z_i)\geq 0$, $x_i\geq 0$, $z_i\geq 0$ for $i\in\{1,2\}$. (Edit: i.e. the matrix is positive semidefinite). Then $$ f(x_1+x_2,y_1+y_2,z_1+z_2)\geq 0 \tag{$\ast$} $$ (something that can be formulated as convexity of the cone of symmetric $2 \times 2$ positive semidefinite (p.s.d.) matrices). Indeed, $(*)$ can be seen by viewing $f$ as the discriminant of the quadratic polynomial $$F(a,b,c)(t)=at^2+2bt+c \tag{**}$$ with $a\geq 0$, $c\geq 0$, observing that then $f(a,b,c)\geq 0$ iff $F(a,b,c)(t)\geq 0$ for all $t$, and seeing that $$F(x_1+x_2,y_1+y_2,z_1+z_2)(t)=F(x_1,y_1,z_1)(t)+F(x_2,y_2,z_2)(t)\geq 0$$ The question is whether there is a (more) direct proof of $(*)$, and what kinds of generalisations are known. E.g. I am interested in the situation where $a,b,c$ are multivariate polynomials, and under which conditions $f(a,b,c)$ is a sum of squares (s.o.s.) of polynomials---with potential applications to efficient s.o.s. decompositions. Edit: As well, is there an analogue of $(**)$ for higher order positive semidefinite matrices? REPLY [3 votes]: Here is a direct proof: Simple computations show that $$f\left(x_1+x_2,y_1+y_2,z_1+z_2\right) = f\left(x_1,y_1,z_1\right) + f\left(x_2,y_2,z_2\right) + \left(x_1z_2+x_2z_1-2y_1y_2\right).$$ It thus remains to check that all three addends on the right hand side are nonnegative. For $f\left(x_1,y_1,z_1\right)$ and $f\left(x_2,y_2,z_2\right)$, this follows straight from the assumptions. For $x_1z_2+x_2z_1-2y_1y_2$, we have to prove that $x_1z_2+x_2z_1 \geq 2y_1y_2$. But thanks to the nonnegativity of $x_i$ and $z_i$, we have $x_1 z_1 \geq y_1^2$ (since $x_1 z_1 - y_1^2 = f\left(x_1, y_1, z_1\right) \geq 0$) and $x_2 z_2 \geq y_2^2$ (similarly), so that the AM-GM inequality yields $x_1z_2+x_2z_1 \geq 2 \sqrt{x_1z_2 \cdot x_2z_1} = 2 \left(\underbrace{x_1 z_1}_{\geq y_1^2} \underbrace{x_2 z_2}_{\geq y_2^2} \right)^{1/2} \geq 2 \left( y_1^2 y_2^2 \right)^{1/2} = 2 \left| y_1 y_2 \right| \geq 2 y_1 y_2$, which is precisely what we needed to prove. Note that this argument might not help you with your question about sum-of-squares decompositions. I am not actually sure what exactly constitutes a sum-of-squares decomposition for an inequality that only holds under assumptions...<|endoftext|> TITLE: The minimum of a sum of absolute values of inner products in $\mathbb{R}^d$ QUESTION [14 upvotes]: Consider a collection of unit vectors $v_1, \ldots, v_n$ in $\mathbb{R}^d$ (we think of $n$ being much larger than $d$). I would like to minimize the sum: $$\sum_{i\neq j}|\langle v_i,v_j\rangle|.$$ Clearly, if $n=d$, the minimum is attained by taking $v_i=e_i$. Could it be that for $n>d$ in order to minimize the latter expression it is still best to take the vectors $v_i=e_{i\, \text{mod}\, d}$? REPLY [4 votes]: In fact, the minimizers of the sum $\sum \langle v_i, v_j \rangle^2$ are precisely tight frames, i.e. sets such that for some $C>0$and for each $x\in \mathbb R^d$, one has $\| x \|^2 = C \sum \langle x, v_j \rangle^2$. This was proved in Benedetto, Fickus "Finite Normalized Tight Frames", Adv. Comp. Math., 18 (2003), pp. 357-385. Thus for $n$ not a multiple of $d$, the set $v_i = e_{i\mod d}$ is not a minimizer, since it is not a tight frame. Although, I believe that for $\ell^1$ the answer may still be correct.<|endoftext|> TITLE: Category theorists stance on deductive systems QUESTION [12 upvotes]: Lambek & Scott: Introduction to higher order categorical logic says that "For example, categorists may be unhappy when we treat categories as special kinds of deductive systems and logicians may be unhappy when we insist that deductive systems need not be freely generated from axioms and rules of inference" I am curious about the latter: why a deductive system need not be freely generated from axioms and rules of inference? In computer science, derivable true formulae are the ones derivable from axioms by a finite series of using inference rules, that is, an inductively defined set. Why is this doubtful for a category theorist? REPLY [5 votes]: I think the idea should pretty much like this: once you drop the requirement for the deductive system to be freely generated from the axioms by the inference rules (i.e. you accept the existence of non free deductive systems) you can start to study morphisms from the free deductive systems into not free ones (what Lambek and Scott call functors, if memory serves me well). In this way you can hopefully prove properties of the free systems by looking to the not free ones. For instance for proving that two proofs (i.e. arrows) are different in a free deductive system you could try to find a nice functor that sends the said proofs in different arrows. Here by nice I mean that such functor sends the proofs in a different system where it is easier to prove the inequality of the two proofs. As a example (probably a not really interesting one) you can consider the functor that goes from a syntactic-freely generated deductive system to the deductive system $\mathbb N$ (the additive monoid of natural numbers, seen as a category, hence as a deductive system), that sends every object (proposition) to the only object of the one-object category $\mathbb N$ and every morphism/proof to its length (which is indeed a natural number, that is a morphism in $\mathbb N$). This functor allows one to distinguish between two proofs because they have different length. I suppose that someone with much more knowledge on the subject than me can provide some more interesting examples.<|endoftext|> TITLE: What are the points of simple algebraic groups over extensions of $\mathbb{F}_1$? QUESTION [26 upvotes]: The "field with one element" $\mathbb{F}_1$ is, of course, a very speculative object. Nevertheless, some things about it seem to be generally agreed, even if the theory underpinning them is not; in particular: $\mathbb{F}_1$ seems to have a unique extension of degree $m$, generally written $\mathbb{F}_{1^m}$ (even if this is slightly silly), which is thought to be in some sense generated by the $m$-th roots of unity. See, e.g., here and here and the references therein. If $G$ is a (split, =Chevalley) semisimple linear algebraic group, then $G$ is in fact "defined over $\mathbb{F}_1$", and $G(\mathbb{F}_1)$ should be the Weyl group $\mathcal{W}(G)$ of $G$. For example, $\mathit{SL}_n(\mathbb{F}_1)$ should be the symmetric group $\mathfrak{S}_n$. (See also this recent question.) I think this is, in fact, the sort of analogy which led Tits to suggest the idea of a field with one element in the first place. These two ideas taken together suggest the following question: What would be the points of a semisimple (or even reductive) linear algebraic group $G$ over the degree $m$ extension $\mathbb{F}_{1^m}$ of $\mathbb{F}_1$? My intuition is that $\mathit{GL}_n(\mathbb{F}_{1^m})$ should be the generalized symmetric group $\mu_m\wr\mathfrak{S}_n$ (consisting of generalized permutation matrices whose nonzero entries are in the cyclic group $\mu_m$ of $m$-th roots of unity); and of course, the adjoint $\mathit{PGL}_n(\mathbb{F}_{1^m})$ should be the quotient by the central (diagonal) $\mu_m$; what $\mathit{SL}_n(\mathbb{F}_{1^m})$ should be is already less clear to me (maybe generalized permutation matrices of determinant $\pm1$ when $m$ is odd, and $+1$ when $m$ is even? or do we ignore the signature of the permutation altogether?). But certainly, the answer for general $m$ (contrary to $m=1$) will depend on whether $G$ is adjoint or simply connected (or somewhere in between). I also expect the order of $G(\mathbb{F}_{1^m})$ to be $m^r$ times the order of $G(\mathbb{F}_1) = \mathcal{W}(G)$, where $r$ is the rank. And there should certainly be natural arrows $G(\mathbb{F}_{1^m}) \to G(\mathbb{F}_{1^{m'}})$ when $m|m'$. (Perhaps the conjugacy classes of the inductive limit can be described using some sort of Kac coordinates?) Anyway, since the question is rather speculative, I think I should provide guidelines on what I consider an answer should satisfy: The answer need not follow from a general theory of $\mathbb{F}_1$. On the other hand, it should be generally compatible with the various bits of intuition outlined above (or else argue why they're wrong). More importantly, the answer should be "uniform" in $G$: that is, $G(\mathbb{F}_{1^m})$ should be constructed from some combinatorial data representing $G$ (root system, Chevalley basis…), not on a case-by-case basis. (An even wilder question would be if we can give meaning to ${^2}A_n(\mathbb{F}_{1^m})$ and ${^2}D_n(\mathbb{F}_{1^m})$ and ${^2}E_6(\mathbb{F}_{1^m})$ when $m$ is even, and ${^3}D_4(\mathbb{F}_{1^m})$ when $3|m$.) REPLY [7 votes]: Not an answer - just opposite - some additional requirments which good answer may satisfy and further comments. 0) Parabolic subgroups should also be defined. I mean good definition should also come with definition of parabolic subgroups. For example for $S_n$, it seems natural to consider $S_{d_1}\times ... \times S_{d_k}$ as parabolics. 1) Element count should be compatible with "known" (= widely agreed) zeta-functions of P^n (and also Grassmanians, Flags). I mean one can define P^n, Grassmanians, Flags in a standard way just by quotients of G/Parabolic. So one should get number of points for such manifold over $F_{1^l}$, hence one can write a standard Weil's zeta function. For example for S_n for Grassamnian $|S_n/(S_k\times S_{n-k}) | = n!/(k!(n-k)!)$. Requirment: zeta should agree with the "known" one (for example s(s-1)...(s-n+1) for P^n). 2) Structure of representation theory of such groups should be similar to G(F_q). For example for GL(F_1^l) one may expect that similar structure as in Zelevinsky theorem (see MO272686), i.e. all representations (for all "n" GL(n)) organized into Hopf algebra using induction and restriction should be isomorphic to several copies of the same Hopf algebra for S_n, each copy for each cuspidal representation. Interesting question how many cuspidals will one have for fixed GL(n,F_1^l) ? 3) Alvis-Carter duality should work. It is defind as roughly speaking as follows - take character - restrict it to parabolic and than back induction, with summation over all parabolics with appropriate sign. The statement is that: it has order two and isometry on generalized characters. Steinberg representation is dual to trivial. For S_n it is transposition of Young diagram (if I rember correctly). 4) Frobenius action and Lang-Steiberg theorem. One knows that for F_q there is Frobenius and Lang-Steinberg theorem (e.g. G.Hiss page 15) holds that map $G^{-1}F(G)$ is surjection. How to define Frobenius ? 5) "Good" bijection between irreducible representation and conjugacy classes (toy Langldands correspondence). Similar to MO270916 one may expect "good" bijection between conjugacy classes - which might be considered as a toy model for local Langlands correspondence for field with one element. For example for S_n there is well-known "good" bijection between the two sets via Young diagrams. OP made intersting proposal for an $GL(F_{1^n})$ - monomial matrices with roots of unity entries. It would be very intersting to check whether the properties above holds true for such proposal. What is most unclear for me - what should be Frobenius map ?<|endoftext|> TITLE: 2x2-determinantal representations of cubic curves QUESTION [7 upvotes]: Let $H_d:=\mathbb{C}_d[x,y,z]$ denote the space of homogeneous degree $d$ polynomials in $x$, $y$, $z$ with complex coefficients. I'd like to show that every $f\in H_3$ can be represented as $$ f=\det\begin{pmatrix}q_1 & q_2\\ \ell_1 & \ell_2\end{pmatrix}, q_i\in H_2, \ell_i\in H_1, i=1,2. $$ This is different from the usual determinantal representations, where each matrix entry is a linear form. I suspect that the answer is true, and it can be proved using technique from A.Beauville's "Determinantal hypersurfaces" - which is not easy to read. Is there an easier argument? REPLY [9 votes]: Let $(\ell_1,\ell_2)$ be the ideal of a point in $V(f)$, so $f \in (\ell_1,\ell_2)$.<|endoftext|> TITLE: Is every transitive ZF-model of inaccessible height a truncation of an inner model? QUESTION [19 upvotes]: Let $\kappa$ be an inaccessible cardinal and let $M \subseteq V_{\kappa}$ be an inner model of $V_{\kappa}$, i.e., a transitive model of $\mathsf{ZF}$ containing all the ordinals up to $\kappa$. My question is whether such a model is always a rank-truncation of an inner model of $\mathbf V$ (defined using $M$ as parameter). Equivalently this question can be phrased as follows: If $M \subseteq V_{\kappa}$ is an inner model of $V_\kappa$ is then $\mathbf L(M) \cap V_\kappa = M$, where $\mathbf L(M)$ is the minimal inner model of $\mathbf V$ containing $M$ ? If the answer to this question is negative, I am curious whether there can be a counterexample $M$ which is definable in $V_\kappa$ without parameters and/or which is a model of $\mathsf{ZFC}$. If the answer to this question is positive, I am curious whether it is still positive if we weaken the assumption that $\kappa$ is inaccessible to $\kappa$ being worldly. REPLY [10 votes]: Theorem: Let $\kappa$ be strongly inaccessible in $V$, such that $V \models ZFC$. If $M\models ZF$, then $L(M) \cap V_\kappa = M$. Proof: Let us prove by induction on $\alpha < \kappa$ that $L(M) \cap V_\alpha = M \cap V_\alpha$. Let $x \in L(M) \cap V_{\alpha + 1}$, so $x \subseteq M \cap V_\alpha$. Let $\gamma$ be an ordinal such that $x\in L_{\gamma}(M)$. Let us find $X \prec L_\gamma(M)$, such that: $x\in X$, $V_\alpha \subseteq X$, $X \cap \kappa \in \kappa$ $X \cap M$ is transitive and equal to $M \cap V_\beta$ for some $\beta$ This is possible by the strong inaccessiblity of $\kappa$: We define, by induction of $n < \omega$, $X_n$. Let $X_0 = M\cap V_{\alpha} \cup \{x\}$. Extend $X_0$ to $X_0'$ such that $X_0'\prec L_\gamma(M)$ and $|X_0'| = |X_0| < \kappa$. Now, take $X_1 \supseteq X_0'$ such that assumptions 2 and 3 hold ($M \cap X_1$ is transitive and $X \cap \kappa$ is ordinal). $|X_1| < \kappa$, by the inaccessiblity of $\kappa$. Let $X_1 \subseteq X_1' \prec L_\gamma(M)$, and so on. Let $X$ be $\bigcup X_n$. Let $\bar{X}$ be the transitive collapse of $X$. $\bar{X} = L_{\bar{\gamma}}(\bar{M})$ where $\gamma < \kappa$ and $x\in \bar{X}$. Since $M \cap X$ is transitive, $\bar{M} = M \cap X$. Since $M\cap X = M \cap V_\zeta$ for some $\zeta$, $\bar{M} = M \cap V_\zeta \in M$ (as $\zeta \leq \bar{\gamma} \in \kappa$). In particular, $M$ can compute the model $L_{\bar{\gamma}}(\bar{M})$ and compute $x$. We conclude that $x \in M$. Remark: Without the inaccessibility assumption, one can get a model $M$ of ZFC such that $L(M) \cap V_{\omega + 1} \neq M \cap V_{\omega + 1}$. Let us start with $V = L$ and let $\kappa$ be a worldly cardinal of countable cofinality. Let $\langle \alpha_n \mid n < \omega\rangle$ be the first cofinal sequence of singular cardinals with limit $\kappa$, in the canonical well order of $L$. Let us consider the following class forcing in $L_\kappa$: for every singular cardinal $\mu\in L_{\kappa}$, let $\lambda = \mu^{++}$, and force with the lottery sum of $Add(\lambda, \lambda^{++})$ and the trivial forcing. Take the $<\kappa$ support product of those forcing notions. This forcing does not collapse cardinals. Let $c$ be a Cohen real. Let $G$ be an $L_\kappa$-generic filter such that $L_\kappa[G] \models 2^{\alpha_n^{++}} = \alpha_n^{+4}$ iff $c(n) = 1$. This is possible by splitting the forcing into class forcing that only decides for which cardinal $\lambda$ we force $2^\lambda = \lambda^{++}$ (this class forcing does not add sets) and the second step in which we change the value of the corresponding $2^\lambda$-s. In $L[c]$ one can find a generic for the first step that codes $c$ as above. Let $M = L_\kappa[G]$. $c \notin M$ (as $M$ and $L$ share the same reals), while $c \in L(M)$.<|endoftext|> TITLE: Cohomological dimension of $G \times G$ QUESTION [16 upvotes]: $\DeclareMathOperator\cd{cd}$A question that I have already posted in the Mathematics section, but which seems to be too delicate for that section (see here and here): Let $\cd(G)$ denote the cohomological dimension of a group $G$, i.e. the minimal length of a projective resolution of Z over $\mathbb{Z} G$. Under which conditions on a group $G$ of finite cohomological dimension is it true that $$\cd(G \times G) > \cd(G) \; ?$$ On the one hand, there are examples of groups with $G \times G \cong G$, though I am not aware of examples of finite cohomological dimension. On the other hand, an affirmative answer follows from the Kuenneth formula for group cohomology as discussed in Chris Gerig's detailed answer to this question. Nevertheless, this line of argument is only applicable if we find a local coefficient system $A$ for which $H_*(G;A)$ is finitely generated and for which there exists $k\in \mathbb{N}$ with $2k>\cd(G)$ and $H^k(G;A)\neq 0$. Under which conditions does such $A$ exist? REPLY [10 votes]: If $G$ is a non-trivial group with $G\times G\cong G$, then $G$ has infinite cohomological dimension as @YCor suggests. If $H\leq G$, then $cd(H)\leq cd(G)$ because $\mathbb ZG$ is a free $\mathbb ZH$-module and so any $\mathbb ZG$-free resolution of the trivial module is a $\mathbb ZH$-free resolution. If $G$ has an element of finite order, then $G$ has infinite cohomological dimension because finite cyclic groups do. If $G$ has an element of infinite order, then $G\cong G\times G$ implies $\mathbb Z^n$ is a subgroup of $G$ for all $n>0$. Since $cd(\mathbb Z^n)=n$, this means $G$ that $cd(G)\geq n$ for all $n>0$ and hence is infinite.<|endoftext|> TITLE: Lattices in semisimple Lie groups QUESTION [5 upvotes]: I am interested in the following question: Can semisimple Lie group of real rank $\geq 2$ contain an abelian lattice? REPLY [14 votes]: No, and it has nothing to do with the higher-rank assumption. Probably the easiest way to see this is by the Borel Density Theorem, which says that if $G$ is a semisimple Lie group which has no compact factor and it is algebraic, and if $\Gamma$ is a lattice in $G$, then $\Gamma$ is Zariski dense in $G$. Now, if you have a non-compact semisimple Lie group and a lattice in it, you can mod up the center and get an algebraic group and a lattice in it. Further, the algebraic group is a product of factors and by modding out compact ones you are in the situation described above. Another easy way to see that a lattice cannot be abelian is by showing that it cannot be amenable. If it was then the enveloping group would be amenable as well, but it is not. Eg it contains a closed free group.<|endoftext|> TITLE: Which manifolds are sensitive to the cocycle in the Dijkgraaf-Witten model? QUESTION [10 upvotes]: Often, TQFTs are defined in families, parametrised by some algebraic data. For example, the Turaev-Viro-Barrett-Westbury TQFTs are parametrised by spherical fusion categories, the Crane-Yetter TQFTs are parametrised by ribbon fusion categories, and the $n$-dimensional Dijkgraaf-Witten theory is parametrised by a finite group $G$ and an $n$-cocycle $\omega$. Instead of regarding TQFTs with a fixed datum as an invariant of manifolds (and cobordisms), one can also fix a manifold and regard a family of TQFTs as invariant of the parametrising data. For example, the Crane-Yetter invariant of $\mathbb{CP}^2$ is the "Gauss sum" $\sum_X d(X)^2 \theta_X$ of the ribbon fusion category, where $X$ ranges over simple objects and $\theta$ is the twist eigenvalue. (I'm thanking Ehud Meir for making me appreciate this viewpoint.) From this viewpoint, my question is: In the Dijkgraaf-Witten TQFT, which manifolds give invariants that are sensitive to the cocycle? In details, let us define the following invariant: $$ DW_{G,\omega}(M) = \sum_{\phi\colon \pi_1(M) \to G} \int_M \phi^* (\omega)$$ Here, $M$ is an $n$-manifold, $\omega \in H^n(G,U(1))$ is an element of the $n$-th group cohomology, and $\phi^*\colon H^n(G,U(1)) \to H^n(M,U(1))$ is induced by the flat $G$-connection $\phi$. I'm looking for a manifold $M$ such that $DW_{G,\omega}(M) \neq DW_{G,\omega'}(M)$ for some $\omega \neq \omega'$. Ideally, the example would be in 3 or 4 dimensions. REPLY [9 votes]: The example $G = \mathbb Z/2$ and $M = \mathbb{RP}^3$ works. The inclusion $\mathbb Z/2\to\{\pm 1\}\subset\mathrm U(1)$ induces an isomorphism $H^3(B\mathbb Z/2, \mathbb Z/2)\to H^3(B\mathbb Z/2, \mathrm U(1))$, so we can pull the cocycles back to $\mathbb Z/2$ cohomology and evaluate on the $\mathbb Z/2$ fundamental class. $\mathbb{RP}^3$ has two isomorphism classes of principal $\mathbb Z/2$-bundles, the trivial bundle $\varepsilon$ and the connected double cover $\xi$. Each determines a classifying map to $B\mathbb Z/2$, and hence a map in cohomology $\phi^*\colon H^*(B\mathbb Z/2; \mathbb Z/2)\cong \mathbb Z/2[\alpha]\to H^*(\mathbb{RP}^3; \mathbb Z/2)\cong \mathbb Z/2[x]/(x^4)$. For $\varepsilon$, this is the zero map; for $\xi$, this is the ring homomorphism induced by $\alpha\mapsto x$. Since $H^3(B\mathbb Z/2;\mathbb Z/2)\cong\mathbb Z/2$, there are two cohomology classes of cocycles. Let $\omega$ be a coboundary, so that $\phi^*\omega = 0$ for all principal $\mathbb Z/2$-bundles, and hence the Dijkgraaf-Witten partition function is $$\mathrm{DW}_{\mathbb Z/2, 0}(\mathbb{RP}^3) = \underbrace{\frac{e^{i\pi(0)}}{2}}_{\text{from } \varepsilon} + \underbrace{\frac{e^{i\pi(0)}}{2}}_{\text{from } \xi} = 1.$$ Let $\omega$ be a cocycle in the other cohomology class (in the notation above, $\alpha^3$). Then, $\varepsilon$ pulls it back to $0$, but $\xi$ pulls it back to $x^3$, and $\langle x^3, [\mathbb{RP}^3]\rangle = 1$, so $$\mathrm{DW}_{\mathbb Z/2, \alpha^3}(\mathbb{RP}^3) = \underbrace{\frac{e^{i\pi(0)}}{2}}_{\text{from } \varepsilon} + \underbrace{\frac{e^{i\pi(1)}}{2}}_{\text{from } \xi} = \frac 12 - \frac 12= 0.$$<|endoftext|> TITLE: "Extra Euler factors" in one definition of the L-function of a twist of a modular form QUESTION [9 upvotes]: Let $(\rho_{f,\lambda})_\lambda$ be the system of Deligne's $\ell$-adic representations attached to a modular newform $f$ (where $\lambda$ runs over the finite places of the number field $K$ generated by the Fourier coefficients of $f$), and take a character $\chi\colon\mathrm G_{\mathbb Q}\rightarrow\overline{\mathbb Q}^\times$ of finite order (assume for simplicity it takes values in $K$). Fix embeddings of $\overline{\mathbb Q}$ into each $\overline{\mathbb Q}_\ell$. Then we can look at the system of $\lambda$-adic realizations $(\rho_{f,\lambda}\otimes\chi)_\lambda$. Let me call a system $(\rho_\lambda)_\lambda$ of $\ell$-adic representations compatible if for each prime $p$ the polynomial $$ P_p(T)=\begin{cases}\det(1-\mathrm{Frob_p}T,\rho_\lambda^{I_p}), & \lambda\not\mid p,\\ \det(1-\varphi T,\mathrm D_{\mathrm{cris}}(\rho_\lambda)), & \lambda\mid p\end{cases} $$ does not depend on the choice of a place $\lambda$ and has coefficients in $\overline{\mathbb Q}$. It is known that the system $(\rho_{f,\lambda})_\lambda$ is compatible in this sense (using results of Scholl and Saito), and it seems to be well-known that this holds also for the system $(\rho_{f,\lambda}\otimes\chi)_\lambda$. Probably this can be seen somehow using Weil-Deligne representations, but I haven't yet worked out the details (any hints or references on this are welcome!). But my actual question is the following. If we define the $L$-function of such a compatible system as usual as the Euler product $$ L((\rho_\lambda)_\lambda,s)=\!\!\!\prod_{p\text{ any prime}}\!\!\! P_p(p^{-s}),$$ then the $L$-function associated to $(\rho_{f,\lambda})_\lambda$ is just the $L$-function of $f$. But what is the $L$-function of $(\rho_{f,\lambda}\otimes\chi)_\lambda$? If we view $\chi$ as a Dirichlet character of $(\mathbb{Z}/N)^\times$ for the minimal possible $N$ via class field theory, then one often considers the $L$-function defined by $$ L(f,\chi,s)=\sum_{n=1}^\infty \chi(n)a_nn^{-s}, $$ where the $a_n$ are the Fourier coefficients. But in general the $L$-function of the system $(\rho_{f,\lambda}\otimes\chi)_\lambda$ differs from this one. Indeed, D. Loeffler's answer to my question How large is Dcris of certain twists of modular forms? shows that we get at least an extra Euler factor at $p$ if $f$ is $p$-ordinary and the $p$-part of $\chi$ cancels the $p$-part of the nebentype of $f$. Can we in general tell which additional Euler factors (compared to $L(f,\chi,s)$) this $L$-function has? REPLY [4 votes]: If $f(z)=\sum_{n \geq 1} a_n q^n$ is a newform of level $\Gamma_1(N)$ and $\chi$ is a Dirichlet character modulo $m$, then the naïve twist of $f$ by $\chi$ is the modular form $f_\chi(z) = \sum_{n \geq 1} a_n \chi(n) q^n$. As was already pointed out $f_\chi$ is not always a newform, but there is a unique newform $f \otimes \chi$ sharing the same Hecke eigenvalues at primes $p$ not dividing $m$. If $N$ and $m$ are coprime then $f_\chi = f \otimes \chi$ is a newform, but this is not always the case in general. A criterion for $f_\chi$ being a newform (equivalently $f_\chi = f \otimes \chi$) has been worked out by Atkin--Li in their article Twists of newforms and pseudo-eigenvalues of $W$-operators (see Corollary 3.1). The question of determining the Euler factor of $f \otimes \chi$ at $p$ is clearly a local one, so we may assume that $\chi$ is a primitive Dirichlet character of conductor $p^\alpha$ with $\alpha \geq 1$, and that $p$ divides $N$. In general the Euler factor of $f \otimes \chi$ at $p$ can be determined from the local automorphic representation associated to $f$, as explained by Peter Humphries. There is however a special case which is easy, namely when $f$ is $p$-primitive, meaning that $f$ has minimal level among its twists by characters of $p$-power conductor. If $f$ is $p$-primitive and $a_p \neq 0$ then we have the formula $$L_p(f \otimes \chi,s)^{-1} = 1- \bar{a}_p \cdot (\psi \chi)_0(p) p^{-s}$$ where $\psi$ is the Nebentypus character of $f$, and $(\psi \chi)_0$ is the primitive Dirichlet character associated to $\psi \chi$. This is explained in Merel's article Symboles de Manin et valeurs de fonctions $L$ (Section 2.6).<|endoftext|> TITLE: When n! = some Fibonacci number QUESTION [13 upvotes]: I'm trying to work out following problem: Let $m$ be some Fibonacci number and $n$ some integer. Find all non-trivial pairs $(n,m)$, where $n! = m$. Trivial pair is pair $(1,1)$. I know that for every number $a$ there exists some Fibonacci number that's divisible by $a$. I also tried to bruteforce my way trough, but I didn't succeed. Here is my sub-optimal code: a = 1 b = 1 mylist = [here are factorials up to 50!] for i in range(0,100): c = a+b a = b b = c print(b) if b in mylist: print("found") I tried many other things including googling it, but I didn't find anything that I can even start with. I have highschool level of math knowledge. REPLY [21 votes]: There are only three terms of the Fibonacci sequence which are equal to a factorial: $$F_{1}= F_{2} = 0!=1! \quad \mbox{and} \quad F_{3}=2!$$ What is more, F. Luca proved in this paper (published in 1999) that the only nontrivial solutions of the diophantine equation $$F_{n} = m_{1}! \cdots m_{t}!$$ are $F_{3}=2!$, $F_{6}=(2!)^{3}$, and $F_{12} = (2!)^{2}(3!)^{2} = 3! 4!$. REPLY [14 votes]: On Fibonacci numbers that are factorials, Surajit Rajagopal and Martin Griffiths (2014) This is behind a paywall, but thanks to SJR the paper can be accessed here. The "simple proof" that for any $n\geq 4$ the Fibonacci number $F_n$ is not a factorial starts on page 14.<|endoftext|> TITLE: Counting some binary trees with lots of extra stucture QUESTION [7 upvotes]: While working on some computations on Hilbert schemes, I came across the following combinatorial problem. Let $D(k,n)$ be the weighted number of binary trees (children are left/right) with $n$ internal nodes (internal node = node with children), with each node labeled by an integer (labels need not be unique, negatives are allowed), with the root labeled $k$, and the label of each internal node is equal to the sum of the labels of the children, and each leaf is labeled either $1$ or $-1$. The weight (which may be negative) assigned to such a tree is the product of the labels of the internal nodes. Furthermore, there is an ordering of the internal nodes such that a parent always comes before a child. For example, $D(1,2) = 4$, with the trees 1 2 -1 1 1 and 1 -1 2 1 1 each counted with weight 2 (and each of these has only one possible ordering of the internal nodes). In the end, I'm actually only interested in the sequence $D(1,2i)$. This sequence appears to be http://oeis.org/A151403, which counts a certain type of lattice paths in the first quadrant, and is equal to $4^i$ times the $i^{th}$ Catalan number. But I defined $D(k,n)$ because then you can get the recurrence relation $$ D(k,n) = k \sum_{a+b=k} \sum_{c+d=n-1} \binom{n-1}{c} D(a,c) D(b,d) $$ where $c,d \ge 0$ but $a,b$ are any integers. We have initial conditions $D(\pm1,0) = 1$ and $D(k,0) = 0$ otherwise. It is easy to see that $D(k,n) = 0$ if $|k| > n+1$, so the sum here is finite. (Each child node starts a new tree, the factor of $k$ takes care of the weight, and the $\binom{n-1}{c}$ is the number of ways of "interweaving" the ordering.) With some computation, I was able to guess the formula for $D(k,n)$. We have $$ D(k,k-1) = 2 (2k)^{k-2} $$ and if $n-k+1 \neq 0$, then $$ D(k,n) = (2k)^n \binom{n}{\frac{n-k-1}{2}} \frac{2}{n-k+1} $$ (where the binomial coefficient is 0 if $\frac{n-k-1}{2}$ is not an integer or is negative). How can I prove the formula for $D(1,2i)$? (1) I could try to use induction and the recurrence formula, but that seems messy. (2) It seems more nice to provide a bijection between these funny trees and the lattice paths or something on the A151403 page. But this seems tricky with the weights (especially since some are negative). (3) I can turn the recurrence relation into a two variable generating function and get a PDE, but it doesn't look like I can get a solution explicit enough to be useful. I though I would ask in case this is well known to someone with more experience in tree/lattice path counting. REPLY [6 votes]: I was able to find an interesting generalization of your formula, but I'm having trouble finding a reference in the literature. Let's call a 0-1-2 tree, a rooted tree where every vertex can have no child, a left child, a right child, or both. Let's call such a tree increasing if we also have a total ordering on the vertices where each parent is less than their child. We will denote by $B_n$ the set of increasing 0-1-2 trees on $n$ vertices. There is an obvious way to append $n+1$ new leaves to a tree $T\in B_n$ to make it a full binary tree where the internal vertices correspond exactly to the vertex set of $T$. Let's call this binary tree $\tilde{T}$. Suppose we associate a variable $x_1,x_2,\dots,x_{n+1}$ to each leaf. Then every vertex in $\tilde{T}$ has a weighted hook-length given by the sum of the variables associated to each leaf it covers (i.e., to each leaf that is a descendant of this vertex). The weight of $\tilde{T}$ is the product of these hook-lengths over all internal vertices. Let's define a polynomial $\tilde{T}(x_1,x_2,\dots,x_{n+1})$ to be the sum of the weight of $\tilde{T}$ over all permutations of $x_1,\dots,x_{n+1}$. For example starting with the tree in $B_2$ 1 2 we can extend it to 1 1 1 1 1 1 2 z , 2 z , 2 y , 2 y , 2 x , 2 x x y y x x z z x y z z y so its polynomial is $$\tilde{T}(x,y,z)=\sum_{\text{Sym}}(x+y)(x+y+z)=4(x+y+z)^2.$$ We can define $$F_n(x_1,x_2,\cdots,x_{n+1})=\sum_{T\in B_n}\tilde{T}(x_1,x_2,\dots,x_{n+1}),$$ so that, for example, $F_2(x_1,x_2,x_3)=8(x_1+x_2+x_3)^2$, by the above. We also regard the empty graph as a 0-1-2 tree, with weight $1$. Now we can state the result in general: Theorem: For all $n\in \mathbb N$ we have the following identity $$F_n(x_1,x_2,\dots,x_{n+1})=2^n n!(x_1+x_2+\cdots+x_{n+1})^{n}.$$ Proof: Let $\mathcal P'$ be the set of subsets of $\{x_1,x_2,\dots,x_{n+1}\}$ except for the empty set and the full set. First we notice that the root of a binary tree $\tilde{T}$ always has hook-length $(x_1+\cdots +x_{n+1})$. Next notice that two increasing 0-1-2 trees of sizes $a,b$, respectively, combine together to form another 0-1-2 tree of size $a+b+1$, but the ordering can be extended in $\frac{(a+b)!}{a!b!}$ ways. This gives us the recurrence $$F_n(x_1,x_2,\cdots,x_{n+1})=(x_1+x_2+\cdots+x_{n+1})\sum_{S\in \mathcal P'}\binom{n-1}{|S|-1}F_{|S|-1}(S)F_{n-|S|}(S^{c}).$$ This will allow us to prove our theorem inductively. From here we will make use of the weighted Cayley formula for labeled trees on a set $S={1,2,\dots,p}$, which can be obtained from the matrix tree theorem. It says that $$y_1y_2\cdots y_p(y_1+y_2+\cdots+y_p)^{p-2}=\sum_{T}y_1^{\deg 1}\cdots y_p^{\deg p}$$ where the sum on the right runs over all labeled trees and keeps track of the degree of each vertex. If instead we enumerated rooted labeled trees and thought of the root as having a "half edge" (so its degree is increased by 1) then we get $y_1y_2\cdots y_p(y_1+y_2+\cdots+y_p)^{p-1}$. Now a tree on the vertex set $\{1,2,\dots,n+1\}$ can be constructed by making a rooted tree on vertex sets $S\in \mathcal P$ and $S^{c}$ and then joining the roots. This way a tree is constructed exactly $2n$ times: because there are $n$ edges to choose for the splitting, and then ordering the two subsets. So by Cayley's formula we get $$\sum_{S\in \mathcal P}\left(\prod_{i\in S}x_i\left(\sum_{i\in S}x_{i}\right)^{|S|-1}\cdot\prod_{i \in S^c}x_i\left(\sum_{i\in S^{c}}x_i\right)^{n-|S|}\right)=2nx_1x_2\cdots x_{n+1}(x_1+\cdots+x_{n+1})^{n-1}$$ $$\iff \sum_{S\in \mathcal P}\left(\left(\sum_{i\in S}x_{i}\right)^{|S|-1}\cdot\left(\sum_{i\in S^{c}}x_i\right)^{n-|S|}\right)=2n(x_1+\cdots+x_{n+1})^{n-1}$$ which tells us $$(x_1+x_2+\cdots+x_{n+1})\sum_{S\in \mathcal P'}\binom{n-1}{|S|-1}F_{|S|-1}(S)F_{n-|S|}(S^{c})=2^nn!(x_1+x_2+\cdots+x_{n+1})^n$$ giving us the induction step. $\blacksquare$ Now to get to your problem we will put $\frac{n+k+1}{2}$ of the variables equal to $1$, and the remaining ones to $-1$. This is equivalent to enforcing the root to be labeled $k$ in your notation. However my function $$F_n(1,\dots,1,-1,\dots,-1)=2^nn!k^n$$ enumerates the variables as distinct so we have to divide by $\left(\frac{n+k+1}{2}\right)!\left(\frac{n-k+1}{2}\right)!$ in order to make the 1's indistinguishable, and similarly for the -1's. Finally we can say for your function $$D(k,n)=\frac{(2k)^nn!}{\left(\frac{n+k+1}{2}\right)!\left(\frac{n-k+1}{2}\right)!}$$ as you had predicted.<|endoftext|> TITLE: A simplified version of an old problem about the generating function for unrestricted partitions QUESTION [9 upvotes]: This is my latest attempt to simplify an old problem of mine so much that the simplified problem can actually be answered. Starting with the generating function for unrestricted partitions: $$(1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)\ldots$$ Change some of the plus signs in the leftmost expression in parentheses to minus signs. Is it possible that the resulting series has coefficients all of which are $1$, $-1$, or zero? I believe that the answer is no, but I'm not convinced by my computer aided search. REPLY [3 votes]: REVISED No. Let $$A=(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)\ldots$$ and call a polynomial $f=1+\sum_1^da_ix^i$ feasible if all the $a_i \in \{{-1,1\}}$ and all the coefficients of $fA$ up to that of $x^d$ are in $\{{-1,0,1\}}.$ Finally, call $f$ maximal if it is feasible but neither of $f+x^{d+1}$ nor $f-x^{d+1}$ is feasible. There are no feasible polynomials of degree 31. There are $40$ maximal polynomials. The largest degree maximal polynomials are ${x}^{30}+{x}^{29}+{x}^{28}-{x}^{27}-{x}^{26}+{x}^{25}-{x}^{24}-{x}^{23 }-{x}^{22}-{x}^{21}-{x}^{20}+{x}^{19}+{x}^{18}+{x}^{17}-{x}^{16}-{x}^{ 15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{9}+{x}^{8}+{x}^{ 7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-x+1 $ ${x}^{30}+{x}^{29}-{x}^{28}-{x}^{27}-{x}^{26}+{x}^{25}-{x}^{24}+{x}^{23 }-{x}^{22}+{x}^{21}-{x}^{20}+{x}^{19}-{x}^{18}-{x}^{17}+{x}^{16}-{x}^{ 15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x}^{10}-{x}^{9}+{x}^{8}+{x}^{ 7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}-x+1 $ and $-{x}^{25}-{x}^{24}-{x}^{23}+{x}^{22}+{x}^{21}+{x}^{20}+{x}^{19}-{x}^{ 18}+{x}^{17}-{x}^{16}-{x}^{15}+{x}^{14}-{x}^{13}+{x}^{12}-{x}^{11}+{x} ^{10}-{x}^{9}+{x}^{8}+{x}^{7}-{x}^{6}+{x}^{5}-{x}^{4}+{x}^{3}-{x}^{2}- x+1. $ the degrees of the maximal polynomials are $30, 30, 25, 24, 24, 24, 24, 24, 24, 24, 22, 21, 21, 20, 19, 18, 18, 18, 18, 18, 18, 17, 17, 16, 15, 15, 14, 14, 14, 14, 14, 13, 10, 10, 9, 8, 8, 7, 7, 6.$<|endoftext|> TITLE: Chromatic number of the plane and phase transitions of Potts models QUESTION [13 upvotes]: There is a simple connection between ground states of antiferromagnetic Potts models and colorings of the plane: if the unit distance graph of the plane ($G=(\mathbb R^2,\{\{x,y\},d_2(x,y)=1\})$) is admissibly colored with $q$ colors, then this is a ground state of the $q$-state Potts model, and conversely, a ground state without frustration (therefore with trivially computed energy and ground state entropy) yields a coloring of $G$. But the recent advances in our understanding of the ferromagnetic Potts models let us hope that a related statistical physics connection could yield insight into the chromatic number of the plane. Duminil-Copin et al. have proved that the phase transition for the Potts model is continuous up to and including $q=4$ states. And it is discontinuous/1st order for $q\ge 26$ and conjectured to be for 5 and more states. These results are for the square lattice but they have obtained related results on isoradial subgraphs of the plane, many such graphs are subgraphs of $G$. It seems on page 5 they conjecture that for all planar graphs the nature of the transition is the same given a number of states $q$. It seems to me that this should be interpreted as a hint that the chromatic number of the plane is 5. I think that Erdos and other experts never really knew what to believe about that number, which is 4, 5, 6, or 7. The usual example for a non-3-colorable subgraph is Moser's spindle. While a hexagonal tiling with different colors on adjacent tiles give an admissible 7-coloring. The rationale for phase transitions at $q>4$ yielding admissible colorings could be, I am guessing, that if $G$ is $q$-colorable it is possible with $q$ states to have an ordered phase, with an infinite cluster (thus not admissible), at $T=T_t$ (say adiabatically continued raising temperature from $T4$. Furthermore, this is expected to hold for all infinite "2D" graphs (the "universality" property). However, the phase transitions in the 2D AF Potts model are believed to be rather different. It is expected that for some $q_c$ which depends on the lattice, when $q>q_c$ there are no phase transitions at all -- the system is disordered even at zero temperature. When $q TITLE: Source for analysis of identification of structures in learner's mind and mathematical structures? QUESTION [13 upvotes]: Concerning the structure of the learner's mind, psychologist Piaget claimed that There exists, as a function of the development of intelligence as a whole, a spontaneous and gradual construction of elementary logico-mathematical structures and that these 'natural' ('natural' the way that one speaks of the 'natural' numbers) structures are much closer to those being used in 'modern' mathematics than to those being used in traditional mathematics. (p. 79 in Piaget 1973). Piaget appears to postulate an affinity between, on the one hand, the structures of the mind and, on the other, the structures of modern mathematics (mainly following Bourbaki). The essay in question is Piaget, J. "Comments on Mathematical Education," in A. G. Howson, ed., Developments in Mathematical Education: Proceedings of the Second International Conference on Mathematical Education, 79--87, Cambridge: Cambridge University Press, 1973. Piaget's postulated affinity has apparently been challenged by some scholars in the context of the New Math controversy. Question. Is there a source that provides a detailed analysis of such a postulation of a connection between the structures in the learner's mind on the one hand, and Bourbaki-style structures in the foundations of mathematics, on the other? (Note that I am not looking for general sources on the New Math/Modern Math controversy, nor am I particularly interested in Piaget's work in general, but rather for an analysis of this particular identification of Piaget's psychological structures and Bourbaki's mathematical structures). I just came across a book that might be relevant: All Positive Action Starts with Criticism: Hans Freudenthal and the Didactics of Mathematics. By Sacha la Bastide-van Gemert. Springer, 16 Jan 2015 Here the author quotes Freudenthal as follows on page 211: It thus did not begin with the Sputnik shock. It had already begun in the early 1950s. They had even managed to convince Piaget, who did not understand anything of it except for the fact that the word "structure" appealed to him. With Piaget's name on the billboard they felt confident of the support of psychology. What now, psychology! Mathematics is ruled by a logical order and he who teaches mathematics is easily seduced to sacrifice the psychological, the educational order to the logical order. I have done my utmost to avoid this and in my 'fragment Rekendidactiek' of 1942, if not earlier, I wanted to warn others. But what was now happening before my very eyes? A logical order brought to ecstasy, a systematic of mathematics as a whole--that is how mathematics should be taught. It is clear from this that Freudenthal was sceptical of these developments but unfortunately he does not elaborate the details of his objections. REPLY [6 votes]: You say that Freudenthal does not elaborate his objections, but in a sense he did just that in his book Mathematics as an educational task (1973), Appendix I, starting: “The somewhat summary criticism I administered on several occasions to Piaget's work demands more detailed argument.” However, this addresses egregious mathematical misconceptions and methodological flaws (ill-chosen questions) more than the (“ontogeny parallels phylogeny”?) postulation you ask about. Against that, his argument seems much shorter: (p. 46): Bourbaki. How convincing this organization of mathematics is! So convincing that Piaget could rediscover Bourbaki's system in developmental psychology. Poor Piaget! He did not fare much better than Kant, who had barely consecrated Euclidean space as "a pure intuition" when non-Euclidean geometry was discovered! (...) Mathematics is never finished – anyone who worships a certain system of mathematics should take heed of this advice. (p. 192): Impressed by Cantor's analysis he turned to studying the development of the number concept under this aspect. (...) Piaget believed that the concept of natural number could be entirely derived from potencies. (...) This may have seduced him to believe that it is also psychologically true; it was one of his ideas to trace in developmental psychology the system of mathematics he happened to be acquainted with. Or, as reported by F. Goffree in The legacy of Hans Freudenthal (1993): (p. 36): Piaget thought that the cognitive development in children took place from poor to rich structures whereas HF thought it was the other way round. Geometry showed something similar. In the Erlanger Program Klein had given a hierarchy of geometrical structures: topological, projective, affine and euclidean, in HF's terms from poor to rich. Children start by drawing irregular circles, anyone can see that. This was sufficient reason for Piaget to presume that their geometric development began with the topological structure. HF remarked condescendingly that those very same children were quite capable of distinguishing between correctly drawn circles and other figures. (Note added: These last remarks, not sourced by Goffree, are in Revisiting mathematics education (1991), p. 27. So far as I can tell, the criticized “postulation” is most detailed in Épistémologie des mathématiques. Partie II. Chap. 8 & 11 (1961, transl. 1966).)<|endoftext|> TITLE: A simple but curious determinantal inequality QUESTION [14 upvotes]: Let $A$ and $B$ be $n\times n$ Hermitian positive definite matrices and $k>0$ real. Then $A^k$ is well-defined and experimentally, we have $$\det(A^k+BABA^{-1})\geqslant \det(A^k+BA^{-1}BA),$$or equivalently $$\det(A^k+A^{-1}BAB)\geqslant \det(A^k+ABA^{-1}B).$$ Note that this would imply the inequality $\color{green}{\det (A^2+BAAB)\geqslant \det(A^2+ABBA)}$ found (but presumably unproved) by M. Lin, which is mentioned in a comment here. One striking fact is that $B$ (or $A$, same outcome) can be multiplied by a positive factor, which means that the inequality remains valid for any "proportionality" between the two terms of each sum. Indeed, for any $\lambda>0$, we have equivalently $$\det(A^k+\color{red}\lambda BABA^{-1})\geqslant \det(A^k+\color{red}\lambda BA^{-1}BA).$$ Any ideas how to prove this conjecture? REPLY [3 votes]: Let $C = A^{-(k+1)/2} B A B A^{-(k+1)/2}$ and $D = A^{-(k-1)/2} B A^{-1} B A^{-(k-1)/2}$ . We have to show that $det(I+C) \ge det(I+D)$ . Now my goal is to apply equation (5.21) in Ando, Majorizations and Inequalities in Matrix Theory, http://ac.els-cdn.com/0024379594903417/1-s2.0-0024379594903417-main.pdf?_tid=1af72ca2-58c4-11e7-a518-00000aab0f01&acdnat=1498298674_bf9275307496eb46505ba4b5b84a7dcc : Choose $E = A^{-(k+1)/2} B A^{-(k+1)/2}$ and $F = A^{k/2+1}$. Then $C^{1/2} = \vert F E \vert$ and $D^{1/2} = \vert F^{a_1} E^{b_1} F^{a_2} E^{b_2} \vert$ , where $a_1 = (k/2)/(k/2+1)$, $b_1 = 1$, $a_2 = 1/(k/2+1)$, $b_2 = 0$. Since $0 \le a_1 \le b_1$ and $0 \le a_2$ and $0 \le b_2$ and $a_1 + a_2 = b_1 + b_2 = 1$ it follows that $D^{1/2} \prec_{\log} C^{1/2}$ and therefore $D \prec_{\log} C$. Since $log(1+e^x)$ is convex in $x$ it follows that $I+C$ weakly log majorizes $I+D$ and therefore $det(I+C) \ge det(I+D)$ .<|endoftext|> TITLE: Are triangulated equivalence detected at compact level? QUESTION [15 upvotes]: Suppose that $D$ and $E$ are compactly generated triangulated categories, even algebraic (i.e. equivalent to derived categories of small dg categories) if we want, and asume that their subcategories $D^c$ and $E^c$ of compact objects are triangle equivalent. Are $D$ and $E$ triangle equivalent?. By Theorem 9.2 of Keller's 'Deriving dg categories', we know that the answer is 'yes' when either $D$ or $E$ is the derived category of (the category of modules over) a small $k$-linear category, but I do not know the general answer. Any help would be highly appreciated. REPLY [6 votes]: Consider the case where $D$ is the ordinary stable homotopy category of spectra, and $E$ is assumed to have all coproducts (which is clearly a necessary condition), and we also assume that $D^c\simeq E^c$ as tensor triangulated categories. It is an old conjecture of Margolis that this forces $D$ to be equivalent to $E$, and I believe that this is still open in full generality. In both $D$ and $E$ one can define a square-zero ideal of phantom maps. Hovey, Palmieri and I proved that $D/\text{Ph}_D\simeq E/\text{Ph}_E$ and that $\text{Ph}_D\simeq\text{Ph}_E$ as a module over this quotient. Schwede and Shipley proved in https://arxiv.org/abs/math/0108143 that $D\simeq E$ provided that $E\simeq\text{Ho}(E_0)$ for some Quillen model category $E_0$.<|endoftext|> TITLE: Spiral lattice random walk QUESTION [6 upvotes]: Execute a random walk from the origin on the integer lattice, but bias the four compass-direction probabilities from $\frac{1}{4}$ each to prefer to step in a spiraling direction. Calling the four step vectors $c_0,c_1,c_2,c_3$, with $c_i=(\cos (i \, \pi/2), \sin (i \, \pi/2))$, adjust the probabilities as follows. Let $v$ be the vector from the origin to the last point on the path, and $n$ the unit normal to $v$, counterclockwise $90^\circ$ to $v$. Then select step $c_i$ with probability $\frac{1}{4} (1+ c_i \cdot n)$.                     $\theta=60^\circ$. $\frac{1}{4}(1+c_2 \cdot n) = (1+\sqrt{3}/2)/4 \approx 0.47$. Unsurprisingly, the random walks spiral around the origin:             $2000$-step random walks. Origin: green. Last point: red. Three examples, followed by five examples at reduced scale. Q. Does Pólya's recurrence theorem hold for these walks? Do the walks return to the origin with probability $1$? All but one of the above examples (the penultimate) returned to the origin, but usually rather quickly. REPLY [4 votes]: Here is the theorem cited in Serguei Popov's answer: and here are the three assumptions:<|endoftext|> TITLE: Computing a determinant involving roots of unity QUESTION [23 upvotes]: Let $d \geq 2$ be an integer and $\xi=\exp(\frac{2\pi i}{d})$. I am trying to compute the determinant of the matrix $$ (\xi^{ij}-1)_{1 \leq i, j \leq d-1}. $$ Let me call it $\Delta(d)$. For small values of $d$ I get:     $\Delta(2)=-2$     $\Delta(3)=-3\sqrt{3}i$     $\Delta(4)=-16i$ But I can't seem to find a general formula. How can I do this? REPLY [8 votes]: (Too long for a comment) Replacing $\xi$ with a variable $x$, Mathematica gives the following: \begin{align*} \Delta(2)&=-1+x\\ \Delta(3)&=(-1+x)^3 x (1+x)\\ \Delta(4)&=(-1+x)^6 x^4 (1+x)^2 \left(1+x+x^2\right)\\ \Delta(5)&=(-1+x)^{10} x^{10} (1+x)^4 \left(1+x^2\right) \left(1+x+x^2\right)^2\\ \Delta(6)&=(-1+x)^{15} x^{20} (1+x)^6 \left(1+x^2\right)^2 \left(1+x+x^2\right)^3 \left(1+x+x^2+x^3+x^4\right) \end{align*} etc... There is a nice pattern: for $d\ge3$, $\Delta(d)$ contains as factors exactly the cyclotomic polynomials of degree 1 to $d-1$, and $x$, raised to some powers. Edit: I checked up to $d=12$ and, as Benjamin remarks, the exponents seems to be: triangular numbers for the factor $x-1$ (see https://en.wikipedia.org/wiki/Triangular_number) tetrahedral numbers for the factor $x$ (see https://en.wikipedia.org/wiki/Tetrahedral_number) I searched in the OEIS for other exponent sequences, but it returns a lot of possibilities. For example, the exponents over $1+x$ go like this: 1,2,4,6,9,12,16,20,25,30,..., and the OEIS gives 16 results: https://oeis.org/search?q=1%2c2%2c4%2c6%2c9%2c12%2c16%2c20%2c25%2c30 Another edit: Problem solved, see at the end of T. Amdeberhan's answer.<|endoftext|> TITLE: Graded and projective (but not bounded below) module that is not graded-projective? QUESTION [8 upvotes]: Let $A = \bigoplus_{n = 0}^\infty A_n$ be a graded algebra over a field $k$ that is locally finite: each $A_n$ is a finite-dimensional $k$-vector space. We say that a graded left $A$-module $P = \bigoplus_{n \in \mathbb{Z}} P_n$ is graded-projective if there is a graded free module $F = \bigoplus A(l_i)$ (for some shifting parameters $l_i \in \mathbb{Z}$) and a graded module $Q$ such that $F \cong P \oplus Q$ as graded modules. We say that $P$ is bounded below if $P_n = 0$ for $n \ll 0$. In section 2 of The structure of AS Gorenstein algebras by Minamoto and Mori, it is shown that if $P$ is a left $A$-module that is graded, projective (as an "ungraded" module), and bounded below, then it is graded-projective. Nakayama's lemma seems crucial to their methods, so I expect something to go wrong if $P$ is not bounded below. Question: Is there a locally finite graded algebra $A$ and a graded left $A$-module $P$ that is projective, not bounded below, and not graded-projective? If something indeed does go wrong, I would already expect it to occur in the case where $A$ is connected, meaning that $A_0 = k$. (This question arose earlier on Math StackExchange.) REPLY [6 votes]: The answer to your question is no. More generally, the properties of being graded-projective or projective are equivalent in settings far more general than yours. See C. Nastasescu, F. Van Oystaeyen, Graded Ring Theory, North-Holland Mathematical Library 28, 1982.<|endoftext|> TITLE: Simply connected slices QUESTION [17 upvotes]: Assume $\Omega$ is an open set in $\mathbb R^3$ such that the intersection of $\Omega$ with any horizontal plane is simply connected. Can you prove that $\Omega$ is simply connected? (Note that by the definition, simply connected set can not be empty.) Comments. The proof given by Tom Goodwillie below is done with bare hands. I would prefer to find ready to use tool for answering this and similar questions. REPLY [15 votes]: Yes, I think so. Let's show that every compact set $K\subset \Omega$ is contained in some compact contractible subset of $\Omega$. We use the fact that in a simply connected open subset of the plane every compact set is contained in some compact contractible set. Denote by $P_t$ the plane $\mathbb R^2\times t$, and define the set $\Omega_t\subset\mathbb R^2$ by $\Omega_t\times t=\Omega\cap P_t$. Define $K_t$ likewise. For each $t$ choose a compact contractible set $C_t\subset \Omega_t$ such that $K_t\subset C_t$. There must be an interval $J_t$ containing $t$ such that for every $t'\in J_t$ we have $K_{t'}\subset C_t\subset \Omega_{t'}$. The set of all $t$ such that $K_t$ is nonempty can be covered by finitely such intervals. Thus for some $a$ there are real numbers $s_0<\dots TITLE: What two ordinals are these (based on definable ordinals)? QUESTION [6 upvotes]: Let $D$ be the set of definable ordinals. An ordinal s is definable if there is a predicate $p$ (in the language of (first-order) set theory), such that $p(x) \iff x=s$ for all $x$. This is definitely a set (not a proper class), since the list of all syntactically valid predicates is countable, and $D$ is $\le$ to that in size. In particular, $D$ is countable (although most of its elements are not). We can define two ordinals based on $D$: The ordinal $\delta$ defined as the least ordinal not in $D$. This ordinal is countable (since $D$ is countable, and so can not contain every countable ordinal), and has the property that an ordinal less than $\delta$ iff it is recursively definable (i.e., every ordinal $x<\delta$ is definable and every $y TITLE: Very weak square and good points QUESTION [7 upvotes]: This is probably well known but I'll appreciate pointers to references: Is there any model where for a singular cardinal $\kappa$ of cofinality $\omega$, Very Weak Square holds at $\kappa$ but every scale on $\kappa$ has stationarily many bad points of cofinality $\omega_1$? Definitions: Very Weak Square at $\kappa$: there is a sequence $\langle C_\alpha: \alpha<\kappa^+\rangle$ where for a club of $\alpha\in \kappa^+$ $C_\alpha\subset \alpha$ is unbounded for any bounded $x\in [C_\alpha]^\omega$, there is $\beta<\alpha$ such that $x=C_\beta$. Scales at $\kappa:$ For some $\langle \kappa_i: i\in \omega\rangle$ increasing cofinal regular cardinals, $\langle f_\alpha\in \prod_{i\in \omega} \kappa_i: \alpha<\kappa^+\rangle$ is a scale if it is increasing and cofinal in $<^*$ (increasing mod finite). $\alpha<\kappa^+$ is a good point if $cf(\alpha)>\omega$ and there is $A\subset \alpha$ cofinal and $m\in \omega$ such that $\{f_\alpha\restriction_{>m}: \alpha\in A\rangle$ is pointwise increasing. $\alpha$ is bad if it's not good. It is known that the assertion is false at $\aleph_\omega$ (Foreman-Magidor) as VWS at $\aleph_\omega$ implies there is a club $C\subset \aleph_{\omega+1}$ such that all $\alpha\in C\cap cof(\omega_1)$ is good. But the proof goes through an intermediate principle (approachability) and VWS does not in general imply AP at larger singular cardinals. REPLY [8 votes]: Though Very Weak Square at $\kappa$ does not in general imply $\mathrm{AP}_\kappa$, it does imply that $\kappa^+ \cap \mathrm{cof}(\omega_1)$ is in the approachability ideal on $\kappa^+$. In fact, the Very Weak Square sequence $\langle C_\alpha \mid \alpha < \kappa^+ \rangle$ is itself a witness to this, provided that we have $\mathrm{otp}(C_\alpha) = \omega_1$ for all $\alpha \in \kappa^+ \cap \mathrm{cof}(\omega_1)$, which can always be arranged. It then follows from the usual arguments, as in the Foreman-Magidor paper on Very Weak Square, that, in a scale on $\kappa$, the set of bad points of cofinality $\omega_1$ is non-stationary.<|endoftext|> TITLE: Properties of coefficients in expansion of $E_6/E_4$ and $E_8/E_6$ QUESTION [9 upvotes]: Let $a(n)$ and $b(n)$ be define by the following; $E_6/E_4 = 1 - 744q + 159768q^2 - 36866976q^3 + 8507424792q^4 - 1963211493744q^5 + \cdots = \Sigma a(n)q^n,$ $E_8/E_6 = 1 + 984q + 574488q^2 + 307081056q^3 + 164453203992q^4 + 88062998451984q^5 + \cdots = \Sigma b(n)q^n.$ (Please see A288261( https://oeis.org/A288261 ) and A288840 ( https://oeis.org/A288840 ).) Are $\frac{a(5^m * n) - a(n)}{3000}$ and $\frac{b(5^m * n) - b(n)}{3000}$ integers for all $m, n >= 0$? (For example) $\frac{a(5) - a(1)}{3000} = \frac{-1963211493744 - (-744)}{3000} = -654403831$. $\frac{b(5) - b(1)}{3000} = \frac{88062998451984 - 984}{3000} = 29354332817$. REPLY [14 votes]: The congruence is true for all $m,n$, and Manyama's calculations for $n \leq 500$ are more than enough to prove it. Similar congruences hold for the coefficients of other quotients of modular forms; e.g. the $q^n$ and $q^{5n}$ coefficients of $E_6/E_8$ are also congruent $\bmod 3000$ for all $n$, while those of $E_4/E_6$ are also congruent $\bmod 120$. As David Loeffler noted already in a comment, once we know the $m=1$ congruences $$ a(5n) \equiv a(n) \bmod 3000, \qquad b(5n) \equiv b(n) \bmod 3000 $$ for all $n$, the congruences for $m \geq 2$ follows by induction. So we need only prove it for $m=1$. The key is the following observation: for any power series $F(q) = \sum_n c_n q^n$ and any integer $r>0$, the series $F^{[r]}(q) := \sum_n c_{rn} q^n$ equals $r^{-1} \sum_{q_1^r = q} F(q_1)$, the sum extending over the $r$ complex roots $q_1$ of $q$. In our setting, $F$ is a meromorphic modular form for $\Gamma(1) = {\rm SL_2}({\bf Z})$, so $F^{[r]}$ is a meromorphic modular form of the same weight for the congruence subgroup $\Gamma_0(r)$ --- indeed it is a linear combination of $F$ with the image of $F$ under the Hecke operator $T_r$. Thus the same is true of $F^{[r]} - F$. So, to show that all the coefficients of $F^{[r]} - F$ are multiples of $M$, it is enough to recognize $(F^{[r]} - F) / M$ as a meromorphic modular form with integer coefficients. This is essentially routine with a package such as gp once we account for the poles of $F$ and $F^{[r]}$. Both $F = E_6/E_4$ and $F = E_8/E_6$ are meromorphic forms of weight $2$, and the desired congruence has $r=5$ and $M=3000$. The modular curve $X_0(5)$ associated to $\Gamma_0(5)$ is rational, parametrized by the "Hauptmodul" $$ h = (\eta_1/\eta_5)^6 = \frac1q \prod_{k=1}^\infty \left(\frac{1-q^k}{1-q^{5k}} \right)^{\!6} = q^{-1} - 6 + 9q + 10q^2 - 30q^3 + 6q^4 - 25q^5 \cdots; $$ and weight-2 meromorphic forms are rational functions of $h$ times the Eisenstein series $$ e_2 = -q \frac{dh/dq}{h} = 1 + 6q + 18q^2 + 24q^3 + 42q^4 + 6q^5 + 72q^6 + 48q^7 + \cdots. $$ Now if $F = E_6/E_4$ then $F$ has simple poles at the zeros of $j=0$, which are the $\Gamma$ orbit of the cube root of unity $\rho = (-1 + \sqrt{-3})/2$; so $F^{[5]}$ has simple poles at the $\Gamma$ orbits of $(\rho+a)/5$, and it is known that those are the points where $j$ is a root of some quadratic polynomial, namely $X^2 + 654403829760 X + 5 \cdot 101376^3 =: Q(X)$. So $j \, Q(j) (F^{[5]}-F) / e_2$ is a modular function on $X_0(5)$ with no poles except at the cusps (the zeros of $e_2$ turn out to cancel automatically); since $F^{[5]}-F$ is regular at the cusps, multiplying by a cubic in $j$ yields a linear combination of $\sum_{t=-15}^3 \alpha_t h^t$, which we can determine by computing the first $19$ coefficients of $j Q(j) (F^{[5]}-F) / e_2$ (we actually computed twice as many coefficients for a sanity check). We find that the integers $\alpha_n$ are all multiples of $3000$, which proves the congruence $a(5n) \equiv a(n) \bmod 3000$ because $h$, and thus also each $h^t$ with $t \in \bf Z$, has integer coefficients and leading coefficient $1$. The case $F = E_8/E_6$ is treated similarly; here the simple poles are at roots of $j-1728$, which are the $\Gamma$ orbit of $i = \sqrt{-1}$, and $F^{[5]}$ has simple poles at the $\Gamma$ orbits of $(i+a)/5$, where $j$ is either $1728$ again (for $a \equiv \pm 2 \bmod 5$) or a root of the quadratic $Q_1(X) := X^2 - 44031499226496X - 6635376^3$. Again we calculate that $$ (j-1728) Q_1(j) (F^{[5]}-F) / e_2 = \sum_{t=-15}^3 \beta_t h^t $$ with each $\beta_t \in 3000 \bf Z$, which proves the congruence $b(5n) \equiv b(n) \bmod 3000$.<|endoftext|> TITLE: Adams spectral sequence and short exact sequences. Some clarifications QUESTION [8 upvotes]: as the title suggests I'm looking for some clarifications in the computations of the ext charts of some $A(1)$-modules arising as extensions of other modules. In particular, I've the following example I'd like to consider: $$0 \to \Sigma^3D \to \Bbb R P^{\infty} \to M \to 0$$ where $D$ is the $A(1)$-module described in Freed and Hopkins' paper at page 75 (table on the right), $M$ is the Moore space (whose $A(1)$-resolution can be found here, and $\Bbb R P^{\infty}$ is the infinite projective space. Here I draw the s.e.s with the usual convention for the action of $Sq^1$ and $Sq^2$: And here you can find what (according to me) is the situation at the level of ASS: The green arrows should be the map representing the connecting homomorphisms in the l.e.s. of ext-groups induced by the above s.e.s. Since we know what should be the stable page of the ASS for $\Bbb R P^{\infty}$, and there are no differentials for dimensions reason, we see that the only non-trivial green arrows are the one starting in position $(5,1)$ and $(6,2)$. This is what I don't understand: If the arrows represent the maps induced by the l.e.s. of the ext's then at least the first one $$\delta \colon \hom_{A(1)}(\Sigma^3D, \Bbb F_2) \to Ext^1_{A(1)}(M, \Bbb F_2)$$ shouldn't be trivial, since the s.e.s. above is non-split. In particular this would be the green arrow starting from position $(1,0)$, which we observed that has to be zero. So what am I missing? I apologise in advance if the question is stupid. REPLY [4 votes]: The red dot in (3,0) comes from a map $\Sigma^3 D \to \mathbb{F}_2$, and this map is the image of a map $\mathbb{R}P^\infty \to \mathbb{F}_2$, so it goes to zero under the coboundary map. This agrees with your first claim. I think the correct statement about splitting is: the s.e.s. is split if and only if, for all $A$-modules $N$, the coboundary map $\textrm{hom}_A(\Sigma^* D, N) \to \textrm{Ext}_A^1(M, N)$ is zero. (Is that right?) In particular, the coboundary map may be zero for some choices of $N$ even if the original s.e.s. is not split.<|endoftext|> TITLE: Invertibility of the Schur Complement QUESTION [5 upvotes]: Suppose that $$ M = \begin{bmatrix}A & B \\ C & D\end{bmatrix}. $$ I know that if $D$ and $M\setminus D$ (where $M\setminus D$ is the Schur Complement of $D$ in $M$) are invertible, then $M$ is invertible. However, suppose that we know that $M$ and $D$ are invertible, but we know nothing about the invertibility of $A$. Can we say that $M\setminus D$ is invertible? REPLY [3 votes]: Yes, if you take the determinants, you obtain with $$\operatorname{det}(M)=\operatorname{det}(M/ D)\cdot\operatorname{det}(D) $$ therefore if $\operatorname{det}(M)$ is non-zero then $\operatorname{det}(M/ D)$ non-zero.<|endoftext|> TITLE: Research-only permanent positions worldwide QUESTION [28 upvotes]: Most academic jobs involve some amount of teaching. Post-docs generally do not, but they are only short-term positions. Question: in which countries can one obtain a research-only permanent position in mathematics? Please provide a link to a relevant website if possible. Please mention only one country per answer, and since there is obviously no best answer, this is a community-wiki question. REPLY [4 votes]: Italy: CNR. Calls for positions are published online here (in Italian only and difficult to search for discipline). :( Also, Scuola Normale Superiore di Pisa, Scuola Internazionale Superiore di Studi Avanzati, and the Gran Sasso Science Institute are not research-only, but they have only honors and PhD courses with a very small number of students.<|endoftext|> TITLE: Reference for Mori program QUESTION [6 upvotes]: Are there some introductory and self-contained books or lecture notes for Mori-program? It would be much better to have enough examples. REPLY [6 votes]: Two good introductions to the Mori program and the geometry of higher dimensional varieties are Kenji Matsuki, Introduction to the Mori program, Universitext. New York, NY: Springer (ISBN 0-387-98465-8/hbk). xxiii, 478 p. (2002). ZBL0988.14007. Olivier Debarre, Higher-dimensional algebraic geometry, Universitext. New York, NY: Springer. xiii, 233 p. (2001). ZBL0978.14001.<|endoftext|> TITLE: optimal estimate for generalized Kloosterman sum QUESTION [7 upvotes]: Let $p$ be an odd prime. Denote $e(x):=e^{2\pi i\frac{x}{p}}$. Let $n\ge 2$ be an integer. Consider the exponential sum $$ S(f,g)=\sum_{g(x_1,\dots,x_n)=0, x_i\in\mathbb{F}_p}e(f(x_1,\dots,x_n)), $$ where $f$ and $g$ are polynomials of $n$ variables. Such $S(f,g)$ is common. For example, Kloosterman sum can be written in this form. I am seeking conditions on $f$ and $g$ so that we have the optimal estimate \begin{equation} S(f,g)\ll p^{(n-1)/2}. \end{equation} My thoughts so far: if from $g(x_1,\dots,x_n)=0$ we can write $x_1$ in terms of other variables in a polynomial way, then $S(f,g)$ becomes a sum over the full $(n-1)$-dim space $\mathbb{F}_p^{n-1}$ and $f$ becomes a polynomial of $n-1$ variables. Then a theorem of Deligne says that we just need to check whether the homogeneous leading term of $f$ defines a smooth projective hypersurface. But I do not know how to deal with general (or other) $g$. For example, what about the case $g(x_1,\dots,x_n)=h(x_1)+\dots+h(x_n)$ for some single-variable polynomial $h$? If $h$ is quadratic, then we can't write $x_1$ in terms of other variables in a polynomial way at all. It seems to me that Deligne's paper already considers a much more general sum than $S(f,g)$. But the Deligne's criterion to get the optimal estimate is abstract and hard to check for people who do not know algebraic geometry. I hope that in the simpler case of $S(f,g)$ there is an easy-to-check criterion on $f$ and $g$. REPLY [5 votes]: The work of Katz (https://web.math.princeton.edu/~nmk/sommesexp.pdf, 5.1.1, see also the beginning of https://web.math.princeton.edu/~nmk/ESES.pdf for an English exposition) provides an analogue of Deligne's criterion. Katz's $X$ here will be the projective closure of the solution set of $g$, his $L$ will be the hyperplane at $\infty$, and his $H$ will be the vanishing locus of $f$. One demands that $L$ is transverse to $X$, which is equivalent to saying that the leading term of $g$ defines a smooth projective hypersurface, that the degree of $f$ is prime to $p$, and that $H$ is transverse to $X \cap L$, which is equivalent to saying that the leading terms of $f$ and $g$ together define a smooth codimension $2$ subvariety in projective space. Of course, if these smoothness assumptions don't hold, then in general you may need to use monodromy, Fourier transform, vanishing cycles, etc. to solve the problem. (This criterion doesn't even apply to the classical Kloosterman sum for $n>2$.) Essentially the boundary between the cases that admit square-root cancellation and those that don't should be very complicated in general, and no general black-box result is expected to cover a significant fraction of the cases where square-root cancellation can be proved.<|endoftext|> TITLE: How do we construct the Gödel’s sentence in Martin-Löf type theory? QUESTION [27 upvotes]: In Martin-Löf dependent type theory (MLTT), under the proposition-as-types correspondence, we sometimes say that a proposition $A$ is true if the type $A$ is inhabited. However, there is no doubt that MLTT meets the minimum criteria of expressiveness of arithmetic required by Gödel’s first incompleteness theorem to apply, which then would imply the existence of "true but improvable propositions". I am aware that the first, intuitionistic, notion of 'provability' (the one inherent to MLTT under the proposition-as-types) refers to provability in general and that the term 'improvable' in the incompleteness theorems refers to non-derivability in a given formal system. Even so, I find this all very confusing. Especially because if there are true but improvable propositions in MLTT, then there is a type $G$ that is true, but the judgment $g : G$ is not derivable within the system for any term $g$! My doubts are further aggravated by the fact that most material on Gödel's incompleteness theorems that can be found on the internet typically only cover first-order logical systems. To be precise, I am looking for type theory what has been done for category theory in this proof (§2) in this nLab entry on the incompleteness theorems (I am also looking for something slightly more accessible, even though I have a basic understanding of category theory). In any case, I will be very glad if anyone could shed some light in these issues. In particular: How can we construct the Gödel’s sentence $G$ in MLTT? How come there is no contradiction if $G$ is true but it is not inhabited? What interpretation are we referring when we say that $G$ is true? Any help is highly appreciated! REPLY [4 votes]: In CTT truth of a proposition, indeed, is expressed as the instantiation of its set of proofs: $$ A \text{ is true } = \mathsf{Proof}(A) \text{ exists} \tag{1} $$ where the existence is the Brouwer-Weyl constrcutive notion of existence, defined by the assertion condition: $$a \text{ is an } \alpha \implies \alpha \text{ exists} $$ Consider now what we may call the "arithmetical fragment" of CTT, given in the language that comprises the connectives and quantifiers, and the natural numbers, plus the appropriate constructors, with the usual introdcution and elimination rules for the connectives and quantifiers, and the definition by recursion as elimination rule for the set $\mathbb{N}$. Question 1: Gödel's theorem then takes the form: there is an an arithmetical proposition $G$, for which we can construct a proof-object, but this cannot be done using just the constructors pertaining only to the arithmetical fragment. This takes care of question 1. Question 2: The set $\mathsf{Proof}(G)$ is instantiated by a proof-object $t$, but this cannot be given just by using the "arithmetical constructors". Question 3: the standard CTT notion of propositional truth given by $(1)$. I have dealt with these matters in some detail in my 'Antirealism and the Roles of Truth' in the Handbook of Epistemology. The article can be found here and §8, pp. 456-459, spells out the Gödel issues for CTT.<|endoftext|> TITLE: A closed formula for $A_n(X)=\sum\limits_{i=0}^n X^{i^2}$ QUESTION [9 upvotes]: I want to know if there exists a closed formula for sum $A_n(X)=\sum \limits_{i=0}^n X^{i^2}$. I have found if n is odd then $(X^n+1)\text{ | } A_n(X)$, but I don't have found a closed formula. REPLY [6 votes]: Not at all an answer, just observations. Firstly, a conjecture, based on experiment. CONJECTURE $A_{2n}$ is irreducible for all $n.$ (empirically true for all $n\leq20.$) CONJECTURE 2 $A_{2n+1}/(x^{2n+1}+1)$ has at most one non-cyclotomic factor. (empirically true for all $n\leq 20$). Finally, the roots of these things cluster around the unit circle (see the graphic for $A_{14}$). Is it true that the zeros of the infinite series are on the unit circle? (oops, this is not supposed to be a question).<|endoftext|> TITLE: A weak version of high dimensional Abhyankar's conjecture QUESTION [5 upvotes]: I am encountering the following situation which is similar to the Abhyankar's higher dimensional conjecture on étale fundamental groups, but with much stronger assumptions: Let $S$ be a finitely generated subring of $\mathbb{C}$, let $X$ be a smooth affine variety over $S$, and let $G$ be a finite group such that the following holds. For any large enough prime $p$ and a base change $S\to k$ to an algebraically closed field of characteristic $p$, the variety $X_{k}$ admits a Galois covering with the Galois group $G$. Does this imply that $X_{\mathbb{C}}$ also admits a Galois covering with the group $G?$ I believe the answer is "yes" if $X$ is a curve, or is a complement of divisors with normal crossings in a projective space (by a result of Abhyankar). Any suggestions or references would be greatly appreciated. REPLY [7 votes]: Here's a sketch of a possible way: Since $p$ is large enough, we can assume that $G$ is of order prime to $p$. If $X$ was projective, then the prime-to-$p$ completions of $\pi_1(X_{\mathbb{C}})$ and $\pi_1(X_{\bar k})$ would be isomorphic by the results of SGA1, and we would be done. To "reduce" to this case, we might assume that we have a smooth compactification $\bar X$ of $X$ such that $\bar X\setminus X$ is a divisor with simple normal crossings. In this case we can use Abhyankar's lemma to extend the covering of $X_{\bar k}$ to a tamely ramified covering of $\bar X_{\bar k}$. The magic of log geometry should allow one to lift this covering to characteristic zero exactly as in the case without ramification. In particular, a single $p$ (satisfying some good reduction hypotheses) should be enough. EDIT. Let me add some details. First, a definition (the terminology is not standard). Definition. (1) (cf. SGA1 Exp. XIII, 2.1) Let $\pi:X\to S$ be a smooth morphism of schemes, $D\subseteq X$ an effective Cartier divisor on $X$. We say that $D$ has normal crossings relative to $S$ if locally on $X$, there exists an etale $S$-morphism $g:X\to \mathbf{A}^n_S$ such that $D=g^*(D(x_1\cdot\ldots\cdot x_r))$ for some $r$. We will call $(X, D)$ a log smooth pair over $S$. (2) We say that a finite morphism $f:Y\to X$ over $S$ is a tame cover of a log smooth pair $(X, D)$ if etale locally on $Y$, there exists a morphism $g:X\to \mathbf{A}^n_S$ as in (1), integers $e_1 ,\ldots, e_r$ invertible on $S$, and an isomorphism over $X$ between $Y$ and the pullback of $(x_1, \ldots, x_n)\mapsto (x_1^{e_1}, \ldots, x_r^{e_r}, x_{r+1}, \ldots, x_n):\mathbf{A}^n_S\to \mathbf{A}^n_S$ along $g$. Then Abhyankar's lemma (see SGA1 Exp. XIII, Appendice I, Proposition 5.5) can be phrased as follows: Lemma 1. Assume that $S={\rm Spec}\, k$, and let $(X, D)$ be a log smooth pair over $S$. Set $U=X\setminus D$. Then the restriction functor $$ \left( \text{tame covers of }(X,D)\right) \longrightarrow \binom{\text{finite etale covers of }U}{\text{tame along }D} $$ is an equivalence. The "log magic" I referred to above is hidden in the proof of the following lemma: Lemma 2. Let $S\to \widetilde{S}$ be a nilpotent closed immersion, let $(\widetilde X, \widetilde D)$ be a log smooth pair over $\widetilde S$, $(X, D)$ its base change to $S$. Then the restriction functor $$ \left(\text{tame covers of }(\widetilde X, \widetilde D)\right) \longrightarrow\left(\text{tame covers of }(X, D)\right)$$ is an equivalence. This can be deduced from Kato's first article on log geometry "Logarithmic structures of Fontaine-Illusie", section 3. (Probably there is an easier way of seeing this directly.) The point is that "tame covers" are log etale if $X$ and $Y$ are given their natural log structures. If we replace $\Omega^1_{Y/X}$ with its variant with log poles, we get the zero sheaf, which is why we obtain a unique extension property for tame covers. From this we obtain the result you need. More precisely, suppose that $R$ is a henselian dvr with residue field $k$ and that $(X, D)$ is a log smooth pair over $S={\rm Spec}\, R$ such that $X$ is proper over $S$. Let $U=X\setminus D$. If $V_k\to U_k$ is a connected $G$-torsor, where $G$ is of order invertible in $k$, then Lemma 1 provides an extension to a connected $G$-tame cover $Y_k$ of $(X_k, D_k)$. Applying Lemma 2 to all ${\rm Spec}\,k\to {\rm Spec}\, R/m^n$, we obtain a connected $G$-tame cover $\mathcal{Y}$ of the formal completion $(\mathcal{X}, \mathcal{D})$. By Grothendieck's existence theorem applied to $f_*\mathcal{O}_{\mathcal{Y}}$, this comes from a connected $G$-tame cover $Y$ of $(X, D)$, which stays connected on the geometric generic fiber $(X_{\bar K}, D_{\bar K})$. Restricting to $U_{\bar K}$, we obtain a desired $G$-cover of your affine scheme $U_{\bar K}$.<|endoftext|> TITLE: Terminology for this notion of "$\sigma$-algebra" in a topos QUESTION [11 upvotes]: Let $\mathcal{E}$ be a Grothendieck topos. I want to define a sort of "$\sigma$-algebra" for it, and I'm asking about what it should be—or already is—called. I know from nlab that Cheng spaces are an accepted constructive notion of $\sigma$-algebras but I don't yet see why I might find them interesting. So one may answer this question by instead telling me why my notion (see below) may not be well-behaved, or what's so nice about Cheng spaces. But let's leave that question aside for now. In the proposed definition below, I'll use the internal language of $\mathcal{E}$. I'll use the convention that $A\Rightarrow B\Rightarrow C$ means $A\Rightarrow(B\Rightarrow C)$. I'll use $\lambda$ notation for terms of exponential objects; for example given types $X$ and $P$ and a term $p:P$, I'll write $\lambda(x:X)\ldotp p$ to denote the function $X\to P$ that is constant at $p$. Definition: Let $\mathcal{E}$ be a Grothendieck topos, $\top\colon 1\to\Omega$ its subobject classifier, and $\mathbb{N}$ its natural numbers object. A pseudo-$\sigma$ algebra in $\mathcal{E}$ consists of an object $X$ and a morphism $\Sigma\colon \Omega^X\to\Omega$ satisfying the following axioms: $\forall(\omega:\Omega)\ldotp\Sigma\big(\lambda(x:X)\ldotp\omega\big)\;\;\;\;$ ["constant propositions are measurable"] $\forall(\phi,\psi:X\to\Omega)\ldotp \Sigma(\phi)\Rightarrow\Sigma(\psi)\Rightarrow\Sigma(\phi\wedge\psi)\;\;\;\;$ ["closed under $\wedge$"] $\forall(\phi,\psi:X\to\Omega)\ldotp \Sigma(\phi)\Rightarrow\Sigma(\psi)\Rightarrow\Sigma(\phi\vee\psi)\;\;\;\;$ ["closed under $\vee$"] $\forall(\phi,\psi:X\to\Omega)\ldotp \Sigma(\phi)\Rightarrow\Sigma(\psi)\Rightarrow\Sigma(\phi\Rightarrow\psi)\;\;\;\;$ ["closed under $\Rightarrow$"] $\forall(\phi:\mathbb{N}\to\Omega^X)\ldotp\big(\forall(n:\mathbb{N})\ldotp\Sigma(\phi(n))\big)\Rightarrow\Sigma\big(\forall(n:\mathbb{N})\ldotp\phi(n)\big)\;\;\;\;$ ["closed under countable $\forall$"] $\forall(\phi:\mathbb{N}\to\Omega^X)\ldotp\big(\forall(n:\mathbb{N})\ldotp\Sigma(\phi(n))\big)\Rightarrow\Sigma\big(\exists(n:\mathbb{N})\ldotp\phi(n)\big)\;\;\;\;$ ["closed under countable $\exists$"] For any $\phi:X\to\Omega$, say that $\phi$ is measurable if $\Sigma(\phi)=\top$. Question: What might be a more common, more useful, or better terminology for what I've called pseudo-$\sigma$ algebras? I guess it's different than just a sub-Heyting algebra of $\Omega^X$ because of the "countable quantification". REPLY [4 votes]: This is an answer to the new question formulated in the comments. Point-set notions of topological spaces are a poor fit for arbitrary toposes because constructing points typically requires the axiom of choice, whereas constructing the lattice of open sets does not. This is the reason why locales are used instead of topological spaces in this context. This can be concisely illustrated by the following fact: in the presence of the axiom of choice, locally compact locales as well as compact regular locales are spatial, i.e., come from topological spaces. For example, open sets in the Zariski spectrum of a commutative ring are radical ideals, whereas points are prime ideals, and proving that a ring has sufficiently many prime ideals requires the axiom of choice, whereas proving that it has sufficiently many radical ideals does not. Similar statements can be made about C*-algebras and other types of rings. The situation for measure spaces is similar. First, if one looks at typical theorems in measure theory, it becomes clear that the data of a σ-algebra alone is never sufficient to formulate an interesting theorem in measure theory; one needs to know at least what the sets of measure 0 are (for more details, see Is there an introduction to probability theory from a structuralist/categorical perspective?). If we adopt this new notion of measurable space (i.e., a set X, a σ-algebra M of measurable subsets, and a σ-ideal N of negligible sets, equivalently, a measure class), the resulting category of measurable spaces is equivalent to the category of measurable locales (see https://ncatlab.org/nlab/show/measurable+locale): given a measurable space (X,M,N), send it to the poset (a frame, in fact) M/N, i.e., equivalence classes of measurable sets modulo negligible sets. One can then illustrate the above points (pun intended) for measurable spaces by considering the equivalence between commutative von Neumann algebras, measurable locales, and measurable spaces. A measurable locale can be extracted from a von Neumann algebra in a canonical fashion: it is simply the complete Boolean algebra of its projections. To extract a measurable space, however, one needs the axiom of choice: the points of the measurable space are maximal C*-ideals in the von Neumann algebra. One must mention that measure theory in a topos has not yet been developed to the same extent that general topology is. For instance, the equivalence between commutative C*-algebras and compact regular locales is proven in a very satisfactory form in arXiv:1411.0898v1. There is no similar exposition for commutative von Neumann algebras and measurable locales in the literature. (I had some correspondence with Simon Henry about this, and the conclusion was that it seems like it can be done, but no details can be found in the literature.)<|endoftext|> TITLE: A question about the generalization of filters QUESTION [7 upvotes]: Let $W$ be any non-empty set. We call a subset $A$ of $\mathscr{P}(W)$ a 3-fold filter on $W$ if the following hold: (i) $W\in A$ and $\varnothing\notin A$; (ii) If $a\in A$ and $a\subset b$ then $b\in A$; (iii) If $a,b,c\in A$ then $(a\cap b)\cup(b\cap c)\cup(c\cap a)\in A$. Now let $A,B$ be two 3-fold filters on $W$. Let $X=\{a\times W\mid a\in A\}$ and $Y=\{W\times b\mid b\in B\}$. Prove that $X\cup Y$ can be extended to a 3-fold filter on $W\times W$, i.e., $\varnothing$ is not contained in the closure of $X\cup Y$ under the operation $(a\cap b)\cup(b\cap c)\cup(c\cap a)$. REPLY [5 votes]: First, a bit of notation: if $a_1,\dots,a_n\subseteq W$, let $\chi_{a_1,\dots,a_n}\colon W\to\{0,1\}^n$ denote the “$n$-fold characteristic function” $$\chi_{a_1,\dots,a_n}(u)=([u\in a_1],\dots,[u\in a_n]),\qquad u\in W,$$ where $[\ ]$ is the Iverson bracket. For a Boolean function $f\colon\{0,1\}^n\to\{0,1\}$, let $S_f\colon\mathcal P(W)^n\to\mathcal P(W)$ be the corresponding set operator $$S_f(a_1,\dots,a_n)=\{u\in W:f(\chi_{a_1,\dots,a_n}(u))=1\}.$$ In this notation, a 3-fold filter is a nonempty upper subset $A\subseteq\mathcal P(W)$ closed under $S_m$, where $m(x,y,z)$ is the ternary majority function. (I am allowing $\varnothing\in A$ here; I will call a 3-fold filter proper if $\varnothing\notin A$.) I will also need a little information on Boolean clones, see Post’s lattice. The majority function $m(x,y,z)$ generates the clone of all monotone self-dual Boolean functions. It follows that for a nonempty set $X\subseteq\mathcal P(W)$, the 3-fold filter generated by $X$ consists of sets $a$ such that $$a\supseteq S_f(a_1,\dots,a_n)$$ for some $a_1,\dots,a_n\in X$, and a monotone self-dual function $f\colon\{0,1\}^n\to\{0,1\}$. It will be actually more convenient to use a slightly different characterization. Since 3-fold filters are closed upwards, they are closed under $S_f$ not only for monotone self-dual $f$, but also for Boolean functions $f$ such that $f(x_1,\dots,x_n)\ge g(x_1,\dots,x_n)$ for some monotone self-dual function $g$. Now, the collection of all such functions $f$ forms the clone denoted $T_1^2$ on the linked page; it can be described intrinsically as the set of all functions $f\colon\{0,1\}^n\to\{0,1\}$ such that $$\alpha\lor\beta=\mathbf1\implies f(\alpha)\lor f(\beta)=1$$ for all $\alpha,\beta\in\{0,1\}^n$, where $\alpha\lor\beta$ is defined coordinatewise, and $\mathbf1=(1,\dots,1)$. (That is, $f$ is polymorphism of the relation $\{(1,1),(1,0),(0,1)\}$.) So, the 3-fold filter generated by $X\subseteq\mathcal P(W)$ is $$\langle X\rangle=\{a\subseteq W:\exists a_1,\dots,a_n\in X\,\exists f\in T_1^2\,a\supseteq S_f(a_1,\dots,a_n)\}.$$ Lemma: If $A\subseteq\mathcal P(W)$ is a proper 3-fold filter, and $a_1,\dots,a_n\in A$, there exist $u,u'\in W$ such that $\chi_{a_1,\dots,a_n}(u)\lor\chi_{a_1,\dots,a_n}(u')=\mathbf1$. Proof: Define a function $f\colon\{0,1\}^n\to\{0,1\}$ by $$f(\alpha)=0\iff\exists u\in W\,\chi_{a_1,\dots,a_n}(u)=\alpha.$$ By definition, $S_f(a_1,\dots,a_n)=\varnothing\notin A$, hence $f\notin T_1^2$. Thus, there are $\alpha,\beta$ such that $\alpha\lor\beta=\mathbf1$, and $f(\alpha)=f(\beta)=0$. The latter means there are $u,u'\in W$ such that $\alpha=\chi_{a_1,\dots,a_n}(u)$ and $\beta=\chi_{a_1,\dots,a_n}(u')$. QED Now, let $A$, $B$, $X$, and $Y$ be as in the question, and assume for contradiction that $\varnothing\in\langle X\cup Y\rangle$. Thus, there are $a_1,\dots,a_n\in A$, $b_1,\dots,b_m\in B$, and a function $f\in T_1^2$ such that $$S_f(a_1\times W,\dots,a_n\times W,W\times b_1,\dots,W\times b_m)=\varnothing.$$ Using the lemma, there are $u,u',v,v'\in W$ such that $$\begin{align} \chi_{a_1,\dots,a_n}(u)\lor\chi_{a_1,\dots,a_n}(u')&=\mathbf1,\\ \chi_{b_1,\dots,b_m}(v)\lor\chi_{b_1,\dots,b_m}(v')&=\mathbf1. \end{align}$$ Put $w=(u,v)$ and $w'=(u',v')$. Then $w,w'\in W\times W$ satisfy $$\chi_{a_1\times W,\dots,a_n\times W,W\times b_1,\dots,W\times b_m}(w)\lor\chi_{a_1\times W,\dots,a_n\times W,W\times b_1,\dots,W\times b_m}(w')=\mathbf1,$$ thus $$f(\chi_{a_1\times W,\dots,a_n\times W,W\times b_1,\dots,W\times b_m}(w))\lor f(\chi_{a_1\times W,\dots,a_n\times W,W\times b_1,\dots,W\times b_m}(w'))=1$$ as $f\in T_1^2$. However, this means that $w$ or $w'$ is an element of $S_f(a_1\times W,\dots,a_n\times W,W\times b_1,\dots,W\times b_m)$, a contradiction.<|endoftext|> TITLE: Categories whose auto-equivalences are naturally isomorphic to the identity QUESTION [5 upvotes]: Are there any useful characterisation of categories whose auto-equivalences are all naturally isomorphic to the identity? For example, I read in this thread, What are the auto-equivalences of the category of groups? , that the category of groups is one such category. Is there some nice general property that I can use to check whether or not a category admits of auto-equivalences that are not naturally isomorphic to the identity? If not, it'd be useful just to find some more interesting examples of categories with this property. More specifically, what examples (if any) are there of auto-equivalences that send every object to an isomorphic object but are not naturally isomorphic to the identity? For instance, I know that any auto-equivalence of SETS must send every object to an isomorphic object, but is it true that any such equivalence is naturally isomorphic to the identity? If so, is this a general property of toposes or just a special case? REPLY [4 votes]: Since you mention toposes, it may be worth pointing out a simple example of a topos that admits a nonidentity automorphism: $\mathrm{Set}\times\mathrm{Set}$ has an automorphism that swaps the two copies. On the other hand, the proof that every automorphism of $\mathrm{Set}$ is the identity can be internalized to show that every indexed automorphism of any topos $\mathcal{E}$ is the identity. Here an "indexed automorphism" consists of automorphisms $\mathcal{E}/X \simeq \mathcal{E}/X$ for all $X\in\mathcal{E}$ that commute appropriately with the pullback functors $f^*:\mathcal{E}/Y \to \mathcal{E}/X$. Informally, since every topos "thinks that it is the category of sets", it also "thinks that every automorphism of itself is the identity". The argument is basically the same: in the world of $\mathcal{E}$-indexed categories, $\mathcal{E}$ itself is the free cocompletion of 1. Similarly, if $\mathcal{V}$ is a complete and cocomplete closed symmetric monoidal category, then every $\mathcal{V}$-enriched automorphism of $\mathcal{V}$ is the identity. The proof is again the same: in the world of $\mathcal{V}$-enriched categories, $\mathcal{V}$ itself is the free cocompletion of 1.<|endoftext|> TITLE: Partitioning a rectangle into different isosceles triangles QUESTION [7 upvotes]: After all the discussion raised by this old question, I am wondering about a somewhat complementary one: For any given rectangle, does there exist a finite set of pairwise different isosceles triangles which tile it? It is easy to tile e.g. a $1\times a$ rectangle for $1 TITLE: Who computed the third stable homotopy group? QUESTION [24 upvotes]: I have spend some time with the geometric approach of framed cobordisms to compute homotopy classes, due to Pontryagin. He computed $\pi_{n+1}(S^n)$ and $\pi_{n+2}(S^n)$. After surveying the literature (not too deeply) I was under the impression that the computation of $\pi_{n+3}(S^n)\cong \mathbb{Z}/24\mathbb{Z}$ for $n\rightarrow \infty$ with similar methods is due to Rohlin in the following paper: MR0046043 (13,674d) Reviewed Rohlin, V. A. Classification of mappings of an (n+3)-dimensional sphere into an n-dimensional one. (Russian) Doklady Akad. Nauk SSSR (N.S.) 81, (1951). 19–22. 56.0X It came a bit of a surprise to me that in the review of this paper on Mathscinet, Hilton states that the results in this paper are incorrect. Does the error only concern the unstable groups? Is it fair to cite this paper for the first computation of the third stable homotopy group of spheres, or should I cite papers by Barrat-Paechter, Massey-Whitehead and Serre? As I understand it these methods are much more algebraic and further removed from the applications that I have in mind. REPLY [23 votes]: The error is that Rokhlin claimed that $\pi_6(S^3)=\mathbb{Z}/6$, but Hilton, in his review, points out that the paper instead shows that $\pi_6(S^3)/\pi_5(S^2) = \mathbb{Z}/6$. The error lies in a prior calculation (reviewed here) that Rokhlin claimed showed $\eta^3=0$, but in fact this element is 2-torsion. Rokhlin corrects his mistake and calculates the stable homotopy group $\pi_3^s$ in Rohlin, V. A. MR0052101 New results in the theory of four-dimensional manifolds. (Russian) Doklady Akad. Nauk SSSR (N.S.) 84, (1952). 221–224. The review states that this result "agrees with, and were anticipated by, results of Massey, G. W. Whitehead, Barratt, Paechter and Serre." Serre's CR note Sur les groupes d'Eilenberg-MacLane. C. R. Acad. Sci. Paris 234, (1952). 1243–1245 (BnF) found the correct $\pi_6(S^3)$ by homotopical means. Barratt and Paechter found an element of order 4 in $\pi_{3+k}(S^k)$ when $k\geq 2$. The reference to Massey-Whitehead is a result presented at the 1951 Summer Meeting of the AMS at Minneapolis; all we have is the abstract in the Bulletin of the AMS 57, no. 6 If one wants to analyse 'dates received' to establish priority, then by all means. REPLY [15 votes]: The mistake is corrected in [Rohlin, V. A. New results in the theory of four-dimensional manifolds. (Russian) Doklady Akad. Nauk SSSR (N.S.) 84, (1952). 221–224, MR0052101]. This and other matters are discussed in the monograph [À la recherche de la topologie perdue. (French) [Remembrance of topology past] I. Du côté de chez Rohlin. II. Le côté de Casson. [I. Rokhlin's way. II. Casson's way] Edited by Lucien Guillou and Alexis Marin. Progress in Mathematics, 62. Birkhäuser Boston, Inc., Boston, MA, 1986, MR0900243] where Rohlin's four papers are reproduced with comments. This book was also translated into Russian (I own a copy) but it seems not into English.<|endoftext|> TITLE: Examples of $(\infty,1)$-topoi that are not given as sheaves on a Grothendieck topology QUESTION [24 upvotes]: An $(\infty,1)$-topos according to Lurie is defined as (accessible) left exact localization of a presheaf $(\infty,1)$-category $\text{P}(\mathcal C)$. Those $(\infty,1)$-topoi $\text{Sh}(\mathcal C)$ arrising from a site $C$ correspond precisely to topological left exact localizations of $\text{P}(\mathcal C)$. What is an example of an $(\infty,1)$-topos not given as $\text{Sh}(\mathcal C)$ - i.e. an $(\infty,1)$-topos arrising from a non-topological localization? Should I think of them as pathological or as useful to have? REPLY [20 votes]: Marc's examples are good ones, but let me add two more (which are closely related to each other): 1) Let $\mathcal{C}$ be an accessible $\infty$-category which admits small filtered colimits, and let $\mathcal{X}$ be the $\infty$-category of functors from $\mathcal{C}$ to $\mathcal{S}$ which preserve small filtered colimits. Then $\mathcal{X}$ is an $\infty$-topos, but I do not know how to realize $\mathcal{X}$ as an $\infty$-category of sheaves on a site. 2) Let $X$ be a locally compact topological space, and let $\mathcal{X}$ be the $\infty$-topos of sheaves on $X$. Then $\mathcal{X}$ is exponentiable in the setting of $\infty$-topoi: that is, for every $\infty$-topos $\mathcal{Y}$, there exists another $\infty$-topos $\mathcal{Y}^{\mathcal{X}}$ such that geometric morphisms from $\mathcal{Z}$ into $\mathcal{Y}^{\mathcal{X}}$ are the same as geometric morphisms from $\mathcal{Z} \times \mathcal{X}$ into $\mathcal{Y}$ (where the product is formed in the $\infty$-category of $\infty$-topoi). I do not know if $\mathcal{Y}^{\mathcal{X}}$ can be realized as an $\infty$-topos of sheaves on a site, even if it is assumed that such a description is known for $\mathcal{Y}$.<|endoftext|> TITLE: Proofs of Beilinson-Bernstein QUESTION [27 upvotes]: The Beilinson-Bernstein localization theorem states roughly that the category of $D$-modules on the flag variety $G/B$ is equivalent to the category of modules over the universal enveloping algebra $U\mathfrak{g}$ with zero Harish-Chandra character. Here $G$ is any semisimple algebraic group over $\mathbb{C}$, $\mathfrak{g}$ its Lie algebra, and $B$ a Borel in $G$. (Really I should probably generalize to modules over a twist $D_\lambda$ of the ring of differential operators and other Harish-Chandra characters here.) I think it's fair to say that this is one of the most important theorems in geometric representation theory. One sign of this is the huge number of generalizations to different settings (arbitrary Kac-Moody, affine at negative and critical levels, characteristic p, quantum, other symplectic resolutions...), each of which plays an key role in the corresponding representation theory. A lot of the papers in this area seem to contain phrases like "Here we are generalizing a proof of the original BB theorem", but they seem to be all referring to different proofs. Unfortunately, I only know one proof, more or less the one that appears in the original paper of Beilinson and Bernstein, "Localisation de $\mathfrak{g}$-modules." So my question is What other approaches are there to proving the localization theorem? To be a little more precise, let's separate BB localization into three parts: The computation of $R\Gamma(G/B,\mathcal{D}_\lambda)$ (for $\lambda$ with $\lambda+\rho$ dominant) The exactness of $R\Gamma$ (true for all weights $\lambda$ with $\lambda+\rho$ dominant) The conservativity of $R\Gamma$ (true for $\lambda$ with $\lambda+\rho$ dominant regular). I'm mainly interested in proofs of parts $2$ and/or $3$ assuming part $1$ (which I know many proofs of.) Standard disclaimer: I'm not an expert and all of this could be wrong. REPLY [12 votes]: I have sketched several possible proofs in this document: http://math.mit.edu/~bezrukav/localization_notes.pdf Please let me know if there are comments or questions.<|endoftext|> TITLE: The definition of amalgamated free product for general von Neumann algebras QUESTION [5 upvotes]: When discussing the amalgamated free product von Neumann algebras, people often assume that the algebras are $\sigma$-finite. I am wandering if there is any literature on the amalgamated free product for general (non $\sigma$-finite) von Neumann algebras? Thank you in advance. REPLY [4 votes]: Let's just consider the case of amalgamation over $\mathbb C$ as the general case is similar. Almost all definitions of free product of von Neumann algebras call for faithful normal states. This condition is equivalent to the algebras being $\sigma$-finite by Takesaki (volume I, II.3.19). However, one can still do the construction using non-faithful normal states but with the realization that your original von Neumann algebras may fail to be embedded in the free product. The only instance of this I could find was in a few of Dykema's papers where he looks at von Neumann algebras with normal states whose GNS representations are faithful, a condition that is weaker than having a faithful normal state.<|endoftext|> TITLE: The mean value of $y \log{y}$ over the ordinates of the CM points QUESTION [9 upvotes]: Let $-D < 0$ be a negative fundamental discriminant and let $y$ range over the values $y = y_Q = \frac{\sqrt{|D|}}{2a}$, as the values $(a,b,c)$ run through the reduced binary quadratic forms $Q = aX^2+bXY+cY^2$ of discriminant $b^2-4ac = -D$ (thus $\mathrm{gcd}(a,b,c) = 1$, $-a < b \leq a \leq c$). Those are the ordinates of the CM points $z_Q = \frac{b + \sqrt{-D}}{2a}$ in the standard fundamental domain for $\mathcal{H} / \mathrm{PSL}(2,\mathbb{Z})$. By Duke's theorem, the $z_Q$ are equidistributed in the measure $\frac{3}{\pi} \frac{dx \, dy}{y^2}$. The highest lying point has ordinate $k := \sqrt{|D|}/2$, and if $\varphi: [1,\infty) \to \mathbb{R}$ is any compactly supported or fast enough decaying function, it follows that the mean value of $\varphi(\frac{\pi}{3}y_Q)$ (over all reduced forms $Q$ if the given discriminant $-D$) is equal to $\int_1^k \varphi(t) \frac{dt}{t^2} + o_{\varphi}(1)$, as $D \to \infty$. The asymptotic $\mathrm{Avg}_Q \varphi(\frac{\pi}{3}y_Q) \sim \int_1^k \varphi(t) \frac{dt}{t^2}$ turns out to be also satisfied for $\varphi(t) = t$ (assuming $L(s,\chi_D)$ satisfies the Riemann hypothesis), for a very special reason: the Kronecker-Chowla-Selberg limit formula expresses $$ \mathrm{Avg}_Q \big(\frac{\pi}{3} y_Q - \log{y_Q} \big) = \log{k} + \frac{L'}{L}(1,\chi_D) + O(1), $$ with an absolutely bounded $O(1)$ term. On the other hand it is plain that $\mathrm{Avg}_Q \varphi(\frac{\pi}{3}y_Q) \sim \int_1^k \varphi(t) \frac{dt}{t^2}$ fails to hold with $\varphi(t) = t^{1+\epsilon}$, for any fixed $\epsilon > 0$: already the principal form (with $a = 1$) contributes an $\sim k^{\epsilon} \frac{k}{h}$ to this average ($h$ denoting the class number), a contribution that can be as big as $k^{\epsilon} \log{\log{k}}$. I am interested in what happens for $\varphi(t) = t\log{t}$. Question. Assume GRH, and possibly some other standard analytic hypotheses. What can be said about the mean value of $\frac{\pi}{3} y_Q \log{y_Q}$, asymptotically as $D \to \infty$? Evidently this mean value lies somewhere between $\log{k}$ and $(\log{k})^2$. Should the limit $\lim_{D \to \infty} \mathrm{Avg}_Q(y_{Q}\log{y_Q}) \big/ (\log{k})^2$ even exist? [Of course, the same question could be asked for the positive discriminants (indefinite binary quadratic forms and closed geodesics on the modular surface). ] REPLY [10 votes]: Let $g : \Gamma \backslash \mathbb{H} \to \mathbb{C}$ be any bounded continuous function. Duke's theorem states that \[\frac{1}{h(D)} \sum_{A \in \mathrm{Cl}_K} g(z_A) = \frac{1}{\mathrm{vol}(\Gamma \backslash \mathbb{H})} \int_{\Gamma \backslash \mathbb{H}} g(z) \, d\mu(z) + o_g(1)\] as $D \to -\infty$ through negative fundamental discriminants, where $h(D) = \# \mathrm{Cl}_K$ the class number of $K = \mathbb{Q}(\sqrt{D})$, $\mathrm{Cl}_K$ the class group, and $d\mu(z) = \frac{dx \, dy}{y^2}$ the $\mathrm{SL}_2(\mathbb{R})$-invariant measure on the upper half-plane $\mathbb{H}$. The proof uses the spectral decomposition \[g(z) = \frac{\langle g, 1\rangle}{\mathrm{vol}(\Gamma \backslash \mathbb{H})} + \sum_{f \in \mathcal{B}_0(\Gamma)} \langle g, f \rangle f(z) + \frac{1}{4\pi} \int_{-\infty}^{\infty} \left\langle g, E\left(\cdot, \frac{1}{2} + it\right)\right\rangle E\left(z, \frac{1}{2} + it\right) \, dt,\] where $\mathcal{B}_0(\Gamma)$ denotes an orthonormal basis of Hecke-Maass cusp forms and $E(z,s)$ denotes the real-analytic Eisenstein series. We want to take $g(z) = y \log y$, but this is of course not square-integrable with respect to $d\mu(z)$. The trick is to let $g_s(z) = y^s \log y$ and $G_s(z) = g_s(z) - \frac{\partial}{\partial s} E(z,s)$ with $\Re(s) > 1/2$ and $s \neq 1$. Then $G_s(z)$ is square-integrable, with \[g_s(z) = \frac{\partial}{\partial s} E(z,s) + \frac{\langle G_s, 1\rangle}{\mathrm{vol}(\Gamma \backslash \mathbb{H})} + \sum_{f \in \mathcal{B}_0(\Gamma)} \langle G_s, f \rangle f(z) + \frac{1}{4\pi} \int_{-\infty}^{\infty} \left\langle G_s, E\left(\cdot, \frac{1}{2} + it\right)\right\rangle E\left(z, \frac{1}{2} + it\right) \, dt.\] It is clear that $\langle G_s, f\rangle = \langle g_s, f\rangle$. Using Zagier's method of renormalisation of integrals, one can show that \[\langle G_s, 1\rangle = \int_{\sqrt{3}/2}^{1} y^{s - 2} \log y \left(1 - 2\sqrt{1 - y^2}\right) \, dy + \frac{1}{(s - 1)^2},\] and that \[\left\langle G_s, E\left(\cdot, \frac{1}{2} + it\right)\right\rangle = 0.\] (Both of these are somewhat nontrivial to show, however.) It follows by analytic continuation that \[g(z) = \lim_{s \to 1} \left(\frac{\partial}{\partial s} E(z,s) + \frac{1}{\mathrm{vol}(\Gamma \backslash \mathbb{H}) (s - 1)^2}\right) + C + \sum_{f \in \mathcal{B}_0(\Gamma)} \langle g, f \rangle f(z),\] where \[C = \frac{1}{\mathrm{vol}(\Gamma \backslash \mathbb{H})} \int_{\sqrt{3}/2}^{1} y^{-1} \log y \left(1 - 2\sqrt{1 - y^2}\right) \, dy \approx -0.00181698.\] Now we let $z = z_A$ and sum over all $A \in \mathrm{Cl}_K$. Then the sum over $f \in \mathcal{B}_0(\Gamma)$ is $O((-D)^{1/2 - \delta})$ for some $\delta > 0$ via the period formula for Hecke-Maass cusp forms at Heegner points and subconvexity bounds for $L(1/2,f \otimes \chi_D)$. Next, we have that \[\sum_{A \in \mathrm{Cl}_K} E(z_A,s) = \frac{w_K}{2} \frac{\Lambda_K(s)}{\Lambda(2s)},\] where $w_K = \# \mathcal{O}_{K,\mathrm{tors}}^{\times}$ (which is equal to $2$ for $D \leq -4$), $\Lambda_K(s) = (2\pi)^{-s} \Gamma(s) \zeta(s) L(s,\chi_D)$, and $\Lambda(2s) = \pi^{-s} \Gamma(s) \zeta(2s)$. We may rewrite this as \[\frac{w_K \sqrt{-D}}{4} \frac{\exp\left((s - 1) \log \frac{\sqrt{-D}}{2}\right) \zeta(s) L(s,\chi_D)}{\zeta(2s)}.\] This has a simple pole at $s = 1$ with residue \[\frac{3}{\pi} \frac{w_K \sqrt{-D} L(1,\chi_D)}{2\pi} = \frac{h(D)}{\mathrm{vol}(\Gamma \backslash \mathbb{H})}\] by the class number formula. It follows that if $C' = C'(D)$ denotes the coefficient of $s - 1$ in the Laurent expansion of this, then \[\sum_{A \in \mathrm{Cl}_K} g(z_A) = C' + C h(D) + O((-D)^{1/2 -\delta}).\] The main term from $C'$ is \[\frac{3}{\pi} \frac{w_K \sqrt{-D} L(1,\chi_D)}{2\pi} \frac{1}{2} \left(\log \frac{\sqrt{-D}}{2}\right)^2 = \frac{h(D)}{\mathrm{vol}(\Gamma \backslash \mathbb{H})} \frac{1}{2} \left(\log \frac{\sqrt{-D}}{2}\right)^2.\] To prove that the remaining terms in $C'$ are smaller requires bounds of the form \[\frac{L'}{L}(1,\chi_D) = o(\log (-D)), \quad \frac{L''}{L}(1,\chi_D) = o((\log (-D))^2).\] Such bounds certainly follow from the generalised Riemann hypothesis, though I don't believe they are known unconditionally. I should mention that this method works more generally for $g(y) = y (\log y)^m$ with $m$ a nonnegative integer, in which case you need to take $m$-th derivatives with respect to $s$, or even more generally for $g(y) = y^s (\log y)^m$ with $s \neq 1$, in which case you no longer need to take limits, but $h(D)$ will no longer naturally appear in the main term, so the answer may fluctuate wildly depending on the size of $L^{(k)}(s,\chi_D)/L(1,\chi_D)$ for $0 \leq k \leq m + 1$.<|endoftext|> TITLE: Growth of a linear recurrent sequence QUESTION [8 upvotes]: Consider the sequence defined by $a_0=a_1=1$ and $a_n=2a_{n-1}-3a_{n-2}$ for $n\geq 2$. This is the sequence https://oeis.org/A087455. I would like to prove that $|a_n|>100$ when $n>10$. How can we do it? (Also, is there an explicit non-trivial lower bound for the sequence $|a_n|$?) The first few terms: 1, 1, -1, -5, -7, 1, 23, 43, 17, -95, -241, -197, 329, 1249, 1511, -725, -5983, -9791, -1633, 26107, 57113, 35905, -99529, -306773, -314959, 290401, 1525679, 2180155, -216727, -6973919, -13297657, -5673557, 28545857, 74112385, 62587199, -97162757, -382087111, -472685951, 200889431, 1819836715, 3037005137, 614500129, -7882015153, -17607530693, -11569015927, 29684560225, 94076168231, 99098655787, -84031193119, -465358353599, -678623127841, 38828805115, 2113526993753, 4110567572161, 1880554163063, -8570594390357, -22782851269903, -19853919368735, 28640715072239, 116843188250683, 147764231284649, -55001102182751, -553294898219449, -941586489890645, -223288285122943, 2378182899426049, 5426230654220927, 3717912610163707, -8842866742335367, -28839471315161855, -31150342403317609, 24217729138850347 Explicit formulas: $$\begin{align*}a_n&=\frac{(1+i\sqrt2)^n+(1-i\sqrt2)^n}{2}\\&=(\sqrt{3})^n\cdot \cos (n\cdot \theta)\end{align*}$$ where $\theta=\tan^{−1}(\sqrt2)$. Thank you. REPLY [8 votes]: Besides the archimedean technique from Baker's theorem in another answer, this type of question can be handled both qualitatively and quantitatively using $p$-adic methods. This was illustrated, for the sequence you asked about, on https://math.stackexchange.com/questions/873147/finding-non-negative-integers-m-such-that-1-sqrt-2m-has-real-part/873529 with the task being a determination of when $a_n = \pm 1$. I wrote up an account of this in another $p$-adic field at http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/strassmannapplication.pdf.<|endoftext|> TITLE: How to translate a graph coloring problem to algebraic or geometric language and solve it? QUESTION [5 upvotes]: I want to know whether there are ways to use algebraic methods for solving graph theory problems (graph coloring problems). For example, is it possible to prove the four-color theorem purely with algebraic or geometric methods? REPLY [4 votes]: See Sebastian A. Csar, Rik Sengupta, and Warut Suksompong, On a subposet of the Tamari lattice; "Relying on work of Whitney in [13], Kauffman reformulated the Four Color Theorem using the vector cross product in [5]. More recently, in [ 1], Cooper, Rowland and Zeilberger transformed the Four Color Theorem into a question about another binary function on $T_n$: the size of the set ParseWords$(T_1, T_2)$ consisting of all words $w \in \{\,0, 1, 2\,\}^n$ which are parsed by both $T_1$ and $T_2$. Here, a word $w$ is parsed by $T$ if the labeling of the leaves of $T$ by $w_1,w_2,\dots,w_n$ from left to right extends to a proper 3-coloring with colors $\{\,0, 1, 2\,\}$ of all $2n − 1$ vertices in $T$ , such that no two children of the same vertex have the same label and such that no parent and child share the same label. The Four Color Theorem is equivalent to the statement that for all $n$ and all $T_1,T_2 \in T_n$, one has $|{\rm ParseWords}(T_1 , T_2 )| \ge 1$. Tamari offers a similar reformulation of the Four Color Theorem in [12]." Perhaps this will go some way to answering Gordon Royle's question in the comments. The Kauffman reference is to Kauffman, Louis H., Map coloring and the vector cross product, J. Combin. Theory Ser. B 48 (1990), no. 2, 145–154, MR1046751 (91b:05078). Here is an excerpt from the review by François Jaeger: This paper presents a nice algebraic reformulation of the four color theorem. Consider an orthonormal basis $i,j,k$ of ${\bf R}^3$ endowed with the usual vector cross product written multiplicatively (so that $ij=k=−ji$, $i^2=0$, and so on). Let $L$ and $R$ be two expressions formed from a word $X_1\cdots X_n$ by correct insertion of parentheses (i.e. $L$ and $R$ represent well-defined products). A sharp solution of the equation $L=R$ is an assignment of values $i,j,k$ to each variable $X_r$ such that the resulting values for $L$ and $R$ (using the vector cross product) are equal and nonzero. For instance a sharp solution to the equation $X_1(X_2X_3)=(X_1X_2)X_3$ is given by $X_1=i$, $X_2=k$, $X_3=i$. Then the four color theorem is equivalent to the statement that every equation of the form $L=R$ has a sharp solution. Another paper worth a mention is Miranda, Rick, Colorings of planar maps and residues of 1-forms, Methods in module theory (Colorado Springs, CO, 1991), 237–247, Lecture Notes in Pure and Appl. Math., 140, Dekker, New York, 1993, MR1203811 (94b:05086). The review by Steve Fisk says, This paper interprets coloring in terms of algebraic geometry. A cubic map on the sphere corresponds to a set of lines in a projective space, considered as a variety. A coloring with $q$ colors corresponds to a 1-form over ${\bf F}_q$ with all nonzero residues. Fisk also writes, There are many reformulations of coloring and the four-color problem [see T. L. Saaty, Amer. Math. Monthly 79 (1972), 2–43; MR0295965; B. T. Datta, in Graph theory and related topics (Waterloo, ON, 1977), 121–131, Academic Press, New York, 1979; MR0538040; F. Jaeger, J. Combin. Theory Ser. B 26 (1979), no. 2, 205–216; MR0532588; D. W. Barnette, Map coloring, polyhedra, and the four-color problem, Math. Assoc. America, Washington, DC, 1983.... so perhaps it would be worthwhile tracking down those references. Continuing to follow my nose through the internet, I come across Matiyasevich, Yuri, Some arithmetical restatements of the four color conjecture, Weak arithmetics. Theoret. Comput. Sci. 257 (2001), no. 1-2, 167–183, MR1825093 (2002f:03107), where the summary reads, in part, The four colour conjecture is reformulated as a statement about non-divisibility of certain binomial coefficients.<|endoftext|> TITLE: Harmonic function properties on $\mathbb R^3$ QUESTION [11 upvotes]: Let $X$ be the set of all harmonic functions external to the unit sphere on $\mathbb R^3$ which vanish at infinity, so if $V \in X$, then $\nabla^2 V(\mathbf{r}) = 0$ on $\mathbb R^3 - S(2)$ and $\lim_{r \rightarrow \infty} V(r) = 0$. Now consider a function $f: X \rightarrow \mathbb R$, defined by $$ f(V)(\mathbf{r}) = || \nabla V(\mathbf{r}) ||^2 $$ For some given $V \in X$, I am looking for all functions $W \in X$ which satisfy $$ f(V) = f(W) $$ Certainly $W = \pm V$ will satisfy the condition. Can anyone find nontrivial solutions for $W$? My approach so far: The condition on $V$ and $W$ is $$ \nabla V \cdot \nabla V = \nabla W \cdot \nabla W $$ By defining $\phi = V + W$ and $\psi = V - W$, this is equivalent to $$ \nabla \phi \cdot \nabla \psi = 0 $$ I then tried expanding $\nabla \phi$ and $\nabla \psi$ in a basis of vector spherical harmonics and plugging into the above formula. This step makes use of the fact $\nabla^2 \phi = \nabla^2 \psi = 0$ and leads to the following condition on the expansion coefficients: $$ \nabla \phi \cdot \nabla \psi = \sum_{nm,n'm'} \phi_{nm} \psi_{n'm'} \left( \frac{1}{r} \right)^{n+n'+4} \left( (n+1)(n'+1) Y_{nm} Y_{n'm'} + \partial_{\theta} Y_{nm} \partial_{\theta} Y_{n'm'} + \frac{1}{\sin^2{\theta}} \partial_{\phi} Y_{nm} \partial_{\phi} Y_{n'm'} \right) $$ Its not clear to me how to proceed from here, or whether this is even the correct approach to take. I could get rid of the sum over $n',m'$ by integrating both sides over a unit sphere and using the orthogonality relations for the spherical harmonics. Doing this gives: $$ \sum_{nm} (n+1)(2n+1) \phi_{nm} \psi_{nm} = 0 $$ though I'm not sure that yields any additional insight. I would appreciate any ideas. REPLY [3 votes]: This problem has an important background in geomagnetism. When planning the MAGSAT satellite mission (1979/80) to determine the spherical harmonic coefficients of the Earth's magnetic field from space, Backus (JGR, 1970) showed that a measurement of the total field intensity $||\nabla V||$ on a spherical shell is in general not sufficient to uniquely determine $V$ (not regarding trivial non-uniqueness due to gauge and sign). He did this by explicitly constructing some counterexample by means of similar arguments as used in the question and comments here. As a consequence, MAGSAT became the first mission carrying a vector magnetometer instead of the much simpler absolute field sensors. In relation to the problem as posed here, Backus (Quart. Journ. Mech. and Applied Math., 21, 195-221 , 1968) proved the following theorem: THEOREM 5: Suppose $\phi$ and $\phi'$ are harmonic outside some open bounded set $W$ in $\mathbf{R}^n$ and vanish at infinity. Suppose that $| \nabla \phi| = | \nabla \phi'|$ outside some sphere which contains $W$. If $n \geq 3$ then one of the two functions $\phi -\phi'$ and $\phi +\phi'$ vanishes identically outside $W$.<|endoftext|> TITLE: What is the shortest polynomial divisible by $(x-1)(y-1)(x^2y-1)$ QUESTION [19 upvotes]: I am interested in polynomials with few terms ("short polynomials", "fewnomials") in ideals. A simple to state question is Given an ideal $I\subset k[x_1,\dots,x_n]$, what is the shortest polynomial in $I$? There are answers for "Does an ideal contain a monomial/binomial?", but the general question seems to be hard. Let's try an easy specific question: How few terms can a polynomial in $\langle(x-1)(y-1)(x^2y-1)\rangle \subset k[x,y]$ have? The generator has 8 terms so this is an upper bound for the minimum. A lower bound for the minimum is 6 which can be seen as follows: Let $f$ be the generator. The Newton polytope of $f$ is a hexagon with two interior points. Any polynomial in $I$ is of the form $fg$ for some $g\in k[x,y]$ and its Newton polytope is $\text{Newton}(f) + \text{Newton}(g)$. Since the number of vertices cannot decrease under the Minkowski sum, $fg$ has at least six terms. Are there polynomials with six or seven terms in $I$? REPLY [8 votes]: Your general question for $n=1$ (univariate polynomials) is the subject of "Computing sparse multiples of polynomials" by Mark Giesbrecht, Daniel S. Roche, Hrushikesh Tilak: We consider the problem of finding a sparse multiple of a polynomial. Given f in F[x] of degree d over a field F, and a desired sparsity t, our goal is to determine if there exists a multiple h in F[x] of f such that h has at most t non-zero terms, and if so, to find such an h. When F=Q and t is constant, we give a polynomial-time algorithm in d and the size of coefficients in h. When F is a finite field, we show that the problem is at least as hard as determining the multiplicative order of elements in an extension field of F (a problem thought to have complexity similar to that of factoring integers), and this lower bound is tight when t=2. There are some talk slides from 2010 by Roche here which state on the last slide the multivariate case as an open problem.<|endoftext|> TITLE: Sheaf (Gieseker) compactification of moduli space of vector bundles QUESTION [8 upvotes]: I am given to understand that the moduli space $M_k^G$ of $G$ vector bundles with second Chern class $c_2=k$ over an algebraic curve/variety (for me a Riemann surface is enough/projective space for varieties) is a non-compact space. I am familiar with the Uhlenbeck compactification $$ \widetilde{M}_k^G = M_{k}^G \cup (M_{k-1}^G \times \mathbb{R}^4) \cup (M_{k-2}^G \times \mathrm{Sym}^2\mathbb{R}^4) \cup \ldots \cup \mathrm{Sym}^k\mathbb{R}^4 $$ Apparently, there is another kind of compactification using sheaves. Vector bundles are locally free sheaves so I am trying to understand, in other words, the moduli space of locally free sheaves. Now, if we include more general sheaves, not necessarily locally free, we get the moduli space of coherent (torsion free) sheaves. Is this statement correct? What is the precise idea behind this compactification, what is its relation to the Uhlenbeck one and is it true that this compact moduli space is now singular? And what if we have a smooth surface, e.g. $\mathbb{P}^n$ or any of the Hirzebruch surfaces? I would appreciate very much any reference on this "sheaf" compactification. P.S. I think this is referred to as Gieseker compactification of the moduli space of sheaves but a Google search wont give me much info and Huybrecht's book is way to hard and lengthy. REPLY [5 votes]: This is perhaps more of a comment, but it's too long. I'm coming from the algebraic side of things, but it seems to me that there are several issues here. Be warned that when I say curve I mean a (smooth) algebraic curve/Riemann surface, and when I say surface I mean a (smooth) complex surface. All my vector bundles are algebraic/holomorphic. Don't you need to mention some type of stability (Gieseker or slope, say) to get some reasonable moduli space? On a curve, a torsion-free sheaf is automatically locally free. (Basically by the structure theorem for modules over a PID.) So, in the curve case moduli of torsion-free sheaves = moduli of vector bundles. Moduli of stable bundles are typically not compact, but if you allow (S-equivalence classes of) semi-stable bundles then the resulting spaces are compact. This will typically introduce singularities. Everything with Gieseker semistability is really concerning bundles on a surface or higher dimensional variety. In the curve case, Gieseker semistability reduces to ordinary slope-semistability. You're mentioning that the 2nd Chern class is fixed. Bundles on curves don't have 2nd Chern classes; you would fix the degree(/first Chern class) if you care about curves/Riemann surfaces. A more gentle book than Huybrechts/Lehn would be Le Potier's "Lectures on Vector Bundles."<|endoftext|> TITLE: Vanishing of $w_2$ for orientable 3-manifolds QUESTION [6 upvotes]: Let $M$ be oriented manifold: this happens if and only if $w_1(M)=0$ (the first Stiefel Whitney class, being an element in $H^1(M,\mathbb{Z}_2)$. There is a result that if $M$ is three dimensional then from the vanishing of $w_1$ automatically follow that also $w_2=0$. This should be rather easy consequence of properties of Wu classes and Steenrod squares however I don't see how to proceed in the calculation. I expressed the total Stiefel Whitney class $w$ of $M$ as $Sq(v)$ where $v$ is the total Wu class. After evaluating $Sq(v)$ I arrived, using the fact that $M$ is 3 dimensional (if I'm not mistaken) at the following: $1+w_1+w_2+w_3=1+v_1+v_2++v_1 \cup v_1$. From $w_1=0$ we infer $v_1=0$ therfore also $v_1 \cup v_1=0$ and the question boils down to showing $v_2=0$ but I don't see how to prove this. REPLY [9 votes]: From the relation $Sq(v)=w$ we get that $v_2=w_2+w_1^2$. More specifically By Thm 11.14 (which is just an extended form of the above equality) in M-S you have $w_1=v_1$ and $w_2=Sq^1(v_1)+v_2=v_1^2+v_2$. Plug them together you obtain my first claim. Recall we have a nice characterisation of the (second) Wu-class, as $$Sq^{2}(x)=v_2\smile x$$ for any $x \in H^{n-k}(M;\Bbb Z_2)$ (See Milnor Stasheff page 132). Since in our case $n=3$, we get that for any $x\in H^1(M;\Bbb Z_2)$, $0=v_2\smile x$, which gives you $v_2=0$. To see this last equality, assume $v_2 \neq 0$. We would have $$1=\langle v_2, (v_2)_*\rangle=\langle v_2, [M]\frown x\rangle=0$$ Where $(v_2)_* \in H_2(M;\Bbb Z_2)$ is the dual of $v_2$ and Poincaré Duality (mod 2) tells us that there exists an element $x \in H^1(M; \Bbb Z_2)$ such that $[M]\frown x=(v_2)_*$ Therefore you have $0=w_2+w_1^2$, which proves your claim<|endoftext|> TITLE: Algebraic surfaces with no deformations QUESTION [6 upvotes]: Is very well known that the only algebraic curve which admits no deformations is the projective line. Q. What are "rigid" smooth algebraic surfaces? Is there a sensible classification? REPLY [10 votes]: There are several different notions of "rigidity" (local rigidity, global rigidity, infinitesimal rigidity, étale rigidity and strong rigidity) and it is possible to provide examples for each of them. This topic is discussed in the paper by I. Bauer and F. Catanese On rigid compact complex surfaces and manifolds, you can have a look at it for more details and references to the literature.<|endoftext|> TITLE: An example of a non-geometric $C^\infty(M)$-module QUESTION [10 upvotes]: Let $M$ be a smooth manifold and let $Q$ be an arbitrary $C^\infty(M)$-module. $Q$ is called geometric if $$\bigcap_{p\in M}\mu_pQ=0,$$ where $\mu_p$ is an ideal in $C^\infty(M)$ of functions vanishing at point $p.$ From the definition, $q\in\mu_pQ$ iff $q=\sum_{i=1}^Nf_iq_i$ for some $f_i$'s from $\mu_p$ and $q_i$'s from $Q.$ Request. Construct a non-geometric $C^\infty(M)$-module. For any vector bundle $E\to M$ the $C^\infty(M)$-module $\Gamma(E)$ is geometric. This is because, if $\xi\in\mu_p\Gamma(E),$ then $\xi(p)=0.$ Hence we have to look for non-geometric modules in a different way. I came across this notion in the following Jet Nestruev's book: Nestruev, Jet, Smooth manifolds and observables, Graduate Texts in Mathematics. 220. New York, NY: Springer. xiv, 222 p. (2003). ZBL1021.58001. Remark about the notion: I investigated more or less Vinogradov's bibliography and I guess this notion first appeared in the following book: Krasil’shchik, I.S.; Lychagin, V.V.; Vinogradov, A.M., Geometry of jet spaces and nonlinear partial differential equations. Transl. from the Russian by A. B. Sosinskij, Advanced Studies in Contemporary Mathematics, 1. New York etc.: Gordon and Breach Science Publishers. xx, 441 p. (1986). ZBL0722.35001. REPLY [4 votes]: What about $\Gamma(M,\mathcal{C}_{M}^{\infty}/\mathcal{I}_{x}^{2})$ where $\mathcal{I}_x$ is the ideal sheaf of a point $x\in M$?<|endoftext|> TITLE: Are there primitive quartic CM fields whose norms of units give all totally positive units of the real quadratic subfield? QUESTION [5 upvotes]: Let $K$ be a primitive (i.e. not biquadratic) quartic CM-field. That is we have $[K:\mathbb{Q}]=4$ and let $K_0=\mathbb{Q}(\sqrt{d})$ be the totally real quadratic subfield, here $d> 1$, that is we have $[K:K_0]=[K_0:\mathbb{Q}]=2$. Denote by $O_K^{\times}$ the set of units in $K$ and by $(O_{K_0}^{\times})^{+}$ the set of totally positive units in $K_0$. Then $N_{K/K_0}(O_K^{\times})$ is a subgroup of $(O_{K_0}^{\times})^{+}$ $\textbf{Question:}$ Can we classify all such fields $K$ with the property $r=[(O_{K_0}^{\times})^{+}:N_{K/K_0}(O_K^{\times})]=1$? a) If $\epsilon_0$ denotes the fundamental unit of $K_0$ and we have $N_{K_0/\mathbb{Q}}(\epsilon_0)=-1$ then $r=1$. So these are all $K$ with the property that $K_0$ has a (fundamental) unit of norm -1. Are these exactly the fields $\mathbb{Q}(\sqrt{t^2+4})$ for some $t\in \mathbb{Z}$? b)What happens if $N_{K_0/\mathbb{Q}}(\epsilon_0)=1$? Is it possible that there is a unit $\epsilon\in K$ such that $\epsilon_0=N_{K/K_0}(\epsilon)$? Then we also should get $r=1$. So are the there such fields with a unit $\epsilon$ such that $N_{K/K_0}(\epsilon)$ is a fundamental unit of $K_0$ with norm 1? Can we classify all such $K$? c) Are there any other $K$ with $r=1$? REPLY [2 votes]: Except for the finitely many fields where nontrivial roots of unity are present, the fundamental unit $\varepsilon$ of the quadratic subfield $k$ is the norm of a unit from $K$ if and only if $K = k(\sqrt{\pm \varepsilon})$. If $N(\varepsilon) = +1$, then $K$ is biquadratic, and if $N(\varepsilon) = -1$, then $K$ is not CM (by simple Galois theory). Thus the fundamental unit in $K$ is the same as in $k$ in your cases, and you have $r = 1$ if and only if $N(\varepsilon) = -1$. In this case, $k$ has the form $k = {\mathbb Q}(\sqrt{t^2+4})$ for trivial reasons: write $\varepsilon = (T + U \sqrt{m})/2$; then $T^2 - mU^2 = -4$, hence $T^2 + 4 = mU^2$ and ${\mathbb Q}(\sqrt{m}\,) = {\mathbb Q}(\sqrt{mU^2}\,) = {\mathbb Q}(\sqrt{T^2+4}\,)$. But I would not regard this as a classification.<|endoftext|> TITLE: Non-compact manifolds of positive/non-negative Ricci curvature QUESTION [11 upvotes]: Consider a non-compact complete Riemannian manifold $(M, g)$ with smooth compact boundary $\partial M$. Suppose also that $M \setminus \partial M$ has positive/non-negative Ricci curvature. My question is, is it possible to remove $\partial M$, and replace it by another compact manifold $N$ with boundary $\partial N = \partial M$, and $M \cup N$ is a complete manifold (without boundary) of positive/non-negative Ricci curvature? Let us assume the maximal volume growth on $M$, and let us also assume for convenience that $M \setminus K$, where $K$ is compact, has only one connected unbounded component. The naive mental picture I am having is "fitting a spherical cap to an end of a cylinder". But I am not sure how true such a heuristic is in general. Thanks in advance for any suggestions! REPLY [7 votes]: There are ways to glue while preserving positive Ricci curvature, but of course they require some geometric control near $\partial M$. Let me mention two relevant papers: On the moduli space of positive Ricci curvature metrics on homotopy spheres by D. Wraith. Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers by G. Perelman. For example, Perelman shows how to glue two positive Ricci curvature manifolds with isometric boundaries to get a positively Ricci metric, which requires (roughly speaking) that the normal curvatures at one boundary is greater than the negative of the normal curvature at the other boundary when the normals are chosen correctly, like in filling a cylinder with a cap. There are also obstructions to such gluings. For example in [MR1216628, Greene, R. E., Wu, H. "Non-negatively curved manifolds which are flat outside a compact set". Differential geometry: Riemannian geometry (Los Angeles, CA, 1990), 327–335, Proc. Sympos. Pure Math., 54, Part 3, Amer. Math. Soc., Providence, RI, 1993] one find the following: If $M$ is a complete Riemannian manifold of nonnegative Riccie curvature which is flat outside a compact set and is simply-connected at infinity , then $M$ is flat. For example, $\mathbb R^n$ is simply-connected at infinity when $n>2$, so if you replace a round disk in $\mathbb R^n$ with a compact manifold with sphere boundary, and hope to have nonnegative Ricci curvature on the result, then the result must be flat, and in particular, be finitely covered by the product of a Euclidean space and a torus. Note that a capped $2$-dimensional cylinder is not simply-connected at infinity, while a capped $n$-dimensional cylinder ($n>2$) is not flat outside the cap.<|endoftext|> TITLE: Given two elements $g,h$ in a finite group $G$ of the same order, does there exist a finite group $G'$ containing $G$ where they are conjugate? QUESTION [8 upvotes]: Let $G$ be a finite group, and let $g,h\in G$ be two elements of the same order. Does there exist a larger finite group $G'\ge G$ such that $g,h$ are conjugate in $G'$? If not, what is known about groups which have this property? REPLY [2 votes]: Since the question was reopened in spite of a duplicate in MathSE, I also "duplicate" my answer from there (a little improved anyway), so as to provide a self-contained answer. I leave it cw since it relies on ideas form other answers and comments. (a) The answer to the OP's question is yes, and more generally one has: Fact. Consider a finite group $F$ and two subgroups $A,B$ with an isomorphism $t:A\to B$. Then one can embed $F$ in a larger finite group in which $A$ and $B$ are conjugate (the conjugation inducing the isomorphism $t$ from $A$ to $B$). This applies in particular to the case when $A,B$ are cyclic subgroups of the same order. Proof of the fact. $A$ has two free actions on $F$, one being given by $a\cdot g=ag$, the other by $a\cdot g=t(a)g$. Writing $k=|F/A|$, find $k$ points $x_1,\dots,x_k$, one in each orbit of the first action, and $k$ points $y_1,\dots,y_k$, one in each orbit of the second action. Then extend the assignment $x_i\mapsto y_i$ to a permutation $\sigma$ of $F$, by the assignment $\sigma(ax_i)=t(a)y_i$. This is well-defined by freeness of the action. Then $$\sigma(abx_i)=t(ab)y_i=t(a)t(b)y_i=t(a)\sigma(bx_i)$$ for all $a,b\in A$ and $i$, and thus $\sigma(ag)=t(a)\sigma(g)$ for all $a\in A$ and $g\in F$. In other words, $\sigma\circ L_a=L_{t(a)}\circ \sigma$. So in the permutation group of $F$, where $F$ is identified to its image through left multiplication, the isomorphism $t:A\to B$ is realized by conjugation by $\sigma$. $\Box$ This is essentially Will's argument, which was only written for $A,B$ cyclic. (b) In the same setting, the universal group in which $A$ and $B$ are conjugate through $t$, is the HNN extension $H$ of $(F,A,B,t)$, defined by the relative presentation $$\langle u,G\mid uau^{-1}=t(a),\forall a\in A\rangle.$$ It is well-known that this HNN-extension is virtually free and hence residually finite. Thus, there exists a finite quotient of $H$ in which $F$ is mapped injectively, and in this quotient $A$ and $B$ are indeed conjugate (by $t$). Of course this argument (mentioned by HJRW in the comments to Will's answer), which looks somewhat immediate at first glance, is less elementary than the one in (a) since it relies on residual finiteness of HNN extensions of finite groups; anyway it's a very natural proof since the construction is universal.<|endoftext|> TITLE: Adjoint functor theorems on 2-categories QUESTION [5 upvotes]: Adjoint functor theorems are theorems stating that under certain conditions a functor that preserves limits is a right adjoint, and a functor that preserves colimits is a left adjoint. (from the nLab.) Now, the condition that a functor is a left/right adjoint can be stated without any further assumption on a generic 2-category. The condition that a functor preserves co/limits can be stated provided the 2-category we live in has a Yoneda structure. Of course, the co/limit must be weighted. Is there an adjoint functor theorem that can be stated in a 2-category with a Yoneda structure? In particular, is there something resembling a "solution set condition" that can be tested formally? Is such a question meaningful? What can be said, and what is an interesting open problem? REPLY [2 votes]: There is actually such a theorem in the original Street-Walters paper on Yoneda structures (p372): any colimit-preserving functor from a total object to an admissible object has a right adjoint. (This is a "special" adjoint functor theorem, i.e. it imposes "reasonability" conditions on the categories rather than on the functor.)<|endoftext|> TITLE: A correlation matrix problem QUESTION [6 upvotes]: I have a linear algebraic question about an arbitrary correlation matrix. If $$ \lambda_{\min} \begin{pmatrix} 1 & \rho_1 & \rho_2 \\ \rho_1 & 1 & \rho_3 \\ \rho_2 & \rho_3 & 1 \end{pmatrix} \ge \alpha $$ does the following hold? $$ (\rho_1,\rho_3) \begin{pmatrix} 1 & \rho_2 \\ \rho_2 & 1 \end{pmatrix}^{-1} \begin{pmatrix} \rho_1 \\ \rho_3 \end{pmatrix} \le 1-\alpha $$ Thanks so much. REPLY [4 votes]: If $\rho_1=\rho_3=0$ it is obvious, so we assume $\rho_1^2+\rho_3^2>0$. First note that $$ A= \begin{pmatrix} 1 & \rho_2 \\ \rho_2 & 1 \end{pmatrix}^{-1}=\frac{1}{\rho_2^2-1}\begin{pmatrix} -1 & \rho_2 \\ \rho_2 & -1 \end{pmatrix}$$ with eigenvalues $\lambda_1=\frac{1}{1-\rho_2}$ and $\lambda_2=\frac{1}{1+\rho_2}$. Since $A$ is symmetric, we know that the Rayleigh quotient of $A$ is upper bounded by $\max\{\lambda_1,\lambda_2\}$, that is $$\frac{\langle Ax,x\rangle}{\langle x,x\rangle} \leq \max\{\lambda_1,\lambda_2\} \qquad \forall x\neq 0.$$ In particular for $z=(\rho_1,\rho_3)^\top$ we obtain the relation $$\langle Az,z\rangle \leq \max\{\lambda_1,\lambda_2\}(\rho_1^2+\rho_3^2)=\max\Big\{\frac{\rho_1^2+\rho_3^2}{1-\rho_2},\frac{\rho_1^2+\rho_3^2}{1+\rho_2}\Big\}.$$ So, we need to prove that $$\max\Big\{\frac{\rho_1^2+\rho_3^2}{1-\rho_2},\frac{\rho_1^2+\rho_3^2}{1+\rho_2}\Big\}\leq 1-\alpha \iff \min\Big\{1-\frac{\rho_1^2+\rho_3^2}{1-\rho_2},1-\frac{\rho_1^2+\rho_3^2}{1+\rho_2}\Big\}\geq \alpha \tag{1} $$ Now, the matrix $$M =\begin{pmatrix} 1 &\rho_1 &\rho_2 \\ \rho_1&1 & \rho_3 \\ \rho_2 & \rho_3 &1 \end{pmatrix}$$ is also symmetric and thus, again by Rayleigh quotient argument we have $$\frac{\langle Mu,u\rangle }{\langle u,u\rangle}\geq \alpha \qquad \forall u\neq 0.\tag{2}$$ The idea is now to prove (1) by plugging in (2) some smart choice of $u$. These choices exist but I could not find nicely elegant ones. Here are some choices obtained with Mathematica: Take $u_{\pm} = (\alpha_{\pm},0,1)$ with $$\alpha_+=\frac{\sqrt{4 \left(\rho_2^2+\rho_2\right)^2-4 \left(\rho_1^2+\rho_3^2\right)^2}-2 \rho_2^2-2 \rho_2}{2 \left(\rho_1^2+\rho_3^2\right)}$$ and $$\alpha_-=\frac{\sqrt{-\rho_1^4-2 \rho_1^2 \rho_3^2+\rho_2^4-2 \rho_2^3+\rho_2^2-\rho_3^4}+\rho_2^2-\rho_2}{\rho_1^2+\rho_3^2}$$<|endoftext|> TITLE: A very basic question about projections in formal PDE theory QUESTION [7 upvotes]: I am learning formal PDE theory for my research and I am currently struggling to have a basic understanding of the operations involved in completing a (say, linear) PDE system to an involutive one (Cartan-Kuranishi procedure) in terms of the partial differential operators involved. First, let us put some context in order to make the question a bit more interesting to a larger audience and to fix notation. In what follows, all manifolds are smooth, Hausdorff, paracompact, connected and oriented, and all maps between any two of them are assumed to be smooth. Given two vector bundles $\pi:E\rightarrow M$, $\pi':E'\rightarrow M$ over the base manifold $M$, a (linear) partial differential operator of type $E\rightarrow E'$ and order $k\geq 0$ is a linear map $$P:\Gamma(\pi)\rightarrow\Gamma(\pi')$$ ($\Gamma(\pi^{(}{}'{}^{)})=\{\phi\in C^\infty(M,E^{(}{}'{}^{)})\ |\ \pi^{(}{}'{}^{)}\circ\phi=\text{id}_M\}$ is the space of sections of $\pi^{(}{}'{}^{)}$) of the form $$P=\Phi_P\circ j^k\ ,$$ where $j^k$ stands for the $k$-th order jet prolongation of sections of vector bundles and $$\Phi_P:J^kE\rightarrow E'$$ is a vector bundle map covering $\text{id}_M$, i.e. $\pi'\circ\Phi_P=\pi\circ\pi^k$, where $\pi^k:J^k E\rightarrow E$ is the projection map of the $k$-th order jet bundle $J^k E$ of $\pi$ onto the base $E$ (with the convention $J^0 E=E$, $\pi^0=\text{id}_E$, $j^0=\text{id}$). The (homogeneous linear) PDE system associated to $P$ is given by $$\mathcal{R}_k=\ker\Phi_P\ .$$ The $q$-th order prolongation of $P$ ($q\geq 0$) is the partial differential operator $\rho_q P$ of type $E\rightarrow J^q F$ and order $k+q$ given by $$\rho_q P\doteq j^q\circ P=\rho_q(\Phi_P)\circ j^{k+q}\ ,$$ so that $\rho_0 P=P$. The vector bundle map $\rho_q(\Phi_P):J^{k+q}E\rightarrow J^q E'$ covering $\text{id}_M$ is uniquely determined by the second equality above (particularly, $\rho_0(\Phi_P)=\Phi_P$). The PDE system associated to $\rho_q P$, called the $q$-th order prolongation of the PDE system $\mathcal{R}_k$, is then denoted by $$\mathcal{R}_{k+q}=\ker\rho_q(\Phi_P)\ .$$ Therefore, the operation of prolongation of PDE systems has a clear, global interpretation in terms of prolongation of the associated partial differential operators. In what follows, we assume in addition that $P$ is regular, that is, $\rho_q(\Phi_P)$ has constant rank for all $q\geq 0$ - this is demonstrably equivalent to assuming that $\mathcal{R}_{q+k}$ is a vector sub-bundle of $J^{k+q}E$ for all $q\geq 0$. Conversely, if $\mathcal{R}_k$ is a vector sub-bundle of $J^kE$, it is locally the kernel of $\Phi_P$ for some partial differential operator $P$ or order $k$. Finally, let $$\pi^{r+s}_r:J^{r+s}E\rightarrow J^r E\ ,\quad r,s\geq 0$$ be the natural jet projection maps, so that $$\pi^r_0=\pi^r\ ,\quad\pi^{r+s}_r\circ\pi^{r+s+t}_{r+s}=\pi^{r+s+t}_r$$ for all $r,s,t\geq 0$. It can be shown that $\pi^{k+q}_{k+r}(\mathcal{R}_{k+q})\subset\mathcal{R}_{k+r}$ for all $0\leq r\leq q$. If $P$ is regular and $\pi^{k+q}_{k+r}:\mathcal{R}_{k+q}\rightarrow\mathcal{R}_{k+r}$ have constant rank for all such $r,q$, we say that $P$ is sufficiently regular. Particularly, if $P$ is regular and the above maps are surjective for all such $r,q$, we say that $P$ is formally integrable. The other key operation employed in the Cartan-Kuranishi completion algorithm besides prolongation is projection. Both operations together unveil the hidden integrability conditions of $P$. In terms of the PDE system $\mathcal{R}_k$ associated to $P$ and its prolongations $\mathcal{R}_{k+q}$ of order $q>0$, projection is a simple operation to describe (unlike prolongation, which is more straightforwardly described in terms of $P$ as done above). Namely, the projection $\mathcal{R}^{(r)}_{k+q}$ of order $r\geq 0$ of $\mathcal{R}_{k+q+r}$ is given by $$\mathcal{R}^{(r)}_{k+q}=\pi^{k+q+r}_{k+q}(\mathcal{R}_{k+q+r})\subset\mathcal{R}_{k+q}\ .$$ For the purposes of the Cartan-Kuranishi algorithm, it suffices to consider $r$-th order projections of PDE systems with $r=1$. Now we have reached a point where I can ask my Question: Suppose that $P$ is sufficiently regular. Is there a simple, global formula for a partial differential operator to which the projected PDE system $\mathcal{R}^{(1)}_{k+q}=\pi^{k+q+1}_{k+q}(\mathcal{R}_{k+q+1})$ is associated? (I understand that such an operator should not be unique) I simply could not find such a formula in the standard literature on the subject (e.g. the 1964 thesis of Quillen and the books of Pommaret, Bryant et al. and Seiler). My conjecture is that composing $\rho_{q+1}P=\rho_{q+1}(\Phi_P)\circ j^{k+q+1}=j^{q+1}\circ\Phi_P\circ j^k$ with the natural projection $\psi_{q+1}$ onto the cokernel of its principal symbol $\sigma(\rho_{q+1} P)$ would provide such an operator and hence a positive answer to my question. In other words, Subquestion: Is there a vector bundle map $$\rho^{(1)}_q(\Phi_P):J^{k+q}E\rightarrow\text{coker}(\sigma(\rho_{q+1} P))$$ covering $\text{id}_M$ such that $$\rho^{(1)}_q(\Phi_P)\circ\pi^{k+q+1}_{k+q}=\psi_{q+1}\circ\rho_{q+1}(\Phi_P)\ ?$$ If that is the case, is it true that $\mathcal{R}^{(1)}_{k+q}=\ker(\rho^{(1)}_q(\Phi_P))$? (Remark: if $P$ is sufficiently regular, then $\sigma(\rho_{q+1} P)$ has constant rank and therefore both $\ker(\sigma(\rho_{q+1} P))$ and $\text{coker}(\sigma(\rho_{q+1} P))$ are vector bundles) If true, this would provide a nice formula for the partial differential operator $D$ produced by the Cartan-Kuranishi algorithm such that $DP$ is involutive and equivalent to $P$, in the same spirit as the compatibility complex associated to an involutive (or, more generally, a formally integrable) partial differential operator, construced by Quillen in his thesis. Moreover, a variation of this procedure would also possibly work with quasilinear partial differential operators. REPLY [4 votes]: Your conjecture is almost correct. But consider the extreme case when your equation has no integrability conditions, so that $\operatorname{coker}(\sigma(\rho_{q+1} P)) = 0$, whence $\rho^{(1)}_{q+1}(\Phi_P) = 0$, which definitely doesn't give you the the equation $\mathcal{R}^{(1)}_{k+q}$ that you wanted. What you need to do instead is to augment the $q$-prolonged equation to include the integrability conditions obtained from the projection: $$ \rho^{(1)}_q(\Phi_P)\colon J^{k+q}E \to J^qE' \oplus_M \operatorname{coker}(\sigma(\rho_{q+1} P)) , $$ so that $\rho^{(1)}_q(\Phi_P) = \rho_q(\Phi_P) \oplus \Phi_{Q\circ P}$, where $\Phi_Q\colon J^{k+q}E \to \operatorname{coker}(\sigma(\rho_{q+1} P))$ is some differential operator whose principal symbol is precisely the projection $$ S^qT^*\otimes_M E' \xrightarrow{\sigma(Q)} \operatorname{coker}(\sigma(\rho_{q+1} P)) . $$ Of course, the lift of a symbol $\sigma(Q)$ to a differential operator is not unique. But it can be chosen globally, for instance, using an auxiliary connection like you did in your earlier question here.<|endoftext|> TITLE: F-sigma subset of plane meeting every circle at 3 points QUESTION [6 upvotes]: Is there an $F_{\sigma}$-set (countable union of closed subsets of plane) $S \subseteq \mathbb{R}^2$ that meets every circle at 3 points? REPLY [5 votes]: There is not. If there were an $F_\sigma$ set meeting every circle in three points, then a suitable Moebius transformation would turn it into an $F_\sigma$ set meeting every line in three points. Bouhjar, Dijkstra, and Maudlin proved that no set meeting every line in exactly three points can be $F_\sigma$, extending an old result of Larman that says the same thing but for sets meeting every line in two points.<|endoftext|> TITLE: Reconstruction of hyperbolic curves using the fundamental group QUESTION [5 upvotes]: In the paper Curves and their Fundamental Groups written by Gerd Faltings, Mochizuki's proof of Grothendick's conjecture on anabelian curves is explained. In the proof, he shows that for two hyperbolic curves $X$ and $Y$, if there exists an isomorphism between the algebraic fundamental groups $\pi_1(X)$ and $\pi_1(Y)$ then these curves are isomorphic. My question is: With only the fundamental group $\pi_1(X)$, can the hyperbolic curve $X$ be reconstructed? Specifically can the differential sheaf $\omega_X$ be reconstructed using only the algebraic fundamental group $\pi_1(X)$? REPLY [2 votes]: The way to reconstruct the differential sheaf from the fundamental group is, of course, $$\pi_1(X)\curvearrowright X\curvearrowright \omega_X$$ But I don't think you can get to the differential sheaf directly, without passing through the underlying curve at some point. At least not with Mochizuki's original result alone. Perhaps $\omega_X$ can be carved out of $\pi_1(X)$ with the techniques of mono anabelian reconstruction (see Topics in absolute anabelian geometry III), but that's a different story. Not sure if that helps. I think the same should apply to any other piece of relevant information from $X$. Ultimately, only $\pi_1(X)$ is anabelian enough.<|endoftext|> TITLE: In which sense are Euler-Lagrange PDE's on fiber bundles quasi-linear? QUESTION [8 upvotes]: In what follows, all manifolds are smooth, Hausdorff, paracompact, connected and oriented, and all maps between any two of them are assumed to be smooth. Let $\pi:E\rightarrow M$ be a fiber bundle over the base manifold $M$, which we assume to have dimension $n$. Given a bundle map $$\mathscr{L}:J^kE\rightarrow\wedge^nT^*M$$ over $M$, where $J^kE$ is (the total space of) the jet bundle of order $k$ of $\pi$ - i.e. $\mathscr{L}$ is a Lagrangian density of order $k$ - we define the action functional $S_{\mathscr{L}}:C^\infty_c(M)\times\Gamma(\pi)\rightarrow\mathbb{R}$ associated to $L$ (here $C^\infty_c(M)$ is the space of smooth real-valued functions on $M$ with compact support, and $\Gamma(\pi)=\{\phi\in C^\infty(M,E)\ |\ \pi\circ\phi=\text{id}_M\}$ is the space of smooth sections of $\pi$) as $$S_{\mathscr{L}}(f,\phi)=\int_M f(j^k\phi)^*\mathscr{L}\ .$$ The Euler-Lagrange operator $E(\mathscr{L})$ associated to $\mathscr{L}$ is the (usually nonlinear) partial differential operator of order $\leq 2k$ defined by the formula $$\int_M E(\mathscr{L})[\phi](\vec{\phi})=\int_M\left.\frac{\partial}{\partial t}\right|_{t=0}f(j^k\phi_t)^*\mathscr{L}\ ,\quad\vec{\phi}\in\Gamma_c(\phi^*VE\rightarrow M)\ ,f|_{\text{supp}\vec{\phi}}\equiv 1\ ,$$ where $\phi_t=\Phi(t,\cdot)$ for any $\Phi\in C^\infty(I\times M,E)$ with $0\in I\subset\mathbb{R}$ an open interval, $\phi_0=\phi$, $\left.\frac{\partial\Phi}{\partial t}\right|_{t=0}=\vec{\phi}$ and $\phi_t\in\Gamma(\pi)$, $\phi_t|_{M\smallsetminus\text{supp}\vec{\phi}}=\phi|_{M\smallsetminus\text{supp}\vec{\phi}}$ for all $t\in I$. The above definition of $E(\mathscr{L})$ can be shown to be independent of the particular choice of $f$ and $\Phi$ under the above conditions. There are other possible ways to define $E(\mathscr{L})$ in the literature, all leading to the same object. One can see from the above definition that $E(\mathscr{L})[\phi]\in\Gamma(\phi^*V^\circledast E\rightarrow M)$, where $\pi_{V^\circledast E}:V^\circledast E=\pi^*(\wedge^nT^*M)\otimes V^*E\rightarrow E$ is the so-called twisted dual of the vertical bundle $VE=\ker T\pi\rightarrow E$ of $E$. Since $$\phi^*V^\circledast E=\{(p,q)\in M\times V^\circledast E\ |\ \phi(p)=\pi_{V^\circledast E}(q)\}\subset M\times_M V^\circledast E$$ for all $\phi\in\Gamma(\pi)$, it is clear that $\phi^*V^\circledast E$ is a sub-bundle of $M\times_M V^\circledast E$ over $M$. As such, we can write $$E(\mathscr{L})=\rho(E(\mathscr{L}))\circ j^{2k}\ ,$$ where $$\rho(E(\mathscr{L})):J^{2k}(E)\rightarrow M\times_M V^\circledast E$$ is a bundle map covering $\text{id}_M$ ($M$ is seen above as a fiber bundle over itself with singleton fibers and $\text{id}_M$ as the projection map). When one writes a local formula for $E(\mathscr{L})[\phi]$ using a local trivialization of $E$ and a local chart for the typical fiber $Q$ of $\pi$, one sees that $E(\mathscr{L})[\phi]$ is always an affine function of the highest-order derivatives of $\phi$ if the order of $E(\mathscr{L})$ happens to be nonzero (which we assume to be the case), so it would make sense to say that $E(\mathscr{L})$ is a "quasi-linear" partial differential operator. However, $M\times_M V^\circledast E$ is generally not a vector or affine bundle over $M$ (unless $E$ itself is), despite $\phi^*V^\circledast E$ being a (different) vector bundle over $M$ for each $\phi\in\Gamma(\pi)$, so $E(\mathscr{L})$ does not fit into the usual way to globally define quasi-linear partial differential operators (see e.g. the discussion in Section IX.2, pp. 393ff of the book Exterior Differential Systems by R.L. Bryant et al.), which requires the target bundle of the operator to be a vector (or at least an affine) bundle. Question: in which global (i.e. coordinate-independent) way one may define quasi-linear partial differential operators in order to encompass Euler-Lagrange operators $E(\mathscr{L})$ acting on smooth sections of general fiber bundles? REPLY [2 votes]: Igor gave a coordinate independent definition of quasilinear equations which the Euler-Lagrange equations satisfy. Still missing is a definition of quasilinear differential operator, which the Euler-Lagrange operator satisfies. You already observed that $M\times_M V^\circledast E$ is generally not a vector or affine bundle over $M$. hence the definition of quasilinear operator found in Bryant, Chern et. al. on p. 397 (which agrees with the definition found in Krasil'shchik, Lychagin, Vinogradov, Geometry of jet spaces and nonlinear partial differential equations p.160) cannot be applied directly, since they assume vector bundles over $M$. But $V^\circledast E$ is a vector bundle over $E$, and the fix might be to expand the definition of nonlinear DO: Definition: Let $V\to E$ be a vector bundle over $E$ and $\psi:J^k E \to V$ a morphism of fiber bundles over $E$ (not neccessarily linear, since we don't assume a linear structure on $E\to M$), then the map $\psi \circ j^k$ is called a nonlinear DO. Call such an operator quasilinear if $\psi$ is affine, when restricted to any fiber of $J^k E \to J^{k-1} E$. I haven't checked, but suspect that the E-L-operators satisfies this definition of quasilinear DO (with $V=V^\circledast E$). Observe that the zero set of a quasilinear $\psi$ is a quasilinear PDE $\mathcal{E}\subset J^k E$, as in Igors answer. Observe also that a nonlinear DO in the more restrictive sense of: a map of fiber bundles $\rho: J^k E \to W$ over $M$ is a special case of the previous definition by pulling back $W$ to $E$. Note also that such a generalization of DO is unavoidable in cases like minimal surfaces, where we don't have a fiber bundle structure on the ambient space $E$. (In fact, in the case of jets of submanifolds we probably need to allow vector bundles $V$ over $J^1 E$ instead of over $E$, since the vertical bundle $VE$ doesn't exist and is replaced by a "normal" bundle.)<|endoftext|> TITLE: Atiyah-Singer style index theorem for elliptic cohomology? QUESTION [27 upvotes]: In 1994, Mike Hopkins wrote a paper called Topological Modular Forms, the Witten Genus, and the Theorem of the Cube. As usual, the introduction was fantastic, explaining the power of various cobordism invariants and connections between homotopy theory and other fields. He states "it is believed that there is an "index" theorem relating analysis on loop space to elliptic cohomology. So far, a satisfying mathematical theory is lacking." What is the status of this question? Has such a theory been developed? REPLY [21 votes]: The status of this question is OPEN. This theory has NOT been developed yet. That being said, the evidence is as compelling as ever, I don't know of any obstructions to making this work, and I'm convinced that there's an awesome theory out there, waiting to be discovered. Roughly 8 years ago, I wrote an unsuccessful ERC proposal, where I outlined a program. This proposal can be found on my Utrecht website here and here (warning: that website will probably stop existing one year from now, so the links will become broken -- but the linked material will still be there to be found on whatever new website I end up having in the future). There are small bits and pieces of what one might call progress, which I've made available on my website: Here's one. In this draft, I take a compact simply connected Lie group $G$ of dimension $d$, and I consider the map $p:G\to \{pt\}$. I construct, geometrically, the $TMF$-pushforward $p_!(1)\in TMF^{-d}(\{pt\})=\pi_d(TMF)$ of the element $1\in TMF^0(G)$ along the map $G\to \{pt\}$. Here's another one. In this draft, I show that there's a new type of 2-equivariance for $TMF$, where the group of equivariance gets replaced by a fusion category.<|endoftext|> TITLE: Another graph characteristic QUESTION [5 upvotes]: This question concerns a method of drawing graphs and a graph characteristic about which I want to learn more. Consider a connected directed graph with at least one node with in-degree 0 and one node with out-degree 0 (let's call it input-output-graph). Draw the input nodes equally spaced on layer 0: Each node has a well-defined minimal distance $d_{in}$ to the input layer 0. Draw the nodes with $d_{in}=n$ on layer $n$: Now draw all the edges that connect adjacent layers: And finally, draw all the other edges: Note, that none of these layers can be interpreted as the output layer, but based on a prescribed output layer the same construction can be made (in the opposite direction). Each edge $(v_0,v_1)$ has a unique and well-defined length $\lambda$ with respect to the layers it connects. Let $v_0$ be a node on layer $n_0$ and $v_1$ a node on layer $n_1$. $$\lambda((v_0,v_1)) = n_1 - n_0$$ Note, that by construction there cannot be an edge with $\lambda(e)>1$ The characteristic I have in mind is nothing but the distribution $d$ of the lengths $\lambda$: $$d(l) = \text{number of edges $e$ with $\lambda(e)=l$}$$ Note, that the corresponding distribution with respect to the output layer may look quite different. It is obvious that for strictly layered input-output-graphs (esp. trees) $d(1) = \text{number of edges}$ and $d(l)=0$ für $l\neq 1$. But what about small-world or random input-output-graphs? Would the characteristic be a good one to distinguish them? Question: Has this characteristic been defined before, and under which name? REPLY [2 votes]: This seems to be related to crossing numbers. Crossing numbers are minima over all possible drawings of the graph with certain restrictions; the least restricted version is where the only condition on arc images is that an arc image does not cross any 3rd vertex, and two arcs intersect transversally (if at all). One talks about rectilinear crossing number if each arc image must be a line segment. Book crossing number for a $k$-page book arises where one places all the vertices on a "book spine" (a straight line $B$) and each arc is drawn on one of the $k$ "pages" (i.e. half-planes attached to $B$). There are probably many more variations of this.<|endoftext|> TITLE: Isomorphism between a filtered vector space and its associated graded QUESTION [9 upvotes]: $\DeclareMathOperator\gr{gr}$Let $ V $ be a vector space with a decreasing filtration $$ V = F_0 V \supseteq F_1 V \supseteq F_2 V \supseteq\dotsb .$$ We define the associated graded of $ V $ to be $$ \gr V := \bigoplus_{k=0}^\infty F_k V / F_{k+1} V. $$ Of course $ \gr V $ can also be regarded as a filtered vector space and we have a canonical isomorphism $\gr (\gr V) = \gr V $. We say that $ V $ “admits an expansion” if there is an isomorphism of filtered vector spaces between $ \gr V $ and $ V $, which becomes the identity map after applying $ \gr $ to both $ \gr V $ and $ V $. This condition is equivalent to the existence of subspaces $ W_k \subset F_k V $ such that $ F_k V = W_k \oplus F_{k+1} V $ and $ V = \bigoplus_k W_k $. Note that not every filtered vector space admits an expansion. For example, the vector space $ V = \mathbb C[[x]] $ with the filtration $ F_k V = x^k \mathbb C[[x]]$ does not admit an expansion. On the other hand, $ V = \mathbb C[x] $ with the same filtration does admit an expansion. Here are my questions: Does this property have a different name in the literature? Let $V$, $W $ be two filtered vector spaces which admit expansions. Suppose that I have a filtration-preserving map $ \phi : V \rightarrow W $ such that $ \gr \phi : \gr V \rightarrow \gr W $ is an isomorphism. Can I conclude that $ \phi $ is an isomorphism? REPLY [2 votes]: Too long for a comment, I was wondering about the cost of completely unfolding Darij's excellent argument. In fact, we have a characterisation of that sort of perturbations of identity that are isomorphisms. The statement is as follows Lemma. Let $k$ be a field and $x$ an indeterminate. For every $Q\in k[x]$, let $f_Q$ be the morphism of $k[x]$ sending $x$ to $x+x^2Q$. Then $f_Q$ is an isomorphism iff $Q=0$. Proof. One way being obvious (if $Q=0$ $f_Q=Id$), it is sufficient to prove that, if $Q\not=0$ then $x\notin Im(f_Q)=f_Q(k[x])$. It is not difficult to see that $Im(f_Q)$ is a subalgebra as follows $$Im(f_Q)\subset k\oplus (x+x^2Q)k[x]$$ which does not contain $x$ (indeed $P\in Im(f_Q)$ implies that $P(x)-P(0)$ can be divided by $(x+x^2Q)$ which is true for $P=x$ iff $Q=0$).<|endoftext|> TITLE: Is there a bound for Lipschitz constant in terms of second differences? QUESTION [7 upvotes]: It is easy to show that if $f\colon[0,1]\to\mathbb R$ and $|f|\leq A$ and $|f''|\leq B$ then~$|f'|\leq 4A+B$. Indeed, by Taylor formula with remainder $f(x)=f(c)+(x-c)f'(c)+\frac12(x-c)^2f''(d)$ where $d$ is between $x$ and $c$. Therefore $f'(c)=\frac{f(x)-f(c)-\frac12(x-c)^2f''(d)}{x-c}$. Now let $c$ be a maximum of $|f'|$. We can assume without loss of generality that $c\geq\frac12$ (otherwise apply the same argument to the function $f(\frac12-x)$ to get the same bound). Then \begin{aligned} |f'(c)|&\leq2\left|f(0)-f(c)-\frac12(0-c)^2f''(d)\right| \\&\leq2|f(0)|+2|f(c)|+c^2 |f''(d)| \\&\leq4A+B \end{aligned} as required. In particular, $f$ is $(4A+B)$-Lipschitz. The question is, can one obtain similar bounds for the Lipschitz constant of $f$ if one does not require differentiability but only assumes a bound on second differences: $|f(x)-2f(x+h)+f(x+2h)|\leq Bh^2$ ? (asked unsuccessfully at MSE.) REPLY [9 votes]: Smoothifying by convolution as Pietro Majer suggests is pretty ok, but if you prefer more direct argument, you may use a standard Lemma. If a bounded function $f$: $[0,1]\to \mathbb{R}$ satisfies $f(\frac{x+y}2)\leqslant \frac{f(x)+f(y)}2$, then $f$ is convex. Proof. At first, we prove that $f$ is continuous on $(0,1)$. If not, there exists a $c\in (0,1/2)$ and a sequence of points $x_n\in [c,1-c]$ and $\delta_n\rightarrow 0$ such that $f(x_n+\delta_n)-f(x_n)\geqslant c$. We get $f(x_n+k\delta_n)\geqslant f(x_n)+kc$ whenever $0\leqslant x_n+k\delta_n\leqslant 1$, this contradicts to the assumption that $f$ is bounded when $\delta_n$ tends to 0. The rest is easy: we get $f(\alpha x+(1-\alpha)y)\leqslant \alpha f(x)+(1-\alpha)f(y)$ for $x\ne y\in [0,1]$ and $\alpha\in \{1/2^n,2/2^n,\dots,(2^n-1)/2^n\}$ by induction in $n$. For arbitrary $\alpha\in (0,1)$ approximate it by such numbers and use continuity (at a point $\alpha x+(1-\alpha)y\in (0,1)$). Thus your condition implies that the function $g(x)=f(x)-Bx^2/2$ is concave and $h(x)=f(x)+Bx^2/2$ is convex. Fix $0 TITLE: Where did Zermelo first model the natural numbers by iterates of the singleton operator, and have the definitions been compared by himself? QUESTION [13 upvotes]: E. Zermelo is widely said to have modelled the (axioms of the) natural numbers by iterating the singleton operation $\{\cdot\}\colon \mathsf{Set}\rightarrow\mathsf{Set}$, $S\mapsto\{S\}$, whence the technical term Zermelo model. J. von Neumann modelled the (axioms of the) natural numbers by iterating the successor operation ${\cdot}^+\colon \mathsf{Set}\rightarrow\mathsf{Set}$, $S\mapsto S\cup\{S\}$, whence the usual technical term von Neumann ordinal. Incidentally, they both seem to have started at $S=\emptyset$, with von Neumann only writing `0' in his letter (the use of the Norvegian vowel started about a generation later). Questions. (0) Do you know a reference to original work of Zermelo's where the definition via iterates of $\{\cdot\}$ appears? (1) Are there scholarly references (or possibly, testimonies from people having attended those lectures) about whether Zermelo defined the natural numbers this way in his lectures in Göttingen and Zürich? (It of course seems unlikely that he did, having been much involved in axiomatic set theory, and having received what is now the standard definition from von Neumann.) EDIT: This question now has been copied to a more focused satellite question, which is where it should be answered. (2) Are there scholarly references about whether Zermelo ever publicly discussed, in lectures or in writing, von Neumann's definition, possibly comparing his definition to his own? EDIT: This question now has been copied to a more focused satellite question, which is where it should be answered. (3) Are there mathematico-historical treatments in specialized journals of these two models? (This is not asking for a discussions, from a modern point of view, of properties of the two models in this thread, see remark below.) EDIT: If one is content with one example, and if a thesis counts as a "specialized journal", this question has been more or less answered by a reference kindly provided by Francois Ziegler in one of the comments. Remarks. Motivation for this question comes partly from research, partly from an expository writing project. I can read German, did have a look at some original publications of Zermelo's (though not for long...) prior to writing this question, and also did more than just scan for patterns like $\{\{\}\}$, but did not find any trace of the eponymous Zermelo model so far. (As I said, I did not read in his papers for very long.) Von Neumann's definition of the finite ordinals appears at least in a handwritten letter, dated 15. VIII. 1923, that von Neumann sent to Zermelo. This letter is reproduced in facsimile in a German mathematico-historical book by H. Meschowski, partially visible online. So the analogous question appears to be answerable about as nicely as one could possibly imagine. (Of course, it would be somewhat interesting learn that even in this case, Stigler's law of eponomy is validated, but it seems that it is not, and von Neumann ordinals really were first published by von Neumann.) Zermelo's definition appears to deemed technically inferior to von Neumann's in various well-known aspects (non-transitivity of Zermelo's sets, cardinality of Zermelo's sets being only 0 and 1, non-suitability of Zermelo's sets for defining limit-ordinals, etc). This is an interesting topic in and of itself, possibly fit for another MO-thread, but a bottomless pit and not the topic of this question, which is rather historical and reference-requestish; may the only reflection in this thread of theoretical considerations be through Zermelo himself, see question (3) above. REPLY [13 votes]: Another go at the original publications turned up an easy, and probably widely-known, answer to question (0): Zermelo's model appears, even typographically identical to the usual form given today, at least on p. 267 of Math. Ann. Vol.65, No. 2 (1908). For completeness, here is the relevant paragraph: Translation: Now if $Z'$ any other set of the kind required by the axiom, then there is a smallest subset $Z_0'$ of $Z'$ with the property under consideration, which corresponds to $Z'$ in exactly the same way as $Z_0$ corresponds to $Z$. But also the intersection $Z_0\cap Z_0'$, which is a common subset of $Z$ and $Z'$, must have the properties of $Z$ and $Z'$, and, being a subset of $Z$, must contain the subset $Z_0$, and, being a subset of $Z'$, must contain the subset $Z_0'$. It thus follows by I that $Z_0\cap Z_0'=Z_0=Z_0'$, and that thus $Z_0$ is a subset of all possible sets which have the properties of $Z$, even though those sets need not be the elements of any set. The set $Z_0$ contains the elements $0$, $\{0\}$, $\{\{0\}\}$ and so forth. Let $Z_0$ be called the "sequence of numbers", because its elements can be used instead of number-symbols. This set is the simplest example of a "countably infinite" set (Nr. 36). As further background, since it appears not too off-topic, here is part of von Neumann's 1923 letter: (source is the book H. Meschkowski: Denkweisen [...] Vieweg, 1990, ISBN-13: 978-3-322-85074-4. DOI: 10.1007/978-3-322-85073-7 accessed with the help of a known search engine) Transcript of the above part of the letter. (I am slighly unsure about the reading of some of the words, but will not record these instances of doubt, since there seems little point in doing so. The writing is sometimes a bit curious ("Grund Idee", "0Menge"). Also, von Neumann consistently (well, consistently-to-degree-two) uses five dots for what one might call the set of all finite ordinals except the first four finite ordinals, but uses fewer dots (namely four dots) for the larger set of what one might call the set of all finite ordinals except the three first finite ordinals .The difference in number of dots equals the difference in number of initial finite ordinals left out, somewhat suggesting that he did this intentionally. He is inconsistent, though, in that the number of horizontally aligned dots below $3$ is unequal to the number of dots behind $3$ within the curly bracket. Incidentally, this letter shows that "naive set theory" seems to have been a usual phrase far before Halmos' textbook. Moreover, at least here, in this somewhat introductory writing, von Neumann defines limit-ordinals purely syntactically, not in the modern first-define-ordinal-as-transitive-and-well-ordered-by-the-membership-relation-then-divide-the-class-of-sets-so-defined-into-two-classes-the-successors-and-the-non-successors way. I do not know whether he used the latter more axiomatic definition in the full version of the work he is here introducing to Zermelo. I also do not know what is the reference for the "naïve" treatment that von Neumann is announcing at the end of the quoted passage. So here is what seems an accurate transcription, to the point of doubly-dotting two of the i's: In den beiden ersten Teilen [?] der Arbeit wird diese ganze Axiomatik auseinandergesetzt, und die Herleitung der bekannten Mengenlehre durchgeführt. Das geschieht, schon der klaren Auseinandersetzung der angewandten Methode wegen, ziemlich weitgehend und detaillirt. Da es sich größtenteils nur um die formalistische Herleitung bekannter Sätze handelt, mußte dabei viel triviales behandelt werden. Neu sind (von einigen Kleinigkeiten abgesehen) in dieser Darstellung wohl nur die folgenden Punkte: Die Theorie der Ordnungszahlen (zweiter Teil, zweites Kapitel). Es gelang mir die Ordnungszahlen auf Grund der Mengenlehre Axiome allein auf zu stellen. Die Grund Idee war die folgende: Jede Ordnungszahl ist die Menge aller vorhergehenden. So wird: (0 die 0Menge) $0=0$, $1=\{0\}$, $2=\{0,\{0\}\}$, $3=\{0,\{0\},\{0,\{0\}\}\}$, . . . . . . $\omega= \{ 0,\{0\},\{0,\{0\}\}, \{0,\{0\},\{0,\{0\}\}\},..... \}$, $\omega+1= \{ 0,\{0\},\{0,\{0\}\}, .... , \{0,\{0\},....,\{0,\{0\}\}\},..... \}\}$, . . . . . . . . . . (Für die positiven endlichen Zahlen lautet also die Regel so: $x+1 = x \overset{\cdot}{+}\{x\}.)$ Die Theorie hat auch im Rahmen der "naïven Mengenlehre" Sinn. (Sie wird, naïv behandelt, demnachst in der Zeitschrift der Szegediner Universität erscheinen.) For good measure, here is a translation of the above transcript. In the first two parst [word in the autograph is illegible but it is probably the German word "Teilen", an inflected form of "Teil"=part] of the work the axioms are explained in their entirety, and the deduction of the elements of the known set theory is carried out. This is done, if only to clearly explain the method applied, in a rather lengthy and detailed fashion. ["auseinandergesetzt" is, in this way of using it, (a participle of) a by now obsolete German verb, meaning "to explain" or "to analyse"; similar remark for "Auseinandersetzung" which in contemporary German only means "row" or "conflict" but used to mean "analysis"] Since for the most part this amounts to a formalistic derivation of known theorems, many trivialities had to be treated. Only (disregarding a few small things) the following aspects are probably new: The theory of ordinal numbers (part two, chapter two). I succeeded in founding the ordinal numbers on the axioms of set theory alone. The basic idea is the following: Each ordinal number is the set of all preceding ordinal numbers. This implies (writing $0$ for the $0$set) [see above for the formulas] (For the positive finite numbers the rule therefore is: $x+1=x\overset{\cdot}{+}\{x\}$.) This theory also makes sense within the framework of "naive set theory". (It will, treated naively, soon be published in the journal of the university of Szeged.) Remarks. It might help when trying to read Zermelo's paper to be told that Zermelo uses the somewhat unusual convention of closing a section introducing an axiom with that axiom's "name". I.e., he does not use "Axiom des Unendlichen" (i.e. "axiom of infinity") as a section-title under which he describes the axiom, but rather first describes the axiom and then, tombstone-like, closes the paragraph with "(Axiom des Unendlichen)". By "the axiom" in the above translation, Zermelo refers to the axiom of infinity. The "kind required by the axiom" is that this be a set with the two properties that (0), it contains 0 as an element, and (1), it is closed w.r.t. applying the singleton-operator. By "with the property under consideration" he to all appearances means two properties, namely, the two properties required by the axiom of infinity. The sets $Z$ and $Z_0$ in "as $Z_0$ corresponds to $Z$" he refers to two sets discussed one paragraph earlier (which I won't explain here, to keep a lid on things). The grammatical construction "in genau derselben Weise wie $Z_0$ dem $Z$ eine [...]" is one of the apparently rather rare examples where mathematical German is more concise than mathematical English, the latter lacking (for all I know) dative-inflected article (such as "dem"). Translating "möge" as "Let" is the closest I can think of; it is (what I would call) an optative mood (though someone more versed in the theory of grammar may disagree-there seem to be subtypes of "optative"). "Nr. 36" refers forward in the paper, to Section 36 on p. 280, where a proof is given that $Z_0$ is an "infinite" set in the sense that it has the property that it is isomorphic to a proper subset of itself, and that every set with this property must contain an isomorphic image of $Z_0$. (Zermelo appears not to mention Dedekind in the context of such considerations.) Incidentally (I will not dwell on this since this is not the topic of this thread), the introduction of the above-cited paper contains an explicit mention of Zermelo's, to the effect that he could not yet rigorously prove the consistency of his axiom system, although he considers this question "sehr wesentlich" (and hints at deeper future work in that direction). So, somewhat interestingly, the consistency of ZFC has been called into question already in what arguably is its founding document. Questions (1),(2),(3) of the OP, which are basically about the reception of von Neumann's definition, remain open(-ended).<|endoftext|> TITLE: No kernel of the form $\lvert x - y\rvert^{-1}$ on tempered distributions? QUESTION [5 upvotes]: Does there exist a continuous bilinear form $\mathcal{B}$ on $\mathcal{S}(\mathbb{R})\times \mathcal{S}(\mathbb{R})$ such that \begin{equation} \mathcal{B}(\varphi_1, \varphi_2) =\int_{\mathbb{R}\times\mathbb{R}} \frac{\varphi_1(x) \varphi_2(y)}{\lvert x - y \rvert} \mathrm{d}x\mathrm{d}y \end{equation} for every $\varphi_1, \varphi_2 \in \mathcal{S}(\mathbb{R})$ with disjoint supports? Context: A bilinear form $\mathcal{B}$ comes together with a kernel $K \in \mathcal{S}'(\mathbb{R}\times \mathbb{R})$ such that $\mathcal{B}(\varphi_1 ,\varphi_2) = \langle K , \varphi_1 \otimes \varphi_2 \rangle$ (by the kernel theorem). The restriction to functions with disjoint supports allows us to avoid possible diagonal problems for the kernel $K$. It is possible to define a valid kernel of the form $K(x,y) = \lvert x - y \rvert^{-\lambda}$ outside the diagonal for $0<\lambda<1$. My guess is here that this is impossible for $\lambda = 1$, and therefore that it is impossible to find a bilinear form $\mathcal{B}$ satisfying the equation above when restricted to test functions with disjoint supports. Is this correct and if yes, how can one prove it? REPLY [2 votes]: You can try $\mathcal B(\varphi_1,\varphi_2)=\int \int\varphi'_1(x)\ \varphi_2'(y) (|x-y|\ln|x-y|)\ dx\ dy$ , i.e. $K$ is the second derivative $\partial_x\partial_y$ of (the distribution defined by) the locally integrable function $H(x,y)=|x-y|\ln|x-y|$ . This extension is clearly non unique (since any distribution applied to $\varphi_1\varphi_2$ vanishes if the two functions have disjoint supports) but a somewhat standard solution (different from the one above) would use Schwartz's pseudo-function Pf. $r^m$ defined e.g. in Théorie des distributions, formula (II,3;4), a well-defined distribution whose restriction to $\{0\}^c$ coincides with the locally summable function $|x|^m$. This leads (via the Fourier transform of Pf. $r^{-1}$, formula (VII,7;18)) to $$\mathcal B(\varphi_1,\varphi_2)=-2[\int \hat{\varphi_1}(\xi)\ \hat{\varphi_2}(-\xi)\ln|\xi|\ d\xi\ +\ (\mathcal C+\ln(2\pi))\int\varphi_1\int\varphi_2]$$where $\mathcal C$ is Euler's constant. Or equivalently (through the definition of Pf. $r^{-1}$)$$\mathcal B(\varphi_1,\varphi_2)=\lim_{\varepsilon\to0}[\int\int_{|x-y|>\varepsilon}\frac{\varphi_1(x)\varphi_2(y)}{|x-y|}\ dx\ dy\ -\ I(\varepsilon)]$$where $I(\varepsilon)\propto \ln\varepsilon$ , i.e. the finite part of the possibly diverging integral.<|endoftext|> TITLE: Probabilities of small balls with convergent center points under Gaussian measure QUESTION [6 upvotes]: I'm in the following situation: Consider a centred Gaussian measure $\mu_0$ on a separable Hilbert space $X$ with covariance operator $Q \in \mathcal{L}(X)$ (positive definite, self-adjoint, trace class). Denote the Cameron-Martin space of $\mu_0$ by $E := Q^\frac12(X)$ with $\|\cdot\|_E := \|Q^{-\frac12}\cdot\|$. Let $(u_n)_{n\in\mathbb{N}}$ be a sequence in $X$ that is unbounded with respect to $\|\cdot\|_E$, but converges towards $\bar u \in E$ with respect to $\|\cdot\|_X$. Moreover, let $\varepsilon_n \to 0$ with $\varepsilon_n > 0$ for all $n \in \mathbb{N}$. Now I want to show, that $\lim\sup_{n \to \infty} \frac{\mu_0(B_{\varepsilon_n}(u_n))}{\mu_0(B_{\varepsilon_n}(\bar u))} \le 1$, where $B_\varepsilon(u) := \{x \in X: \|x - u\|_X \le \varepsilon\}$. Are there ideas how to prove this or can anyone recommend me according literature? Remark: We have $\lim_{n\to\infty} \frac{\mu_0(B_{\varepsilon_n}(0))}{\mu_0(B_{\varepsilon_n}(\bar u))} = e^{\frac12\|\bar u\|_E^2}$, so we can consider $\frac{\mu_0(B_{\varepsilon_n}(u_n))}{\mu_0(B_{\varepsilon_n}(0))}$ alternatively. REPLY [2 votes]: Masoumeh Dashti kindly provided the following proof, which I reproduce here in my words and notation: Let $u \mapsto W_u$ denote the unique linear extension of $E \to L^2(X,\mu_0)$, $u \mapsto \widetilde{W}_u = (Q^{-\frac12}u,\cdot)_X$, to $X$. We first note that $Z := Q(X)$ is dense in $E = Q^{\frac12}(X)$, and that for every $w \in Z$ the linear functional $W_{Q^{-\frac12}w} = (Q^{-1}w,\cdot)_X$ is continuous. Now by the Cameron-Martin Theorem and Anderson's inequality, $$ \begin{align*} \mu_0(B_{\varepsilon_n}(u_n)) &= \int_{B_{\varepsilon_n}(u_n - w)} \exp\left(-\|w\|_E^2 + W_{Q^{-\frac12}w}(v)\right) \mu_0(dv) \\ &\le e^{-\frac12\|w\|_E^2} \sup_{u \in B_{\varepsilon_n}(u_n - w)} \left\{\exp((Q^{-1}w,v)_X)\right\} \mu_0(B_{\varepsilon_n}(u_n - w)) \\ &\le e^{-\frac12\|w\|_E^2} \sup_{u \in B_{\varepsilon_n}(u_n - w)} \left\{\exp((Q^{-1}w,v)_X)\right\} \mu_0(B_{\varepsilon_n}(0)) \end{align*} $$ holds for all $w \in Z$ and $n \in \mathbb{N}$. On the other hand, the symmetry of $B_{\varepsilon_n}(0)$ implies $$ \begin{align*} \mu_0(B_{\varepsilon_n}(\bar u)) &= e^{-\frac12\|\bar u\|_E^2} \int_{B_{\varepsilon_n}(0)} \exp\left(W_{Q^{-\frac12}\bar u}(v)\right) \mu_0(dv) \\ &= e^{-\frac12\|\bar u\|_E^2} \int_{B_{\varepsilon_n}(0)} \frac12 \left(\exp\left(W_{Q^{-\frac12}\bar u}(v)\right) + \exp\left(-W_{Q^{-\frac12}\bar u}(v)\right)\right) \mu_0(dv) \\ &\ge e^{-\frac12\|\bar u\|_E^2} \mu_0(B_{\varepsilon_n}(0)). \end{align*} $$ Using the continuity of $(Q^{-1}w,\cdot)_X$ and the convergence $u_n \to \bar u$ in $X$, we obtain $$ \begin{align*} {\lim\sup}_{n \to \infty} \frac{\mu_0(B_{\varepsilon_n}(u_n))}{\mu_0(B_{\varepsilon_n}(\bar u))} &\le e^{\frac12\|\bar u\|_E^2 - \frac12\|w\|_E^2} \exp((Q^{-1}w, \bar u - w)_X) \\ &= e^{\frac12\|\bar u\|_E^2 - \frac12\|w\|_E^2} \exp((w, \bar u - w)_E) \end{align*} $$ for all $w \in Z$. In particular, if we consider a sequence $\{w_j\}_{j\in\mathbb{N}} \subset Z$ with $w_j \to \bar u$ in $E$ as $j \to \infty$, the previous estimate leads to $$ {\lim\sup}_{n \to \infty} \frac{\mu_0(B_{\varepsilon_n}(u_n))}{\mu_0(B_{\varepsilon_n}(\bar u))} \le 1. $$<|endoftext|> TITLE: Equation $x=\phi(x)+\phi(x+1)-1$ QUESTION [11 upvotes]: Does anybody have any ideas how to solve the equation $x=\phi(x)+\phi(x+1)-1$, where $x$ is a natural number and $\phi$ is Euler's totient function? I failed even to figure out whether this equation has finite number of solutions or infinite. Any help will be appreciated. REPLY [18 votes]: This is OEIS Sequence A067798. Nothing else seems to be known about it; at any rate OEIS gives no references to the literature, only a link to a list of further such $x$ from Giovanni Resta that extends it from the 43rd solution, $722015$, to the 76th, $1103806594815$, which is just a a bit larger than $2^{40}$. (Thanks to Robert Israel for noting this link in his comment.) It seems very hard to do anything with the problem other than make a few elementary observations (e.g. all $x>2$ will be odd), do some heuristics (biases mod $3$, $4$, $5$, and other small moduli; expected asymptotics), and use yet more computing power to find a handful of further solutions.<|endoftext|> TITLE: Legendre's Constant QUESTION [7 upvotes]: In a couple of web pages, I see that Legendre's constant is defined to be $\lim_{n \to \infty} (\pi(n) - (n/\log(n)))$ (for example, here and here). Actually the first uses $\lim_{n \to \infty} (\log(n) - (n/\pi(n)))$, but I have the same question in either case. To be honest, I just assumed that this was a typo for $\lim_{n \to \infty} (\pi(n) / (n/\log(n)))$ (that is, the prime number theorem). However, in both pages it looks to me as if the claim really is that $\lim_{n \to \infty} (\pi(n) - (n/\log(n)))$ exists and is $1$. Really? I don't know whether to be more surprised that the limit exists or that its value is 1 Does anyone else find it surprising? REPLY [10 votes]: 1. The prime number theorem in the form $$\pi(n)=\mathrm{li}(n)+O(ne^{-c\sqrt{\log n}})$$ combined with the approximation $$\mathrm{li}(n)=\frac{n}{\log n}+\frac{n}{\log^2 n}+O\left(\frac{n}{\log^3 n}\right)$$ shows that $$\pi(n)-\frac{n}{\log n}=\frac{n}{\log^2 n}+O\left(\frac{n}{\log^3 n}\right).$$ So the left hand side tends to infinity quite rapidly, it has no finite limit. 2. The correct definition of Legendre's constant is $$A:=\lim_{n\to\infty}\left(\log n-\frac{n}{\pi(n)}\right),$$ and the third display above shows that it equals $1$: \begin{align*}\log n-\frac{n}{\pi(n)} &=\log n-\frac{n}{\frac{n}{\log n}+\frac{n+o(n)}{\log^2 n}}\\ &=\log n-\frac{\log n}{1+\frac{1+o(1)}{\log n}}\\ &=(\log n)\left(1-\frac{1}{1+\frac{1+o(1)}{\log n}}\right)\\ &=(\log n)\left(1-1+\frac{1+o(1)}{\log n}\right)\\ &=(\log n)\frac{1+o(1)}{\log n}\\ &=1+o(1).\end{align*}<|endoftext|> TITLE: Behaviour at natural boundary QUESTION [7 upvotes]: Suppose I have a holomorphic function $f$ in a domain $\Omega$ with natural boundary $\partial \Omega.$ Let $p \in \partial \Omega.$ Is it true that there is some analogue of Picard's little theorem - that is, by choosing an appropriate sequence $x_1, \dotsc, x_n, \dotsc \in \Omega$ converging to $p$ we can have the limit of $f(x_i)$ be (almost) any complex number? EDIT As divined by Noam Elkies, this was inspired by the recent discussion of $\sum_{i=0}^n z^{i^2},$ and the fact that its zeros seem to cluster near its natural boundary. Indeed, consider the function (a slight variant of that occurring in both answers) $$ \sum_{i=0}^n \frac{z^{2^i-1}}{(i+1)^2}. $$ Here is the graph of its zeros($n=10$): This would tend to imply that you don't have to work very hard to tend to zero when approaching the boundary, and I assume that zero is not a particularly exceptional value. Indeed, if you plot the $1$s of the function (preimages of the value $1$), you get an identical plot: Which would tend to indicate that my conjecture has at least a grain of truth in it. REPLY [5 votes]: There is, indeed, a grain of truth in your conjecture. One possible formalization of it is as follows. Suppose that there is a sequence of analytic in the unit disk $\mathbb D$ functions $f_n$ such that $f_n$ converge pointwise on some set $E\subset \mathbb D$ having an accumulation point inside $\mathbb D$. If $f_n$ omit two fixed different values in $\mathbb D$, then $f_n$ converge uniformly inside $\mathbb D$. This is just the usual mumbo-jumbo about normal families (Montel's theorem, to be exact) plus the uniqueness theorem (any limit of a subsequence has prescribed values on $E$). If you replace $\mathbb D$ by a disk around $p$ and take $f_n$ to be Taylor polynomials of $f$, say, you'll see that either $f$ can be analytically extended to a neighborhood of $p$, or you have the effect you observed for $a$-points of $f_n$ with almost every $a\in \mathbb C$. Alas, more often than not, the zeroes on your picture will lie outside $\Omega$, so you cannot extract too much information about $f$ itself from that picture.<|endoftext|> TITLE: The lattice handshake number ("nearly kissing" number)? QUESTION [11 upvotes]: Update: I'm happy to say that this question has been made essentially obsolete by the breakthrough result of Serge Vlăduţ, who showed that the kissing number is exponentially large: https://arxiv.org/abs/1802.00886 ! Recall that the kissing number in $n$ dimensions is the maximal number $M$ such that there exist $x_1, \ldots, x_M \in \mathbb{R}^n$ with $\|x_i\| = 1$ and $\|x_i - x_j\| \geq 1$ for $i\neq j$. Equivalently, the kissing number is the largest number of non-overlapping spheres of fixed radius that are all tangent to some center sphere with the same radius. It is presumably called the kissing number because the spheres are all "kissing" (i.e., tangent to) the center sphere. The lattice kissing number is the same thing, but we restrict to the case when the $x_i$ span a lattice. I.e., the lattice kissing number is the maximal number of points of length $\lambda_1(L)$ in a lattice $L \subset \mathbb{R}^n$, where $\lambda_1(L)$ is the length of the shortest non-zero lattice vector. It is a long-standing open problem to find a lattice with kissing number $2^{C n}$ for some positive constant $C$. The current best constructions are quite far from this, giving a lower bound of only $n^{\Theta(\log n)}$. (In contrast, if we don't restrict to lattices, then we know that the kissing number is exponential.) My question is what happens when we relax the requirement that the spheres kiss. I.e., what if some of them only shake hands? Formally, for $\epsilon > 0$, we call the $\varepsilon$-lattice handshake number the maximal number of non-zero lattice points in a ball of radius $(1+\varepsilon) \cdot \lambda_1(L)$. It is relatively easy to see that this value is at least $(1+\varepsilon)^n$. (This is true in expectation for a random lattice sampled from the Haar measure.) So, it's exponential when $\varepsilon$ is constant, but what happens when $\varepsilon = \varepsilon(n) = o(1)$ is a function of $n$ that goes to zero? Are there known families of lattices that achieve exponential handshake number for such $\varepsilon$? I have a particular application in mind, but one can imagine how a family of "near-kissing" lattices would be quite useful in general. Edit: The application is now up on the arXiv, https://arxiv.org/abs/1712.00942 . In particular, we need a family of lattices whose $\varepsilon$-handshake number is at least $(1+\varepsilon+\delta)^n$ for any constants $\varepsilon, \delta > 0$ in order to show a certain quantitative hardness result for the lattice Shortest Vector Problem. REPLY [5 votes]: I'm happy to say that this question has been answered by the breakthrough result of Serge Vlăduţ, who showed that the kissing number is exponentially large: https://arxiv.org/abs/1802.00886 ! I.e., there exists a constant $C > 0$ and a sequence of lattices $L_n \subset \mathbb{R}^n$ such that $L_n$ has at least $2^{C n}$ non-zero vectors of exactly minimal length. This is far stronger than what I wanted, which was a sequence with exponentially many vectors of approximately minimal length, and it in particular more than suffices for the application that I mentioned in the question. (It was also a tremendous surprise (at least to me)! I started thinking about the handshake number specifically because I did not expect a result like Vlăduţ's in the near future.) More specifically, if we define the handshake number as $$N_{\varepsilon}(n) := \max_{L\, \subset\, \mathbb{R}^n} \;\#\; \{ y \in L \ : \ 0 < \|y\| \leq (1+\varepsilon) \lambda_1(L) \}$$ for positive integer $n$ and $\varepsilon \geq 0$, then Vlăduţ's result immediately yields the correct asymptotics of $N_{\varepsilon}(n)$ for all $\varepsilon$ [1]: $$\max\{ 2^{\Omega(n)},\ (1+\varepsilon)^n\} \leq N_{\varepsilon}(n) \leq 2^{O(n)} (1+\varepsilon)^n \; .$$ Indeed, the lower bound of $2^{\Omega(n)}$ is Vlăduţ's amazing result. The other bounds are more-or-less trivial: the upper bound follows from a packing argument. The lower bound of $(1+\varepsilon)^n$ is true in expectation for a random lattice in the sense of Siegel (equivalently, this is true in expectation for $\{ z \in \mathbb{Z}^n \ : \ \langle a, x \rangle \equiv 0 \bmod p\}$ with $a \in \mathbb{Z}_p^n$ chosen uniformly at random for a sufficiently large prime $p$). So, "all" that remains is to settle the constant in the exponent. (The question is also interesting in other norms. It will be interesting to see whether Vlăduţ's result extends to other $\ell_p$ norms, for example.)<|endoftext|> TITLE: Shift invariant subspaces of $l^1$ QUESTION [10 upvotes]: There is a simple characterization of shift-invariant closed subspaces of $l^2$: for any measurable subset $S$ of $\mathbb{T} = \mathbb{R}/2\pi\mathbb{Z}$, the set of elements of $l^2$ whose Fourier transform is supported on $S$ is a shift-invariant closed subspace, and every such subspace has this form. This is pretty easy. Is the analogous statement true of shift-invariant closed subspaces of $l^1$? I.e., for any such subspace $E$ there is a measurable subset $S \subseteq \mathbb{T}$ such that $E = \{f \in l^1: \hat{f}$ is supported on $S\}$? If so, my next question would be which measurable subsets of $\mathbb{T}$ arise in this way, as the "support" of a shift-invariant closed subspace of $l^1$? REPLY [4 votes]: This is not a full answer, as my memory / personal bookshelf is not good enough. But it should give some hints. As the comments suggest, we can reduce this to a question about the Fourier algebra $A(\mathbb T)$. Indeed, the steps are: We can turn $\ell^1(\mathbb Z)$ into a commutative Banach algebra for the convolution product. Closed, shift-invariant subspaces of $\ell^1(\mathbb Z)$ are the same as closed ideals of $\ell^1(\mathbb Z)$. The Fourier transform $\ell^1(\mathbb Z) \rightarrow \mathbb A(\mathbb T)$ is an isometric algebra homomorphism. So we wish to classify closed ideals of $A(\mathbb T)$. All this works for a general locally compact abelian group. Under a suitable identification, $\mathbb T$ is the spectrum of $\ell^1(\mathbb Z)$ and the Fourier transform is nothing but the Gelfand transform. Let $I$ be a closed ideal of $A(\mathbb T)$. The hull of $I$ is $$ \nu(I) = \{ h\in \mathbb T : f(h)=0 \ (f\in I) \} $$ a closed subset of $\mathbb T$. Conversely, if $N\subseteq\mathbb T$ is a closed subspace then the kernel of $N$ is $$ \iota(N) = \{ f\in A(\mathbb T) : f(h)=0 \ (h\in N) \} $$ a closed ideal in $A(\mathbb T)$. Clearly $I \subseteq \iota(\nu(I))$ but sadly we do not always have equality. We say that an ideal $I$ has spectral synthesis if we have $I = \iota(\nu(I))$. If $G$ is a discrete abelian group then every closed ideal in $A(G)$ has spectral synthesis. If $G$ is a non-discete abelian group, then a theorem of Malliavin shows that there is a closed ideal without spectral synthesis. In conclusion, closed subsets of $\mathbb T$ do not classify closed ideals of $\ell^1(\mathbb Z)$. Theorem: If the boundary of $\nu(I)$ does not contain a non-empty perfect set then $\iota(\nu(I))=I$ so $I$ is of spectral synthesis. Theorem: If $f\in A(G)$ vanishes on a neighbourhood of $\nu(I)$ then $f\in I$. This 2nd result hints at the connection between spectral synthesis and the ability (or not) to be able to approximate elements in a closed ideals by elements which vanish on a slightly large set than the hull. (This is all in Folland's book "A course in Abstract Harmonic Analysis". You should find a lot more in Hewitt and Ross Vol 2, or Rudin's book, etc.)<|endoftext|> TITLE: What are the potential applications of perfectoid spaces to homotopy theory? QUESTION [44 upvotes]: This year's Arizona Winter School was on perfectoid spaces, and there were quite a few homotopy theorists in the audience. I'd like to get a "big list" of reasons homotopy theorists might care about perfectoid spaces. Of course, a general answer is that homotopy theorists know by now to pay attention whenever number theorists get excited about something, but I'm looking for specific examples. Here are a few examples I know of: $THH$ and $TC$ are intimately related to the $A\Omega$ cohomology theory introduced by Bhatt-Morrow-Scholze. This is my favorite example. Perfectoid spaces feature heavily in Scholze-Weinstein's paper on moduli of $p$-divisible groups, and homotopy theorists really like $p$-divisible groups. I don't know if there are specific ideas yet on how to apply their work in homotopy theory, though. I've talked to some people interested in applying them to the study of Lubin-Tate space and the Gross-Hopkins period map. There's some speculation that it could help with $TAF$. I don't know much about any of these examples past the first, so please feel free to expand on these examples in addition to providing new ones. I'd also be happy to hear about applications of related things like $p$-adic Hodge theory in general or the pro-étale site, even if they don't explicitly mention perfectoid spaces. REPLY [3 votes]: As Peter points out, it is more reasonable to look for connections between prisms and homotopy theory. I'll answer my own queston with some recent observations/speculations on the relation between prisms and equivariant homotopy theory, most of which are recorded in my paper A slice refinement of Bökstedt periodicity. Let $A=\mathbb Z[q]$, acted on by Adams operations $\psi^k(q)=q^k$, and write $B^\bullet$ for the multiplicative monoid underlying a semiring $B$. Then $n \mapsto ([n]_q, \psi^n)$ defines a monoid map $$\operatorname{End}({\mathbb T}) \ge \mathbb N^\bullet \to A^\bullet \rtimes \operatorname{End}_{\mathrm{Ring}}(A).$$ For an oriented prism $(A,d)$ we can do something similar with the submonoid $p^{\mathbb N}\le\mathbb N^\bullet$ by sending $p^k\mapsto (d\dotsm\phi^{k-1}(d), \phi^k)$. The prism condition $\delta(d)\in A^\times$ is equivalent to $FV=p$ (Corollary 3.28), which is a special case of the Mackey functor condition $\operatorname{res}(\operatorname{tr}(x)) = \sum_{g\in G/H} g\cdot x$. $q$-divided powers give lifts of the Norm map (Propositions 3.32 and 3.33), which is part of the Tambara functor structure of $\mathrm{THH}$. Conversely, the fact that $q$-divided powers descend to a map $W_1(R) \to W_2(R)$ is essentially equivalent to the "existence of higher $q$-divided powers" lemma (Lemma 16.7 of Prisms and prismatic cohomology) In view of the previous point, the fact that the Norm scales slice filtration is closely related to a lemma needed for the convergence of the $q$-logarithm (Remark 1.4 of my paper and Proposition 4.9 of The $p$-completed cyclotomic trace in degree 2) So this suggests it might be possible to define a notion of $G$-prism for a compact Lie group $G$, recovering prisms for $G=\mathbb Q_p/\mathbb Z_p$, presumably related to Dress-Siebeneicher's $G$-Witt vectors, such that $\pi^G_0$ of a nice $G$-$E_\infty$-ring spectrum is a $G$-prism.<|endoftext|> TITLE: Upper bound on the critical points of a rational function with zeros and poles on the disk QUESTION [5 upvotes]: Let $f$ be a rational function with $j$ zeros and $k$ poles, all of which reside in the closed unit disk (excepting of course the zeros or poles at $\infty$ when $j\neq k$). What is the smallest number $R>0$ such that all the (finite) critical points of $f$ must lie in the closed disk centered at the origin with radius $R$? When $j\neq k$, a lower bound for the answer is $\dfrac{j+k}{|j-k|}$, as can be seen by inspecting the example $f(z)=\dfrac{(z-1)^j}{(z+1)^k}$. I suspect that $R=\dfrac{j+k}{|j-k|}$ is the answer in general (again assuming $j\neq k$). When $j=k$, I am not sure what I expect the answer to be. NOTE: This question was originally posted on MSE, without receiving any answers. REPLY [7 votes]: Your conjecture is correct. We can assume that the largest critical point occurs at a positive $z=R>0$. If we denote the zeros and poles by $a_n$ and $b_n$, respectively, then the condition for a critical point is $$ \sum \frac{1}{R-a_n} = \sum \frac{1}{R-b_n} . $$ In particular, the real parts of both sides must be equal to one another, and if $R>1$, then, given an $|a|\le 1$, we can find a real $-1\le\alpha\le 1$ such that $\textrm{Re}\: 1/(R-a) = 1/(R-\alpha)$. (To do this, I just confirmed that $1/(R+1)\le\textrm{Re}\: 1/(R-a)\le 1/(R-1)$ by a slightly tedious, but elementary calculation, but maybe there's a better way to see this.) This means that our search for optimal functions can be restricted to those with real zeros and poles, but then it's clear that your example gives the largest possible $R$, by just looking at the effect of changing a single $a_n$ or $b_n$; notice that $F(R)=LHS-RHS$ will have the same sign as $j-k$ for $R$ larger than the largest zero. For $j=k$, there is no bound: consider $\frac{z^2}{(z-a)(z-b)}$. This has its critical point at $z=2ab/(a+b)$, which we can make arbitrarily large.<|endoftext|> TITLE: Rights to edit "Archives Grothendieck" QUESTION [16 upvotes]: The Archives Grothendieck are now "Archives en accès libre", I would like to know if one can edit them... I would like to take an excerpt (split some pages of the archive i.e. from page 10-30 of motifs) of an archive and join its tex version in a single archive. Are there legal issues with editing the material..? At the end my intention is to begin to type som parts of the archives publicly. Mentions légales REPLY [12 votes]: Contact me at michele.bolognesi@umontpellier.fr and we shall discuss the details. In the meanwhile I'll hear from the staff of the university how to deal with this. Best, Michele<|endoftext|> TITLE: Identity with binomial coefficients and k^k QUESTION [13 upvotes]: In process of doing some computations on Hilbert schemes, I stumbled across the following identity, for $k \ge 2$: $$ k^{k-3} = \frac{1}{2} \sum_{i=1}^{k-1} \binom{k-2}{i-1} i^{i-2} (k-i)^{k-i-2} $$ which can be verified computationally for small values of $k$ (I did it for the first 500). This is actually a special case of a bigger problem, so I'm hoping to get ideas about what methods might be applicable to prove this kind of identity. I understand that there are good methods for automatically proving hypergeometric identities, but am I correct in thinking that this is not hypergeometric, due to the $k^{k-3}$ type terms? EDIT Here is a link to the bigger question that I had that this is a special case of: Counting some binary trees with lots of extra stucture REPLY [2 votes]: In my (first research) paper, A note on evaluation of Abel's sums, (that appeared second, in 1979) I described another way to evaluate Abel's sum of the form $$A_n(x,y;p,q)=\sum_{k=0}^n {{n} \choose {k}}(k+x)^{k+p}(n-k+y)^{n-k+q},$$ via (simpler) auxiliary sums defined by $$F_n(x,y;p,q)=\sum_{k=0}^n {{n} \choose {k}}(-1)^k (k+x)^{p}(n-k+y)^{q}.$$ The $F_n$ functions have simple evaluations in terms of difference operators. I was inspired by Riordan's book Combinatorial Identities and later corresponded with Riordan, and met him in 78.<|endoftext|> TITLE: Rankin-Selberg integral for GL(3) form with Odd Maass form on GL(2) QUESTION [7 upvotes]: Let $F$ be a Hecke-Maass cusp form for $SL_3(\mathbb Z)$. Let $u$ be a Hecke-Maass cusp form for $SL_2(\mathbb Z)$. The following integral $$\mathcal L(s,F\times u)=\int_{{SL}(2,\mathbb{Z})\backslash GL_+(2,\mathbb R)/SO(2)}F\left(\begin{pmatrix}z&\\&1\end{pmatrix}\right)u(z)\det(z)^{s- \frac 1 2}{d^*z}$$ produces $$\text{gamma factors }\times\; L(s, F\times u), $$ where $L(s,F\times u)$ is the Rankin-Selberg $L$-functions and $u$ is an even Maass form. However, when $u$ is an odd Maass form, $\mathcal L(s,F\times u)$ is identically zero because of trivial reason. But still $L(s,F\times u)$ is well defined. Question: What integral (maybe similar to $\mathcal L$) will give $L(s,F\times u)$ when $u$ is an odd Maass form? Reference: Proposition 4.4 of Li, Xiaoqing, and Matthew P. Young. "The $L^2$ restriction norm of a $GL_3$ Maass form." Compositio Mathematica 148, no. 3 (2012): 675-717. REPLY [3 votes]: Like Peter Humphries suggests, you need to replace u with $\Lambda_2 u$ and replace $F$ with something that transforms appropriately on the upper-left copy of SO(2). Thinking of the weight 2 raise $\tilde{u}=\Lambda_2 u$ as a function on $\Gamma\backslash G$, and provided I have my signs correct, this function transforms as $$ \tilde{u}\left(\begin{pmatrix}y&x\\&1\end{pmatrix}\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}\right) = \tilde{u}\left(\begin{pmatrix}y&x\\&1\end{pmatrix}\right) e^{-2i\theta}. $$ So you need some raise of $F$ that transforms by $e^{2i\theta}$. A little background: The higher-weight structure of $GL(3)$ is attached to the representations of $SO(3,\Bbb{R})$, and there is exactly one of these (up to isomorphism) for every odd dimension (taking the trivial representation at dimension 1), these are Wigner-D matrices. The spherical forms have images in every dimension except 3. We like to collect the $K$-finite scalar-valued forms into (row) vector-valued forms which transform by the Wigner D-matrix. For $d \ge 2$, if $\tilde{F}=(\tilde{F}_{-d}, \ldots, \tilde{F}_d)$ is a vector-valued form that transforms by the $(2d+1)$-dimensional Wigner D-matrix (that is, $\tilde{F}(gk)=\tilde{F}(g)\mathcal{D}^d(k)$), then the entry at -2 has what you need, i.e. $$ \tilde{F}_{-2}\left(g\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\&&1\end{pmatrix}\right) = \tilde{F}_{-2}(g) e^{2i\theta}. $$ In the notation of this paper, there are two vector-valued forms at $d=2$ which might work: $Y^2 F$ and $Y^0 Y^2 F$. I would guess that you want $$ (Y^2 F)_{-2} = \tilde{Y}^{0,2}_{-2,-2} F = Z_{-2} F=(2y_2 \partial_{y_2}-y_1\partial_{y_1}-2iy_2\partial_{x_2})F, $$ and this guess matches up nicely with Peter Humphries', as well.<|endoftext|> TITLE: Steenrod squares as power operations vs. as cohomomology operations QUESTION [17 upvotes]: There are already several excellent questions and answers on MO regarding Steenrod squares, understanding them in various ways and relating them to power operations and I think I get this. Still, I am confused about the sense in which the power operation perspective coincides with the stable cohomology operation perspective. Namely, there seems to be two very different constructions that for the Eilenberg-MacLane spectrum $H\mathbb{F}_2$ yield (almost) the same answer. 1) Given a spectrum $E$, we can consider the graded (non-commutative) ring $\mathcal{A}_E =[E,E]_*$ with multiplication induced by composition. For every spectrum $X$, the ring $\mathcal{A}_E$ acts on the graded abelian group $E^*(X)=[X,E]_*$. For $E=H\mathbb{F}_2$, we get the ordinary Steenrod algebra which is generated by the $Sq^i$ operations with $i\ge 0$. This can be calculated explicitly as $$\mathcal{A}_2^*=\lim_nH^{*+n}(K(\mathbb{F}_2,n);\mathbb{F}_2)$$ and the Borel-Serre computation of the cohomology of Eilenberg-MacLane spaces (I mention this to emphasis that, as far as I understand, the presentation with $Sq^i$-s is obtained by computation) 2) Given an $E_\infty$-ring spectrum $R$ we can consider the homotopy groups of the endomorphism spectrum of the forgetful functor from $E_\infty$-algebras over $R$ to spectra. Less formally, the thing that naturally acts on the underlying spectrum of every $R$-algebra. This includes the power operations for example. For $R = H\mathbb{F}_2$, we get the big Steenrod algebra (more precisely, a certain completion of it. See Lurie's notes), which is generated (topologically) by all Steenrod squares $Sq^i$ with $i\in\mathbb{Z}$. The ordinary Steenrod algebra can be obtained from it by imposing the single relation $Sq^0 = Id$. In general, the two constructions take different things as input and produce an algebra that in general acts on different things. Yet for $H\mathbb{F}_2$, we get closely related algebras. Question: Is there a conceptual explanation of the relation between (1) and (2)? Is there anything more general we can say about this relation for (ring) spectra other then $H\mathbb{F}_p$? Following Lurie's notes cited above, it seems that the main step in the comparison is the fact that the $\mathbb{F}_2$-cochains of $K(\mathbb{F}_2,n)$ as an algebra over $H\mathbb{F}_2$ is almost the free algebra on one generator of degree $n$, where the only relation we need to impose is precisely $Sq^0=Id$. But again, this is verified in a very computational way and from this perspective seems to me like a miracle. Another point is that power operations for other cohomology theories (like K-theory) are unstable in general, so this also seems rather special. REPLY [8 votes]: Here are some things that one can say for more general ring spectra. I choose to work with even periodic spectra, where $\pi_1(E)=0$ and $\pi_2(E)$ contains an invertible element, so that one can generally concentrate on $E_0(X)$ and $E^0(X)$. Key examples are $E=KU$ and $E=HP=\bigvee_{i\in\mathbb{Z}}\Sigma^{2i}H$ and $$ E=MP=\bigvee_{i\in\mathbb{Z}}\Sigma^{2i}MU=\text{ the Thom spectrum for } \mathbb{Z}\times BU. $$ (Here $H$ will be the Eilenberg-MacLane spectrum for mod $p$ (co)homology, for some fixed prime $p$.) Stable operations $F^0(X)\to E^0(X)$ are controlled by $E^0F$, which in good cases is the $E_0$-linear dual of $E_0F$. This is most naturally described in terms of the scheme $\text{spec}(E_0F)$. Associated to $E$ we have a base scheme $S_E=\text{spec}(E_0)$ and a formal scheme $G_E=\text{spf}(E^0BS^1)$ which should be thought of as a bundle of groups over $S_E$. We have similar data for $F$, using which we construct the scheme $\text{Iso}(G_E,G_F)$ consisting of triples $(a,b,u)$ with $a\in S_E$ and $b\in S_F$ and $u$ an isomorphism between the corresponding fibres of $G_E$ and $G_F$. There is a natural map $\text{spec}(E_0F)\to\text{Iso}(G_E,G_F)$ which is an isomorphism when one of $E$ and $F$ is Landweber exact. In the case $E=F=HP$ we find that $\text{Iso}(G_E,G_F)$ is the polynomial part of the dual Steenrod algebra, but this picture loses the exterior part. Unstable operation $F^0(X)\to E^0(X)$ (for spaces $X$) are controlled by $E^0(\Omega^\infty F)$. This has a natural Hopf algebra structure, and the additive unstable operations are the primitive elements. In good cases $E^0(\Omega^\infty F)$ is dual to $E_0(\Omega^\infty F)$, and the primitive elements are dual to the indecomposables in $E_0(\Omega^\infty F)$. Here $\Omega^\infty F$ is a ring up to homotopy, and the two ring operations give rise to two different products on $E_0(\Omega^\infty F)$. We take indecomposables with respect to the addition product, and the multiplication product passes to the quotient to give a ring structure on $\text{Ind}(E_0(\Omega^\infty F))$. This is again best described in terms of schemes. We can define $\text{Hom}(G_E,G_F)$ in the same style as $\text{Iso}(G_E,G_F)$ and we get a map $$\text{spec}(\text{Ind}(E_0(\Omega^\infty F))) \to \text{Hom}(G_E,G_F) $$ which is an isomorphism when $F$ is Landweber exact. The stabilization map $\Sigma^\infty\Omega^\infty F\to F$ gives rise to a map $\text{spec}(E_0F)\to\text{spec}(\text{Ind}(E_0\Omega^\infty F))$ which is compatible with the inclusion $\text{Iso}(G_E,G_F)\to\text{Hom}(G_E,G_F)$. In the case $E=F=HP$ this picture again sees the polynomial part of the cohomology of Eilenberg-MacLane spaces, but not the exterior part. Now let $\text{Sub}(G_E)$ denote the scheme of pairs $(a,A)$ where $a\in S_E$ and $A$ is a finite flat subgroup scheme of $(G_E)_a$. Put $DS^0=\bigvee_nB\Sigma_n$. The group $E^0DS^0$ can be made into a ring using the transfer maps associated to the inclusions $\Sigma_i\times\Sigma_j\to\Sigma_{i+j}$. There is a natural map $\text{spf}(\text{Ind}(E^0DS^0))\to\text{Sub}(G)$, which is an isomorphism in good cases. In the case $E=HP/p$ this encodes the standard relationship between $H^*(B\Sigma_{p^d})$ and the Dickson invariant subalgebra of $H^*(BC_p^d)$, again ignoring the exterior parts. (The indecomposables are zero in $H^*(B\Sigma_m)$ if $m$ is not a power of $p$.) If $E$ is an $H_\infty$ ring spectrum, we have the following picture. For each point $(a,A)\in\text{Sub}(G_E)$ there is a naturally associated point $b\in S_E$ and an isogeny $q_A\colon(G_E)_a\to(G_E)_b$ with kernel $A$. This creates a relationship between power operations and a particular subclass of isogenies. Stable operations give an action of all isomorphisms, and one can combine this with the action of special isogenies to get an action of all isogenies, but this is slightly artificial. Going back to the case of $HP$, the stories with stable operations, unstable operations and power operations for $\Sigma_{p^d}$ give coaction maps \begin{align*} HP^0(X) &\to HP^0(X)[\zeta_0,\zeta_1,\zeta_2,\dotsc][\zeta_0^{-1}] \\ HP^0(X) &\to HP^0(X)[\zeta_0,\zeta_1,\zeta_2,\dotsc] \\ HP^0(X) &\to HP^0(X)[\zeta_0,\zeta_1,\zeta_2,\dotsc,\zeta_d]/(1-\zeta_d) \end{align*} These are all compatible in an obvious way. Indeed, because $G_E=\text{spf}(E^0BS^1)$ the claim is true by construction when $X=BS^1$. The class $\mathcal{X}$ of spaces where it is true is closed under products and coproducts. Moreover, if $f:X\to Y$ induces a surjection in cohomology, and $Y\in\mathcal{X}$, then $X\in\mathcal{X}$. In the case $E=HP$ we can now use the relationship between $BS^1$, $BC_p=K(\mathbb{Z}/p,1)$ and $K(\mathbb{Z}/p,n)$ to deduce that all spaces lie in $\mathcal{X}$. If $E$ is Landweber exact then one can make a similar argument using the spaces in the $MU$ spectrum instead of $K(\mathbb{Z},n)$.<|endoftext|> TITLE: Who was Guillaume Maran? QUESTION [26 upvotes]: This week I visited the Capitole de Toulouse, where probably the most famous statue for us math people is Pierre de Fermat. But I found another statue of a mathematician: Guillaume Maran. He must be important, at least at that time, to be memorialized in the city hall. But all information I can find online says that he was a lawyer in Toulouse. I don't know if this is the same person as the statue. I can't find his date of birth/death or any of his work. Do you know anything about him? UPDATE: I went again and took a picture of Emmaneul Maignan. But I don't understand the description under his name: $ $ REPLY [18 votes]: For an authoritative answer, I contacted the professor of history at the University of Toulouse, Jacques Krynen, who has written a biography of Guillaume Maran. His response [*] to the puzzling inscription on Maran's statue in de Salles des Illustres strengthens the case that it was an error. My (unsubstantiated) guess is that this happened when the 17th century statues were reconstructed in 1892 after having been destroyed by a fire a few years earlier (as described here). [*] Il y a bien longtemps que je ne me suis pas rendu salle des illustres... en tout état de cause Guillaume Maran était professeur de droit. Pas mathématicien! (I have not been to the hall of celebrities for quite some time... in any case, Guillaume Maran was professor of law. No mathematician!)<|endoftext|> TITLE: Virtual Motives Infinitely Divisible by Lefschetz Motive QUESTION [6 upvotes]: Let $K_0(Var_k)$ be the grothendieck group of the category of $k$-varieties, and call its elements virtual motives. $\mathbb{L}:=[\mathbb{A}^1_k]$ is called the Lefschetz motive. I think that if a virtual motive is divisible by arbitrarily high powers of $\mathbb{L}$, then it must be $0$. Is this true? If so, is there a proof of this? REPLY [6 votes]: This is an open problem. It's probably quite hard.<|endoftext|> TITLE: Axiomatizations of arithmetical parts of theories QUESTION [7 upvotes]: For common theories that talk about something more general than first-order arithmetic (e.g. set theories and subsystems of second-order arithmetic), are there nice axiomatizations of their arithmetic parts? I know that the arithmetic part of $ACA_0$ is $PA$ and the arithmetic parts of $RCA_0$ and $WKL_0$ are both $I\Sigma^0_1$, and I'm mostly interested in stronger theories (like $\Delta^1_1-CA$, $ATR_0$, $\Pi^1_1-CA$, $Z_2$, $KP$, $ZFC$). I'm also aware that Craig's theorem provides a recursive axiomatization of the arithmetical part of any recursively enumerable theory, but I'm interested in "nice" axiomatizations (basically meaning finite lists of axioms and schemas, where you should be able to tell whether a sentence satisfies a schema by doing some rudimentary syntax-checking instead of anything fancy), instead of using cheap tricks like from the proof of Craig's theorem. REPLY [7 votes]: A somewhat general method has been explained by Azriel Lévy [Axiomatization of induced theories, Proc. Am. Math. Soc. 12, 251-253 (1961); ZBL0178.31603; MR0122702] The method does not give axioms that are very intuitive, but I think they fit your "rudimentary syntax-checking" criterion as I understand it. In short, the arithmetical part of ZF is axiomatized by the Peano axioms and all arithmetical sentences which say "if $\Phi \vdash \widehat{\sigma}$ then $\sigma$" where $\Phi$ ranges over finite subsets of ZF, $\sigma$ ranges over arithmetical sentences, $\widehat{\sigma}$ is the sentence in the language of set theory that says that $(\omega;0,1,{+},{\cdot}) \vDash \sigma$. Lévy proves that the same is true for any essentially reflexive theory (a theory that proves the consistency of any finite fragment of itself). Any recursive extension of ZF has this property and I think Z2 also has this property, but finitely axiomatizable theories will not have this property.<|endoftext|> TITLE: A connected reductive algebraic group over a separably closed field is a rational variety QUESTION [6 upvotes]: I need either a proof or a reference in modern (scheme-theoretic) language. According to Sansuc, this result can be gleaned from Borel's book on linear algebraic groups, but the old-style algebraic geometry language makes my head spin. Isn't there a modern proof of the above statement somewhere, for goodness sake? REPLY [3 votes]: How's about these notes by Gille? And these notes by Colliot-Thelene. And these notes by Brian Conrad (Prop 7.2.3) REPLY [2 votes]: This is 16.56 + 21.56 of J.S. Milne, Algebraic Groups: The theory of algebraic group schemes over a field, Cambridge U. P., 2017 (available September)<|endoftext|> TITLE: Must an inverse limit of simply connected groups be simply connected? QUESTION [9 upvotes]: While the fundamental group $\pi_1$ preserves products, it is not true in general that an inverse limit of simply connected topological spaces is simply connected. I would like to know if similar things can happen with topological groups. Let $G=\varprojlim_{n}(G_n,p_{n+1,n}:G_{n+1}\to G_n)$ be the inverse limit of an inverse sequence of 1-connected topological groups $G_n$ and continuous homomorphisms. If $e$ is the identity element of $G$, must $\pi_1(G,e)$ be trivial? What if each $G_n$ is a Lie group? REPLY [5 votes]: It seems that the inverse limit of simply connected Lie groups is simply connected. The argument uses the well-known fact that the second homotopy group of any Lie group is trivial (see e.g. Homotopy groups of Lie groups). Applying the long exact sequence of homotopy groups to the bonding homomorphism $p_{n+1,n}:G_{n+1}\to G_n$ between the simply connected topological groups, we conclude that its kernel $K_{n+1}=p_{n+1,n}^{-1}(e)$ has trivial homotopy group $\pi_1(K_{n+1})=\pi_2(G_n)=0$. Using this fact, and also the fact that a locally trivial bundle over a contractible space is trivial, for any loop $\gamma:\partial\mathbb D\to G$ defined on the boundary of the unit disk $\mathbb D$ we can inductively construct a sequence of maps $\bar\gamma_n:\mathbb D\to G_n$ such that $p_{n+1,n}\circ \bar\gamma_{n+1}=\bar\gamma_n$ and $\bar\gamma_n|\partial\mathbb D=p_{\infty,n}\circ\gamma$ for all $n$. Here $p_{\infty,n}:G\to G_n$ denotes the limit projection. Then the maps $\bar \gamma_n$ determine a map $\bar \gamma:\mathbb D\to G$ extending the map $\gamma$ and witnessing that the topological group $G$ is simply connected. I hope that this argument is correct (this is a question to specialists in Algebraic Topology).<|endoftext|> TITLE: Functions from a lattice to $[0,1]$ QUESTION [6 upvotes]: Let $\mathcal L$ be a bounded modular lattice. Suppose also that $\mathcal L$ is complete and upper-continuous (i.e. for any directed subset $\{x_i:i\in I\}$ of $\mathcal L$ and any $x\in \mathcal L$, $x\wedge\bigvee_{i\in I}x_i=\bigvee_{i\in I}(x\wedge x_i)$). Is there any known sufficient condition for $\mathcal L$ to admit an order-preserving function $f\colon\mathcal L\to [0,1]$, such that: $f(x)=0$ if and only if $x=0$; if $a\leq b$ in $\mathcal L$, $f(a)=f(b)$ implies $a=b$; $f$ commutes with (possibly infinite) joins of directed families. Is there any known structure theorem for lattices admitting such a map? Of course, the obvious case is when $\mathcal L$ is an atom, in which case $\mathcal L\cong\{0<1\}$. I am much more interested in situations where the image of $f$ is a dense subset of $[0,1]$. Motivating Example. Let $D$ be an Archimedean valuation domain, and denote by $v:D^*\to \mathbb R_{\geq0}$ the valuation on $D$. It is possible to define a function $L_v:Mod(D)\to \mathbb R_{\geq0}\cup\{\infty\}$ via the following steps: given an ideal $I$ of $R$, let $v(I):=\inf\{v(d):d\in I\}$; given a finitely generated right $D$-module $F$, let $\{x_1,\dots,x_n\}$ be a generating set with the minimum possible cardinality. Define inductively $F_0=0$; $F_1=x_1D$; if $F_k$ is defined for some $k TITLE: When a Kähler manifold is isometric to $\mathbb C^n$ QUESTION [7 upvotes]: Let $(X,\omega)$ be an $n$-dimensional complete Kähler manifold. Then when it is isometric to complex Euclidean space $\mathbb C^n$ REPLY [12 votes]: The story of characterization of the isometry class of $\mathbb C^n$ equipped with the flat metric comes back to around 80 of the nice work of Burns in Annals of Mathematics and also others I know the following theorem, it may help in the foliation language. Theorem: An $n$-dimensional Kähler manifold $(X,\omega)$ is isometric to $\mathbb C^n$ if and only if the Kähler metric equals $\omega= \partial\bar\partial\tau$, where $ \tau:M\to [0,∞)$ is an $\mathbb C^\infty$ strictly plurisubharmonic (psh) exhaustion function which satisfies the following Monge-Ampère foliation $$(\partial\bar\partial\log\tau)^n=0$$ See Burns, Dan, Curvatures of Monge-Ampère foliations and parabolic manifolds. Annals of Mathematics. (2) 115 (1982), no. 2, 349–373.<|endoftext|> TITLE: Category theory from MK class theory perspective? QUESTION [10 upvotes]: I'm looking for a text that treats category theory from the perspective of MK class theory. MK is already very well-designed and equipped for the type of abstraction that occurs in category theory, and I suspect that this together with its ability to discuss proper classes with relative ease would make a treatment in that context very elegant and concise. EDIT: It is also my understanding that many category theorists are less interested in viewing set theory as a foundation for mathematics, and more interested in it as simply a base language with which we can construct collections of objects in an intuitive manner and then manipulate them. It seems to me that a good option in this case is to simply pick an absurdly strong set theory to use as the underpinning 'language', and MK is more powerful than other widely studied set theories without the existence of inaccessible cardinals. REPLY [19 votes]: Morse-Kelley set theory doesn't seem adequate for all the things one would like to do in category theory. It provides a nice treatment of proper classes, so it can deal with large categories like the category of sets or the category of groups. It can deal with a functor between two such categories, like the forgetful functor from groups to sets or the "free group" functor from sets to groups. But the category of all functors between two large categories (or even from a large category to a small one) is in general a collection of classes, and that's beyond what MK can handle directly. For a smooth development of things like functor categories, Kan extensions, and related concepts, especially if you want to handle iterations, like the category of functors between two categories of functors between large categories, you'll want infinitely many levels of the cumulative hierarchy of sets beyond your large categories. In other words,you'll want not only classes but also super-classes (i.e., collections of classes), super-duper classes, etc. That sort of set theory is stronger than MK. It's still considerably weaker than an inaccessible cardinal, but the underlying intuition is pretty close to an inaccessible cardinal. Furthermore, once you've introduced all these super-duper-etc.-classes, your original large categories, like the category of sets, don't look so comprehensive any more; you might want a category of all sets, classes, super-classes, etc. Unless I put some serious constraints on my wishful thinking, I'm likely to end up just where Grothendieck did, in a universe full of inaccessible cardinals (while set theorists are looking down at these pitifully "small large cardinals").<|endoftext|> TITLE: Intuition for the Drift Term of the Laplace-Beltrami Operator QUESTION [8 upvotes]: In coordinates, the Laplace-Beltrami operator on a Riemannian manifold $(M,g)$ can be written as: $$ \Delta_g = g^{ij}\partial_{ij} - g^{jk}\Gamma^\ell_{jk}\partial_\ell $$ The second term: $$ \mu^\ell = - g^{jk}\Gamma^\ell_{jk} $$ can be viewed as the "convection term" in the Riemannian heat equation or the drift term of the (Ito) stochastic differential equation defining Brownian motion on $(M,g)$. Question: What is the geometric meaning of this $\mu$? I would really like some intuition as to how the geometry generates this term (i.e. how to interpret it geometrically). Any other insights into intuitively understanding this term would be appreciated as well (e.g. other places where it appears). (Note: this is a refinement of this question). REPLY [3 votes]: You really should think of $\Delta$ as an $L^2$ self-adjoint, elliptic operator in it's own right irrespective of coordinates. $-\Delta$ has positive spectrum, with countable eigenvalues accumulating at $\infty$ etc. It's the average of the Hessian $\nabla^2 f$ in the sense that contracting the $(0, 2)$ tensor $\nabla^2 f$ with the metric $g$ produces a $(1, 1)$ tensor that you can trace (which is an averaging operation). In this sense there is no drift term. On the other hand, there is a way think of the extra term $\mu$ coming from the connection. For a smooth function $f : M \to \mathbb{R}$, from the smooth structure alone, you can always form the differential $Df \in \Gamma^{\infty}(T^{\ast} M)$ as smooth section of the cotangent bundle. Without a connection, you can think of $Df : TM \to \mathbb{R}$ as a map between smooth manifolds, differentiate again, and this gives $$ D^2 f : TTM \to \mathbb{R} $$ I used $D$ for the differential so as not to confuse this with the exterior derivative $d$ which would have $d^2 = 0$. Now if you have a connection, you can split $TTM$ as $$ TTM \simeq VTM \oplus HTM $$ into vertical and horizontal sub-bundles. In this particular case both $VTM$ and $HTM$ are isomorphic to $TM$. $VTM$ is the kernel of the map $d\pi : TTM \to TM$ where $\pi: TM \to M$ is the bundle projection and this is isomorphic to $TM$. The connection allows you to "split" the map $d\pi : TTM \to TM$ obtaining an injective bundle morphism $TM \to TTM$ complementary to $VTM \simeq TM$ whose image I'll denoted $HTM$. The vertical bundle consists of elements of $TTM$ that are tangent to the fibres of $TM$ while the horizontal bundles consists of elements of $TTM$ that are tangent to the base $M$. To help clarify, for a general vector bundle $\pi : E \to M$, the kernel $VE$ of $d\pi : TE \to TM$ is isomorphic to $E$ and a connection gives a splitting $TM \to TE$ whose image is denoted $HE$ and such that $TE \simeq VE \oplus HE \simeq E \oplus TM$. Now, what this all has to do with "drift" is that under the identification $TTM \simeq TM \oplus TM$, the $g^{ij} \partial_{ij}$ term comes from $VE$ - it's the term tangent to the fibre and corresponds to $\partial_{ij} f = D^2 f$. The other term - the one with the connection coefficients $\Gamma$ corresponds to the part that is tangent to the base $M$. In other words, the $g^{ij} \partial_{ij}$ part arises from differentiating while moving along the fibres of $TM$ and the $\Gamma$ part arises from moving along the base $M$. Thus the drift is measuring the change in the fibres of the bundle $TM$ as measured by the connection as the basepoint $x \in M$ varies.<|endoftext|> TITLE: Nori's Mixed Motives and Realisation Functors QUESTION [5 upvotes]: The conjecture 51 in Levine's Mixed Motives in Handbook of K-theory http://www.math.illinois.edu/K-theory/handbook/1-429-522.pdf states that the functor induced by $hs:\text{ECM} \rightarrow \text{MHS}$, denoted by, \begin{equation} \mathfrak{hs}:\text{NMM}(k)_{\mathbb{Q}} \rightarrow \text{MHS}_{\mathbb{Q}} \end{equation} where $\text{NMM}(k)_{\mathbb{Q}}$ is Nori's mixed motives over a field $k$ (which admits an embedding $\sigma: k \rightarrow \mathbb{C}$) tensored by $\mathbb{Q}$. I am not clear with the construction of the functor $\mathfrak{hs}$ (I guess this is the Hodge realisation functor of Nori's mixed motives), could anyone give me some references? I also have two questions about the properties of $\mathfrak{hs}$. Is $\mathfrak{hs}$ (conjectured to be) exact or not? Has the exactness of it been proved? If $\mathfrak{hs}$ is exact, then it has a derived functor, \begin{equation} D\mathfrak{hs}:D^b(\text{NMM}(k)_{\mathbb{Q}}) \rightarrow D^b(\text{MHS}_{\mathbb{Q}}) \end{equation} In Harrer's thesis, https://arxiv.org/abs/1609.05516 he constructs a functor, \begin{equation} R_n:\text{DM}_{gm}(k,\mathbb{Q}) \rightarrow D^b(\text{NMM}(k)_{\mathbb{Q}}) \end{equation} The composition of it with $D\mathfrak{hs}$ is a functor \begin{equation} D\mathfrak{hs} \circ R_n :\text{DM}_{gm}(k,\mathbb{Q}) \rightarrow D^b(\text{MHS}_{\mathbb{Q}}) \end{equation} Is this functor equivalent to the Hodge realisation functor? Any references? REPLY [3 votes]: Did you read http://home.mathematik.uni-freiburg.de/arithgeom/preprints/buch/buch-v1.pdf? About 1: since the forgetful functor $MHS_{\mathbb{Q}}\to \mathbb{Q}-Vect$ is exact and faithful, it suffices to verify that the singular cohomology functor from Nori motives is exact. The latter statement should be an immediate consequence of the definition of Nori motives.<|endoftext|> TITLE: About the existence of a convergent sequence QUESTION [6 upvotes]: Let $(A_n)$ be a set sequence in a Banach space wheresuch that $A_n$ is nonempty, closed and convex for every $n=1,2\dots$. Assume that $\displaystyle\lim_{n,m\to \infty} d(A_n,A_m)=0$ where d is the Hausdorff distance between two sets. My question is whether there exists a convergent sequence $(x_n)$ satisfying $x_n\in A_n$ for every $n$? I asked this question on MSE, but haven't got answers. REPLY [3 votes]: Fedor Petrov's answer to my more general question implies a positive answer to this question, as Banach spaces are complete metric spaces.<|endoftext|> TITLE: Hausdorff distance and Cauchy sequences QUESTION [7 upvotes]: This is a generalization of an older question. Let $(X,d)$ be a metric space and let $(A_n)_{n\in\mathbb{N}}\subseteq X$ be a sequence of non-empty closed subsets such that for all $\varepsilon > 0$ there is $N\in\mathbb{N}$ such that for all $m,n\geq N$ we have $d_H(A_m,A_n)<\varepsilon$ (where $d_H(\cdot,\cdot)$ denotes the Hausdorff distance). Can we pick $x_n\in A_n$ for all $n\in\mathbb{N}$ such that $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence? REPLY [6 votes]: Yes. Choosing a subsequence $n_1m$ and a point $x_m\in A_m$ on the distance at most $d_H(A_m,A_{n_i})+1/m$ from $x_{n_i}$. The whole sequence $(x_1,x_2,\dots)$ is still Cauchy.<|endoftext|> TITLE: Harmonic analysis on nilpotent Lie groups and the Campbell-Hausdorff formula QUESTION [5 upvotes]: I am trying to understand the non-commutative analysis for nilpotent Lie groups, so I've been reading Corwin's and Greenleaf's book on the representation theory of nilpotent groups and going through examples of the orbit method. My problem is that I would like to apply the Fourier transform to the Lie algebra generators, and so basically I only need the unitary representations of the Lie algebra. We can simply differentiate the Lie group representations of one parametric subgroups, but as I was trying to go into more difficult examples, the calculations became quiet tedious very fast, because they rely (at least the way it's done in the book) on the Campbell-Hausdorff formula to construct the induced representations. So my question is: are there any constructions of irreducible unitary representations of nilpotent Lie groups/algebras, that do not use the Campbell-Hausdorff formula or that are more friendly for high-dimensional cases? Thanks. REPLY [2 votes]: One friendly class comes from parabolic induction. So in case your nilpotent algebra is nilradical of a parabolic subgroup, there is a general formula that is straightforward to apply. See Theorem 1.3 of Algebraic analysis on scalar generalized Verma modules of Heisenberg parabolic type I.: $A_n$-series. This case deals with Verma modules which are unitarizable only in very special case (where the nilradical is actually Abelian), but more or less the same combinatorics should work even in the real case. Indeed, it seems that this approach can be generalised considerably, see later works of the first author.<|endoftext|> TITLE: Injectivity implies surjectivity QUESTION [85 upvotes]: In some circumstances, an injective (one-to-one) map is automatically surjective (onto). For example, Set theory An injective map between two finite sets with the same cardinality is surjective. Linear algebra An injective linear map between two finite dimensional vector spaces of the same dimension is surjective. General topology An injective continuous map between two finite dimensional connected compact manifolds of the same dimension is surjective. Are you aware of other results in the same spirit? Is there a general framework that somehow encompasses all these results? REPLY [2 votes]: There is an example of this in stable homotopy theory, related to the famous `Generating Hypothesis,' conjectured by Freyd (and still open as far as I know!). For finite complexes $X$ and $Y$, the map from the stable homotopy classes of (pointed) maps from $X$ to $Y$ can be related to the algebraic maps between stable homotopy groups via $$ [ X, Y]^S \to \mathrm{Hom}(\pi_*^S(X), \pi_*^S(Y) ). $$ The Generating Hypothesis says that this function is always injective finite complexes $X$ and $Y$; and it is a theorem of Freyd that if it is true, then the map is always surjective, too. It has been long enough since I read up on this that I can't remember if the surjectivity follows on a pointwise basis, or whether you need the full conjecture to get the implication for a particular $X$ and $Y$.<|endoftext|> TITLE: Finding a 1-form adapted to a smooth flow QUESTION [43 upvotes]: Let $M$ be a smooth compact manifold, and let $X$ be a smooth vector field of $M$ that is nowhere vanishing, thus one can think of the pair $(M,X)$ as a smooth flow with no fixed points. Let us say that a smooth $1$-form $\theta$ on $M$ is adapted to this flow if $\theta(X)$ is everywhere positive; and The Lie derivative ${\mathcal L}_X \theta$ is an exact $1$-form. (By the way, I'd be happy to take suggestions for a better name than "adapted". Most adjectives such as "calibrated", "polarised", etc. are unfortunately already taken.) Question. Is it true that every smooth flow with no fixed points has at least one $1$-form adapted to it? At first I was sure that there must be counterexamples (perhaps many such), but every construction of a smooth flow I tried to make ended up having at least one adapted $1$-form. Some examples: If the flow is isometric (that is, it preserves some Riemannian metric $g$), one can take $\theta$ to be the $1$-form dual to $X$ with respect to the metric $g$. If the flow is an Anosov flow, one can take $\theta$ to be the canonical $1$-form. If $M$ is the cosphere bundle of some compact Riemannian manifold $N$ and $(M,X)$ is the geodesic flow, then one can again take $\theta$ to be the canonical $1$-form. (This example can be extended to a number of other Hamiltonian flows, such as flows that describe a particle in a potential well, which was in fact the original context in which this question arose for me.) If the flow is a suspension, one can take $\theta$ to be $dt$, where $t$ is the time variable (in local coordinates). If there is a morphism $\phi: M \to M'$ from the flow $(M,X)$ to another flow $(M',X')$ (thus $\phi$ maps trajectories to trajectories), and the latter flow has an adapted $1$-form $\theta'$, then the pullback $\phi^* \theta'$ of that form will be adapted to $(M,X)$. Some simple remarks: If $\theta$ is adapted to a flow $(M,X)$, then so is $(e^{tX})^* \theta$ for any time $t$, where $e^{tX}: M \to M$ denotes the time evolution map along $X$ by $t$. In many cases this allows one to average along the flow and restrict attention to cases where $\theta$ is $X$-invariant. In the case when the flow is ergodic, this would imply in particular that we could restrict attention to the case when $\theta(X)$ is constant. Conversely, in the ergodic case one can almost (but not quite) use the ergodic theorem to relax the requirement that $\theta(X)$ be positive to the requirement that $\theta(X)$ have positive mean with respect to the invariant measure. The condition that ${\mathcal L}_X \theta$ be exact implies that $d\theta$ is $X$-invariant, and is in turn implied by $\theta$ being closed. For many vector fields $X$ it is already easy to find a closed $1$-form $\theta$ with $\theta(X) > 0$, but this is not always possible in general, in particular if $X$ is the divergence of a $2$-vector field with respect to some volume form, in which case the integral of $\theta(X)$ along this form must vanish when $\theta$ is closed. However, in all the cases in which this occurs, I was able to locate a non-closed example of $\theta$ that was adapted to the flow. (But perhaps if the flow is sufficiently "chaotic" then one can rule out non-closed examples also?) REPLY [11 votes]: For a slightly different perspective: The Reeb foliation of an annulus (https://en.wikipedia.org/wiki/Reeb_foliation) can be doubled to give a flow on a torus, generated by a nowhere zero vector field $X$, such that all but one trajectory spirals towards a closed loop in forward time. This flow has the property that (i) all but 2 trajectories are nonrecurrent and (ii) the foliation has no closed transversal (it is the standard example of a "non-taut" foliation, see e.g. Thurston, "A Norm on the Homology of a 3-Manifold"). It follows that there is no $X$-invariant 1-form $w$ with $w(X)>0$. Indeed, if $dw \neq 0$ then $|dw|$ gives a smooth X-invariant measure, contradicting (i), and if $dw = 0$ then $w$ is a closed, nowhere zero 1-form, and the leaves of the measured foliation defined by $Ker(w)$ are recurrent and transverse to the leaves of $X$, giving a closed transversal, contradicting (ii). This is the same as Bryant's example; as he observes, one can uses the additional fact that every $X$-invariant function is constant to even rule out $w$ with $L_X(w) = df$.<|endoftext|> TITLE: Triangles whose vertices and center have all the same color QUESTION [6 upvotes]: A plane is colored with two colors. It's an easy exercise to prove that it's always possible to find an equilateral triangle whose vertices have all the same color. Does anyone know any proof or reference for the following problem? A plane is colored with two colors. Is it always possible to find a triangle whose vertices and center C have all the same color? Consider four different cases: i) C is the incenter ii) C is the circumcenter iii) C is the orthocenter iv) C is the barycenter. Any help would be appreciated. REPLY [3 votes]: Actually, for any finite subset $P$ of the plane, there are $2^{\aleph_0}$ monochromatic scaled copies of $P$: https://arxiv.org/abs/1304.3154 (see also the question Monochromatic point sets in two-colored plane)<|endoftext|> TITLE: Finding the nearest matrix with real eigenvalues QUESTION [17 upvotes]: In this thread on MATLAB Central, I found a discussion on finding the nearest matrix with real eigenvalues. The first hypothesis was to simply truncate the complex part of the eigenvalues. So, given matrix $A$, the closest matrix to $A$ in some norm would be $$A' = V \; \mathrm{real}(D)\; V^{-1}$$ where $A= V D V^{-1}$ is the eigendecomposition of $A$ (assuming $A$ is diagonalizable). This has been found to be false, however, with the counterexample $$A = \begin{bmatrix} 1& 1& 0 \\ -1& 0& 0 \\ 0& 0& 0 \\ \end{bmatrix}$$ The procedure above would produce $$A'=\begin{bmatrix} 0.5& 0& 0 \\ 0& 0.5& 0 \\ 0& 0& 0 \\ \end{bmatrix}$$ but there is another matrix $$A''=\begin{bmatrix} 0.9& 0& 0 \\ -2& -0.1& 0 \\ 0& 0& 0 \\ \end{bmatrix}$$ that has real eigenvalues and is closest to $A$ both in spectral and Frobenius norm. Now, let $ A = U T U' $ be the real Schur decomposition of $A$, with $U$ orthogonal and $T$ quasi-upper triangular (there are 2-by-2 blocks on the diagonal corresponding to complex eigenvalues). Consider $ A_s = U \; \mathrm{triu}(T) \; U' $ where triu returns the upper triangle part. By numerical simulations $A_s$ it is always closer to random $A$ than $A'$ and on the counterexample it is also closest to $A$ than $A''$ (in Frobenius and spectral norm). My conjecture is that $A_s$ is the closest matrix with real eigenvalues to $A$ in Frobenius (or spectral) norm. Unfortunately I could not prove or disprove this conjecture. Can anybody? A starting point can be $$||A-A_s||_F = ||T-\mathrm{triu}(T)||_F$$ REPLY [4 votes]: Vanni Noferini and I have published an arxiv e-print, Nearest $\Omega$-stable matrix via Riemannian optimization, in which we address also this problem (in Section 7.5). Quick summary of our findings: There does not seem to be a simple closed-form solution. In particular, you can't get the solution by just truncating the Schur form or the eigendecomposition of $A$. We suggest an interesting reformulation of the problem. Namely, let $\operatorname{triu}$ be the operator that returns the upper triangular part of a matrix (diagonal included) $$ \operatorname{triu}(A)_{ij} = \begin{cases} A_{ij} & i\leq j,\\ 0 & i > j \end{cases} $$ and $\operatorname{tril}(A) = A - \operatorname{triu}(A)$ the one that returns its lower triangular part (diagonal excluded). Then the nearest matrix to $A$ with real eigenvalues (returned already in Schur decomposition) is $B = Q\operatorname{triu}(Q^*AQ) Q^*$, where the orthogonal matrix $Q$ solves $$ Q = \arg\min_{Q \in O_n} \|\operatorname{tril}(Q^*AQ)\|_F^2. $$ In this reformulation, the objective function is a nice simple smooth function: it just the sum of squares of the strictly lower triangular entries of $Q^*AQ$. In addition, this formulation has a smaller dimension (the manifold of orthogonal matrices has dimension $\frac{n(n-1)}{2}$ rather than $n^2$) and simpler constraints ("$Q$ is orthogonal" is easier to deal with numerically than "all eigenvalues of $B$ are real"). Unfortunately this problem (in both formulations, the original one and the alternative formulation) is non-convex and has multiple local minimizers. In similar problems, the number of minimizers grows exponentially with the dimension $n$, and we believe that this holds also for this problem. However, code for optimization on matrix manifolds can be used to compute a (local) minimizer numerically in a very fast way, at least for small matrices (up to $n\approx 50$). Our Matlab code is available. None of the minimizers suggested by the previous answers for the $3\times 3$ and $4\times 4$ examples in this thread are global minima; we can beat them all. For matrices that are so small, repeating the minimization procedure with various starting points one gets only few local minima, so numerical evidence suggests that the matrices that we compute in Section 7.5 are global minimizers. For instance, for the matrix $$ A = \begin{bmatrix} 1 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} $$ in the original question, our method computes the minimizer $$ B \approx \begin{bmatrix} 1.2110 & 0.7722 & -0.0962\\ -0.7722 & -0.2416 & -0.0962\\ -0.0962 & 0.0962 & 0.0306\\ \end{bmatrix}, $$ with $\|A-B\|_F = 0.4946$. It is not hard to prove that for each (local) minimizer $B$, $\operatorname{Tr} B = \operatorname{Tr} A$; otherwise one could construct a closer matrix $B + (\operatorname{Tr} A - \operatorname{Tr} B)\frac{1}{n}I$. Matrices with real distinct eigenvalues have a neighbourhood formed of matrices with real distinct eigenvalues: hence a (local) minimizer $B$ cannot have distinct eigenvalues. Numerically, often the minimizers have eigenvalues with high multiplicities. For instance, in the example above $B$ has a triple eigenvalue in $1/3$. It does not seem easy to extend the results to other matrix norms. We started our research by thinking about this problem, but then we were happily surprised to find out that the technique can be applied also to a harder problem that had already been studied in the literature on numerical linear algebra and control theory, that of finding the nearest Hurwitz stable matrix. And, more generally, it can be extended to solve numerically the problem of finding $$ B = \arg \min_{S_\Omega} \|B-A\|_F, $$ where $S_\Omega$ is the set of all (real or complex) matrices with all eigenvalues in a given closed set $\Omega$. Another nice example of how procrastinating and thinking about Mathoverflow questions can lead to useful research sometimes. :)<|endoftext|> TITLE: A conjecture on planar graphs QUESTION [37 upvotes]: I don't know the following is a known result, but it would be very useful to me in my research if it were true. Conjecture: Let $G$ be a planar graph. The sum $$ \sum_{\{x,y\} \in E(G)}{\min(\deg(x),\deg(y))} $$ is at most linear in the number of vertices. What I know about this problem: This conjecture would be false if one replaces the minimum by the average - the star graph is a counterexample, in which the sum is quadratic. I can prove an upper bound of $O(n \log(n))$ as follows. Let $A_i = \{v : 2^i \leq \deg(v) < 2^{i+1}\}$, and let $$ E_i = \{\{x,y\} \in E(G): x \in A_i ,y \in \cup_{j \geq i}{A_j}\}. $$ Now, $E(G)$ is the union of the $E_i$'s, and the contribution of an edge from $E_i$ is at most $2^{i+1}$. On the other hand, as $G$ is planar the size of $E_i$ is at most $3|\cup_{j \geq i} A_i |$. Now, as the average degree in a planar graph is at most 6, the number of vertices whose degree is at least $2^i$ is at most $6n/2^i$. Therefore $|E_i| \leq 18 n/ 2^i$. We have $$ \sum_{\{x,y\} \in E(G)}{\min(\deg(x),\deg(y))} \leq \sum_{i=0}^{\log_2(n)} \sum_{\{x,y\} \in E_i}{\min(\deg(x),\deg(y))} $$ $$ \leq\sum_{i=0}^{\log_2(n)} |E_i| \cdot 2^{i+1} \leq \sum_{i=0}^{\log_2(n)} (18 n/ 2^i) \cdot 2^{i+1} = 36 n \log_2(n). $$ REPLY [10 votes]: This is just an elaboration on Brendan McKay's beautiful answer, but too long for a comment. The crucial idea is to simplify the problem by generalising it, introducing a maximisation on the indices of the sum, detached from the original planar graph $G$. For $x = (x_1, \ldots, x_n) \in \mathbb{R}_{\geq 0}^n$ and a multi-graph $H$ with $V(H) \subseteq \{ 1, \ldots, n \}$, let $$ L_H(x) := \sum_{ij \in E(H)} \min \{ x_i, x_j \} . $$ For a natural number $d$, let $\mathcal{H}_d$ be the class of all finite multi-graphs $H$ with $V(H) = \{ 1, \ldots, n \}$ (for any $n$) such that $e(H[A]) \leq d(|A|-1)$ holds for any $A \subseteq V(H)$ (in other words, graphs of arboricity at most $d$). Theorem: For any $x \in \mathbb{R}_{\geq 0}^n$ and any $H \in \mathcal{H}_d$ we have $L_H(x) \leq d(\sum_{i=1}^n x_i - \max_i x_i)$. Proof: We may assume that $x_i >0$ holds for every $i$. Permute $\{1, \ldots, n\}$ so that $x_1 \geq x_2 \geq \ldots \geq x_n$. Note that $$ L_H(x) = \sum_{ij \in E(H) \atop i \, < \, j} x_j $$ Extend the class $\mathcal{H}_d$ slightly by requiring only that $e(H[A]) \leq d(|A| - 1)$ holds when $A = \{ 1, \ldots, k \}$ for some $k$ (that is, $A$ is an initial segment of $\{ 1, \ldots, n\}$). Call this new class $\mathcal{H}_d'$. Choose $H \in \mathcal{H}_d'$ with $V(H) = \{ 1, \ldots, n \}$ so that $L_H(x)$ is maximum and, subject to this, so that $$ R(H) := \sum_{ij \in E(H)} i + j$$ is minimum. We claim that $H$ is a star with center 1. Indeed, if $ij \in E(H)$ and $1 < i < j \leq n$, then we could replace $ij$ by the edge $1j$ and obtain a graph $H'$ with $L(H') = L(H)$ and $R(H') < R(H)$. Trivially $H' \in \mathcal{H}_d'$ as well, hence contradicting our choice of $H$. Moreover, every $j \in \{ 2, \ldots, n \}$ has degree exactly $d$ in $H$. Otherwise, choose $j$ minimum with $d_H(j) \neq d$. If $d_H(j) > d$, then for $A := \{ 1, \ldots, j \}$ we have $e(H[A]) > d(|A| - 1)$, a contradiction to $H \in \mathcal{H}_d'$. Hence $d_H(j) \leq d-1$. Add an edge $1j$ to $H$. If there is some $k > j$ with $d_H(k) > d$, take a minimum such $k$ and delete one edge $1k$. If there was no such $k$, then the resulting graph $H'$ satisfies $L(H') > L(H)$. If there was such a $k$, then $L(H') = L(H)$ and $R(H') < R(H)$. Either way, $H'$ is easily seen to lie in $\mathcal{H}_d'$ and thus contradicts our choice of $H$. Now that $H$ is explicitly given, it follows that $$ L_H(x) = \sum_{j = 2}^n d x_j . $$ This finishes our proof. The original statement now follows by taking as $x$ the degree-sequence of a planar graph and noting that $G \in \mathcal{H}_d$ for $d = 3$.<|endoftext|> TITLE: Can a Laplace eigenfunction have a level set with a cusp like singularity? QUESTION [10 upvotes]: Let $\Omega$ be a precompact open subset of ${\mathbb R}^2$ with piecewise smooth boundary. Let $u$ be an eigenfunction of the Laplacian on $\Omega$ with either Dirichlet or Neumann conditions on $\partial \Omega$. Is it possible that some level set of $u$ has a `cusp'? By cusp, I mean a critical point $z$ of $u$ such that the level sets of $u$ topologically foliate a neighborhood of $z$. The prototypical example of a cusp is the critical point $(0,0)$ of the function $f(x,y)=x^2+y^3$. A Mathematica generated sketch of the topological foliation is provided here: S. Y. Cheng observed many years ago that if a critical point lies on the zero level set of a planar eigenfunction, then near the critical point, the zero level set is akin to that of a harmonic polynomial. It follows that the answer to the question is `no' if restricted to critical points that lie in the zero level set. REPLY [3 votes]: The question has been answered affirmatively by Olti Myrtaj and Kelly Chen as part of a Research Experiences for Undergraduates project. See here<|endoftext|> TITLE: Distribution of distances of successive zeros of $f(x)={\rm cos}(x)+{\rm cos}(\alpha x)+{\rm cos}(\beta x)$ QUESTION [6 upvotes]: Let $\alpha$ and $\beta$ be incommensurate real numbers. Consider the function $f(x)={\rm cos}(x)+{\rm cos}(\alpha x)+{\rm cos}(\beta x)$ and its positive zeros $x_k(\alpha,\beta)$. Fix $\alpha$ and $\beta$, and consider the distances between successive zeros $z_k=x_{k+1}-x_k$. The concrete distribution of the zeros depends on $\alpha$ and $\beta$, but the distribution always shows singularities. Here are some examples for the first 100k zeros for various $\alpha$ and $\beta$: The position of the singularities does obviously depend on the continued fraction representation of $\alpha$ and $\beta$, as higher rational approximations to $\alpha$ and $\beta$ will display similar distributions. My question is: Can one write down explicit approximations for the positions of these singularities in these distributions of the distances between the zeros in terms of the continued fraction expansions of $\alpha$ and $\beta$? (The distances $z_k$ and $z_{k+1}$ are strongly correlated.) The function $f(x)$ is almost periodic. In the literature about almost periodic functions I could not locate information about the distribution of the distances between the zeros. Mathematica code to generate the images: https://drive.google.com/open?id=0B649LNvIOdYnaDctRElFYjQ2RlU REPLY [3 votes]: Here is how I think of this problem from a ergodic theory point of view. Because the system is ergodic for long time the point $(t [2\pi] , \alpha t [2\pi], \beta t [2\pi] )$ should span $[2\pi]^3 $ with equidistribution . Therefore the distribution of the distance $t$ between two consecutive zeros should be given by the measure of the set defined by $$\begin{cases} \cos(x)+\cos(y)+\cos(z)=0 \\ \cos(x+t)+\cos(y+\alpha t)+\cos(z+\beta t)=0\end{cases} $$ Let $(u,v,w)$ an orthogonal basis with $u=(1,\alpha,\beta)$ and let $(a,b,c)$ the coordinate of $(x,y,z)$ in this basis. The previous equations give us $$\begin{cases} a_1=g_1(b,c) \\ a_2=g_2(b,c) \\ t=a_2-a_1\end{cases} $$We can think of $a_1$ and $a_2$ as the fist and second time we cross the set $\cos(x) +\cos(y)+\cos(z) =0 $ starting from the point $(0,b,c)$. On the $\nabla (g_2-g_1)\neq 0$, the solution is a smooth curve,therefore the singalarities can only appear when $\nabla (g_2-g_1) = 0$. As a conclusion, the singularities should only exists for $t$ such that exists $(x,y,z)$ solution of $$\begin{cases} \cos(x)+\cos(y)+\cos(z)=0 \\ \cos(x+t)+\cos(y+\alpha t)+\cos(z+\beta t)=0 \\ \nabla(g_2-g_1)=0 \end{cases}$$ This explain the singularity at the boundaries of the set on your first diagram (extremum $\Rightarrow \nabla (g_2 -g_1)=0$ ). I try to solve the system numerically for $\alpha=\sqrt{2},\beta=\sqrt{3}$ and it gives me $t \approx 2.275$. Does it make sense with your last diagram?<|endoftext|> TITLE: A naive diophantine approximation question QUESTION [6 upvotes]: Let $\alpha$ be a positive real number (bigger than one, and irrational if it matters) (the one I am secretly thinking of is $\varphi,$ the golden ratio). I want to know, given an $\epsilon>0,$ whether there always exist $n, m \in \mathbb{N}$ with $n > 0$, such that $|\alpha^n - m|< \epsilon.$ If yes, is there any way to estimate/find them? REPLY [11 votes]: The answer is no. Start with $\alpha$ between 3.25 and 3.75; take squares of those 2 to see that one can restrict $\alpha^2$ to being between 11.25 and 11.75; take 3/2 powers of those to see that one can restrict $\alpha^3$ to being between 38.25 and 38.75; take 4/3 powers of those etc. One gets the idea: since $\alpha>3$ a 0.5 wide interval is magnified to an interval of width >1.5 by moving to the next power. Such interval can always be narrowed to $[n+0.25,n+0.75]$ for some integer $n$.<|endoftext|> TITLE: Strong duality for a particular moment problem QUESTION [5 upvotes]: Reading the paper in this Link (see pag 13) with the objective of understanding a topic related to stochastic optimization I came across a problem in demonstrating one of the theorems. The situation is the following: The context: Let $(\Xi,\mathcal{E})$ be a measurable space where $\Xi\subseteq \mathbb{R}^{m}$, $\widehat{\xi}_{i}\in \Xi$ for $i=1,\ldots,N$ and $\ell(\xi):=\max_{k\leq K}\ell_{k}(\xi)$ where $\ell_{k}:\Xi\rightarrow \mathbb{R}$ are functions. We denote by $\mathcal{M}(\Xi)$ the set of probability distributions supported on $\Xi$. We consider the infinite optimizatión program \begin{align} & \left\{\begin{array}{cl} {\displaystyle\sup_{\mathbb{Q}_{i}\in\mathcal{M}(\Xi)} }&{\displaystyle \frac{1}{N}\sum_{i=1}^{N}\int_{\Xi}\ell(\xi)\mathbb{Q}_{i}(d\xi)} \\ \mbox{s.t.} & {\displaystyle \frac{1}{N}\sum_{i=1}^{N}\int_{\Xi}\left\|\xi-\widehat{\xi}_{i}\right\|\mathbb{Q}_{i}(d\xi)\leq \varepsilon.} \end{array} \right. \tag{P} \end{align} We know that the dual problem is \begin{align} & \left\{\begin{array}{cl} {\displaystyle\inf_{\lambda} }&{\displaystyle \sup_{\mathbb{Q}_{i}\in\mathcal{M}(\Xi)}\left(\frac{1}{N}\sum_{i=1}^{N}\int_{\Xi}\ell(\xi)\mathbb{Q}_{i}(d\xi) +\lambda \varepsilon -\lambda \frac{1}{N}\sum_{i=1}^{N}\int_{\Xi}\left\|\xi-\widehat{\xi}_{i}\right\|\mathbb{Q}_{i}(d\xi) \right)} \\ \mbox{s.t.} & {\displaystyle \lambda \geq 0.} \end{array} \right. \tag{D} \end{align} We denote $\mathrm{Val(P)}$ and $\mathrm{Val(D)}$ the optimals values of $\mathrm{(P)}$ and $\mathrm{(D)}$ respectively. Assumption: The paper I am reading establishes the following assumption: Assumption 1. [Convexity] The uncertainty set $\Xi\subseteq \mathbb{R}^{m}$ is convex and closed, and the negative constituent functions $-\ell_{k}$ are proper, convex and lower semicontinuous for all $k\leq K$. Moreover, se assume that $\ell_{k}$ is not identically $-\infty$ on $\Xi$ for all $k\leq K$. The problem: If the convexity Assumption 1 holds, I need to show that we have strong duality, that is, I need to show that $\mathrm{Val(P)}=\mathrm{Val(D)}.$ Remark: According to the paper I am reading this is a consequence of a extended version of well-known strong duality result for moment problems, in this sense, the paper cite the following work: A. Shapiro, On duality theory of conic linear problems, in Semi-Infinite Programming, M. A. Goberna and M. A. L´opez, eds., Kluwer Academic Publishers, 2001, pp. 135–165 In this work, in Proposition 3.4, they show strong duality for the optimization problem $$ \left\{\begin{array}{cl} \max_{\mu\in\mathcal{C}} & \left\langle \varphi, \mu \right\rangle \\ \mbox{s.t.} & \mathcal{A}\mu-b \in K \end{array} \right. \tag{1} $$ where the context of $\mathrm{(1)}$ is the following: Is (P) a particular case of (1)? If the answer is yes, then how can I make this evident? REPLY [2 votes]: Yes. $\mathcal{M}(\Xi)$ is the space of all probability measures defined on the measure space $\Xi$ with Levy distance and superscript means usual direct product of spaces. ${\displaystyle \frac{1}{N}\sum_{i=1}^{N}\int_{\Xi}\left\|\xi-\widehat{\xi}_{i}\right\|\mathbb{Q}_{i}(d\xi)\leq \varepsilon}$ can be rewritten as $\mathcal{A}(\mathbb Q)=\mathcal{A}(\mathbb{Q}_i)_{i=1,\cdots,N}=\frac{1}{N}\sum_{i=1}^N\mathcal{A}_{\hat\xi_i}\mathbb{Q}_{i}$ You can verify that $$\mathcal{A}_{\hat\xi_i}:\mathcal{M}(\Xi)^N\rightarrow \mathbb R, \mathbb{Q}\mapsto \int_{\Xi}\left\|\xi-\widehat{\xi}_{i}\right\|\mathbb{Q}_i(d\xi)$$ is linear in $\mathbb{Q}$ and the sum of linear operators is still linear. So $\mathcal{A}:\mathcal{M}(\Xi)^N\rightarrow\mathbb R$ is a linear operator and the constraint is $\mathcal{A}(\mathbb{Q}_i)_{i=1,\cdots,N}-0\in \{r\in\mathbb{R} : 0\le r\le K\}$ Now the inner product is just the dual product $\langle\varphi,(\mathbb{Q})_{i=1,\cdots N}\rangle=\int_{\Xi}\varphi (d\mathbb{Q_1}+\cdots+d\mathbb{Q_N})$. Here $\varphi=(\varphi_1(\omega)\cdots\varphi_N(\omega))\in L(\Xi)$ the space of all integrable real functions defined on $\Xi$, but with the setting of (P), you are just choosing a special case that $\varphi(\omega)=(\varphi_1(\omega)\cdots\varphi_N(\omega))=(\frac{1}{N}l(\omega),\cdots,\frac{1}{N}l(\omega))$. In short the inner product is defined on $L(\Xi)\times \mathcal{M}(\Xi)^N=L(\Xi)\times \oplus_{i=1}^N \mathcal{M}(\Xi)$<|endoftext|> TITLE: Is this BBP-type formula for $\ln 257$ and $\ln 65537$ true? QUESTION [28 upvotes]: We have the known BBP(Bailey–Borwein–Plouffe)-type formulas, $$\ln3 = \sum_{n=0}^\infty\frac{1}{2^{2n}}\left(\frac{1}{2n+1}\right)$$ $$\ln5 = \frac{1}{2^2}\sum_{n=0}^\infty\frac{1}{2^{4n}}\left(\frac{2^2}{4n+1}+\frac{2^2}{4n+2}+\frac{1}{4n+3}\right)$$ However, I noticed that if we define the function, $$R\big(a,b\big) = \sum_{n=0}^\infty\frac{1}{(2^a)^n}\left(\sum_{j=1}^{a-1}\frac{2^{a-1-j}}{an+j}+\sum_{k=1}^{a/b-1}(-1)^{k+1}\frac{2^{a-1-bk}}{an+bk}\right)\tag1$$ then it seems BBP-type formulas for Fermat numbers $2^{2^m}+1$ with $m>0$ have a common form, $$\ln 5 = \frac1{2^{2}}R\big(2^2,2^1\big)$$ $$\ln 17 = \frac1{2^{13}}R\big(2^4,2^2\big)$$ $$\ln 257 = \frac1{2^{252}}R\big(2^8,2^3\big)$$ $$\color{brown}{\ln 65537 \overset{?}= \frac1{2^{65531}}R\big(2^{16},2^4\big)}$$ Q: Is the formula for $p=65537$ true? I've used Mathematica to verify the $p=5,\,17,\,257$ to hundreds of decimal digits, and also $p=65537$ using its initial terms, but how do we rigorously prove that $(1)$ is true for Fermat numbers $>3$? P.S. In this paper, the authors were not(?) able to find $p=65537$ and is missing in the list. REPLY [35 votes]: Yes, this actually holds for all Fermat numbers! Let's start with the identity $$\log 2=\sum_{k=1}^{\infty}\frac{1}{k2^k}$$ and try to work out an expression for $$\log (2^{2^s}+1)=2^{s}\log 2+\log\left(1-\frac{1}{2^{2^{s+1}}}\right)-\log\left(1-\frac{1}{2^{2^s}}\right)$$ $$=2^s\left(\sum_{k=1}^{\infty}\frac{1}{k2^k}\right)-\left(\sum_{k=1}^{\infty}\frac{1}{k2^{k2^{s+1}}}\right)+\left(\sum_{k=1}^{\infty}\frac{1}{k2^{k2^{s}}}\right)$$ $$=\sum_{n=0}^{\infty}\frac{1}{2^{n2^{2^s}}}\left(\sum_{j=1}^{2^{2^s}}\frac{2^{s-j}}{n2^{2^s}+j}-\sum_{h=1}^{2^{2^s-s-1}}\frac{2^{-h2^{s+1}}}{n2^{2^{s}-s-1}+h}+\sum_{l=1}^{2^{2^s-s}}\frac{2^{-l2^s}}{n2^{2^s-s}+l}\right)$$ $$=\sum_{n=0}^{\infty}\frac{1}{2^{n2^{2^s}}}\left(\sum_{j=1}^{2^{2^s}}\frac{2^{s-j}}{n2^{2^s}+j}-\sum_{h=1}^{2^{2^s-s-1}}\frac{2^{-(2h)2^{s+1}}}{n2^{2^{s}-s}+2h}+\sum_{l=1}^{2^{2^s-s}}\frac{2^{-l2^s}}{n2^{2^s-s}+l}\right)$$ $$=\sum_{n=0}^{\infty}\frac{1}{2^{n2^{2^s}}}\left(\sum_{j=1}^{2^{2^s}}\frac{2^{s-j}}{n2^{2^s}+j}+\sum_{l=1}^{2^{2^s-s}}\frac{(-1)^{l+1}2^{-l2^s}}{n2^{2^s-s}+l}\right)$$ here we see that the terms at $j=2^{2^s}$ and $l=2^{2^s-s}$ cancel out, so after factoring out $\frac{1}{2^{2^{2^s}-s-1}}$ we are left with the defining expression of $R$ with $a=2^{2^s}, b=2^s$: $$\log(2^{2^s}+1)=\frac{1}{2^{2^{2^s}-s-1}}R(2^{2^s},2^s).$$<|endoftext|> TITLE: A reference to a theorem on the equivalence of ideals of measure zero in the Cantor cube QUESTION [12 upvotes]: I am looking for a reference of the following (true) fact: Theorem. For any two continuous strictly positive Borel probability measures $\mu,\lambda$ on the Cantor cube $2^\omega$ there exists a homeomorphism $h:2^\omega\to 2^\omega$ such that $h(\mathcal N_\mu)=\mathcal N_\lambda$. Here by $\mathcal N_\mu=\{B\subset 2^\omega:\mu(B)=0\}$ we denote the $\sigma$-ideal of sets of $\mu$-measure zero in the Cantor cube; and $h(\mathcal N_\mu)=\{h(B):B\in\mathcal N_\mu\}$. A measure $\mu$ on a topological space $X$ is called continuous if $\mu(\{x\})=0$ for all $x\in X$; and strictly positive if $\mu(U)>0$ for any non-empty open set $U\subset X$. I believe that such a useful (and not very difficult) fact should be already known and published somewhere. Right? Remark. Two measures on the Cantor cube need not be homeomorphic. For example, the Haar measures on the zero-dimensional compact topological groups $(\mathbb Z/2\mathbb Z)^\omega$ and $(\mathbb Z/3\mathbb Z)^\omega$ are not homeomorphic as they have different sets of values on closed-and-open subsets (and this set of values is a topological invariant of a measure). REPLY [4 votes]: Finally I have found a good reference to this theorem, which can be easily derived from the following Theorem 2.12 in this paper of Akin: Theorem (Akin, 1999). For any strictly positive continuous measures $\mu,\nu$ on the Cantor cube $2^\omega$ and any $\varepsilon>0$ there exists a homeomorphism $h$ of $2^\omega$ such that $$(1-\varepsilon)\mu(B)\le \nu(h(B))\le(1+\varepsilon)\mu(B)$$for any Borel subset $B$ of $2^\omega$.<|endoftext|> TITLE: Are there any known approaches of generalizing functions that do not have a limit at infinity to values at infinity? QUESTION [5 upvotes]: Let's consider the affinely extended real line. The functions that have a limit on positive or negative infinity $\lim_{x\to+\infty} f(x)$ or $\lim_{x\to-\infty} f(x)$ can be generalized to the values at infinity $f(+\infty)$, $f(-\infty)$ based on these limits. But what about functions that do not have such limits? For instance, if we take these integrals following Borel (or Abel) generalization, we get: $\int_0^\infty \cos x=0$ and $\int_0^\infty \sin x=1$ Since integral of $\cos x$ is $\sin x$ and integral of $\sin x$ is $-\cos x$, we can probably generalize these functions to infinity (assuming the fundamental theorem of calculus should still hold): $\sin (\infty)=\cos(\infty)=0$ (notice, this would apparently break the main trigonometric identity $\sin^2 x+\cos^2 x =1$ at this point but one of the books linked in the comments resolves the paradox by claiming that $(\sin \infty)^2\ne \sin^2 \infty$ and the same for cosine). Still, I wonder whether there were other attempts at such generalization? Possibly, something along Cesaro lines (limit of mean value of the function) or sometyhing else? REPLY [2 votes]: Many people developed a theory of limits of distributions at finite or infinite values in the 50's and 60's and for these all of your formulae hold. A particularly elementary version was developed by J. Sebastião e Silva and a complete description can be found online by googling has name and "limits of distributions at infinity". There is also a text book version by J. Campos Ferreira et al.---"Introduction to the theory of distributions".<|endoftext|> TITLE: Ehresmann's theorem for singular varieties QUESTION [7 upvotes]: Let $X$ and $Y$ be two algebraic varieties over $\mathbb{C}$ and $f\colon X\rightarrow Y$ be a proper map. Assume that $Y$ is smooth. I am interested in sufficient and necessary conditions for $f$ to be topologically locally trivial on the target, i.e., whether for each $y\in Y$ there exists an open neighborhood $V$ such that $f$ factors through a homeomorphism $f^{-1}(V){\cong} V\times f^{-1}(y)\rightarrow V$. If $X$ is a smooth variety, the answer comes from Ehresmann's theorem: $f$ is topologically locally trivial on $Y$ if and only if it is a proper submersion. This also implies that $f$ is a smooth proper algebraic map. But what if $X$ is singular? I could not find a reference. Surely, $f$ must be topologically locally trivial on the $source$. Then I would like to know which tools may be used to patch together these local trivializations along a fiber, and to have a reasonable understanding of when this is not possible. REPLY [4 votes]: It seems that you are looking for a good notion of "equisingularity." About this I know quite little and I hope someone else will give a nice answer, but my guess is that a sufficient and necessary condition would be the existence of a horizontal Whitney stratification, at least locally on the source. Perhaps the book "Stratified Morse Theory" by Goresky and MacPherson can be of some help. Another (necessary but perhaps insufficient) condition (at least if $Y$ is the germ of a curve) is the vanishing of vanishing cycles. Regarding the second question (local to global topological triviality), see Corollary 6.14 in L. C. Siebenmann "Deformations of Homeomorphisms on Stratified Sets" Commentarii Mathematici Helvetici 47 (1972): a proper separated topological submersion whose fiber is stratifiable is in fact a fiber bundle.<|endoftext|> TITLE: Number of points of elliptic curve over $\mathbb{F}_p$ with CM by $\sqrt{-2}$ when $p\equiv 1\bmod 8$ QUESTION [8 upvotes]: I was trying to calculate the number of points of the curve $E:y^2 = x^3 + 4x^2 + 2x$ over $\mathbb{F}_p$ for $p\equiv 1\bmod 8$ (In order to have $\sqrt{-2}\in\mathbb{F}_p$) but I did not succeed. This curve is mentioned in Silverman's Advanced Topics in the Arithmetic of Elliptic curves (Proposition 2.3.1) to have multiplication by $\sqrt{-2}$. Over these primes $p\equiv 1\bmod 8$ the curve $E$ has full $2$-torsion so $E(\mathbb{F}_p)\cong \mathbb{Z}/(2)\times \mathbb{Z}/(k)$. In this case my conjecture is that the size will be related to the factorization of $p=(a+b\sqrt{-2})(a-b\sqrt{-2})$ over $\mathbb{Z}[i\sqrt{2}]$, that is $p=a^2 + 2b^2$. Hence, $\#E(\mathbb{F}_p)=p+1\pm 2a$ where $a$ is odd (and the sign I do not know how to choose it yet). Calculating this reminds me to the proof of the Last Entry of Gauss Tagebuch. I would like to have an elliptic curve with CM by $\sqrt{-2}$ such that I can know the number of points in terms of $p$. Does anybody has a suggestion? or maybe another curve? Thanks REPLY [7 votes]: Sorry for self-advertisement: the question of the "sign" that one must choose (equivalently which $\pi$ or $\overline{\pi}$) is entirely answered in my book GTM239, Section 8.5.2, some of the results being due to Mark Watkins (unpublished I believe). The main idea is to show that any CM curve is equivalent in an evident sense to a "basic" CM elliptic curve with a specific equation, giving a relation between the number of points of the initial curve and the "basic" curve, and then Theorem 8.5.8 determines explicitly the number of points of that basic curve. I did not do the exercise for your specific example, but it is completely algorithmic.<|endoftext|> TITLE: The Picard group of a semisimple algebraic group in positive characteristic QUESTION [13 upvotes]: Let $k$ be a field of positive characteristic and let $G$ be a connected semisimple algebraic group over $k$ with fundamental group $\mu$. Note that $\mu$ can be non-smooth. It is stated in Sansuc's 1981 Crelle paper Groupe de Brauer et arithmétique des groupes algébriques..., Lemma 6.9(iii), p.41, that ${\rm Pic}\, G=\mu^{*}(k)$ is the group of $k$-characters of $\mu$. Sansuc refers the reader to the Fossum-Iversen paper On Picard groups of algebraic fiber spaces, Proposition 4.6, for this. Now, it is not clear to me that the latter paper really works at the required level of generality. To confuse matters even more (for me), Raynaud, in his thesis (Lecture Notes in Math. 119), discusses in particular the canonical map $\mu^{*}(k)\to {\rm Pic}\, G$ (Proposition VII.1.5, p.106). By examining Raynaud's proof, all I can see is that the latter map is an injection whose cokernel is a torsion group annihilated by some power of the characteristic of $k$. So, my question is: has it really been proved in the literature that Sansuc's statement ${\rm Pic}\, G=\mu^{*}(k)$ is true in positive characteristic even when $\mu$ is not smooth? REPLY [10 votes]: I'm not sure if there is really a literature gap, or maybe it is addressed in some reference that just hasn't come to mind. Anyway, let $G$ be a perfect smooth connected affine group over a field $k$, so $G_K$ has no nontrivial $K$-homomorphism to ${\rm{GL}}_1$ for any extension field $K/k$, and hence $K[G]^{\times} = K^{\times}$ by Rosenlicht's Unit Theorem. If $L'$ is a line bundle on ${\rm{Pic}}(G_{k'})$ for a finite Galois extension $k'/k$ such that the isomorphism class of $L'$ is ${\rm{Gal}}(k'/k)$-invariant, it therefore follows via Hilbert 90 that the isomorphisms to the Galois-twists can be arranged to satisfy the cocycle condition, so the natural map $${\rm{Pic}}(G) \rightarrow {\rm{Pic}}(G_{k'})^{{\rm{Gal}}(k'/k)}$$ is surjective, and hence by passing to the direct limit on such $k'/k$ we see that the natural map $${\rm{Pic}}(G) \rightarrow {\rm{Pic}}(G_{k_s})^{{\rm{Gal}}(k_s/k)}$$ is surjective. But this latter map is also injective. Indeed, again by limit considerations it suffices to show if $L$ is a line bundle on $G$ such that $L_{k'}$ admits a nowhere-vanishing global section $s'$ over $G_{k'}$ for a finite Galois extension $k'/k$ then $L$ admits a nowhere-vanishing global section. For each $\gamma \in {\rm{Gal}}(k'/k)$, the canonical $\gamma$-equivariant action on $L_{k'}(G_{k'}) = k' \otimes_k L(G)$ satisfies $\gamma(s') = c_{\gamma} s'$ for a unique $c_{\gamma} \in k'[G]^{\times} = {k'}^{\times}$, and $\gamma \mapsto c_{\gamma}$ is a 1-cocycle. Thus, by Hilbert 90 there exists $u' \in {k'}^{\times}$ such that $c_{\gamma} = \gamma(u')/u'$ for all $\gamma$, so $s'/u'$ is a nowhere-vanishing global section that is Galois-invariant and hence corresponds to a trivialization of $L$ over $G$. Now assume $G$ is a connected semisimple $k$-group. Letting $\widetilde{G} \rightarrow G$ be the simply connected central cover over $k$, with central kernel denoted $\mu$, we have ${\rm{Pic}}(\widetilde{G}) = 1$ by the above injectivity applied to $\widetilde{G}$ because ${\rm{Pic}}(\widetilde{G}_{k_s})=1$. To see this latter triviality, note that for any split connected reductive group $H$ over a field $F$, the "open cell" relative to a choice of split maximal $F$-torus and Borel $F$-subgroup containing it provides a dense open subscheme that is $F$-isomorphic to a direct product of copies of $\mathbf{A}^1_F$ and $\mathbf{A}^1_F - \{0\}$. Using such an open subscheme enables one to show that the group ${\rm{Pic}}(H)$ is identified with the group of central extensions $E$ of $H$ by ${\rm{GL}}_1$ as $F$-groups by adapting the proof of the Weil-Barsotti formula describing the dual abelian variety in terms of extensions by ${\rm{GL}}_1$ -- see section 4 in the paper "Local Properties of Algebraic Group Actions" by Knop, Kraft, Luna, and Vust for that argument (presented in the setting over algebraically closed fields, but via completely general methods that work for a split connected reductive group over any field). If the $F$-split $H$ is connected semisimple and simply connected then there are no such nontrivial extensions $E$ since necessarily $E$ is connected reductive and hence $\mathscr{D}(E) \rightarrow H$ is a central isogeny (so an isomorphism when $H$ is simply connected). This shows ${\rm{Pic}}(H)=1$ in such cases, so ${\rm{Pic}}(\widetilde{G}_{k_s})=1$ as claimed (and from the consequence ${\rm{Pic}}(\widetilde{G})=1$ we obtain the non-obvious fact that $k[\widetilde{G}]$ is a UFD). The result VII 1.5 that you refer to in Raynaud's thesis provides an injective homomorphism $\mu^*(k_s) \rightarrow {\rm{Pic}}(G_{k_s})$ (this map assigns to any $k_s$-homomorphism $\chi:\mu_{k_s} \rightarrow {\rm{GL}}_1$ the $\chi$-pushout of the central extension $1 \rightarrow \mu_{k_s} \rightarrow \widetilde{G}_{k_s} \rightarrow G_{k_s} \rightarrow 1$), and it is Galois-equivariant by design. Hence, as long as this is an equality, which is purely a counting question (due to its injectivity), it follows by the preceding arguments that passage to ${\rm{Gal}}(k_s/k)$-invariants gives that the natural map $\mu^*(k) \rightarrow {\rm{Pic}}(G)$ is an equality. In this way, we see that the result asserted without proof in Remark VII 1.7(a) of Raynaud's thesis (basically the affirmative answer to the question posed in the split case) gives the desired result over $k_s$ and hence over $k$. This doesn't answer the question of providing a literature reference with a complete proof, but it "confirms" that a complete proof should have been known to experts in Paris since the 1960's.<|endoftext|> TITLE: McKay conjecture for finite groups in the simplest case G=GL(2,F_p) ( puzzle: Borel knows about cuspidals) QUESTION [21 upvotes]: The McKay conjecture and related (Alperin, Issacs-Navarro) are one of the "main problems in the representation theory of finite groups" (G.Navarro pdf). Statement of the McKay conjecture is quite simple: McKay conjecture: for any finite group $G$ and prime $p$ the following holds: The number of irreducible representations of G of dimension not divisible by $p$ is equal to the number of irreducible representations of the normalizer of the Sylow $p$-subgroup $P$ of $G$ of dimension also not divisible by $p$. i.e. $$ |\mathrm{Irr}_{p'}(G) | = | \mathrm{Irr}_{p'}(N_G(P)) |. $$ (all representations here are over complex numbers). Notation $\mathrm{Irr}_{p'}(H)$ denotes set of irreducible repsentations of some group $H$ with dimensions not divisible by $p$. The conjecture states the equality of the two numbers, but, of course, one may hope that there should be bijection between underlying sets of irreducible representations. So: Question: What can be a natural bijection in the simplest case $G=\mathrm{GL}(2,\mathbf{F}_p)$ ? (For $p$ same as above). It is known that there is no choice-free bijection, but, nevertheless there should be some natural family of bijections in order to explain this somehow. What is puzzling (see more details below) - it is very very straightforward to get HALF of the representations of $G$, from representations of $N_G(P)$ (which is just the Borel subgroup ): it is just the induction from the Borel (=$N_G(P)$) to $G$. But how to get the other half? The other half are the so-called cuspidal representations, which by definition are those which ARE NOT INDUCED from the Borel (=$N_G(P)$) subgroup. Moreover when you do induction you get gluing of irreducible representation of the Borel subgroup by the Weyl group action, so one must make some choice of representatives in the Weyl group orbits to "unglue". More details on puzzle: Let me remind the classification of irreducible representation for $GL(2,F_p)$ (one may look at Garrett page 11, Etingof&K page 69 or refrences in MO271389). And of the Borel (=$N_G(P)$) subgroup. The puzzling outcome will be that: there should be bijection between (p-1)^2 characters of standard torus (diagonal matrices) and the main part (including cuspidals) of representaions of $GL(2,F_p)$. But that contradicts standard viewpoint - cause cuspidal irreps correspond to characters of the non-split torus, while McKay somehow predicts them from split torus. There seems to be no known way (to me) to get cuspidals from the standard-split torus (=diagonal matrices). Representations of $\mathrm{GL}(2,\mathbf{F}_p)$ are well-known to come in 4 series: 1) Series 1 - "det" - count = $(p-1)$, dimension = 1 1-dimensional representations factoring through the determinant, 2) Series 2 - "regular principal series" - count = $(p-1)(p-2)/2$, dimension = $p+1$ those induced from the Borel subgroup to $G$ from character of Borel which is "regular" meaning that character has different values on the two generators of the torus, 3) Series 3 - "cuspidal" - count = $(p)(p-1)/2$, dimension = $p-1$ those that are not induced from Borel and hard to get them 4) Series 4 - "special", count = $(p-1)$, dimension = $p$ , actually those are "irregular principal series". So the 4th case is not interesting to us in McKay conjecture since, its dimension is $p$ and divisible by $p$. From cases 2 and 3 we get $(p-1)^2$ irreducible representations and adding those from case 1, we get $(p-1)^2 + (p-1) = p(p-1)$. Representations of Borel (=$N_G(P)$ - normalizer of Sylow p-subgroup) Borel = semidirect product of 2-torus and abelian subgroup of unipotent matrices. Representations of semi-direct products easy to describe (see e.g. Etingof&K page 76). So for the particular case of Borel in GL(2): We have $(p-1)^2$ irreps factoring through the torus (they are one-dimensional), and we have $(p-1)$ non-trivial $(p-1)$-dimensional irreps that are induced from non-trivial characters of the abelian subgroup of unipotent matrices. So in total we have $(p-1)^2+(p-1) = p(p-1)$. So we get numerical coincidence $$ p(p-1) = |\mathrm{Irr}_{p'}(GL(2,\mathbf{F}_p) | = | \mathrm{Irr}_{p'}(Borel(2,\mathbf{F}_p)) | = |N_G(P)|. $$ - which is is predicted by the McKay conjecture. But I do not see how the bijection between the two sets can be made ! (Except it is easy to propose that the $(p-1)$ of type 1 for $\mathrm{GL}(2)$ should correspond to $(p-1)$ non-one-dimensional in irreps of Borel). The problem is that we somehow should get cuspidal from the charactetrs of Borel = characters of the standard torus (diagonal matrices), but there seems to be no known way to do it and moreover cuspidals always correspond to characters of NON-split torus. From high level that is Deligne-Lusztig theory, from down-to-earth considerations that is seen by looking on conjugacy classess as in references above. REPLY [6 votes]: Here is an answer of sorts. In this case, by Brauer's First Main Theorem, there is a bijection between $p$-blocks of $G = {\rm GL}(2,p)$ with defect group $P$ and $p$-blocks of $B = N_{G}(P)$ with defect group $P$. Choose a block $\beta$ of $G$ with defect group $P$ and let $\gamma$ be the unique Brauer correspondent of $\beta$ for $B$. The numerical invariants of $\beta$ and $\gamma$ are determined by the theory of blocks of defect one (already developed by Brauer, and later generalized to the cyclic defect group theory by Dade and Green). In this case, it is (almost) immediate that $B$ has $p-1$ $p$-blocks with defect group $P$ (one for each class of scalar matrices in $B$), and that in each case the all-important invariant, the inertial index $e$ , is $p-1$, being $[N_{G}(P):C_{G}(P)]$ in each case. In fact, one could say that the reason the McKay conjecture holds in this case is because this inertial index is independent of the particular block (of full defect)- possibly together with the fact (also due to Brauer) that in a group whose order is divisible by $p$ but not by $p^{2}$, every irreducible character of degree divisible by $p$ lies in a $p$-block of defect zero. The general theory of blocks of defect one now gives that the block $\beta$ contains $p$ irreducible characters, all necessarily of degree prime to $p$ as noted above, due to the fact that there are $e + \frac{p-1}{e}$ irreducible characters in a block of defect one, where $e$ is the inertial index of the block. The same holds within $B = N_{G}(P)$, so that $\gamma$ also contains $p$ irreducible characters, all of degree prime to $p$. Hence in this case, Brauer's First Main Theorem, and the theory of blocks of defect one "explains" why the McKay conjecture holds for $G$. In some sense it also explains why there is no "canonical" bijection. In a general $p$-block of defect one with inertial index $e$, there are $\frac{p-1}{e}$ "exceptional" irreducible characters, and $e$ "non-exceptional characters". When $e < p-1,$ there is a reasonably natural way to pair the exceptional irreducible characters for the whole group and the exceptional characters for the normalizer of the defect group because of irrationality of character values on non-identity $p$-elements. But when $e = p-1,$ the irrationalities disappear and the distinction between exceptional and non-exceptional characters appears somewhat artificial.<|endoftext|> TITLE: Maximum number of perfect matchings in a planar graph? QUESTION [8 upvotes]: What is the maximum number of perfect matchings a planar $k$-partite $|V|$ number of vertices simple graph can have where $k=2,3,4$ ($k>4$ is impossible for a planar graph)? Since number of matchings cannot exceed $2^{\alpha_k |V|}$ where $\alpha_k>0$ is an absolute constant that depends only on $k$ in essence 'what is the best estimate of $\alpha_k$?' is the query. REPLY [9 votes]: In order to get an upper bound on $\alpha_4$ (which is probably far from being a tight bound), you can use the Kahn-Lovász Theorem (a generalisation of Brègman's Theorem). The Kahn-Lovász Theorem says that the number of perfect matchings in any graph $G$ is at most $$\prod_{v\in V(G)}(d(v)!)^{1/2d(v)}.$$ Its not immediately clear to me how to obtain an upper bound of the form $2^{c|V(G)|}$ from this, but it looks like it might be possible since planar graphs have at most $3|V(G)|-6$ edges (and therefore the sum of the vertex degrees is at most linear in $|V(G)|$). See the edit below. Like I said, this is unlikely to be tight, but at least it might give you a bound. The Kahn-Lovász Theorem is tight for general graphs, but I think that the unique tight example is a disjoint union of balanced complete bipartite graphs, which is not planar unless every component is isomorphic to $K_2$ or $K_{2,2}$. Edit: If the sum of the degrees is fixed, then the product in the upper bound here is maximized when all of the degrees are as similar as possible. This can be derived, e.g., from Karamata's Inequality. In a planar graph, the average degree is at most $6$. So, the number of perfect matchings is bounded above by $$(6!)^{n/12} = (1.73026\dots)^n.$$<|endoftext|> TITLE: When is every orbit closure uniquely ergodic? QUESTION [10 upvotes]: Given a topological dynamical system $(X,T)$ (so that $T$ is a homeomorphism of the compact metric space $X$) and a point $x\in X$ we call the set ${\mathcal O}(x):=\overline{\{T^nx:n\in\mathbb Z\}}$ the orbit closure of $x$. Question 0: Is there a name for systems with the property that the orbit closure of every point is uniquely ergodic (i.e., supports a unique invariant measure)? It is well known that nilsystems have this property but not all distal systems do. Question 1: Is it true that if $(X,T)$ and $(Y,S)$ are uniquely ergodic, then $(X\times Y,T\times S)$ has the property that every orbit closure is uniquely ergodic? Question 2: What if in addition $(X,T)$ and $(Y,S)$ are distal? REPLY [5 votes]: I think the two-sided Morse system is a counterexample to both questions the first question. Consider the two points $x^{(1)}=\ldots 0110100110010110\cdot0110100110010110\ldots$ and $x^{(2)}=\ldots 1001011001101001\cdot0110100110010110$. The orbit closure of $(x^{(1)},x^{(2)})$ supports exactly two ergodic invariant measures, the diagonal measure on the Morse system and the anti-diagonal measure. By the way, I have a paper with Emmanuel Lesigne and Máté Wierdl establishing lots of strange properties of the Morse system (such as for any point, $x=(x^{(1)},x^{(2)},x^{(3)})$ of $M^3$, $x$ is generic for some ergodic invariant measure, but there exist points of $M^4$ that are not generic for any invariant measure).<|endoftext|> TITLE: Motivations to study the cohomology of the moduli space of curves QUESTION [10 upvotes]: Could anyone give some interesting motivations to understand the cohomology of $\mathcal{M}_g$? What I know: I have read the various approaches to construct $\mathcal{M}_g$ via orbit spaces for group actions, period maps, and Teichmüller theory. I learned from literature that the cohomology of $\mathcal{M}_g$ is an important object to study, though I have not started reading some classical papers on the topic, such as Mumford's Towards an enumerative geometry of the moduli space of curves, Miller's The homology of the mapping class group, or Morita's Characteristic classes of surface bundles. What aspects am I interested in: Some notes mentioned that the cohomology of moduli space is an important source of motives. Although I do not really understand what is a motive, any elaboration in this direction would be appreciated. Besides that, I am more interested in any concrete applications in algebraic geometry, differential geometry, or topology to motivate the study of cohomology of $\mathcal{M}_g$. REPLY [16 votes]: As the title to Mumford's famous paper "Toward an enumerative geometry..." suggests, knowing the cohomology / cycle theory of the moduli space of curves allows one to answer enumerative geometry questions for curves. Here is an (very concrete) example that came up in real life for a student of mine. He had a family of genus 2 curves over $\mathbb{P}^2$ constructed by taking the family of lines in $\mathbb{P}^2$ and for each line taking the double cover of the line branched at the six points given by the intersection of the line with a fixed (generic) sextic curve (equivalently, this is the defining linear system on a generic genus 2 $K3$ surface). He needed to know how many of those genus two curves admit a degree $d$ map to a fixed elliptic curve $E$. This is a calculation in the cohomology ring of $\overline{M}_2$: we are looking for the number of points in the intersection of the two cycles in $\overline{M}_2$ given by the two families of curves: the 2-cycle given by the family over $\mathbb{P}^2$ and the 1-cycle given by the locus of genus 2 curves admitting a degree $d$ map to $E$ (since the dimension of $\overline{M}_2$ is 3, we expect these cycles to intersect in points). This intersection problem is dual to a cup product computation in the cohomology ring of $\overline{M}_2$. Using Mumford's complete description of this cohomology ring, one can determine how to express each of our cycles in terms of Mumford's generators and then compute the cup product. There are some minor issues to fuss with involving the orbifold structure and some general position issues, but the problem is essentially solved with the cohomology of the moduli space. This was a rather specific example, but maybe you get the general idea: if we want to count the "number of curves satisfying this that and the other", we try to formulate the question as a problem in the cohomology ring of the moduli space.<|endoftext|> TITLE: Heat Equation with an integral boundary condition QUESTION [6 upvotes]: I have been struggling with following Heat equation IBVP, \begin{equation} \frac{\partial v\left(x, t\right)}{\partial t} = \alpha \frac{\partial^2 v\left(x, t\right)}{\partial x^2}, \quad t \in \left(0, T\right], x \in \left(- \infty, 1\right]. \label{heatEqn} \end{equation} with the following initial and boundary conditions \begin{equation} v\left(x, 0\right) = \delta\left(x\right), \quad v\left(-\infty, t\right) = 0, \quad \int_{-\infty}^{1}v\left(x, T\right) dx = \beta T^{\alpha-1} \end{equation} where $\alpha \in \left(0, 1\right)$ and $\beta$ is some positive constant. Some interesting observation about this problem, if we consider the semi-infinite plane $x \in \left(- \infty, 1\right]$ $\times$ $t \in \left[0, T\right]$, we know line integrals along three of the boundaries of this rectangle, as in $\int_{-\infty}^{1}v\left(x, 0\right) dx = 1$, $\int_{0}^{T}v\left(-\infty, t\right) dt = 0$ and the boundary condition with the integral equation above. Also if we can deduce $v\left(1, t\right)$ and $\frac{\partial v\left(x, t\right)}{\partial x}|_{x = 1}$ then we can attack the problem using laplace transform. I am wondering if there is version of Greens or Divergence Theorem which can be applied to tackle this problem. Any help or guidance will be greatly appreciated. Thank you. REPLY [2 votes]: Here is the answer to your modified question via potential approach but simpler than proposed in the comments. Denote $Z(x,t)=(4\alpha\pi t)^{-1/2} e^{-x^2/(4\alpha t)}$ the fundamental solution for the heat equation. I'll write $t$ instead of $T$ in the question. Subtracting $Z(x,t)$ from $v$ reduces the initial condition to zero and the integral to $$ \beta t^{\alpha-1}-\int_{-\infty}^1Z(x,t)\,dx= \beta t^{\alpha-1}-\frac{1}{2} \left(\text{erf}\left(\frac{1}{2 \sqrt{\alpha t}}\right)+1\right):=g(t). $$ Shifting the side boundary to $x=0$ (for convenience) the problem becomes $$ \left\{ \begin{array}{rccl} u_t&=&\alpha u_{xx},&t>0,\ x<0,\\ u|_{t=0}&=&0,\\ \int_{-\infty}^0u(x,t)\,dx&=&g(t),&t>0. \end{array} \right. $$ Its solution can be written down explicitly as a potential: $$ u(x,t)=2\alpha\int_{-\infty}^0Z_{xx}(x,t-\tau)g(\tau)\,d\tau. $$ To see that consider the double layer potential $$ Wg(x,t)=\int_0^tZ_x(x,t-\tau)g(\tau)\,d\tau. $$ It satisfies the heat equation for $x<0$ and for continuous densities $g$ has a property $$ \lim_{x\to0-}Wg(x,t)=\frac{g(t)}{2\alpha}. $$ Actually it's the jump relation for the double layer potential in this particular case. For $\varepsilon>0$ we have $$ \int_{-\infty}^{-\varepsilon}u(x,t)\,dx= 2\alpha\int_0^t\int_{-\infty}^{-\varepsilon}Z_{xx}(x,t-\tau)g(\tau)\,dx d\tau= $$ $$ 2\alpha\int_0^tZ_{x}(-\varepsilon,t-\tau)g(\tau)\,d\tau= 2\alpha Wg(-\varepsilon,t). $$ Taking $\varepsilon\to0+$ gives the result. The solution of the initial problem is $v(x,t)=u(x-1,t)+Z(x,t)$.<|endoftext|> TITLE: Uniformisation for non simple closed curves QUESTION [7 upvotes]: Given a simple closed curve in the plane, there is a homeomorphism from the unit open disk to the interior of the curve. The homeomorphism can be taken conformal, this is the uniformisation theorem. Is there a version of this result for closed curves that are not simple? For example, given a $C^1$ closed curve $\gamma: S^1 \rightarrow {\bf R}^2$ with finitely many self-intersections, all of them transverse, is there a continuous map (maybe even conformal) from the unit open disk to the plane, such that the number of preimages of any point in ${\bf R}^2\backslash \gamma(S^1)$ is equal to the absolute value of the number of turns the curve makes around the point? $\hskip 1in$ REPLY [3 votes]: It sounds like, to some extent, what you are asking is whether an immersed closed curve in the plane can be extended to an immersion of a disk. This would yield the multiplicities that you seek; however, such an extension is not always possible. Necessary and sufficient conditions were first given in the thesis of Blank in 1967. For instance see this paper for more on this topic and references.<|endoftext|> TITLE: Which groups can occur as the quotient of a group by its centre? QUESTION [40 upvotes]: Given a group $G$, its centre $Z(G)$ is a normal subgroup of $G$ and one can consider the quotient $G/Z(G)$. Which groups $H$ can occur as $G/Z(G)$ for some group $G$? For example, it is easily proved that $G/Z(G)$ cannot be cyclic unless it is trivial. A certain amount of general literature survey and google search did not give me any answer. But since I do not work in algebra, it could be the case that I missed something easy. It can be shown that $G/Z(G)$ is isomorphic to the group of inner automorphisms $\mathrm{Inn}(G)$ of $G$. REPLY [51 votes]: In the literature the following terminology is common. If $H$ is a group such that $H \cong G/Z(G)$ for some $G$, we say that $H$ is capable. As you mention, non-trivial cyclic groups are not capable. Another example is the quaternion group of order $8$, which is not capable. The earliest paper on this seems to be the one by Baer: "Groups with preassigned central and central quotient group. Trans. Amer. Math. Soc. 44 (1938), no. 3, 387–412." He gives a complete classification of finite abelian $H$ that are capable. In general the description of groups that are isomorphic to some $G/Z(G)$ is not known and seems to be a difficult problem. But you will find many papers on the subject for example by looking up "capable groups" on google, or by looking up on MathSciNet which papers cite the 1938 paper of Baer.<|endoftext|> TITLE: Representations of $SL_1(D),$ where $D$ a division algebra over a local field QUESTION [8 upvotes]: Let $k$ be a local field of residue characteristic $p$, and let D be a central division algebra over $k$ of index $n>2$. How to determine the irreducible complex representations of the group $SL_1(D)$? Suggest some reference regarding this. REPLY [3 votes]: An article by Shai Shechter recently appeared on Math ArXiv: "Characters of the Norm-One Units of Local Division Algebras of Prime Degree" http://front.math.ucdavis.edu/1512.02448<|endoftext|> TITLE: What is Atiyah's topological formulation of the odd order theorem? QUESTION [49 upvotes]: Here is a quote from an article by Daniel Gorenstein on the history of the classification of finite simple groups (available here). During that year in Harvard, Thompson began his monumental classification of the minimal simple groups. He soon realized that he didn't need to know that every subgroup of the given subgroup was solvable, but only its local subgroups, and he dubbed such groups N-groups. However, the odd order theorem was still fresh in his mind. One afternoon I ran into him in Harvard Square and noticed he had a copy of Spanier's book on algebraic topology under his arm. "What in the world are you doing with Spanier?" I asked. "Michael Atiyah has given a topological formulation of the solvability of groups of odd order and I want to see if it provides an alternate way of attacking the problem," was his reply. What is this topological formulation of the solvability of groups of odd order? REPLY [30 votes]: In an email correspondence with Atiyah, I brought this up. The comments are meant to provide background for Thompson's remark. "When I first heard of FT I thought there should be a simpler proof using fixed point theorems and K-theory and I propagated the idea. The problem was that fixed point theorems could only deal with fixed points of elements or cyclic groups. So I knew we needed a theory that would cover fixed points of a whole group. We could then apply it to the action of a finite group on the projective space of the reduced regular representation." He went on further to say, "It was only recently that I realized we had to use equivariant K-theory and not its completion at the identity." Then he mentioned his completion theorem of Brauer induction (with Segal), and how Snaith had given a topological proof of Brauer induction. Although not definitive, Atiyah may have a shortened proof of the Feit-Thompson theorem, and if so it would be presented in the near future. In case it is helpful, I see that the quote from Gorenstein's paper was around 1960. Atiyah had published the finite group version of the Atiyah-Segal completion in 1961 ("Characters and cohomology of finite groups"). So, based on this recent email correspondence, these 1961 ideas make their appearance in a formulation of FT.<|endoftext|> TITLE: What function is this? -Counterexample found: it is not Lipschitz- QUESTION [9 upvotes]: THE FRAMEWORK Let $0<\lambda\le1$ and consider $$ \Psi:(\Bbb R[X]_0,||\cdot||_{\lambda})\longrightarrow(\mathcal C^{\lambda}[0,1],||\cdot||_{\lambda}) $$ defined as $$ \Psi(p):=\sup_{0\le u\le\cdot}p(u) $$ in the sense that the polynomial $p$ is sent by $\Psi$ to the function $t\mapsto\sup_{0\le u\le t}p(u)$. Here: $\Bbb R[X]_0$ denotes the space of one variable polynomials with real coefficients which vanish at $0$; $\mathcal C^{\lambda}[0,1]$ is the space of $\lambda-$ Holder continuous functions $f:[0,1]\to\Bbb R$; $||\cdot||_{\lambda}$ denotes the usual $\lambda-$Holder seminorm, i.e. $$ ||f||_{\lambda}:=\sup_{0\le s TITLE: Has anyone seen a nice map of multiplicative cohomology theories? QUESTION [26 upvotes]: I have seen lots of descriptions of this map in the literature but never seen it nicely drawn anywhere. I could try to do it myself but I really lack expertise, hence am afraid to miss something or do it wrong. Let me just provide some glimpses, and maybe somebody can nicely tie them together. At the "initial end" there is the stable (co)homotopy, corresponding to the sphere spectrum. At the "terminal end" there is the rational cohomology (maybe extending further to real, complex, etc.) From that terminal end, chains of complex oriented theories emanate, one for each prime; the place in the chain corresponds to the height of the formal group attached. Now here I am already uncertain what to place at each spot - Morava $K$-theories? Or $E$-theories? At the limit of each chain there is something, and again I am not sure whether it is $BP$ or cohomology with coefficients in the prime field. Next, there is complex cobordism mapping to all of those (reflecting the fact that the complex orientation means a $MU$-algebra structure). But all this up to now only happens in the halfplane. There are now some Galois group-like actions on each of these, with the homotopy fixed point spectra jumping out of the plane and giving things like $KO$ and $TMF$ towards the terminal end and $MSpin$, $MSU$, $MSp$, $MString$, etc. above $MU$. Here I have vague feeling that moving up from $MU$ is closely related to moving in the plane from the terminal end (as $MString$, which is sort of "two steps upwards" from $MU$, corresponds to elliptic cohomologies which are "two steps to the left" from $H\mathbb Q$) but I know nothing precise about this connection. As you see my picture is quite vague and uncertain. For example, I have no idea where to place things like $H\mathbb Z$ and what is in the huge blind spot between the sphere and $MU$. From the little I was able to understand from the work of Devinatz-Hopkins-Smith, $MU$ is something like homotopy quotient of the sphere by the nilradical. Is it correct? If so, things between the sphere and $MU$ must display some "infinitesimal" variations. Is there anything right after the sphere? Also, can there be something above the sphere? How does connectivity-non-connectivity business and chromatic features enter the picture? What place do "non-affine" phenomena related to algebroids, etc. have? There are also some maps, like assigning to a vector bundle the corresponding sphere bundle, which seem to go backwards, and I cannot really fit them anywhere. Have I missed something essential? Or all this is just rubbish? Can anyone help with the map, or give a nice reference? REPLY [4 votes]: An informal reference could be the diagram on page 2 of the lecture notes from my September 2000 lecture in Oberwolfach, where I discussed the chromatic red-shift conjecture.<|endoftext|> TITLE: About pointwise Kan extension QUESTION [11 upvotes]: Suppose that you want to look at the left Kan extension of a functor $F : \mathcal{C} \to \mathcal{A}$ along a functor $K : \mathcal{C} \to \mathcal{B}$. It is widely known that if the colimit of the canonical diagram $$ \mathrm{colim}(K \downarrow d \to \mathcal{C} \overset{F}{\rightarrow} \mathcal{A})$$ exists in $\mathcal{A}$ for all object $d$ of $\mathcal{D}$, then the wanted left Kan extension exists and its value on $d$ is given by this colimit. However, to establish that result the previous colimit has to exist for all object $d$ of $\mathcal{D}$. This raises the following question. Suppose that the left Kan extension of $F$ along $K$ exists, call it $L : \mathcal{D} \to \mathcal{A}$, and suppose that $$ \mathrm{colim}(K \downarrow d \to \mathcal{C} \overset{F}{\rightarrow} \mathcal{A})$$ exists for a particular object $d$ of $\mathcal{D}$, but not necessarily for all objects of $\mathcal{D}$ (that's the point). Is it still true that $$L(d) \cong \mathrm{colim}(K \downarrow d \to \mathcal{C} \overset{F}{\rightarrow} \mathcal{A} )$$ ? REPLY [2 votes]: The answer is no (I think -- non-pointwise Kan extensions are a pain and I may have messed something up!). I wouldn't lose too much sleep over this, though -- in practice, you never know that some functor is a Kan extension without knowing it's a pointwise one. The non-pointwise Kan extension I gave here works for a counterexample. See there for more details. The Kan extension looks like this: (The best intuition I can offer is that it was actually constructed to have $F$ fully faithful, but the comparison 2-cell $\eta$ non-invertible, which forces it to be a non-pointwise Kan extension.) $\require{AMScd}$ \begin{CD} \mathbf{B}\mathbb{N} @>G>> C\\ @VFVV \nearrow L \\ \mathbf{B}\mathbb{N}_+ \end{CD} Here $\mathbf{B}\mathbb{N}$ is the monoid $\mathbb{N}$ regarded as a one-object category, $\mathbf{B}\mathbb{N}_+$ is the same with an initial object adjoined, $F$ is the natural inclusion, and $C$ looks like this: $G\bullet \overset{(\eta_n)_{n \in \mathbb{N}}}{\overset{\to}{\to}} L\bullet \overset{L!}{\leftarrow} L\emptyset$ Here $G\bullet$ is the fully faithful image of $G$ (so it's a copy of $\mathbf{B}\mathbb{N}$), and $L\emptyset \overset{L!}{\to} L\bullet$ is the fully faithful image of $L$ (so it's a copy of $\mathbf{B}\mathbb{N}_+$ where $L\emptyset$ is the initial object). $C(G\bullet, L\bullet)$ is generated by an arrow $\eta = \eta_0$ under the equation $\eta_n = \eta \circ Gn = Ln \circ \eta$. So $\eta$ constitutes a comparison natural transformation $G \implies LF$. Why it's a left Kan extension: An exhaustive analysis of the functors $H: \mathbf{B}\mathbb{N}_+ \to C$ (there are 5 families of them, depending on where the objects are sent) reveals that the only one which admits a natural transformation $G \implies HF$ or $L \implies H$ is $L$ itself, and this diagram is in fact a left Kan extension $L = \mathrm{Lan}_F G$. Why it's a counterexample: You can check that $Hom(G\bullet,-)$ and $Nat(Hom(F,\bullet),Hom(G,-))$ are isomorphic; they map $L\emptyset \mapsto \emptyset$, $L\bullet \mapsto \mathbb{N}$, $G\bullet \mapsto \mathbb{N}$. So $G\bullet$ "should" be the value of $\mathrm{Lan}_F G(\bullet)$. even though the value in the actual left Kan extension $L$ is $L\bullet$.<|endoftext|> TITLE: Oscillation of the summatory Möbius function QUESTION [9 upvotes]: Let the Mertens function $$M(x) = \sum_{n \le x} \mu(n)$$ I assume (perhaps foolishly) that it is known that $M(x)$ changes sign infinitely often. If that's true, the question is a quantitative version : How many sign changes of $M(x)$ are there between $1$ and $y$ (asymptotically) ? **ADDITION* GH from MO cites a result which gives a logarithmic number of changes. This, while better than nothing, is not (empirically the truth): for $N=1000000,$ you get around $5500$ sign changes, for $N=10000000,$ around $12000,$ and here is the graph of the total number of sign changes.This looks square-rootish. Now, what is even more interesting is that for a symmetric random walk, the number of returns to the origin is asymptotic to $\frac{2}{\pi} \sqrt{N},$ which is much smaller than this. (see https://math.stackexchange.com/questions/1338097/expected-number-of-times-random-walk-crosses-0-line for deviation). The difference is even more striking, if you remember that a lot of numbers are non-square-free. REPLY [3 votes]: You may also find quantitative results concerning error term from prime number theorem in the article S. B. Stechkin, A. Yu. Popov, “The asymptotic distribution of prime numbers on the average”, Uspekhi Mat. Nauk, 51:6(312) (1996), 21–88, english translation is published in Russ. Math. Surv. 51 (1996), pp. 1025-1092.<|endoftext|> TITLE: Lie algebra of a p-group QUESTION [6 upvotes]: Given a p-group P, the first hochschild cohomology of the group algebra (over a field of characteristic p) of P is a nonzero Lie algebra. Is it known what Lie algebra results depending on P? I have no experience with this, so not sure if this is a too easy question. One can replace p-group algebra by local selfinjective finite dimensional algebra if it is too easy. Can the simple lie algebras realised as first hochschild cohomology of such (or other) algebras? REPLY [7 votes]: I'm not an expert but it looks like some interesting simple Lie algebras do appear this way, yet one does not know how many. If the base field $k$ is $\mathbb{F}_p$, where $p$ is odd, and $G=\mathbb{Z}_p$ is a cyclic group of order $p$, then $HH^*(kG,kG)$ is recently described along with all relevant operations (even as a BV-algebra). It seems that the Lie algebra $HH^1(\mathbb{Z}_p,\mathbb{Z}_p)$ is isomorphic to the Witt algebra $W(1,\underline{1})$ which is the derivation algebra of the truncated polynomial ring $k[X]/(X^p)$. The Witt algebra belongs to a large family of finite dimensional simple Lie algebras called Lie algebras of Cartan type (there are four types: $W, S, H$ and $K$). Markus Linckelmann showed recently that some Cartan type algebras of type $W$ can be realised as $HH^1(B,B)$ where $B$ is a block of a finite group algebra. In the more general setting of a self-injective finite dimensional local algebra, the answer is unknown.<|endoftext|> TITLE: Independence over ZFC + CH QUESTION [6 upvotes]: Acknowledging Woodin's result on $\Sigma^2_1$-absoluteness for forcing-models satisfying the continuum hypothesis (CH), it is natural to ask: Are there examples of statements $\phi$ in "the usual realm of mathematics" that cannot de decided in ZFC + CH? Surely modern mathematicians (oblivious of metamathematics) could have naturally come up with problems of complexity beyond $\Sigma^2_1$ that are provably undecidable in ZFC + CH. A ZFC example to contrast with would be Borel's conjecture. But what are the ZFC + CH examples if any? REPLY [8 votes]: I wanted to add a couple of examples in addition to those mentioned in the comments. In algebra, Shelah's work on the Whitehead Problem established that the statement ``All Whitehead groups are free'' is independent of ZFC + CH. He originally proved the statement's independence from ZFC, and then developed forcing techniques for getting independence from ZFC + CH. In set-theoretic topology, I've done a lot of work on building "interesting" models of ZFC+CH using iterated forcing. For example, the following statement is not decided by ZFC + CH: "a first countable countably compact (=every infinite set has a point of accumulation) space is either compact or contains a subspace homeomorphic to $\omega_1$" Counterexamples to this can be built from $\diamondsuit$; constructions due to Ostaszewski and Fedorcuk are the most well-known. Independence from ZFC was established in a flush of results from the late 1980s involving Balogh, Dow, Fremlin, and Nyikos (and others as well). In particular, the statement is a consequence of PFA. We established that the statement is consistent with ZFC + CH in joint work with Nyikos, so taken with the $\diamondsuit$ results we have the required independence. The best result we have along these lines (done with Alan Dow) is that ZFC + CH is consistent with "compact countably tight spaces are sequential''. Here a space $X$ is countably tight if whenever $x$ is in the closure a set $A\subseteq X$ if and only if $x$ is in the closure of a countable subset of $A$. A space $X$ is sequential if a subset $A$ of $X$ is closed if and only if it is sequentially closed. Roughly speaking, the statement "compact countably tight spaces are sequential" says that if the topology of a compact space is determined by its countable subsets, then it is determined by its convergent sequences". Again, this statement is a consequence of PFA (a famous result of Balogh), and in the other direction $\diamondsuit$ provides us with counterexamples -- in fact, Fedorcuk showed that $\diamondsuit$ entails the existence of a compact countably tight hereditarily separable space of size $2^{\aleph_1}$ with no convergent sequences at all. The work we did with Dow proves the statement is consistent with (and therefore independent from) ZFC + CH. REPLY [6 votes]: Is there a definable non-measurable set? (Here I mean definable without parameters -- it is cheating if you define it from an ultrafilter, or a well-ordering of $\mathbb R$, without first defining those too.) The answer to this question is independent of ZFC+CH. The answer is yes if $V=L$, but it becomes no if you add a Cohen real or a random real in certain forcing extensions. I'm not sure whether or not you consider this question to lie within the realm of "usual mathematics" -- but it is a question that I would imagine a wide variety of mathematicians have asked at some point (not just set theorists or logicians). EDIT: Over the weekend I realized that I'd made a mistaken claim in this post, namely that adding a Cohen or random real to a model of ZFC+V=L produces a model with no parameter-free-definable non-measurable sets. (I think I might have made the substitution ''ultrafilter'' $\rightarrow$ ''non-measurable set'' in the back of my mind; adding a Cohen or random real does produce a model without parameter-free-definable ultrafilters.) For adding a random real, what I claimed is definitely false. If $r$ is random over $L$, then $L[r]$ admits a non-measurable set with a particularly simple definition (only three symbols!), namely $L \cap \mathbb R$. This can be deduced from Example 26.51 in Jech's book (and Jech attributes the example to Solovay). I don't know whether adding a single Cohen real to $L$ results in a model without parameter-free-definable non-measurable sets. In his famous paper in which he proved that there is a model of ZF+DC in which every set of reals is measurable, Solovay shows that there is a forcing extension of $L$ where every non-measurable set of reals fails to be definable (in fact, he shows much more: no non-measurable set of reals is definable from a countable sequence of ordinals). His forcing construction uses an inaccessible cardinal, so all I can say for certain is that ``there exists a definable non-measurable set'' is independent of ZFC+CH assuming the consistency of ZFC+"there is an inaccessible cardinal".<|endoftext|> TITLE: What is the growth rate of the number of unoriented cobordism classes? QUESTION [7 upvotes]: Let $\Omega_n^O$ denote the abelian group of cobordism classes of closed, unoriented manifolds of dimension $n$, and let $d(n) := \lvert\Omega_n^O\rvert$. What are the asymptotics of $d(n)$? It's possible to express $\log_2 d(n)$ as a modified partition function: by work of Thom, $\Omega_n^O$ is an $\mathbb F_2$-vector space with a basis of the form $$\{x_{\ell_1,\dotsc,\ell_j}\mid \ell_1 + \dotsb + \ell_j = n, \ell_i \ne 2^r - 1\},\qquad\qquad (*)$$ so $\dim_{\mathbb F_2}\Omega_n^O = \log_2 d(n)$ is equal to the number of partitions of $n$ into numbers that are not one less than a power of 2. Hence if $P(n)$ denotes the usual partition function, $d(n) = O(2^{P(n)})$. Is this sharp? What about lower bounds? REPLY [10 votes]: If you denote by $P'(n)$ the number of partitions of $n$ with parts not of the form $2^k-1$, it holds that $$\log P'(n)\sim \pi \sqrt{\frac{2n}{3}}$$ just like for the usual partition function. This will be true whenever you allow the parts to come from a set of integers of asymptotic density 1. See for example Nathanson's paper: "Asymptotic density and the asymptotics of partition functions".<|endoftext|> TITLE: transitive subgroups of $S_n$ QUESTION [8 upvotes]: For a subset $X$ of $S_n$ (the symmetric group of degree $n$) define $C(X)$ to be the union of all conjugates of elements of $X.$ Question 1 What is that called? Question 2 Over all transitive subgroups $H$ of $S_n,$ which one has the smallest $C(H)?$ REPLY [4 votes]: Since nothing else came up, I'm posting this as an answer. For $n$ a power of a prime $p$, the answer is any group of exponent $p$, acting regularly. Indeed, if $n$ is a power of $p$, then a Sylow $p$-subgroup $P$ of $H$ is transitive, and any element of order $p$ in the center of $P$ will be fixed-point-free, so $H$ always contains an element of cycle type $(p,\ldots,p)$. In a group of exponent $p$ acting regularly, every non-trivial element has this cycle type, so this is best possible. For other small values of $n$, I obtained the following answers computationally: $n=6$ : dihedral group, acting regularly. $n=10$ : dihedral group, acting regularly. $n=12$ : $A_4$, acting regularly. $n=14$ : dihedral group, acting regularly (edging out by less than 0.1% a group isomorphic to $AGL(1,8)$). $n=15$ : $A_5$, acting on the cosets of a Sylow $2$-subgroup. $n=18$ : generalised dihedral group over an elementary abelian group, acting regularly. $n=20$ : a group of the form $C_2^4:C_5$ (edging out another group by about 0.01%). I'm not sure why there are so many extremely close calls. I'm guessing it has to do with the wide range of size of conjugacy classes in $S_n$ (so two groups might contain the same large classes, but differ in some small classes). Perhaps some other cases can be done, like $n$ twice a prime but, to me, the general problem seems hard, even intractable.<|endoftext|> TITLE: Hochschild cohomology of certain local algebras QUESTION [6 upvotes]: Let $K$ be a field and $A$ the algebra $K\langle x_1,...,x_n\rangle/J^m$ for $n \geq 1$ and $m \geq 2$, where $K\langle x_1,...,x_n\rangle$ is the non-commutative polynomial ring in $n$ variables over $K$ and $J$ the ideal generated by $x_1,\dotsc,x_n$. What is the Lie algebra of the first Hochschild cohomology of $A$ (or at least its dimension)? What is the Hochschild cohomology ring of that algebra? The same questions, with the non-commutative polynomial ring replaced by a commutative one. REPLY [3 votes]: The right term to look for is "truncated quiver algebras". There are two relevant references which I believe lead to a complete answer to your question in the non-commutative case. First, Section 8.2 of "Comparison morphisms and the Hochschild cohomology ring of truncated quiver algebras" by Guillermo Ames, Leandro Cagliero, and Paulo Tirao (http://www.sciencedirect.com/science/article/pii/S0021869309002907) suggests that the product in cohomology is zero in positive degrees (for $n>1$). Second, Theorem 1.3 of the article "The Lie algebra structure of the first Hochschild cohomology group for monomial algebras" by Claudia Strametz (http://www.sciencedirect.com/science/article/pii/S1631073X02023464) has formulas for the Lie bracket on the first Hochschild cohomology in the general case of a monomial algebra (using combinatorics of Anick chains / Bardzell resolution). It should be possible to unravel these formulas completely in your case.<|endoftext|> TITLE: Looking for the name or reference regarding a bipartite graph parameter QUESTION [6 upvotes]: I'm writing a paper about a math puzzle and the thing I'm studying ends up equivalent to finding the following parameter of a bipartite graph G with parts X and Y: The largest $k$ such that any $k$ vertices in $X$ all have a common neighbor. I'm trying to figure out if this parameter has a name and if the corresponding decision problem is NP-complete (which I strongly suspect it is). My best lead so far has been this restatement of the parameter: the largest $k$ such that a set family is $k$-wise intersecting. But none of the papers I've looked at seem interested in finding $k$ -- generally it is given and other results follow. Does anyone know the name of this parameter (if it exists), or if it is NP-complete? REPLY [3 votes]: The problem is equivalent to the set cover problem and thus NP-hard. The set formulation you gave is: Given a finite set $\mathcal{X}$, a family $\mathcal{F} \subseteq \mathcal{P}(\mathcal{X})$, what is the largest $k$ such that the intersection of every $k$-subset of $\mathcal{F}$ is nonempty? A decision version of this is: Given a finite set $\mathcal{X}$, a family $\mathcal{F} \subseteq \mathcal{P}(\mathcal{X})$, and integer $k$, is there a $k$-subset of $\mathcal{F}$ whose intersection is empty? Taking the complement of each set in $\mathcal{F}$, we get the decision version of the set cover problem: Given a finite set $\mathcal{X}$, a family $\mathcal{F} \subseteq \mathcal{P}(\mathcal{X})$, and integer $k$, is there a $k$-subset of $\mathcal{F}$ whose union is $\mathcal{F}$? In a similar fashion, we can take the complement (of only edges between $X$ and $Y$) in the graph formulation, and then $k+1$ is the minimum number of vertices in $X$ needed to cover $Y$. But after a quick search I couldn't find this kind of covering number used in literature.<|endoftext|> TITLE: Action of SL(2,Z) on upper triangular primitive integer matrices of determinant N, from the right. Is it transitive? QUESTION [13 upvotes]: I am porting this question across from StackExchange, since it has received no answers and perhaps is sufficiently deep to fit here. I am considering the set of upper triangular matrices $$D_N=\left\{\begin{pmatrix}a&b\\0&d\end{pmatrix}:a,b,d\in\mathbb{Z}, ad=N, \gcd(a,b,d)=1, 0\leq b TITLE: Naive equivariant transfer QUESTION [7 upvotes]: Given a group $G$, a $\mathbb Z$-graded cohomology theory $E^*_G$, and a $n$-sheeted covering $p\colon X \to Y$, I would like a transfer map $$p_!\colon E^*_G X \to E^*_G Y$$ satisfying $$\require{cancel}\xcancel{p_! p^* = n \cdot \mathrm{id}.}$$ [this formulation was wrong; see the comments] such that when $n$ is inverted, $p_! p^*$ becomes an isomorphism (what I really need is that $p^*$ is an injection to a direct factor). I'm aware such transfers exist for arbitrary fibrations $p$ (the number $n$ becomes the Euler characteristic of the fiber) under the additional assumption $E^*_G$ is $RO(G)$-graded, and that the $RO(G)$-grading is necessary for that result, but I only need them for covering maps, and I'd like to avoid the additional hypothesis. So, are there transfer maps in this generality? REPLY [4 votes]: Let me try to turn my comments into something like an answer (but the 'tldr' version is "I don't know.") When $G=*$, a space $X$ represents a functor with homotopy coherent transfers if and only if $X$ is a $\Gamma$-space (and hence equivalent to an $E_{\infty}$-space). Actually, you might take 'being a $\Gamma$-space' as the definition of having homotopy coherent transfers. One could (and Quillen did) ask whether this is equivalent to just asking that $[-,X]$ has functorial transfers for finite covers. That was called "the transfer conjecture". This turns out to be the same as asking that $X$ be an $H_{\infty}$-space. And so one asks "Is every $H_{\infty}$-space an $E_{\infty}$-space?" The answer is no, and a counterexample was provided by Kraines and Lada. For finite $G$, a $G$-space $X$ represents a functor with homotopy coherent transfers for finite covers (of $G$-spaces) if and only if it admits the structure of an equivariant Segal space (and hence is equivalence to a $G-E_{\infty}$-space. So, up to group completion, $X$ is the zeroth space of a genuine $G$-spectrum (and so represents an $RO(G)$-graded cohomology theory). If $X$ is a $G$-space representing a functor with homotopy coherent transfers for finite covers fibered in trivial $G$-sets, then this is like saying $X$ is an $E_{\infty}$-space in $G$-spaces. In particular, up to group completion, it admits ordinary deloopings and represents a $\mathbb{Z}$-graded cohomology theory for $G$-spaces. In your situation, you seem to have a space $X$ as in (3) and are asking if it's possibly to ask that $[-,X]$ admit transfers for all covers without requiring that $X$ admit $RO(G)$-deloopings, i.e. without requiring that $X$ be equivalent to a $G-E_{\infty}$-space. My guess is that this is possible, but that any example would be manufactured (like the counterexample of Kraines and Lada). However, I should point out that this situation is not precisely parallel to that in (1) because you have already placed an $E_{\infty}$-structure on $X$, so it's at least plausible that this homotopy coherence together with some weak notion of more exotic transfers could be enough... but again, I doubt it. In your last comment you mention that you are only interested in transfers for covers with cyclic structure group. Even for those, I think I stand by my intuition from (4), but I could be wrong. One imagines that whatever obstruction is responsible for establishing the conjectured example in (4) would be known to cyclic covers, especially after localizing at a prime. I've gone this whole answer without saying $N_{\infty}$-operad. Consider it said. (It's relevant to this business of asking for fewer transfers, somewhere between "fibered in trivial $G$-sets" and "fibered in arbitrary $G$-sets". Blumberg-Hill is the place to learn about these.)<|endoftext|> TITLE: Gray product on $(\infty,2)$-categories QUESTION [5 upvotes]: In the book of Gaitsgory and Rozenblyum Ch A.1 Basics of $(\infty,2)$-categories, 3.2, they define the Gray tensor product of $(\infty,2)$-categories. Here are my questions: What is an explicit description of $\mathrm{Seq}_\bullet ([m]\otimes [n])$, where $\otimes$ is the Gray product, and $\mathrm{Seq}_\bullet: \text{2-Cat}\rightarrow(\text{1-Cat})^{\Delta^{op}}$ is their model of $(\infty,2)$-categories? In Lurie's paper $(\infty,2)$-Categories and the Goodwillie Calculus I, he defines several models for the homotopy theory of $(\infty,2)$-categories. I was wondering whether it is possible to define Gray product on these model categories, e.g. $\mathrm{Set}_\Delta^{\mathrm{sc}}$, $\mathrm{Fun}(\Delta^{op}, \mathrm{Set}_\Delta^+)$, and whether there are explicit definitions? Thanks in advance! REPLY [5 votes]: For question (2), there is actually a left Quillen bifunctor $$ \times_{\mathrm{gr}}: \mathrm{Set}_\Delta^{\mathrm{sc}} \times \mathrm{Set}_\Delta^{\mathrm{sc}} \to \mathrm{Set}_\Delta^{\mathrm{sc}} $$ which models the Gray tensor product. If $(X,T),(Y,S)$ are scaled simplicial sets then $(X,T) \times_{\mathrm{gr}} (Y,S) = (X \times Y,R)$ where the set $R$ of thin triangles consists of those triangles $\sigma: \Delta^2 \to X \times Y$ such that: 1) The image of $\sigma$ in both $X$ and $Y$ is thin. 2) Either the images of $\sigma$ and $\sigma|_{\Delta^{\{1,2\}}}$ in $X$ are degenerate or the images of $\sigma$ and $\sigma|_{\Delta^{\{0,1\}}}$ in $Y$ are degenerate. For example, $\Delta^1 \times_{\mathrm{gr}} \Delta^1$ is a square in which one of the two triangles is thin and the other is not. This data is essentially the same as a square which commutes up to a non-invertible $2$-cell. Note however that even if $(X,T)$ and $(Y,S)$ are both fibrant then $(X,T) \times_{\mathrm{gr}} (Y,S)$ will generally not be fibrant (because it can have invertible $2$-cells which are not designated as thin). This construction features prominently in a forthcoming preprint of mine concerning limits and colimits in $(\infty,2)$-categories, which I will hopefully put on the arXiv soon.<|endoftext|> TITLE: Was there ever proposed a theory where the value of Dirac Delta at zero had meaning on itself? QUESTION [5 upvotes]: Was there ever proposed a theory where $\delta(0)$ has a meaningful value or used in a formal way outside integrals? Particularly, following Fourier transforms, we can formally obtain $$\pi\delta(0)=\int_0^\infty dx=\int_{0^+}^\infty \frac1{x^2}dx$$ Were there attempts to treat these integrals as some quantities? REPLY [2 votes]: This works fine in Robinson's framework by choosing for example the Cauchy distribution (in the probability sense, not the Schwartz sense) with an infinitesimal value of the parameter. The "infinitely tall, infinitely narrow" delta-function this obtained returns the value (of a test function at the point when integrated against it) up to an infinitesimal. Robinson in his 1966 book proves the existence of delta functions which return the value of the test functions "on the nose" but these are a bit harder to define.<|endoftext|> TITLE: When does the determinant distribute over addition? QUESTION [7 upvotes]: When does $\det(A+B)=\det(A)+\det(B)$ hold? I actually wonder if there is an easy answer for when $Per(A+B)=Per(A)+Per(B)$. REPLY [15 votes]: let me assume $A$ is invertible, then you ask when $$\det(1+X)=1+\det X,\;\;X=A^{-1}B $$ so if $X$ has eigenvalues $x_i$, $i=1,2,\ldots n$, you would need $$\prod_{i}(1+x_i)=1+\prod_i x_i$$ basically you can take arbitrary values for $x_1,x_2,\ldots x_{n-1}$ and then the only requirement is that $$x_n=\frac{1-U}{U-V},\;\;U=\prod_{i=1}^{n-1}(1+x_i),\;\;V=\prod_{i=1}^{n-1}x_i$$<|endoftext|> TITLE: Mapping space from a quotient space QUESTION [10 upvotes]: For $X/{\sim}$ a quotient space, $$ Map(X/{\sim},Y)\subset Map(X,Y). $$ But is this inclusion always a homeomorphism on its image? (Assuming compact-open topology on the mapping spaces.) If not what would be the most general setting to make it true? We can also assume that $X$ and $Y$ are compactly generated. A related question: if $q\colon X\to X/{\sim}$ is a quotient map and $X/{\sim}$ is compact, does always exist a compact $Y\subset X$ such that $q(Y)=X/{\sim}$. REPLY [3 votes]: @Victor has pointed out to my error in the pre-edited version--thank you, Victor. Now everything IS under control and FIXED. Here, after @TarasBanakh, there is another example of a quotient map $\ q: X\rightarrow X/{\sim} $ such that $X$ is compact but there is no compact subset $Y\subseteq X$ such that $f(Y)=f(X)$. Let $Q\subset\mathbb R$ be the set of all rational numbers. Let $\ J:=[0;1]:=\{x\in\mathbb R: 0\le x\le 1\},\ $ Define $$ X\ :=\ \{(x\ y)\in J^2\,:\, |\{x\ y\}\cap Q| = 1\} $$ And let $\ p:X\rightarrow J\ $ be the projection $\ p(x\ y)\ := x.\ $ Then $p$ is onto, and for every $A\subseteq J$ we have: $p^{-1}(A)$ is open in $X$ when $A$ is open in $J$ because $p$ is induced by the Cartesian projection; $p^{-1}(A)$ is not open in $X$ when $A$ is not open in $J$ because $p^{-1}(x)$ is dense in $\ \{x\}\times J\ $ for every $\ x\in J.\ $ Thus, $\ p\ $ is topologically equivalent to the respective quotient map. More than this, $p$ is an open map. Indeed, sets $$ B_{abcd}\ :=\ ((a;b)\times(c;d))\,\cap\, X$$ form a topological base of $X$, and $\ p(B_{abcd}) = (a;b)\cap J.\ $ Thus $p$ is an open map.   Let $\ Y\subseteq X\ $ be a compact subset such that $\ p(Y)=[0;1]\ $ (a proof by contradiction). Then $\ Y\ $ is a countable union of its compact subsets $\ C_a:=(\{a\}\times\mathbb R)\cap Y\ $ and $\ D_a:=(\mathbb R\times\{a\})\cap Y,\ $ where $\ a\ $ runs over rational numbers. Then sets $\ p(C_a)\ $ and $\ p(D_a)\ $ are compact and they cover $\ [0;1].\ $ Thus, by Baire's theorem one of these projections must have a non-empty interior in $\ [0;1]\ $ -- a contradiction. Thus such $\ Y\ $ does not exist.<|endoftext|> TITLE: Modern sources on Lie algebraic methods in combinatorics QUESTION [8 upvotes]: I am looking for modern textbooks, expository papers and journal articles regarding the use of Lie algebras in combinatorics. In particular, I am interested to know the extent to which people still care about this. For example, the representation theory of $\mathfrak{sl}_2(\mathbf{C})$ can be used to show that certain sequences are unimodal and symmetric. This and similar material can be found in Richard Stanley's papers (e.g. 41, 57, 62, 72, 84). In the textbooks I've looked at (e.g. Aigner's A Course in Enumeration, Stanley's EC1/2) you can find material on things like posets, Young tableaux, and the character theory of $S_n$ and $\mathrm{GL}(n)$. But it seems like Lie algebras such as $\mathfrak{sl}_2(\mathbf{C})$ have been left out. REPLY [6 votes]: A lot of the literature on crystals (in the modern sense of algebraic combinatorics) uses the theory of crystal bases (which is part of quantum group theory) as a black box. For one example of such a use, see Mark Shimozono, Crystals for dummies, which proves various properties of semistandard Young tableaux using crystals. Vertex algebras can be used to prove partition identities. For example: Debajyoti Nandi, Partition identities arising from the standard $A_2^{(2)}$-modules of level $4$, thesis or Mirko Primc, Vertex Algebras and Combinatorial Identities and S. Capparelli, On Some Representations of Twisted Affine Lie Algebras and Combinatorial Identities. Do you count vertex algebras as part of Lie-algebra theory? Jean-Louis Loday, Todor Popov, Parastatistics Algebra, Young Tableaux and the Super Plactic Monoid obtains a generalized Cauchy identity for hook Schur functions (a super-version of Schur functions) from representations of Lie superalgebras.<|endoftext|> TITLE: Open problems in mathematical physics QUESTION [36 upvotes]: What are good, still unsolved problems in mathematical physics that are in vogue? I always get the same answers: reference to Millennium Problems by the Clay Institute, or "there's still a lot to do in this or that..." or the three body problem. Could you be more specific? REPLY [6 votes]: Exact formulas, or better approximations in minimum times in quantum control, a.k.a., the quantum speed limit, is largely unsolved. Expressed mathematically: Given $a,b \in \mathfrak{su}(n)$ which are bracket generating and some $G \in SU(n)$, consider a system obeying the equation $\dot{U}_t = (a + f(t)b)U_t$. What is the minimum time $T=T^*$, over all controls $f$ (where all functions, or even delta functions are permitted), to achieve $U_T = G$.<|endoftext|> TITLE: Postnikov-type tower for a map between spaces QUESTION [5 upvotes]: Part I Let $f: X\to Y$ be an arbitrary map between topological spaces/simplicial sets $X$ and $Y$. Then we can associate with $f$ a tower of fibrations $$ \dots \to F_i\to\dots \to F_1\to F_0=Y $$ constructed in a following way: 0.let $F_0 = Y$ and we take $C_0$ as homotopy cofiber of $f$ 1.define $F_1$ as homotopy fiber of $Y\to C_0$ and since composition $X\to Y\to C_0$ is null-homotopic we have a map $X\to F_1$. $C_1$ is a homotopy cofiber of this map. 2.$F_2$ is a homotopy fiber of $Y\to C_1$ and so on. Therefore we got a tower of principle (under some assumptions on $\pi_1$-action I believe) fibrations $$ \Omega C_i\to F_{i+1}\to F_i $$ together with collection of maps $X\to F_i$. Many questions about this construction can be asked. What does this tower tell us about $f$? About $X$ ? Under what conditions on $f$ $X\to \lim F_i$ is an equivalence ? etc Q1: I'm sure such construction have been studied before, where I can find any information about it? Part II Actually, Postnikov tower for a space $X$ is constructed in a very similar way, with one more additional step: instead of taking homotopy fiber straight away, we kill homotopy groups of cofiber using fundamental class. In particular, we start with $X\to K(\pi_1 X,1)$, take its homotopy cofiber $C_0$ and define next step of a tower as a homotopy fiber of composition $K(\pi_1 X, 1)\to C_0\to K(\pi_3 C_0, 3)$. Here $C_0$ is $2$-connected and $\pi_3 C_0=\pi_2 X$. Q2 What is the ``phylosophical'' meaning of taking this additional step $C_n\to K(\pi_{n+2} X,n+3)$ ? What will happen with Postnikov tower if we will not kill homotopy groups of cofiber, apart from losing nice description of $k$-invariants as cohomology classes ? P.S. In general, I would like to compare two constructions (without killing homotopy groups of cofiber and with it) and to understand, which one is more ``canonical''. Conjecturally, taking different morphisms $f$ (for example $X\to \Omega^{\infty}\Sigma^{\infty} X$) we will get towers, that a related to already known constructions (such as Goodwillie tower, for example). EDIT Actually (if I'm not wrong), if $X$ is simple, we can start the Postnikov tower for $X$ straight away from a map $X\to *$. Fundamental class of $\Sigma X=\mathrm{hocofib\{X\to *\}}$ is then represented by $\Sigma X\to K(\pi_1 X, 2)$ and $X_1=\mathrm{hofib}\{*\to K(\pi_1 X,2)\}=K(\pi_1 X,1)$. If we apply my construction to a map $X\to *$, then as $F_1$ we will have $\Omega\Sigma X$, instead of $X_1=K(\pi_1 X,1)$ in a usual Postnikov tower. In general, comparing two constructions, we have a map $F_i\to X_i$, in the example above in the level one this is $\Omega\Sigma X\to K(\pi_1 X,1)$, which is an isomorphism on $\pi_1$. REPLY [3 votes]: Here is an answer to your "Part I." The construction you outlined (and its dual) was considered in detail in the papers of Ganea. In particular: Ganea, T. Induced fibrations and cofibrations. Trans. Amer. Math. Soc. 127 1967 442–459. (See Section 3 of the above for your case.) For the dual question see my answer to: Fibrations and Cofibrations of spectra are "the same"<|endoftext|> TITLE: Extending ground model ultrafilters QUESTION [5 upvotes]: Work over a model of $\sf GCH$. Suppose that $\kappa$ is some regular cardinal, and $U$ is a uniform ultrafilter on $\kappa^+$. Does $U$ have some canonical extension after forcing with $\operatorname{Add}(\kappa,\kappa)$? (Alternatively, replace $\kappa^+$ by an arbitrary regular $\lambda>\kappa$.) REPLY [4 votes]: At Asaf's suggestion, I'm posting my earlier comment as an answer; I'll also append something about Asaf's subsequent question. I can't imagine any of the extensions deserving the name "canonical". Each of the subsets $A$ of $\kappa$ that you've added produces a subset of $\kappa^+$, by chopping the ordinal $\kappa^+$ into blocks of length $\kappa$ and putting $A$ on each block. An extension of $U$ has to choose between this set and its complement, and I see no way to make that choice canonical. Perhaps a more rigorous argument would be that no extension of $U$ can be invariant under the automorphisms of the forcing, not even the automorphisms that interchange the generic sets with their complements. Asaf also asked about "slightly more canonical ultrafilters" versus just using AC to extend the co-small subsets filter. The only improvement I can see is to start with a richer filter then the co-small subsets, for example the club filter. But in the end, one still needs to extend it in what looks to me like a hopelessly non-canonical application of AC.<|endoftext|> TITLE: Does Kechris' conjecture contradict both parts of Martin's conjecture, or just part 1? QUESTION [15 upvotes]: By Kechris' conjecture (KC) I mean the assertion that Turing equivalence $\equiv_T$ is a universal countable Borel equivalence relation. On the other hand, Martin's conjecture (MC) is a long-lasting conjecture about the set of Turing invariant functions as preordered by $$ f\le_m g\iff f(x)\le_T g(x) \text{ on a Turing cone}. $$ Roughly speaking, MC says that this structure is as simple as possible. For an introduction to MC, see e.g. https://arxiv.org/pdf/1109.1875.pdf MC and KC are usually presented as completely contraposed conjectures; in fact, it is well known that if Turing equivalence is universal, then there are Borel Turing-invariant functions which are neither constant on a cone nor increasing on a cone (see https://arxiv.org/pdf/math/0001173.pdf at page 4). So, it is provable in ZF + DC that if KC is true, then part 1 of MC is false, even in its weaker "Borel" formulation. My question is: does anyone know of any contradiction arising from KC and the second part of MC? REPLY [2 votes]: This doesn't answer the question, but it cannot be that KC is true and $\leq_m$ is a prewellordering for all the Borel Turing invariant which are not constant a cone (not just the increasing ones). This is because as we show below, this implies there is a $\Delta^1_2$ wellordering of $\mathbb{R}$ (contradicting AD or AC+large cardinals). We prove the claim: suppose $\equiv_T$ was a universal countable Borel equivalence relation, let $=_\mathbb{R}$ be the equality relation on $\mathbb{R}$, and consider the relation $=_\mathbb{R} \times \equiv_T$ which must be Borel reducible to $\equiv_T$ via some Borel reduction $f$. For each $x \in \mathbb{R}$, let $f_x$ be the Borel Turing invariant function where $f_x(y) = f((x,y))$. These $f_x$ are not constant one a cone and if $x \neq x'$, then for all $y$, $f_x(y) \not \equiv_T f_{x'}(y)$ (because $f$ is a Borel reduction). Hence, the functions $f_x$ are all distinct under $\leq_m$ which wellorders them. So the ordering $x \leq x'$ iff $f_x \leq_m f_{x'}$ is a $\Delta^1_2$ wellordering of $\mathbb{R}$.<|endoftext|> TITLE: “Totally transitive” polytopes which are not regular QUESTION [8 upvotes]: Is it possible to have an abstract polytope which is vertex-transitive, edge-transitive, face-transitive, etc. (individually transitive on faces of each particular dimension) and yet not flag-transitive? [I had thought this might be an easy curiosity for Math StackExchange to readily answer for me, but I had no engagement there, and thus I now cross-post it here…] REPLY [3 votes]: An illustration for the answer by John Machacek: Identifying opposite sides of the picture, one gets a 3-polytope with five vertices, ten edges and five square faces. (This is the well known torus map $\{4,4\}_{(1,2)}$.)<|endoftext|> TITLE: A Naive Question about Nekovar's Paper on Beilinson's Conjecture QUESTION [5 upvotes]: I post this naive question on Math stackexchange, but have got no reply, so I decide to bother the mathoverflow community. https://math.stackexchange.com/questions/2350436/rational-structures-of-cokernel-of-linear-maps In Nekovar's paper on Beilinson Conjecture, http://math.stanford.edu/~conrad/BSDseminar/refs/BeilinsonintroII.pdf I got stuck with the rational structures on cokernel, see sections(2.1) and (2.2), and the question could be stated as follows, $M_B^+$ and $M_{dR}/F^0$ are both rational vector spaces and $\alpha_M$ is an injetive linear map, \begin{equation} \alpha_M: M_B^+ \otimes_{\mathbb{Q}} \mathbb{R} \rightarrow (M_{dR}/F^0) \otimes_{\mathbb{Q}} \mathbb{R} \end{equation} From the paper the rational structures on $M_B^+$ and $M_{dR}/F^0$ define a natural rational structure on $\text{det(Coker}~\alpha_M)$ where $\text{det}(V)$ means the highest exterior power of the vector space $V$, but I don't know how? If $\alpha_M$ is an isomorphism, then $\text{Coker}~\alpha_M$ is trivial, from the paper $\text{det(Coker}~\alpha_M)$ now is canonically isomorphic to $\mathbb{R}$, and the rational structure is $\text{det}(\alpha_M)^{-1}\mathbb{Q}$, which seems very confusing, could someone clarify this? REPLY [4 votes]: For every short exact sequence of $k$-vector spaces \begin{equation*} 0 \to V_1 \xrightarrow{\alpha} V \xrightarrow{\beta} V_2 \to 0 \end{equation*} there is an isomorphism $\det(V) \cong \det(V_1) \otimes \det(V_2)$. Looking at the special case $V_2=0$, it is natural to define $\det(0)=k$. Now if $k_0$ is a subfield of $k$ and if $V_1$ and $V$ are equipped with a $k_0$-rational structure then $\det(V_1)$ are $\det(V)$ are also equipped with a $k_0$-rational structure and it is natural to equip $\det(V_2)$ with the rational structure $\det(\alpha)^{-1} \cdot k_0$. In this way the isomorphism above respects the rational structures. Actually when formulating the Beilinson conjecture for a motive $M$, under the assumptions of Nekovar's paper, it is convenient to introduce the fundamental $\mathbf{Q}$-line \begin{equation*} \Xi(M) = \det(H^1_f(M))^{-1} \otimes \det(H^1_f(M^*(1))^*) \otimes \det(M_B^+)^{-1} \otimes \det(M_{\mathrm{dR}}/\mathrm{Fil}^0) \end{equation*} (see The equivariant Tamagawa number conjecture by Flach). The conjecture says that there is a canonical isomorphism $\Xi(M) \otimes \mathbf{R} \cong \mathbf{R}$ and that the $\mathbf{Q}$-rational structure $\Xi(M)$ corresponds to the leading term of the $L$-function $L^*(M)$. Note that the rational structure defined by $\Xi(M)$ is equal to $\det(\alpha_M)$ times the determinant of the height pairing, and that the problem of defining $\det(0)$ does not appear anymore.<|endoftext|> TITLE: Function and Fourier transform vanish on an interval QUESTION [11 upvotes]: I'm no expert on these things (and this may not be cutting edge research level; it's really motivated by this MSE question), but it seems that there are non-zero measures (and also functions (?), I would think) on $\mathbb R$ such that both the function and its Fourier transform vanish on an interval. This is usually mentioned in the context of Beurling's work on spectral gaps. (1) Is there a nice snappy way to see this? (2) How regular can such an $f$ be? Could it be a Schwartz function? REPLY [8 votes]: Over at MSE (see linked question), user1952009 (= reuns on MO) has given a very elegant construction of a Schwartz function with this property. I am reproducing his/her answer here. Let $g,h\in C_0^{\infty}$ with support contained in $[1/3,1/2]$, and write $\widehat{h}_n=\int_0^1 h(x)e^{-2\pi inx}\, dx$ for the discrete Fourier coefficients of $h$, viewed as a $1$-periodic function $h_p$. Let $$ f(x)=\sum_{n=-\infty}^{\infty} \widehat{h}_n g(x-n) . $$ Then $$ \widehat{f}(\xi)= \sum \widehat{h}_n e^{2\pi in\xi}\widehat{g}(\xi) = h_p(\xi)\widehat{g}(\xi) . $$ Thus $f$ is as desired; this calculation also confirms that $f\in\mathcal S$.<|endoftext|> TITLE: Asymmetric $A \iff B$ proofs QUESTION [12 upvotes]: When proving that conditions $A$ and $B$ are equivalent, it is often an arbitrary choice whether to first prove $A\implies B$ or $B\implies A$. Are there examples where the second implication uses the first in a nontrivial way, and becomes significantly more difficult without invoking the former? REPLY [5 votes]: This is perhaps overly elementary since it is something that might appear in an undergraduate course, but it's what came immediately to my mind. For $\Omega \subseteq \mathbb{C}$ simply connected and open, a continuous function $f\colon \Omega \to \mathbb{C}$ is holomorphic if and only if $$\int_\gamma f(z)\,dz = 0$$ for any closed curve $\gamma$ in $\Omega$. The forward direction is Cauchy's theorem. As a consequence once obtains the famous integral formula and from that the fact that holomorphic functions are analytic. The reverse direction is a particular case of Morera's theorem, which can be easily proven as follows: observe that the condition on integrals implies $f$ has an antiderivative, which is necessarily holomorphic and therefore analytic. Since $f$ is the derivative of an analytic function it has a derivative itself.<|endoftext|> TITLE: Examples of the large sieve inequality where a constant larger than 1 is needed QUESTION [7 upvotes]: Let $S(x) = \sum_{n=0}^{N-1} a_n e^{2 \pi i n x}$ be a trigonometric polynomial of length $N$. The analytic/harmonic large sieve inequality in its sharpest form states that $$ \sum_{r=1}^R |S(x_r)|^2 \leq (N + \delta^{-1}-1) \int_{0}^{1} |S(x)|^2 dx$$ where $x_1,x_2,\ldots,x_R$ are $\delta$ separated points. This can be thought of as a discretization of Parseval's identity on the circle. In many of the applications of this inequality to number theory one takes $\delta = 1/N$ in which case the right hand side of the inequality has a factor $2N$. The factor of $2$ here leads to some very unfortunate inefficiencies in sieve theoretic applications, such as the $2$ in the Brun-Titchmarsh inequality. It seems that improving the factor of $2$ in these applications is related to both the parity problem in sieve theory and the Siegel zero problem. See, for instance, this paper of Maynard. However it is easy to see in some cases, such as when $x_1,x_2, \ldots, x_R$ are equally spaced (using Fourier analysis on the group of residues mod $N$), that the inequality indeed holds with a constant $1$ in place of the constant $2$. Are there examples of trigonometric polynomials and $\delta = 1/N$ separated points known where the constant in the analytic large sieve inequality is required to be greater than $1$? REPLY [3 votes]: Some comments: 1) Montgomery's survey paper https://projecteuclid.org/euclid.bams/1183540922 page 559, Theorem 3 and equation (19). It essentially says that Selberg gave an example, with the comment that this is the only situation where this is sharp. See also page 563 for lower bounds on $\Delta$, and also upper bounds, when one is sieving modulo some subset of the integers $q\leq Q$. This paper gives links to the rich literature of that time. 2) Similarly, also some comments after Theorem 4.7 of Tenenbaum's Introduction to analytic and probabilistic number theory, (3rd edition). Two more recent items, maybe not so well known: 3) Vandermonde Matrices with Nodes in the Unit Disk and the Large Sieve, C\'eline Aubel and Helmut Bölcskei https://arxiv.org/abs/1701.02538 4) Refinements of Selberg's sieve, Sara Elizabeth Blight, PhD Thesis, Rutgers, 2010. Especially chapter 4 https://rucore.libraries.rutgers.edu/rutgers-lib/27420/<|endoftext|> TITLE: Detecting/Characterising positive elements in free groups QUESTION [9 upvotes]: Let $X$ be a set, and let $F(X)$ be the free group generated by $X$. I will say that an element of $F(X)$ is positive if it is in the monoid generated by all the conjugates in $F(X)$ of every member of $X$. For example, if $X=\left\{a,b\right\}$ then $a, b, a^2 b a^{-1}$ are positive, but $aba^{-1}b^{-1}$ is not positive. Is there any algorithm that can detect, given an element of $F(X)$, whether is it positive? REPLY [6 votes]: Yes, there is an algorithm. This is based on the following simple fact: Any positive element can be reached (but in non-reduced form usually) by only applying the operations right multiplication by a generator $R_g(x)=xg$ and conjugation by a generator $C_g^{\pm}(x)=g^{\pm 1}xg^{\mp 1}$. To see this, just rewrite $$ xax^{-1}yby^{-1} \ldots wpw^{-1}zqz^{-1} = zz^{-1}ww^{-1} \ldots yy^{-1} xax^{-1}yby^{-1} \ldots wpw^{-1}zqz^{-1} $$ (here $a,b, \ldots ,p,q\in X$, while $x,y,\ldots$ are general elements of $F(X)$). Note that this at most doubles the length of our word. Conversely, it's easy to see that any word reached by a combination of these operations will be positive. This gives the following procedure: (1) Given a word $W$, list all non-reduced words of length $\le 2|W|$ that represent the same element; (2) for each such word $W'$, find all $W''$ (if any) with $W'=O(W'')$, with $O$ one of the operations from above; (3) repeat with these new words $W''$ etc. $W$ is positive precisely if the empty word ever shows up in this list. The algorithm terminates because undoing a multiplication or conjugation reduces the length of a word.<|endoftext|> TITLE: Does the forgetful functor $\mathbf{Comp} \rightarrow \mathbf{Top}$ have a left-adjoint? QUESTION [7 upvotes]: The forgetful functor $\mathbf{CompHaus} \rightarrow \mathbf{Top}$ from compact Hausdorff spaces to topological spaces famously has a left-adjoint, the Stone-Cech compactification. Question. Does the forgetful functor $$\mathbf{Comp} \rightarrow \mathbf{Top}$$ from compact spaces to topological spaces have a left-adjoint? This should be well-known, but I haven't been able to find any relevant information. REPLY [14 votes]: No, it does not. If it did, then $\mathbf{Comp}$ would be a reflective subcategory of the total category $\mathbf{Top}$, and hence would be total itself. Now, total categories admit all small limits, and there is the rub: $\mathbf{Comp}$ does not have all small limits. In fact, $\mathbf{Comp}$ does not have equalizers (of course it admits small products, by the Tychonoff theorem). Here is a simple example to show why not. Consider Sierpinski space $\mathbf{2} = \{0, 1\}$ where the point $1$ is open. This is of course compact. Take two continuous maps $f, g: [0, 1] \rightrightarrows \mathbf{2}$ where $f^{-1}(1) = [0, 1/2)$ and $g^{-1}(1) = [0, 1]$. Suppose there is an equalizer $i: E \to [0, 1]$ of $f, g$ in $\mathbf{Comp}$. Since $\hom(1, -): \mathbf{Comp} \to \mathbf{Set}$ would preserve this equalizer, we see the underlying set of $E$ would have to be $[0, 1/2)$. In order for $i: E \to [0, 1]$ to be continuous, the topology on $E = [0, 1/2)$ would have to be equal to or finer than the subspace topology on $[0, 1/2)$. But since the subspace topology on $[0, 1/2)$ is not compact, no finer topology on $[0, 1/2)$ can be either; contradiction.<|endoftext|> TITLE: The axiom of choice as a consequence of a stronger semantics? QUESTION [16 upvotes]: I've never had a problem with the axiom of choice, but it has often confused me how many authors find full choice so much different from finite choice. In my head they seem quite similar. We are picking elements in either case--why treat the infinite case as fundamentally different from the finite one? [Note: This is a rhetorical question, which I answer later, to motivate the remainder of the post.] Their answer was to emphasize that finite choice was provable in ZF, but the full axiom of choice is independent. Okay, so then I'd delve into the proofs of finite choice, and get even more confused. They'd start with the trivial claim that an empty set has a choice function, the empty function. So far so good. Then they'd work by induction, sometimes skipping all the details. My problem was that this was exactly the point where I (and apparently others, see this previous question) got confused. What axiom of set theory allows someone to pick a single element from a single non-empty set? The answer to this question is probably obvious to most logicians, but it took me a lot of delving to find it. The answer is: It isn't an axiom of set theory that allows this choice. It is the principle of "existential elimination" from first order logic. In other words, singleton choice is a meta-logical assumption. It says that we interpret the sentence $$ \exists x(x\in y) $$ as asserting the ability to pick (non-constructively) an element $x$ inside $y$. [I'm being a bit informal here, but hopefully you get the idea.] So in my mind that begs the question of why our semantics does not treat the sentence $$ \forall y \exists x (x\in y) $$ as asserting the ability to pick (non-constructively) for each $y$ an element $x_y\in y$. [The subscript here is merely to emphasize that the choice of $x$ depends on $y$.] Thus my question: Is there a natural semantics/language where the sentence $\forall y \exists x\ P(x,y)$ is interpreted as the ability to pick for each $y$ some $x_y$ such that $P(x_y,y)$ holds? It seems like this is somewhat strengthening the usual interpretation of $\forall$, by allowing infinitely many operations (one for each $y$) simultaneously. Edited to add: Motivated by Goldstern's comment below, an alternate way to frame my question might be: Is ZF+GC (Zermelo-Fraenkel set theory with global choice) a sufficiently strong theory in which the semantics I proposed above is sound? REPLY [16 votes]: I like this question very much. I find the difference between the two readings of $\forall x\exists y\ P(x,y)$ to be similar to the issues often brought up by questions of uniformity in mathematical existence assertions. So this may be a good keyword leading to further discussion. For example, the discussion of uniformity in my answer to the question Can a problem be simultaneously NP and undecidable?. To explain a little, in the context of computability theory, say, in the ordinary mathematical reading of $\forall x\exists y\ P(x,y)$, there is no reason to think that there should be a computable function mapping each $x$ to such a $y$. But if we assert that the existence claim is uniform, then this means something much closer to the semantics about which you inquire, namely, that there should be a computable function allowing us to pass from $x$ to a witness $y$. In ZFC set theory, the axiom of choice can be interpreted as the assertion that all existence claims of the form $\forall x\in u\exists y\ P(x,y)$ are uniform, so that there is a function $f$ with domain $u$ so that $P(x,f(x))$ for each $y\in u$. In class theories such as KM, the global axiom of choice is the assertion that all assertions of the form $\forall x\exists y\ P(x,y)$ are uniform, so that there is a class function $F$ with $P(x,F(x))$ for all $x$. Since the issue of the uniformity of existence assertions is often central in many mathematical domains, I propose that this is the answer to your question.<|endoftext|> TITLE: Kneser graph with overlap QUESTION [7 upvotes]: Consider a graph with the vertices being all subsets of size $n$ of a set of size $2n$. Two vertices are connected if their overlap has size at most one. What is the chromatic number of this graph? If we have two vertices connected when the overlap is empty, then this corresponds to the Kneser graph, for which we know the chromatic number for any $n,k$. In this special case $k=n/2$, the chromatic number of the Kneser graph is trivially two. In our case, we have the generalized Kneser graph $KG(2n,n,1)$. It seems that the chromatic number is always $6$ for any $n\geq 2$. Can we prove this, or does it follow from some known result? As domotorp wrote in the comments, the graph is also called the discrete Borsuk graph. Is something known about its chromatic number? REPLY [2 votes]: The graph you are interested in is the generalized Kneser graph $KG(2n,n,1)$. The generalized Kneser graph $KG(n,r,s)$ has the same vertex set as a Kneser graph $KG(n,r)$ but with two vertices joined by an edge if their intersection has no more than $s$ elements. A web search for chromatic numbers of such graphs generates many hits including a paper of Frankl and one of Tort in which the chromatic number is computed when $s = 1$ under some restrictions on $n$.<|endoftext|> TITLE: Sequential addition of points on a circle, optimizing asymptotic packing radius QUESTION [11 upvotes]: Suppose I have to put $N$ points $x_1, x_2, \ldots, x_N$ on the circle $S^1$ of length 1 so as to achieve the largest minimum separation (packing radius). The optimal solution is the equally spaced arrangement $x_k = k/N \bmod 1$, which achieves $\min_{i0$, write $n = 2^q+r$ with $0 \leq r <2^q$ and put $x_n$ at $\log_2\left(\tfrac{2n+1}{2^{q+1}}\right)$. So the first few points are $0$, $\log_2(1+1/2)$, $\log_2(1+1/4)$, $\log_2(1+3/4)$, $\log_2(1+1/8)$, $\log_2(1+3/8)$, $\log_2(1+5/8)$, $\log_2(1+7/8)$, ... When point $x_n$ is inserted, the smallest interval will either be the one just to the right of $x_n$ or else the furthest right interval. Approximating $\log$ by its linearization, these have lengths roughly $\tfrac{2^{q+1}}{(2n+1) \log 2} \tfrac{1}{2^{q+1}}=\tfrac{1}{(2n+1) \log 2}$ and $\tfrac{1}{2 \log 2} \tfrac{1}{2^q}=\tfrac{1}{2^{q+1} \log 2}$ respectively. Multiplying these by $n+1$ (the number of intervals) gives $\tfrac{n+1}{(2n+1) \log 2}$ and $\tfrac{n+1}{2^{q+1} \log 2}$. The former approaches $\tfrac{1}{\log 4}$ while the latter oscillates between $\tfrac{1}{\log 2}$ and $\tfrac{1}{\log 4}$.<|endoftext|> TITLE: Two set of axioms for Stiefel-Whitney classes QUESTION [5 upvotes]: Let $E \to X$ be a vector bundle. We can associate to $E$ several invariants: among them are the Stiefel-Whitney classes $w_i(E) \in H^i(X;\mathbb{Z}_2)$. These classes may be defined using the axioms: 0. $w_0(E)=1$ and $w_i(E) \in H^i(X;\mathbb{Z}_2)$. 1. $w(f^*E)=f^*w(E)$ for continuous maps $f$ (here $w=1+w_1+w_2+\cdots$ is the total class) 2. $w(E \oplus F)=w(E) \cup w(F)$ 3. $w_1(\gamma_1) \neq 0$ for the tautological bundle $\gamma_1$ over $\mathbb{R}P^{\infty}=BO(1)$ In particular, we can rephrase these axioms to recognize the first Stiefel Whitney class: 1'. $w_1(f^*E)=f^*w_1(E)$ for continuous maps $f$ 2'. $w_1(E \oplus F)=w_1(E) + w_1(F)$ 3'. $w_1(\gamma_1) \neq 0$ for a tautological bundle $\gamma_1$ over $\mathbb{R}\mathbb{P}^{\infty}=BO(1)$. However, in Lawson-Michelsohn's book "Spin Geometry" it is stated that in order for a cohomology class $v_1$ to equal $w_1$ we only need to check: 1''. $v_1(f^*E)=f^*v_1(E)$ for continuous maps $f$ 2''. $v_1(\gamma_n) \neq 0$ for the tautological bundle $\gamma_n$ over $BO(n)$ for every natural number $n$. How do we prove that these two sets of axioms are equivalent (and therefore characterize $w_1$)? Concerning higher Stiefel-Whitney classes, does a set of axioms like 1''-2'' define $w_k$ for each $k$? REPLY [6 votes]: The naturality condition 1'' reduces the question of whether $v_1=w_1$ to the special case of the tautological bundles $\gamma_n$. In this case both $v_1$ and $w_1$ lie in $H^1(BO(n);{\mathbb Z}_2)$. This group is just ${\mathbb Z}_2$, as shown in the book of Milnor and Stasheff for example, or just from the fact that $\pi_1BO(n)=\pi_0O(n)={\mathbb Z}_2$. So if $v_1(\gamma_n)$ is nonzero it must equal the nonzero class $w_1(\gamma_n)$. This argument does not work for $w_2$ since $H^2(BO(n);{\mathbb Z}_2)= {\mathbb Z}_2 \times {\mathbb Z}_2 $ with basis $w_2$ and $w_1^2$, assuming $n>1$. Thus to characterize $w_2$ one needs to know more than just that $w_2(\gamma_n)$ is nonzero since $w_1^2$ also has this property, as does $w_1^2+w_2$. On the other hand, if one considers only oriented vector bundles then $H^2(BSO(n);{\mathbb Z}_2)={\mathbb Z}_2$ when $n>2$ so in this situation $w_2$ is characterized by naturality and being nonzero for the tautological bundle over $BSO(n)$.<|endoftext|> TITLE: A Collatz-like problem on prime numbers QUESTION [30 upvotes]: Consider the function $f$ on the prime numbers defined by $$ f(p):= \text{ the greatest prime factor of } 2p+1.$$ The iteration of $f$ from any prime $p<10^8$ converges to the cycle $$(3,7,5,11,23,47,19,13)$$ Question: Is it true for any prime $p$? Let $\ell(p)$ be the number of iterations needed to join the cycle from $p$. The prime number $p$ will be called champion if for any prime $q p:=1;; bb:=0;; while p<100000000 do p:=NextPrimeInt(p); a:=p; b:=0; while a<>3 and a<>7 and a<>5 and a<>11 and a<>23 and a<>47 and a<>19 and a<>13 do b:=b+1; L:=PrimeDivisors(2*a+1); l:=Length(L); a:=L[l]; od; if b>bb then Print([p,b]); bb:=b; fi; od; [ 2, 1 ][ 29, 3 ][ 41, 4 ][ 53, 6 ][ 79, 7 ][ 89, 9 ][ 311, 10 ][ 1223, 11 ][ 1889, 12 ][ 2833, 13 ][ 3821, 14 ][ 18149, 16 ][ 63521, 17 ][ 222323, 18 ][ 779111, 19 ][ 2167289, 20 ][ 7585511, 21 ][ 19487999, 23 ] We can compute a prime $p$ such that $\ell(p) = r$ (see below for $r=30$). gap> p:=17;; r:=30;; while r>0 do q:=3;; while not IsPrime((q*p-1)/2) do q:=NextPrimeInt(q); od; r:=r-1; Print([q,p]); p:=(q*p-1)/2; od; [ 7, 17 ][ 13, 59 ][ 61, 383 ][ 7, 11681 ][ 13, 40883 ][ 373, 265739 ][ 61, 49560323 ][ 13, 1511589851 ][ 157, 9825334031 ][ 199, 771288721433 ][ 13, 76743227782583 ][ 31, 498830980586789 ][ 1423, 7731880199095229 ][ 163, 5501232761656255433 ][ 823, 448350470074984817789 ][ 79, 184496218435856252520173 ][ 1171, 7287600628216321974546833 ][ 1663, 4266890167820656516097170721 ][ 193, 3547919174542875893134797454511 ][ 733, 342374200343387523687507954360311 ][ 1627, 125480144425851527431471665273053981 ][ 61, 102078097490430217565502199699629413543 ][ 127, 3113381973458121635747817090838697113061 ][ 163, 197699755314590723869986385268257266679373 ][ 277, 16112530058139143995403890399362967234368899 ][ 2437, 2231585413052271443363438820311770961960092511 ][ 2731, 2719186825804192753738350202549892917148372724653 ][ 193, 3713049610635625205229717201581878778366102955513671 ][ 1453, 358309287426337832304667709952651302112328935207069251 ][ 1609, 260311697315234435169341091280601170984606971427935810851 ] We can generalize the problem to any function $f_k$, where $kp+1$ replaces $2p+1$ above. For $k=3$, it is the cycle $(2,7,11,17,13,5)$. For $k=4$, it is $(5,7,29,13,53,71,19,11)$. For $k=5$, it is $(2,11,7,3)$. For $k=6$, there are two cycles:$(47,283,1699,2039,2447,14683,8009,1373,107,643,227)$ and $(13,19,23,139,167,59,71,61,367,2203,13219,547,67,31,17,103,619,743)$. For $k=7$, it is $(3,11,13,23)$. For $k=8$, it is $(11,89,31,83,19, 17,137,1097,131,1049,109,97,37)$. For $k=9$, two cycles: $(13,59,19,43,97,23)$ and $(37,167,47,53,239,269,173,41)$. Everything is checked for $p<10^6$. Bonus question 1: Does the iteration of $f_k$ converges to finitely many possible cycles, for any $k \ge 2$? We can generalize the problem to any polynomial function $f \in \mathbb{N}[X]$ splitting on $\mathbb{Q}$ with $f(0)=1$. For $f(p)=(p+1)(2p+1)$, it is the cycle $(5,11,23,47,19,13,7)$. For $f(p)=(2p+1)(3p+1)$, it is the cycle $(31,563,23,47,71,107,43,29,59,89,179,359,719,1439,2879,617,463,139)$. For $f(p)=(3p+1)(4p+1)(5p+1)$, it is the cycle $( 71, 107, 67, 269, 673, 2693, 6733, 1171, 937, 163, 653)$ Everything is checked for $p<10^6$. Bonus question 2: Can we extend to that case? It seems that it can't be extended to the non splitting case. For $f(p)=p^2+1$ and from $p=2$, we get the (probable) sequence: 2, 5, 13, 17, 29, 421, 401, 53, 281, 3037, 70949, 1713329, 1467748131121, 37142837524296348426149, 101591133424866642486477019709, 1650979973845742266714536305651329, 78343914631785958284737, 4029445531112797145738746391569, 350080544438648120162733678636001, 26208090024628793745288451837610346882122253572537, ... For $f(p)=p^2+3p+1$ and from $p=2$, we get the (probable) sequence: 2, 11, 31, 211, 821, 135301, 3809941, 742299251, 2894402701, 11096115237403051, 13495491562451, 5906592644484061, 3006276317783130610918295261, 680868245636686686301066879953955425558991, 859331554798594732550606265780004082746150814706504421, 13431381921273506538508334090334652350875716299550588398947479075941548746770801901,... Bonus question 3: Is it true that for any polynomial function $f \in \mathbb{N}[X]$ with $f(0)=1$ and without rational root, a sequence starting from any prime number $p$ never reach a cycle? Note that we have seen a class of polynomials for which we expect the convergence to finitely many possible cycles, and a class for which we expect no cycle at all. We are wondering about an intermediate case: Is there a polynomial with a convergence to infinitely many cycles? REPLY [3 votes]: On bonus question 3, for $f_k(p)=kp^2+1$, I have found several values for $k$ which have a 2-cycle. I can't show that all $p$ will fall into these cycles or that these are the only cycles. Working backwards from these cycles, we can find which numbers do fall into them. Edit update: In addition to the examples below, I google searched looking for other information, and here someone found for $k=1$ we have a 5-cycle $(19121, 10753313, 1241761, 3817193, 107837)$. This confirms part of my speculation that my computer search was limited to the computational low-hanging fruit. /Edit For $k=1$ we have a cycle $(89,233)$; $k=5$ has cycle $(43,67$); $k=11$ has cycle $(23,97)$; $k=23$ has cycle $(3,13)$; $k=41$ has cycle $(67,409)$; $k=71$ has cycle $(5,37)$; $k=85$ has cycle $(383,769)$; $k=113$ has cycle $(509,1021)$; $k=143$ has cycle $(7,73)$; $k=215$ has cycle $(257, 467)$; $k=311$ has cycle $(67,277)$; $k=359$ has cycle $(11,181)$; $k=863$ has cycle $(17,433)$; $k=951$ has cycle $(67,73)$; $k=1238$ has cycle $(13,41$); $k=1427$ has cycle $(167,293)$; $k=1808$ has cycle $(107,271)$; $k=1811$ has cycle $(17,61)$; $2089$ has cycle $(31,241)$; $k=2501$ has cycle $(167,251)$; $k=4199$ has cycle $(59,101)$; $k=5903$ has cycle $(61,151)$; $k=6119$ has cycle $(61,179)$; $k=6269$ has cycle $(29,67)$; $k=7679$ has cycle $(109,151)$; $k=9407$ has cycle $(7,97)$. I found one 3-cycle at $k=7$ which goes $(79,127,1283)$. As well, for $g_k(p)=kp^2-1$, aside from perfect square $k's$, I found $k=11$ has cycle $(239,3307)$; $k=19$ has cycle $(157,233)$; $k=23$ has cycle $(929,2711)$; $k=61$ has cycle $(127,503)$; $k=103$ has cycle $(857,1283)$; $k=1639$ has cycle $(127,197)$. $k=123$ has a 3-cycle $(443,449,823)$; $k=409$ also has a 3-cycle $(71,317,137)$. At first glance I notice how many of the $k's$ are prime, but not all. Now, why so many 2-cycles? I think it is because of my computational limitations since the numbers get large quickly. I found these using unsigned long integer data type; I predict arbitrary precision calculations would uncover larger cycles. For $k=23$, we have $23*3^2+1=2^4*13$, and $23*13^2+1=2^4*3^5$, and you can see that these large numbers must be very composite to have such low greatest prime factors. Another pleasing example is $k=143$; $143*73^2+1=2^6*3^5*7^2$. That explains why these are so rare.<|endoftext|> TITLE: How is a universal deformation ring in a 1-dimensional DM stack related to the complete etale local ring of coarse moduli scheme? QUESTION [5 upvotes]: Let $\mathcal{M}$ be a smooth 1-dimensional Deligne-Mumford stack with finite diagonal, and let $M$ be its coarse moduli scheme (all over some field $k$). Suppose $M$ is also smooth over $k$, and that all local deformation problems for $\mathcal{M}$ are pro-representable (I think this is already implied by the smooth + DM conditions?) Let $x_0 : \text{Spec }\bar{k}\rightarrow\mathcal{M}$ a geometric point, and let $\bar{x_0}\in M$ be its image in $M$. Let $\mathcal{R}$ be the completion of the etale local ring at $x_0$ (colimit of global sections of scheme etale neighborhoods of $x_0$), and let $R$ be the completion of the etale local ring of $M$ at $\bar{x_0}$. This yields a natural map $R\rightarrow\mathcal{R}$ (both are isomorphic to a 1-variable power series ring over $\bar{k}$), whence a map $$p : \text{Spec }\mathcal{R}\rightarrow\text{Spec }R$$ I believe the infinitesimal lifting property of etale morphisms then implies that $\mathcal{R}$ is also the universal deformation ring of the object corresponding to $x_0$ (am I missing any conditions here?). Let $X_0/\bar{k}$ be the object corresponding to $x_0$, and let $X^\text{univ}/\mathcal{R}$ be the universal deformation of $X_0$ over $\mathcal{R}$. I would like to say something like: $p$ is finite flat totally ramified with ramification index equal the order of the group of automorphisms of $X_0$ modulo the subgroup of those automorphisms which extend to an automorphism of $X^\text{univ}/\mathcal{R}$. Is this true? What is a good reference? If it's false, what is the relation between the universal deformation ring of $x_0$ and the complete etale local ring of $x_0$? Actually, some calculations of mine seem to indicate that may be false, though it's possible I've made a mistake somewhere else... REPLY [2 votes]: I believe that this is true - that is, "$p$ is finite flat totally ramified with ramification index equal to the order of the group of automorphisms of $X_0$ modulo the subgroup of those which extend to an automorphism of $X^\text{univ}/\mathcal{R}$." Specifically, let $G$ be the automorphism group of the object $x_0$. Let $\mathcal{R}$ be the etale local ring of $\mathcal{M}$ at $x_0$. Then there is a natural action of $G$ on $\mathcal{R}$. Let $R$ be the etale local ring of $\overline{x_0}\in M$. Let $\mathcal{M}_{(x_0)} := \mathcal{M}\times_M\text{Spec }R$. From the proof of Theorem 11.3.1 of Olsson's book "Algebraic Spaces and Stacks", we find that $$\mathcal{M}_{(x_0)} \cong [\text{Spec }\mathcal{R}/G]$$ and moreover the composition $$\text{Spec }\mathcal{R}\rightarrow[\text{Spec }\mathcal{R}/G]\rightarrow \text{Spec }R$$ is finite. Since the map $\text{Spec }R\rightarrow M$ is flat, the projection $[\mathcal{R}/G]\cong\mathcal{X}_{(x_0)}\rightarrow\text{Spec }R$ identifies the target with the coarse moduli scheme of $[\mathcal{R}/G]$ (Theorem 11.1.2 of the same book). Thus, we have $R = \mathcal{R}/G = \mathcal{R}^G$. Since $M,\mathcal{M}$ are smooth and 1-dimensional, $R,\mathcal{R}$ are discrete valuation rings, and the map $R\rightarrow\mathcal{R}$ being a finite map between regular schemes must be flat (see Katz-Mazur's book, end of p507). At this point standard ramification theory tells us that if $K$ is the kernel of the action of $G$ on $\mathcal{R}$, then since we're working with strictly henselian DVRs, the extension $\mathcal{R}/R$ is totally ramified with ramification index $|G/K|$. Let $X_0$ be the object over $\overline{k}$ corresponding to $x_0$, and let $X/\mathcal{R}$ be the object corresponding to the canonical map $\text{Spec }\mathcal{R}\rightarrow\mathcal{M}$. It remains to show that $K$ is precisely the subgroup of automorphisms of $X_0$ which extend to automorphisms of $X^\text{univ}/\mathcal{R}$. For this, we use the interpretation of $\mathcal{R}$ as the universal deformation ring which pro-represents the deformation functor $F_{X_0}$ which to every artinian local $\overline{k}$-algebra $A$ associates the set of isomorphism classes $$\{(X/A,\;\;\varphi : X_{\overline{k}}\cong X_0)\}/\cong$$ Then group $G$ acts naturally on these sets by acting on the isomorphism $\varphi$, ie $g\in G$ sends $(X,\varphi)$ to $(X,g\circ\varphi)$, and hence $G$ acts naturally on the functor $F_{X_0}$, and hence on the prorepresenting object $\mathcal{R}$. From this it is clear that an element $g\in G$ fixes $\mathcal{R}$ if and only if for every deformation $X/A$ over Artinian local $\overline{k}$-algebras $A$, there exists an extension of $g$ to an automorphism of $X/A$. Since $\mathcal{M}$ is assumed Deligne-Mumford, the diagonal is unramified (hence the automorphism group schemes of objects are unramified - in particular, formally unramified), so such extensions of $g$, if they exist, are unique. This implies that whenever $g$ extends to every $X/A$, the various extensions are compatible with each other, and hence gives an automorphism of $X^\text{univ}/\mathcal{R}$. Conversely, it's clear that automorphism of $X^\text{univ}/\mathcal{R}$ restricts to an automorphism $g\in G$ of $X_0$ which acts trivially on the deformation functor, and hence on $\mathcal{R}$.<|endoftext|> TITLE: Positivity of coefficients of a polynomial derived from Schubert polynomials QUESTION [5 upvotes]: Let $W=\bigcup_{n=1}^\infty S_n$ be the union of all symmetric groups $S_n$. For an element $w\in W$, denote by $\mathfrak{S}_w$ the Schubert polynomial associated to $w$, and by $\partial_w$ the divided difference operator associated to $w$. Question: Let $u,v\in W$. Why is $$ \partial_u(\mathfrak{S}_u\mathfrak{S}_v)-\mathfrak{S}_v $$ always a positive polynomial, in the sense that all coefficients are non-negative integers? Preliminary thoughts: It is clear that $\partial_u(\mathfrak{S}_u\mathfrak{S}_v)$ is positive. One may use the twisted Leipniz rule and induction on the length of $u$ to reduce the problem to an apparently simpler one. Indeed, let $s_i$ be the rightmost factor in a reduced expression of $u$. Then it suffices to show that $\partial_{us_i}(s_i\mathfrak{S}_u\partial_i\mathfrak{S}_v)$ is positive. (Here, $W$ acts on $\mathbb{Z}[x_1,x_2,x_3,\ldots]$ by permuting the variables.) I was thinking to realize the polynomials in question as characters of representations, in the spirit of the work of M. Watanabe on KP modules, but had not much success. Any help would be greatly appreciated. I computed many examples and cannot find a counter example. In fact, I have a stronger "positivity conjecture" which implies this one, and even to this stronger conjecture, I was not able to find neither a counterexample nor a proof. REPLY [2 votes]: You might want to try to prove Conjecture 1 in Kirillov's paper when the length difference between w and v is 2, starting from Liu's formula for skew divided difference operators in terms of /partial_{ij}.<|endoftext|> TITLE: Is there a simple proof of the following Identity for $\sum_{k=m-1}^l(-1)^{k+m}\frac{k+2}{k+1}{\binom l k}\binom{k+1}m$? QUESTION [7 upvotes]: While studying the behaviour of umbilic points on Weingarten surfaces I discovered that the following combinatorial identity must be true. For all $l,m\in{\mathbb N}$ with $l\geq m-1\geq0$ the following holds: $ \sum_{k=m-1}^l(-1)^{k+m}\frac{k+2}{k+1}{l \choose k}{k+1 \choose m}= \left\{\begin{array}{ccl} 0&if& l>m\\ 1&if& l=m\\ -1-\frac{1}{l+1} &if& l=m-1 \end{array} \right. $ Improbable as it at first appears, it is easy to check the second and third options are true, and I have computer-checked and found it is true for all $l,m\leq 60$. Perhaps it is well-known or has an easy proof. Either would be good to know. REPLY [6 votes]: We obtain for $l,m\in\mathbb{N}$ with $0\leq m-1 \leq l$: \begin{align*} \color{blue}{\sum_{k=m-1}^{l}}&\color{blue}{(-1)^{k+m}\frac{k+2}{k+1}\binom{l}{k}\binom{k+1}{m}}\\ &=\frac{1}{m}\sum_{k=m-1}^l(-1)^{k+m}(k+2)\binom{l}{k}\binom{k}{m-1}\tag{1}\\ &=\frac{1}{m}\binom{l}{m-1}\sum_{k=m-1}^l(-1)^{k+m}(k+2)\binom{l-m+1}{k-m+1}\tag{2}\\ &=\frac{1}{m}\binom{l}{m-1}\sum_{k=0}^{l-m+1}(-1)^{k+1}(k+m+1)\binom{l-m+1}{k}\tag{3}\\ &=\frac{m+1}{m}\binom{l}{m-1}\sum_{k=0}^{l-m+1}(-1)^{k+1}\binom{l-m+1}{k}\\ &\qquad+\frac{l-m+1}{m}\binom{l}{m-1}\sum_{k=1}^{l-m+1}(-1)^{k+1}\binom{l-m}{k-1}\tag{4}\\ &=-\frac{m}{m+1}\binom{l}{m-1}[[l=m-1]]\\ &\qquad+\frac{l-m+1}{m}\binom{l}{m-1}\sum_{k=0}^{l-m}(-1)^k\binom{l-m}{k}\tag{5}\\ &=\left(-1-\frac{1}{m}\right)[[l=m-1]]+\frac{l-m+1}{m}\binom{l}{m-1}[[l=m]]\tag{6}\\ &\color{blue}{=\left(-1-\frac{1}{m}\right)[[l=m-1]]+1[[l=m]]} \end{align*} and the claim follows. Comment: In (1) we use the binomial identity $$\binom{p+1}{q+1}=\frac{p+1}{q+1}\binom{p}{q}$$ In (2) we use the binomial identity $$\binom{p}{q}\binom{q}{r}=\binom{p}{r}\binom{p-r}{q-r}$$ In (3) we shift the index of the sum to start with $k=0$. In (4) we split the sum and work similarly as in (1). In (5) we do some simplifications regarding $(1-1)^{l-m+1}$ using Iverson brackets. We also shift the index of the sum again. In (6) we do a similar job as in (5).<|endoftext|> TITLE: Monoidal tensor product which preserves directed limits QUESTION [8 upvotes]: Given a symmetric monoidal category $Q$, is there a construction of a (preferably full and faithful strong) monoidal embedding of $Q$ into some symmetric monoidal closed category $M$ which has all filtered colimits for which the functor $A\otimes -$ preserves cofiltered limits (i.e., limits with a directed poset as shape, I do not mean colimits here) for each fixed $A\in M$? It would be nice if $M$ is both complete and cocomplete, but not essential. However, the existence of filtered colimits and the monoidal closure of $M$ are essential. A possible solution might be considering the Yoneda embedding of $Q$ into $M:=[Q^\mathrm{op},\mathbf{Set}]$, which becomes symmetric monoidal closed when equipped with a tensor product via the Day Convolution. Could it be that this tensor product preserves cofiltered limits? The Day convolution can be defined in terms of a colimit, which does not commute with limits in general. But could we have an escape route, since we require the limits to have a special shape, namely of a directed poset? Alternatively, is there some other construction of an embedding into a symmetric monoidal closed category $M$ where the functor $A\otimes -$ for fixed $A\in M$ has a left adjoint, hence preserves all limits? The reason why we are interested in this question is because we aim to model circuit programming languages, where the circuits are represented by morphisms in the category $Q$. In principle, we would like to have a construction for arbitrary symmetric monoidal categories $Q$, but maybe there is some construction that only works if we impose some extra conditions on $Q$? A specific choice for $Q$ might be a free symmetric monoidal category over a finite monoidal category/signature. Another specific choice for $Q$ might be the category of finite-dimensional C*-algebras with completely positive subunital maps. But a construction for an arbitrary monoidal categories $Q$ would be preferable. REPLY [5 votes]: Public Service Announcement! It's very confusing -- I'd daresay incorrect -- to say "directed limit" to mean "limit indexed by a cofiltered diagram". Actually, historically the term "directed limit" has been used to mean "colimit indexed by a filtered diagram" or even just "colimit"-- this usage predates the introduction of the terms "limit" and "colimit", I believe. The term you're looking for is "inverse limit", or if you want to be 100% clear, say "cofiltered limit" (because "inverse limit" has sometimes been used to just mean "limit"). Answer to Question: Now to your question. As mentioned in the comments, if $C$ is a symmetric monoidal category then the Day convolution is a universal way to turn $C$ into a cocomplete category with a symmetric monoidal product that preserves colimits in each variable. It's straightforward to check that the Day convolution restricts to a symmetric monoidal structure on the Ind-completion of $C$, denoted $Ind(C)$, with $\otimes$ preserving filtered colimits in each variable. Here $Ind(C) \subset [C^\mathrm{op},\mathsf{Set}]$ is the full subcategory of presheaves which are filtered colimits of representables. It is the universal category with filtered colimits generated by $C$, and so the Day convolution makes it the universal way to turn $C$ into a symmetric monoidal category with filtered colimits preserved in each variable by $\otimes$. To get a similar universal construction for cofiltered limits, just dualize everything. The dual Day convolution is a symmetric monoidal structure on $[C,\mathsf{Set}]^\mathrm{op}$ preserving limits in each variable, and restricts to a symmetric monoidal structure the Pro-completion of $C$, $Pro(C) = Ind(C^\mathrm{op})^\mathrm{op} \subset [C,\mathsf{Set}]^\mathrm{op}$, which preserves cofiltered limits in each variable. $Pro(C)$ is the universal category with cofiltered limits on $C$, and it consists of those copresheaves on $C$ which are filtered colimits of representables -- with morphisms being natural transformations in the opposite direction. Caveat emptor: The $Ind$ and $Pro$ completions really are universal. In particular, the inclusion $C \to Ind(C)$ does not preserve filtered colimits that might happen to exist in $C$ / dually the inclusion $C \to Pro(C)$ does not preserve cofiltered limits that might happen to exist in $C$. So if you are interested in the cofiltered limits that already exist in $C$, then $Pro(C)$ will be unsatisfactory for you because the Pro construction simply ignores all existing cofiltered limits and freely adds in new ones. (This is just like how if $G$ is a group, then the free group $F(G)$ on the underlying set of $G$ will completely ignore the existing multiplication on $G$.) However, if $C$ already has cofiltered limits, then the inclusion $C \to Pro(C)$ will have a right adjoint which takes an object of $Pro(C)$, which you can think of as a formal cofiltered limit, and evaluates that limit in $C$ (in the group analogy, if $G$ is a group then there is a canonical group homomorphism $F(G) \to G$ which multiplies together formal words in the group language over $G$). Then you can start asking questions like how this functor interacts with the Day convolution. Of course, if you don't care about cofiltered limits that might happen to be present in $C$, then $Pro(C)$ is probably exactly what you want. It's nice to know, though, that $C \to Ind(C)$ does preserve finite colimits / $C \to Pro(C)$ does preserve finite limits. Supplementary note: If you're not familiar with the $Ind$ and $Pro$ completions, I should probably tell you that $Ind(C)$ can alternately be described as follows. An object consists of a filtered category $I$ and a diagram $X: I \to C$. The homsets are $Hom_{Ind(C)}(X: I \to C, Y: J \to C) = \varprojlim_{i \in I} \varinjlim_{j \in J} Hom_C(X_i,Y_j)$. Dually, an object in $Pro(C)$ consists of a cofiltered category $P$ and a diagram $Z: P \to C$. The homsets are $Hom_{Pro(C)}(Z: P \to C, W: Q \to C) = \varprojlim_{q \in Q} \varinjlim_{p \in P} Hom_C(Z_p, W_q)$. This is a quite explicit sense in which $Ind(C)$ consists of "formal filtered colimits" / $Pro(C)$ consists of "formal cofiltered limits" -- the objects literally are the diagrams you want to take the (co)limit of!<|endoftext|> TITLE: Generating the topology of a manifold QUESTION [5 upvotes]: Let $X$ be a topological manifold of dimension $d$, and let $F$ be a collection of continuous maps from $X$ into $\mathbf{R}^d$ such that: $F$ separates points of $X$, i.e. for any two distinct points $x,y\in X$ there is $\varphi\in F$ such that $\varphi(x)\ne\varphi(y)$; for every $x\in X$ there is an open neighbourhood $U$ of $x$ and $\varphi\in F$ such that $\varphi|_U$ is a homeomorphism of $U$ onto an open set $W$ in $\mathbf{R}^d$. Does it follow that $F$ generates the topology of $X$, i.e. the original topology of $X$ is the minimal topology in which all elements of $F$ are continuous? Since $F$ consists of continuous maps, it generates the topology weaker than the original one, and since $F$ separates points, the generated topology is Hausdorff. It seems that this topology is also locally Euclidean, but I cannot articulate the right reasoning. Then we get that the identity is a continuous bijection between the original and generated topologies, and since they both are locally Euclidean it follows that this map is open, and therefore a homeomorphism. However, I feel concerned about the very last argument, since it does not use Hausdorff property, without which the statement is wrong. PS You may assume that all the objects considered are actually smooth or holomorphic, but I don't think it is relevant here. REPLY [2 votes]: This is just a variation of Corbennick's comment. The statement in the question is wrong for any non-compact manifold: take $x\in X$ and consider all maps into $\mathbf{R}^{d}$, such that their limit at infinity equals their value at $x$. This collection generates the topology of the one point compactification of $X$ factorized by gluing $\infty$ with $x$, which is not the original one. One has to tweak this argument for the holomorphic case, however, because holomorphic functions cannot have limit at infinity unless they are constants. Let $X=D$ be the unit disk and let $F'$ be the set of all functions $f$ on $\overline{D}$ which are continuous on $\overline{D}$, holomorphic on $D$ and such that $f(0)=f(1)$. Let $F$ be the set of restrictions of the members of $F'$ on $D$. Since $F$ contains functions $z^2-z$ and $z^3-z$, it follows that at every point there is a local biholomorphism. However, $F$ generates the topology of the "folded" disk, not equivalent to the original one.<|endoftext|> TITLE: Intersections of quadratic planes as elliptic curves QUESTION [5 upvotes]: An elliptic curve defined over a field $k$ is a smooth projective curve of genus $1$, plus a $k$-rational point. Every elliptic curve can be written in a Weierstrass form, i.e. as a plane cubic curve of the form $y^2 + a_1 x y + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$. Now let $A$ and $B$ be two symmetric $4 \times 4$ matrices. Consider the curve $C$ in $\mathbb{P}^3$ defined by: $^tx A x = 0$, $^tx B x = 0$. If the homogeneous polynomial $\det(\lambda A + \mu B) \in k[\lambda, \mu]$ does not have multiple fators, then the curve $C$ is of genus $1$. Thus if there is a rational point on $C$, then one can define an elliptic curve. My main question is: how can one deduce a Weierstrass form for such a curve? By "how", I mean an algorithm which accepts $A$ and $B$ as inputs and gives the $a_i$'s as outputs. For other models such as cubic plane curves, or hyperelliptic curves of degree $4$, I am aware of such algorithms. But I cannot find reference for this intersection model. A secondary question is: is it possible to do "computations" directly on the curve $C$? For example, if an elliptic curve is given as a cubic plane curve, i.e. by a homogeneous polynomial of degree $3$: $F(X, Y, Z) = 0$, then the addition law can be described easily by intersecting projective lines with the curve: if $O$ is the zero point and $P$, $Q$ are two points on the curve, then one can find the sum $P + Q$ as follows: intersect the line $PQ$ with the curve, call the third intersection $R$, then intersect the line $OR$ with the curve, and the third intersection is just $P + Q$. Is there a similar procedure in the model of intersection of quadric surfaces? REPLY [6 votes]: Regarding your main question, this is done in Cassels, Lectures on elliptic curves, $\S$ 8 (iv) p. 36. We may assume that the common rational point of the quadrics is $(X:Y:Z:T)=(0:0:0:1)$. Then the quadrics have the shape \begin{align*} Q_1 & = TL + R\\ Q_2 & = TM + S \end{align*} where $L,M$ (resp. $R,S$) are linear (resp. quadratic) in $X,Y,Z$. The polynomials $L,M$ are necessarily linearly independent, and eliminating $T$ gives the cubic equation \begin{equation*} P(X,Y,Z) = LS-RM=0. \end{equation*} So we are reduced to put a cubic into canonical form, which you know how to do. Regarding your second question, it is indeed possible to work out the group law directly on the intersection $C$ of two quadric surfaces. This is explained in Husemöller, Elliptic curves, Introduction, $\S$ 8, p. 20. The idea is as follows: given two points $P,Q$ on $C$, consider the plane containing $O,P,Q$, and intersect it with $C$. By Bézout's theorem there will be 4 points of intersection, $O$, $P$, $Q$ and a fourth point $-(P+Q)$. The map $P \to -P$ is defined similarly.<|endoftext|> TITLE: zeros on the circle of convergence QUESTION [6 upvotes]: In this question some experiments were used to conjecture that the zeros of partial sums of a series converging to a function with natural boundary on the unit circle were (weakly) converging to the unit circle. Well, nothing is new under the Sun, and this is a Theorem of Jentsch, from his recent PhD thesis: R. Jentsch, Untersuchungen zur Theorie der Folgen analytischer Funktionen, Inauguraldissertation Berlin, 1914,39 s. Actually, Jentsch's theorem does not need the circle to be the natural boundary, but simply the circle of convergence, so the only thing it is lacking is the stronger statement that the empirical measure of the zeros converges to the Lebesgue measure on the unit circle. EXACT STATEMENT Let $$f(z) = \sum_{i=0}^\infty a_n z^n,$$ with radius of convergence $1.$ Then every point of the unit circle is an accumulation point of zeros of partial sums of $f.$ END EXACT STATEMENT The question is: does anyone here know of a more accessible place where the proof of this result could be found (or maybe there is a 10 line argument someone could just post)? REPLY [3 votes]: Disclaimer: I learned the trickery below from N.K.Nikolskii, who was giving us a special topics course in complex analysis when I was a fourth year undegraduate student. I have no idea whether it can also be traced back 100 years but, certainly, it has been used by many people on many occasions and is worth teaching to our graduate students (IMHO). Main Lemma: Let $r<1 TITLE: Is the "surface-minor" ordering of plane graphs a well-quasi-ordering? QUESTION [7 upvotes]: A plane graph is a finite simple graph with a fixed embedding into the two-sphere. The embedding induces an embedding on a minors of a plane graph (i.e. a graph obtained by successive removal of vertices, removal of edges, and contraction of edges). In other words, we may consider minors of plane graphs to be plane graphs themselves. These minors are called surface minors. The surface-minor relation is an ordering on plane graphs, which is finer than the minor relation: a plane graph may be a minor of another plane graph, without being a surface minor (indeed, if a graph has several non-isomorphic embeddings into the two-sphere, then these are examples of that behavior). The question is: Is the surface-minor ordering of plane graphs a well-quasi-ordering? That is to say, is there among any infinite collection of plane graphs a pair of two plane graphs, one of which is a surface minor of the other? This seems a very natural question to me, yet I couldn't find an answer in the literature. One would assume that this question must be answered in one of Robertson and Seymour's papers, since both the notion of well-quasi-orderings and the notion of surface minors are pretty central in them. If one restricts oneself to plane trees, the answer is positive: this is a version of Kruskal's Tree Theorem. The answer is also positive if one restricts oneself to 3-connected plane graphs, since Whitney's Theorem says that they have a unique embedding, and so the surface-minor and the minor relation coincide. Of course, one can ask this question not only about the two-sphere, but about any other fixed (not necessarily orientable) closed surface. Please note this question is not a direct corollary of the Robertson-Seymour theorem that the minor ordering on finite graphs is a well-quasi-ordering; and it does not have a direct relation to Kuratowski's Theorem, which gives the two forbidden minors for planar graphs (not to be confused with the plane graphs this question is about). REPLY [3 votes]: This is a partial answer, for the case when the given sequence $G_1,G_2,\dots$ of plane graphs has unbounded treewidth. In such a case, for every $n$ there is an $i$ such that $G_i$ contains the $n\times n$ grid as a minor, and thus also as a plane minor. The rest follows from the simple fact that $G_1$ is a plane minor of a sufficiently large plane grid.<|endoftext|> TITLE: flat/crystalline cohomology of abelian variety QUESTION [8 upvotes]: Let $A/k$ be an abelian variety over an algebraically closed field and $\ell \neq \mathrm{char}\,k$. In http://jmilne.org/math/articles/1986b.pdf, Theorem 15.1(b) it is proved that $$H^r_{et}(A, R) = \bigwedge^rH^1_{et}(A,R)\quad\text{for $R = \mathbf{Z}_\ell,\mathbf{Q}_\ell,\mathbf{F}_\ell$.}$$ The proof uses $$H^1_{et}(A,\mathbf{Z}_\ell) = \mathrm{Hom}_{cont}(\pi_1^{et}(A,0),\mathbf{Z}_\ell).$$ For $R = \mathbf{Z}/p\mathbf{Z}$ one has $$H^r_{et}(A,\mathbf{Z}/p\mathbf{Z}) = H^r_{fppf}(A,\mathbf{Z}/p),$$ since $\mathbf{Z}/p\mathbf{Z}$ is a smooth quasi-projective commutative group scheme. Using induction, the five lemma and short exact sequences $$0 \to \mathbf{Z}/p \to \mathbf{Z}/p^{n+1} \to \mathbf{Z}/p^n \to 0,$$ one gets $$H^r_{et}(A,\mathbf{Z}/p^n) = H^r_{fppf}(A,\mathbf{Z}/p^n)\quad\text{ for all $n \geq 0$,}$$ and hence $$H^r_{et}(A,\mathbf{Z}_p) = H^r_{fppf}(A,\mathbf{Z}_p).$$ One has $$\pi_1^{et}(A,0) = \prod_{\ell} T_\ell(A)$$ with $T_p(A) = \varprojlim_nA[p^n]^r$ with $A[p^n] = A[p^n]^0 \times A[p^n]^r$, with $A[p^n]^0$ local and $A[p^n]^r$ reduced (Mumford, Abelian Varieties, Chapter IV.18) for $p = \mathrm{char}\,k$. By Lei Fu, Étale cohomology theory, Proposition 5.7.20, one has $$H^1_{et}(X,G) = \mathrm{Hom}_{cont}(\pi_1^{et}(X),G)$$ for a finite group $G$ and $X$ connected Noetherian. Is there a similar isomorphism for crystalline cohomology or flat cohomology with $R = \mathbb{Z}_p,\mathbb{Q}_p,\mathbb{F}_p$, $p = \mathrm{char}\,k$? Edit (26.04.2018): The answer below settles the question for $W(k)$ and $\mathrm{Quot}(W(k))$ coefficients and crystalline cohomology. What is the analogue of $H^q_\mathrm{et}(\bar{A},\mu_{\ell^n}) = \mathrm{Hom}(\bigwedge^qA[\ell^n],\mu_{\ell^n})$ in crystalline cohomology? REPLY [2 votes]: In Illusie, Complexe de de Rham-Witt et cohomologie cristalline, p. 651, (7.1.1) it is proved that $H^*_{cris}(A/W) = \bigwedge^*H^1_{cris}(A/W)$. Is $H^1_{cris}(A/W) = T_pA$? I have found in http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.295.7431&rep=rep1&type=pdf, p. 202, Theorem 5.7.1: For $A/W(k)$, $k$ algebraically closed of characteristic $p > 0$, there is a short exact sequence of $W = W(k)$-modules: $$0 \to H^1_{et}(A_k,\mathbf{Z}_p)\otimes W \to H^1_{cris}(A_k/W) \to \mathbf{D}(\hat{A}/W) \to 0$$ Edit (26.04.2018): This settles the question for $W(k)$ and $\mathrm{Quot}(W(k))$ coefficients. What is the analogue of $H^q_\mathrm{et}(\bar{A},\mu_{\ell^n}) = \mathrm{Hom}(\bigwedge^qA[\ell^n],\mu_{\ell^n})$ in crystalline cohomology?<|endoftext|> TITLE: Donaldson and DT invariants QUESTION [14 upvotes]: Let $X$ be a smooth projective surface. Then, using the compactified moduli space of anti self-dual connections or torsion free sheaves we can construct Donaldson invariants of $X$. Similarly, one can take a CY3-fold and by slanting elements of the universal sheaf with elements of the Chow group of the 3fold, construct the DT (Donaldson-Thomas invariants) - at least morally I think this is the idea. If CY3 is the canonical bundle of $X$ do the DT invariants of $K_X$ and the Donaldson invariants of $X$ relate to each other in any sense? More generally, is there any short of relation between Donaldson and Donaldson-Thomas invariants? REPLY [5 votes]: In some sense this is the topic of Vafa-Witten theory for complex surfaces; see the many recent papers of Göttsche-Kool on the subject. In DT theory the virtual dimension is 0, so you don't usually use insertions (or the slant product) -- you just get one number. It is (a virtual version of) the Euler characteristic of the Donaldson moduli space, and Göttsche-Kool have shown that can be expressed in terms of classical Donaldson invariants. Then there are refined invariants, which recover (a virtual version of) the Hirzebruch $\chi^{\ }_y$-genus of the Donaldson moduli space. I do not know if that can be expressed in terms of Donaldson invariants, but it seems reasonable to expect it can.<|endoftext|> TITLE: Is every $GL_2(\mathbb{Z}/n\mathbb{Z})$-extension contained in some elliptic curve's torsion field? QUESTION [16 upvotes]: I suppose this question could be phrased in terms of Galois representations, but I'm asking it this way. Let $n>1$ be an integer. If $K$ is a number field with $\operatorname{Gal}(K/\mathbb{Q}) \cong GL_2(\mathbb{Z}/n\mathbb{Z})$ (edit) and containing the $n$-th roots of unity (/edit), must there exist an elliptic curve $E$ defined over $\mathbb{Q}$ such that $K \subseteq \mathbb{Q}(E[n])$? It's not necessary to produce the curve, although it would be cool. If not, is there a good way to test if this holds for a given $K$? (If you want to restrict to the case where $n$ is a prime power, that's fine.) REPLY [4 votes]: Two comments. First, there is a trick used by Serre, Ribet and others to prove, using an old result of Goursat, that under certain assumptions such as absolute irreducibility, two mod p Galois representations cut the same Galois extension if and only if they are twists of each other. I guess using this combined with the reference in a previous answer should be enough to answer your question. Secondly, if you take n to be a prime power with exponent at least 2, it is not even clear that a mod n representation can be lifted at all to any p-adic representation, even if you admit coefficients in extensions of the p adic field. Further assumptions such as finite ramification set make the question even harder. My guess is that in general you should not expect existence of any geometric p adic lift, in particular this rules out elliptic curves.<|endoftext|> TITLE: On representing Turing Degrees relative to iterates of the jump QUESTION [5 upvotes]: Consider the following definition. We define the 'index' of a Turing degree $A$, (denoted $i(A)$) as the smallest ordinal $\alpha\in \omega_1$ such that $A \ngeq 0^{(\alpha)}$. [is this concept found anywhere in the literature, and if so, under what name?] I have 2 questions relating to this concept: 1) Does the following property hold: $i(A\vee B) = \max\{i(A), i(B)\}$? 2) If $A$ is a Turing degree, and $i(A)=\alpha+1$ is a successor ordinal, then can we always write $A = X \vee 0^{(\alpha)}$ where $X$ is a Turing degree satisfying $i(X) = 1$? Is there a natural way to extend this idea to limit ordinals? I'm suspecting that the first question is false, and the second is true, but I don't know where to start. REPLY [5 votes]: First, note that "$0^{(\alpha)}$" doesn't make sense unless $\alpha$ is a computable ordinal. We can do a bit better, especially via mastercodes, but that still only reaches up to $\omega_1^L$ (which might be vastly smaller than $\omega_1$). So let's restrict attention to those Turing degrees which are not above every set of the form $0^{(\alpha)}$ for $\alpha$ a computable ordinal (that is, those Turing degrees not bounding every hyperarithmetic degree). The answer to your first question is extremely "no:" for any computable $\alpha$, we can find sets $A, B$ neither of which computes $0'$ but such that $A\oplus B$ computes $0^{(\alpha)}$. Proof: let $\mathbb{P}$ be the set of pairs $(p, q)$ of finite partial maps $\omega\rightarrow 2$ with the same domain $D$, such that $\{a\in D: p(a)=q(a)\}\subseteq 0^{(\alpha)}$ and $\{a\in D: p(a)\not=q(a)\}\subseteq \overline{0^{(\alpha)}}$. Now taking a sufficiently generic filter through $\mathbb{P}$ yields a pair of sets which agree precisely on $0^{(\alpha)}$ (hence whose join computes $0^{(\alpha)}$) but neither of which computes $0'$ (since each of the two sets will be sufficiently Cohen-generic). REPLY [4 votes]: Even if you fix a meaning for $0^{(\alpha)}$ for every countable ordinal $\alpha$, by fixing a code for $\alpha$, say, then it still isn't necessarily the case in ZFC that every real is below one of them, and so you haven't really defined a degree notion on the degrees. The reason is that this would imply the continuum hypothesis, as there are only $\omega_1$ many countable ordinals $\alpha$, and each $0^{(\alpha)}$ would compute only countably many reals. In this sense, the existence of a backbone sequence of length $\omega_1$ cofinal in the Turing degrees is equivalent to CH.<|endoftext|> TITLE: Bounded-width Konig's lemma in reverse math QUESTION [6 upvotes]: We define $\mathsf{BWKL}$ as follows: Every infinite binary tree of bounded width has an infinite path. This obviously follows from $\mathsf{WKL}$. Is this principle true in $\mathsf{RCA}_0$? If not, what does it need? I've found a paper that addresses a related principle by the same name and shows it follows from $\mathsf{I\Sigma}^0_2$ ($\Sigma^0_2$-induction)... but it appears to assume that the tree has finite width, not merely bounded. (Also, I don't think it gives any lower bound on the principle's strength.) EDIT: After some off-site discussion with Prof. Kołodziejczyk (one of the authors of this paper), I'd like to correct two minor confusions on my part. The bounded-width Konig's lemma discussed by Kołodziejczyk, Michalewski, Pradic, and Skrzypczak is precisely what I was asking about (which appears in Simpson & Yokoyama's work, as mentioned in Denis Hirschfeldt's answer). This is already pointed out in the journal version of the paper, though not in the preprint I'd linked. On the other hand, it is true that the paper does not address lower bounds, so it did not already answer my question. Also: of course the distinction I was drawing between finite & bounded does not apply in this context; since we're talking about sets IN the model, it's straightforward to see that every bounded set (with maximum less than some first-order object) is finite (in the sense of having a canonical index). My only excuse: I always get bogged down in the details of first-order concerns, so I erred on the overly-cautious side. REPLY [5 votes]: If I understand your question correctly, this principle is indeed provable from either WKL or I$\Sigma^0_2$, but not in RCA$_0$, or even WWKL$_0$. See slides 12 and 13 in this talk by Yokoyama. Based on this paper by Nies, Triplett, and Yokoyama, it seems that these results will appear in a paper by Simpson and Yokoyama. It is interesting that, as noted in both these links, if we bound the size of prefix-free sets rather than just that of levels, the principle becomes provable in RCA$_0$.<|endoftext|> TITLE: Leray-Serre spectral sequence for algebraic groups QUESTION [6 upvotes]: Let $G$ be a semisimple, simply-connected, complex algebraic group. Fix a Borel subgroup $B$ and let $P$ be a parabolic subgroup properly containing $B$. If $M$ is a $B$-module, then we have the Leray-Serre spectral sequence corresponding to the fibration $P/B\to G/B\to G/P$ $E^{p,q}_2=H^p(G/P, H^q(P/B,M))\implies H^{p+q}(G/B,M)$ The notation $H^•(G/B, M)$ means the sheaf cohomology of the sections of the bundle $G\times_B M\to G/B$ and similarly with $G,B,M$ replaced by another group $G$ with closed subgroup $B$ and $B$-module $M$. My question is, how can the module on the right be seen as a graded version of the module on the left? What is the filtration? REPLY [2 votes]: In the vein of @MarkGrant's comment: Since this is a first quadrant spectral sequence, for a given pair $(p,q)$ the individual terms $E_2^{p,q},E_3^{p,q},E_4^{p,q},\ldots$ occurring on each subsequent page of the spectral sequence will eventually reach a stable value $E_\infty^{p,q}$. (The larger $p+q$ is, the more pages of the spectral sequence it may take for $E_r^{p,q}$ to reach its stable value.) In general $E_\infty^{p,q}$ will only be a subquotient of $E_2^{p,q}$. Then the statement that the spectral sequence converges to $H^*(G/B,M)$ means that $H^{p+q}(G/B,M)$ admits a decreasing filtration $F^* H^{p+q}(G/B,M)$ with $$F^p H^{p+q}(G/B,M) / F^{p+1} H^{p+q}(G/B,M) = E_\infty^{p,q}.$$ If you happen to know that the spectral sequence collapses at the $E_2$-page, and hence for all $p,q$ that $E_2^{p,q} = E_\infty^{p,q}$, then this tells you how to think of the $H^p(G/P,H^q(P/B,M))$ as the filtration layers of $H^{p+q}(G/B,M)$.<|endoftext|> TITLE: String cobracket from TFT QUESTION [10 upvotes]: Let $M$ be a closed oriented manifold. Chas and Sullivan (https://arxiv.org/abs/math/0212358) introduced a Lie bialgebra structure on $H_\bullet^{S^1}(LM, M)$, $S^1$-equivariant homology of the loop space $LM=\mathrm{Map}(S^1, M)$ relative to constant loops $M\rightarrow LM$. It seems an "explanation" of the string topology operations comes from topological field theories. I will switch to chains for simplicity and ignore some shifts. Then $C_\bullet(\Omega M)$ is a (smooth, not proper) Calabi-Yau dg algebra, so it defines a positive boundary oriented 2d TFT (this construction is reviewed at the end of https://arxiv.org/abs/0905.0465). This gives rise to some operations on $CH_\bullet(C_\bullet(\Omega M)) = C_\bullet(LM)$ which are supposed to be the string topology operations. Here $CH_\bullet(-)$ are Hochschild chains which is the value of the TFT on $S^1$. For instance, you get a structure of an $E_2^{fr}$-algebra on $C_\bullet(LM)$, where $E_2^{fr}$ is the operad of framed little disks. Passing to $S^1$-coinvariants you obtain a gravity algebra (https://arxiv.org/abs/math/0605080); in particular, you get a Lie bracket on $C_\bullet^{S^1}(LM)$. Similarly, the TFT picture seems to give a non-counital $E_2^{fr}$-coalgebra structure on $C_\bullet(LM)$. Again passing to $S^1$-coinvariants I would expect to see a Lie cobracket on $C^{S^1}_\bullet(LM)$. However, it is known that to define the cobracket one has to pass to reduced chains $C_\bullet^{S^1}(LM, M)$. A similar construction exists in the algebraic context, e.g. in https://arxiv.org/abs/0804.4748 it is shown that Hochschild homology is a BV algebra and cyclic homology is a gravity algebra (in particular, has a Lie bracket) while only reduced Hochschild homology is a BV coalgebra and reduced cyclic homology is a gravity coalgebra (in particular, has a Lie cobracket). (Note that the paper deals with the Koszul dual situation of a cyclic coalgebra.) Question: how can one see that the string cobracket exists only on reduced cyclic homology and not on the unreduced one from the TFT perspective? How does reduced Hochschild homology appear in the TFT? REPLY [7 votes]: The string cobracket you are referring to is not part of the TQFT structure of string topology given by the smooth Calabi Yau algebra structure on $C_*(\Omega M)$ but rather associated to an action of the chains of certain compactification of the moduli space of Riemann surfaces as explained in Sullivan's survey "String Topology: background and present state". It is an operation arising from a homotopy at the chain level. In fact, this cobracket arises from a secondary coproduct on $C_*(LM)$. Given a closed manifold $M$ of degree $d$ we may define two coproducts of degree $-d$ in ordinary singular chains $C_*(LM)$ that pass to homology: namely we may consider self intersections at $t=0$ and split into two loops or at $t=1$ and split into two loops. Note that these are really coproducts factoring through maps $C_*(LM) \to C_{*-d}(M \times LM)$ and $C_*(LM) \to C_{*-d}(LM\times M)$, respectively. These two coproducts are chain maps and moreover they are chain homotopic, so they define the same coproduct of degre $-d$ in homology, which is in fact part of the TQFT structure of string topology on $H_*(LM)$ (the upside down pair of pants coproduct). However, there is a canonical chain homotopy between the two coproducts at the chain level given by a map $\vee: C_*(LM) \to C_{*+1-d}(LM \times LM)$ which may be defined as a one parameter family of self intersections (the transversality assumptions are rather subtle). The map $\vee$ of course is not a chain map, however, it is a chain map if we work modulo constant loops so it defines a coproduct in relative homology! It is in this sense that $\vee$ is a secondary operation. Then the string cobracket in $S^1$-equivariant homology modulo constant loops is induced by $\vee$ and the maps relating ordinary and equivariant homology in the long exact sequence. Thus, there are two coproducts, one of them of degree $-d$ defined on $H_*(LM)$ and is trivial if we work modulo constant loops (also trivial if Euler characteristic is zero, actually this coproduct is rather boring since it is essentially multiplication by the Euler characteristic) and the other one of degree $1-d$ defined on $H_*(LM,M)$ which shows a more interesting behavior. In particular, the latter induces a cobracket of degree $2-d$ equivariant homology modulo constant loops and it defines a involutive Lie bialgebra structure together with the string bracket. The Hochschild story is completely analog to what I have described above. An interesting subtlety on the algebraic Hochschild theory of Frobenius algebras (Koszul dual to the one with $C_*(\Omega M)$) is that "reduced" Hochschild complex only makes sense for commutative dg algebras (or cocommutative dg coalgebras, if you are working with coalgebras) while the algebraic versions of the loop product make sense in the associative case. In a recent paper of Zhengfeng Wang and myself (https://arxiv.org/abs/1703.03899) we actually describe a way to combine the algebraic loop product and the algebraic coproduct (the one of degree $1-d$) in an extended version of the Hochschild complex without having to assume commutativity. Essentially when there is a problem at the algebraic version of the "constant loops" we change the differential of the underlying complex instead of killing these "constant loops".<|endoftext|> TITLE: Galois action on $p$-adic Tate module of Abelian variety over finite field semisimple? QUESTION [6 upvotes]: Let $A,B$ be positive dimensional Abelian varieties over a finite field and $p$ be an arbritrary prime. By Zarhin, Homomorphisms of abelian varieties over finite fields http://www.math.nyu.edu/~tschinke/books/finite-fields/final/10_zarhin.pdf, Theorem 10.2, one has an isomorphism $$\mathrm{Hom}(A,B) \otimes \mathbf{Z}_p \to \mathrm{Hom}(A(p),B(p))$$ with $A(p)$, $B(p)$ the $p$-divisible groups of $A$ and $B$ (a generalisation of Tate's Endomorphisms of Abelian Varieties over Finite Fields to $p = \mathrm{char}(K)$). Is the Galois action on the $p$-adic Tate module also semisimple as it is for $\ell \neq p$? REPLY [9 votes]: You're just asking whether Frobenius acts semsimply on the $p$-adic Tate module. We know from Tate's theorem that Frobenius acts semisimply on the $\ell$-adic Tate module, and hence satisfies some squarefree polynomial (its minimal polynomial). Now because the map from the ring of endomorphisms to the endomorphisms of the Tate module is injective, it also satisfies this squarefree polynomial in the ring of endomorphisms of the abelian variety. Hence it satisfies a squarefree polynomial, and thus is semisimple, when acting on the $p$-adic Tate module as well.<|endoftext|> TITLE: What did Zermelo say he was hoping for on the consistency of set theory? QUESTION [6 upvotes]: Question. What precise things are known about what Zermelo is hinting at in the below citation? What are scholarly references on Zermelo's own attempts at proving consistency of his axioms? What did Zermelo hope for? Most concretely: are there other publications of Zermelo's on consistency of set theory? Did he lecture on this and if yes, what did he say? I'm looking for more than "Well, he was hobnobbing with Hilbert in Göttingen and Hilbert's optimism, still unfazed at the time, was giving Zermelo hope for some sort of absolute consistency proofs ...", in particular, looking for a dedicated discussion of how much of the relative-turn (i.e. from a hope of some absolute consistency-proof, in some sense, to the more modest notion of consistency relative to another formal system) was *already recognizable in the documents from the time around 1900, in particular, whether the conjecture kindly provided by Ed Dean in one of the answers below, i.e. whether Zermelo was hoping or planning to work out a relative consistency proof modelled on Hilbert's "Foundations of Geometry", i.e., did Zermelo write about this method of Hilbert's anywhere? Citation. In p. 262 of Math. Ann. Vol.65, No. 2 (1908) one can read: For convenience, I provide an unidiomatic literal translation: "[...] of these principles may remain undiscussed here. Even the--certainly very essential--"contradictionlessness" of my axioms I have not yet been able to rigorously prove; rather I have had to restrict myself to occasional remarks that those "antinomies" known today all disappear, if the principles proposed here are adopted. With this [work] I want to at least offer useful preparations to future investigations into such deeper problems." Remarks. The emphasis on today is mine; evidently Zermelo here is referring to the mundane phenomenon of absence of known problems, the known unknowns, as they say. This question seems appropriate here given the comments of two not entirely unknowledgeable mathematicians to this question. I expect the novelty in all of this to be nil. I do not have illusions that there is anything mathematically new to come of looking into this historical issue. Nor do I want to create a Zermelo-myth along the lines of "Zermelo took a proof of inconsistency of ZFC into the grave" or something like that. Consistency questions are arguably the most studied topic in logic and set-theory, Zermelo's work has been thoroughly digested, and in particular his Math. Ann. 65 paper appears in English translation in volume 1 of his collected works edited by H.-D.Ebbinghaus and A. Kanamori, and Zermelo's mentioning consistency proofs is emphasized, with a (translated) citation here in the first Section. I even expect this very question to have been treated somewhere, but did not search for it. (Isn't this---within reason of course---what Q&A sites are for?) According to the usual narrative, "Hilbert's program" still lay about ten years in the future when this was published. A general recapitulation of the basics on ZFC (in particular the second incompleteness theorem) should perhaps be kept out of this thread. There are many good references on this, on this site and elsewhere. This thread is rather meant to EDIT to clarify this passage: thread is meant to focus on giving a picture of what proof-theory and consistency-proofs meant to mathematicians around 1908, which is more than 20 years before Gödel published the second incompleteness theorem. In particular, are there dedicated historical/mathematical articles on precursors to relative consistency? While I am actively working on something related to (variants of) models of ordinals, whence this question, consistency of ZFC is not an (active) interest of mine; I resolved to ask this nevertheless, because recency can make up for non-novelty, and there seems to be some demand for such a question, and because it sometimes is good to be reminded of, or served with known things, and to complement this question, and to provide a new generation of mathematicians with an occasion to have a (relevant) discussion here. And who knows, maybe something new comes of it? REPLY [5 votes]: The following remarks may not speak to what you're really after, but given your explicit reference to Hilbert's Program still being years away when wondering what Zermelo might have in mind, they may be somewhat useful. It's true that at the point of Zermelo's 1908 passage, Hilbert had not yet adopted the more formal approach (along the lines of Frege or Russell-Whitehead) to logical deduction that informed the quest in Hilbert's Program for consistent and complete foundations of mathematics. He didn't have available a notion of syntactic completeness, for instance. But even so, already in Hilbert's 1899 Foundations of Geometry (pdf here) one finds notions of relative consistency proofs that today's logicians would recognize, despite the informal-by-today's-standards surrounding logical framework, wherein e.g. he essentially shows the independence of an axiom $\varphi$ from a set of axioms $S$ by giving a model for $S+\{\neg\varphi\}$. So it would be well within the realm of possibility that Zermelo in 1908 hoped to devise a relative consistency proof for his axiomatization along similar lines, thus going beyond providing the mere appearance of having addressed the known antinomies that Zermelo references, such as Russell's paradox to which Frege's Grundgesetze had succumbed in 1903. As to exactly what sort of construction (and relative to what background theory) Zermelo might have envisioned, I can't say (and maybe that's the only part that you're really asking about). But it's also not clear, just from the cited passage at least, that his thoughts even progressed far beyond the point of recognizing the bare desire for such a consistency proof. While not directly relevant to your particular questions, you might enjoy Reck and Awodey's "Completeness and Categoricity, Part I: Nineteenth-century Axiomatics to Twentieth-century Metalogic" (pdf here) for its general discussions of some logical concepts in the era preceding that of Hilbert's Program. Its discussion of Foundations of Geometry has some bearing on my remarks at least.<|endoftext|> TITLE: product of power sets QUESTION [9 upvotes]: For a set $X$, let $\mathcal P(X)$ denote its power set and let $\mathcal P(X)\otimes\mathcal P(X)$ denote the product $\sigma$-algebra in $X^2$. When $|X|\leq\aleph_0$ then $\mathcal P(X)\otimes\mathcal P(X)=\mathcal P(X^2)$ but when $|X|>2^{\aleph_0}$ this equality is known to fail. What happens when $\aleph_0<|X|\leq 2^{\aleph_0}$? REPLY [6 votes]: The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Lebesgue measure defined on all sets of reals. Here, $X=\mathbb R$ and $\Sigma=\mathcal P(\mathbb R)$. Since $\nu$ extends Lebesgue measure, the space satisfies the assumptions of the result just stated, and $\mathcal P(\mathbb R)\otimes\mathcal P(\mathbb R)$ is not $\mathcal P(\mathbb R^2)$. By the way, there is a recent article in the Monthly dealing precisely with this problem and discussing how $\mathsf{CH}$ implies that $\mathcal P(\mathbb R)\otimes\mathcal P(\mathbb R)$ is $\mathcal P(\mathbb R^2)$ while the existence of extensions of Lebesgue measure gives a negative answer: MR3626256 Avilés, Antonio; Plebanek, Grzegorz. A little ado about rectangles. Amer. Math. Monthly 124 (2017), no. 4, 345–350.<|endoftext|> TITLE: Compact operators on $\ell^1$ QUESTION [5 upvotes]: Let $T$ be a compact symmetric operator on $\ell^2$ and $T\vert_{\ell^1}$ be bounded on $\ell^1$. Are there any non-trivial conditions that $T\vert_{\ell^1}$ is compact as well (for example would $T$ belonging to some Schatten-class on $\ell^2$ be sufficient)? The obvious proof estimating $$\left\lVert \sum_{i=0}^{\infty} \lambda_i \langle \cdot ,\varphi_n \rangle \varphi_n - \sum_{i=0}^{k} \lambda_i \langle \cdot ,\varphi_n \rangle \varphi_n \right\rVert_{L(\ell^1)} $$ does not work as the eigenvectors, we get from the $\ell^2$ representation, are not necessarily bounded in $\ell^1$ norm. Recall also that a set $M \subset \ell^1$ is compact if it is bounded, closed and $\lim_{n \rightarrow \infty} \operatorname{sup}_{x \in M} \sum_{k=n}^{\infty} \left\lvert x_k \right\rvert=0$ EDIT: Since I received an answer that outlined to me that it does not work in the non-symmetric case, I thought it would be good to explain why I think that symmetry could help. In this case, I can show that $\sigma(T) \subset \sigma(T_{\vert_{\ell^1}})$ and the point spectra coincide. Moreover, also all finite-dimensional eigenspaces are the same for each element of the point spectrum. In other words, the spectrum of $T\vert_{ \ell^1}$ also contains the spectrum that one would assume to have for a compact operator. However, I could so far not exclude any continuous spectrum for $T\vert_{ \ell^1}$ which would be a necessary condition for compactness. REPLY [5 votes]: Here's an example showing that $T$ can be trace-class but $T|_{\ell^1}$ is not compact. Let $(x_n)$ be a sequence of vectors in $\ell^2$ with disjoint supports, $\sum_n \|x_n\|_2 \leq 1$ and $\|x_n\|_1=1$ for all $n$. Define $$ T(\xi) = \sum_n \xi_n x_n \qquad (\xi\in\ell^2). $$ Then $\| T(\xi) \|_1 \leq \sum_n |\xi_n| \|x_n\|_1 \leq \|\xi\|_1$ so $T$ is bounded on $\ell^1$. However, $(T(e_n)) = (x_n)$ has no convergent subsequence so $T$ is not compact on $\ell^1$. As $\sum_n \|x_n\|_2\leq 1$, $T$ is trace-class. An example of such a sequence is as follows. Let $N(n)$ be a rapidly increasing sequence of integers, and choose $x_n$ to be the sequence $(0,\cdots,0,N(n)^{-1},\cdots,N(n)^{-1},0,\cdots)$ where we repeat $N(n)^{-1}$ exactly $N(n)$ times, and we place the non-zero terms so that the $x_n$ have disjoint support. Then $\|x_n\|_1=1$ but $\|x_n\|_2 = N(n)^{-1/2}$ so as long as $N(n)$ increases fast enough that $\sum_n N(n)^{-1/2} \leq 1$ we're done. This $T$ is not self-adjoint, but notice that $T^*$ is trace-class, and $T^*$ is still bounded on $\ell^1$. Furthermore, $S=T+T^*$ is seen to still be such that $S(e_n)$ has no convergent subsequence in $\ell^1$.<|endoftext|> TITLE: Fermat's Last Theorem in finite fields QUESTION [7 upvotes]: Consider the finite field $\mathbb{F}_q$. Schur (1916) proved that, given $n$, when the field is sufficient large, this equation, $$x^n+y^n= z^n$$ always has a nontrivial solution. What conditions does the number of solutions satisfy? I. Schur, Über die Kongruenz $x^{m} + y^{m} \equiv z^{m} \pmod{p}$, Jahresber. Deutschen Math. Verein. 25 (1916), 114–117. REPLY [8 votes]: It is more or less easy to obtain, via exponential sums, an asymptotic formula for the number $J$ of solutions of the congruence $$x^{n}+y^{n} \equiv z^{n} \pmod{p}$$ where $$1 \leq x, y, z \leq p-1.$$ Indeed, since \begin{eqnarray*} J &=& \sum_{x=1}^{p-1} \sum_{y=1}^{p-1} \sum_{z=1}^{p-1} \frac{1}{p}\sum_{a=0}^{p-1}e^{2 \pi i \frac{a(x^{n}+y^{n}-z^{n})}{p}}\\ &=& \frac{(p-1)^{3}}{p}+ \frac{1}{p} \sum_{a=1}^{p-1} \sum_{x=1}^{p-1}\sum_{y=1}^{p-1} \sum_{z=1}^{p-1}e^{2\pi i \frac{a(x^{n}+y^{n}-z^{n})}{p}}\\ &=& \frac{(p-1)^{3}}{p} + \frac{1}{p}\sum_{a=1}^{p-1}\left|\sum_{x=1}^{p-1}e^{2\pi i \frac{ax^{n}}{p}}\right|^{2}\sum_{y=1}^{p-1}e^{2\pi i \frac{ay^{n}}{p}}\\ \end{eqnarray*} and $$ \sum_{a=0}^{p-1} \left| \sum_{x=1}^{p-1} e^{2\pi i \frac{ax^{n}}{p}} \right|^{2} \leq n\,p\,(p-1)$$ and $$\left|\sum_{y=1}^{p-1}e^{2\pi i \frac{\alpha y^{n}}{p}}\right| \leq n\sqrt{p}$$ for any integer $\alpha$ which is not divisible by $p$, it follows that $$J = \frac{(p-1)^{3}}{p}+O\left(n^{2}(p-1)\sqrt{p}\right).$$<|endoftext|> TITLE: Dense and co-dense subsets in connected $T_2$-spaces QUESTION [7 upvotes]: Is there a connected $T_2$-space $(X,\tau)$ with more than 1 point and with the following property? Whenever $D\subseteq X$ is dense, $X\setminus D$ is not dense. REPLY [4 votes]: There is such a thing as a submaximal topology, in which every dense subset is open. These obviously satisfy your condition. Take any connected Hausdorff space $(X,\tau)$. Let $\mathscr F$ be an ultrafilter of $\tau$-dense sets. Let $\tau'$ be the topology generated by $\tau\cup \mathscr F$. Then $(X,\tau')$ is submaximal Hausdorff connected.<|endoftext|> TITLE: an identity related to the pentagonal numbers QUESTION [6 upvotes]: How can I prove the following? $$1-x+x^2+x^5-x^7-x^{12}+x^{15}-x^{22}-x^{26}+x^{35}-x^{40}+\dots \\= \prod_{i=1}^{\infty} [(1 - x^{8 i - 7}) (1 + x^{8 i - 6}) (1 + x^{8 i - 5}) (1 + x^{8 i - 4}) (1 + x^{8 i - 3}) (1 + x^{8 i - 2}) (1 - x^{8 i - 1}) (1 - x^{8 i})]$$ It doesn't seem to follow from the Triple Product formula and I haven't been able to come up with a combinatorial proof. REPLY [14 votes]: This is an instance of Watson's quintuple product identity (also Macdonald identity for $BC_1$): $$\prod_{n\geq 1}(1-s^n)(1-s^nt)(1-s^{n-1}t^{-1})(1-s^{2n-1}t^2)(1-s^{2n-1}t^{-2})=\sum_{n\in \mathbb Z}s^{\frac{3n^2+n}{2}}(t^{3n}-t^{-3n-1}).$$ By plugging in $t=x^{-1}$ and $s=-x^{4}$ this becomes exactly your identity. Somewhat amusingly, the quintuple product identity can be proven directly from the triple product identity. See this article by Carlitz, or this article of Foata and Han.<|endoftext|> TITLE: Jordan curves admitting only acyclic inscriptions of squares QUESTION [22 upvotes]: The (recently solved) inscribed square problem or Toeplitz conjecture posits that every closed, plane continuous (Jordan) curve ${\it \Gamma}$ in $\mathbb{R}^2$ contains all vertices of some square. It appears this theorem was just proven! Most examples resemble the one on the left, in which the natural continuous parameterization along ${\it \Gamma}$ intersects the square's points in sequence. Let's call such an inscription cyclic. However, some Jordan curves contain the points of a square in non-sequential order, such as shown at the right. Let's call such an inscription acyclic. Questions Are there Jordan curves that admit only acyclic inscriptions (and not also cyclic inscriptions)? Given the inscribed square problem was just answered in the affirmative, can one prove whether acyclic-only curves exist? Alternatively, or additionally: can one provide an example of such an acyclic-only curve? My conjecture is that there are no such acyclic-only Jordan curves. My first approach has been to assume that there is a given acyclic square inscription for a Jordan curve and then prove--invoking continuity assumptions and topological methods--that there must also be a cyclic inscription. Alas, such a proof for even this partial case has been elusive. REPLY [6 votes]: I am not a member here and so could not provide a comment. The claim that it has recently been solved is inaccurate. Green and Lobb solve the $\textbf{smooth}$ version of the Rectangle Peg Problem. For the square case, that’s been known since about a 100 years now (I suppose Schnirelman was the first) or at least some decades now. Indeed, it is true for all continuously differentiable curves. I suppose you can check a few results of Richard Schwarz, in the last couple of years, which may answer your questions regarding the (a)cyclicality.<|endoftext|> TITLE: Interesting geometric application of Hitchin Fibration QUESTION [10 upvotes]: Let $X$ be a smooth complex projective variety. Let $M_{Higgs}(X, P)$ be the coarse moduli which universally corepresents the functor: $$M^{\#}_{Higgs}(X, P): Sch/\mathbb{C}\longrightarrow \mathcal{Set}$$ which assigns to any scheme $S$ the set of isomorphism classes of semi-stable Higgs bundles $(E, \theta)$ over $X\times S$ with Hilbert poly $P$. Here $P=nP_0$, with $P_0$ being the Hilbert polynomial of $\mathcal{O}_X$ with respect to a fixed polarization $\mathcal{O}_X(1)$. Then we have the proper Hitchin map $$H_1: M_{Higgs}(X, P) \rightarrow \bigoplus_{i=1}^nH^0(X, sym^i\Omega^1(X))$$ which maps $(E,\theta)$ to the characteristic polynomial of $\theta$. (Take the coefficients.) Consider the Dolbeault Moduli space $M_{Dol}(X, P)$, which universally corepresents $$M^{\#}_{Dol}(X, P): Sch/\mathbb{C}\longrightarrow \mathcal{Set}$$ which assigns to any scheme $S$ the set of isomorphism classes of semi-stable Higgs bundles $(E, \theta)$ over $X\times S$ with Hilbert polynomial $P$ for which the Chern classes of $E$ vanish. Since $M_{Dol}(X, P)$ is disjoint union of some of the connected components of $M_{Higgs}(X, P)$, we should also have the following proper Hitchin map $$H_2: M_{Dol}(X, P) \rightarrow \bigoplus_{i=1}^nH^0(X, sym^i\Omega^1(X))$$ Can any one give some interesting, even small, geometric applications of the Hitchin map $H_1$ or $H_2$? Any answers and references are appreciated. REPLY [2 votes]: If $X$ is a smooth projective curve with genus $g \geq 2$ and $M_{Higgs}(X,P)$ is smooth, the map: $$ H_1 : M_{Higgs}(X,P) \longrightarrow \bigoplus_{i} H^0(X, Sym^i \Omega_X)$$ gives $M_{Higgs}(X,P)$ a structure of complete algebraic integrable system. I think this result is due to Beauville. See : https://projecteuclid.org/download/pdf_1/euclid.acta/1485890603. As far as I know, it is quite difficult to provide examples of complete algebraic integrable systems. EDIT : as Beauville comments below, the algebraic complete integrability of this moduli space was proved by Hitchin himself.<|endoftext|> TITLE: Non-vanishing of the Borel classes in the cohomology of $\operatorname{SL}_n(\mathbb Z)$ QUESTION [9 upvotes]: $\DeclareMathOperator\SL{SL}$The stable real cohomology of $\SL_n(\mathbb Z)$ was computed by Borel: it is given by $\mathbb R[z_i\mid i=5,9,13,\dotsc]$ with $z_i$ in degree $i$. One may wonder whether the pull back of the stable class $z_i$ on $\SL(\mathbb Z)$ to $\SL_n (\mathbb Z)$ for some finite $n$ is non-zero. That is, whether $\iota_n^* z_i = 0 \in H^i(\SL_n(\mathbb Z),\mathbb R)$ for $\iota_n \colon \SL_n(\mathbb Z) \to \SL(\mathbb Z)$. Borel's proof also gives a real cohomological stability result and hence gives a range of $n$'s for which this is true, but in 1978 Ronnie Lee announced a proof in On unstable cohomology classes of $\SL_n(\mathbb Z)$ that $z_i$ is non-zero as long as $i<2n-3$, a much larger range of $n$'s than what can be deduced from Borel's proof. His paper says Because the proofs of Theorems 1 and 2 require careful geometric construction, it is planned to present the detailed proof later, and we will indicate here some of the ideas involved in the case when $n$ is odd. but as far as I know no detailed proof has appeared. My questions are: Has any proof of Ronnie Lee's theorem appeared in the literature? If not, what is the best known range for non-vanishing of the stable Borel classes? REPLY [2 votes]: You can indeed read this off from the work of Franke, as was done in Section 4.3 of Characteristic classes of bundles of K3 manifolds and the Nielsen realization problem by Jeffrey Giansiracusa, myself, and Bena Tshishiku. In particular, Lee's result is true.<|endoftext|> TITLE: Conditions under which inequality holds for all triplets of non-negative reals QUESTION [5 upvotes]: I have the following inequality: $$ \left(\frac{1}{3} \left( \left(\frac{a+b}{2}\right)^3 + \left(\frac{a+c}{2}\right)^3 + \left(\frac{b+c}{2}\right)^3 \right) \right)^\frac{1}{3} \leq \left(\frac{a^p+b^p+c^p}{3}\right)^\frac{1}{p}$$ which is true $\forall a,b,c \in \mathbb{R}^+\cup\{0\} $. I would like to show that this inequality above holds if and only if $ p \geq \frac{3}{2} $. i.e. the inequality holds for all non-negative reals $a,b,c $ as long as $ p \geq \frac{3}{2}$. Any help whatsoever would be appreciated, so please comment even if you don't have a full solution (even proving 1 direction of the implication would be very helpful). P.S. I'm not sure exactly what to tag this, I've tagged it as convex-optimisation for now, as one of the things I've tried unsuccessfully is maximising/minimising one the sides subject to the constraint of the other side being held constant. If this is tagged wrong, please let me know the right tag(s). REPLY [5 votes]: it would be quite a nice result, but unfortunately it's proven quite difficult. I'm not sure about "nice" (after all, one can invent infinitely many inequalities for 3 positive numbers) but it is certainly not difficult. WLOG, $a+b+c=3$. Write $a=1+A, b=1+B, c=1+C$ with $A,B,C\ge -1, A+B+C=0$. Then the LHS equals $$ \left[\frac{(1-\frac A2)^3+(1-\frac B2)^3+(1-\frac C2)^3}3\right]^{1/3} $$ Part 1: $p$ must be $\ge 3/2$ Consider the case when $A,B,C\to 0$ and look at both sides up to the second order. We have $$ LHS=1+\frac{A^2+B^2+C^2}{12}+\text{higher order terms}\,; \\ RHS=1+\frac{p-1}6(A^2+B^2+C^2)+\text{higher order terms} $$ whence $p-1\ge \frac 12$ is necessary. Part 2: $p=3/2$ works (and, thereby, any larger $p$ works as well by Holder). We need to prove that $$ \sqrt{\frac{(1-\frac A2)^3+(1-\frac B2)^3+(1-\frac C2)^3}3} \le \frac{(1+A)^{3/2}+(1+B)^{3/2}+(1+C)^{3/2}}3 $$ Opening the parentheses, recalling that $A+B+C=0$, and using the inequality $\sqrt{1+X}\le 1+\frac X2$, we estimate the LHS from above by $$ 1+\frac{A^2+B^2+C^2}8-\frac{A^3+B^3+C^3}{48} $$ Thus, it would suffice to show that $$ (1+A)^{3/2}\ge 1+\frac 32A+\frac 38A^2-\frac 1{16}A^3 $$ However the RHS here is just the cubic Taylor polynomial of $A\mapsto (1+A)^{3/2}$ at $A=0$ and the fourth derivative is positive.<|endoftext|> TITLE: growth of a free group automorphism is same for finite index subgroups? QUESTION [7 upvotes]: Let $X=\{x_1,\dots,x_N\}$ and $F=F(X)$ be a free group generated by $X$. Let $\phi\colon F\to F$ be an automorphism of $F$. Define a growth function of $\phi$ as: $$ \operatorname{gr}_{\phi,X}(n)=\operatorname{max}_{1\le i\le N}\{\|\phi^n(x_i)\|_X\}, $$ where $\|.\|_X$ denotes the word length with respect to $X$. We consider these functions up to equivalence $f\simeq g$ defined in the following way. For functions $f,g\colon [0,+\infty)\to[0,+\infty)$ we say that $f \preceq g$ if there exist $C>0$ such that for all $n\in [0,+\infty)$: $$ f(n)\le Cg(Cn+C)+C. $$ We say that $f\simeq g$ if $f\preceq g$ and $g\preceq f$. We extend this relation to functions $\mathbb N\to [0,+\infty)$ by assuming them to be constant on each interval $[n,n+1)$. [EDIT 07/28/17: Since the function $n\mapsto \|\phi^n(x_i)\|_X$ is not monotone (think of a finite order automorphism modeled on some subset of $X$), it is better to use a slightly different definition of $f\preceq g$: we say that $f\preceq g$ if there exist constants $A,B>0$, $C,D\geq0$ such that for all $n\in [0,+\infty)$: $$ f(n)\le Ag(Bn+C)+D.] $$ It can be shown that, viewed up to $\simeq$ equivalence, functions $\operatorname{gr}_{\phi,X}(n)$ are independent of generating set $X$ (so can be denoted just $\operatorname{gr}_\phi(n)$). And it looks plausible that: If $H\le F$ is a subgroup of finite index, invariant under $\phi$, then $\operatorname{gr}_{\phi|_H}(n)\simeq \operatorname{gr}_{\phi}(n)$. Question: Is the detailed proof of this statement written somewhere in the literature? (I somehow find it difficult to prove that $\operatorname{gr}_{\phi|_H}(n)\succeq \operatorname{gr}_{\phi}(n)$, for arbitrary automorphism $\phi$.) REPLY [3 votes]: An alternate way to address point 5 in Lee Mosher's above answer, using only that $\phi$ and $\phi|_H$ have quasi-isometric mapping tori, is to use a result of Macura (Macura, Nataša, Detour functions and quasi-isometries., Q. J. Math. 53, No. 2, 207-239 (2002). ZBL1036.20033.). Indeed, let $G=F\rtimes_\phi\mathbb Z$ and let $G'=H\rtimes_\phi\mathbb Z$. Suppose that $\phi$ has polynomial growth of order $r$. Then, as Lee explained above, $\phi|_H$ also has non-exponential, and thus polynomial, growth. But Macura's result (Theorem 1.1 in the above) says that $G$ has detour function a polynomial of degree exactly $r+1$, and another application of the same theorem shows that $G'$ has detour function a polynomial of degree $r'+1$, where $r'$ is the degree of the polynomial growth function of $\phi|_H$. But $G$ is obviously quasi-isometric to $G'$, and Macura shows that the order of the detour function is a quasi-isometry invariant, so $r=r'$. (The detour function is very similar to the perhaps more familiar divergence function.)<|endoftext|> TITLE: Blow-ups, pullbacks and proper transforms QUESTION [10 upvotes]: Let $X$ be a smooth projective variety, $Z$ a smooth subvariety of $X$, and let $f:\widetilde{X}\to X$ be the blow-up of $X$ along $Z$. Then for a subvariety $V\subset X$, we have two cohomology classes on $\widetilde{X}$: The pullback $f^*[V]$ and $[\widetilde{V}]$, the class of the proper transform of $V$. My question is the following: Assuming $V\not\subseteq Z$ (but not that the intersection $V\cap Z$ is proper!), is the class $f^*[V]$ effective (i.e., a non-negative sum of classes of subvarieties)? If the intersection $V\cap Z$ is transversal, then $f^*[V]=[\widetilde V]$. More generally, Fulton's book (Chapter 6) gives a formula relating $f^*[V]$, $[\widetilde{V}]$ and the Segre classes of the intersection $V\cap Z$, but it is not clear (to me at least) whether this gives an effective expression for $f^*[V]$. REPLY [3 votes]: I guess you are right... :) It seems to me that this indeed fails quite often. Let $d=\mathrm{codim}_XZ$ and $P=f^{-1}Z\subseteq \widetilde X$ the exceptional divisor. By the assumptions $P\to Z$ is a $\mathbb P^{d-1}$-bundle. First note that as long as $d\leq 2$, then the statement is probably true for the simple reason that if $V\neq X$, then $\dim (V\cap Z)\leq \dim V-1$ and hence the largest dimensional part of the pull-back of the Segre class $s(V\cap Z, V)$ has dimension at most $\dim V$ and hence that's the only component that makes a meaningful appearance in the formula for $f^*[V]$. So, let's go to $d=3$. In this case $P\to Z$ is a $\mathbb P^{2}$-bundle. Now let $V\subseteq X$ be a surface such that $C=V\cap Z\subseteq V$ is a Cartier divisor in $V$. Then the Segre class in question is $$ s(C,V)=[C]-[C]^2. $$ Since we have a nice blow up, the "mysterious" excess normal bundle is just the relative tangent bundle $T_{P/Z}$, but this actually doesn't really matter, all we need is that there is a short exact sequence, $$ 0\to \mathscr O_P(-1) \to f^*N \to E\to 0, $$ where $N=N_{Z/X}$ is the normal bundle of $Z$ in $X$. So we need the class $\left(c(E)\cdot_P f^*([C]-[C]^2)\right)_2$. Observe that $f^*[C]$ has dimension $3$ and $f^*[C]^2=f^*([C]\cdot_V[C])$ has dimension $2$, so we need to intersect the first with $c_1(E)=c_1(f^*N)-c_1(\mathscr O_P(-1) )$ and the second with $c_0(E)=1$. Using that $$ c_1(f^*N)\cdot_P f^*[C]= f^*(c_1(N)\cdot_Z [C]), $$ we get that $$ \left(c(E)\cdot_P f^*([C]-[C]^2)\right)_2= f^*(c_1(N)\cdot_Z [C]-[C]^2)+ c_1(\mathscr O_P(1))\cdot_P f^*[C]. $$ So, it seems that if the intersection $C=V\cap Z$ is such that $C^2>0$ on $V$, and $c_1(N)\cdot_Z C<0$ on $Z$, then the above class will not be effective. Note that these choices make the $f^*$ part of the class negative effective while the other class is "horizontal", so it is unlikely to cancel the "vertical" classes, but in any case one can probably play around with $V$ and $Z$ to make this more precise. It seems that if the codimension of $Z$ is even higher and the intersection $V\cap Z$ is still large in $V$, then there is even more that can go wrong.<|endoftext|> TITLE: What kind of non-cuspidal automorphic representation are not isobaric sums? QUESTION [13 upvotes]: Let's say $\pi$ is an automorphic representation on $GL_3(A_{\mathbb Q})$ (or $GL_n(A_{\mathbb Q})$). If $\pi$ is not cuspidal, what $\pi$ can be other than isobaric sums? If there is such a thing, what's their $L$-functions? Do $L$-functions decompose like those of isobaric sums? REPLY [13 votes]: EDIT. A colleague wrote to me to point out that my original answer to this question was actually completely wrong: I had confused "isobaric" representations with "pure" representations (which are not at all the same thing). The correct answer, according to my colleague, is this. Consider the one-dimensional representation $\sigma = |\cdot|^{1/2} \boxtimes|\cdot|^{-1/2}$ of $(GL_1 \times GL_1)(\mathbf{A})$. Then we can parabolically-induce this up to a representation $I(\sigma)$ of $GL_2(\mathbf{A})$. The representation $I(\sigma)$ is very highly reducible: its irreducible factors are precisely the representations of the form $\pi = \prod_v \pi_v$ where all but finitely many $\pi_v$ are the one-dimensional trivial representation of $GL_2(\mathbf{Q}_v)$, and the remainder are the Steinberg representation. All of these $\pi$'s are automorphic, but only the globally trivial representation (with no Steinberg places) is isobaric in Langlands' sense. Notice that any two representations occurring in $I(\sigma)$ are isomorphic locally almost everywhere, so there is no "strong multiplicity one" for non-cuspidal automorphic reps. The point of isobaric representations seems to be to cut down from all automorphic reps to a smaller class in which there is some hope of multiplicity-one results being true. Original answer: You can already see this for $GL_2$, so let me give an answer there, rather than for $GL_3$. Consider the function $L(s) = \zeta(s - r +\tfrac{1}{2}) \zeta(s + r - \tfrac{1}{2})$, for an integer $r \ge 2$. This is the $L$-function of a non-cuspidal automorphic representation of $GL_2$ (given by the holomorphic Eisenstein series $E_{2r}$). However, it is not isobaric -- on the Galois side, it corresponds to a direct sum of two $\ell$-adic Galois representations which are each pure, but with different weights, while isobaric representations have all components pure of the same weight. ("Same weight" is the literal meaning of the word "isobaric".)<|endoftext|> TITLE: Field with one element look at counting index-$n$ subgroups in terms of Homs to $S_n$, generalization to $F_{1^k}$? QUESTION [5 upvotes]: Main idea shortly: As we discussed recently MO272045, there is beautiful fomula which counts index-n subgroups in terms of homomorphisms to $S_n$. Let me give "field with one element" interpretation of that formula, and propose a formula which can be seen as as $F_{1^k}$-generalization ($S_n$ changed to $\mu_k \wr{S}_n=GL(n,F_{1^k})$, subgroups enriched by pairs: subgroup + character). My question (see details below): whether such generalization is true / known ? Details. 1. The formula: Let $G$ be any finitely generated group, the following is true (see MO272045, Qiaochu Yuan's blog1(proof), blog2, blog3, R. Stanley EC2 Exercise 5.13): $$ \sum_{n \ge 0} \frac{| \mathrm{Hom}(G,S_n) | }{n! } z^n = \exp\left( \sum_{n \ge 1} \frac{|\text{Index}~n~\text{subgroups of}~ G|}nz^n \right) . $$ Here $| \mathrm{Hom}(G,S_n) | $ is the number of homomorphisms from $G$ to the symmetric group $S_n$ (pay attention that we count homomorphisms themselves - not up to conjugation). $|\mathrm{Index}~n~\text{subgroups of}~ G|$ is the number of index $n$ subgroups in $G$ (pay attention subgroups themselves, not up to conjugation). 2. Field with one element interpretation: Although field with one element is quite mysterious, but there are simple and widely agreed heuristics about it: $S_n$ is $GL(n,F_1)$ (see e.g. MO272498) Respectively $Hom(G,S_n)$ or an action on n-points set is n-dimensional representation of $G$ Respectively transitive action on finite set is IRREDUCIBLE representation, i.e. irreducible representations comes from index n-subgroups of $G$ via action on $G/H$ So we come to : $$ \sum_{n \ge 0} \frac{| n\text{-dim}~F_1 ~ \text{representations of}~ G ~\text{with choice of basis} | }{|Automorphisms| } z^n = \exp\left( \sum_{n \ge 1} \frac{|n\text{-dim}~F_1~\text{IRREDUCIBLE representations of}~ G|}{|Automorphisms|} z^n \right) . $$ That interpretation seems clarifying for me, it puts the formula into general framework of "exponential formulas", counting all representations as exponential of irreducible ones. Analogies above hopefully explain the interpretation. Let me comment why $n$ in the denominator at the RHS is interpreted as |Automorphism|. Indeed for n-point set $M$ with transitive action of $G$, we can send any point "x" to any other point "y" and extend it to map $ \phi:M->M$ by $\phi(gx)=g\phi(x)$ in a unique way, thus getting automophism of $M$ which commutes with action of $G$. So we have $n$ such automorphisms. 3. Question on generalization to $F_{1^k}$ Kapranov and Smirnov 1995 (see also MO272698) propose to consider algebraic extension $F_{1^k}$ of the $F_1$ for each $k$ and $GL(n,F_{1^k})$ to be generalized symmetric group $\mu_k\wr {S}_n$ (consisting of generalized permutation matrices whose nonzero entries are in the cyclic group $\mu_k$ of $k$-th roots of unity) So, proposal: $$ \sum_{n \ge 0} \frac{| \mathrm{Hom}(G,\mu_k \wr{S}_n) | }{n! } z^n = \exp\left( \sum_{n \ge 1} \frac{|\text{Index}~n~G\text{-subgroups and their characters to k-th roots of unity }|}nz^n \right) . $$ The idea is that $Ind^G_H V_{trivial} = G/H$ as representation, so we just consider $Ind^G_H V_{character}$ - it gives representation like $G/H$ but twisted by roots of unity. That seems the way to get all irreducible representation of $G$ over $F_{1^k}$. So we arrive to the formula motivated by $F_1$, but its formulation is completely independent of anything related to mysteries of $F_1$. QUESTION: Is the formula true/known ? REPLY [7 votes]: Yes this is known. The paper "On Wohlfahrt series and wreath products" by Y. Takegahara contains slightly more general results and builds on previous work (Muller's paper). Corollary 2.7 says: $$\sum_{n=0}^{\infty}\frac{|\operatorname{Hom}(A,G\wr S_n)|}{|G|^nn!}x^n=\exp\left(\sum_{B\subset A}\frac{|\operatorname{Hom}(B,G)|}{|G||A:B|}x^{|A:B|}\right)$$ where $A$ is any group that has finitely many subgroups of each index, $G$ is finite and $|A:B|$ is the index of $B$ as a subgroup of $A$. REPLY [5 votes]: Thomas M\"uller has done a very careful and thorough study of problems of this type. While I don't know that the formula you write is a special case of Theorem 1 in his paper ``Enumerating Representations in Finite Wreath Products", which appeared in Advances in Math in 2000, you might well find something useful therein.<|endoftext|> TITLE: Endomorphism ring of bimodules QUESTION [6 upvotes]: Given an algebra $A$ with a right $A$-module $M$ with $End_A(M) \cong A$. Then we can view $M$ as a natural $A$-bimodule. When is $M$ as a bimodule indecomposable and what is its endomorphism ring as a bimodule? In fact in my examples $M$ was always indecomposable and the endomorphism ring was isomorphic to the center of $A$ (but I did not calculate many examples, since my computer can not do that). REPLY [8 votes]: The endomorphism ring of such a bimodule is indeed always the center: If $f$ is a bimodule-endomorphism of $M$, then in particular $f\in End({_A M})$ so that by assumption $f(m)=ma$ for some $a\in A$. If this is also a right-module homomorphism, then $\forall b,m: mab = f(m)b = f(mb) = mba$. Because the right action is faithful, this means $\forall b: ab=ba$. This also shows that $M$ is indecomposable iff $Z(A)$ has no non-trivial idempotents iff $A$ is indecomposable as an algebra.<|endoftext|> TITLE: Does Ribet's construction of class fields give us eigenspaces of rank 1? QUESTION [8 upvotes]: Ribet's paper on the Herbrand-Ribet theorem constructs a representation $\rho: Gal(\overline{\Bbb Q}/\Bbb Q) \to GL_2(\mathbb F_q)$ where $q = p^r$ of the specific form: $ \begin{bmatrix} 1 & *\\ 0 & \chi \end{bmatrix}$ where $\chi$ is a power of the cyclotomic character mod $p$. In particular, if we let $K$ be the kernel of $\chi$, the matrix is of the form $ \begin{bmatrix} 1 & *\\ 0 & 1 \end{bmatrix}$ and our representation looks like $\rho: Gal(\overline{\Bbb Q}/K) \to \mathbb F_q$. The image is a subgroup (under addition) of $\mathbb F_q$ and if $B$ is the field fixed by the kernel of this map, then Ribet shows that $B/K$ is an unramified extension with abelian Galois group of the form $(p,\dots,p)$ on which $Gal(K/\Bbb Q)$ acts by $\chi^{-1}$. I believe it is conjectured that $B/K$ should be a cyclic extension. Morever, by the main conjecture of Iwasawa theory, we can deduce that the degree of $B/K$ is less than the p-adic valuation of a particular Bernoulli number and most of the time, this is just $1$. This suggests that $B/K$ should always be cyclic - can we prove this directly? REPLY [3 votes]: I don't think we know how to prove this directly. Indeed, recent works by Wake and Wake–Erickson show that this cyclicity is equivalent to a conjectured improvement of Mazur–Wiles' result to the effect that a suitable localization of the Hecke algebra $\mathfrak{H}_\mathfrak{p}$ is Gorenstein. The conjuction of the two papers shows how Greenberg's conjecture that $\lambda^+=0$ implies the cyclicity you mention. This is quite remarkable, in the sense that Greenberg's conjecture is known to imply the Main Conjecture (at least for abelian fields), but so far none has been able to deduce it from the Main Conjecture. It thus seems that the Main Conjecture lies somehow shallower than Greenberg's, in turn equivalent to the cyclicity of $B/K$. Let me finally mention Kurihara's paper where he directly studies the equivalence between the statement that Ribet's extension attached to $\omega_\mathrm{cyc}^{(1-k)}$ be cyclic and the existence of an element of precise order $p^n$ (where $n=\operatorname{ord}_p(\zeta(1-k))$ inside $\mathfrak{I}/\mathfrak{I}^2$ where $\mathfrak{I}$ is an Eisenstein Ideal.<|endoftext|> TITLE: the criteria for 3-dim manifolds diffeomorphic to $\mathbb{R}^3$ QUESTION [15 upvotes]: In Schoen and Yau's paper "Complete three-dimensional manifolds with positive Ricci curvature and scalar curvature", they mentioned “If $M^3$ is contractible, to prove it is diffeomorphic to $\mathbb{R}^3$, from a topological result by Stallings "Group Theory and Three DimensionalManifolds", it suffices to prove that $M^3$ is simply connected at infinity and irreducible.” I checked Stallings’ book, did not find the exact result they referred to. Can anyone tell me which theorem in Stallings’ book they referred to or any other place to find the detailed proof of the above statement? Thanks. REPLY [10 votes]: Husch and Price is the right reference, but I thought I could give a quick sketch of a proof. Let $X^3$ be simply connected at infinity and contractible. For every compact $C \subset X$, there exists a compact $D$ containing $C$ such that $π_1(X-D)\to π_1(X-C)$ is trivial. Taking a regular neighborhood, we may assume that $D$ is a compact submanifold with boundary. Compress $∂D$ as much as possible in the complement of $C$ to obtain a compact incompressible surface $S \subset X-C$. If $S$ is not a union of 2-spheres, then each non-trivial component is $π_1$-injective in $X-C$ by the loop theorem. But the fundamental group of this surface is in the image of $π_1(∂D) \to π_1(X-C)$, which is trivial, a contradiction. Thus, $S$ must consist of a collection of 2-spheres. Since $X$ is contractible, each such sphere bounds a contractible submanifold, which is a 3-ball by the Poincaré conjecture (and I suppose van Kampen's theorem). Hence $C$ is contained in a 3-ball. Thus, $X$ is a nested union of 3-balls, and is therefore homeomorphic to $R^3$.<|endoftext|> TITLE: What is the relation between BRST quantization and gauge fixing quantization QUESTION [7 upvotes]: To quantize gauge field, one usually use gauge-fixing procedure and then plus ghost field, my question is what the relation between BRST quantization and gauge fixing quantization is? Because it seems the latter is a special case of the former. Thanks! REPLY [6 votes]: I do not think that it makes sense to say that the gauge-fixing is a special case of BRST quantization: $\rightarrow$ The gauge-fixing procedure is actually a normalization technique and it is utilized in order to eliminate unphysical degrees of freedom (gauge directions) in the path integral quantization of the various gauge theories. This -rather naturally- destroys the gauge invariance. A standard method to restore the lost gauge invariance -and thus to recover the physics- is the Batalin-Vilkovisky (BV) or antifield formalism: $\rightarrow$ after introducing some extra fields in the theory (ghost fields and conjugate antifields), an extended action is constructed involving all these variables. This action is invariant under the BRST symmetry operator $s$ (usually called BRST-differential), which now replaces the original gauge symmetry. Its cohomology essentially contains the physics, in the following sense: $\rightarrow$ The BRST-differential is nilpotent $s^2=0$ and thus its cohomological groups $H^n(s)$ can be constructed. $H^0(s)$ consists of the gauge invariant functions (that is: the observables). You can find more details in these notes and the references therein. Edit: The question: What is the BRST-anti-BRST formalism? and the answer there, are also related and may be of some further help.<|endoftext|> TITLE: Locales in constructive mathematics QUESTION [16 upvotes]: It is well known that locales are much more well behaved in a constructive setting than topological spaces. Nevertheless, many authors develop the theory of locales in classical mathematics. Are there any textbooks in which the theory is treated constructively? I'm actually interested in specific questions. First, let us define a map of locales to be an open (resp., closed) embedding if it is a pullback of the open (resp., closed) point of the Sierpiński locale. We say that a locale $X$ is discrete if the diagonal $X \to X \times X$ is open. We say that $X$ is Hausdorff if the diagonal is closed. Now, my questions are: Are these definitions reasonable in a constructive setting? Is it true that the locale corresponding to the frame of subsets of a set is discrete? I think I can prove this for discrete sets (that is, sets with decidable equality). Also, are there some implications in the converse direction? My guess is that the answer to both questions is "no", but I don't have any counterexamples. How discrete, Hausdorff, and spatial locales are related? Again, I cannot prove this, but I guess there are no implications between these notions. REPLY [6 votes]: PreScriptum. Having almost finished the answer below I saw the answer by Simon Henry which largely subsumes mine. I still decided to post it as it contains some details/proofs, so may be viewed as an addendum to that answer. Concerning 2.: First note that $f:X\to Y$ is an open embedding in your sense iff $X$ is up to isomorphism an open of $Y$. (More precisely this means that there is an $U_X\in\operatorname{Opens}(Y)$ and an isomorphism $i:\operatorname{Opens}(X)\cong\{U\in\operatorname{Opens}(Y)\mid U\subseteq U_X\}$ such that $if^*(\_)=U_X\cap\_$ for the frame homomorphism $f^*:\operatorname{Opens}(Y)\to\operatorname{Opens}(X)$ determining $f$. Proof - if the latter holds then the needed map $Y\to$ Sierpiński corresponds to the frame homomorphism $\{\varnothing,\{o\},\{o,c\}\}\to\operatorname{Opens}(Y)$ sending $\{o\}$ to $U_X$, since the pushout of this map along the homomorphism determining the embedding of the open point into Sierpiński is the quotient of $\operatorname{Opens}(Y)$ by the smallest frame congruence identifying $U_X$ and $Y$; conversely, given such a map $f:Y\to$ Sierpiński, take $U_X=f^{-1}(o)$.) Next observe that if $\operatorname{Opens}(Y)=\operatorname{Subsets}(S)$, then for any subset $T\subseteq S$ the frame homomorphism $T\cap\_:\operatorname{Subsets}(S)\to\operatorname{Subsets}(T)$ determines an open embedding in $Y$ of an $X$ with $\operatorname{Opens}(X)=\operatorname{Subsets}(T)$. This in particular applies to $T\subseteq T\times T$ and we are done given that for $\operatorname{Opens}(X)=\operatorname{Subsets}(T)$ and $\operatorname{Opens}(X')=\operatorname{Subsets}(T')$ one has $\operatorname{Opens}(X\times X')=\operatorname{Subsets}(T\times T')$. For the converse direction (and this also addresses 1. a bit), there is an additional necessary condition: if $\operatorname{Opens}(X)=\operatorname{Subsets}(T)$, then not only the diagonal $X\to X\times X$ but also $X\to\text{point}$ is open (i. e. the image of any open of $X$ under $X\to\text{point}$ is an open subset of the single point locale). Such locales are usually called $\textit{overt}$. If the logic is not classical, there might exist non-overt locales with open diagonal. For example, if a singleton $\{*\}$ has a non-complemented subset $S\subset\{*\}$, then $S$ determines an open sublocale $U$ of the single point locale with $\operatorname{Opens}(U)=\operatorname{Subsets}(S)$; then this open sublocale has the complementary closed sublocale $C$, and if $C$ would be open too then $S$ would be complemented in $\{*\}$. Now $C$ is an example of a non-overt locale with open diagonal since ($C$ being a sublocale of the single point locale) the image of $C\to\text{point}$ is $C$ and the diagonal $C\to C\times C$ is an isomorphism. In particular, it follows that there is no set $T$ with $\operatorname{Opens}(C)$ isomorphic to $\operatorname{Subsets}(T)$, since otherwise $C$ would be overt. (Actually $\operatorname{Opens}(C)$ is isomorphic to $\{S'\subseteq\{*\}\mid S'\supseteq S\}$.)<|endoftext|> TITLE: Extending $\Bbb N$ to a semiring with isomorphic additive and multiplicative structure QUESTION [21 upvotes]: Seen $(\Bbb N,+,\cdot)$ as a semiring, is it possible to extend it to a semiring $(R,+,\cdot)$ so that the additive and multiplicative monoids become isomorphic? This means there is some monoid-isomorphism $$\varphi:(R,\cdot)\cong(R,+)$$ and $\Bbb N$ is a sub-semiring of $R$. Here, $\Bbb N$ is meant to include $0$. I do not think that there is such an extension, but I cannot find a contradiction. I also wonder if the problem becomes easiert when asking for an isomorphism $$\varphi:(R\setminus\{0\},\cdot)\cong(R,+)$$ instead? Observations The multiplication will be commutative. Also, there will be many new "numbers", e.g. a unique additively absorbing element $\eta:=\varphi(0)$, i.e. $\eta+x=\eta$ for all $x\in R$. For $n\in\Bbb N^+$ we have $$n\cdot \eta=\underbrace{\eta+\cdots+\eta}_n=\eta.$$ Therefore we have further elements $\tilde\eta=\varphi(\eta)$ that absorbe some numbers when added to them, e.g. $\varphi(n),n\in\Bbb N^+$, but not all (there can be only one universally absorbing element). REPLY [18 votes]: There is an extension $R$: take the closure of $\mathbb N$ by the operations $\text{L}$ (or $\varphi$ in the OP) and its inverse $\text{E}$, which are the logarithm and exponential in base $1.2$. Notice that the choice of base (see below) implies that all new numbers generated by repeated applications of the 4 operations ($+$, $\cdot$, $\text{L}$ and $\text{E}$) are $\gt 1$, except for infinities, so that there is never a need to apply $\text{L}$ to negative numbers. For infinities impose the following rules: $0\cdot x=0$ $\text{L}(0)+x=\text{L}(0)$ $\text{L}(0)\cdot x=\text{L}(0)$ if $x\ne0$ $\text{L}(\text{L}(0))+x=\text{L}(\text{L}(0))$ if $x\ne \text{L}(0)$ $\text{L}(\text{L}(0))\cdot x=\text{L}(\text{L}(0))$ if $x\ne \text{L}(0),0$ $\text{L}(\text{L}(\text{L}(0)))+x=\text{L}(\text{L}(\text{L}(0)))$ if $x\ne \text{L}(\text{L}(0)),\text{L}(0)$ ... These rules obey the semiring axioms and cover all possibilities for mixing reals and infinities. Notice also that $\text{L}(0)$, $\text{L}(\text{L}(0))$, $\text{L}(\text{L}(\text{L}(0)))$ etc. are then the only non-reals in $R$. As for the choice of a base $\lt\sqrt{2}$, notice that if $a=1.2577...$ is the solution to $1.2^a=a$, then both $\text{L}(x)>a$ and $\text{E}(x)>a$ if $x>a$ and therefore the only finite elements in this model $R$ that can be $ TITLE: Categories which are both monadic and comonadic over another category QUESTION [10 upvotes]: I heard a professor say that $\lambda$-rings are both monadic and comonadic over commutative rings. Remark 2.11(a) on the nlab page says the same. What does it mean, intuitively, that a category is both monadic and comonadic over another category? What follows from this formally? What are some additional examples of such phenomena? REPLY [6 votes]: Here is my own rule of thumb about this. One important example of a monadic category over $C$ is one obtained by adjoining to $C$ a system of $n$-ary operations satisfying some relations. For example, the category of rings with involution is monadic over sets: it can be given by two binary operations ($+,\times$), two $0$-ary operations ($0,1$), and two $1$-ary operations (negation and the involution). Perhaps if you adopt a suitably enlightened definition of operations with relations then all monadic functors would arise in this way. Now a $1$-ary operation is just a morphism, so you can also view it as a $1$-ary operation in the opposite category. So if you're adding only $1$-ary operations, then you should get a category which is both monadic and comonadic over the original category. Indeed, the usual way of defining lambda-rings is by adjoining a bunch of $1$-ary operations to the category of commutative rings. The monad is the free lambda-ring functor, and the comonad is the big Witt vector functor (also called the co-free lambda-ring functor). Another example is the category formed from objects of the original category $C$ together with an action of your favorite group (or monoid) $G$. In representation theory, the monad is often called the induced representation functor, and the comonad is called the co-induced representation functor. I'd expect that, as above, with suitably enlightened definitions, all examples would arise in this way. Some things that follow formally from this set up are that the new category has all the same kinds of limits and colimits that the original category $C$ has, and the forgetful functor preserves them. Beck's theorem says that some kind of converse is also true.<|endoftext|> TITLE: What Turing degree would allow you to "compute" the axioms of ZFC in some countable model of ZFC? QUESTION [7 upvotes]: It is established in this post that you there is no computable model of ZFC, yet it can be computed in by a PA-degree oracle machine. Note that when we see "compute a model", we just mean that membership is decidable. My question is what Turing degree do you need if you want more than that? In particular, we require the following: A computable set $S$, representing the sets. A decidable relation $\in_S$, such that $(S,\in_s)$ is a model for ZFC. That $=_s$ is computable (i.e. equality between sets in the model) Given a nonempty set $x$, we can compute a disjoint set $y$ such that $y \in_s x$ (axiom of regularity) For any formula $\phi(x)$ and given a set $x$, we may compute the set $\{x \in_s z : \phi(x)\}$ (axiom schema of specification) Given $x$ and $y$, we can compute $\{x,y\}$ (axiom of pairing) Given $x$, we can compute $\bigcup x$ (axiom of union) Given $x$, we can compute $P(x)$ (axiom of powerset) Given $x$, we can compute a well order for $x$ (Well-ordering theorem) Note that we do need to worry about the axiom of infinity, or any of the axioms of replacement, since they assert that a certain set exists, and we can just "hard code" those constants as output. Of course, this is based only just one axiomatization of ZFC. I could have asked about others, or I could just ask that for any provable statement $\exists x:\phi(x,y)$, given a $y$ we can compute a $x$ but this is just about getting a feel for what Turing degrees we will need for some axiomatization of ZFC. What Turing degree is needed to compute all of the above. (My best guess is that any PA degree would be sufficient, since there is a PA degree computing a model of Morse-Kelly set theory, but I'm not sure.) REPLY [2 votes]: Any PA degree is still sufficient. (Emil answered this in the comments, I'm just expanding on how to do it.) To see this, let's remember the characterization of PA degrees: ${\bf d}$ is PA iff for every computable theory $T$, ${\bf d}$ computes a model of $T$. Now by "model," as you observe, I just mean the atomic diagram (e.g. for ZFC, the underlying set and the membership relation). However, the point is that I can improve this by appropriately changing the theory. In this case, let's consider an expanded language. First, consider the language $$L^-=\{\in, c_{inf}, f_{reg}, f_{un}, f_{pow}, f_{wo}, g_{pair}\}$$ where $c_{inf}$ is a constant symbol, $f_{reg}, f_{un}, f_{pow},f_{wo}$ are each unary function symbols and $g_{pair}$ is a binary function symbol. Now the language $L$ we want consists of $L^-$ together with appropriate-arity function symbols $h_{sep}^\varphi$ and $h_{rep}^\varphi$ for each formula $\varphi$ in the usual language of set theory (you left Replacement out of your list of axioms, but I assume you want it there). Now I'm going to build an $L$-theory, call it "ZFC'," which says (in addition to ZFC) that the additional function symbols we've added witness the corresponding ZFC axioms. This is basically partial Skolemization. E.g., amongst the axioms of ZFC', we'll have: $c_{inf}$ is $\omega$. $f_{pow}(a)$ is the powerset of $x$. $f_{wo}(a)$ is a well-ordering of $x$. $g_{pair}(a,b)=\{a,b\}$. For $\varphi(x, y, z)$ a formula of three variables, $h_{sep}^\varphi(a, b, c)=\{d\in a: \varphi(d, b, c)\}$. And so on. Note that each of these is in fact expressible as an $L'$-sentence; in particular, since $\omega$ has a definition in the language $L$, we can use it as an abbreviation for that definition here. Now the key facts are: ZFC' is computable. If $M\models$ ZFC,' then we can turn it into a model of ZFC of the kind you want (that is, "annotated" appropriately). And now we're done, by invoking the PA degree. Note that this trick is far more general than just applying to ZFC. In fact, for any computable theory $T$ and any PA degree ${\bf d}$, from ${\bf d}$ we can compute a model of $T$ together with its complete theory and all of its Skolem functions. The one change we need to make is that now we don't know what sentences we want to be true, so we need to use conditional Skolemization: e.g. for each formula $\psi(x, y)$ we add a function symbol $f_\psi$ and an axiom stating $$\forall x(\exists y(\psi(x, y))\implies \psi(x, f_\psi(x))).$$ We didn't need to do this in the ZFC case since the formulas we wanted to Skolemize were ones we wanted to hold everywhere.<|endoftext|> TITLE: Manifolds with polynomial transition maps QUESTION [21 upvotes]: Title edited I thank მამუკა ჯიბლაძე and Corbennick for their suggestion on the title of this question. I changed the title based on the suggestion of Corbennick. What is an example of a manifold $M$ which does not admit an atlas $\mathcal{A}$ with the following property?: For every two charts $(\phi,U)$ and $(\psi,V)$ in $\mathcal{A}$, $\psi \circ \phi^{-1}$ is a polynomial map.(Its components are polynomial functions). Motivation: The motivation for this question comes from the concept "Affine manifolds". I am indebted to Mike Cocos, for learning this concept and its related problem, that is Chern conjecture. REPLY [21 votes]: Brief sketch of slight simplification of Bryant's answer: Without loss of generality $\mathcal A$ is maximal with respect to the condition that all transition functions are polynomial. Now make a space $\tilde M$ by gluing together $U$ and $V$ whenever $(\phi, U)$ and $(\psi,V)$ are charts such that $\phi$ and $\psi$ agree in $U\cap V$. This is a smooth manifold (or rather each of its connected components is), because polynomials are so rigid. It is equipped with both a covering projection to $M$ and an immersion in $\mathbb R^n$. So if $M$ is simply connected then $M$ immerses in $\mathbb R^n$. If $M$ is also compact and $n>0$ then this is impossible.<|endoftext|> TITLE: Existence of a "diagonal" set in certain set systems QUESTION [7 upvotes]: Let $\kappa\geq\aleph_0$ be an infinite cardinal, and suppose that ${\cal A}$ is a collection of subsets of $\kappa$ such that for all $A\in {\cal A}$ we have $|A| = \kappa$ and for $A,B\in {\cal A}$ with $A\neq B$ we have $|A\cap B|<\kappa$. Is there $D\subseteq \kappa$ such that for all $A\in {\cal A}$ we have $A\cap D \neq \emptyset$, and $A \not\subseteq D$ ? REPLY [8 votes]: This is false in general when $\kappa=\omega$. Let $\mathcal{A}=\langle A_\alpha:\alpha<\mathfrak{c}\rangle$ be an almost disjoint family of size continuum, and let $\langle D_\alpha:\alpha<\mathfrak{c}\rangle$ list all subsets of $\omega$. For each $\alpha<\mathfrak{c}$, one of $A_\alpha\cap D_\alpha$ and $A_\alpha\setminus D_\alpha$ is infinite, so we can shrink $A_\alpha$ to an infinite subset $B_\alpha$ such that either $B_\alpha\subseteq D_\alpha$ or $B_\alpha\cap D_\alpha=\emptyset$. The family $\langle B_\alpha:\alpha<\mathfrak{c}\rangle$ is still almost disjoint, and for any subset $D$ of $\omega$ there is an $\alpha$ such that either $B_\alpha\subseteq D$ or $B_\alpha\cap D=\emptyset$. REPLY [4 votes]: At least, this is true if $|\mathcal A|=\aleph_0$, and also if $|\mathcal A|<\kappa$. In the former case ($|\mathcal A|=\aleph_0$) we can write $\mathcal A=\{A_1,A_2,\ldots\}$. Choose now elements $a_1\in A_1$, $a_2\in A_2$, $a_3\in A_3,\ldots$ so that $a_2\notin A_1$ (this is possible as $|A_2|>|A_1\cap A_2|$), then $a_3\notin A_1\cup A_2$ (possible in view of $|A_3|>|(A_1\cup A_2)\cap A_3|$), etc. Taking $D:=\cup_i\{a_i\}$, we get $\{a_i\}\subseteq D\cap A_i\subseteq\{a_1,\ldots,a_i\}\subsetneq A_i$ for each $i\ge 1$. In the latter case ($|\mathcal A|<\kappa$), for each $A\in\mathcal A$ we have $|\cup_{B\ne A}(A\cap B)|<|A|$, and therefore there exists $a\in A$ with $a\notin \cup_{B\ne A} B$. Thus, $\mathcal A$ admits a system of distinct representatives, and taking $D$ to be the union of all these representatives, we get $D\cap A=\{a\}$, for every $A\in\mathcal A$.<|endoftext|> TITLE: Connected algebraic subgroup of $PGL_3$ and $PGL_2 \times PGL_2$ QUESTION [5 upvotes]: I am looking for a reference regarding the maximal proper connected algebraic subgroups of $PGL_3$ and $PGL_2 \times PGL_2$ respectively when the base field is any algebraically closed field (of characteristic $\neq 2$). For instance in the case $G=PGL_2 \times PGL_2$, if $H$ is a maximal proper connected algebraic subgroup of $G$, then I believe that $H$ is as follows: $PGL_2 \times B$, where $B$ is a Borel subgroup of $PGL_2$; $B \times PGL_2$; or conjugate to the diagonal embedding of $PGL_2$ in $G$. I know a quite simple proof of this result over a field of characteristic zero but it uses the correspondence between Lie subalgebras and algebraic subgroups. Thank you in advance for your help! REPLY [4 votes]: Here is another answer in case $G = \mathrm{PGL}_2 \times \mathrm{PGL}_2$, just using elementary calculations. So let $H \subset G$ be a connected maximal algebraic subgroup. For $i = 1, 2$, denote by $p_i \colon G \to \mathrm{PGL}_2$ the projection to the $i$-th factor. If $p_1(H) \neq \mathrm{PGL}_2$, then $p_1(H)$ is a maximal (connected) subgroup of $\mathrm{PGL}_2$ and thus $H = B \times \mathrm{PGL}_2$ for some Borel subgroup $B$ of $\mathrm{PGL}_2$. Analogously, if $p_2(H) \neq \mathrm{PGL}_2$, then $H = \mathrm{PGL}_2 \times B$. Now, assume $p_1(H) = p_2(H) = \mathrm{PGL}_2$. Let $N \subseteq H$ be a normal subgroup of $H$. Since $\mathrm{PGL}_2$ is simple, $p_i(N) = 1$ for $i =1, 2$ and hence $N$ lies in the kernel of $p_1$ and of $p_2$. Thus $N = 1$ and therefore $H$ is simple. Since $H$ is a subgroup of $\mathrm{PGL}_2 \times \mathrm{PGL}_2$, it must be isomorphic to $\mathrm{PGL}_2$. Thus, $p_i \colon H \to \mathrm{PGL}_2$ is an isomorphism for $i=1, 2$ and hence H is conjugate to the diagonal embedding (here one uses the fact that all automorphisms of $\mathrm{PGL}_2$ are inner). [ADDED] There is a more direct argument in case $p_1(H) = p_2(H) = \mathrm{PGL}_2$: the kernel of $p_1 |_H \colon H \to \mathrm{PGL}_2$ is the normal subgroup $H \cap \{ 1 \} \times \mathrm{PGL}_2$ of $H$. Since $p_2(H) = \mathrm{PGL}_2$, one can easily see that $H \cap \{ 1 \} \times \mathrm{PGL}_2$ is normal in $\{ 1 \} \times \mathrm{PGL}_2$. Since $H \subsetneq \mathrm{PGL}_2 \times \mathrm{PGL}_2$, this implies that $H \cap \{ 1 \} \times \mathrm{PGL}_2$ is trivial and thus $p_1 |_H$ is an isomorphism. Analogously one sees that $p_2 |_H$ is an isomorphism. Hence $H$ is conjugate to the diagonal embedding (here one uses the fact that all automorphisms of $\mathrm{PGL}_2$ are inner). [ERRATUM] The direct argument above only shows that $p_1 |_H$ and $p_2 |_H$ are bijective.<|endoftext|> TITLE: Probability that biggest area stays greater than 1/2 in a unit square cut by random lines QUESTION [21 upvotes]: The square $[0,1]^2$ is cut into some number of regions by $n$ random lines. We can chose these random lines by randomly picking a point on one of the four sides, picking another point randomly from any of the other three sides and then connecting the dots. We do this $n$ times. What is the probability after $n$ lines that the largest region has area $1/2$ or greater? (A follow-up question: Is the circle in the square best at avoiding random lines?) REPLY [11 votes]: Aaron and fedja have pointed out that the problem is equivalent to finding the convex region in the plane with area $1/2$ with the highest probability that a random line does not intersect it. The optimal convex region $\Delta$ has boundary a union of eight segments, each satisfying a differential equation from a certain one-parameter family, that hence are smooth. Pick a corner of the square, choose coordinates so that that corner is point $(0,0)$, and consider the segment of the boundary $C$ of $\Delta$ whose tangent lines touch the two sides of the square adjacent to that point. If we write this segment of $C$ as the graph of a decreasing function $y(x)$, then the tangent line at the point $(x,y)$ connects the points $\left(x- \frac{y}{\dot{y}},0\right)$ and $\left(0, y+ \dot{y}x \right)$ on these two sides, where $\dot{y}$ is the derivative with respect to $x$. So if we plot the region in the $a,b$ plane consisting of those $(a,b)$ such that the line connecting $(a,0)$ and $(b,0)$ does not intersect $\Delta$, the boundary of that region is the parameteric curve $\left(x- \frac{y}{\dot{y}}, y- \dot{y}x \right)$ and thus the area of the region is $$ - \int \left(x- \frac{y}{\dot{y}} \right) \frac{d}{dx} \left(y - \dot{y}x \right) dx$$ $$= - \int \left(x- \frac{y}{\dot{y}} \right) \left(\frac{dy}{x} - \ddot{y}x - \dot{y} \right) dx$$ $$= \int x \ddot{y} \left(x- \frac{y}{\dot{y}} \right) dx$$ the negative sign being because $\left(y- \dot{y}x \right)$ is a decreasing function of $x$ by convexity. So we are optimizing $$ \int x \ddot{y} \left(x- \frac{y}{\dot{y}} \right) dx$$ subject to an upper bound on $ \int y dx$ which by Lagrange multipliers is equivalent to optimizing $$ \int x \ddot{y} \left(x- \frac{y}{\dot{y}} \right) dx - \lambda \int y dx$$ for some $\lambda>0$. By calculus of variations, if we set $F(y, \dot{y}, \ddot{y}) = x \ddot{y} \left(x- \frac{y}{\dot{y}} \right) - \lambda y$, then the optimal value of $y$ satisfies $$\frac{dF}{dy} - \frac{d}{dx} \left( \frac{dF}{d \dot{y}} - \frac{d}{dx} \left(\frac{dF}{d \ddot{y}} \right) \right) =0$$ We can evaluate $$\frac{dF}{d \ddot{y}} = x^2 - \frac{xy}{\dot{y}}$$ $$\frac{d}{dx} \left(\frac{dF}{d\ddot{y}} \right)= 2x-\frac{y}{\dot{y}} - x + \frac{xy \ddot{y}}{\left(\dot{y}\right)^2}$$ $$ \frac{dF}{ d\dot{y}} =\frac{xy \ddot{y}}{\left(\dot{y}\right)^2}$$ $$ \frac{dF}{d \dot{y}} - \frac{d}{dx} \left(\frac{dF}{d \ddot{y} }\right) = \frac{y}{\dot{y}} -x $$ $$ \frac{d}{dx} \left( \frac{dF}{d \dot{y}} - \frac{d}{dx} \left(\frac{dF}{d \frac{dy^2}{dx^2}}\right) \right) = 1 - \frac{y \ddot{y}}{\left( \dot{y}\right)^2} -1$$ $$\frac{dF}{dy} = - \lambda - x \frac{\ddot{y}}{\dot{y}}$$ so the differential equation is $$ - \lambda - x \frac{\ddot{y}}{\dot{y}} + \frac{y \ddot{y}}{\left( \dot{y}\right)^2} =0$$ or $$ \lambda \left( \dot{y}\right)^2 +( x \dot{y} -y )\ddot{y} =0$$ If we let $t= \dot{y} \frac{x}{y}$ be the dimensionless derivative, then $\dot{y} = t\frac{y}{dx}, \ddot{y} = \frac{d}{dx}\left( t\frac{y}{x}\right)= \dot{t}\frac{y}{x} + t^2 \frac{y}{x^2} - t \frac{y}{x^2} =\frac{y}{x^2} \left( \frac{dt}{d\log x} +t^2-t\right)$ so we can write the equation (ignoring factors of $y$ or $x$) as $$ \lambda t^2 + (t-1) \left( \frac{dt}{d\log x} + t^2-t \right) =0$$ $$\frac{dt}{d \log x} = - \lambda \frac{t^2}{t-1} + t -t^2 $$ so either we have $t$ a constant solution of $(t^2-t)(t-1) + \lambda t^2 =0$ with $y$ a constant times $x^t$ or we can express $\log x$ and $\log y$ as integrals of rational functions of $t$. $$\log x = \int \frac{1}{ - \lambda \frac{t^2}{t-1} + t -t^2} dt$$ $$\log y = \int \frac{t}{ - \lambda \frac{t^2}{t-1} + t -t^2} dt$$ Matt F. in the comments did the integrals and found that the formulas, while explicit, are quite nasty. Perhaps this can be fixed by changing the parameter, but this seems unlikely. It should be possible to do similar calculations for the other kind of segment, but the next step would be to calculate the different ways these segments can be stitched together, which amounts to solving an equation involving eight of these explicit solutions. That seems difficult unless the solutions are really nice - although I'm sure it can be done with the aid of a suitable computer algebra system.<|endoftext|> TITLE: Proof of the Schauder Lemma QUESTION [6 upvotes]: Schauder's Lemma in functional analysis states the following: Let $E$ and $F$ be metrizable locally convex topological vector spaces, and let $E$ be Fréchet. Then if the linear continuous map $A:E\to F$ is nearly open, that is to say, for any neighborhood of the origin $U\subset E$ we have that $\emptyset\ne\operatorname{Int}(\overline{A(U)})\subset F$, then $A$ is surjective and open. However, I've been unable to find a reference for a proof of this, and I haven't managed to figure out why it is true either. Either a proof or a reference to a proof would be much appreciated. The most I've been able to show is that $\overline{A(E)}=F$. REPLY [6 votes]: The result has not much to do with the linear structure. In the book Introduction to Functional Analysis of Meise and Vogt you find a version for metric spaces (Lemma 3.9): Let $X$ and $Y$ be metric spaces; $X$ be complete. Let $f:X\to Y$ be continuous and assume that for every $\varepsilon>0$ there exists a $\delta>0$, such that for all $x\in X$, $\overline{f(U_\varepsilon(x))} \supset U_\delta(f(x))$. Then the map $f$ is open. As far as I remember this can also be found in Bourbaki.<|endoftext|> TITLE: Complete residue system modulo n (permutation of numbers 0 to n-1) such that QUESTION [6 upvotes]: I have a task: Find all $n\ \epsilon \ N, \ n > 1$ for which a permutation $a_1,\ a_2,\ ...,\ an\ $ of numbers $ 0,1, ..., n - 1$ exists such that $a_1,\ a_1+a_2,\ ...,\ a_1+a_2+\ ...\ +an\ $ form a CRS $mod\ n$. So far I've come to the conclusion that $a_1$ must be $0$ because otherwise there would be two equal numbers from the listed above (so it won't be a CRS) and I think that odd numbers don't form CRS because $n*(n-1)/2 \ \ (mod\ n)$ is also $0$ so again won't form CRS. My assumption is that all even numbers can form CRS and satisfy the conditions above since I have found such permutations for numbers 2,4,6,8, but I don't know how to prove it and if it is right at all. Would appreciate some help: first - whether I have made any mistakes so far, and second - with proof if I am right. Thanks in advance! REPLY [11 votes]: Your question is about sequenceable groups, introduced in 1961 by Basil Gordon. A finite group is called sequenceable if its elements can be written as a sequence $(g_1,g_2,\dotsc,g_n)$ so that all the partial products $g_1,g_1g_2,\dotsc,g_1g_2\dotsb g_n$ are pairwise distinct. Gordon has shown that a finite abelian group is sequenceable if and only if it contains exactly one involution (element of order $2$). It is easily seen that this condition is necessary, and Gordon proved that it is also sufficient. A great review of this subject can be found in this paper by Matt Ollis; see also more recent papers by the same author, such as this one.<|endoftext|> TITLE: Properties of right rejective subcategories QUESTION [5 upvotes]: I am reading this paper finiteness of representation dimension, on page 1012 there is a place I don't understand: Why $\mathcal{C}(, \mathbb{F}(X)) \rightarrow [\mathcal{C'}](,X)$ is an isomorphism? How to get $\mathcal{C''}$ is a right rejective subcategory of $\mathcal{C}$ by $\mathcal{C'}$ a right rejective subcategory of $\mathcal{C}$ and $\mathcal{C''}$ a right rejctive subcategory of $\mathcal{C'}$? REPLY [2 votes]: In Peter LeFanu Lumsdaine's answer, he answers most of the question, leaving only the question of why the composition of counits for a chain of right rejective subcategories should be mono. In fact, this is not necessarily true, and is corrected in a later paper of the same author, Iyama, "Representation Dimension and Solomon Zeta Function". In 2.3.1 of that paper, and the footnote on the same page, he adds the condition that the counit of the adjunction for $\mathcal{C}'$ and $\mathcal{C}''$ is pointwise mono in $\mathcal{C}$, not just in $\mathcal{C}'$, which I think solves the problem that Peter identified. He also points out that this doesn't affect any of the applications in the first paper, where all the categories were full subcategories of a module category, and the relevant maps were even monic in the module category. Here's an example to show that the correction is necessary. Fix a field $k$, and let $\mathcal{C}$ be the category of representations of the quiver $\bullet\to\bullet$ over $k$, so an object is a diagram $U\stackrel{\alpha}{\longrightarrow}V$ of vector spaces. Let $\mathcal{C}'$ be the full subcategory of representations where $\alpha$ is surjective, and let $\mathcal{C}''$ be the full subcategory of representations where $\alpha$ is an isomorphism. The inclusion of $\mathcal{C}'$ into $\mathcal{C}$ has a right adjoint which takes $U\stackrel{\alpha}{\longrightarrow}V$ to $U\stackrel{\alpha}{\longrightarrow}\text{im}(\alpha)$, with the counit given by the obvious inclusion. So $\mathcal{C}'$ is a right rejective subcategory of $\mathcal{C}$. The inclusion of $\mathcal{C}''$ into $\mathcal{C}'$ also has a right adjoint taking $U\stackrel{\alpha}{\longrightarrow}V$ (where $\alpha$ is surjective) to $U\stackrel{\text{id}}{\longrightarrow}U$, with the counit given by the obvious surjection. But this surjection is monic in $\mathcal{C}'$, since its kernel in $\mathcal{C}$ is $0\longrightarrow\ker(\alpha)$, which has no nonzero maps from objects of $\mathcal{C}'$. So $\mathcal{C}''$ is a right rejective subcategory of $\mathcal{C}'$. However the fact that this surjection is not monic in $\mathcal{C}$ means that $\mathcal{C}''$ is not a right rejective subcategory of $\mathcal{C}$.<|endoftext|> TITLE: Does this product have analytic continuation? QUESTION [11 upvotes]: The product $$ F(s)=\prod_{p}\frac1{(1-p^{-s})^p}, $$ converges for $\mathrm{Re}(s)>2$, when $p$ runs over all primes. Does it admit analytic continuation beyond the line $\mathrm{Re}(s)=2$? Any papers where it has been studied? REPLY [8 votes]: $$P(s) = \sum_p p^{-s}, \qquad \log F(s) = \sum_{p^k} \frac{p^{1-sk}}{k} = \sum_{k\ge 1} \frac{P(sk-1)}{k}$$ $P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ and $P_N(s) = \sum_{n=N+1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ is analytic for $\Re(s) > \frac{1}{N+1}$ so that $$e^{N! P(s)} = e^{N! P_N(s)}\prod_{n=1}^{N-1} \zeta(ns)^{\mu(n) \frac{N! }{n}}$$ is meromorphic for $\Re(s) > \frac{1}{N+1}$ providing the analytic continuation of $P(s)$ : $P(s)$ has a branch point at $\frac{\rho}{N}$ for each $N\ge 1$ and non-trivial zero $\rho$ of $\zeta$. Therefore $P(s)$ has a natural boundary on $\Re(s) = 0$ and no analytic continuation exists beyond there. For the same reason $F(s)^{N!}$ is meromorphic for $\Re(s) > 1+\frac{1}{N+1}$ and $\log F(s)$ has a branch point at $1+\frac{\rho}{N}$ for every $N\ge 1,\rho$ and hence a natural boundary on $\Re(s) = 1$ and no analytic continuation exists beyond there.<|endoftext|> TITLE: Bounding weight multiplicities by number of certain Coxeter elements QUESTION [12 upvotes]: This question concerns lower bounds of certain weight multiplicities in finite dimensional representations of algebraic groups (or Lie groups, Lie algebras). Let's say $G$ is a simple algebraic group of rank $r$ over $\mathbb{C}$ and $\lambda$ a dominant integral weight. Consider the irreducible finite dimensional representation $V_\lambda$ with highest weight $\lambda$ and let $\mu$ be a weight occuring in $V_\lambda$. Let $W$ be the Wely group of $G$. Suppose that $\lambda$ is regular (lies in interior of Weyl chamber), $\mu$ is dominant, and $\lambda-\mu$ lies in the interior of the positive root cone (the "wide cone"), then it seems to be true that $m_{\lambda\mu}\ge 2^{r-1}$ Here $m_{\lambda\mu}$ is the dimension of $\mu$ weight space in $V_\lambda$. Question: Does anyone know an elementary or straightforward proof (or maybe counterexample) of this lower bound? For example using some multiplicity formula or combinatorial models. I came to this possible lower bound in a very indirect way when studying geometry of certain affine Springer fibers. But I guess there should be a straightforward way to see this, for example by some weight multiplicity formula, which I'm not quite familiar with. It would also be great if this could be seen by using some combinatorial models of algebraic representations (crystals, path models, MV polytopes etc). But since I'm not familiar with all these, any comments or suggestions are welcome. Remarks on the number $2^{r-1}$: Recall $G$ is simple with rank $r$. Fix a set of simple reflections $s_1,...,s_r$ in the Weyl group $W$, then $2^{r-1}$ is the number of elements in $W$ that can be written as products of these $r$ simple reflections, each occurring precisely once. Previously I have abused terminology and called this the "number of Coxeter elements", which lead to some confusion. As far as I know, this abuse of terminology is also present in literature, so it's better to pay attention to the context when seeing these words. Thanks to Jim Humphreys for clarifying this in his answer below. Simply speaking, $2^{r-1}$ only counts those Coxeter elements that can be expressed as products of a given set of simple reflections, which is in general smaller than the number of all Coxeter elements. For $W=S_n$, we have $2^{r-1}=2^{n-2}$ while the Coxeter elements in $S_n$ are the $n$-cycles, so there are $(n-1)!$ of them in total. Update (Nov 1st, 2017): The very indirect way I found this inequality is in here. See Corollary 4.5.2. REPLY [5 votes]: I have a full proof in type $A$, and most of a proof in the other types. Notation: $\alpha_1$, $\alpha_2$, ..., $\alpha_n$ are the (positive) simple roots, $\Phi^{+}$ is the set of all positive roots, $\rho$ is determined by the condition $\langle \alpha^{\vee}_i, \rho \rangle =1$ for all $i$. The element $\sigma$ is defined as $\sum_i \alpha_i$ I'll write $\alpha \geq \beta$ to indicate that $\alpha -\beta = \sum c_i \alpha_i$ for $c_i \in \mathbb{Z}_{\geq 0}$. So the condition that $\lambda-\mu$ is in the interior of the positive cone and occurs in $V_{\lambda}$ states that $\mu \leq \lambda - \sigma$. Our strategy is to show that $$K_{\lambda \mu} \geq K_{\lambda (\lambda - \sigma)} = \# (\mbox{number Coxeter elements}).$$ The gap in this argument is to give a proof in all types of The Annoying Lemma: Suppose that $\beta \geq \delta$ and both $\beta$ and $\delta$ are dominant. Then there is a sequence of dominant weights $\gamma_0$, $\gamma_1$, ..., $\gamma_N$ such that $\beta = \gamma_0$, $\gamma_N=\delta$ and $\gamma_i - \gamma_{i+1} \in \Phi^{+}$. Note that the Annoying Lemma is false if we ask for $\gamma_i - \gamma_{i+1}$ to be a simple root. For example, write dominant $GL_3$ weights as partitions in the usual way and take $\beta = (3,2,1)$ and $\delta = (2,2,2)$. Then the Lemma is true, because $\beta - \delta = (1,0,-1)= \alpha_1 + \alpha_2$ is a positive root. However, either of the two sequences $(\beta, \beta-\alpha_1, \delta)$ or $(\beta, \beta-\alpha_2, \delta)$ has a non-dominant element in the middle (namely, $(2,3,1)$ and $(3,1,2)$ respectively.) Proof of $K_{\lambda \mu} \geq K_{\lambda (\lambda - \sigma)}$ assuming the Annoying Lemma: We immediately reduce to the case that the Dynkin diagram is connected. Also, the $SL_2$ case is immediate, so we assume we are not in it. With those reductions made, $\rho-\sigma$ is dominant. Since $\lambda$ is regular dominant, we have that $\lambda - \rho$ is domininant, so $\lambda - \sigma$ is dominant. We may therefore apply the Annoying Lemma to obtain a chain $\gamma_0 = \lambda - \sigma$, $\gamma_1$, $\gamma_2$, ..., $\gamma_N = \mu$ where $\gamma_i - \gamma_{i+1} \in \Phi^+$. It is enough to show that $K_{\lambda \gamma_{i+1}} \geq K_{\lambda \gamma_{i}}$. Restrict to the $SL_2$ corresponding to $\pm (\gamma_i - \gamma_{i+1})$. Then the weights $\gamma_i$ and $\gamma_{i+1}$ lie on the same $SL_2$ string, and the condition that they are both dominant says that $\gamma_{i+1}$ lies nearer the center than $\gamma_i$, so $K_{\lambda \gamma_{i+1}} \geq K_{\lambda \gamma_{i}}$ as desired. $\square$ Proof that $K_{\lambda (\lambda - \sigma)} = \#(\mbox{number of Coxeter elements})$. We recall the BGG resolution $$0 \leftarrow V_{\lambda} \leftarrow M_{\lambda} \leftarrow \bigoplus_{i} M_{\lambda - (\langle \alpha_i^{\vee}, \lambda \rangle +1) \alpha_i} \leftarrow \cdots$$ where $M_{\kappa}$ is the Verma module with highest weight $\kappa$. Since $\lambda$ is regular dominant, $\langle \alpha_i^{\vee}, \lambda \rangle \geq 1$ and thus $\lambda - \sigma = \lambda - \sum_j \alpha_j$ is not a weight of $M_{\lambda - (\langle \alpha_i^{\vee}, \lambda \rangle +1) \alpha_i}$. So the multiplicity of $\lambda - \sigma$ in $V_{\lambda}$ is the same as in $M_{\lambda}$; that is to say, it is the multiplicity of $-\sigma$ as a weight of the universal enveloping algebra of $\mathfrak{n}_-$. Let $e_1$, $e_2$, ..., $e_n$ be the Chevalley generators of $\mathfrak{n}_-$. Then $U(\mathfrak{n}_-)$ is generated by the $e_i$ modulo the Chevalley-Serre relations. We see that a monomial in the $e_i$ is of degree $\sigma$ if and only if it uses each $e_i$ once, and the only Serre relations in such low degree are those of the form $e_i e_j = e_j e_i$ when $A_{ij} =0$. So the multiplicity of $-\sigma$ as a weight of $U(\mathfrak{n}_+)$ is the number of permutations of $e_1$, ..., $e_n$, up to interchanging commuting elements; this is precisely the description of the Coxeter elements. (It is also not hard to directly prove, using for example PBW bases, that the number is $2^{n-1}$, but I thought this was more fun.) The annoying Lemma in Type A This paper by Matthew Fayers asserts the Annoying Lemma in type A as Proposition 2.3 but leaves the proof to the reader. Prop 1.2 of this paper is the same application that I intended -- showing that if $\mu \geq \nu$ and $\mu$ and $\nu$ are both dominant then $K_{\lambda \mu} \leq K_{\lambda \nu}$. I have completed Fayer's exercise but am delaying posting the solution in hopes that I'll find a less messy one that works for all types.<|endoftext|> TITLE: limits of stable theories QUESTION [10 upvotes]: Say that a complete theory $T$ is a limit of stable theories if for every $\phi \in T$ there is a stable completion of $\{\phi\}$. (Equivalently, $T$ is the ultraproduct of stable theories.) Question: Is every simple theory a limit of stable theories? (We allow that theories of finite structures are stable.) As motivation, all the simple unstable theories I am aware of are random in various ways (e.g. random graph) and the independence property only comes from the full random schema. Example: any pseudo-finite theory is a limit of stable theories. This includes a lot of of simple theories, e.g. the theory of the random graph. It also includes pseudo-finite linear orders, which are SOP. Example: suppose $T$ is a completion of ACFA (the model companion of algebraically closed fields with an automorphism), so $T$ is simple. Suppose $\phi \in T$. Then $\phi$ is consistent with ACFA, so for arbitrarily large powers of primes $p^n$, $(F, x \mapsto x^{p^n}) \models \phi$, where $F$ is some (any) algebraically closed field of characteristic $p$. But these structures are stable. A strong counterexample would be a finitely axiomatizable simple unstable theory. REPLY [6 votes]: No, not every simple theory is a limit of stable theories. For example, let $K$ be a pseudoalgebraically closed field with a small, nontrivial Galois group not isomorphic to $\widehat{\mathbb{Z}}$. Possibly after replacing $K$ with a finite algebraic extension, there will be some natural number $n$ coprime to the characteristic for which the $n^\text{th}$ power map is not onto. By smallness of the Galois group, the group $K^\times/(K^\times)^n$ is finite, say of size $m$. That the absolute Galois group is not isomorphic to $\widehat{\mathbb{Z}}$ is reflected by there being some natural number $\ell$ for which either there is no extension of degree $\ell$ or there are two distinct extensions of degree $\ell$. The theory of $K$ is supersimple, but the sentence $\phi$ which asserts that $K$ is a field, that there is some $x \in K$ which is not an $n^\text{th}$ power, that there are $m$ elements $y_1, \ldots, y_m$ of $K$ so that every nonzero element of $K$ may be expressed as $x_i z^n$ for some $z \in K$ and $1 \leq i \leq m$, and that either there are no extensions of degree $\ell$ or there are two distinct such extensions holds in $K$ but cannot hold in any stable structure: any such structure is a field (by the first item) and finite fields cannot satisfy the fourth item while infinite stable fields have connected multiplicative groups and hence cannot satisfy the conjunction of the second and third items.<|endoftext|> TITLE: Decomposition of Henstock-Kurzweil-integrable functions QUESTION [9 upvotes]: Let $f:[a,b]\to\mathbb R$ be a Henstock-Kurzweil-integrable function (short: HK-integrable). Can $f$ always be written as a sum of a Lebesgue-integrable function and a function which has a classical primitive, i.e. are there $f_1\in L^1([a,b])$ and an everywhere differentiable function $F:[a,b]\to\mathbb R$ such that $f=f_1+F'$? Since every integral function of HK-integrable functions is $ACG_*$, the following question is equivalent: Can every $ACG_*$-function be written as a sum of an absolutely continuous function and an everywhere differentiable function? For me it is sufficient to (dis)prove this "only" for HK-integrable functions $f$ which are in $L^1([c,b])$ for each $c\in(a,b)$. Since HK-integration is brand new to me, I don't really have a feeling whether this is true or not. By reading a couple of books I've gathered a bunch of nice properties of HK-integrable functions, but none of them helped. I would be very grateful if anyone who is more familiar with this type of integration theory can at least have a look at this problem. Thank you very much in advance! REPLY [4 votes]: The answer is negative. Consider the space $\Delta'=\Delta'([a,b])$ of functions, which are classical derivatives of functions $[a,b]\to\mathbb R$. Assume the claim was true. Then any HK-integrable $f:[a,b]\to\mathbb R$ had a representation $f=f_1+f_2$, where $f_1\in L^1([a,b])$ and $f_2\in\Delta'$. Let $g$ be a multiplier for $\Delta'$, i.e. a function $[a,b]\to\mathbb R$ with the property that $hg\in\Delta'$ whenever $h\in\Delta'$. Then it is easy to see that $g$ is bounded and hence belongs to $L^\infty([a,b])$. Then $f_1g\in L^1([a,b])$ and hence $f_1g$ is HK-integrable, and $f_2g\in\Delta'$ and thus $f_2g$ is HK-integrable. But then $fg$ where HK-integrable. Since $f$ was arbitrary, $fg$ is HK-integrable whenever $f$ is HK-integrable. But then it is well-known that $g$ must agree almost everywhere with a function of bounded variation, and since $g$ has a primitive, it can be shown that $g$ must be continuous and of bounded variation. This would imply that every multiplier for $\Delta'$ is continuous and of bounded variation, which is not true in general. There are examples of such multipliers which are neither.<|endoftext|> TITLE: What homotopy classes can attaching an $E_n$-cell kill? QUESTION [15 upvotes]: Let $A$ be a connected $E_{n+1}$-ring spectrum and let $\alpha\in\pi_k(A)$. I am having trouble showing that attaching an $E_n$-cell along $\alpha$ will necessarily not kill an element $\beta\in\pi_k(A)$ unless $\beta$ is a multiple of $\alpha$. Perhaps this is not true, but it seems plausible to me since attaching an $E_n$-cell seems like everything above the cell that kills $\alpha$ itself should be in higher dimensions. To be precise: we can "attach an $E_n$-$A$-cell" to $A$ along $\alpha$ by taking the pushout of the following span in $E_n$-$A$-algebras: $$A\overset{\overline{0}}\leftarrow F_{E_n}(\Sigma^kA)\overset{\overline{\alpha}}\to A $$ where $F_{E_n}$ is the free $E_n$-$A$-algebra functor, $\overline{0}$ is the adjoint of the zero map $\Sigma^kA\to A$ and $\overline{\alpha}$ is the adjoint of the $A$-module map $\Sigma^kA\to A$ induced by $\alpha$. It seems to me that there should be a "bottom layer" of $F_{E_n}(\Sigma^kA)$ that kills of $\alpha$, but that everything else (used to kill off the powers of $\alpha$ in a homotopy coherent way) should happen in higher dimensions, so cannot kill $\beta$. Is this true? It may be useful to notice that, according to Lemma 4.4 of this paper of Antolin-Camarena and Barthel, that the above pushout is equivalent to $Ind_0^n(cof(\Sigma^k A\to A))$ where $Ind_0^n$ is the left adjoint to the forgetful functor from $E_n$-$A$-algebras to $E_0$-$A$-algebras (where $E_0$-$A$-algebras here just means unital $A$-modules). I should also mention that this is the pretty clearly the best we can do in general, since as Tyler pointed out in the comments it's relatively easy to attach a structured cell along something in degree $k$ and kill something in degree $k+1$. And his example is not rare either. I can construct an infinite family of ring spectra in which this occurs for arbitrarily large degrees, when only attaching $E_1$-$A$-cells. REPLY [5 votes]: This is not an answer but rather extended commentary on Tom Bachmann's answer. I hope to address some of Jon's comments on that answer. Bar constructions with respect to coproducts compute pushouts First let me address Tom's point (3), that an instance of the two-sided bar construction computes pushouts. This is probably well-known but I had to think about it for a bit, so I'll include an explanation in case it helps someone else. Also, this explanation helps later with coherence issues for maps between simplicial objects. The two-sided bar construction can be defined very generally, for a monoid $M$ for some monoidal structure $\otimes$ together with left and right $M$-modules $L$ and $R$. Now consider the special case when $\otimes$ is given by the coproduct. In this case, the monoidal structure on $M$ is unique, given by the fold $M \sqcup M \to M$, and the module structures on $L$ and $R$ are given simply by morphisms $M \to L$, $M \to R$. And the claim is that the geometric realization of the bar construction is simply the pushout of $R \leftarrow M \to L$. One way to see that is to consider the functor $b : \mathrm{Span} \to \Delta^\mathrm{op}$ where $\mathrm{Span} = \{ r \leftarrow m \to l \}$ and $b(m \to l) = (d_0 : [1] \to [0])$ and $b(m \to r) = (d_1 : [1] \to [0])$. It's not hard to check that the coproduct-based bar construction $\mathrm{Fun}(\mathrm{Span}, \mathcal{C}) \to \mathrm{Fun}(\Delta^\mathrm{op}, \mathcal{C})$ is given by left Kan extension along this functor $b$. The usual argument about left Kan extending along the composite $\mathrm{Span} \to \Delta^\mathrm{op} \to \ast$, shows that a span and its corresponding bar construction have the same colimit. Defining the relevant map of simplicial $A$-modules Instead of trying to check that Tom's maps are compatible with faces and degeneracies, let's try to build simplicial maps wholesale. I'll use $F : \mathrm{Mod}_A \to \mathrm{Alg}^{E_n}_{A}$ for the free $E_n$-$A$-algebra functor defined on $A$-module spectra, and $U : \mathrm{Alg}^{E_n}_{A} \to \mathrm{Mod}_A$ for the corresponding forgetful functor. The key ingredient in Tom's argument is a simplicial $A$-module $B_{\bullet}$ whose geometric realization is $U(A/\!/\alpha)$, the underlying $A$-module of the $E_n$-$A$-algebra in Jon's question. By the previous section, the simplicial $E_n$-$A$-algebra $\mathcal{B}_{\bullet} := \mathrm{Lan}_b(A\overset{\overline{0}}\leftarrow {E_n}(\Sigma^kA)\overset{\overline{\alpha}}\to A)$ has geometric realization given by the pushout of that span, $A /\!/ \alpha$. Since the forgetful functor $U$ preserves geometric realizations, the simplicial $A$-module $B_{\bullet} := U \circ \mathcal{B}_{\bullet}$ will have $U(A /\!/ \alpha)$ as geometric realization. Now let $C_{\bullet} := \mathrm{Lan}_b(0 \leftarrow \Sigma^k A \to A)$ be the coproduct-based bar construction in $\mathrm{Mod}_A$. We want to define a morphism of simplicial $A$-modules $C_{\bullet} \to B_{\bullet} = U \circ \mathcal{B}_{\bullet}$, or equivalently a morphism of simplicial $E_n$-$A$-algebras $F \circ C_{\bullet} \to \mathcal{B}_{\bullet}$. Now, since $F$ is a left adjoint it preserves the left Kan extension defining $C_{\bullet}$, so $F \circ C_{\bullet}$ is the bar construction for the span $F(0) \leftarrow F(\Sigma^k A) \to F(A)$. The following diagram is a natural transformation between that span and the one defining $\mathcal{B}_{\bullet}\require{AMScd}$: $$\begin{CD} F(0) @<{F(0)}<< F(\Sigma^k A) @>{F(\alpha)}>> F(A) \\ @V{\mathrm{id}}VV @V{\mathrm{id}}VV @V{\mu_A}VV \\ A @<{\bar{0}}<< F(\Sigma^k A) @>{\bar{\alpha}}>> A \\ \end{CD}$$ This map of spans induces a simplicial map between their respective bar constructions. The rest of the argument I think that now the rest of Tom's argument runs fine: let $B'_{\bullet}$ be the cofibre in simplicial $A$-modules of $C_{\bullet} \to B_{\bullet}$. Now, without worrying about compatibility with faces and degeneracies, you can identify for each $n$ the map $C_n \to B_n$ to describe $B'_n$ and see that $B'_n$ is $2k$-connective.<|endoftext|> TITLE: The number of commuting m-tuples is divisible by order of group: Improvements? QUESTION [19 upvotes]: The number of commuting pairs of elements in finite group G is equal to the product $k(G)*|G|$ (see MO271757 ) where $k(G)$ is the number of conjugacy classes. Thus it is is divisible by $|G|$ (the number of elements of $G$). That divisibility also follows from a theorem by L. Solomon, stated below. Question 0: It seems that the number of commuting $m$-tuples $c_m(G)$ is also divisible by $|G|$ for any $m$; is this correct? It appears to follow from a result cited in Klyachko, Mkrtchyan (details below). Question 1: Does the ratio $c_m(G)/|G|$ have some group theoretic interpretation for $m>2$? (When $m=2$, this is the number of conjugacy classes). If the group is abelian, then obviously $c_m(G) = |G|^m$, so it is divisible by a very high power of $|G|$. Question 2: Is there any improvement possible for this type of divisibility by $|G|$ for nilpotent or $p$-groups ? Remark: Any improvement cannot contradict the analogues of the 5/8 bound for general $m$: $c_{m+1}(G) \leq \frac{3 \cdot 2^m - 1}{2^{2m+1}} |G|^{m+1}$ by Lescot (see MO108392). Reminder Let me state theorems cited in Klyachko & Mkrtchyan 2012 (found in MO98639), and apply it to our situation. Solomon theorem [1969]. In any group, the number of solutions to a system of coefficient-free equations is divisible by the order of this group if the number of equations is less than the number of unknowns. Application Consider the equation $xy=yx$ in a group---one equation, two unknowns---thus the number of solutions should be divisible by the order of the group. Hence the number of commuting pairs is divisible by the order of group. Note that we cannot apply that theorem for the number of commuting triples, $m$-tuples, since the number of equations exceeds the number of unknowns. So one needs a refinement, and that seemed to be known: Gordon,Rodriguez-Villegas theorem arXiv:1105.6066. In any group, the number of solutions of a system of coefficient free equations is divisible by the order of this group if the rank of the matrix composed of exponent sums of $i$th unknown in $j$th equation is less than the number of unknowns. (It is presented like this by Klyachko & Mkrtchyan 2012, it is not immediately clear (to me) how to extract this formulation from the original paper). Application Consider equations defining commuting $m$-tuples: $x_ix_jx_i^{-1}x_j^{-1} = 1$. The sums the of exponents is ZERO! Thus the rank of the matrix is zero and hence the theorem ensures that the number of solutions is divisible by order of $G$. I hope that this correct, and that an expert can confirm it. Motivation I hope (see MO271752) that there should be a nice generating function, $$ \sum_{n\geq 0} \frac{|\mathrm { commuting~} m\mathrm{-tuples~ in~ GL}(n,F_q)|}{|GL(n,F_q)|} x^n = ??? $$ This is similar to the known, $$ \sum_{n\geq 0} \frac{|\mathrm { commuting~} \mathrm{pairs~ in~ GL}(n,F_q)|}{|GL(n,F_q)|} x^n = \prod_{j\geq 1} \frac{1-x^j}{1-qx^j} $$ which would imply divisibility results at least for $GL(n,F_q)$ (and some other groups too), so it is nice to have support for such a belief. About other groups, it would be quite interesting for me to know especially about the group $UT(n,q)$ (unitriangular matrices over a finite field)--- by what power of $q$ the numbers $c_m(G)$ are divisible. REPLY [7 votes]: Another strategy is to use induction on $m,$ there being nothing to do if $m=1.$ Note that $(x_{1},x_{2},\ldots,x_{m})$ is a commuting $m$-tuple if and only if $(x_{2},\ldots,x_{m})$ is a commuting $(m-1)$-tuple of $C_{G}(x_{1}).$ By induction then, for every choice of $x_{1},$ the number of completions of $x_{1}$ to a commuting $m$-tuple with first component $x_{1}$ is divisible by $|C_{G}(x_{1})|.$ It follows easily that given a conjugacy class $C$ of $G,$ the number of commuting $m$-tuples whose first component is an element of $C$ is divisible by $|G|.$ Since the conjugacy classes partition $G,$ the number of commuting $m$-tuples is indeed divisible by $|G|.$ Notice also that if $p$ is a prime which does not divide $|G|$, then the number of mutually commuting $p$-tuples of elements of $G$ is congruent to $|G|$ (mod $p$), since the cyclic permutation of order $p$ acts on such commuting $p$-tuples with $|G|$ fixed points (the ordered $p$-tuples with all components equal).<|endoftext|> TITLE: Is there a classification of pointed nodal genus 1 curves? QUESTION [6 upvotes]: Any pointed nodal (ie, proper semistable with a specified rational point lying in the smooth locus) curve of arithmetic genus 1 over a field $k$ must be irreducible and has precisely 1 node, which must be rational over the base field, and its normalization must be $\mathbb{P}^1_k$. Over an algebraically closed field, all such curves are isomorphic to the compactification of the plane curve $y^2 = x^3 + x^2$, say with the marked point at $P = (0,0)$. Over a general field $k$, is there a classification of pointed nodal curves of arithmetic genus 1 (such curves would essentially be twists of $y^2 = x^3 + x^2$)? REPLY [3 votes]: I am just posting my comment as an answer. The closed substack $\Delta$ of $\overline{\mathcal{M}}_{1,1}$ parameterizing pointed, nodal, stable curves of arithmetic genus $1$ is naturally equivalent to the classifying stack $BC_2$, where $C_2$ is the cyclic group of order $2$. Thus, for every scheme $S$ (or algebraic space), the set of equivalence classes of $1$-morphisms from $S$ to $\Delta$ is bijective to the set of isomorphism classes of $C_2$-torsors over $S$, i.e., finite, degree $2$, étale covers of $S$ (possibly disconnected). For a field $k$ of characteristic $\neq 2$, the set of isomorphism classes is bijective to $k^\times/(k^\times)^2$.<|endoftext|> TITLE: Internal hom in $(\infty,2)$-categories QUESTION [7 upvotes]: Let $X,Y$ be two $(\infty,2)$-categories, viewed as two fibrant objects in $\mathrm{Fun}(\Delta^{op},\mathrm{Set}_\Delta)$ with the complete Segal model structure (one uses the Joyal model structure on $\mathrm{Set}_\Delta$ to define this). It has been proved that the $(\infty,1)$-category 2-$\mathrm{Cat}$ of $(\infty,2)$-categories is Cartesian closed. The category of bisimplicial sets $\mathrm{Fun}(\Delta^{op}, \mathrm{Set}_\Delta)$ is also Cartesian closed. My question is: Does $Y^X$ in 2-$\mathrm{Cat}$ coincide with $Y^X$ in $\mathrm{Fun}(\Delta^{op}, \mathrm{Set}_\Delta)$, i.e. is the latter one fibrant? If not, is there a reference about the relation between the two (more specific than saying the first $Y^X$ is a fibrant replacement of the latter $Y^X$)? REPLY [5 votes]: Yes because the model structure on Fun(Δ^op, sSet) is a cartesian model structure. This follows from the fact that the Joyal model structure is cartesian, and §10, §11 in Rezk's paper “A model for the homotopy theory of homotopy theory”.<|endoftext|> TITLE: Notable examples of syntactic proofs whose existence is guaranteed by completeness, but having been found later than a semantic proof? QUESTION [7 upvotes]: Question. What are examples (preferably documented and explicitly commented on from this perspective in the literature, preferably in an article dedicated to this aspect alone) of the following well-known aspect of the usual completeness theorem for first-order logic (summarizing it here slightly flippantly, for brevity)? If $\mathbb{T}$ is a theory in first-order logic (with $=$) over a language $L$, and if $\sigma$ is any $L$-sentence, and if $\vDash$ denotes entailment w.r.t. a given semantics for $L$, and if $\vdash$ denotes existence of a finite proof w.r.t. a given usual proof system, and if you prove $\mathbb{T}\vDash\sigma$ with no holds barred, then you know the existence-statement $\mathbb{T}\vdash\sigma$ to be true. Remark. Motivation for the question is partly expository writing, partly my working on an open problem about triangle-free graphs whose statement is one first-order sentence in a relational language. It seems especially interesting, in particular for expository purposes, to have notable examples of a first-order syntactic proof having to exist by the completeness theorem but not yet having been found so far (and researchers in the field being aware of that and deploring it), and there being some hope that the shortest syntactic proof is short enough to be found in future (and possible even appreciably simple). Even though my research-motivation is about a statement which does not even use function-symbols, my exposition does not make any such restriction to purely relational languages, and I would also appreciate examples involving first-order statements which do use function symbols. In expositions (I will not give examples here since such mentionings would be rather negativistic, to the effect of "Look, Author A does not give an example.") of the usual completeness theorem for first-order logic, one sometimes encounters a discussion pointing out the above consequence of the completeness theorem, but I have not seen any notable example being given in such expositions. It is easy to devise very artificial examples. The metaphor "with no holds barred" in the above refers to any mathematical theory or logic being allowed to prove that each model of $\mathbb{T}$ is a model of $\sigma$. This MO thread is similar in spirit, but technically quite different. All famous examples (that I can think of, that is) of ("elementary" here in an informal sense) elementary statements first being proved by non-elementary methods and later being given an elementary proof do not qualify as examples, for one technical reason or the other. (For example, it would be too much of a stretch to pass off e.g. the elementary proofs given by P. Erdős and A. Selbert of the theorem on the distribution of the primes as an example. The statement each Robbins algebra is a Boolean algebra fits the logical bill, but there the first-proved-by-semantic-non-elementary-methods-bit is totally lacking: the first proof found for this was syntactic. Some other examples I tried do not fit for similar reasons.) REPLY [9 votes]: There's a good reason that concrete examples are going to be rare. Existence of a first-order proof (at least in a countable language) is a $\Sigma_1$ property in the language of arithmetic, and it's a general metamathematical principle that proofs of $\Sigma_1$ statements should be constructive. More precisely, if you have a proof that such a proof exists via the completeness theorem, one should be able to formalize this proof in some strong enough theory of arithmetic and then extract a syntactic proof (for instance, using the functional (or "Dialectica") interpretation). That doesn't mean there can't be such examples---the process of extracting a syntactic proof could be tedious, or simply not done yet---but it's unusual and unlikely to last for long if there's interest in obtaining a syntactic proof.<|endoftext|> TITLE: Property between trace class and Hilbert-Schmidt QUESTION [7 upvotes]: Consider the following condition on a bounded operator $T$ on a Hilbert space: $\ \ \ \ \ $(A) there exists an orthonormal basis $(e_j)$ with $\sum_j\parallel Te_j\parallel<\infty$. We have the implications $\ \ \ \ \ $(trace class) $\ \Rightarrow\ $ (A) $\ \Rightarrow\ $ (Hilbert-Schmidt) But what about the converse directions? Are they both false or is one of them true? REPLY [5 votes]: Condition (A) is equivalent to $T$ being trace class. I will use the usual notation $|T|:=(T^\ast T)^{1/2}$. 1. Assume that condition (A) holds. Then $$\sum_j\langle|T|e_j,e_j\rangle\leq\sum_j\||T|e_j\|=\sum_j\|Te_j\|<\infty.$$ The left hand side is finite, hence $T$ is trace class by definition. 2. Assume that $T$ is trace class. Let $(e_j)$ be an orthonormal eigenbasis of $|T|$. Then $$\sum_j\|Te_j\|=\sum_j\||T|e_j\|=\sum_j\langle|T|e_j,e_j\rangle<\infty.$$ The left hand side is finite, hence condition (A) holds. Note that this answers the question completely, since trace class is a strictly stronger property than Hilbert-Schmidt.<|endoftext|> TITLE: Are there infinitely many primes N congruent to 1 mod 8 with h(-N) and h(-2N) both powers of 2? QUESTION [5 upvotes]: The title says it all. For N less than 1000, if I've looked up the tables correctly, when the condition holds N can only be 17, 41, 73, 113, or 257. Motivation: Let N be an odd positive integer, f in Z[[q]] a modular form of level Gamma_0 (N), and p a prime not dividing 2N. I'll say N satisfies the mod 2 local nilpotence condition * if: (*) For every choice of f and p, some power of the Hecke operator T_p takes f into 2Z[[q]]. Suppose in particular that N is prime. If N is 3,5 or 7 mod 8 and satisfies * then N is 3,5 or 7. Suppose however N is 1 mod 8. If h(-N) is not a power of 2 one may use an ideal-class-character whose order is not a power of 2 to construct first a weight 1 modular form of level 4N with quadratic character, and then from this form a mod 2 eigensystem of level N other than the identically zero eigensystem, thereby showing that * does not hold. A similar argument applies when h(-2N) is not a power of 2. So a necessary (but not sufficient) condition for * to hold is that h(-N) and h(-2N) are powers of 2. (The LMFDB database can be used to show that * fails when N is 73,113 or 257. I'm not sure what happens when N is 17 or 41) REPLY [5 votes]: Up to more than 10^8, the list is 17 41 73 113 257 1153 1217 2657 4481 4993 5569 57649 1164817 140618353 I would guess this goes on.<|endoftext|> TITLE: Can one define quantized universal enveloping algebras in a basis-free way? QUESTION [13 upvotes]: (For the background, I am learning about quantum groups — essentially in order to understand crystal/global/canonical bases in the context of this question — from the books by Jantzen and by Hong&Kang and the 1995 paper "On Crystal Bases" by Kashiwara, along with a few others.) The definition given of a quantized universal enveloping algebra (at least in the sources mentioned in the above parenthesis, or here on Wikipedia) is an explicit construction by generators and relations. What I would like to understand is whether this is merely convenient (simply construct the objects we care about and then work with them) or if there is some deeper reason: Is there an alternative definition of the quantized universal enveloping algebra of a semisimple Lie algebra $\mathfrak{g}$ that does not involve giving an explicit construction with generators and relations? (I am willing to restrict myself to finite dimensional $\mathfrak{g}$ if that helps.) This could mean, for example, a combination of one of the following ideas that come to my mind: Abstractly defining a quantum deformation of the universal enveloping algebra of a semisimple Lie algebra. Defining a particular condition on Hopf algebras (along the lines of "the category of modules is semisimple") and then classifying the algebras satisfying this condition using root systems just like one does for finite-dimensional semisimple Lie algebras. Describing the quantized universal enveloping algebra as solution to a universal problem (or representing some functor). Perhaps only in the classical ($A_n$, $B_n$, $C_n$, $D_n$) cases, constructing the algebra starting from a "standard representation" that itself can be obtained from basis-free data (such as a vector space perhaps with a quadratic form attached to it, or something). Using the "canonical basis" to define the algebra in the first place. (Maybe some of these ideas are completely stupid. I merely list them in order to explain the sort of thing I'd be happy to see.) Even a construction that still involves generators and relations but avoids choosing a basis of the root system would be interesting to see. As things stand, I don't even understand to what extent the $e,f,k$ generators of the algebra can be recovered from the algebra itself, or what choices have to be made for that (this is admittedly a different question, but I imagine it is strongly related), so answers along that line are also welcome. REPLY [13 votes]: For complex simple $\mathfrak g$, Drinfeld (1986, p. 807) already characterized his $\mathrm U_h\mathfrak g$ as the unique (up to equivalence and change of parameter) deformation of $\mathrm U\mathfrak g$ admitting a “quantum Cartan involution”, i.e. an algebra automorphism and coalgebra antiautomorphism extending the Cartan involution and for which comultiplication is cocommutative on the Cartan subalgebra. (This exact statement is from Shnider (1991, Cor. 3), more details and a proof in Shnider-Sternberg (1993, p. 285).) REPLY [4 votes]: I do not know the answer in general. But towards the end of the OP you say: "Even a construction that still involves generators and relations but avoids choosing a basis of the root system would be interesting to see" Motivated from this remark, I was wondering whether this description might be of interest for your purposes. If yes, you can find more details at C. Kassel's book on Quantum groups, ch. $VI$, sect. $2$, p. 125-126 (Prop. 2.1, 2.2).<|endoftext|> TITLE: Concrete description of an exceptional minuscule variety QUESTION [9 upvotes]: Let $G$ be a complete reductive Lie group. A simple root $\alpha$ is said to be minuscule if the multiplicity of the coroot $\alpha^\vee$ in $\beta^\vee$ is at most $1$ for all positive roots $\beta$. Associated to a minuscule root $\alpha$ is a maximal parabolic subgroup $P_\alpha$ of $G$, and the quotient $X = G/P_\alpha$ is called a minuscule variety and has various nice properties (see the textbook of Billey and Lakshmibai for details). In the case $G= SL_n$, all the simple roots are minuscule and the corresponding minuscule varieties are complex Grassmannians, which can be thought of as parameter spaces for $k$-dimensional linear subspaces of an $n$-dimensional complex vector space. In the other classical types, we can also understand the minuscule varieties $X$ that arise as some sort of parameter space, where for example the points of $X$ correspond to subspaces that are isotropic with respect to some bilinear form. That leaves exactly two minuscule varieties of exceptional type. One is the projective plane over the octonions, which is not too bad. The other comes from Lie type $E_7$ and I don't know how to interpret it as a parameter space. What is this space? Is there a way to describe it (even non-rigorously) that avoids Lie theory? Sometimes the name Freudenthal is attached to this space, though I don't know that he studied it explicitly. I have also seen it given the notation $G_\omega(\mathbb{O}^3, \mathbb{O}^6)$, which is suggestive; unfortunately, I've never seen this notation explained anywhere, so I don't know how to interpret it. REPLY [7 votes]: One description is $\smash{X=E_7\left/E_6\times S^1\right.}$ (quotient of compact groups, of real dimension 133 – 79 = 54), as removal of the root $\alpha$ in question leaves an $E_6$ diagram. Another is, by Borel-Weil-Tits, $X= G$-orbit of the highest weight line in $\smash{\mathbf{CP}^{55}}$, projective space of the fundamental representation attached to $\alpha$. This is a projective variety whose complex dimension (27), degree (13110), and equations are already in Cartan (1894, p. 144; 1932, p. 160): see Hirzebruch (1957, p. 100). But the “parameter space description” you want is probably the one given by Tits (1955, p. 138): Generally if $\beta$ is another simple root, Tits knows that mapping each $x\in G/P_\alpha$ to the “$\alpha$-plane” $\{y\in G/P_\beta: y $ meets $x$ (as cosets inside $G$)$\}$ bijects $G/P_\alpha$ with the set of all $\alpha$-planes in $G/P_\beta$. That makes $X$ a parameter space, and becomes “concrete” if we take for $\beta$ the highest root (at the other end of the $E_7$ diagram): then, he shows, $G/P_\beta$ can be viewed as the “antihermitian quadric” $$ \left\{(x_0,x_1,x_2,y_0,y_1,y_2)\in\mathbf O^6:\sum_{i=0}^2x_i\bar y_i - y_i\bar x_i=0\right\}, $$ projectivized and complexified, and its set of $\alpha$-planes as the set of suitably defined “$\mathbf o$-planes” $\mathbf{OP}^2$ in it. (See p. 89, where these models of $G/P_\beta$ and $G/P_\alpha$ are denoted $\tilde{\mathrm Q a\mathrm h}^5_{\mathbf o}$ and $\tilde{\varGamma a\mathrm h}^{5;2}_{\mathbf o}$.) Edit 1: As you suggest, Freudenthal (1954) already had essentially the same models of $G/P_\alpha$ and $G/P_\beta$ (denoted $\mathfrak M$ and $\mathfrak N$). For a recent exposition see e.g. Omoda (2000, §2.3), where $G$ is denoted $\smash{G^{[2]}}$. Earlier, Freudenthal (1953) also corrected Cartan’s equations of $X$ in $\smash{\mathbf{CP}^{55}}$. Edit 2: $X$ also occurs as the special case $\smash{LGr_{\mathbf{CO}}(3)}$ in the “realization of all compact symmetric spaces as ((Double) Lagrangian) Grassmannians” by Huang-Leung (2011) or Eschenburg-Hosseini (2013). Formally this consists of subspaces $\smash{(\mathbf C\otimes\mathbf O)^3\subset(\mathbf H\otimes\mathbf O)^3}$, or the similar thing projectively, but as they discuss, this is subtle to make sense of, as non-associativity makes $\smash{\mathbf O^n}$ not an $\mathbf O$-module.<|endoftext|> TITLE: Polynomials non-negative on the integers QUESTION [28 upvotes]: Let $P$ be a real polynomial of exact degree $2n$ ($n \geq 1$) whose zeros are real numbers and such that \begin{equation*} P(j) \geq 0 \quad \text{for any} \quad j \in \mathbb{Z}. \end{equation*} Does there exist non-negative real numbers $\alpha_0,\alpha_1,\ldots,\alpha_n,$ with at least one of the $\alpha_i$ non-zero, such that the polynomial \begin{equation*} Q(x) = \sum_{k=0}^{n} \alpha_k P(x+k) \end{equation*} is non-negative on the whole real line, i.e.; $Q(x) \geq 0$ for any $x \in \mathbb{R}$ ? I would like to add that this question is not merely a mathematical curiosity but pops up naturally while working on spectral transformations of discrete measures. REPLY [29 votes]: Occasionally I wish somebody could give me a good whack on the head to keep my brains running and the older I get, the more frequently I need it. The problem is actually trivial. I will prefer to think that $P$ is non-negative on some disjoint with integers arithmetic progression $\Lambda$ with step $1$ . Then we need to show that $$ \sum_{k=0}^n {n\choose k}^2 P(k)\ge 0\,. $$ Let $\lambda$ be a number such that $\Lambda=\{x:\cos(\pi x+\lambda)=0\}$ and put $Q(x)=x(x-1)\dots(x-n)$. Consider the meromorphic function $$ F(z)=\frac{\tan(\pi z+\lambda)-\tan\lambda}{Q(z)^2}P(z)\,. $$ Note that the poles of $F$ are simple and $F(z)$ decays like $|z|^{-2}$ on big circles between the poles of the tangent, so the sum of the residues converges to $0$. Now the residue at the zero $k$ of $Q$ is $\frac{\pi}{(n!)^2\cos^2\lambda}{n\choose k}^2P(k)$ while the residues at the poles $x\in\Lambda$ of the tangent are $-\frac 1{\pi Q(x)^2}P(x)\le 0$. The end.<|endoftext|> TITLE: Non conformally geodesible vector field QUESTION [5 upvotes]: What is an example of a smooth vector field $V$ on an open set of the plane which is a geodesible vector field but there is no a conformal metric $g$ such that $V$ is geodesible vector field with respect to $g$. A geodesible vector field is a non vanishing vector field for which there is a Riemannian metric $g$ such that all trajectories of $V$ are (unparametrized) geodesics with respect to $g$. The motivation for this question is described here: Limit cycles of quadratic systems and closed geodesics REPLY [8 votes]: Here is how one can construct an example: Consider the smooth, nonvanishing $1$-form $$ \omega = y^3(1{-}y)^2\,\mathrm{d}x + \big(y^3-2(1{-}y)^2\bigr)\,\mathrm{d}y. $$ Note: This $\omega$ came from Exercises 5 and 6 of Section 16 of Chapter XVIII of Volume IV of Dieudonné's Treatise on Analysis. These exercises show that $\omega$ cannot be written globally in the form $g\,\mathrm{d}f$ for two smooth functions on $\mathbb{R}^2$. (In other words, $\omega$ has no global 'integrating factor'.) I am going to use Dieudonné's $1$-form to show that the nonvanishing vector field $$ V = y^3(1{-}y)^2\,\frac{\partial\ }{\partial x} + \big(y^3-2(1{-}y)^2\bigr)\,\frac{\partial\ }{\partial y}, $$ while geodesible, is not geodesible with respect to any conformal metric on the plane, i.e., a metric of the form $g = \mathrm{e}^{2u(x,y)}(\mathrm{d}x^2+\mathrm{d}y^2)$. First, I will show that $V$ is geodesible. To do this, it suffices to find a closed $1$-form $\phi$ such that $\phi(V)>0$. I construct $\phi$ as follows: Let $\rho \approx 0.639$ be the unique real root of $\rho^3-2(1{-}\rho)^2 = 0$. Now set $$ f(y) = \frac{\rho^3(1{-}\rho)^2-y^3(1{-}y)^2}{y^3-2(1{-}y)^2}, $$ and note that $f(y)$ is a rational function of $y$ with a quadratic denominator that never vanishes. Hence $f(y)$ is a smooth function of $y$. Now set $$ \phi = \mathrm{d}x + f(y)\,\mathrm{d}y. $$ Then $\phi$ is closed and computation yields $\phi(V) = \rho^3(1{-}\rho)^2>0$. Thus, $V$ is geodesible. To see that $V$ not geodesible with respect to any metric of the form $g= \mathrm{e}^{2u(x,y)}(\mathrm{d}x^2+\mathrm{d}y^2)$, write $$ V = h(y)\left(\cos\theta(y)\,\frac{\partial\ }{\partial x} + \sin\theta(y)\,\frac{\partial\ }{\partial y}\right) $$ where $h(y)>0$, and consider, for any function $u(x,y)$, the two $1$-forms $$ \eta_1 = \mathrm{e}^{u(x,y)}\bigl(\cos\theta(y)\,\mathrm{d}x{+}\sin\theta(y)\,\mathrm{d}y\bigr) \ \ \text{and}\ \ \eta_2 = \mathrm{e}^{u(x,y)}\bigl(-\sin\theta(y)\,\mathrm{d}x{+}\cos\theta(y)\,\mathrm{d}y\bigr). $$ If $V$ is to be geodesic with respect to the conformal metric $$ {\eta_1}^2+{\eta_2}^2 = \mathrm{e}^{2u(x,y)}(\mathrm{d}x^2+\mathrm{d}y^2), $$ then one must have $\mathrm{d}\eta_1 = 0$, which would imply, since $\mathbb{R}^2$ is simply connected, that $\eta_1 = \mathrm{d}v$ for some function $v=v(x,y)$ on $\mathbb{R}^2$. However, by construction, $$ \mathrm{d}v = \eta_1 = \mathrm{e}^{u(x,y)}\,\frac{\omega}{h(y)}, $$ so $\omega = \mathrm{e}^{-u(x,y)}h(y)\,\mathrm{d}v$, which is impossible by Dieudonné's Exercises. Thus, $V$ is not geodesible with respect to any conformal metric on $\mathbb{R}^2$.<|endoftext|> TITLE: A theorem of M. Artin QUESTION [16 upvotes]: If I got it right, there is a theorem due to M. Artin that on a projective complex manifold any point has a Zariski open neighbourhood which is a $K(\pi, 1)$ space. I have two questions about it. Is there a proof of this result written in English? (I am really bad at French.) Is it true that a point has an arbitrarily small neighbourhood with this property (i.e. smaller than a given one)? REPLY [24 votes]: Quick answer. I. See M. Olsson On Faltings’ method of almost etale extensions, chapter 5. He discusses there a version of this fact over a dvr, but I think you can easily extract what you want, if you really don't want to look at SGA. II. Yes. One possible precise statement is: a smooth scheme $X$ over a field $K$ of characteristic zero admits a basis for the Zariski topology consisting of affine open subsets $U\subseteq X$ which are $K(\pi, 1)$ for the etale topology. Moreover, if $K=\mathbb{C}$, then each $U(\mathbb{C})$ is a $K(\pi, 1)$ as a topological space, and the fundamental group $\pi_1(U(\mathbb{C}))$ is a good group (in the sense of Serre). Longer answer. Since I learned some of this from multiple sources for my thesis, I figured I can try to explain what you need to know, especially since some of it is not explicitly stated in SGA. Definition. An elementary fibration is a map of schemes $f:X\to S$ which admits a factorization (too lazy to draw a diagram here!) as an open immersion $j:X\to \overline X$ followed by a smooth projective $\overline{f}:\overline{X}\to S$ such that the geometric fibers of $\overline{f}$ are integral curves, and such that there exists a complementary to $j$ closed immersion $i:Y\to \overline{X}$ such that $Y\to S$ is finite etale and surjective. Simply speaking, an elementary fibration is just a fibration in smooth affine curves. Definition. Let $k$ be a field. An Artin neighborhood over $k$ is a scheme $U/k$ admitting a chain of elementary fibrations $$ U = U_d\to U_{d-1}\to \ldots \to U_1\to U_0 = {\rm Spec}\, k. $$ Artin's theorem consists of two parts: If $X$ is smooth over an infinite field $k$, then $X$ can be covered by Zariski open Artin neighborhoods. (This uses Bertini, and works in any characteristic). If $U$ is an Artin neighborhood over a field $k$ of characteristic zero, then $U$ is a $K(\pi, 1)$ for the etale topology. If moreover $k=\mathbb{C}$, then $U(\mathbb{C})$ with the complex topology is a $K(\pi, 1)$ space and $\pi_1(U(\mathbb{C}))$ is a good group. To understand (2), let us recall the definitions: Definition. A connected topological space (resp. a connected qcqs scheme) $X$ with a point (resp. geometric point) $x\in X$ is a $K(\pi, 1)$ if for every local system (resp. every locally constant constructible abelian etale sheaf) $\mathscr{F}$ on $X$, the natural map $$ H^*(\pi_1(X, x), \mathscr{F}_x) \longrightarrow H^*(X, \mathscr{F}) $$ is an isomorphism. In good cases ($X$ a CW complex or locally noetherian), this is equivalent to the vanishing of $\pi_q(X)$ for $q>1$ (in the etale setting, these are the etale homotopy groups of Artin and Mazur). Examples of $K(\pi, 1)$'s are smooth curves not isomorphic to $\mathbb{P}^1$ or spectra of fields. Definition. A group $G$ is a good group if for every finite $G$-module $M$, the natural maps $$ H^*(\hat G, M)\longrightarrow H^*(G, M) $$ are isomorphisms (here $\hat G$ denotes the profinite completion of $G$). Finite groups, free groups, and extensions of good groups are good - this will be important in a second. A nice example of a group which is not good is ${\rm Sp}(2g, \mathbb{Z})$ for $g>2$ (I don't remember the exact bound). This implies that the moduli space of principally polarized complex abelian varieties with suitable level structure is a $K(\pi, 1)$ in the topological sense but not for the etale topology. To understand why (2) is true, it is enough to know the following: Lemma. Let $f:X\to S$ be an elementary fibration over a regular noetherian $\mathbb{Q}$-scheme $S$. Then (a) If $S$ is a $K(\pi, 1)$ then so is $X$, (b) If $S$ is of finite type over $\mathbb{C}$ and $S(\mathbb{C})$ is a $K(\pi, 1)$, then $X(\mathbb{C})$ is a $K(\pi, 1)$, (c) If $S$ is of finite type over $\mathbb{C}$, then $\pi_1(X(\mathbb{C}))$ is an extension of $\pi_1(S(\mathbb{C}))$ by a free group. In particular, if $\pi_1(S(\mathbb{C}))$ is good, so is $\pi_1(X(\mathbb{C}))$. Assertions (b) and (c) follow easily from the (not entirely obvious but intuitively clear) fact that $f:X(\mathbb{C})\to S(\mathbb{C})$ is a locally trivial fibration whose fiber is an open Riemann surface, whose homotopy type is that of a wedge of circles. Assertion (a) can be proved directly using Abhyankar's lemma (this is where we need characteristic zero, otherwise the proof fails due to wild ramification problems) and the Leray spectral sequence, cf. Lemma 5.5 in Olsson's article. Alternatively, assertion (a) can be deduced from (b) and (c) if $S$ is of finite type over $\mathbb{C}$ using the comparison theorems between singular and etale cohomology, but that would be a wrong thing to do since Artin's original motivation was to prove that comparison! Which brings us to why Artin proved this theorem in the first place. Application. Let $X$ be smooth and of finite type over $\mathbb{C}$ and let $\mathscr{F}$ be an locally constant constructible sheaf on $X$. Then the natural maps $$ H^*(X, \mathscr{F})\longrightarrow H^*(X(\mathbb{C}), \mathscr{F}) $$ are isomorphisms. If $X$ is an Artin neighborhood, then this follows from the fact that the above maps can be factored as (sorry, I forgot how to draw a commutative square on MO): $$ H^*(X, \mathscr{F}) \cong H^*(\pi_1(X, x), \mathscr{F}_x) \cong H^*(\pi_1(X(\mathbb{C}), x), \mathscr{F}_x) \cong H^*(X(\mathbb{C}), \mathscr{F}). $$ Here the first isomorphism comes from the fact that $X$ is a $K(\pi, 1)$ as a scheme, the second from the fact that $\pi_1(X(\mathbb{C}),x)$ is a good group (combined with the theorem from SGA1 which says that $\pi_1(X, x)$ is the profinite completion of $\pi_1(X(\mathbb{C}), x)$), and the last one from the fact that $X(\mathbb{C})$ is a $K(\pi, 1)$ space. If $X$ is not necessarily an Artin neighborhood, it can be covered by such, and their pairwise intersections can be covered by such, etc., and on both sides we can write a spectral sequence of a hypercovering. In fact, this application is sort of a red herring, as later Artin proved a much more general result (with $X$ finite type but not necessarily smooth and $\mathscr{F}$ an arbitrary constructible etale sheaf) using reduction to curves. Still, Artin's result has had some influence on algebraic geometry, for example it showed up in Faltings' work on $p$-adic Hodge theory (cf. Olsson's paper!). A sketch of Artin's construction (added Aug 6, 2017). I found the following short explanation of Artin's construction of elementary neighborhoods on my hard drive. It is enough to prove: Lemma. Let $X$ be a smooth variety over an infinite field $k$, let $x\in X(k)$ be a point. Then there exists a Zariski neighborhood $U$ of $x$ and an elementary fibration $\pi:U\to S$. Proof (sketch). We can assume $X$ is affine, embed $X$ into a normal projective $\bar X\subseteq \mathbf{P}^N$, let $Y = \bar X - X$ be the complement. By a Bertini-type argument, after passing to a higher Veronese embedding, there exists a linear subspace $L$ of codimension $\dim X - 1$ containing $x$, avoiding the singularities of $\bar X$ and $Y$ and transverse to $\bar X$ and $Y$. We can furthermore find a subspace $C\subseteq L$ of codimension 1, not containing $x$, disjoint from $Y$ and meeting $X\cap L$ transversely. The projection $\mathbf{P}^N\setminus C\to \mathbf{P}^{\dim X - 1}$ along $C$ is a rational map which extends to the blow-up of $\mathbf{P}^N$ along $C$. Taking the proper transform $\bar X'$ of $\bar X$ in this blow-up, we get a proper surjective map $\bar f: \bar X'\to \mathbf{P}^{\dim X - 1}$ whose fiber at $\bar f(x)$ is the smooth curve $L\cap \bar X$. Then $\bar f$ defines an elementary fibration as needed.<|endoftext|> TITLE: More pseudoholomorphic curves than complex valued functions QUESTION [6 upvotes]: A lecture I heard had a remark - "There is a rich class of pseudohoplomorphic curves to a symplectic manifold with an almost complex structure (tamed by the symplectic structure). On the other hand, there are very few complex valued functions on a symplectic manifold". Can someone explain why? Or point me to some reference? I always wondered why people do not study complex valued functions -that should be equally insightful. REPLY [10 votes]: I think that, instead of complex-valued functions $f:M\to\mathbb{C}$ you mean $J$-holomorphic functions, i.e., complex-valued functions $f:M\to\mathbb{C}$ that satisfy $f'(x)(Jv) = i\,f'(x)(v)$ for all $v\in T_xM$. The complex-valued functions don't have anything to do with the almost-complex structure $J$, while the $J$-holomorphic functions clearly do. The reason these latter aren't usually interesting on an almost-complex manifold is that, while the equations for pseudoholomorphic curves form a determined elliptic system (and so have plenty of local solutions), the equations for $J$-holomorphic functions form an overdetermined PDE system when the domain has (real) dimension greater than $2$, and the integrability of the underlying almost complex structure is exactly the condition that this overdetermined system be 'maximally compatible'. In fact, for most almost-complex manifolds $(M^{2n},J)$ of real dimension $2n>2$, the sheaf of $J$-holomorphic functions on $M$ is just the sheaf of constant functions, so there is nothing special to say about it.<|endoftext|> TITLE: Does a certain points and lines configuration exist? QUESTION [16 upvotes]: For which $n$ we may mark $n$ red and $n$ blue points on the Euclidean plane, not all on a line, so that any line which passes through two points of different colour contains another point? For $n=1991$ this was proposed in a not-up-to-date edition of Prasolov's problem book on planimetry, but the suggested solution actually solves a different problem (in the newest edition this is fixed.) The following example for $n=6k$ is communicated by M. Belozerov: take a regular $4k$-gon, colour its vertices alternatively and infinite points of the sides arbitrarily. REPLY [12 votes]: Using a Pappian-ish configuration, we can get $n = 5$ and $n = 7$. For $n = 7$, put points at $(x,0)$ and $(x,1)$ for integer $x$ with $0 \le x \le 3$, at $(x,1/2)$ for integer or half-integer $x$ with $1/2 \le x \le 5/2$, and at horizontal infinity. Color them like this: For $n = 5$, remove the rightmost two blue points and the rightmost two red points, turn the rightmost blue point on the central horizontal line red, and make one of the red points on the middle vertical line blue. This construction breaks down for $n > 7$ since you start to get lines of shallow enough slope to just connect two points lying entirely in one "half" of the diagram. Edited following the suggestion of Jan Kyncl: A modification of the above in fact gives a configuration for any $n \ge 4$. For $n = 3k+1$ with $k \ge 1$, put points at $(x,0)$ and $(x,1)$ for integer $0 \le x \le 2k-1$, at $(x,1/2)$ for integer or half-integer $(k-1)/2 \le (3k-1)/2$, and at horizontal infinity. Then color the points on the horizontal lines $y = 0$ and $y = 1$ red if $x < k$ and blue if $x \ge k$, color the points on the horizontal line $y = 1/2$ blue if $x < k$ and red if $x \ge k$, and color horizontal infinity red. (This directly extends the $n = 7$ case drawn above) For $n = 3k$ with $k \ge 2$, just omit the leftmost and rightmost point on the central horizontal line. For $n = 3k + 2$ with $k \ge 1$, put points at $(x,0)$ and $(x,1)$ with integer $0 \le x \le 2k$, at $(x,1/2)$ with $k/2 \le x \le 3k/2$, and at horizontal infinity. On the outer lines $y = 0$ and $y = 1$, color points red if $x < k$ and blue if $x > k$, and on the central line $y = 1/2$ color points red if $x > k$ and blue if $x < k$. On the line $x = k$, color $(k,1)$ red, $(k,1/2)$ and $(k,0)$ blue, and color horizontal infinity red.<|endoftext|> TITLE: Local limit theorems for positive random walks QUESTION [5 upvotes]: Let $\xi_1,\xi_2,\ldots$ be i.i.d. positive random variables with infinite mean $\mathbb E[\xi_i]=\infty$, and consider the random walk $$T_n=\sum_{i=1}^n\xi_i,\qquad n\in\mathbb N.$$ Here's an example that illustrates why I'm interested in such questions. Example. If $(S_n)_{n\in\mathbb N}$ is a simple symmetric random walk on $\mathbb Z$, then $\xi_1,\xi_2,\ldots$ could represent the time intervals between consecutive returns to zero of $S_n$. Q. Does there exist general local limit theorems for the random walk $T_n$? In the case of the example I have above, everything can be computed explicitly quite nicely, and it can be argued that $$n\cdot \mathbb P\{T_{\sqrt{n}x}=n\}\asymp xe^{-x^2/2}.$$ I'm interested in more general statements of this form. REPLY [6 votes]: Yes, there is a general result, see Chapter 9 of Gnedenko-Kolmogorov book. The theorem says that if $\xi_i$ are i.i.d with values in $\mathbb{Z}$ such that $$\text{gcd}\{s-s':\mathbb{P}(\xi_1=s)>0,\;\mathbb{P}(\xi_1=s')>0\}=1,$$ and $b_n^{-1}(T_n-a_n)$ converges in distribution to a stable law, then also the local limit theorem holds, namely, $$ b_n\mathbb{P}(T_n=k)-g(b_n^{-1}(k-a_n))\to 0 $$uniformly in $k$, where $g$ is the density of the stable law. There is also a necessary and sufficient condition for the existence of $b_n$ and $a_n$ such that $b_n^{-1}(T_n-a_n)$ converges in distribution, see e. g. Feller's or Durrett's book.<|endoftext|> TITLE: Eigenvalues of a matrix with binomial entries QUESTION [9 upvotes]: I am trying to determine the eigenvalues and eigenvectors of the following matrix: $$M_{ij} = 4^{-j}\binom{2j}{i}$$ where it is understood that the binomial coefficient $\binom{m}{k}$ is zero if $k<0$ or $k>m$. The indices $i,j$ traverse a discrete finite range, $i,j \in \{a, a+1, \dots, b\}$, from $a$ to $b$, where $a,b$ are non-negative integers with $0\le a\le b$. Therefore the matrix $M_{ij}$ has dimensions $(b-a+1) \times (b-a+1)$. I would content myself with an approximate asymptotic expression (if there is no exact analytical result), valid for large $a,b$ (say for each fixed ratio $a/b$). I am mostly interested in the largest eigenvalue (not in absolute value, but the largest positive eigenvalue) and the corresponding eigenvector. Also a recurrence relation would be useful. Anything that helps... To be explicit, we want to solve the following eigenquation: $$\sum_{j=a}^b 4^{-j} \binom{2j}{i} x_j = \lambda x_i,\quad i=a, a+1, ..., b$$ for the eigenvalues $\lambda$ and eigenvectors $x_i$. Update: Numerical experiments suggest that if $a,b\rightarrow \infty$ with a fixed ratio $a/b$, the largest eigenvalue $\rightarrow 1$ always, irrespective of the value of the ratio $a/b$. There is a leading order correction proportional to $1/\sqrt{b}$, and the value of the proportionality constant depends on the value of the ratio $a/b$. I do not know how to prove any of these statements, and I cannot be sure they are correct. It would be nice if we could compute the value of leading order coefficient. REPLY [7 votes]: I don't have an answer, but it appears that the eigenvalues are always real. I don't have a proof, but have checked this using Sturm sequences for $1 \le a \le b \le 30$. You're unlikely to get "an exact analytic result", as the characteristic polynomial seems to be irreducible over the rationals unless $a=0$ (in which case $\lambda - 1$ is a factor). Thus for $a=1, b=5$, the characteristic polynomial is $${\lambda}^{5}-{\frac {437\,{\lambda}^{4}}{256}}+{\frac {29823\,{ \lambda}^{3}}{32768}}-{\frac {85687\,{\lambda}^{2}}{524288}}+{\frac { 15115\,\lambda}{2097152}}-{\frac{1}{32768}} $$ which has Galois group $S_5$.<|endoftext|> TITLE: Is the moduli space of unorientable Riemann surfaces with $pin^+$ structure orientable? QUESTION [28 upvotes]: By a non-orientable Riemann surface ${\cal C}$, I mean a compact non-orientable two-manifold without boundary that is endowed with a conformal structure. Such objects have a moduli space that is somewhat like the more familiar moduli space of oriented Riemann surfaces, but this moduli space is itself non-orientable. (See corollary 2.3 of https://arxiv.org/abs/1309.0383, where this is proved with a simple explicit example.) My question is this: Suppose that ${\cal C}$ is endowed with a $pin^+$ structure. (This is one of the two possible analogs of a spin structure in the non-orientable case, the other being a $pin^-$ structure.) I've come to suspect that the moduli space of non-orientable Riemann surfaces with a $pin^+$ structure is itself orientable. I wonder if this is a known result and where it might be found. REPLY [4 votes]: This is not an answer to the question but rather a comment related to the motivation behind it that might be of interest. A more natural point of view from the perspective of GW theory is to consider symmetric Riemann surfaces i.e. orientable RS with an orientation reversing involution. The moduli of such curves is not orientable but the moduli of equivariant J-hol. maps to a symplectic manifold with an anti-symplectic involution of dimension 2(2n+1) often are and give rise to real GW invariants (in particular this is the case for target curves and CY 3-folds arXiv). The calculations in the target P^1 case suggest that these are related to matrix models for SO/Sp - I hope I'll be able to say something more precise soon.<|endoftext|> TITLE: Primes that are "almost" multiples of each other QUESTION [6 upvotes]: Let $P$ denote the set of primes in $\mathbb{N}$. For $k\in \mathbb{N}, k\geq 2$ set $$M_k = \big\{p\in P: \{kp-1, kp+1\}\cap P \neq \emptyset\big\}.$$ Is there $k\in \mathbb{N}, k\geq 2$ such that $M_k$ is infinite? REPLY [12 votes]: You're asking whether, for some $k \ge 2$, there are infinitely many primes $p$ such that either $kp-1$ or $kp+1$ is prime. That would mean there are infinitely many primes such that $kp-1$ is prime or infinitely many primes such that $kp+1$ is prime. Of course, $k$ had better be even. If $k$ is even, Dickson's conjecture would say that in both cases the answer is yes. But Dickson's is still a conjecture, and there are no $k$ for which the answer has been proven to be yes.<|endoftext|> TITLE: Density of Cliques in Random Graphs QUESTION [5 upvotes]: I've found and read a considerable amount about the density of cliques on non-random graphs, notably the paper "The Clique Density Theorem" by Reiher. I was wondering if there was any analogous work on the density of cliques on random graphs? I would love to learn about any results out there. REPLY [2 votes]: If by "density" you do not strictly restrict your question to quantitative results (in the sense of results about the numerical value of the clique density), but rather also more qualitative/structural/topological results, then the work of Matthew Kahle is very relevant to your question. In particular, Kahle investigated the clique complex of $G(n,p)$. (Synonyms: flag complex, Vietoris--Rips complex). This is the abstract finite simplicial complex whose $d$-dimensional independent sets are precisely the $(d+1)$-cliques in the graph. However, if you really mean "analogous work", in particular, analogous to theorems of Reiher, and of Razborov, in particular "work discovering a globally-convex-yet-piecewise-concave-dependency between $K^2$-density and $K^r$-density", then I think there simply does not exist anything analogous seen through the lens of (the measure of) $G(n,p)$. The dependency between $p$ and $K^r$-density appears convex. Cf. e.g. M. Kahle: Sharp vanishing thresholds for cohomology of random flag complexes. Annals of Mathematics 179 (3), pp. 1085-1107<|endoftext|> TITLE: What is the precise relation between Kolmogorov complexity and Shannon's entropy? QUESTION [8 upvotes]: Consider the discrete case: Shannon's entropy is $H(x)=-\sum\limits_i^n p(x_i) log\space p(x_i)$. Probability based on prefix-free Kolmogorov complexity is $R(x_i)=2^{-K(x_i)}$ where $K(x_i)$ is the prefix-free Kolmogorov complexity of $x_i$. What is the relation between $R(x_i)$ and $p(x_i)$? Are they equal? I remember vaguely that a book has discuss on the relation without any precise result. REPLY [9 votes]: What follows, which is all standard stuff (and should be in, say, Li and Vitanyi's monograph, or Downey and Hirschfeldt perhaps), might be the sort of relationship you vaguely recall seeing. Your $R(x):= 2^{-K(x)}$ is not itself a probability measure, but it is a lower semi-computable semi-measure; and it is optimal among such in the following sense: for any lower semi-computable semi-measure $p$ on strings, there is some constant $c_p >0$ such that $2^{c_p} \cdot R(x) \ge p(x)$ for all strings $x$. Moreover, then, $K(x)\leq c_p - \log_2 p(x)$ for all $x$. Also, it's known that $c_p = K(p)+O(1)$. For a fixed such $p$, then, we have that the expected value (w.r.t. $p$) of the Kolmogorov complexity of a string is $$ \sum_x p(x) K(x) \leq K(p) + H(p) + O(1), $$ where I'm using the definition $H(p) := - \sum_x p(x) \log_2 p(x)$ for the Shannon entropy of $p$. On the other hand, Shannon's source coding theorem yields $H(p)\leq \sum_x p(X) K(x)$, so in total we have $$ 0 \leq \left(\sum_x p(x)K(x)\right) - H(p) \leq K(p)+O(1). $$ Thus the Shannon entropy is close to the expected value of Kolmogorov complexity for low-complexity $p$.<|endoftext|> TITLE: Section of an $n$-dimensional convex polytope by $2$-dimensional plane QUESTION [7 upvotes]: Consider an $n$-dimensional convex polytope with $k$ vertices. In the worst case the number of faces is exponential in $n$ and $k$. Consider a $2$-dimensional plane which intersects this polytope, i.e., it intersects only a subset of all faces. Can I bound the number of such intersected faces? In the worst case, will this number be a polynomial in $n$ and $k$ or still exponential? Thanks in advance. REPLY [2 votes]: I found a paper wich shows that in the worst case it still can be exponential: "Shadows and slices of polytopes, Nina Amenta et al., 1996" https://dl.acm.org/citation.cfm?id=237228<|endoftext|> TITLE: An inequality inspired by the isoperimetric inequality QUESTION [5 upvotes]: Let us consider the simplest isoperimetric inequality. Consider a smooth simple closed curve given by $r=\rho(\theta)$ in polar coordinates, where $\rho(\theta)>0$ can be regarded as a smooth periodic function with period $2\pi$. As $dA=rdrd\theta$, the area of the region $\Omega$ enclosed by the closed curve is \begin{align*} A=\iint_{\Omega}rdrd\theta =\frac{1}{2}\int_0^{2\pi}\rho(\theta)^2 d\theta. \end{align*} On the other hand, it is an easy exercise in calculus that the length of the curve is given by \begin{align*} L=&\int_{0}^{2\pi}\sqrt{\rho(\theta)^2+\rho'(\theta)^2}d\theta. \end{align*} By the isoperimetric inequality, we have $L^2\ge 4\pi A$, so we must have \begin{equation} 2\pi \int_{0}^{2\pi} \rho(\theta)^2 d\theta \le \left(\int_{0}^{2\pi}\sqrt{\rho(\theta)^2+\rho'(\theta)^2}d\theta\right)^2. \end{equation} This looks similar to the Wirtinger's inequality, which states that \begin{equation} \begin{split} \int_{0}^{2\pi}\rho'(\theta)^2d\theta \ge \int_{0}^{2\pi} \left(\rho(\theta)-\overline \rho\right)^2 d\theta = \int_{0}^{2\pi} \rho(\theta)^2 d\theta-\frac{1}{2\pi} \left(\int_{0}^{2\pi}\rho(\theta)d\theta\right)^2. \end{split} \end{equation} So \begin{align*} 2\pi \int_{0}^{2\pi} \rho(\theta)^2 d\theta \le& \left(\int_{0}^{2\pi}\rho(\theta)d\theta\right)^2+2\pi\int_{0}^{2\pi}\rho'(\theta)^2d\theta, \end{align*} which isn't quite what I want. It doesn't seem right to me that we can apply Wirtinger's inequality directly to prove it because the equality case in Wirtinger's inequality holds when $\rho(\theta)=C+a\cos \theta+ b\sin \theta$, but the equality in our inequality can only hold when $\rho$ is constant the equality in our inequality doesn't hold in this case. (However, by geometric consideration, the equality does hold for, say, $\rho(\theta)=\sin \theta$, but only if we restrict $\theta$ to $[0, \pi]$.) This is where I am stuck. So my question is: Can we show this inequality without using the isoperimetric inequality (say by Fourier analysis or using Wirtinger's inequality more carefully)? Can it be used to prove (at least a simple case of) the isoperimetric inequality? If not, why? (If nothing else, at least I obtain an inequality on circle :-) ) To elaborate further, it is well-known that we can apply Wirtinger's inequality (or Fourier type argument) to prove the isoperimetric inequality on the plane. Indeed, the Wirtinger's inequality and the isoperimetric inequality are equivalent (e.g. Osserman's paper on isoperimetric inequality). Usually, these kinds of proofs involve shifting the center of mass to $0$, applying the Green's theorem and the Wirtinger's (or Cauchy-Schwarz) inequality on some combination of two functions (say $x(s), y(s)$). So as a subquestion, why is there no such argument involving only a single function (say $\rho(\theta)$)? REPLY [3 votes]: I will illustrate the Bellman function approach to prove Wirtinger's inequality which, of course, is simpler than the original problem. The advantage of the approach is that it does not use any Fourier analysis (which apparently is the best thing to do for this particular problem since $f$ is periodic), Cauchy--Schwarz, or variational calculus such as Euler--Lagrange equation. If somebody finds the approach interesting you can try to use it to prove the original problem (or maybe I will try to do it myself but later), and as Paul Bryan noticed, unfortunately it will only give you isoperimetric inequality for the sets whose boundary has a nice 1-1 parametrization in polar coordinate systems (for example, star shaped sets). Consider the function of 4 variables $$ M(t,x,y,z):=\frac{2tx^{2}(\cos(t)-1)-y^{2}\sin(t) +z^{2}(t\cos(t)-\sin(t))+2(1-\cos(t))(2xy+xzt-yz)}{2-2\cos(t)-t\sin(t)} $$ defined in the domain $(0,2\pi)\times \mathbb{R}^{3}$. In what follows $M_{t}, M_{x}, M_{y}, M_{z}$ denote partial derivatives of $M$. Claim 1: For any $v \in \mathbb{R}$ and all $(t,x,y,z)\in (0,2\pi)\times \mathbb{R}^{3}$ we have $$ x^{2}-v^{2}\leq M_{t}+v M_{x}+xM_{y}+vM_{z}. \qquad (*). $$ Proof: Optimize the inequality over all $v$. The optimal value is attained when $v=-\frac{M_{x}+M_{z}}{2}$. Therefore it is enough to have $$ x^{2}\leq -\left(\frac{M_{x}+M_{z}}{2}\right)^{2}+M_{t}+xM_{y}. \qquad (**) $$ After straightforward calculations one notices that the inequality $(**)$ is in fact equality! The details are left to the reader. Claim 2: For any $f\in C^{1}([0,2\pi])$ we have $$ \int_{0}^{2\pi} f^{2}-(f')^{2}\leq \limsup_{t\to 2\pi}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right) - \liminf_{t\to 0}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right). $$ Proof: Notice that $$ f^{2}(t)-(f'(t))^{2}\leq \frac{d}{dt} M\left(t, f(t), \int_{0}^{t}f, \int_{0}^{t}f' \right). \qquad (***) $$ Indeed, after taking the derivative the inequality $(***)$ simplifies to $$ f^{2}(t)-(f'(t))^{2}\leq M_{t}+M_{x}\, f'(t)+M_{y}\,f(t)+M_{z}\,f'(t). $$ The latter follows from $(*)$ where we take $v=f'(t), \;x=f(t), \;y=\int_{0}^{t} f$ and $z=\int_{0}^{t}f'$. Finally we just integrate $(***)$ in $t$ from $t_{1}$ to $t_{2}$ where $t_{1}, t_{2}\in (0,2\pi)$, and take the lower and upper limits $t_{1}\to 0$ and $t_{2} \to 2\pi$. Claim 3: Let $f\in C^{1}([0,2\pi])$ be such that $f(0)=f(2\pi)$ and $\int_{0}^{2\pi}f=0$. Then $$ \int_{0}^{2\pi} f^{2}-(f')^{2}\leq 0 $$ Proof: Indeed, using Claim 2 it will be enough to show that \begin{align*} &\limsup_{t\to 2\pi}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right)= \lim_{t\to 2\pi}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right) =0\\ &\liminf_{t\to 0}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right)=\lim_{t\to 0}\, M\left(t, f(t), \int_{0}^{t} f, \int_{0}^{t}f'\right)=0. \end{align*} These inequalities roughly speaking follow from the following observations \begin{align*} &\lim_{\delta \to 0}\; M(2\pi-\delta, f(0)-\delta f'(0),-\delta f(0), -\delta f'(0))=0\\ &\lim_{\varepsilon \to 0}\; M(\varepsilon, f(0)+\varepsilon f'(0), \varepsilon f(0), \varepsilon f'(0))=0 \end{align*} for any $f(0)$ and $f'(0)$. The end.<|endoftext|> TITLE: Is there a stronger form of recursion? QUESTION [15 upvotes]: I'm wondering if there are any recursion principles more general than the following, first given by Montague, Tarski and Scott (1956): Let $\mathbb{V}$ be the universe, and $\mathcal{R}$ be a well-founded relation such that for all $x\in Fld\mathcal{R}$, $\{y:y\mathcal{R}x\}$ is a set. Further, let $\mathbb{F}$ be a function with $dmn\mathbb{F}=Fld\mathcal{R}\times\mathbb{V}$. Then there exists a unique function $\mathbb{G}$ such that $dmn\mathbb{G}=Fld\mathcal{R}$ and for all $x\in Fld\mathcal{R}$, $$\mathbb{G}(x)=\mathbb{F}(x,\mathbb G\restriction\{y:y\mathcal{R}x\}).$$ Specifically, I would like to be able to drop the requirement that $\{y:y\mathcal{R}x\}$ be a set. I am attempting to define by recursion a function on $O_n\times O_n$ ordered lexicographically, which is a well ordering and consequently a well-founded relation, however $\{y:y<(0,\alpha)\}$ is a proper class for all $\alpha>0$. The proof of the above theorem in the context of MK class theory (and even its statement) rely pretty explicitly on this not happening, however I am aware that category theorists often work in situations where they need to aggregate together many proper classes and manipulate them/construct morphisms between them. Is there a (perhaps large-cardinal based) strengthening of this theorem that would allow one to legitimately make such a definition by recursion in a class-theoretical context? REPLY [18 votes]: Yes, there are such principles. In fact, there is a natural hierarchy of such class-theoretic recursion principles, which form a hierarchy of strength transcending Gödel-Bernays set theory GBC. These principles have become important in the emerging field known as the reverse mathematics of second-order set theory, which seeks to classify various natural second-order set theoretic principles into a hierarchy over GBC. The principle of elementary transfinite recursion asserts that for any well-ordered class relation $\Gamma$ on a class $I$, not necessarily set-like, and any first-order property $\varphi$, allowing class parameters $Z$, there is a solution $S$, meaning a class $S\subset I\times V$, such that $S_i=\{x\mid \varphi(x,S\upharpoonright i,Z)\}$, where $S\upharpoonright i=\{(j,x)\in S\mid j<_\Gamma i\}$ is the part of the solution below $i$. Thus, we define a class by recursion along $\Gamma$, where at each stage we define a new class in terms of the solution previous to that stage. One can formalize your function-based recursion using this kind of formalism. In my recent paper V. Gitman and J. D. Hamkins, Open determinacy for class games, in Foundations of Mathematics, Logic at Harvard, Essays in Honor of Hugh Woodin’s 60th Birthday, A. E. Caicedo, J. Cummings, P. Koellner, and P. Larson, Eds., American Mathematical Society, 2016. (arxiv:1509.01099) we proved that ETR is equivalent to the principle of clopen determinacy for class games. Meanwhile, open determinacy is a little stronger, and below $\Pi^1_1$-comprehension. In my recent paper V. Gitman, J. D. Hamkins, P. Holy, P. Schlicht, and K. Williams, The exact strength of the class forcing theorem, under review, (arxiv:1707.03700) we proved that the principle of $\text{ETR}_{\text{Ord}}$, which allows recursions only of length $\text{Ord}$, is equivalent to a list of 12 natural statements, including the class forcing theorem, the existence of truth predicates of various kinds for infinitary logic, the existence of iterated truth predicates for first-order logic, the existence of Boolean set-completions of partial orders, and others. The picture that is emerging from this analysis is that the natural second-order set theories are ranked by the amount of recursion they support.<|endoftext|> TITLE: Partitions of the reals such that closures of partition elements are saturated QUESTION [5 upvotes]: Suppose we have a partition $P$ of a set $S$ and a unary set operation $u:\mathscr{P}(S)\to\mathscr{P}(S)$ such that for each $A\in P$ the set $uA$ is saturated with respect to $P$ (a union of elements in $P$). Can anyone suggest any references on this topic? So far the closest thing I have found is a 2013 paper by Christian Ronse entitled "Closures on partial partitions from closures on sets." Here is the specific question I am curious about: Conjecture. Let $X$ be a connected finite topological space. Let $u$ be the closure operation on $\mathbb{R}$ under the usual topology. Then there exists a partition $P$ of $\mathbb{R}$ satisfying the condition stated above, such that the quotient $\mathbb{R}/P$ is homeomorphic to $X$. Even if this an intractable problem, it can be fun to search for counterexamples. Certain spaces are difficult to find a corresponding real partition (of the type above) for; some have had me thinking they were counterexamples for sure, then suddenly I find a partition that works. Can anyone suggest any leads? Added 22 July 2017 Taking into account the path (described below) that led me to the conjecture, instead of restricting to just one unary operation $u$ that is arbitrarily defined on $\mathscr{P}(S)\setminus P$, it might be better to allow any given collection of operations (of whatever arity), some or all of which may be required to satisfy various axioms (such as for example, closure axioms), with the partition condition applying to each. But the meat of this post is supposed to be the conjecture, so if the above framework is too general to be of any use (it may be), then let's just scrap it. I arrived at the conjecture as follows. Back in grad school in the mid-1980s, the Kuratowski closure-complement-intersection problem (a good discussion can be found here) attracted my attention. I noticed that one could easily write a program to generate families of sets of reals under the usual topology (from a single seed set) by partitioning and working solely within the partition—provided it satisfies the above condition for the closure operation on $\mathbb{R}$ (the condition holds trivially for the complement operation). In 2009 I posted a Javascript app here that lets users vary the seed set in a ten-point quotient of the reals to see what family it generates under closure and complement. About a year ago, I became interested in the actual space “under the hood” of this app. A natural question arose: since connected seven-point spaces exist that contain Kuratowski 14-sets, might it be possible to replace my ten-point quotient with a seven-point one? (The answer is yes.) Further exploration led me to the conjecture, which I am currently about halfway finished verifying for all 94 connected five-spaces (it holds for all smaller ones). This is where I stop...there are too many connected six-spaces! As one would guess, Cantor sets play a prominent role in many of the partitions (roughly half). Added 5 August 2017 Professors who like to occasionally throw curveballs at students on Ph.D. quals might find a few in my answer below. While some partitions are trivial and most are easy to find, others are more challenging. For example, here is a moderately difficult one: Problem. Give an example of a partition $\mathscr{P}=\{F_1,F_2,G_1,G_2,G_3\}$ of $\mathbb{R}$ under the usual topology such that $\varnothing\not\in\mathscr{P}$, each $F_i$ is closed, each $G_i$ is open, and $\overline{G_i}=G_i\cup F_1\cup F_2$ for $i=1,2,3.$ Solution. See space number 19 in the list of five-point spaces in my answer. All of the evidence so far points to the truth of the conjecture. Short of finding a proof, it might be interesting to look for counterexamples in weakened spaces, for example in some other infinite connected space besides $\mathbb{R}.$ Or it might be interesting to replace $\mathbb{R}$ with finite spaces. One more question. Many spaces in my answer seem to require the “middle thirds” structure of the Cantor construction in their associated real partitions. Many other spaces clearly do not. Assuming that some do in fact require it, what is it about them that makes this so? REPLY [2 votes]: Every nonempty connected space of five or fewer points satisfies the conjecture. $\def\R{\mathbb{R}}\def\hc{\hfill\cr}\def\sm{\setminus}\def\lw{\leftarrow}\def\ts{\textstyle}\def\sp#1#2#3#4{\matrix{#1\hc #2\hc #3\hc #4\hc}}\def\nq{\hfill\newline\quad}\def\nqq{\hfill\newline\quad\quad}$ We say a nonempty partition of a topological space is compatible with the topology (or just compatible) if the closure of each partition element is saturated. (Note that we get an equivalent definition using interior instead of closure.) For each nonempty connected space $X$ such that $|X|\leq5$ we present a compatible partition $P$ of $\R$ such that $\R/P$ is homeomorphic to $X$ (partitions of finite intervals will always be defined so they extend to $\R$ by adjoining translations). Spaces are ordered so it is clear that no two are homeomorphic. The list is thus self-contained, assuming one knows how many non-homeomorphic connected spaces exist of cardinalities one through five: 1, 2, 6, 21, 94 (sequence A001928). A list of all topological spaces of four or fewer points (up to homeomorphism) appears here and a list all five-point spaces appears here. Per the usual Cantor set construction, for any non-degenerate closed interval $J,$ let $G_1(J),G_2(J),\ldots$ denote the sequence of (unions of) open middle thirds of $J.$ Thus, for each $n\geq1$ the set $G_n(J)$ is a disjoint union of $2^{n-1}$ finite open intervals. Moreover $G_m(J)\cap G_n(J)=\varnothing$ for $m\neq n.$ Let $M_0(J)$ denote the set of endpoints of $J$ and $M_n(J)$ the set of endpoints of intervals in $G_n(J).$ Let $\sp{M(J)}{G(J)}{F(J)}{F'(J)}\sp{=}{=}{=}{=}\sp{\bigcup_{n=0}^\infty M_n(J)}{\bigcup_{n=1}^\infty G_n(J)}{J\sm G(J)}{F(J)\sm M(J)}\sp{\rm\ (endpoints),}{\rm\ (open\ middle\ thirds),}{{\rm\ (Cantor\ set),\ and}}{\rm\ (non\mbox{-}endpoints).}$ For any disjoint union $U$ of like intervals and partition $P$ of this same type of interval, the expression “$U\{P\}$” shall refer to the partition of $U$ that puts a congruent copy of $P$ in each interval of $U.$ For example, the notation $G_2([0,1])\{A\lw(0,1);B\lw[1,2)\}$ represents $$\ts A\supset({2\over18},{3\over18})\cup({14\over18},{15\over18})\quad{\rm and}\quad\ts B\supset[{3\over18},{4\over18})\cup[{15\over18},{16\over18}).$$ The arrow symbol $\leftarrow$ serves to remind that sets within braces are generally not members of the final partition. Given a non-degenerate interval $I$ and discrete set $S=\{s_1,s_2,\ldots\}\subset I,$ let $h_S^+(I)$ denote an arbitrary discrete set $\{t_{i,j}\}\subset I\sm S$ such that for each $i=1,2,\ldots$ we have $t_{i,j}\nearrow s_i.$ Define $h_S^-(I)$ similarly with $t_{i,j}\searrow s_i.$ Let $H^+([a,b))$ denote the following “halving decomposition” of $[a,b)$: $$H^+([a,b))=\bigcup_{n=0}^\infty\big[b-{1\over2^n}(b-a),\,\,b-{1\over2^{n+1}}(b-a)\big).$$ Define $H^-((a,b])$ similarly. Given a disjoint union $V$ of non-degenerate intervals, let $W_1(V),W_2(V),\ldots$ denote an arbitrary finite disjoint family of subsets of $V,$ each dense in $V,$ whose union is all of $V.$ For any given partition, the size of this family (for any given $V)$ will simply be the largest subscript that appears. Finite topological spaces appear as ordered lists of closures of singletons. For example the list ace bde c d means “the topological space $\{a,b,c,d\}$ such that $\overline{\{a\}}=\{a,c,e\},$ $\overline{\{b\}}=\{b,d,e\},$ $\overline{\{c\}}=\{c\},$ and $\overline{\{d\}}=\{d\}.$” (Note that a space is $T_0$ iff this list contains no repetitions.) These lists appear in increasing order of (a) their total length, (b) for any given length in (a), their number of one-point elements, (c) for any given number in (b), their number of two-point elements, etc. Partition elements are denoted $A,B,\ldots$ where the homeomorphism sends $a$ to $A,$ etc. Lastly, note that since quotients preserve connectedness, the conjecture cannot be satisfied by any disconnected space. One-point connected space a $A=\R$ Two-point connected spaces $\def\sy#1{#1&=\ts}\def\sk#1{#1&\supset\ts}\def\sp#1#2{\begin{align}\sy{A}#1&\\\sy{B}#2&\\\end{align}}\def\ss#1#2{\begin{align}\sk{A}#1&\\\sk{B}#2&\\\end{align}}$ ab b $\sp{\R\sm B}{\{0\}}$ ab ab $\sp{W_1(\R)}{W_2(\R)}$ Three-point connected spaces $\def\sy#1#2{\lower#1pt\hbox{$#2$}&=\ts}\def\sk#1#2{\lower#1pt\hbox{$#2$}&\supset\ts}\def\sp#1#2#3#4#5#6{\begin{align}\sy{#4}{A}#1&\\\sy{#5}{B}#2&\\\sy{#6}{C}#3&\\\end{align}}\def\ss#1#2#3#4#5#6{\left\{\begin{align}\sk{#4}{A}#1&\\\sk{#5}{B}#2&\\\sk{#6}{C}#3&\\\end{align}\right.}$ ac bc c $\sp{(-\infty,0)}{(0,\infty)}{\{0\}}{0}{0}{0}$ abc b c $[0,2)\ss{(0,1)\cup(1,2)}{\{0\}}{\{1\}}{0}{0}{0}$ abc bc c $[0,1)\ss{(0,1)\sm B}{h_{\{0\}}^-([0,1))}{\{0\}}{.3}{.3}{.3}$ abc bc bc $[0,1]\ss{G([0,1])}{M([0,1])}{F'([0,1])}{0}{0}{0}$ abc abc c $\sp{W_1(\R\sm C)}{W_2(\R\sm C)}{\{0\}}{.2}{.2}{.2}$ abc abc abc $\sp{W_1(\R)}{W_2(\R)}{W_3(\R)}{0}{0}{0}$ Four-point connected spaces $\def\sy#1#2{\lower#1pt\hbox{$#2$}&=\ts}\def\sk#1#2{\lower#1pt\hbox{$#2$}&\supset\ts}\def\sp#1#2#3#4#5#6#7#8{\begin{align}\sy{#5}{A}#1&\\\sy{#6}{B}#2&\\\sy{#7}{C}#3&\\\sy{#8}{D}#4&\\\end{align}}\def\ss#1#2#3#4#5#6#7#8{\left\{\begin{align}\sk{#5}{A}#1&\\\sk{#6}{B}#2&\\\sk{#7}{C}#3&\\\sk{#8}{D}#4&\\\end{align}\right.}$ ad bd cd d $[0,1]\ss{\bigcup_{n=0}^\infty G_{3n+1}([0,1])}{\bigcup_{n=0}^\infty G_{3n+2}([0,1])}{\bigcup_{n=0}^\infty G_{3n+3}([0,1])}{F([0,1])}{.3}{.3}{.3}{.3}$ acd bd c d $[0,3)\ss{(0,1)\cup(1,2)}{(2,3)}{\{1\}}{\{0,2\}}{0}{0}{0}{0}$ abcd b c d $[0,3)\ss{(0,1)\cup(1,2)\cup(2,3)}{\{0\}}{\{1\}}{\{2\}}{0}{0}{0}{0}$ acd bd cd d $[0,2)\ss{(0,1)\sm C}{(1,2)}{h_{\{0\}}^-([0,1))\cup h_{\{1\}}^+((0,1])}{\{0,1\}}{.2}{.2}{.2}{.2}$ acd bcd c d $[0,2)\ss{(0,1)}{(1,2)}{\{0\}}{\{1\}}{0}{0}{0}{0}$ abcd bd c d $[0,2)\ss{(0,1)\cup(1,2)\sm B}{h_{\{1\}}^+((0,1])}{\{0\}}{\{1\}}{0}{0}{0}{0}$ acd bcd cd d $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1);\nq B\lw(1,2);\nq C\lw\{1\}\}\hc \lower.2pt\hbox{$D$}\supset F([0,1])\hc}\right.$ ad bcd bcd d $[0,2)\ss{(0,1)}{W_1((1,2))}{W_2((1,2))}{\{0,1\}}{0}{0}{0}{0}$ abcd bd cd d $[0,1)\ss{(0,1)\sm(B\cup C)}{h_{\{0\}}^-([0,1))\sm C}{h_{\{1\}}^+((0,1])}{\{0\}}{0}{0}{0}{0}$ abcd b cd cd $[0,2]\ss{G([0,2])\sm B}{\{1\}}{M([0,2])}{F'([0,2])}{0}{0}{0}{0}$ abcd bcd c d $[0,2)\ss{(0,1)\cup(1,2)\sm B}{h_{\{0\}}^-([0,1))\cup h_{\{1\}}^-([1,2))}{\{0\}}{\{1\}}{.3}{.3}{.2}{.2}$ acd bcd cd cd $[0,1]\ss{\bigcup_{n=0}^\infty G_{2n+1}([0,1])}{\bigcup_{n=0}^\infty G_{2n+2}([0,1])}{M([0,1])}{F'([0,1])}{.3}{.3}{.3}{.3}$ abcd bcd cd d $[0,1)\ss{(0,1)\sm(B\cup C)}{h_C^-((0,1))}{h_{\{0\}}^-([0,1))}{\{0\}}{.3}{.3}{.3}{.2}$ abcd abcd c d $[0,2)\ss{W_1((0,1)\cup(1,2))}{W_2((0,1)\cup(1,2))}{\{0\}}{\{1\}}{0}{0}{0}{0}$ abcd bcd cd cd $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup(1,2);\nq B\lw\{1\}\}\hc \lower.2pt\hbox{$C$}\supset M([0,1])\hc \lower.2pt\hbox{$D$}\supset F'([0,1])\hc}\right.$ abcd bcd bcd d $[0,1]\ss{G([0,1])}{\bigcup_{n=1}^\infty M_n([0,1])}{F'([0,1])}{M_0([0,1])}{0}{0}{0}{0}$ abcd abcd cd d $[0,1)\ss{W_1((0,1))\sm C}{W_2((0,1))\sm C}{h_{\{0\}}^-([0,1))}{\{0\}}{0}{0}{0}{0}$ abcd abcd cd cd $[0,1]\ss{W_1(G([0,1]))}{W_2(G([0,1]))}{M([0,1])}{F'([0,1])}{0}{0}{0}{0}$ abcd bcd bcd bcd $[0,1]\ss{G([0,1])}{\bigcup_{n=0}^\infty M_{2n}([0,1])}{\bigcup_{n=0}^\infty M_{2n+1}([0,1])}{F'([0,1])}{.3}{.3}{.3}{.3}$ abcd abcd abcd d $\sp{W_1(\R\sm D)}{W_2(\R\sm D)}{W_3(\R\sm D)}{\{0\}}{.2}{.2}{.2}{.2}$ abcd abcd abcd abcd $\sp{W_1(\R)}{W_2(\R)}{W_3(\R)}{W_4(\R)}{0}{0}{0}{0}$ Five-point connected spaces $\def\sy#1{#1&=\ts}$ $\def\sk#1{#1&\supset\ts}$ $\def\sp#1#2#3#4#5{\begin{align}\sy{A}#1&\\\sy{B}#2&\\\sy{C}#3&\\\sy{D}#4&\\\sy{E}#5&\\\end{align}}$ $\def\ss#1#2#3#4#5{\left\{\begin{align}\sk{A}#1&\\\sk{B}#2&\\\sk{C}#3&\\\sk{D}#4&\\\sk{E}#5&\\\end{align}\right.}$ ae be ce de e $[0,1]\ss{\bigcup_{n=0}^\infty G_{4n+1}([0,1])}{\bigcup_{n=0}^\infty G_{4n+2}([0,1])}{\bigcup_{n=0}^\infty G_{4n+3}([0,1])}{\bigcup_{n=0}^\infty G_{4n+4}([0,1])}{F([0,1])}$ ade bd ce d e $[0,4)\ss{(0,1)\cup(2,3)}{(1,2)}{(3,4)}{\{1,2\}}{\{0,3\}}$ ade be ce d e $[0,2]\ss{\bigcup_{n=0}^\infty G_{3n+1}([0,2])\sm D}{\bigcup_{n=0}^\infty G_{3n+2}([0,2])}{\bigcup_{n=0}^\infty G_{3n+3}([0,2])}{\{1\}}{F([0,2])}$ ace bde c d e $[0,4)\ss{(0,1)\cup(1,2)}{(2,3)\cup(3,4)}{\{1\}}{\{3\}}{\{0,2\}}$ acde be c d e $[0,4)\left\{\matrix{H^-((0,1])\{\nq A\lw(0,1);\nq B\lw(1,2);\nq E\lw\{1,2\}\}\hc\hskip-2pt\begin{align}A&\supset(1,2)\cup(2,3)\cup(3,4)&\\ C&\supset\{2\}&\\ D&\supset\{3\}&\\ E&\supset\{0\}&\end{align}\hc}\right.$ abcde b c d e $[0,4)\ss{(0,1)\cup(1,2)\cup(2,3)\cup(3,4)}{\{0\}}{\{1\}}{\{2\}}{\{3\}}$ ade be ce de e $[0,1]\left\{\matrix{G_{3n+1}([0,1])\{\nq A\lw(0,1)\cup(1,2);\nq D\lw\{1\}\}{\rm\ for\ }n=0,1,2,\ldots\hc\hskip-1pt\begin{align}\sk{B}\bigcup_{n=0}^\infty G_{3n+2}([0,1])&\\\sk{C}\bigcup_{n=0}^\infty G_{3n+3}([0,1])&\\\sk{E} F([0,1])&\end{align}\hc}\right.$ ade bde ce d e $[0,1]\left\{\matrix{G_1([0,1])\{\nq A\lw(0,1);\nq B\lw(1,2);\nq D\lw\{1\}\}\hc\hskip-1pt\begin{align}\sk{A}\bigcup_{n=0}^\infty G_{3n+2}([0,1])&\\\sk{B}\bigcup_{n=0}^\infty G_{3n+3}([0,1])&\\\sk{C}\bigcup_{n=1}^\infty G_{3n+1}([0,1])&\\\sk{E} F([0,1])&\end{align}\hc}\right.$ ace bde ce d e $[0,1]\left\{\matrix{G_{2n+1}([0,1])\{\nq A\lw(0,1)\cup(1,2);\nq C\lw\{1\}\}{\rm\ for\ }n=0,1,2,\ldots\hc G_2([0,1])\{\nq B\lw(0,1)\cup(1,2);\nq D\lw\{1\}\}\hc\hskip-1pt\begin{align}\sk{B}\bigcup_{n=1}^\infty G_{2n+2}([0,1])&\\\sk{E} F([0,1])&\end{align}\hc}\right.$ acde be ce d e $[0,1]\left\{\matrix{G_{2n+1}([0,1])\{\nq A\lw(0,1)\cup(1,2);\nq C\lw\{1\}\}{\rm\ for\ }n=1,2,\ldots\hc G_1([0,1])\{\nq A\lw(0,1)\cup(1,2);\nq D\lw\{1\}\}\hc\hskip-1pt\begin{align}\sk{B}\bigcup_{n=0}^\infty G_{2n+2}([0,1])&\\\sk{E} F([0,1])&\end{align}\hc}\right.$ acde bd ce d e $[0,3)\ss{(0,1)\cup(1,2)\sm C}{(2,3)}{h^-_{\{1\}}([1,2))}{\{0,2\}}{\{1\}}$ acde bde c d e $[0,3)\ss{(0,1)\cup(1,2)}{(2,3)}{\{1\}}{\{0\}}{\{2\}}$ abcde be c d e $[0,3)\ss{(0,1)\cup(1,2)\cup(2,3)\sm B}{h^-_{\{2\}}([2,3))}{\{0\}}{\{1\}}{\{2\}}$ ade bde ce de e $[0,1]\left\{\matrix{G_{2n+1}([0,1])\{\nq A\lw(0,1);\nq B\lw(1,2);\nq D\lw\{1\}\}{\rm\ for\ }n=0,1,2,\ldots\hc\hskip-2pt\begin{align}\sk{C}\bigcup_{n=0}^\infty G_{2n+2}([0,1])&\\\sk{E}F([0,1])&\end{align}\hc}\right.$ ace bde ce de e $[0,1]\left\{\matrix{G_{2n+1}([0,1])\{\nq A\lw(0,1)\cup(1,2);\nq C\lw\{1\}\}{\rm\ for\ }n=0,1,2,\ldots\hc G_{2n+2}([0,1])\{\nq B\lw(0,1)\cup(1,2);\nq D\lw\{1\}\}{\rm\ for\ }n=0,1,2,\ldots\hc\hskip-2pt\begin{align}\sk{E}F([0,1])&\end{align}\hc}\right.$ ae be cde cde e $[0,1]\left\{\matrix{\lower.2pt\hbox{$A$}\supset\bigcup_{n=0}^\infty G_{3n+1}([0,1])\hc\lower.2pt\hbox{$B$}\supset\bigcup_{n=0}^\infty G_{3n+2}([0,1])\hc G_{3n+3}([0,1])\{\nq C\lw W_1((0,1));\nq D\lw W_2((0,1))\}{\rm\ for\ }n=0,1,2,\ldots\hc \lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ acde be ce de e $[0,1]\left\{\matrix{G_{2n+1}([0,1])\{\nq A\lw(0,1)\cup(1,2)\cup(2,3);\nq C\lw\{1\};\nq D\lw\{2\}\}{\rm\ for\ }n=0,1,2,\ldots\hc\hskip-2pt\begin{align}\sk{B}\bigcup_{n=0}^\infty G_{2n+2}([0,1])&\\\sk{E}F([0,1])&\end{align}\hc}\right.$ acde bc c de de $[0,4)\ss{(0,1)\cup G([1,2])\cup(2,3)}{(3,4)}{\{0,3\}}{M([1,2])}{F'([1,2])}$ ade bde cde d e $[0,4)\ss{\bigcup_{n=0}^\infty \big(G_{3n+1}([0,1])\cup G_{3n+1}([2,3])\big)\cup(1,2)\cup(3,4)}{\bigcup_{n=0}^\infty \big(G_{3n+2}([0,1])\cup G_{3n+2}([2,3])\big)}{\bigcup_{n=0}^\infty \big(G_{3n+3}([0,1])\cup G_{3n+3}([2,3])\big)}{F([0,1])}{F([2,3])}$ abe b cde cde e $[0,2]\left\{\matrix{\lower.2pt\hbox{$A$}\supset\bigcup_{n=0}^\infty G_{2n+1}([0,2])\sm B\hc\lower.2pt\hbox{$B$}\supset\{1\}\hc G_{2n+2}([0,2])\{\nq C\lw W_1((0,1));\nq D\lw W_2((0,1))\}{\rm\ for\ }n=0,1,2,\ldots\hc \lower.2pt\hbox{$E$}\supset F([0,2])\hc}\right.$ acde bde ce d e $[0,2)\ss{(0,1)\sm C}{(1,2)}{h_{\{0\}}^-([0,1))}{\{1\}}{\{0\}}$ acde be cde d e $[0,3)\ss{(0,1)\cup(1,2)\sm C}{(2,3)}{h^-_{\{0\}}([0,1))\cup h^+_{\{1\}}((0,1])\cup h^+_{\{2\}}((1,2])}{\{1\}}{\{0,2\}}$ acde bde c de e $[0,4)\left\{\matrix{G([0,1])\{\nq A\lw(0,1);\nq B\lw(1,2);\nq D\lw\{1\}\}\hc\hskip-2pt\begin{align}\sk{A}(1,2)\cup G([2,3])\cup(3,4)&\\\sk{C}F([2,3])\\\sk{E}F([0,1])&\end{align}\hc}\right.$ abcde be ce d e $[0,2)\ss{(0,1)\cup(1,2)\sm(B\cup C)}{h^+_{\{1\}}((0,1])}{h^-_{\{1\}}([1,2))}{\{0\}}{\{1\}}$ abcde bd ce d e $[0,2)\ss{(0,1)\cup(1,2)\sm(B\cup C)}{h^-_{\{0\}}([0,1))}{h^-_{\{1\}}([1,2))}{\{0\}}{\{1\}}$ abcde b c de de $[0,4)\ss{(0,1)\cup G([1,2])\cup(2,3)\cup(3,4)}{\{0\}}{\{3\}}{M([1,2])}{F'([1,2])}$ acde bcde c d e $[0,4)\ss{(0,1)\cup(2,3)}{(1,2)\cup(3,4)}{\{0\}}{\{1,3\}}{\{2\}}$ abcde bde c d e $[0,3)\ss{(0,1)\cup(1,2)\cup(2,3)\sm B}{h^-_{\{1\}}[1,2)\cup h^+_{\{2\}}(1,2]}{\{0\}}{\{1\}}{\{2\}}$ ade bde cde de e $[0,2)\left\{\matrix{H^-((0,1])\{\nq A\lw\bigcup_{n=0}^\infty G_{3n+1}([0,1]);\nq B\lw\bigcup_{n=0}^\infty G_{3n+2}([0,1]);\nq C\lw\bigcup_{n=0}^\infty G_{3n+3}([0,1]);\nq D\lw F([0,1])\sm\{0\}\}\hc\hskip-2pt\begin{align}\sk{A}(1,2)&\\\sk{E}\{0\}&\end{align}\hc}\right.$ abe be cde cde e $[0,2]\left\{\matrix{G_{2n+1}([0,2])\{\nq A\lw(0,1)\cup(1,2);\nq B\lw\{1\}\}{\rm\ for\ }n=0,1,2,\ldots\hc G_{2n+2}([0,2])\{\nq C\lw W_1((0,1));\nq D\lw W_2((0,1))\}{\rm\ for\ }n=0,1,2,\ldots\hc \lower.2pt\hbox{$E$}\supset F([0,2])\hc}\right.$ acde bde ce de e $[0,3]\left\{\matrix{G([0,3])\{\nq A\lw(0,1)\cup(1,2);\nq B\lw(2,3);\nq C\lw\{1\};\nq D\lw\{2\}\}\hc \lower.1pt\hbox{$E$}\supset F([0,3])\hc}\right.$ acde be cde de e $[0,1]\left\{\matrix{G_{2n+1}([0,1])\{\nq A\lw(0,1)\cup(1,2)\sm C;\nq C\lw h^-_{\{1\}}([1,2));\nq D\lw\{1\}\}{\rm\ for\ }n=0,1,2,\ldots\hc \lower.1pt\hbox{$B$}\supset\bigcup_{n=0}^\infty G_{2n+2}([0,1])\hc \lower.1pt\hbox{$E$}\supset F([0,1])\hc}\right.$ acde bde c de de $[0,2]\ss{\bigcup_{n=0}^\infty G_{2n+1}([0,2])\sm C}{\bigcup_{n=0}^\infty G_{2n+2}([0,2])}{\{1\}}{M([0,2])}{F'([0,2])}$ abcde be ce de e $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup(1,2)\cup(2,3)\cup(3,4);\nq B\lw\{1\};\nq C\lw\{2\};\nq D\lw\{3\}\}\hc \lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ abcde bc c de de $[0,3)\ss{G([0,1])\cup(1,2)\cup(2,3)\sm B}{h^-_{\{2\}}([2,3))}{\{2\}}{M([0,1])}{F'([0,1])}$ acde bde cde d e $[0,4)\left\{\matrix{\big(G_{2n+1}([0,1])\cup G_{2n+1}([2,3])\big)\{\nq A\lw(0,1)\cup(1,2);\nq C\lw\{1\}\}{\rm\ for\ }n=0,1,2,\ldots\hc \begin{align}\sk{A}(1,2)\cup(3,4)&\\\sk{B}\bigcup_{n=0}^\infty\big(G_{2n+2}([0,1])\cup G_{2n+2}([2,3])\big)&\\\sk{D}F([0,1])&\\\sk{E}F([2,3])\end{align}\hc}\right.$ acde bcde ce d e $[0,1]\left\{\matrix{G_1([0,1])\{\nq A\lw(0,1);\nq B\lw(1,2);\nq D\lw\{1\}\}\hc G_n([0,1])\{\nq A\lw(0,1);\nq B\lw(1,2);\nq C\lw\{1\}\}{\rm\ for\ }n=2,3,\ldots\hc\lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ ae bcde bcde d e $[0,3)\ss{(1,2)}{W_1((0,1)\cup(2,3))}{W_2((0,1)\cup(2,3))}{\{0\}}{\{1,2\}}$ abcde bde ce d e $[0,2)\ss{(0,1)\cup(1,2)\sm(B\cup C)}{h^-_{\{0\}}([0,1))\cup h^+_{\{1\}}((0,1])}{h^-_{\{1\}}([1,2))}{\{0\}}{\{1\}}$ abcde bde c de e $[0,2)\ss{(0,1)\cup(1,2)\sm(B\cup D)}{h^-_D((1,2))}{\{0\}}{h^-_{\{1\}}([1,2))}{\{1\}}$ abcde bcde c d e $[0,3)\ss{(0,1)\cup(1,2)\cup(2,3)\sm B}{h^-_{\{0\}}([0,1))\cup h^-_{\{1\}}([1,2))\cup h^-_{\{2\}}([2,3))}{\{0\}}{\{1\}}{\{2\}}$ ade bde cde de de $[0,1]\ss{\bigcup_{n=0}^\infty G_{3n+1}([0,1])}{\bigcup_{n=0}^\infty G_{3n+2}([0,1])}{\bigcup_{n=0}^\infty G_{3n+3}([0,1])}{M([0,1])}{F'([0,1])}$ abcde bc bc de de $[0,4)\ss{G([0,1])\cup(1,2)\cup G([2,3])\cup(3,4)}{M([0,1])}{F'([0,1])}{M([2,3])}{F'([2,3])}$ abe abe cde cde e $[0,1]\left\{\matrix{G_{2n+1}([0,1])\{\nq A\lw W_1((0,1));\nq B\lw W_2((0,1))\}{\rm\ for\ }n=0,1,2,\ldots\hc G_{2n+2}([0,1])\{\nq C\lw W_1((0,1));\nq D\lw W_2((0,1))\}{\rm\ for\ }n=0,1,2,\ldots\hc\lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ acde bde cde de e $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\sm C;\nq B\lw(1,2);\nq C\lw h^+_{\{1\}}((0,1]);\nq D\lw\{1\}\}\hc\lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ acde be cde cde e $[0,1]\left\{\matrix{G_{2n+1}([0,1])\{\nq A\lw(0,1)\cup G([1,2])\cup(2,3);\nq C\lw M([1,2]);\nq D\lw F'([1,2])\}{\rm\ for\ }n=0,1,2,\ldots\hc\begin{align}\sk{B}\bigcup_{n=0}^\infty G_{2n+2}([0,1])&\\\sk{E}F([0,1])\end{align}\hc}\right.$ acde bcde ce de e $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup(2,3);\nq B\lw(1,2);\nq C\lw\{1\};\nq D\lw\{2\}\}{\rm\ for\ }n=0,1,2,\ldots\hc\lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ acde bcde c de de $[0,4)\ss{\bigcup_{n=0}^\infty\big(G_{2n+1}([0,1])\cup G_{2n+1}([2,3])\big)\cup(1,2)\cup(3,4)}{\bigcup_{n=0}^\infty\big(G_{2n+2}([0,1])\cup G_{2n+2}([2,3])\big)}{F([0,1])}{M([2,3])}{F'([2,3])}$ ae bcde bcde de e $[0,2)\ss{(0,1)}{W_1((1,2))\sm D}{W_2((1,2))\sm D}{h^-_{\{1\}}([1,2))\cup h^+_{\{2\}}((1,2])}{\{0,1\}}$ abcde bde ce de e $[0,1)\ss{(0,1)\sm(B\cup C\cup D)}{h^+_D((0,1))}{h^-_{\{0\}}([0,1))\sm(B\cup D)}{h^+_{\{1\}}((0,1])}{\{0\}}$ abcde bde c de de $[0,3)\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup(1,2);\nq B\lw\{1\}\}\hc\hskip-1pt\begin{align}\sk{A}(1,2)\cup(2,3)&\\\sk{C}\{2\}&\\\sk{D}M([0,1])&\\\sk{E}F'([0,1])\end{align}\hc}\right.$ acde bcde cde d e $[0,4)\left\{\matrix{G([0,1])\{\nq A\lw(0,1);\nq B\lw(1,2);\nq C\lw\{1\}\}\hc G([2,3])\{\nq A\lw(0,1);\nq B\lw(1,2);\nq C\lw\{1\}\}\hc\hskip-1pt\begin{align}\sk{A}(1,2)\cup(3,4)&\\\sk{D}F([0,1])&\\\sk{E}F([2,3])\end{align}\hc}\right.$ ade bcde bcde d e $[0,2)\ss{(0,1)}{W_1((1,2))}{W_2((1,2))}{\{0\}}{\{1\}}$ abcde bde cde d e $[0,2)\ss{(0,1)\cup(1,2)\sm(B\cup C)}{h^-_{\{0\}}([0,1))\cup h^-_{\{1\}}([1,2))\sm C}{h^+_{\{1\}}((0,1])\cup h^+_{\{2\}}((1,2])}{\{0\}}{\{1\}}$ abcde b cde cde e $[0,2]\left\{\matrix{\lower.2pt\hbox{$A$}\supset\bigcup_{n=0}^\infty G_{2n+1}([0,2])\sm B\hc\lower.2pt\hbox{$B$}\supset\{1\}\hc G_{2n+2}([0,2])\{\nq A\lw(0,1)\cup G([1,2])\cup(2,3);\nq C\lw M([1,2]);\nq D\lw F'([1,2])\}{\rm\ for\ }n=0,1,2,\ldots\hc\lower.2pt\hbox{$E$}\supset F([0,2])\hc}\right.$ abcde bcde ce d e $[0,2)\ss{(0,1)\cup(1,2)\sm(B\cup C)}{h^-_{\{0\}}([0,1))\cup h^-_{C}((1,2))}{h^-_{\{1\}}([1,2))}{\{0\}}{\{1\}}$ abcde abcde c d e $[0,3)\ss{W_1((0,1)\cup(1,2)\cup(2,3))}{W_2((0,1)\cup(1,2)\cup(2,3))}{\{0\}}{\{1\}}{\{2\}}$ acde bde cde de de $[0,1]\left\{\matrix{G_{2n+1}([0,1])\{\nq A\lw(0,1)\cup(1,2);\nq C\lw\{1\}\}{\rm\ for\ }n=0,1,2,\ldots\hc\begin{align}\sk{B}\bigcup_{n=0}^\infty G_{2n+2}([0,1])&\\\sk{D}M([0,1])&\\\sk{E}F'([0,1])\end{align}\hc}\right.$ acde bcde cde de e $[0,2)\left\{\matrix{H^-((0,1])\{\nq H^-((1,2])\{\nqq A\lw(0,1);\nqq B\lw(1,2);\nqq C\lw\{1,2\}\};\nq A\lw(0,1);\nq D\lw\{1\}\}\hc\lower.2pt\hbox{$A$}\supset(1,2)\hc\lower.2pt\hbox{$E$}\supset\{0\}\hc}\right.$ ade bcde bcde de e $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1);\nq B\lw W_1((1,2));\nq C\lw W_2((1,2));\nq D\lw\{1\}\}\hc\lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ abcde bde cde de e $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup(1,2)\sm(B\cup C);\nq B\lw h^+_{\{1\}}((0,1]);\nq C\lw h^-_{\{1\}}([1,2));\nq D\lw\{1\}\}\hc\lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ abcde be cde cde e $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup(1,2)\cup G([2,3])\cup(3,4);\nq B\lw\{1\};\nq C\lw M([2,3]);\nq D\lw F'([2,3])\}\hc\lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ abcde bcde ce de e $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup(1,2)\cup(2,3)\sm B;\nq B\lw h^-_{\{1\}}([1,2))\cup h^-_{\{2\}}([2,3));\nq C\lw\{1\};\nq D\lw\{2\}\}\hc\lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ abcde bcde c de de $[0,1]\left\{\matrix{G_1([0,1])\{\nq A\lw(0,1)\cup(1,2)\sm B;\nq B\lw h^-_{\{1\}}([1,2));\nq C\lw\{1\}\}\hc G_n([0,1])\{\nq A\lw(0,1)\cup(1,2);\nq B\lw\{1\}\}{\rm\ for\ }n=2,3,\ldots\hc\lower.2pt\hbox{$D$}\supset M([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ abcde bcde cde d e $[0,2)\ss{(0,1)\cup(1,2)\sm(B\cup C)}{h^-_{C\,\cap\,(0,1)}((0,1))\cup h^-_{C\,\cap\,(1,2)}((1,2))}{h^-_{\{0\}}([0,1))\cup h^-_{\{1\}}([1,2))}{\{0\}}{\{1\}}$ abcde abcde ce d e $[0,2)\ss{W_1((0,1)\cup(1,2))\sm C}{W_2((0,1)\cup(1,2))\sm C}{h^-_{\{1\}}([1,2))}{\{0\}}{\{1\}}$ acde bcde cde de de $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1);\nq B\lw(1,2);\nq C\lw\{1\}\}\hc\lower.2pt\hbox{$D$}\supset M([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ ade bcde bcde de de $[0,1]\left\{\matrix{G_{2n+2}([0,1])\{\nq B\lw W_1((0,1));\nq C\lw W_2((0,1))\}{\rm\ for\ }n=0,1,2,\ldots\hc\lower.2pt\hbox{$A$}\supset\bigcup_{n=0}^\infty G_{2n+1}([0,1])\hc\lower.2pt\hbox{$D$}\supset M([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ abcde bde cde de de $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup(1,2)\cup(2,3);\nq B\lw\{1\};\nq C\lw\{2\}\}\hc\hskip-1pt\begin{align}\sk{D}M([0,1])&\\\sk{E}F'([0,1])\end{align}\hc}\right.$ acde bcde cde cde e $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup\bigcup_{n=0}^\infty G_{2n+1}([1,2])\cup(2,3);\nq B\lw\bigcup_{n=0}^\infty G_{2n+2}([1,2]);\nq C\lw M([1,2]);\nq D\lw F'([1,2])\}\hc\lower.2pt\hbox{$E$}\supset F([0,1])\hc}\right.$ abcde b cde cde cde $[0,3)\ss{G([0,1])\cup(1,2)\cup(2,3)}{\{2\}}{\bigcup_{n=0}^\infty M_{2n}([0,1])}{\bigcup_{n=0}^\infty M_{2n+1}([0,1])}{F'([0,1])}$ ae bcde bcde bcde e $[0,2)\ss{(0,1)}{W_1((1,2))}{W_2((1,2))}{W_3((1,2))}{\{0,1\}}$ abcde bcde cde de e $[0,1)\ss{(0,1)\sm(B\cup C\cup D)}{h^-_C((0,1))\sm D}{h^-_D((0,1))}{h^-_{\{0\}}([0,1))}{\{0\}}$ abcde abcde ce de e $[0,1)\ss{W_1((0,1))\sm(C\cup D)}{W_2((0,1))\sm(C\cup D)}{h^-_{\{0\}}([0,1))\sm D}{h^+_{\{1\}}((0,1])}{\{0\}}$ abcde abcde c de de $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw W_1((0,1));\nq B\lw W_2((0,1))\}\hc\lower.2pt\hbox{$C$}\supset M_0([0,1])\hc\lower.2pt\hbox{$D$}\supset\bigcup_{n=1}^\infty M_n([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ abcde bcde bcde d e $[0,4)\left\{\matrix{\left(H^+([0,1))\cup H^+([2,3))\right)\{\nq A\lw G([0,1])\cup(1,2);\nq B\lw M([0,1]);\nq C\lw F'([0,1])\}\hc\begin{align}\sk{A}(1,2)\cup(3,4)&\\\sk{D}\{1\}&\\\sk{E}\{3\}\end{align}\hc}\right.$ abcde abcde cde d e $[0,2)\ss{W_1((0,1)\cup(1,2))\sm C}{W_2((0,1)\cup(1,2))\sm C}{h^-_{\{0\}}([0,1))\cup h^-_{\{1\}}([1,2))}{\{0\}}{\{1\}}$ abcde bcde cde de de $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup(1,2)\sm B;\nq B\lw h^-_{\{1\}}([1,2));\nq C\lw\{1\}\}\hc\lower.2pt\hbox{$D$}\supset M([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ abcde bcde cde cde e $[0,2)\left\{\matrix{H^+([1,2))\{\nq G([0,1])\{\nqq A\lw(0,1)\cup(1,2);\nqq B\lw\{1\}\};\nq A\lw(1,2);\nq C\lw M([0,1]);\nq D\lw F'([0,1])\}\hc\lower.2pt\hbox{$A$}\supset(0,1)\hc\lower.2pt\hbox{$E$}\supset\{0\}\hc}\right.$ abcde bcde bcde de e $[0,2)\left\{\matrix{H^+([1,2))\{\nq A\lw G([0,1])\cup(1,2);\nq B\lw\bigcup_{n=1}^\infty M_n([0,1]);\nq C\lw F'([0,1]);\nq D\lw M_0([0,1])\}\hc\lower.2pt\hbox{$A$}\supset(0,1)\hc\lower.2pt\hbox{$E$}\supset\{0\}\hc}\right.$ abcde abcde cde de e $[0,1)\ss{W_1((0,1))\sm(C\cup D)}{W_2((0,1))\sm(C\cup D)}{h^-_D((0,1))}{h^-_{\{0\}}([0,1))}{\{0\}}$ acde bcde cde cde cde $[0,1]\ss{\bigcup_{n=0}^\infty G_{2n+1}([0,1])}{\bigcup_{n=0}^\infty G_{2n+2}([0,1])}{\bigcup_{n=0}^\infty M_{2n}([0,1])}{\bigcup_{n=0}^\infty M_{2n+1}([0,1])}{F'([0,1])}$ abcde bcde bcde de de $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw (0,1)\cup G([1,2])\cup(2,3);\nq B\lw M([1,2]);\nq C\lw F'([1,2])\}\hc\lower.2pt\hbox{$D$}\supset M([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ abcde abcde cde de de $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw W_1((0,1)\cup(1,2));\nq B\lw W_2((0,1)\cup(1,2));\nq C\lw\{1\}\}\hc\lower.2pt\hbox{$D$}\supset M([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ abcde abcde cde cde e $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw W_1((0,1)\cup G([1,2])\cup(2,3));\nq B\lw W_2((0,1)\cup G([1,2])\cup(2,3));\nq C\lw M([1,2]);\nq D\lw F'([1,2])\}\hc\lower.2pt\hbox{$D$}\supset M([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ abcde abcde abcde d e $[0,2)\ss{W_1((0,1)\cup(1,2))}{W_2((0,1)\cup(1,2))}{W_3((0,1)\cup(1,2))}{\{0\}}{\{1\}}$ abcde bcde cde cde cde $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw(0,1)\cup(1,2);\nq B\lw\{1\}\}\hc\lower.2pt\hbox{$C$}\supset\bigcup_{n=0}^\infty M_{2n}([0,1])\hc\lower.2pt\hbox{$D$}\supset\bigcup_{n=0}^\infty M_{2n+1}([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ abcde bcde bcde bcde e $[0,1]\ss{G([0,1])}{\bigcup_{n=0}^\infty M_{2n+1}([0,1])}{\bigcup_{n=0}^\infty M_{2n+2}([0,1])}{F'([0,1])}{M_0([0,1])}$ abcde abcde abcde de e $[0,1)\ss{W_1((0,1))\sm D}{W_2((0,1))\sm D}{W_3((0,1))\sm D}{h^-_{\{0\}}([0,1))}{\{0\}}$ abcde abcde cde cde cde $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw W_1((0,1));\nq B\lw W_2((0,1))\}\hc\lower.2pt\hbox{$C$}\supset\bigcup_{n=0}^\infty M_{2n}([0,1])\hc\lower.2pt\hbox{$D$}\supset\bigcup_{n=0}^\infty M_{2n+1}([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ abcde abcde abcde de de $[0,1]\left\{\matrix{G([0,1])\{\nq A\lw W_1((0,1));\nq B\lw W_2((0,1));\nq C\lw W_3((0,1))\}\hc\lower.2pt\hbox{$D$}\supset M([0,1])\hc\lower.2pt\hbox{$E$}\supset F'([0,1])\hc}\right.$ abcde bcde bcde bcde bcde $[0,1]\ss{G([0,1])}{\bigcup_{n=0}^\infty M_{3n}([0,1])}{\bigcup_{n=0}^\infty M_{3n+1}([0,1])}{\bigcup_{n=0}^\infty M_{3n+2}([0,1])}{F'([0,1])}$ abcde abcde abcde abcde e $\sp{W_1(\R\sm\{0\})}{W_2(\R\sm\{0\})}{W_3(\R\sm\{0\})}{W_4(\R\sm\{0\})}{\{0\}}$ abcde abcde abcde abcde abcde $\sp{W_1(\R)}{W_2(\R)}{W_3(\R)}{W_4(\R)}{W_5(\R)}$<|endoftext|> TITLE: Multiples in sets of positive upper density QUESTION [11 upvotes]: Suppose we are given $A \subseteq \mathbb{N}$ with $\lim\sup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} > 0$. For $k\in \mathbb{N}, k\geq 2$ we set $$M_A(k) = \{a\in A: ka \in A\}.$$ Does there exist $k\in \mathbb{N}, k\geq 2$ such that $M_A(k)$ is infinite? REPLY [15 votes]: Not necessarily: you can in fact have $M_k(A)=\varnothing$ for all integer $k\ge 2$. This was shown by Besicovitch ("On the density of certain sequences of integers", Math. Ann. 110 (1935), no. 1, 336–341) who has constructed a set (of positive integers) of positive upper density such that none of the elements of the set is divisible by any other element. REPLY [10 votes]: Besicovitch's result mentioned by Seva is quite hard, while in your question you may simply take $A=\cup [n!+1,2n!]$.<|endoftext|> TITLE: Books and resources on PDEs that use Mathematica and Matlab QUESTION [7 upvotes]: Can you recommend some reference books that use software like MATLAB and Mathematica to deal with the basic topics in analysis of PDE (the ones you can find in Strauss' book Partial Differential Equations: An Introduction) and numerical analysis of PDE? REPLY [2 votes]: 4 reference books for the study of PDE with MATLAB: Coleman, Matthew P. An introduction to partial differential equations with MATLAB. Second edition. Chapman & Hall/CRC Applied Mathematics and Nonlinear Science Series. CRC Press, Boca Raton, FL, 2013. Thorough treatment of PDEs and their applications, includes numerous problem-solving exercises, MATLAB code on the author's website, first edition from 2005. Li, Jichun; Chen, Yi-Tung Computational partial differential equations using MATLAB. With 1 CD-ROM (Windows, Macintosh and UNIX). Chapman & Hall/CRC Applied Mathematics and Nonlinear Science Series. CRC Press, Boca Raton, FL, 2009. Provides standard finite difference and finite elements, novel techniques, such as high-order compact finite difference and meshless methods, applications from the fields of mechanical and electrical engineering, presents both theoretical numerical analysis and practical implementations, many computer projects and problems, includes a CD-ROM with MATLAB source code. Stanoyevitch, Alexander Introduction to numerical ordinary and partial differential equations using MATLAB. Pure and Applied Mathematics (New York). Wiley-Interscience, Hoboken, NJ, 2005. Thorough coverage of analytic concepts, geometric concepts, programs and algorithms, and applications. Extensive chapter on the finite element, including mesh generation, enables one to numerically solve general elliptic boundary value problems, FTP site that includes downloadable files. Cooper, Jeffery Introduction to partial differential equations with MATLAB. Applied and Numerical Harmonic Analysis. Birkhauser Boston, Inc., Boston, MA, 1998. Includes analytical and numerical methods, separation of variables technique, together with transform methods associated with Fourier, Green's function, variational methods, theory and applications of finite elements for boundary value problems, method of finite differences is also considered. and three books for PDE with Mathematica: Adzievski, Kuzman; Siddiqi, Abul Hasan Introduction to partial differential equations for scientists and engineers using Mathematica. CRC Press, Boca Raton, FL, 2014. Provides fundamental concepts, ideas, and terminology related to PDEs, discusses separation of variable method, studies the solution of the heat equation using Fourier and Laplace transforms, examines the Laplace and Poisson equations of different rectangular circular domains, and discuss finite difference methods. Kythe, Prem K.; Puri, Pratap; Schäferkotter, Michael R. Partial differential equations and boundary value problems with Mathematica. Second edition. Chapman & Hall/CRC, Boca Raton, FL, 2003. Theory and applications for solving initial and boundary value problems involving, in general, the first-order partial differential equations, and in particular, the second-order partial differential equations of mathematical physics and continuum mechanics. Vvedensky, Dimitri Partial differential equations with Mathematica. Physics Series. Addison-Wesley Publishing Company, Wokingham, 1993. Covers linear and nonlinear partial differential equations with exemplar examples, inspired by the symbolic software Mathematica.<|endoftext|> TITLE: Do minimal submanifolds minimize area locally? QUESTION [10 upvotes]: A few days ago I asked this question on math.stackexchange: Consider $(\tilde{M},g)$ a riemannian manifold and $M \subset \tilde{M}$ riemannian submanifold. Is it true that if $M$ is a minimal submanifold of $\tilde{M}$ then for every $p \in M$ there exists a neighborhood $W$ of $p$ in $\tilde{M}$ such that $V=W\cap M$ has least area among every $\Omega \subset W$ with $\partial \Omega = \partial V$? I've been thinking about it, I think it is true but I don't know how to prove. If it's true, how should I go about proving it? Link: Do minimal submanifolds minimize area locally? As asked there, we say a submanifold is minimal if the mean curvature vanishes identically, or equivalently, it's a critical point of the area functional. Thanks in advance! REPLY [21 votes]: Yes, this is true, but finding an explicit general proof in the literature seems to be a challenge. I think that many people just believe it without having seen an actual proof. One reason is that it's very easy to prove that small pieces of minimal submanifolds are strongly stable, i.e., small nontrivial perturbations of a minimal submanifold supported in sufficiently small domains must strictly increase the area. While this is pretty convincing, it doesn't actually prove local minimizing, though, because you don't know that everything with the same boundary is a perturbation of the thing you start with. For a proof that works when the ambient manifold $\tilde M$ is Euclidean space, see, for example, Theorem 2.1 in Curvy slicing proves that triple junctions locally minimize area, by Gary Lawlor and Frank Morgan, J. Diff. Geom 44 (1996), 514–528. I think that standard techniques can be used to extend their proof to cover general ambient manifolds, since all you want is the local result. If you are willing to assume that the competitor $\Omega$ is an oriented submanifold, there is an easier proof by the method of calibrations. The strategy is this: With $p\in M^n$ given, one constructs a closed $n$-form $\phi$ on $\tilde M$ with the property that $|\phi(e_1,\ldots,e_n)|\le 1$ for every orthonormal $n$-tuple $e_1,\ldots,e_n\in T_x\tilde M$ for every $x\in M$ and such that $\phi$ pulls back to equal the area form on $M\cap V$ for some open convex $p$-neighborhood $V\subset\tilde M$. [This construction of the calibration $\phi$ is a little tricky, since the first thing you would think of is to try to do this by pulling back the volume form on $M$ to a tubular neighborhood of $M$ by the 'closest point on $M$' projection, which works when $M$ has dimension $1$ or codimension $1$ but not otherwise, even when the ambient space is flat. Of course, you will need to use the fact that $M$ is a minimal submanifold (i.e., its mean curvature vanishes) in order to construct $\phi$.] Anyway, once $\phi$ is constructed, the rest is easy: If $\Omega\subset \tilde M$ is $n$-dimensional and oriented and satisfies $\partial\Omega = \partial(M\cap V)$ (in the oriented sense), then because $\Omega\subset V$ and $V$ is convex (which implies that $\Omega \cup (-(M\cap V)$ is homologous to $0$), we have $$ \mathrm{vol}(\Omega) \ge \int_\Omega \phi = \int_{M\cap V} \phi = \mathrm{vol}(M\cap V). $$ (The first inequality follows because, by construction, $\phi$ pulls back to $\Omega$ to be no more than the volume form on $\Omega$. The second equality follows by Stokes' Theorem, since $\mathrm{d}\phi=0$ and $\Omega \cup (-(M\cap V))$ has no boundary and is null-homologous. The third equality follows because $\phi$ pulls back to $M\cap V$ to be its volume form.)<|endoftext|> TITLE: Monochromatic point sets in two-colored plane QUESTION [9 upvotes]: Which are the configrations $P\subset \mathbb{R}^2$ of points, such that the following property holds: Property M (for Monochromatic): Every two-coloring of $\mathbb{R}^2$ contains a monochromatic copy of $P$. (By "copy" we allow Euclidean motions and also scaling by a positive factor. By "monochromatic" we understand, as the name suggests, that all points in the copy have the same color) Context: In the question "Triangles whose vertices and center have all the same color", the point configurations "equilateral triangle" and "equilateral triangle with its center" are considered. The question "Finding monochromatic rectangles in a countable coloring of $\mathbb{R}^2$" considered a slightly different problem: the colorings are countable and the rectangle is axis aligned (but its aspect ratio is not fixed). If a complete classification of point configurations which satisfy Property M is too much to ask for, here are some perhaps simpler questions: What is the smallest example of a point configuration, which does not satisfy Property M? Can point configurations with Property M be arbitrarily large, perhaps even infinite (countable or not)? Given a point set $P$, is there an algorithm to decide whether it has Property M? Has this question been studied somewhere? (I am aware of some work, where one does not allow the scaling of the point set, e.g. Monochromatic triangles in two-colored plane REPLY [5 votes]: Not an answer, just an illustration: Permit me to mention Ron Graham's challenge from 2003:     (Figure from Computational Geometry Column 46.) The citations here overlap with those referenced by Jan Kyncl.<|endoftext|> TITLE: Expectation involving maximum of Gaussian variables QUESTION [7 upvotes]: Let $X\sim N(0, I_d)$ be a $d$-dimensional Gaussian random vector. Let $W_1, \ldots, W_k \in \mathbb{R}^d$ be $k$ fixed vectors in general positions. It is clear that $w_i^\top X, \ldots, w_k^\top X$ are jointly Gaussian random variables. Let $$ Y = \max _{i \in [k]} W_i^\top x ,$$ my question is how to compute $\mathbb{E} [ Y\cdot X]$. My idea is to use a smooth approximation of the max function and the Stein's identity. Let $$ f(y, \alpha ) = \alpha^{-1} \log \left [ \exp(\alpha y_1) + \ldots + \exp( \alpha y_k ) \right ],$$ it is known that $ \left | \max_i y_i - f(y, \alpha) \right | \leq \log k / \alpha$. Then by Stein's identity, we consider \begin{align} \mathbb{E} \left [ f(WX , \alpha) \cdot X \right ] = \mathbb{E} [ \nabla_{X} f(W X , \alpha) ] = \mathbb{E} \left [ \frac{ \sum_{i \in [k]} \exp ( \alpha \cdot W_i ^\top X ) \cdot W_i }{ \sum_{i \in [k]} \exp(\alpha \cdot W_i ^\top X )} \right ]. \end{align} If I naively take $\alpha \rightarrow +\infty$, I would get \begin{align} &\lim_{\alpha\rightarrow +\infty} \mathbb{E} \left [ f(WX , \alpha) \cdot X \right ] = \mathbb{E} \left[ \sum_{j\in[k]} I\left\{ j =\arg\max _{i\in[k]} W_i^\top X \right \} \cdot W_j \right ] \\ &\quad = \sum_{j\in [k]} \mathbb{P}\left (j =\arg\max _{i\in[k]} W_i^\top X \right ) \cdot W_i. \end{align} I wonder whether my naive derivation gets the correct answer and whether such derivation could be made rigorous. REPLY [2 votes]: In the $d=2$ case the result is explicit and nice. Let $x$ and $y$ be i.i.d standard normal, and let $(t,u)$ and $(v,w)$ be fixed vectors. Let $M = \begin{pmatrix} t & u \\ v & w \\ \end{pmatrix}, \text{ and } X = \begin{pmatrix} x \\ y \\ \end{pmatrix} $. We want $E\big[\max(MX)\,X\big]$. The first coordinate is \begin{align} &E\big[\max(tx+uy,vx+wy)x\big] \\ ={}&E\big[(tx+uy+vx+wy)x+\left|tx+uy-vx-wy\right|x\big]\ /\ 2 \\ ={}&E\big[(tx+uy+vx+wy)x\big]\ /\ 2 \\ ={}&(t+v)\ /\ 2 \end{align} The first equation uses the identity $\max(a,b)=(a+b+|a-b|)/2$. (This is the part specific to $d=2$: are there any analogs with more than two variables being maximized?) The second equation uses the symmetry of $(x,y)$ and $(-x,-y)$ in the distribution. The third equation uses the standard normal variance and independence of $X$ and $Y$. So using similar reasoning for the second coordinate, $$ E\big[\max(MX)\,X\big]= \frac{1}{2} \begin{pmatrix} t+v\\ u+w\\ \end{pmatrix}. $$<|endoftext|> TITLE: Number of tilting modules QUESTION [7 upvotes]: Let $A=A_n$ be the algebra of upper triangular matrices over a field $K$ with $n$ simple modules. It is a nice result that there are $C_{n+1}=1,2,5,14,...$ (Catalan numbers for $n \geq 1$) tilting $A_n$-modules, where a tilting module $T$ is a module with $n$ indecomposable summands (we assume all modules are basic) and projective dimension 1 and $Ext^{1}(T,T)=0$. Let $J$ be the Jacobson radical of $A_n$ and $B_{n,l}:=A_n /J^l$ for some $n-1 \geq l \geq 2$. Computer experiments with small n and l suggest the following generalisation: The number of tilting $B_{n,l}$ modules equals $C_l$. Is this true? If yes, there should be a simple reason, which I do not see at the moment. If no, what is the correct number of $B_{n,l}-$tilting modules? My guess goes as follows: Let $e$ be the idempotent of $B=B_{n,l}$ such that $eB$ is minimal faithful projective-injective. Then $eB$ is a summand of any $B$-tilting module. Thus any tilting module is of the form $T=eB \oplus X$ and $X$ corresponds to a tilting-module of $B/BeB$ (why?) which can be identified with $A_{l-1}$. Thus there are as many tilting $B$-modules as $A_{l-1}$ tilting modules which is $C_l$. REPLY [6 votes]: Let $e$ be the idempotent in $B$ such that $Be$ is the direct sum of the $n-l$ indecomposable projective-injectives which do not have projective proper submodules. Then the two-sided ideal $BeB=Be$ is projective as a left $B$-module, so $B \to B/BeB$ is a homological epimorphism, see for instance Koenig, Steffen; Nagase, Hiroshi, Hochschild cohomology and stratifying ideals., J. Pure Appl. Algebra 213, No. 5, 886-891 (2009). ZBL1181.16009. This means that $\operatorname{Ext}^*_{B/BeB}(M,N) \cong \operatorname{Ext}^*_B(M,N)$ for all $B/BeB$-modules $M$ and $N$. We also have an isomorphism of algebras $B/BeB \cong A_l$. Claim: If $X$ is a tilting $B/BeB$-module, then $Be \oplus X$ is a tilting $B$-module. Proof: Suppose $X$ is tiling $B/BeB$-module, and let $T=Be \oplus X$. 1) $X$ has $l$ indecomposable summands, so $T=Be \oplus X$ has $n$ indecomposable summands. 2) For $n \geq 2$, we have $$\operatorname{Ext}^n_B(T,B) \cong \operatorname{Ext}^n_B(X,B/BeB) \cong \operatorname{Ext}^n_{B/BeB}(X,B/BeB)=0,$$ so the projective dimension of $T$ is at most $1$. 3) $\operatorname{Ext}^1_B(T,T) \cong \operatorname{Ext}^1_B(X,X) \cong \operatorname{Ext}^1_{B/BeB}(X,X)=0. $ Claim: There are no other tilting $B$-modules. Proof: Suppose $T=Be \oplus Y$ is a tilting $B$-module and $Y$ has an indecomposable direct summand $Y'$ with $(BeB)Y' \neq 0$. Let $f \colon P(Y') \to Y'$ be the projective cover. Then $P(Y')$ is an indecomposable direct summand of $Be$, so $\ker f$ is not projective. Hence the projective dimension of $Y'$ is at least $2$. Contradiction. Conclusion: The number of tilting $B$-modules is equal to the number of tilting $A_l$-modules.<|endoftext|> TITLE: Are trivial zeros of the zeta function important? QUESTION [8 upvotes]: Non-trivial zeros play an important (main) role in the distribution of prime numbers. Are there theorems in which trivial zeros play an important (main) role? REPLY [5 votes]: A conjecture of Quillen-Lichtenbaum $$ \lim_{s \to n} (n-s)^{-\mu_n} \zeta_F (-s) = \pm \frac{\mid{K_{2n} (O_F)_{tor} }\mid}{\mid{K_{2n+1} (O_F)_{tor}}\mid} R_{F,n} * 2^{?}\\ $$ where $F$ is the number field like $\mathbb{Q}$ and $O_F$ is the integers therein generalizing $\mathbb{Z}$ in the case of Riemann zeta. $\mu_n$ is the multiplicity of the zero there. That makes sure you don't just get $0$. $K_\bullet (O_F)_{tor} $ means the torsion subgroup of algebraic K theory of that ring. $R_{F,n}$ is a so-called regulator and $2^?$ is for an unknown power of $2$. So you see the left hand side is leading information about when you have zeroes at negative integers. This is proven in a bunch of cases via Voevodsky. In the $O_F=\mathbb{Z}$ case you just get the numerators and denominators of the Bernoulli numbers. That recovers the $-\frac{B_{n+1}}{n+1}$ above up to signs, powers of 2 and regulators. So if you are interested in computing anything in the RHS, zooming in on the zeroes of the associated $\zeta_F$ are the most important.<|endoftext|> TITLE: Can one prove the elementary divisor theorem for PIDs by elementary matrix operations? QUESTION [23 upvotes]: The elementary divisor theorem was originally proved by a calculation on integer matrices, using elementary (invertible) row and column operations to put the matrix into Smith normal form. That is the matrix is zero off the diagonal, and on the diagonal each entry divides the one below it. This calculation immediately generalizes to matrices over any Euclidean ring. The key point is that in a Euclidean ring the GCD of two elements can be found by a series of steps where, in each step, one of the arguments is not multiplied by any non-unit. Specifically you reduce an entry $B$ modulo an entry $A$ by adding some (generally non-unit) multiple of A to B to get a result with smaller Euclidean norm than $A$. But $B$ is not multiplied by anything in this step, and so certainly not multiplied by any non-unit. This does not generalize directly to every PID. In a PID elements $A,B$ have a GCD which is a linear combination of them, But that linear combination may require non-unit multiples of each (even in the integers). And it is not obvious that in every PID the process can be broken into steps where one or the other argument enters the linear combination with no multiplier (or at worst some unit multiplier). Is there either some way to do it that I have not seen, or a proof that in some PIDs it cannot be done? Can one calculate Smith normal forms over PIDs by elementary matrix operations? The motivation for this question is to understand exactly what facts Emmy Noether faced when she gave her abstract algebra proof of the elementary divisor theorem for PIDs. Her proof was one striking confirmation that the Noetherian condition on rings is a powerfully useful idea. The discussion by მამუკა ჯიბლაძე and Luc Guyot reveals the interesting point that, while the answer to my question is simply no for the case of the integers in $\mathbb{Q}(\sqrt{-19})$, you actually can get Smith normal forms for that case by essentially the same kind of arithmetic, if you also allow yourself to also add new columns to the matrix as new `work space.' But algebraic $K$-theory shows the answer remains negative for other cases even when you allow that. As to the history, it seems Noether could well have seen elementary matrix calculations as hopeless for the problem, but she could not have proved they cannot work. She is not known to have ever done things like concrete calculations on $\sqrt{-19}$. REPLY [13 votes]: The answer is no: it is not possible, in general, to reduce a matrix over a principal ideal domain (PID) to a diagonal (or trigonal) matrix by means of elementary row and column operations. (This topic has been discussed in this MO post). The following example is a result proved by P. M. Cohn [1, Theorem 6.1 and subsequent discussion] in 1966 and doesn't rely on some $SK_1$ obstruction. Let $R = \mathbb{Z} [\frac{1 + \sqrt{-19}}{2}]$ and let $\theta \in R$ be a root of $X^2 - X + 5$. Then $R$ is a PID (see, e.g., Hardy and Wright) and the $1$-by-$2$ matrix $( 3 - \theta,\, 2 + \theta)$ cannot be reduced to a trigonal form $(*, \, 0)$. Edit. The following lines aim at comparing P. M. Cohn's example with those presented by მამუკა ჯიბლაძე in his answer. Here is a first remark which applies to both examples of PIDs. Because the Bass stable rank of $R$ is at most $2$ (use Bass Cancellation Theorem), it easily follows that any $1$-by-$n$ matrix over $R$ with $n \ge 3$ can be reduced to a matrix of the form $(*, \, 0, \dots,\,0)$ using elementary transformations. This certainly contrasts with the above example of $1$-by-$2$ matrix. This second remark only holds for P. M. Cohn's example. Let $E_n(R)$ denote the subgroup of $SL_n(R)$ generated the elementary matrices over $R$, i.e., those matrices which differ from the identity by a single off-diagonal entry. Then we observe that $SL_n(R)/E_n(R) \simeq SK_1(R)$ for every $n \ge 3$ by the classical stability theorems [2, Corollary 11.19] and $SK_1(R) = 1$ by the Bass-Milnor-Serre Theorem [2, Theorem 11.33]. For P. M. Cohn's matrix [1, Theorem 6.1], from which we borrowed the first row, this means that $\begin{pmatrix} 3 - \theta & 2 + \theta \\ - 3 - 2\theta & 5 - 2\theta \end{pmatrix} \in E_3(R) \setminus E_2(R)$. Eventually, in the case of the D. Grayson's PID, i.e., $R = S^{-1} \mathbb{Z}[T]$, with $S$ the multiplicative subset generated by $T$ and the polynomials $T^m - 1$, one can find $A \in SL_2(R)$ such that $A \notin E_n(R)$ for every $n \ge 2$ (I am unable to provide an explicit matrix at the moment). [1] "On the structure of the $GL_2$ of a ring", P. M. Cohn, 1966. [2] B. Magurn, "An algebraic introduction to K-theory", 2002.<|endoftext|> TITLE: Continuity of mapping sending a function to its (brouwer) fixed point QUESTION [8 upvotes]: Let $f:[0,1]^n \rightarrow [0,1]^n$ be a continuous mapping. Brouwer's fixed point theorem says that $f$ has a fixed point, i.e., some $x$ such that $f(x) = x$. Suppose we have a continuous family, i.e., a continuous function $f:[0,1]^n \times [0,1] \rightarrow [0,1]^n$. Then for each $r \in [0,1]$ we have that there is a point $x_r$ such that $f(x_r,r) = x_r$. It is known that, in general, there is no continuous mapping $r \mapsto x_r$. The following counter-example exists in the case $n=1$: take $f(x,r) = 2rx$ if $r \leq 1/2$ and $f(x,r) = (2r-1) + (2-2r)x$ - for $r < 1/2$ we have the only fixed point as 0, and $r > 1/2$ we have the only fixed point as $1$. My question is "what happens if we demand there are unique fixed points?" i.e., if every $f(\cdot,r)$ has a unique fixed point, is the mapping $r \mapsto x_r$ continuous? In the case $n=1$ my intuition (which is very possibly wrong!) seems to indicate that if each of the functions $f(\cdot,r)$ has a unique fixed point, then the mapping sending such a function to its fixed point is continuous. I'm sure this has been studied before by someone, however searching for it is a little difficult. Does anyone have any references, or know what happens for other $n$? I have little to no geometric insight as to what happens for $n \geq 2$. REPLY [4 votes]: Yes, it is indeed continuous. Consider the correspondence (set-valued map) $F:[0,1]\to 2^{[0,1]^n}$ given by $$F(x)=\{y\in[0,1]^n:f(y,x)=y\}.$$ Obviously, $F$ has closed values in a compact Hausdorff space and has a closed graph. Therefore, $F$ is upper hemicontinuous. Now if $F$ is single-valued, and therefore essentally a function, being upper hemicontinuous coincides with being continuous.<|endoftext|> TITLE: Is the circle in the square best at avoiding random lines? QUESTION [20 upvotes]: This question is inspired by a recent one (and takes a great deal from the answers there). Given a convex subset $\Delta$ of the unit square, let $p(\Delta)$ be the probability that a random line does not intersect $\Delta.$ The exact model for the selection of the lines is that a random point is chosen uniformly on the boundary and then a second is chosen randomly from the union of the other three sides. Consider the circle $\Delta_O$ of radius $\frac1{\sqrt{2\pi}}$ centered at $(\frac12,\frac12),$ I can show that $p(\Delta_0)=0.34470989\dots.$ Let $S$ be the family of convex subsets of the square with area $\frac12.$ What are the best upper and lower bounds you can give on $\sup_{S}p(\Delta)?$ In particular, is there a specific $\Delta \in S$ with $p(\Delta) \gt p(\Delta_O)?$ Here is a certain octagon $\Delta_8$ along with the circle $\Delta_0$ described above. I find that the octagon is the best of its kind but it is not quite as good as the circle. I get that $p(\Delta_8)=0.31984\dots.$ The lower left corner is at about $(0.32606,0.11670).$ Will Sawin in a comment to the question mentioned above observes that an optimal region will have no straight sides and also no corners. Any set $t\Delta_O+(1-t)\Delta_8$ with $0 \lt t \lt 1$ will be in $S.$ I don't know if these are worth investigating very deeply since some other octagon might be better for the purpose. Such a convex combination would still have $8$ corners but the segments would be curved. To deal with the corners on this, or some other $\Delta$ with corners, we could, for a very small positive $\epsilon,$ take all points within $\epsilon$ of a point of $\Delta.$ This would enlarge the area a bit so we would then shrink the smoothed figure to restore the correct area. This smoothing will fix corners but not straight sides (of length exceeding $\epsilon.$) It seems clear that an optimal region has the same $8$ lines of reflective symmetry as the square. I'm not claiming to have a proof of that, nor am I asking for one. In an answer to the linked question Will Sawin also makes an impressive start on a proof on finding the optimal set using the Calculus of Variations. I recommend reading it. The question here is more humble, just beat the circle. Perhaps there is a $16$-gon that succeeds or gets close. If so, it could perhaps be massaged to do even better. An upper bound is $p(\Delta) \lt \frac23.$ It is given by Christian Remling. He calls it crude, and I do think it is pretty far from optimal, but it is also the record that I know of so far and the proof sketched is very nice. REPLY [2 votes]: Indeed it seems that the circle is not optimal although the improvements I've found only raise the probability slightly and are visually hard to distinguish from a circle. Here is the circle in red and a region $\Delta_h$ bounded by $4$ hyperbolas. Recall that the circle has $p(\Delta_0)=0.34470989\dots.$ It turns out that $p(\Delta_h)=0.34494\dots.$ It was suggested that the optimum might be a shape bounded by $4$ hyperbolas near the corners and $4$ parabolas near the middles of the sides. The best I could do with that is $\Delta_{hp}$ shown here with $p(\Delta_{hp})=.34544686\dots.$ The parabola at the bottom is roughly $y=1.221614(x-\frac12)^2+.1045232.$ The transition from the parabola to the hyperbola to its left occurs between $(0.26957, 0.16938)$ and $(0.267287, .170719)$ (the respective endpoints of the polygonal approximations) with the middle of the hyperbola at $(0.213614,0.213614).$ Some disclaimers: I am not certain this is correct. I expected that the contact point of a tangent line rolling around would transition from hyperbola to parabola exactly when its intersection of the line with the square passes a corner and begins to intersect opposite sides. I believe this would maintain a constant area between the tangent and the square.However I don't see that happening here. Also, The later digits may not be exact, I tried to truncate before a small calculated digit. These are actually polygons obtained by sampling (at equally radially spaced points). The area may come out to be less than $\frac12$ in which case the figure is dilated appropriately. This will map parabolas to parabolas but deform hyperbolas. However replacing the deformed sections by a best hyperbola approximation changes the area only slightly. Iterating a few times stabilizes. I did not do this adjusting to $\Delta_h$ so it may be cheated out of a bit. I don't know if this is really the optimal shape. I could investigate small perturbations of the polygon but have not. I am pretty sure that something similar to a quarter circle in a corner OR a parabola along with most of the top OR the top, with the upper portions of the two sides connected by the arc of a conic could not do as well. I'm not saying it would be hard to rule those out. I just haven't rigorously done so.<|endoftext|> TITLE: What is homology anyway? QUESTION [85 upvotes]: Disclaimer: I don't feel qualified to ask this question and yet it's been troubling me for some time now and I lost my patience and decided to ask to get some kind of answer. If there are any stupid mistakes please treat them as such and try to focus on the main issue raised if at all possible. As the title suggests I'm struggling with the meaning of "Homology". In particular how are "Homology" and "Cohomology" related. By the end of my question I hope it will be clear what I mean. Let me start with some of the possible interpretations I'm (somewhat) familiar with, and after that let me say what troubles me. (All categories and functors are $\infty$ unless stated otherwise) Cohomology $\sim \operatorname{Hom}$ — Homology $\sim \otimes$ To make this precise consider the suspension $\infty$-functor sending spaces to their suspension spectra $\Sigma^{\infty}_+ :\mathrm{Spaces} \to \mathrm{Sp}$. The category of spectra is a symmteric monoidal $\infty$-category so for every space $X$ and spectrum $E$ one can define the $E$-homology of $X$ as the homotopy groups of the smash product $E_*X\mathrel{:=}\pi_*(\Sigma^{\infty}_+X \otimes_{\mathbb{S}} E)$. The $E$-cohomology of $X$ in this picture is the homotopy groups of the mapping spectrum $E^*X\mathrel{:=}\pi_*(\operatorname{Map}(\Sigma^{\infty}_+X,E))$. Homology $\sim$ Abelianization To make this precise one can consider the tangent category to $\mathrm{Spaces}$ which is the fiberwise stabilization of the codomain fibration $\mathrm{Spaces}$. The fiber over a space $X$ will be the category spectra parametrized by $X$. Then one can define the Homology of $X$ as the image of the identity map $X \to X$ under the stabilization procedure. This is the "absolute cotangent complex" $L_X$. One has a kind of shriek pushforward for these parametrized spectra which for the case $X \to \mathrm{pt}$ sends $L_X$ to $\Sigma^{\infty}_+X$ and one recovers some of the above from this viewpoint (I'm not so sure about this statement suddenly, is this true?). In a sense this is the relative setting for the above. Cohomology $\sim \mathrm{limits}$ - Homology $\sim \mathrm{colimits}$ To make this precise start with a local system over a space $X$. Let's take as a definition for a local system a functor from $X$ considered as an infinity groupoid to some category of coefficients (say spectra). Take this local system $L:X \to \mathrm{Sp}$ and define $L$-cohomology of X to be $\operatorname{Lim} L$ (this coincides with the sheaf cohomology definition) and $L$-homology to be $\operatorname{Colim} L$ (giving the same answer as 1 for the case of a constant functor $L=E$). Homology $\sim$ dual to Cohomology This is the most cheeky definition. There are many flavors of this I believe the basic archetype being the Poincaré duality for oriented manifolds $H^i_{\mathrm c}(M) \cong H_{n-i}(M)$. The main idea is to define homology in such a way that one gets "Poincaré duality". For example in Verdier duality for locally compact (sufficiently nice) spaces one can define homology with coefficients in a sheaf $F$ as the compactly supported cohomology with coefficients in the Verdier dual of $F$. For example on a manifold if $F= \mathbb{Z}$ is the constant sheaf then the Verdier dual will be $\operatorname{OR}_M$ the orientation sheaf (perhaps shifted depends on one's conventions). The point is that this definition is concocted so that one always has a duality between homology and cohomology. This can be done in any cohomology theory which has good duality properties (i.e. six functors). Why am I not satisfied? Here are my concerns. Some of the interpretations above answer some of the concerns but none of them answer all of the concerns in a satisfactory way: Lack of convenient relative framework: For sheaf cohomology one has a very convenient framework for working in a relative situation (push/pull) in any context no matter how general. All one needs is a site and one immediately can ask questions about how cohomology behaves in this site, what kind of properties does it satisfy? Does it have 6 functor formalism? If not maybe at least 5 or 4? Does it have any interesting dualities? etc.… For Homology one seems to run into several persistent problems when trying to translate the above interpretations into a relative general setting like this. Using duality as a crutch: As much as I like dualities sometimes I feel like we're being a bit unfair to "Homology" treating it like a deformed creature which only has a right to exist as a dual to cohmology when in fact homology is the older brother of the two! Asymmetry between co/homology: In cohomology one has sheaves, sections, resolutions etc.… What do we have in homology? I'm kind of wishing that all the homology business is part of a bigger story Cosheaf Homology — Sheaf Cohomology. Unfortunately I have no idea what the words in the left hand side mean or even what they should mean. I just wish there was some way to put homology and cohomology on an equal footing. Only locally constant data: This is related to the above point. Why is there no "Constructible Homology" or "Coherent Homology"? Why doesn't Homology deserve these variants? I hope by now I've made it clear what's my "problem" with my current understanding of Homology. As I said I don't feel like I'm qualified to ask this question so if anyone has any suggestion for an edit or a revision please don't even ask permission just edit away! REPLY [20 votes]: For a long time (and still today), I very much shared the confusion of the OP. I think Jacob Lurie gives a very clear take on the standard perspective, but Mike Shulman does have a very valid contrasting point. The issue is that in the types of $6$-functor formalisms that are usually studied say in étale cohomology, there just is no left adjoint to pullback, and cohomology is not the dual of anything. For example, if $S$ is a profinite set, then $H^0(S,\mathbb F_\ell)$ are the locally constant functions $S$ with values in $\mathbb F_\ell$, which is often a countably dimensional $\mathbb F_\ell$-vector space, and hence cannot be the dual of anything. I've highlighted this in some other MathOverflow questions before, but I wanted to do it here again: In Chapter VII of Geometrization of the local Langlands correspondence, we define for any "small v-stack" $X$ a closed symmetric monoidal category $D_{\blacksquare}(X,\mathbb Z_\ell)$ in such way that for any map $f: Y\to X$, the pullback $f^\ast$ admits a left adjoint $f_\natural$ in addition to the right adjoint $Rf_\ast$. (We do not use the notation $f_!$, as the latter means compactly supported cohomology, which is different from homology.) Moreover, completely general base change and projection formulas hold true. So this is a version of a $6$-functor formalism, allowing constructible coefficients, that puts homology and cohomology on equal footing, and arguably homology is now more fundamental again: For example, it satisfies a projection formula (which cohomology doesn't in general), and cohomology is always the dual of homology, but not the other way around. What is a small v-stack? One source comes from analytic adic spaces, so for example from rigid spaces over $p$-adic fields, and this is the perspective taken in our work. But actually any condensed set, or in fact condensed groupoid (and we might as well allow a condensed anima), defines a small v-stack, over a fixed geometric point $\mathrm{Spa} C$. So the above applies in particular to say compactly generated weak Hausdorff spaces $X$ (regarded as a condensed set, regarded as a diamond over $\mathrm{Spa} C$, which is a special kind of small v-stack). This formalism notably allows profinite sets $S$ as above. The problem gets resolved by allowing homology to be a topological group (rather, condensed group); in that case, the measures on $S$ with $\mathbb F_\ell$-coefficients, which is naturally a profinite $\mathbb F_\ell$-vector space, whose (continuous) dual are the locally constant functions $H^0(S,\mathbb F_\ell)$. I'm in a state of perpetual confusion over how expressive this formalism is; many things usually expressed in terms of a $6$-functor formalism do not have an obvious translation, but usually have some translation into this picture. For example, if $f: Y\to X$ is proper and smooth, then the dualizing complex is given by the inverse of $$R\pi_{1\ast} \Delta_{f\natural} \mathbb Z_\ell,$$ where $\Delta_f: Y\to Y\times_X Y$ is the diagonal and $\pi_1: Y\times_X Y\to Y$ the projection. Concretely, the fibre of this at a closed geometric point $y\in Y$ is given by the homology of $Y$ with coefficients in $i_{y\natural} \mathbb Z_\ell$, where $i_y: \{y\}\to Y$ is the closed immersion. Here $$i_{y\natural} \mathbb Z_\ell = \varprojlim_{U\ni y} j_{U!} \mathbb Z_\ell$$ where $U\to Y$ runs over étale neighborhoods of $y$, so the homology of $Y$ with coefficients in $i_{y\natural} \mathbb Z_\ell$ is the limit of the compactly supported cohomologies $R\Gamma_c(U,\mathbb Z_\ell)$. This should be thought of as the compactly supported cohomology of a small ball around $y$, which should indeed be dual to the dualizing complex at $y$. (While I tend to think about analytic adic spaces for $X$ and $Y$ here, this actually also works if $X$ and $Y$ are manifolds. In that case, one can even replace $\mathbb Z_\ell$ by $\mathbb Z$.) Also, one can give a very similar treatment replacing analytic adic spaces with schemes. Some things are a bit different though, for example the base change results are not quite as general. (I find it very curious that analytic adic spaces are better than schemes here, and have no good intuitive explanation.)<|endoftext|> TITLE: Tensor product space with projective norm is incomplete QUESTION [6 upvotes]: Ryan says in his book "Introduction to Tensor Products of Banach Spaces"(pg. 17) that for Banach spaces $X$ and $Y$, $X\otimes Y$ equipped with projective norm is not complete unless $X$ and $Y$ are finite dimensional. First I want the example of this. Second, is there any sources about the proof of this statement? Thanks. REPLY [4 votes]: It wasn't clear to me, immediately, how to do the general case, but I think you can use Biorthogonal Systems. Indeed, let $(x_n)\subseteq X$ and $(f_n)\subseteq X^*$ satisfy that $f_n(x_m) = \delta_{n,m}$. Let $(y_n) \subseteq Y$ be a linearly independent sequence. Set $$\tau = \sum_{n=1}^\infty \|x_n\|^{-1} \|y_n\|^{-1} 2^{-n} x_n\otimes y_n \in X \widehat\otimes Y. $$ Suppose, towards a contradiction, that $\tau \in X\otimes Y$. Now, $\tau$ induces a bounded linear map $T:X^*\rightarrow Y$, and by our assumption, $T$ is finite rank. However, clearly $T(f_n) = \|x_n\|^{-1} \|y_n\|^{-1} 2^{-n} y_n$ and so $(T(f_n))$ is a linearly independent sequence, contradicting $T$ being finite rank. To construct $(x_n), (f_n)$ we can follow an old argument due to Markushevich. The argument is not hard, but is a touch long to type out here. The best link I could find was: Biorthogonal Systems in Banach Spaces (Google books). Search for "M-basis".<|endoftext|> TITLE: A property of real numbers concerning integer parts of multiples QUESTION [11 upvotes]: For a given positive real number $\alpha$, define the set $T_\alpha$ by $T_\alpha = \{ [n\alpha] \mid n = 1,2,\dots \}$. What is a necessary and sufficient condition (in terms of $\alpha$ and $\beta$) to have $T_\alpha \subseteq T_\beta$ ? REPLY [6 votes]: A partial answer. It is clear that we should consider only $\alpha,\beta>1$. Suppose we want to find $n$ such that $n\in T_\beta$, $n\notin T_\alpha$. It is sufficient to find integers $l$ and $k$ such that $$n\le l\beta TITLE: Collections of examples and counterexamples in (real, complex, functional) analysis, ODEs and PDEs QUESTION [6 upvotes]: What books collect examples and counterexamples (or also "solved exercises", for some suitable definition of "exercise") in real analysis, complex analysis, functional analysis, ODEs, PDEs? The only book of this kind that I know of is the famous Counterexamples in Analysis by Bernard R. Gelbaum and John M. H. Olmsted. REPLY [2 votes]: I coincidentally read this question, and despite it is not currently active, I believe it is right to share with the forum my personal answer. I think that reference 1 is very nice book for what pertains to the last two points of your question. It is not a list of simple exercises like other books on the same topics, but a collection of many carefully chosen problems (many of them at the research level) and examples illustrating many more aspects of ODE (in the first part of the book) and PDE theory (in the second part), with answer or hints for solution. For example you can find the Garabedian-Grushin and the Lewy examples, examples in semigroup theory and in the theory of functions of several complex variables, examples pertaining the Gilbarg-Serrin theorem (you may notice that I am more accustomed to the second part of the book). In order to get a more precise idea of the aim of the book see the main portion of the Author's preface below and also its Zbl review. 1 Biler, P., Nadzieja, T. Problems and Examples in Differential Equations, Pure and Applied Mathematics 164, New York:Marcel Dekker, Inc., ISBN: 0-8247-8637-8, viii+244 (1992), MR1198886, Zbl 0760.34001.<|endoftext|> TITLE: Relation between affine flag and Grassmannian Steinberg variety QUESTION [10 upvotes]: Let $\mathcal{K}=\mathbb{C}((t))$ be the field of formal Laurent series over $\mathbb{C}$, and by $\mathcal{O}=\mathbb{C}[[t]]$ the ring of formal power series over $\mathbb{C}$. Given a semi-simple Lie group $G$, affine Grassmannian $Gr_G$ is defined by the coset space $G(\mathcal K)/G(\mathcal O)$ wheareas the affine flag variety is defined by $Fl_G=G(\mathcal K)/I$ where $I$ is the Iwahori subgroup which is the preimage of a Borel subgroup $B$ under the map $G(\mathcal O)\to G$. Bezrukavnikov-Finkelberg-Mirković showed that the $G(\mathcal{O})$-equivariant $K$-theory of affine Grassmannian $Gr_G$ is the coordinate ring of the following phase space $$ \textrm{Spec}\; K^{G(\mathcal{O})}(Gr_G)=(T\times T^\vee)/W $$ where $T$ is the maximal torus and W is the Weyl group. My question is as follows: if you replace affine Grassmannian $Gr_G$ by affine flag variety $Fl_G$, then is the corresponding space $$ \textrm{Spec}\; K^{G(\mathcal{O})}(Fl_G) $$ just the direct product with the contangent bundle $T^*(G/B)$ of the flag variety (or nil-cone $\mathcal N$) $$ ((T\times T^\vee)/W)\times (T^*(G/B))~~~? $$ Or is $\textrm{Spec}\; K^{G(\mathcal{O})}(Fl_G)$ a non-trivial $T^*(G/B)$-bundle (or some another bundle) over $(T\times T^\vee)/W$? REPLY [3 votes]: To clarify the issue you're having: you can look at the spaces $$\mathcal{\tilde{R}}=\{(x,g)\in \mathfrak{g}(\mathcal{O})\times G((t)) \mid \mathrm{Ad}_{g^{-1}}(x)\in \mathfrak{g}(\mathcal{O})\}$$ $$\mathcal{\tilde{Z}}=\{(x,g)\in \mathrm{Lie}(I)\times G((t)) \mid \mathrm{Ad}_{g^{-1}}(x)\in \mathrm{Lie}(I)\}$$ For former has left and right $G(\mathcal O)$ actions via $$g'\cdot (x,g) \cdot g''=(\mathrm{Ad}_{g'}(x), g'gg'')$$ and the same formulae define left and right actions of $I$ on the latter. To get an algebra, you should mod out by the same group on both sides, so you get $$K\big(G(\mathcal{O})\backslash \mathcal{\tilde{R}}/G(\mathcal{O})\big) \cong \mathbb{C}(T\times T^{\vee})^W$$ If instead, you consider $K\big(I\backslash \mathcal{\tilde{R}}/I\big)$, then you get the above algebra tensored with matrices on the vector space $H^*(G/B)$, and this is always hold true; the bimodules $K\big(G(\mathcal{O})\backslash \mathcal{\tilde{R}}/I\big)$ and $K\big(I\backslash \mathcal{\tilde{R}}/G(\mathcal{O})\big)$ induce the Morita equivalence. So, in order to get something more interesting, you need to change the underlying space to $\mathcal{\tilde{Z}}$.<|endoftext|> TITLE: Compact quaternionic Kahler manifolds of negative curvature: examples QUESTION [12 upvotes]: There is a well known problem of LeBrun-Salamon: are there any non-symmetric compact quaternionic-Kahler manifolds of positive scalar (and Ricci) curvature? It is hard and still unsolved: Quaternionic-Kahler metrics whose universal covers have only discrete isometry groups? The symmetric compact quaternionic-Kahler manifolds ("Wolf spaces") are understood and classified. However, for each Wolf space there is the dual symmetric space, say, $G/H$, which is quaternionic-Kahler of negative scalar (and Ricci) curvature. For any lattice $\Gamma\subset G$, the double quotient $\Gamma\backslash G/H$ is a locally symmetric quaternionionic-Kahler orbifold of negative curvature and finite volume. However, it can have cusp points, and then it is non-compact. Are there any compact locally symmetric quaternionionic-Kahler orbifolds? Manifolds? I could not find a reference to either existence or non-existence results. REPLY [8 votes]: Any (Riemannian) symmetric space admits a cocompact lattice. This is due to A. Borel, Compact Clifford-Klein forms of symmetric spaces, Topology 2, 1963, pp.111-122. The quaternionic hyperbolic space is symmetric and quaternionic-Kahler.<|endoftext|> TITLE: Polynomials leaving invariant the Gaussian integers QUESTION [16 upvotes]: It can easily be shown that if a complex polynomial $P$ leaves invariant $\mathbb{Z}$ ($P(\mathbb{Z}) \subseteq \mathbb{Z}$) then it must be a linear combination (with integer coefficients) of Hilbert polynomials $H_k$, i.e. polynomials of the form : $$ H_k(X) : = \frac{X(X-1)\cdots (X-k+1)}{k!} $$ Now, what happens when $P$ stabilizes an entire lattice, say the Gaussian integers $\mathbb{Z}[i] $, i.e $P(\mathbb{Z}[i] ) \subseteq \mathbb{Z}[i] $ ? Pretty clearly, the set $\mathcal{A}$ of all such polynomials is an additive sub-group of $\mathbb{Q}[i][X]$ (and even a $\mathbb{Z}[i]$ module). Such polynomials can be expressed as linear combinations (with coefficients in $\mathbb{Z}[i]$) of the polynomials $H_k$ (same proof as in the case of $\mathbb{Z}$). It is also straightforward to see that $\mathbb{Z}[i] [X]$ is contained in $\mathcal{A}$. However equality does not arise, since the polynomial $ \widetilde{H}_2(X): = (1+i)\frac{X(X-1)}{2} = (1+i)H_2(X)$ does not lie in $\mathbb{Z}[i] [X]$ and yet verifies the property of leaving invariant the Gaussian integers. Question : Can we give a precise description of such polynomials, say an explicit basis of $\mathcal{A}$ (seen as a module) ? REPLY [13 votes]: Your question is related to the study of (generalized) numerical polynomials: If $R$ is an integral domain and $K$ the field of fractions of $R$, then the set ${\rm Int}(R) := \{f \in K[x]: f(R) \subseteq R\}$ is a subdomain of $K[x]$, whose elements are called the numerical polynomials over $R$ (in one variable $x$). The domain ${\rm Int}(R)$ has been the subject of a great deal of research: Original work in the area was entirely centered on the case where $R$ is the ring of integers, and was motivated by interpolation problems in the early days of calculus. It was only in 1919 that A. Ostrowski and G. Pólya first considered numerical polynomials in their own right, though focused on the case where $R$ is the ring of integers of a number field. In particular, they could show, in this context, that ${\rm Int}(R)$ has a regular basis $(f_k)_{k \ge 0}$ as an $R$-module if and only if the products of prime ideals of $R$ of every given norm are principal, which is certainly true if $R$ is a PID ("regular" means that $\deg f_k = k$ for all $k$): Their proof is constructive, so the answer to your question ("Can we give a precise description of such polynomials etc.?") is yes. For further details and results, you may want to have a look to P.-J. Cahen and J.-L. Chabert's monograph, Integer-valued polynomials, Math. Surveys Monogr. 48, Amer. Math. Soc., 1997. More specifically, see Remark II.1.5(ii) for an "explicit basis".<|endoftext|> TITLE: Is this closed subspace of Fréchet space complemented QUESTION [6 upvotes]: In the hope of completing the rich tapestry of complemented (or not) topological vector subspaces, I would like to know (maybe it is immediate for specialists) whether the space of analytic functions is complemented within the space of infinitely differentiable ones. I begin with the one-variable case ... and make this precise. Let $\Omega\subset \mathbb{C}$ be an open subset. We consider $$ H(\Omega)=C^\omega(\Omega;\mathbb{C})\subset C^\infty(\Omega;\mathbb{C}) $$ the large one being endowed with the standard topology defined by the seminorms $$ p_{\,n,B}=sup_{\ 0\leq |\alpha|\leq n\atop t\in B}|D^\alpha(f)[t]|\ . $$ where $n\in \mathbb{N}, \alpha\in \mathbb{N}^2$, $B$ is a relatively compact open subset of $\Omega$ and the bi-indexed derivative is $$ D^\alpha:=(\frac{\partial}{\partial x})^{\alpha[1]}(\frac{\partial}{\partial y})^{\alpha[2]}\ . $$ I know that the subspace $H(\Omega)=C^\omega(\Omega;\mathbb{C})$ is complete and then closed for this (standard) topology. My question is the following Q) Is there a known closed complement of it i.e. a decomposition $$ C^\infty(\Omega;\mathbb{C})=C^\omega(\Omega;\mathbb{C})\oplus W=H(\Omega)\oplus W $$ where $W$ is closed ? (maybe the projector is an integro-differential operator ?) at least for some particular domains $\Omega$ ? Remark i) This question is a reformulation of this one in MSE where it did not receive a complete answer. ii) With the given topology, $C^\infty(\Omega;\mathbb{C})$ and $H(\Omega)=C^\omega(\Omega;\mathbb{C})$ are m-convex Fréchet algebras. Maybe (if possible) $W$ could have some algebraic structure (ideal ?). REPLY [6 votes]: $H(\Omega) $ is not complemented in $C^\infty (\Omega)$ e.g. for the unit disc in $\mathbb C $. This follows from the structure theory of Frechet spaces: The space of smooth functions is isomorphic to $s^\mathbb N$ and has a certain property (DN$_{loc}$) of Vogt. If $H (\Omega) $ were complemented it would also have this property and hence even property (DN) because it has continuous norms. But this is not true for power series spaces of finite type. I don't have any literature at hand. The book of Meise and Vogt is a good starting point. The local condition is in a more recent article of Vogt.<|endoftext|> TITLE: Has anybody studied strict/pseudo morphisms of monads? QUESTION [6 upvotes]: There is a notion of morphism from a monad $T:\mathscr C\to \mathscr C$ to another one $T':\mathscr C'\to \mathscr C'$. It arose here on MO e. g. in "Functors between monads": what are these really called? or Distributive law between Kleisli triples. Such morphisms have been considered by Pumplün, Street and probably others, they are functors $F:\mathscr C\to\mathscr C'$ together with a transformation $T'F\to FT$ with some coherence conditions generalizing those for a distributive law. The idea is to have an induced functor between Eilenberg-Moore algebras. While this is certainly "the" correct notion, I've recently encountered a situation when it is not. With several colleagues we are studying varieties like meet-semilattices-with-a-nucleus; "correct" morphisms $(M,j)\to(M',j')$ must be $f:M\to M'$ which preserve meets and satisfy $j'f=fj$, not just $j'f\leqslant fj$ as it would be if one would view $j$ and $j'$ as monads and take monad morphisms in the above sense. The natural "categorified" version of such morphisms would then be natural isomorphisms $T'F\cong FT$; that is, a $\textit{pseudo}$ version, while Pumplün-Street notion would then be the $\textit{lax}$ version. Question: can one distinguish such "strict"/"pseudo" morphisms of monads among more general ones by some abstract-nonsensical properties? Could be something like having functors both between Eilenberg-Moore and Kleisli categories, compatible in certain way. Somehow the precise formulation escapes me. Is anybody aware of some work on these? REPLY [2 votes]: I don't know of very much work about these specifically, or a characterization of them in terms of how they act on Eilenberg-Moore or Kleisli categories. But there is a precise sense in which they are the "pseudo" (or "strong") to Street's "lax" morphisms. Namely, there is a 2-monad $M$ on $Cat$ whose algebras are categories equipped with a monad, and these are the pseudo and lax $M$-morphisms respectively. By the way, I've found uses for this kind of monad morphism too; see for instance Definition 11.5 and Lemma 11.10 of this paper.<|endoftext|> TITLE: Classification of finite type structures leads to Dynkin diagrams? QUESTION [14 upvotes]: Classification of finite type structures in mathematics often lead to the Dynkin diagrams (Example: representation-finite hereditary algebras, simple Lie algebras, Cluster algebras,... and I have read that there are nearly 50 other such structures connected with the Dynkin diagrams). Questions: Are there classification results of such "finite type" structures where the answer surprisingly does not correspond to exactly the Dynkin diagrams, but to Dynkin diagrams with maybe a finite number of other diagrams and/or some simply laced diagrams missing (like for example $E_6$)? REPLY [7 votes]: I would mention the classification of finite-dimensional Nichols algebras $\mathfrak{B}(V_1\oplus\cdots\oplus V_\theta)$, $\theta\geq2$, over decomposable Yetter-Drinfeld modules $V_1\oplus\cdots\oplus V_{\theta}$ over groups. In the case $\theta\geq4$ the classification is essentially based on Dynking diagrams of finite type (with labels). In the cases $\theta\in\{2,3\}$ there are exceptions, most of them appearing in the case $\theta=2$. References: https://arxiv.org/abs/1412.0857 and https://arxiv.org/abs/1311.2881.<|endoftext|> TITLE: homogeneous surface in $\mathbb{R}^4$ QUESTION [5 upvotes]: It is well known that the only homogeneous surfaces in $\mathbb{R}^3$ are the spheres, cylinders or planes. My question is about other examples in dimension $4$. Such a surface should have "constant curvature" but it seems that there is no good scalar invariant such as mean curvature or Gauss curvature...except perhaps the square norm of the second fundamental form. In another codimension 2 situation, namely curves in $\mathbb{R}^3$, there are with lines and circles,also the helix. Is there a homogeneous surface in $\mathbb{R}^4$ which is not $S^2$, $S^1\times \mathbb{R}$ or $\mathbb{R}\times \mathbb{R}$? Edit: and $S^1\times S^1$ but connected REPLY [8 votes]: I'm rearranging my answer a little bit because I realized that I overlooked an apparent possibility (that turns out not to occur), and I didn't want my answer to be misleading: If the surface in Euclidean $\mathbb{R}^4$ has positive Gauss curvature and is homogeneous, it will be complete and hence compact. Hence the group of ambient symmetries will have to preseve its center of mass, which we can take to be $0\in\mathbb{R}^4$. Thus, the group of symmetries will lie in $\mathrm{O}(4)$. However, the identity component of the symmetries of a compact surface of positive curvature must be a quotient of $\mathrm{SU}(2)$, so we are looking at a subgroup of $\mathrm{O}(4)$ whose identity component must be either $\mathrm{SU}(2)$ or $\mathrm{SO}(3)$. The former can only be represented one way, up to conjugacy, in $\mathrm{O}(4)$, and that action has no $2$-dimensional orbits (they are either the origin or $3$-spheres). Thus, it must be $\mathrm{SO}(3)\subset\mathrm{O}(4)$, uniquely up to conjugacy. The two dimensional orbits of $\mathrm{SO}(3)$ are round $2$-spheres lying in $3$-planes, so this is all of the possible homogeneous surfaces with positive curvature. (Note that in $\mathbb{R}^5$, though, there is a homogeneous $\mathbb{RP}^2$ that sits linearly fully in $\mathbb{R}^5$.) There can't be any homogeneous surface of negative Gauss curvature whose ambient symmetries act as the full isometries of the surface, since the identity component of the symmetry group of its simply-connected cover would be $\mathrm{PSL}(2,\mathbb{R})=\mathrm{SO}(2,1)^\circ$, which has no nontrivial homomorphism into the group of isometries of $\mathbb{R}^4$ (for Lie algebra reasons). Thus, the action of the identity component of the ambient symmetries of the surface would have to be a 2-dimensional subgroup of $\mathrm{SO}(2,1)^\circ$, and these are all conjugate to each other. In particular, the ambient symmetries would preserve a foliation by geodesics and, consequently, an orthonormal coframing on the surface. Now one can set up the structure equations for such an ambient coframing and easily derive that this case cannot occur. Thus, homogeneous negatively curved surfaces do not exist in $\mathbb{R}^4$. Finally, there remain only the homogeneous surfaces with zero Gauss curvature in $\mathbb{R}^4$ to classify, and a calculation with the structure equations shows that, up to isometry, there are only two types: First, there is the product of a homogeneous curve (i.e., line, circle or circular helix) in $\mathbb{R}^3\subset\mathbb{R}^4$ with an orthogonal line; second, there are the products of two circles in orthogonal 2-planes in $\mathbb{R}^4$. In higher dimensions, of course, the number of types of homogeneous surfaces increases.<|endoftext|> TITLE: Relation between Legendre and Chebyshev polynomials QUESTION [5 upvotes]: Where I could find relationships between Legendre and Chebyshev polynomials? For example I found with maple $$ P_n(\cos\theta)=\sum_{k=0}^n(-1)^{n+k}\frac{2-\delta_{k0}}{4^n} \binom{n-k}{\frac{n-k}{2}}\binom{n+k}{\frac{n+k}{2}}\cos(k\theta)$$ The sum runs over $n+k$ even, and $\delta_{k0}=1$ if and only if $k=0$. (And $\cos(k\theta)$ are the Chebyshev polynomials) But would like to know how its proved, and what the inverse relationship is. Are there any papers or books with these types of relationships? REPLY [2 votes]: Both the Legendre and Chebyshev polynomials are particular cases of Jacobi polynomials $P_n^{(\alpha,\beta)}(x)$. A general connection formula of the type $$P_n^{(\gamma,\delta)}(x)=\sum_{k=0}^nc_{n,k}^{\gamma,\delta;\alpha,\beta}P_k^{(\alpha,\beta)}(x)$$ can be found on page 256 of the book [Mourad E.H. Ismail, Classical and quantum orthogonal polynomials in one variable, Encyclopedia of Mathematics and its Applications 98, Cambridge University Press, Cambridge, 2005].<|endoftext|> TITLE: "Unimodality" of the positive eigenvector of a non-negative irreducible matrix? QUESTION [6 upvotes]: Consider an eigenvalue / eigenvector problem for a matrix $A$ that is known to be non-negative and irreducible (so the Perron-Frobenius theorem applies): $$\sum_j A_{ij} x_j = \lambda x_i$$ Here $\lambda$ is the largest eigenvalue and $x_i$ is an eigenvector with all entries positive. I say that the eigenvector $x_i$ is unimodal if there exists an $i^*$ such that $x_{i+1} \ge x_i$ whenever $i < i^*$ and $x_{i+1} \le x_i$ whenever $i \ge i^*$. What conditions must be placed on the matrix $A$ such that $x_i$ unimodal in $i$? I don't know if this kind of problem has been studied before. Perhaps it has been addressed for some specific matrices. If there is some terminology that could help me find relevant papers, or some papers that treat something similar, please let me know. REPLY [2 votes]: Let $A$ be as in the question and let $n$ denote the size of $A$, i.e. $A \in \mathbb{R}^{n \times n}$. Troughout, fix $i^* \in \{1,\dots,n\}$. We call a vector $x \in \mathbb{R}^n$... ... $i^*$-unimodal if $x_{1} \le x_2 \le \dots \le x_{i^*}$ and $x_{i^*} \ge x_{i^*+1} \ge \dots \ge x_n$. ... strictly $i^*$-unimodal if $x$ is $i^*$-unimodal and if, in addition, all the above inequalities are strict. Theorem. Let $T \in \mathbb{R}^{n \times n}$ be the ($i^*$-dependent) matrix given by formula $(*)$ below. If all entries of $T^{-1}AT$ are nonnegative, then $A$ has an eigenvector $x$ for the eigenvalue $\lambda$ such that $x$ is $i^*$-unimodal and such that each entry of $x$ is strictly positive. This result is actually the content of Theorem 5 below. For the proof, we need a bit of preparation. Let $\mathbb{R}^n_+ := \{x \in \mathbb{R}^n: \; x_k \ge 0 \text{ for all } k \}$ be the standard cone in $\mathbb{R}^n$ and let $K := \{x \in \mathbb{R}^n: \; x \text{ is } i^*\text{-unimodal and } x_1 \ge 0\}$. Then $K$ is a closed convex cone in $\mathbb{R}^n$, we have $K \cap (-K) = \{0\}$ and $K$ has non-empty interior (the interior points of $K$ being exactly the vectors which are strictly $i^*$-unimodal and whose first component is strictly positive); hence, the cone $K$ is generating in $\mathbb{R}^n$, i.e. $K-K = \mathbb{R}^n$. Lemma 1. Let $x$ be an eigenvector of $A$ for the eigenvalue $\lambda$. If $x \in K$, then all entries of $x$ are strictly positive. Proof. By the Perron-Frobenius theorem the irreducibility of $A$ implies that the eigenspace $\ker(\lambda - A)$ is spanned by a vector $y$ whose entries are all strictly positive. Thus, $x = \alpha y$ for some non-zero real number $\alpha$. We have $0 \le x_1 = \alpha y_1$, so $\alpha > 0$ since $y_1 > 0$. This proves the lemma. We point out that the lemma implies in particular that $x_n > 0$, although we did not even assume $x_n$ to be nonnegtive in the definition of $K$. Now we apply the Krein-Rutman Theorem to the ordered Banach space $(\mathbb{R}^n,K)$. This immediately yields the following result: Theorem 2. If $AK \subseteq K$, then the eigenspace $\ker(\lambda - A)$ contains a vector $x \in K$ (whose entries are all strictly positive according to Lemma 1). So our next task is to find sufficient conditions for the property $AK \subseteq K$. Fortunately, the cone $K$ is quite well-behaved - in fact, the ordered space $(\mathbb{R}^n,K)$ is isomorphic to $(\mathbb{R}^n,\mathbb{R}^n_+)$ as the following lemma shows: Lemma 3. Let $T: \mathbb{R}^n \to \mathbb{R}^n$ be given by \begin{align*} Tx = \begin{pmatrix} x_1 \\ x_1 + x_2 \\ \vdots \\ x_1 + x_2 + \dots + x_{i^*} \\ x_1 + x_2 + \dots + x_{i^*} - x_{i^*+1} \\ \vdots \\ x_1 + x_2 + \dots + x_{i^*} - x_{i^* + 1} - \dots - x_{n} \end{pmatrix}. \qquad (*) \end{align*} Then $x \in \mathbb{R}^n_+$ if and only if $Tx \in K$. Proof. Straightforward. Corollary 4. We have $AK \subseteq K$ if and only if $T^{-1}AT (\mathbb{R}^n_+) \subseteq \mathbb{R}^n_+$ if and only if every entry of the matrix $T^{-1}AT$ is nonnegative. Remark. Note that the matrix $T^{-1}$ can be explicitely computed (and it actually has a more "sparse" structure than $T$ itself). By combining Theorem 2 and Corollary 4 we arrive at the following result which gives a sufficient condition for the desired property (and which has already been stated at the beginning of this answer): Theorem 5. Assume that all enntries of $T^{-1}AT$ are nonnegative. Then $A$ has an eigenvector $x$ for the eigenvalue $\lambda$ such that $x$ is $i^*$-unimodal and such that all entries of $x$ are strictly positive. Outlook. It is important to note that the condition in Theorem 5 is not an equivalence; this is, in some sense, due to the fact that the spectral properties asserted by the Perron-Frobenius Theorem (or by the Krein-Rutman Theorem) do not, conversely, imply nonnegativity of the matrix. However, it might be possible to give a characterization result which is rather close to the spirit of Theorem 5, if one imposes a stronger condition on the eigenvector (and also on the dual eigenvector) and if one employs the following result from the theory of eventually positive matrices which can be found in [D. Noutsos: On Perron–Frobenius property of matrices having some negative entries (2006), Theorem 2.2]: Theorem 6. Let $B \in \mathbb{R}^{n\times n}$. Then the following assertions are equivalent: (i) The spectral radius $r := r(B)$ of $B$ is a strictly dominant and algebraically simple eigenvalue of $B$, and each of the eigenspaces $\ker(r - B)$ and $\ker(r - B')$ is spanned by a vector whose entries are all strictly positive. (ii) There exists a number $k_0 \ge 1$ such that every entry of $B^k$ is strictly positive for every $k \ge k_0$. Here, $B'$ denotes the transposed matrix of $B$, and a strictly dominant eigenvalue of a matrix is an eigenvalue is modulus is strictly larger than the modulus of any other eigenvalue of the given matrix. Remarks 7. (a) At first glance the definition of the notion strong Perron-Frobenius property in [op. cit.] (which is used to state [op. cit., Theorem 2.2]) does not appear to include the algebraic simplicity of the spectral radius. However, one can see from the proofs in [op. cit.] that it's the author's understanding that the notion strong Perron-Frobenius property includes the property that the spectral radius is an algebraically simple eigenvalue. This can also be seen from other papers of the same author, see for instance [D. Noutsos, M. J. Tsatsomeros: Reachability and holdability of nonnegative states (2008), Definition 2.3]. (b) Given the other properties in Theorem 6(ii), algebraic simplicity of the eigenvalue $r$ of $B$ is actually equivalent to geometric simplicity. This is for instance proved in [D. Daners, J. Glueck, J. Kennedy: Eventually positive semigroups of linear operators (2016), Proposition 3.1]. Remark 8. If one whishes to employ Theorem 6 in order to turn Theorem 5 into a characterisation result one probably has to take the following phenomena into account: (a) One needs to use a strictly unimodal eigenvector instead of a merely unimodal eigenvectors, now. (b) For the transposed matrix $A'$ one probably needs some kind of dual notion to unimodality since the eigenvector of $A'$ will be of the form $(T^{-1})'y$ for some vector $y$ whose entries are all strictly positive. I'm not sure how (and if) vectors of the form $(T^{-1})'y$ for strictly positive $y$ can be easily described. (c) One needs the assumption that the spectral radius of $A$ is a strictly dominant eigenvalue. Note on an EDIT made 2018-09-01: I've removed the former Theorem 6 from the former version of the post since it contained several mistakes (it did not take Remark 8(b) and (c) into account). Instead I've added some information from the literature in Theorem 6 and Remark 7 which might be useful when trying to characterise the existence of strictly unimodal eigenvectors. However, I haven't though in detail about how such a characterisation result might look precisely.<|endoftext|> TITLE: About small $\omega$-orthogonality classes and Gabriel-Ulmer duality QUESTION [5 upvotes]: I am reading the paper http://www.numdam.org/article/CTGDC_2001__42_1_51_0.pdf fixing the implication $(ii)\Rightarrow (i)$ of Theorem 1.39 of Adamek-Rosicky's book. The correct statement is: if $\mathcal{K}$ is a reflective subcategory of a LFP category $\mathcal{L}$ closed under filtered colimits and such that the theory of the embedding $\mathcal{K}\subset \mathcal{L}$ is a quotient, then $\mathcal{K}$ is a small $\omega$-orthogonality class of $\mathcal{L}$. I can "prove" that the theory of the embedding $\mathcal{K}\subset\mathcal{L}$ is always a quotient, which is obviously wrong. Where am I wrong in what follows ? Let $r:\mathcal{L} \to \mathcal{K}$ be the left adjoint of the inclusion functor $i:\mathcal{K}\subset \mathcal{L}$. I am using the terminology of https://ncatlab.org/nlab/show/Gabriel-Ulmer+duality: $\text{Lex}$ is the category of finitely complete (essentially) small categories together with the finite limit-preserving functors. The theory of the embedding $\mathcal{K}\subset \mathcal{L}$ is the $\text{Lex}$-functor $r^{op}:\mathcal{L}_{fp}^{op} \to \mathcal{K}_{fp}^{op}$. To prove that it is a quotient, we have to prove two things: 1) that every object $B$ of $\mathcal{K}_{fp}^{op}$ is isomorphic to an object $r^{op}A$: indeed, $r^{op}i^{op}B\cong B$ since $\mathcal{K}$ is a full subcategory of $\mathcal{L}$; 2) that every map $b:X\to r^{op}A$ factors as a composite $X\to r^{op}A' \to r^{op}A$ where the left-hand map is an isomorphism and the right-hand map is $r^{op}f$ for some $f:A'\to A$, which seems to be a consequence of $\mathcal{K}^{op}_{fp}(X,r^{op}A) \cong \mathcal{K}(rA,X) \cong \mathcal{L}(A,A') \cong \mathcal{L}^{op}_{fp}(A',A)$ with $A'=iX$. REPLY [3 votes]: I understand where is the mistake. $r$ does preserve finite presentability (the proof is straightforward and it is due to the fact that it is a left adjoint of a functor preserving filtered colimits). But there is no reason for $A'=iX$ to be finitely presentable.<|endoftext|> TITLE: Is there any relativistic interpretation on considering Kaehler-Einstein metrics and Calabi-Yau manifolds? QUESTION [6 upvotes]: Perhaps I sould ask this question on a physics forum, but I am curious about answers coming from mathematicians. Calabi-Yau manifolds are examples of Ricci-flat Kaehler manifolds. As we know, in the semi-riemannian case, Ricci flat metrics describes solutions for Einstein field equations on vacuum. So my question is: by considering the additional structure "Kaehler", and consequently, the holomorphic structure on $M$, does there any physical interpretation for such manifolds? More generally, what is the physical meaning of Einstein-Kahler manifolds? Is this related yet with Einstein field equations? What does the holomorphic sctructure on $M$ offers in addition to real manifolds? REPLY [5 votes]: First, a purely mathematical remark: it is not so easy to construct Riemannian Ricci-flat metrics on compact manifolds. Ricci flat Kähler (= Calabi-Yau) metrics give a large class of examples and are "easy" to obtain: by Yau's theorem, it is enough to check a complex geometric condition (vanishing first Chern class) on a compact Kähler manifold to obtain a Ricci-flat metric. The point is that the equation Ricci curvature vanishes is easier to control with the additional Kähler requirement: a Kähler metric is locally determined by one complex valued function, whereas a general metric is locally determined by d(d-1)/2 real valued functions in (real) dimension d. In physics, most of appearances of complex geometry are related in one way or the other to supersymmetry. Given a supersymmetric theory on a Minkowski spacetime $\mathbb{R}^{1,D}$, you might want to consider the same theory on $\mathbb{R}^{1,d} \times M$, where $M$ is a compact manifold of dimension $D-d$. To obtain a supersymmetric theory on $\mathbb{R}^{1,d}$, the simplest possbility is to ask for $M$ to admit a covariantly constant spinor. Compact Riemannian manifolds admitting covariantly constant spinors are automatically Ricci-flat and necessarily have reduced holonomy. Calabi-Yau manifolds give a large class of such examples (but there are others, not necessarily obviously related to complex geometry, as 7-manifolds of G2 holonomy for example). A "dual" point of view. Physically, it is natural to think to a particle probing spacetime and to study the worldline theory of this particle. For any Riemannian manifold, you can always consider a free particle, whose trajectories are geodesics. You can also consider a spinning particle, whose worldline theory is supersymmetric, and you can ask to which condition on the metric is the supersymmetry of the worldline theory bigger than expected. You find that Kähler manifolds are a natural class of examples where the worldline theory of the spinning particle has such an extended supersymmetry. (It is possible to go much further along these lines, e.g. replacing a point by a higher dimensional probing object as a string or a membrane and/or considering a quantum probing object, and to recover/interpret many facts about special holonomy metrics from such physics point of view).<|endoftext|> TITLE: Double Series involving Gamma function QUESTION [6 upvotes]: Does anyone have any ideas on howto verify $$\sum_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)}\cdot \frac{1}{3^{n+m+3x-1}} = \Gamma(x)$$ for $x>0$? I posted this question also on math.stackexchange. This is not an exercise from a book, but arises due to my research in the study of a probability density. REPLY [4 votes]: This problem can be reduced at least formally to a compact double integral, which might be easier to solve. Starting with the integral representation for the Gamma function, we write the double sum as an integral of the square of the confluent hypergeometric function ${}_1F_1$, then apply analogue of Euler's transformation formula: \begin{align} &\sum_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)}\cdot \frac{1}{3^{n+m+3x-1}}\\ &=3\int_0^\infty e^{-3t}t^{3x-1}\sum_{n,m=0}^\infty \frac{t^{n+m}}{\Gamma(n+1+x)\Gamma(m+1+x)}dt\\ &=\frac3{\Gamma^2(x+1)}\int_0^\infty e^{-3t}t^{3x-1}~\left[{}_1F_1(1,x+1;t)\right]^2dt\\ &=\frac3{\Gamma^2(x+1)}\int_0^\infty e^{-t}t^{3x-1}~\left[{}_1F_1(x,x+1;-t)\right]^2dt\\ &=\frac3{\Gamma^2(x+1)}\int_0^\infty e^{-t}t^{3x-1}~\sum_{n,m=0}^\infty\frac{(x)_n(x)_m}{(x+1)_n(x+1)_m}\frac{(-t)^{n+m}}{n!m!}dt\\ &=\frac{3x^2}{\Gamma^2(x+1)}\int_0^\infty e^{-t}t^{3x-1}~\sum_{n,m=0}^\infty\frac{1}{(x+n)(x+m)}\frac{(-t)^{n+m}}{n!m!}dt\\ &=\frac{3}{\Gamma^2(x)}\int_0^\infty e^{-t}t^{3x-1}\int_0^1 du\int_0^1 dv\sum_{n,m=0}^\infty\frac{(-t)^{n+m}u^{n+x-1}v^{m+x-1}}{n!m!} dt\\ &=\frac{3}{\Gamma^2(x)}\int_0^\infty e^{-t}t^{3x-1}\int_0^1 du\int_0^1 dv~ e^{-t(u+v)}(uv)^{x-1}dt\\ &=\frac{3\Gamma(3x)}{\Gamma^2(x)}\int_0^1\int_0^1 \frac{u^{x-1} v^{x-1}}{(u+v+1)^{3 x}} dudv. \end{align} This means that the initial problem is equivalent to $$\int_0^1\int_0^1 \frac{u^{x-1} v^{x-1}}{(u+v+1)^{3 x}} dudv=\frac{\Gamma^3(x)}{3\Gamma(3x)}. $$<|endoftext|> TITLE: Sets with no "full" projection on sufficiently large subset of coordinates QUESTION [5 upvotes]: Let $\ M\ $ be a finite set. Call set $A\ \ d$-fully shattering ($d\le|M|$) $\ \Leftarrow:\Rightarrow\ $ the following are true: $A \subset \{0,1\}^M$. $\forall_{\, S \in\binom Md}\ \pi_S(A)\ne\{0,1\}^S,\,\ $ where $\,\ \pi_S:\{0,1\}^M\rightarrow\{0,1\}^S\ $ is the canonical projection. I'm interested in the upper bounds on the size of $d$-fully shattering sets especially in the regime $d=|M|^\alpha$. I'm pretty sure this object was studied before but I don't know under what name. Any references or hints would be greatly appreciated. REPLY [2 votes]: This is a comment about shattering. Let $U$ be a finite set. If $S\subseteq [n]$ has cardinality $d$, then call $U^n$ an $n$-dimensional cube and call $U^S$ the $d$-dimensional face indexed by $S$. A subset $A\subseteq U^n$ shatters the face $U^S$ if the projection of $U^n$ onto $U^S$ maps $A$ onto $U^S$. The Shattering Lemma, due to Vapnik-Chervonenkis and also to Perles–Sauer–Shelah, concerns the case $U=\{0,1\}$. It is the statement that the maximal size of a subset $A\subseteq \{0,1\}^n$ that shatters no $d$-dimensional face is $\sum_{k1$) there is some $A\subseteq U^n$ that shatters every $d$-dimensional face, which satisfies the bound $$ |A|\leq \lceil d\log_b(n)+\log_b(u^d/d!) \rceil, $$ where $u=|U|$ and $b=u^d/(u^d-1)$. For our purposes, $u, d, b$ are fixed, but $n$ is allowed to vary. Although this $\sim \log_b(n)$ bound may not be optimal, we know that there is no $A\subseteq U^n$ of size $<\log_u(n)$ that shatters every $d$-dimensional face, so $\log(n)$ is the right growth order. The thing that might interest readers of this thread is: we proved the existence of a fully-shattering set of $\log(n)$-size with a probabilistic argument. We computed the expected number of unshattered $d$-dimensional faces for a randomly chosen subset $A\subseteq U^n$ of size $\lceil d\log_b(n)+\log_b(u^d/d!) \rceil$ and found it to be less than $1$, so at least one of the sets of this size unshatters zero faces (i.e. shatters every face). We don't have an explicit description of any $\log(n)$-size fully-shattering sets. [This combinatorial problem came up while investigating the minimal sizes of generating sets of powers $\mathbb U^n$ of a finite algebraic structure $\mathbb U$.]<|endoftext|> TITLE: For which abelian groups $G$ does the monoid of zero-sum sequences over $G$ embed into a ring as a divisor-closed subsemigroup? QUESTION [7 upvotes]: Let $K$ be a multiplicatively written semigroup (either commutative or not) and $H$ a subsemigroup of $K$. We say that $H$ is divisor-closed (in $K$) if $x \in H$ for all $x, y \in K$ such that $x \mid_K y$ (i.e., $y = uxv$ for some $u, v \in K$) and $y \in H$. Accordingly, we say that a semigroup $S$ is annular (bear with me, I don't have a better word for the moment) if it embeds as a divisor-closed subsemigroup into the multiplicative monoid of a ring. So here is my question: Q. Given an abelian group $G$, denote by $\mathscr B(G)$ the monoid of zero-sum sequences over $G$, that is, the submonoid of $\mathscr F(G)$, the free abelian monoid with basis $G$, given by the inverse image of $0_G$ under the canonical (monoid) epimorphism $\mathscr F(G) \to G$. For which $G$ is $\mathscr B(G)$ annular? This is a very special case of a more general question (namely, when does a semigroup embeds into a ring etc.?), which, however, is also much harder and beyond the scope of this post. Indeed, my motivation is simply the following: Factorization theory (that is, the theory of non-unique factorization) grew up out of algebraic number theory, and a turning point in its history was when the theory, until then focused on integral domains, was reforged in the language of monoids, based at least in part on the consideration that the latter provide "models" (for studying various phenomena of interest) that wouldn't have been available otherwise, with monoids of zero-sum sequences being probably the most representative of these models. So, the whole point of my question is that I'd like to understand to which extent the statement in bold is well-grounded. Update #1. A fruitful exchange with Benjamin Steinberg in the comments below has eventually shown that if $H$ is a non-trivial monoid with trivial group of units (as in the case of interest here) embedding as a divisor-closed subsemigroup into the multiplicative monoid of a unital ring $R$, then the characteristic of $R$ is $2$, simply because the condition of divisor-closedness implies $R^\times \cong H^\times = \{1_H\}$. REPLY [4 votes]: Figured it out (sorry for answering my own question). I'll prove the following: Lemma. Let $H$ be a linearly orderable monoid and $R$ a domain whose group of units is trivial. Then $H$ embeds as a divisor-closed submonoid into the multiplicative monoid of the monoid ring $R[H]$. This will show that the answer to the question in the OP is "For all $G$", since $\mathscr B(G)$ is a torsion-free, cancellative, commutative monoid, and hence is linearly orderable. Proof of the lemma. Let $\preceq$ be a total order on $H$ for which $(H, \preceq)$ is a linearly ordered monoid, that is, $xz \prec yz$ and $zx \prec zy$ for all $x, y, z \in H$ with $x \prec y$ (as usual, $u \prec v$ means $u \preceq v$ and $u \ne v$). Moreover, let $\delta$ be the canonical embedding $H \to R[H]$, so that $\delta(x)$ is, for every $x \in H$, a Kronecker delta $H \to R$ centered at $x$, which I'll rather denote by $\delta_x$. Now pick $x \in H$, and assume $\delta_x = f\ast g$ for some $f, g \in R[H]$, with $\ast$ being the multiplication in $R[H]$. I claim that $|{\rm supp}(f)| = |{\rm supp}(g)| = 1$. Indeed, it is clear that $f$ and $g$ cannot be identically zero, so the support of each of them is non-empty. Accordingly, let $y_f$ and $z_f$ be, respectively, the minimum and the maximum of $S_f := {\rm supp}(f)$ relative to the order $\preceq$ (which exist because $\preceq$ is total and $S_f$ is not only non-empty, but also finite); define $y_g$ and $z_g$ in a similar way (only with $g$ in lieu of $f$). Then set $y := y_f y_g$ and $z := z_f z_g$. We have $$ \delta_x(y) = \sum_{uv=y} f(u) g(v) = f(y_f) g(y_g) \ne 0_R.$$ To see this, note that, if $u \prec y_f$ or $v \prec y_g$, then $f(u) g(v) = 0_R$, and on the other hand, if $y_f \preceq u$ and $y_g \preceq v$, then $y = y_f y_g \preceq uv$, with equality iff $u = y_f$ and $v = y_g$ (by the assumption that $H$ is linearly ordered by $\preceq$). Then the rest is trivial, because $y_f \in {\rm supp}(f)$ and $y_g \in {\rm supp}(g)$ give $f(y_f) g(y_g) \ne 0_R$ (by the fact that $R$ is a domain). To wit, we have shown that $y$ is in the support of $\delta_x$, and so is $z$ by an analogous argument. It follows $x = y = z$, which is, however, possible only if $y_f = z_f$ and $y_g = z_g$ (by construction, $y_f \preceq z_f$ and $y_g \preceq z_g$, and we can't have $y_f \neq z_f$ or $y_g \ne z_g$, otherwise $H$ being linearly ordered by $\preceq$ would yield a contradiction). In other words, $f = a\delta_u$ and $g = b \delta_v$ for some $a, b \in R$ and $u, v \in H$ such that $ab= 1_R$. But $R^\times$ is trivial (by hypothesis), and therefore $a = b = 1_R$. So $f, g \in \delta(H)$, and we are done. [] In the last line of the proof, we use that domains are, of course, Dedekind-finite rings, so that a product is a unit iff so are all the factors (see here).<|endoftext|> TITLE: Cohomology of certain arithmetic groups QUESTION [7 upvotes]: This is a question on literature about cohomology of arithmetic groups. Let $M$ denote a quaternion algebra over $\mathbb Q$ and assume it is non-split over $\mathbb R$. Fix a maximal order $\Lambda$ in $M$ and for any ring $R$ let $$ M(R)=\Lambda\otimes_{\mathbb Z}R,\qquad G(R)=M(R)^\times/R^\times. $$ Let $p$ be a prime number at which $M$ splits and set $$ \Gamma_p=G({\mathbb Z}[1/p]) $$ I need to know how the first Betti-number $$ \dim H^1(\Gamma_p,{\mathbb C}), $$ depends on $p$. I would be grateful if you could point me towards some literature on this. REPLY [6 votes]: We can write $G(\mathbb Q_p)/ G(\mathbb Z_p) =GL_2(\mathbb Q_p)/GL_2(\mathbb Z_p)$ as the set of vertices of a tree (the Bruhat-Tits tree) of degree $p+1$. The group $\Gamma_p$ acts on this tree with finite stabilizers. Hence the cohomology of the quotient $\Gamma_p \backslash G(\mathbb Q_p)/ G(\mathbb Z_p) $ is equal to the group cohomology of $\Gamma_p$. That's some graph with $n$ vertices and at most $n(p+1)/2$ edges, where $n$ is the cardinality of $\Gamma_p \backslash G(\mathbb Q_p)/ G(\mathbb Z_p) $, and thus Euler characteristic at least $-n (p-1)/2$ (and at most $1$). $n$ is bounded independently of $p$ because any class in $G(\mathbb Q_p)/G(\mathbb Z_p)$ defines a locally free left $M(\mathbb Z)$-submodule inside $M(\mathbb Z[1/p])$, with two modules $\Gamma_p$-conjugate if and only if they are isomorphic, and because there are finitely many isomorphism classes of locally free left $M(\mathbb Z)$-modules.<|endoftext|> TITLE: Inverse of a matrix with binomial entries QUESTION [9 upvotes]: This is closely related to this question: Eigenvalues of a matrix with binomial entries. We consider the matrix: $$M_{ij} = 4^{-j}\binom{2j}{i}$$ where it is understood that the binomial coefficient $\binom{m}{k}$ is zero if $k<0$ or $k>m$. The indices $i,j$ traverse a discrete finite range, $i,j \in \{a, a+1, \dots, b\}$, from $a$ to $b$, where $a,b$ are non-negative integers with $0\le a\le b$. Therefore the matrix $M_{ij}$ has dimensions $(b-a+1) \times (b-a+1)$. Can we find the inverse matrix, $M^{-1}$? Numerical computations of the determinant suggest that this is a non-singular matrix (for all $0 \le a < b$). A close-form expression for the determinant could be useful. Why this question is not a duplicate: Although in principle the eigenvalues and eigenvectors of a matrix are enough to invert it, the other question focuses on the largest positive eigenvalue alone. Moreover the eigenvalues/eigenvectors (not even the largest alone) have not been solved, so maybe finding the inverse of this matrix turns out to be easier. So, if the other question suddenly received a complete response and all the eigenvalues / eigenvectors were found, then yes, this question would be automatically solved. But that does not seem likely to happen. REPLY [6 votes]: Let's refer everything to square matrices indexed from $0$ to $h$, that I will denote as $$ {\bf M}_{\,h} = \left\| {\;f(n,m)\;} \right\|_{\,h} $$ with $n$ being the row index and $m$ the column index. I will then denote by $$ \left( {f(n) \circ {\bf I}_{\,h} } \right) $$ the diagonal matrix whose entries are equal to $f(n)$. So I write the matrix you proposed as $$ \bbox[lightyellow] { {\bf M}_{\,h} (a) = \left\| {\;4^{ - n - a} \left( \matrix{ 2n + 2a \cr m + a \cr} \right)\;} \right\|_{\,h} = \left( {4^{ - n - a} \circ {\bf I}_{\,h} } \right)\;\left\| {\;\left( \matrix{ 2n + 2a \cr m + a \cr} \right)\;} \right\|_{\,h} }$$ where $h=b-a$. That premised, consider that in general $$ \eqalign{ & \left( \matrix{ r\,n + t \cr m + q \cr} \right) = {{\left( {r\,n + t} \right)^{\,\underline {\,m + q} } } \over {\left( {m + q} \right)!}} = {{\left( {r\,n + t} \right)^{\,\underline {\,q} } \left( {r\,n + t - q} \right)^{\,\underline {\,m} } } \over {\left( {m + q} \right)^{\,\underline {\,q} } \;m^{\,\underline {\,m} } }} = \cr & = \left( {r\,n + t} \right)^{\,\underline {\,q} } \left( \matrix{ r\,n + t - q \cr m \cr} \right){1 \over {\left( {m + q} \right)^{\,\underline {\,q} } }} \cr} $$ where $x^{\,\underline {\,a} } $ denotes the falling factorial ($x^{\overline {\,a\,} } $ the rising) and where, for the present problem, we consider $q$ to be a non-negative integer, while $r$ and $t$ could be real (or even complex). Then we have that we can write the binomial as $$ \begin{gathered} \left( \begin{gathered} r\,n + s \\ m \\ \end{gathered} \right) = \frac{1} {{m!}}\left( {r\,n + s} \right)^{\,\underline {\,m\,} } = \frac{1} {{m!}}\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right)} {\left( \begin{gathered} m \\ k \\ \end{gathered} \right)s^{\,\underline {\,m - k\,} } \left( {r\,n} \right)^{\,\underline {\,k\,} } } = \hfill \\ = \frac{1} {{m!}}\sum\limits_{\left\{ \begin{subarray}{l} \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right) \\ \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \;h} \right) \end{subarray} \right.} {\left( \begin{gathered} m \\ k \\ \end{gathered} \right)s^{\,\underline {\,m - k\,} } \left( { - 1} \right)^{\,k - j} \left[ \begin{gathered} k \\ j \\ \end{gathered} \right]r^{\,j} n^{\,j} } = \hfill \\ = \sum\limits_{\left\{ \begin{subarray}{l} \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right) \\ \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \;h} \right) \end{subarray} \right.} {n^{\,j} r^{\,j} \left( { - 1} \right)^{\,k - j} \left[ \begin{gathered} k \\ j \\ \end{gathered} \right]\frac{1} {{k!}}\left( \begin{gathered} s \\ m - k \\ \end{gathered} \right)} \hfill \\ \end{gathered} $$ Then in the last line we can replace $n^m$ with $$ n^{\,m} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \;h} \right)} {\left\{ \matrix{ m \cr k \cr} \right\}n^{\,\underline {\,k\,} } } = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \;h} \right)} {\left( \matrix{ n \cr k \cr} \right)k!\left\{ \matrix{ m \cr k \cr} \right\}} $$ Thus we arrive finally to $$ \bbox[lightyellow] { \eqalign{ & {\bf M}_{\,h} (a) = \left\| {\;4^{ - n - a} \left( \matrix{ 2n + 2a \cr m + a \cr} \right)\;} \right\|_{\,h} = \left( {4^{ - n - a} \circ {\bf I}_{\,h} } \right)\;\left\| {\;\left( \matrix{ 2n + 2a \cr m + a \cr} \right)\;} \right\|_{\,h} = \cr & = \left( {4^{ - n - a} \circ {\bf I}_{\,h} } \right)\left( {\left( {2\,n + 2a} \right)^{\,\underline {\,a} } \circ {\bf I}_{\,h} } \right)\;\left\| {\;\left( \matrix{ 2n + a \cr m \cr} \right)\;} \right\|_{\,h} \left( {{1 \over {\left( {n + a} \right)^{\,\underline {\,a} } }} \circ {\bf I}_{\,h} } \right) = \cr & = \left( {4^{ - n - a} \circ {\bf I}_{\,h} } \right)\left( {\left( {2\,n + 2a} \right)^{\,\underline {\,a} } \circ {\bf I}_{\,h} } \right)\;{\bf B}_{\,h} \left( {n! \circ {\bf I}_{\,h} } \right)\;\overline {{\bf St}_{{\bf 2}\,h} } \left( {2^{\,n} \circ {\bf I}_{\,h} } \right)\;\overline {{\bf St}_{{\bf 2}\,h} } ^{\,{\bf - }\,{\bf 1}} \left( {n! \circ {\bf I}_{\,h} } \right)^{\,{\bf - }\,{\bf 1}} \left( {{\bf I}_{\,h} + \overline {{\bf E}_{\,h} } } \right)^{\,{\bf a}} \left( {{1 \over {\left( {n + a} \right)^{\,\underline {\,a} } }} \circ {\bf I}_{\,h} } \right) \cr} \tag{1} }$$ with $$ \eqalign{ & {\bf B}_{\,h} = \;\left\| {\;\left( \matrix{ n \cr m \cr} \right)\;} \right\|_{\,h} \quad {\bf St}_{{\bf 2}\,h} = \;\left\| {\;\left\{ \matrix{ n \cr m \cr} \right\}\;} \right\|_{\,h} \quad {\bf I}_{\,h} + {\bf E}_{\,h} = \;\left\| {\;\left( \matrix{ 1 \cr n - m \cr} \right)\;} \right\|_{\,h} \cr & \overline {\bf X} = transpose({\bf X}) \cr} $$ After that the determinant follows easily, since the matrices other than the diagonal ones have unitary determinant $$ \bbox[lightyellow] { \left| {\,{\bf M}_{\,h} (a)\,} \right| = \left( {\prod\limits_{0\, \le \,n\, \le \;h} {{{\left( {2\,\left( {n + a} \right)} \right)^{\,\underline {\,a} } } \over {2^{\,n + 2a} \left( {n + a} \right)^{\,\underline {\,a} } }}} } \right) = \left( {\prod\limits_{0\, \le \,n\, \le \;h} {{{\left( \matrix{ 2\,\left( {n + a} \right) \cr a \cr} \right)} \over {2^{\,n + 2a} \left( \matrix{ n + a \cr a \cr} \right)}}} } \right) \tag{2}}$$ Some notes concerning the inversion of identity (1), and further analysis you might possibly want perform on that. For the Binomial $$ {\bf B}_{\,h} ^{\,{\bf r}} = \;\left\| {\;r^{\,n - m} \left( \matrix{ n \cr m \cr} \right)\;} \right\|_{\,h} = \left( {r^n \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} \;\left( {r^n \circ {\bf I}_{\,h} } \right)^{\, - \,{\bf 1}} \quad \;\left| {\;r \in R,C} \right. $$ where the second expression for $r=0$ is understood to be taken in the limit. So $$ {\bf B}_{\,h} ^{\, - \,{\bf 1}} = \left( {\left( { - 1} \right)^n \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} \;\left( {\left( { - 1} \right)^n \circ {\bf I}_{\,h} } \right)^{\, - \,{\bf 1}} = \left( {\left( { - 1} \right)^n \circ {\bf I}_{\,h} } \right){\bf B}_{\,h} \;\left( {\left( { - 1} \right)^n \circ {\bf I}_{\,h} } \right) $$ For the Stirling Numbers, 1st and 2nd kind are related by $$ {\bf St}_{{\bf 2}\,h} ^{\, - \,{\bf 1}} = \;\left\| {\;\left( { - 1} \right)^{\,n - m} \left[ \matrix{ n \cr m \cr} \right]\;} \right\|_{\,h} = \left( {\left( { - 1} \right)^n \circ {\bf I}_{\,h} } \right)\;{\bf St}_{{\bf 1}\,h} \;\left( {\left( { - 1} \right)^n \circ {\bf I}_{\,h} } \right) $$ ${\bf E}$ is the "shift", "first off-diagonal", .. matrix, i.e: $$ {\bf E}_{\,h} = \left\| {\;\left[ {1 = n - m} \right]\;} \right\|_{\,h} = \left\| {\;\left( \matrix{ 0 \cr n - m - 1 \cr} \right)\;} \right\|_{\,h} $$ (where $[P]$ is the Iverson bracket) then $$ \eqalign{ & {\bf E}_{\,h} ^{\,{\bf q}} = \left\| {\;\left[ {q = n - m} \right]\;} \right\|_{\,h} = \left\| {\;\left( \matrix{ 0 \cr n - m - q \cr} \right)\;} \right\|_{\,h} \quad \;\left| {\;0 \le q \in Z} \right. \cr & \left( {{\bf I}_{\,h} + {\bf E}_{\,h} } \right) = \left\| {\;\left[ {0 \le n - m \le 1} \right]\;} \right\|_{\,h} = \left\| {\;\left( \matrix{ 1 \cr n - m \cr} \right)\;} \right\|_{\,h} \cr & \left( {{\bf I}_{\,h} + {\bf E}_{\,h} } \right)^{\,{\bf r}} = \sum\limits_{0\, \le \,k\,\left( { \le \;h} \right)} {\left( \matrix{ r \cr k \cr} \right){\bf E}_{\,h} ^{\,{\bf k}} } = \left\| {\;\left( \matrix{ r \cr n - m \cr} \right)\;} \right\|_{\,h} \quad \;\left| {\;r \in R,C} \right. \cr} $$ and finally that ${\bf B}$ and ${\bf {I+E}}$ are tied by the similarity $$ {\bf B}_{\,h} \; = \left( {{\bf St}_{\,{\bf 2}\,h} \left( {n! \circ {\bf I}_{\,h} } \right)} \right)\left( {{\bf I}_{\,h} + {\bf E}_{\,h} } \right)\left( {{\bf St}_{\,{\bf 2}\,h} \left( {n! \circ {\bf I}_{\,h} } \right)} \right)^{{\bf - 1}} $$ and by a bunch of other relations, among which $$ \left( {{\bf I}_{\,h} + {\bf E}_{\,h} } \right)^{\,{\bf - q}} \quad \left| {\;0 \le {\rm integer }q} \right.\quad = \left( {{\bf B}_{\,h} \left( {n^{\,\underline {\, - q\,} } \circ {\bf I}_{\,h} } \right)} \right)^{\,{\bf - 1}} \;\left( {\left( {n^{\,\underline {\, - q\,} } \circ {\bf I}_{\,h} } \right)\;\;{\bf B}_{\,h} } \right) $$<|endoftext|> TITLE: Is there a higher dimensional analogue of the Dirac belt trick? QUESTION [12 upvotes]: The Dirac belt trick produces a nice 3-dimensional geometric object with symmetry group $Spin(3) = SU(2)$: a 2-sphere with a properly embedded framed ray (usually presented by using orientations to reduce the framing of the ray to a single normal vector field, then integrating this to give a "belt" of finite width), with the ray or "belt" regarded up to smooth isotopies fixing the sphere and trivial at infinity. A naive generalization to higher dimensions using a (2-dimensional) "belt" does not work to produce an object with symmetry group $Spin(n)$ since the "belt" can always be untwisted. My question: Is there any properly embedded subspace of ${\mathbb R}^n$ with boundary on $S^{n-1}$, possibly equipped with an auxiliary structure like a framing of its normal bundle, such that when the subspace is regarded up to isotopies fixing $S^{n-1}$ and trivial at infinity, the symmetry group of the "sphere and subspace" is $Spin(n)$? I am particularly interested in the case of $n=4$. REPLY [6 votes]: If I understand the question correctly, the same "belt" construction works in higher dimensions. The belt should have a full framing of its normal bundle ($n−1$ normal vectors; using the orientation of $\mathbb R^n$, you can get by with $n−2$ normal vectors; adding the orientation tangent to the belt gives $n$ vectors total). $Spin(n)$ is the simply connected double cover of $SO(n)$, and one explicit model of this double cover is the set of pairs $(f, p)$, where $f\in SO(n)$ and $p$ is a homotopy class of path from $f$ to a base point in $SO(m)$. The map of the sphere into $\mathbb R^n$ determines an element $f\in SO(n)$, and the framings along the belt determine a path from $f$ to a base point (that base point being the fixed framing near infinity on the belt). I actually use a construction very similar to this when doing fermionic TQFT calculations. I usually call the framed arc a "Dirac belt".<|endoftext|> TITLE: Is $\beta \mathbb{N}$ homeomorphic to its own square? QUESTION [23 upvotes]: Let $\mathbb{N}$ be the set of natural numbers and $\beta \mathbb N$ denotes the Stone-Cech compactification of $\mathbb N$. Is it then true that $\beta \mathbb N\cong \beta \mathbb N \times \beta \mathbb N $ ? REPLY [5 votes]: An indirect argument: Since the Banach space of continuous functions $C(\beta\mathbb{N})$ is isomorphic to $\ell_\infty$, it contains no complemented copies of $c_0$. Since $C(\beta\mathbb{N}\times\beta\mathbb{N})$ is isomorphic to $C\big(\beta \mathbb{N},C(\beta\mathbb{N})\big)$, it contains a complemented copy of $c_0$. See [P. Cembranos. $C(K,E)$ contains a complemented copy of $c_0$. Proc. Amer. Math. Soc. 91 (1984), 556-558.]<|endoftext|> TITLE: IMO 2017/6 via arithmetic geometry QUESTION [55 upvotes]: The (very nice) final problem of IMO 2017 asked contestants to show: If $S$ is a finite set of lattice points $(x,y)$ with $\gcd(x,y)=1$, then there is a nonconstant homogeneous polyonmial $f \in \mathbb Z[x,y]$ such that $f(x,y) = 1$ for all $(x,y) \in S$. It's claimed in this forum post that the above IMO problem is a special case of Lemma 7.3 of arXiv:16040.01704. The former post phrases the lemma as follows: If $X$ is a finite scheme over $\operatorname{Spec} \mathbb Z$ then $\operatorname{Pic}(X)$ is finite. Being unknowledgable as I am, I do not see how to deduce the IMO problem from the lemma. Can someone make the connection explicit? REPLY [33 votes]: The set $S$ gives rise to a subscheme (which let's also denote by $S$) of $\mathbb{P}^1_{\mathbb{Z}},$ because relatively a prime pair $(x,y)$ corresponds to a section of $\mathbb{P}^1_{\mathbb{Z}}\rightarrow\operatorname{Spec}\mathbb{Z}$, and we take the union of these divisors in $\mathbb{P}^1_\mathbb{Z}$. Now, we have a map $\mathcal{O}(d)_{\mathbb{P}^1}\rightarrow\mathcal{O}(d)_S,$ and it will suffice to show that for some $d$ some element in the image of the induced map on global sections is nowhere zero. (If one composes this map with the map $\mathcal{O}(d)_S\rightarrow\mathcal{O}(d)_{\operatorname{Spec}\mathbb{Z}}$ corresponding to a section $(x,y)$, one gets at the level of global sections the evaluation map $\mathbb{Z}[x,y]_d\rightarrow\mathbb{Z}$ on degree $d$ homogeneous polynomials.) For this, it suffices that one can find $d$ so that this map is surjective on global sections and so that $\mathcal{O}(d)_S$ is trivial. The map will be surjective on global sections for large enough $d$ by ampleness of $\mathcal{O}(1)$, and the triviality of a power of $\mathcal{O}(1)_S$ follows from the finiteness of $\operatorname{Pic}(S)$. It's maybe slightly inaccurate to say just that it's a case in Lemma 7.3 in the above paper; rather, this exact argument is ran in the proof of Corollary 1.3 (which immediately follows the aforementioned lemma) to prove a much more general result.<|endoftext|> TITLE: What is the formula for Fourier Transform or Mellin transform of Riemann Zeta $\zeta(s)$ QUESTION [7 upvotes]: I have seen the Fourier and Mellin transform for Riemann $\Xi (t)=\xi ({\frac 12}+it)$ where: $\xi (s)={\tfrac {1}{2}}s(s-1)\pi ^{{-s/2}}\Gamma \left({\tfrac {1}{2}}s\right)\zeta (s)$ Fourier transform of $\Xi(t)$ is: $\Xi (t) = \int_{-\infty}^\infty\Phi(u)e^{iut}\,du$ Where: $\Phi(u) = \sum_{n=1}^\infty (4\pi^2n^4e^{9u/2} - 6n^2\pi e^{5u/2} ) exp(-n^2\pi e^{2u})$ But I had not seen something similar for Fourier or Mellin transform for Riemann $\zeta(s)$ itself. Is there a "closed form" formula for Fourier or Mellin transform for Riemann $\zeta(s)$ itself ? Thank you. REPLY [8 votes]: $\zeta(s)=\sum_{n=1}^\infty n^{-s}$ is the Laplace transform of the distribution $S(u) = \sum_{n=1}^\infty \delta(u-\log n)$. $\zeta(s) = \mathcal{L}[S(u)](s) = \int_{-\infty}^\infty S(u) e^{-su}du$ converges for $\Re(s) > 1$. For $\Re(s) \in (0,1)$ it becomes the bilateral Laplace transform $\zeta(s) = \mathcal{L}[S(u)-e^u](s) =\int_{-\infty}^\infty (S(u)-e^u) e^{-su}du$ For $\Re(s) \in (-1,0)$ it is $\zeta(s) = \mathcal{L}[S(u)-e^u+\frac{1}{2}](s) =\int_{-\infty}^\infty (S(u)-e^u+\frac{1}{2}) e^{-su}du$ For $\Re(s) \in (-K,-K+1)$ it is $\zeta(s) = \mathcal{L}[S(u)-\sum_{k=0}^K \frac{B_k}{k!}e^{(1-k)u}](s) =\int_{-\infty}^\infty (S(u)-\sum_{k=0}^K \frac{B_k}{k!}e^{(1-k)u}) e^{-su}du$ where $B_k$ are the Bernouilli numbers. Thus for $\sigma \in (-K,-K+1)$, the inverse Fourier transform (in the sense of distributions) of $\hat{f}(\xi)=\zeta(\sigma+2i \pi \xi)$ is $f(u)= e^{-\sigma u}(S(u)-\sum_{k=0}^K \frac{B_k}{k!}e^{(1-k)u})$.<|endoftext|> TITLE: An identity involving hook-lengths QUESTION [8 upvotes]: I am reading Macdonlad's book on "symmetric functions and Hall polynomials" and I have difficulty figuring out an identity which involves hook-lengths. I would like to ask for a hint. Let $\lambda=(\lambda_1,\dots, \lambda_k)$ be a partition and define $\mu_i=\lambda_i+k-i$, for $1\leq i\leq k$. The hook-length of a box $x\in \lambda$ in the Young diagram is denoted by $h(x)$. One can show the following identity: $$ \sum_{x\in \lambda}t^{h(x)}+\sum_{1\leq i TITLE: Zariski open subset on family of Kaehler-Einstein manifolds QUESTION [10 upvotes]: Let $\pi:\mathcal X\to B$ be a family of Kaehler manifolds then if we take $B'\subset B$ be the set of parameters such that $X_b$ admit Kaehler-Einstein metric(with zero, negative, or positive Ricci curvature) , then $B'$ is Zariski open subset of $B$ always? REPLY [10 votes]: The answer depends on diameter bounds of the fibers $(X_b,\omega_b)$ and the $L^\infty$ norm of the Ricci curvature. (You can derive the answer by using a nice paper of J. Cheeger-A.Naber. See the nice survay paper of Donaldson.) When $b\to 0$, the diameter of $X_b$ may blow up near the central fiber. If such an assumption is satisfied then $B'$ is a Zariski open subset of $B$. When fibers $X_b$ are Calabi-Yau manifolds, then, fixing holomorphic $n$-forms $\Omega_b$ on $X_b$, $b\neq 0$, which vary holomorphically in $b$, we may define the Weil-Petersson pseudometric on $Δ^×$ by Tian's formula: $$\omega_{WP}=-\sqrt{-1}\partial_b\bar\partial_b\log\left((-1)^{n^2/2}\int_{X_b}\Omega_b\wedge\overline {\Omega_b}\right),$$ which is a smooth semipositive definite Hermitian form on $Δ^×$, well-defined independent of the choice of $\Omega_b$. (It would be nice to find the same formula when dimension of base is bigger than one.) Then boundedness of diameter of $X_b$ corresponds to boundedness of $$\int_{X_b}\Omega_b\wedge\overline {\Omega_b} TITLE: Relatively hyperbolic knot groups QUESTION [6 upvotes]: I am looking for references on the geometry of knot groups. For instance, I am interested in the following question: When is a knot group relatively hyperbolic? Hyperbolic knots are known to have a group which is relatively hyperbolic, but is the reciprocal true? Is the answer easier when restricted to alternating knots? And what about acylindrically hyperbolic knot groups? REPLY [9 votes]: The following theorem, which can be deduced from a combination theorem of Dahmani, is stated as Theorem 7.2.2 in our book 3-manifold groups . Theorem: Let $N$ be a compact, orientable, irreducible 3-manifold with empty or toroidal boundary. Let $M_1,\ldots,M_k$ be the maximal graph-manifold pieces of the JSJ decomposition of $N$, let $S_1,\ldots,S_l$ be the tori in the boundary of $N$ that adjoin a hyperbolic piece and let $T_1,\ldots,T_m$ be the tori in the JSJ decomposition of $N$ that separate two (not necessarily distinct) hyperbolic pieces of the JSJ decomposition. The fundamental group of $N$ is hyperbolic relative to the set of parabolic subgroups $$ \{H_i\}=\{\pi_1M_p\}\cup\{\pi_1 S_q\}\cup\{\pi_1 T_r\}~. $$ In particular, such a 3-manifold $M$ is non-elementarily relatively hyperbolic if and only if it is not a graph manifold (including the possibility of being Seifert fibred or a Sol manifold). I believe that, historically, knots whose complements are graph manifolds were called hose knots, so the statement is that a knot complement group is non-elementarily relatively hyperbolic if and only if it's not a hose knot. You also ask about acylindrical hyperbolicity, a weaker notion which merely asks for some acylindrical action on a hyperbolic space. It turns out that non-geometric graph manifold groups always act acylindrically on the Bass-Serre tree of the JSJ decomposition (see Lemma 2.4 here), and so are acylindrically hyperbolic. So the fundamental group of a knot complement will be acylindrically hyperbolic unless the knot is Seifert fibred. Finally, the remaining Seifert fibred knot complements (except for the trivial knot) all admit $\mathbb{H}^2\times\mathbb{R}$ geometry. In particular, after quotienting out a central infinite-cyclic subgroup, they act properly on the hyperbolic plane. The central $\mathbb{Z}$ acting trivially is the reason the corresponding action of the knot group on the hyperbolic plane fails to be acylindrical -- so even in this case, the groups are "nearly" acylindrically hyperbolic.<|endoftext|> TITLE: Fourth cohomology of the modular group QUESTION [6 upvotes]: Is $H^4(PSL(2,\mathbb{Z}),\mathbb{Z})$ known? I ask this in response to the recent calculation of the same cohomology group for $\mathrm{Co}_0$ and $\mathrm{Co}_1$. REPLY [6 votes]: From Jeremy Rickard's comments, the group cohomology (with coefficients a module with trivial action) of a free product of (discrete) groups is sent to direct sum (eg Proposition 1.3.16.3 in C. Löh, Group Cohomology & Bounded Cohomology (pdf)), so $$ H^n(PSL(2,\mathbb{Z}),\mathbb{Z}) \simeq H^n(C_2\ast C_3,\mathbb{Z})\simeq H^n(C_2,\mathbb{Z})\oplus H^n(C_3,\mathbb{Z}), $$ and positive, even-degree cohomology of cyclic groups is $H^{2k}(C_n,\mathbb{Z}) = \mathbb{Z}/n$ ($k\gt 0$) and $0$ in odd degree, hence $H^4(PSL(2,\mathbb{Z}),\mathbb{Z}) = \mathbb{Z}/2 \oplus \mathbb{Z}/3$.<|endoftext|> TITLE: Best provable and unconditional lower and upper bounds for Brun's constant QUESTION [8 upvotes]: Question: What are the currently known best provable and unconditional lower and upper bounds for Brun's constant $B$, corresponding to the sum of the reciprocals of the twin primes? Remark. According to Dominic Klyve's thesis "Explicit bounds on twin primes and Brun's Constant" (2007, p.23) the best provable unconditional bounds known at the time were given by $1.830484424658 < B < 2.347$ the lower bound being obtained by computation of the sum up to $10^{16}$. Hence, apparently, not even the first digit was known at the time i.e. the question $B<2$ was open. Was there any significant progress since then? Note that the sharper estimates that one usually sees, as that of Nicely $B=1.90216 05823 \pm 0.00000 00008$ are in fact conjectured but not rigorously proved (95% "confidence interval" according to its author). REPLY [7 votes]: In March 7, 2018 (several months after you asked this question!), Dave Platt and Tim Trudgian published a paper called Improved bounds on Brun’s constant, where they show the following improved bounds: $$1.840503 < B < 2.288513$$<|endoftext|> TITLE: Endomorphism algebras of restricted representations QUESTION [5 upvotes]: Let $G$ be a group, and $$\rho:G\to \mathrm{GL}(V)$$ be an absolutely irreducible, finite-dimensional representation over a characteristic $0$ field $k$. For each finite index subgroup $H\le G$, let $$\mathrm{End}_H(V) = \{\phi\in\mathrm{End}(V) :g\cdot\phi(v) = \phi(g\cdot v)\;\;\forall v\in V, \ g\in H\}$$ be the endomorphism ring of $\rho|_H$. Fix some finite index normal subgroup $N\lhd G$. Then there is a natural map $$\begin{align}\begin{Bmatrix}\text{Subgroups $H$ with }\\N\le H\le G\end{Bmatrix}&\to\begin{Bmatrix}\text{Subalgebras $A$ with } \\\mathrm{End}_G(V)\subseteq A\subseteq\mathrm{End}_N(V)\end{Bmatrix}\\ \\H&\mapsto \mathrm{End}_H(V).\end{align}$$ Can we determine which subalgebras $A$ are in the image of this map? I'm particularly interested in the case where $\dim V = 4$ and $\mathrm{End}_N(V) = M_2(k)$. In this case, is there a sugbroup $H$ such that $\mathrm{End}_H(V) = k\times k$? Edit: My motivation is as follows. Suppose I have a representation (in my case, a Galois representation) $$\rho:G\to\mathrm{Aut}(V)\cong\mathrm{GL}_4(k)$$ and I know that for some normal subgroup $N$, $$\rho|_N = \sigma\oplus\sigma$$ for some representation $\sigma:N\to\mathrm{GL}_2(k)$. A priori, I have no information about $N$. Under what circumstances can I find an $H$ such that $$\rho|_H=\sigma_1\oplus\sigma_2,$$ where the $\sigma_i:H\to\mathrm{GL}_2(k)$ are distinct? Since I know nothing about $N$, I'm hoping for a condition that is intrinsic to the representation in some way. Further edit: Let $G$ act on $\mathrm{End}_N(V)$ by $$g\cdot \phi = \rho(g)\circ\phi\circ\rho(g^{-1})$$ for $g\in G$, $\phi\in \mathrm{End}_N(V)$. Enlarging $N$ if necessary, we can assume that $N$ is the kernel of this action, and we get an injection $$G/N\hookrightarrow \mathrm{Aut}(M_2(k))=\mathrm{PGL}_2(k).$$ Since $G/N$ is finite, it is either cyclic, dihedral, $A_4$, $S_4$ or $A_5$. As Johannes points out in his answer, it is only the $A_5$ case which presents a difficulty. Are there any facts that I could use about $\rho$, without knowing what $N$ is, that would enable me to rule out this case? REPLY [4 votes]: EDIT: Finiteness isn't necessary for this argument. Instead I use that $V$ is semisimple over $N$, $|G:N|<\infty$ and $char(k)=0$. $Res_N^G(V)$ is semisimple because it is the restriction of a simple module to a normal subgroup. And it having $k^{2\times 2}$ as endomorphism ring is equivalent to it being the sum of two isomorphic two-dimensional, absolutely simple $N$-modules by the Wedderburn theorem. In particular, every $v\in V\setminus\{0\}$ generates a two-dimensional, simple $N$-module and all of those are isomorphic. Now note that $Res_H^G(V)$ is also semisimple if $N\leq H\leq G$. This is because every $H$-invariant $U\leq V$ is also $N$-invariant and therefore has a $N$-invariant complement. Averaging over $H/N$ gives a a $H$-invariant complement. Therefore $Res_H^G(V)$ can have the following endomorphism rings: $k$ (iff the restriction is still absolutely simple), a quadratic extension of $k$ (iff the restriction is simple, but not absolutely simple), $k\times k$ (iff the restriction decomposes into two different irreducibles) and $k^{2\times 2}$ (iff the restriction decomposes into two isomorphic irreducibles). Theorem: Let $G$ is a group and $V$ a absolutely simple $k[G]$-module, $N\unlhd G$ a normal subgroup such that $End(Res_N^G(V))=k^{2\times 2}$. Then the following are equivalent: There exists a subgroup $N\leq H\leq G$ such that $End_k(Res_H^G(V)) = k\times k$. There exists a $g\in G$ such that $Res_{\langle g\rangle N}^G(\chi_V)$ is not divisible by two in the character ring $Ch(\langle g\rangle N)$. Proof: If $g\in G$ is arbitrary and $v\in V$ an eigenvector of $g$, then $U:=span\{nv \mid n\in N\}$ is a $g$-invariant subspace. Therefore we get a decomposition $V=U\oplus U'$ into $\langle g,N\rangle$-invariant subspaces. It follows that $\chi_V(g) = \chi_U(g) + \chi_{U'}(g)$. $1.\implies 2.$ Now $Res_H^G(V) = U_1 \oplus U_2$ for two irreducible, non-isomorphic $H$-modules and $Res_{H'}^H(U_i)$ is still irreducible for all $N\leq H'\leq H$. If we choose $g\in H$ such that $\chi_{U_1}(g) \neq \chi_{U_2}(g)$ and an eigenvector $v\in U_1$, then our construction gives $U=U_1$ and $U' \cong U_2$, $Res_{\langle g\rangle N}(\chi_V) = Res_{\langle g\rangle N}^H(\chi_{U_1}) + Res_{\langle g\rangle N}^H(\chi_{U_2})$ and these two summands are different irreducible characters. Therefore $Res_{\langle g\rangle N}(\chi_V)$ can not be divisible by two. $2.\implies 1.$ Conversely if such a $g$ exists, then $H:=\langle g\rangle N$ satisfies the conditions. Corollary: If there is a $g\in G$ such that $2 \nmid \chi_V(g)$, then $H=\langle g,N\rangle$ satisfies $End(Res_H^G(V)) = k\times k$. Here is an example where the Corollary is actually applicable: Let $G=S_3\times S_3$ and $\rho=\sigma\otimes\sigma$ where $\sigma$ is the two-dimensional character of $S_3$. Over $N:=S_3\times 1$ this representation restricts to $\sigma+\sigma$. The element $g:=(c,c)$ where $c\in S_3$ satisfies $\chi_V(g)=1$. Corollary 2: If $G/N$ is supersolvable, $H$ exists with the desired properties. Being supersolvable means that there exists a normal series $N=M_0 < M_1 < \ldots < M_k = G$ such that every quotient is cyclic. We have just demonstrated that $\sigma$ can be extended to a (automatically irreducible) character $\widehat{\sigma}$ of $\langle g,N\rangle$ which is also a constituent of the restriction of $\rho$, no matter what $g\in G$ is. In particular it can be extended to $\sigma_1\in Irr(M_1)$. Let $\sigma_2 := Res_{M_1}^G(\rho) -\sigma_1$. If $\sigma_1\neq\sigma_2$, we are done. If not, we're back in the original situation: We have a normal subgroup $M_1\unlhd G$ such that $Res_{M_1} ^G(\rho)$ is the sum of two isomorphic $M_1$-modules so that we can proceed by induction. QED<|endoftext|> TITLE: Is there a universal bound for this ratio of expectations? QUESTION [6 upvotes]: Let $X$ and $Y$ be two zero-mean independent and identically distributed random variables. Is there a bound for the following ratio, $$\frac{\mathbb{E}[|X+Y|]}{\mathbb{E}[|X|+|Y|]}=\frac{\mathbb{E}[|X+Y|]}{2\mathbb{E}[|X|]} ,$$ where $\mathbb{E}$ and $|.|$ are the expectation and absolute value operations, respectively? REPLY [5 votes]: This is Jensen's inequality. Applied to the distribution of Y one gets $E|X+Y| \ge |X + E(Y)|$<|endoftext|> TITLE: What was Burroni's sketch for topological spaces? QUESTION [11 upvotes]: In a 1981 talk, René Guitart cites Albert Burroni as having given "A first interesting example of a mixed sketch...for the category of topological spaces" in 1970. This was apparently done in Burroni's Université de Paris thesis, and I don't see any evidence of it having ever been published more, well, publically. It seems that this must be a large sketch, since categories sketchable by a small sketch are accessible, and in any case topological spaces would be locally presentable if they were accessible, and thus sketchable by a limit sketch. Regarding large sketches, one has the large limit sketch of Spanier giving quasi-topological spaces. Burroni's thesis title includes the word "quasi-topologies," although it's not clear to me whether that is meant in Spanier's sense. Could Burroni have done something similar, adding some colimit cones to Spanier's topology on compact Hausdorff spaces to restrict to precisely ordinary spaces? In any case, does anyone know his construction, or another sketch for spaces, or a reference in which one is written down? The only comparable thing I can think of is (EDIT:) Barr's theorem on spaces as relational $\beta$-algebras, ($\beta$ being the ultrafilter monad) and I doubt there's going to be some theorem relating relational algebras and large-sketchable categories. REPLY [2 votes]: Perhaps Albert's thesis, Esquisses des catégories à limites et des quasi-topologies is still available. If you go to this Cahiers page and follow the link on the left hand page to ESQUISSES MATHEMATIQUES, it says Most of them are still available and can be obtained by sending an e-mail to the email address given there. Albert's thesis is no. 5 on the list.<|endoftext|> TITLE: Do Abelian varieties isomorphic to all their conjugates descend? QUESTION [5 upvotes]: Suppose $A$ is an abelian variety over $\overline{\mathbb{Q}}$ of dimension $g$, such that $A$ is isomorphic to all of its Galois conjugates. Note that I'm not including any polarization data. Can I conclude that $A$ descends to $\mathbb{Q}$, or at least to a field $K$ of bounded degree over $\mathbb{Q}$? Thanks! REPLY [3 votes]: 1.Theorem (Shimura 1971, "On the Field of Rationality for an Abelian Variety"): If $g$ = dim $A$ is odd and $Aut(A)=\{\pm1\}$ then $A$ has a model over its field of moduli. Proof: $\{1\} = H^1(\overline{\mathbb{Q}}^*) \rightarrow H^2(\pm1) \rightarrow H^2(\overline{\mathbb{Q}}^*)$ and $\{\pm1\}$ acts faithfully on $H^0(A_{\overline{\mathbb{Q}}}, \Omega^g) = \overline{\mathbb{Q}}$ since $g$ is odd, so the image of the obstruction in $H^2(Aut(H^0(A_{\overline{\mathbb{Q}}}, \Omega^g))) = H^2(\overline{\mathbb{Q}}^*)$ can be identified as the obstruction to the descent of a one-dimenstional $\overline{\mathbb{Q}}$ vector space to $\mathbb{Q}$, which is therefore trivial. 2.Of interest: Maus 1973, "On the Field of Moduli of an Abelian Variety with Complex Multiplication" Shimura 1982, "Models of an Abelian Variety with Complex Multiplication Multiplication over Small Fields" Shioda 1977, "Some Remarks on Abelian Varieties"<|endoftext|> TITLE: generalisation of umbilic surfaces QUESTION [10 upvotes]: It is well known that if you have a complete surface in $\mathbb{R}^3$ with umbilic points, that is to say $k_1=k_2$ everywhere, where $k_1$ and $k_2$ are the principal curvatures, that is to say the eigenvalues of the second fundamental form, then is a plane or a sphere. Is something known for $k_1-k_2=constant$? Of course the cylinder provides an example, but is there some example with $k_1$ non constant? Finally I wonder if it may be easier to have example in higher codimension? For instance for the case of surfaces in $\mathbb{R}^d$. REPLY [4 votes]: ..a small addition to Professor Bryant's answer For this class of surfaces, where $k1-k2 = constant$, we have introduced the name "Costant Skew Curvature Surfaces" (CSkC-surfaces) and we have studied an aspect concerning the Bonnet-surfaces. It is well know that the famous question that Bonnet asked was: "When does there exist an isometric embedding $x:M^2 \rightarrow R^3$ such that the mean curvature function of the immersion is $H$?" Our work was born from the question: Can a surface be CSkC and Bonnet at the same time, and, if that is the case, what does it represent? We showed that the CSkC-surfaces with principal curvatures ($k_1$ and $k_2$) nonconstant cannot contain any Bonnet-surfaces, so if and only if $k_1$ and $k_2$ are both constant the class of CSkC-surfaces can admit Bonnet-surfaces. This means that the only CSkC-surfaces for which exists a nontrivial isometric deformation preserving the mean curvature $H$ are (patch of) a circular cylinders. see (Alexander Pigazzini, Magdalena Toda): https://projecteuclid.org/euclid.jgsp/1518577293 Another aspect of the CSkC-surfaces is that if we setting $k_1-k_2=constant$ renders the shape equation for an elastic membrane equivalent to the Schrodinger equation for a particle on the same surface.. then the same equations have the same solutions... see (Victor Atanasov, Rossen Dandoloff): https://iopscience.iop.org/article/10.1088/0143-0807/38/1/015405/pdf By relating the two works we can for example say that: "if you want to look for an isometric deformation that preserve this relation (shape equation equivalent to the Schrodinger equation) then you can get the result only if the surface is (patch of) circular cylinder, and then search through its isometric deformations".<|endoftext|> TITLE: Sets in constructive mathematics QUESTION [5 upvotes]: It is not completely clear how Bridges, Richman and Youchuan treated sets in their paper. Example is in the following lemma (Lemma 7 on p. 7): Let $U$ and $V$ be (inhabited to mean $\exists u \in U, \exists v \in V$) sets of a Banach space such that $U \cup V$ is dense. Then, the following holds: If $u_0 \in U$ and $v_0 \in V$, then $\rho([u_0, v_0], \bar U \cap \bar V) = 0$ $\rho(x, \bar U \cap \bar V) = \rho(x, U) \wedge \rho(x, V)$ ... To do this, choose $w$ in $U \cup V$ within ... If a set $X$ is inhabited, then there is (at least) one point that can be constructed. How can other points be constructed? REPLY [4 votes]: Suppose we know that there exists $x \in A$ such that $\phi(x)$, and we want to prove $\psi$. Then the elimination rule for existential quantifiers allows us to argue as follows: We know that $\exists x \in A . \phi(x)$, and so we may assume to have $a \in A$ such that $\phi(a)$. [Insert argument using $a$ and the fact that $\phi(a)$ here.] Therefore $\psi$, as required. There is a technical condition, namely that $a$ must not appear in $\psi$. This is precisely how we always use knowledge that something exists. Many authors use the word "choose", as follows: We know that $\exists x \in A . \phi(x)$, and so choose $a \in A$ such that $\phi(a)$. [Insert argument using $a$ and the fact that $\phi(a)$ here.] Therefore $\psi$, as required. This has nothing to do with choice! It is still just elimination of existential quantifiers, but the word "choose" confuses many into thinking we're applying the axiom of choice. All of the above holds equally well classically and constructively. But people worry about constructive math, as if somehow there existence is more special, so let me address this as well. If we have the assumption $\exists x \in A . \phi(x)$ then we need not "construct" an element $a \in A$ such that $\phi(x)$. The assumption gives us some $a \in A$ such that $\phi(a)$. We do not know which $a$ it gives us, but it gives us one. We are thus allowed to use such an $a \in A$, keeping in mind that all we know about it is $\phi(a)$. Let us apply this to density. Suppose you know that $U \cup V$ is dense in $A$. The definition of density is: for every $\epsilon > 0$ and $x \in A$ there exists $y \in U \cup V$ such that $d(x,y) < \epsilon$. So, given $x_0 \in A$ and $\epsilon_0 > 0$, we may conclude that there exists $y \in U \cup V$ such that $d(x_0, y) < \epsilon_0$. Therefore, we may say: there is $a \in U \cup V$ such that $d(x_0, a) < \epsilon_0$. There is no need to "construct" $a$. The fact that $a$ is there is precisely the existential assumption!<|endoftext|> TITLE: Hirzebruch Surface F2 QUESTION [5 upvotes]: Can the Hirzebruch Surface $F_2:=\mathbb{P}(\mathcal{O}\oplus \mathcal{O}(2))$ be obtained by some GIT quotient of $\mathbb{P}^4$ (or $\mathbb{C}^4$)? REPLY [6 votes]: Depending on the interpretation of your question, the answer is Yes. In fact the Hirzebruch surface $\mathbb{F}_n =\mathbb{P}(\mathcal{O}\oplus \mathcal{O}(n))$ is the quotient of $X = \mathbb{A}^2 \setminus \{0\} \times \mathbb{A}^2 \setminus \{0\}$ with respect to the group action $$ \mathbb{G}_m^2 \times X \to X, \quad (\lambda,\mu) \cdot (s,t;x,y) = (\lambda s, \lambda t; \mu x, \lambda^{-n} \mu y).$$ From a highbrow perspective, such a realisation exists as $\mathbb{F}_n$ is toric. (Every toric variety is a quotient of an open subset of an affine space by the action of some multiplicative group, by the theory of Cox rings).<|endoftext|> TITLE: Critical values of L-functions and weights of Eisenstein Series QUESTION [7 upvotes]: I have been reading Serre's paper on p-adic modular forms and there seems to be a connection between critical values of L-functions and weights of Eisenstein series in the following sense: For the Riemann zeta function, the values of zeta functions at positive even integers $2k$ correspond to the Eisenstein series $$G_{2k} = 2\zeta(2k) + c_k\sum_{n\geq 1}\sigma_{2k-1}(n)q^n$$ and we see that the constant term is more or less the evaluation of $\zeta$ at the weight. For the Kubota-Leopoldt p-adic L function $L_p(s,\omega^{1-u})$ where $(s,u) \in X = \mathbb Z_{p}\times \mathbb Z/(p-1)$ and $\omega$ the Teichmüller character, Serre constructs an Eisenstein series $E_k$ for $k = (s,u)\in X$ such that $$E_{k} = 2L_p(s,\omega^{1-u}) + \sum_{n\geq 1}\sigma^*_{k-1}(n)q^n.$$ Once again, the value of the L- function is visible as the constant term. For $K$ a totally real number field of degree $r$, let $\zeta_K$ be it's Dedekind zeta function. For $m=2k$ an even integer, we can define a modular form of weight $k=rm$: $$f_m = 2^{-r}\zeta_K(1-m) + \sum_{\mathfrak a}\sum_{v\in \mathfrak d^{-1}\mathfrak a}(N\mathfrak a)^{m-1}q^{Tr(v)}$$ where $\mathfrak d$ is the discriminant and the sum is over integral ideals $\mathfrak a$. Yet again, one can see the value of the zeta function in the constant term. (I don't know this example very well so I might have made some mistakes here - please correct me). I only have these examples so far but is there a general conjecture that explains this pattern? Can one always associate an Eisenstein series of some form to special values of L functions? Is this a part of the Langland's conjectures/philosophy? Can one extend this to L-functions coming from modular forms or groups other than $GL_1$? REPLY [5 votes]: (I disclaim expertise on p-adic L-function and p-adic automorphic forms things, though I did make some earlier contributions to rationality statements and local archimedean computations that helped p-adic developments in recent years.) Although by now the catch-phrase "Langlands Philosophy" is often taken as an umbrella for almost anything having to do with automorphic forms, L-functions, and representation theory, in any case I think this is an insufficient explanation... especially as Langlands' conjectures and remarks came after many very interesting examples, and, further, the documented literal conjectures do not go quite so far as to explain every possible aspect of automorphic forms and such. For example, the third illustrative example in the question is (modulo typos and details) the restriction of a (holomorphic) Hilbert modular Eisenstein series along the diagonal to an elliptic modular form. This was used by H. Klingen in the early 1960s to prove the appropriate rationality properties of (abelian) L-functions of totally real number fields. (G. Shimura talked about this in a course in the mid-1970s, and it was very striking to me. My old book on Hilbert modular forms gives this example, too.) In that same vein, much of G. Shimura's work in the early 1970s, and later, proved special value results about various L-functions by using rationality properties of (holomorphic discrete series) automorphic forms. This line of thinking was involved in my own thesis under Shimura in the mid-70s, and affected my thinking subsequently: suitable pullbacks of holomorphic modular forms (Siegel, hermitian, ...?) are holomorphic, and have (numerical) Fourier coefficients in the same field as the original form. I. Satake's work on hermitian imbeddings of hermitian symmetric spaces in the mid-60s assured that these things work as best as could be hoped, even quite generally (but, sadly, not interacting much with exceptional groups E6 and E7...) Another thread with old antecedents is the 1939 Rankin-Selberg integral representation, which was immediately aimed at giving analytic results to approach Ramanujan's conjectures (best known estimates, I think, prior to P. Deligne's 1974 proof of the last bit of the Weil conjectures), but also re-used by Shimura c. 1973 to prove essentially the full range of special values for suitable Hecke L-functions (after work of Damerell by different methods). In the early 1980s, Piatetski-Shapiro and Rallis, myself, M. Harris, and a few others pursued the general idea (mentioned by P.-S. at the Budapest conference c. 1971) that certain configurations of subgroups could produce Euler products when various things (e.g., Eisenstein series) were integrated against each other. (My paper in the Shimura-conference volume, AMS 66, also on my web page, attempts to consider a sort of general case of this.) The Euler-product feature can easily occur without any special-value results. For that matter, apparently E. Hecke and H. Maass were aware that certain $GL(1,k)$ periods of $GL(2)$ Eisenstein series $E_s$ (with quadratic extension $k$) produced essentially $\zeta_k(s)/\zeta(2s)$, and such things. In the last 10-20 years, many people have demonstrated increasingly-sophisticated ways to evaluate periods of Eisenstein series (not just constant terms, as in Langlands-Shahidi) to produce Euler products. E. Lapid, O. Offen, G. Chinta, and many others (my apologies for not making a bigger list: lazy). H. Jacquet's relative trace formula sometimes helps in converting a not-quite-Euler-product situation to a visibly Euler-product situation. The most-authentic "Langlands conjecture" version of such things is that (literally) (... and via a suitable form of "the fundamental lemma", see Loeser, Ngo, et al) the most general automorphic L-function (whatever this means exactly) should be equal to a/the standard L-function attached to a cuspform on $GL(n)$. There is no immediate conjecture about special values. (Deligne's 1978 ICM conjectures incorporated things known at the time, mostly due to Shimura, but also with some prescient interpolation.) I don't think Langlands made conjectures about p-adic L-functions... but it is entirely believable to me that appealing additions to the original conjectures could be made that did refer to p-adic stuff. Given my limited appreciation of these things, I don't see that the long-ago conjectures really suggested so much about many recent developments, but those recent developments (Wiles, Wiles-Taylor, Taniyama-Shimura-etal...) are compatible with the vague general idea that many things are related to automorphic forms, and in non-obvious ways.<|endoftext|> TITLE: Homology of a limit of semidirect products QUESTION [7 upvotes]: Suppose I have two families of groups $A_k$ and $B_k$ indexed by the natural numbers and suppose $B_k$ acts on $A_k$. Suppose there are groups homomorphisms $A_{k+1} \rtimes B_{k+1} \to A_k \rtimes B_k$ extending maps $A_{k+1} \to A_k$ and $B_{k+1} \to B_k$. Also suppose that $\lim_k H_i(A_k)=0$ for $i>0$. Can I conclude that $\lim_k H_i(A_k \rtimes B_k) \cong \lim_k H_i(B_k)$? I am happy to make any finiteness assumptions about these homology groups and their limits. REPLY [6 votes]: No, because it is not even true for constant families: let $A$ be an acyclic group, so $H_i(A)=0$ for $i>0$, and $B$ be a group which $A$ acts on interestingly, e.g. $B= F(A)$ is the free group on the underlying set of $A$. Then $H_1(B \rtimes A) = \mathbb{Z}$ but $H_1(B) = \mathbb{Z}^A$.<|endoftext|> TITLE: Surreal number: trying to construct complete ordered fields QUESTION [9 upvotes]: Let $R$ be a subring of $\mathbf{No}$, the set of surreal number. We try to construct $\tilde{R}$, the Cauchy completion of $R$, just like the ordinary Cauchy completion for metric space. In the following we only consider the sequences in $R$ indexed by (i.e. with length equal) $\mathrm{Cf}(R)$, the cofinality of $R$. For any Cauchy sequences $(x_{\alpha})$ and $(y_{\alpha})$, we define an equivalent relation: $$x\sim y\;\;\mathrm{ iff }\;\; |x_{\alpha}-y_{\alpha}|\rightarrow0.$$ Let $\tilde{R}$ be the set of all equivalent classes of Cauchy sequences. On $\tilde{R}$ we define addition $[x]+[y]=[x+y]$ and multiplication $[x][y]=[xy]$ of classes. It is standard to check that these operations are well defined, $\tilde{R}$ becomes a ring and is Cauchy complete, and that $R$ is dense in $\tilde R$. For each ordinal number $\alpha$, denote $O_{\alpha}$ the set of surreal numbers with birthday $<\alpha$. It is known that if $\alpha=\omega^{\omega^{\beta}}$ for some ordinal $\beta$ then $O_{\alpha}$ is a ring, and if $\alpha=\epsilon_{\beta}$ for some ordinal $\beta$ then $O_{\alpha}$ is a field. It is easy to check that in the latter case, $\tilde{O_{\alpha}}$ is not only ring but also a field. The question is, in the case $\alpha=\omega^{\omega^{\beta}}$, is $\tilde{O_{\alpha}}$ actually a field? It is worth pointing out that if $\beta=0$ then $O_{\alpha}$ is the set of dyadic fraction and hence $\tilde{O_{\alpha}}=\mathbf R$, the set of reals, and is certainly a field. Apparently the difficult part is about the existence of multiplicative inverse. REPLY [12 votes]: In Fields of surreal numbers and exponentiation (Fund. Math. 167 (2001), pp. 173-188, doi:10.4064/fm167-2-3), Lou van den Dries and I show that $O_\alpha$ is an ordered field if and only if $\alpha$ is an epsilon number (see Corollary 4.9). Moreover, for epsilon $\alpha$, $O_\alpha$ is never Cauchy Complete in the familiar generalized sense you have in mind. On the other hand, for epsilon $\alpha$, $O_\alpha$ has a Cauchy completion consisting of $O_\alpha$ together with all the surreal numbers of tree rank $\alpha$ that fill the Dedekind gaps in $O_\alpha$ having breadth $0$ (where the breadth of a Dedekind cut $(X,Y)$ of an ordered abelian group $G$ is the largest convex subgroup $G'$ of $G$ for which $x+|g'|\in X$ for all $x\in X$ and all $g' \in G'$). Edit: Suppose $\alpha > \omega$. Then the Cauchy Completion of $O_\alpha$ is an ordered field if and only if $\alpha$ is an epsilon number. In particular, for the case you have in mind, consider the following. By Lemma 4.8 of the aforementioned paper, we have: If $\beta >1$ is not an epsilon number, then the tree rank of $ \omega^{-\beta} < \omega^{\beta}$. Accordingly, if $\beta >1$ is not an epsilon number, then $\omega^{-\beta}\in O_{\omega^{\beta}}$ but $\omega^{\beta}$ is not in $O_{\omega^{\beta}}$, so $O_{\omega^{\beta}}$ is not an ordered field. Moreover, since $\omega^{\beta}$ does not fill any gap in $O_{\omega^{\beta}}$ of breadth 0, $\omega^{\beta}$, which is the multiplicative inverse of $\omega^{-\beta}$, is not in the Cauchy completion of $O_{\omega^{\beta}}$, and hence the Cauchy completion of $O_{\omega^{\beta}}$ is not an ordered field.<|endoftext|> TITLE: What are the current breakthroughs of Geometric Complexity Theory? QUESTION [16 upvotes]: I've read from Wikipedia about Geometric Complexity Theory (GCT) which (if I understood correctly) is a program for coping with the $ P=NP $ problem using algebraic methods. That program seems promising, but I wonder what the current results of it are. So what are the current breakthroughs of Geometric Complexity Theory? REPLY [18 votes]: Let $X=(X_{ij})_{1\le i,j\le n}$ be a generic $n\times n$ matrix and $F_1(X)={\rm det}(X)$ the degree $n$ homogeneous polynomial given by the determinant. Let $$ F_2(X)=(X_{nn})^{n-m}\times {\rm perm}\left[(X_{ij})_{1\le i,j\le m}\right] $$ which takes the permanent of an $m\times m$ submatrix and multiplies by one's favorite linear form in order to make another homogeneous polynomial of degree $n$ (one could also use the entry $X_{11}$ instead of $X_{nn}$). This modification is called padding. Then define the number $$ c(m)=\min\{\ n\ |\ n\ge m\ \ {\rm and}\ \ \overline{G\cdot F_2}\subset \overline{G\cdot F_1}\ \} $$ where $G$ is $GL(n^2)$ acting on the space $V$ of dimension $n^2$ where $X$ lives and therefore also on the space of degree $n$ polynomial functions of $X$. The $\overline{G\cdot F_i}$ are Zariski closures of orbits. The big conjecture in the area or Valiant's Hypothesis (a complex version of ${\rm P}\neq{\rm NP}$) is that $c(m)$ grows faster than any polynomial in $m$. Now if $\overline{G\cdot F_2}\subset \overline{G\cdot F_1}$, then one has a surjective $G$-equivariant map $$ \mathbb{C}[\overline{G\cdot F_1}]_d\longrightarrow \mathbb{C}[\overline{G\cdot F_2}]_d $$ between degree $d$ parts of the coordinate rings of these orbit closures. So the game is to try to show that this does not happen, for $n$ insufficiently large relative to $m$, by proving the existence of a multiplicity obstruction, i.e., an irreducible representation $\lambda$ for which multiplicities satisfy $$ {\rm mult}_{\lambda}(\mathbb{C}[\overline{G\cdot F_1}]_d)<{\rm mult}_{\lambda}(\mathbb{C}[\overline{G\cdot F_2}]_d) $$ or at the level of ideals $$ {\rm mult}_{\lambda}(I[\overline{G\cdot F_1}]_d)>{\rm mult}_{\lambda}(I[\overline{G\cdot F_2}]_d)\ . $$ An optimistic approach is to try to show there exist occurrence obstructions, i.e., $\lambda$'s such that ${\rm mult}_{\lambda}(\mathbb{C}[\overline{G\cdot F_1}]_d)=0$ and ${\rm mult}_{\lambda}(\mathbb{C}[\overline{G\cdot F_2}]_d)>0$. This hope has been squashed in the work of Bürgisser, Ikenmeyer and Panova mentioned by Timothy. However, the possibility of multiplicity obstructions is still open. I think the approach by Mulmuley is to try prove the existence of such multiplicity obstructions by leveraging all the tools available from representation theory for the computation of these multiplicities. Personally, I have never been a fan of this approach. Having studied 19th century invariant theory in some depth, it seems more natural to me to approach the orbit separation problem using the explicit tools from that era. This article by Grochow seems to also point in a similar direction (I suspect the third article mentioned by Joseph is in the same vein). In classical language (see Turnbull or Littlewood), one has to explicitly construct a mixed concomitant which vanishes on $F_1$ but not on $F_2$. One also has to do this infinitely often (in $m$) in order to establish the superpolynomial growth property. Such a (degree $d$) concomitant is the same as a specific $G$-equivariant map from your favorite model for the irreducible representation $\lambda$ to ${\rm Sym}^d({\rm Sym}^n(V))$ (Grochow calls that a separating module). Invariant theorists from the 19th century had two methods for generating such objects: elimination theory and diagrammatic algebra. A very baby example where $F_1$ and $F_2$ are binary quartic forms under the action of $G=SL(2)$ (see this MO question) is say $$ F_1(x,y)=x^4+8x^3y+24x^2y^2+32xy^3+16y^4 $$ and $$ F_2(x,y)=16x^4-24x^3y+12x^2y^2-2xy^3\ . $$ A separating concomitant (here in fact a covariant) is the Hessian of a generic binary quartic $F$ $$ H(F)(x,y)=\frac{\partial^2 F}{\partial x^2}\frac{\partial^2 F}{\partial y^2}-\left( \frac{\partial^2 F}{\partial x\partial y} \right)^2\ . $$ It vanishes (identically in $x,y$) for $F=F_1$ but not for $F=F_2$. In this case, the Hessian can be seen as an equivariant map form the irreducible given by the second symmetric power (of the fundamental two-dimensional representation) into the coordinate ring for the affine space of binary quartics. So a possible superoptimistic "plan" for GCT involves the following sequence of steps. Find a way to generate tons of concomitants. Identify some explicit candidates for the vanishing on $F_1$ and prove that property. Show they don't vanish on $F_2$. Step 1) is in principle solved by the First Fundamental Theorem for $GL(n^2)$ but there is a mismatch: the determinant is a natural object in the invariant theory for $GL(n)\times GL(n)$ (acting on rows and columns) rather than $GL(n^2)$. One could try to repair the mismatch by expressing the basic building block for the invariant theory of $GL(n^2)$ in terms of the one for $GL(n)\times GL(n)$ (see this MO question for a similar reduction problem from $SL(n(n+1)/2)$ to $SL(n)$). Guessing the right candidates for Step 2) looks hard to me. Knowing beforehand that some multiplicities ${\rm mult}_{\lambda}(I[\overline{G\cdot F_1}]_d)$ are nonzero would definitely help. Although, one could procrastinate and defer the proof of nonidentical vanishing of the concomitant to Step 3) which should show more than that anyway. If one has such right candidates, showing they vanish on $F_1$ may be easy by arguments one could call Pauli's exclusion principle (contracting symmetrizations with antisymmetrizations), high chromatic number property, or simply "lack of space". However, I think the most difficult part is Step 3). For example, in my paper "16,051 formulas for Ottaviani's invariant of cubic threefolds" with Ikenmeyer and Royle, the guessing was done by computer search, but with the right candidate in hand, the vanishing on $F_1$ was relatively easy to explain (it's a rather pretty example of chromatic number due to the global properties of the graph rather than some big clique). The analogue of Step 3) in our article was done by brute force computer calculation and we still don't have a clue for why it is true. The paradigmatic problem related to Step 3) is the Alon-Tarsi conjecture (see this MO question and this one too). In my opinion, one needs to make progress on that kind of question (the Four Color Theorem is of this type too, via a reduction due to Kauffman and Bar-Natan) before Valiant's Conjecture. Since the question is about breakthroughs in GCT. I think this article by Landsberg and Ressayre also deserves some attention since it suggests that a reasonable guess for the exact value of $c(m)$ is $$ \left(\begin{array}{c}2m\\ m \end{array}\right)-1\ . $$ Note that a proof of concept for the explicit "Step 1),2),3) approach", on a much simpler problem, was given by Bürgisser and Ikenmeyer in this article. Finally, for more information on GCT, I highly recommend the review "Geometric complexity theory: an introduction for geometers" by Landsberg. PS: I should add that my pessimism is specific to the Valiant Hypothesis which is the "Riemann Hypothesis" in the field. Of course, one should not throw the baby with the bath water and denigrate GCT because it so far failed to prove this conjecture. There are plenty of more approachable problems in this area where progress has been made and more progress is expected. See in particular the above-mentioned article by Grochow and review by Landsberg.<|endoftext|> TITLE: How can I calculate the period matrix of this Riemann surface? QUESTION [13 upvotes]: I am attempting to calculate the period matrix of the Riemann surface associated to the zero set of a complex curve: $y^3 = x^4 -1$. Background: It is my understanding that the period matrix of a Riemann surface is defined as $\int_{\gamma_i} \omega_j$, where $\omega_1, ..., \omega_g$ are a choice of basis of the invariant differentials of the surface, and $\gamma_1, ..., \gamma_{2g}$ are a choice of basis for $H_1(S)$. I have calculated an invariant differential basis as $dx/y, dx/y^2, xdx/y^2$. To compute the period matrix, I am struggling mightily to actually write down the line integrals necessary. Q1. How can I write down these line integrals? How can I view the generators of $H_1(S)$ as living on $\mathbb{C}$? Toward coordinatizing $S$, I have a presentation of $S$ as 3 sheeted cover of $S^2$. However, naively gluing these octagons together gives me no insight on how to visualize the $\gamma_i$ as loops on these octagons, much less write the necessary associated line integral. In attempting to understand how to write $\gamma_i$ as line integrals on these octagons, it seems that I must understand how to tile a 3-torus with octagons (since every gluing of my octagons I seem to try gives me a horribly embedded 3-torus). So, I am attempting to mimic Klein's tiling of $x^3y + y^3z + z^3x = 0$ by heptagons in order to get a tiling of a 3-torus by octagons. In other words, finding a part of the {8, 3} hyperbolic plane tiling which glues together to give a 3-torus. In conclusion, I restate the same question: Q2. How can I understand the loops that form a basis of $H_1(S)$ as loops on the {8, 3} tiling? What part of the {8, 3} tiling forms a 3-holed torus? REPLY [6 votes]: You can gain infinite enlightenment by reading the very cool paper: Gianni, Patrizia; Seppälä, Mika; Silhol, Robert; Trager, Barry, Riemann surfaces, plane algebraic curves and their period matrices, J. Symb. Comput. 26, No.6, 789-803 (1998). ZBL0964.14047. (note that a lot of the paper is devoted to producing the equation, but you already have that).<|endoftext|> TITLE: Projective bundle QUESTION [8 upvotes]: Let X be a variety. Suppose $\mathcal{E_1}$ and $\mathcal{E_2}$ are two vector bundles on X. Is there an example such that $\mathbb{P}(\mathcal{E_1})$ and $\mathbb{P}(\mathcal{E_2})$ are isomorphic as varieties but not as $\mathbb{P}^n$-bundles over X? REPLY [8 votes]: Here is one method of constructing many such examples. Claim. Let $(X,H)$ be a projective variety, let $\phi \colon X \stackrel\sim\to X$ be an automorphism, and let $\mathscr L$ be a line bundle on $X$ with $P_H(\mathscr L,n) \neq P_H(\mathcal O_X,n)$ such that $\phi^* \mathscr L \not \cong \mathscr L$ and $\phi^* \mathscr L \not \cong \mathscr L^{-1}$. Define $\mathscr E_1 = \mathcal O_X \oplus \mathscr L$ and $\mathscr E_2 = \mathcal O_X \oplus \phi^* \mathscr L$. Then $\mathbb P(\mathscr E_1) \cong \mathbb P(\mathscr E_2)$ as varieties, but not as $\mathbb P^1$-bundles over $X$. Example. Let $X = \mathbb P^1 \times \mathbb P^1$, $\phi$ the coordinate swap, and $\mathscr L = \mathcal O(1,0)$. Example. Let $X = E$ be an elliptic curve, $\phi = [-1]$ the inversion, $P$ a point that is not $2$-torsion, and $\mathscr L = \mathcal O_X(P)$. Proof of claim. Clearly $\phi$ can be enhanced to a commutative diagram (in fact, a pullback square) $$\begin{array}{ccc}\mathbb P(\mathcal O_X \oplus \phi^* \mathscr L) & \to & \mathbb P(\mathcal O_X \oplus \mathscr L)\\ \downarrow & & \downarrow \\ X & \stackrel \phi \to & X\end{array}$$ whose horizontal arrows are isomorphisms. This shows that $\mathbb P(\mathscr E_1) \cong \mathbb P(\mathscr E_2)$ as varieties. Suppose that $\mathbb P(\mathscr E_1) \cong \mathbb P(\mathscr E_2)$ as $\mathbb P^1$-bundles over $X$ (equivalently, as varieties over $X$). The long exact sequence $$\ldots \to H^1_{\operatorname{\acute et}}(X,\mathbb G_m) \to H^1_{\operatorname{\acute et}}(X,\operatorname{GL}_2) \to H^1_{\operatorname{\acute et}}(X,\operatorname{PGL}_2) \to \ldots$$ shows that two vector bundles give the same $\operatorname{PGL}_2$-bundle (equivalently, the same projective bundle) if and only if they differ by a line bundle. Hence, there exists a line bundle $\mathscr M$ such that $$\mathcal O_X \oplus \mathscr L \cong \mathscr M \otimes \left(\mathcal O_X \oplus \phi^* \mathscr L\right) = \mathscr M \oplus (\mathscr M \otimes \phi^* \mathscr L).$$ By assumption, the Hilbert polynomials of $\mathscr L$ and $\mathscr O_X$ are not the same, so the factors in the decomposition $\mathcal O_X \oplus \mathscr L$ have different Gieseker slopes. Uniqueness of the Harder–Narasimhan filtration implies that $\mathscr M = \mathcal O_X$ and $\mathscr M \otimes \phi^* \mathscr L = \mathscr L$, or $\mathscr M = \mathscr L$ and $\mathscr M \otimes \phi^* \mathscr L = \mathscr O_X$. Both contradict our hypotheses. $\square$ I haven't thought about what conditions give $\mathscr L_1 \oplus \mathscr L_2 \cong \mathscr L_3 \oplus \mathscr L_4 \Rightarrow (\mathscr L_1, \mathscr L_2) = (\mathscr L_3, \mathscr L_4)$. My current proof uses stability, but there may well be an easier argument, perhaps dropping the hypothesis on Hilbert polynomials.<|endoftext|> TITLE: Universal homeomorphism of stacks and etale sites QUESTION [9 upvotes]: A morphism between schemes is a universal homeomorphism if it is integral, surjective, universally injective. For morphism between algebraic stacks, this notion also make sense. It is well know that a universal homeomorphism between schemes induces: (1) homeomorphism between their underlying topological spaces; and (2) equivalence between their etale sites. I have seen the proof of these statements some time ago but didn't really understand well. So first a somewhat vague question: Question 1: Is there a good conceptual reason why for schemes, homeomorphism between underlying topological spaces should imply equivalence between etale sites? For me, at least psychologically, the underlying topological space and the etale site of a scheme seems quite unrelated. So I would prefer an intuitive explanation of this (seemingly?) coincidence. Now for universal homeomorphism of stacks, one can also define its underlying topological space and its etale site and as far as I know, (1) is still true. It seems (2) may not be true, but something weaker is true, i.e. it induces equivalence of etale topoi, see for example Remark 4.26 of Behrend's book here. In particular, if $\mathcal{X}$ is a (reasonable) Deligne-Mumford stack, the Keel Mori theorem says that it has a coarse moduli space $X$ and the natural morphism $\mathcal{X}\to X$ is a proper universal homeomorphism. (see the second paragraph in Conrad's paper). Thus the etale topos of $\mathcal{X}$ is isomorphic to the etale topos of $X$. I'm trying to understand this isomorphism in the following special case: Let a finite group $G$ acts on a ring $A$. Then my understanding is that the stack quotient $[(Spec A)/G]$ has a coarse moduli space $Spec(A^G)$, i.e. the GIT quotient. Then the etale topos of $[(Spec A)/G]$ is isomorphic to that of $Spec(A^G)$. In particular, when $A$ is an algebraically closed field with trivial $G$ action, then I reached a suspicious conclusion that the etale topos of $BG$ is isomorphic to etale topos of a point. Question 2: Is this correct? If yes, what is the key (commutative algebra) ingredient behind this isomorphism of topos in this special case? Otherwise any comments on where I made mistakes are welcome. REPLY [4 votes]: (1) is by definition. The standard proof of (2) is via descent of étale morphisms along universal submersions (see SGA1, Exp IX, Thm 4.10, or http://stacks.math.columbia.edu/tag/04DY, or my paper "Submersions and effective descent of étale morphisms", Thm 5.21). The point is that given a universal homeomorphism $f\colon X'\to X$ and an étale morphism $E'\to X'$, it comes equipped with a unique descent datum along $f$ because the diagonal of $f$ is a nil-immersion. Since $f$ is a universal submersion, the descent datum is effective: $E'\to X'$ is the pull-back of an étale morphism $E\to X$. I believe that one could prove that we have an equivalences of étale topoi without this result. Namely, it is enough to prove that the unit and counit of the adjunction $(f^*,f_*)$ are natural isomorphisms which is easy using the description of $f_*$ of a finite (or integral) morphism. This uses the description of finite morphisms over strictly henselian local rings. Concerning Question 1, the site-equivalence is more subtle than the topoi-equivalence, essentially due to the difference between schemes and algebraic spaces. The topoi-equivalence is more conceptual but does depend on the properties of finite morphisms over henselian rings as we will see. The equivalence of étale sites/topoi fails as you have stated it for algebraic spaces and stacks. It is true for a separated homeomorphism of algebraic spaces and hence for representable and separated homeomorphisms of algebraic stacks. It is false for a non-separated homeomorphism of algebraic spaces (e.g., see my paper cited above). It is also false for non-representable separated universal homeomorphisms in general. This highlights that the argument on finite/integral morphisms in the topoi-argument is crucial and hence perhaps not completely "intuitive". Concerning Question 2, Behrend has a non-standard, stricter, definition of universal homeomorphism (Def 4.18). In that definition, it is not sufficient that $f\colon X\to Y$ is a homeomorphism after arbitrary base change. The definition says that there should be a representable separated universal homeomorphism $X'\to X$ such that $X'\to X\to Y$ is a representable separated universal homeomorphism (Behrend says "schematic" but this is equivalent to separated in this context). Example: In Behrend's notation: $BG\to *$ is a universal homeomorphism if $G$ is an infinitesimal group (that is, $G\to *$ is a homeomorphism, e.g., $G=\mu_p$ or $G=\alpha_p$ in characteristic $p$) but not if $G$ is étale (e.g., $G=\mathbb{Z}/n\mathbb{Z}$). Since representable separated universal homeomorphism give rise to equivalences of étale topoi, it is obvious that Behrend's definition does as well. When it comes to a coarse moduli map $\pi\colon \mathcal{X}\to X$, it is a proper (hence separated) universal homeomorphism, but not representable, and it typically does not give rise to an equivalence of étale sites or topoi (it does for $BG\mu_p\to *$). The correct statement is the following: Theorem: There is an equivalence of stabilizer-preserving étale morphisms $\mathcal{E}\to \mathcal{X}$ and étale morphisms $E\to X$ (this also holds for étale morphisms that are representable by algebraic spaces, hence also for étale sheaves). Here stabilizer-preserving can be taken to mean pointwise stabilizer-preserving: for every $e\colon \operatorname{Spec} k\to \mathcal{E}$ with image $x\colon \operatorname{Spec} k\to \mathcal{X}$, the homomorphism $\mathrm{stab}(e)\to \mathrm{stab}(x)$ is an isomorphism of $k$-groups. You can find a proof (for quasi-separated étale morphisms) in my paper "Existence and properties of geometric quotients", Thm 6.12 (via 6.6-6.7).<|endoftext|> TITLE: Independent vector fields $X,Y$ on $S^3$ with $[X,Y]=Y$ QUESTION [7 upvotes]: Are there two smooth independent vector fields $X,Y$ on $S^3$ with $[X,Y]=Y$? REPLY [4 votes]: You can also use this result which appears in the following paper of E. Ghys see corollary I.1.2: Let $G$ be a Lie group homeomorphic to $\mathbb{R}^n$ which acts locally freely on a $n+1$-dimensional manifold $M$. Then the universal cover of $M$ is diffeomorphic to $\mathbb{R}^{n+1}$. The vector field $X,Y$ define an action of $Aff(\mathbb{R})$ on $S^3$, but the universal cover of $S^3$ is $S^3$ and not $\mathbb{R}^3$. http://perso.ens-lyon.fr/ghys/articles/actionslocalement.pdf<|endoftext|> TITLE: Continuous right inverse to the Laplacian operator on $C^\infty$ QUESTION [8 upvotes]: For each $f\in C^\infty(\mathbb R^n)$, there exists $u\in C^\infty(\mathbb R^n)$ such that $\Delta u=f$. This, I guess, has been known well before the more general Malgrange-Ehrenpreis theorem that says the same for a general differential operator with constant coefficients. My question is twofold: 1°) Who proved this first? 2°) Does there exist a continuous operator $S:C^\infty(\mathbb R^n)\to C^\infty(\mathbb R^n)$ such that $\Delta Sf\equiv f$ ? REPLY [5 votes]: Assuming Question 2 asks about a continuous linear right inverse, the answer is no. In section 3, Theorem 3.3 of the article "Some results on continuous linear maps between Fréchet spaces" by Dietmar Vogt it is proved that no hypoelliptic operator on $C^{\infty}(\Omega)$ admits a continuous linear right inverse for any open $\Omega \subset \mathbb{R}^n$ with $n>1$.<|endoftext|> TITLE: Tangent space of the space of smooth sections of a bundle QUESTION [7 upvotes]: Let $E\to M$ be a real vector bundle of finite rank over a closed differentiable manifold $M$. Let $C^{\infty}(E)$ denote the space of smooth sections of $E$ and let $e\in C^{\infty}(E)$ be a section. I often see statements of the type: "The tangent space to $C^{\infty}(E)$ at any given section $e\in C^{\infty}(E)$ is isomorphic to $C^{\infty}(E)$ itself, namely $T_{e}C^{\infty}(E)\simeq C^{\infty}(E)$." I wonder what is the precise formulation of the statement above. I know that if we only take smooth sections, $C^{\infty}(E)$ admits the structure of an infinite dimensional Frechet manifold with respect to the appropriate topology. The statement should be then that the isomorphism $T_{e}C^{\infty}(E)\simeq C^{\infty}(E)$ holds understanding $C^{\infty}(E)$ as a Frechet manifold and taking $T_{e}C^{\infty}(E)$ to be the tangent space of $C^{\infty}(E)$ as a Frechet manifold? Or rather, should it be understood by assuming that we have implicitly Sobolev-completed $C^{\infty}(E)$ into $H_{s}(E)$ using some appropriate Sobolev norm as to make $H_{s}(E)$ into a smooth Hilbert manifold and then the isomorphism that actually holds is $T_{e}H_{s}(E)\simeq H_{s}(E)$? My second question is: can one make sense of an isomorphism of the type $T_{e}C^{\infty}(E)\simeq C^{\infty}(E)$ if the base is non-compact? And lastly, let $Q\to M$ be a smooth fiber bundle over a closed manifold $M$ with typical fiber $F$ given by a smooth manifold. How should be understood the tangent space at a point of the space of sections of $Q$? References are welcome. REPLY [10 votes]: $C^\infty(E)$ is a Frechet VECTOR space. Thus its tangent space at each point equals $C^\infty(E)$ via its affine structure. Added: This is also true if $M$ is not compact. However, for a fiber bundle $Q\to M$ one has to be more careful with the topology (if $M$ is not compact). See 10.10 of Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980) (pdf) for an answer. In principle, the tangent space is the space of sections of the vertical bundle of $Q$ restricted to the the image of a section. There is also Sections 42, 43, .. of Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997 (pdf) where a more easily handable notion of differentiability is developped and then used. These methods also work for Sobolev spaces of sections. See MR3135704 Reviewed Inci, H.; Kappeler, T.; Topalov, P. On the regularity of the composition of diffeomorphisms. (English summary) Mem. Amer. Math. Soc. 226 (2013), no. 1062, vi+60 pp. for the basics of these.<|endoftext|> TITLE: What is the Turing degree associated with an ultrafilter $U$? QUESTION [8 upvotes]: I asked Turing degree of a turing machine with access to an (arbitrary) nonstandard integer, not thinking about the possiblity that this could depend on the model used. The question was not formulated very well, but I did not want to delete it since I got some good answers. Now I will ask the question I originally had in mind. Let $U$ be a ultrafilter on $\mathbb N$ and $f$ a partial computable function of two arguments, which can either accept or reject its input (or not halt) (I'm using a finite range so that the output is guaranteed to be standard, which should make things slightly simpler). We say that $f$ accepts $n$ with help from $U$ if $\{i \in \mathbb N : f(n,i)\text{ accepts}\} \in U$, and rejects or does not halt likewise (due to the definition of the ultrafilter, exactly one of these possibilities will be true). This can be seen as defining running $f$ in a model defined by a ultrapower of $\mathbb N$ with input $n$ (which is standard) and $H$=$(0,1,2,\dots)$. My question is, what is the turing degree associated with such functions, for a given ultrafilter $U$. If the ultrafitler $U$ is principal, then its turing degree will just be $0$, so we will mostly consider nonprincipal ultrafilters. For a nonprincipal ultrafilter $U$ (we will assume all ultrafilters are nonprincipal from this point on), we can construct a function $f(n,i)$ which interprets $n$ as a standard turing machine, and runs this machine for $i$ steps. If the machine halts, it will after a standard number of steps, and so $f$ will accept. If the machine has not halted after $i$ steps, it never will, and so $f$ will reject. $f$ has solved the halting problem for standard turing machines, and so the Turing degree of $U$ is at least $0'$. (Note that if the machine associated with $n$ is one that looks for a contradiction in, say, $ZFC$, $f$ is still guaranteed to give the right answer (according to whether or not $ZFC$ is consistient), even if the length of shortest contradiction is nonstandard in whatever metatheory we are using. That's because the ultraproduct model layers another layer of "nonstandardness" over what the metatheory already has.) We can also note that $U$'s turing degree must be at least $0'$, since it can be used to model an eventually correct machine. On the other hand, for any set $S \subseteq \mathbb N$, we can construct an ultrafilter $U$ such that, with its help, we can compute $S$. Choose a computable encoding of finite sets of natural numbers.. Now, let $a$ be the set of encodings of subsets of $S$ of the form $S_{ TITLE: How many commutative local algebras are there? QUESTION [5 upvotes]: Is there a computer programm that can list all finite dimensional commutative local (nonsemisimple) k-algebras $K[x_1,x_2,...,x_n]/I$ over a finite field k (of order $q$) with $J^l=0$ for a given $q, n$ and $l$ , when $J=\langle x_1,...,x_n\rangle$ denotes the Jacobson radical and $I \subseteq J^2$? One way might be as follows: Let $A$ be the algebra $K[x_1,x_2,...,x_n]/J^l$, then one just has to find all ideals $I$ (submodules) of $A$ and look at the quotients. Is there an estimate how many there are (up to isomorphism or not)? How and in what time can a computer check if two such algebras are isomorphic? REPLY [5 votes]: The paper https://arxiv.org/pdf/1201.3529.pdf shows there are an astronomical number of commutative semigroups satisfying the product of any 3 elements is 0. The contracted semigroup algebra (the zero of the semigroup is identified with the zero of the algebra) is then an algebra of your sort. So there is no way to enumerate them all. They also show there is still an astronomical number of these semigroups up to isomorphism. I would be very surprised if two such semigroups can have isomorphic algebras without being isomorphic since $J^3=0$ doesn't give you much room but I have not tried to prove this.<|endoftext|> TITLE: References on the obstacle problem for the heat equation QUESTION [5 upvotes]: Can you point out some references that deal with the obstacle problem for the heat equation? $$(OP) \quad\begin{cases} \max\{\Delta u -\partial_t u, \varphi - u \} = 0 & \text{ in } (0,T)\times \mathbb{R}^n \\ u(0,\cdot) = \varphi(0,\cdot) & \text{ in } \mathbb{R}^n \end{cases}.$$ Works on elliptic obstacle problems appear to be much easier to find (see Wikipedia, for instance). Since a bounty has been offered for this question, I'll write down what I feel is missing in the current (very nice) answer and that I'd like to see: complete argument for the existence (with references too) further details on the representation of solutions using the heat kernel references about numerical analysis of the problem (and Matlab/Mathematica codes) references on physical motivations for the problem. REPLY [4 votes]: I will get you started, but there are lots of blanks to fill. We are interested in the PDE \begin{align*} \min\left\{ -u_{t}+\Delta u,u-\varphi\right\} & =0 & \text{in }(0,T]\times\mathbb{R}^{n}\\ u(0,\cdot)-\varphi(0,\cdot) & =0 & \text{in }\mathbb{R}^{n} \end{align*} where $\varphi$ is a smooth function of polynomial growth. Existence Let $u^{0}$ be a classical solution of \begin{align*} -u_{t}^{0}+\Delta u^{0} & =0 & \text{in }(0,T]\times\mathbb{R}^{n}\\ u^{0}(0,\cdot)-\varphi(0,\cdot) & =0 & \text{in }\mathbb{R}^{n} \end{align*} that is unique in an appropriately picked space of functions of polynomial growth. Define inductively $u^{k}$ as a classical solution of \begin{align*} -u_{t}^{k}+\Delta u^{k}+k\min\left\{ u^{k-1}-\varphi,0\right\} & =0 & \text{in }(0,T]\times\mathbb{R}^{n}\\ u^{k}(0,\cdot)-\varphi(0,\cdot) & =0 & \text{in }\mathbb{R}^{n} \end{align*} that is unique in an appropriately picked space of functions of polynomial growth. Now you would have to show that you can pass to limits ($k\rightarrow\infty$) to obtain a solution of the original PDE. The limiting solution is not, in general, twice differentiable in space. But, you should probably be able to establish sufficient conditions for it to be a once differentiable viscosity solution. This technique is called a penalty method and is at least useful for establishing existence. Uniqueness Uniqueness can be handled in the space of viscosity solutions using a comparison principle argument. I wrote about this in an expository post, but I'm sure this is available in different forms elsewhere. The setting of the post proves uniqueness in the space of bounded functions, but you should be able to generalize the arguments. Heat Kernel You can use smooth pasting to write the solution with the heat kernel, but you will probably not be able to find a "nice expression" for the free boundary, as this is believed to be hard. References Van Moerbeke, Pierre. "On optimal stopping and free boundary problems." Archive for Rational Mechanics and Analysis 60.2 (1976): 101-148. Pham, Huyên. "Optimal stopping, free boundary, and American option in a jump-diffusion model." Applied mathematics & optimization 35.2 (1997): 145-164.<|endoftext|> TITLE: Maximal Cisinski model structure on simplicial sets QUESTION [7 upvotes]: This is a very simple question coming from the observation that every (pre)sheaf category has the maximal Cisinski model structure on it. This is the Cisinski model structure with the smallest class of weak equivalences possible. Now, it is natural to ask: what is the maximal Cisinski model structure on the most canonical category in this setting, namely the category of simplicial sets? Is it larger than the Joyal model structure? Assuming that the answer is "yes", can we give an explicit description of its weak equivalences and fibrant objects? REPLY [7 votes]: (Edit: after 2.5 years, I've finally typed up the details as a paper! https://arxiv.org/abs/2201.13400 ) I think I've worked out a relatively nice description of the fibrant objects. I won't be able to include the full proofs here, but I will give the description and explain a bit about why it works. (I'm planning to include the details in a longer note, to which I'll post a link here once I make it available, but that will probably take at least a couple more months.) Let me first recap some of Andrea's answer and the comments following it: the fibrant objects in the minimal model structure are those with lifts against $\textbf{An}_{\mathfrak{L}}$. The class $\textbf{An}_{\mathfrak{L}}$ is generated by the pushout-products of the boundary inclusions $\partial\Delta[n]\hookrightarrow\Delta[n]$ with the maps $\{\varepsilon\}\hookrightarrow L$ for $\varepsilon=0,1$, where $L$ is the subobject classifier in simplicial sets. The first thing to notice is that we can actually use $J$, the nerve of the groupoid with two objects and an isomorphism between them, instead of $L$. One way to see why is to work out what $L$ is as a simplicial set, and see that $J$ is actually a retract of $L$. I talked to Alex Campbell, however, and he told me that if you trace through Cisinski's arguments in his book, you can actually see that everything he says about $L$ would apply to a separating cylinder which is trivially fibrant, such as $J$. So, at any rate, now we know the fibrant objects are those with lifts against $\textbf{An}_{J}$, which is generated by the pushout-products of the boundary inclusions $\partial\Delta[n]\hookrightarrow\Delta[n]$ with the map $\{0\}\hookrightarrow J$. With this description already, there is some nice intuition: having lifts against the pushout-product $(\partial\Delta[n]\times J) \cup (\Delta[n]\times \{0\})\hookrightarrow \Delta[n]\times J$ tells you that any time you have an $n$-simplex and an "isomorphism of boundaries" which starts at the boundary of that $n$-simplex, then it extends to an "isomorphism of $n$-simplices" which starts at that $n$-simplex. The description so far, which is not new, was not satisfying to me. I wanted something which felt more like the horn extension definition of Kan complexes and quasi-categories, so I played around a bit and found the following: Define an "isoplex," denoted $\mathfrak{D}_i[n]$, to be the nerve of the category $c_0\rightarrow c_1 \rightarrow \ldots c_{i-1} \rightarrow c_i \leftrightarrow c_{i+1} \rightarrow c_{i+2} \rightarrow \ldots \rightarrow c_n$, i.e., where the morphism $c_i\rightarrow c_{i+1}$ is an isomorphism. Think of this as analogous to $\Delta[n]$, which is the nerve of the category $c_0\rightarrow c_1 \rightarrow \ldots \rightarrow c_n$. In particular, we can define faces of $\mathfrak{D}_i[n]$ in a similar manner, where $d_j\mathfrak{D}_i[n]$ is the maximal subcomplex not containing the $j$ vertex. Then we can define the iso-horn of $\mathfrak{D}_i[n]$, denoted $\mathbb{V}_i[n]$, to be the union of all of its faces except the $i$th face. Let $\text{IsoHorn}$ be the set of all iso-horn inclusions $\mathbb{V}_i[n]\hookrightarrow \mathfrak{D}_i[n]$. It turns out that $\text{IsoHorn}$ generates the class $\textbf{An}_J$. In particular, the fibrant objects in the minimal model structure are the simplicial sets with lifts against $\text{IsoHorn}$. That $\overline{\text{IsoHorn}}\subseteq \textbf{An}_J$ is not so bad to check, because each iso-horn extension is a retract of a generator of $\textbf{An}_J$. The other direction follows because each of the generators of $\textbf{An}_J$ can be built out of iso-horn extensions (by transfinite composition of pushouts), which I think is pretty intuitive, but which did take a couple pages of combinatorics for me to check carefully. As for the minimal model structure being different from Joyal's, in addition to Andrea's properness argument, we can also just see directly that the horn $\Lambda^1[2]$ is fibrant in the minimal model structure. But not only is the minimal model structure different from Joyal's, I'm confident there will be multiple interesting Cisinski model structures between them. In particular, my thesis project is to find a Cisinski model structure where the fibrant objects model up-to-homotopy versions of 2-Segal sets (which is why I was thinking about the minimal model structure in the first place).<|endoftext|> TITLE: What is this sequence? QUESTION [16 upvotes]: This is again a question that I asked at Stack Exchange, but got no answer so far, so I am trying here. Let: $$ a_n=\sum_{k\ge0}(k+1) {n+2\brack k+2}(n+2)^kB_k$$ $B_k$ is the Bernoulli number. ${n\brack {k}}\;$ is the unsigned Stirling number of first kind, $\left( {0\brack {0}}\;=1 \text{ and }{{n}\brack {k}}\;=(n-1){{n-1}\brack {k}}\;+{{n-1}\brack {k-1}}\;\right)$. From $n=0$, the first terms are: $ \ \ 1\ ,\ 0\ ,\ -5\ ,\ 0\ ,\ 238\ ,\ 0\ ,\ -51508\ ,\ 0\ ,\ 35028576\ , ..$ The $a_n$ are all integers, and the odd-indexed $a_{2n+1}$ vanish. A generating function should be even, but I could not find it. Also, any possible combinatorial interpretation (when removing the sign)? I would welcome any help or indication on this. Thank you in advance. EDIT(01/09/18): Actually, this is a particular case of: $$ a_{n,h}=\sum_{k\ge0}{k+h-1\choose k} {n\brack h+k}n^k B_k$$ Here is a table for $a_{n,h}$, for $1\le n,h \le 9$. \begin{matrix} n&|&a_{n,1}&a_{n,2}&a_{n,3}&a_{n,4}&a_{n,5}&a_{n,6}&a_{n,7}&a_{n,8}&a_{n,9}\\ -&&---&---&---&---&---&---&---&---&---\\ 1&|&1&0&0&0&0&0&0&0&0\\ 2&|&0&1&0&0&0&0&0&0&0\\ 3&|&-1&0&1&0&0&0&0&0&0\\ 4&|&0&-5&0&1&0&0&0&0&0\\ 5&|&24&0&-15&0&1&0&0&0&0\\ 6&|&0&238&0&-35&0&1&0&0&0\\ 7&|&-3396&0&1281&0&-70&0&1&0&0\\ 8&|&0&-51508&0&4977&0&-126&0&1&0\\ 9&|&1706112&0&-408700&0&15645&0&-210&0&1\\ \end{matrix} I know now how to show that $a_{n,h}$ is integer, and that it is zero when $n-h$ is odd. But the proofs that I have found are quite lengthy, technical, and not really enlightening about the mathematical signification of these numbers. Any idea? REPLY [5 votes]: I've got that $a_{n,h}$ is the coefficient of $x^{n-1}$ in \begin{split} &(-1)^{n+h+1} \frac{n!}{2\cdot (h-1)!}\frac{\log(1+x)^{h}\left(\coth(-\frac{n}2\log(1+x))-1\right)}{1+x} \\ =\ &(-1)^{n+h} \frac{n!}{(h-1)!}\frac{\log(1+x)^{h}}{(1-(1+x)^{-n})(1+x)}. \end{split} It's yet unclear if the dependency on $n$ can be eliminated to turn this expression into a generating function.<|endoftext|> TITLE: For how many residues $x \pmod{p}$ is $x^{-1} \in [1, (p-1)/2]$? QUESTION [9 upvotes]: Given a prime $p$, let $N_p$ denote the number of $x \in [1, (p-1)/2]$ such that $$x^{-1} \pmod{p} \in [1, (p-1)/2]$$ What is the limiting distribution of $$S_n := \left \{\frac{N_p}{(p-1)/2}\, : \, 1\le \text{prime} \, p \le n \right\}$$ as $n\to\infty$? Numerical evidence suggests this quantity is distributed like a normal random variable with mean $1/2$ and standard deviation given by $f(n)$, where the graph of $f$ on $3\le n \le 300$ looks like the below plot: REPLY [13 votes]: Let me try to explain why this problem is hard (concurring with Alexey Ustinov here). Fourier analysis shows that $$ N_p = \frac{1}{p^2} \sum_{m=0}^{p-1} \sum_{n=0}^{p-1} \left( \sum_{x=1}^{p-1} e_p( nx + mx^{-1} )\right) \left(\sum_{x=1}^{(p-1)/2} e_p(-nx) \right) \left(\sum_{x=1}^{(p-1)/2} e_p(-mx) \right) $$ where $ \left( \sum_{x=1}^{p-1} e_p( nx + mx^{-1} \right)$ is a Kloosterman sum $K(mn;c)$ and $ \left(\sum_{x=1}^{(p-1)/2} e_p(-nx) \right) \left(\sum_{x=1}^{(p-1)/2} e_p(-mx) \right)$ is large for $n$, $m$ close to zero and small elsewhere. The term $n=0, m=0$ gives the main term, where $n=0$ and $m \neq 0$ or vice versa gives a small and easily computable term, and the interesting error term comes from the range where $n$ and $m$ are nonzero. This can be represented as a linear combination of Kloosterman sums $K(mn;p)$, with the largest contributions coming from $mn = \pm 1$. We expect the sums $K(a;p)/\sqrt{p} $ for any $a$ to be equidistributed as random variables with a Sato-Tate (semicircular) distribution. It's reasonable to also conjecture independence over different $a$, which would enable calculating the distribution of any finite sum of Kloosterman sums. (In fact this follows from a conjecture I made in my PhD thesis). This is an infinite sum of Kloosterman sums, but it is probably reasonable to make a similar independence statement, which should give a Gaussian distribution or one very similar to it. However, almost nothing is known about the distribution of $K(a;p)$ for fixed $a$ and varying $p$. (The case of fixed $p$ and varying $a$ is well-known by work of Deligne and Katz). Even the first moment is not known, although we do have a bound for the average of $K(a;c)$ over all numbers $c$, proved using the Kusnetsov formula. This sum of Kloosterman sums looks much harder to analyze than the distribution of a single Kloosterman sum, so it is probably also too hard for current techniques.<|endoftext|> TITLE: The $32$-deg polynomial for the tetrahedron inscribed in the icosahedron? QUESTION [7 upvotes]: This MO answer discusses this table involving the maximal side lengths of the five Platonic solids $T,C,O,D,I$ inscribed in the other solids, This table is also found in Moritz Firsching's paper. I noticed that almost all are roots of equations with solvable Galois groups. For example, $d = 0.162631\dots$ is a root of, $$4096d^{16} - 3701760d^{14} + 809622720d^{12} - 17054118000d^{10} + 79233311025d^8 - 94166084250d^6 + 31024053000d^4 - 3236760000d^2 + 65610000=0$$ The discriminant of $F(\sqrt{d})=0$ is a perfect square and Magma says the above has a solvable Galois group. The exception seems to be $t=1.3474429\dots$ (the tetrahedron in icosahedron case) which is a root of, $$5041 t^{32} - 1318386 t^{30} + 60348584 t^{28} - 924552262 t^{26} + 5246771058 t^{24}-15736320636 t^{22} + 29448527368 t^{20} - 37805732980 t^{18} + 35173457839 t^{16}-24298372458 t^{14} + 12495147544 t^{12} - 4717349124 t^{10} + 1256858478 t^8 - 217962112 t^6 + 21904868 t^4 - 1536272 t^2 + 160801 = 0$$ The discriminant of $F(\sqrt{t})=0$ is not a perfect square (it is divisible by the seemingly random prime $466369383062945371$), and Magma says $F(\sqrt{t})$ has group $16T1952$, order $2^{15}\cdot3^4\cdot5^2\cdot7^2$, hence is not solvable. Q: Is there an a priori reason why the T in C case is not a radical, unique among the other cases? (Or is there something wrong with the polynomial?) REPLY [10 votes]: I guess it would be difficult to prove that the answero your question is "no", since proving that "no a priori reason exists" might be hard. More modestly, I can say that I don't really know a good reason. However, here are some pointers: One essential fact is that four of the five platonic solids are centrally symmetric; all but the tetrahedron. Croft observed: in an optimal inclusion two concentric polytopes will share a center. Therefore, an optimal inclusion for two platonic solids, both of which are not the tetrahedron will result in a concentric situations. This makes these configurations considerably easier. This (at least hopefully somewhat) explains that the degree of the algebraic numbers in the 12 cases not involving the tetrahedron. Here is a table with the exact values: Let's look at the cases involving the tetrahedron, that is the first row and the first columns of the table above. The pairs $(T\subset C)$ and $(O\subset T)$ are again concentric, and so are the pairs $(T\subset D)$ and $(I\subset T)$. The case $(T\subset O)$ is particulary simple to describe, since $T$ and $O$ share a face in the optimal solution. What remains are really just $(C\subset T)$, $(D\subset C)$ and $(T\subset I)$. And indeed those are the one with solutions that are more algebraically involved. (The case $(C\subset T)$ not so much, since in the optimal configuration edges align and the constraints to begin with are "only" coming from $T$ and $C$ and not, as in the other two cases from the more complicated $D$ and $I$.) to summarize: There are really only two cases where one might expect more involved solutions, one of which is radical, the other not. To address the second part of the question: "Or is there something wrong with the polynomial?": I hope not. Let me give you a few more details, how one can calculate the polynomial in question. We assume we know the incidences, i.e. what vertex of the tetrahedron lies on what face of the icosahedron and deduce the optimal solution from here. The incidences are shown in the following picture: one vertex of T coincides with a vertex of I, one lies in an edges, the two remaining in two seperate faces of I. I fix an icosahedron with edge-length $1$ and vertices $$\begin{align} v_{0} & = \left(0,\,\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4}\right) \\ v_{1} & = \left(0,\,-\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4}\right) \\ v_{2} & = \left(\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0\right) \\ v_{3} & = \left(\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0\right) \\ v_{4} & = \left(\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0,\,\frac{1}{2}\right) \\ v_{5} & = \left(\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0,\,-\frac{1}{2}\right) \\ v_{6} & = \left(-\frac{1}{2},\,\frac{1}{4} \sqrt{5} + \frac{1}{4},\,0\right) \\ v_{7} & = \left(-\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0\right) \\ v_{8} & = \left(-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0,\,\frac{1}{2}\right) \\ v_{9} & = \left(0,\,\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4}\right) \\ v_{10} & = \left(0,\,-\frac{1}{2},\,-\frac{1}{4} \sqrt{5} - \frac{1}{4}\right) \\ v_{11} & = \left(-\frac{1}{4} \sqrt{5} - \frac{1}{4},\,0,\,-\frac{1}{2}\right) \\ \end{align}$$ The four points of the tetrahedron are then given as follows: $$\begin{align}p_0 =& v_{11}\\p_1 =& f_0v_0 + f_1v_2 + (1-f_1-f_0)v_6 \\ p_2 =& g_0v_1 + g_1v_3 + (1-g_0-g_1)v_7 \\p_3 =& e_0v_5 + (1-e_0)v_{10}\end{align}$$ For some positive variables $f_0, f_1, g_0, g_1$ and $e_0$. For each $(i,j)\in\binom{[4]}{2}$, we have $(p_i-p_j)^2 = s^2$ for some real variable $s$. Putting it all together, we obtain the following system of 6 equations with 6 variables: $$\begin{align} 0 =& f_{0}^{2} + f_{0} f_{1} + f_{1}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} f_{0} + \frac{1}{2} \, \sqrt{5} f_{1} - \frac{1}{2} \, f_{0} - \frac{1}{2} \, f_{1} + 1 \\ 0 =& g_{0}^{2} + g_{0} g_{1} + g_{1}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} g_{0} + \frac{1}{2} \, \sqrt{5} g_{1} - \frac{1}{2} \, g_{0} - \frac{1}{2} \, g_{1} + 1 \\ 0 =& e_{0}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} e_{0} - \frac{1}{2} \, e_{0} + 1 \\ 0 =& -\frac{1}{2} \, \sqrt{5} f_{0} g_{0} + f_{0}^{2} + f_{0} f_{1} + f_{1}^{2} - \frac{1}{2} \, f_{0} g_{0} - f_{1} g_{0} + g_{0}^{2} - f_{0} g_{1} - 2 \, f_{1} g_{1} + g_{0} g_{1} + g_{1}^{2} - s^{2} - f_{0} - g_{0} + \frac{1}{2} \, \sqrt{5} + \frac{3}{2} \\ 0 =& -\frac{1}{2} \, \sqrt{5} e_{0} f_{1} + e_{0}^{2} - e_{0} f_{0} + f_{0}^{2} - \frac{1}{2} \, e_{0} f_{1} + f_{0} f_{1} + f_{1}^{2} - s^{2} - e_{0} - f_{1} + \frac{1}{2} \, \sqrt{5} + \frac{3}{2} \\ 0 =& -\frac{1}{2} \, \sqrt{5} e_{0} g_{0} - \frac{1}{2} \, \sqrt{5} e_{0} g_{1} + e_{0}^{2} - \frac{1}{2} \, e_{0} g_{0} + g_{0}^{2} - \frac{1}{2} \, e_{0} g_{1} + g_{0} g_{1} + g_{1}^{2} - s^{2} + \frac{1}{2} \, \sqrt{5} e_{0} + \frac{1}{2} \, \sqrt{5} g_{0} - \frac{1}{2} \, e_{0} - \frac{1}{2} \, g_{0} - g_{1} + 1 \\ \end{align}$$ This systems happens to have the solution $$\begin{align} f0=& 0.356785524577257 \text{..., zero of } 5751x^{16} + 54216x^{15} - 434286x^{14} - 466374x^{13} + 9452306x^{12} - 19323022x^{11} - 33022460x^{10} + 206565938x^{9} - 384738484x^{8} + 362774804x^{7} - 180708354x^{6} + 47907122x^{5} - 9497814x^{4} + 61518x^{3} + 895260x^{2} - 144318x + 401 \\ f1=& 0.352452740635196 \text{..., zero of } 5751x^{16} + 24732x^{15} - 16650x^{14} - 342012x^{13} - 901126x^{12} - 959712x^{11} + 743040x^{10} + 4747162x^{9} + 2877417x^{8} - 2704036x^{7} + 2767104x^{6} + 2220422x^{5} - 977942x^{4} - 2032764x^{3} - 504834x^{2} + 243738x + 64251 \\ g0=& 0.595049283356260 \text{..., zero of } 5751x^{16} + 46926x^{15} - 623124x^{14} + 3139674x^{13} - 7786156x^{12} + 2057864x^{11} + 49657348x^{10} - 189659290x^{9} + 400533046x^{8} - 535383644x^{7} + 467013828x^{6} - 298170008x^{5} + 167384576x^{4} - 74703516x^{3} + 17034188x^{2} - 698296x + 164 \\ g1=& 0.0729071548475811 \text{..., zero of } 5751x^{16} - 46224x^{15} - 194904x^{14} + 1789782x^{13} - 3522208x^{12} - 3388962x^{11} + 50328778x^{10} - 43454770x^{9} - 123966325x^{8} + 205675176x^{7} - 80109508x^{6} - 35043358x^{5} + 51387632x^{4} - 18740400x^{3} + 1053588x^{2} + 9072x - 369 \\ e0=& 0.645495309223693 \text{..., zero of } 71x^{16} + 972x^{15} - 2730x^{14} - 19898x^{13} + 64290x^{12} - 61466x^{11} + 46202x^{10} - 95276x^{9} + 136499x^{8} - 92310x^{7} + 51560x^{6} - 48088x^{5} + 35854x^{4} - 11920x^{3} - 2804x^{2} + 2406x - 41 \\ s=& 1.34744285033120 \text{..., zero of } 5041x^{32} - 1318386x^{30} + 60348584x^{28} - 924552262x^{26} + 5246771058x^{24} - 15736320636x^{22} + 29448527368x^{20} - 37805732980x^{18} + 35173457839x^{16} - 24298372458x^{14} + 12495147544x^{12} - 4717349124x^{10} + 1256858478x^{8} - 217962112x^{6} + 21904868x^{4} - 1536272x^{2} + 160801 \\ \end{align}$$ While the solution was found by using Newton's method combined with integer relations algorithms, the solution can be checked using exact calculations in $\mathbb{A}$ or in a smaller numberfield, which contains all the relevant numbers. Indeed one can take the number field $F$ with defining polynomial any of the ones defining the solutions (substituting s by sqrt(s)). They are all isomorphic and have discriminant $10637699079912558734361600000000 = 2^{16} \cdot 3^{4} \cdot 5^{8} \cdot 11 \cdot 466369383062945371$, divisible by the seemingly random prime.<|endoftext|> TITLE: Hitchin fibration and Springer resolution QUESTION [11 upvotes]: Let C be a curve and let us assume $G=GL_N$ and $\mathfrak{g}=\mathfrak{gl}_N$ for simplicity. The moduli space $\mathcal{M}_H(C,G)$ of $G$-Higgs bundles admits the Hitchin fibration $\pi: \mathcal{M}_H(C,G) \to \mathcal{A}=\bigoplus_{i=1}^{N} H^0 (C,K_C^i)$ where $K_C$ is the canonical bundle of $C$. The preimage $\pi^{-1}(0)$ of zero under the Hitchin fibration is called global nilpotent cone. The Hitchin fibration is a completely integrable system and a generic fiber is the Jacobian of the corresponding spectral curve of $C$ which is a Lagrangian complex tori. However, the global nilpotent cone is generally a singular fiber. The first question is whether there exist generally loci (not only zero) in $\mathcal{A}$ on which Hitchin fibers are singular. If so, in these loci, does the spectral curve of $C$ degenerate into a nodal curve and is the singular fiber a compactified Jacobian? Is there any good reference on geometry of singular fibers of the Hitchin map $\pi$? The second question: why is $\pi^{-1}(0)$ called the global nilpotent cone? Is there any relation to the nilpotent cone $\mathcal{N}$ which is the subset of nilpotent elements of $\mathfrak{g}$? Instead it looks to me that a Springer fiber under the Springer resolution $\mu:T^*(G/B)\to \mathcal{N}$ is similar to the global nilpotent cone. (Note that $B$ is a Borel subgroup and $G/B$ is the complete flag variety.) However, the Springer resolution is not a completely integrable system. The third question: how can we connect the Hitchin fibration to the Springer resolution? REPLY [5 votes]: I'll try to answer the second and third questions. My preferred way to organize this circle of ideas is to think of the following ladder of theories : Representation theory of $\mathfrak{g}$ (or) the group $G$ Representation theory of affine Lie algebra $\hat{\mathfrak{g}}$ (or) the loop group $\hat{G}$ "Global" Representation theory of Hitchin systems $M_H(G,C)$ There is non-trivial interaction (and inclusions) between all three of them. The usual nilpotent cone in $\mathfrak{g}$ plays a prominent role in the Rep theory of $\mathfrak{g}$. I'll define it in a way that makes the analogy with Global/Hitchin case obvious. Recall the adjoint quotient map (in $GL_n$, this is the map to the coefficients of the Charachteristic polynomial) $h : \mathfrak{g} \rightarrow \mathfrak{h} /W $ where $\mathfrak{h}$ is the Cartan subalgebra and $W$ is the Weyl group. Now, inverse image of the adjoint quotient map over zero, $h^{-1}(0)$ is the nilpotent cone, $\mathcal{N}_\mathfrak{g} = h^{-1}(0)$. The fully honest way to say this involves schemes/stack theoretic inverse image since $\mathfrak{h}/W$ and (the Nilcone) have many strata. So, the $"0"$ is not just a point. But, the basic picture should be clear even in the absence of the more careful terminology. Now, in the Hitchin case, the Hitchin map $\mu : \mathcal{M}_H \rightarrow \mathcal{B} $ where $\mathcal{M}_H$ is the total space of the Hitchin system and $\mathcal{B}$ is the Hitchin Base. The Hitchin base is nothing but the globalized version of the charachteristic polynomial of the Higgs field. So, one can think of the Hitchin map as a global version of the adjoint quotient map $h$. Fibers of the Hitchin map are interesting for various reasons. The most interesting (and difficult to study) fiber is, arguably, the one over $"0"$, $\mathcal{N}_{global}= \mu^{-1}(0)$. By analogy with the adjoint quotient case, this is called the Global Nilpotent Cone. On the relation to Springer Theory, To get an exact connection, my understanding is that you have to simplify to special cases of the Hitchin system. For $C$ being a cuspidal elliptic curve, D. Nadler has shown how to obtain Springer Theory from the Hitchin Fibration. More precisely, he relates some natural A-branes of the Hitchin System to certain perverse sheaves appearing in Springer Theory. This case is simpler than the general situation because $Bun_G$ is a lot simpler in this case. In the more general case, one could, with good reason, view a study of the Hitchin fibration as defining a Global version of Springer Theory (as is done, for example, in the works of Z. Yun).<|endoftext|> TITLE: What is the total space of a stack after all? QUESTION [10 upvotes]: From my general experience I think for myself of what follows as some kind of taboo question for some reason: in my imagination, everybody wants an answer to this but somehow thinks it shall not be asked. OK among many ways to present a stack, I choose this one: we are given a Grothendieck topos $\mathbf X$ represented by sheaves of sets on a site $(\mathbb C,J)$, and then we have "a large category $\mathscr C$ in the world of $\mathbf X$", that is, a presheaf $\mathbb C^{\mathrm{op}}\to\text{Categories}$ satisfying the (three-step) glueing conditions w. r. t. $J$. In this question, by stacks are meant such $\mathscr C$'s. More precisely, we speak about stacks on $\mathbf X$, or stacks on $(\mathbb C,J)$. Given that, we may consider the notion of "presheaf of $\mathbf X$-world sets on $\mathscr C$ in the $\mathbf X$-world". Again, there are several ways to define this, for example, as another gadget $\mathscr E$ of the same kind as $\mathscr C$ together with a "functor in the $\mathbf X$-world" $\mathscr E\to\mathscr C$ which is a discrete fibration. All such discrete fibrations form a category which I will denote by $\operatorname{Sets}(\mathbf X)^{\mathscr C^{\mathrm{op}}}$, since, I believe, it can be appropriately described as the category of contravariant functors, in the $\mathbf X$-world, from $\mathscr C$ to $\operatorname{Sets}(\mathbf X)$, the latter being yet another gadget of the same kind as $\mathscr C$, with the "underlying" $\mathbb C^{\mathrm{op}}\to\text{Categories}$ sending $c\in\mathbb C$ to $\mathbf X/a(h_c)$ (slice over the associated sheaf of $h_c:=\hom_{\mathbb C}(-,c)$). The question now is simply this: under what conditions does it happen that there is another Grothendieck topos $\mathbf Y$ such that the category $\operatorname{Sets}(\mathbf X)^{\mathscr C^{\mathrm{op}}}$ is equivalent to $\mathbf Y$? Remarks I am primarily interested in the case when $\mathscr C$ is the associated stack of an internal category of $\mathbf X$. I believe in this case several things simplify. Since $\mathscr C$ is in general not small (i. e. not the externalization, in a known way, of an internal category of $\mathbf X$), there is in general no well-defined geometric morphism $\operatorname{Sets}(\mathbf X)^{\mathscr C^{\mathrm{op}}}\to\mathbf X$, but even if there is no such morphism, I believe it is still natural to call $\mathbf Y$, when it exists, the total space of the stack $\mathscr C$. Whereas if there is such a geometric morphism, it still might be different from the one with inverse image "constant presheaf" and direct image "$\varprojlim_{\mathscr C}$". Or it does coincide but is not bounded. Or further, although not bounded, is $\textit{locally}$ bounded. Hence subquestion: can such things happen? There is a variation which might be needed to have more natural examples - when $\mathscr C$ comes naturally equipped with its own "$\mathbf X$-world Grothendieck topology" which one cannot ignore, i. e. one has to consider the $\mathbf X$-world $\textit{sheaves}$ rather than $\operatorname{Sets}(\mathbf X)^{\mathscr C^{\mathrm{op}}}$ to obtain something sensible. Finally, the natural reverse questions are - which geometric morphisms $f:\mathbf Y\to\mathbf X$ are of this form? For those which are - what, if any, additional data on $f$ enable to recover the stack $\mathscr C$? REPLY [3 votes]: Suppose $\mathscr{C}$ is the stackification of an internal category $C$ in $\mathbf{X}$. In this case, since $\mathrm{Sets}(\mathbf{X})$ is a stack, morphisms of stacks $\mathscr{C}^{\mathrm{op}} \to \mathrm{Sets}(\mathbf{X})$ are equivalent (by the universal property of stackification) to morphisms $C^{\mathrm{op}} \to \mathrm{Sets}(\mathbf{X})$, which in turn are equivalent to $\mathbf{X}$-internal discrete fibrations over $C$. (Probably I am using here the fact that by the "comparison lemma", stacks over $\mathbb{C}$ are equivalent to stacks over $\mathbf{X}$ with its canonical topology.) The category of such is the "$\mathbf{X}$-indexed functor category" $[C^{\mathrm{op}},\mathbf{X}]$, which is a Grothendieck topos equipped with a bounded geometric morphism to $\mathbf{X}$; see sections B2.3 and B3.2 of Sketches of an Elephant. Since $[C^{\mathrm{op}},\mathbf{X}]$ is the free cocompletion of $C$ in the $\mathbf{X}$-world, by internalizing the usual arguments it determines $C$ up to "Morita equivalence", i.e. equivalence in the bicategory of $\mathbf{X}$-internal profunctors. This can equivalently be stated as "up to internal weak Cauchy completion" in $\mathbf{X}$, i.e. the equivalence relation generated by internal functors that are "fully faithful" and "surjective up to splitting idempotents" in the internal language of $\mathbf{X}$. Upon passage to associated stacks, internally "fully faithful and essentially surjective" functors get inverted, so this becomes up to "Cauchy completion in the world of stacks", i.e. simultaneously splitting idempotents and stackifying. If we additionally consider sheaves for an internal Grothendieck topology on $C$, we obtain another Grothendieck topos $\mathrm{Sh}_{\mathbf{X}}(C)$ that also comes with a bounded geometric morphism to $\mathbf{X}$, and the internal version of Giraud's theorem says that every bounded geometric morphism to $\mathbf{X}$ is of this form; see sections B3.3 and C2.4 of Sketches of an Elephant.<|endoftext|> TITLE: Manifold of probability measures: connections between two types of metrics QUESTION [37 upvotes]: The space of probability measures could be viewed as an infinite-dimensional manifold, equipped with two possible types of metrics — (1) Wasserstein and (2) Fisher-Rao. Metric (1) is connected with optimal transport, and metric (2) is connected with information geometry. Question: What are the connections between these two metrics? We know that the Fisher-Rao metric is characterized by the Fisher information matrix, but what is the corresponding characterization for the Wasserstein metric? Any references are much appreciated. REPLY [7 votes]: Just a quick follow-up: Very recently (well, actually in 2015) three teams came up independently and almost simultaneously with the same construction of a new "optimal-transport-like" distance on the space of Radon measures $\mathcal M^+$, which somehow interpolates continuously between Wasserstein and Fisher-Rao. This distance now goes by the name of Wasserstein-Fisher-Rao (WFR) metrics, sometimes also Hellinger-Kantorovich (HK) distance, and gave rise to a whole new topics generally referred to as unbalanced optimal transport. As pointed out by @GabeK in his excellent answer the Wasserstein and Fisher-Rao structures interact in an interesting way and lead to unsuspected behaviour (well, at least for me). The underlying structure possesses several rich and geometric underlying formulations. In a nutshell, the WFR distance can be seen as an infimal-convolution of Wasserstein and Fisher-Rao distances. For recent developments see the citations of the original trhee papers below (I belong to the first team) [1] Kondratyev, S., Monsaingeon, L., & Vorotnikov, D. (2016). A new optimal transport distance on the space of finite Radon measures. Advances in Differential Equations, 21(11/12), 1117-1164. [2] Chizat, L., Peyré, G., Schmitzer, B., & Vialard, F. X. (2018). An interpolating distance between optimal transport and Fisher–Rao metrics. Foundations of Computational Mathematics, 18(1), 1-44. [3] Liero, M., Mielke, A., & Savaré, G. (2018). Optimal entropy-transport problems and a new Hellinger–Kantorovich distance between positive measures. Inventiones mathematicae, 211(3), 969-1117.<|endoftext|> TITLE: Do varieties without rational curves contain sub-polynomially many rational points? QUESTION [8 upvotes]: Suppose $V$ is a projective algebraic variety over $\mathbb{Q}$ which does not contain any rational curves. Is it true/provable that the number of rational points on $V$ of Weil height at most $T$ grows slower than $T^{\delta}$ for any $\delta>0$? I'm not sure what is implied by all the variations of Lang's conjectures out there, so clarifications on this would also be much appreciated. Thanks!! REPLY [2 votes]: I found it in an a paper of Mckinnon: https://arxiv.org/pdf/1011.5825.pdf Apparently this is known as the "rational curve conjecture" and Mckinon almost proves that it follows from Vojta's conjectures. Specifically, assuming Vojta's conjecture he proves that any variety $X$ of non-negative Kodaira-dimension has an open subset $U\subset X$ containing sub-polynomially many points. If one then assumes that all varieties with negatvie Kodaira dimensio are ruled( apparently a conjecture of the minimal model program) one almost gets the result (you get rational curves, but perhaps not defined over $\mathbb{Q}$.<|endoftext|> TITLE: Steinberg representation for sporadic simple groups? QUESTION [13 upvotes]: The Steinberg representation is a remarkable irreducible representation of a reductive algebraic group over a finite field or local field, or a group with a BN-pair. It is analogous to the 1-dimensional sign representation ε of a Coxeter or Weyl group that takes all reflections to –1. The Steinberg representation is the Alvis–Curtis dual of the trivial 1-dimensional representation. Question: What are the analogs of the Steinberg representation for sporadic simple groups ? (If there are any ideas for other finite groups beyond standard - it also welcome). Wikipedia suggests: Some of the sporadic simple groups act as doubly transitive permutation groups so have a BN-pair for which one can define a Steinberg representation, but for most of the sporadic groups there is no known analogue of it. However googling does not lead me to an answer even about that "some of sporadic ... " . Further question (cohomological construction) : At Mathoverflow D.Pasechnik answering S.Lentner question: (weak?) BN-Pair / Tits System for Sporadic Groups ? writes: Instead of starting from a weak BN-pair, one can weaken Tits' axioms from his "Local approach to buildings" to develop a theory dealing with sporadics. That may give a way to answer my question since it is known that Steinberg representation is realized in the cohomology group of the Bruhat–Tits building. Question 2 Are there similar cohomological constructions of "Steinberg" representation for sporadic groups ? Further question (mod-p reduction): In the remarkable (must&pleasure to read - imho) survey "The Steinberg representation" BAMS 1987 J. E. Humphreys takes modular representation point view on the Steinberg representation. The point is: for G(F_p) "St" can be reduced mod "p". Moreover the reduction is quite remarkble - it preserves irreducibility and some other properties (quote: it is also a "principal indecomposable" representation determining a "block" by itself. Moreover, the character of the representation vanishes at all elements of G having order divisible by p) - that follows from Brauer-Nesbit theory. Question 3 If there any mod-p properties of "St" for sporadic ? The positive answer would have a strange conclusion - that some sporadics might be considered as kind of groups defined over "F_p" for some "p". Motivation (one of): As described here: MO271067 I would hope to have some bijection(s) between irreducible representations of sporadic groups (and other groups also) and their conjugacy classes. Distinguishing some representations as "Steinberg like" would be quite helpful. For example for Mathiew group M11 looking on the character table (e.g. here page 3 or Magma): Class | 1 2 3 4 5 6 7 8 9 10 Size | 1 165 440 990 1584 1320 990 990 720 720 Order | 1 2 3 4 5 6 8 8 11 11 X.1 + 1 1 1 1 1 1 1 1 1 1 X.2 + 10 2 1 2 0 -1 0 0 -1 -1 X.3 0 10 -2 1 0 0 1 Z1 -Z1 -1 -1 X.4 0 10 -2 1 0 0 1 -Z1 Z1 -1 -1 X.5 + 11 3 2 -1 1 0 -1 -1 0 0 X.6 0 16 0 -2 0 1 0 0 0 Z2 Z2#2 X.7 0 16 0 -2 0 1 0 0 0 Z2#2 Z2 X.8 + 44 4 -1 0 -1 1 0 0 0 0 X.9 + 45 -3 0 1 0 0 -1 -1 1 1 X.10 + 55 -1 1 -1 0 -1 1 1 0 0 Z1 = i * sqrt(2) Z2 = (-1+i * sqrt(11))/2 One can easily guess to map pair of conjugacy classes of order 8 to pair of irreps X3,X4 and classes of order 11 to X.6 X.7 because 1) that are only 4 complex classes/irreps - that distiguishes 4 classes and 4 irreps 2) orders 8, 11 of classes correspond to degree of rationality of those irreps - thus we get: two classes of order 8 <-> two irreps X3,X4; two classes of order 11 <-> two irreps X6,X7; Bonus Question What irrep is Steinberg for M11 and what conjugacy class might correspond to it ? More generally one may ask is there an analog of Alvis-Curtis duality for sporadics ? Is there splitting of classes and characters to semisimple/unipotent for sporadics ? Analogs of parabolic subgroups ? Etc... (i.e. can one extend properties from Lie groups to sporadics ? ) EDIT: From the comments it becomes clear that there is lot of research generalizing the Steinberg representation. Still it is not clear for the direction suggested in Wikipedia - from double transitive action to BN-pair , and hence to Steinberg reprsentation. REPLY [15 votes]: The approach that I have taken to generalizing the Steinberg module to finite groups other than groups of Lie type is that, in general, the object we should consider is a chain complex, rather than just a module. For every finite group G and for every prime p there is a finite chain complex of projective modules, canonically defined up to isomorphism, that in the case of a group with a split (B,N) pair in characteristic p, is the Steinberg module, shifted in degree so that it appears in degree equal to the dimension of the building. I like to call this canonically defined complex the `Steinberg complex' of G at the prime p, and it exists for every finite group and every prime. The Steinberg complex is shown to exist and is defined for the first time (without calling it by that name) in P.J. Webb, A split exact sequence of Mackey functors, Commentarii Math. Helv. 66 (1991), 34-69, doi: 10.1007/BF02566635. It arises from Theorem 2.7.1 there, applied to Brown's simplicial complex of non-identity p-subgroups of G, or to any complex G-homotopy equivalent to this complex. This includes Quillen's complex of non-identity elementary abelian p-subgroups, Bouc's complex of p-radical subgroups, as well as Benson's complex which has been discussed recently. Benson's complex was introduced in an exercise at the end of section 6.6 of his book Representations and Cohomology II, and it is discussed by Smith in his book Subgroup Complexes. My theorem states that over a $p$-local ring (such as a field of characteristic p) the augmented chain complex of Brown's complex is the direct sum of a contractible complex and a complex of projective modules. This implies that if we remove all contractible summands from the augmented chain complex of Brown's complex we obtain a complex of projective modules, unique up to isomorphism After my initial paper, other proofs have been given of my result. The following papers both contain proofs: P. Symonds, The Bredon cohomology of subgroup complexes, J. Pure Appl. Algebra 199 (2005), 261-298, doi: 10.1016/j.jpaa.2004.12.010. S. Bouc, Resolutions de foncteurs de Mackey, Proc Symposia in Pure Math. 63 (1998), 31-83. The fact that the Steinberg complex exists has as a consequence the method of computing group cohomology described in my paper, A local method in group cohomology, and used quite extensively by various people, including Adem and Milgram. Work on the structure of the Steinberg complex in the situation where its homology is not projective was done by Dan Swenson in his thesis. You can download a copy at http://www-users.math.umn.edu/~webb/PhDStudents/index.html In 1986 I wrote a survey with the title Subgroup Complexes that appeared in vol 47 of Proc. Symposia in Pure Math, before I had proved my theorem about the Steinberg complex. In that survey I gave values of the Lefschetz module of the Steinberg complex for certain groups, including some sporadic groups. This is the alternating sum of the projective modules in the Steinberg complex, so it is a virtual projective module, and it could be called the generalized Steinberg module. I just want to point out that there are many errors in that article. The value I gave for the generalized Steinberg module of M_12 is completely wrong, and the value for A_7 should have a minus sign. These and other errors are listed on my web site, at http://www-users.math.umn.edu/~webb/Publications/SubgroupComplexes.errors I hope this helps.<|endoftext|> TITLE: Is every map of rank smaller than r dominated by a constant rank map? QUESTION [5 upvotes]: Let $\, f:M \to N$ be a smooth map, with rank $df \le r$ everywhere. Does there exist a smooth map $\tilde f:M \to N$ of constant rank $r$, such that each level set of $\tilde f$ is contained in some level set of $f$? Is it true at least locally? (It is easy to see rank $df \le r$ is a necessary condition for the existence of such $\tilde f$). Motivation: I am trying to visualize maps of small rank. In the case of constant rank maps , this is easy since the level sets form a foliation of the domain. The intuition is that a map of rank $\le r$ is "more constant" than a map of rank exactly $r$. REPLY [7 votes]: No. The simplest case is $M = S^1$ and $N = \mathbb{R}$. Then any nonconstant map $f:M\to N$ has rank at most 1, but there is no smooth map from $M$ to $N$ that has constant rank $1$. The question would be more interesting if you were considering $\tilde f: M\to \tilde N$ of rank at most $r$ instead of $\tilde f: M \to N$. (The answer would still be 'no', but the 'counterexamples' would be more interesting. For example, any nonconstant smooth map $f:S^2\to \mathbb{R}$ has rank at most $1$, but there is no smooth map from $S^2$ to any manifold that has constant rank $1$, since the tangent bundle of $S^2$ is irreducible.)<|endoftext|> TITLE: Find a metric so that a given submanifold is totally geodesic QUESTION [8 upvotes]: Let $M$ be a smooth manifold, and $N$ is a submanifold. My question is simply that, does there exist a metric $g$ on $M$ so that $N$ is totally geodesic? In general the answer might be 'no'. But it might be possible that, when we assumed additional conditions on $M$ or $N$, the answer became 'yes'. This question is kind of related to a previous question here. I'll appreciate it if you know any reference on this issue. Thanks. REPLY [3 votes]: I happened to think of an alternative proof long after this question was answered, but I decided to post it anyway now. As mentioned in the comments, if $N$ is properly embedded and has trivial normal bundle then taking any product metric on a tubular neighborhood and extending with a partition of unity does the trick. If the normal bundle $E$ of $N$ is nontrivial, let $E'$ be an inverse bundle so that $E \oplus E' \cong N \times \mathbb{R}^n$. Again giving $N\times \mathbb{R}^n$ any product metric renders $N\times \{0\}$ totally geodesic in $N \times \mathbb{R}^n$. A glance at the second fundamental form reveals that $N\times \{0\}$ is also totally geodesic in $E$ (identifying $E$ with its image in $N \times \mathbb{R}^n$). Now pull back the induced metric on $E$ to a tubular neighborhood $U$ of $N$, and extend to $M$ with a partition of unity.<|endoftext|> TITLE: Alternative way to prove the functional equation for Eisenstein series? QUESTION [6 upvotes]: Let $E(z,s):=\pi^{-s}\Gamma (s) \sum_{(m,n)=1}\frac{y^s}{|mz+n|^{2s}}$ be the real-analytic Eisenstein series. It satisfies the functional equation $E(z,s)=E(z,1-s)$ with two poles at $s=0,1$. The method I know to prove this is to calculate the Fourier coefficients individually. They are either divisor functions or Riemann zeta function. Each satisfies the functional equation. However, is there more conceptual way to prove this function equation without looking at individual coefficients? REPLY [6 votes]: As usual, the functional equation on the Dirichlet series side comes from a theta function on the modular side. Using the poisson summation formula we show $$\vartheta_z(x) = \sum_{(c,d) \in \mathbb{Z}^2} \exp(-\pi x \frac{|cz+d|^2}{|\Im(z)|}) = x^{-1} \vartheta_z(1/x)$$ Then let $$E_z(s) = \sum_{\gamma \in SL_2(\mathbb{Z})} \Im(\gamma(z))^{s}=\sum_{(c,d) \in \mathbb{Z}^2, gcd(c,d)=1} (\frac{|cz+d|^{2}}{\Im(z)})^{-s}$$ (with $gcd(0,d) = d$). It comes from a Mellin transform $$\Gamma(s)\pi^{-s}\zeta(2s)E_z(s) = \Gamma(s)\pi^{-s}\sum_{c,d \in \mathbb{Z}^2 \setminus(0,0)} (\frac{|cz+d|^2}{|\Im(z)|})^{-s}\\ = \int_0^\infty (\vartheta_z(x)-1) x^{s-1}dx = \int_1^\infty (\vartheta_z(x)-1) x^{s-1}dx+\int_1^\infty (\vartheta_z(1/x)-1) x^{-s-1}dx \\ = \frac{1}{s-1}+\frac{1}{-s }+\int_1^\infty (\vartheta_z(x)-1) (x^{s-1}+x^{-s})dx$$ Which proves the functional equation. Let $h(u) = e^{-\pi \|u\|^2}, u \in \mathbb{R}^2$ which is its own Fourier transform. For some self-adjoint matrix $B \in GL_2(\mathbb{R})$ let $g(u)= e^{-\pi u^T B u} = h(B^{1/2}u)$ then $\widehat{g}(u) = \frac{1}{|\det(B)|^{1/2}} h((B^{-1/2})^Tu)$. Then apply the Poisson summation formula to $$\theta_B(x) = \sum_{n \in \mathbb{Z}^2} e^{-\pi x (u^T Bu)}= \sum_{n \in \mathbb{Z}^2} h((x^{1/2} I)B n)\\=\sum_{n \in \mathbb{Z}^2} \frac{x^{-1}}{|\det(B)|^{1/2}} h((x^{-1/2}I)B^{-T/2}n)= \frac{x^{-1}}{|\det(B)|^{1/2}}\theta_{B^{-1}}(1/x)$$ Finally $|cz+d|^2 = (c,d)B{\scriptstyle\begin{pmatrix} c \\ d \end{pmatrix}}$ where $B = {\scriptstyle\begin{pmatrix} |z|^2 & \Re(z) \\ \Re(z) & 1\end{pmatrix}}, \det(B) = \Im(z)^2, B^{-1} = \frac{1}{\Im(z)^2} {\scriptstyle\begin{pmatrix} 1 & -\Re(z) \\ -\Re(z) & |z|^2\end{pmatrix}}$ so that $(c,d)B^{-1}{\scriptstyle\begin{pmatrix} c \\ d \end{pmatrix}} = \frac{|c-dz|^2}{\Im(z)^2}$ and $\theta_{B^{-1}}(x) =\theta_B(\frac{x}{\Im(z)^2})$ and $\vartheta_z(x) = \theta_B(\frac{x}{|\Im(z)|})$. (it is very possible there are some typos)<|endoftext|> TITLE: A Compact Manifold with odd Euler characteristic whose tangent bundle admits a field of lines QUESTION [12 upvotes]: I understand that the top Stiefel Whitney class is an obstruction for the tangent bundle of a manifold to have a trivial line sub-bundle. I am looking for a counterexample when removing the word "trivial", i.e: A compact manifold $M$ of dimention n such that $w_n(TM)\neq0$ (or equivalently $\chi(M)$ is odd) and there exists a sub-bundle $\xi\subset TM$ with $\operatorname{rank}(\xi)=1$. After some thought I understand why one cannot find an example using surfaces, odd-dimensional manifolds or any of $S^n$,$T^n$,$\mathbb{R}P^n$,$\mathbb{C}P^n$. I also understand why manifolds with $H^1(M;\mathbb{Z}_2)=0$ will not work, and why such a manifold (if it exists) cannot be null-cobordant. My intuition on 4-manifolds (or god forbid anything higher dimensional) is not good enough to know where to look. Any insight as to where one should look for such a creature (or why the hell it should not exist) will be greatly appreciated. REPLY [16 votes]: I believe there is no example satisfying all your constraints. If I recall (my memory is a little foggy on this) the result likely goes back to Hopf, and one of his variations on the Poincare-Hopf index theorem. This question might be addressed in the Milnor and Stasheff text. Here is one way to argue the point. Say the tangent bundle of the manifold $N$ admits a field of lines. Then (at worst) some 2:1 cover of $N$ admits an everywhere non-vanishing vector field. Since the Euler characteristic is the obstruction to such a vector field existing (Poincare-Hopf index theorem) the Euler characteristic of this covering space is zero. But $\chi N$ is a multiple of the euler characteristic of the cover. i.e. your assumption that the Euler characteristic is odd excludes the possibility of a 1-dimensional sub-bundle. REPLY [2 votes]: Let $\gamma$ be the canonical (real) line bundle over $RP^2$. In other words, $w_1(\gamma)\neq 0$. The total Stiefel-Whitney class of the Whitney sum $\gamma\oplus\gamma$ is $(1+w_1(\gamma))^2=1+w_1(\gamma)^2$. So $w_2(\gamma\oplus\gamma)=w_1(\gamma)^2$ which is nonzero.<|endoftext|> TITLE: Does MCMC overcome the curse of dimensionality? QUESTION [6 upvotes]: I want to compute an integral like this $$\frac{\int_y g(y) e^{-\beta f(y)} \text{d} y } {\int_y e^{-\beta f(y)} \text{d} y}$$ where $f(y)$ is not necessarily convex and the dimension $d$ of $y$ is large. This problem can be viewed as to integrate a function with respect to a density function whose normalized factor is unknown. It seems MCMC is a good choice? However for this general problem I didn't find any literature showing the convergence rate w.r.t the dimension $d$. Remark: there are two ways I know to implement MCMC algorithms. One is by Metropolis–Hastings algorithm; another is by Metropolis-adjusted Langevin algorithm, which simulates a SDE $dX(t)=-\nabla f(x(t))dt + \sqrt{2\beta^{-1}} dB(t) $. In the latter case we also need to consider discretization errors of the SDE. I guess without convexity assumptions on $f(y)$ the convergence rate could be $\mathcal{O}(e^{-d})$ slow. If this is true, it may suffer from the curse of dimensionality? The reason why I came up with this problem is that we can just use simple Monte Carlo method, i.e. just sample uniformly distributed random variable to compute the integral on numerator and denominator separately. And as we know this time the variance of our estimate is not related to $d$. I am very confused about the role MCMC plays in high dimension problems. Does someone there can help me figure out this? Any references are much appreciated. REPLY [5 votes]: You need a global convexity to enjoy the optimal convergence rate, otherwise even local convexity will almost surely(not in probabilistic sense) lead to the worst rate you pointed out. MCMC(Markov Chain Monte Carlo) does not overcome the curse of dimensionality. Quite the contrary, Bayesians are working very hard in two directions to solve the problems that caused by the high dimension of the parameter space. (1) The ABC(Approximate Bayesian Computation) scheme. This originated from Laplace's approximation of an exponential integral and its main idea is to discard those samples from each MH step that is "dissimilar from existing samples". But this method suffers from failure of detecting outliers and over-concentration of the posterior. On the other hand even if this method works well during sampling, we do not have any consistency guarantee since when nuisance parameters $\theta_2$ consist of the majority of the parameter space $\boldsymbol{\theta}=(\theta_1,\theta_2)$, Neyman-Scott example will nullify the information about the parameter $\theta_1$ of concern brought in by the samples. That is also a reason why regularization methods are so popular in high dimension inference problems, they simply produce a weighted norm that cannot be washed out by samples. (2) The scalable methods(For example). These are sometimes referred to as divide-and-conquer problems. The Bayesian model, or more precisely the parameter space is decomposed into subspaces, and the high dimension issue is divided into many low dimension issues and the $O(n^d)$ problem becomes $O(n^{d/k}(n/k))$ problems. But the problem is also obvious, that is decomposing and combining parameter spaces will artificially delete and add correlations between parameter spaces, which makes the prior information about $\boldsymbol \theta$ lose.<|endoftext|> TITLE: When are counits monic? QUESTION [7 upvotes]: The counit of an adjunction is split monic precisely when the right adjoint is full. Can anything nice be said about when the counit is monic (ideally in terms of the induced comonad)? REPLY [7 votes]: As noted in a comment (in the dual case), if the left adjoint is fully faithful then this property has a name: a coreflective subcategory whose counits are mono is called monocoreflective (the dual of "epireflective"). This is of course a property of the induced comonad (since a coreflection is the same as an idempotent comonad). I don't think it has any more concrete characterization, but you can google it and read more about it. However, we can say that the general case reduces to this special one. Namely, in an arbitrary adjunction if the counits are monic, then since $\epsilon \circ F G \epsilon = \epsilon \circ \epsilon F G$ by naturality, monicity of $\epsilon$ gives $F G \epsilon = \epsilon F G$, and therefore the adjunction is idempotent. Thus it factors as a composite of a reflection $F_1\dashv G_1$ (with invertible counit) and a coreflection $F_2 \dashv G_2$ (with invertible unit), in which case monicity of $\epsilon$ is equivalent to monicity of $\epsilon_2$. Since an adjunction is also idempotent if and only if its induced comonad is idempotent, we can say that for a general adjunction the counit is monic if and only if the induced comonad is idempotent and corresponds to a monocoreflective subcategory.<|endoftext|> TITLE: Reference for exponential Vandermonde determinant identity QUESTION [24 upvotes]: I recently stumbled upon the following identity, valid for any real numbers $\alpha_1,\dots,\alpha_n$ and $\lambda_{n1} \leq \dots \leq \lambda_{nn}$: $$ \mathrm{det}( e^{\alpha_i \lambda_{nj}} )_{1 \leq i,j \leq n} = V(\alpha) \int_{GT_\lambda} \exp( \sum_{i=1}^n \sum_{j=1}^i \lambda_{ij} (\alpha_{n+1-i}-\alpha_{n-i}))$$ where $V(\alpha)$ is the Vandermonde determinant $$ V(\alpha) := \prod_{1 \leq i < j \leq n} (\alpha_j - \alpha_i),$$ $GT_\lambda$ is the Gelfand-Tsetlin polytope of tuples $(\lambda_{ij})_{1 \leq j \leq i < n}$ obeying the interlacing relations $\lambda_{i+1,j} \leq \lambda_{i,j} \leq \lambda_{i,j+1}$ and with the usual Lebesgue measure, and one has the convention $\alpha_0 := 0$. Thus for instance when $n=1$ one has $$ e^{\alpha_1 \lambda_{11}} = \exp( \lambda_{11} \alpha_1 )$$ when $n=2$ one has $$ \mathrm{det} \begin{pmatrix} e^{\alpha_1 \lambda_{21}} & e^{\alpha_1 \lambda_{22}} \\ e^{\alpha_2 \lambda_{21}} & e^{\alpha_2 \lambda_{22}} \end{pmatrix} $$ $$= (\alpha_2 - \alpha_1) \int_{\lambda_{21} \leq \lambda_{11} \leq \lambda_{22}} \exp( \lambda_{11} (\alpha_2-\alpha_1) + \lambda_{21} \alpha_1 + \lambda_{22} \alpha_1 )\ d\lambda_{11}$$ and when $n=3$ one has $$ \mathrm{det} \begin{pmatrix} e^{\alpha_1 \lambda_{31}} & e^{\alpha_1 \lambda_{32}} & e^{\alpha_1 \lambda_{33}} \\ e^{\alpha_2 \lambda_{31}} & e^{\alpha_2 \lambda_{32}} & e^{\alpha_2 \lambda_{33}} \\ e^{\alpha_3 \lambda_{31}} & e^{\alpha_3 \lambda_{32}} & e^{\alpha_3 \lambda_{33}} \end{pmatrix} $$ $$ = (\alpha_2 - \alpha_1) (\alpha_3 - \alpha_1) (\alpha_3 - \alpha_2) \int_{\lambda_{31} \leq \lambda_{21} \leq \lambda_{32}} \int_{\lambda_{32} \leq \lambda_{22} \leq \lambda_{33}} \int_{\lambda_{21} \leq \lambda_{11} \leq \lambda_{22}}$$ $$ \exp( \lambda_{11} (\alpha_3-\alpha_2) + \lambda_{21} (\alpha_2-\alpha_1) + \lambda_{22} (\alpha_2-\alpha_1) + \lambda_{31} \alpha_1 + \lambda_{32} \alpha_1 + \lambda_{33} \alpha_1)$$ $$ d \lambda_{11} d\lambda_{22} d\lambda_{21},$$ and so forth. The identity can be proven easily by induction. I first discovered it by starting with the Schur polynomial identity $$ \mathrm{det}( x_j^{a_i} )_{1 \leq i,j \leq n} = V(x) \sum_T x^{|T|}$$ where $0 \leq a_1 < \dots < a_n$ are natural numbers in increasing order, $T$ ranges over column-strict Young tableaux of shape $a_n-n+1, \dots, a_2-1, a_1$ with entries in $1,\dots,n$, and $x^{|T|} := x_1^{c_1} \dots x_n^{c_n}$ where $c_i$ is the number of occurrences of $i$ in $T$, and taking a suitable "continuum limit" as the $a_i$ go to infinity and the $x_j$ go to one in a particular fashion. It can also be derived from the Duistermaat-Heckmann formula for the Fourier transform of Schur-Horn measure, combined with the fact that this measure is the pushforward of Lebesgue measure on the Gelfand-Tsetlin polytope under a certain linear map. Note that the identity also provides an immediate proof that any $n$ distinct exponential functions on $n$ distinct real numbers are linearly independent. I am certain that this formula already appears in the literature, and perhaps even has a standard name, but I was unable to locate it with standard searches. So my question here is if anyone recognizes the formula and can supply a reference for it. REPLY [17 votes]: Write $\beta = \lambda_n$, the top row of your GT patterns. It's a theorem of [Baryshnikov] that if we choose a uniformly random point in the polytope GT${}_\lambda$, it's equivalent to choosing a Haar-random Hermitian matrix with spectrum $\beta$ and then taking its "principal minors". (I've also seen this fact credited to Weyl, and others.) More precisely, let $B = \mathrm{diag}(\beta)$, and form a matrix $X = U B U^\dagger$, where $U$ is a random unitary. Then let $\lambda_{11}$ be the top-left entry of $X$, let $\lambda_{21}, \lambda_{22}$ be the eigenvalues of the top-left $2 \times 2$ submatrix of $X$, ..., and let $\lambda_{n1}, \dots, \lambda_{nn}$ be the eigenvalues of the top-left $n \times n $ submatrix of $X$ (namely, $\beta$). Then $\lambda$ is uniformly random in the polytope GT${}_\lambda$. This probability distribution on $\lambda$ is basically your integral, but we have to divide by the volume of the polytope, which is $V(\lambda)/[(n-1)! (n-2)! \cdots 2! 1!]$. I guess this is standard? If not, it's also in Baryshnikov. Having done so, your identity is the Harish-Chandra--Itzykson--Zuber identity applied to the matrices $A = \mathrm{diag}(\alpha)$ and $B$. This follows by inferring the diagonal entries of $X$ from the Gelfand-Tsetlin pattern $\lambda$, which you can do because the Gelfand--Tsetlin pattern gives you the traces of all the top-left submatrices. (By the way, I think the [Faraut] paper referenced below has a good exposition of some related things.) Baryshnikov, Yu., GUEs and queues, Probab. Theory Relat. Fields 119, No.2, 256-274 (2001). ZBL0980.60042. Faraut, Jacques, Rayleigh theorem, projection of orbital measures and spline functions, Adv. Pure Appl. Math. 6, No. 4, 261-283 (2015). ZBL1326.15058. REPLY [14 votes]: This looks like a special case of a formula by Samson Shatashvili related to the HCIZ integral as mentioned in Ryan's answer. Compare, in particular the two ways of computing $\langle 1\rangle$ given by Equations 3.2 and 3.4 in "Correlation Functions in The Itzykson-Zuber Model" (thanks to Leonid Petrov for letting me know about this reference in his answer to this MO question).<|endoftext|> TITLE: Poincare's argument for maximizing the Coulomb energy QUESTION [8 upvotes]: For $\Omega\subset \mathbb{R}^3$ a region with $|\Omega| = |B_1|$, let $$ C(\Omega) = \int_\Omega\int_\Omega \frac{dxdy}{|x-y|} $$ denote the Coulomb (or gravitational, etc) energy. Poincaré is credited with an incomplete proof that (*) $C(\Omega) \leq C(B_1)$. I know the (now standard) proof using symmetrization, but my impression is that this is not Poincaré's approach. My question is: What was Poincaré's argument for (*)? REPLY [6 votes]: H. Poincaré, Sur une théorème de M. Liapounoff relatif a l’équilibre d’une masse fluide, Comptes Rendus de L’Academie des Sciences 104, 622–625 (1887). As discussed by G.C. Evans, Poincaré assumes tacitly that there do exist one or more bodies of a given volume, with smooth boundaries, which provide relative minima for the electrostatic capacity $C$ with respect to neighboring forms; and his treatment amounts to a proof that among these the sphere provides the absolute minimum. (Note that the Coulomb energy $U=Q^2/2C$, so minimal $C$ corresponds to maximal $U$ for given total charge $Q$.) A complete proof, without this assumption, was given by G. Szegö, Über einige Extremalaufgaben der Potentialtheorie (1930).<|endoftext|> TITLE: Is integration on manifolds unique? QUESTION [17 upvotes]: I asked this question on math.stackexchange.com, but haven't received an answer, so I thought I'd brave the waters here. Suppose $M$ is a smooth oriented compact connected $m$-dimensional manifold and let $A^m(M)$ denote the set of smooth exterior differential $m$-forms on $M$. We can integrate members of $A^m(M)$, and the mapping $\phi\mapsto\int_M\phi$ has several properties: It's not identically $0$. It's linear. It's symmetric. By "it's symmetric", I mean that if $f:M\to M$ is an orientation-preserving diffeomorphism, then $\int_M \phi = \int_M f^*\circ \phi$. I think that expresses the idea of symmetry I have in mind. Basically, all points on $M$ are equivalent as far as integration is concerned. My question is: Do these properties uniquely determine the space $A^m(M)$ and the usual definition of integration? More precisely, is there some smooth compact connected $m$-manifold $M$, a space $S$ of smooth sections of the tensor bundle of $M$, a large family $F$ of diffeomorphisms of $M$, and a non-trivial linear functional $\lambda:S\to\mathbb{R}$ such that $\lambda( \phi) = \lambda( f^*\circ \phi)$ for all $\phi\in S$ and $f\in F$, but is not just integration of exterior $m$-forms, up to constant scale factor? My motivation for asking this question: I suspect almost everyone who's ever learned about integration of differential forms on manifolds has wondered why those particular definitions were chosen. The answer is probably some variant of, "Because those definitions work." But must it be this way? Are they the only definitions that make integration work? REPLY [11 votes]: Here's a somewhat abstract but pretty satisfying way to see that integration of compactly supported densities on a not-necessarily orientable and not-necessarily compact $n$-manifold $M$ is unique. This point of view is explained in Remark 3.3.10 in Kashiwara & Schapira's Sheaves on Manifolds. Let $t: M \to 1$ be the terminal map to a point, and recall that the category of sheaves of $\mathbb{R}$-vector spaces on a point is just the familiar category of vector spaces. Recall also that the pushforward with proper supports along the terminal map $t$ is the compactly supported global sections functor; i.e. $$t_!=\Gamma_c(M; -).$$ The counit $\epsilon: Rt_! t^! \Rightarrow 1$ of the Poincare-Verdier duality adjunction $Rt_! \dashv t^!$ is a generalization of integration. Since adjoints, like other universal objects, are unique up to unique isomorphism, it follows that integration is unique in this sense. To see how the counit here generalizes integration, apply it to the real numbers (or complex numbers) and take the $0$th cohomology to recover integration. In a bit more detail, notice that $$H^0 Rt_! t^! \mathbb{R}=H^0 Rt_! \mathrm{or}_M[n] =H^0 R \Gamma_c(M; \mathrm{or}_M[n]) = H^n_c(M; \mathrm{or}_M).$$ So the counit mentioned above induces the integration map in compactly supported sheaf cohomology: $$\int_M: H^n_c(M; \mathrm{or}_M) \to \mathbb{R}.$$ The densities (or $n$-forms if you pick an orientation for an orientable manifold) appear when you take the usual de Rham resolution of the orientation sheaf $\mathrm{or}_M$, thus using de Rham cohomology to compute sheaf cohomology. For $p: E \to M$ a topological submersion, we can use Poincare-Verdier duality, the relative orientation sheaf $\mathrm{or}_{E/M}$, and a somewhat similar approach to define fibre integration of "relative" densities.<|endoftext|> TITLE: Braid groups on topological spaces QUESTION [5 upvotes]: The configuration space $C_n(M)$ of $n$ particles in some connected graph $M$ (thought of as the topological realisation of a one-dimensional CW-complex) is $$M^n \backslash \{ (x_1, \ldots, x_n) \mid x_i=x_j \ \text{for some} \ i \neq j\},$$ and the corresponding unordered configuration space $UC_n(M)$ is the quotient $C_n(M)/ \mathfrak{S}_n$ where $\mathfrak{S}_n$ acts freely on $C_n(M)$ by permuting the coordinates. The braid group $B_n(M)$ is the fundamental group of $UC_n(M)$ (which is connected if $M$ is connected itself). Is it true that $B_{m}(M)$ admits an injective homomorphism into $B_n(M)$ if $m \leq n$? Is it true at least if $m \leq 2$? It is worth noticing that the question is not trivial if we replace the graph $M$ with a sphere according to this question. REPLY [2 votes]: I think the answer is yes if $M$ has a leaf. Subdivide $M$ sufficiently so that Abram's locally CAT(0) model $X_n(M)$ of $B_n(M)$ (called the reduced braid group, $RB_n(M)$ is this paper of Crisp and Wiest) can be used. We can map $X_n(M)$ to $X_{n+1}(M')$ by mapping a configuration $(p_1, \dots, p_n)$ to $(p_1, \dots, p_n, p_{n+1})$. Here $M$ is the subdivided graph and $M'$ is $M$ with an additional subdivided edge glued onto the degree one vertex of the distinguished leaf of $M$. The point $p_{n+1}$ is fixed and corresponds to the degree one vertex of the distinguished (extended) leaf of $M'$. The map $X_n(M) \to X_{n+1}(M')$ is a cubical map and it seems that the image of the link of any vertex in $X_n(M)$ is a full subcomplex of the link of the image point in $X_{n+1}(M')$. So, this map is $\pi_1$-injective. But, $X_{n+1}(M')$ is isomorphic to $X_{n+1}(M)$, so we have an injective homomorphism $B_n(M) \to B_{n+1}(M)$. Here's some additional detail/background, though Crisp & Wiest (and Abrams in his thesis) say it much better if you read their work. Given a subdivided graph $M$, the carrier of a point $x$ in $M$ is the closed cell (vertex or edge) containing it. The space $Y_n(M)$ consists of configurations of $n$ ordered points in $M$ such that the carriers of points are mutually disjoint. $X_n(M)$ is the quotient under the action of the symmetric group. The cubical structure comes from the fact that a configuration of $n$ points where exactly $k$ of these points have edges as their carriers corresponds to a $k$-cube.<|endoftext|> TITLE: Geometric or conceptual way to understand supersymmetry algebra QUESTION [10 upvotes]: Is there any geometric or more direct conceptual way to understand a supersymmetry algebra, rather than starting from a Lagrangian including boson and fermion fields, deriving all the expressions ensuring the supersymmetry invariance and then writing down the supersymmetry algebra? For a geometric or algebraic way, I mean to (at least partially) derive the supersymmetry algebra from pure geometry (spinors, spin group, etc.). REPLY [3 votes]: In this paper about a global theory of supermanifolds, Alice Rogers develops the theory of supermanifolds as they underly supersymmetric field theories from a rigorous but physicist-friendly differential-geometric point of view from topological scratch, and also puts some additional structures such as vector fields and tangent spaces on them. She also compares the $G^{\infty}$ or deWitt supermanifolds to the algebro-geometric approach of for example Kostant or Leites. Alice Roger's 2007 textbook explains the supermathematics needed for doing superphysics (including Grassmann algebras, super Lie groups such as the super Poincare group, etc) and contains more applications to different physics topics such as $N=1$ supersymmetry, supergravity, some aspects of string theory, or Brownian motion from the same nice differential-geometric point of view.<|endoftext|> TITLE: Characterize constant objects in the internal language of a topos? QUESTION [16 upvotes]: A Grothendieck topos $\mathcal{E}$ is equivalent to the category of sheaves on some site $Q$. We say a sheaf $X\colon Q^{\text{op}}\to\mathsf{Set}$ is constant if it is the sheafification of a constant presheaf, i.e. one that factors through the terminal map $Q^{\text{op}}\to \{*\}$. But what if we forget the site $Q$ and consider $X$ as an object in the topos? Can we characterize the property of $X\in\mathcal{E}$ being constant? More specifically, two questions: Is the property of $X$ being constant dependent on the choice of site $Q$? Is there a way to characterize constant objects in the internal language of $\mathcal{E}$? REPLY [5 votes]: The answer to your second question is also no: only local properties can be characterised in the internal language of a topos, and it is possible to have locally constant (pre)sheaves that are not constant. A simple, and well-known, example runs as follows: let $\mathbb C$ be the poset with four elements $n$, $e$, $s$, and $w$, satisfying $w TITLE: Spectral sequences in $K$-theory QUESTION [21 upvotes]: There is an algebraic analogue of the Atiyah-Hirzebruch spectral sequence from singular cohomology to topological $K$-theory of a topological space. For a field $k$, let $X$ be smooth variety $X$ over $k$. The following spectral sequence will be referred to in the sequel as the motivic spectral sequence: $$E_2^{i,j} := H^{i-j}(X, \mathbf{Z}(-j)) \Rightarrow K_{-i-j}(X).$$ See: the Bloch-Lichtenbaum motivic spectral sequence in [BL], and the generalizations by Levine [L] and Friedlander-Suslin [FS] to smooth varieties over $k$. the Voevodsky motivic spectral sequence [V]. the Grayson motivic spectral sequence [G]. For $k$ and $X$ as in the foregoing, we may form the étale hypercohomology of the Bloch complex $z^{j}(X,\bullet)$ ([B]) on $X_{\rm\acute{e}t}$, denoted $H^{\bullet}_{L}(X, \mathbf{Z}(j))$ and usually called Lichtenbaum cohomology. Questions: Is an "étale analogue" of the motivic spectral sequence from the foregoing, i.e.: $$E_2^{i,j} := H_L^{i-j}(X, \mathbf{Z}(j))\Rightarrow K_{-i-j}^{\rm\acute{e}t}(X)$$ available? If the answer to $(1)$ is "yes", what is the currently known generality? If the answer to $(1)$ is "yes", references? References: [BL] S. Bloch, S. Lichtenbaum, A spectral sequence for motivic cohomology, K-theory, 1995. [L] M. Levine, Techniques of localization in the theory of algebraic cycles, 2001. [FS] E. M. Friedlander, A. Suslin, The spectral sequence relating algebraic K-theory to motivic cohomology, 2002. [V] V. Voevodsky, A possible new approach to the motivic spectral sequence for algebraic K-theory, 2002. [G] A. Suslin, On the Grayson spectral sequence, 2003. [B] S. Bloch, Algebraic cycles and Higher $K$-theory, 1986. REPLY [5 votes]: Yes. Firstly, I apologize for the self-referencing. But the references for this is a joint paper with Marc Levine (basically finishing up what he had sketched in aforementioned paper), Markus Spitzweck and Paul Arne Ostvaer. It is currently under revision but has been available on the arxiv for sometime: https://arxiv.org/abs/1711.06258 The spectral sequence itself is available over regular schemes since it is an incarnation of the slice spectral sequence in motivic homotopy theory. Over more general bases, we are computing the etale version of Weibel's homotopy K-theory The paper also addresses convergence issues in extensive, if not terse, details.<|endoftext|> TITLE: A quick algorithm for calculating the $\ell_1$-distance between two finite sets on the real line? QUESTION [6 upvotes]: For two non-empty finite sets $A,B$ in the real line define the $\ell_1$-distance $d_1(A,B)$ between $A$ and $B$ as the smallest Lebesgue measure of a closed subset $\Gamma\subset \mathbb R$ such that $A\cup B\subset\Gamma$ and each connected component of $\Gamma$ intersects both sets $A$ and $B$. It is clear that such minimal set $\Gamma$ is a disjoint union of closed intervals with ends in the set $A\cup B$. As was noted by @fedja in his comment, the number of such unions is $\le 2^{n-1}$ where $n$ is the cardinality of the set $A\cup B$. So, a brute force algorithm for calculating the distance $d_1(A,B)$ has exponential complexity. Problem. Is there a polynomial-time algorithm for calculating $d_1(A,B)$? Or this problem is NP-hard? NP-complete? Remark. A similar problem for the plane seems to be NP-hard, see Remark 2 here. REPLY [3 votes]: While it doesn’t seem to be possible to beat the $O(n\log n)$ bound for general input, the problem can be solved in time $O(n)$ if $A\cup B$ is given sorted. This builds on the ideas in fedja’s and Aaron Meyerowitz’s answers. As implicit in the other posts, I am assuming unit-cost real RAM or a similar model, so that arithmetic operations on real numbers are exact and take constant time. Let $A\cup B=\{x_1,\dots,x_n\}$ with $x_1 TITLE: Algebraic cycles, Chow spaces and Hilbert-Chow morphisms QUESTION [10 upvotes]: In the sequel, let $S$ be a scheme, and $X$ a locally of finite type algebraic space over $S$. In his thesis ([R1-R4]), David Rydh introduces, among several others, the notion of relative cycles on $X\to S$, and, for integers $r\ge 0$, he defines the functor: $$\text{Chow}_r(X/S) : (\text{Sch}/S)^{\rm opp}\to\text{Set}$$ by: $$\text{Chow}_r(X/S)(T) := \text{Cycl}^{\rm prop}_r(X\times_ST/T)$$ where the right side is the set of equidimensional relative cycles of dimension $r$ on $X\times_ST\to T$, with proper support (see [R4]). $\text{Chow}_r(X/S)$ is an fppf sheaf, ultimately because the fppf sheaf of divided powers of $X\to S$ is representable (see [R1]). In [R4], representability of $\text{Chow}_r(X/S)$ is shown only in those cases when $\text{Chow}_r(X/S)$ can be proved to be isomorphic as a functor to some other functor known to be representable (e.g., Angéniol's Chow space (see [An]), when $S$ is of pure characteristic zero, and $X\to S$ is smooth and separated). It appears that the best representability result one may extrapolate from [R4] is, therefore, as follows: Theorem 1. Let $S$ be a scheme of pure characteristic zero, $X\to S$ a smooth and separated algebraic space, $r\ge 0$ an integer. Then the fppf sheaf $\text{Chow}_r(X/S)$ is represented by a separated algebraic space over $S$, locally of finite type. The algebraic space representing $\text{Chow}_r(X/S)$ will be called Chow space, in the sequel. It is easy to show, using representability of the Hilbert functor when $X\to S$ is projective, that the following holds: Theorem 2. Let $S$ be a scheme of pure characteristic zero, $X\to S$ a smooth and projective algebraic space, $r\ge 0$ an integer. Then: $\text{Chow}_r(X/S)$ is proper. the Hilbert-Chow morphism: $$\text{Hilb}_r(X/S)\to\text{Chow}_r(X/S)$$ constructed in [R4], is proper. Questions: Is it true that, under the assumptions in Theorem 2, $\text{Chow}_r(X/S)$ in $(1)$ is projective? Under the assumptions in Theorem 2, what are the currently known properties of the Hilbert-Chow morphism $\text{Hilb}_r(X/S)\to\text{Chow}_r(X/S)$ in $(2)$? Is it surjective/surjective on geometric points/birational? Upon inspecting Angéniol's proof of representability of his Chow functor, it appears to me the smoothness assumption can be removed, upon appropriately exploiting the deformation theory in vol. 1 of Illusie's thesis. Has this been done anywhere in the literature? What can one say, to this day, about the case when $S$ is of pure characteristic $p>0$? Around question $(2)$. Is representability of $\text{Chow}_r(X/S)$ known at least in the case when $S = \text{Spec}(k)$, for $k$ a field of characteristic $p>0$, and $X\to S$ a smooth projective $S$-scheme? References: [An] B. Angéniol. Familles de cycles algébriques. Springer. [R1] D. Rydh. Families of zero-cycles and divided powers: I. Representability. [R2] D. Rydh. Families of zero-cycles and divided powers: II. The universal family. [R3] D. Rydh. Hilbert and Chow schemes of points, symmetric products and divided powers. [R4] D. Rydh. Families of cycles. REPLY [10 votes]: Mathoverflow answer In my thesis [R4], I gave an ad hoc definition of a Chow functor ($\mathrm{Chow}_r$ above) that was meaningful also in characteristic p and close to Barlet's and Angéniol's definitions. It is "ad hoc" because the definition involves specifying zero-cycles over every suitable projection to an r-dimensional smooth space and imposing compatibility conditions, whereas Barlet (reduced, char 0) and Angéniol (char 0) have a single object representing the family of cycles. Theorem 1: I could only show that $\mathrm{Chow}_r$ coincides with Barlet/Angéniol's functor when X/S is smooth and separated for (a) reduced parameter schemes, (b) families of "multiplicity-free" cycles (the multiplicity of every component is 1), and (c) families of Weil divisors, see [R4, Thm 16.2]. In particular, Thm 1 above is, as far as I know, not known in general. It is possible that my ad-hoc definition needs to be modified slightly to agree in general (see introduction and discussion after 5.4). Theorem 2: $\mathrm{Chow}_r(X/S)$ is not proper: it is (typically an infinite) disjoint union of projective schemes $\mathrm{Chow}_{r,d}(X/S)$ where you fix the degree d of the cycles. The morphism $\mathrm{Hilb}_r \to \mathrm{Chow}_r$ is also not proper for similar reasons, but $\mathrm{Hilb}_P\to \mathrm{Chow}_r$ is projective where $P$ is a Hilbert polynomial of degree $r$ and leading coefficient $dr!$. Q1: In characteristic zero, it is known that $(\mathrm{Chow}_r)_\mathrm{red}$ coinicides with the reduction of Angéniol's functor (see above) and for projective schemes the latter is known to coincide with the classical Chow variety [R4, Cor 16.3], which is projective after fixing the degree $d$. There is also an explicit map $\mathrm{CH}\colon \mathrm{Chow}_{r,d}(X/S)\to \mathrm{Div}^d(G/S)$ where $G$ is a suitable Grassmannian [R4, Def 17.4]. Since $\mathrm{CH}_\mathrm{red}$ is a closed immersion, $\mathrm{CH}$ is at least finite and hence $\mathrm{Chow}_{r,d}$ is projective (if representable, otherwise Angéniol's variant). Q2: As Jason mentioned, you need a tiny bit to ensure that the Hilb-Chow morphism is surjective: a geometric $k$-point of Chow is simply a cycle and such can be lifted to a subscheme if it has at least codimension $1$ (in the local ring of every generic point of the support of the cycle, choose a closed subscheme with length equal to the multiplicity and take the closure). The restriction of $\mathrm{Hilb}_r(X/S)\to \mathrm{Chow}_r(X/S)$ to the open subscheme parametrizing families of normal equidimensional subschemes is an open immersion [R4, Cor 12.9]. At least in characteristic zero: I think there is a characteristic zero assumption missing in Cor 12.9 (12.8 and 12.9 implicitly rely on Thm 9.8) but with the "correct" definition of $\mathrm{Chow}$ this should work out (Thm 9.8 should then hold in any characteristic). The restriction of $\mathrm{Hilb}_r(X/S)\to \mathrm{Chow}_r(X/S)$ to the open subscheme parametrizing reduced equidimensional subschemes is a monomorphism [R4, Cor 9.10]. The point is that the flat subscheme $Z$ can be recovered from its associated family of cycles. Although [R4, Cor 9.10] assumes characteristic zero (via Cor 9.9 which for any family of multiplicity-free cycles gives a unique potentially non-flat $Z$), one can in any characteristic prove [R4, Cor 9.10] directly from [R4, Prop 9.2] ($Z$ is in fact unique). I do not know if $\mathrm{Hilb}_P(X/S)\to \mathrm{Chow}_r(X/S)$ is an immersion after restricting to reduced equidimensional subschemes or even integral subschemes. This is claimed, without proof, in [R4, Rmk 9.11] (it also does not follow from the second sentence of the remark as claimed even if we knew that Chow_r was representable). Indeed, it could very well be false as Jason indicates (I don't know if Mumford's example gives a counter-example though). One would obtain a counter-example if one has a non-flat family $Z\to T$ with $Z$ and $T$ reduced such that the fibers are irreducible and generically reduced and their reductions all have the same Hilbert polynomial. Q3: Angéniol requires $X\to S$ to be smooth (which I think can be generalized to smoothly embeddable). He heavily relies on using cotangent sheaves but one could perhaps replace these with dualizing complexes but I think a lot of work would be required (IIRC, the most involved part is how the cohomology class representing the cycle behaves under change of projections and this could be difficult to generalize). Q4: Unfortunately, even now I can say nothing about the representability of $\mathrm{Chow}_r$ in positive characteristic. The key would be to understand the deformation theory of families of cycles and at the time I wrote my thesis I knew very little about deformation theory. I thought, and still think, that it can be done with a reasonable amount of work for (1) multiplicity-free cycles and (2) relative Weil-divisors such that every irreducible component meets the smooth locus (e.g., families of semi-log canonical pairs). Q5: No, not as far as I know. I think this case is as difficult as the general case.<|endoftext|> TITLE: What can be the applications of a theory of schemes à la Grothendieck to the category of groups? QUESTION [8 upvotes]: Baumslag, Miasnikov, and Remeslennikov have developed in [1] a theory similar to the classical theory of algebraic geometry in the category of groups. Let $G$ be a group, a $G$-group is defined by a couple $(H,f_H)$ where $f_H:G\rightarrow H$ is an (injective) morphism. For every element $x\in H$, we denote by $G(x)$, the subgroup of $H$ generated by $\{f_H(g)xf_H(g)^{-1}, g\in G\}$. An element $x\neq 1$ of $H$ is a divisor of zero if there exists an element $y\neq 1$ such that $[G(x),G(y)]=1$. A $G$-group $H$ is a domain if there does not exist a divisor of zero in $H$. An ideal $I$ of $(H,f_H)$ is just a normal subgroup of $H$ distinct of $H$. If $I$ of $H$, $(H/I,f_{H/I})$ is a $G$-group where $f_{H/I}$ is the composition of $f_H$ and the quotient morphism $H\rightarrow H/I$. The ideal $P$ is a prime ideal if and only if $(H/I,f_{H/I})$ is a domain. This is equivalent to saying that for every $x,y\in H$, $[G(x),G(y)]\subset P$ implies that $x\in P$ or $y\in P$. Let $G[X_1,...,X_n]$ be the free product of $G$ and the free group generated by $n$ elements. Every element $f$ of $G[X_1,...,X_n]$ defines a map $f_P:H^n\rightarrow H$ such that $f_P(h_1,..,h_n)$ is obtained by replacing $X_i$ by $h_i$ in the expression of $f$. Such a function is called a polynomial function. For every subset $S$ of $G[X_1,...,X_n]$, $V_H(S)=\{x\in H^n:\forall f\in S, f_P(x)=1\}$. The Zariski topology of $H^n$ is the topology whose set of subsets is generated by $\{S\subset G[X_1,...,X_n],V_H(S)\}$. I have started to adapt the point of view of Grothendieck to this theory. Here I consider the comma category $C(G)$ whose objects are morphisms $f_H:G\rightarrow H$.Let $Spec_G(H)$ be the set of prime ideals of $H$. For every normal subgroup $I$ of $H$, $V_H(I)$ is the set of prime which contains $I$. I have noticed that $V_H([I,J])=V_H(I)\bigcup V_H(J)$ and for a family of normal subgroup $(I_a)_{a\in A}$, let $E_A$ be the group generated by the family $(I_a)_{a\in A}$, $V(E_a)=\cap_{a\in A}V(I_a)$. This endows $Spec_G(H)$ with a topology. Let $U$ be an open subset of $Spec_G(H)$, we denote by $O_H(U)$ the set of functions defined on $U$ such that for every $f\in O_H(U)$, for every $P\in U$, there exists an open subset $V$ of $U$ which contains $P$, an element $h_V\in H$ such that for every $Q\in V$, $f(Q)=l_Q(h_V)$ where $l_Q:H\rightarrow H/Q$ is the quotient map. The correspondence $U\rightarrow O_H(U)$ defines a sheaf on $Spec_G(H)$. A $G$-scheme $(X,O_X)$ is a topological space $X$ endowed with a sheaf $O_X$ such that there exists an open cover $(U_i)_{i\in I}$ of $X$, for every $i\in I$ a $G$-group $H_i$ such that there exists an isomorphism of $G$-spaces between $(U_i,{O_X}_{U_i})$ and $(Spec_G(H_i),O_{H_i})$. See the second reference for more details. We can also define another notion of prime by saying that an ideal $I$ is a prime if and only if for every $x,y\in H$ such that $G(x)\cap G(y)\subset I$, $x\in I$ or $y\in I$ and develop a similar theory. Question. The theory of schemes has very important applications in algebraic geometry, what are the problems in group theory which can be solved by using this framework ? Let $G$ be a group, we can study the spectrum $Spec_G(G)$. Suppose that $G$ is the free group generated by $n$ elements. A closed subset of $Spec_G(G)$ is defined by a normal subgroup $I$ of $G$ which is the presentation of a finitely generated group. What can be the interpretation of the topology of $Spec_G(G)$ in this case ? Is anyone familiar with this construction ? Baumslag, G, Miasnikov, A. Remeslennikov, V.N. Algebraic geometry over groups I. Algebraic sets and ideal theory. J. Algebra. 1999, 219, 1679. Tsemo Aristide. Some properties of G-schemes. http://xxx.lanl.gov/pdf/1708.00359v1 REPLY [10 votes]: The topology you discuss is closely related to the Gromov--Grigorchuk topology on the space of marked groups. You might be interested in this nice paper of Champetier--Guirardel. In recent years, most of the work on this topic has been aimed at the case when $G$ is free. (Indeed, I think this is what Baumslag et al. had in mind.) The culmination of this work is Zlil Sela's solution to Tarski's problems on the elementary theory of the free group, and the parallel project of Kharlampovich--Myasnikov. As I think you'll see in the Champetier--Guirardel paper, for specific groups $G$, the "soft" Grothendieck-ian point of view encompassed by the topology on $\mathrm{Spec}(G)$ doesn't get you very far -- I think the point is that non-commutative groups are sufficiently "hard" that even basic properties don't hold in general, and so are inaccessible to "soft" techniques. To give you some idea of the techniques involved, Sela's work hinges entirely on the fact that a free group $F$ acts freely on a tree. He needs to analyse infinite sequences of homomorphisms $f_i:H\to F$ (I hope my $H$ is the same as yours, but am not sure) and he always does so by passing to a limiting action of $H$ on a real tree, and analysing the dynamics of this action. All very un-Grothendieck-ian! Kharlampovich--Myasnikov's techniques are superficially more combinatorial, but morally amount to much the same thing.<|endoftext|> TITLE: Kazhdan-Lusztig theorem for composition factors of Verma modules QUESTION [8 upvotes]: The Kazhdan-Lusztig Conjecture (which is actually a theorem) gives the character of some irreducible modules of a (say) simple complex Lie algebra $\mathfrak{g}$ in terms of characters of Verma modules. More precisely, following the notations of Representations of Semisimple Lie Algebras in the BGG Category O, by Humphreys, we define for $w$ in the Weyl group $M_w(\lambda)$ to be the Verma module with highest weight $w(\lambda + \rho) - \rho$ $L_w(\lambda)$ to be the irreducible module with the same highest weight Then if we choose $\lambda = -2 \rho$, we have (see Conjecture 8.4 in Humphreys's book) $$\mathrm{ch}\, L_w (\lambda) = \sum\limits_{x \leq w} (-1)^{\ell (x,w)} P_{x,w} (1) \, \mathrm{ch}\, M_x (\lambda) \, . $$ In this formula, the polynomials $P_{x,w}$ are the Kazhdan-Lusztig associated to the Weyl group, and we use the Bruhat ordering. Is this formula still valid for (antidominant) weights $\lambda \neq -2 \rho$ ? REPLY [4 votes]: It's valid for integral weights of the form $\lambda-2\rho$, with $\lambda$ antidominant (these are the anti-dominant weights that have a dominant weight $w_0\lambda$ in their orbit under the dot action). The proof is that if you tensor the simples $L_w(-2\rho)$ and Vermas $M_w(-2\rho)$ with the simple represenation of lowest weight $\lambda$, and then consider the summand where the center of the universal enveloping algebra acts with the expected character, you get the simples $L_w(\lambda-2\rho)$ and Vermas $M_w(\lambda-2\rho)$, on the nose. If $\lambda$ is not integral, then the formula is wrong; also if we choose a weight that doesn't have a dominant weight in its dot orbit (a singular block), then the answers are different as well.<|endoftext|> TITLE: Locus of roots of all convex combinations of two monic polynomials QUESTION [7 upvotes]: Let $p,q$ be monic polynomials in $\mathbb{C}[t]$ and for $\alpha \in [0,1]$, let $c_\alpha := \alpha p + (1-\alpha)q \in \mathbb{C}[t]$. Since the roots of a polynomial vary continuously with respect to its coefficients, following the fundamental theorem of algebra, the locus $$ L(p,q) := \{ z \in \mathbb{C}\mid c_\alpha(z)=0,~\alpha\in[0,1]\} $$ consists of $n$ (not necessarily distinct) continuous paths. A priori determination of the endpoints of the paths is ostensiblty diffcult, but experimental evidence suggests the following. Conjecture. Let $P = \{ \lambda_1,\dots,\lambda_n\}$ and $Q=\{\mu_1,\dots,\mu_n\}$ denote the roots of $p$ and $q$, respectively. For $1 \leq i,j\leq n$, let $d(i,j):= |\lambda_i- \mu_j|$. If $$\text{argmin}_{(i,j)}(d) =\{(k,\ell)\},$$ then there is a path from $\mu_\ell$ to $\lambda_k$. I am interested in knowing whether the above conjecture is known or if it can be established from what is known from complex analysis and the geometry of polynomials. Note that it is not clear what happens when there is a tie. Notice that if $p$ and $q$ share a simple root $\lambda$, then there is a (degenerate) path from that root to itself (this corresponds to the case when the distance is zero). The picture below contains a typical example generated from the following MATLAB code p=[1 randn(1,5)+i*randn(1,5)] q=[1 randn(1,5)+i*randn(1,5)] hold on scatter(real(roots(p)),imag(roots(p)),'x','r') scatter(real(roots(q)),imag(roots(q)),'o','b') for k=0:.01:1 c=k*p+(1-k)*q; scatter(real(roots(c)),imag(roots(c)),'.','m'); end Update: Per Christian's answer below, the conjecture, as stated above, is not true; however, I am still interested in a priori determination of the paths (new question posted). REPLY [4 votes]: Whatever the truth of the conjecture is (well, Christian's answer tells us...) the zeros of linear combinations of polynomials have been studied, see for example: On the zeros of a linear combination of polynomials (Pacific Journal, 1966m Robert Vermes).<|endoftext|> TITLE: Schubert calculus expressed in terms of the cotangent space of the Grassmannians QUESTION [14 upvotes]: Let $T^*_{\mathbb{C}}(Gr_{n,r})$ denote the cotangent space of the Grassmannian of $r$-planes in $\mathbb{C}^n$. Moreover, let $\Lambda^\bullet$ denote the exterior algebra of $T^*_{\mathbb{C}}(Gr_{n,r})$. Condsidering $Gr_{n,r}$ as the homogeneous space $U_n/(U_r \times U_{n-r})$, we have a unique representation of $U_n/(U_r \times U_{n-r})$ on $\Lambda^{\bullet}$ for which the associated homogeneous vector bundle is the direct sum $\bigoplus_{k \in \mathbb{N}} \Omega^k$. (i) Just as for any homogeneous space, every de Rham cohomology class of $Gr_{n,r}$ has a $G$-invariant representative. Moreover, every $G$-invariant element must be harmonic, and so, gives by Hodge decomposition a cohomology class. Is it correct to conclude from this that the cohomology group $H^\bullet$ is isomorphic as a vector space to the space of $U(r) \times U(n-r)$-invariant elements in $\Lambda^\bullet$? (ii) With respect to a standard weight basis of $T^*(Gr_{n,r})$, what do the $U(r) \times U(n-r)$-invariant elements look like, and how does this presentation of Schubert calculus relate to the partition presentation given in this question? REPLY [5 votes]: Regarding your second question, I think the answer is in the famous Kostant "Lie Algebra Cohomology and the Generalized Borel-Weil Theorem" or rather its second part.<|endoftext|> TITLE: When is "metric dimension" well defined? QUESTION [9 upvotes]: A subset $B$ of a metric space $(M,d)$ is called a metric generating set if and only if $$[\forall b \in B, d(x,b)=d(y,b)] \implies x = y \,. $$ A metric generating set $B$ is called a metric basis if it is minimal with respect to inclusion among metric generating sets (in obvious analogy to vector spaces). The metric dimension of $(M,d)$ is the cardinality of any metric basis. Question: For which metric spaces is metric dimension well-defined? When can we be sure that any metric basis for a metric space has the same cardinality? Sufficient criteria will suffice for answers, as will necessary criteria, although of course the holy grail of answers would be a non-trivial necessary and sufficient criterion. Note: This is a follow-up to my previous question. There, the accepted answer pointed out that the notion of metric dimension does not make sense in arbitrary metric spaces. In a matroid, any basis has the same cardinality, but there are metric spaces with metric generating sets of minimal yet non-equal cardinalities. Nevertheless, it does seem possible that metric dimension may make sense for certain classes of metric spaces, e.g. Euclidean spaces (Murphy, A Metric Basis Characterization of Euclidean Space, 1975) or graphs (Ramirez-Cruz, Oellermann, Rodriguez-Velazquez, The Simultaneous Metric Dimension of Graph Families, 2015). It is unclear to me what property common to these two types of metric spaces allows the definition to be well-formed/well-defined for them. In the case of Euclidean spaces, it seems intuitively clear that this notion should be related to that of affine independence, but coordinate-free definitions of affine independence (solely in terms of the metric) are rare (e.g. section 2.6 here), so I am still working on the algebra to show the connection. REPLY [4 votes]: Metric dimension is well-defined The usual definition of metric dimension (and the one initially given in the OP) is the smallest cardinality of any metric basis. This generalizes the notion for metric dimension in graphs and is well-defined for any metric space (finite or not). To see this let $(M,d)$ be a metric space and $\mathcal{G}$ be the collection of metric generating sets. There are two natural posets one can define on $\mathcal{G}$: let $P_1 = (\mathcal{G}, |\cdot|)$ be defined by $G \prec G'$ if $|G| < |G'|$ and let $P_2 = (\mathcal{G}, \subset)$ be defined by $G \prec G'$ if $G \subset G'$. Note that the minimal elements of $P_1$ are the metric bases $\mathcal{B}(M,d)$. Also notice that the metric bases are contained in the minimal elements of $P_2$ and that this containment is generally strict. Metric Bases and Matroids Fix a metric space $(M,d)$ where $M$ is a finite set. Let $r$ be the metric dimension of $(M,d)$. Since the metric bases $\mathcal{B}(M,d)$ of $(M,d)$ all have the same cardinality one can ask if and when $\mathcal{B}$ is the set of bases of a matroid. Recall that a collection $\mathcal{B}$ of (finite) sets is the set of bases of a matroid if it satisfies the following exchange axiom: for every $B,B' \in \mathcal{B}$ and every $e \in B$ there is some $f \in B'$ such that the set $B \setminus \{e\} \cup \{f\}$ is also in $B$. First let's see that there are some finite metric spaces whose metric bases are the bases of a matroid. Let $G = (V,E)$ be an undirected connected graph and let $d: V \times V \to \mathbb{N}$ be the map that takes a pair of vertices to the length of the shortest path between them. Then $M(G) := (V,d)$ is a metric space. A simple computation shows the following fact. Fact: Let $G = (V,E)$ be a simple connected graph with $|V| \le 4$. Then the metric bases of $M(G)$ are the bases of a matroid. This fact does not extend to all graphs with $|V|=5$. To see this consider the graph $G = ([5], \{13,14,15,24,25,35\})$. Then $M(G)$ has 22 metric generating sets and six metric bases $$\mathcal{B}(M(G)) = \{12, 15, 23, 24, 25, 34\}.$$ Notice that for $B = 12$, $e = 2$, and $B'= 34$ there is no element $f \in B'$ such that $B \setminus 2 \cup f$ is also a metric basis. So the metric bases of $M(G)$ are not the bases of any matroid. This graph is unique among simple connected graphs on five vertices in that it is the only one whose metric bases are not matroidal. Let $f(n)$ be the number of the simple connected graphs on $n$ vertices and let $g(n)$ be the number of such graphs whose metric bases are not matroidal. We used these Macaulay2 scripts to compute $f(n)$ and $g(n)$ for $n \le 7$. n = 1 2 3 4 5 6 7 f(n) = 1 1 2 6 21 112 853 g(n) = 0 0 0 0 1 18 323 More on when the metric bases of a graph are matroidal can be found in these two papers: BC2011 and B2013. Another example Finally let's return to Example 1 of this answer to the previous question. In that example we have $M = \{0,1,2,3\}$ and $d$ given by $$d(x,y) = \begin{cases} 2 \text{ if } x,y \neq 0 \\ 2 + \frac{1}{y} \text{ if } x = 0. \end{cases}$$ The metric generating sets of $(M,d)$ consist of all subsets of $\{0,1,2,3\}$ other than the singletons $\{i\}$, where $i \in \{1,2,3\}$. In particular, $\{0\}$ is a metric generating set of cardinality one. So the metric dimension of $(M,d)$ is one and $\{0\}$ is the only metric basis. It follows that the set of metric bases $\mathcal{B}(M,d)$ is matroidal with the corresponding matroid on four elements being isomorphic to the uniform matroid $U_{1,1}$ together with three loops.<|endoftext|> TITLE: Can the group of holomorphic automorphisms of an open subset of the complex plane be isomorphic to the additive group of real numbers? QUESTION [12 upvotes]: We can construct open sets in the complex plane $\mathbb{C}$ whose automorphism group is isomorphic to $\mathbb{Z}$, but is there an open set whose automorphism group is isomorphic to $\mathbb{R}$? REPLY [16 votes]: No, we cannot. Let $S$ be a connected noncompact Riemann surface whose group of conformal automorphisms is nondiscrete. Then (this follows from the uniformization theorem) $S$ is either conformal to the cylinder or to the annulus or to the once punctured disk or to the disk. Now, it is elementary to observe that the (connected components of the identity, which have index at most 2 in the full automorphism group) groups of conformal automorphisms of these surfaces are respectively ${\mathbb C}^\times$, $S^1$, $S^1$ and $PSL(2, {\mathbb R})$. Edit. In the case $S$ is disconnected one can argue as follows. If two components of $S$ are conformal to each other then $Aut(S)$ contains order 2 elements, while ${\mathbb R}$ is torsion-free. Now, assume that none of the components are conformal to each other. You have the connected components $S_j, j\in J$. Then $$ Aut(S)\cong \prod_{j\in J} Aut(S_j).$$ Then, none of the exceptional cases $Aut_0(S_j)\cong {\mathbb C}^\times, \cong S^1, \cong PSL(2,R)$, can occur (since they all contain nontrivial elements of finite order), hence, each $S_j$ is a hyperbolic surface with nonabelian fundamental group. Then $Aut(S_j)$ is discrete and, moreover, contains no infinitely divisible infinite order elements, i.e. infinite order elements $g$ such that $g^{1/n}$ exists for infinitely many natural numbers $n$ (for this you need to know a little bit about Fuchsian groups, I can explain if you are interested). This means that $Aut(S)$ also has no infinitely divisible elements. However, the entire group ${\mathbb R}$ is divisible. A contradiction. REPLY [14 votes]: It cannot be $R$. Consider two cases: a) Your region $D$ is hyperbolic Then $D=H/G$, where $H$ is the upper half-plane, and $G$ a discrete group. Let $\Gamma$ be the pullback of your group of automorphisms. Then we have $\gamma g=\beta(g)\gamma$ for every $\gamma\in\Gamma$ and every $g\in G$ and some automorphism $\beta:G\to G$. As $\Gamma$ is not discrete, but $G$ is discrete, we conclude that $\beta=\mathrm{id}$, so $G$ and $\Gamma$ commute. But the centralizer of any element $h$ of $SL(2,R)$ is the one-parametric subgroup passing through $h$. We conclude that $G$ is contained in a one-parametric group. As $G$ is discrete, and infinite (unless it is trivial), it must be isomorphic to $Z$, or be trivial. Then $D$ is a punctured disc or a ring or the disk if $G$ is trivial. In all three cases the full group of automorphisms is not isomorphic to $R$. b) If $D$ is parabolic, it is either $C$ or $C^*$. In both cases the full group of automorphisms is not isomorphic to $R$.<|endoftext|> TITLE: Higher Chow groups revisited QUESTION [6 upvotes]: Let $X$ be an algebraic variety over a field $k$. Bloch defines the "algebraic singular complex" using the algebraic simplices $$\Delta^n = \text{Spec}(k[x_0,\dots,x_n]/(x_0+x_1+\dots+x_n=1) \subset \mathbf{A}^{n+1}_k$$ that are easily arranged into a cosimplicial scheme $\Delta_k^{\bullet}$. Then one defines: $$z^i(X, j)= \mathbf{Z}[C_{ij}]$$ the free abelian group on $C_{ij}$, where $C_{ij}$ is the set of integral closed subschemes in $X\times_k\Delta^j$ of codimension $i$ and in "good position", ie. intersecting every face of $\Delta_k^j$ in $X\times_k\Delta_k^j$ in codimension $\ge i$. See https://www.uni-due.de/~bm0032/publ/CycleComplexes.pdf for a quick intro. The cosimplicial structure on $\Delta_k^{\bullet}$ makes $z^i(X,\bullet)$ into a simplicial abelian group, and the Zariski hypercohomology of the associated complex (take the differential to be the alternating sum of the degeneracy maps) is denoted $H^{\bullet}(X, \mathbf{Z}(j))$ because if $X$ is smooth it agrees with motivic cohomology. Let us construct a similar complex, using, instead of $\Delta_k^{\bullet}$, rather the cosimplicial scheme $\Gamma_k^{\bullet}$ where $\Gamma_k^n$ is the projective closure of $\Delta_k^n$ in $\mathbf{P}^{n+1}_k$, i.e.. the hyperplane $x_0 + \dots + x_n = x_{n+1}$ in $\mathbf{P}^{n+1}_k$. As before indeed $\Gamma_k^{\bullet}$ forms a cosimplicial scheme in a way that is completely analogous to $\Delta_k^{\bullet}$. Then as before we define $Z^i(X, j)$ to be the free abelian group on integral closed subschemes in $X\times_k\Gamma_k^{j}$ intersecting every face in $\Gamma_k^j$ (included the one at $\infty$) in the product, in codimension $\ge i$. It looks like the last condition still ensures that pullbacks between the abelian groups $Z^i(X,j)$ are defined, and hence we still get a simplicial abelian group, and then a complex as before. There's an obvious map between complexes of abelian groups $z^i(X,\bullet)\to Z^i(X,\bullet)$. My question is. Is the above map a quasi-isomorphism? After all, if we analyze $Z^i(X,\bullet)$ in degree zero and one, we get the usual presentation for $\text{CH}^i(X)$ where rational equivalence is defined comparing cycle classes in $X\times_k\mathbf{P}_k^1$ and not in $X\times_k\mathbf{A}^1_k$. Related question: What do higher Chow groups mean? REPLY [6 votes]: Good point. Can provide details as needed, but first let me suggest the answer in the form of an exercise. Take $X$ to be a smooth projective curve over a number field $k$. Use de Franchis' Lemma to show your proposed complex has finite cohomology group in degree $3$ and weight $1$, while Bloch's complex typically won't. It seems there are reasons from $K$-theory for why using $\Gamma_k^{\bullet}$ vs $\Delta_k^{\bullet}$ doesn't always work. Another intuition should probably come from the Dold-Thom thm in topology, whose analogue should be seen as the quasi-isomorphism between the Bloch complex and the Suslin complex, this latter morally being "algebraic singular cohomology" of $\text{Sym}_k(X) = \varinjlim_{n\ge 0}\text{Sym}_k^n(X)$. Using $\Gamma_k^{\bullet}$, $k$-morphisms $\Gamma_k^i\to\text{Sym}^n_k(X)$ are "too few".<|endoftext|> TITLE: What condition makes unitary reductive group unramified? QUESTION [5 upvotes]: I am a little bit confused with the definition of an unramified unitary group. Let $F$ be a local field of characteristic zero whose residue field is finite field of characteristic $p$. Then for a connected reductive group $G$ defined over $F$, recall that $G$ is unramified if it is quasi-split and split over maximal unramified extension of $F$. Let $E/F$ be an unramified quadratic field extension and $V$ is a hermitian space over $E$. Then unitary group $U(V)$ which is a subgroup of $GL_E$ perserving hermitian form of $V$. Then I am wondering whether $U(V)$ is unramified reductive group. If not, what conditions ensures $U(V)$ unramified? Any comments will be highly appreciated. REPLY [3 votes]: As Mikhail Borovoi explained in a comment, the question reduces to "when is a unitary group over a non-Archimedean local field quasi-split"? The answer does not distinguish between ramified or unramified separable quadratic extensions. Let $E/F$ be a separable quadratic extension of non-Archimedean local fields. Isomorphism classes of pairs $(V,h)$ where $V$ is a finite-dimensional vector space over $E$ and $h$ is a non-degenerate hermitian form on $V$ are classified by $(\dim V, \mathrm{disc}(h))$ where $\mathrm{disc} (h) = \det H \mod N_{E/F}(E^{\times})$ with $H$ the matrix of $h$ in some basis of $V$ over $E$. This is the analogue of the classification of quadratic forms (see Jacobson, A note on hermitian forms https://projecteuclid.org/euclid.bams/1183502551 ). In positive dimension every discriminant in $F^{\times} / N_{E/F}(E^{\times}) = \mathbb{Z} / 2 \mathbb{Z}$ occurs. As Mikhail Borovoi pointed out, in odd dimension these two forms $h_1,h_2$ have the same unitary group, in fact $h_2 \simeq \lambda h_1$ where $\lambda \in F^{\times} \smallsetminus N_{E/F}(E^{\times})$. In even dimension only one of the two forms gives rise to a quasi-split unitary group, i.e. Mikhail Borovoi's counterexample generalizes to arbitrary even dimension (although in even dimension greater than 2 the group is not anisotropic). There are probably several ways to see this. For example you can show that Borel subgroups of $U(V,h)$ defined over $F$ correspond bijectively to flags $V_1 \subset \dots \subset V_n$ where $\dim_E V = 2n$, $\dim_E V_1 = 1$, $\dim_E V_{i+1}/V_i = 1$ for $1 \leq i \leq n-1$, and $V_n$ is totally isotropic. Or you can argue by taking Galois cohomology of the short exact sequence of algebraic groups over $F$ (for the étale topology) $$ U(1) \rightarrow U(V,h) \rightarrow PSU(V,h) $$ and using $H^2(F, U(1)) = 0$ (a consequence of Tate-Nakayama) and the fact that the only inner form of a quasi-split group that is also quasi-split is the trivial one, which follows from the existence of a pinning defined over $F$. In fact all of this can be understood in the context of Galois cohomology of reductive groups. For example the classification of hermitian forms is essentially equivalent to the Hasse principle (vanishing of $H^1$ of a special unitary group).<|endoftext|> TITLE: Representation theory over any field QUESTION [5 upvotes]: I understand that representation theory of complex reductive groups is essentially combinatorial. By general principles, I imagine Galois theory then determines the theory over any field. For example, over a separably closed field, it works exactly the same way as over the complex numbers. Is that an accurate summary? Is there a reference where this exercise is worked out? REPLY [8 votes]: As already mentioned, the representation theory of a reductive depends heavily on the characteristic of the ground field. In particular, in positive characteristic it is of a very non-combinatorial nature. Nevertheless, the determinantion of the irreducible representations is quite uniform over all characteristics. In this case, Galois theory can be used to determine irreducible representations over any field. For connected reductive groups, this "exercise", as you call it, has been worked out by Tits in his paper Tits, J. Représentations linéaires irréductibles d'un groupe réductif sur un corps quelconque. J. Reine Angew. Math. 247 (1971) 196–220. He even considers irreducible representations over division fields. This exercise is much less trivial than it sounds since it involves, at a minimum, the classification of reductive groups over any field. At the end, the results are very pretty, though. Apart form highest weight theory they involve a homomorphism from the character group of the center of the group $G$ to the Brauer group of the ground field $k$.<|endoftext|> TITLE: Infinite dimensional symplectic geometry QUESTION [20 upvotes]: Could anyone comment on possible references concerning infinite dimensionsal symplectic manifolds?. I am mainly concerned with hilbert spaces, so i am not interested in the convenient analysis approach to the topic given, for instance, by Michor and collaborators, More spectifically, I am interested in an infinite-dimensional analogue of the Marsden-Weinstein reduction and applications. I am specially concerned with its possible uses in quantum mechanics. Thank you in advance. REPLY [9 votes]: A standard way of going from a symplectic manifold to quantum mechanics is Fedosov quantization. Now if you want to do this in infinite dimension, that means you are interested in quantum field theory. A good place to start is the thesis by Giovanni Collini (a former student of Stefan Hollands) which develops such a Fedosov quantization in the QFT context.<|endoftext|> TITLE: Wiener's axiomatization of the group law based on division QUESTION [9 upvotes]: Gian-Carlo Rota wrote that [*]: Wiener axiomatized the group law by taking $xy^{-1}$ as the basic operation, and his axiomatization is quite different from any of the other axiom systems for groups. Does anyone know what axiomatization Rota is referring to? [*] In "The Barber of Seville, or The Useless Precaution", from Indiscrete Thoughts. REPLY [6 votes]: One surprisingly nice axiomatization for this operator is: $$x/x=1$$ $$x/1=x$$ $$(x/z)\,/\,(y/z)=x/y$$ The proof is straightforward, and most easily dealt with by writing it out in detail: We translate sentences from group theory with $(x^{-1})^*=1/x$, $(xy)^*=x/(1/y)$. The first two axioms of group theory then are \begin{align} a1=a&: a\,/\,(1/1)=a/1=a\\ 1a=a&: 1\,/\,(1/a)=(a/a)\,/\,(1/a)=a/1=a\\ aa^{-1}=1&: a\,/\,(1/(1/a))=a/a=1\\ a^{-1}a=1&: (1/a)\,/\,(1/a)=1.\\ \end{align} Associativity is $$(ab)c=a(bc): (a/(1/b))\,/\,(1/c)=a\,/\,(1/(b/(1/c)))$$ which is a special case of \begin{align} (a/(1/b))\,/\,d &=(a/(1/b))\,/\,(d/1)\\ &=(a/(1/b))\,/\,((d/b)/(1/b))\\ &=a\,/\,(d/b)\\ &=a\,/\,((d/d)/(b/d))\\ &=a\,/\,(1/(b/d)). \end{align}<|endoftext|> TITLE: Pairs of roots of unity whose real part satisfies a polynomial identity QUESTION [7 upvotes]: Some motivation: The matrix $M$ is Butson Hadamard if the entries of $M$ are $k^{\textrm{th}}$ roots of unity (for some $k$), and distinct pairs of rows are orthogonal under the usual Hermitian inner product. I am interested in classifying the Butson Hadamard matrices for which some power is a real scalar matrix. (This is equivalent to the corresponding unitary matrix having finite multiplicative order.) In the case of $2\times 2$ matrices, after some reductions, the problem is equivalent to classifying the pairs of roots of unity $(\zeta_{1}, \zeta_{2})$ for which $$\Re( \zeta_{1} ) = \sqrt{2} \Re(\zeta_{2}) $$ This can be restated as $2\Re(\zeta_{2})^{2} - \Re(\zeta_{1})^{2} = 0$. The solutions we have found, up to negation and complex conjugation, are $(i, i)$, $(1, \omega_{8})$ and $(\omega_{8}, \omega_{6})$. We ran computer searches which suggest that any further solution involves a $k^{\textrm{th}}$ root of unity for some $k \geq 33$. Q1: Is this list of solutions to the displayed equation complete? Q2: More generally, what techniques exist for finding all solutions to a polynomial equation in real parts of roots of unity? REPLY [10 votes]: We have $$ \frac{\zeta_1+\frac{1}{\zeta_1}}{\zeta_2+\frac{1}{\zeta_2}}=\sqrt{2}. $$ By applying Galois group automorphisms we can assume that $\zeta_2=e^{2\pi i/n}$ for some $n$, where we can take either value of $\sqrt{2}$. Solving for $\zeta_1$ gives $$ \sqrt{2}\zeta_1=\zeta_2+\frac{1}{\zeta_2} \pm\sqrt{\zeta_2^2+\frac{1}{\zeta_2^2}}. $$ If $n>8$ then $\zeta_2^2+\frac{1}{\zeta_2^2}>0$. This implies that $\zeta_1$ is real, a contradiction. Thus there are no further solutions.<|endoftext|> TITLE: Does the clique-coclique bound hold for all walk-regular graphs? QUESTION [8 upvotes]: The clique-coclique bound is said to hold for a simple graph $G$ on $n$ vertices if $\lvert \omega(G) \rvert \lvert \alpha(G) \lvert \leq n$, letting $\omega(G)$ and $\alpha(G)$ denote its clique and coclique (independent set) numbers respectively. It is known, in particular, that the clique-coclique bound holds for all vertex-transitive graphs and distance-regular graphs - two families of walk-regular graphs. The clique-coclique also appears to hold for all of the examples of walk-regular graphs that I know of that are neither vertex-transitive nor distance-regular. It is also apparent that the clique-coclique bound holds for some other families of walk-regular graphs, namely semi-symmetric graphs. Could it be possible that the clique-coclique bound actually holds for all (connected) walk-regular graphs? By informal reasoning in head, it feels plausible to me that this could be the case? I wonder what would might be a good approach to take to try to prove or disprove this? REPLY [4 votes]: This is an old thread and already answered, but in case it helps anyone, here is a smaller walk-regular graph that does not satisfy the clique/coclique bound. (It may not be the smallest.) It has 18 vertices, and its automorphism group has two orbits on vertices (the pink and the blue in the picture). Graph6: QGEENDpMGwpppowomEHp`FBEKMG It has six maximum cliques of size $5$, with representative $\{0, 6, 7, 12, 13\}$ and six maximum cocliques of size $4$, with representative $\{0, 1, 9, 16\}$. As $\alpha \times \omega = 4 \times 5 = 20 > 18$, this violates the clique-coclique condition.<|endoftext|> TITLE: Are continuous rational functions arc-analytic? QUESTION [5 upvotes]: Let $X\subseteq\mathbb{R}^n$ be a smooth semi-algebraic set (for simplicity we can assume $X=B(0,r)$ is a small ball around the origin). A function $f:X\rightarrow \mathbb{R}$ is called a continuous rational function if it is continuous (w.r.t the euclidean topology) and there exists a Zariski open dense subset $U\subseteq \overline{X}^{zar}$ and a regular function $g$ on $U$ such that $f\restriction _{U\cap X}=g\restriction _{U\cap X}$. A function $f:X\rightarrow \mathbb{R}$ is called arc-analytic if $f\circ \gamma:(-\epsilon,\epsilon)\rightarrow \mathbb{R}$ is analytic for every analytic arc $\gamma :(-\epsilon,\epsilon)\rightarrow X$. Trying to get a hold of the hierarchy of functions on semi-algebraic sets I stumbled upon the following questions: 1) Is every continuous rational function arc-analytic? if not, what is the simplest counterexample? 2) Is there any natural homomorphism from the ring of arc-analytic functions on $X$ (or the stalk of the correspoing (pre-?) sheaf at a point $x\in X$) to some sort of "algebraic" ring? say the algebraic closure of $\mathbb{R}((x_1,...,x_n))$ or something like that? or are these inherently "non algebraic" functions (unlike all of their analytic friends)? REPLY [2 votes]: If $X$ is all of $\mathbf R^n$ then the answer to the first question is "yes", I think. Indeed, for any continuous rational function $f$ on $\mathbf R^n$ there is a stratification of $\mathbf R^n$ in Zariski-locally closed subsets such that the restriction of $f$ to any stratum is regular (Théorème 4.1 of the paper "Fonctions régulues"). It implies that $f\circ \gamma$ is meromorphic and continuous, hence analytic. As for the second question, by a Theorem of Bierstone-Milman (Theorem 1.1 of the paper "Arc-analytic functions"), any semi-algebraic arc-analytic function $f$ on $\mathbf R^n$ is blow-Nash. This means that there is a finite sequence of blow-ups with smooth algebraic centers $\pi\colon X\rightarrow\mathbf R^n$ such that $f\circ \pi$ is Nash. Hence, such functions are algebraic over $\mathbf R(x_1,\ldots,x_n)$, I suppose.<|endoftext|> TITLE: What is the intuition for the trace norm (nuclear norm)? QUESTION [34 upvotes]: I will word this question in terms of linear operators acting on $\mathbb{C}^n$ for simplicity. Feel free to provide an answer in terms of more general Hilbert spaces if you think it makes more sense that way. The standard norm induced by the inner product on $\mathbb{C}^n$ is the Euclidean norm $ \sqrt{\langle x, x\rangle} = \| x \|_2 = \sqrt{\sum_i |x_i|^2}$. Similarly, endowed with the inner product $\langle A, B\rangle = \text{trace}(A^* B)$, the space of $n \times n$ complex matrices forms an inner product space with the induced Frobenius norm $ \|A\|_2 = \sqrt{\text{trace}(A^* A)}$. However, there is a different norm that frequently comes up: it is the trace norm $\|A\|_1 = \text{trace}\left(\sqrt{A^* A}\right)$. There is a sense in which this norm is induced by the inner product on $\mathbb{C}^n$, since if $A = xx^*$, then $\|A\|_1 = \|x\|_2^2$. However, what is the "meaning" of the trace norm? The Euclidean and Frobenius norms have an intuitive meaning in a geometric sense, as the length of a vector. Why do we care about the trace norm? Is it precisely because it is induced by the "natural" Euclidean norm on $\mathbb{C}^n$? Additionally, if we express the trace norm in terms of the singular values of $A$, it corresponds to the L1 norm (i.e. sum of absolute values) of the singular values. Does the L1 norm on $\mathbb{C}^n$, i.e. $\|x\|_1 = \sum_i |x_i|$, have any interpretation, and does it share any similarity with the trace norm on $\mathbb{C}^{n\times n}$? (I apologise if these questions are basic, I have asked this question on Math StackExchange and received no responses, and I could not find any information about the intuition for these norms and their interrelations.) REPLY [7 votes]: If you are interesting in geometrical intuition, here is the possible one: Any matrix (operator) $A$ transforms a unit ball to an ellipsoid. Singular values of $A$ correspond to the lengths of ellipsoid's axis, which orientation is defined by singular vectors. So the nuclear norm represents the sum of the lengths of ellipsoid's axis. You can also play with link between $l_1$ norm and sparsity. Having genetal approximation problem : $$ \beta^* = argmin_{\beta}\|E(\beta)\|_1 = argmin_{\beta}\|A-M(\beta)\|_1 $$ the error matrix $E(\beta^*)$ will probably have some singular values equal to 0, which is not the case with $l_2$ norm. I don't know if all these facts can be useful for you, but at least it gives some intuition about this norm.<|endoftext|> TITLE: How to get this integral's asymptotics? QUESTION [8 upvotes]: Consider the following integral $$ \int_0^{\infty}\frac{e^{-x}-1}{x^{2+\frac{A}{\log b-5/6}}}\frac{1}{\log(b/x)-i\pi/2}\,dx $$ where $A>0$ and $b>0$. I am interested in the small $b$ asymptotics of this integral. Any ideas on how to proceed? REPLY [3 votes]: The integral in question is \begin{equation*} I:=\int_0^\infty f(x)\,dx, \tag{1} \end{equation*} where \begin{equation*} f(x):=\frac{e^{-x}-1}{x^{2-\varepsilon}}\frac{1}{\ln(b/x)-i\pi/2},\quad \varepsilon:=-\frac{A}{\ln b-5/6}\downarrow0. \end{equation*} Let \begin{equation*} C:=C_\varepsilon:=A-\tfrac56\,\varepsilon, \end{equation*} so that \begin{equation*} b=e^{-C/\varepsilon}. \end{equation*} We have \begin{equation*} \int_{b^{1/2}}^\infty f(x)\,dx \sim\int_{b^{1/2}}^\infty \frac{e^{-x}-1}{x^{2-\varepsilon}}\frac{dx}{\ln(b/x)} =\frac{J_1+J_2}{b^{1-\varepsilon}}, \end{equation*} where \begin{equation*} J_1:=\int_{\varepsilon^{1/2}/b}^\infty \frac{1-e^{-by}}{y^{2-\varepsilon}}\frac{dy}{\ln y} \le\int_{\varepsilon^{1/2}/b}^\infty \frac{dy}{y^{2-\varepsilon}}\frac{1}{\ln(\varepsilon^{1/2}/b)} \ll\varepsilon^{1/2}b^{1-\varepsilon}=o(b^{1-\varepsilon}) \end{equation*} and, with $u=\varepsilon\ln y$, \begin{equation*} J_2:=\int_{b^{-1/2}}^{\varepsilon^{1/2}/b} \frac{1-e^{-by}}{y^{2-\varepsilon}}\frac{dy}{\ln y} \sim b \int_{b^{-1/2}}^{\varepsilon^{1/2}/b} \frac{dy}{y^{1-\varepsilon}}\frac{1}{\ln y} = b \int_{C/2}^{C+\varepsilon\ln(\varepsilon^{1/2})}\frac{du}u\,e^u \sim b(\text{Ei}(A)-\text{Ei}(A/2)). \end{equation*} It follows that \begin{equation*} \int_{b^{1/2}}^\infty f(x)\,dx \to e^{-A}[\text{Ei}(A)-\text{Ei}(A/2)]. \tag{2} \end{equation*} Also, for \begin{equation*} g(x):=-\frac1{x^{1-\varepsilon}}\frac{1}{\ln(b/x)-i\pi/2}, \end{equation*} we have \begin{equation*} \int_0^{b^{1/2}} |f(x)-g(x)|\,dx \ll\int_0^{b^{1/2}} \frac{x^2}{x^{2-\varepsilon}}\frac{dx}{\pi/2}\to0. \tag{3} \end{equation*} Then, using the substitution $t=\varepsilon\ln(b/x)$, we have \begin{equation*} \int_0^{b^{3/2}} g(x)\,dx\sim -\int_0^{b^{3/2}}\frac{dx}{x^{1-\varepsilon}}\frac{1}{\ln(b/x)} =-e^{-C}\int_{C/2}^\infty \frac{dt}t\,e^{-t}=-e^{-C}\text{E}_1(C/2)\to -e^{-A}\text{E}_1(A/2). \tag{4} \end{equation*} Further, \begin{equation*} \int_{b^{3/2}}^{b^{1/2}} g(x)\,dx= \int_{b^{3/2}}^b g(x)\,dx+\int_b^{b^{1/2}} g(x)\,dx =\int_{b^{1/2}}^1 g(bu)\,b\,du+\int_{b^{1/2}}^1 g(b/u)\,\frac{b\,du}{u^2} =b\times\int_{b^{1/2}}^1 h(u)\,du, \end{equation*} where \begin{equation*} h(u):=g(bu)+g(b/u)/{u^2} =-2i e^{C(1/\varepsilon-1)}e^t\,\frac{2\pi\cosh(\varepsilon t)+4i t \sinh(\varepsilon t)}{\pi^2+4t^2} \end{equation*} and $t:=-\ln u$. (The latter identity is the key: note that $\Re g(x)$, as well as $\Re f(x)$, changes sign at $x=b$.) So, \begin{equation*} \int_{b^{3/2}}^{b^{1/2}} g(x)\,dx= -2i b e^{C(1/\varepsilon-1)}\int_0^{C/(2\varepsilon)}dt\,\frac{2\pi\cosh(\varepsilon t)+4i t \sinh(\varepsilon t)}{\pi^2+4t^2}. \tag{5} \end{equation*} Next, \begin{equation*} \int_0^{C/(2\varepsilon)}dt\,\frac{2\pi\cosh(\varepsilon t)}{\pi^2+4t^2} \to\int_0^\infty dt\,\frac{2\pi}{\pi^2+4t^2}=\frac\pi2 \tag{6} \end{equation*} and \begin{equation*} \int_0^{C/(2\varepsilon)}dt\,\frac{4i t \sinh(\varepsilon t)}{\pi^2+4t^2} =\int_0^{C/2}dz\,\frac{4i z \sinh z}{\varepsilon^2\pi^2+4z^2} \to\int_0^{A/2}dz\,\frac{4i \sinh z}{4z}=i\,\text{shi}\frac A2, \tag{7} \end{equation*} where $\text{shi}$ is the hyperbolic sine integral function. Collecting the pieces (1)--(7), we have \begin{equation*} I\to e^{-A}(\text{Ei}(A)-\text{Ei}(A/2))-e^{-A}\text{E}_1(A/2)-2i e^{-A} (\tfrac\pi2+i\,\text{shi}\tfrac A2) =e^{-A}(\text{Ei}(A)-i\pi), \end{equation*} which is the same expression as the one obtained by Carlo Beenakker.<|endoftext|> TITLE: Terminology: Lost in translation or multiple-meanings QUESTION [14 upvotes]: I was reading Uniformization of Riemann Surfaces by Henri Paul de Saint Gervais (not a real person, but a group of French mathematicians), and the translator kindly points out that the name of "the greatest theorem of the 19th century" comes from the French word uniforme, meaning single-valued (as opposed to multi-valued). It may still seem obscure, but in the text (the introduction) it was explained pretty well. For a lesser example, in the same introduction it talks about a group $\Gamma$ acting on the upper half plane (fixed point) freely and properly. That casual parenthetical remark cleared up the meaning of "free action" that I could never make sense of and had to look up its definition repeatedly. I hope it is okay to ask for more examples of this sort. (You are welcome to rephrase the question.) It may not strictly be the original intention, but may have contributed to its wide acceptance but has since been forgotten (e.g. What is the naming reason of poles in complex analysis?). As other Terminology questions and answers show (especially in algebra), this may be quite tortuous to explain properly, and may not be more helpful than "just learn the definition." For starters: what is so proper about a proper map? Updated: To avoid being too vague that any terminology can have a story behind it (meromorphic, homology, etc), here are some guiding criteria: 1) Due to translation/importing (most often from French and German), or the multitude of meanings of the English word itself, the original meaning of the terminology has been lost in most texts on the subject; 2) it has been generalized out of the original context; 3) and that the original meaning helps in the understanding of the terminology, without having to give a long explanation. REPLY [6 votes]: Poincaré is usually said to be the first one to use homologous. He uses the term in his Analysis situs to mean the relation between manifolds that we nowadays refer to as cobordism or bordism. I have always interpreted his usage to be an extension of the relation of homology (or perspectivity) between triangles in geometry: one can think of a bordism from a manifold M to another N as a (very flexible!) perspectivity from one to the other.<|endoftext|> TITLE: K-theory of an elliptic curve QUESTION [10 upvotes]: Given an elliptic curve $E$ over $\mathbb{Q}$, I have read somewhere (But I can't remember exactly where) that the Beilinson conjecture asserts that: The rank of the albelian group $K_{2}(E)$ (the second algebraic K-theory) is equal to the rank of the abelian group of the rational points $E(\mathbb{Q})$. Conjecture: $\textrm{rk } K_{2}(E)= \textrm{rk } E(\mathbb{Q})$ Question What are some evidences of such conjecture? Is it verified in some known cases? REPLY [6 votes]: I would like to add a couple of references to the ones provided in Timo Keller's comment. First of all, the conjecture is not correctly stated. According to the conjectures, the rank of the curve (which is related to the rank of $K_0(E)$) should equal the order of vanishing of the Hasse-Weil L-function of $E$ at $s=1$, but the rank of $K_2(E)$ should be the order of vanishing of the L-function at $s=0$. The correctly stated conjecture on the rank of $K_2(E)$ is that the rank of $K_2(E/\mathbb{Q})$ should be $1+S$ where $S$ is the number of places with multiplicative/semistable reduction. Alternatively, the rank of $K_2(\mathcal{E})$ should be 1, where $\mathcal{E}$ is a regular proper model over ${\rm Spec}\mathbb{Z}$ of the curve. More generally, for $E/F$ an elliptic curve over a number field $F$, the rank of $K_2(E/F)$ should be $[F:\mathbb{Q}]+S$. This is discussed in the article of Bloch and Grayson linked in Timo Keller's comment: Bloch, S. and Grayson, D., $K_2$ and $L$-functions of elliptic curves: computer calculations, Applications of algebraic $K$-theory to algebraic geometry and number theory, Part I, II (Boulder, Colo., 1983), Contemp. Math., 55 Amer. Math. Soc., Providence, RI, 1986, 79--88. As far as I know, there is not a single curve for which the conjecture is known. I don't think the full Beilinson conjectures are known for any scheme other than rings of $S$-integers, where they are the classical theorems of algebraic number theory. It seems that this is the only evidence we have. That and the marvelous beauty of the conjectures. Anyway, the conjecture explains the ranks of K-theory resp. motivic cohomology in terms of some analytic invariants which we can actually compute, namely Deligne cohomology. Beilinson's conjectures basically state that Deligne cohomology tells us everything about rational motivic cohomology. (For more precise formulations, look at survey articles on Beilinson's conjectures, such as the ones by Nekovar or Deninger-Scholl.) Ok, some partial results are known: the work of Goncharov-Levin has produced elements with non-trivial regulator, so we know that the rank must be at least one. Goncharov, A. B. and Levin, A. M., Zagier's conjecture on $L(E,2)$, Invent. Math. 132 (1998), no.2, 393--432. Analogous results are known in the function field case by the work of Kondo and Yasuda. Again, we know that the rank is at least the predicted rank, by explicit generators. S. Kondo and S. Yasuda. On the second rational K-group of an elliptic curve over global fields of positive characteristic. Proc. Lond. Math. Soc. (3) 102 (2011), no. 6, 1053–1098. There are further partial results, via K-theory with finite coefficients. A paper of Soulé establishes some results for p-adic K-theory. In particular, he gives examples where the rank of the $p$-adic K-theory is in fact equal to the rank predicted by Beilinson's conjecture. The method is to relate the $p$-adic K-theory to étale cohomology and then compute the rank of the latter. C. Soulé. $p$-adic K-theory of elliptic curves. Duke Math. J. 54 (1987), 249--269. There is also a paper of Kato. It's mostly concerned with generalizations of the Hasse principle, but there is some application to K-theory of curves over number fields. If you dive into the notation a bit, his Corollary 0.10 states that assuming finiteness of K-groups the conjectures on the ranks of $K_1$ and $K_2$ are equivalent. K. Kato. A Hasse principle for two-dimensional global fields. J. reine angew. Math. 366 (1986), 142-180. No upper bounds are known, the key problem is the yet unknown finite generation of K-groups...<|endoftext|> TITLE: Are free ultrafilters as posets product-irreducible? QUESTION [7 upvotes]: Let $\kappa\geq \aleph_0$ be a cardinal, and suppose that ${\cal U}$ is a non-principal ultrafilter on $\kappa$. We regard ${\cal U}$ as a poset $({\cal U}, \subseteq)$. Suppose that there are posets $P, Q$ such that ${\cal U} \cong P\times Q$. Does this imply one of $P, Q$ consists of one point only? REPLY [11 votes]: No. Every nonprincipal ultrafilter $U$, considered as a partial under $\subseteq$, is a nontrivial product order. To see this, suppose that $U$ is a nonprincipal ultrafilter on $\kappa$. Partition $\kappa=A\sqcup B$ into two sets with $A\in U$ and $B$ nonempty. Every $X\in U$ can be written as $X=(X\cap A)\sqcup (X\cap B)$, and furthermore, $X\subseteq Y$ just in case $(X\cap A)\subseteq (Y\cap A)$ and $(X\cap B)\subseteq (Y\cap B)$. Let $P=U\upharpoonright A=\{ X\subseteq A\mid X\in U\}$ and $Q=P(B)=\{X\mid X\subseteq B\}$. These are both nontrivial and $\langle U,\subseteq\rangle$ is isomorphic to the product order $\langle P,\subseteq\rangle\times\langle Q,\subseteq\rangle$ by the map $X\mapsto (X\cap A,X\cap B)$. Indeed, you don't even need the ultrafilter to be non-principal, provided $\kappa\geq 3$. The reason is that if $\kappa\geq 3$, then you can partition $\kappa=A\sqcup B$ where $A\in U$, $B$ is nonempty and $A$ has at least two points. In this case, both $P$ and $Q$ again will have at least two elements each, and the rest of the argument is as before. (Meanwhile, if $\kappa=1$ or $\kappa=2$, then $U$ has only one or two elements, respectively, and so it is not a nontrivial product.)<|endoftext|> TITLE: About primitively recursively recognizable ordinals QUESTION [12 upvotes]: Preliminary: I believe the notion of primitive recursive functions on ordinals is standard and unproblematic (the main difference with the finite case is that one needs to introduce a $\sup$ or $\limsup$ in definition of primitive recursion). If there is any doubt, I refer to the notion defined in either one of the following papers: Stephen G. Simpson, “Short Course on Admissible Recursion Theory”, in: Fenstad, Gandy & Sacks (eds.), Generalized Recursion Theory II (Oslo 1977), North-Holland (1978), p. 355–390, esp. §2. Jeremy Avigad, “An Ordinal Analysis of Admissible Set Theory Using Recursion on Ordinal Notations” (J. Math. Log. 2 (2002), 91–112; preprint version here), esp. §3–4. (If they are not equivalent, then there is something seriously wrong with my understanding of the universe.) Definition: Say that an ordinal $\alpha$ is primitively recursively recognizable (or p.r.-recognizable for short) if the ordinal function taking the value $1$ on $\alpha$ and $0$ on every other ordinal is primitive recursive (without parameters, of course). Remarks: Obviously every finite ordinal is p.r.-recognizable. Also, $\omega$ is p.r.-recognizable (because the predicate “$\alpha$ is finite” is p.r., for example it can be tested as $1+\alpha > \alpha$ and addition of ordinals is p.r.). Less obviously, I think the function taking the value $1$ on the admissible ordinals and $0$ otherwise is primitive recursive (I don't have a satisfactory reference, but Hinman, Recursion-Theoretic Hierarchies, Persp. Math. Logic. 9, Springer 1978, states something slightly weaker in corollary VIII.2.19, and I think the proof he gives actually yields the statement I wrote), so the $n$-th admissible ordinal (and in particular, the Church-Kleene ordinal) is p.r.-recognizable for every finite $n$; the same should also be true of the $n$-th recursively inaccessible ordinal (and much more). On the other hand, not every ordinal is p.r.-recognizable (because there are only countably many p.r. functions [without parameters]). More precisely, I think that if $L_\gamma \mathrel{\preceq_1} L_\beta$ with $\gamma<\beta$ (where $\preceq_1$ means “is a $1$-elementary submodel”) then no ordinal $\alpha$ such that $\gamma\leq\alpha<\beta$ can be p.r.-recognizable, because p.r. functions are absolute for the $L_\beta$ (right?), so if $L_\beta \models \exists \alpha.\varphi(\alpha)$ with $\varphi$ a p.r. predicate recognizing an ordinal, then $L_\gamma \models \exists \alpha.\varphi(\alpha)$ and $\alpha<\gamma$. [Edit: After reading the answer by Philip Welch, I now realize why this reasoning is incorrect: in writing $L_\beta \models \exists \alpha.\varphi(\alpha)$ I implicitly assumed that $\beta$ is p.r.-closed so that no value higher than $\beta$ is used in computing $\varphi(\alpha)$.] Question: What is the smallest non p.r.-recognizable ordinal? More precisely, how does it compare with the smallest $\alpha$ such that $L_\alpha \mathrel{\preceq_1} L_{\alpha+1}$? [Edit: After reading the answer by Philip Welch, I realize that the ordinal I should be asking for comparison with is the smallest $\alpha$ such that $L_\alpha \mathrel{\preceq_1} L_{\varphi(\omega,\alpha+1)}$.] Further comments: The smallest $\alpha$ such that $L_\alpha \mathrel{\preceq_1} L_{\alpha+1}$ is the smallest $\Pi^1_0$-reflecting ordinal, meaning $\Pi_n$-reflecting for every $n$: see Richter & Aczel, “Inductive Definitions and Reflecting Properties of Admissible Ordinals”, in: Fenstad & Hinman (eds.), Generalized Recursion Theory (Oslo 1972), North-Holland (1974), p. 301–381, specifically theorem 1.18 on p. 313&333. For some reason, I had gotten into my head (based on §3 of the aforementioned Richter&Aczel paper) that an ordinal is p.r.-recognizable if and only if, for some (first-order, i.e., $\Pi^1_0$) statement $T$ of the language of set-theory, it is the smallest $\alpha$ such that $L_\alpha \models T$ (this would solve the above question). But there's something seriously wrong, here [edit: no there isn't], because $\alpha$ is p.r.-recognizable iff $\alpha+1$ is, and the statement “there exists a largest ordinal $\gamma$ and $L_\gamma \mathrel{\preceq_1} L$” is first-order and the first $\beta$ such that $L_\beta$ satisfies it is precisely the first $\alpha+1$ such that $L_\alpha \mathrel{\preceq_1} L_{\alpha+1}$… so I run into a contradiction and there must be something seriously wrong with what I wrote. My main goal here is to understand the source of my confusion and dispel it. Because of this confusion, I'm also not sure whether the p.r.-recogizable ordinals are an initial segment of the ordinals. So let's make this into an: Extra question: Are the p.r.-recognizable ordinals an initial segment of the ordinals? If not, what is the smallest ordinal which is p.r.-recognizable but which is greater than at least one non-p.r.-recognizable ordinal? REPLY [6 votes]: Call $\alpha\in On$ p.r.closed, if every p.r. set function $f$ is total on $L_{\alpha}$. (Note (1) If $\alpha^\ast$ is the least p.r. closed ordinal $> \alpha$ then it is the next image of a point in the $\omega$'th Veblen function, so in any case is much smaller than the next admissible. Hence (2) the p.r.closed ordinals form a c.u.b. class, and in fact are c.u.b. below any admissible ordinal $>\omega$. (3) By Jensen-Karp [1] the p.r. functions on ordinals to ordinals are all the restrictions of the p.r. set functions to $On$. Hence it is easier to reason with the latter class.) Definition Let $\delta$ be p.r. reflecting if for all p.r. functions $F:On\rightarrow On$, if $F(\delta)\neq 0$ then $\exists \alpha < \delta (F(\alpha)\neq 0 )$. Then Claim $\delta$ not p.r. reflecting iff $\delta$ is p.r. recognizable. Now let $\beta$ be the least p.r. closed ordinal with a $\beta_0<\beta$ satisfying $L_{\beta_0}\prec_{\Sigma_1}L_\beta$. (So larger than the least $\Pi^1_0$-reflecting ordinal, but less than the least $\Pi^1_1$-reflecting ordinal.) By p.r. closure, and $\Sigma_1$-elementarity, the totality of the p.r. functions on $L_\beta$ ensures: Lemma The least p.r. reflecting ordinal is less than or equal to $\beta_0$. This gives an upper bound to the first p.r. reflecting, and so non-p.r. recognizable, ordinal, and will answer the first part of the Extra Quest. negatively, as we shall see below there are p.r. recognisable ordinals $>\beta$. As an imprecise lower bound one can take the least $\Pi^1_0$-reflecting ordinal mentioned above. [Edit: This is fully answered below at (C).] (If $\gamma$ is this ordinal, and so we have $L_\gamma \prec_{\Sigma_1}L_{\gamma +1}$, it is easy to see all ordinals $\tau < \gamma$ are p.r. recog. as for any such $\tau$ there is some sentence $B$ so that $L_\tau$ is the least level of the $L$-hierarchy where $B$ is true. But $\gamma+1$ is itself p.r. recog. as the first $\tau$ where $L_\tau$ has a proper $\Sigma_1$-substructure, etc., etc. (This answers the second part of the original Question) Similar statements hold for larger stretches of the ordinals $[0,\gamma']$ for $\gamma + 1 < \gamma'$.) To find the least upper bound of the p.r. recog. ordinals, one can reason as follows: If the $\Sigma_1$ sentence $A$ is first true at some stage $L_{\delta}$, then $\delta$ is p.r. recog. Consequently the p.r. recog. ordinals will be cofinal in $\sigma_1$ the least $\Sigma_1$-stable ordinal, i.e. the least ordinal with $L_{\sigma}\prec_{\Sigma_1}V $ (as new $\Sigma_1$-sentences become true cofinally in $\sigma_1$). Thus for example if $\delta$ is least so that $L_\delta$ is a $ZF^-$ model, it will be p.r. recog. Note then that no ordinal $\geq \sigma_1$ can be p.r.recongizable, as otherwise this would be a new $\Sigma_1$ fact true at a stage beyond $\sigma_1$. Thus: Lemma The supremum of the p.r. recognisable ordinals is $\sigma_1$. If $\tau_0$ is the least p.r. reflecting ordinal, and $\tau >\tau_0$ the least p.r. closed ordinal above that, then the answer to the second question of the Extra Question, is to take $\tau +1$. The $\Sigma_1$ statement that "there exists such a pair $\tau_0<\tau$ " can be used to recognise this ordinal. Comment: if the definition of recognizable is adjusted (call it recognizable*) so that $\alpha$ is recognizable* if for some p.r. function $F$, $F$ is everywhere $0$ except that for some (unique) $\tau$ $F(\tau)=\alpha$, (and thus $\alpha$ is the sole non-zero value that $F$ takes) then the recognizable* ordinals are precisely those $< \sigma_1$. [1] R.B. Jensen and C. Karp Primitive Recursive set functions, in Proceedings of Symposia in Pure Mathematics,vol.13 Part 1, "Axiomatic Set Theory", Ed. D. Scott, AMS, 1971, pp 143-167. Edit added to address Gro-Tsen's queries (see comment below) and intended to complete both the original Question and its second reformulation: (A) The function $F(\delta)=L_\delta$ is p.r. (See Devlin, "Constructibility" but I think it is in [1] anyway.) Satisfaction is likewise p.r., hence the function: $G(\delta)=1$ if $L_\delta \vDash $ '' $A\wedge \forall\delta' L_{\delta'} \vDash \neg A$''; $G(\delta)=0$ otherwise is p.r. and gives that the least $\delta$ with $L_\delta\vDash A$ is p.r. recognizable. (B) Let $\beta<\sigma_1$ . We want that $\beta$ is p.r. recognizable*. Let $A$ be a $\Sigma_1$ sentence that is first true at some $\gamma \in (\beta, \sigma_1)$. By standard arguments the $<_L$-least bijection $f_\gamma:\omega\leftrightarrow \gamma$ is definable over $L_\gamma$. Suppose $f_\gamma(n)=\beta$. By the kind of argument in (A) we see that $ \gamma$ is p.r. recog*. But then so is $L_{\gamma+1}$ and $f_\gamma$. Put these facts together to build a p.r. function $G$ whose only non-zero value is $G(\gamma)=\beta$. (C) We answer this (and, using the first Lemma above, finish the original Question) by: Lemma Let $\alpha < \beta_0$. Then $\alpha$ is p.r. recog. Hence the least p.r. reflecting ordinal is $\beta_0$ which in turn is the least non-p.r. recog. ordinal. Proof: Say that $\alpha$ begins a gap if $\exists \delta ( L_\alpha \prec_{\Sigma_1} L_\delta)$. We say that $[\alpha,\delta]$ is a gap, if $\alpha$ begins a gap, and $\delta$ is maximal with $L_\alpha \prec_{\Sigma_1} L_\delta$. $\bullet$ If $\omega < \delta < \beta_0$ is not in any gap, then we are sufficiently low down in the $L$-hierarchy where there is a $\Sigma_1$ sentence $A=A(\delta)$ so that $L_{\delta+1}\vDash $ ``$A\wedge\forall \delta’<\delta L_{\delta’ + 1}\vDash \neg A$ ‘’. As argued above at (A), this makes $\delta+1$ and so $\delta$ p.r. recog. We just need the ordinals of gaps $[\alpha,\delta]$ with $\alpha <\beta_0$ to be p.r. recognizable. So let $[\alpha,\delta]$ be such a gap. We show that $\alpha$ is p.r. recog. and variants of the argument suffice for the other ordinals in the gap. Then $\alpha < \delta < \alpha^* <\beta_0$. Using (un)/pairing functions (which are p.r.) etc. one has that the closure of $\{\alpha\}$ under p.r. functions includes all ordinals up to $\alpha^*$. So let $F$ be a p.r. function, with $F(\alpha)=\delta’$ for some $\delta’\in (\delta,\alpha^*)$. Let $A=A(\delta’)$. Let: $G(\xi) = 1$ if $L_{F(\xi)+1}\vDash A$; $G(\xi) = 0$ otherwise. Then $G$ witnesses that $\alpha$ is p.r. recognizable, so we are done. Q.E.D. (The definition of $\beta_0$ and $\beta$ is that given in Gro-Tsen's second formulation of question, in terms of the Veblen function (if I understand the terms correctly).)<|endoftext|> TITLE: A Schur-like product theorem on groups QUESTION [5 upvotes]: Let $G$ be a finite group, and consider the composition $X * Y$ on $\mathbb{C}G$ defined by $$(\sum_g \alpha_g u_g) * (\sum_g \beta_g u_g) = \sum_g \alpha_g \beta_g u_g.$$ This composition can be reformulated as a convolution using the Fourier transform. Question: Let $X,Y \in \mathbb{C}G$ be positive elements. Is $X * Y$ also positive? Remark: It is an application of Theorem 4.1 in http://dx.doi.org/10.1090/tran/6582 to finite groups theory, but we are here interested in a purely group-theoretic reference or proof. It is like to Schur product Theorem because $X * Y$ is an entrywise product like the Schur product. Bonus question: Can we extend to any locally compact group? REPLY [4 votes]: The answer is positive. For ease of notation, let me write $a=\sum_g \alpha_g u_g$ and $b = \sum_g \beta_g u_g$. Let's equip $\mathbb{C}G$ with the positive inner product induced by the standard basis. In terms of the usual normalized trace on a group algebra, this inner product is given by $(x,y)\mapsto\mathrm{tr}(x^*y)$. So when you decompose $\mathbb{C}G$ into a direct sum of matrix algebras, this is just the usual Hilbert-Schmidt inner product. In particular, the cone of positive elements is nicely self-dual: we have $y\geq 0$ if and only if $\mathrm{tr}(xy)\geq 0$ for all $x\geq 0$. Hence it is enough to show that $\mathrm{tr}\left(x(a \ast b)\right)\geq 0$ for all $x\geq 0$. Now your composition operation $*:\mathbb{C}G\otimes\mathbb{C}G\to\mathbb{C}G$ is the adjoint of the usual comultiplication $\Delta:\mathbb{C}G\to\mathbb{C}G\otimes\mathbb{C}G$, meaning that $$ \mathrm{tr}\left(x(a \ast b)\right) = \mathrm{tr}(\Delta(x)(a\otimes b)). $$ Finally, we have $\Delta(x)\geq 0$ since $\Delta$ is a $*$-homomorphism, and $a\otimes b\geq 0$ since $a\geq 0$ and $b\geq 0$ by assumption. I don't know about the generalization to the locally compact case.<|endoftext|> TITLE: Fixed-point-free group action on a finite, contractible, 3-dimensional simplicial complex QUESTION [11 upvotes]: Let $K$ be a finite simplicial complex with an admissible action of a finite group $G$. (Terminology: By an action of a group $G$ on $K$ I mean an action by simplicial automorphisms. The action is called admissible if for all $g\in G$ and all simplices $\sigma\in K$ the equality $g\sigma=\sigma$ implies that $gv=v$ for all vertices $v\in \sigma$. By $K^G$ I denote the subcomplex formed by stationary points of the action (the set of stationary points being a subcomplex is guaranteed by admissibility of the action). Note that the admissibility condition is not really relevant, since for any action of $G$ on $K$ the action induced on the barycentric subdivision is admissible.) Q1: Can $K^G$ be empty if $\dim(K)=3$ and $K$ is contractible? Q2: What if $K$ is additionaly collapsible (in the sense of Whitehead's simple homotopy theory)? Is suspect the answer is that $K^G$ can be empty. However, I wasn't able to find an example in the literature. I know that: If $\dim(K)=2$ and $K$ is collapsible, then $K^G$ is non-empty and collapsible [6]. If $\dim(K)=2$ and $K$ is contractible, then it is conjectured [8,9] that $K^G$ is non-empty (this, as far as I know, is open even for infinite $K$, see [7]). If $\dim(K)=2$ and $K$ is acyclic, then $K^G$ may be empty, and such actions have been classified [5]. If $\dim(K)=3$ and $K$ is collapsible, then $K^G$ is either empty or acyclic [6]. (My question 2 is whether the case $K^G=\emptyset$ can really occur.) If we allow $K$ to be infinite, then for contractible $K$ of dimension $3$ there exist finite group actions with $K^G=\emptyset$. [2,Corollary II.7.4]. There are fixed-point-free group actions on finite, contractible complexes $K$ of higher dimension (see [3] or [4]). Using [1] one can make these examples collapsible by considering products $K\times [0,1]^n$ for $n$ large enough. References: [1] K. Adiprasito, B. Benedetti, Subdivisions, shellability, and collapsibility of products; also available on arXiv. [2] A. H. Assadi, Finite Group Actions on Simply-Connected Manifolds and CW Complexes. [3] E. E. Floyd, R. W. Richardson, An action of a finite group on an n-cell without stationary points [4] R. Oliver, Fixed-point sets of group actions on finite acyclic complexes. [5] R. Oliver, Y. Segev, Fixed point free actions on Z-acyclic 2-complexes. [6] Y. Segev, Some remarks on finite 1-acyclic and collapsible complexes. [7] J. M. Corson, On Finite Groups Acting on Contractible Complexes of Dimension Two. [8] M. Aschbacher, Y. Segev, A Fixed Point Theorem for Groups Acting on Finite 2-Dimensional Acyclic Simplicial Complexes [9] C. Casacuberta, W. Dicks, On finite groups acting on acyclic complexes of dimension two REPLY [4 votes]: EDIT (2021-02-26): The authors of https://arxiv.org/abs/2102.11458, https://arxiv.org/abs/2102.11459 claim that they've proven that every action of a finite group on a finite and contractible 2-complex has a fixed point. It follows (see the reasoning below) that the answer to Q2 is negative (i.e. every action on a finite, collapsible 3-complex has a fixed point). After a second thought and re-reading Segev's proof that $K^G$ is either empty or acyclic for $3$-dimensional, collapsible simplicial complexes, I conclude that Q2 is open. Details below. Let $K$ be a collapsible, $3$-dimensional simplicial complex equipped with an admissible action of $G$. By the proof of [5, Theorem (5.1)] (reference numbering as in the question) we know that $K$ collapses to a $2$-dimensional, $G$-invariant subcomplex $L$, and $K^G$ collapses to $L^G$. If $K^G=\emptyset$, then $L^G=\emptyset$. But $L$ is a contractible, $2$-dimensional complex. This means that having such an action would answer in the negative the Casacuberta-Dicks-Aschbacher-Segev conjecture mentioned in the question: If $dim(K)=2$ and $K$ is contractible, then it is conjectured [8] that $K^G$ is non-empty (this, as far as I know, is open even for infinite $K$, see [7]). So if we know an answer to Q2, then it must be that a fixed-point-free action on a $3$-dimensional, collapsible simplicial complex does not exist. However, any paper proving this would probably cite [5], and I can't find such an article. Also note that if a fixed-point-free action on a collapsible $3$-complex does not exist and Zeeman's collapsibility conjecture is true (which is a very strong assumption), then we have a positive solution to the Casacuberta-Dicks-Aschbacher-Segev conjecture. Proof: Let $G$ act on a contractible $2$-complex $K$. For $n$ large enough, by Zeeman's conjecture and [1, Proposition II.2.1] the $n$-th barycentric subdivision of $K\times I$ is a collapsible $3$-complex, with $G$ acting non-trivially on the first coordinate. Thus, a fixed point of this action has to correspond to an element of $K^G$. Update: The answer to Q1 is positive - there exist fixed point free group actions on contractible $3$-complexes. Let $G$ be a finite group acting without fixed points on a $2$-dimensional, acyclic simplicial complex $X$. Then $G\times \mathbb{Z}_2$ acts without fixed points on the suspension $SX$, which is contractible. A very similar idea was applied in the construction of a fixed point free action on a disk in the 1959 paper by E. Floyd and R. Richardson An action of a finite group on an n-cell without stationary points. I've learned this from section 1 of A. Adem's Finite group actions on acyclic 2 -complexes.<|endoftext|> TITLE: Rice's theorem in type theory QUESTION [6 upvotes]: From the formula $$\forall f\colon A\to A\,\exists x\colon A\,(f(x)=x)$$ we can get the scheme $$\forall x\colon A\,(\phi(x)\vee\neg\phi(x))\Rightarrow\forall x\colon A\,\phi(x)\vee\forall x\colon A\,\neg\phi(x)$$ From the scheme $$\forall x\colon A\,\exists y\colon A\,\phi(x,y)\Rightarrow\exists x\colon A\,\phi(x,x)$$ we can get $$\forall x\colon A\,(\phi(x)\vee\psi(x))\Rightarrow\neg\neg(\forall x\colon A\,\phi(x)\vee\forall x\colon A\,\psi(x))$$ Where can I find these results? Give me a link, please. P.S. Sorry, the second scheme must be $$\forall x\colon A\,(\neg\phi(x)\vee\neg\psi(x))\Rightarrow\neg\neg(\forall x\colon A\,\neg\phi(x)\vee\forall x\colon A\,\neg\psi(x))$$ The results are not quite difficult, but I never saw it in articles or books. REPLY [6 votes]: I'll prove the first scheme; you can find a link by using the cite button below this answer. First, we show that $$\forall x,y : A\,(\phi(x) \lor \lnot\phi(y))\tag{*}\label{eq}.$$ To see this, given $x,y : A$, define the function $$f_{x,y}(z) = \begin{cases} x & \text{if $\phi(z)$,} \\ y &\text{if $\lnot\phi(z)$.} \end{cases}$$ By hypothesis, $f_{x,y}$ has must have a fixed point. Since $$\forall z: A\,(f_{x,y}(z) = x \lor f_{x,y}(z) = y)$$ that fixed point must either be $x$ or $y$. Thus $$\forall x, y: A\,(f_{x,y}(x) = x \lor f_{x,y}(y) = y).$$ If $f_{x,y}(x) = x$ then $\phi(x)$ and if $f_{x,y}(y) = y$ then $\lnot\phi(y)$, so we conclude that \eqref{eq} is indeed true. This is almost what we need. For the final stretch, first note that $A$ is inhabited since the identity function must have a fixed point. So fix $z:A$. We know that $\phi(z) \lor \lnot\phi(z)$. If $\phi(z)$ then $\phi(x) \lor \lnot\phi(z)$ is equivalent to $\phi(x)$ for any $x:A$, and so the instance of \eqref{eq} with $y = z$ gives $\forall x: A\,\phi(x)$. If $\lnot\phi(z)$ then $\phi(z) \lor \lnot\phi(y)$ is equivalent to $\lnot\phi(y)$ for any $y:A$, and so the instance of \eqref{eq} with $x = z$ gives $\forall y:A\,\lnot\phi(y).$ Thus, we conclude that $$\forall x:A\,\phi(x) \lor \forall y:A\,\lnot\phi(y),$$ which is an alphabetic variant of your desired conclusion.<|endoftext|> TITLE: Online References for Cartan Geometry QUESTION [15 upvotes]: I would like to learn more about Cartan Geometry ("les espaces généralisés de Cartan"). I ordered Rick Sharpe's book "Differential Geometry: Cartan's generalization...", which would take a long time to arrive though. In the mean time, can someone recommend possibly some online lecture notes, or some online papers containing an introduction to Cartan Geometry, with I hope several examples worked out? I kind of get what it is. When you model it on Euclidean geometry, it yields Riemannian Geometry. When you model it on affine space, it yields a manifold with an affine connection. When you model it on G/H, it gives a kind of curved space, which looks infinitesimally like G/H. Ok, this is the rough idea, but I would like to learn a bit more. Does anyone know of a few online resources on the topic by any chance? Edit 1: I thank everyone who replied. I have learned a lot from various people, and I thank you all. Edit 2: Check out the very nice and short introduction to Cartan geometry by Derek Wise (it is very well written and concise): https://arxiv.org/abs/gr-qc/0611154 A hilarious point in the explanation, is the image of a hamster rolling inside a sphere tangent to the manifold (followed by expressions such as "hamster configurations" etc). It was really funny, and it explained the idea very well. I just realized that it is a shortened version of Derek Wise's thesis, which Tobias Fritz had already suggested as a reference in the comments below (thank you!). Edit 3: R. Sharpe's book has arrived. I find it interesting that R. Sharpe's motivation for writing the book was this question "why is Differential Geometry the study of a connection on a principal bundle?". He wrote that he kept bugging differential geometers with this question, and that attempting to answer this question eventually led him to write his book on Cartan geometry! I appreciate anecdotes like this one. REPLY [10 votes]: There is a series of four recorded lectures by Rod Gover introducing conformal geometry and tractor calculus. Tractor bundles are natural bundles equipped with canonical linear connections associated to $(\mathfrak{g}, H)$-modules. Tractor connections play the same role in general Cartan geometries that the Levi-Civita connection plays in Riemannian geometry; for general Cartan geometries the tangent bundle does not have a canonical linear connection. There's also a set of introductory notes on conformal tractor calculus written by Rod Gover and Sean Curry. If you have the book in your library, I would also suggest having a look at Cap & Slovak's Parabolic Geometries text. This is the modern bible on Cartan geometry, and parabolic geometries in particular. It is more terse than Sharpe, but also covers much more. Parabolic geometries are Cartan geometries modelled on $(\mathfrak{g}, P)$ where $\mathfrak{g}$ is semisimple and $P$ is a parabolic subgroup. Parabolic geometries include conformal, projective geometry, CR geometry, and many more geometries of interest. In the parabolic setting representation-theoretic tools are often used to construct invariant differential operators. For instance, there are so-called BGG sequences of operators associated to irreducible representations, which in the flat case compute the same sheaf cohomology groups as the twisted de Rham sequence. REPLY [10 votes]: You can download some shorter text dealing with conformal geometries from Slovák's homepage One could download his book with Čap from the infamous Russian server (which appears to be down at the moment). I'm not sure whether the Sharpe's book is there as well. I think a really good introductory text is the book Cartan for beginners by Ivey and Landsberg which doesn't really deal with Cartan geometries per se but rather teaches the Cartan method which, in a sense, is precisely the machinery that really makes the Cartan geometries work. Tractor connections and tractor bundles are not really part of Cartan geometries but rather an independent (and in many cases equivalent) approach to study geometrical problems. In conformal geometry they were discovered by T. Y. Thomas in the mid twenties. See Thomas's structure bundle for conformal, projective and related structures. for details.<|endoftext|> TITLE: Does Dyer's Thesis prove that the sheaf/fibration equivalence fails in dimension n>2? QUESTION [11 upvotes]: I've been recently looking into extending the Grothendieck construction to the case of a strict $\omega$-functor $F$ from a strict $\omega$-category $X$ to $\omega-\operatorname{Cat}$. By some recent developments (a paper of Dimitri Ara and Georges Maltsiniotis on defining the lax/oplax join and slice for strict $\omega$-categories), it has become pretty much a breeze to define the Grothendieck construction for such an $\omega$-functor as the oplax colimit of $F$ in $\omega-\operatorname{Cat}$, as it is simply defined to be the initial object of the oplax slice of $\omega-\operatorname{Cat}$ under $F$. However, in a private conversation, I was informed of Scott Dyer's thesis, where he argues that in dimensions higher than $2$, things go rather bad and we end up in a situation without enough Cartesian lifts. Here's the trouble: Dyer's thesis is pretty abstruse, and he uses a lot of nonstandard notation. It's also quite long. I'm wondering if anyone here might be familiar enough with it to save me some time reading through it only to find out that I'm barking up totally the wrong tree: 1.) Is Dyer's Grothendieck integral exactly the same construction as the aforementioned oplax colimit? 2.) Assuming the first question is true, has anyone come up with a loosening of either the definition of cartesianness for higher cells or the definition (as far as one exists) of a strict n-fibration. If fibrations in dimension n>2 are not the correct characterization of the image of the Grothendieck construction, has anyone come up with an alternative one since then? Edit: Here's the paper REPLY [9 votes]: On glancing at the paper, I think that Dyer is just confused. If anything, the place where something new and different happens is $n=2$, not $n=3$, but that case was explained by Hermida. Dyer's proposition 5.3.2 says (in the case $k=n=2$ that he proves) that every 2-cell factors as a "composition" of a cartesian arrow with a vertical one (replacing his terminology "canonical" and "flat" with the standard words "cartesian" and "vertical"). However, this is not a literal single composition in a hom-category, which would be ill-typed for the same reasons he complains about in the $k=3$ case. As can be seen from his proof of Prop. 5.3.2, the true statement is that every 2-cell factors as a "pasting composite" of a cartesian arrow and a vertical one; more specifically it is an ordinary composite of a whiskering of a cartesian arrow and a whiskering of a vertical one. The whiskering of the cartesian arrow is in fact another cartesian arrow, but the whiskering of the vertical arrow is no longer vertical. It seems to me that the same thing is happening in the case $k=3$. In the pictures at the bottom of his p98, if you whisker $(b,\chi)$ by $\tilde{l}$ on the bottom, and whisker $\tilde{\mu}$ by $(b,x)$ on the top, then their domains and codomains will match up, and their composite should be $(\mu,\chi)$ as desired.<|endoftext|> TITLE: Is there a simple proof of the following binomial Identity (part 2)? QUESTION [5 upvotes]: This is a related question to the one I posted on MO earlier: Is there a simple proof of the following Identity for $\sum_{k=m-1}^l(-1)^{k+m}\frac{k+2}{k+1}{\binom l k}\binom{k+1}m$? It arose in the same context: the degeneracy of umbilic points on Weingarten surfaces. For all $l,m\in{\mathbb N}$ with $l\geq m\geq0$ the following identities appear to hold: \begin{eqnarray} &(1-(2m+1)(m+1)){\textstyle{{l+1 \choose m}}}\nonumber\\ +\sum\limits_{k=m+1}^{l+1}&(-1)^{\scriptstyle{{k+m}}}{\textstyle{{l+1 \choose k}}}\left[(1-(2k+1)(m+2)){\textstyle{\frac{2m+2}{2k+1}{k \choose m+1}}}+(1-(2k+1)(m+1)){\textstyle{\frac{2m+1}{2k+1}{k \choose m}}}\right]\nonumber\\ &= \left\{\begin{array}{ccl} 0&if& l>m\\ 2(l+1)(l+2) &if& l=m \end{array}. \right.\nonumber \end{eqnarray} Obviously the $l=m$ case is trivial (I include it for completeness). So, any suggestions for a proof of this? REPLY [3 votes]: Following the hint @darijgrinberg stated in the comment section with respect to the beauty inside the square brackets we focus on the sum and we obtain \begin{align*} \color{blue}{\sum_{k=m+1}^{l+1}}&\color{blue}{(-1)^{k+m}\binom{l+1}{k} \left[(1-(2k+1)(m+2))\frac{2m+2}{2k+1}\binom{k}{m+1}\right.}\\ &\qquad\qquad\qquad\qquad\quad \color{blue}{\left.+(1-(2k+1)(m+1))\frac{2m+1}{2k+1}\binom{k}{m}\right]}\\ &=\sum_{k=m+1}^{l+1}(-1)^{k+m}\binom{l+1}{k}[m-2k(m+2)]\binom{k}{m}\tag{1}\\ &=\binom{l+1}{m}\sum_{k=m+1}^{l+1}(-1)^{k+m}\binom{l+1-m}{k-m}[m-2k(m+2)]\tag{2}\\ &=\binom{l+1}{m}\sum_{k=1}^{l+1-m}(-1)^{k}\binom{l+1-m}{k}[-2k(m+2)-m(2m+3)]\tag{3}\\ &=-2(m+2)\binom{l+1}{m}\sum_{k=1}^{l+1-m}(-1)^{k}\binom{l+1-m}{k}k\\ &\qquad-m(2m+3)\binom{l+1}{m}\left([[l+1=m]]-1\right)\tag{4}\\ &=-2(m+2)\binom{l+1}{l+1-m}(l+1-m)\sum_{k=1}^{l+1-m}(-1)^{k}\binom{l-m}{k-1}\\ &\qquad-m(2m+3)\binom{l+1}{m}\left([[l+1=m]]-1\right)\tag{5}\\ &=2(m+2)(l+1)\binom{l}{m}\sum_{k=0}^{l-m}(-1)^{k}\binom{l-m}{k}\\ &\qquad-m(2m+3)\binom{l+1}{m}\left([[l+1=m]]-1\right)\tag{6}\\ &\color{blue}{=2(l+1)(l+2)[[l=m]]}\\ &\qquad\color{blue}{-(1-(2m+1)(m+1))\binom{l+1}{m}\left([[l+1=m]]-1\right)}\tag{7}\\ \end{align*} in accordance with OPs claim. Comment: In (1) we use @darijgrinbergs simplified bracketed beauty. In (2) we use the binomial identity $$\binom{p}{q}\binom{q}{r}=\binom{p}{r}\binom{p-r}{q-r}$$ In (3) we shift the index to start with $k=1$. In (4) we split the sum and do some simplifications regarding $(1-1)^{l+1-m}$ using Iverson brackets. In (5) and (6) we use the binomial identity $$\binom{p+1}{q+1}=\frac{p+1}{q+1}\binom{p}{q}$$ and we shift the index to start with $k=0$. In (7) we do some final simplifications and adaptions to better see the relationship with OPs identity.<|endoftext|> TITLE: Brownian motion and its maximum and its minimum QUESTION [6 upvotes]: Let $W_u, 0\leq u \leq t$ be Brownian motion. Let $m_t= min_{0\leq u\leq t} W_u$ and $M_t = max_{0 \leq u \leq t} W_u$. The fact that $(M_t , W_t)$ is absolutely continuous with respect to Lebesgue measure on $\mathbb{R}^2$ is known in some stochastic calculus book. By symmetry of Brownian motion, $(m_t,W_t)$ is absolutely continuous with respect to Lebesgue measure on $\mathbb{R}^2$. I want to know whether $(m_t, M_t , W_t)$ is absolutely continuous with respect to Lebesgue measure on $\mathbb{R}^3$. Since $P(x\leq m_t \leq M_t \leq y, W_t \in dx) = \sum_{k=-\infty}^{\infty} \bigg{[} \phi \big( \frac{x-2k(b-a)}{\sqrt{t}}) - \phi \big( \frac{x-2b-2k(b-a)}{\sqrt{t}}) \bigg{]}dx$, If $(m_t,M_t,W_t)$ is absolutely continuous to Lebesgue measure on $\mathbb{R}^3$, I think I can calculate joint density of $m_t,M_t,W_t$. Could you help me? REPLY [4 votes]: It appears that a proof is given in Choi, ByoungSeon; Roh, JeongHo, On the trivariate joint distribution of Brownian motion and its maximum and minimum, Stat. Probab. Lett. 83, No. 4, 1046-1053 (2013). ZBL1266.60142.<|endoftext|> TITLE: What is the need for MathSciNet reviewing again the journal versions of conference papers? QUESTION [5 upvotes]: Suppose that a conference paper already has a MathSciNet review. If later that paper appears with the same abstract in a journal, what is the point of making a new review? Also, wouldn't it make more sense to first ask the person who made the review for the conference version? (In a way, of course it's better to have different opinions, but with this mentality everything could be reviewed twice, so this is irrelevant here.) REPLY [17 votes]: At the moment most of those decisions come from me, at least for computer science papers (those with a 68 class as primary). The practice of having proceedings and final versions of papers is not exclusive to computer science, but this is where it is most common. I've found more often than not that the journal version is significantly different from the proceedings (not just in the straightforward sense that it contains proofs omitted from the first version due to page restrictions) so, if only one is reviewed, the journal version is the obvious choice. Unfortunately there are many wrinkles. There are papers that appear in proceedings and a final version never makes it to a journal. Sometimes we do not receive the proceedings version until after the journal version has been reviewed. Sometimes the title changes or for some other reason we fail to identify that one corresponds to the other. And yes, sometimes there is no discernible difference between both versions, and I do not notice this, or I notice it and feel it may be something a reviewer may want to point out explicitly. Etc. Lately, with a few exceptions, I've been trying to only have the journal version reviewed. This has been traditionally how this situation is handled. If there seems to be a need for a review of both, I usually try to contact first whoever did the first review; I agree this is the natural choice of reviewer. However, this person is not always available. And sometimes, a different reviewer makes more sense once the details of proofs reveal connections not highlighted before. Anyway, please feel free to email me (to my AMS account) if there are specific papers that concern you in this regard. And any comments or feedback you may have would be most appreciated as well.<|endoftext|> TITLE: Are σ-sets preserved by Borel isomorphisms? QUESTION [7 upvotes]: Recall that a $\sigma$-space is a topological space such that every $F_{\sigma}$-set is $G_{\delta}$-set. $X$ - $\sigma$-set, if $X$ is a $\sigma$-space and it is subset of real line $R$. Let $F$ be a Borel isomorphism (Baire isomorphism order $\alpha$) from a $\sigma$-set $X$ onto a topological space $Z$. Does it follow from this that $Z$ is a $\sigma$-space ? REPLY [5 votes]: Under some set-theoretic assumptions the answer to this question is negative. Namely, if there exists a $Q$-set $X$, then $X$ is a $\sigma$-set which is Borel isomorphic to a (hereditarily normal compact) topological space $Y$ which is not a $\sigma$-space. A topological space $X$ is called a $Q$-space if each subset of $X$ is of type $F_\sigma$. A $Q$-space is called a $Q$-set if $X$ is a subset of the real line. It is well-known that uncountable $Q$-sets exist under MA+$\neg$CH (more precisely, under MA each subset $A\subset\mathbb R$ of cardinality $|A|<\mathfrak c$ is a $Q$-set). Now take any uncountable $Q$-set $X$ and let $Y$ be the one-point compactification of a discrete space of cardinality $|X|$. Each subset of $Y$ is Borel (more precisely, open or closed). Then any bijective map $f:X\to Y$ is a Borel isomorphism. But $Y$ is not a $\sigma$-space since for the unique non-isolated point $y$ of $Y$ the singleton $\{y\}$ is not of type $G_\delta$ in $Y$. But this answer is a bit unfair. It would be much interesting to know if a $\sigma$-set can be Borel isomorphic to a subset of the real line which is not a $\sigma$-set. This question is related to an old open problem of Miller who asked in 1979 if for any ordinals $2\le \alpha<\beta<\omega_1$, a $Q_\alpha$-set $A$ and a $Q_\beta$-set $B$ we have $|A|<|B|$. A counterexample to this Miller's problem for $\alpha=2$ would give a $Q$-set (and hence $\sigma$-set) $A$ which is Borel isomorphic (by any bijection) to a $Q_\beta$-set which is not a $\sigma$-set. But this problem remains open since 1979, so seems to be difficult. Concerning the problem of preservation of $\sigma$-sets by Borel isomorphisms, let us remark that Serpinski sets (which are $\sigma$-sets) are preserved by Borel isomorphisms.<|endoftext|> TITLE: If $G$ is a finite abelian group, and $R$ a ring, then does every automorphism of $G$ induce the identity on the group $R[G]^\times/G$? QUESTION [5 upvotes]: Let $G$ be a finite abelian group, and let $R$ be a ring (commutative with 1). In particular I'm interested in the case where $R$ is $\mathbb{Z}/n\mathbb{Z}$ for some $n$. By functoriality, every automorphism $\alpha\in\text{Aut}(G)$ induces an automorphism of the group algebra $R[G]$, and hence an automorphism of the group of units $R[G]^\times$, which stabilizes the subgroup $G\subset R[G]^\times$. I'd like to understand in general the units of $R[G]$ which don't come from $G$. In particular, must $\alpha$ induce the identity on $R[G]^\times/G$? REPLY [6 votes]: No. Let $G=\mathbb{Z}_8=\langle w\rangle$ and set $v=2-w^4+(1-w^4)(w+w^{-1})$. This is a (normalized) unit in $\mathbb{Z} G$ by 10.8 in S. K. SEHGAL, Units in Integral Group Rings (with an appendix by A. Weiss), Pitman Monographs and Surveys in Pure and Applied Math.69, Longman Scientific & Technical, Harlow, 1993. Consider $\phi\in\text{Aut}(G)$ given by $\phi(w)=w^3$. Expanding we have $$v = 2 - w^4 + w + w^{-1} - w^{-3}-w^3.$$ Therefore $$\phi(v) = 2-w^4+w^3+w^{-3}-w^{-1}-w.$$ It follows that there is no $x\in G$ such that $x\phi(v) = v$, and so $\phi$ does not give the identity on $(\mathbb{Z}G)^\times/G$.<|endoftext|> TITLE: The letter $\wp$; Name & origin? QUESTION [23 upvotes]: Do you think the letter $\wp$ has a name? It may depend on community - the language, region, speciality, etc, so if you don't mind, please be specific about yours. (Mainly I'd like to know the English names, if any, but other information is welcome.) If yes, when and how did you come to know it? When, how, and how often do you mention it? (See below.) What's the origin of the letter? In computing, various names, many of which are bad, have been given to $\wp$. See my answer. Background: (Sorry for being a bit chatty.) Originally I raised a related question at Wikipedia. The user Momotaro answered that in math community it's called "Weierstrass-p". Momotaro also gave a nice reference to the book The Brauer-Hasse-Noether Theorem in Historical Perspective by Peter Roquette. The author's claim supports Momotaro. (The episode in the book about the use of $\wp$ by Hasse and Emmy Noether is very interesting - history amuses - but it's off topic. Read the above link to Wikipedia. :) However I'm not completely sure yet, because the occasions on which the letter's name becomes a topic must be quite limited. For example perhaps in the classroom a professor draws $\wp$, and students giggle by witnessing such a weird symbol and mastery of handwriting it; then the professor solemnly announces "this letter is called Weierstrass-p", like that? And "Weierstrass-p" is never an alias of the p-function? After reading Momotaro's comment, I think I've read somewhere that the letter was invented by Weierstrass himself, but my memory about it is quite vague. Does anyone know something about it? Is it a mere folklore, or any reference? I don't think mathoverflow is a place for votes, but if it were, I'd like one: "Have you ever heard of the name of the letter $\wp$? Slightly off-topic, about the p-function's name in Japanese; In Japanese, the names of the Latin alphabets are mostly of English origin, エー, ビー, シー... (eh, bee, cee, etc.) But $\wp$-function is called ペー (peh), indicating its German origin. See e.g. 岩波 数学公式 III, p34, footnote 2 I don't know the name of the letter in Japanese. (In fact, most non-English European languages read "p" as "peh"...) EDIT Typography in some early literature (off-topic, but interesting): First see the excellent comment below by Francois Ziegler $\wp$ that looks like the original (?) and today's glyph: Elliptische Functionen. Theorie und geschichte. (sic) (1890) by Alfred Enneper with Felix Müller, p60. The first ed (1876) by Enneper alone does not seem to mention $\wp$-function. A Course of Modern Analysis (1902) by Whittaker, 1st ed, p322. (Famous Whittaker & Watson, but the 1st ed was by Whittaker alone.) Similar to Kurrent/Sütterlin (see the answer below by Manfred Weis) lowercase p. All were published by the publisher Gauthier-Villars in Paris, in French: Traité des fonctions elliptiques et de leurs applications (1886) by Georges Henri Halphen, p 355. Éléments de la Théorie des Fonctions Elliptiques (1893) by Jules Tannery and Jules Molk, vol 1, page 156. Principes de la théorie des fonctions elliptiques et applications (1897) by Emile Lacour and Paul Appell, p 22. BTW Abramowitz & Stegun uses $\mathscr{P}$. Wow. See p 629. REPLY [6 votes]: My own answer, about the name of the letter $\wp$. Mostly in computing, with little math. (Lengthy) Summary In computing, the letter $\wp$ has been plagued with a plethora of inappropriate names. I guess "Weierstrass p" is one of such names that some standard created, and has been copied without much scrutiny. I suspect that ISO-8879 in 1986 was the root of all evil. Names and references that survive in latest standards are: \wp: TeX script capital p (wrong) and weierstrass elliptic function (alias): Unicode ℘: HTML (and XML?) Unicode Unicode is not directly related to my question, but it is important behind the scenes, so a good point to start. Basics In Unicode the letter $\wp$ is given the codepoint U+2118 in the block "letterlike symbols", named "script capital p". But in fact it's lowercase. There's also an official alias "weierstrass elliptic function". Unicode Technical Note #27 regrets this misnomer: Should have been called calligraphic small p or Weierstrass elliptic function symbol, which is what it is used for. It is not a capital "P" at all. A formal alias correcting this to WEIERSTRASS ELLIPTIC FUNCTION has been defined. The first version of Technical Note #27 in 2006 was a bit more emphatic Should have been called calligraphic small p or perhaps even Weierstrass elliptic function symbol, which is what it is used for. It's not a capital "P" at all. (Boldification is done by me.) Unicode 1.0 (1991) Unicode 1.0 (1991) defined the letter, and the name was "script p". It was also called "weierstrass elliptic function". (See Footnote) Unicode 2 (1996) screwed it up They gave wrong names. As seen later, this change matters for us. From the file Nameslist: 2118 SCRIPT CAPITAL P = per = power set = Weierstrass elliptic function (All of them were legitimate names. See also the file index.) Alas, power set; $\wp$ is lowercase. (But what's "per"? According to Wiktionary's entry "per" the preposition "per" used to be written sometimes with a script letter "p". An example can be found in 1911 Encyclopædia Britannica. See Wikisource or archive.org) Teminology note: The file "NamesList" maps from the codepoint (letter itself) to their names, the main one and aliases, if any. "Index" is the inverse mapping. (Paranoiac detail: In Unicode 2.1, they changed its writing direction from "Other neutral" to "Left-to-right" See here. For writing directionos see this. Unicode 3.0 (1999) In 3.0, two wrong names "per" and "power set" can't be seen in Nameslist, but they still exist in Index. Wow. They still remained in 3.2 Index But not sure if this removal was officially announced Unicode never changes the main names once they are given, so "script capital p" will stay for ever. Unicode 4.0 (2003) In Unicode 4.0, it seems the index file was abolished. So it's finally settled to be "script capital p" and "weierstrass elliptic function". Unicode 6.1 (2012) Finally in Unicode 6.1, the meaning of aliases were clarified. "weierstrass elliptic function" of U+2118 was defined to be "correction". Knuth's book (1986) Computer Modern Typeface (google books link) by Don Knuth was published in 1986. In p 26, there's a line there are special symbols like Weierstrass's ‘p’ (℘). However, in the index in p 580, there's only the entry "Weierstrass, Karl Theodor Wilhelm", pointing to pp 26, 233 and 235, and the entry "Weierstrass'p" is not there. So I guess Knuth did not mean that "Weierstrass'p" was the name of the symbol. Old ISO, MathML, XML... true culprit? Section summary: "Weierstrass p" was introduced sometime, but not sure exactly when. It has becoming obsolete, but not completely yet. In HTML, you can write $\wp$ by escaping as ℘. Dunno other markup languages. In MathML 1.01 specification (1998), $\wp$ is given the name "weierp", of which description is "Weierstrass p" and the alias "wp". See here. The letter derived from "ISOamso", but I can't be sure if the name came from ISOamso, too. The status of the latest MathML, 3.0, will be stated at the end of this section. ISOamso ("Added Math Symbols: Ordinary", has nothing to do with AMS :) is a part of SGML = ISO 8879:1986 in 1986, according to this. When the first MathML was under construction, ISO 10646 (roughly speaking Unicode itself, but its counterpart in ISO) were apparently not there. See this. They had to gather letters from various standards. HTML is worse: I couldn't find any entity specification about $\wp$ in HTML 1 - 3. (In very rough translation, entity = letter. See this intro) In HTML4 (1999), it is defined as Because HTML4 has been used so long (it's still in use), this wrong sepcification has spreaded widely. See below. In HTML5 (2014), the word "entity" was abolished. "Character entity references" was change to "Character references", and the term "Named entities" was replaced with "Named character references". See this. "Weierp" seems to have lost peculiarity up to HTML4; it's simply U+2118, and that's all. See this (Work on HTML5 started back in 2004, but it took long time until final publication.) There's also "XML entity definitions". Although XML is rather old, the official specification of entities seems to have had been lacking for long. In its latest version (2nd ed) in 2014, $\wp$ is nothing but "weierp". See chap 2. ISOamso is now one of legacy entity sets. In its first version in 2010 however, it's classified as ISOamso Anyway in ISOamso pages (ver 2 and ver 1) its description is "/wp - Weierstrass p", with the alias "wp". It seems the slash in "/wp" corresponds to LaTex backslash. (Cf. /hbar, /ell, /Re etc) Unfortunately in MathML 3.0, (the latest, released in 2010) it still depends on XML's "legacy" entity sets, and it refers to ISOamso. "Weierstrass'p" is not yet dead. "Weierstrass-p" spreaded The word "Weierstrass p" appears in e.g. Encyclopedia Machintosh (1990), MacUser magazine (1992) I googled for "weierstrass's p" - "weierstrass's p function" limiting to pre-1986 instances. There had not been any to mean the symbol $\wp$. The result of dropping "'s" is similar. Date limitation in Google search is not so reliable thus it does not prove anything. Wrong HTML4's specification is now popular. Microsoft's document HTML Character sets reads "Office 2003", although its exact date and scope are not clear. It says $\wp$ is: "script capital P, =power set, =Weierstrass p, U2118 ISOamso". XML in a Nutshell (2002) and Beginning HTML, XHTML, CSS, and JavaScript (2011) say Weierp is "Script capital P, power set, Weierstrass p" Acknowledgments The books "Encyclopedia Machintosh", "MacUser magazine", "Computer Modern Typeface" and "XML in a Nutshell" were pointed out in Wikipedia Users Momotaro and The Man in Question. I thank them. Footnote In Unicode 1.0.0 (1991), it was NOT "script capital p", the current name. In Unicode Name Index (pdf), names "script p" and "weierstrass elliptic function" are found. See pdf pages 22 and 27. This did not change in Unicode 1.0.1 (1992). Unicode 1.0 implicitly, but clearly meant it was capital. All lowercase letters were named "small". It seems that in Unicode 1.1 (1993) the name was changed to "script capital p" See UnicodeData-1.1.5<|endoftext|> TITLE: Young-Fibonacci tableaux, content, and the Okada algebra QUESTION [9 upvotes]: Using the French convention, the content of the $i \times j$ box in the Young diagram of a partition $\lambda$ is $i-j$. Now if $\lambda$ is partition of $n$ and $\sigma_\lambda: S_n \longrightarrow V_\lambda$ is the corresponding irreducible representation of the symmetric group $S_n$ then the sum of contents of all boxes in the young diagram of $\lambda$ equals \begin{equation} {\text{tr} \, \sigma_\lambda \big( t \big) \over {\text{dim} V_\lambda}} \cdot \big| T \big| \end{equation} where $T$ is the conjugacy class consisting of all transpositions and $t$ is any choice of transposition. Moreover each Young tableau $Y_\lambda$ encodes an eigenvector in $V_\lambda$ for the operator $\sigma_\lambda \big( J_k \big)$ with eigenvalue $c_k$ where \begin{equation} J_k \, := \, \sum_{i=1}^{k-1} \, \big(i,k \big) \, \in \Bbb{C} \big[ S_n \big] \end{equation} and $c_k$ is the content of the box in $Y_\lambda$ labeled by $k$. Consider now the Young-Fibonacci lattice whose elements consist of words $w = a_1 \cdots a_d$ taken from the alphabet $\{1,2 \}$ which can be visualised by stacking boxes into adjacient vertical columns going from left to right such that the number of boxes in the $i$-th column is $a_i$. The rank $|w|$ of a word is simply the number of boxes in such a picture; equivalently $|w|$ equals $a_1 \, + \, \cdots \, + \, a_d$. I won't describe the covering relations that give the lattice structure --- but suffice it to say each word $w$ or rank $n$ encodes an irreducible representation $V_w$ of the Okada algebra $\mathcal{A}_n$ and each complete chain $Y_w$ ending at $w$ indexes a basis vector in $V_w$. Question: (1) Are there pairwise commuting operators $\tilde{J_1}, \dots, \tilde{J_n}$ within the Okada algebra $\mathcal{A}_n$ for which each complete chain $Y_w$ (viewed as a basis vector in $V_w$) is a simultaneous eigenvector and (2) is there a notion of content (a value for each covering relation in the Young-Fibonacci lattice) so that the $k$-th content $c_k$ of $Y_w$ (viewed as a complete chain) is the eigenvalue of $\tilde{J_k}$ corresponding to $Y_w$ and (3) will the sum of such contents along any complete chain $Y_w$ be constant ? regards, A. Leverkühn REPLY [2 votes]: Errors have been fixed In this second response let me try to address what might play the role of the operator sum $\tilde{J_1} + \cdots + \tilde{J_n}$ which is the analogue of $\sum_{t \in T} \, t \ \in \Bbb{C}[S_n]$ where $T$ is the conjugacy class consisting of all transpositions. Given two sets of generic parameters $x_1, \dots, x_{n-1}$ and and $y_1, \dots, y_{n-2}$ the associated Okada algebra $\mathcal{F}_n$ has a presentation given by generators $E_1, \dots, E_{n-1}$ subject to the defining relations \begin{equation} \begin{array}{rll} E_i^2 &= \ x_i \, E_i & \\ E_i \, E_j &= \ E_j \, E_i &\text{whenever $|i-j| >1$} \\ E_j \, E_i \, E_j &= \ y_i \, E_j &\text{whenever $j = i+1$} \end{array} \end{equation} Given a word $w$ the Okada power-symmetric functions $p_w$ is defined by the recursion \begin{equation} p_w \ = \ \left\{ \begin{array}{ll} \displaystyle x_1 \cdots \, x_k &\text{if $\, w = 1^k \, $ with $\, k \geq 0$} \\ \displaystyle q_k \cdot p_u \big[ + (k + 2) \big] &\text{if $\, w= u 2 1^k \,$ with $\, k \geq 0$} \end{array} \right\} \end{equation} where \begin{equation} q_k \ := \ (x_1 \cdots \, x_k) \, \Big( x_{k+1} x_{k+2} \, - \, (k+2) \,y_{k+1} \Big) \end{equation} and $p_u \big[ +l \big]$ means perform the substitutions $x_i \mapsto x_{i+l}$ and $y_i \mapsto y_{i+l}$ for all $i \geq 1$. For the purposes of this discussion, let's provisionally define the Okada power-symmetric function in terms of the Okada-Schur functions and the Okada characters values: \begin{equation} p_u \, = \ \sum_{|v| = |u|} \, X^v_u \, s_v \end{equation} Now let's define $\mathcal{F_n}$-valued versions $\Bbb{s}_w$ and $\Bbb{p}_w$ of the Okada-Schur and power-symmetric functions. For $n \geq k \geq 1$ define the following $k \times k$ tri-diagonal determinants whose values are in the Okada algebra $\mathcal{F_n}$ \begin{equation} \begin{array}{ll} \mathcal{P}_k \, := &\det \, \begin{pmatrix} x_1 & x_2 E_1 & 0 & \cdots \\ 1 & x_2 & x_3 E_2 & & \\ 0 & 1 & x_3 & & \\ \vdots & & & \ddots \end{pmatrix} \\ \\ \\ \mathcal{Q}_{k-1} \, := &\det \, \begin{pmatrix} x_2 E_1 & x_1 x_3 E_2 & 0 & \cdots \\ 1 & x_3 & x_4 E_3 & & \\ 0 & 1 & x_4 & & \\ \vdots & & & \ddots \end{pmatrix} \end{array} \end{equation} where $\mathcal{P}_0 := 1$. Clearly an order must be observed when tabulating the determinant --- following the conventions of Kerov and Goodman, the $l$-th factor in the expansion will always be selected from the $l$-th column. The values of the determinants $\mathcal{P}_k$ and $\mathcal{Q}_{k-1}$ are in fact independent of the order in which products are taken in the Laplace expansion: This is because indices of the generators $E_1, \dots, E_{n-1}$ which participate in any given monomial in the expansion of such a tri-diagonal determinant will always differ by at least two. Employ the same recursion above being mindful to place the accumulating $\mathcal{Q}$-factors to the left and in order: \begin{equation} \Bbb{s}_w \, := \ \left\{ \begin{array}{ll} \mathcal{P}_k &\text{if $w= \, 1^k \, $ and $k \geq 0$} \\ \\ \mathcal{Q}_k \, \big[+ |v| \big] \cdot \Bbb{s}_v &\text{if $w= \, 1^k \, 2 \, v \, $ and $k \geq 0$} \end{array} \right\} \end{equation} Once again $\mathcal{Q}_k \big[ + |v| \big]$ means shift all indices by $|v|$. Play the same game and define the $\mathcal{F}_n$-valued power-symmetric functions by the expansion: \begin{equation} \Bbb{p}_u \, := \ \sum_{|v| = |u|} \, X^v_u \, \Bbb{s}_v \end{equation} I want to make use of Okada's trace functional $\text{Tr}: \mathcal{F}_n \longrightarrow \Bbb{C}$ which is defined for an element $a \in \mathcal{F}_n$ using the Okada-Schur values by \begin{equation} \text{Tr}(a) \ = \ {1 \over {(x_1 \cdots \, x_n)}} \, \sum_{|v|=n} \, s_v \cdot \text{tr} \Big[ \sigma_v(a) \Big] \end{equation} where $\sigma_v : \mathcal{F}_n \longrightarrow \text{End}(V_v)$ is the irreducible representation of $\mathcal{F}_n$ associated to a word $v$ with $|v|=n$ and where $\text{tr} \big[ \cdot \big]$ denotes the usual trace. Okada proves in his paper that \begin{equation} \begin{array}{ll} \displaystyle \text{Tr} \, (1) &\displaystyle = \, 1 \\ \displaystyle \text{Tr} \, \big( ab \big) &\displaystyle = \, \text{Tr} \, \big(ba \big) \\ \displaystyle \text{Tr} \, \big( aE_i \big) &\displaystyle = \, {y_i \over {x_{i+1}}} \, \text{Tr} \, (a) \quad \text{when $a \in \mathcal{F}_i$} \quad \left( { \scriptstyle \begin{array}{l} \text{As far as I can tell there seems to be} \\ \text{a missing $x_{i+1}$ in the denominator of} \\ \text{part (4) of Proposition 2.7 in Okada's} \\ \text{paper which I have tried to correct here.} \end{array} }\right) \end{array} \end{equation} Using these multiplicative properties, a simple induction on the length $|v|$ reveals that $\text{Tr} \, \big( \Bbb{s}_v \big) \, = \, s_v$ and consequently $\text{Tr} \, \big( \Bbb{p}_u \big) \, = \, p_u$. Moreover \begin{equation} \begin{array}{c} \displaystyle \text{tr} \, \Big[ \pi_v \big( \Bbb{p}_u \big) \Big] \, = \, \big( x_1 \cdots x_n \big) \, X^v_u \\ \displaystyle \text{--- and ---} \\ \displaystyle \text{tr} \, \Big[ \pi_v \big( \Bbb{s}_u \big) \Big] \, = \, \big(x_1 \cdots x_n \big) \, \delta_{u,v} \end{array} \end{equation} Isn't this observation an indication that $\Bbb{p}_u$ is an analogue of a characteristic function of a conjugacy class in the group setting ? However, pursuing this analogy, it's not immediately clear whether the elements $\mathcal{p}_u$ are central in $\mathcal{F}_n$. yours, Ines. p.s. In fact the order of products taken in the expansion of the $\mathcal{F}_n$-valued determinants $\mathcal{P}_k$ and $\mathcal{Q}_{k-1}$ is irrelevant --- this is because all monomials which occur involve $E$-generators whose subscripts differ by at least two.<|endoftext|> TITLE: Does $GL_2(\widehat{\mathbb{Z}})$ contain a dense finitely generated subgroup? QUESTION [9 upvotes]: It's well known that $SL_2(\widehat{\mathbb{Z}})$ contains $SL_2(\mathbb{Z})$ as a dense and finitely generated subgroup. However, $GL_2(\mathbb{Z})$ is not dense in $GL_2(\widehat{\mathbb{Z}})$, since $GL_2(\mathbb{Z})$ is contained in the closed subgroup of matrices with determinant $\pm 1$, which is very far from the entirety of $GL_2(\widehat{\mathbb{Z}})$ (whose determinant map surjects onto $\widehat{\mathbb{Z}}^\times$) Is there a finitely generated dense subgroup of $GL_2(\widehat{\mathbb{Z}})$? REPLY [21 votes]: The answer is no. If $GL_2(\widehat{\mathbb Z})$ is topologically finitely generated, then so is the quotient $\widehat {\mathbb Z}^*$ (quotient via the determinant map). The latter has quotient $\widehat {\mathbb Z}^*/ (\widehat{\mathbb Z}^*)^2=\prod _p {\mathbb Z}_p^*/({\mathbb Z}_p^*)^2$, where the product is over all primes. This group has as quotient $\prod _p {\mathbb Z}/2{\mathbb Z}$. The latter is an infinite dimensional vector space over the finite field ${\mathbb F}_2$ of two elements and is hence not topologically finitely generated.<|endoftext|> TITLE: What are the important geometric-topological consequences of 4-dimensional version of Gauss-Bonnet-Chern theorem? QUESTION [13 upvotes]: The Gauss–Bonnet theorem in differential geometry is an important statement about surfaces which connects their geometry to their topology and has very important applications to Riemann surface theory. The generalized Gauss–Bonnet theorem (Gauss-Bonnet-Chern) in dimension $n=4$, for a compact oriented manifold states that $$\chi(M)=\frac{1}{32\pi^2}\int_M\left(|Rm|^2-4|Rc|^2+r^2\right)d\mu,$$ where $Rm$ is the full Riemann curvature tensor, $Rc$ is the Ricci curvature tensor, $r$ is the scalar curvature. My question is What are the important local-global results of the 4-dimensional version of Gauss-Bonnet-Chern theorem similar to the 2-Dimensional case? Update:(2,September,2017) This update is just an additional information. If $W$ denote the Weyl curvature of $(M,g)$ then $${32\pi^2}\chi(M)=\int_M(|W|^2+ 8Q_g)d\mu,$$ where $$Q_g:=-\frac{1}{12}(\Delta_gr-r^2+3|Rc|^2),$$ is the Paneitz $Q$ curvature introduced by Branson. REPLY [4 votes]: A good paper in this direction is "Some implications of the generalized Gauss-Bonnet theorem" written by Bishop and Goldberg which proved the following two theorems Theorem 1.1. A compact and oriented Riemannian manifold of dimension $4$ whose sectional curvatures are non-negative or nonpositive has non-negative Euler-Poincare characteristic. If the sectional curvatures are always positive or always negative, the Euler-Poincare characteristic is positive. Theorem 1.2. In order that a $4$-dimensional compact and orientable manifold M carry an Einstein metric, i.e., a Riemannian metric of constant Ricci or mean curvature $R$, it is necessary that its Euler-Poincare characteristic be non-negative. and a corollary Corollary 1.2. If $V$ is the volume of $M$, $$\chi(M)\geq\frac{VR^2}{12\pi^2}$$ equality holding if and only if $M$ has constant curvature.<|endoftext|> TITLE: A "paradox" regarding Voronin's universality theorem QUESTION [8 upvotes]: The famous and remarkable Voronin's universality theorem states: Theorem (Voronin 75): let $00$, there exists a positive real number $\tau$ such that: $$max_{\vert s \vert \leq r} \vert \zeta(s+3/4+i \tau)-g(s) \vert <\epsilon. $$ Which practically means that $g(s)$ could be approximated by $\zeta(s)$ for some "high enough" values of $\tau$ - on the right hand side of the critical strip. The thing is that there is a certain numerical "mystery" with respect to this theorem - when it comes to the question "how high is high enough"? Of course - there are effective analytic studies in various cases. But its worthwhile to pay attention to direct calculation via computer - it turns out to be really "hard" to verify Voronin numerically (in my eyes, the illustration makes the theorem even more impressive). For instance - let us take the constant function $g(z)=e^{3}$. Question: Is there any estimate on $\tau$ for which Voronin's approximation works for $g(s)=e^{3}$? It is important to point out and compare, for instance, the following graph of $\log \vert \zeta(0.75+e^{0.0001 \tau} i) \vert$ for $\tau = 0,...,250000$ in this case: As you can see - for quite big values - the function still doesn't seem to cross the bound $\pm 3$. So when does Voronin's theorem start to kick in this case? (It is interesting to note also some implications to zeros of zeta (RH), for instance.) REPLY [4 votes]: Consider $e^3$ and $e^{-3}$ separately. For $e^3$: Think about the order of $\zeta(s)$ in the critical strip: Titchmarsh Chapter V. Define $\mu(\sigma)$ as the lower bound of the numbers $\xi$ such that $$ \zeta(\sigma+it)=O(|t|^\xi). $$ The Lindelof Hypothesis (which is a consequence of the Riemann Hypothesis) is that $\mu(\sigma)=1/2-\sigma$ for $\sigma\le 1/2$ and $\mu(\sigma)=0$ for $\sigma>1/2$. In this case $\zeta(3/4+it)=O(|t|^\xi)$ is true for every $\xi>0$, so the zeta function grows very slowly on this vertical line. It's simply going to take a long time to reach $e^3$. For $e^{-3}$: $\zeta(s)$ is near $e^{-3}$ when $1/\zeta(s)$ is near $e^3$. But on the Riemann Hypothesis, $1/\zeta(s)=O(t^\epsilon)$ for $\sigma>1/2$ and every positive $\epsilon$. (Titchmarsh (14.2.6)). So again, the function $1/\zeta(s)$ can grow only very slowly on the $3/4$ line. Update: "Can you give an estimate...?" The point of my answer was to show why $\tau$ will likely be quite large for the function you chose, beyond the range where computation is easy. Your comment below indicates you're not really asking about Voronin's Theorem; you're asking about making effective the constants implied by Titchmarsh's use of O. Many of these can be made effective by paying careful attention to the proofs. It depends how badly you need them whether it's worth the effort.<|endoftext|> TITLE: Homotopy groups of even-dimensional spheres QUESTION [7 upvotes]: I need to understand the structures of $\pi_{4n}(S^{2n})$ and $\pi_{r}(S^{2n};\mathbb{Z}_k) (r\geq 4n-1)$, where the latter group is the homotopy group with coefficient defined as $[P^r(k), S^{2n}]$. Does anyone know their computations and any references are available? REPLY [14 votes]: The $2$-primary parts of the first few groups are as follows: $$\begin{array}{ll} n & \pi_{4n}(S^{2n}) \\ 1 & \mathbb{Z}_2 \\ 2 & \mathbb{Z}_2^2 \\ 3 & \mathbb{Z}_2\\ 4 & \mathbb{Z}_2^4\\ 5 & \mathbb{Z}_2\oplus\mathbb{Z}_4\\ 6 & \mathbb{Z}_2^2\\ 7 & \mathbb{Z}_2\oplus\mathbb{Z}_8\\ 8 & \mathbb{Z}_2^4\\ 9 & \mathbb{Z}_2\oplus\mathbb{Z}_4\oplus\mathbb{Z}_8. \end{array}$$ These are all in Toda's book "Composition methods in the homotopy groups of spheres". As you can see, there is no obvious pattern, and no general description is known. The difficulty of describing $\pi_i(S^j)$ is primarily controlled by $i-j$; the fact that $i=2j$ and $j$ is even is not going to help very much.<|endoftext|> TITLE: Extension of a vector field to an orthonormal frame for a flat metric QUESTION [8 upvotes]: Assume that $U$ is an open set in the plane and $X$ is a non vanishing vector field on $U$. Is there a non vanishing vector field $Y$ on $U$ such that the pair $\{X,Y \}$ plays the role of an orthonormal frame for a flat Riemannian metric on $U$? REPLY [2 votes]: This is by no means a complete proof, but I will try at least to describe an approach. The main idea is to look at integral curves, as suggested by Tom Goodwillie in the comments. We will construct the desired metric by perturbing $U$ together with $X$ in such a way that the $X$-trajectories will have constant velocity $1$. This will guarantee that $X$ has norm $1$. Then $Y$ can be defined simply by rotating $X$ by $90$ degrees. I will assume that the vector field is bounded from above, by rescaling let us assume that $|X|<\varepsilon$. Also I assume that $U$ can be covered by small 'eyes' that look like $X$ flows upwards. For a point $x\in U$ denote by $\Phi_t(x)$ the trajectory of $x$ flowing along $X$ so that $\Phi_0(x)=x$. Assume a point $a$ on the lower side of the eye ends up in time $t(a)$ on another side of the eye. Then the whole eye is parametrized by pairs $(a,s)$ such that $a$ is a point on the lower side of the eye and $0\leq s\leq t(a)$. Suppose I construct a different flow inside the eye, call it $\Psi_t(x)$ such that $\Psi_{t(a)}(a)=\Phi_{t(a)}(a)$ and such that velocity of a point flowing along $\Psi$ is $1$. Then we can define an isomorphism from the eye to itself by $\alpha \Phi_s(a) = \Psi_s(a)$. Then we define a metric inside the eye by pulling back the standard metric via $\alpha$. The pullbacks of the trajectories of $\Psi$ will be the trajectories of $X$, so $X$ will have unit length. How do we construct such a flow? Note that $$ |\Phi_{t(a)}(a)-a|\leq \varepsilon t(a). $$ We can assume that the projection of $X$ on the vertical direction is at least $\varepsilon'$ so we have $$ t(a)\leq \frac{1}{\varepsilon'} |\Phi_{t(a)}(a)-a|. $$ So we can move from $a$ to $\Phi_{t(a)}(a)$ by following a zig-zag line as shown on the picture, so that the length of the zig-zag line is $t(a)$. I think it is possible to organize the zig-zags for different points $a$ into a smooth flow. Also we may assume that each zig zag follows $X$ for a little while in its beginning and in the end. Then close to the boundary of the eye $\alpha$ is just stretching in the $X$ direction, so that the maps $\alpha$ for different eyes glue together nicely. Also there can be a problem to make $\alpha$ smooth.<|endoftext|> TITLE: The approximation property for some spaces of holomorphic functions QUESTION [6 upvotes]: I am reading a circle of papers which use arguments based on Fredholm determinants of nuclear operators to compute numerical quantities associated to real-analytic and holomorphic dynamical systems. In order to understand these papers I am learning about Grothendieck's theorems on traces and determinants of nuclear operators. (My reference for these results is the book Traces and Determinants of Linear Operators by Goh'berg, Goldberg and Krupnik, which I am finding easier to read than the original papers of Grothendieck.) The operators are typically defined on classical Banach spaces of holomorphic functions, sometimes of several variables. In order for Grothendieck's results on the existence of traces and Fredholm determinants to be applied it seems that the Banach space $\mathfrak{X}$ on which the operators are to be studied must satisfy the approximation property: for every compact subset $K$ of $\mathfrak{X}$ and every $\varepsilon>0$ we must be able to find a finite-rank operator $F$ such that $\|x-Fx\|<\varepsilon$ for all $x \in K$. I was not previously familiar with the approximation property (my background is mainly in ergodic theory and dynamical systems) and am trying to better understand the scope of this particular hypothesis. Unfortunately the approximation property does not seem to be very explicitly treated in the papers I am reading and I have not yet discovered any useful references. I wonder if anyone can point me to a reference for the following: Let $D \subset \mathbb{C}^d$ be nonempty, open and bounded. Is it known whether or not the Banach space of bounded holomorphic functions $D \to \mathbb{C}$ with continuous extensions to $\overline{D}$, equipped with the uniform norm, has the approximation property? Some qualifications: I would be happy with a reference which states that the problem is open, as well as with a definite answer either way. I'm interested both in cases where $D$ is connected, and in cases where it is not connected. Any information about either would be welcome. I understand that the approximation property for $H^\infty(D)$ is unknown when $D\subset \mathbb{C}$ is the unit disc. Can anyone give me a reference stating that this problem is open? REPLY [2 votes]: A classical reference for the case of the unit disk: Bourgain, J. and Reinov, O. On the approximation properties for the space $H^\infty$. Math. Nachr. 122 (1985), 19–27. (Link) As recently as 2013, the question is still mentioned as open in: Brudnyi, A. Banach-valued holomorphic functions on the maximal ideal space of $H^\infty$. Invent. Math. 193 (2013), no. 1, 187–227. (Link)<|endoftext|> TITLE: Height of associated primes in regular rings QUESTION [8 upvotes]: Let $I$ be an ideal in a regular ring $R$. Suppose $I$ can be generated by $n$ elements. Let $P$ be an associated primes of $I$. Is it true that the height of $P$ is bounded above by $n$? Remark: (1) The question has an affirmative answer when $P$ is a minimal associated prime of $I$ by Krull's principal theorem. (2) The question is false for general Noetherian rings Indeed, suppose $I = 0$ and $R$ has an embedded associated prime ideal. REPLY [7 votes]: Just a couple of remarks: a) You can simplify Jason's example slightly with $I=(u^2,v^2,xu-yv)$. The annihilator of $uv$ is the maximal ideal. b) One can in fact find an example of a three-generated $I$ with an associated prime of height $N$ for any $N$. Take $R$ to be a regular local ring of dim $N$ and maximal ideal $m$. A theorem by Bruns [1] states that the second syzygy of any ideal is the second syzygy of a three-generated ideal. So take $M = syz^2(m)$, then $M=syz^2(I)$ where $I$ has $3$ generators. Counting depth gives $depth(I)=1$, so $R/I$ has $m$ as an associated prime. [1] http://www.sciencedirect.com/science/article/pii/0021869376900478<|endoftext|> TITLE: Is Girard's LU just an embedding of classical and intuitionistic logic into linear logic? QUESTION [12 upvotes]: This question is about Girard's system LU, presented in his paper On the unity of logic. Girard starts by giving a "modal" sequent calculus with two zones of both hypotheses and consequents, $\Gamma;\Gamma'\vdash \Delta';\Delta$, of which $\Gamma'$ and $\Delta'$ are treated classically (with contraction and weakening) and $\Gamma$ and $\Delta$ are treated linearly (without contraction and weakening). He gives rules in this calculus for the usual connectives $\otimes,\oplus,\&,⅋,\neg,\multimap,!,?,\bigwedge,\bigvee$ of classical linear logic, and explains how the result is bi-interpretable with ordinary classical linear logic. At this point it seems clear that he has just given a different presentation of the "same" classical linear logic. Then he introduces new "chimeric" connectives $\wedge,\vee,\Rightarrow,\supset,\forall,\exists$. It appears as though each of these connectives is defined in terms of the linear ones, although not by a uniform definition but rather by a case analysis on whether the input formulas are "positive, negative, or neutral". For instance, $A\wedge B$ is $A\otimes B$ if $A$ and $B$ are both positive, but it is $A\& B$ if $A$ and $B$ are both negative, while if $A$ is positive and $B$ is negative then $A\wedge B$ is $A\otimes !B$, and so on. He gives explicit rules for these new connectives as well, but a brief perusal of these rules suggests to me that they are derived from the definitions of the new connectives in terms of the linear ones. Finally he considers several fragments of the resulting system that embed classical and intuitionistic nonlinear logics: classical logic uses $\wedge,\vee,\neg,\Rightarrow,\forall,\exists$ while intuitionistic logic uses $\wedge,\vee,\supset,\bigvee,\exists$. It seems to me, therefore that LU is just a rephrasing of linear logic together with new embeddings of classical and intuitionistic logic into it that are defined by case analysis on the polarity of formulas rather than "uniformly" as the more common embeddings are. (This "just" is not meant to be disparaging; I find LU very interesting, I'm just trying to understand it better.) But Girard doesn't seem to view it this way; and moreover in Vauzeilles's proof of cut-elimination for LU it takes $3\frac12$ extra pages to extend the result to the chimeric connectives, which doesn't seem like it would be the case if they were simply defined in terms of the linear ones. So I must be missing something; what is it? REPLY [6 votes]: Sorry if my answer comes so late, maybe you already figured it out by yourself in the meantime, I hope this helps anyway. I think the main misunderstanding is that the "non-chimeric" fragment of $\mathbf{LU}$ is almost a different presentation of linear logic, but not quite (of course it does not help that Girard calls this the linear fragment of $\mathbf{LU}$...). There is a subtle discrepancy given by the presence of polarities: in $\mathbf{LU}$, structural rules on the right (resp. on the left) are allowed for every negative (resp. positive) formula, including the atoms. This is justified by the fact (which Girard mentions in passing at some point) that structural rules (on the right) are derivable in "usual" linear logic for negative formulas, which are those defined by $$M,N ::= \bot \mathrel{|} \top \mathrel{|} M\& N \mathrel{|} M⅋N \mathrel{|} \forall x.N \mathrel{|} ?A,$$ where $A$ is an arbitrary formula. The generalized promotion rule $$\frac{\vdash\mathcal N,A}{\vdash\mathcal N,!A}.$$ is also derivable when $\mathcal N$ is composed of negative formulas only. There are now two observations to make: the negative formulas defined above do not contain atoms, so $\mathbf{LU}$ goes slightly beyond linear logic in that respect. Indeed, "usual" linear logic really corresponds to the linear fragment of $\mathbf{LU}$ restricted to neutral atoms. Unfortunately, Girard fails to mention this subtlety. The structural rules and "promotion-with-negative-context" rule are derivable, but not cut-free derivable. This means that the obvious way of embedding the linear fragment of $\mathbf{LU}$ in "usual" linear logic (mapping each negative (resp. positive) atom $X$ to $?X$ (resp. $!X$)) is not "computationally transparent". The second observation justifies the need to make sure that cut-elimination holds in $\mathbf{LU}$ as a logical system of its own (this had essentially already been pointed out by Andrej in his comment). That being done, one may study computationally meaningful embeddings of classical or intuitionistic logic in $\mathbf{LU}$, which turn out to be slightly more efficient than the ones directly in linear logic (there are less !'s and ?'s to take care of, because of the more liberal structural rules). Such embeddings were studied more thoroughly by Olivier Laurent in his Ph.D. thesis. In particular, he defined polarized linear logic ($\mathbf{LLP}$), which is essentially the linear fragment of $\mathbf{LU}$ without neutral atoms (and, therefore, without neutral formulas in general), and showed how the $\lambda\mu$-calculus (Parigot's calculus for classical logic) may be nicely translated in it. You may look on his web page for more. Laurent's simplified system is what has been used in practice, so $\mathbf{LU}$ kind of fell into oblivion (in fact, "On the unity of logic" is one of the rare papers by Girard on linear logic which I had never read!).<|endoftext|> TITLE: A refinement of Serre's finiteness theorem on unstable homotopy groups of spheres QUESTION [26 upvotes]: Serre's finiteness theorem says if $n$ is an odd integer, then $\pi_{2n+1}(S^{n + 1})$ is the direct sum of $\mathbb{Z}$ and a finite group. By looking at the table of homotopy groups, say on Wikipedia, one empirically observes that if $n \equiv 3 \pmod 4$, then we in fact have $$ \pi_{2n+1}(S^{n+1}) \cong \mathbb{Z} \oplus \pi_{2n}(S^n). $$ This holds for all the cases up to $n = 19$. On the other hand, for $n \equiv 1 \pmod 4$ (and $n \neq 1$), the order of the finite part drops by a factor of $2$ when passing from $\pi_{2n}(S^n)$ to $\pi_{2n+1}(S^{n+1})$. From the EHP sequence, we know that these two are the only possible scenarios. Indeed, we have a long exact sequence $$ \pi_{2n+2}(S^{2n+1}) \cong \mathbb{Z}/2\overset{P}{\to} \pi_{2n}(S^n) \overset{E}{\to} \pi_{2n+1}(S^{n+1}) \overset{H}{\to}\pi_{2n+1}(S^{2n+1}) \cong \mathbb{Z}. $$ Since $H$ kills of all torsion, one sees that the map $E$ necessarily surjects onto the finite part of $\pi_{2n+1}(S^{n+1})$. So the two cases boil down to whether or not $P$ is the zero map. What we observed was that it is zero iff $n \equiv 3 \pmod 4$, up to $n = 19$. Since $\pi_{2n+2}(S^{2n+1}) \cong \mathbb{Z}/2\mathbb{Z}$ has only one non-zero element, which is the suspension of the Hopf map, it seems like perhaps one might be able to check what happens to this element directly. However, without a good grasp of what the map $P$ (or the preceeding $H\colon\pi_{2n+2}(S^{n+1}) \to \pi_{2n+2}(S^{2n+1})$) does, I'm unable to proceed. Curiously, I can't seem to find any mention of this phenomenon anywhere. The closest I can find is this MO question, but this phenomenon is not really about early stabilization, since for $n = 3, 7$, the group $\pi_{2n-1}(S^n)$ is not the stable homotopy group, but something smaller. I'd imagine either this pattern no longer holds for larger $n$, or there is some straightforward proof I'm missing. Note: Suggestions for a more descriptive title are welcome. REPLY [27 votes]: $P(\eta) = [i_n,i_n] \circ \eta$, where $[i_n,i_n]: S^{2n-1} \rightarrow S^n$ is the Whitehead product of the identity map with itself. So you are asking if this composite is null. I don't know if this is an easy problem. One sufficient condition is that $[i_n,i_n]$ be divisible by 2, but, ha, ha, this is now known to only rarely happen, thanks to the Hill-Hopkins-Ravenel theorem on the Kervaire invariant. (See [HHR, Thm 1.5].) But it does for $n=63$, so $P$ is zero in that case. I'd search the old literature - papers of Mahowald, Barratt, James and their collaborators - for any general results, if they exist. Mahowald, in particular, has many papers with examples of families of elements on which $H$ acts nonzero. Added 30 minutes later: [Mahowald, Some Whitehead products in $S^n$, Topology 4, 1965, Theorem 1.1.2(a)] answers your question. $P(\eta)$ is zero if $n \equiv 3 \mod 4$ and is nonzero in basically all other cases.<|endoftext|> TITLE: Is $V=L$ equivalent to there being a $\Sigma_1$ well-ordering of the universe? QUESTION [9 upvotes]: Working in ZF, it's well-known that for any $n \ge 2,$ the claim that there is a $\Sigma_n$ well-ordering of the universe is equivalent to the axiom $V=HOD.$ It seems natural to believe there should be a similar theorem for $n=1,$ perhaps that there being a $\Sigma_1$ well-ordering of the universe is equivalent to $V=L.$ I can't find any counterexamples to this proposition, having checked various generic extensions of $L$ and canonical inner models like $L[U],$ so I'm guessing this is true. It would suffice to check that if there a $\Sigma_1$ well-ordering of $V,$ then all sets of ordinals are in $L.$ This is true for subsets of $\alpha$ for any $\alpha$ countable in $V,$ by applying Mansfield's theorem that if there is a $\Sigma_2^1$ well-ordering of $\mathbb{R},$ then $\mathbb{R} \subset L.$ Then the first non-trivial case would be to show this holds for subsets of $\omega_1,$ but I haven't been able to do this (the proof of Mansfield's theorem doesn't seem to extend to this more general case). So, are there any known results similar to what I'm asking? REPLY [11 votes]: No, see the paper On $ Σ_1$ Well-Orderings of the Universe by Harrington and Jech.<|endoftext|> TITLE: Existence of relative Dirichlet density of primes starting with 1 QUESTION [12 upvotes]: This question is a duplicate of an existing MO question, but that other MO question has an accepted answer that does not actually answer the question, and I'm not sure how to fix that other than by re-asking the question. On page 76 of Serre's book A Course in Arithmetic, he writes: [T]here exist sets having an analytic density but no natural density. It is the case, for example, of the set $P^1$ of prime numbers whose first digit (in the decimal system, say) is equal to 1. One sees easily, using the prime number theorem, that $P^1$ does not have a natural density and on the other hand Bombieri has shown me a proof that the analytic density of $P^1$ exists (it is equal to $\log_{10}2 = 0.301029995\ldots$). There is a slight misstatement here because literally speaking, $P^1$ has natural density zero, but clearly the intent is to speak of the relative density of $P^1$ inside the set $P$ of all primes. In other words, Bombieri's result is a kind of "Benford's law for primes": $$\lim_{s\to1} {\sum_{m\in P_1} m^{-s} \over \sum_{p\in P} p^{-s}} = \log_{10}2.$$ My question is, how does one prove that the above limit (which goes by various names—relative analytic density, relative Dirichlet density, relative zeta density) exists? Serre does not say anything about this. The accepted answer to the duplicate MO question cites two papers, one by Cohen and Katz, and one by Raimi. But the paper by Cohen and Katz simply restates what Serre says without giving a proof. The paper by Raimi cites a paper by R. E. Whitney (Initial digits for the sequence of primes, Amer. Math. Monthly 79 (1972), 150–152) but Whitney's paper considers logarithmic density rather than Dirichlet density: $$\lim_{N\to\infty} {\sum_{m\in P_1, m\le N} 1/m \over \sum_{p\in P, p\le N} 1/p}.$$ It's not clear to me that this implies Bombieri's result. In a comment to a now-deleted MO question, KConrad suggested looking in the book Prime Numbers by Ellison and Ellison, but I did not find the answer there either. REPLY [6 votes]: The answer by so-called friend Don indicates why the existence of logarithmic density implies the existence of Dirichlet density, with the same value. Below is an argument explaining why the set of primes with a specified initial digit has a logarithmic density (of Benford type), and thus a Dirichlet density. In my experience, the number of times authors cite Serre on this point is much greater than the number of references that provide an actual proof, so another write-up of a proof is probably helpful. The paper by R. E. Whitney mentioned in the OP uses the Mertens formula $\sum_{p \leq x} 1/p = \log \log x + M + O(1/(\log x)^2)$, where the $O$-term is stronger than the remainder term typically found in treatments of this estimate, which is $O(1/(\log x))$. The argument below avoids this. We start with a useful technical lemma that will be applied a few times. Lemma. For $0 < c_1 < c_2$ and $b > 1$, $$ \sum_{c_1b^k \leq p \leq c_2b^k} \frac{1}{p} \sim \frac{\log_b(c_2/c_1)}{k} $$ as $k \to \infty$, where the sum on the left runs over primes. Proof. Our argument is based on p. 7 here, which is the paper cited at the end of the Wikipedia page about Dirichlet density. That paper proves a much more general result than what we need. Using partial summation, for $x \geq 1$ $$ \sum_{p \leq x} \frac{1}{p} = \frac{\pi(x)}{x} + \int_1^x \frac{\pi(y)}{y^2}\,dy. $$ Therefore \begin{align*} \sum_{c_1b^k \leq p \leq c_2b^k} \frac{1}{p} & = \sum_{p \leq c_2b^k} \frac{1}{p} - \sum_{p \leq c_1b^k} \frac{1}{p} + O\left(\frac{1}{b^k}\right) \\ & = \frac{\pi(c_2b^k)}{c_2b^k} - \frac{\pi(c_1b^k)}{c_1b^k} + \int_{c_1b^k}^{c_2b^k} \frac{\pi(y)}{y^2}\,dy + O\left(\frac{1}{b^k}\right). \end{align*} To show this is asymptotic to $\log_b(c_2/c_1)/k$ as $k \to \infty$, we will show $$ \frac{\pi(c_2b^k)}{c_2b^k} - \frac{\pi(c_1b^k)}{c_1b^k} = o\left(\frac{1}{k}\right), \ \ \ \int_{c_1b^k}^{c_2b^k} \frac{\pi(y)}{y^2}\,dy \sim \frac{\log_b(c_2/c_1)}{k}. $$ By the Prime Number Theorem, $\pi(c_2b^k) \sim c_2b^k/\log(c_2b^k) \sim c_2b^k/(k\log b)$ and $\pi(c_1b^k) \sim c_1b^k/(k\log b)$ as $k \to \infty$, so $$ k\left(\frac{\pi(c_2b^k)}{c_2b^k} - \frac{\pi(c_1b^k)}{c_1b^k}\right) \to \frac{1}{\log b} - \frac{1}{\log b} = 0. $$ Thus $\pi(c_2b^k)/c_2b^k - \pi(c_1b^k)/c_1b^k = o(1/k)$ as $k \to \infty$. To estimate $\int_{c_1b^k}^{c_2b^k} (\pi(y)/y^2)\,dy$, pick $\varepsilon > 0$. For large $y$, say $y \geq y_\varepsilon$, we have $1-\varepsilon \leq \pi(y)/(y/\log y) \leq 1+\varepsilon$, so $(1-\varepsilon)/(y\log y) \leq \pi(y)/y^2 \leq (1+\varepsilon)/(y\log y)$. For $k$ large enough that $c_1b^k \geq y_\varepsilon$, \begin{equation}\label{int-long} (1-\varepsilon)\int_{c_1b^k}^{c_2b^k}\frac{dy}{y\log y} \leq \int_{c_1b^k}^{c_2b^k} \frac{\pi(y)}{y^2}\,dy \leq (1+\varepsilon)\int_{c_1b^k}^{c_2b^k}\frac{dy}{y\log y}, \end{equation} and as $k \to \infty$, \begin{align*} \int_{c_1b^k}^{c_2b^k}\frac{dy}{y\log y} = \log\log(c_2b^k) - \log\log(c_1b^k) & = \log\left(\frac{k\log b + \log c_2}{k\log b + \log c_1}\right) \\ & = \log\left(1 + \frac{\log_b(c_2/c_1)}{k + \log_b(c_1)}\right) \\ & = \frac{\log_b(c_2/c_1)}{k + \log_b(c_1)} + O\left(\frac{1}{k^2}\right) \\ & = \frac{\log_b(c_2/c_1)}{k} + O\left(\frac{1}{k^2}\right). \end{align*} Therefore $$ 1-\varepsilon \leq \varliminf_{k \to \infty} \frac{\int_{c_1b^k}^{c_2b^k} (\pi(y)/y^2)\,dy}{\log_b(c_2/c_1)/k} \leq \varlimsup_{k \to \infty} \frac{\int_{c_1b^k}^{c_2b^k} (\pi(y)/y^2)\,dy}{\log_b(c_2/c_1)/k} \leq 1 + \varepsilon. $$ Taking $\varepsilon \to 0^+$ shows $\int_{c_1b^k}^{c_2b^k} (\pi(y)/y^2)\,dy \sim \log_b(c_2/c_1)/k$, so we're done. QED For $d \in \{1, \ldots, 9\}$, let $A_d$ be the set of primes with leading digit $d$. For example, \begin{align*} A_1 & = \{11, 13, 17, 19, 101, 103, 107, 109, 113, 127, 131, 137, 139, \ldots\}, \\ A_2 & = \{2, 23, 29, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, \ldots\}. \end{align*} We can describe the primes in $A_d$ in terms of inequalities: $$ A_d = \{p : d \cdot 10^k \leq p < (d+1)10^k \ {\rm for \ some } \ k \geq 0\}. $$ Theorem. For each $d$, $A_d$ has logarithmic density $\log_{10}((d+1)/d)$. Since existence of logarithmic density implies existence of Dirichlet density with the same value, $A_d$ has Dirichlet density $\log_{10}((d+1)/d)$. Proof. Instead of estimating $$ \frac{\sum_{p \in A_d, p \leq x} 1/p}{\log \log x} $$ for large $x$, we want to restrict $x$ to powers of $10$. We will first check the ratio changes negligibly when $x$ runs between consecutive powers of $10$. For $n \geq 1$, if $10^n \leq x < 10^{n+1}$ then $$ \log n + \log \log 10 \leq \log\log x < \log(n+1) + \log \log 10, $$ so $\log \log x \sim \log n$ as $x \to \infty$. We will show $$ \frac{\sum_{p \in A_d, 10^n \leq p < 10^{n+1}} 1/p}{\log n} \to 0 \ {\rm as } \ n \to \infty, $$ so $$ \lim_{x \to \infty} \frac{\sum_{p \in A_d, p \leq x} 1/p}{\log \log x} = \lim_{n \to \infty} \frac{\sum_{p \in A_d, p < 10^n} 1/p}{\log n}, $$ in the sense that if the limit on the right exists then so does the limit on the left and they agree. If $d \not= 1$ then $\{p \in A_d : 10^n \leq p < 10^{n+1}\} = \emptyset$. If $d = 1$ then $\{p \in A_d : 10^n \leq p < 10^{n+1}\} = \{p : 10^n \leq p \leq 2 \cdot 10^{n}\}$ and $$ \sum_{10^n \leq p \leq 2 \cdot 10^{n}} \frac{1}{p} \sim \frac{\log_{10} 2}{n} $$ by the lemma, so $$ \frac{\sum_{p \in A_1, 10^n \leq p < 10^{n+1}} 1/p}{\log n} \sim \frac{\log_{10} 2}{n\log n} \to 0. $$ It remains to show $$ \lim_{n \to \infty} \frac{\sum_{p \in A_d, p < 10^n} 1/p}{\log n} = \log_{10}\left(\frac{d+1}{d}\right). $$ Break up the numerator into sums between consecutive powers of $10$: $$ \sum_{p \in A_d, p < 10^n} \frac{1}{p} = \sum_{0 \leq k \leq n-1} \left(\sum_{d \cdot 10^k \leq p < (d+1)10^k} \frac{1}{p}\right). $$ For $k \geq 1$, $$ \sum_{d \cdot 10^k \leq p < (d+1)10^k} \frac{1}{p} = \sum_{d \cdot 10^k \leq p \leq (d+1)10^k} \frac{1}{p} \sim \frac{\log_{10}((d+1)/d)}{k} $$ by the lemma. Therefore $$ \sum_{p \in A_d, p < 10^n} \frac{1}{p} \sim \sum_{1 \leq k \leq n-1} \frac{\log_{10}((d+1)/d)}{k} \sim \log_{10}\left(\frac{d+1}{d}\right)\log n, $$ so $A_d$ has logarithmic density $\log_{10}((d+1)/d)$. QED Remark. In a similar way, if $m \geq 2$ and $1 \leq a \leq m-1$, then the set of primes whose number of digits (in base $10$, say) is congruent to $a \bmod m$ has logarithmic density $1/m$ (it has no natural density). For example, the set of primes with an odd number of digits has logarithmic density $1/2$.<|endoftext|> TITLE: Which elements of $1+(x_1,x_2)\subset\mathbb{Z}_p[[x_1,x_2]]^\times$ are in $\langle 1+x_1,1+x_2\rangle$? QUESTION [9 upvotes]: It's a classical fact that the commutative power series ring $\mathbb{Z}_p[[x_1,x_2]]$ is isomorphic to the completed group algebra $\mathbb{Z}_p[[\mathbb{Z}_p\times\mathbb{Z}_p]]$, the isomorphism sending generators $a_1,a_2$ of $\mathbb{Z}_p\times\mathbb{Z}_p$ to $1+x_1,1+x_2$. In $\mathbb{Z}_p[[x_1,x_2]]$, we have the coset $1+(x_1,x_2)$ of the ideal $(x_1,x_2)$, which is a closed and open subgroup of its group of units, which certainly contain $1+x_1,1+x_2$, and hence contains the closed subgroup $\langle 1+x_1,1+x_2\rangle$ generated by $1+x_1,1+x_2$, which is isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$, but is strictly smaller than $1+(x_1,x_2)$. Is it possible to describe the image of the closed subgroup $\langle 1+x_1,1+x_2\rangle$ in $\mathbb{Z}_p[[x_1,x_2]]$? Ie, suppose we are given a power series $f\in 1+(x_1,x_2)\subset\mathbb{Z}_p[[x_1,x_2]]$. How can we determine if $f$ lies in the closed subgroup $\langle 1+x_1,1+x_2\rangle$? (I know this question isn't really well-defined. I'm just looking for certain criteria, ideally linked to algebraic properties of the power series, which can be used to identify the image of $\langle 1+x_1,1+x_2\rangle$). It would also be nice if one could describe the quotient $(1+( x_1,x_2))/\langle 1+x_1,1+x_2\rangle$. REPLY [5 votes]: The usual power series for $\log(1+x)$ determines an injection from $1+(x_1,x_2)\subset\mathbb{Z}_p[[x_1,x_2]]$ into $\mathbb{Q}_p[[x_1,x_2]]$. A power series $f\in 1+(x_1,x_2)$ is in $\langle 1+x_1,1+x_2\rangle$ if and only if there are $\alpha_1$, $\alpha_2\in\mathbb{Z}_p$ such that \begin{align*} \log f(x)&=\alpha_1\log(1+x_1)+\alpha_2\log(1+x_2)\\ &=\alpha_1\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}x_1^n+\alpha_2\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}x_2^n. \end{align*} Note that $\alpha_1$ and $\alpha_2$ can be read off from $\log f(x)$: they are the coefficients of $x_1$ and $x_2$, respectively. So if $$ \log f(x)=\sum_{i,j\geq 0}a_{ij}x_1^{i}x_2^{j}, $$ a necessary and sufficient condition for $f(x)\in\langle 1+x_1,1+x_2\rangle$ is that $a_{ij}=0$ if $i$, $j\geq 1$, and $$ a_{n,0}=\frac{(-1)^{n-1}}{n}a_{1,0}, $$ $$ a_{0,n}=\frac{(-1)^{n-1}}{n}a_{0,1}. $$ REPLY [2 votes]: The group algebra $\mathbb{Z}_p[[\mathbb{Z}_p\times \mathbb{Z}_p]]$ has the structure of a complete Hopf algebra, where $\mathbb{Z}_p\times \mathbb{Z}_p$ consists of precisely the group-like elements. Via the given isomorphism, we get a comultiplication on $\mathbb{Z}_p[[x_1,x_2]]$, where $1+x_1$ and $1+x_2$ are group-like. The formula for the comultiplication is $\Delta x_i=x_i\otimes 1+1\otimes x_i+x_i\otimes x_i$ for $i\in\{1,2\}$. A power series $f\in 1+(x_1,x_2)$ is in $\langle 1+x_1,1+x_2\rangle$ if and only if $f$ is group-like, i.e. if and only if there is an equality of formal power series $$ f(x_1+y_1+x_1y_1,x_2+y_2+x_2y_2) = f(x_1,x_2)f(y_1,y_2)\in\mathbb{Z}_p[[x_1,x_2,y_1,y_2]]. $$<|endoftext|> TITLE: Gap in Przytycki's computation of the skein module of links in a handlebody? QUESTION [13 upvotes]: I am reading the paper [1], where the author proves that the skein module of links in a handlebody $F\times I$ has a free basis given by products $D_1 \cdots D_n$ where each $D_i$ is the closure of $v_i\in \pi_1(F)$, $v_i$ is the smallest length lexicographically smallest representative of its conjugacy class and $v_1\leq v_2\leq\cdots\leq v_n$. Besides several instances where the author makes a vague argument which presumably can be fixed using diamond lemma, I am stuck in the point where he proves that such products span the skein module. The only place devoted to this issue is the second paragraph of "Proof of Theorem 2.11", p. 333. He says that, as an algebra both the skein module $\mathcal{S}(F\times I)$ and the symmetric algebra over the span of conjugacy classes of the fundamental group $S(R\hat{\pi}^\circ)$ are generated by the representatives of conjugacy classes of $\pi_1(F)$, hence the claim. But the map between these is not an algebra homomorphism, so this argument does not apply. I don't know how to fix this gap. Naively one would think that induction on the number of crossings would work. But here is the problem. We need to be able to do two things: 1) switch the order of components, 2) change a representative of a component (i.e. conjugate the element $v_i\in\pi_1(F)$ to make it lexicographically smaller). Now the problem with these two operations is that they behave well with respect to different invariants. 1) works up to diagrams with smaller number of crossings, 2) works up to diagrams whose components have words of smaller lengths (see Lemma 1.7). Clearly 1) messes up the maximal length, indeed, usually the left-overs will have words whose lengths are sums of lengths of what we start with. On the other hand, 2) involves performing some isotopies. Even if the number of self-crossing of the component does not go up under such isotopies, the number of crossings with other components may increase. So we are chasing our own tail here. UPDATE: First note that the question is about modules over the polynomial ring $\mathbb{Z}[v,v^{-1},h]$, not over the field of fractions. It turns out the question is not that easy. I did some experiments and it turns out the statement is actually false in almost all cases, except as stated by the author. For instance, if the surface is not planar, i.e. for the punctured torus it is false. If it is planar, i.e. a disk with several disks removed, and if the order of generators is different, or the choice of generators is different, it is false. The order of generators plays a role in saying what's lexicographically smaller. For instance, take a disk with $2$ disks removed. Suppose the generators of the fundamental group are $x$ going around one puncture, and $y$ going around both punctures. If the order on the generators is $x TITLE: Has anyone seen these posets before? QUESTION [8 upvotes]: I've come across a family of posets which appear to have a couple of remarkable enumerative properties, and I'm wondering whether anyone has seen these before. Take $n\geqslant3$, and let $\preccurlyeq$ be a partial order on the set $\{1,\dots,n\}$. Say that $\preccurlyeq$ has property M if: There is a unique way to write $\{1,\dots,n\}$ as the union of two $\preccurlyeq$-chains, and $\preccurlyeq$ is maximal with this property, i.e. any refinement of $\preccurlyeq$ destroys the uniqueness in (1). For example, if $n=5$, there are four partial orders (up to isomorphism) with property M. Given as sets of covers, these are as follows: $$\{(1,2),(2,3),(3,4)\},\{(1,2),(2,3),(2,5),(4,5)\},\{(1,3),(3,4),(2,3),(2,5)\},\{(1,3),(3,5),(2,4),(1,4),(2,5)\}$$ Now here are the (apparent) remarkable properties: Up to isomorphism, there are exactly $2^{n-3}$ partial orders on $\{1,\dots,n\}$ with property M. If $\preccurlyeq$ has property M, then the size of $\preccurlyeq$ (i.e. the number of pairs $i\preccurlyeq j$) is $\binom n2+1$. I've checked these properties for $n\leqslant8$. Does anyone know whether they are true generally? (I also have a conjectural construction of all partial orders with property M, coming from representation theory, but I can't prove anything!) REPLY [8 votes]: I claim: A poset with property M has precisely two maximal elements. Precisely one maximal element is greater than every non-maximal element. Call this the supermaximal element. Removing the supermaximal element (if $n>3$) leaves a poset with property M. So there is a unique way to build any example from the $n=3$ example by repeatedly adding a new supermaximal element. For an $n=k$ example, there are two ways to add a new supermaximal element (choose which of the existing maximal elements remains maximal), each of which increases the "size" by $k$. Now the two properties you observed follow by induction.<|endoftext|> TITLE: A homeomorphism with a prescribed action on the fundamental group - decidable or not? QUESTION [9 upvotes]: I am curious if the following topological problem is decidable. Let $M,N$ be two closed manifolds. Given a group isomorphism $p: \pi_1(M)\to \pi_1(N)$, is there a homeomorphism $\phi: M\to N$ such that $\phi_*=p$? EDIT. (Thanks for helpful comments.) There are some relevant results: "Algorithmic aspects of homeomorphism problems" by Nabutovsky and Weinberger. There is a preprint version https://arxiv.org/abs/math/9707232 (click ps, pdf is a mess!), and a published version http://www.ams.org/books/conm/231/. A sort of review of their methods was written by R.I.Soare, "Computability theory and differential geometry" (Bulletin of Symbolic Logic, 2004). According to the paper, the problem is decidable for simply connected manifolds of dimension at least 5. They also construct a counterexample for a non-simply connected case (Proposition 0.1 in the published paper). However, I would like to point out that this counteraxample does not answer the question. Even if algorithm for solving a problem formulated above exists, it cannot be used unless you have an explicit isomorphism between fundamental groups. In the situation of Proposition 0.1, it leads to no contradiction. Interestingly, the results of Nabutovsky and Weinberger may be taken as a hint that the answer to the question is positive. At least, this possibility is not excluded for all I know. EDIT. Actually, the example in Proposition 0.1 does show that the problem is undecidable. (So, I got it wrong. Thanks to Achim Krause for clarifying the details.) REPLY [7 votes]: This is undecidable. For a construction, see the first page of Nabutovsky and Weinberger, "Algorithmic aspects of homeomorphism problems", Rothenberg Festschrift 1998.<|endoftext|> TITLE: Cartan's Structure Equations VS Cartan's Method of Equivalence QUESTION [5 upvotes]: There have been a number of posts on related questions, such as: Geometric interpretation of Cartan's structure equations, What is the geometric significance of Cartan's structure equations? and Maurer-Cartan structure equation derivation. While my question is related, it is I hope a bit more specific. I am trying to learn some of Cartan's methods, and getting a little confused, because they all seem to be closely related, so that differentiating between them is difficult for me. I would like to think of Cartan's structure equations this way. You start with an adapted co-frame and apply d, then express the results in terms of old data (the co-frame), but while respecting the Lie algebra. This gives 2 new things, the connection 1-form, and torsion. We then apply d again, and express the results in terms of "old" data (maybe while respecting an underlying Lie algebra). I am trying to make sense of this. It seems that this is the same process than the one used in Cartan's equivalence method. Am I right? So torsion is the first "invariant", which could be used to compare 2 different geometric structures locally, while curvature would be the next "invariant". But what is the relevant EDS perhaps? It seems that we are building something recursively, so I am a little confused. Perhaps we need to go to infinity, to see the whole structure, right? As in, using infinity-structures, like Urs Schreiber seems to be suggesting, in the second link above. Can someone please comment or answer my questions? Edit: after some thinking, and reading a good chunk of Olver's book "Equivalence, Invariants and Symmetry", here is my current understanding of Cartan's structure equations. Let's say you have a Riemannian manifold $(M,g)$, and let $(\theta^i)$, for $1 \leq i \leq m$, with $m=\dim M$, be a smooth local orthonormal coframe. Applying $d$ to the coframe gives our first set of "invariants" (or perhaps I should write $O(m)$-invariants). That the first set of "invariants" is nothing but the Levi-Civita connection is the meaning of the first structure equations. We then apply $d$ a second time, and get a second set of "invariants". That this second set of invariants can be broken in 2 parts, the first quadratic in the Levi-Civita connection and the second one nothing but the curvature of $g$, is the content of Cartan's second structure equation. REPLY [3 votes]: You are right that structure equations are the result of applying the method of equivalence. You don't start with an adapted coframe. (Some people might say that you do, but that is not quite what the method does.) You start with the bundle of all adapted coframes. You then construct on it the bundle of all adapted pseudoconnections, i.e. coframes on the total space of the first bundle. And so on. As you go, at each step you reveal one more collection of structure equations. You keep going until (1) you do not have a geometric hypothesis at hand which can justify the next order frame adaptation or (2) you reach involution in the sense explained in Robert Gardner's book. Since the entire procedure is invariant under local isomorphisms of G-structures, the resulting structure equations "respect" the Lie algebra of symmetries. The relevant EDS is that of the isomorphisms of the G-structures, as explained in Gardners' Lecture 7.<|endoftext|> TITLE: Definition of the Bergman fan QUESTION [5 upvotes]: Let $M$ be a matroid on the set $\{1,\dots,m\}$. The Bergman fan $\tilde{B}(M)$ is defined in literature to be the set in $\mathbb{R}^{m}$ consisting of the vectors $v=(w_1,\dots,w_m)$ such that for every circuit $c\in C(M)$, the minimum $\min_{i\in c}w_i$ is attained at least twice. It is then mostly left implicit what the fan structure on this set should be. Sometimes people mention equivalence classes formed by $v_1\sim v_1$ iff $M_{v_1}=M_{v_2}$, where $M_v$ is the matroid with as bases those bases $B$ of $M$ attaining the minimum $\min_{B\in B(M)} \sum_{b\in B}v_b$. For what I gather it is implied that these should define a fan structure on $\tilde{B}(M)$, but simple examples show that it doesn't: Let $M=U_{3,3}$. Then there are no circuits, so the support of $\tilde{B}(M)$ is the whole space. Since there is only a single basis, there is only a single equivalence class. So the equivalence classes do not define a fan structure on $\tilde{B}(M)$. What am I not getting here? REPLY [5 votes]: From Matroid polytopes, nested sets and Bergman fans, Feichtner and Sturmfels, just above Proposition 2.5 on the bottom of page 4: Two vectors $w,w' \in \mathbb{R}^n$ are considered equivalent for the matroid $M$ if $M_w = M_{w'}$. The equivalence classes are relatively open convex polyhedral cones. These cones form a complete fan in $\mathbb{R}^n$. This fan is the normal fan of $P_M$. If $\Gamma$ is a cone in the normal fan of $P_M$ and $w \in \Gamma$ then we write $M_\Gamma = M_w$. The following proposition shows that the Bergman fan $\tilde{\mathcal{B}}(M)$ is a subfan of the normal fan of the matroid polytope $P_M$. For $M = U(3,3)$ the one equivalence class is $\mathbb{R}^3$, which is indeed open, convex and a cone (it is the positive hull of $\{\pm e_1, \pm e_2, \pm e_3\}$). The set $\{\mathbb{R}^3\}$ is a fan: every face of a cone in the fan is in the fan (check: $\mathbb{R}^3$ is the only face) the intersection of any two cones in the fan is a face of each (check: $\mathbb{R}^3$ is the only intersection) I tried working out another example but I'm still new to this so bear with me. For another example, let $M = U(1,3)$. Then the equivalence classes can be parameterized in terms of subsets of the bases of $M$: $M_v$ has only $1$ as a basis iff $v = (a,b,b)$ with $a > b$. $M_v$ has $1,2$ as bases iff $v = (a,a,b)$ with $a > b$ $M_v$ has $1,2,3$ as bases iff $v = (a,a,a)$ and we have similar conditions for permutations of the variables. Therefore the equivalence classes are $$[1,0,0], [0,1,0], [0,0,1], [1,1,0], [1,0,1], [0,1,1], [1,1,1]$$ where $[1,0,0] = -[0,1,1], [0,1,0] = -[1,0,1], [0,0,1] = -[1,1,0]$. The line $[1,1,1]$ is a face of every cone. The intersecton of any two cones $C_1, C_2$ is either the line $[1,1,1]$ or if $C_1 = C_2$ then $C_1 \cap C_2 = C_1$.<|endoftext|> TITLE: An Averaged Version of Gromov's Waist Inequality QUESTION [5 upvotes]: Gromov's waist inequality for unit n-sphere $\mathbb{S}^{n}$ says: For any continuous function $f: \mathbb{S}^{n} \rightarrow \mathbb{R}^{m} $, there is some $y \in \mathbb{R}^{m}$ s.t. $Vol_{n-m}(f^{-1}(y)) \geq Vol_{n-m}(\mathbb{S}^{n-m}) $. I'm wondering if there is an averaged version of the inequality, comparing the averaged fiber volume and some averaged $\mathbb{S}^{n-m}$ volume. For example, is it true that for some constant $C(n, m)$ depending only on dimensions: $ \int_{f(\mathbb{S}^{n})} Vol_{n-m}(f^{-1}(y)) \geq C(n, m) Vol_{m}(f(\mathbb{S}^{n})) Vol_{n-m}(\mathbb{S}^{n-m}) $ It is, of course, interesting to consider the tubular version: $ \int_{f(\mathbb{S}^{n})} Vol_{n}(f^{-1}(y) + \epsilon ) \geq C(n, m) Vol_{m}(f(\mathbb{S}^{n})) Vol_{n}(\mathbb{S}^{n-m} + \epsilon ) $, as well as similar inequalities for ball, cube, etc. in place of sphere. REPLY [2 votes]: It is more natural to estimate the total area of large fibres, but it also does not work:https://arxiv.org/abs/1402.2856<|endoftext|> TITLE: Combinatorics problem related to Motzkin numbers with prize money I QUESTION [19 upvotes]: Here a combinatorics problem. I offer 30 euro for a proof and 100 bounty points for a counterexample: Let $n \geq 2$. An $n$-Kupisch series is a list of $n$ numbers $c:=[c_1,c_2,...,c_n]$ with $c_n=1$, $c_i \ge 2$ for $i \neq n$ and $c_i-1 \leq c_{i+1}$ for all $i=1,...,n-1$ and setting $c_0:=c_n$. The number of such $n$-Kupisch series is equal to $C_{n-1}$ (Catalan numbers). The CoKupisch series $d$ of $c$ is defined as $d=[d_1,d_2,...,d_n]$ with $d_i:= \min \{k | k \geq c_{i-k} \} $ and $d_1=1$. One can show that the $d_i$ are a permutation of the $c_i$. A number $a \in \{1,...,n \}$ is a descent if $a=1$ or $c_a >c_{a-1}$. Define a corresponding set, indexed by descents: $X_1 := \{1,2,...,c_1-1 \}$, and $X_a := \{ c_{a-1}, c_{a-1}+1 ,..., c_a -1 \}$ for descents $a > 1$. A $n$-Kupisch series is called $2-$Gorenstein if it satisfies the following condition: for each descent $a$, and each $b \in X_a$: either $c_{a+b} \geq c_{a+b-1}$ or $d_{a+b-1} = d_{a+b + c_{a+b}-1} - c_{a+b}$ is satisfied. Conjecture: The number of $n$-Kupisch series that are $2$-Gorenstein for $n \geq 2$ equals 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, ... which are the Motzin numbers. https://oeis.org/A001006 My computer says it is true for $n=2,3,...,14$. (also for n=1 if [1] is the unique 1-Kupisch series, by a representation-theoretic interpretation) Here an example which visualises the c and d sequences in a picture of a Dyck path: https://drive.google.com/file/d/0B9hKtvQe-4-bQlpjcURfYnNzUGs/view Background: This has a representation theoretic background (see http://www.sciencedirect.com/science/article/pii/0022404994900442 )and would give a certain classification result together with a new categorification of the Motzkin numbers. There is a very natural bijection of n-Kupisch series to Dyck paths from (0,0) to (2n-2,0) and probably the 2-Gorenstein algebras among them might give a new combinatorial interpretation of Motzkin paths as subpaths of Dyck paths. I found in fact many such things and translated them into elementary problems, but I have no talent in complicated combinatorics :( example for n=5: All 5-Kupisch series (number is 14): [ [ 2, 2, 2, 2, 1 ], [ 3, 2, 2, 2, 1 ], [ 2, 3, 2, 2, 1 ], [ 3, 3, 2, 2, 1 ], [ 4, 3, 2, 2, 1 ], [ 2, 2, 3, 2, 1 ], [ 3, 2, 3, 2, 1 ], [ 2, 3, 3, 2, 1 ], [ 3, 3, 3, 2, 1 ], [ 4, 3, 3, 2, 1 ], [ 2, 4, 3, 2, 1 ], [ 3, 4, 3, 2, 1 ], [ 4, 4, 3, 2, 1 ], [ 5, 4, 3, 2, 1 ] ] All 2-Gorenstein 5-Kupisch series (number is 9): [ [ 2, 2, 2, 2, 1 ], [ 3, 2, 2, 2, 1 ], [ 2, 3, 2, 2, 1 ], [ 4, 3, 2, 2, 1 ], [ 2, 2, 3, 2, 1 ], [ 3, 2, 3, 2, 1 ], [ 3, 3, 3, 2, 1 ], [ 2, 4, 3, 2, 1 ], [ 5, 4, 3, 2, 1 ] ] Representation theoretic conjecture/background: Conjecture: The number of 2-Gorenstein algebras that are Nakayama algebras with n simple modules with a linear oriented line as a quiver is equal to the sequence of Motzkin numbers. Background/explanations: Here the $c_i$ are the dimension of the indecomposable projective modules which determine the algebra uniquely (assuming that the algebras are connected quiver algebras over a field). The $d_i$ are the dimension of the indecomposable injective modules at point $i$. 2-Gorenstein means that the dual of the regular module $D(A)$ has a projective presentation $P_1 \rightarrow P_0 \rightarrow D(A) \rightarrow 0$ with $P_1$ having injective dimension bound by 1. ($P_0$ is always projective-injective for Nakayama algebras) Here a link for the program I used to test things (copy all programs, and the finaltest program does the job then. 1 means the Kupisch series is 2-Gorenstein and 0 means it is not): https://docs.google.com/document/d/1U9mriuvCEE9FeXY1TfY_yJ1mi05TCdHRfppwCSwi1S8/pub edit: Since we have 2 people with proofs now, I award 30 Euro each to them in case their proof is correct (I still have to check) and then no more money for new proofs. I award additionally 200 bounty points to the nicest proof (which might be one of the two first posted proofs or another not yet posted proof), which can be decided by the community in terms of upvotes near the end when the bounty expires (16.08.) edit 2: Now at 16.08. I give the bounty points to Anton, since it is the most complete answer and the details of findstats answer are not posted yet. I will give 30 Euro to each after checking findstats proof (when it is posted and correct). edit 3: Since findstat updated the answer and now has a full proof of an extension of the conjecture just a little time after I awarded the first bounty, I also will give a bounty to them and accept their answer (second bounty had to be 400 or higher and can be awarded after 24 hours). REPLY [3 votes]: Here is a sketch of the proof. For me Dyck paths of length $n$ are paths from $(0,0)$ to $(n,n)$ consisting of North and East steps staying weakly above the diagonal. North means $(0,1)$, East means $(1,0)$. A path is called primitive if it is non-empty and touches the diagonal only at the endpoints. A number of the form $a+b-1$ for a descent $a$ and $b\in N_a$ will be called double rise. A number $i$ such that $c_{i+1} TITLE: Is the set of points with smallest stabilizer open? QUESTION [5 upvotes]: Let $G$ be a complex reductive group acting linearly on a complex affine variety $X$, and let $K$ be the kernel of the action, i.e. $$K:=\{g\in G:g\cdot x=x\text{ for all }x\in X\}.$$ Is $$X_K:=\{x\in X:\mathrm{Stab}_G(x)=K\}$$ Zariski-open in $X$? REPLY [4 votes]: The answer is affirmative if $G$ is finite or abelian since in that case subgroups are rigid. Otherwise, $X_K$ may not even be dense, let alone open. The standard example is due to Luna from his slice paper: Let $G=SL(2,\mathbb C)$ act on the space $X=S^3\mathbb C^2$ of binary cubics. The action is effective so $K=1$. Every non-degenerate cubic $f$ is the product of $3$ distinct linear factors. Then $Stab_G(f)$ contains the cyclic permutation of these factors which shows that $X_K$ is not dense. It is not empty either, since $f=x^2y$ has trivial stabilizer. In my opinion, the question is ill-posed since $X_K$ may be open for the trivial reason that it is empty. Let, e.g., $G$ be absolutely simple (e.g. $G=SO(2n+1,\mathbb C)$) and $H$ a proper reductive subgroup. Then $X=G/H$ is affine with $K=1$ but $Stab_G(x)\ne K$ for all $x\in X$. The way to go is to look at the stabilizers of only the closed orbits. Their behavior is much more regular which gives rise to the Luna stratification.<|endoftext|> TITLE: What are traces? QUESTION [27 upvotes]: Let $A$ be a Noetherian commutative ring and Let $A\rightarrow B$ be a finite flat homomorphism of rings. We can thus form the so called "trace" $\mathrm{Tr_{B/A}}:B\rightarrow A$, which is a homomorphism of $A$ - modules defined as follows: Every $b\in B$ acts on $B$ (when viewed as an $A$ - module) by multiplication. Since $B$ is finite flat over $A$ and $A$ is Noetherian, $B$ is a locally free $A$ - module and hence multiplication by $b$ is given locally (on a principal open subset $\mathrm{Spec}(A_s)\subseteq \mathrm{Spec}(A),s\in A$ and under some isomorphism $B_s\cong A_s^n$) by multiplication by a matrix. We define $\mathrm{Tr_{B/A}}(b)$ to be the trace of this matrix. Since the trace of a matrix is independent of the choice of basis this homomorphism of $A$ - modules glues nicely and is well defined. In the case $A\rightarrow B$ is finite etale one can even show that this morphism is nondegenerate, i.e.: Induces an isomorphism $B\overset{\sim}\rightarrow \mathrm{Hom}_A(B,A)$ by adjunction (this is a well known claim of Galois theory in the case where $A\rightarrow B$ is a finite seperable field extension). My (rather ill-formulated) questions are the following: 1) Are there any other algebraic/geometric constructions I should think of as similar to this one? 2) Is there a deeper reason for the existence of such trace morphisms (for example some categorical phenomena that this is a special case of)? what's so special about finite flat homomorphisms that makes this happen? It seems to me pretty mysterious that such a homomorphism even exists, and I do not seem to completely grasp it's geometric meaning. 3) What is the geometric intuition behind the cannonical isomorphism of $A$ - modules $B\overset{\sim}\rightarrow \mathrm{Hom}_A(B,A)$ in case $A\rightarrow B$ is finite etale? [edit] Let me try to be abit more specific about what bothers me: Given a ring homomorphism $\phi :A\rightarrow B$ there's an obvious adjunction: $\mathrm{Forget}:\mathsf Mod_B \substack{\longrightarrow\\\perp \\\longleftarrow \\}\mathsf Mod_A:\mathrm{Hom}_A(B,-)$ which, by evaluating the counit at $A$, gives a map $\mathrm{Tr}_\phi:\mathrm{Hom}_A(B,A)\rightarrow A$ in $\mathsf{Mod}_A$. Also, given a proper map $f:X\rightarrow Y$ between reasonable schemes (for example essentially finite type schemes over a field) we have the adjunction given by Grothendieck duality $Rf_*:\mathsf D^b_c(X) \substack{\longrightarrow\\\perp \\\longleftarrow \\}\mathsf D^b_c(Y):Rf^!$ which induces (again by evaluating the counit at $\mathcal{O}_Y$) a morphism $\mathrm{Tr}_f:Rf_*Rf^!\mathcal{O}_Y\rightarrow\mathcal{O}_Y$ in $\mathsf{D}_c^b(Y)$ What bothers me is that my original trace, unlike the two trace maps I just mentioned, does not seem to come as naturally from some adjunction or anything like that. Where is it coming from? What is it? REPLY [8 votes]: It seems worth mentioning the geometric setting: Let $A$ and $B$ be the rings of functions on varieties $X$ and $Y$ over an algebraically closed field $k$, so we have a map $\phi: Y \to X$. Let $d$ be the degree of $\phi$ and let $x \in X$ be a closed point over which $\phi$ is etale, so $\phi^{-1}(x) = \{ y_1, \ldots, y_d \}$. Let $x$ correspond to the maximal ideal $\mathfrak{p} \subset A$ and let $y_i$ correspond to the maximal ideal $\mathfrak{q}_i \subset B$. By the Chinese Remainder Theorem, $$B \otimes_A A/\mathfrak{p} = \bigoplus B/\mathfrak{q}_i.\quad (\ast)$$ Let $g \in B$. Then $\mathrm{Tr}(g)$, evaluated at the point $x$, is the same as the trace of the map $\bar{g} :B \otimes_A A/\mathfrak{p} \to B \otimes_A A/\mathfrak{p}$ induced by multiplication by $g$. Under the isomorphism $(\ast)$, $\bar{g}$ acts on $B/\mathfrak{q}_i$ by $g(y_i)$. In particular, this action is diagonal! So, for $x$ as above, we have $$\mathrm{Tr}(g)(x) = \sum g(y_i).$$ The nifty fact is that this extends to a regular function on the locus where $\phi$ is branched.<|endoftext|> TITLE: Which groups can have $GSp(4)$ as local component? QUESTION [6 upvotes]: In some cases the relations between a global group $G$ (over the adeles $\mathbb{A}$ of a field $F$) and its local components $G_v$ (where $v$ are the places of $F$) are well known. Obviously a group determines its local components, so the question aims at, given a family of local groups , understanding whether or not it can arises as the local components of global group. Some examples: the local components of a quaternion algebra are $GL_2(F_v)$ almost everywhere, and an even number of local division quaternion algebras $B_v$ the local components of a unitary group are $GL_2(F_v)$ at the split places, and any collection of local unitary groups at the other places, maybe safe one (Hasse invariants condition, empty if the number of variables is odd) Now here is the question: what natural global group can have $GSp(4)$ as (one, many or every) local component(s)? REPLY [3 votes]: In addition to the relatively boring extension/restriction of scalars for $GSp(2n,\widetilde{k})$ for an extension field $\widetilde{k}$ of the ground field $k$... : The Galois twist groups often denoted by $Sp^*(p,q)$ (or $GSp^*(p,q)$...) over ground field $\mathbb R$, defined via non-degenerate quaternion-symmetric forms with quaternion algebras over the ground field (or over extensions...) almost everywhere locally become $Sp(p+q,k_v)$, because the quaternion algebra splits almost everywhere locally. (Throwing in the similitudes is minor...) That is, with quaternion (division) algebra $B$ over ground field $k$ (a global field), with non-degenerate $B$-valued, symmetric form on a $B$-vectorspace $V$ of $B$-dimension $n$, the isometry group $G$ has $k_v$-points isomorphic to $Sp(2n,k_v)$ locally almost everywhere. EDIT: it may be worth adding that this sort of Galois twisting allows easy creation of compact quotients analogous to Shimura curves for $SL(2)=Sp(2)$, by choosing a symmetric form whose signature at one real place is $(0,q)$ or $(p,0)$, so that (by a standard, if not widely understood, reduction theory result) the arithmetic quotient is compact. But/and the representation theory is just that of (locally, split) $Sp(n)$ almost everywhere.<|endoftext|> TITLE: $GSp(4)$ vs $PSp(4)$ QUESTION [6 upvotes]: After some months wandering through examples of algebraic groups in the theory of automorphic forms and number theory, I wonder why so many efforts are spent in understanding $GSp(4)$ (local newforms, liftings, correspondences, etc.) without any mention to $PSp(4)$. Explicit examples and computable settings are rare and valuable, and it seems far easier to deal with $PSp(4)$ than with $GSp(4)$ at first. Is there any reason for not having anything in that direction? Are the results and theories involved straightforward from other known cases? REPLY [6 votes]: If we think about classical modular forms, one typically works on SL(2). One could also work on PSL(2), but one would like to write down congruence subgroups in terms of matrices, so one often phrases things for SL(2). Moreover, if one wants to deal with nebentypus, one is forced to work with SL(2). However, if we want to study things representation theoretically, one usually passes to automorphic representations on GL(2), rather than SL(2). One loses nothing by doing so, and the representation theory is somewhat easier. E.g., there is one archimedean representation which is discrete series of weight $k$ for GL(2), rather than one of weight $k$ and one of weight $-k$ for SL(2). Globally we have strong multiplicity 1 for GL(2), but not for SL(2) and this is because the sizes of local $L$-packets for GL(2) are 1, whereas they are of size 1 or 2 for SL(2). Because of this, multipicity 1 is significantly harder to prove for SL(2) than for GL(2). Now if we move to higher rank, in some sense the closest analogue of GL(2) is GSp(4). You still have discrete series (unlike GL(3) or GL(4)), and we have the exceptional isomorphism PGSp(4) $\simeq$ SO(5) analogous to PGL(2) $\simeq$ SO(3). (Here SO means the split special orthogonal group.) One could also work with other groups with have compact forms (and often does, e.g., unitary groups) but GSp(4) is of interest because it treats the classical theory of Siegel modular forms of degree 2. (Note working on representations of GSp(4) with trivial central character is equivalent to working on PGSp(4).) Working classically with such Siegel modular forms, you work on Sp(4). Like SL(2), this is slightly more convenient and more general than PSp(4). On the other hand, working representation theoretically, GSp(4) is simpler than Sp(4), similar to the GL(2)/SL(2) situation. E.g., local $L$-packets no longer have size 1, but can have size 1 or 2 on GSp(4), but on Sp(4) they can have size up to 16---see work of Gan and Takeda. So most people who use representation theory find GSp(4) or PGSp(4) easier, and if one is going to work with the more difficult group Sp(4), there is no reason to restrict to PSp(4). Finally, I don't think that examples are any harder to come up with on GSp(4) rather than PSp(4), due for instance to the connection with SO(5) and theta series. In addition, work on PGSp(4) or GSp(4) may generalize to SO$(2n+1)$.<|endoftext|> TITLE: Topologies with no minimal $T_2$ topologies above them QUESTION [6 upvotes]: Let $(X,\tau)$ be a topological space. With $T_2(\tau)$ we denote the collection of $T_2$-topologies on $X$ that contain $\tau$. Is there an example of a topology $\tau$ such that the partially ordered set $(T_2(\tau),\subseteq)$ contains no minimal members? REPLY [4 votes]: There exists a topology $\tau$ with two non-isolated points on a countable set $X$ such that the poset $T_2(\tau)$ does not have minimal elements. To construct such topology $\tau$, take any Hausdorff $(\omega_1,\omega_1)$-gap on $\omega$, which is a pair $\big((A_\alpha)_{\alpha\in\omega_1},(B_\alpha)_{\alpha\in\omega_1}\big)$ of families of infinite subsets of $\omega$ satisfying the following two conditions: (H1) for any $\alpha<\beta<\omega_1$ we have $A_\alpha\subset^* A_\beta\subset^* B_\beta\subset^* B_\alpha$; (H2) for any set $C\subset\omega$ one of the sets $\{\alpha\in \omega_1:A_\alpha\subset^* C\}$ or $\{\alpha\in\omega_1:C\subset^* B_\alpha\}$ is at most countable. Here the notation $A\subset^* B$ means that the complement $A\setminus B$ is finite. It is well-known that Hausdorff $(\omega_1,\omega_1)$-gaps do exist in ZFC. Let $\tau$ be the topology on the countable set $X=\omega\cup\{-\infty,+\infty\}$ consisting of sets $U\subset X$ such that $\bullet$ if $-\infty\in U$, then there exists $\alpha\in\omega_1$ such that $B_\alpha\subset^* U$; $\bullet$ if $+\infty \in U$, then there exists $\alpha\in\omega_1$ such that $\omega\setminus A_\alpha\subset^* U$. It is clear that $-\infty$ and $+\infty$ are unique non-isolated points of the topological space $(X,\tau)$. We claim that no topology $\sigma$ in the poset $T_2(\tau)$ is minimal. Since $\sigma$ is Hausdorff, the points $-\infty,+\infty$ have disjoint neighborhoods $U_-,U_+\in\sigma$. By the condition (H2), one of the sets $A=\{\alpha\in\omega_1:A_\alpha\subset^* U_-\}$ or $B=\{\alpha\in\omega_1:U_-\setminus\{-\infty\}\subset^* B_\alpha\}$ is countable. If the set $A$ is countable, then we can find a countable ordinal $\alpha$ such that $A_\alpha\not\subset^*U_-$. Consider the topology $\sigma'$ on $X$ consisting of sets $W\subset X$ satisfying two conditions: $\bullet$ if $-\infty\in W$, then there exists $U\in\sigma$ such that $-\infty\in U\cup A_\alpha\subset^* W$; $\bullet$ if $+\infty\in W$, then there exists $U\in\sigma$ such that $+\infty\in U\subset^* W$. It is clear that $\tau\subset\sigma'\subset\sigma$ and $\sigma'\ne\sigma$ as $U_-\in\sigma\setminus\sigma'$. The topology $\sigma'$ is Hausdorff since $U_-\cup A_\alpha$ and $U_+\setminus A_\alpha$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively. If the set $B$ is countable, then we can find a countable ordinal $\alpha$ such that $U_-\setminus\{-\infty\}\not\subset^* B_\alpha$ and hence $\omega\setminus B_\alpha\not\subset^* U_+$. In this case we can consider the topology $\sigma'$ on $X$ consisting of sets $W\subset X$ satisfying two conditions: $\bullet$ if $-\infty\in W$, then there exists $U\in\sigma$ such that $-\infty\in U\subset^* W$; $\bullet$ if $+\infty\in W$, then there exists $U\in\sigma$ such that $+\infty\in U\cup(\omega\setminus B_\alpha)\subset^* W$. It is clear that $\tau\subset\sigma'\subsetneq\sigma$. The topology $\sigma'$ is Hausdorff since $U_-\cap B_\alpha$ and $U_+\cup(\omega\setminus B_\alpha)$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively. In both cases we have constructed a strictly weaker topology $\sigma'\subset \sigma$ in $T_2(\tau)$ witnessing that the topology $\sigma$ is not minimal.<|endoftext|> TITLE: Prime race modulo $12$. When is the first sign change? QUESTION [5 upvotes]: Define $\pi(x;q,a)$ as the number of primes less than or equal to $x$ which are congruent to $a$ modulo $q$. Up to $x=10^{11}$ we have $\pi(x;12,1) \le \pi(x;12,-1)$. What is the smallest $x$ for which $\pi(x;12,1) > \pi(x;12,-1)$? Assuming GRH the sign changes infinitely often. REPLY [11 votes]: The comments above point to the paper Chebyshev's Bias by Rubinstein and Sarnak, and the paper of Martin. For context, let's consider first primes mod $3$. Assuming you're familiar with the terminology of Rubinstein and Sarnak, they compute the bias of primes to be $2$ mod $3$ over $1$ mod $3$ to be $0.9990\ldots$ (This is $\delta(P_{3;N;R})$ in their notation). In 1978, Bays and Hudson computed that $608,981,813,029$ was the smallest integer $x$ such that $\pi_{3,2}(x)<\pi_{3,1}(x)$. The paper of Martin shows the bias of primes to be $11$ mod $12$ over $1$ mod $12$ is much stronger: $\delta_{12;11,1}=0.999977$ (in Martin's slightly different notation.) Since this is a logarithmic scale, the 'extra' $0.000977$ means that the bias is almost two orders of magnitude stronger. Meanwhile, Ford and Hudson showed in that your $x$ will be less than $10^{84}$.<|endoftext|> TITLE: Uniquely ordered commutative rings QUESTION [7 upvotes]: I am wondering whether there are reasonable necessary and/or sufficient conditions to dedice whether a commutative ring can be uniquely ordered (like for instance $\mathbb{Z}$) or not. In the field case, for example, we know that every real closed field has a unique order (namely the sums of squares). I was hoping that the notion of "real closed rings" does the same, but this is not the case. In a more specific context, if I have a valued ring $(R,v)$, I'd like to know under which conditions the residue ring $Rv := R_v/I_v$ is uniquely ordered, where $R_v := \{x \in R: v(x) \geq 0\}$ and $I_v := \{x \in R: v(x) > 0\}$ REPLY [11 votes]: In a field, an element can be made negative in some order iff it is not a sum of squares. Thus, $F$ is uniquely ordered iff for every $a\in F^\times$, $a$ or $-a$, but not both, is a sum of squares. An order on a domain $R$ extends uniquely to its fraction field, thus $R$ is uniquely ordered iff its fraction field is. Thus: Proposition: The following are equivalent for any domain $R$: $R$ is uniquely ordered. $0$ is not a sum of nonzero squares, and for every $a\in R$, there is a nonzero $b\in R$ such that $ab^2$ or $-ab^2$ is a sum of squares.<|endoftext|> TITLE: Can a game be an option of itself? QUESTION [5 upvotes]: My question is, can a game contain itself as an option? and can it be a surreal number? For example $A=\{A|\}$ or $B=\{C|B\}$ where $C$ is a surreal number. from the point of view of games, it is equivalent to having the player play by doing nothing an effectively passing his turn to the second player. But I am not sure if this is okay from the point of view as surreal numbers. But as a number it is a problem, since for example to proof that $A=\{A|\}$ is a number, we  will need to prove that all of its options are number, which means that we need to prove that $A$ is a number! So is it even possible to have a game with itself as an option? or will I have a "set which contains itself" paradox or something like that. Thanks! REPLY [9 votes]: A surreal number cannot have itself as an option, because numbers are defined inductively: if $L$ and $R$ are sets of numbers, then $\{L|R\}$ is a number. Notice that $L$ and $R$ have to already contain things that are numbers before we can say that $\{L|R\}$ is a number. I suppose there would nothing a priori wrong with allowing games containing themselves as options, but it would obviously limit one's ability to analyze them as generalizations of surreal numbers. This is probably why in ONAG Conway writes "...we adopt the convention that in no game is there an infinite sequence of positions each of which is an option of its predecessor." (p.72) So although games containing themselves as options might conceivably model something interesting, they are explicitly ruled out by definition.<|endoftext|> TITLE: Program analysis for Turing machines QUESTION [5 upvotes]: What is considered the state-of-the-art on program analysis (static and dynamic) for Turing machines? What references can I consult for this problem? I am thinking of things like determining whether a state in the transition graph is reachable, which strings of consecutive symbols on the tape are possible or impossible, and so on. Of course, this is undecidable in general, but I'm looking for analyses that cover a wide range of cases. REPLY [2 votes]: This is a bit beyond what you are seeking, but nevertheless you might be interested in the work of Adam Yedidia and Scott Aaronson. They have constructed three different Turing machines, $G$, $R$, and $Z$, that have these properties: $G$: Halts iff Goldbach's conjecture is false. $R$: Halts iff the Riemann hypothesis is false. $Z$: Cannot be proved in ZFC to not halt. If it halts, it proves ZFC is inconsistent. "$Z$ is a Turing machine for which the question of its behavior (whether or not it halts when run indefinitely) is equivalent to the consistency of ZFC." Their work is intricately related to the Busy Beaver function, and certainly involves "state-of-the-art program analysis." "A Relatively Small Turing Machine Whose Behavior Is Independent of Set Theory." arXiv:1605.04343. 2016. "The 8000th Busy Beaver number eludes ZF set theory: new paper by Adam Yedidia and me." Scott Aaronson blog Their appendix contains a precise description of the $7,918$-state TM $Z$, a snippet of which looks like this:           To attempt respond to the original intent of your question more directly, "which strings of consecutive symbols on the tape are possible or impossible, and so on" You probably know that: (1) Finding a TM that accepts (say) all strings of exactly length $3$ is undecidable. This can be established via Rice's theorem. (2) Detecting "dead code," i.e., unreachable states of a TM, is undecidable. (3) Deciding whether a TM accepts any word at all, is undecidable.<|endoftext|> TITLE: Does every compact metric space have a canonical probability measure? QUESTION [19 upvotes]: Edit: Shortly after this post it was rightly pointed out by @AntonPetrunin that the measure $\mu$ may not be unique. @R W then showed how one can construct a metric space where the limiting measure is not unique. Hence @R W answered the following question in the negative: Does the following construction always produce a unique limiting measure? Therefore the construct below doesn't produce a canonical probability measure for every compact metric space (as was claimed below). Thank you @AntonPetrunin and @R W for catching the error and providing an example. Original Post: My coauthor Sean Li and I recently ran across the (seemingly not well-known) fact that every compact metric space is endowed with a canonical probability measure. The construction has been used to prove the existence of Haar measures on compact groups (see for instance page 3 of Assaf Naor's notes). However, we're not able to find a reference for a general compact metric space. Does anyone have a reference? Let $(X,d)$ be a compact metric space, let $\epsilon>0$, and let $Y_{\epsilon}\subset X$ be an $\epsilon$-net of minimal cardinality. That is, $\displaystyle\cup_{y\in Y_{\epsilon}}B(y,\epsilon)=X$ where $B(y,\epsilon)$ is the ball of radius $\epsilon$ centered at $y$, and $\# Y_{\epsilon}$ is minimal with respect to all such sets. Let $\mu_{\epsilon}$ be the Borel probability measure on $X$ defined by $\mu_{\epsilon}(E)=(\# E\cap Y_{\epsilon})/\# Y_{\epsilon}$. By compactness it follows that there exists a Borel probability measure $\mu$ on $X$ such that $\mu$ is the weak limit (up to taking a subsequence) of $\mu_{\epsilon}$ as $\epsilon\to 0$. Theorem The measure $\mu$ is unique. That is, it doesn't depend on the choice of minimal $\epsilon$-nets $Y_{\epsilon}$. A proof is supplied in the notes of Naor cited above on page 4 and 5, during the course of his proof of existence of Haar measure. According to Naor, he got this proof from Milman--Schechtman (page 1 and 2), and Milman and Schechtman cite the paper "Abstrakte fastperiodische Funktionen" of Maak (Abh. Math. Sem. -- 1936). Naor seems to be the first to point out that such a unique measure didn't rely on the underlying homogeneous structure of the metric space (see the comments in his notes). Can anyone point us to a reference (besides the notes of Naor) of this fact? Also, we're not able to track down a copy of Maak's paper to check if there is any mention of any of this there. REPLY [16 votes]: As Anton has already mentioned, one can only claim that if the sequence $\mu_{\epsilon_n}$ associated to a certain sequence of minimal $\epsilon_n$-nets $Y_n$ converges, then it will also converge to the same limit for for any other sequence of $\epsilon_n$-nets. However, a different sequence $(\epsilon_n)$ may produce a different limit measure. For the simplest counterexample let $T$ be the (genealogical) tree constructed in the following way: the progenitor $o$ has two first generation descendants $a,b$. Further, in the branch starting from $a$ (resp., $b$) everyone in even generations (counted with respect to $o$) has 4 (resp., 2) descendants, and everyone in odd generations has 2 (resp., 4) descendants. Let now $X=\partial T$ be the boundary of this tree ($\equiv$ the set of infinite geodesic rays issued from $o$ $\equiv$ the set of all inifnite lines of descendants starting from $o$) endowed with the metric $$ d(x,x') = 2^{-(x|x')} \;, $$ where $(x|x')$ denotes the confluence of the rays $x$ and $x'$ (i.e., the length of their common part). Then OP's construction produces two limit measures corresponding to the sequences $\epsilon_n=2^{-2n}$ and $\epsilon_n=2^{-2n+1}$. In this case the two limit measures are equivalent, but one can easily obtain examples with singular limit measures, with inifnitely many limit measures, etc, etc.<|endoftext|> TITLE: When can a 'polynomial' have an infinite number of zeroes QUESTION [5 upvotes]: As indicated in the title the intended meaning of 'polynomial' here is slightly nonstandard, and if there is a better word please feel free to edit. For our purposes: A polynomial $\mathfrak{p}$ will be a member of some integral monoid ring $\mathbb{Z}(T^\mathbb{M})$, so for each $\mathfrak{p}\in\mathbb{Z}(T^\mathbb{M})$ there is some $\alpha\in \mathbf{On}$, the class of ordinals, and a unique pair of sequences $\{z_i\}_{i<\alpha}\subset\mathbb{Z}$ and $\{m_i\}_{i<\alpha}\subset\mathbb{M}$ (an additive monoid) such that $$\mathfrak{p}=\sum_{i<\alpha}z_iT^{m_i}.$$ Addition and multiplication behave as expected in usual polynomial multiplication with $T$ as an arbitrary indeterminate, where the addition in $\mathbb{M}$ may or may not be commutative. Further, $\alpha$ may be finite or transfinite, and in the latter case I would like to allow a transfinite number of nonzero terms in the polynomial. I believe this is equivalent to looking at Hahn series or a slight generalization of them if we require that $\mathbb{M}$ be an ordered monoid, however I am only interested in an algebraic property of $\mathfrak{p}$ in general so I cast the question in terms of more algebraic objects. My question is: Under what general conditions will $\mathfrak{p}$ have a unique* transfinite prime factorization in $\mathbb{Z}(T^\mathbb{M})$, in the sense that there exists a sequence $\{\mathfrak{p}_i\}_{i<\lambda}\subset\mathbb{Z}(T^\mathbb{M})$ for $\lambda\geq\omega$ such that $\forall i\big(\mathfrak{p}_i\neq1\big)$ and $\mathfrak{p}=\Pi_{i<\lambda}\mathfrak{p}_i$, while no other sequence produces $\mathfrak{p}$ as its product? *possibly with stipulations about the uniqueness of the factorization If this question is perhaps too general, I would still be very much interested in the special case where $\mathbb{M}$ is discretely ordered, or the addition in $\mathbb{M}$ is commutative, or we restrict $\alpha$ to be countable or $\omega$. This came up during some research I'm doing pertaining to non-Archimedean ordered rings, which can be viewed as integral monoid algebras with a judicious choice of discretely ordered monoid and allowed 'polynomial' length, and in this setting polynomial factorization corresponds to factorization of elements in the ring. It seems like all the monoids I'm looking at produce unique finite factorizations for $\alpha<\omega$, but it feels like there should be some simple generalizations which will produce integral monoid algebras with the above property. REPLY [3 votes]: $\DeclareMathOperator{\Ord}{Ord}$$\DeclareMathOperator{\Noo}{No}$It seems to me that you are redefining existing notions: this is exactly the Hahn series structure except the ordered group is now the positive part of a discretely ordered group, while $\mathbb{Z}_{\infty}$ is the Grothendieck group (/ring) associated to the monoïd (/semi-ring) structure of $\Ord$, and $\mathbb{Q}_{\infty}$ is its fraction field. I don't think the latter's real closure can be densely embedded in $\Noo$ for it contains no element strictly between $\mathbb{N}$ and $X^{\frac{1}{\mathbb{N}+1}}$. I understand why you might want to see $\mathbb{Z}_{\infty}$ and $\mathbb{Q}_{\infty}$ as "versions" of integers and rationnal numbers, but do they really share their properties to the extent that such vocabulary and notations would be relevant? Apart form that, a proof that those rings have a kind of euclidean division (mind that the term "euclidean domain" is reserved for domains with an euclidean function) would be very nice indeed. Do you have someone checking your proof? A note on how to add cuts to ordered fields: Given an ordered field $F$, and a cut $(L,R)$ in it, that is subsets (or subclasses here) with $L < R$, $F = L \cup R$ and $L/R$ have no maximum/minimum, there is a canonical way to add it to $F$ and thus obtain a simple extension denoted $F(L \ | \ R)$ henceforth. Let's say that $(L,R)$ is of algebraic type if there is some polynomial $P \in F[X]$ such that there is neigborhood of the cut, that is, a convex subset $C$ of $F$ intersecting $L$ and $R$ if they are not empty, such that $P$ is strictly negative on $C \cap L$ and strictly positive on $C \cap R$. Otherwise, say that $(L,R)$ is of transcendantal type. Since $F$ can be embedded in a real closure, polynomials in $F[X]$ have less sign shifts than their degree, and this implies that: $(i)$: If $(L,R)$ is of transcendental type, then for each polynomial $P \in F[X]$, there is a neighborhood $C$ of $(L,R)$ such that $P$ has constant sign on $C$. $P$ is defined to be strictly positive in $F[X]$ if this sign is strictly positive. This defines a positive cone on $F[X]$ that is naturally extended to $F(L \ | \ R):= F(X)$, and in the resulting field, $L < X < R$. $(ii)$: If $(L,R)$ is of algebraic type, chose $P$ with minimal degree satisfying the conditions of the definition. $P$ is then irreducible for having constant sign on neighborhoods is stable by product. For $S = PQ + T \in F[X] / (P)$ with $\deg(T) < \deg(P)$, we find a neighborhood $C$ of $(L,R)$ such that $T$ has constant sign on $C$ and define this as the sign of $S$. To see that this defines a positive cone on $F[X] /(P)$, the non trivial part is to see that if $S > 0$ then $S^2 > 0$. Towards this, write $T^2 = PQ_1+ R_1$ with $\deg(R_1) < \deg(P)$, note that $\deg(Q_1) = 2\deg(T) - \deg(P)< \deg(P)$, so $Q_1$ has constant sign on a neighborhood $C_1$ of $(L,R)$ and by choosing a small enough one we get that $T,R_1$ have constant sign as well. Since $R_1 = T^2 - PQ_1$ is strictly positive on either $L$ or $R$, the constant sign is $1$ so $S^2 > 0$. In the resulting ordered field $F( L \ | \ R):= F[X] / (P)$, $L < X + (P) < R$ (and of course $X + (P)$ is a root of $P$). There is an equivalent formulation with pseudo-Cauchy sequences but it only works with cuts that are sufficiently spaced so that valuation theoretic arguments can apply. With this one as with the valued field version, there is a subtlety in the algebraic type version in that if $F(a)$ is a proper simple ordered field extension of $F$ and $(L,R)$ denotes the cut in $F$ defined by $a$, that is $L = F \cap ]-\infty;a[$ and $R := F \cap ]a;+\infty[$, then $a$ may not be algebraic over $F$ even if $(L_a,R_a)$ has algebraic type. When the cut is "Dedekind" in the sense that $\{r-l \ | \ (l,r) \in L \times R\}$ is coinitial in $F^{>0}$, the resulting ordered field $F(L \ | \ R)$ is a dense extension, and adding the cut is similar to adding a Cauchy sequence. It is possible to add all such cuts at once and define Dedekind-like operations on the set of Dedekind cuts in $F$. (note that those cuts can have algebraic or transcendent type) You then get a dense extension without proper dense extensions (=maximal dense extension), which is a nice generalization of completeness for ordered fields. It is also possible to add all Cauchy sequences at once or even to take a uniform completion of the field to get the same result. Adding all agebraic type cuts inductively gives a real closure. In fact, an ordered field is real closed iff it has no algebraic type cut. Adding all cuts inducitvely (unlike adding enough pseudo-limits) is impossible for an ordered field $F'$ always has $(F',\varnothing)$ as a cut (and as soon as $F'$ is non-archimedean, it has the cut defined by $\mathbb{N}$). I suppose it is possible under global choice to add all set-sized cuts, and get Field which may be saturated and may under global choice be isomorphic to $\Noo$, but I guess this is a bit artificial for your taste. So this works because one can evaluate polynomials in $F[X]$ at points of $F$. The problem is that it is hard to see how to evaluate elements such as $X^{\omega}$ in $R:=\mathbb{Q}_{\infty}[X^{\mathbb{Z}_{\infty}^+}]$ (I think the notation $\mathbb{Q}_{\infty}((X^{\mathbb{Z}_{\infty}^+}))$, which speaks for itself when one knows Hahn series, should be prefered here), and that there is no clear way, even with an euclidean-like function, of specifying a condition for an element in the polynomial ring $R[\mathbb{Z}_{\infty}^+]$ of polynomials with indeterminates in the monoïd $\mathbb{Z}_{\infty}^+$ and coefficients in $R$, so that given an irreducible element $a$ of $R[\mathbb{Z}_{\infty}^+]$, the elements in $R[\mathbb{Z}_{\infty}^+]$ with less degree cannot for instance change signs over one cut. (The problem here being that the degree takes values in $\mathbb{Z}_{\infty}^+$ which is not well-ordered.) You could try to embed $R$ in a nice way in $\Noo$, find out the cut defined by $a$ in $\Noo$, intersect it with $R$ (fortunately, the maps $0 TITLE: Does every non-amenable group contain a 2-generated non-amenable subgroup? QUESTION [9 upvotes]: It is known that there are non-amenable groups not containing $F_2$, the free group on two generators; for example, Olshanskii's group. But does every non-amenable group contain a 2-generated non-amenable subgroup? REPLY [15 votes]: The answer is "no". There are non-amenable Golod-Shafarevich groups where every 2-generated subgroup is finite. See my answer here.<|endoftext|> TITLE: Cut-free proofs in ZFC QUESTION [11 upvotes]: If a statement $P$ has a ZFC proof of length $n$, must it also have a cut-free ZFC proof of length polynomial in $n$? By a cut-free ZFC proof, I mean a proof in sequent calculus without cut rule of some sequent "finite set of ZFC axioms $\implies P$ ". This is arguably nonstandard and different from a cut-free proof in sequent calculus augmented with ZFC axioms in that we essentially allow a single cut on ZFC axioms as the final step. Background: Every statement provable in predicate calculus is provable in sequent calculus without the cut rule, but sometimes with an iterated exponential increase in proof size. However, ZFC is not finitely axiomatizable, and for every finite fragment $T$ of ZFC and statement $P$, it proves "if $T$ proves $P$ , then $P$ " (this property also holds for many other theories such as $\mathrm{PA}$ and $\mathrm{Z}_2$, and one can also ask this question for those systems), and the question is whether this can be used to circumvent the slowdown from cut elimination by changing a proof of "finite fragment of ZFC $\implies P$ " into a cut-free proof of "a bigger finite fragment of ZFC $\implies P$ ". An answer may give us some insight about to what extent polynomial length cut-free proofs are a well-behaved quantity, avoiding short proofs that every cut includes $2^k_n$ but working somewhat well otherwise. REPLY [13 votes]: Yes, one can eliminate cuts polynomially for proofs in ZFC, PA, and similar theories that have nontrivial axiom schemata allowing arbitrary formulas. For PA, replace a cut $$\color{blue}{\dfrac{\Gamma\implies A,\Delta\qquad\Gamma,A\implies\Delta}{\Gamma\implies\Delta}}$$ with a cut-free subderivation $$\dfrac{\dfrac{\color{blue}{\Gamma,A\implies\Delta}}{\Gamma,\forall x\,A\implies\Delta}\qquad\dfrac{{\atop\displaystyle\strut\color{blue}{\Gamma\implies A,\Delta}}\qquad\dfrac{\dfrac{\dfrac{}{\Gamma,A\implies A,\Delta}}{\Gamma\implies A\to A,\Delta}}{\Gamma\implies\forall x\,(A\to A),\Delta}}{\Gamma\implies A\land\forall x\,(A\to A),\Delta}}{\dfrac{\Gamma,A\land\forall x\,(A\to A)\to\forall x\,A\implies\Delta}{\color{blue}{\Gamma,\color{red}{\forall\vec y\,(A\land\forall x\,(A\to A)\to\forall x\,A)}\implies\Delta}}},$$ where $\vec y$ is the list of free variables of $A$, and $x$ is a fresh variable. The extra red formula is now a (dummy) instance of an induction axiom, and we let it pass through to the final sequent. For ZFC, we may abuse replacement or separation axioms in a similar way.<|endoftext|> TITLE: Probability of good colorings in randomly-colored graphs QUESTION [8 upvotes]: Each vertex in a graph is randomly and independently colored either red or blue with equal probability. A coloring is called $r$-good, for some fraction $r\in[0,1]$, if at least a fraction $r$ of the edges touch at least one red vertex. Define $p(G,r)$ as the probability that a graph $G$ is $r$-good. Obviously $p(G,r)$ is decreasing with $r$. What is the largest $r$ such that $p(G,r)>1/2$ for all finite graphs $G$? There is an upper bound of $2/3$ (as noted by Kevin P. Costello) and a lower bound of $1/2$. For the upper bound, let $G$ be the clique with $3$ vertices. It has eight different colorings: one is 0-good, three are 2/3-good and four are 1-good. Therefore, $p(G,2/3)= 7/8$, but for every $r>2/3$, $p(G,r)\leq 1/2$. For the lower bound, consider an arbitrary graph $G$. For every coloring of $G$, either it or its opposite colorng is $1/2$-good; therefore $p(G,1/2)\geq 1/2$. To prove that $p(G,1/2)>1/2$, it is sufficient to prove the existence of a coloring that both it and its opposite are $1/2$-good. To construct such a coloring, assign a unique number to each vertex of $G$, and assign to each edge $(u,v)$ the number $(u+v)/2$. Let $D$ be the median of all edges' numbers. Color all vertices smaller than $D$ red and all other vertices blue. Both this coloring and its opposite are $1/2$-good. Hence $p(G,1/2)>1/2$. Intuitively, the 3-clique seems to be the worst case, since when there are many vertices, the expected number of good edges is $3/4$. So my conjecture is that the real value is 2/3. In other words: Conjecture. For every finite graph $G$, $p(G,2/3)>1/2$. Is this true? Alternatively, can you prove tighter bounds? REPLY [4 votes]: For $r > 1/2$, consider the star graph $K_{1,n}$ for large $n$. Half of the time, the central vertex will be red, and the graph will certainly be $r$-good. The other half of the time, the central vertex will be blue, and the graph will be $r$-good only if a fraction $r$ of the other vertices are red. The probability of this event approaches $0$ as $n \rightarrow \infty$. So the probability of a large star graph being $r$-good (for $r > 1/2$) is only slightly greater than $1/2$. You have defined $p(r)$ to be the smallest probability that a graph is $r$-good. In this case, there may not be a smallest probability; the probabilities approach a limit of $1/2$ (unless of course there is some other graph I have not considered which has a smaller probability of being $r$-good). So, if it is fair to alter the statement of the question to define $p(r)$ as the infimum of all such probabilities, then $p(r) \leq 1/2$ for all $r > 1/2$.<|endoftext|> TITLE: Is it possible to describe the image of the $p$-adic logarithm on $1+\mathfrak{m}$, where $\mathfrak{m}$ is the maximal ideal of a $p$-adic field? QUESTION [10 upvotes]: Let $R$ be the ring of integers of some finite extension of $\mathbb{Q}_p$. In particular, I'm interested in the case $R = \mathbb{Z}_p[\zeta_{p^k}]$ (the totally ramified extension of $\mathbb{Z}_p$ of degree $(p-1)p^{k-1}$) Let $\pi$ be a uniformizer of $R$, and $\mathfrak{m} = (\pi)$ the maximal ideal. The $p$-adic logarithm on $1+\mathfrak{m}$ is the function: $$\log_p : 1+\mathfrak{m}\rightarrow \mathfrak{m}$$ given by $$\log_p(1+x) = \sum_{n\ge 1}\frac{(-1)^{n-1}}{n}x^n$$ which clearly converges on $1+\mathfrak{m}$, and whose image lies in $\mathfrak{m}$. If the ramification index of $R/\mathbb{Z}_p$ is $e$, then $$\log_p|_{1+\mathfrak{m}^r} : 1+\mathfrak{m}^r\longrightarrow\mathfrak{m}^r$$ is an isomorphism for any $r > e/(p-1)$, but is in general neither injective nor surjective on $\mathfrak{m}$. Is it possible to describe in general the image $\log_p(1+\mathfrak{m})\subset\mathfrak{m}$ of $\log_p$? (or at least in the case of $R = \mathbb{Z}_p[\zeta_{p^k}]$? REPLY [4 votes]: It is certainly not true that the logarithm maps into the maximal ideal of the integers, when the ramification index is large. For instance, if $p=2$ and $\lambda^6=2$, you get $v_2(\log(1+\lambda))=-5/3$, where $v_2$ is the $2$-adic valuation normalized so that $v_2(2)=1$. Indeed, for all $z$ with $v_p(z)$ unequal to any number $\frac1{(p-1)p^m}$, $v_p(\log(1+z))$ is given by the (function whose graph is the boundary of the) Newton copolygon. The copolygon encodes the same information as the polygon, but in a way that may be more useful for some purposes. Starting with a convergent power series $f=\sum_nc_nx^n$, you take, in the right-hand Cartesian half-plane, the intersection of all the sets given by $\eta\le n\xi+v(c_n)$. The result is a closed convex set with polygonal boundary. The vertices have for their $\xi$-coordinates the negative slopes of the Newton polygon’s segments, and the segments have for their slopes the first coordinates of the Newton polygon’s vertices. Call $\psi$ the function whose graph is the boundary of the copolygon. Now, if $v(\lambda)$ is unequal to the $\xi$-coordinate of any vertex of the copolygon of $f$, then $v(f(\lambda))=\psi(v(\lambda)$, as is easily seen. The fact that the copolygon of the logarithmic series $x-x^2/2+x^3/3-\cdots$ descends infinitely far in the open right-hand half-plane shows that the possible valuations of $\log(1+\lambda)$ are all real numbers. But if you don’t like copolygon talk, you can prove directly by looking at the polygon of $\log(1+x)-\mu$, for a $\mu$ in an algebraic closure of $\Bbb Q_p$, or even $\mu\in\Bbb C_p$, that the logarithm maps onto the specified algebraically closed field.<|endoftext|> TITLE: Result attribution for eigenvalues of a matrix of Pascal-type QUESTION [6 upvotes]: A few years ago, I wanted to cite a result in a paper, for which I could not find a reference. I ended up not using the full strength of it, and the part that I needed could be easily proved. Still, I'd like to know where the full version appears. The result is as follows: The eigenvalues of the matrix $$\left[\binom{i+j}{i}\binom{2n-i-j}{n-i}\right]_{0\le i,j\le n}$$ are $$\binom{2n+1}{k}, \quad 0\le k\le n.$$ If a formula for the corresponding eigenvectors also appears somewhere, that would be helpful, too. (I only needed the fact that $\binom{2n+1}{n}$ has eigenvector $[1,1,\dots,1]^T$, and that's easy to see.) Thanks. REPLY [5 votes]: I don't know a reference. One way to show the eigenvalues starts from the observation (which can be proved using generating functions) that $\sum_{i=0}^n {i\choose k} A_{i,j}={2n+1 \choose n-k} {j+k \choose k}={2n+1 \choose n-k}\,\sum_{\ell=0}^k {k\choose \ell} { j \choose \ell}$ (where $A$ is the matrix above). With the row vectors $\mathbf{v}_k$ with coordinates $\mathbf{v}_k(i)={i \choose k}$ that is $$\mathbf{v}_k A={2n+1 \choose n-k}\left(\sum_{\ell=0}^k {k \choose \ell} \mathbf{v}_\ell\right).$$ The rest is routine. ADDED: (for the record) With some patience one finally finds that $$\mathbf{e}_k=\sum_{j=0}^k (-1)^j{n-j \choose k-j}{k+j \choose j}\mathbf{v}_j$$ is an eigenvector to the eigenvalue ${2n+1 \choose n-k}$.<|endoftext|> TITLE: Commuting with self-adjoint operator QUESTION [5 upvotes]: Let $T$ be an (unbounded) self-adjoint operator. Assume that there is a bounded operator $S$ such that $TS=ST.$ For which kind of $f$ do we have that $f(T)S=Sf(T)?$ My thought was that using a strategy exploiting Stone's formula for the resolvent and then the Stone-Weierstrass theorem one can show this for $f \in C_0(\mathbb{R}).$ (continuous functions tending to zero). REPLY [4 votes]: Any bounded Borel function $f: \mathbb{R} \to \mathbb{R}$. If $TS = ST$ then (taking adjoint of both sides) $S^*T = TS^*$. Therefore both ${\rm Re}(S) = \frac{1}{2}(S + S^*)$ and ${\rm Im}(S) = \frac{1}{2i}(S - S^*)$ commute with $T$, and since they are self-adjoint it follows from standard spectral theory that they commute with $f(T)$. Taking linear combinations, $S$ commutes with $f(T)$. Edit: here is a possibly more direct proof. Suppose $TS = ST$, meaning that $S$ preserves the domain of $T$ and they commute on this domain. Then the same is true with $T + iI$ in place of $T$ (since the domain doesn't change), and that operator is invertible. Multiplying both sides of $(T + iI)S = S(T + iI)$ by $(T + iI)^{-1}$ (no problems here, $(T + iI)^{-1}$ is bounded and takes everything into the domain of $T + iI$) yields $S(T + iI)^{-1} = (T + iI)^{-1}S$. Now $(T + iI)^{-1}$ is normal (its adjoint is $(T - iI)^{-1}$), so by standard spectral theory $S$ commutes with every bounded Borel function of $(T + iI)^{-1}$. But every bounded Borel function of $T$ is also a bounded Borel function of $(T + iI)^{-1}$, QED.<|endoftext|> TITLE: Set family $\mathcal{F}$ such that for all $A,B,C \in \mathcal{F}$ both $A \cap B \not \subseteq C$ and $C \not \subseteq A \cup B $ QUESTION [13 upvotes]: This question initially arose out of a question in asymptotic matroid theory. The matroid question has since been answered in a different way, but the extremal set theory question remains unanswered and may be of interest in its own right. Question. Let $\mathcal{F}$ be a set family such that for all distinct sets $A,B,C \in \mathcal{F}$ both $A \cap B \not \subseteq C$ and $C \not \subseteq A \cup B$. Let $f(n)$ be the maximum size of such a set family on ground set $[n]$. What are the best known upper and lower bounds on $f(n)$? Here is a simple argument that shows at least that $f(n)$ goes to infinity. Just draw a Venn diagram and make sure that all the cells contain a point. I am hoping the extremal combinatorics people already know the answer. For example, there is this nice answer of Sergey Norin to this MO question on poisoned wines which describes the related concept of strongly union-free families (which just means that all pairwise unions are distinct). My condition implies that $\mathcal{F}$ is both strongly union-free and strongly interection-free, but the converse does not hold. For example, the set family consisting of $\{1,2\}, \{2,3\}$, and $\{3,4\}$ is both strongly union-free and strongly interection-free, but does not satisfy my condition. REPLY [3 votes]: Just by the fact that your family needs to be both strongly union-free and intersection-free you can improve the upper-bound. First notice that a family $\mathcal{F} \subseteq 2^{[n]}$ is strongly union-free, iff $\overline{\mathcal{F}}= \{[n] \setminus A|A \in \mathcal{F}\}$ is strongly-intersection free. Let's denote by $f_k(n)$ The size of the largest strongly union- free k-uniform family $\mathcal{F}_k \subseteq {[n] \choose k}$, and by $g_k$ the same for strongly intersection-free. Then $g_k(n)=f_{n-k}(n)$. If you demand that your family is both strongly union-free and intersection-free, then the number of sets in the family in the $k$th layer is at most $\min(f_k(n),f_{n-k}(n))$. So the size of a family $\mathcal{F}$ fulfilling your constraints is at most $n\cdot \max_{k \in [n]}\min(f_k(n),f_{n-k}(n))$. Multiplying by $n$ of course becomes negligible as $n$ grows if you just want bounds on the base of the exponent. If you look at this paper of Coppersmith and Shearer, they actually give a bound for $k$-uniform strongly union-free families, and then see for which $k$ this bound is largest, and multiply by $n$. I haven't done the calculations, but just because now you have the minimum of $f_k(n)$ and $f_{n-k}(n)$, their method should give you a better upper-bound.<|endoftext|> TITLE: Number of connected components of the intersection of two maximal tori QUESTION [10 upvotes]: Let $G$ be a connected complex semisimple Lie group and $S$, $T$ two maximal tori in $G$. Is there a known upper bound on the number of connected components of $S\cap T$? For example, is it bounded by the cardinality of the centre $Z_G$: $$|\pi_0(S\cap T)|\leq|Z_G|?$$ REPLY [8 votes]: Summary: Let $X = \mathrm{Hom}(T,\mathbb{G}_m)$ be the weight lattice, $\Phi \subset X$ the root system. Define a sublattice $L$ of $X$ to be a "root sublattice" if $L$ is generated as an abelian group by $L \cap \Phi$. Then the possible component groups of $S \cap T$ are the torsion subgroups of $X/L$, as $L$ ranges over root sublattices. This will follow from: Theorem: Let $Z$ be a subgroup of $T$. Then there exists a maximal torus $S$ with $S \cap T = Z$ if and only if there is a connected subgroup $H$ with $T \subseteq H \subseteq G$ such that $Z = Z(H)$. We first make some comments about connected groups $H$ with $T \subseteq H \subseteq G$. Write $\mathfrak{g} = \mathfrak{t} \oplus \bigoplus_{\beta \in \Phi} \mathfrak{g}_{\beta}$, where $\mathfrak{g}$ and $\mathfrak{t}$ are the Lie algebras of $G$ and $T$ and $\mathfrak{g}_{\beta}$ are the root spaces. Connected subgroups are determined by their Lie subalgebras, and a subalgebra containing $\mathfrak{t}$ must be of the form $\mathfrak{t} \oplus \bigoplus_{\beta \in I} \mathfrak{g}_{\beta}$ for some subset $I$ of $\Phi$. More specifically, $\mathfrak{t} \oplus \bigoplus_{\beta \in \Phi} \mathfrak{g}_{\beta}$ will be a Lie-sub-algebra if and only if, for $\beta_1$ and $\beta_2 \in I$, if $\beta_1+\beta_2 \in \Phi$ then $\beta_1 + \beta_2 \in I$. For such an $I$, we write $H_I$ for the corresponding connected subgroup. We note that there are only finitely many $H_I$, since there are only finitely many subsets of $\Phi$. For any such $I$, set $J = I \cap (-I)$. Then $H_J$ is a reductive subgroup of $G$, and we have a short exact sequence $0 \to N_{I \setminus J} \to H_I \to H_J \to 0$ where $N_{I \setminus J}$ is the unipotent group corresponding to $\bigoplus_{\beta \in I \setminus J} \mathfrak{g}_{\beta}$. This sequence is semidirect. Lemma: The centralizer of $T$ in any $H_I$ is $T$. Proof: Let $J = I \cap (-I)$ and consider the above semidirect sequence $0 \to N_{I \setminus J} \to H_I \to H_J \to 0$. Let $\pi$ be the map $H_I \to H_J$. Let $Z_{H_I}(T)$ be the centralizer of $T$ in $H_I$. Then $\pi(Z_{H_I}(T)) \subseteq Z_{H_J}(T) = T$, where the latter inequality is standard. On the other hand, $T$ clearly does centralize $T$. So we have $T \subseteq Z_{H_I}(T) \subseteq \pi^{-1}(T)$ and thus we have a short exact sequence $0 \to N_{I \setminus J} \cap Z_{H_I}(T) \to Z_{H_I}(T) \to T \to 0$. But the Lie algebra of $N_{I \setminus J}$ is a direct sum of weight spaces for $T$ with nonzero character, so no element of $N_{I \setminus J}$ centralizes $T$. We deduce that $N_{I \setminus J} \cap Z_{H_I}(T)$ is trivial, so $Z_{H_I}(T) = T$. $\square$ Corollary: The center of $H_I$ is contained in $T$. Proof: Clearly, the center of $H_I$ centralizes $T$. $\square$. We can now show that $S \cap T$, for any maximal torus $S$, is of the form $Z(H_I)$ for some $I$. Let $H$ be the Lie-sub-group generated by $S$ and $T$. Clearly, $T \subseteq H$ and $H$ is connected, so $H$ is of the form $H_I$ for some $I$. By the corollary, $Z(H_I) \subseteq T$ and similarly $Z(H_I) \subseteq T$. This shows $Z(H_I) \subseteq S \cap T$. On the other hand, $S$ and $T$ commute with every element of $S \cap T$, so $H_I$ commutes with every element of $S \cap T$ and we have $S \cap T \subseteq Z(H_I)$. We have proven both containments. We now know that all intersections are of the form $Z(H_I)$. We want to show, in reverse, that any group of the form $Z(H_I)$ occurs as $S \cap T$. Given $I$, let $L \subseteq X$ be the lattice generated by $I$. Let $K = L \cap \Phi$. Then $Z(H_I)$ is the subgroup of $T$ on which the characters of $L$ vanish, and we thus deduce that $Z(H_I) = Z(H_K)$. So it is enough to show that $Z(H_K)$ is of the form $S \cap T$. The group $H_K$ is reductive, so all we need is Lemma: Let $H_K$ be as above. There is a maximal torus $S$ in $H_K$ such that $S \cap T = Z(H_K)$. Proof: Let $Y = \bigcup_{K' \subsetneq K} H_{K'}$. Then $Y$ is a union of finitely many subgroups of lower dimension, so the complement of $Y$ is Zariski dense. Let $s$ be a regular element in $H_K \setminus Y$, and let $S$ be the connected centralizer of $s$. We claim that $S \cap T = Z(H_K)$. We know that every maximal torus in $H_K$ contains $Z(H_K)$. Suppose, for the sake of contradiction, that $t \in T \setminus Z(H_K)$ and $t \in S$. Let $Z(t)$ be the centralizer of $t$ in $H_K$; since $t$ is not central, $Z(t)$ is not $H_K$. Let $Z(t)_0$ be the connected component of the identity of $Z(t)$. So $Z(t)_0$ is a connected subgroup of $H_K$ containing $T$, and must be of the form $H_{K'}$ for some $K' \subsetneq K$. Also, since $t \in S$, we have $S \subseteq Z(t)_0$. So $s \in H_{K'}$, contrary to the choice of $s$. We have obtained a contradiction, and deduce that $S \cap T = Z(H_K)$. We have now proven the theorem. As we noted above, $Z(H_I)$ is the subgroup of $T$ where the characters in $I$ vanish. We deduce that $Z(H_I)$ is the dual group to $X/\mathrm{Span}_{\mathbb{Z}} I$ and the component group of $Z(H_I)$ is the torsion subgroup of $X/\mathrm{Span}_{\mathbb{Z}} I$. What remains is combinatorics. As nfdc23 suggests, it is convenient to work with the adjoint form of the group, in which case $X = \mathrm{Span}_{\mathbb{Z}} \Phi$. For the general case, multiply all bounds by $|X / \mathrm{Span}_{\mathbb{Z}} \Phi| = |Z(G)|$. I'll list the root sublattices and state the largest one. Proofs will be provided if requested. I've chosen In $A_n$, the root sublattices are $A_{n_1} \oplus A_{n_2} \oplus \cdots A_{n_r}$ for $\sum n_i = n$ and $Z(H_I)$ is trivial. In $B_n$, we obviously have $B_{n_1} \oplus B_{n_2} \oplus \cdots \oplus B_{n_r}$. Each of these $B_m$'s, in turn, contain $D_m$ and $A_{m-1}$. The largest index comes from $D_2^{\lfloor n/2 \rfloor}$ giving index $2^{\lfloor n/2 \rfloor}$. Here $\lfloor x \rfloor$ means $x$ rounded down, and $D_2 = \{ \pm e_1 \pm e_2 \} \subset B_2 = \{ \pm e_1 \pm e_2, \pm e_1, \pm e_2 \}$. In $C_n$, we obviously have $C_{n_1} \oplus C_{n_2} \oplus \cdots \oplus C_{n_r}$ and we also have $A_{m-1} \subset C_m$. The largest index comes from $C_1^{\oplus n}$, that is to say, from $\{ \pm 2 e_i \}$ inside $C_n = \{ \pm 2 e_i, \pm e_i \pm e_j \}$, with index $2^n$. In $D_n$, we obviously have $D_{n_1} \oplus D_{n_2} \oplus \cdots \oplus D_{n_r}$ and we also have $A_{m-1} \subset D_m$. The largest index comes from $D_2^{\lfloor n/2 \rfloor}$, giving $2^{\lfloor n/2 \rfloor -1}$. The exceptional types seem like a pain, but they definitely harbor some surprises! Both $A_8$ and $D_8$ are root sublattices of $E_8$, with index $3$ and $2$ respectively.<|endoftext|> TITLE: Partition of [3n] into summoids QUESTION [12 upvotes]: Let $ [n] $ be the set $ \{1,2,\ldots n\}$. A summoid is a subset $ A \subset [n] $ of the form $ \{a,b,a+b\} $ (you can choose a better name, if it doesn't exist already). Now, I developed by accident this simple result: There is no partition of $ [9] $ into (disjoint) summoids. I want to ask the following questions: Is it true for general $ [3k] $ when $ k > 1 $? According to a computer program I found that it is true for $ [12] $, but it seems my method for $ [9]$ can't be applied to the general case (maybe my program for $[12]$ isn't trustworthy). According to the comment below by R. van Dobben de Bruyn, there is no such partition when $ k \equiv 2,3 \pmod 4$. According to the comment below by Gerhard Paseman, there is a counterexample for $ k = 5 $ that extends also to $ n = 3 \cdot 4^k $ and to $ n = 3 \cdot (1+4^k) $. Is it useful for something? Is there any study of such results? REPLY [9 votes]: I have the following results: N = 12: (1, 11, 12) (3, 7, 10) (4, 5, 9) (2, 6, 8) N = 15: (1, 14, 15) (3, 10, 13) (4, 8, 12) (5, 6, 11) (2, 7, 9) N = 24: (1, 23, 24) (2, 20, 22) (5, 16, 21) (6, 13, 19) (7, 11, 18) (8, 9, 17) (3, 12, 15) N = 27: (1, 26, 27) (2, 23, 25) (4, 20, 24) (7, 15, 22) (8, 13, 21) (9, 10, 19) (6, 12, 18) (3, 14, 17) (5, 11, 16) and so on. My guess would be that there are always (a lot of) solutions for any $k \equiv 0, 1\mod 4$. In my opinion, this is some kind of Goldbach-ish problem. It is likely to get some probabilistic heuristic, but may be difficult to prove. For reference, here I add the number of different solutions for each $k$. N = 12: 8 solutions; N = 15: 21 solutions; N = 24: 3040 solutions; N = 27: 20505 solutions; and so on. EDIT: Thanks to OEIS, we now have some reference. In this paper: http://oeis.org/A104429/a104429.pdf, the existence of solutions is discussed in detail in section I.3 (starting from page 22).<|endoftext|> TITLE: Integral involving Laguerre, Gaussian and modified Bessel function QUESTION [5 upvotes]: I am trying to prove that the integral \begin{align} \int_{0}^{\infty } e^{-\frac{r^2}{2B}} r^{l-n} L_n^{l-n}\left(\frac{r^2}{C}\right) I_{l-n}\left(\rho r \right) r dr \end{align} has the form \begin{align} B^{l+1} e^{\frac{B}{2}\rho^2} \rho^{l-n} L_n^{l-n}\left(\frac{\rho^2}{C} \right), \end{align} where $L_n^{l-n}$ is the generalised Laguerre function, and $I_{l-n}$ is the modified Bessel function of the first kind. Expanding the modified Bessel function into an infinite sum, and then using Eq. 7, section 7.414 (pg. 809) from Tables of Integrals, Series & Products (Ed. 7) (by I.A. Gradshteyn & I.M.Ryzhik), which is \begin{align} \int_{0}^{\infty} e^{-st} t^{\beta} L_n^{\alpha}(t) dt = \frac{\Gamma(\beta+1) \Gamma(\alpha+n+1)} {n! \Gamma(\alpha+1) s^{\beta+1}} {}_1F_2\left(-n, \beta+1; \: \alpha+1; \: \frac{1}{s} \right), \end{align} I can get close, but not close enough! Does anyone know how to do this? Please let me know if you need any further information. I haven't done this very often, but the few times I have I always seem to forget something pertinent! REPLY [4 votes]: Mathematica can evaluate the integral$^\ast$ $$I_{n,l}=\begin{align} \int_{0}^{\infty } e^{-\frac{r^2}{2B}} r^{l-n} L_n^{l-n}\left(\frac{r^2}{C}\right) I_{l-n}\left(\rho r \right) r dr \end{align},\;\;B,C,\rho>0,$$ for any integer $n\geq 0$ as a function of $l>n-1$. The results are consistent with $$I_{n,l}=\begin{align} B^{l+1} (1/B-2/C)^ne^{\frac{B}{2}\rho^2} \rho^{l-n} L_{n}^{l-n}\left(\frac{B^2\rho^2}{C-2B}\right),\;\;l+1>n\geq 0, \end{align}$$ Not quite what the OP suggested, but similar. $^\ast$ The evaluation of the integral for given $n$ follows from the general formula $$\int_0^\infty r^p e^{-a r^2} I_q(r)\,dr=\frac{\Gamma \left(\frac{1}{2} (p+q+1)\right)}{2^{q+1}a^{\frac{1}{2} (p+q+1)}\,\Gamma (q+1)} \, _1F_1\left(\tfrac{1}{2} (p+q+1);q+1;\frac{1}{4 a}\right)$$ (for $p>-1$, $p+q>-1$, $a>0$)<|endoftext|> TITLE: A cohomology theory for fusion categories QUESTION [13 upvotes]: It is well known that for a finite group $G$, the associator of the fusion category of $G$-graded $k$-vector spaces is given by an element of $H^3(G,k^*)$, up to equivalence of categories. ($k^*$ is the multiplicative group of units in $k$.) A crucial step when showing this is the fact that in $G$-graded vector spaces, all simple objects are invertible, and therefore the tensor product of two simples is simple again. Denote the simple object generating the $g$-graded subcategory, $g \in G$, by $k_g$. Then $k_{g_1} \otimes (k_{g_2} \otimes k_{g_3}) = k_{g_1 g_2 g_3}$ is again simple, and thus is automorphisms are given by an element of $k^*$, for each triple of group elements. Two categories of $G$-graded vector spaces may be monoidally equivalent via a monoidal functor $(F, F^2, F^0)$, but if so, the coherence morphism $F^2_{g,h}\colon Fk_g \otimes Fk_h \to F(k_g \otimes k_h)$ is again given by a number in $k^*$ for every tuple $(g,h)$ of group elements, and it's in fact the coboundary for two representatives of the same cohomology class. We might want to generalise this result and start with a based fusion ring, corresponding to the Grothendieck ring of our future fusion category, with the simples as chosen basis. For the previous example, the fusion ring is $\mathbb{C}[G]$ (the group ring), and the chosen basis is $G$ (as a subset of the group ring). Of course, most fusion categories don't have all simples invertible. This means, e.g. that the associators and the monoidal functor coherences live in entirely different spaces, and it's not so obvious how to repeat the cohomology construction. Is it still possible to classify associators as elements of some cohomology? How about other data, such as pivotal structures and braidings? Ideally, the cohomology theory could just be formulated given the based fusion ring. Additionally, what's the relation to already known homotopical and (co)homological data typically associated to fusion categories? I.e. is such a cohomology related to the cohomology of the Brauer-Picard groupoid? In which way is its "tangent cohomology" the Davydov-Yetter cohomology (as discussed here? REPLY [9 votes]: There is no such cohomology theory known (in particular, this is not related to Davydov-Yetter cohomology which is about deformations and vanishes for finite groups). In my mind this is a very important open problem in the field which could have some major applications to classification of Izumi categories. One can speculate about what the other small cohomology groups should be: If A is the fusion ring of some C, then $H^1(A)$ should give the tensor natural isomorphisms of the identity functor. If A is the fusion ring of some C, then $H^2(A)$ should give the "gauge automorphisms" of C (i.e. the tensor autoequivalences whose underlying functor is the identity modulo natural isos). $H^3(A)$ should classify possible C with fusion ring A up to tensor equivalences whose underlying functor is the identity. I think one can even write down some kind of definitions that make the above work, but the problem is to understand in what sense it's a "cohomology theory." For example does one ever get spectral sequences or any other sort of "algebraic topology" computational tools? It's possible that fusion rings are not the right input. In particular, you might be better off thinking of "Ocneanu cells" as an $H^3$ not for fusion rings but for certain kinds of graphs related to module tensor categories over a fixed braided tensor category. But again I don't know any meaningful sense in which these form a "cohomology theory." I spent some time discussing trying to build such a cohomology theory with Chris Schommer-Pries and Chris Douglas, but we never made any substantial progress.<|endoftext|> TITLE: What are Homotopy rings good for? QUESTION [14 upvotes]: In his paper, Note on quasi-Lie rings, P. J. Hilton defines the (non-associative) Homotopy ring of a pointed space $X$ as$$\bigoplus_{n>1}\pi_n(X)$$ where the Whitehead product $\pi_m(X)\times\pi_n(X)\to\pi_{m+n-1}(X)$ is the product. I suppose in a naïve fashion, one would hope for this to sprout some interesting structure in a similar way to cohomology rings. However it seems to be the case the non-associativity (stemming from the Whitehead product) kills this dream. As far as I can tell, no interesting ring theory can be done on this object, not to mention the fact that it is highly non-trivial for most $X$. I failed to find anymore literature on this subject, partly due to terminology changing from one source to the next. I suppose the correct terminology would be to call this a quasi-Lie algebra, but this just doesn't have the same ring to it. Is there anything more to be said about this construction? Is it wrong to try to force a structure on homotopy groups in this way? Are there similar constructions that have the same feel as cohomology rings but for homotopy? REPLY [22 votes]: The rationalization of this ring can be understood in a very nice way, as follows. Suppose for simplicity that $X$ is simply connected. Then we can define its rational homotopy groups $$\pi_n(X, \mathbb{Q}) \cong \pi_n(X) \otimes \mathbb{Q}.$$ What can we say about their structure? One observation is that $\pi_n(X) \cong \pi_{n-1}(\Omega X)$, and $\Omega X$ has a loop space structure. This means that the rational homology $H_{\bullet}(\Omega X, \mathbb{Q})$ has the structure of a graded Hopf algebra, with product given by the Pontryagin product. Furthermore, there is a rational Hurewicz map $$\pi_{n+1}(X, \mathbb{Q}) \cong \pi_n(\Omega X, \mathbb{Q}) \to H_n(\Omega X, \mathbb{Q}).$$ Theorem 1: These maps are injective. Theorem 2: The image of the rational Hurewicz map consists precisely of the primitive elements in the Hopf algebra $H_{\bullet}(\Omega X, \mathbb{Q})$. As in the ungraded case, the primitive elements of a (graded) Hopf algebra naturally have the structure of a (graded) Lie algebra, under the commutator bracket. This is precisely (the rationalization of) the Whitehead bracket. Furthermore: Theorem 3: $H_{\bullet}(\Omega X, \mathbb{Q})$ is the (graded) universal enveloping algebra of the (graded) Lie algebra $\pi_{\bullet}(\Omega X, \mathbb{Q}) \cong \pi_{\bullet-1}(X, \mathbb{Q})$. This is one way to think about the shift in degree necessary to define the Whitehead bracket. These theorems have many nice applications. For example, you can use them to compute the rational homotopy groups of spheres, by computing the rational homology groups of the based loop spaces of spheres. There is also a nice connection to Koszul duality; the fact that the rational homotopy groups form a graded Lie algebra while the rational cohomology groups form a graded commutative ring reflects the fact that the Lie operad is Koszul dual to the commutative operad.<|endoftext|> TITLE: Besicowitch distance between sets of invariant measures, ergodic vs non-ergodic QUESTION [7 upvotes]: When working with Dominik Kwietniak and Jakub Konieczny, the question appeared: Let $X$ and $Y$ be two subshifts on the same alphabet, $M(X)$, $M(Y)$ the sets of shift-invariant measures on $X$ and $Y$, respectively, $M_e(X)$ and $M_e(Y)$ be the set of the ergodic ones. The Besicovitch measure $\bar{d}$ considered on the set of all shift-invariant measures can be lifted in a way the Hausdorff distance is established and we can define the distance between the sets of ergodic measures, or all invariant measures, as follows: $$\bar{d}(M_e(X),M_e(Y))=\max\left(\sup_{\mu\in M_e(X)}\inf_{\nu\in M_e(Y)}\bar{d}(\mu,\nu),\sup_{\nu\in M_e(Y)}\inf_{\mu\in M_e(X)}\bar{d}(\mu,\nu)\right),$$ $$\bar{d}(M(X),M(Y))=\max\left(\sup_{\mu\in M(X)}\inf_{\nu\in M(Y)}\bar{d}(\mu,\nu),\sup_{\nu\in M(Y)}\inf_{\mu\in M(X)}\bar{d}(\mu,\nu)\right).$$ If we are not wrong (if someone disagree I can add the details), it is true that $$\bar{d}(M_e(X),M_e(Y))\le \bar{d}(M(X),M(Y)).$$ The question is whether $$\bar{d}(M(X),M(Y))\le \bar{d}(M_e(X),M_e(Y)).$$ The reason why we "claim" it is the observation that it is true when on the left side are the convex hulls of the ergodic measures instead of all invariant measures. The idea of the proof is to take for given $\mu\in M(X)$ its ergodic decomposition $\mu=\int_{M_e(X)}\mu'd\lambda(\mu')$. Every ergodic $\mu'$ assign with an ergodic measure $\nu'$ on $Y$ that satisfies $$\bar{d}(\mu',\nu')<\bar{d}(M_e(X),M_e(Y))+\varepsilon.$$ It defines a map $f$ from $M_e(X)$ to $M_e(Y)$, $f(\mu')=\nu'$. Now integrate $$\nu=\int_{M_e(X)}f(\mu') d\lambda(\mu')$$ and prove that $\bar{d}(\mu,\nu)<\bar{d}(M_e(X),M_e(Y))+\varepsilon.$ But I do not know how to show measurability of the map $f$. The definition of the mapping is not very constructive and is not unique at all. Any comment, suggestion would be appreciated! REPLY [3 votes]: 2 Proof based on approximations by convex combinations This proof is much simpler and does not involve the advanced notions and theorems from the descriptive set theory. Since the role $X$ and $Y$ is interchangeable with respect to the assumptions, it is satisfactory to prove that for every $D>\bar{d}(M_e(X),M_e(Y))$ and $\mu\in M(X)$, there is $\nu\in M(Y)$ such that $\bar{d}(\mu,\nu)\leq D$. Let $\mu_n=\sum^{k_n}_{i=1}a_{n,i}\mu_{n,i}$, $n\in\mathbb{N}$, be a sequence of convex combinations of ergodic measures $\mu_{n,i}$ on $X$ that converges to $\mu$ in the weak$^\star$ topology. By the assumption on $D$, for every $\mu_{n,i}$ there exists an ergodic measure $\nu_{n,i}$ on $Y$ such that $\bar{d}(\mu_{n,i},\nu_{n,i})\bar{d}(M_e(X),M_e(Y))$ and put $$P=\left\{\xi\in M_e(X\times Y)\mid \int_{X\times Y}1_{x_0\neq y_0} d\xi(x,y)\leq D\right\}.$$ It is a closed subset in the Polish space $M_e(X\times Y)$ equipped with the weak$^*$ topology. Hence, $P$ itself with the restriction of Borel structure $M_e(X\times Y)$ is a standard Borel space as well as the space $M_e(X)$ where the weak$^*$ topology is considered. By the assumption on $D$, the standard projection $\pi_X:P\to M_e(X)$ is onto and continuous, in particular Borel. The Jankov-von Neumann theorem and its corrolary in the Exercise 18.3 in the Kechris' book ensures that there exists a function $f:M_e(X)\to P$ such that $\pi_X(f(\mu))=\mu$ and which is measurable with respect to the $\sigma$-algebra on $M_e(X)$ generated by analytic sets. By Lusin theorem, $f$ is universally measurable since $M_e(X)$ is standard Borel (see Theorem 21.1 in Kechris' Book). It implies, that for every $\mu\in M(X)$ and its ergodic decomposition $\mu=\int_{\mu'\in M_e(X)}\mu'\lambda(\mu')$, the Pettis integral $$\xi=\int_{\mu'\in M_e(X)}f(\mu')\lambda(\mu')$$ is well defined. It is easy to see that $\xi$ is an $\sigma\times\sigma$-invariant measure on $X\times Y$ that projects on $\mu$. Moreover, $$\int_{X\times Y}1_{x_0\neq y_0}d\xi(x,y)\leq D.$$ It implies, that for the projection $\nu=\pi_Y(\xi)$, $\bar{d}(\mu,\nu)\leq D$. It concludes the proof.<|endoftext|> TITLE: Upper bound for $p_{n^2} - p_{(n-1)^2}$? QUESTION [7 upvotes]: What is the best unconditional upper bound for $p_{n^2}-p_{(n-1)^2}$ such that $p_n$ is the $n$-th prime number? Asymptotics suggest it's somewhere near $4 n \ln n$, but how to prove this? Edit: it's known that there is a prime between $[x,x+x^{13/23}]$, so one can bound $p_{n^2}-p_{(n-1)^2} \leq 2 n n^{13/23}$, but are there any better bounds? REPLY [12 votes]: By the results of Baker, R.C.; Harman, G.; Pintz, J., The difference between consecutive primes. II, Proc. Lond. Math. Soc., III. Ser. 83, No.3, 532-562 (2001). ZBL1016.11037. the number of primes in the interval $[x, x+x^{0.525}]$ is $\gg x^{0.525}/\log x$ for $x$ large enough (see the final inequality of the paper); in particular, $p_{(n-1)^2} + p_{(n-1)^2}^{0.525} \geq p_{n^2}$ for $n$ large enough. This gives a bound of the form $O( n^{1.05} \log^{0.525} n )$. Any improvement on this bound would likely improve the Baker-Harman-Pintz bound on large prime gaps, so I doubt one can do much better using what is in the literature. Of course, the situation is much better on RH or even on LH (the Lindelöf Hypothesis). (Also, one can show the expected bound of $O(n \log n)$ for almost all $n$, with a fairly small exceptional set, using the known results on primes in short intervals on average.)<|endoftext|> TITLE: A strange two-variable recursion QUESTION [5 upvotes]: In some work I was doing with a colleague the following function of two natural number variables, defined by a recursion, came up and we have no clue how to solve it. Any suggestions or improvements on the upper bound given below would be appreciated. Asymptotics are also interesting. The boundary conditions are $$f(m,1)=2^m-1\ \text{and }\,\, f(1,n)=1.$$ The recursion for $m,n\geq 2$ is given by $$f(m,n)= f(m-1,2^{m-1}-1)+f(m,n-1)+1.$$ We can show that $f(m,n) \leq n\cdot 2^{\binom{m}{2}+1}$. For the cases of interest to us $n\leq 2^m-1$. REPLY [6 votes]: First, we note that $f(m,n) - f(m,n-1)$ does not depend on $n$, so for a fixed $m$ we can write $f(m,n) = \alpha_m n + \beta_m$ for some $\alpha_m,\beta_m$ which depend only on $m$. Imposing the boundary conditions lets us write $$ f(m,n) = a_m (n-1) + (2^m-1) $$ Next, substituting that into the defining equation and simplifying gives $$ a_m = a_{m-1}(2^{m-1}-2) + 2^{m-1} $$ For example, it is easy to see that $a_2 = 2$ and $a_3 = 8$, with $$ f(2,n) = 2(n-1)+3\\ f(3,2) = 8(n-1)+7 $$ The usual trick for such recursions is to let (starting at $m=3$) $$ a_m = x_m\prod_{i=2}^{m-1}(2^i-2) $$ which gives $x_3 = a_3/(2^2-2) = 4$ and for $m>3$ $$ x_m = x_{m-1} + \frac{2^{m-1}}{\prod_{i=2}^{m-1}(2^i-2)} \\ x_m = x_3 + \sum_{k=4}^m \frac{2^{k-1}}{\prod_{i=2}^{k-1}(2^i-2)} $$ $x_m$ very rapidly approaches (from above) $\prod_{s=0}^m (1+2^{-s})$. So $a_m$ approaches (from above) $$\frac13 \prod_{i=2}^m \frac{2^i-2}{1+2^{-i}} \rightarrow \frac13\prod_{i=2}^m (2^i-3)$$. THis not only verifies (and slightly tightens) your relation, but also gives a handle on a lower limit, which comes from taking only one $-3$ in just one term of the product (and summing over which term uses the $-3$). (You can actually get $x_m$ and $a_m$ in closed form, but it involves Q-Pochammer symbols and elliptic functoins so that is not at all illuminating.)<|endoftext|> TITLE: Finding the closest matrix to $\text{SO}_n$ with a given determinant QUESTION [26 upvotes]: $\newcommand{\GLp}{\operatorname{GL}_n^+}$ $\newcommand{\SLs}{\operatorname{SL}^s}$ $\newcommand{\dist}{\operatorname{dist}}$ $\newcommand{\Sig}{\Sigma}$ $\newcommand{\id}{\text{Id}}$ $\newcommand{\SOn}{\operatorname{SO}_n}$ $\newcommand{\SOtwo}{\operatorname{SO}_2}$ $\newcommand{\GLtwo}{\operatorname{GL}_2^+}$ I am trying to find the Euclidean distance between the set of matrices of constant determinant and $\SOn$, i.e calculating $$ F(s)= \min_{A \in \GLp,\det A=s} \dist^2(A,\SOn). $$ Since the problem is $\SOn$-invariant we can effectively work with SVD; Using geometric reasoning, we can reduce the problem to diagonal matrices with at most two distinct values for its entries: Indeed, denote by $\SLs$ the submanifold of matrices with determinant $s$; Let $\Sig \in \SLs$ be a closest matrix to $\SOn$. By orthogonal invariance, we can assume $\Sig$ is positive diagonal. Then its unique closest matrix in $\SOn$ is the identity. Consider the minimizing geodesic between $I,\Sig$: $$ \alpha(t) =\id+t(\Sig-\id). $$ Since a minimizing geodesic to a submanifold is orthogonal to it, we have $$\dot \alpha(1) \in (T_{\Sig}SL^{s})^{\perp}=(T_{(\sqrt[n]s)^{-1}\Sig}SL^{1})^{\perp}=\big((\sqrt[n]s)^{-1}\Sig T_{\id}SL^{1}\big)^{\perp}=\big(\Sig \text{tr}^{-1}(0)\big)^{\perp}.$$ Since $\Sig^{-1} \in \big(\Sig \text{tr}^{-1}(0)\big)^{\perp} $ is a basis for $\big(\Sig \text{tr}^{-1}(0)\big)^{\perp}$, we deduce $$ \Sig-\id=\dot \alpha(1)=\lambda \Sig^{-1} \, \, \text{for some} \, \, \lambda \in \mathbb{R}, \, \text{i.e}$$ $$ \sigma_i-1=\frac{\lambda}{\sigma_i} \Rightarrow \sigma_i^2-\sigma_i-\lambda=0.$$ We see from the equation that if $\sigma_i$ is a solution, then so it $1-\sigma_i$, so if we denote by $a$ one root, the other must be $1-a$. We just proved $\{\sigma_1,\dots,\sigma_n\} \subseteq \{a,1-a \}$. Moreover, if the closest matrix $\Sigma$ does indeed have two distinct diagonal values, then they must be of the form $a,1-a$; Since both are positive, this implies $0 \frac{1}{n}(1-\frac{1}{n})^{n-1}$ the minimizer is conformal (the other candidate "$a,1-a$" does not exist). Tim also showed that if $s \le (\frac{1}{2})^n$ then the minimizer is not conformal. It remains to determine what happens when $(\frac{1}{2})^n 0,\quad b>0,\quad \alpha \in [0,1], \quad n\alpha \in \mathbb{N}$$ where $H(a,b,\alpha)$ is $$H(a,b,\alpha) = \alpha \log (a) + (1-\alpha) \log (b).$$ We can perform the method of Lagrange multipliers in the $a$ and $b$ coordinates. This will give us that either $b=a$ or $b = (1-a)$. A more direct path to this result is to observe in your equation $$\sigma^2 - \sigma - \lambda = 0 \text{ for some } \lambda \in \mathbb{R}$$ we have that if $\sigma$ is a solution, so is $(1-\sigma)$. The easy competitor Now if $a = b$ then $\alpha$ is irrelevant, and we have the first competitor to the minimization $$a = b = e^L, \quad \alpha = \text{anything}, \quad dist = (e^L-1)^2$$ The harder competitor The case $b = 1-a$ is harder to analyze. First note that for this solution we must restrict to $a < 1$. Given that $b = 1-a$ we can rewrite our optimization as minimizing $$\alpha (a-1)^2 + (1-\alpha)a^2 $$ over the set \begin{equation}\tag{1} \alpha \log(a) + (1 - \alpha) \log(1-a) = L \end{equation} \begin{equation} \tag{2} a \in (0,1/2), \quad a \leq \min(e^L, 1-e^L), \quad n\alpha \in \mathbb{N} \end{equation} The first constraint above comes from assuming (WLOG) $a$ is smaller than $b$. The second is from observing that $L$ is a convex combination of $\log a$ and $\log (1-a)$ so $\log (a) \leq L \leq \log(1-a)$. To simplify, we can solve the constraint $\alpha \log a + (1-\alpha)\log (1-a)$ for $\alpha$: \begin{equation} \tag{3} \alpha = \frac{L - \log(1-a)}{\log(a) - \log(1-a)} \end{equation} and now rewrite our minimization as: minimize $$f(a) = \frac{(L - \log(1-a))(1-a)^2 + (\log(a)-L)a^2}{\log(a)-\log(1-a)}$$ over $$a \in (0, \min(e^L, 1-e^L)), \quad n\alpha \in \mathbb{N}$$ Now we claim the following three facts: $f(a)$ as zero or one critical points $\lim_{a \to 0}f(a) = 0$ $f(e^L) = f(1-e^L) = (e^{L}-1)^2$ Given these three facts we see that if we forget the condition $n \alpha \in \mathbb{N}$ from our constraints we see that the infimum of $f$ is $0$ which is not realized for $a > 0$. The function $f(a)$ either looks like or (These are L = -1/5 and -3.) The red line in these pictures is the value from taking the matrix of diagonals, (e^L - 1)^2. If we now enforce the discrete condition $n \alpha \in \mathbb{N}$ we see that it suffices to check the smallest $a$ possible against $(e^L -1 )^2$. Checking these facts I checked items 2 and 3 with a CAS. For item 1 I did the following. First implicitly differentiate the constraint (1) with respect to $a$ to find $$ \frac{d\alpha}{da} \left(\log a - \log(1-a) \right) = - \left(\frac{\alpha}{a} - \frac{1-\alpha}{1-a} \right) $$ Then differentiate $f$ and set it to zero to find $$ 2 (a - \alpha) \left(\log a - \log(1-a) \right) = (-2a + 1) \left( \frac{\alpha}{a} - \frac{1-\alpha}{1-a} \right) $$ Multiply by $a(1-a)$ to find $$ 2a(1-a)(a-\alpha) \left( \log a - \log(1-a) \right) = (-2a+1) \left(\alpha - a \right) $$ If $\alpha \neq a$ we can divide by $\alpha-a$ and find (checking that there's only one solution) $a = 1/2$. This is disallowed by our constraints (2). On the other hand, we have a solution if $\alpha = a$. Then, check that there is at most one solution to (1) with $\alpha = a$. Add-on: I will write what I know about $f(a)$ more precisely. For $L \geq \log (1/2)$ there is a solution to the constraint $\alpha \log a + (1-\alpha) \log (1-a) = L$ with $a= \alpha$. Equivalently, for $1/2 \leq y < 1$ there is a solution to $\frac{a^a}{(1-a)^{1-a}} = y$. There are actually two solutions on $[0, 1]$ related by $a_1 = 1-a_2$, so since we enforce $a<1/2$ there is only one solution on our domain for $a$. For $L < \log (1/2)$ there is no solution with $a = \alpha$. This means that $f$ as a function of $a$ has a critical point if $L \geq \log(1/2)$, namely where $a = \alpha$. If $L \leq \log(1/2)$ there is no critical point. Edit: Here are some pictures for why the smallest value of $a$ below does always corresponds to the smallest value of $\alpha$. I think it's helpful to visualize first. Here is $\alpha$ as a function of $a$ for $L = \log(.99)$, $L = \log(.51)$, $L = \log(1/2)$, and $L = \log(.49)$. $L = \log(.99)$ $L = \log(.51)$ $L = \log(.5)$ $L = \log(.49)$ If $L \leq \log(1/2)$ then $\alpha$ as a function of $a$ has no critical points and is increasing (similar reasoning as with $f(a)$). Therefore taking the smallest possible $\alpha$ gives you the smallest possible $a$. If $L > \log(1/2)$ then for $a \in [0, 1-e^{L}] = [0, min(e^L, 1-e^L)]$, $\alpha$ has a single critical point, a maximum, and is zero at the endpoints. Therefore, we can (mistakenly) choose $\alpha = 1/1000$ and then take $a$ to be large. But, you can visualize all allowable values of $a$ by drawing a discrete collection of horizontal lines on this picture (here's $L = log(.51), n=10$): The possible choices of $a$ in our discrete set are given by intersections of the red lines with our blue curve, by the nature of the function each red line has two intersections with the blue curve. The lowest red line has both the smallest and largest value of $a$. This picture also illustrates how to find the counterexample to your stackexchange question.<|endoftext|> TITLE: A conjectured trace inequality for some products of powers of matrices QUESTION [7 upvotes]: Let $B, R\in M_{n}(\mathbb{C})$ hermitian and $B$ positive semidefinite. Let $s,t \in \mathbb{R}$ and $s,t \ge 0$ . Does then hold $Tr[B^s (B R^2 B)^t] \ge Tr[B^s (R B^2 R)^t]$ ? REPLY [5 votes]: I think the inequality is false. Consider for instance the choices \begin{equation*} B = \begin{bmatrix} 5& 6& -2\\ 6 & 13 & 2\\ -2 & 2 & 5\end{bmatrix},\quad R = \begin{bmatrix} -8 & 4 & 4\\ 4 & -2 & -1\\ 4 & -1 & 0 \end{bmatrix},\quad s=5,\ t=3. \end{equation*} Then, we have (computed using Mathematica) lhs = Tr[MatrixPower[b, s].MatrixPower[b.r.r.b, t]] which yields 415274500333934, whereas rhs = Tr[MatrixPower[b, s].MatrixPower[r.b.b.r, t]] yields 450223588494254, so that lhs-rhs = -34949088160320.<|endoftext|> TITLE: Reference - motives of curves QUESTION [12 upvotes]: There is a really interesting comment in this question that I was unable to find a reference... Under the "Tate conjectures, then every motive belongs to the tensor category generated by motives of curves." I would like to know where I should go to learn more about this... Many thanks, REPLY [20 votes]: This phenomenon is specific to the fields $\mathbb F_q$ (in this case you need the Artin motives as well) and $\bar{\mathbb F}_q$. Roughly, the outline is as follows. If the Tate conjecture is true and the Frobenius action is semisimple, then the homological standard conjecture is true. Then Jannsen's Motives, numerical equivalence, and semi-simplicity implies that the category of motives with homological equivalence is semisimple. Moreover, the Tate conjecture gives a description of the simple objects in terms of the characteristic polynomial of Frobenius. Then Honda–Tate theory constructs all such characteristic polynomials inside the cohomology of an abelian variety. These are always dominated by Jacobians, so all desired Frobenius eigenvalues occur in the $H^1$ of a curve. A great general reference for motives is the AMS Proceedings of Symposia in Pure Mathematics Vol. 55.1: Motives. See here for the AMS ebook collection version (you might be able to access this from within your institution). See in particular Proposition 2.6 and Remark 2.7 of Milne's chapter Motives over finite fields (preprint here). Remark. As I understand it, the same is not expected to hold over basically any other type of field. For example, it should be "easy" to write down a Hodge structure that is not in the full abelian tensor subcategory generated by the Hodge structures of weight $1$.<|endoftext|> TITLE: Harmonic oscillator discrete spectrum QUESTION [8 upvotes]: Let us act intentionally stupid and assume we do not know that we can solve for the spectrum of the harmonic oscillator $$-\frac{d^2}{dx^2}+x^2$$ explicitly. Is there an abstract argument why the spectrum of this operator is discrete and tends to infinity? It almost looks like a result that would follow from the compactness of some operator involved, yet we are here on an unbounded domain $\mathbb{R}.$ So does it follow from spectral theory only that the spectrum is discrete? REPLY [4 votes]: The physical idea is that $-\Delta+q$ for a sufficiently-growing "confining potential" $q$ on $\mathbb R^n$ should have compact resolvent, which then proves discreteness of its spectrum. In many explicit situations, one can prove this directly by imitating proofs of Rellich's compactness lemma(s). Not an abstract version, but maybe non-trivial, is illustrated for automorphic forms in http://www.math.umn.edu/~garrett/m/v/simplest_afc_schrodinger.pdf<|endoftext|> TITLE: Wiener Measure measure on functions? QUESTION [5 upvotes]: I know that the Wiener measure for the Brownian motion $\{B_t\}_{t\ge 0}$ on the probability space $(\Omega, \mathscr{F},P)$ can be defined as $\mu=P\circ B^{-1}$ acting on the sigma-algebra generated by the cylindrical sets: $$Cly:=\{x\in C([0,\infty),\Bbb{R}^d)|x(t_1)\in A_1,...,x(t_n)\in A_n\}$$ Here $A_i$ are Borel sets of $\Bbb{R}^d$. I also know the action of $\mu$ on $A\in Cly$ is given by: $$\mu[A] =\prod^n_{i=1} (2\pi(t_i-t_{i-1}))^{-d/2} \int_{A_1}...\int_{A_n} \exp\left( -\frac{1}{2} \sum^n_{i=1} \frac{|x_i-x_{i-1}|^2}{t_i-t_{i-1}}\right) dx_n...dx_1$$ My question is how do we go from this to the integral of a function $f(\omega)$ to (as indicated in (1)): $$\lim_{n\rightarrow \infty}\left(\frac{n}{4D\pi t}\right)^{nd/2}\int_{\Bbb{R}^d}...\int_{\Bbb{R}^d}\hat f(x_1,...,x_n)\exp\left( - \sum^n_{i=1} \frac{|x_i-x_{i-1}|^2}{4Dt/n}\right) dx_n...dx_1$$ (1) also explains that the existence of the measure is to do with the Riesz Representation Theorem - how is this the case? Sources (1) Nathanson, E.S., 2014. Path integration with non-positive distributions and applications to the Schrödinger equation. The University of Iowa. pdf pg$\sim$27 actual pg$\sim$17 REPLY [4 votes]: The trick is to regard the Wiener measure as a random sample function $f(x,t)$ where $x\in (\Omega, \mathscr{F},P)$ and $t\in \mathscr{T}$ is the time index set. Then the whole stochastic process can be regarded as a curve/function when $x$ is fixed. Along with probability measure $P$ this consists of a random function. When you look at the curve of path generated by a Wiener measure, it is actually a realization of a function when the time index $t$ varies, see [1]. The reason why we need to invoke Riesz representation theorem is that when the path is $L^2$ integrable(and for a Wiener measure this is the case because its path is smooth with probability $P$ 1) we have a nice duality between function space and the time index $\mathscr{T}$,which is neatly explained in [2]. When the sample path $g_x(t):=f(x,t)$ is not $L^2$ integrable this perspective will not hold, that is why we need some conditions on covariance function of $\mu$ to make the sample path smooth with probability $P$ 1. Reference [1]Cramér, Harald. "Stochastic processes as curves in Hilbert space." Theory of Probability & Its Applications 9.2 (1964): 169-179. [2]Enchev, Ognian, and Daniel W. Stroock. "Towards a Riemannian geometry on the path space over a Riemannian manifold." Journal of Functional Analysis 134.2 (1995): 392-416. REPLY [2 votes]: It seems that what you are looking for is the notion of "abstract Wiener space", which provides a rigorous setting for putting a white noise on various spaces of functions. Long story short, you define the white noise as a random distribution via a larger Banach space of functions; this distribution makes sense once taken against elements of the dual of this Banach space (hence the use of the Riesz representation theorem for seeing this dual space as a subspace of your initial Hilbert space of functions). A nice introduction can be found in Stroock's "Abstract Wiener space, revisited".<|endoftext|> TITLE: Minimal subcoverings of a cover whose sets intersect in at most $1$ point QUESTION [6 upvotes]: Suppose we are given a nonempty set $X$. Let ${\cal U}$ be a set of subsets of $X$ such that $\bigcup {\cal U} = X$, and $U_1\neq U_2\in {\cal U}$ implies $|U_1\cap U_2| \leq 1$. Is there a subcollection ${\cal U}_0\subseteq {\cal U}$ such that $\bigcup {\cal U}_0 = X$, and if $U\in{\cal U}_0$ then $\bigcup \big({\cal U}_0 \setminus \{U\}\big) \ne X$ ? REPLY [4 votes]: Edit 08/21/2017 Péter Komjáth detected a gap in my original argument, so I will edit it now to limit what is claimed. I claim that if (i) $|X|=\omega$, or if (ii) $|X|=\kappa>\omega$ and $\mathcal U$ is a locally finite cover of $X$, then the question has an affirmative answer. Here I say that $\mathcal U$ is a locally finite cover of $X$ if each element of $X$ is contained in only finitely many elements of $\mathcal U$. From this point until the last paragraph, the only edits I intend to make involve replacing some $U$'s with ${\mathcal U}$'s. (These were typos in my original writeup.) I will explain how to construct ${\mathcal U}_0$. Let $(x_{\lambda})_{\lambda<\kappa}$ be an enumeration of $X$. I will examine the elements of $X$, roughly in order, and decide which elements of ${\mathcal U}$ to put into ${\mathcal U}_0$. To keep track of things, I will write $(X,\emptyset)$ and $({\mathcal U},\emptyset)$ to indicate the starting state and $(\emptyset,X)$ and $(\emptyset, {\mathcal U}_0)$ to indicate the ending state. Middle states will be of the form $(X',X'')$ and $({\mathcal U}', {\mathcal U}'')$ where $\{X',X''\}$ is a partition of $X$ into two cells, where the elements of $X''$ have been `handled' and the elements of $X'$ have yet to be handled. As we handle the elements of $X$ we move them from $X'$ to $X''$ and either discard sets from ${\mathcal U}'$ or else we move some sets from ${\mathcal U}'$ to ${\mathcal U}''$. Throughout the process we will arrange that $X'$ is contained in the union of the sets in ${\mathcal U}'$, while ${\mathcal U}''$ is a minimal cover of $X''$. The process will end with a minimal cover ${\mathcal U}_0$ of $X$. As I move through the process I will want to maintain an additional assumption. As mentioned above, I will assume at each stage $(X',X'')$ and $({\mathcal U}', {\mathcal U}'')$ that $X'$ is contained in the union of the sets in ${\mathcal U}'$, and ${\mathcal U}''$ is a minimal cover of $X''$. (This includes the assumption that $\bigcup_{u\in {\mathcal U}''}u=X''$.) For each $v\in {\mathcal U}''$ the set ${\mathcal U}''-\{v\}$ does not cover $X''$, so there is an element $x_v\in X''$ contained in $v$ and in no other element of ${\mathcal U}''$. Call $x_v$ an anchor for $v$. The assumption that ${\mathcal U}''$ is a minimal cover of $X''$ means exactly that every $v\in {\mathcal U}''$ has an anchor in $X''$. The additional assumption I intend to maintain throughout the process is: no set in ${\mathcal U}'$ contains an anchor $x_v\in X''$ of an element $v\in {\mathcal U}''$. The additional assumption holds in the starting state, since ${\mathcal U}''=\emptyset$ at the beginning. The construction really starts now. I allow two ways for the process to move forward. A move of type 1. We are at stage $(X',X'')$ and $({\mathcal U}', {\mathcal U}'')$, where (i) $X'$ is contained in the union of the sets in ${\mathcal U}'$, (ii) ${\mathcal U}''$ is a minimal cover of $X''$, and (iii) no set in ${\mathcal U}'$ contains an anchor $x_v\in X''$ of an element $v\in {\mathcal U}''$. I define a directed graph with vertex set $X'$. Write $x\to y$ if the implication [$x\in u$ implies $y\in u$] holds for every $u\in {\mathcal U}'$. There will be loops on every $x\in X'$, but they are unimportant. If $x\to y$ is a nonloop, then any $u\in {\mathcal U}'$ that contains $x$ must also contain $y$. Therefore there cannot be two sets $u, v\in {\mathcal U}'$ containing $x$, since this would yield $u\cap v \supseteq \{x,y\}$, contradicting $|u\cap v|\leq 1$. Thus each $x\in X'$ that is the tail vertex of a nonloop is contained in a unique $u_x\in {\mathcal U}'$. I intend to (i) move all tail vertices $x$ of nonloops from $X'$ to $X''$, move all associated sets $u_x$ from ${\mathcal U}'$ to ${\mathcal U}''$, (iii) treat $x$ as the anchor of $u_x$, and (iv) move from $X'$ to $X''$ all other elements covered by the $u_x$'s. I claim that after these steps we again have a stage $(\overline{X}',\overline{X}'')$ and $(\overline{\mathcal U}', \overline{\mathcal U}'')$, where the assumptions are satisfied. (I overlined sets $X', X'', {\mathcal U}', {\mathcal U}''$ to indicate the state of the set after the step is completed.) To see that $\overline{X}'\subseteq \bigcup_{u\in \overline{\mathcal U}'} u$ it suffices to note that $\overline{X}'\cup \overline{X}''=X= {X}'\cup {X}''$, $\overline{\mathcal U}'\cup \overline{\mathcal U}''= {\mathcal U}'\cup {\mathcal U}''$ (which covers $X$), and $\overline{X}''=\bigcup_{u\in \overline{\mathcal U}''} u$. To see that $\overline{\mathcal U}''$ is a minimal cover of $\overline{X}''$ it suffices to note that $\overline{X}''=\bigcup_{u\in \overline{\mathcal U}''} u$ and each $u\in \overline{\mathcal U}''$ has an anchor in $\overline{X}''$. Finally, no set in $\overline{\mathcal U}'$ contains an anchor element in $\overline{\mathcal U}''$ since (i) $\overline{\mathcal U}'\subseteq {\mathcal U}'$, so no set in $\overline{\mathcal U}'$ contains an old anchor element (one that existed in $X''$), and (ii) no set in $\overline{\mathcal U}'$ contains one of the new anchors $x$, since each such $x$ was a tail of a nonloop. The preceding step accomplishes something useful only if the associated graph is not discrete. If it is, then we need A move of type 2. Suppose now we are at a stage $(X',X'')$ and $({\mathcal U}', {\mathcal U}'')$ where the associated directed graph is discrete. If we are not yet to the stage where $X'=\emptyset$, then choose the element $x\in X'$ that has least index in our earlier enumeration $(x_{\lambda})_{\lambda<\kappa}$, and add it to $X''$. I would like it to be an anchor element, so choose any $u\in {\mathcal U}'$ that contains $x$ and add $u$ to ${\mathcal U}''$. Also, add all elements of $u$ to $X''$. To ensure that $x$ remains an anchor element for the rest of the construction, delete all $v\in {\mathcal U}'$ containing $x$. Now the move is complete. I claim that after these steps we again have a stage $(\overline{X}',\overline{X}'')$ and $(\overline{\mathcal U}', \overline{\mathcal U}'')$, where the assumptions are satisfied. To see that $\overline{X}'\subseteq \bigcup_{u\in \overline{\mathcal U}'} u$ and also that $\overline{\mathcal U}'$ contains no anchor elements in $X''$, it suffices to observe that, since at the start of the step the associated graph was discrete, if $y\in \overline{X}'\subseteq X'$, then there is a $v\in {\mathcal U}'$ such that $y\in v$ and $x\notin v$. The set $v$ contains none of the old anchor elements, by induction, and $v$ does not contain the new anchor element $x$, so $v\in\overline{\mathcal U}'$ and $v$ contains no anchors from $\overline{X}''$. To see that $\overline{\mathcal U}''={\mathcal U}''\cup\{u\}$ is a minimal cover of $\overline{X}''=X''\cup u$ it is enough to note that every set in $\overline{\mathcal U}''$ has an anchor in $\overline{X}''$. If we alternate between moves of these two types we will eventually exhaust $X$, because moves of type 1 cannot occur twice in a row and moves of type 2 progress through the enumeration of $X$. We will end with $(\emptyset,X)$ and $(\emptyset, {\mathcal U}_0)$ where ${\mathcal U}_0$ is a minimal cover of $X$. Last paragraph As Péter Komjáth pointed out, the inclusion $X'\subseteq \bigcup {\mathcal U}'$ might not be preserved at nonzero limit stages for general $X$, $\mathcal U$. This is not a problem if $|X|=\omega$, since there are no nonzero limit stages. I claim that it is also not a problem if $|X|=\kappa>\omega$ and $\mathcal U$ is a locally finite cover of $X$. Without offering the full details, the reason is that: (i) the inclusion $X'\subseteq \bigcup {\mathcal U}'$ is preserved at successor stages, and (ii) when $\mathcal U$ is a locally finite covering, a point of $X'$ can only be uncovered at a successor stage if our operations only allow moving sets from ${\mathcal U}'$ to ${\mathcal U}''$ or deleting sets from ${\mathcal U}'$.<|endoftext|> TITLE: If $x_{n+1}= \frac{nx_{n}^2+1}{n+1}$ then $x_{n}=1$ QUESTION [24 upvotes]: I asked this question at MSE, but I think it's more appropriated to MO. Let $x_{n}$ be a sequence, such that $x_{n+1}= \dfrac{nx_{n}^2+1}{n+1}$ and $x_n>0$ for all $n$. There is a positive integer $N$ such that $x_n$ is integer for all $n>N$. Does it follow that $x_n=1$ for all positive integers $n$? I tried to prove that $x_1 \equiv 1 \text{(mod p)}$ for all prime numbers $p$ but I couldn't make any progress. Does anyone know if this sequence has ever been studied? I'm looking for a proof or any reference of this result. Any help would be appreciated. REPLY [9 votes]: I did the following experiment: Let $p$ be a prime number. Then a necessary condition for the sequence to remain in $\mathbb{Z}$ is that $x_{p-1} \equiv \pm 1 \mod p$. So for every starting value $x_1$, I calculated $x_{p-1} \mod p$ for the first several prime numbers $p$ to see if there are obstructions. It turns out that for every choice of $x_1$ between $2$ and $100000$, there are always obstructions. The first obstruction (i.e. the smallest prime $p$ such that $x_{p-1}$ is not congruent to $\pm 1$ modulo $p$) for $2 \leq x_1 \leq 100$ is listed below. x[1] = 2: obstruction at 2 x[1] = 3: obstruction at 5 x[1] = 4: obstruction at 2 x[1] = 5: obstruction at 5 x[1] = 6: obstruction at 2 x[1] = 7: obstruction at 5 x[1] = 8: obstruction at 2 x[1] = 9: obstruction at 23 x[1] = 10: obstruction at 2 x[1] = 11: obstruction at 7 x[1] = 12: obstruction at 2 x[1] = 13: obstruction at 5 x[1] = 14: obstruction at 2 x[1] = 15: obstruction at 5 x[1] = 16: obstruction at 2 x[1] = 17: obstruction at 5 x[1] = 18: obstruction at 2 x[1] = 19: obstruction at 11 x[1] = 20: obstruction at 2 x[1] = 21: obstruction at 7 x[1] = 22: obstruction at 2 x[1] = 23: obstruction at 5 x[1] = 24: obstruction at 2 x[1] = 25: obstruction at 5 x[1] = 26: obstruction at 2 x[1] = 27: obstruction at 5 x[1] = 28: obstruction at 2 x[1] = 29: obstruction at 13 x[1] = 30: obstruction at 2 x[1] = 31: obstruction at 7 x[1] = 32: obstruction at 2 x[1] = 33: obstruction at 5 x[1] = 34: obstruction at 2 x[1] = 35: obstruction at 5 x[1] = 36: obstruction at 2 x[1] = 37: obstruction at 5 x[1] = 38: obstruction at 2 x[1] = 39: obstruction at 7 x[1] = 40: obstruction at 2 x[1] = 41: obstruction at 11 x[1] = 42: obstruction at 2 x[1] = 43: obstruction at 5 x[1] = 44: obstruction at 2 x[1] = 45: obstruction at 5 x[1] = 46: obstruction at 2 x[1] = 47: obstruction at 5 x[1] = 48: obstruction at 2 x[1] = 49: obstruction at 7 x[1] = 50: obstruction at 2 x[1] = 51: obstruction at 19 x[1] = 52: obstruction at 2 x[1] = 53: obstruction at 5 x[1] = 54: obstruction at 2 x[1] = 55: obstruction at 5 x[1] = 56: obstruction at 2 x[1] = 57: obstruction at 5 x[1] = 58: obstruction at 2 x[1] = 59: obstruction at 7 x[1] = 60: obstruction at 2 x[1] = 61: obstruction at 11 x[1] = 62: obstruction at 2 x[1] = 63: obstruction at 5 x[1] = 64: obstruction at 2 x[1] = 65: obstruction at 5 x[1] = 66: obstruction at 2 x[1] = 67: obstruction at 5 x[1] = 68: obstruction at 2 x[1] = 69: obstruction at 11 x[1] = 70: obstruction at 2 x[1] = 71: obstruction at 11 x[1] = 72: obstruction at 2 x[1] = 73: obstruction at 5 x[1] = 74: obstruction at 2 x[1] = 75: obstruction at 5 x[1] = 76: obstruction at 2 x[1] = 77: obstruction at 5 x[1] = 78: obstruction at 2 x[1] = 79: obstruction at 29 x[1] = 80: obstruction at 2 x[1] = 81: obstruction at 7 x[1] = 82: obstruction at 2 x[1] = 83: obstruction at 5 x[1] = 84: obstruction at 2 x[1] = 85: obstruction at 5 x[1] = 86: obstruction at 2 x[1] = 87: obstruction at 5 x[1] = 88: obstruction at 2 x[1] = 89: obstruction at 13 x[1] = 90: obstruction at 2 x[1] = 91: obstruction at 7 x[1] = 92: obstruction at 2 x[1] = 93: obstruction at 5 x[1] = 94: obstruction at 2 x[1] = 95: obstruction at 5 x[1] = 96: obstruction at 2 x[1] = 97: obstruction at 5 x[1] = 98: obstruction at 2 x[1] = 99: obstruction at 11 x[1] = 100: obstruction at 2 Up to $x_1 = 100000$, the biggest "first obstruction" appears at: x[1] = 13589: obstruction at 103 Even if one allows $x_1$ to be $\sqrt{k}$ for some integer $k$, the results are similar - one just starts from $x_2$, and for $2 \leq x_2 \leq 10000$ there are always obstructions at (small) prime numbers. These results seem to support the original conjecture. EDIT Following this idea, I further calculated, for a given prime $p$, the residue classes of $x_1 \mod p$ that will lead to an obstruction at $p$. Let us call them "bad" residues. The result seems to be interesting for its own sake. p bad residues x[1] mod p mod 2: 0 mod 3: mod 5: 0 2 3 mod 7: 0 3 4 mod 11: 0 3 5 6 8 mod 13: 0 2 3 5 8 10 11 mod 17: 2 4 5 12 13 15 mod 19: 0 6 8 11 13 mod 23: 3 7 8 9 11 12 14 15 16 20 mod 29: 0 3 5 6 7 8 10 11 13 14 15 16 18 19 21 22 23 24 26 mod 31: 0 2 5 8 10 13 15 16 18 21 23 26 29 mod 37: mod 41: mod 43: 0 2 3 4 5 6 7 8 9 10 11 14 16 18 19 20 21 22 23 24 25 27 29 32 33 34 35 36 37 38 39 40 41 mod 47: 0 7 8 9 10 11 12 13 14 15 18 21 23 24 26 29 32 33 34 35 36 37 38 39 40 mod 53: 0 9 12 15 17 18 20 23 30 33 35 36 38 41 44 mod 59: 3 5 8 13 14 15 16 18 21 26 33 38 41 43 44 45 46 51 54 56 mod 61: 0 2 5 8 11 12 15 17 19 20 21 22 23 24 26 29 32 35 37 38 39 40 41 42 44 46 49 50 53 56 59 mod 67: mod 71: 5 6 15 19 20 24 25 31 33 35 36 38 40 46 47 51 52 56 65 66 mod 73: 9 23 27 28 29 44 45 46 50 64 mod 79: mod 83: mod 89: 0 2 3 4 5 16 17 22 23 24 27 30 31 32 35 40 49 54 57 58 59 62 65 66 67 72 73 84 85 86 87 mod 97: 0 2 3 8 11 14 15 17 21 23 24 28 29 30 35 38 39 44 47 50 53 58 59 62 67 68 69 73 74 76 80 82 83 86 89 94 95 And here is the table which counts the number of bad residues modulo $p$: p number of bad residues x[1] mod p mod 2: 1 mod 3: 0 mod 5: 3 mod 7: 3 mod 11: 5 mod 13: 7 mod 17: 6 mod 19: 5 mod 23: 10 mod 29: 19 mod 31: 13 mod 37: 0 mod 41: 0 mod 43: 33 mod 47: 25 mod 53: 15 mod 59: 20 mod 61: 31 mod 67: 0 mod 71: 20 mod 73: 10 mod 79: 0 mod 83: 0 mod 89: 31 mod 97: 37 mod 101: 50 mod 103: 35 mod 107: 29 mod 109: 20 mod 113: 30 mod 127: 22 mod 131: 93 mod 137: 33 mod 139: 115 mod 149: 121 mod 151: 59 mod 157: 6 mod 163: 111 mod 167: 85 mod 173: 111 mod 179: 98 mod 181: 127 mod 191: 0 mod 193: 83 mod 197: 4 mod 199: 130 mod 211: 85 mod 223: 34 mod 227: 77 mod 229: 57 mod 233: 85 mod 239: 137 mod 241: 56 mod 251: 140 mod 257: 79 mod 263: 0 mod 269: 44 mod 271: 129 mod 277: 20 mod 281: 26 mod 283: 231 mod 293: 171 The most noticeable thing, to me, is those primes with "$0$" bad residues. Here are they: 3, 37, 41, 67, 79, 83, 191, 263, 347, 353, 373, 379, 421, 449, 463, 509, 557, 619, 647, 661, 673, 719, 733, 757, 787, 823, 839, 911 This sequence is not found in OEIS. Let us call those primes "exceptional". If one excludes those exceptional primes, then the proportion of bad residues (i.e. [number of bad residues mod $p$] divided by $p$) seems to distribute uniformly on the interval $[0, 1)$. This suggests that the exceptional primes may of particular interest. EDIT To illustrate the distribution of the proportion of bad residues, here I add the statistical data (for primes $p < 5000$): "bad proportion" number of primes 0 77 (i.e. number of exceptional primes) (0.00, 0.02] 10 (0.02, 0.04] 15 (0.04, 0.06] 13 (0.06, 0.08] 18 (0.08, 0.10] 7 (0.10, 0.12] 11 (0.12, 0.14] 8 (0.14, 0.16] 14 (0.16, 0.18] 21 (0.18, 0.20] 14 (0.20, 0.22] 16 (0.22, 0.24] 11 (0.24, 0.26] 17 (0.26, 0.28] 15 (0.28, 0.30] 13 (0.30, 0.32] 11 (0.32, 0.34] 15 (0.34, 0.36] 15 (0.36, 0.38] 17 (0.38, 0.40] 17 (0.40, 0.42] 19 (0.42, 0.44] 13 (0.44, 0.46] 15 (0.46, 0.48] 23 (0.48, 0.50] 20 (0.50, 0.52] 16 (0.52, 0.54] 11 (0.54, 0.56] 15 (0.56, 0.58] 16 (0.58, 0.60] 14 (0.60, 0.62] 12 (0.62, 0.64] 6 (0.64, 0.66] 13 (0.66, 0.68] 20 (0.68, 0.70] 14 (0.70, 0.72] 8 (0.72, 0.74] 9 (0.74, 0.76] 9 (0.76, 0.78] 6 (0.78, 0.80] 11 (0.80, 0.82] 12 (0.82, 0.84] 7 (0.84, 0.86] 6 (0.86, 0.88] 5 (0.88, 0.90] 6 (0.90, 0.92] 3 (0.92, 0.94] 3 (0.94, 0.96] 2 (0.96, 0.98] 0 (0.98, 1.00] 0 There is clearly a concentration on $0$, i.e. on the exceptional primes. The "average proportion", calculated as $\frac{\sum_p proportion_p}{\sum_p 1}$, is about $0.37551$.<|endoftext|> TITLE: Applications of cluster algebras QUESTION [14 upvotes]: Why are so many algebraists nowadays interested in cluster algebras? (This is a rewording of one half of the closed question Cluster algebras and teichmuller theory.) REPLY [10 votes]: One reason is that cluster algebras have motivated many recent developments in the representation theory of associative algebras. There is a lot one can say about this, so I will try to just give an overview of some of the key ideas, and suggest further reading. I recommend Keller's survey article (https://arxiv.org/abs/0807.1960) as a good source for a more detailed discussion. A simple case of a cluster algebra is the cluster algebra without frozen variables defined by a skew-symmetric matrix. A skew-symmetric matrix is essentially the same data (modulo some technicalities) as a quiver (directed graph) without loops or $2$-cycles, e.g. $$\begin{pmatrix}0&-2\\2&0\end{pmatrix}\ \longleftrightarrow\ 1\Rightarrow 2$$ where the '$\Rightarrow$' represents a pair of arrows. The matrix here is the skew-symmetrisation of the adjacency matrix of the quiver, the assumptions on loops and $2$-cycles ensuring that there is no cancellation when computing this matrix, so that the process is invertible. (For simplicity, to avoid algebras breaking up into direct products, I consider only the pairs in the above correspondence such that the quiver is connected; this is most important in the classification statements below.) Such quivers, when they have no oriented cycles, also define associative algebras; they correspond to the finite dimensional basic hereditary algebras over a field $k$ via $Q\leftrightarrow kQ$ (I take $k$ to be algebraically closed, just in case). The algebra $kQ$ has basis given by the paths of $Q$ (including the length $0$ paths at each vertex) with multiplication given by concatenation of paths when the result is a path and $0$ otherwise, extended via linearity. A classical result of Gabriel says that the algebra $kQ$ has finitely many isomorphism classes of indecomposable modules if and only if $Q$ is an orientation of one of the simply laced Dynkin graphs, i.e. those of type $\mathsf{A}_n$, $\mathsf{D}_n$ or $\mathsf{E}_{6,7,8}$. Suggestively, Fomin–Zelevinsky prove in their second paper on cluster algebras that the cluster algebras (without frozen variables, defined by a skew-symmetric matrix) with finitely many cluster variables are precisely those from matrices corresponding to these quivers. It turns out that, whenever $Q$ is an acyclic quiver, there is in fact a bijection between most of the cluster variables (precisely, those not appearing in the initial seed) of the cluster algebra $\mathcal{A}_Q$ given by $Q$, and the indecomposable representations of $kQ$. To make this connection stronger, one can introduce the 'cluster category' $\mathcal{C}_Q$, defined by Buan, Marsh, Reineke, Reiten and Todorov (https://arxiv.org/abs/math/0402054), which is a kind of extension of the category of $kQ$-modules by adding in some extra indecomposable objects corresponding to the initial cluster variables of $\mathcal{A}_Q$, so that each cluster variable $x$ of $\mathcal{A}_Q$ now corresponds to an indecomposable object $M_x$ of $\mathcal{C}_Q$. Various properties of the cluster variables may now be translated into properties of these objects. For example, two cluster variables $x$ and $y$ are compatible (appear in the same cluster) if and only if the corresponding objects have the homological property $\operatorname{Ext}^1_{\mathcal{C}_Q}(M_x,M_y)=0$. The seeds of $\mathcal{A}_Q$ correspond to basic cluster-tilting objects of $\mathcal{C}_Q$, which are direct sums $T$ of some of the $M_x$s such that $\operatorname{Ext}^1_{\mathcal{C}_Q}(T,M_x)=0$ if and only if $M_x$ appears as a summand of $T$. The quiver of the seed corresponding to such an object $T$ is the ordinary quiver of the endomorphism algebra $\operatorname{End}_{\mathcal{C}_Q}(T)$, i.e. the unique (up to isomorphism) quiver $Q_T$ such that $\operatorname{End}_{\mathcal{C}_Q}(T)\cong kQ_T/I$ for some ideal generated by linear combinations of paths of length at least $2$. The category $\mathcal{C}_Q$ is a '$2$-Calabi–Yau triangulated category', so its cluster-tilting objects have a mutation theory by work of Iyama–Yoshino (https://arxiv.org/abs/math/0607736) which turns out to correspond to mutations of seeds. There is considerably more work in this direction, including cluster categories for cluster algebras corresponding to non-acyclic quivers by Amiot (https://arxiv.org/abs/0805.1035), and for cluster algebras with frozen variables by, e.g. Geiß–Leclerc–Schröer (https://arxiv.org/abs/math/0609138), Jensen–King–Su (https://arxiv.org/abs/1309.7301), Demonet–Iyama (https://arxiv.org/abs/1503.02362) and myself (https://arxiv.org/abs/1510.06224, https://arxiv.org/abs/1702.05352). One can also develop some theory for skew-symmetrizable matrices, see e.g. Demonet (https://arxiv.org/abs/0909.1633), Labardini-Fragoso–Zelevinsky (https://arxiv.org/abs/1306.3495) and Geiß–Leclerc–Schröer again (https://arxiv.org/abs/1410.1403). While construction of these kinds of representation-theoretic models is useful for understanding cluster algebras, it also suggests lots of interesting representation theory that may not otherwise have been considered. This can make a precise connection difficult to pin down: I am not really aware of cluster algebras being used directly to solve open representation-theoretic problems, but they have suggested lots of new lines of representation-theoretic inquiry (e.g. the second reference to Geiß–Leclerc–Schröer above, which I think is not really about cluster algebras at all, a priori). One example of this is Adachi–Iyama–Reiten's $\tau$-tilting theory (https://arxiv.org/abs/1210.1036). I am far from an expert on this theory, but my understanding is that this is really a purely representation-theoretic theory that could have been developed entirely without cluster algebras (and, I am told, was almost discovered by Auslander and co-authors in the 80s!) but in the end it was, at least in part, thinking about cluster categories that provided the inspiration. I note that one slightly confusing part of this connection is that one quite rarely studies the cluster algebra itself as an algebra! There are some exceptions to this, such as work of Lampe (https://arxiv.org/abs/1210.1502).<|endoftext|> TITLE: A combinatorial identity QUESTION [15 upvotes]: I hope this is a suitable MO question. In a research project, my collaborator and I came across some combinatorial expressions. I used my computer to test a few numbers and the pattern was suggesting the following equation for fixed integers $K\geq n>0$. $$\dfrac{K!}{n!K^{K-n}}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}=\displaystyle {K-1\choose n-1}.$$ We tried to think of a proof but failed. One can probably move these $K!, n!$ to the right and rewrite the RHS, or maybe move $K!$ into the summation to form combinatorial numbers like $K\choose k_1,k_2,\dotsc,k_n$. We don't know which is better. The questions are: Anyone knows a proof for this identity? In fact the expression that appears in our work is $\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \sigma_p(k_1,\dotsc,k_n) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}$, where $p$ is a fixed integer and $\sigma_p(\dotsc)$ is the $p$-th elementary symmetric polynomial. The equation in the beginning simplifies this expression for $p=0,1$. Is there a similar identity for general $p$? ----------Update---------- Question 2 is perhaps too vague, and I'd like to make it a bit more specific. Probably I should have written this down in the beginning, but I feared this is too long and unmotivated. But after seeing people's skills, I'm very tempted to leave it here in case somebody has remarks. In fact, question 2 partly comes from the effort to find a proof for the following (verified by computer). $$ \frac{1}{K!} \prod_{r=1}^{K} (r+1 -x)= \sum_{n=1}^K \frac{(-1)^n}{n!} \left[ \sum_{p=0}^n K^{n-p} \prod_{r=1}^p (x +r-4) \left( \sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!} \right) \right], $$ Where $x$ is a fixed number (in our case, an integer). REPLY [6 votes]: Here is a generating-function proof of your conjectured identity (and an answer to question 2). The main ingredient is a formula for the appearing symmetric sums. Let $T(z)$ (the ``tree function'') be the formal power series satisfying $T(z)=z\,e^{T(z)}$. If $F$ is a formal power series the coefficients of $G(z):=F(T(z))$ are given by (Lagrange inversion) $$[z^0]G(z)=[z^0] F(z) \mbox{ , } [z^k]G(z)=\tfrac{1}{k} [y^{k-1}] F^\prime(y)\,e^{ky} =[y^k](1-y)F(y)\,e^{ky}\mbox{ for } k\geq 1\;.$$ In particular (as is well known) $$T(z)=\sum_{n\geq 1}\frac{n^{n-1}}{n!}z^n \;\mbox{ and }\; \frac{T(z)}{1-T(z)}=\sum_{n\geq 1}\frac{n^{n}}{n!}z^n$$ Thus $T(z)\left(1+\frac{t}{1-T(z)}\right)=\sum_{n\geq 1} \frac{(1+tn)n^{n-1}}{n!}$. Therefore \begin{align*}S_{p,n}(K):&=\sum_{{k_1+\ldots +k_n=K \atop k_i\geq 1}} \sigma_p(k_1,\ldots,k_n)\prod_{i=1}^n \frac{k_i^{k_i-1}}{k_i!}\\ &=[t^p]\sum_{k_1+\ldots +k_n=K \atop k_i\geq 1} \prod_{i=1}^n \frac{(1+tk_i) k_i^{k_i-1}}{k_i!}\\ &=[t^p z^K]\, T(z)^n \left(1+\frac{t}{1-T(z)}\right)^n \end{align*} and $$S_{p,n}(K)={n \choose p} [z^K]\frac {T(z)^n}{(1-T(z))^p}={n \choose p}[y^K]\,y^n\, \frac{(1-y)}{(1-y)^p}\,e^{Ky}\;\;\;\;\;(*)$$ Now consider the sum ($m:=x-4$) $$R(K,m):=\sum_{n=0}^K\frac{(-1)^n}{n!}\bigg[\sum_{p=0}^n K^{n-p}\,\left(\prod_{r=1}^p (m+r)\right) S_{n,p}(K)\bigg]$$ Since $K\geq 1$ the sum remains unchanged if we start the summation at $n=0$ (all $S_{0,p}(K)$ are $0$). Using that and $(*)$ gives \begin{align*} R(K,m):&=\sum_{n=0}^K\frac{(-1)^n}{n!}\bigg[\sum_{p=0}^n K^{n-p}\,\left(\prod_{r=1}^p (m+r)\right) S_{n,p}(K)\bigg]\\ &=[y^K]\,(1-y) \sum_{n=0}^K\frac{(-1)^n}{n!}\sum_{p=0}^n K^{n-p}\,p!{m+p \choose p}{n \choose p}\frac{y^n}{(1-y)^p}\,e^{Ky}\\ &=[y^K]\,(1-y) \sum_{p=0}^K\sum_{n=p}^K\frac{(-1)^n}{(n-p)!} K^{n-p}{m+p \choose p}\frac{y^n}{(1-y)^p}\,e^{Ky}\\ &=[y^K]\,(1-y) \sum_{p=0}^K {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\bigg[\sum_{n=p}^K\frac{(-1)^{n-p}}{(n-p)!} K^{n-p}y^{n-p}\,e^{Ky}\bigg]\\ &=[y^K]\,(1-y) \sum_{p=0}^K {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\bigg[1 +\mathcal{O}(y^{K-p+1})\bigg]\\ \end{align*} where $\mathcal{O}(y^{K-p+1})$ denotes a formal power series which is a multiple of $y^{K-p+1}$. Taking into account the respective factors $y^p$ the terms in these series do not contribute to the coefficient $[y^K]$. Therefore \begin{align*} R(K,m)&=[y^K]\,(1-y) \sum_{p=0}^K {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\\ &=[y^K]\,(1-y) \sum_{p\geq 0} {m+p \choose p}\frac{(-1)^p y^p}{(1-y)^p}\\ &=[y^K]\,(1-y) \left(\frac{1}{1+\tfrac{y}{1-y}}\right)^{m+1}\\ &=[y^K]\,(1-y)^{m+2}\\ &=(-1)^K\,{m+2 \choose K},\,\mbox{ as desired }. \end{align*} REPLY [4 votes]: Here is the answer to Question 2. It may be probably simplified. Denote $y=3-x$, then we rewrite your identity as $$\binom{y+K-2}K=\frac{(y-1)y(y+1)\dots (y+K-2)}{K!}=c_0\binom{y}0+c_1\binom{y}1+\dots+c_K\binom{y}K,$$ where $$c_p=p!\sum_{n=p}^K(-K)^{n-p}\frac1{n!}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}.$$ On the other hand, by Vandermonde--Chu identity we have $$\binom{y+K-2}K=\sum_{i=2}^{K}\binom{y}i\binom{K-2}{K-i},$$ so your identity is equivalent to the formula $$ \sum_{n=p}^K(-K)^{n-p}\frac{K!}{n!}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!}=\frac{K!}{p!}\binom{K-2}{K-p}, $$ I multiplied both parts by $K!/p!$. Note that $$ \frac{K!}{n!}\sum\limits_{ \begin{subarray}{c} k_1+\dotsb+k_{n}=K \\ k_i \geq 1 \end{subarray}} \sigma_p(k_1,\dotsc,k_{n}) \prod\limits_{i=1}^n \dfrac{k_i^{k_i-2}}{(k_i-1)!} $$ is a number of the trees $T$ on $\{0,1,\dots,K\}$ such that degree of 0 equals $n$ and $p$ vertices in different components of $T\setminus\{0\}$ are marked. Indeed, if these components $A_1,\dots,A_n$ are enumerated (this corresponds to the multiple $n!$) and $i$-th component $A_i$ has $k_i$ vertices, then we have $\frac{K!}{k_1!\dots k_n!}$ ways to choose $A_i$, $\sigma_p(k_1,\dotsc,k_{n}) $ ways to mark $p$ vertices in different components, $k_i^{k_i-1}$ ways to make a tree on $A_i$ and choose a vertex in $A_i$ joined with 0. Note that each (out of $\binom{K}p$ sets) set of $p$ marked vertices makes the same contribution to the sum. So, we may suppose that the marked set is $\{1,2,\dots,p\}$ and we have to prove that the sum of $(-K)^{n-p}$ over admissible trees (where the tree $T$ is admissible if $1,2,\dots,p$ are in different components of $T\setminus \{0\}$) equals $\frac1{\binom{K}p}\frac{K!}{p!}\binom{K-2}{K-p}=(p-1)p\dots (K-2)$. We start to prove this from the case $p=0$, $p=1$, where the restriction that $1,2,\dots,p$ are in different components of $T\setminus \{0\}$ disappears. Then the sum $z_0^{n-1}z_1^{d_1-1}\dots z_K^{d_K-1}$, $d_i=\deg(i)$, over all trees on $\{0,\dots,K\}$ equals, as is well known and easy to prove, to $(z_0+\dots+z_K)^{K-1}$. Substituting $z_0=-K$, $z_1=\dots=z_K=1$ we get the result. Now we deal with the more involved case $p\geqslant 2$. Denote $K=p+m$ and consider the variables $z_0,z_1,\dots,z_p,z_{p+1},\dots$ (infinitely many for simplicity of notations). Denote $s=z_0+z_1+\dots$, write $\sigma_i$ for the $i$-th elementary symmetric polynomial of $z_{p+1},z_{p+2},\dots$. Denote $\varphi_0=1$, $\varphi_m=s\varphi_{m-1}+(p-1)p\dots (p+m-2)\sigma_m$ for $m\geqslant 1$. I claim that the sum of $z_0^{n-p}z_1^{d_1-1}\dots z_{p+m}^{d_{p+m}-1}$ over all admissible trees equals $\varphi_m(z_0,z_1,\dots,z_{p+m},0,0,\dots)$. Note that this implies our claim, as follows from the substitution $z_0=-K=-p-m,z_1=\dots=z_{p+m}=1$. The proof is on induction in $m$. Base $m=0$ is clear. For the induction step, look at coefficients of any specific monomial $z_0^{n-p}z_1^{d_1-1}\dots z_{p+m}^{d_{p+m}-1}$. Consider two cases: 1) $d_i=1$ for a certain index $i\in \{p+1,\dots,p+m\}$, without loss of generality $i=p+m$. This corresponds to the case when $p+m$ has degree 1, such a vertex may be joined with any of other vertices, and removing corresponding edge we get a tree (it remains admissible) on $\{0,1,\dots,K-1\}$. This corresponds to the summand $s\varphi_{m-1}$: namely, $z_j\varphi_{m-1}$ corresponds to the edge between $p+m$ and $j$; $j=0,1,\dots,p+m-1$. 2) $d_{p+1},\dots,d_{p+m}$ are greater than 1. Then they are all equal to 2, since the degree of the whole monomial equals $m$. In this case there are $p(p+1)\dots (p+m-1)$ admissible trees (well, they are all admissible for such a choice of degrees and we may either apply the above formula for all trees, or prove it by induction, or as you wish). It remains to prove that the coefficient of $z_{p+1}\dots z_{p+m}$ in the function $\varphi_m$ equals $p(p+1)\dots (p+m-1)$. Since $\varphi_m=s\varphi_{m-1}+(p-1)p\dots (p+m-2)\sigma_m$, it is equivalent to proving that the coefficient of $z_{p+1}\dots z_{p+m}$ in $s\varphi_{m-1}$ equals $p(p+1)\dots (p+m-1)-(p-1)p\dots (p+m-2)=mp(p+1)\dots(p+m-2)$. We should take some $z_j$, $p+1\leqslant j\leqslant p+m$, from the multiple $s=\sum z_i$, and for each choice of $j$ we have a coefficient of $z_j^{-1}\cdot z_{p+1}\dots z_{p+m}$ in $\varphi_{m-1}$ equal to $p(p+1)\dots(p+m-2)$ - by induction (base $m-1=0$ is clear).<|endoftext|> TITLE: Do de Rham cohomologies commute with direct limits? QUESTION [10 upvotes]: Given a smooth manifold $M$ and a relatively compact exhaustion $M=\bigcup_{n\in\mathbb N} M_n$ with open and relatively compact $M_n\subseteq M_{n+1}$ (hence $M=\lim\limits_{\to}M_n$) do we always have $$H^k(M)=\lim\limits_{\leftarrow} H^k(M_n)?$$ This looks so natural that the answer should be known in which case I would like to get a reference. EDIT (inspired by Weibel's book and an article of Milnor from 1962). For fixed $k$ the projective spectrum $\Omega^k(M_n)$ of $k$-forms on $M_n$ satisfies the (strict) Mittag-Leffler condition so that the derived functor ${\lim\limits_{\leftarrow}}^1 \Omega^k(M_n)$ vanishes. We thus get an exact sequence $0\to \Omega^\ast(M)\to \prod_n \Omega^\ast(M_n)\to \prod_n \Omega^\ast(M_n)\to 0$ of cochain complexes (the last map is the difference map $(\omega_n)_n\mapsto (\omega_n-\rho_{n+1}(\omega_{n+1}))_n$ with the restriction operator $\rho_{n+1}$). The long cohomology sequence thus yields an exact sequence $$\cdots \to H^k(M)\to \prod_nH^k(M_n) \to \prod_nH^k(M_n) \to H^{k+1}(M)\to \cdots.$$ From this one gets short exact sequences $$0\to {\lim\limits_{\leftarrow}}^1 H^{k-1}(M_n) \to H^k(M)\to \lim\limits_{\leftarrow}H^k(M_n)\to 0.$$ It seems to me that the spectrum $H^0(M_n)$ of the spaces of locally constant functions again satisfies the strict Mittag-Leffler condition so that ${\lim\limits_{\leftarrow}}^1H^0(M_n)=0$. This means that $H^1(M)=\lim\limits_{\leftarrow} H^1(M_n)$ is always true. Another consequence is $H^k(M)=\lim\limits_{\leftarrow} H^k(M_n)$ if all cohomologies $H^{k-1}(M_n)$ are finite dimensional (because this implies a Mittag-Leffler condition). However, in general the question remains open (and I would rather expect a counterexample). REPLY [7 votes]: $\def\RR{\mathbb{R}}\def\Hom{\mathrm{Hom}}$This seems too simple, so please tell me what I'm missing. We first prove the corresponding result in homology. Let $C_i(X)$ be the group of singular $i$-chains in $X$. Then $C_i(M) = \lim_{j \to \infty} C_i(M_j)$ (because $i$-simplices are compact, so the image of any finite number of them must lie in some $M_j$). Homology commutes with direct limits, so we deduce that $H_i(M) = \lim_{j \to \infty} H_i(M_j)$. Now, the universal coefficient theorem tells us that $H^i(M, \RR) = \Hom(H_i(M), \RR)$ and the same for $M_j$. So the question is whether $\Hom( \ , \RR)$ turns direct limits into inverse limits, and the answer is yes. Both the universal coefficient theorem, and the identification of singular and de Rham cohomology, work for general manifolds. I also have a direct proof of Mittag-Leffler for $H^q(M_0) \leftarrow H^q(M_1) \leftarrow \cdots$. Lemma Let $M$ be a manifold (PL or better) and let $K$ be a compact subset of $M$. Then $H^q(M) \to H^q(K)$ has finite dimensional image. Proof Take a PL triangulation of $M$. For each face $\sigma$ of the triangulation, let $U_{\sigma}$ be the union of the relative interiors of those faces $\tau$ containing $\sigma$. So the $U_{\sigma}$ form an open cover of $M$. As $K$ is compact, it can be covered by finitely many $U_{\sigma}$. Let $N$ be the union of the closures of those $U_{\sigma}$. So $K \subset N$ and $H^q(M) \to H^q(K)$ factors through $H^q(N)$. But $N$ is a finite simplicial complex, so $H^q(N)$ is finite dimensional. $\square$ Remark Actually, the union of the $U_{\sigma}$ also has finite cohomology, without taking the closure, but I didn't see a one line way of saying it. Now, let $M_0 \subset M_1 \subset M_2 \subset \cdots$ be an ascending chain of manifolds, with the closure $\overline{M_i}$ (in $M_{i+1}$) compact. Then $H^q(M_{i+1}) \to H^q(M_i)$ factors through $H^q(\overline{M}_i)$, so it has finite dimensional image. For each $j \geq i+1$, the image of $H^q(M_j)$ lies in that finite dimensional image. So the Mittag-Leffler condition holds and $\lim_{\infty \leftarrow j}^1 H^q(M_j)=0$.<|endoftext|> TITLE: Intuition for symplectic groups QUESTION [26 upvotes]: My question essentially breaks down to How do you, a working mathematician, think about (real) symplectic groups? How do you visualize symplectic (linear) transformations? What intuition do you have? For example, we can think of rotations by thinking, "Okay, let's just restrict to a plane fixed by the transformation and see what it does there." This doesn't quite work for the symplectic group, though, since on the plane it is indistinguishable from the area-preserving linear transformations. The next smallest example is for a 4-dimensional vector space, which isn't particularly easy for me to visualize. REPLY [14 votes]: In a different direction from Victor Protsak's answer, I will focus on your comment The next smallest example is for a 4-dimensional vector space, which isn't particularly easy for me to visualize. It's true that the definition of the Lie group $\mathrm{Sp}(4,\mathbb{R})$ involves a $4$-dimensional vector space, but actually there are two nice $3$-dimensional homogeneous spaces on which this group acts. The first, most obvious one, is the projective space $\mathbb{RP}^3$ of $1$-dimensional subspaces in $\mathbb{R}^4$.You can picture it as a compactification of $\mathbb{R}^3$ where you add a point at infinity for each possible direction of a line. The symplectic group acts on this space because it acts linearly on $\mathbb{R}^4$. The difference between the symplectic group and the full group of projective automorphisms $\mathrm{PGL}(4,\mathbb{R})$ is that $\mathrm{Sp}(4,\mathbb{R})$ does not act transitively on lines. Some lines are special, because they come from projectivizing Lagrangian planes. Conversely, any continuous transformation of $\mathbb{RP}^3$ which preserves lines and Lagrangian lines is an element of $\mathrm{Sp}(4,\mathbb{R})$ (this is a version of the fundamental theorem of projective geometry). The second is the Lagrangian Grassmannian, the subset of the Grassmannian of $2$-planes consisting of planes where the symplectic form vanishes. It is a $3$-dimensional space again, which is a bit more topologically complicated. It's homeomorphic to $(\mathbb{S}^2\times \mathbb{S}^1)/\langle\iota\rangle$, where $\iota$ acts as the antipodal map on both spaces. This space admits an invariant conformal Lorentzian metric. This means that at each point (Lagrangian) $L$, there is a light cone, and in this case it consists of all Lagrangians intersecting $L$ in a line. Again, any continuous transformation of this space which preserves light cones comes from something in $\mathrm{Sp}(4,\mathbb{R})$. The Lorentzian structure is an accident coming from the isomorphism $\mathrm{Psp}(4,\mathbb{R})\cong \mathrm{SO}^0(3,2)$ and so does not exist for higher symplectic groups, but the incidence structure of Lagrangians is still present. You can imagine this space of Lagrangians as a compactification of $\mathbb{R}^{2,1}$, the $(2+1)$-dimensional Minkowski space, preserving its light cone structure.(see for instance https://arxiv.org/pdf/0706.3055.pdf for details) If you get familiar with one or both of these spaces, it should help your intuition of how this particular group works. Edit: Here is an additional description to answer R. van Dobben de Bruyn's comment There is actually a more intuitive description due to Sophus Lie, the "Lie quadric". The space of Lagrangians in $\mathbb{R}^4$ is identified with the space of oriented circles on the $2$-sphere, where zero-radius circles ("point circles") are also allowed. The way to set it up is to use $\mathbb{C}^2$ as your real 4-dimensional vector space, fix a complex-valued symplectic form $\omega$ and use $Im(\omega)$ as your real-valued symplectic form. Then, the projection $(\mathbb{C}^2\backslash\{0\})\rightarrow\mathbb{CP}^1$ sends Lagrangian planes to circles 2-to-1 (and so Lagrangians to oriented circles 1-to-1). Two Lagrangians intersect if and only if the corresponding circles are tangent. The group $\mathrm{Sp}(4,\mathbb{R})$ acts on oriented circles and points as tangency-preserving transformations. These transformations take a bit of time to get used to, since they can send a point to a circle... You just have to remember that, unlike in projective geometry, the transformations on the "space of circles" do not preserve points. The transformations which do preserve the "point circles" are the same as the Moebius transformations acting on $\mathbb{CP}^1$. This is a $6$-dimensional subgroup of the $10$-dimensional $\mathrm{Sp}(4,\mathbb{R})$. There is also a "increase radius by $r$" operation which takes every circle in $\mathbb{C}\subset\mathbb{CP}^1$ to a circle centered at the same point, of radius $r$ bigger. Circles with negative orientation are thought of as having "negative radius". If two circles are externally tangent, they have opposite orientations, and if you add the same amount of "signed radius" to them, they stay tangent. These two types of transformations (Moebius and radius addition) together generate the full group of tangency-preserving transformations. (for more details see https://en.wikipedia.org/wiki/Lie_sphere_geometry)<|endoftext|> TITLE: A variant of $\ell^2$-cochains QUESTION [12 upvotes]: Suppose $X$ is an infinite countable CW complex which satisfies the following property: for all $k$-cells $e$, the number of $(k+1)$-cells incident to $e$ is at most $c_k$, where the latter is some number that depends on $k$. Let $X_k$ be the set of $k$-cells. Let $\ell^2_k(X)$ be the set of functions $a : X_k \to \Bbb R$, such that the series $$ \sum_{e \in X_k} a(e)^2 $$ converges (this implicitly makes use of the counting measure on $X_k$). Then the incidence bound assumption implies that coboundary operator $$ \delta: \ell^2_k(X) \to \ell^2_{k+1}(X) $$ is defined (this uses the same formula that arises when defining the cellular cochain complex of $X$). When $\dim X =1$ this construction was introduced by Dodziuk and Kendall in Dodziuk, J.(1-CUNYG); Kendall, W. S.(4-STRA) Combinatorial Laplacians and isoperimetric inequality. From local times to global geometry, control and physics (Coventry, 1984/85), 68–74, Pitman Res. Notes Math. Ser., 150, Longman Sci. Tech., Harlow, 1986. Questions Has this construction been investigated in the generality described above? How is the cohomology of this complex related to the usual cellular cohomology of $X$? Is there a set of reasonable conditions on $X$ which guarantee that this cohomology is finite dimensional? How does the above relate to other notions of $L^2$-cohomology? REPLY [3 votes]: I'm not sure if this should be a comment rather than an answer, since I'm not attempting to answer the hard questions (2) and (3). But it's too long to be a comment. That said: First of all, I just want to clarify that presumably you want incidence to be bounded with multiplicity, that is, the coefficients of the boundary map to also be bounded. For (4), take a look at this technical report by Attie and Block: https://pdfs.semanticscholar.org/cfd5/297840f262e255f86777e112757dbffdf9e5.pdf Theorem 4 is a de Rham theorem for $L^p$ cohomology -- for a manifold with bounded geometry, $L^p$ cohomology is equal to simplicial $L^p$ cohomology with real coefficients, on any given bounded triangulation. (I suppose you can also define a "piecewise de Rham" cohomology and then the isomorphism holds for any simplicial complex, in standard and $L^p$ flavors.) Now if you have a manifold with a nice enough CW structure, then you could replace each cell with a bounded number of simplices up to homotopy equivalence, and you would have an isomorphism from your theory to the "usual" de Rham version. The pullback to the universal cover of a CW structure on a compact manifold is of course nice enough, but a CW complex of bounded incidence is not. E.g. take an $\mathbb{R}$ and at the $n$th lattice point attach an $S^2$ which in turn has a 4-cell attached to it via a map with Hopf invariant $n$. This has bounded incidence, since only the 1-cells are incident to anything at all, but an infinite number of homotopy types of attaching maps and so a simplicial complex version would have to get more and more complicated as you go further out. I don't know of anyone studying the $L^2$ version in this level of generality. I studied the $L^\infty$ analogue at some length in my thesis (see sections 5 and 6 of https://arxiv.org/abs/1410.3368), but I was only concerned with universal covers of compact spaces, which avoids a lot of technical issues.<|endoftext|> TITLE: Dimensions of irreducible representations of $GL(n,F_q)$ are polynoms in q having roots ONLY at roots of unity and zero? QUESTION [16 upvotes]: Consider the group $GL(n,F_q)$ for finite field $F_q$, consider its irreducible representations over complex numbers. Questions Is my understanding correct that the dimensions of all such irreps are polynomials in $q$ with integer coefficients having zeros only at roots of unity and zero ? I understand that the answer should be contained in Green's 1955 paper THE CHARACTERS OF THE FINITE GENERAL LINEAR GROUPS, but as for me paper is difficult for extracting information. The case $q=2$ should be treated with care as demenstrated in comment from user148212. Questions 2 If anwer is yes - is there any conceptual/nice reason for it ? (R. Stanley comment below perfectly explains the part about roots). Questions 3 If someone can give some nice formula for such dimensions that would be quite helpful. Questions 4 For other finite groups of Lie type is there any similar phenomena ? Questions 5 From the perspective of $F_1$ it would be nice to know is there always a manifold such that number of $F_q$ points is given by these polynomials ? (Flags are of that type). Let me give some examples known to me supporting the positive answer to the question: For $GL(2,F_q)$ dimensions are : $1$ (det-like irreps) , $q+1$ (principal series), $q-1$ (cuspidal), $q$ (Steinberg = irregular principal series). See e.g. MO273764, MO271389. In general "regular princinpal series" - irreps induced from non-trivial characters of the Borel subgroup will have dimension $[n]_q!$. Just because $GL/Borel = Flag$ manifold has such number of points. Cuspidal irreps: the degree of a cuspidal character of $GL(n, q)$ is $(q − 1)(q^2 − 1)· · ·(q^{n-1}-1)$ (see page 135 Corollary 5.4.5. of very nice thesis "Character Tables of the General Linear Group and Some of its Subroups" containing huge amount of concrete information). For the so-called unipotent irreps there is q-analogue of "hook formula". The degrees of the unipotent characters are “polynomials in q”: $ q^{d(λ)} \frac{(q^n − 1)(q^{n−1} − 1)· · ·(q − 1)}{ \prod_{h(λ)}(q^h − 1) }$ with a certain d(λ) ∈ N, and where h(λ) runs through the hook lengths of λ. See nice survey by G. Hiss FINITE GROUPS OF LIE TYPE AND THEIR REPRESENTATIONS (top page 26, section 3.2.6). From above source - section 3.2.7: The degrees of the unipotent characters of $GL(5,q)$ for table $(5)$ dim = $ 1 $, for table $(4, 1)$ dim = $q(q + 1)(q^2 + 1)$ for table $(3, 2)$ dim = $q^2(q^4 + q^3 + q^2 + q + 1)$ for table $(3, 1^2)$ dim = $q^3(q^2 + 1)(q^2 + q + 1)$ for table $ (2^2, 1)$ dim = $q^4(q^4 + q^3 + q^2 + q + 1)$ for table $(2, 1^3)$ dim = $q^6(q + 1)(q^2 + 1)$ for table $(1^5)$ dim = $ q^{10}$ characters for GL(3), GL(4) has been computed by R. Steinberg The representations of GL(3,q), GL(4,q), PGL(3,q), PGL(4,q) Canad. J. Math. 3(1951), 225-235. Which "This paper is part of a Ph.D. thesis written at the University of Toronto under the direction of Professor Richard Brauer". The degrees of the irreducible characters of GL3(q): $(q − 1)^2(q + 1)$, $(q − 1)(q^2 + q + 1)$, $ (q + 1)(q^2 + q + 1)$, $q^2 + q + 1$, $q(q^2 + q + 1)$, $q(q + 1)$, $q^3$, $1$. See e.g. G.Hiss quoted above section 3.3.6 page 28. REPLY [5 votes]: As Jim has noted, for $\mathrm{GL}_n$ the formulas were given long ago by Green, Springer, and Macdonald and it was shown that they are polynomials in q with integer coefficients. Since the question asked for an explicit formula, I will now produce their formula. First, let us discuss the parameterisation of the irreducible representations. Let $\Phi$ denote the orbits of the Frobenius automorphism $x\mapsto x^q$ on $\overline{\mathbb{F}_q}^\times$. For an element $\gamma\in \overline{\mathbb{F}_q}^\times$, we let $\{\gamma\}$ denote the corresponding orbit, and let $d(\gamma)$ denote the size of this orbit. Let $\mathcal{P}$ denote the set of partitions. The irreducible complex representations of $G_n:=\mathrm{GL}_n(\mathbb{F}_q)$ are parameterised by the set of maps $$\Lambda: \Phi\rightarrow \mathcal{P}$$ such that $\Lambda$ is constant on the orbits and the sum of all the partitions in the image of $\Lambda$ equals $n$. We let $d(\Lambda)$ denote the dimension of the corresponding irreducible representation. Then showing $d(\lambda)\in \mathbb{Z}[q]$ is more or less equivalent to showing that $|G_n|/d(\lambda)\in \mathbb{Z}[q]$. (One can use Gauss's Lemma, primitive polynomials, etc.). Now we state the formula for $|G_n|/d(\lambda)$ and explain why it is a polynomial in $\mathbb{Z}[q]$. For each partition $\lambda$, define the normalised Hook polynomial of $\lambda$ by $$ H_\lambda(q):=q^{-\frac{1}{2}<\lambda, \lambda>} \prod(q^h-1). $$ Here, the product is taken over the boxes in the Young diagram of $\lambda$ and $h$ is the hook length of the box. Moreover, $<\lambda, \lambda>:=\sum_{i} (\lambda'_i)^2$, where $\lambda'$ is the dual partition. Next, we defined the normalised hook polynomial of $\Lambda$. Let $$ m_{\lambda, d}:=\#\{\{\gamma\} \, | \, d(\gamma)=d, \Lambda(\gamma)=\lambda\} $$ and set $$H_\Lambda(q):= \prod_{d,\lambda} H_\lambda(q^d)^{m_{d,\lambda}}. $$ Finally, we have the formula $$|G_n|/d(\Lambda')=(-1)^n q^{\frac{1}{2}n^2} H_{\Lambda},$$ where $\Lambda'$ is obtained from $\Lambda$ by dualizing all the partitions; see also this paper, equation 3.1.5. The only thing remaining is to see why this expression, which is seemingly a polynomial in $q^{\frac{1}{2}}$, is actually a polynomial in $q$. This boils down to showing that $$ n^2- \sum_{d,\lambda} m_{\lambda,d} d\sum_{i}(\lambda_i')^2 $$ is even. But note that since a partition and its dual have the same size, we have $$ n=|\Lambda|=\sum_{d,\Lambda'} m_{d,\lambda} d|\lambda| = \sum_{d,\lambda} m_{d,\lambda} d (\sum_i \lambda_i'). $$ Thus, modulo $2$, we have $$ n^2- \sum_{d,\lambda} m_{\lambda,d} d\sum_{i}(\lambda_i')^2 = \sum_{d,\lambda} \left((m_{d,\lambda} d)^2 (\sum_i (\lambda_i')^2) - (m_{\lambda,d} d)\sum_{i}(\lambda_i')^2\right)=0, $$ where the last equality follows from the fact that $x^2-x$ is always even.<|endoftext|> TITLE: A sum involving Euler totient function QUESTION [6 upvotes]: I wanted to know if we know any asymptotic formula/bound for the following sum $$ \sum_{1 \leq a < n \\(a,n) = 1 } \phi(a) \ .$$ A trivial upper bound could be $$ \sum_{a=1}^{n} \phi(a) = \frac{3}{\pi^2} n^2 + O(n^{1+\epsilon}), $$ but I wanted to know if we could do better than that. Thanks in advance for the help! References https://en.wikipedia.org/wiki/Euler%27s_totient_function#Other_formulae REPLY [12 votes]: The answer here is $$\frac{3}{\pi^2}n^2\prod_{p\mid n} \frac{p}{p+1}+O(nd(n)\log\log n+ n\log n),$$ where $d(n)$ is a divisor function and product is over prime values of $p$. To prove this, note first that $$\sum_{\substack{a\leq n \\ (a,n)=1}} \varphi(a)=\sum_{a \leq n} \varphi(a)\sum_{d \mid (a,n)} \mu(d)=\sum_{d \mid n} \mu(d)\sum_{bd\leq n} \varphi(bd).$$ Now, using the identity $$\varphi(a)=a\sum_{d \mid a}\frac{\mu(d)}{d}$$ we get $$\sum_{bd \leq n} \varphi(bd)=\sum_{b \leq \frac{n}{d}}bd\sum_{c\mid bd} \frac{\mu(c)}{c}.$$ Changing order of summation, we obtain $$\sum_{bd \leq n}\varphi(bd)=d\sum_{c \leq n}\frac{\mu(c)}{c}\sum_{\substack{b \leq n/d \\ \frac{c}{(c,d)} \mid b}}b.$$ As for any $x$ and $y$ the equality $$\sum_{\substack{b \leq x \\ y \mid b}} b=\frac{x^2}{2y}+O(x)$$ holds, the inner sum equals $$\sum_{\substack{b \leq n/d \\ \frac{c}{(c,d)} \mid b}}b=\frac{n^2(c,d)}{2cd^2}+O(\frac{n}{d}).$$ Therefore, $$\sum_{bd \leq n} \varphi(bd)=\frac{n^2}{d}\sum_{c \leq n} \frac{\mu(c)(c,d)}{2c^2}+O(n\log n).$$ Let us now compute the sum in the RHS of this identity: $$\sum_{c \leq n} \frac{\mu(c)(c,d)}{c^2}=\sum_{c=1}^{+\infty} \frac{\mu(c)(c,d)}{c^2}-\sum_{c>n} \frac{\mu(c)(c,d)}{c^2}=S_1-S_2$$ (both series are obviously convergent) Now we should compute $S_1$ and estimate $S_2$. Note that the function $\frac{\mu(c)(c,d)}{c^2}$ is multiplicative due to the multiplicativity of $\mu(c), c^2$ and $(c,d)$. Thus, $$S_1=\prod_p (1-\frac{(c,p)}{p^2})=\prod_p (1-1/p^2)\prod_{p \mid d}\frac{p}{p+1}=\frac{6}{\pi^2}\prod_{p \mid d} \frac{p}{p+1}.$$ To obtain the estimate for the $S_2$, observe that $$|S_2|\leq \sum_{c>n} \frac{(c,d)}{c^2}\leq \sum_{t \mid d} \sum_{ft>n} \frac{t}{(ft)^2}=O\left(\frac{d(n)}{n}\right).$$ So, $$\sum_{bd \leq n} \varphi(bd)=\frac{3n^2}{\pi^2d}\prod_{p \mid d} \frac{p}{p+1}+O(nd(n)/d+n\log n).$$ Summing this over all $d$, we finally get $$\sum_{\substack{a\leq n \\ (a,n)=1}} \varphi(a)=\frac{3n^2}{\pi^2}\sum_{d \mid n} \frac{\mu(d)}{d}\prod_{p \mid d} \frac{p}{p+1}+O(nd(n)\log\log n+n\log n)=\frac{3n^2}{\pi^2}f(n)+O(nd(n)\log\log n+\log n).$$ To get a bit more closed-form expression for $f(n)$ note that $f$ is again multiplicative and for any prime $p$ and positive $\alpha$ we have $$f(p^\alpha)=\frac{\mu(1)}{1}+\frac{\mu(p)}{p}\frac{p}{p+1}=1-\frac{1}{p+1}=\frac{p}{p+1},$$ therefore $$f(n)=\prod_{p \mid n} \frac{p}{p+1},$$ as needed.<|endoftext|> TITLE: Complexity of induction formulas in proof theoretic ordinals QUESTION [10 upvotes]: According to The Art of Ordinal Analysis, the proof theoretic ordinal of a theory $T$ is the least ordinal $\alpha$ such that: $${\bf ERA}+TI(\alpha,ECP)\vdash Con(T)$$ In above definition, $ECP$ stands for Elementary computable predicates and $TI(\alpha, A)$ stands for transfinite induction up to $\alpha$ for predicates in complexity class $A$. Q. Is it possible to reduce the complexity of predicates for transfinite induction in above definition to a smaller complexity class? For example, Is $Con(T)$ provable from ${\bf ERA}+TI(\beta,P)$ for some ordinal $\beta$ ?($P$ stands for polynomial time predicates.) REPLY [3 votes]: By a padding argument, for reasonable notation systems, an elementary time computable predicate $P$ in $\mathrm{TI}(β,ECP)$ can be chosen to be polynomial time computable. For example, for limit $α<β$, set $P'(α+(2^n+1) 2^{\mathrm{code}(α)}) ⇔ P(α+n)$ with $P'$ true for ordinals that are not in that form ('+' refers to ordinal addition; n∈ℕ). With this padding, we get an order preserving bijection between counterexamples to $P$ and counterexamples to $P'$. We have $\mathrm{ERA}⊢\mathrm{TI}(β,P')⇔\mathrm{TI}(β,P)$, and if $P$ is exponential time computable, then $P'$ is polynomial time computable (as a predicate on codes for ordinals using the notation system in the $\mathrm{TI}$). A caveat is that we require the coding of $α+n$ to be well-behaved in relation to the coding of $α$, but this is satisfied by reasonable notation systems. Note, however, that there are recursive ordinal notation systems where, for example, $α→α+1$ is not recursive. Also, the reason elementary time computability was used over polynomial time computability is that reasonable representations of ordinals appear to be elementary time isomorphic (at least for ordinals for which ordinal analysis is well-understood), but polynomial time reducibility distinguishes between, for example, unary and binary numbers. Also, the form in the question is just one of a number of different ways to define the proof ordinal of a theory. Since your definition uses Con(T), it is a form of $Π^0_1$ ordinals, but like $Π^0_2$ ordinals and unlike a more fine-grained notion of $Π^0_1$ ordinals, it does not distinguish between say PA and PA+Con(PA). For 'natural' theories (and reasonable ordinal representations), the different definitions lead to the same ordinal, but that is not the case in general. An Extension There are 'pathological' representations of $ε_0$ such that (1) ordinal comparison is exponential time computable, (2) ERA constructively proves that the representation is elementary time equivalent to Cantor Normal Form, and (3) ERA proves transfinite induction for polynomial time predicates. Essentially, given a polynomial time $P$ that holds for some $n > 2 \, \mathrm{code}(P)$, make sure that the least such $n$ codes a finite ordinal, with comparison of finite ordinals agreeing with comparison of their codes. However, if ERA constructively proves that $ωβ=β$ (using an order-preserving computable injection $β×ω→β$) and that ordinal comparison is polynomial time computable, then $\mathrm{ERA}⊢\mathrm{TI}(β,Π^0_1) ⇔ \mathrm{TI}(β,\mathrm{P})$ (P (without italics) means polynomial time). Essentially (using a slightly stronger assumption), given a $Π^0_1$ $P$, if $¬P(α)$, then set $¬P'(α')$ for every $α'$ with $ωα≤α'<ω(α+1)$ with $\mathrm{code}(α')$ sufficiently large for us to have time to refute $P(α)$ (and to compute $ωα$ and $ω(α+1)$).<|endoftext|> TITLE: Long identity-free sequences of permutations QUESTION [5 upvotes]: Let $N = [n]$ and for any subset $A \subseteq N$, let $S_A$ denote the subgroup of the symmetric group $S_n$ that fixes all objects outside $A$. Say that a sequence $A_1, \dots, A_k \subseteq N$ is "identity free" if the equivalence $$s_1 s_2 \dots s_k = \mathop{id} \qquad \text{where } s_i \in S_{A_i} \text{ for all } i$$ has no solutions, except for the trivial one where $s_1 = s_2 = \dots = s_k = \mathop{id}$. Have these been studied under any name that I can search for? While I would be interested in any discussion at all, I am particularly interested in the question of the extremal length $L$ of the longest identity-free sequence when all $A_i$ are constrained to have some fixed size $\alpha$. The only upper bounds on $L$ I have been able to observe so far come from the easy observation that no two $A_i$ may overlap on more than one element (and then apply Cauchy-Schwarz). REPLY [5 votes]: Construct a bipartite graph $G$ where one part is $[n]$ and the other is $[k]$ such that there is an edge between $i\in[n]$ and $j\in[k]$ iff $i\in A_j$. Then $A_1, \dots, A_k$ are identity free if $G$ is acyclic. The converse may not hold though (as pointed out by Jan Kyncl). If $|A_j|=\alpha>1$ for all $j\in[k]$, then $G$ has $\alpha k$ edges. In this case, $G$ may be acyclic only if $\alpha k \leq n+k-1$, i.e. $$k \leq \frac{n-1}{\alpha-1}.$$ This inequality does not have to hold if $G$ is not acyclic.<|endoftext|> TITLE: Unique words in dihedral groups QUESTION [11 upvotes]: Suppose $x$ is a word over the alphabet $\{0,1\}$. Let $a$, $b$ be elements of the group Dih$_k$ for some $k$. Let $\varphi=\varphi_{a,b,k}$ be the map from words over $\{0,1\}$ to elements of the dihedral group Dih$_k$ (having $2k$ elements) such that $\varphi(0)=a$, $\varphi(1)=b$, and $\varphi$ takes concatenation to multiplication: $\varphi(xy)=\varphi(x)\varphi(y)$. Let's say that $x$ is dihedrally simple if there is some $\varphi_{a,b,k}$ such that $\varphi(x)\ne\varphi(y)$ for all $y\ne x$ of the same length as $x$. Computer experimentation suggests the Conjecture: The dihedrally simple words form a regular language, namely $S\cup T$ where $$ S=\bigcup_{n=0}^\infty \{0^n,\quad 0^{n-1}1,\quad (01)^{n/2},\quad 01^{n-1},\quad 01^{n-2}0\} $$ where $(01)^{t+\frac12}=(01)^t0$, and $T$ is obtained from $S$ by interchanging 0 and 1. My question is: Is the Conjecture similar to anything in the literature? Or do you see how to prove it? Edit: to answer @LucGuyot's question below: this definition arises in my model of quantum security from a recent UCNC'17 paper (see also arXiv version) except that there is an additional constraint there, that we map the locked state $|0\rangle$ to the unlocked state $|1\rangle$. The relevance of dihedral groups is that Dih$_k$ is representable as a subgroup of the projective unitary group $\mathrm{PU}(2, \mathbb C)$. The interpretation then is that $x$ is a secret code which should be punched into a quantum device with buttons labeled 0 and 1 and which trigger unitary operations $U_A$, $U_B$. Any code of the same length as $x$ but different from $x$ will not have the same effect (say, unlocking the device). Any attempt to inspect the state of the device during entering of the code, as one might do with a simple padlock, will constitute a measurement of the device and therefore reset the quantum superposition. REPLY [9 votes]: The conjecture holds true and I don't know of anything similar. Given a word $w$ over $\{0, 1\}$, we denote by $\overline{w}$ the word obtained from $w$ by interchanging $0$ and $1$. Let us show first that the elements of $S \cup T$, where $T = \overline{S}$, are dihedrally simple. Claim 1. A word $w$ over $\{0, 1\}$ is dihedrally simple if and only if $\overline{w}$ is. Proof. Consider $\overline{\varphi}_{a,b,k} \Doteq \varphi_{b, a, k}$. Because of Claim 1, it suffices to show Claim 2. The elements of $S$ are dihedrally simple. The following will come soon in handy. Lemma 1. Let $a \in \text{Dih}_k$ be either central or of order $k > 2$ and let $b \in \text{Dih}_k$ be a non-central element of order $2$. Let $w$ be a word over $\{0, 1\}$ of length $n$. If $w$ contains at least two $1$'s (resp. more than two $1$'s), then $\varphi_{a, b, k}(w)$ is either of the form $a^i$ or $a^ib$ with $\vert i \vert \le n - 2$ (resp. $\vert i \vert < n - 2$). If $w$ is of the form $0^i 1 0^j$, then $\varphi_{a, b, k}(w) = a^{i - j}b$. Proof. Write $w = 0^{i_1}1^{j_1} \cdots 0^{i_s}1^{j_s}$ with $i_t, j_t \ge 0$. It follows from the identities $b^2 = 1$ and $bab = a^{-1}$ that $\varphi_{a, b, k}(w) = a^{\sum_{t = 1}^s \pm i_t}b^{\varepsilon}$ with $\varepsilon \in \{0, 1\}$ and $\varepsilon \equiv \sum_{t = 1}^s j_t\mod 2$. If $w$ contains at least two $1$'s (resp. more than two $1$'s), then $\sum_{t = 1}^s i_t \le n - 2$ (resp. $\sum_{t = 1}^s i_t < n - 2$), which implies the first statement. Consider now the second statement with $w = 0^i 1 0^j$. Then $\varphi_{a, b, k}(w) = a^iba^j = a^{i - j}b$. We are now in position to prove Claim 2. Proof of Claim 2. If $w \in S$ is of length $n = 0$, then $w$ is certainly dihedrally simple. Let us assume that $n > 0$ and let $k > 4n$. Let $a \in \text{Dih}_k$ be an element of order $k$ and let $b \in \text{Dih}_k$ be a non-central element of order $2$. Then $\varphi_{a,b,k}(0^n) = a^n$. Since $\text{Dih}_k = \langle a \rangle \rtimes \langle b \rangle$, it follows from Lemma 1 that the image by $\varphi_{a,b,k}$ of any other word of length $n$ is distinct from $a^n$. Thus $0^n$ is dihedrally simple. Since $\varphi_{a,b,k}(0^{n - 1}1) = a^{n - 1}b$, it also follows from Lemma 1 that $0^{n - 1}1$ dihedrally simple. The case of $01^{n - 1}$ is very similar but relies on $\overline{\varphi}_{a,b,k}$ instead; we will omit it. Let us now consider $\varphi_{a,b,k}(10^{n -2}1) = a^{-(n - 2)}$. We infer from Lemma 1 that $10^{n -2}1$, and hence $01^{n -2}0$, is dihedrally simple. Eventually, let us consider $\varphi_{b, ba, k}((01)^{n/2})$. This is $a^{n/2}$ if $n$ is even and $a^{(n - 1)/2}b$ if $n$ is odd. Since $b^2 = (ba)^2 = 1$, we have $\varphi_{b, ba, k}(00) = \varphi_{b, ba, k}(11) = 1$. Therefore the image of any word of length $n$ by $\varphi_{b, ba, k}$ is the image of a word of the form $(01)^{i/2}$ or $(10)^{i/2}$ with $0 \le i \le n$, that is, a group element $g$ of the form $a^{\lfloor i/2 \rfloor}b^{\varepsilon}$ or respectively $a^{-\lfloor i/2 \rfloor}(ba)^{\varepsilon}$ with $\varepsilon \in \{0, 1\}$. It is easily checked that such an element $g$ coincides with $\varphi_{b, ba, k}((01)^{n/2})$ if and only if $i = n$, which shows that $(01)^{n/2}$ is dihedrally simple. The following claim will settle the conjecture. Claim 3. Let $w$ be a dihedrally simple word. Then $w \in S \cup T$. Proof. Let $w$ be dihedrally simple word of length $n$. Since $S \cup T$ contains all words of length at most three, we can assume that $n \ge 4$. By definition, we can find $k \ge 2$, $a,b \in \text{Dih}_k$ such that $\varphi_{a, b, k}(w) \neq \varphi_{a, b, k}(v)$ for every word $v \neq w$ of length $n$. We can split our reasoning into three cases. The group elements $a$ and $b$ commute. If $w$ contains a subword of the form $01$ or $10$, then we can interchange these two subwords, producing a word $v \neq w$ such that $\varphi_{a, b, k}(v) = \varphi_{a, b, k}(w)$, a contradiction. Therefore $w$ is of the form $0^n$ or $1^n$, and hence lies in $S \cup T$. The group elements $a$ and $b$ are non-commuting elements of order $2$. We have then $\varphi_{a, b, k}(00) = \varphi_{a, b, k}(11) = 1$. Thus $w$ cannot contain any of the subword $00$ or $11$, since interchanging them would yield a contradiction. Therefore $w$ is of the form $(01)^{n/2}$ or $(10)^{n/2}.$ Thus $w$ lies in $S \cup T$. The group elements $a$ and $b$ are non-commuting elements of order $2$ and $k > 2$ respectively. We have then $\varphi_{a, b, k}(00) = 1$. If $w$ contains at least one $1$, i.e., $w = w'1w''$, then it cannot contain a subword of the form $00$, since otherwise moving this subword from $w'$ to $w''$, or vice versa, would yield a word $v \neq w$ with the same image. As we also have $\varphi_{a, b, k}(101) = \varphi_{a, b, k}(000)$, the word $w$ cannot contain a subword of the form $101$. As a result, $w$ is of the form $0^n$, or $1^{n - 1}0$, or $01^{n - 2}0$. Therefore $w$ lies in $S \cup T$. The very last case consists in interchanging the orders of $a$ and $b$, but this is too similar to case (3).<|endoftext|> TITLE: Curious decomposition between two sets QUESTION [10 upvotes]: Suppose $X,Y$ are sets, and $f:X\to Y$ and $g: Y \to X$. Then there are disjoint subsets $X_1,X_2 \subseteq X$ with $X_1\cup X_2= X$ and disjoint subsets $Y_1,Y_2 \subseteq Y$ with $Y_1\cup Y_2= Y$ such that $f(X_1) = Y_1$, and $g(Y_2) = X_2$. (This curious result is a consequence of the Knaster-Tarski fixed point theorem.) Can this statement be generalized to partial functions, or even binary relations $R_1, R_2 \subseteq X\times Y$? REPLY [5 votes]: Yes, I think the exact same proof goes through for binary relations $R: X \nrightarrow Y$ and $S: Y \nrightarrow X$ (which of course includes the partial function case). Each induces a monotone operation between power sets, e.g. $\exists R: PX \to PY$ takes $A \subseteq X$ to $\{y: \exists_{x \in A} R(x, y)\}$. Letting $\neg_X$ denote complementation on subsets $A \subseteq X$, we obtain a covariant (i.e. monotone) operation $$\neg_X \circ \exists S \circ \neg_Y \circ \exists R: PX \to PX$$ which by the Knaster-Tarski theorem has a fixed point $X_1 \in PX$. Put $Y_1 = (\exists R)(X_1)$ and $Y_2 = \neg_Y Y_1$ and $X_2 = (\exists S)(Y_2)$. Then $X_2 = \neg_X X_1$ by the fixed point equation.<|endoftext|> TITLE: Why do we use hexagons in percolation? QUESTION [12 upvotes]: In some cases, hexagons are used in percolation. Why do we use hexagons in percolation? REPLY [13 votes]: Site percolation on the triangular lattice (or, equivalently, face percolation on the hexagonal lattice) is the only case for which conformal invariance of percolation at criticality has been proven. This has to do with very special combinatorial properties of the lattice. For a detailed explanation of this, see this paper by Vincent Beffara.<|endoftext|> TITLE: Is a model of set theory determined by the Cohen reals over it? QUESTION [19 upvotes]: This question concerns the amount of information about a model $M$ that is contained in the collection of all reals Cohen over $M$. Specifically, let $M$ and $N$ be countable transitive models of ZFC and suppose that they have the same collection of Cohen reals, i.e. any real $c\in V$ is Cohen over one of them iff it is Cohen over the other one. Does this tell us anything about $M$ and $N$? Naively one might hope to get that $M=N$, but this is far too much. Already if $M$ and $N$ merely agree on $\mathcal{P}(\omega)$, then they will share the same Cohen reals, since the dense subsets of the Cohen poset are effectively coded by reals. So we can get easy examples of models $M\neq N$ which share the same Cohen reals, e.g. by letting $N$ be a sufficiently closed forcing extension of $M$. This tells us that we cannot hope to recover information about $M$ and $N$ beyond their reals. But can we recover that? If $M$ and $N$ share their Cohen reals, is $\mathcal{P}(\omega)^M=\mathcal{P}(\omega)^N$? REPLY [14 votes]: Unless I'm missing something, don't you get a counterexample if $N=M[s]$ where $s$ is a Sacks real over $M$? The point is that every dense open subset of $2^{<\omega}$ in a Sacks extension includes such a dense open set in the ground model and so a Cohen real over $M$ remains Cohen over $M[s]$. This is a standard fusion argument. One can proceed as follows: Lemma: Suppose $p\Vdash``\text{$\dot D$ is dense open in $2^{<\omega}$}"$, $n<\omega$, and $\sigma\in 2^{<\omega}$. Then there are $q\leq p$ and $\tau\in 2^{<\omega}$ such that $\tau\supseteq \sigma$, $q\upharpoonright\vphantom{B}^n2= p \upharpoonright\vphantom{B}^{n}2$, and $q\Vdash \check{\tau}\in\dot D$ proof: Let $\{a_i:i TITLE: Equivariant cohomology ring is an integer domain QUESTION [5 upvotes]: Let $G$ be a connected compact Lie group and let $V$ be a complex $G$-representation. Denote by $\mathbb{P}(V)$ the projectivization of the vector space $V$. I would like to ask a couple of questions about the equivariant cohomology ring $H^*_G(\mathbb{P}(V),\mathbb{Q})$. Under what conditions on the representation $V$ the ring $H^*_G(\mathbb{P}(V),\mathbb{Q})$ is an integer domain? The second question is more delicate. Suppose that $i\colon X \to \mathbb{P}(V)$ is a closed $G$-equivariant embedding of a $G$-submanifold $X$ (for example, $X$ is a closed $G$-orbit in $\mathbb{P}(V)$). Denote by $i_!1\in H^*_G(\mathbb{P}(V),\mathbb{Q})$ the image of $1\in H^0_G(X,\mathbb{Q})$ under the equivariant pushforward map $i_!\colon H^*_G(X,\mathbb{Q}) \to H^*_G(\mathbb{P}(V),\mathbb{Q})$. I would like to know under what conditions the cohomology class $i_!1$ is not a zero divisor in the ring $H^*_G(\mathbb{P}(V),\mathbb{Q})$. REPLY [7 votes]: It is very rare for these rings to be integral domains. To see this, put $$ f_V(t)=\sum_kc_k(V)t^{\dim(V)-k} \in H^*(BG)[t]. $$ (All cohomology here has rational coefficients.) It is then standard that $H_G^*(PV)=H^*(BG)[x]/f_V(x)$, and also that $f_{V\oplus W}(t)=f_V(t)f_W(t)$. From this it is clear that $H_G^*(PV)$ can only be a domain if $V$ is irreducible. However, irreducibility is far from being sufficient. To see this, let $T$ be a maximal torus with Weyl group $W$. By a "fake representation" of $G$ I will mean a $W$-invariant representation of $T$. The $W$-orbits in $T^*$ give a basis over $\mathbb{N}$ for the fake representations. Because $H^*(BG)=H^*(BT)^W$, a fake representation $V$ still has an associated polynomial $f_V(t)\in H^*(BG)[t]$. Each fake representation is the restriction of a virtual representation of $G$, but not usually a genuine representation of $G$. If a genuine representation $V$ can be decomposed nontrivially as a sum of fake representations, then we get a factorisation of $f_V(t)$, showing that $H^*_G(PV)$ is not a domain. I think it is almost always possible to find such a fake decomposition. To find the list of exceptions is probably an easy exercise for someone more fluent in Lie theory than I am.<|endoftext|> TITLE: For which exact couples do associated spectral sequences degenerate at $E_1$? QUESTION [6 upvotes]: It is well known that a bigraded exact couple of objects of an abelian category yields a spectral sequence (cf. https://ncatlab.org/nlab/show/exact+couple#SpectralSequencesFromExactCouples). My question is:under which conditions does this spectral sequence degenerate at $E_1$ (actually, I am more interested in necessary conditions)? This condition appears to be equivalent to the image of the morphism $f_1:E_1\to D_1$ lying inside all levels of the filtration of $D_1$ by $g_1^i(D_1)$ (here I ignore the upper indices, and $g_1^i$ denotes the $i$th iterate of $g_1:D_1\to D_1$). Is this true; are there any references for this fact? Actually, I would like to conclude that $f_1=0$; does this follow from the degeneration at $E_1$? I am more interested in the bounded case; so an answer to the first part of my question is sufficient for my purposes; yet are there any other conditions ensuring that $f_1=0$ whenever the spectral sequence degenerates at $E_1$? REPLY [3 votes]: Degenerating at $E_1$ is, as you describe, equivalent to the image of the map $f_1: D_1 \to E_1$ being contained in the image of $g_1^i$ for all $i$. Roughly, this is because the definition of the $d_r$-differential on a class $x$ is the equivalence class of any element $h_1 g_1^{1-r} f_1(x)$, where $h_1$ is the map $D_1 \to E_1$. If the spectral sequence degenerates, then $E_1 = E_r$ for all $r$ and so we don't need to worry about equivalence classes: then this asks that $g_1^{1-r} f_1(x)$ is in the kernel of $h_1$, hence the image of $g_1$. Then we repeat. I do not know a place in the literature where this is explicitly stated. It is not necessarily the case that $f_1 = 0$ if the spectral sequence degenerates. For example, if the map $h_1$ is zero so that $0 \to E_1 \xrightarrow{f_1} D_1 \to D_1 \to 0$ is an exact sequence, the spectral sequence still degenerates at the $E_1$-page. This is not uncommon behavior for degenerating spectral sequences in cohomology, rather than homology.<|endoftext|> TITLE: Intuition for Szabo's geometric spectral sequence QUESTION [8 upvotes]: In https://arxiv.org/abs/1010.4252, Szabo defines a link invariant $\hat{H}(L)$ which can be computed combinatorially from a link diagram and shows that there is a spectral sequence from Khovanov homology to $\hat{H}(L)$. Conjecturally, this spectral sequence is isomorphic to the spectral sequence (proved by Ozsvath-Szabo) from Khovanov homology to $\hat{HF}(\Sigma(L))$ (the hat version Heegaard-Floer homology for the double branched cover). In particular, one should have $\hat{H}(L)\cong \hat{HF}(\Sigma(L))$. To define $\hat{H}(L)$, Szabo divides the resolution diagrams into many types and writes down formula for the differential explicitly, case by case. While it is very concrete, I do not see how he obtains these formulas. I understand that he tries to add in differentials corresponding to higher dimensional faces of the resolution cube. But it seems that there are some deeper motivation to help him finding the right formula for these higher differentials. He does not explain much about these motivations in the paper. My question is: Any one could explain the intuition underlying Szabo's construction? REPLY [9 votes]: The first intuition is that the differentials should depend on a choice of orientation of the surgery arcs (the first differential eventually turns out to be independent, anyways). Let's do a split, with a single circle $C$ splitting into two, $C_1$ and $C_2$, by an elementary saddle $S$. Then the monopoles for $C$, $C_1\cup C_2$, and $S$ are $M(C)=S^1$, $M(C_1\cup C_2)=S^1\times S^1$ and $M(S)=S^1$ respectively, and the restriction maps are $M(S)\to M(C)$ is identity and $M(S)\to M(C_1\cup C_2)$ is the diagonal map. The split map differential $d_1\colon H_*(M(C))\to H_*(M(C_1\cup C_2))$ is the pull-push map; at the space level, the pull-push map is not cellular (it is the diagonal embedding), so it you want to get a cellular chain level map, you need to perturb the map off the diagonal. So you need to choose whether you are moving the diagonal towards the $S^1$ factor corresponding to $C_1$ or $C_2$, and that is the choice of an orientation of the surgery arc joining $C_1$ to $C_2$. (Of course the $d_1$ differential doesn't depend on how you chose your perturbation, but you did choose one. If $1$ and $x$ are the top (1-dim) and bottom (0-dim) generators, then the split map becomes $x\mapsto x\otimes x, 1\mapsto 1\otimes x+x\otimes 1$.) The second intuition is what should be homotopy $H$ relating these different choices of orientations (perturbations)? Once again, for the case of a split, there are two cellular perturbations of the diagonal $S^1$ inside $S^1\times S^1$. These two perturbations are homotopic, and the homotopy is the map $1\mapsto 1\otimes 1$. Now you combine these two facts, and try to come up a second differential $d_2$ in bigrading $(2,2)$. It should satisfy the following: $d_2$ depends on the orientations of the surgery arcs. $[d_1:d_2]=0$. $d_2^2$ is null-homotopic. And if we change the orientation of the arcs at a single crossing $c$, then $d_2$ changes as: $d'_2-d_2=[H_c:d_1]$, where $H_c$ is the above homotopy for the crossing $c$. And you will come up with Szabo's $d_2$. And then you proceed. $d_3$ is forced to satisfy $d_3^2=[d_1:d_2]$, so try to come up with such as a $d_3$, and you will get Szabo's $d_3$. And if you are Szabo, at this point, you figure out the pattern.<|endoftext|> TITLE: Largest number of points one can pick in finite projective space without getting three on a line QUESTION [5 upvotes]: Consider the projectivization $\mathbb P\mathbb F_p^n$ of $\mathbb F_p^n$. How large a set $B \subseteq \mathbb P \mathbb F_p^n$ can I pick so that no three points of $B$ lie on the same line? REPLY [10 votes]: The term for such sets is "caps". The problem you ask was posed by Bose ("Mathematical theory of the symmetrical factorial design", Sankhyā 8 (1947) 107–166), and is important in relation to coding theory: see Hill, A first course in coding theory (1986), around figure 14.9. At least at the time of Hill's book from which the following values are taken, the maximal size was known just for $p=2$ (it is $2^n$), for $n=2$ (it is $p+1$ for $p$ odd, $p+2$ for $p$ a power of $2$), for $n=3$ (it is $p^2+1$) and in the further cases $(n,p)=(4,3)$ and $(n,p)=(5,3)$ (where it is $20$ and $56$ respectively). (Some of these results are due to Bose himself.) PS: It might also be worth while to point out that for $n=2$, caps are also known as "arcs", and a theorem of B. Segre states that (for $p$ odd), an arc of maximal size (also, "oval") is necessarily a conic.<|endoftext|> TITLE: For which $n$ is there a regular $n$-simplex with vertices in $\mathbb{Z}^n$? QUESTION [13 upvotes]: For which $n$ is there a regular $n$-simplex with vertices in $\mathbb{Z}^n$ (or equivalently, in $\mathbb{Q}^n$)? Some easy observations: Such an $n$-simplex exists with vertices in $\mathbb{Z}^{n+1}$: just take the $n+1$ points $(0, \ldots, 0, 1, 0, \ldots, 0)$. If $n$ is even and $n+1$ is not a perfect square, then no regular $n$-simplex with vertices in $\mathbb{Z}^n$ exists: on the one hand, the side length $x$ is the square root of an integer, and thus the volume, namely $\frac{\sqrt{n+1}}{n! 2^{n/2}} x^n$, is irrational; on the other hand the volume must be rational by the formula using determinants. For $n=3$ a regular tetrahedron with vertices in $\mathbb{Z}^3$ does exist: $\{(0,0,0),(1,1,0),(1,0,1),(0,1,1)\}$. If there exists a Hadamard matrix of order $n+1$, we can normalize it so the first column is all ones and then the remaining $n$ columns give the coordinates of the $n+1$ vertices of a regular $n$-simplex with vertices in $\{1,-1\}^n$. REPLY [8 votes]: As shown in the accepted answer to this question, the Hadamard matrix condition is necessary and sufficent, so you have answered your own question... EDIT As pointed out by Noam, I misread the linked-to question, and to atone: Robin Chapman gives the answer to the actual OP question, and here is his 1998 answer (slightly reformatted): This is a good problem. It reduces to a question in the theory of rational quadratic forms. Let's ask for which $n$ an $n$-simplex can be embedded $n$-space with integral coordinates. The answer is if and only if $n + 1$ is an odd square, or $n + 1$ is a sum of $2$ odd squares, or $n + 1 \equiv 0 \mod 4.$ It's equivalent to consider rational coordinates, as we can scale, and we can also translate to put one of the vertices at the origin. Let $v_1,\dots, v_n$ be the other vertices. Then for some rational number m we have $v_i\cdot v_i = 2m,$ and $v_i\cdot v_j = m$ for $i \neq j.$ This means that the quadratic forms $Q_1 = x_1^2 + x_2^2 + ... + x_n^2$ and $2m Q_2$ where $ Q_2 = x_1^2 + x_1 x_2 + x_1 x_3 + \dots + x_1 x_n + x_2^2 + x_2 x_3 + \dots + x_n^2$ are equivalent over the rationals. Indeed this is a necessary and sufficient condition. One can use the Hasse-Minkowski theory of rational quadratic forms to determine when an m exists so that this is the case, and doing so leads, after some effort, to the stated condition.<|endoftext|> TITLE: Conflating reals and sets of countable ordinals "nicely" QUESTION [8 upvotes]: It is consistent with ZFC that $2^{\aleph_1}=2^{\aleph_0}$. This can be gotten easily via forcing; more interestingly, it is a direct consequence of forcing axioms (which also set this value at $\aleph_2$). However, just because a bijection exists between the subsets of $\omega$ and the subsets of $\omega_1$ doesn't mean that such a bijection is "nice" in any sense. My question is, roughly: Can we ever have a "nice" injection from $\mathcal{P}(\omega_1)$ to $\mathcal{P}(\omega)$, or at least a "nice" surjection from $\mathcal{P}(\omega)$ to $\mathcal{P}(\omega_1)$? Exactly what "nice" here means is a bit up for grabs; my instinct at first is to lean on generalized descriptive set theory and ask for something like "Borel on $\omega_1^{\omega_1}$," but I'm not at all sure that's a good thing to do. Another take might be along the lines of the following: Is it consistent with ZFC that there is a parameter-freely-definable surjection from $\mathcal{P}(\omega)$ to $\mathcal{P}(\omega_1)$? We can also ask a "local" version of this question: Is it consistent with ZFC that for every $s\subset\omega_1$ there is a $r\subset\omega$ with $s\in L[r]$? While this wouldn't establish the existence of a "nice" correspondence, it would at least establish a more meaningful equivalence than follows immediately from them having the same cardinality. (Note, however, that this would imply that $\omega_1=\omega_1^r$ for some real $r$; since this is contradicted by large cardinals, this seems like it might be the wrong question to ask.) REPLY [6 votes]: The technique of almost disjoint forcing was introduced in MR0289291 (44 #6482). Jensen, R. B.; Solovay, R. M. Some applications of almost disjoint sets. In Mathematical Logic and Foundations of Set Theory (Proc. Internat. Colloq., Jerusalem, 1968), pp. 84–104, North-Holland, Amsterdam, 1970. Fix an almost disjoint family $X=(x_\alpha:\alpha<\omega_1)$ of subsets of $\omega$, so $x_\alpha\cap x_\beta$ is finite for all $\alpha<\beta<\omega_1$. Given $s\subseteq\omega_1$, the poset $\mathbb P_{X,s}$ adds a real $r$ such that $$s=\{\alpha<\omega_1:r\cap x_\alpha\mbox{ is infinite}\}.$$ This shows that $s\in L[X,r]$; if $X\in L$ then in fact $s\in L[r]$. The poset $\mathbb P_{X,s}$ is ccc, and it follows from standard arguments that if $\mathsf{MA}_{\omega_1}$ holds and $\omega_1=\omega_1^L$, then $\mathcal P(\omega_1)=\bigcup_{r\subseteq\omega}\mathcal P(\omega_1)^{L[r]}$. Moreover, by taking as $X$ the first almost disjoint family in the standard enumeration of $L$, we see that $X$ is definable without parameters, and this gives us a simple parameter-free injection of $\mathcal P(\omega_1)$ into $\mathcal P(\omega)$ as long as there is, for instance, a parameter-free definable well-ordering of $\mathcal P(\omega)$. This is consistent with $\mathsf{MA}_{\omega_1}+\omega_1=\omega_1^L$. (In fact, it is a consequence of $\mathsf{BPFA}+\omega_1=\omega_1^L$, see my paper with Friedman listed below and available at my page. For the consistency result just for $\mathsf{MA}$, this is a result of Friedman, see Theorem 8.51 in his class forcing book.) In general, in the context of forcing axioms, to get a nice well-ordering of $\mathcal P(\omega_1)$, it suffices to find a nice well-ordering of $\mathcal P(\omega)$ and apply almost disjoint forcing. If we further arrange that the almost disjoint family we begin with is sufficiently nice, this reflects as well in the complexity and/or parameters of the well-ordering of $\mathcal P(\omega_1)$. The same holds for a nice injection of $\mathcal P(\omega_1)$ into $\mathcal P(\omega)$ and similar variants. MR2895389 (2012m:03123). Caicedo, Andrés Eduardo; Friedman, Sy-David. $\mathsf{BPFA}$ and projective well-orderings of the reals. J. Symbolic Logic 76 (2011), no. 4, 1126–1136. I like the description of the poset $\mathbb P_{X,s}$ in Jech's book: A condition is a function $f\!:d_f\to2$ whose domain $d_f$ is a subset of $\omega$ such that $d_f\cap x_\alpha$ is finite for all $\alpha\in s$ and $\{n\in d_f: f(n)=1\}$ is finite. The order is inclusion. Note that if $p,q$ are incompatible conditions, then in particular $\{n:p(n)=1\}\ne\{n:q(n)=1\}$. This shows that $\mathbb P_{X,s}$ is actually Knaster. For any $p$ and any $\alpha\in\omega_1\smallsetminus s$, we can extend $p$ to a condition $q$ whose domain contains $x_\alpha$ and satisfies $q(n)=0$ for all $n\in x_\alpha\smallsetminus d_p$ (since the $x_\beta$ are almost disjoint). Similarly, for any $\alpha\in s$ and any $k\in\omega$, we can extend any condition to ensure that its domain meets $x_\alpha$ in a set of size at least $k$. From this, standard density arguments give us the result. By the way, it is consistent that $\mathsf{MM}$ holds and there is a parameter-free well-ordering of $\mathcal P(\omega_1)$, which naturally gives us a nice injection of $\mathcal P(\omega_1)$ into $\mathcal P(\omega)$. (I am using "nice" loosely here, but the definition of the well-ordering, and therefore of the injection, is actually not too bad.) See MR2474445 (2009k:03085). Larson, Paul B. Martin's maximum and definability in $H(\aleph_2)$. Ann. Pure Appl. Logic 156 (2008), no. 1, 110–122.<|endoftext|> TITLE: How are motives related to anabelian geometry and Galois-Teichmuller theory? QUESTION [23 upvotes]: In Recoltes et Semailles, Grothendieck remarks that the theory of motives is related to anabelian geometry and Galois-Teichmuller theory. My understanding of these subjects is not very solid at this moment, but this is what I understand: Anabelian geometry tries to ask how much information about a variety is contained in its etale fundamental group. In particular, there exist "anabelian varieties" which should be completely determined by the etale fundamental group (up to isomorphism). The determination of these anabelian varieties is currently ongoing. Galois-Teichmuller theory tries to understand the absolute Galois group $\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ in terms of the automorphisms of the "Teichmuller tower", which is constructed as follows. We begin with the moduli stacks of curves with genus $g$ and $\nu$ marked points. These moduli stacks $\mathcal{M}_{g,\nu}$ have homomorphisms to each other, which correspond to "erasing" marked points and "gluing". The Teichmuller tower $\hat{T}_{g,\nu}$ comes from the profinite fundamental groupoids of these moduli stacks. The theory of motives is some sort of "universal cohomology theory" in the sense that any Weil cohomology theory (which is a functor from smooth projective varieties to graded algebras over a field) factors through it. This is obtained from some process of "linearization" of algebraic varieties (considering correspondences as morphisms, followed by the process of "passing to the pseudo-abelian envelope", and formally inverting the Lefschetz motive). Related to the theory of motives is the concept of a Tannakian category, which provides a kind of higher-dimensional analogue of Galois theory. I think the category of motives is conjectured to be a Tannakian category, via Grothendieck's standard conjectures on algebraic cycles (please correct me if I am wrong about this). So I'm guessing Tannakian categories might provide the link between the theory of motives and anabelian geometry and Galois-Teichmuller theory (which are both related to Galois theory) that Grothendieck was talking about in Recoltes et Semailles, but I'm not really sure. Either way, the ideas are still not very clear to me, and I'd like to understand the connections more explicitly. REPLY [24 votes]: There's a very deep connection between motives and Grothendieck-Teichmüller theory but it isn't well-understood yet. I can't even frame it precisely in higher genus, but at least I can frame a precise conjecture in genus zero. It has to do with motives that are connected to periods of moduli spaces on the one hand, and Grothendieck-Teichmüller being connected to automorphisms of fundamental groups of moduli spaces on the other. So on the one hand, we have the Tannakian category $MTM$ of mixed Tate motives over Z. Goncharov and Manin gave a construction showing how to construct motivic multiple zeta values from the cohomology of the moduli spaces $M_{0,n}$ of genus zero curves with $n$ marked points, giving rise to a Tannakian subcategory of MTM, and F. Brown subsequently proved that the motivic multiple zeta category actually fills all of $MTM$. Take the Tannakian fundamental group of this category $MTM$, and take its (pro)-unipotent radical. The associated graded Lie algebra is known to be freely generated by one generator in each odd weight. Now on the other hand, take the Grothendieck-Teichmüller group, which can be identified with the outer automorphism group of the tower of all the fundamental groups of the moduli spaces $M_{0,n}$ that commute (up to inner automorphisms) with certain standard maps between these fundamental groups coming from erasing points or subsurface inclusion (or doubling the braid strands, if you like to think of the fundamental group of $M_{0,n}$ as a braid group). This is a priori a profinite group, but it has a pro-unipotent version, which has an associated Lie algebra that is isomorphic to the graded Grothendieck-Teichmüller Lie algebra $grt$ defined by three relations that are the exact additive analogues of the three defining relations in the profinite group. Thanks to the result that "the motivic multizeta values satisfy the associator relations", together with Brown's result, we can deduce that the Tannakian fundamental Lie algebra of $MTM$ injects into the Grothendieck-Teichmüller Lie algebra $grt$. The important conjecture is that these two Lie algebras are equal, even though one of them arises from a motivic construction which on the one hand "lifts" up actual real numbers (the real multizeta values, which are periods of the moduli spaces $M_{0,n}$ to mixed Tate motives, which then turn out to generate the full category of mixed Tate motives, and the other arises from the automorphism group of the fundamental groups of the moduli spaces $M_{0,n}$. This answer is not disjoint from Will Sawin's answer above, but expressed differently.<|endoftext|> TITLE: Characterization of Cohen reals QUESTION [16 upvotes]: The following is a well-know fact: Theorem The real $r$ is Cohen over $V$ iff if it does not belong to any meager Borel set coded in $V$. Now suppose that $\kappa$ is an uncountable cardinal and let $(r_i: i<\kappa)$ be a sequence of reals. Question. Is there a characterization theorem as above for the sequence $(r_i: i<\kappa)$ to be $Add(\omega, \kappa)$-generic over $V$, where $Add(\omega, \kappa)$ is the Cohen forcing for adding $\kappa$-many Cohen reals. If there is a known characterization, giving a reference is welcome. Remark. I know there are characterizations of Cohen algebra, for example the one given in the paper Characterizations of Cohen algebras, but I'm interested in a characterization parallel to the one given in the above theorem. Remark 2. The reason I'm asking the question is the following: In the paper Adding many random reals may add many Cohen reals I showed that forcing with $R(\kappa) \times R(\kappa)$ adds a generic filter for $Add(\omega, \kappa)$ (where $R(\kappa)$ is the usual forcing for adding $\kappa$-many random reals), which generalizes the well-known fact that forcing with $R \times R$ adds a Cohen real (where $R$ is the random forcing). The usual proof of the above known result uses characterization of Cohen reals given in the above theorem, while I presented a direct proof without using any characterization result. If there is a characterization as asked above, there might be a different proof parallel to the known one. REPLY [8 votes]: Chapter 20 of the Handbook of Set Theoretic Topology ("Random and Cohen reals" by Ken Kunen, pp 887-911) deals with such questions. Quoting from Truss's review: "Quite a proportion of the paper is devoted to a study of the properties of Cohen extensions of a countable transitive model by $2^I/\mathcal{I}$, where I is an arbitrary index set and $\mathcal{I}$ is the natural "lifting'' to $\mathcal{P}(I)$ of one of the ideals under consideration." More specifically, Lemma 3.8 on page 903 gives a general result akin to what Will Brian suggested in the comments above. 3.8 Lemma Let $\mathcal{I}$ be a reasonable ideal. Let $M$ be a countable transitive model for ZFC with $I\in M$. Let $F\in 2^I$. Then $F$ is $\mathcal{I}$-generic over $M$ if and only if $F$ is not in any $M$-coded Baire set in $\mathcal{I}$. Recall that the Baire sets are defined as belonging to the $\sigma$-algebra generated by the clopen sets. The paper is written from an abstract point of view so there's a lot of notation, but the intent is to lay out what Cohen and Random forcing have in common, and why they can be lifted to large index sets.<|endoftext|> TITLE: Scaling a set of reals to be nearly integers QUESTION [8 upvotes]: A version of this question was previously asked on MSE. I'll mention progress below. A geometric construction I'm exploring leads to a set $R$ of $n$ positive real numbers, for example: $$ R = \{ \pi, e, \sqrt{2} \} \approx \{3.14159, 2.71828, 1.41421\} \;. $$ Given some $\epsilon > 0$, I would like to find the smallest scale factor $s$ so that, for each $x \in R$, $s x$ is within $\epsilon$ of a natural number (i.e., excluding $0$). More precisely, if $[z]$ is $z$ rounded to the nearest integer, then $| sx - [sx] | < \epsilon$. For example, if $\epsilon = 0.08$, then $s=7.018$ works for the above $R$: $$ 7.018 \, R \approx \{22.048, 19.077, 9.925\} \;, $$ and the gaps to the nearest integers are $$ \{0.048, 0.077, 0.075 \} \;, $$ each less than $\epsilon$. But I don't know that $7.018$ is the minimum. Q. What is a general procedure to compute the smallest $s$, given $R$ and $\epsilon$? Ideally I would like an algorithm. Two algorithms were proposed at MSE, both somewhat brute-force searches. One searches through all the integers to which the scaled reals might round. The other searches through rational approximations to pairwise ratios of the reals. In my circumstance, $|R|$ might be $10$ or $20$, or even larger. For fixed $R$, how the minimum $s$ varies with $\epsilon$ is of interest to me, so I understand the tradeoffs between $s$ and $\epsilon$. A comment at MSE suggests the relationship between $s$ and $\epsilon$ could behave "a bit wildly." REPLY [2 votes]: Even though you asked for the optimal solution and others have already answered in this regard, I think it's still worth mentioning the following efficient approach to getting "good" solutions which may not be optimal: Let me first mention a different but related problem: given $(\xi_1,\ldots,\xi_r) \in \mathbb{R}^r$ (all irrational, say), how can we simultaneously approximate the $\xi_i$ by rationals $p_i/q$ having the same denominator $q$? I.e., how can we scale $(\xi_1,\ldots,\xi_r)$ by a $q\in\mathbb{N}_{>0}$ so that $|q\xi_i - p_i|$ are small without $q$ being too large? Dirichlet tells us that there exist $q$ arbitrarily large so that $|q\xi_i - p_i| \leq q^{-1/r}$ where $p_i = \lceil q\xi_i \rfloor$ ("closest integer to"); it doesn't help us find them, unfortunately. But here's how we can obtain a not-quite-so-good approximation in an algorithmically convenient way: for $A>0$ real, consider the image of the $\mathbb{Z}$-linear map $\mathbb{Z}^{r+1} \to \mathbb{R}^{r+1}$ taking $(p_1,\ldots,p_r,q)$ to $(A(q\xi_1-p_1),\ldots,A(q\xi_r-p_r),q/A^r)$. This is a lattice in $\mathbb{R}^{r+1}$, which I just described through a matrix, and we are trying to find short nonzero vectors in it: using the LLL algorithm we can find something like $|q\xi_i - p_i| \leq 2^{r/2}/A$ with $q\leq 2^{r/2} A^r$. As for your original problem, if we are given $(\xi_0,\ldots,\xi_r)$, you can first scale by $\xi_0^{-1}$, say putting $\xi'_i = \xi_i/\xi_0$ so that $\xi'_0 = 1$, forget about this one and apply what I just said to $(\xi'_1,\ldots,\xi'_r)$: this gives you a scale factor $s := q/\xi_0$ such that $s\xi_i \approx p_i$ is close to an integer and $s\xi_0 = q$ is an integer. Given that the problem of finding the shortest nonzero vector in a lattice is hard (in various practical or conjectural ways), I suspect your problem is algorithmically hard if you insist on getting the optimal solution and if $r$ (your $n-1$) is large. But I thought the LLL algorithm deserved at least a mention.<|endoftext|> TITLE: Unprovable cases of $\Pi^0_1$-conservativity QUESTION [5 upvotes]: Let $T$ be a r.e. and consistent extension of PA in the language of artihmetic. Are there examples of arithmetical sentences $\phi, \psi$ such that $T+\phi \equiv_{\Pi^0_1} T+ \psi$ but $T \not \vdash Con_{\tau + \phi} \leftrightarrow Con_{\tau + \psi}$, where $\tau$ numerates the $T$-axioms in $T$ and $\tau + \phi := \tau(x) \vee x = \overline{\phi} $ (respect. $\psi$)? REPLY [5 votes]: I'd say that this situation is indeed possible. In general, the $\Pi_1^0$-conservativity of one theory over another does not guarantee that a proof of their relative consistency can be formalised in $PA$. This observation is due to Guaspari (Partially conservative extensions of arithmetic, Transactions of the AMS, Vol. 254, 1979, pp. 47-68, Theorem 3.3). Recall that $\Pi_1^0$-conservativity and interpretability coincide for essentially reflexive, r.e. theories. Theorem (Guaspari). Let $T$ be r.e. and essentially reflexive. Then there is a $\Sigma_1^0$ sentence $\phi$ such that $T + \phi$ is interpretable in $T$, but $PA \nvdash Con(T) \rightarrow Con(T+\phi)$.<|endoftext|> TITLE: Line bundles and modular forms QUESTION [9 upvotes]: Let $X$ be a modular curve and $\mathcal{L}$ a line bundle on $X$. I found some literature regarding modular forms as global sections on $\mathcal{L}$. Under this context, my question is: For any line bundle $\mathcal{L}$ on $X$ (given modular curve), can any global section of $\mathcal{L}$ be considered as a modular form? If then, could I see the way of interpretation? REPLY [15 votes]: Write your curve as $\Bbb{H}/\Gamma $, with $\Gamma \subset PGL(2,\mathbb{R})$ acting freely on $\Bbb{H}$. The pull back of $\mathcal{L}$ to $\Bbb{H}$ is the trivial line bundle $\Bbb{H}\times \mathbb{C}$; this implies that $\mathcal{L}$ is the quotient of $\Bbb{H}\times \mathbb{C})$ by $\Gamma $ acting by $\gamma \cdot (\tau , z)=(\gamma \tau , e_\gamma (\tau )z)$, where $\gamma \mapsto e_{\gamma }$ is a 1-cocycle of $\Gamma $ with values in $\mathcal{O}(\Bbb{H})^*$ (see the beginning of Mumford's Abelian varieties for a completely analogous analysis). It follows that the sections of $\mathcal{L}$ correspond to functions $f:\Bbb{H}\rightarrow \mathbb{C}$ satisfying $f(\gamma \tau )= e_{\gamma }(\tau )f(\tau )$. These can be considered as modular forms in a generalized sense. If you take $\mathcal{L}=K_X^m$, the $m$-th power of the canonical bundle, we have $e_{\gamma }(\tau )=(c\tau +d)^{2m}$ for $\gamma =\begin{pmatrix} a & b\\ c & d \end{pmatrix}$, and $\ f$ becomes a genuine modular form of weight $2m$ for $\Gamma $.<|endoftext|> TITLE: Continuous maps $f:S^n \to \mathbb{C}P^m$ with $f(x)\perp f(-x) $ QUESTION [7 upvotes]: Question 1: What is a complete classification of all positive integers $m,n$ with the following property: There is a continuous map $f:S^n \to \mathbb{C}P^m$ such that $f$ maps antipodal points to orthogonal lines. Namely for every $x\in S^n$ we have $\;f(x) \perp f(-x)$. Here the later perpendicularity is meant as $f(x)$ is orthogonal to $f(-x)$ with respect to the standard inner product of $\mathbb{C}^{m+1}$ In particular, is it true to say that such map does not exist if $n>2m$? Motivation: One can prove the three dimensional Borsuk Ulam theorem without any explicit or implicit use of homology-cohomology as follows: (An Equivalent formulation of )Borsuk_Ulam in dimension $3$: There is no an odd continuous function $f:S^3\to S^2$. Proof: We identify $S^2$ with $\mathbb{C}P^1$. Then, as I learned from Sebastian Goette via his MO comment, the antipodal points of $S^2$ corresponds to orthogonal lines in $\mathbb{C}P^1$. So we have to prove that there is no a continuous map $f:S^3\to \mathbb{C}P^1$ with the property that $f$ maps antipodal points to orthogonal lines. For the contrary assume that such $f$ exist. Every map $f:S^3 \to \mathbb{C}P^1$ determines a complex line bundle $\ell$ over $S^3$ where $\ell$ is athe pull back of the tautological line bundle over $\mathbb{C}P^1$ so is a subbundle of the trivial bundles $S^3 \times \mathbb{C}^2$. Obviousely every line bundle over $S^3$ is a trivial bundle because the corresponding clutching function $K:S^2 \to GL(1,\mathbb{C})$ is null homotp because any such $K$ has a logarithm by the lifting lemma in the covering space theory. We take a non vanishing section $S:S^3 \to \mathbb{C}^2$ for the line bundle $\ell$. Put $\omega= dx\wedge dy$, the natural determinant $2\_$ form on $\mathbb{C}^2$. Then $\omega(S(x), S(-x)):S^3 \to \mathbb{C} \setminus \{0\}$ is an odd continuous function, a contradiction by the 2 dimensional BU where the later has an elementary proof. Because every continuous function $S^2\to S^1$ has a logarithm so obviously it can not be an odd map $\;\blacksquare$ Remark 1: Instead of identification $S^2$ with $\mathbb{C}P^1$, one can identify $S^2$ with the space of projections of $M_2(\mathbb{C})$. In this case antipodal maps correspond to orthogonal projections. The precise identification is the following: $(x,y,z)\mapsto 1/2\begin{pmatrix} 1-z&x+yi\\x-yi&1+z \end{pmatrix}$ Remark 2 The above proof, which is independent of homology or cohomology, and involves the orthogonality of lines in the projective space or projections of the matrix algebra, not only motivates the question $1$ above but also motivates the following question: Question 2: Can one generalize the above proof to find new proof of the higher dimensional BU, without involving Homology-Cohomology? REPLY [11 votes]: Such a map $S^n\to \mathbb CP^m$ exists if and only if either $n<2m$ or $n=2m=2$. To see this, first note that such a map is the same as a $\mathbb Z/2$-equivariant map from $S^n$ to a certain subspace of $\mathbb CP^m\times \mathbb CP^m$, namely the space of pairs $(L,M)$ such that $L\perp M$. Now note that the latter subspace is an equivariant deformation retract of the space of pairs $(L,M)$ such that $L\neq M$. So the question is, for $X=\mathbb CP^m$, is there a map $f:S^n\to X$ such that $f(-x)\neq f(x)$ for all $x$? If $n<2m$ then the answer is yes because you can embed $$S^n\subset\mathbb R^{n+1}\subseteq\mathbb R^{2m}=\mathbb C^m\subset \mathbb CP^m.$$ To get a negative result when $n=2m$, note that for any map $f:S^n\to N$ to a smooth $n$-manifold there is a mod $2$ obstruction to having $f(x)\neq f(-x)$ for all $x$. It can be defined homologically, or it can be defined more geometrically by first putting $f$ in general position and then counting how many unordered pairs $(x,-x)$ satisfy $f(x)=f(-x)$. This element of $\mathbb Z/2$ depends only on the homotopy class of $f$. If $f$ is homotopic to a constant then you can work out (replacing $N$ by $\mathbb R^n$ if you like) that the obstruction is nontrivial. (This is a way of thinking about Borsuk-Ulam.) When $N=\mathbb CP^m$ then every map $S^{2m}\to X$ is homotopic to a constant unless $m=1$. But there are more homotopy classes of maps $S^2\to \mathbb CP^1$, and in fact for the maps of odd degree that obstruction is trivial. The negative result for $n>2m$ follows from the negative result for $n=2m$ (when $m>1$). The negative result for $S^3\to\mathbb CP^1$ is a case of Borsuk-Ulam, as you observed. Edit: The geometric idea for defining the mod $2$ invariant is this. An equivariant map $F:S^n\to N\times N$ (such as $x\mapsto (f(x),f(-x)$) is always homotopic through equivariant maps to a smooth map such that $F$ is transverse to the diagonal $\Delta_N\subset N\times N$. Then $F^{-1}(\Delta_N)$ is a finite set (zero-dimensional compact manifold) in $S^n$ invariant under $x\mapsto -x$, giving a finite set in $\mathbb RP^n$. The cardinality of this, reduced mod $2$, is the invariant. It is independent of the choice of $F$ within the homotopy class because if $F_0$ and $F_1$, both smooth and transverse to $\Delta_N$, are equivariantly homotopic then the homotopy $H:S^n\times I\to N\times N$ may be chosen to be transverse to $\Delta_N$, and then $H^{-1}(\Delta_N)$ is a cobordism in $S^n\times I$ giving a cobordism in $\mathbb RP^n\times I$, and a compact one-dimensional manifold must have an even number of boundary points.<|endoftext|> TITLE: Did Lurie's model of a homotopy coherent idempotent change? QUESTION [12 upvotes]: In the published version of HTT (Def 4.4.5.2) and on the nlab one finds one definition of a split homotopy coherent idempotent corepresented by a quasicategory $Idem^+_\mathrm{old}$. A few months ago, Gal Dor found an issue in Lurie's treatment of idempotents, and Lurie rewrote the section. The definition of an idempotent in a quasicategory has changed: in the updated version of HTT, it's a map from $Idem^+_\mathrm{new}$, which is simply the nerve of the free split idempotent in $\mathsf{Cat}$. In his question, Dor suggested that these two definitions are the same, but I'm not so sure. I count 8 nondegenerate 3-simplices in $Idem^+_\mathrm{new}$, but 6 nondegenerate 3-simplices in $Idem^+_\mathrm{old}$, so I don't think they're isomorphic. I don't see an obvious bijection between the simplices of the two models. Questions: Are $Idem^+_\mathrm{old}$ and $Idem^+_\mathrm{new}$ equivalent quasicategories? Is there at least a map between them? It seems that all the foundational results about the old definition have been re-proved for the new definition (including a cohrence result -- the new Prop 4.4.5.20 which originally appeared under the old definition in Higher Algebra). Are there any results in the literature which rely on the details of the old construction? EDIT Just for concreteness, here are the non-degenerate simplices in dimension 3: For $Idem^+_\mathrm{old}$: A 3-simplex is an (unlabeled) set of disjoint nonempty subintervals of $[3]$, which is nondegenerate iff each subinterval has exactly one element and each "gap:" between subintervals (including the gaps at the beginning and end) have at most one element. So a nondegenerate 3-simplex can be specified by a set of numbers between 0 and 3 (each representing a singleton interval). They are: $\begin{align*} \{0,1,2\}, \{0,1,3\}, \{0,2,3\}, \{1,2,3\},\{0,2\}, \{1,3\}, \{1,2\}, \{0,1,2,3\} \end{align*}$ for a total of 6 8. For $Idem^+_\mathrm{new}$: A 3-simplex is a sequence of 3 composable morphisms in the free split idempotent $A^{\overset{i}{\to}}_{\underset{r}{\leftarrow}} X \overset{e}{\to} X$, and it's nondegenerate if none of them are the identity. They are: $\begin{align*} (i,r,i), (i,e,r), (i,e,e), (r,i,r), (r,i,e), (e,r,i), (e,e,r), (e,e,e) \end{align*}$ for a total of 8. Side note: I personally find the new definition both much cleaner and much more transparent. I'm still puzzled about the motivation of the old definition. REPLY [6 votes]: A bijection of simplices is not too hard. Just note that a chain of composable non-identity morphisms in $A^{\overset{i}{\to}}_{\underset{r}{\leftarrow}} X \overset{e}{\to} X$ can be recovered from the list of morphisms, or from the list of objects which can be specified as indices of $X$'s. So the 3-simplex example $A \overset{i}{\to} X \overset{e}{\to} X \overset{r}{\to} A$ corresponds to $(i,e,r)$ and $\{1,2\}$.<|endoftext|> TITLE: Model structure on wheeled topological properads QUESTION [5 upvotes]: A wheeled properad is roughly, if I understand correctly, a properad (or PROP) with contraction maps $O_i^j\to O_{i-1}^{j-1}$ which contract an input with an output. There is a book, Infinity Properads and Infinity Wheeled Properads by Hackney, Robertson and Yau, which defines a category of wheeled infinity-properads, from the point of view of a Joyal/Lurie-like combinatorial picture (as simplicial sets with some extra structure). Now the combinatorially defined infinity-category of operads is equivalent to the category of topological operads when the two sides are viewed as model categories or $\infty$-categories obtained from localization. If I understand correctly, a similar statement holds for PROPs. My question is whether one can define a category of topological wheeled properads and a model structure on this category which is equivalent to the definition of Hackney, Robertson and Yau. In fact, I haven't even been able to find a notion of topological wheeled properad defined anywhere - is there a reason why this is nontrivial? REPLY [3 votes]: I can only answer the first half of your question. I defined the notion of a topological wheeled properad in the preprint Dwyer-Kan Homotopy Theory of Algebras over Operadic Collections. As usual the ambient category can be any bicomplete symmetric monoidal closed category. You can also replace wheeled properads with other operad-like structures, such as dioperads, props, and wheeled operads. I showed that there is a Dwyer-Kan model structure on the category of all simplicial wheeled properads. You can replace the ambient category with any convenient model category (see Definition 6.5.2 in that paper). Once again this Dwyer-Kan model structure exists for other operad-like structures, not just enriched wheeled properads. However, I do not know whether this is Quillen equivalent to the combinatorially defined model of infinity wheeled properads in the Hackney-Robertson-Yau book. I also do not know whether this is true for props.<|endoftext|> TITLE: Kolmogorov-Arnold theorem for (just-)functions QUESTION [11 upvotes]: There is famous Kolmogorov-Arnold theorem for continuous functions composition - continuous function of several variables can be composed of continuous functions of two variables. Specialization of such theorem into smooth functions is false: there is no similar composition obeying smoothness - that is there are smooth functions of several variables that cannot be composed of smooth function of 2 variables. Take a look here: Kolmogorov superposition for smooth functions There is an obvious way around: generalization and even grand generalization. Generalization: Is Kolmogorov-Arnold theorem true for just functions ( non-continuous)? If above question is interesting we may ask for Grand Generalization: Let KA(f, X) means the following hypothesis: given object f of some class containing several variables ( function, relation, rules of inference etc), and characteristics X ( continuous, transitive etc) valid for such class of objects, is it possible to compose every object of class f as finite number of objects of the same class with 2 variables with the same characteristic X? Is there something known for Grand Generalization K(f,X) for various f and X? REPLY [9 votes]: The answer is yes for functions $f:[0,1]^n\to\mathbb{R}$: Any such function can be written as $$ f(x_1,\dots,x_n) = g\Big(\sum_{i=1}^n h_i(x_i)\Big). $$ The proof is pretty simple: The $h_i$ should "spread out" the digits of the $x_i$ by $n$ places such that $\sum_{i=1}^n h_i(x_i)$ contains all the digits of all the $x_i$. Thus, the argument that is passed to $g$ contains the same information that the arguments of the function $f$ contain. I've seen this proof in some lecture notes about deep learning - that's one field where Kolmogorov's Theorem was heavily discussed in the 1980s, see Girosi and Poggio's note Representation properties of networks: Kolmogorov's theorem is irrelevant F Girosi, T Poggio - Neural Computation, 1989 - MIT Press Edit: Here is more fleshed out version of the proof. The $h_i$ should "spread out" the digits as follows: Let us denote the inputs by $x^1,\dots,x^n$ and their digits by $$ \begin{split} x^1 & = 0.x^1_1 x^1_2\dots\\ x^2 & = 0.x^2_1 x^2_2\dots\\ \vdots &\\ x^n & = x^n_1 x^n_2\dots \end{split}. $$ Then the $h_i$ work as follows $$ \begin{split} h_1(x^1) & = 0.x^1_1 0 \dots 0 x^1_2 0\dots 0 x^1_3 0\dots\\ h_2(x^2) & = 0.0 x^2_1 0\dots 0 x^2_2 0 \dots 0 x^2_3 0\dots\\ h_3(x^3) & = 0.0 0 x^3_1 0\dots 0 x^3_2 0 \dots 0 x^3_3 0\dots\\ \end{split} $$ with $n-1$ zeros between digits. Then $$ \sum_{i=1}^n h_i(x^i) = 0.x^1_1x^2_1\dots x^n_1 x^1_2 x^2_2\dots x^n_2\dots $$ i.e., this number contains all the digits of all the numbers $x^1,\dots,x^n$. Now define $g$ of this number as $f(x_1,\dots,x_2)$. Since the "interlacing map" $(x^1,\dots,x^n) \mapsto \sum_i h_i(x^i)$ is bijective, such a $g$ exists.<|endoftext|> TITLE: Properties of $\zeta(s)\zeta(2s)\zeta(3s)...$ QUESTION [8 upvotes]: Let's consider the Dirichlet series $f(s)=\sum_{n=1}^\infty a_n n^{-s}$, where $a_n$ is the number of non-isomorphic abelian groups of order $n$. Now $a_n$ is weakly multiplicative and $a_{p^k}=P(k)=$ partition number of $k$, so we get $f(s)=\prod_{p} \sum_{k=0}^\infty P(k) p^{-ks}=\prod_{p} \prod_{k=1}^\infty \frac{1}{1-p^{-ks}}$ because of the generating function of the partition number. So we get $f(s)=\prod_{k=1}^\infty \zeta(k s)$ (where everything converges absolutely). So my question is: what is known about this function? Is there a functional equation or an analytic continuation? Thank you very much. REPLY [15 votes]: As usual when counting objects, it is preferable to assign a weight proportional to the inverse of their automorphism group. While the function $\zeta(s)\zeta(2s)...$ does count the number of non-isomorphism abelian groups of order $n$, if you now count them with the factor $1/|Aut(G)|$ the function becomes $\zeta_\infty(s)=\zeta(s+1)\zeta(s+2)\zeta(s+3)...$ which is MUCH more interesting. What follows is directly taken from the basic paper I wrote with H.W. Lenstra on heuristics for class groups (Springer L.N. 1068), in the special case of abelian groups (one can do the same for finitely generated modules over Dedekind domains). Set $$W(s)=(\Gamma(s/2)^{-1}\Gamma_2(s))^{1/2}\pi^{s^2/4}2^{(s-1)(s-2)/4}\zeta_\infty(s)$$ where $\Gamma_2$ is Barnes's double gamma function, and $$\Lambda(s)=W(s)W(-s)\sin^2(\pi s)/(\pi^2C^2)$$ where $C$ is an easily given normalizing constant. Then $\Lambda(s)$ extends to an entire function of order $2$ (like the Selberg zeta function, but with quite different properties), which is of course even, and its functional equation simply says that it is periodic of period $1$ (morally speaking, it is a gamma factor times $\prod_{n\in\Bbb Z}\zeta(s+n)$), this being equivalent to the functional equation of the zeta function itself. What's fun about this function is that if you graph it on the real line (try it!) you get the constant $1$ (or some other constant if you don't know the value of $C$), which is impossible. In fact, its graph is close to less than $2.10^{-38}$ from the constant $1$, as an easy exercise (or a look at the paper) shows. Just my 2 cents because it's fun.<|endoftext|> TITLE: What is currently known or conjectured about q,t-Kostka polynomials? QUESTION [18 upvotes]: The $q,t$-Kostka polynomials $K_{\lambda,\mu}(q,t)$ appear as the change of basis coefficients between Macdonald polynomials $H_\mu(x;q,t)$ and Schur functions $s_\lambda(x)$: $$H_\mu(x;q,t)=\sum_{\lambda\vdash|\mu|}K_{\lambda,\mu}(q,t)s_\lambda(x)$$ Macdonald conjectured that $K(q,t)\in\mathbb{N}[q,t]$ and this was proved by Haiman as a consequence of his proof of the $n!$ Conjecture. However, this did not provide a combinatorial formula for $K_{\lambda,\mu}(q,t)$. If $q$ is set equal to $0$ then you get the one-variable Kostka polynomial $K_{\lambda,\mu}(t)$ (it says this on p. 7 of http://www.maths.ed.ac.uk/~igordon/pubs/grenoble3.pdf). A lot is known about $K_{\lambda,\mu}(t)$, for example, they come up in describing the cohomology rings of Springer fibers, and they have a closed formula in terms of a statistic called the charge on tableaux on $\lambda$ of weight $\mu$, as explained in another MO post here: How does the grading on the cohomology of a flag variety break up the regular representation of W?. The charge statistic is easy to describe and compute (Disclaimer!!: maybe what I will write isn't exactly right, like maybe it gives the formula for a transpose of $\mu$ or $\lambda$ and/or multiplies the correct poly by a power of $t^{-1}$... sorry, I only first heard of the charge this weekend over a few glasses of wine and then I tried to remember the algorithm with the aid of some googling to make it give the right answer when $q$ is set equal to $0$ in the matrices that appear here: http://garsia.math.yorku.ca/MPWP/qttables/qtkostka2a4.pdf): You put a semistandard Young tableau of weight $\mu=(\mu_1,\mu_2,...,\mu_k)$ on $\lambda$, so fill the boxes of the Young diagram of $\lambda$ with $\mu_1$ $1$'s, $\mu_2$ $2$'s,...,$\mu_k$ $k$'s, in such a way that the entries are nondecreasing in rows and strictly increasing down columns (drawing the Young diagram in the way where the rows get smaller as you go down). Then write a word $w$ which is the sequence of integers you get by reading left to right across each row starting from the bottom row and ending with the top row. So if $\lambda=(3,1)$ and $\mu=(1,1,1,1)$ then there are three possible tableaux $T$ depending on whether you put 4, 3, or 2 in the leg of $\lambda$, and these tableaux give words $w$ that are 4123, 3124, and 2134 respectively. Then to compute the charge of the tableau $T$, you can write $w$ clockwise in a circle and put a dash in between the first letter and last letter of $w$, then starting from the first letter of $w$, going around clockwise, find the first sequence $1,...,k$ in this cyclically written $w$ and remove it from $w$, but in this process, whenever you cross the dash you add $k-j$ to the charge where $j$ was the last letter you plucked out before crossing the dash. Now you have a smaller word $w'$, rinse and repeat on $w'$ but with $k'$ replacing $k$ where $k'$ is the biggest letter of $w'$. And so on. This process ends with the empty word and the charge of $T$. The formula found by Lascoux and Schuetzenberger for $K_{\lambda,\mu}(t)$ is that $$K_{\lambda,\mu}(t)=\sum_{T\hbox{ a semistandard tableau on }\lambda\hbox{ of weight }\mu}t^{\mathrm{charge}(T)}$$ So for $\lambda=(3,1)$ and $\mu=(1^4)$ this produces $K_{\lambda,\mu}(t)=t+t^2+t^3$. So my first question is, is there really no conjectural formula for $K_{\lambda,\mu}(q,t)$ which looks like a natural generalization of the Lascoux-Schuetzenberger formula for $K_{\lambda,\mu}(t)$? All I can find are statements like that finding a formula for $K_{\lambda,\mu}(q,t)$ is an open problem. The best I found was that there are formulas that work when one partition is a hook or the other partition has at most two columns https://math.berkeley.edu/~mhaiman/ftp/jim-conjecture/formula.pdf, but that these formulas are wrong in general https://arxiv.org/pdf/0811.1085.pdf (see Conj 5.5 for exactly how quickly they are supposed to go wrong). Also it seems like there was basically no progress on this topic in the last decade. Is that correct? Recently Carlsson and Mellit proved the Shuffle Conjecture, which refines the combinatorial description of the Frobenius character of the diagonal coinvariant ring. This doesn't seem directly related, but according to http://www.aimath.org/WWN/kostka/schurpos.pdf, at root of both problems are something called LLT polynomials -- is it possible one can work backwards from Carlsson-Mellit, Mellit's works to give a combinatorial description of LLT polynomials and from there to settling the question of a combinatorial description of $K_{\lambda,\mu}(q,t)$? Also, if you look at the tables of Kostka matrices like here http://garsia.math.yorku.ca/MPWP/qtTEXtables.html, you notice that in any column all the entries have the same number of terms, equal to dimension of $\lambda$ -- in fact this was proved by Macdonald, VI.8.18 in Symmetric Functions and Hall Polynomials -- which suggests that maybe the correct formula should be a sum over standard Young tableaux on $\lambda$, but with statistics to give exponents of $q$ and $t$ which depend somehow on $\mu$. Did anyone find a conjectural formula like that? Is it too naive to think the answer would look like that? Surely people tried and there's just no pattern to be found?? My second question is, besides as the Schur coefficients of Macdonald polynomials, is there anywhere else that the $q,t$-Kostka polynomials indepedently have arisen, without reference to Macdonald polynomials, like how the $t$-Kostka polynomials appeared in cohomology of Springer fibers? REPLY [4 votes]: To answer some of your questions - note that it suffices to find the Schur-expansion of certain LLT polynomials, in order to figure out the $qt$-Kostka polynomials. There IS a combinatorial description of LLT polynomials, it is quite easy in fact. However, from the definition, it is not clear that they are Schur positive (only monomial and with some easy work, Gessel positive). Together with G. Panova, in a recent preprint, we connect the LLT polynomials with an open conjecture regarding the $e$-expansion of chromatic symmetric functions (this is also related to the Carlson-Mellit paper). Now, I have a preprint with an explicit conjecture on the $e$-expansion of (vertical-strip) LLT polynomials, with a straightforward combinatorial statistic (and now also a proof). This of course gives the Schur expansion, but there is a slight twist to the problem which introduces signs. My intuition is therefore the following: The Schur-expansion of LLT's is hard, because in reality, the coefficients arise AFTER cancellations in a signed sum (the terms in the sum are easy combinatorial objects).<|endoftext|> TITLE: Square root in complex reductive groups QUESTION [9 upvotes]: Let $G$ be a connected complex reductive linear algebraic group. Does every $g\in G$ have a square root? (That is, some $a\in G$ such that $a^2=g$.) REPLY [11 votes]: This question was raised, both in characteristic 0 and in arbitrary characteristic, in back-to-back 2003 papers by R. Steinberg here and P. Chatterjee here. I recall reviewing these papers together for Math Reviews. The answer is similar for all power maps (not just squares), but the adjective "reductive" is replaced in both papers by the more precise "semisimple". (It's best to treat algebraic tori separately.) Even in characteristic 0 the answer to your question is sometimes negative, but these two papers do give precise criteria for all power maps. REPLY [9 votes]: As the comment shows the answer is negative in general. Perhaps it is worth to mention that for connected compact Lie groups the answer is yes, because its exponential map is surjective. In general, if the exponential map of a Lie group $G$ is surjective, then every group element $g \in G$ has a square root, i.e. an element $a \in G$ with $a^2 = g$, since $\exp(x)$ has $\exp(x/2)$ as a square root for any $x \in \mathfrak {g}=\operatorname{Lie}(G)$.<|endoftext|> TITLE: Volume form induced by a Finsler metric QUESTION [8 upvotes]: I'm interested in knowing more about the volume form canonically induced by a Finsler metric. I've found some reasoning about it in this article http://www.ams.org/journals/bull/1950-56-01/S0002-9904-1950-09332-X/home.html but I was wondering if someone could point out a more recent source with results explained in a clearer way. REPLY [10 votes]: In fact there are very many ways to provide a Finsler manifold with a "canonical" volume. Personally I've gone from thinking that this is a nuisance and trying to pin down which one is really the best to thinking that this is part of the landscape and should be accepted. There is a very good notion of volume that goes by the name of "Holmes-Thompson" volume, but it was also introduced by Dazur and, in my opinion, half-heartedly studied by Busemann. You can find almost all that is known about it in the book Minkowski Geometry by Thompson and in the paper Deane Yang linked to in his comment. In the paper by Busemann that you mention, he claims the Hausdorff measure of the Finsler manifold, viewed as a metric space, is the right notion of volume. It is a very interesting notion, of course, but it has its quirks : totally geodesic submanifolds are not minimal, integral geometry goes out the window, some volume filling results fail, etc. It is also not great for non-reversible Finsler metrics. The Holmes-Thompson is nicer in this respect too because it is more sensitive to non-reversibility. Although it is too long to explain here, my viewpoint has changed and I think that one can and should consider, always with some measure and with a lot of good taste, all the natural notions of volume. Sometimes keeping the same questions and changing the notion of volume opens up new vistas and allows you to tie Finsler geometry to other fields, which is what I think is needed most. Check for example this paper. Behind the paper is the idea that there is a different geometry of numbers for every natural notion of volume. The classical results are for Hausdorff measure and this paper makes the case that for the Holmes-Thompson measure you also get an interesting theory (which was, by the way, forseen by Mahler).<|endoftext|> TITLE: A manifold is a homotopy type and _what_ extra structure? QUESTION [28 upvotes]: Motivation: Surfaces Closed oriented 2-manifolds (surfaces) are "classified by their homotopy type". By this we mean that two closed oriented surfaces are diffeomorphic iff they're homotopy equivalent. This means that somehow the homotopy type of the surface contains essentially all information about the manifold. Let's turn the question around. Let's choose a homotopy type $M$, and ask whether it specifies a manifold. I have to be more precise by what I mean here. First of all, let's fix a dimension, say 2 for now, although we can increase it later. Of course not every homotopy type will correspond to a surface. There a some restrictions on $M$, such as: It has to have cohomological dimension 2, i.e. isomorphisms $H^k(M) \cong 0 \quad \forall k > 2$. We have to specify an isomorphism $\phi\colon \mathbb{Z} \xrightarrow{\cong} H^2(M)$ which will in particular define the fundamental class $[M] := \phi(1)$, corresponding to orientation. Cohomology and homology have to exhibit Poincaré duality, which is to say that the cap product with the fundamental class is an isomorphism: $[M] \cap - \colon H_k(M) \xrightarrow{\cong} H^{2-k}(M)$ Now we're in better shape. Although I don't know a proof and haven't seen this statement anywhere, I'd venture the following conjecture, which should be easy to prove or disprove by anyone who knows more homotopy theory than me: Conjecture Each homotopy type with the extra structure outlined in 1. - 3. corresponds to a closed, oriented surface. In particular, there is an equivalence between the category of homotopy types with extra structure and the category of closed, oriented surfaces. Note also that the cap product is functorial, so a map of surfaces should be a map of the homotopy types preserving all of the structure. The takeaway is this: I've come to believe that surfaces are essentially homotopy types with extra structure on cohomology and homology that comes from the manifold structure. Possibly I haven't captured all structure that is needed. But I guess one could amend the list in that case. Question 1: Am I right so far? Higher dimensions: Topological, PL and smooth structures It gets hairier when we go up dimensions. There are closed 3-manifolds that are homotopy equivalent, but not homeomorphic. On the other hand, simply connected topological 4-manifolds are classified by their intersection form, so they can be completely recovered by the information in 1. - 3. ! For smooth structures, there is of course less luck, although the Kirby-Siebenmann class in 4th cohomology tells you whether there is a PL structure or not, so that sounds like a promising candidate for more extra structure along the lines of what we had so far. Question 2: How far can we carry on the idea and classify higher dimensional (topological, PL, or smooth manifolds) by extra structure on the homotopy type, or its homology and cohomology? Boundaries, noncompact manifolds We could wonder whether it's possible to generalise the story to manifolds with boundaries, or noncompact manifolds. Then the homotopy type will certainly not be sufficient. Already surfaces with boundary are not classified by their homotopy type. (Typical counterexample: The direct sum of two annuli, and the minimal 1-handlebody of a torus.) What is really relevant here is the homotopy type of the boundary inclusion $\partial M \hookrightarrow M$, and the corresponding relative cohomology. (Similarly, for noncompact manifolds what seems to be relevant is compactly supported cohomology, which is related to the compactification of the manifold.) How far can the idea be generalised here? REPLY [20 votes]: Igor is giving a good reference to the topic. For completeness, my education, and satisfaction of other reader's laziness, I'm going to give a rough outline here. A Poincaré complex is, very similarly to what I've conjectured, a finite CW complex together with a chosen fundamental class that induces Poincaré duality. Such a complex of course does not correspond to a unique topological (let alone smooth) manifold, and there is a popular 5-dimensional counterexample. There are also "Poincaré pairs", corresponding to manifolds with boundary, essentially answering half of my last question. Another important construction that one can do with a manifold (indeed, any nice enough vector bundle) is the Thom-Pontryagin construction. It leads to something called the "Spivak normal fibration". Essentially, one takes a vector bundle $p\colon E \to M$, picks out the unit disk bundle and quotients by the unit sphere bundle (thus compactifying each fibre), yielding the Thom space $T(p)$. Now, one can also first embed $M$ (uniquely) into $\mathbb{R}^k$ for $k$ high enough, and compactify the complement of the stable normal bundle $\nu$. Since the 1-point-compactification of $\mathbb{R}^k$ is $S^k$, this gives a map $S^k \to T(\nu)$, the "normal invariant". One can generalise and axiomatise this construction for Poincaré duality spaces, which is then called a "Spivak normal fibration". It consists of a spherical fibration for the CW complex and something generalising the normal invariant. It is a theorem by Spivak that every Poincaré duality space has a Spivak normal fibration, unique up to homotopy equivalence. This is remarkable, since a Poincaré duality space will in general not even have a tangent or normal bundle. When the spherical fibration actually comes from the stable normal bundle of a smooth manifold, it has a reduction to a $O(k)$ structure group. Alternatively, one can think about the classifying space $BG$ of circle bundles, and the classifying space $BO$ of vector bundles, which is mapped into $BG$ by means of the $J$-homomorphism. Only those spherical fibrations in its image can come from a smooth manifold. Similar arguments can be made for topological and $PL$-manifolds. For high dimensions, all this is already pretty close to a full answer. Dimensions 3 and 4 are the hardest, as usual. In 2 dimensions, oriented closed surfaces are classified by Poincaré complexes. (It seems this was proved as late as in the early '80s.) That seems to answer my first question. There is a theorem in $\geq 5$ dimensions by Browder that asserts that a simply connected Poincaré complex is homotopy equivalent to a manifold iff the normal fibration has a $TOP$-reduction. In 3 dimensions, Poincaré complexes were classified by Hendriks and Turaev. But I don't know which ones correspond to manifolds, and what extra data classifies the manifolds. (One could ask e.g. how to distinguish homotopy equivalent Lens spaces. Of course I'm not asking you to directly classify 3-manifolds.) In 4 dimensions, simply connected topological manifolds are classified by the intersection form and the Kirby-Siebenmann invariant. The intersection form is obviously defined for any 4-dimensional Poincaré complex. I think Kirby-Siebenmann does as well, but I'm not so sure about that. To really find a manifold from a Poincaré complex, one needs to study $L$-groups, which are a whole huge subject in itself. (See e.g. Wolfgang Lück's 2004 notes A Basic Introduction to Surgery Theory, Andrew Ranicki's Algebraic $L$-theory and topological manifolds (Cambridge Tracts in Mathematics 102 (1992)), or Wall's Surgery on compact manifolds.) The basic idea seems to be that one studies maps from a manifold into a Poincaré space, and then tries to change the manifold by surgery until the map is a homotopy equivalence. The $L$-groups give obstructions to this.<|endoftext|> TITLE: Entire function that grows along one axis and decays along the other QUESTION [5 upvotes]: I would like to find an entire function on the complex plane that grows with $|\mathrm{Re}(z)|$ no faster than $e^{\alpha |\mathrm{Re}(z)|}$ (for some fixed positive $\alpha$), and decays as fast as possible in both directions along the $\mathrm{Im}(z)$ axis. As an example, we could consider $\cosh(\alpha z)$. This grows like $e^{\alpha |\mathrm{Re}(z)|}$ along the real axis and oscillates along the imaginary axis. By averaging over different $\alpha$'s, we can make it decay like $1/\mathrm{Im}(z)$ in the imaginary direction, $\int_0^\alpha d\alpha' \cosh(\alpha' z) = \frac{\sinh(\alpha z)}{z}.$ And by averaging again over $\alpha$, we can make it decay like $1/\mathrm{Im}(z)^2$, $\int_0^\alpha d\alpha' \frac{\sinh(\alpha' z)}{z} = \frac{\cosh(\alpha z) - 1}{z^2}.$ However, beyond this point the averaging trick doesn't make it decay any faster in the imaginary direction. Is there an entire function that grows at most like $|\mathrm{Re}(z)|$ in the real directions and decays faster than $1/z^2$ in the imaginary directions? REPLY [6 votes]: The answer to this question is contained in the famous theorem of Beurling and Malliavin. It is usually stated with decrease on the real axis (rather than imaginary axis as in your question, but you make the change $z\mapsto iz$ to obtain what you need). First of all, the growth restriction is $$\log|f(z)|\leq (\alpha+o(1))|z|$$ this is weaker than your growth restriction. This is called "exponential type $\alpha$". Then suppose that $|f|$ is bounded on the real axis (or belongs to $L^2$, this can be relaxed). Then it is not difficult to see that $$\int_R\frac{|\log|f(x)||}{1+x^2}dx<\infty.$$ This gives the maximal rate of decrease of $|f|$ on the real axis. Under some additional conditions, this rate is best possible. Beurling and Malliavin give two such conditions. Let $w(x)\geq 1$ be a weight function satisfying $$\int_R\frac{\log w(x)}{1+x^2}dx<\infty.$$ This is called the "logarithmic integral". Then there exists an entire function $f$ of (arbitrarily small) exponential type $\alpha$ such that $fw$ is bounded on the real line, if $w$ satisfies any of the following additional regularity conditions: a) $\log w$ is uniformly continuous, or b) $w$ is itself a function of (some) exponential type. This theorem is difficult. But there is a much simpler result due to Wiener and Paley: it is sufficient that $w$ is even and increasing on the positive ray, and the logarithmic integral is convergent. And of course it is not a problem to construct such function decreasing as $|x|^{-n}$ for every $n$: just divide out as many zeros of $\cos$ as you need. References: MR0147848 Beurling, A.; Malliavin, P. On Fourier transforms of measures with compact support. Acta Math. 107 1962 291–309. MR1430571 Koosis, Paul Leçons sur le théorème de Beurling et Malliavin. (French) [Lessons on the Beurling-Malliavin theorem] Université de Montréal, Les Publications CRM, Montreal, QC, 1996 This book is a very nice discussion of the whole subject, with alternative proofs.<|endoftext|> TITLE: Parallelotope fundamental domains of the n-torus QUESTION [6 upvotes]: The group $\mathbb Z^n$ acts on the topological space $\mathbb R^n$ by translation: if $z = (z_1, \cdots, z_n) \in \mathbb Z^n$ and $x = (x_1, \cdots, x_n) \in \mathbb R^n$, then $z\cdot x := z+x$. The quotient space of this action is the $n$-dimensional torus $\mathbb R^n/\mathbb Z^n$. In this setting, a fundamental domain is a convex set $Z \subset \mathbb R^n$ such that $Z$ surjects onto $ \mathbb R^n / \mathbb Z^n$ via the usual quotient map, and such that this map is injective on the interior of $Z$. My question is: when is a parallelotope in $\mathbb R^n$ a fundamental domain of the $n$-dimensional torus?. By a parallelotope, I mean the $n$-dimensional analogue of a parallelogram. More precisely, a parallelotope is set of the form $\left\{ \sum_{i = 1}^n a_i v_i \mid 0 \leq a_i \leq 1 \right\}$ for some linearly independent set $\left\{ v_1, \cdots, v_n \right\}\subset \mathbb R^n$. I'm fine with assuming that our parallelotopes are rational, meaning $v_i \in \mathbb Q^n$ for all $i$. I'm also inerested more generally in which parallelitopes in $\mathbb R^n$ surject onto $\mathbb R^n / \mathbb Z^n$ via the usual quotient map. My initial guess was that any parallelotope with sufficiently large volume would at least surject onto the $n$-torus, but this dream was quickly crushed by the following example: if $Z$ is the rectangle $[0, 0.9999]\times [0,10000000]$ in $\mathbb R^2$, then $Z$ doesn't surject onto $\mathbb R^2 / \mathbb Z^2$, but a small rotation of $Z$ does surject. On the other hand, it's easy to see that this property is preserved by the action of $SL_n(\mathbb Z)$ on $\mathbb R^n$. Thus we may assume that the matrix $[v_1 v_2 \cdots v_n]$ is in Hermite normal form (and in particular, upper-triangular). Using this trick, I found it's not too hard (though quite messy) to figure out the $n=2$ case by hand, but I'm not sure what to do in higher dimensions. REPLY [5 votes]: Instead of asking which unit-volume n-parallelotopes ($TC$, where $T\in SL_n(\mathbb{R})$, and $C=[0,1]^n$) tile $\mathbb{R}^n$ when translated by $\mathbb{Z}^n$, we can equivalently ask, which unit-determinant lattices $T^{-1}\mathbb{Z}^n$ tile space when applied to the cube $C$ via translations. This is the subject of Hajós's theorem, previously a conjecture of Minkowki, which says that such a lattice has an upper triangular basis with ones on the diagonal.<|endoftext|> TITLE: Any important consequences with presupposition of $\mathbf{P} \neq \mathbf{NP}$ QUESTION [19 upvotes]: As we know, there are lots of consequences with the presupposition of the Riemann Hypothesis. Similarly, are there any important consequences with the presupposition of $\mathbf{P} \neq \mathbf{NP}$ ? An alternative statement of $\mathbf{P} \neq \mathbf{NP}$ is the extended Church-Turing Thesis. So if we have an speedup algorithm of other model than classic Turing Machine, we have to find an new algorithm for Turing Machine with the assumption $\mathbf{P} \neq \mathbf{NP}$ ? that means we have to find new speedup algorithm of factoring and the like. REPLY [17 votes]: Let me take a slightly different angle on this question and interpret it as asking what sorts of hardness results require P ≠ NP only, and which ones require stronger assumptions. By itself, P ≠ NP implies (or more pedantically, is known to imply) less than what many people think. For example, here are some of the things that P ≠ NP is not known to imply: Factoring has no polynomial time algorithm. Graph isomorphism has no polynomial time algorithm. One-way functions exist. There is in general no short certificate that a graph is not Hamiltonian. There is no family of polynomial size circuits for SAT. NP-complete problems have no subexponential time algorithms. NP-complete problems are hard on average. There is no randomized polynomial time algorithm for an NP-complete problem. There is no polynomial time algorithm for an NP-complete problem on a quantum computer. On the other hand, P ≠ NP does imply some results that might seem at first to be stronger than P ≠ NP. A notable class of examples are inapproximability results that follow from the PCP theorem. For example, if P ≠ NP then there is no polynomial time approximation algorithm for the size of the maximum independent set of a graph that is guaranteed to get within a constant (or even logarithmic) factor of the optimum. Another example is Mahaney's theorem that if P ≠ NP then there is no sparse set that is NP-complete.<|endoftext|> TITLE: Relationship between synthetic differential geometry and differential cohesion? QUESTION [10 upvotes]: I'm a big fan of synthetic differential geometry (or smooth infinitesimal analysis), as developed by Anders Kock and Bill Lawvere. It's a beautiful and intuitive geometric theory, which gives justification for the infinitesimal methods used by many of the pioneers of analysis and differential geometry, like Sophus Lie. I have also occasionally come across the notion of differential cohesion, which seems to be another sort of synthetic approach to differential geometry. Unfortunately, while I'm pretty comfortable with 1-category theory, I don't know much homotopy theory or higher category theory, so the language used is pretty foreign to me. There is also a lot of talk of modalities, which I don't really have any intuition for. Question: What is the relationship between differential cohesion and synthetic differential geometry (in the sense of Kock & Lawvere)? REPLY [7 votes]: I'm a co-author on the abstract linked in the comments but I'm coming from the computer science side so I'm not an expert on the models and I know very little classical differential geometry. I had the same question and my current understanding is that Differential Cohesion and Synthetic Differential Geometry have the same "canonical model": the SDG people use the Cahiers topos and Diff Cohesion people use the $\infty$-sheaves on the same site. They are by no means equivalent: for instance Synthetic Differential Geometry axiomatizes a real line object $R$ and the Kock-Lawvere axiom uses it, whereas in differential cohesion the notion of "infinitesimal" distance is more "formal" in that it is just derived from abstract modalities. For instance there are very simple models of DC like this one that have a very strange notion of "infinitesimally close". It's a matter of ongoing research how much differential geometry can be done using just differential cohesion, but clearly if you want to get all of the classical theorems you need to at some point axiomatize the reals (unfortunately any internal definition of the reals via Dedekind cuts will give you the wrong reals because you want to only have the smooth functions). On the other hand Differential Cohesion is more "categorical" which syntactically means we hope to get well-behaved type-theoretic connectives instead of uninterpreted axioms, like how having the identity type in HoTT is much nicer than axiomatizing the n-spheres. The main source for what can be formalized in (solid, differential) cohesion is Urs' book, look for the sections on the differential cohesion modalities to see some comparisons to SDG, but also note that much of that book can be formalized with just a cohesive topos. For some shorter/simpler examples, see Felix Wellen's thesis if you want to see what some of differential geometry looks like using just one of the modalities from differential cohesion. Also maybe check out Mike Shulman's paper for (non-differential) cohesion plus some axioms for a more hybrid style. Also, when it comes to the infinity-stuff, I have been assured by experts that you can just ignore the word infinity when you are looking for intuition. A big takeaway from infinity-category theory and Homotopy Type Theory is that infinity groupoids act a lot like sets.<|endoftext|> TITLE: Norm estimation of identity plus two non-commuting self-adjoint operators QUESTION [6 upvotes]: This is a problem that I have been stuck for a few months. Let $X$ be a Hilbert space and $A:B:X\to X$ be two non-commuting semi-positive self-adjoint bounded linear operators. Is it true that $$\|(I+A+B)^{-1}A\|\le 1.$$ If it is, can you suggest any reference to me? I was able to show $\|(I+A+B)^{-1}A\|\le \|A\|$ and $\|(I+A)^{-1}A\|\le 1$. However, I don't know how to show the desired result without assuming $\|A\|\le 1$ or $AB=BA$. If it helps, I can show $A=C^*D_1^*D_1C$ and $B=C^*D_2^*D_2C$ for some linear bounded $C,D_1,D_2$. REPLY [6 votes]: Though two answers are already posted, let me explain how to understand that it is false from general reasoning. The inequality $\|(I+A+B)^{-1}A\|\leqslant 1$ is equivalent to $\|(I+A+B)^{-1}A x\|\leqslant \|x\|$ for any vector $x$. Take $x=A^{-1}(I+A+B)y$, the inequality rewrites as $\|y\|\leqslant \|y+A^{-1}(I+B)y\|$. We are given only that $A,B$ are positive definite, thus for given vector $z$, $A^{-1}z$ may be any vector which forms an acute angle with $z$. For $z=(I+B)y$ this in general allows the vector to form an obtuse angle with $y$ (unless $y$ is an eigenvector of $B$.) So, if the direction of $A^{-1}(I+B)y$ forms an obtuse angle with $y$ and $A^{-1}$ is small enough, the inequality $\|y\|\leqslant \|y+A^{-1}(I+B)y\|$ fails.<|endoftext|> TITLE: Is there a probabilistic interpretation of McShane's identity? QUESTION [6 upvotes]: Recall McShane's Identity. Take a Brownian Motion on this punctured torus constrained to return to the original point. Now take the "curve" you get this way and try and contract it as much as possible. Maybe the probability you get of ending with a given simple closed geodesic of length $l$ is $2/(1+e^l)$ or something. This would give a probabilistic proof of McShane's identity or something like this. Can one construct this? REPLY [3 votes]: Not that I know of, but there is a Brownian motion argument for the (philosophically related) Basmajian identity: Calegari, Danny, Chimneys, Leopard spots and the identities of Basmajian and Bridgeman, Algebr. Geom. Topol. 10, No. 3, 1857-1863 (2010). ZBL1196.57010. The Brownian motion thing seems to be not really essential, as shown in my preprint. But make of all this what you will.<|endoftext|> TITLE: Does/can threadability imply reflection principles? QUESTION [10 upvotes]: First let me define a few notions to phrase my question simply. Say a regular cardinal $\kappa$ is threadable if the threaded square $\Box(\kappa)$ fails, and $\alpha$-threadable if $\Box(\kappa,\alpha)$ fails, so 1-threadable is simply threadable. Now furthermore say that $\kappa$ is $\alpha$-reflecting if $\text{Refl}(\alpha,\kappa)$ holds; i.e. that for any $\alpha$-sized collection of stationary subsets of $\kappa$ has a common reflection point below $\kappa$. My (perhaps very trivial) question is then what direct implications (can) exist between these cardinals? By Theorem 2.13 in Hayut & Lambie-Hanson (2016) we get that every $<\kappa$-reflecting cardinal is $<\kappa$-threadable, and Theorem 2.8 in the same paper gives us that $2$-reflecting cardinals are $<\omega$-threadable. Theorem 4.9 in the paper also shows that $\textsf{ZFC}\not\vdash 1\text{-reflecting}\Rightarrow \text{threadable}$, assuming a background assumption. A special case: do threadable cardinals admit stationary reflection (i.e. are they 1-reflecting)? REPLY [8 votes]: To address your special case: threadable cardinals are not necessarily 1-reflecting. In fact, the assertion, "$\kappa$ is $\alpha$-threadable for every $\alpha$ such that $\alpha^+ < \kappa$" does not imply that $\kappa$ is 1-reflecting. To see this, suppose that $\kappa$ is a weakly compact cardinal whose weak compactness is preserved by $\kappa$-directed closed forcing (this is overkill, for the sake of concision). Now let $\mathbb{S}$ be the standard forcing notion to add a non-reflecting stationary subset to $\kappa$ (or to $S^\kappa_\omega$, or to your favorite stationary subset of $\kappa$) by initial segments. In $V^{\mathbb{S}}$, let $\mathbb{T}$ be the forcing that shoots a club in $\kappa$ disjoint from this generically added stationary set. The point is that, for all $0 < \beta < \kappa$, the two-step iteration $\mathbb{S} * \dot{\mathbb{T}}^\beta$ (where the second iterand is a full-support product) has a dense $\kappa$-directed closed subset. Now move to $V^{\mathbb{S}}$. In this model, there is a non-reflecting stationary subset of $\kappa$, since we have just explicitly introduced one with $\mathbb{S}$. However, if $\alpha < \kappa$, then $\square(\kappa, \alpha)$ must fail. This is similar to arguments in our paper that you cited, but is basically because forcing with $\mathbb{T}$ would have to add a thread to any $\square(\kappa, \alpha)$-sequence, but such a thread cannot be added by a forcing $\mathbb{Q}$ such that $\mathbb{Q}^{\alpha^+}$ is $\kappa$-distributive. $\kappa$ is inaccessible in this model; similar arguments will work at successors of either singular or regular cardinals. The situation becomes more interesting if you increase the threadability assumption to the failure of $\square(\kappa, < \kappa)$, which is actually equivalent to the tree property holding at $\kappa$. If $\kappa$ is inaccessible, then this is equivalent to $\kappa$ being weakly compact, in which case $\kappa$ is $\alpha$-reflecting for all $\alpha < \kappa$. If $\kappa$ is a double successor cardinal, i.e., $\kappa = \lambda^{++}$, then it is shown in Cummings-Friedman-Magidor-Rinot-Sinapova (2016, preprint) that the tree property can consistently hold at $\kappa$ while there is a non-reflecting stationary subset of $S^\kappa_{<\lambda^+}$. Perhaps somewhat surprisingly, it actually turns out to be rather difficult to simultaneously obtain reflection and the tree property at successors of small singular cardinals. This was first done in Fontanella-Magidor(2017), for $\kappa = \aleph_{\omega^2 + 1}$. The question remains open, and seemingly quite difficult, whether the tree property and stationary reflection can simultaneously hold at $\aleph_{\omega + 1}$. I think this covers the main points regarding implications from threadability to reflection. Implications from full reflection of the form $\mathrm{Refl}(\alpha, \kappa)$ to threadability are covered, I think pretty exhaustively, in the paper with Hayut that you cited. There are some interesting (to me, at least) open questions regarding implications from $\mathrm{Refl}(\alpha, S)$ to threadability, where $S$ is some specific stationary subset of $\kappa$. For example, it is open whether $\mathrm{Refl}(\omega, S^{\omega_2}_{\omega})$ implies the failure of $\square(\omega_2, \omega).$ In a related vein, there is a recent preprint by Fuchs in which he investigates the effect of diagonal stationary reflection hypotheses on threadability.<|endoftext|> TITLE: Rademacher type theorem for Alexandrov spaces QUESTION [6 upvotes]: The classical Rademacher theorem says that any Lipschitz function on a doman in $\mathbb{R}^n$ has the first derivative almost everywhere. I am wondering if this result can be generalized as follows. Let $X$ be an Alexandrov space with curvature bounded below of finite dimension $n$. It is well known that at almost every point $x$ (with respect to the $n$-Hausdorff measure) the tangent cone at $x$ is isometric to the Euclidean space $\mathbb{R}^{n}$. Let us call such points regular. Let $f\colon X\to \mathbb{R}$ be a Lipschitz function. We define its differential (if it exists) in the standard way as follows. Let $X_N$ (resp. $\mathbb{R}_N$) denote the space $X$ (resp. $\mathbb{R}$) with the metric multiplied by $N$. Let $$f_N\colon X_N\to \mathbb{R}_N$$ denote the map $f$ between these rescaled spaces. $f_N$ has the same Lipschitz constant as $f$. Fix a point $x\in X$. Then $(X_N,x)$ converges in the Gromov-Hausdorff sense to the tangent cone $T_xX$ at $x$, and $(\mathbb{R}_N,f(x))$ converges to $(\mathbb{R},0)$. Then if there is a limit map $df_x\colon T_xX\to \mathbb{R}$ we call it the differential of $f$ (automatically it will have the same Lipschitz constant as $f$). Question. Is it true that for almost every regular point $x\in X$ the differential at $x$ of a Lipschitz function $f$ exists and is a linear functional on $\mathbb{R}^n$? REPLY [6 votes]: I assume you are interested in the finite-dimensional case. You can write this function in a distance chart (these charts cover almost all points). Apply the standard Rademacher theorem and notice that the distance chart is differentiable almost everywhere. Hence the result follows.<|endoftext|> TITLE: Bers' constant for compact hyperbolic surfaces with geodesic boundary QUESTION [7 upvotes]: The clasical Bers' theorem about pants decomposition says that any compact Riemann surface of genus $g \geq 2$ has a pants decomposition such that every cutting geodesic in this decomposition is of length $\leq \mathcal{B}_g$, where $\mathcal{B}_g$ is a constant (so-called Bers' constant) depending only on $g$. There are also estimations on Bers' constant. My question is : What is known about Bers' constant on hyperbolic surfaces with boundary? Thank you for your answers! REPLY [6 votes]: Balacheff, Parlier, and Sabourau proved that Bers's theorem also holds for arbitrary complete Riemannian metrics, as long as you rescale appropriately: there's a $C_{g,n}$ such that a complete surface of genus $g$ with $n$ ends and area $A$ has a pants decomposition where each curve has length at most $C_{g,n}\sqrt{A}$. You can apply this to surfaces with boundary by adding a cylinder and a cusp to each boundary component -- i.e., to every boundary component of length $L$, attach a cylinder of height $L$ and circumference $L$, and to the end of that, attach a hyperbolic cusp. Take a pants decomposition. Any curve that goes into the cylinder or the cusp is homotopic to a shorter curve that stays in the original manifold, so a minimal pants decomposition stays in the original surface, and has all its curves of length at most $C_{g,n}\sqrt{A+\sum_i L_i^2+L_i}$. Balacheff, Florent; Parlier, Hugo; Sabourau, Stéphane, Short loop decompositions of surfaces and the geometry of Jacobians, Geom. Funct. Anal. 22, No. 1, 37-73 (2012). ZBL1254.30057.<|endoftext|> TITLE: Decay of positive definite function in $L^p$ QUESTION [6 upvotes]: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous positive-definite function with $f(0)=1$. Positive-definiteness of $f$ means $$ \sum_{i=1}^{n}\sum_{j=1}^{n}f(x_i-x_j)y_i y_j \geq 0 $$ for all $n\geq 1, x,y\in \mathbb{R}^n$. Note that, by Bochner's theorem, $f = \widehat{\mu}$ for some Borel probability measure $\mu$ on $\mathbb{R}$. Question. If $f \in L^{p}$ for some $p < \infty$, must we have $f(x)=O(|x|^{-c})$ for some $c>0$? Edit: Formerly, I had the condition $p>2$ instead of $p < \infty$. Thanks to Christian Remling for pointing out that the condition $p>2$ adds nothing and the condition $p<\infty$ is needed. Context: This is a natural extension of this question: https://math.stackexchange.com/questions/2296804/lp-implies-polynomial-decay I posted this first to MSE a few months ago, got several up-votes, but nothing helpful: https://math.stackexchange.com/questions/2306071/decay-of-positive-definite-functions-in-lp I am following the advice of Cross posts to Math SE regarding cross-posting. REPLY [4 votes]: No, this does not follow. We can take $f=g*g$, with $g\simeq 1$ near $x_n$, with $x_n$ very rapidly increasing. We'll also choose $0\le g\le 1$ as an even continuous function from $L^1$. This will make sure that $\widehat{f}=\widehat{g}^2$ is positive, as required. Moreover, $f\in L^1$ also, but power decay is prevented by just taking the $x_n$ large enough. More specifically, if $g(x)=\sum h(a_n(x-x_n))$, with a compactly supported $h$ and $a_n,|x_n|\to\infty$ and if also $g=1$ near zero, then $f(x_n)\ge \int h(a_nt)\, dt>0$, and this will be $\ge Cx_n^{-\alpha}$ for any given constants $C,\alpha$ if we just take $x_n$ large enough. Notice that it suffices to show that $f$ does not satisfy any of the estimates $f(x)\le N x^{-1/N}$, $x\ge N$, and for each such potential bound, we use one $x_n$ to refute it. Finally, if $\widehat{f}$ is not in $L^1$, then we modify the argument by also multiplying $\widehat{f}$ by a smooth cut-off function $\varphi$ with $\varphi, \widehat{\varphi}\ge 0$ to fix this (as above, we can take $\varphi=\psi*\psi$ to do this). This will change $f$ itself to $\widehat{\varphi}*f$, but this will still be in $L^1$ and fail to satisfy power bounds if the $x_n$ increase rapidly.<|endoftext|> TITLE: Finite-order self-homeomorphisms of $\mathbf{R}^n$ QUESTION [23 upvotes]: Consider the $n$-dimensional euclidean space $\mathbf{R}^n$. A self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ is said to be of finite order if $\phi^m = \mathrm{id}_{\mathbf{R}^n}$ for some positive integer $m$. Question: Does every finite order self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ have a fixed point? What I know about it: If for every divisor $d | m$, the fixed-point set $\left(\mathbf{R}^n\right)^{\phi^d}$ of $\phi^d$ has its cohomology groups $H^*_c\left(\left(\mathbf{R}^n\right)^{\phi^d}, \mathbf{Z}\right)$ finitely generated over $\mathbf{Z}$, then the theorem of "Verdier, Caractéristique d'Euler-Poincaré, 1973" will be applicable. In particular, all the self-homeomorphisms of $\mathbf{R}^n$ of prime order has a fixed-point. In that article Verdier derived a formula of the finite group representation on the alternating sum of the cohomology group with $\mathbf{Q}$-coefficients. This in particular implies a version of Lefschetz trace formula for finite-order self-homeomorphisms. Unfortunately, I don't know if there can be some self-homeomorphism of non-prime order such that a certain power of it has its fixed-point set very complicated. REPLY [24 votes]: There is, naturally, a huge history regarding a basic question like this. This particular problem was figured out between 1930 and the early 1960's. The main names are P.A. Smith, Conner, Floyd. Here is a math review to get you going: MR0130929 (24 #A783) Reviewed Kister, J. M. Examples of periodic maps on Euclidean spaces without fixed points. Bull. Amer. Math. Soc. 67 1961 471–474. 54.80 (57.47) Let T be a homeomorphism of period r on Euclidean space En. The classical result of P. A. Smith is that T has a fixed point if r is prime [Ann. of Math. (2) 35 (1934), 572–578] or a prime power [Amer. J. Math. 63 (1941), 1–8; MR0003199]. The author settles a question of long standing with the following theorem: If r is not a prime power, then there exists a triangulation μ of $E^{9r}$ and a homeomorphism T of period r on $E^{9r}$ without a fixed point, such that T is simplicial over μ. The construction is a modification of an example due to Conner and Floyd [Proc. Amer. Math. Soc. 10 (1959), 354–360; MR0105115].<|endoftext|> TITLE: "Bisecting" a free subgroup with respect to word length QUESTION [12 upvotes]: My broad question is regarding the lengths of (reduced) words in a subgroup of a free group. As motivation, consider the free group $Gp(S)$ where $|S|=n$, that is, a free group of rank $n$. Let $S=\{a_1,a_2,\dots,a_n\}$. Now take a subgroup $N$ of $Gp(S)$ of index $2$. It is automatically normal, so is the kernel of a surjective homomorphism from $Gp(S)$ to $C_2=\{1,-1\}$. This means that every subgroup $N$ of index $2$ can be obtained by starting with a function from $S$ to $\{1,-1\}$ and extending it to a homomorphism of $Gp(S)$. While any such subgroup $N$ has index $2$ (which intuitively can be thought of as roughly half the elements of $Gp(S)$ are in $N$), this intuition does not hold if we work with lengths of reduced words. That is, it is not true that for any given length $k$, the number of reduced words of length $k$ contained in $N$ is exactly half the total number of reduced words of length $k$ in $Gp(S)$. For instance, we could map $a_1 \to -1$ and $a_j \to 1$ for every $2 \leq j \leq n$, and the normal subgroup this gives us, while of index $2$, is certainly not "balanced" with respect to each length. And as MTyson pointed out in his comment, this is not even possible in most cases. For instance, consider $S=\{a_1,a_2\}$, then the mapping $a_1 \to 1$ and $a_2 \to -1$ would map $8$ reduced words of length $2$ to $-1$, and $4$ of them to $1$. More generally, suppose $S=\{a_1,a_2,\dots,a_n\}$ with even $n$. Then on mapping half the number of generators to $-1$ (and the remaining to $1$), we get, for odd $k$, an equal number of reduced words of length $k$ mapped to $1$ and to $-1$. For even $k$, there are slightly more reduced words of length $k$ mapped to $-1$ than to $1$, the precise values of which can be calculated inductively through straightforward combinatorial arguments. My question is whether, and how, this approach can be extended to subgroups of $Gp(S)$. Suppose $H$ is a normal subgroup of $Gp(S)$ of index $m$. Then $H$ itself is a free group (Neilsen's theorem) of rank $mn-m+1$, and so has a basis (of reduced words over alphabet $S$) of cardinality $mn-n+1$. Again, consider an index $2$ subgroup $N$ of $Gp(S)$. The intersection $N \cap H$ is either trivially the whole of $H$ (which we shall ignore), or is an index $2$ subgroup of the group $H$. We can now ask the same question as before for the index $2$ subgroup $N \cap H$ in $H$. Given the normal subgroup $H$ (using a basis), does there exist an index $2$ subgroup $N$ of $Gp(S)$ such that the number of length $k$ reduced words in $N \cap H$ is roughly half the number of length $k$ reduced words in $H$? The main difficulty here arises from the fact that length is still with respect to alphabet $S$ (even though $H$ is itself a free group), and so there can be cancellations making matters difficult. It is hard enough to count the number of reduced words of a given length in $H$ (the word problem), but that is not what we care about (or so I believe)! We just want to "bisect" $H$ at each length using an index $2$ subgroup. The next natural followup question is about quantification: what can be provably achieved? Given the normal subgroup $H$ (using a basis) and an index $2$ subgroup $N$, let $|N \cap H|_{k}$ and $|H|_{k}$ denote the number of reduced words of length exactly $k$ in $N \cap H$ and $H$ respectively. We are then interested in $N$ that minimizes (asymptotically as a function of $n,m,k$) $$\left| \frac{1}{2} - \frac{|N \cap H|_{k}}{|H|_{k}} \right|$$ for a fixed $H$. While this is a rather broad question, I'd be satisfied even with pointers to relevant tools and references that could possibly help. As of now, I am just staring at the wall without ideas on where to even begin. EDIT: MTyson points out that we could ask the same question for a more restricted class of words: cyclically reduced words, and modulo cyclic permutations. Exact balance seems achievable at least for some simple hand-written examples. REPLY [3 votes]: So the right way to look at this is in terms of walks on graphs. Let $G = Gp(S)$, then the Cayley graph of $G/N$ is a $2n$-regular graph on $|G/N|$ vertices. The number of words of length $k$ in $N$ is the same as the number of paths of length $k$ from some fixed point in this graph to itself, where we demand the path never take the same edge in opposite directions in consecutive steps. This is given by $$ \frac{ (\sqrt{n-1})^k }{|G/N| } \operatorname{tr}\left( U_k \left(\frac{A}{2 \sqrt{n-1}}\right)\right)$$ where $A$ is the adjaceny matrix of the Cayley graph and $U_k$ are the Chebyshev polynomials of the second kind. (This can be proved by letting $A_k$ be the matrix whose $i,j$ entry counts backtracking-free paths of length $k$ from $i$ to $j$ and showing that the $A_k$ satisfy the recurrence relation of the Chebyshev polynoimals in $A$. If $N'$ is an index $2$ subgroup of $N$, the same count for $N'$ is $$ \frac{ (\sqrt{n-1})^k }{|G/N'| } \operatorname{tr}\left( U_k \left(\frac{B}{2 \sqrt{n-1}}\right)\right)$$ where $B$ is the adjacency matrix of the Cayley graph of $G/N'$. You want to know the difference between the number of words of length $k$ in $N'$ and the number in $N$ but not in $N'$, which is twice the second count minus the first. If you express this as a sum over eigenvalues, this cancels all the eigenvalues appearing in $A$, as they also appear in $B$, leaving only the "new" spectrum of $B$. As soon as either $G/N'$ is not bipartite or $G/N$ is bipartite, the new spectrum will be contained in the interval $(-2n,2n)$ and therefore, because it is finite, bounded away from $N$, giving a power-savings bound for the error term. This is a group-theoretic nondegeneracy condition which says that the unique isomorphism $N/N' \to \mathbb Z/2$ is not equal on $N$ to the word length mod $2$. If you want the sharpest possible bound, you need the Ramanujan bound: If all the new spectrum of $B$ is in $[-2\sqrt{n-1}, 2\sqrt{n-1}]$, then the Chebyshev polynomial will be bounded by $k+1$, so the difference between the number of length $k$ words in $N'$ and not in $N'$ will be bounded by $(k+1) (\sqrt{n-1})^k$. If we allow $N'$ to be an arbitrary index $2$ subgroup, and not just $N \cap H$ for an index two subgroup $H$ of $G$, then the Cayley graph of $G/N'$ may be an arbitrary $2$-covering of the Cayley graph of $G/N$. The existence of a $2$-covering with all new eigenvalues in that interval was established by Marcus, Spielman, and Srivastava in 2013 (https://arxiv.org/abs/1304.4132), under the additional assumption that $G/N$ is bipartite (i.e. that $N$ is contained in the subgroup of even length words). I don't know of any approach to get a sharp bound, or nearly sharp bound, without using their result, but in many cases it might be possible to get some explicit bound on the power savings by controlling the largest eigenvalue.<|endoftext|> TITLE: Cubic almost-vertex-transitive graphs with given spanning tree QUESTION [8 upvotes]: Consider the infinite 3-regular tree. Pick a vertex $C$, the "center". For any integer $L\ge 1$ consider the closed ball, in the graph distance, of radius $L$ around $C$. Let $T_L$ be the induced subgraph on the set of vertices given by this ball. The finite tree $T_L$ rooted at $C$ has $L+1$ layers or generations. It has $3\times 2^L-2$ vertices and they all have degree $3$ except the $3\times 2^{L-1}$ leaves in the top layer. My question is: Q1: Is it possible to add $3\times 2^{L-1}$ edges between the leaves in order to get a graph $G_L$ with spanning tree $T_L$ such that $G_L$ is vertex-transitive? (Edit: now solved in the negative, see below) For $L=1$ and $L=2$, I can do that which gives me the tetrahedron and the Petersen graph respectively. Does this kind of problem arise in geometric group theory and the study of locally finite graphs? Also, when looking at OEIS A032355, the relevant numbers of vertex-transitive graphs seems abnormally low. Even more surprising, they immediately precede a big surge (look at the values for $n=2,5,11,23,47,95$ in that list). Is there a known explanation for this phenomenon? Edit in view of the awesome answers by Aaron and Gordon: I still have to digest the theory around Moore graphs and bound. My motivation for this question comes from statistical mechanics on the widest possible class of lattices. In particular it has to do with the question of defining "periodic boundary conditions" for given finite subsets. In this regard, an equally useful (for me) weakening of my question is as follows: Q2: Is there a constant $c\in (0,1)$ independent of $L$ such that one can build $G_L$ as above to make the size of the orbit of the center $C$ no smaller than $c\times(3\times 2^{L}-2)$? In other words I want to make this orbit of macroscopic size. The orbit is of course wih respect to the action of the automorphism group of $G_L$. A quick remark is that if one adds no edges, i.e., one takes $T_L$ itself, this orbit is the singleton $\{C\}$ and the automorphims group is a wreath product $\mathbb{Z}_2\wr\cdots\wr\mathbb{Z}_2\wr\mathbb{Z}_3$ with $\mathbb{Z}_2$ appearing $L-1$ times. REPLY [7 votes]: The two you give are the only cubic examples. There are various ways to weaken your requirements which give very small families. One is Moore Graphs which take a $k$-ary tree of diameter $L$ and connect the leaves so that the girth is $2L+1.$ Other than odd cycles $k=2$ and complete graphs (diameter $1$) they must have diameter $2$. There are uniques examples for $k=3,7$ which turn out to be distance transitive. The only other open case is $k=57$ which would be distance regular but could not be distance transitive.<|endoftext|> TITLE: Are there open subgroups of $SL_2(\widehat{\mathbb{Z}})$ which are $GL_2(\widehat{\mathbb{Z}})$-conjugate, but not $SL_2$-conjugate? QUESTION [11 upvotes]: I apologize if this is too obvious, but I figure it must have a quick answer. Are there open subgroups $\Gamma\le SL_2(\widehat{\mathbb{Z}})$ which are conjugate in $GL_2(\widehat{\mathbb{Z}})$, but not conjugate in $SL_2(\widehat{\mathbb{Z}})$? REPLY [11 votes]: Yes, there are indeed such subgroups. Since the open subgroups are exactly the congruence subgroups, it suffices to find two subgroups of $\mathrm{SL}_2(\mathbf{Z}/N\mathbf{Z})$ which are conjugate in $\mathrm{GL}_2(\mathbf{Z}/N\mathbf{Z})$ but not in $\mathrm{SL}_2(\mathbf{Z}/N\mathbf{Z})$. The following Magma code shows that the first example arises for $N=7$: for N:=2 to 10 do G:=GL(2,IntegerRing(N)); H:=SL(2,IntegerRing(N)); Sub:=Subgroups(H); for R1,R2 in Sub do H1:=R1`subgroup; H2:=R2`subgroup; if H1 ne H2 then bool,g:=IsConjugate(G,H1,H2); if bool then print H1,H2,g; end if; end if; end for; end for; whose first output is MatrixGroup(2, IntegerRing(7)) of order 2^3 Generators: [2 1] [2 5] [6 4] [3 1] [6 0] [0 6] MatrixGroup(2, IntegerRing(7)) of order 2^3 Generators: [1 1] [5 6] [6 4] [3 1] [6 0] [0 6] Conjugate by [3 0] [0 1]<|endoftext|> TITLE: Is the space of connections modulo gauge equivalence paracompact? QUESTION [10 upvotes]: I find this question interesting, but need to get it out of my system: is the space of connections (modulo gauge) on a compact four-manifold paracompact, in the Sobolev topology? If so, I believe it would admit partitions of unity, which would surely make life easier in gauge theory. But I haven't seen the experts make use of such a fact. I have also heard that spaces of curves (as used in symplectic geometry) do not always admit partitions of unity. The question occurred to me while reading about the first of the "five gaps" described by the late Abbas Bahri: http://sites.math.rutgers.edu/~abahri/papers/five%20gaps.pdf As Bahri himself points out, this so-called "gap" has been filled in several different ways (via the Freed-Uhlenbeck theorem or via holonomy perturbations; that is, if the original approach is indeed flawed). Still, I think it would be useful for the younger generation to understand the core of the difficulty. I had wondered if it can be summed up in a negative answer to this question. REPLY [7 votes]: Yes, the space of gauge orbits of connections is paracompact (even when you the use Fréchet topology). First, the space of all connections is paracompact since it is an affine space modelled on a nuclear Fréchet space (and/or it is metrisable). Narasimhan & Ramadas (Geometry of SU(2) Gauge Fields) showed that the action of the group of gauge transformation is proper (they actually only consider SU(2) gauge theories over $S^3$ but their argument generalizes to arbitrary structure groups and arbitrary compact base manifolds). For proper group actions, the orbit space is Hausdorff and the canonical projection map is closed. Since the image of a paracompact Hausdorff space under a closed continuous map is also paracompact, we conclude that the orbit space is paracompact. Moreover, the gauge orbit space is stratified by smooth (Fréchet) manifolds (it's the usual orbit type stratification, the top stratum being the space of irreducible connections, see for example The orbit space of the action of gauge transformation group on connections or On the gauge orbit space stratification: a review). Since every stratum is modelled on a nulear Fréchet space, they are even smoothly paracompact (see Convenient Setting of Global Analysis by Kriegl & Michor for further details, especially chapters 14 (Smooth Bump Functions) to 16 (Smooth Partitions of Unity and Smooth Normality)).<|endoftext|> TITLE: Mapping class group and representation of fundamental group of Riemann surfaces QUESTION [9 upvotes]: Let $S$ be a Riemann surface with genus $g>0$. Let $M$ be the mapping class group of $S$. $Hom(\pi_1(S),Gl(n, \mathbb{C}))$ is the representation space of fundamental group of $S$ Question: Is there an integer $n>1$, such that there exists an irreducible representation $\rho \in Hom(\pi_1(S),Gl(n, \mathbb{C}))$ with $\gamma(\rho)$ conjugate to $\rho$ by an element in $Gl(n, \mathbb{C})$, for all $\gamma\in M$ ? REPLY [10 votes]: There are counterexamples as soon as $g > 1$. Let $n$ be the number of surjective homomorphisms $\pi_1(S) \to A_5$, up to $S_5$-conjugacy. (We can see that $n \geq 1$ using the fact that $A_5$ can be generated for two elements.) Then there is a homomorphism $\pi_1(S) \to A_5^n$, where we take the product of all these homomorphisms. I claim this homomorphism is surjective. Proof: Let $H$ inside $A_5^n$ be the image. Its projection onto each factor is surjective, and these surjective morphisms are distinct up to automorphisms of $A_5$. All such subgroups are all of $A_5^n$, by induction on $n$: Given that it is true for $A_5^{n-1}$, the projection of a subgroup $H$ onto the first $n-1$ factors must have image $A_5^{n-1}$, and the kernel is a normal subgroup of $A_5$, hence either $A_5$ or $1$. If $A_5$ then $H= A_5^n$, if $1$ then there is an $n$th homomorphism $H= A_5^{n-1} \to A_5^n$ which is not equal to any of the other ones up to an automorphism, which is false. Now if we take $V$ an irreducible representation of $A_5$ which is stable under the action of $S_5$ (like the standard representation), then $V^{ \otimes n}$ is an irreducible representation of $A_5^n$ and thus by composition an irreducible representation of $\pi_1$. Any element of the mapping class group permutes the homomorphisms $\pi_1(S)$ (and maybe also acts on them by elements of $S_5$) and so preserves $V^{\otimes n}$, so $V^\otimes n$ is invariant under the mapping class group. We could also run this argument with the Mathieu group $M_{11}$, which has no outer automorphisms at all, so any irreducible representation works.<|endoftext|> TITLE: Functorial construction of ("pre"-)spectral sequences? (Or - what is the "higher structure" underlying spectral sequences?) QUESTION [7 upvotes]: Let $\mathcal{C}$ be a stable $\infty$-category. Let $Fun(\mathbb{Z},\mathcal{C})$ be the category of sequences of objects in $\mathcal{C}$. Where the category $\mathbb{Z}$ stands for the nerve of the poset $\mathbb{Z}$. There's a canonical functor: $$Gr:Fun(\mathbb{Z},\mathcal{C}) \to \underset{n \in \mathbb{Z}}{\coprod}{\mathcal{C}} \subset Fun(\mathbb{Z},\mathcal{C})$$ $$X \mapsto \underset{n \in \mathbb{Z}}{\coprod}{Cofib(X(n-1) \to X(n))}$$ Define the category $Fil(\mathcal{C})$ of filtered objects of $\mathcal{C}$ as the localization of $Fun(\mathbb{Z},\mathcal{C})$ w.r.t. morphisms $f: X \to Y$ which induce equivalences on graded pieces $Gr(f):Gr(X) \to Gr(Y)$. (I've taken this definition from Enhancing the filtered derived category by Gwilliam and Pavlov. I have no claims to originality for this or for anything else in this question for that matter). There's an obvious functor on $Fil(\mathcal{C})$ which shifts sequences. For notational convenience we will denote it $T: Fil(\mathcal{C}) \to Fil(\mathcal{C})$ with $T(X)(n):=X(n-1)$ (this convention will be useful later). There's also the suspension functor coming from $\mathcal{C}$ which acts objectswise. We'll denote it as usual by $\Sigma$. Notice that $T$ and $\Sigma$ commute with each other. The functor $Gr$ above sits in the fiber sequence of functors: $$T \to Id \to Gr$$ Or equivalently: $$Id \to Gr \to \Sigma T$$ Precomposing this sequence of functors with the tail $\Sigma T$ we get $$Id \to Gr \to \Sigma T \to Gr \circ \Sigma T \to \Sigma^2 T^2$$ By repeating this process we get an infinite sequence of functors: $$Id \to Gr \to \Sigma T \to Gr \circ \Sigma T \to \Sigma^2 T^2 \to Gr \circ \Sigma^2 T^2 \to \Sigma^3 T^3 \to \dots$$ One can just as easily extend this sequence in the other direction. Notice that inside this sequence sits the sequence: $$\dots \to Gr \to Gr \circ \Sigma T \to Gr \circ \Sigma^2 T^2 \to Gr \circ \Sigma^3 T^3 \to \dots$$ Which is a "complex" in the sense that the composition of any two consecutive natural transformations is null-homotopic. If $\mathcal{C}$ is provided with a t-structure then one should be able to make the sequence above the $E_1$ page of a spectral sequence with values in $\mathcal{C}^{\heartsuit}$. In summary we have managed to make the construction of the $E_1$ page of a filtered object completely functorially and independent of the heart. Here's the (not so well defined) question: How far can we take this? Can one make a functorial construction of an intermediate category between $Fil(\mathcal{C})$ (can be a functor, a sequence of functors etc.) and any target category of spectral which is independent from any t-structure on $\mathcal{C}$ s.t. choosing a t-structure induces a functor to spectral sequence with values in $\mathcal{C}^{\heartsuit}$ (with all the pages and all the differentials)? Or even less precisely: What is the "higher" structure underlying spectral sequences? Edited: As was pointed out to me nothing is wrong with $Fil(\mathcal{C})$ as an answer to this question from this perspective so here's a refinement: The Question: Let $\mathcal{C}$ be a stable $\infty$-category. Can we define a stable $\infty$-category $\mathcal{S(\mathcal{C})}$ associated with $\mathcal{C}$ having the following properties: There's a "natural" functor $Fil(\mathcal{C}) \to \mathcal{S(\mathcal{C})}$ A t-structure on $\mathcal{C}$ induces a t-structure on $\mathcal{S(\mathcal{C})}$ whose heart is the category of spectral sequences with values in $\mathcal{C}^\heartsuit$ (or exact couples). For any t-structure the natural functor from (1) composed with the $\pi_0$ functor on $\mathcal{S(\mathcal{C})}$ gives the canonical functor $Fil(\mathcal{C}) \to SpSeq(\mathcal{C}^{\heartsuit})$ sending a filtered object to its associated spectral sequence (or exact couple). The assignment $\mathcal{C} \mapsto \mathcal{S(\mathcal{C})}$ is functorial. Edited: One potential weakness in the precise version is that one could argue that spectral sequence should morally correspond to $\pi_*$ and not $\pi_0$. I have no idea what to expect. REPLY [7 votes]: Let me propose an answer to the question which isn't quite what you ask for. In fact, for the most part I agree with Denis that the correct object really is $\mathsf{Fil}(\mathcal{C})$. Also, I should mention that a really nice place to look for functorial discussions of spectral sequences is Verdier's thesis- most of what he does there carries over in a straightforward way to whatever fancy setting you'd like. Ok. So the first issue is that the category of exact couples is not abelian, so we can't ask for it to appear as the heart of a t-structure. As a warm-up, let's try to do the analogous thing for long exact sequences, and we'll get an answer which may or may not be satisfying. We'd like some functor from $\mathcal{C}^{[1]}$ to $\mathsf{LES}(\mathcal{C})$ which records `long exact sequences' without choosing a heart. Again, the category of long exact sequences is not abelian, so we can't expect to put a t-structure on $\mathsf{LES}(\mathcal{C})$ and have this be the heart. But whatever, it's sorta clear what we want: we should choose a sequence of pushouts and pullbacks that witness a bunch of triangles tied together. So let $\mathsf{Stair} \subset \mathbb{Z} \times \mathbb{Z}$ be the sub-poset consisting of pairs $(i,j)$ with $j=-i-1, -i, -i+1, -i+2$. An object of $\mathsf{LES}(\mathcal{C})$ will be a functor $X: \mathsf{Stair} \to \mathcal{C}$ such that $X(i, -i-1)$ and $X(i, -i+2)$ are zero objects and every square made up of adjacent vertices is a pushout square. There is a forgetful functor $\mathsf{LES}(\mathcal{C}) \to \mathcal{C}^{[1]}$ recording the value at the arrow $(0,0) \to (1,0)$, and this is a trivial Kan fibration (a section can be obtained by iterated right and left Kan extensions). A section then gives a functor $\mathcal{C}^{[1]} \to \mathsf{LES}(\mathcal{C})$. Any $t$-structure on $\mathcal{C}$ yields a functor $\pi_0: \mathcal{C} \to \mathcal{C}^{\heartsuit}$ and postcomposition gives a functor $\mathsf{LES}(\mathcal{C}) \to \mathrm{Les}(\mathcal{C}^{\heartsuit})$ where the target is the ordinary category of long exact sequences in the heart. Actually, we could have used any homological functor, not just the $\pi_0$ of some heart. Again: this is sort of a waste of fanciness... in the end we're getting a functor to an ordinary category, so this passes through the homotopy category, and we could have just defined things there... but ok. Now for spectral sequences. The indexing category here is more complicated, it's a variation of what Lurie calls "$\mathsf{Gap}(\mathbb{Z}, \mathcal{C})$" in HA.1.2.2. But the intuition isn't so bad. Basically, given a filtered object, we need to record a choice for every 'subquotient' (not just the adjacent ones), and then for every triple we need to build a staircase as above. Maybe I won't try to write this down... we'll call this category $\mathsf{SpObj}(\mathcal{C})$ for 'spectral objects in $\mathcal{C}$'. So an object in here is a functor from some horrible category of quadruples $(i,j,k;s)$ to $\mathcal{C}$ where $i\le j\le k$ and $s$ is in $\mathsf{Stair}$ and a bunch of these values are required to be zero objects, various squares are meant to be pushouts, etc. Basically, if we start with a filtered gadget $X(i)$ then let $X(i,j)$ be a choice of cone for $X(i) \to X(j)$, and then $X(i,j,k;(0,0))$ will be $X(i,j)$, $X(i,j,k;(1,0))$ will be $X(i, k)$, and the rest of the staircase records an extension of the map between these into a 'long exact sequence'. This more elaborate version of taking the associated graded can be turned into a functor $\mathsf{Fil}(\mathcal{C}) \to \mathsf{SpObj}(\mathcal{C})$ which takes a filtered gadget and makes a choice of display like this (it can be built via lots of left and right Kan extensions; though this is no longer an equivalence!) The upshot is that, given any homological functor $H: h\mathcal{C} \to \mathcal{A}$, where $\mathcal{A}$ is an abelian category, we can apply $H$ to every value taken by $X \in \mathsf{SpObj}(\mathcal{C})$ and unwinding we see that for every $i\le j$ we have a sequence of abelian groups that yield $H_n(X(i,j))$ and for every triple we have a chosen and functorial boundary map connecting these. This is precisely the data of what Verdier calls a "spectral object" and is enough to form a spectral sequence (the boundary maps are given, and the pages of the spectral sequence are formed by taking certain images of maps between then $H_{p+q}$'s). I think it's essentially an unrolled exact couple. My personal opinion: this is all more trouble than it's worth...<|endoftext|> TITLE: What is the relation between Coxeter transformations of generalized Cartan matrices and Coxeter transformations of finite-dimensional algebras? QUESTION [7 upvotes]: Note: This question now has a sister :-) The Coxeter transformation of a generalized Cartan matrix: In the paper The spectral radius of the Coxeter transformations for a generalized Cartan matrix, Claus Ringel defines a Coxeter transformation as follows, in several steps: A generalized Cartan matrix of size $n$ is a matrix $A \in M^{n \times n}(\mathbb{Z})$ such that for all $i \neq j$ the following properties are satisfied: $A_{ii} = 2$ $A_{ij} \leq 0$ $A_{ij} \neq 0 \Leftrightarrow A_{ji} \neq 0$ He then goes on to define the reflection $R_i: \mathbb{R}^n \to \mathbb{R}^n$ as the linear map (depending on $A$) which is given on the canonical basis by $e(j) \mapsto e(j) - A_{ji}e(i)$. Now if $\pi: \{1, \dots, n\} \to \{1, \dots, n\}$ is any permutation, he calls the product $C(A, \pi) := R_{\pi(n)} \cdots R_{\pi(1)}$ a Coxeter transformation for $A$. The Coxeter transformation of a finite dimensional algebra: Let $k$ be a field and $A$ be a finite-dimensional $k$-algebra. Let $P(1), \dots, P(n)$ and $I(1), \dots, I(n)$ be the indecomposable projective $A$-modules up to isomorphism and the indecomposable injective $A$-modules up to isomorphism, respectively (in such a way that the orders correspond, i.e. $P(i) = Ae_i$, $I(i) = D(e_iA)$ with the same idempotent $e_i$). Assume that the dimension vectors $\underline{\dim}(P(1)), \dots, \underline{\dim}(P(n))$ form a basis of $\mathbb{R}^n$. Then we can define the Coxeter transformation for $A$ (with respect to the ordering of the projective modules) by $\Phi_{A}: \mathbb{R}^n \to \mathbb{R}^n, \ \underline{\dim}(P(i)) \mapsto -\underline{\dim}(I(i))$, Question: What precisely is the relation between these two notions of Coxeter transformations? How do properties of a generalized Cartan matrix and properties of a finite-dimensional algebra translate via this relation? REPLY [4 votes]: Since I didn't want to think about permutations too much, here is an answer about relating Coxeter transformations of the form $C(A, \text{id})$ to Coxeter transformations of finite-dimensional path-algebras. Let $A$ be a generalized Cartan matrix. Let $A_+$ and $A_-$ be defined by $$\left( A_+\right)_{ij} = \begin{cases} A_{ji}, \ i > j \\ 1, \ i = j \ \ \ \ \ \ \ \ ; \\ 0, \text{else} \end{cases} \ \ \ (A_-)_{ij} = \begin{cases} A_{ij}, \ i > j \\ 1, \ i = j \\ 0, \text{else} \end{cases}$$ Theorem 1: We have $C(A, \text{id}) = -A_{+}^{-1}A_{-}^t$. Proof: This is Theorem $2.9$ in this paper by Sefi Ladkani. Note that I changed the definition of $A_+$ and $A_-$ to match the definition of a Coxeter transformation I gave in my question. $\square$ Now let $H = kQ$ be the path algebra of the finite quiver $Q$ without oriented cycles. Let $S(1), \dots, S(n)$ be the simple modules, ordered in such a way that there is never an arrow in increasing direction, i.e. whenever $1 \leq i \leq j \leq n$ then there is no arrow $i \to j$ in $Q$. This can be achieved by labeling a sink of $Q$ with number $1$, a sink of the remaining quiver after killing $1$ with number $2$ and so on. The homological bilinear form $\left\langle - , -\right\rangle : \mathbb{Z}^n \times \mathbb{Z}^n \to \mathbb{Z}$ is given (since $H$ is hereditary) on dimension vectors by \begin{align*} \left\langle \underline{\dim} S(i), \underline{\dim} S(j)\right\rangle & = \dim_k \text{Hom}_H(S(i), S(j)) - \dim_k \text{Ext}_H^1(S(i), S(j)) \\ & = \delta_{ij} - \#\{\alpha: i \to j\}. \end{align*} It is well known that the Coxeter transformation $\Phi_H$ of $H$ is the unique Coxeter transformation of the homological bilinear form, in the sence of Ladkanis paper (i.e. $\left\langle x,y \right\rangle = - \left\langle y, \Phi_{H}x \right\rangle$). Therefore, if we set $D$ the matrix with entries $(D)_{ij} = \left\langle \underline{\dim}S(i), \underline{\dim}S(j)\right\rangle$ then we get $\Phi_H = -D^{-1}D^t$. Furthermore, the matrix $A = D + D^t$ is a symmetric generalized Cartan matrix, and since $D$ has no nonzero entries at the upper triangle (remember the order of the simple modules), we get $A_+ = D = A_-$. We get the following: Theorem 2: $\Phi_H = C(A, \text{id})$. Proof: By theorem $1$ we have $C(A, \text{id}) = -A_+^{-1}A_{-}^t = -D^{-1}D^t = \Phi_H$. Compare also with Corollary $2.11$ in Ladkanis paper. $\square$ Remark: I'm pretty sure we can generalize this to every permutation. Also note that we can in this way represent every Coxeter transformation of a symmetric generalized Cartan matrix as the Coxeter transformation of a path algebra (since the numbers in the lower triangle show us exactly how many arrows we have to put between the indices). Added later: There is still the question how properties of the algebra $H = kQ$ and properties of the associated Cartan matrix $A$ correspond. Remember that we have $A_{ii} = 2$, $A_{ij} = - \#\{\alpha: i \to j\}$ for $i > j$ and $A_{ij} = A_{ji}$ and that al arrows in $Q$ go in increasing direction. Let $q_A$ be the quadratic form associated to $A$, i.e. $q_A(x) = x^{t}Ax$ for $x \in \mathbb{Z}^n$. Let $q_Q$ be the quadratic form of the quiver $Q$, i.e. $$q_Q(x) = \left\langle x, x \right\rangle = \sum_{i = 1}^{n}x_i^2 - \sum_{\alpha \in Q_1}x_{s(\alpha)}x_{t(\alpha)}.$$ Theorem 3: $q_A = 2q_Q$. Proof: We have \begin{align*} q_A(x) & = x^tAx = \sum_{i,j} x_iA_{ij}x_j \\ & = \sum_{i = 1}^{n}A_{ii}x_{i}^2 + \sum_{i \neq j} A_{ij}x_ix_j \\ & = \sum_{i = 1}^{n}2x_i^2 + \sum_{i > j}2A_{ij}x_ix_j \\ & = 2 \left( \sum_{i = 1}^{n}x_i^2 - \sum_{\alpha \in Q_1} x_{s(\alpha)}x_{t(\alpha)}\right) \\ & = 2q_Q(x), \end{align*} proving the claim. $\square$ Therefore, the quiver $Q$ is Dynkin (i.e. $H$ is representation finite), Euclidean (i.e. $H$ is tame) or wild (i.e. $H$ is wild) iff $A$ is positive definite, positive semidefinite (but not positive definite) or indefinite as a matrix.<|endoftext|> TITLE: Probability that product is a perfect square QUESTION [14 upvotes]: The probability a given integer in $[0,n]$ is a square is $\frac1{\sqrt n}$. What is the probability that if you take two integers uniformly then their product is square? I know the main term is $\frac1n$. I am also looking for correction terms. The difficulty is an average integer has $\omega(\log\log n)$ factors which can go as high as $\frac{\log n}{\log\log n}$ and as low as $2$. Then we are looking at the probability that the prime factors with odd parity in both integers come in pairs. Is there a way to organize the computations and get one precise quantity? In general what is the number of $m$-tuples whose product is a $k$ power where each integer $x_i$ is in $[a_i,b_i]$ where $i\in\{1,\dots,m\}$? REPLY [10 votes]: For the case $m=k=2$, in which we seek the number $N(n)$ of pairs $(x,y) \in [1,n]^2$ for which $xy$ is a square, we give an elementary estimate $$ N(n) = Cn \log n + An + O(n^{2/3}), $$ where $C = 1/\zeta(2) = 6/\pi^2$ and $$ A = \frac{3\gamma-1}{\zeta(2)} - \frac{2\zeta'(2)}{\zeta(2)^2} - 1 = 0.1377775\ldots \, . $$ This agrees with the analytic calculation of Asymptotiac K (and is corroborated by numerical computation up to $n = 2^{30}$), and improves the error term (the contour-integral analysis gave an error estimate equivalent to $n^{21/22+o(1)}$ rather than $n^{2/3}$). Recall that $xy$ is a square iff $(x,y) = (ma^2,mb^2)$ for some positive integers $m,a,b$, and the representation can be made unique by requiring either that $a,b$ be coprime or that $m$ be squarefree, which is the source of the factor $1/\zeta(2)$. Let $M(n)$, then, be the number of triples $(m,a,b)$ of positive integers such that $ma^2\leq n$ and $mb^2\leq n$, without the additional coprimality or squarefree condition. Then Möbius inversion (applied with either condition $\gcd(a,b)=1$ or $\mu(m)^2 = 1$ yields $$ N(n) = \sum_{d=1}^{\lfloor\sqrt{n}\rfloor} \mu(d) M(\lfloor n/d^2 \rfloor). $$ We show: Proposition. $M(n) = n \log n + Bn + O(n^{2/3})$, where $B = 3\gamma - 1 - \zeta(2) = -0.913287\ldots$ (and again $\gamma$ is Euler's constant $0.5772156649\ldots$). Proof: Let $R = \lfloor n^{1/3} \rfloor$. If $m\cdot\max(a,b)^2 \leq n$ then either $m \leq R$ or $\max(a,b) \leq R$. Given $m$, the number of $(a,b)$ pairs is $\lfloor \sqrt{n/m} \rfloor^2 = n/m - O(\sqrt{n/m})$; summing this over $m \leq R$ gives $n H_R - O(n^{2/3})$ where $H_R$ is the harmonic sum $\sum_{m=1}^r 1/m$. In the other direction, each $k \leq R$ occurs $2k-1$ times as $\max(a,b)$ for positive integers $a,b$, each of which accounts for $\lfloor n/k^2 \rfloor$ solutions, for a sum of $$ \sum_{k=1}^R (2k-1) \, (n/k^2 - O(1)) = (2 H_R - \zeta(2)) n - O(n^{2/3}) $$ solutions. For the total count we add this to $n H_r - O(n^{2/3})$, and subtract $R^3 = n - O(n^{2/3})$ which is the number of solutions for which $a,b,m$ are all $\leq R$, finding $(3 H_R - \zeta(2) - 1) n + O(n^{2/3})$. The Proposition then follows from $H_R = \log R + \gamma + O(1/R) = \frac13 \log n + \gamma + O(n^{-1/3})$. $\Box$ The estimate for $N(n)$ then follows from $N(n) = \sum_{d=1}^{\lfloor\sqrt{n}\rfloor} \mu(d) M(\lfloor n/d^2 \rfloor)$; the $2\zeta'(2) / \zeta(2)^2$ comes from the second term of $$ \frac{n}{d^2} \log \frac{n}{d^2} = \frac1{d^2} n \log n - 2n \frac{\log d}{d^2} $$ because $\sum_{d=1}^\infty \mu(d) \log d \, / \, d^2$ is the derivative at $s=2$ of $-1/\zeta(s)$. One can probably use bounds on exponential sums to further reduce the $O(n^{2/3})$ error, both in worst and average case, as is done for the Dirichlet divisor problem and the Gauss circle problem.<|endoftext|> TITLE: Angle estimate in Alexandrov spaces QUESTION [5 upvotes]: I am not sure that this is a research level question. Remark 10.9.4 in the book "A course in metric geometry" by Burago, Burago, Ivanov claims the following. Let $X$ be a finite dimensional Alexandrov space with curvature bounded below. Fix $p\in X$ and $\varepsilon> 0$. Then there exists $r>0$ such that for any two points $x,y$ such that $|px|,|py|0$ and sequences $x_n,y_n\to p$ such that $0<|px_n|\leq |py_n|\to 0$, and \begin{eqnarray}\label{0} \tilde\measuredangle x_npy_n< \measuredangle x_npy_n-\delta. \end{eqnarray} We denote for brevity $$r_n:=|px_n|,\, R_n:=|py_n|,\, \alpha_n:=\measuredangle x_npy_n, \tilde\alpha:=\tilde\measuredangle x_npy_n.$$ Thus $$r_n\leq R_n\to 0,\, \tilde\alpha_n<\alpha_n-\delta.$$ Choosing a subsequence, we may assume that there exists a geodesic $\gamma$ starting at $p$ such that the angle between its direction at $p$ and the direction of $px_n$ is at most a small $\omega$ to be chosen later. Let $z_n\in \gamma$ be such that $|pz_n|=r_n$. We have \begin{eqnarray}\label{1} |x_nz_n|=\sqrt{2r_n^2(1-\cos\tilde\measuredangle x_npz_n)}=2r_n \sin{(\frac{\tilde\measuredangle x_npz_n}{2})}\leq r_n\cdot \omega. \end{eqnarray} Let us show that the angles $\tilde \measuredangle z_npy_n$ and $\tilde \measuredangle x_npy_n$ are very close to each other. We have \begin{eqnarray*} |x_ny_n|^2=r_n^2+R_n^2-2r_nR_n\cos\tilde\measuredangle x_npy_n,\\ |z_ny_n|^2=r_n^2+R_n^2-2r_nR_n\cos\tilde\measuredangle z_npy_n. \end{eqnarray*} Subtracting the two equalities we get \begin{eqnarray*} 2r_nR_n(\cos\tilde\measuredangle z_npy_n-\cos\tilde\measuredangle x_npy_n)=(|z_ny_n|-|x_ny_n|)(|z_ny_n|+|x_ny_n|). \end{eqnarray*} This and (\ref{1}) imply \begin{eqnarray*} |\cos\tilde\measuredangle z_npy_n-\cos\tilde\measuredangle x_npy_n|\leq \omega \frac{|z_ny_n|+|x_ny_n|}{2R_n}\leq \omega \frac{r_n+R_n}{R_n}\leq 2\omega. \end{eqnarray*} Since $|\arccos a-\arccos b|\leq C|a-b|^{1/2}$ for some fixed $C\geq 1$ and any $a,b\in [0,\pi]$, we obtain \begin{eqnarray}\label{2} |\tilde\measuredangle z_npy_n-\tilde\alpha_n|\leq C\sqrt{2\omega}. \end{eqnarray} On the other hand \begin{eqnarray*} \measuredangle z_npy_n-\tilde\measuredangle z_npy_n\geq (\measuredangle x_npy_n-\measuredangle z_npx_n)-(\tilde\alpha_n+C\sqrt{2\omega})\geq\\ \measuredangle x_npy_n-\tilde\alpha_n-(\omega+C\sqrt{2\omega})\geq \delta -(\omega+C\sqrt{2\omega}). \end{eqnarray*} Now let us choose $\omega$ so that $\omega+C\sqrt{2\omega}<\delta/2$. Then we get \begin{eqnarray}\label{3} \measuredangle z_npy_n-\tilde\measuredangle z_npy_n> \delta/2. \end{eqnarray} Thus we reduced the problem to the case when the points $z_n$ lie on the same geodesic $\gamma$ and $|pz_n\leq |py_n|$. Now choose point $w_n\in \gamma$ such that $|pw_n|=|py_n|$. We have $\tilde\measuredangle w_npy_n\leq \tilde\measuredangle z_npy_n$. Hence by (\ref{3}) we have \begin{eqnarray}\label{4} \measuredangle w_npy_n-\tilde\measuredangle w_npy_n> \delta/2. \end{eqnarray} Thus now $|pw_n|=|py_n|$. Like in the previous step we can find another geodesic $\hat\gamma$ such that its direction at $p$ is very close to the directions at $p$ of the geodesics $py_n$. We choose points $t_n\in \hat \gamma$ such that $|pt_n|=|py_n|$. Repeating the argument as above we get for $n\gg 1$ $$\measuredangle w_npt_n-\tilde\measuredangle w_npt_n> \delta/4.$$ But $\measuredangle w_npt_n$ equals the angle between $\gamma$ and $\hat\gamma$, and $\tilde\measuredangle w_npt_n$ converges to this angle by definition. This is a contradiction. QED<|endoftext|> TITLE: Where can we find polynomial's root? QUESTION [5 upvotes]: Let $R$ be a ring (not commutative in general) with identity and let $\Psi(x) = x^m-\sum_{j=0}^{m-1}\psi_jx^j$ be a monic polynomial over $R$. I want to construct a ring extension $K$ of $R$, which contains a root of $\Psi(x)$. One construction follows from non-commutative Hamilton-Caley Theorem. But I also think that there is another construction, which seems to be very interesting. Let $M = R[x]/R[x]\Psi(x)$ be a left module over $R[x]$ and $A = \mathrm{Ann}_{R[x]}M\subset R[x]$ be an annihilator of $M$. I want to prove that $\Psi(x)$ has a root in the ring $$ R[x]/A. $$ The proof is very simple in the case, where $R$ is a commutative ring. So the main interest is when $R$ is not commutative. REPLY [3 votes]: No, $R[x]/A$ does not always have a root. Let $R=\mathbb{Z}\langle a,b\rangle$ be the free ring on two noncommuting elements $a$ and $b$. Let $\Psi=x^2+a\in R[x]$, and let $M$ and $A$ be the objects you defined. In this case, $A=0$. Suppose $f=r_nx^n+\cdots+r_0\in A$. In particular, there is some $g=s_n x^n+\cdots+s_0\in R[x]$ such that $f\cdot b=g\cdot (x^2+a)$. Matching up coefficients, we find that $r_0b=s_0a$, $r_1b=s_1a$, $r_2b=s_2a+s_0$,$\dots$, $r_ib=s_ia+s_{i-2}$. In the free ring $R$ the first two equations force $r_0=s_0=0$ and $r_1=s_1=0$, and by induction the rest of the equations force $f=g=0$. Hence $A=0$. Finally, the ring $R[x]/A\simeq R[x]$ doesn't have any roots of $\Psi$.<|endoftext|> TITLE: Measuring a presheaf's failure to be a sheaf? QUESTION [16 upvotes]: Apologies for the vagueness of question. Background this thread has some nice examples of presheaves failing to be sheaves. Question Is there a generic way to measure "how badly" a presheaf fails at being a sheaf? Something like an invariant that "counts", up to some notion of equivalence, sections that fail to glue or restrict properly? Discussion Can we do this by comparing some invariant of a presheaf $P$ and its sheafification $\tilde P$? The comments on this old math SE thread makes an attempt to argue that the Cech cohomology (taking the cover refinement limit) of the two are equal. But is there something else that we can compare between $P$ and $\tilde P$? REPLY [11 votes]: Let $\mathcal{F}$ be a presheaf on $X$, and suppose $\mathcal{U}=\{U_i\}$ is an open cover of $X$. The Cech complex is the cochain complex whose degree $n$ piece is the direct sum of sections of $\mathcal{F}$ on the $(n+1)$-fold intersections of sets in $\mathcal{U}$. The ordinary Cech complex is supported in degrees $n\geq 0$, but the same definition works for $n=-1$: there is only one $0$-fold intersection, and it is all of $X$. So we get an extended complex supported in degrees $\geq -1$, that starts $$ 0\to \mathcal{F}(X)\to \bigoplus_{i} \mathcal{F}(U_i)\to\bigoplus_{i TITLE: Highest weight representations of Kac—Moody algebras: what is inside the weight spaces? QUESTION [9 upvotes]: Let $V(\lambda)$ be the unique irreducible representation of a Kac—Moody algebra $\mathfrak{g}$ with the highest weight $\lambda$. If $\mathfrak{g}$ is not of finite type, then even for $\lambda$ one of the fundamental weights the weights of $V(\lambda)$ usually have multiplicities $\geqslant2$. Is there a nice way to pick a basis in each of the weight spaces $V(\lambda)_\mu$? Here "nice" means that there is an explicit description of the action of the standard generators of $\mathfrak{g}$, so one could actually write (infinite) matrices for the Lie algebra elements. I have checked some sources including Lie Algebras of Finite and Affine Type by R. Carter and Affine Lie Algebras, Weight Multiplicities and Branching Rules by Kass, Moody, Patera and Slansky, but none contains such details even for the smallest cases like $\tilde{A}_1$. Here is a picture from the second: As you can see, this diagram gives no hint to how exactly the generators act inside the weight spaces. I guess this can be reconstructed from the desription of $V(\lambda)$ as a quotient of the Verma module, but this seems quite computationaly extensive, so I was wondering whether someone has already done it in some cases (say, for the untwisted affine KM algebras for some particular choice of $\lambda$)? REPLY [5 votes]: The key word you're looking for is Littelmann path model, which gives a nice geometric description for the basis of a weight space of a highest-weight representation of a symmetrisable Kac-Moody algebra (see e.g. http://people.bath.ac.uk/lpah20/GeomSemNP.pdf for an introduction), and gives a fairly explicit algebraic description of this basis (see http://www.mi.uni-koeln.de/~littelma/montreal.pdf Theorem 5.2), from which it is possible to calculate the action of the standard generators. While I don't know of a precise reference where this is written up, it is a good exercise to calculate these things for, say, $\tilde A_1$ and $\tilde A_2$.<|endoftext|> TITLE: Is there any published physics article where $q$-mathematics is applied? QUESTION [12 upvotes]: Excuse me for the concern, but I want to ask you a question. In 2002 Professor John Baez had published a few articles on his page regarding the possibility of applying $q$-mathematics in the science of physics (see [1]). The scripts were interesting but so far I could not find any article where $q$-mathematics was applied. Is there any published article where $q$-mathematics is applied? [1] http://math.ucr.edu/home/baez/week183.html REPLY [11 votes]: As another example of the second category in Kostantinos Kanakoglou's answer I think it is fair to mention quantum-integrable systems: this topic in physics was pivotal in the historical development of the notion of quantum groups by the Leningrad group (Faddeev et al), and the Japanese group (Jimbo and Miwa et al). In particular, $U_q(\widehat{\mathfrak{sl}_2})$, the quantum-affine version of $\mathfrak{sl}_2$, naturally arises in the (algebraic Bethe-ansatz) analysis of the partially isotropic (or "XXZ") Heisenberg quantum spin chain, and the closely related six-vertex model in statistical physics. Here the deformation parameter $q$ characterizes the spin chain's partial anisotropy as $\Delta = (q+q^{-1})/2$, with $\Delta=q=1$ the completely isotropic ($\mathfrak{sl}_2$-invariant) point. More, including many references to published articles, can e.g. be found in the books Jimbo and Miwa, Algebraic Analysis of Solvable Lattice Models, esp Chapters 0--3; Gómez, Ruiz-Altaba and Sierra, Quantum Groups in Two-Dimensional Physics, esp Section 2.3 and its references; Chari and Pressley, A Guide to Quantum Groups, esp Section 7.5 and its bibliographical notes. Finally, although I am by no means an expert on the topic, let me point out that the second half of the book by Gómez et al is devoted to quantum groups in conformal field theory, see also Fuchs, Affine Lie Algebras and Quantum Groups: An Introduction, with Applications in Conformal Field Theory.<|endoftext|> TITLE: Adding nonconstructive disjunction to intuitionistic logic QUESTION [10 upvotes]: In constructive mathematics, under realizability interpretations, we can define nonconstructive disjunction $A⅋B$ as follows: A witness for $A⅋B$ gives a candidate witness for $A$ and a candidate witness for $B$, at least one of which is correct (we need not know which one). $A⅋B$ is the strongest connective that always satisfies $(¬C⇒A) ∧ (¬¬C⇒B) ⇒ A⅋B$. It also the strongest monotonic connective such that $¬C⅋¬¬C$ is constructively valid. How do you formalize intuitionistic sequent calculus augmented with nonconstructive disjunction? What kind of semantics is this calculus complete for? The formalization should be conservative over intuitionistic logic. Below is what I know and more precise versions of these questions. Note: I used symbol '⅋' because because it follows the rules of multiplicative disjunction $⅋$ (pronounced 'par') in linear logic (however, $∨_w$ ('weak or') would be another reasonable choice). For example, $A⅋A ⇒ A$ is not constructively valid: We might not know which of the two candidate witnesses is correct. The limitation on contraction is the price for using nonconstructive disjunction in a constructive framework. To axiomatize '⅋' as an addition to intuitionistic logic, let us list its basic properties. Axioms: $A⅋B⇔B⅋A$ $A⅋(B⅋C)⇔(A⅋B)⅋C$ $(B⇒C) ⇒ (A⅋B⇒A⅋C)$ $¬C⅋¬¬C$ $A⅋⊥ ⇔ A$ Is this axiomatization complete for the realizability semantics above? Here, I would like to thank Andrej Bauer for his treatment and general formalization of realizability (his answer below), which includes a semantics for '⅋', though completeness of the axioms remains open. If the axioms are insufficient, we can extend the language with infinitely many predicate variables (without second order quantification), and add inference rule: From $A⅋B⇒D$ infer $(¬C⇒A) ∧ (¬¬C⇒B) ⇒ D$ ($C$ is a predicate variable not used in $A$, $B$, $D$). Conversely, if the axioms above are sufficient, this extension should be conservative over them. Number and Function Realizability Frank Waaldjik's answer and comments note the similarity between $A⅋B$ and $(¬A⇒B)∧(¬B⇒A)$, with the later even satisfying the axioms for $A⅋B$ (unless the axiomatization is incomplete). However, for most realizability interpretations, $A⅋B$ is strictly stronger than $(¬A⇒B)∧(¬B⇒A)$. Two examples are number realizability (with natural numbers coding partial recursive functions) and function realizability (with witnesses being codes for partial continuous functions; constructive validity requires a recursive witness for the universal closure of the formula). In both cases, if $r⊩A$ and $r⊩B$ ('⊩' means 'realizes') are both $Π^0_2$, then $A⅋B$ is equivalent to $∃a,b∈ℝ[ab∉ℚ∧(a∉ℚ→A)∧(b∉ℚ→B)]$, which can be proved using $Π^0_2$-completeness of $a∉ℚ$. For both number and function realizability, we also have the following (without complexity restrictions) $(∃n∈ℕ \, A_0(n)) ⅋ (∃n∈ℕ \, A_1(n)) ⇒ ∃i,n \, (A_i(n)⅋∃n∈ℕ \, A_{1-i}(n))$ (intuitively, run/examine both candidate witnesses until one gives an $n$). Over the domain $R$ of partial functions $ℕ→ℕ$ (treated extensionally, but witnesses enumerate values with an arbitrary order and delay), we even have $(∃a∈R \, A(a)) ⅋ (∃b∈R \, B(b)) ⇔ ∃a,b∈R \, (A(a)⅋B(b))$, which (with the axioms above) suffices to define '⅋' for both number and function realizability (since $¬A(a)⅋¬B(b)$ is $¬(A(a)∧B(b))$). However, these formulas need not hold for other interpretations of '⅋'. While investigating completeness of the axioms, I was stumbled over $(¬E→A∨B)∧(¬¬E→C∨D) → ((¬A→¬¬E)∨(¬B→¬¬E)∨(¬C→¬E)∨(¬D→¬E))$ before realizing that it holds under both number and function realizability but not intuitionistically provable. Sequent Calculus We would also like a reasonable sequent calculus for intuitionistic logic + '⅋', which, if possible, allows cut elimination and subformula property for cut-free proofs. Here we use an idea from linear logic, treating the succedent Δ as a multiset that is interpreted as a ⅋-disjunction of its members. Δ will have exchange and weakening but not contraction. As Damiano Mazza noted below, Girard's LU system ("On the unity of logic") already includes intuitionistic and linear connectives. However, I do not know if its combination of intuitionistic logic and '⅋' works for us, and in any case, LU's generality likely makes it artificially complicated here. Here is my attempt at the sequent calculus. We start with LK sequent calculus (link accessed Aug 23, 2017) and apply these changes: * Remove right contraction. * Strengthen $∨L$ to reflect lack of right contraction: from Γ,$A$⊢Δ and Γ,$B$⊢Δ infer Γ,$A∨B$⊢Δ. * Add $⅋L$: from Γ,$A$⊢Δ$_1$ and Γ,B⊢Δ$_2$ infer Γ,$A⅋B$⊢Δ$_1$,Δ$_2$. * Add $⅋R$: from Γ⊢$A$,$B$,Δ infer Γ⊢$A⅋B$,Δ. * In $→R$, $¬R$, and $∀R$, require formulas in Δ to be Harrop. Harrop formulas are closed under '⅋'. Δ is arbitrary for other rules. Is this the right system for intuitionistic logic with '⅋'? Can we get cut elimination to hold? The changes are straightforward except for the $→R$, $¬R$, and $∀R$ rules. Because intuitionistic logic deals not just with truthfulness but constructiveness, we have to limit side formulas in the succedent, but Harrop formulas are fine since they are determined by their truth values. I do not know whether these restricted rules are sufficient, especially if not using the cut rule. Nonconstructive Existential Quantifier A disjunction can be thought of as an existential quantifier over $\{0,1\}$, and we can generalize nonconstructive disjunction into a weak or nonconstructive existential quantifier (but note that there are also other constructs of varying degrees of constructiveness). A witness for $∃_w x \, A(x)$ is like a witness for $∀x \, A(x)$ except that it need only be valid for one (potentialy unknown) $x$. '$∃_w$' appears to be the strongest connective always satisfying $(¬¬∃x \,C(x)) ∧ ∀x (¬¬C(x)⇒A(x)) ⇒ ∃_w x \, A(x)$. It is also apparently the strongest monotonic connective with $¬¬∃x \,C(x) ⇒ ∃_w x \, ¬¬C(x)$ constructively valid. We can try to axiomatize it with the following basic properties, though the completeness is unclear. $∃_w x ∃_w y \, A(x,y) ⇔ ∃_w y ∃_w x \, A(x,y)$ $∃_w x (A(x) ⅋ B(x)) ⇔ ((∃_w x \, A(x)) ⅋ (∃_w x \,B(x)))$ $∀x (A(x)⇒B(x)) ∧ ∃_w x \, A(x) ⇒ ∃_w x \, B(x)$ $¬¬∃x \, C(x) ⇒ ∃_w x \, ¬¬C(x)$ $(∃_w x \,A(x)) ∧ ∀x (A(x)⇒x=y) ⇒ A(y)$ (assumes equality is identity; presumably ¬¬x=y⇒x=y). If the axioms are insufficient, we can extend the language with infinitely many predicate variables (without second order quantification), and add inference rule: From $∃_w x \, A(x) ⇒ D$ infer $(¬¬∃x \, C(x)) ∧ ∀x (¬¬C(x)⇒A(x)) ⇒ D$ ($C$ is a predicate variable not used in $A$, $D$). Updated (Sep 2): Added relations with both number and function realizability. REPLY [6 votes]: $\newcommand{\par}{\mathbin{⅋}}$ $\newcommand{\rz}{\Vdash}$ $\newcommand{\fst}[1]{\mathsf{fst}(#1)}$ $\newcommand{\snd}[1]{\mathsf{snd}(#1)}$ Let me write $r \rz \phi$ when $r$ realizes $\phi$. If you allow higher-order logic, then $A \par B$ may be expressed as $$\exists X . (\lnot X \Rightarrow A) \land (\lnot\lnot X \Rightarrow B)$$ where $X$ ranges over all propositions. To justify my claim, let us first recall how higher-order logic is interpreted in realizability. Given a partial combinatory algebra $\mathbb{A}$, the propositions are the subsets of $\mathbb{A}$. In the realizability topos $\mathrm{RT}(\mathbb{A})$ the object of all proposition is $\Omega = (\mathcal{P}(\mathbb{A}), =_\Omega)$, where $=_\Omega : \mathcal{P}(\mathbb{A}) \times \mathcal{P}(\mathbb{A}) \to \mathcal{P}(\mathbb{A})$ is the non-standard equality predicate defined by $$(A =_\Omega B) = (A \Rightarrow B) \land (B \Rightarrow A)$$ (Here read $\Rightarrow$ and $\land$ as in the realizability interpretation, i.e., they give sets of realizers.) By unraveling the realizability interpretation of existentials and applying the definition of $\Omega$, we can work out the fact that $$p \rz \exists X . \phi(X)$$ if, and only if, there is $C \subseteq \mathbb{A}$ such that $\fst{p} \in (C =_\Omega C)$ and $\snd{p} \rz \phi(C)$, where $\mathsf{fst}$ and $\mathsf{snd}$ are first and second projection, respectively. Note that $((\lambda x . x), (\lambda x . x)) \in (C =_\Omega C)$ for all $C$. Let us abbreviate $v = ((\lambda x . x), (\lambda x . x))$. Now we may show that $$ A \par B \iff \exists X . (\lnot X \Rightarrow A) \land (\lnot\lnot X \Rightarrow B)$$ is realized. From left to right, the realizer is $$f = \lambda x . (v, ((\lambda y . \fst{x}), (\lambda z . \snd{x})),$$ Indeed, if $(p, q)$ realizes $A \par B$, then $f (p, q) = (v, ((\lambda y . p), (\lambda y . q)))$. We know that $p \rz A$ or $q \rz B$. If $p \rz A$, take $C = \bot = \emptyset$ and observe that $((\lambda y . p), (\lambda y . q))$ realizes $$(\lnot \bot \Rightarrow A) \land (\lnot\lnot \bot \Rightarrow B)$$ because this is equivalent to $$(\top \Rightarrow A) \land (\bot \Rightarrow B).$$ If $q \rz B$, take $C = \top = \mathbb{A}$ and observe that now $((\lambda y . p), (\lambda y . q))$ realizes $$(\lnot \top \Rightarrow A) \land (\lnot\lnot \top \Rightarrow B)$$ because this is equivalent to $$(\bot \Rightarrow A) \land (\top \Rightarrow B).$$ The implication in the other direction is realized by $$g = \lambda x . (\fst{\snd{x}} \star, \snd{\snd{x}} \star)$$ where $\star \in \mathbb{A}$ is some default realizer. Indeed, suppose there is $C \subseteq \mathbb{A}$ and $p \in \mathbb{A}$ such that $\fst{p} \in (C =_\Omega C))$ and $\snd{p} \rz (\lnot C \Rightarrow A) \land (\lnot\lnot C \Rightarrow B)$: If $C = \emptyset$ then $\fst{\snd{p}} \rz \top \Rightarrow A$ and so $\fst{\snd{p}} \star \rz A$. If $C \neq \emptyset$ then $\snd{\snd{p}} \rz \top \Rightarrow B$ and so $\snd{\snd{p}} \star \rz B$. This completes the proof. We could ask whether there is a first-order schema $\Phi(A,B)$, expressed in the propositional calculus, such that $A \par B \iff \Phi(A, B)$ is realized. I do not know, yet. Addendum: without explaining the details, let me just say that we may realize $$A \par B \iff \exists X \in \Omega_{\lnot\lnot} . (X \Rightarrow A) \land (\lnot X \Rightarrow B)$$ where $\Omega_{\lnot\lnot p} = \{p \in \Omega \mid \lnot\lnot p \Rightarrow p\}$. This should be compared to the fact that $$A \lor B \iff \exists X \in 2 . (X \Rightarrow A) \land (\lnot X \Rightarrow B)$$ where $2 = \{p \in \Omega \mid p \lor \lnot p\}$. This gives us the following idea. Consider any object $T = (\{0,1\}, (=_T))$ whose underlying set is $\{0,1\}$, and let us call such objects "binary". (I think up to isomorphism they are the $\lnot\lnot$-dense subobjects of $\Omega_{\lnot\lnot}$.) We can then define $T$-disjunction $\lor_T$ by $$A \lor_T B \iff \exists t \in T . (t =_T 0 \Rightarrow A) \land (t =_T 1 \Rightarrow B).$$ We have a whole spectrum of disjunctions of various constructive strength, two of which were discussed above ($\par$ is $\lor_{\Omega_{\lnot\lnot}}$, and the intuitionistic $\lor$ is $\lor_2$). To give you one in the middle, consider ordinary number realizability and the binary object $\Sigma$ in which $1$ is realized by the elements of the Halting set, and $0$ by the elements of the non-Halting set (this object is known as the Rosolini dominance). This gives a non-symmetric $\Sigma$-disjunction. One might prefer to impose a further restriction on the binary object $T$, namely that the twist map $0 \mapsto 1$, $1 \mapsto 0$ be realized, which would give symmetry of $\lor_T$. And I will leave it as an exercise to figure out what needs to be done for associativity of $\lor_T$. In general $T$-disjunction implies $S$-disjunction iff the identity map $\mathrm{id} : \{0,1\} \to \{0,1\}$ is realized as a morphism $T \to S$. This probably leads into some sort of study of reducibilities that would make a little boring logic paper. Supplemental 2017-08-22: The characterization of $\par$ tells us what its introduction rule should be, namely: $$ \frac{\displaystyle \Gamma, \lnot C \vdash A \qquad \Gamma, \lnot \lnot C \vdash B }{\Gamma \vdash A \par B} $$ This is weird.<|endoftext|> TITLE: Minimal approximations of surfaces by convex polygons QUESTION [9 upvotes]: Suppose you want a collection of convex polygons in $3$-space such that, when you glue them together edge-to-edge, you obtain an orientable surface of genus $g$. What is the fewest number of polygons you need? Is this a known result? I've done some searching, and there's a bunch of literature on polygonal surfaces/polygonal meshes, but I haven't found an answer to my question yet. I'm pretty sure that for $g > 2$, you can do it with $6g$ rectangles, essentially by gluing together a bunch of triangular prisms. Similarly, the best I've found for the torus is $9$ rectangles. Is this the best possible, and is there an easy way to see that? This seems like a natural enough question that I'd be a little surprised if it hasn't been addressed. Does the answer change if we don't require that the polygons be glued together edge-to-edge? Thanks in advance! REPLY [5 votes]: I interpret your question as seeking polyhedra that realize genus-$g$ surfaces and have the fewest number of convex faces. A related goal has seen considerable work: the same problem but minimizing the number of vertices, so-called vertex-minimal triangulations. For example, this paper finds all the $g=3$ and $g=4$ vertex-minimal triangulations, and finds a $12$-vertex $g=5$ example: Hougardy, Stefan, Frank H. Lutz, and Mariano Zelke. "Surface realization with the intersection segment functional." Experimental Mathematics 19.1 (2010): 79-92. (arXiv abs.)           This work has subsequently been extended: Brehm, Ulrich, and Undine Leopold. "Polyhedral Embeddings and Immersions of Many Triangulated 2-Manifolds with Few Vertices." arXiv:1603.04877 (2016). (arXiv abs.)<|endoftext|> TITLE: Upper bound total variation by Wasserstein distance for continuous distance QUESTION [5 upvotes]: I am reading the survey of the relationships between metrics of distributions (see https://arxiv.org/pdf/math/0209021.pdf for the paper). The general results show that for general distributions, we cannot upper bound the total variation by Wasserstein distance. Many answers in MO give the same intuitive counterexample: consider a discrete distribution and a continuous distribution. However, assuming the two underlying distributions are both continuous (for example consider the simpliest when they are both Gaussian mixtures with zeros mean or just Gaussians with zero mean), can we find such an upper bound? Any related suggestions or comments are welcomed! REPLY [4 votes]: If both probability measures have smooth densities, the total variation can be bounded by Wasserstein (even by a weaker metric Levy-Prokhorov), see this paper. More precisely, see Lemma 5.1 and proof of Theorem 2.1. The basic idea is given below. If both $p$ and $q$ are smooth densities, $d_V(p, p_\gamma)$ and $d_V(q, q_\gamma)$ are sufficiently small for every small gamma, where $d_V$ is the total variation and $p_\gamma$ is a convolution of p and the uniform density as defined in the paper. Note that $d_V(p, q) < d_V(p, p_\gamma) + d_V(p_\gamma, q_\gamma) + d_v(q_\gamma, q)$ by the triangle inequality. Lemma 5.1 guarantees that if $p$ and $q$ are close in Levy-Prokhorov metric, then $d_V(p_\gamma, q_\gamma)$ is also small provided that $\gamma$ is much bigger than the Levy-Prochorov distance.<|endoftext|> TITLE: Hausdorff vs Gromov-Hausdorff convergence of convex hypersurfaces QUESTION [5 upvotes]: Let $\{K_i\}$ be a sequence of convex compact $n$-dimensional subsets in a Euclidean space $\mathbb{R}^n$. Assume $\{K_i\}$ converges in the Hausdorff metric to a convex compact set $K$ which is also $n$-dimensional. Consider the sequence of boundaries $\{\partial K_i\}$ equipped with the induced intrinsic (!) metrics. Is it true that the sequence $\{\partial K_i\}$ with these intrinsic metrics converges to $\partial K$ (again, equipped with the intrinsic metric) in the Gromov-Hausdorff sense? A proof or a reference will be helpful. REPLY [10 votes]: The reference is Lemma 10.2.7 in A course of metric geometry by Burago-Burago-Ivanov. They do it in 3d but it does not matter. The main point is that if two convex bodies are Hausdorff close, then one can blow up one of them by slight dilation to contain the other one. Then the nearest point projection lets you project paths on the blown up convex body onto the other one. This easily implies that the intrinsic metrics on the boundaries are close uniformly, and hence GH close.<|endoftext|> TITLE: The dance marathon problem QUESTION [12 upvotes]: In his book, "The Strange Logic of Random Graphs", Joel Spencer describes the "Dance Marathon" problem: Imagine $n$ couples at a Dance Marathon. Each dance each couple remains standing with independent probability one half. A couple that does not remain standing is removed from the competition. A couple wins the prize if at the end of some dance they are the only couple that remained standing. It may happen that none of the couples that began a dance remained standing, in which case the prize is not awarded. What is the probability $f(n)$ that the prize is awarded? The surprising fact is that $\lim_{n\rightarrow \infty} f(n)$ does not exist. We have an exact formula $$f(n) = n\sum_{k=1}^{\infty}(1-2^{-k})^{n-1}2^{-k-1}$$ Imagine the marathon continuing until the winning couple also collapses and let $k$ be the number of dances completed by the winners. There are $n$ choices for the winning couple, $k$ can be any positive integer, the other $n-1$ couples all do not survive the first $k$ dances, and the winning couple survives the first $k$ dances and not the $k+1$-st dance. Now parametrize $n=2^{u}\theta$ with $u$ integral and $\theta\in[1,2)$. For $k=u+i$, with $i$ fixed and $u\rightarrow \infty$, $n(1-2^{-k})^{n-1}2^{-k-2}\sim\theta 2^{-i}e^{-2^{i}\theta}$. Set $$g(\theta)=\sum_{i=-\infty}^{+\infty}\theta 2^{-i}e^{-2^{i}\theta}$$ Some careful analysis gives that for $\theta$ fixed and $n=2^{u}\theta\rightarrow \infty$, $f(n)\rightarrow g(\theta)$ and simple calculation gives that $g(\theta)$ is not a constant. I'm confused by some elementary details (doesn't $g(\theta)$ diverge as $i\rightarrow -\infty$?) and would like to learn more about the "careful analysis". Could anyone either point me to a complete proof of this result or (if it's straightforward) generate one? REPLY [8 votes]: I'll lay out the starting steps; I hope that after that it won't be much work for you to fill in on your own. To be clear, the process is that there are a succession of dances. At the start, $n$ couples are dancing. In each round, each couple is (independently) eliminated with probability $1/2$. So the probability of a couple being eliminated at time exactly $j$ is $2^{-j}$ and the probability that they are eliminated at time $j$ or sooner is $1-2^{-j}$. We want the probability that, eventually, precisely one couple remains. It is convenient to run the process until everyone is eliminated. Then the probability that the first couple is eliminated at time precisely $k+1$, while everyone else is eliminated at time $k$ or earlier, is $2^{-k-1} (1-2^{-k})^{n-1}$. Summing on $k$, the probability that the first couple survives longer than all the others is $\sum_{k=1}^{\infty} 2^{-k-1} (1-2^{-k})^{n-1}$. Then multiply by $n$ to get the probability that some couple survives longer than all the others: $$\sum_{k=1}^{\infty} n 2^{-k-1} (1-2^{-k})^{n-1}.\quad (\ast)$$ Now, fix a positive real number $\theta$ and let $n$ go to infinity through numbers of the form $\theta 2^u$, for $u \in \mathbb{Z}$. So $(\ast)$ becomes $$\sum_{k=1}^{\infty} \theta 2^{u-k-1} \frac{(1-2^{-k})^{\theta 2^u}}{1-2^{-k}}.$$ Putting $\ell = k-u$, that's $$\sum_{\ell=-u+1}^{\infty} \theta 2^{-\ell-1} \frac{(1-2^{-\ell-u})^{\theta 2^u}}{1-2^{-\ell-u}}.$$ As $u \to \infty$, we have $(1-2^{-\ell-u})^{\theta 2^u} \to \exp(-\theta 2^{-\ell})$ and $1-2^{-\ell-u} \to 1$. So, naively taking limits term by term, we get $$g(\theta) = \sum_{\ell=-\infty}^{\infty} \theta 2^{-\ell-1} \exp(-\theta 2^{-\ell}).$$ Note that the sum is convergent: As $\ell \to \infty$, the first factor goes to $0$ and the second to $1$; as $\ell \to -\infty$ the second factor goes to $0$ much faster than the first blows up. Of course, one has to justify taking limits termwise, but I think this is the level of exercise which is reasonable to leave to you. Instead, let me say a few things to make the argument seem plausible. First of all, a sanity check: $g(\theta) = g(2 \theta)$; just change $\ell$ to $\ell+1$. This is as it should be. (I don't know why Spencer insists that $\theta \in [1,2)$.) Second, the effect is numerically quite small. I get a minimum of 0.721342 near $\theta = 1.3$ and a maximum of $0.721354$ near $\theta = 1.9$. Finally, let me try to sketch the heuristic which convinced me the result was reasonable. I don't know if this will make sense to anyone but me. Let's imagine two marathons, which start out with $1.5 \cdot 2^u$ and $2^u$ people, for some large $u$. Very roughly, if we run the first one for a single dance, it will drop down to $1.5 \cdot 2^{u-1}$ people. Then, if we run the second one for one dance, it will drop down to $2^{u-1}$. If we keep going in this manner, the numbers keep jumping over each other while staying in fixed ratio, until we get down to small values, say $12$ and $8$. There is no reason the odds for $n=12$ and $n=8$ should be the same, so why should they be for $1.5 \cdot 2^u$ and $2^u$. More carefully, but still heuristically, if we start with $1.5 \cdot 2^u$ people, the number of people at the next step is a distribution, centered at $1.5 \cdot 2^{u-1}$ but with standard deviation $\sqrt{1.5 \cdot 2^u}$. Think of it as $1.5 \cdot 2^u \cdot 2^{-1} \cdot (1 \pm 2^{-u/2})$. At the next step, $1.5 \cdot 2^u \cdot 2^{-2} \cdot (1 \pm 2^{-u/2})(1 \pm 2^{-(u-1)/2})$. And continuing down $1.5 \cdot 2^{u-k} \cdot (1 \pm 2^{-u/2})(1 \pm 2^{-(u-1)/2}) \cdots$. So the ratio between the two marathons is like $$1.5 {\Big (} (1 \pm 2^{-u/2})(1 \pm 2^{-(u-1)/2}) \cdots {\Big)}^2$$ where we truncate the product at some fixed power of $2$; perhaps at $1 \pm 16^{-1}$. As $u$ grows, that product stays bounded, so the ratio is still staying roughly fixed, not spreading out all the way between $0$ and $\infty$. So I still buy the argument of the paragraph above.<|endoftext|> TITLE: property of convex functions QUESTION [9 upvotes]: I am able to give a proof to the following inequality for convex functions. Most likely this is well known, but I am unable to find a reference. I would appreciate if someone more knowledgeable in the literature of convex analysis could help. Suppose $P$ is an open bounded convex subset of $\Bbb R^n$ and $f: P \to \Bbb R$ is a convex function such that $\int_P f =0$. Then there exists positive constants $\alpha,\beta>0$ (not dependent on $f$) such that $$-\alpha\inf_P f \leq \int_P |f(x)|dx^n \leq -\beta \inf_P f .$$ EDIT: per Fedja's observation, I removed the unnecessary constants from the ineqilities. Also, it is specified that $P$ has to be open and bounded. Also, in lack of a good reference, if someone can furnish a short proof of the first estimate, that would work for me as well. REPLY [10 votes]: Anyway, If you know a 5 line proof for the first inequality please share it with us OK, here goes. Assume that $\inf_P f=-1$ and that it is (nearly) attained at the origin. Let $K=\{x\in P: f(x)\le -\frac 12\}$. Then $E=\{x\in P: f(x)<1\}\subset 4K$ by convexity. Also, $|E|\ge |P|/2$ because otherwise $\int_P f=\int_{P\setminus E}f+\int_E f\ge |P\setminus E|-|E|>0$. Thus, $|K|\ge\frac 12 4^{-n}|P|$ whence $\int_P |f|\ge \int_K|f|\ge 4^{-n-1}|P|$. If you allow 10 lines instead of 5, you can get the sharp constant via a similar argument, but, since it hasn't been requested, I'll stop here :-)<|endoftext|> TITLE: Category of concrete categories QUESTION [9 upvotes]: Consider the following 2-category: • It objects are concrete categories, i.e., categories equipped with a faithful functor to $Set$. • A 1-morphism between $(C_1,U_1)$ and $(C_2,U_2)$ consist of a functor $F:C_1\to C_2$ and a natural transformation $z:U_1\Rightarrow U_2\circ F$. • Its 2-morphisms are the obvious thing. Question: Is there a name for that notion of functor between concrete categories? ... the pair $(F,z)$ is a  [insert adjective]  functor from $C_1$ to $C_2$ ... REPLY [10 votes]: Concrete functor is established in the literature for the related notion where the natural transformation is an isomorhpism (see e.g. Porst 1996 Concrete Categories Are Concretely Equivalent if…) — i.e. the sub-2-category of the slice 2-category of CAT over Set on faithful functors. Your 2-category is similarly the sub-2-category on faithful functors of the colax slice 2-category of CAT over Set. So it seems very natural to call your notion colax concrete functors, though as far as I can find this term hasn’t been used before. A lax concrete functor would be the same thing but with the transformation in the other direction.<|endoftext|> TITLE: Graph isomorphism problem for minimally strongly connected digraphs QUESTION [5 upvotes]: A minimally strongly connected digraph (MSC) is strongly connected (SC), while removal of any arc destroys this. That is, between any two vertices a, b there exists a directed path from a to b, while removal of any arc (a,b) renders b unreachable from a. What is known about graph isomorphism for MSCs? For example, is decision of graph isomorphism between MSCs GI-complete ("graph-isomorphism complete" -- as hard as deciding isomorphism between general digraphs), or is it poly-time decidable? There seems to be no literature on this subject. Another question is about determining UMS(n), the number of isomorphism types of MSCs for n vertices. In the paper Minimal strong digraphs it is mentioned that UMS(n), in other words, the number of unlabeled MSC digraphs with n vertices, is "unknown": does there exist at least a superpolynomial lower bound for UMS(n)? Again, there seems to be nothing in the literature on this subject (so any information is welcome). REPLY [2 votes]: Isomorphism of MSC digraphs is isomorphism-complete. Consider two connected undirected graphs $G,H$ with no vertices of degree 1. It is routine to see that connectivity and minimum degree at least 2 won't help you to determine whether $G$ and $H$ are isomorphic. Now convert $G,H$ to digraphs $G',H'$ by replacing each edge with a directed 4-cycle using two new vertices. Precisely, replace each edge $\lbrace v,w\rbrace$ by the 4-cycle $v\to x_{v,w}\to w\to y_{v,w}\to v$, where $x_{v,w},y_{v,w}$ are new vertices not used anywhere else. Now note that $G',H'$ are MSC digraphs which are isomorphic iff $G,H$ are isomorphic. On the second question I'm not aware of a complete answer. Some small numbers are on OEIS.<|endoftext|> TITLE: Szpiro's conjecture for function fields and Mochizuki's approach to the number field case QUESTION [6 upvotes]: Where can I find more details on the proof of Szpiro's conjecture for function fields, as mentioned in Minhyong Kim's answer to this MO question? I am looking at this in the context of Mochizuki's much-discussed approach to an analogue of the same conjecture for number fields, so I am particularly looking for an emphasis on concepts such as the Gauss-Manin connection and the importance of finding an arithmetic analogue. As for my relevant background, I know some basic algebraic geometry, including cohomology and elliptic curves, but I have very little knowledge of subjects such as deformation theory and Hodge theory (which I am assuming these topics belong to). I know about connections in the context of differential geometry. If there are any relevant prerequisites it would be very helpful to have them enumerated. REPLY [5 votes]: You might want to also consider the geometric ("symplectic") version of the conjecture, since Mochizuki alredy has a paper outlining the relationship between Bogomolov's proof and his own IUT theory. Shinichi Mochizuki, "Bogomolov's Proof of the Geometric Version of the Szpiro Conjecture from the Point of View of Inter-universal Teichmuller Theory" (2016) The original relevant papers are: J. Amorós, F. Bogomolov, L. Katzarkov, T. Pantev, I. Smith, "Symplectic Lefschetz fibrations with arbitrary fundamental groups" (2000) Shouwu Zhang, "Geometry of algebraic points" (2001)<|endoftext|> TITLE: Do the cohomology groups of the structure sheaf of a smooth resolution depend on the resolution? QUESTION [9 upvotes]: Let $X$ be an affine variety. Let $Y$ be smooth and let the map $f\colon Y\rightarrow X$ be proper birational. We will call $Y$ a smooth resolution of $X$. Do the cohomology groups $H^i(Y,\mathcal{O}_Y)$ depend on $Y$? I'm sure that the answer is 'no' if $X$ is normal, but I can't find a reference. Is it true in general (if $X$ is not normal)? REPLY [8 votes]: Edit. This follows from the Elkik-Fujita Vanishing Theorem. There is a more general vanishing theorem due to Elkik and Fujita. One version of this theorem (where I read the theorem) is Theorem 1.3.1 of the following article. MR0946243 (89e:14015) Kawamata, Yujiro; Matsuda, Katsumi; Matsuki, Kenji Introduction to the minimal model problem. Algebraic geometry, Sendai, 1985, pp. 283–360, Adv. Stud. Pure Math., 10, North-Holland, Amsterdam, 1987. The proof in Kawamata-Matsuda-Matsuki is similar in spirit to the proof in my original answer, so I will leave that answer below. In place of the Hodge symmetries, K-M-M use Kawamata-Viehweg vanishing (the original proofs of that theorem use the Hodge decomposition theorem and Kodaira vanishing). In particular, the argument using the Leray spectral sequence and intersecting with general hyperplanes is in the proof in Kawamata-Matsuda-Matsuki (that is probably where I first learned those techniques). There is an extension of Kawamata-Viehweg vanishing to positive characteristic ala Deligne-Illusie (under appropriate hypotheses, of course). So there is probably also an extension of the Elkik-Fujita Vanishing Theorem. To deduce your result from Theorem 1.3.1, apply the theorem to the morphism $g:Z\to Y$ with $E$ equal to the (scheme-theoretic) support of the cokernel of the natural morphism, $$ f^*: f^*\omega_{Y/k} \to \omega_{Z/k}. $$ Take $L$ to be the invertible sheaf $f^*\omega_{Y/k}$ (this will pull through the derived pushforward by the projection formula). Define $D$ and $\widetilde{D}$ to be the empty divisors. Define $\widetilde{L}$ to be the structure sheaf. Original answer. Let me expand a bit on my comment. First of all, for every field $k$, for every birational, proper morphism of $k$-schemes, $g:Z\to Y$, each of the following natural $\mathcal{O}_Y$-module homomorphisms is an isomorphism, $$g^q: \Omega^q_{Y/k} \to g_*\Omega^q_{Z/k}.$$ Since $\Omega^q_{Y/k}$ is locally free, hence torsion-free, and since $g^q$ restricts to an isomorphism on the dense, maximal open $U$ over which $g$ is an isomorphism, the morphism $g^q$ is injective. Next, for every open $V$ of $Y$ and for every section $s$ of $g_*\Omega^q_{Z/k}(V) = \Omega^q_{Z/k}(g^{-1}(V))$, the restriction of $s$ to $U\cap V$ is the image of a section of $\Omega^q_{Y/k}$. The complement of the open subset $U$ has codimension $2$ (by Zariski's Main Theorem, or easier arguments). Since the sheaf $\Omega^q_{Y/k}$ is locally free on a regular scheme, in particular it is $S2$. Thus, the section $s$ extends to a section of $\Omega^q_{Y/k}$ on all of $V$. Therefore $g^q$ is an isomorphism. If $k$ has characteristic $0$, and if $Y$ is projective, then the Hodge symmetries imply that the following maps are also isomorphisms, $$\psi^q:H^q(Y,\mathcal{O}_Y)\to H^q(Z,\mathcal{O}_Z).$$ The claim, to be proved by increasing induction on $m$, is that $\mathcal{O}_Y\to Rg_*\mathcal{O}_Z$ is a quasi-isomorphism in degrees $\leq m$, i.e., $g^\#:\mathcal{O}_Y\to g_*\mathcal{O}_Z$ is an isomorphism and every $R^qg_*\mathcal{O}_Z$ is zero for $00$, and assume that the result is proved for all birational morphisms $g$ of smooth projective varieties for all degrees strictly less than $m$. The claim, to be proved by increasing induction on $d$, is that the support of the sheaf $R^mg_*\mathcal{O}_Z$ has no irreducible component of dimension $d$. For the base case of the induction, assume that the support of $R^mg_*\mathcal{O}_Z$ has an irreducible component of dimension $0$. If there were such a component, then it would give a summand of $R^mg_*\mathcal{O}_Z$ that is a skyscraper sheaf. Thus, $H^0(Y,R^mg_*\mathcal{O}_Z)$ is nonzero. The Leray spectral sequence for the cohomologies $H^m(Z,\mathcal{O}_Z)$ is a stage $2$, first quadrant, cohomological spectral sequence, $$E_2^{p,q} = H^p(Y,R^qg_*\mathcal{O}_Z)\Rightarrow H^{p+q}(Z,\mathcal{O}_Z).$$ By hypothesis, all of the sheaves $R^qg_*\mathcal{O}_Z$ are zero for $00$ be an integer, assume that the vanishing of $R^qg_*\mathcal{O}_Z$ has been proved for all proper birational morphisms from a smooth, proper $k$-scheme to a smooth, projective $k$-scheme for $00,$$ then $R^m(g_H)_*\mathcal{O}_{Z_H}$ equals $R^mg_*\mathcal{O}_Z\otimes_{\mathcal{O}_Y}\mathcal{O}_{Y_H}$. Choose $H$ to be transverse to every $d$-dimensional irreducible component of the support of $R^mg_*\mathcal{O}_Z$ (if there are any). Then the intersection of that component with $H$ gives a $(d-1)$-dimensional irreducible component of the support of $R^m(g_H)_*\mathcal{O}_{Z_H}$. This is a contradiction. Therefore, there are no $d$-dimensional irreducible components of the support of $R^mg_*\mathcal{O}_Z$. By induction on $d$, the support of $R^mg_*\mathcal{O}_Z$ is empty, i.e., this is the zero sheaf. Thus, by induction on $m$, the map $\mathcal{O}_Y\to Rg_*\mathcal{O}_Z$ is a quasi-isomorphism. Now, let $X^o$ be affine and finite type over a field, let $f^o:Y^o\to X^o$ be a projective, birational morphism with $Y^o$ smooth, and let $g^o:Z^o\to Y^o$ be a proper, birational morphism with $Z^o$ smooth. The closure of $X^o$ in a projective space containing affine space is a projective scheme $X$. By the strong resolution of singularities of Hironaka, there exists a projective, birational morphism $f:Y\to X$ with $Y$ smooth and with $f^{-1}(X^o)$ equal to $Y^o$ as an $X^o$-scheme. Similarly, by Nagata compactification and Hironaka, there exists a proper, birational morphism $g:Z\to Y$ with $Z$ smooth and with $g^{-1}(Y^o)$ equal to $Z^o$ as a $Y^o$-scheme. By the previous paragraph, $\mathcal{O}_Y\to Rg_*\mathcal{O}_Z$ is a quasi-isomorphism. By flat base change, also $\mathcal{O}_{Y^o}\to R(g^o)_*\mathcal{O}_{Z^o}$ is a quasi-isomorphism. Thus, applying the derived functor $R(f^o)_*$ also gives a quasi-isomorphism. Since a composition of derived functors of pushforward is the derived functor of pushforward by the composition, taking cohomology modules gives an isomorphism, $$H^q(Y^o,\mathcal{O}_{Y^o})\to H^q(Z^o,\mathcal{O}_{Z^o}).$$ Finally, for any proper, birational morphism $h^o:W^o\to X^o$ with $W^o$ smooth, and for any projective, birational morphism $f^o:Y^o\to X^o$ with $Y^o$ smooth (and such a morphism $f^o$ exists by Hironaka), then again by Hironaka, and by Chow's Lemma, there exists a pair of projective, birational morphisms, $$g^o:Z^o\to X^o, \ \ i^o:Z^o\to W^o,$$ such that $h^o\circ i^o$ equals $f^o\circ g^o$. Covering $W^o$ by open affines and using the previous result with $\text{Id}:W^o\to W^o$ and $i^o:Z^o\to W^o$ in the place of $f^o$ and $g^o$, it follows that $\mathcal{O}_{W^o}\to R(i^o)_*\mathcal{O}_{Z^o}$ is a quasi-isomorphism. Therefore each natural pullback map, $$H^q(W^o,\mathcal{O}_{W^o})\to H^q(Z^o,\mathcal{O}_{Z^o}),$$ is an isomorphism. Therefore every $H^q(W^o,\mathcal{O}_{W^o})$ equals $H^q(Y^o,\mathcal{O}_{Y^o})$.<|endoftext|> TITLE: When can an invertible function be inverted in closed form? QUESTION [21 upvotes]: The Risch algorithm answers the question: "When can a function be integrated in closed form?", see: https://en.wikipedia.org/wiki/Symbolic_integration Is anyone aware of any work that answers the related question, "When can an invertible function be inverted in closed form?" By closed form, I wish to exclude infinite series expansions, unless they describe a special function. I would be happy just to see a short list of some explicitly invertible functions. REPLY [8 votes]: Closed-form functions need the definition which set of functions is allowed to represent the function. Take e.g. the algebraic definition of the Elementary functions by Liouville and Ritt (Wikipedia: Elementary function). There are only few publications that prove that the function given there doesn't have an inverse in closed form. Examples are the proof that the solutions of Kepler's equation are not elementary functions* and the proof that LambertW is not an elementary function**. *) Joseph Liouville Ritt, J. F.: Integration in finite terms. Liouville's theory of elementary methods. Columbia University Press, New York, 1948 **) Why is it that the Lambert W relation cannot be expressed in terms of elementary functions? Dubinov, A.; Galidakis, Y.: Explicit solution of the Kepler equation. Physics of Particles and Nuclei Letters 4 (2007) 213-216 $\ $ 1.) Ritt and Risch Ritt's work Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 treats the topic in general. But it deals only with the elementary functions, and Ritt's method of proof seems to be unfortunately only for the elementary functions. Ritt's theorem is proved also by Risch in [Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759. I assume it is possible to extend this proof and Ritt's theorem to other classes of functions. See my question How to extend Ritt's theorem on elementary invertible bijective elementary functions. Another reference is the last theorem in Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22. Rosenlicht writes: "The preceding theorem is a powerful tool for finding elementary solutions, if such exist, of certain types of transcendental equations, or for proving their nonexistence." Unfortunately, this method is applicable only for certain kinds of equations. But the method is applicable also for some other classes of functions which can be represented by a field. How classes of functions can be represented as a set of functions generated by a field, is treated in differential algebra and is described e.g. in section 1 of Davenport, J. H.: What Might "Understand a Function" Mean. In: Kauers, M.; Kerber, M., Miner, R.; Windsteiger, W.: Towards Mechanized Mathematical Assistants. Springer, Berlin/Heidelberg, 2007, page 55-65. $\ $ It is possible to extend at least one part of Ritt's results directly as follows: a) Let $K_0$ a field, $n\in\mathbb{N}_+$, $X,Y_1,...,Y_n$ sets. $A\colon Y_1\times \cdots\times Y_n \mapsto A(Y_1\times ...\times Y_n), (x_1,...,x_n)\mapsto A(x_1,...,x_n)$ be a function, algebraic over the field $K_0$. $f_1\colon X\to Y_1, x\mapsto f_1(x)$; ...; $f_n\colon X\to Y_n, x\mapsto f_n(x)$ be bijective transcendental functions. $F\colon X\to Y, x\mapsto A(f_1(x),...,f_n(x))$ be a bijective function. $K$ be the extension field of $K_0$ which is generated from $K_0$ only by applying algebraic operations, the identity function and the inverses of $f_1,...,f_n$. If $f_1,...,f_n$ are pairwise algebraically dependent over $K_0$, then $F$ has an inverse in $K$. That means, roughly spoken, each bijective iterated composition of unary univalued functions $f_i$ (maybe this can be extended to $n_i$-ary $n_i$-valued functions) is invertible in the algebraic closure of the differential field which is generated by the $f_i$ and their inverses. This, if formulated as theorem, is important, because each function on an open domain can be made bijective by restriction of its domain. This plays a role at inverting functions and solving equations by partial inverses. I gave an example in the answer to Algebraic solution to natural logarithm equations like $1-x+xln(-x)=0$. $\ $ The following is an unproven assumption of mine: If $trdeg_{K_0}K_0(f_1,f_2,...,f_n)>0$, then $F$ cannot have an inverse in $K$. An indication for this assumption: The defining equations for the inverse of $F$, $F^{-1}(F(x))=x$ and $F(F^{-1}(x))=x$, cannot be solved by applying only functions from $K$. That means, it is not possible to calculate the inverse of $F$ by solving this equations on that way. But Ritt's proof goes further: Ritt $proves$ that a corresponding elementary function $F$ cannot have an inverse in the elementary functions. b) One can easily prove the Theorem: Let $n \in \mathbb{N}_0$, $f_{1},...,f_{n}$ bijective functions, $f=f_{n}\circ f_{n-1}\circ\ ...\ \circ f_{2}\circ f_{1}$ a bijective function, $\phi$ the inverse of $f$. Then $\phi=\phi_{1}\ \circ\ \phi_{2}\ \circ\ ...\ \circ\ \phi_{n-1}\ \circ\ \phi_{n}$, wherein for all $i$ with $1\leq i\leq n$, $\phi_{i}$ is the inverse of $f_{i}$. For inverses in the elementary functions and LambertW, I could apply this theorem for my answer at https://math.stackexchange.com/questions/2309691/equations-solvable-by-lambert-function/2527410#2527410. $\ $ 2.) Lin and Chow As I found by an obvious but possibly new theorem (Proof Check: Nonexistence of the inverse function in a given class of functions), there is a connection between solvability of particular kinds of equations of the form $f(z)=\alpha$ in a given set of numbers and invertibility of functions $f$ in a given set of functions. Nonexistence theorems for solutions of irreducible polynomial equations $P(z,e^z)=0$ in the Elementary numbers and in the Explicit elementary numbers, respectively are given in: Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50 Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448 Are there further nonexistence statements known for solutions of equations in other classes of numbers? $\ $ 3.) Arnold and Khovanskii The topological method of Vladimir Arnold and Askold Khovanskii allow the extension of the problem to a very large class of functions. Khovanskii, A.: Topological Galois Theory - Solvability and Unsolvability of Equations in Finite Terms. Springer 2014, Khovanskii, A.: One dimensional topological Galois theory. 2019: "Definition. A multivalued analytic function of one complex variable is called a $S$-function, if the set of its singular points is at most countable." "Theorem 3 (on stability of the class of $S$-functions). The class $S$ of all $S$-functions is stable under the following operations: differentiation, i. e. if $f∈S$, then $f′ ∈ S$; integration, i. e. if $f ∈ S$ and $g′ = f$, then $g ∈ S$; composition, i. e. if $g$, $f ∈ S$, then $g ◦ f ∈ S$; meromorphic operations, i. e. if $f_i ∈ S$, $i=1,...,n$, the function $F(x_1,...,x_n)$ is a meromorphic function of $n$ variables, and $f=F(f_1,...,f_n)$, then $f ∈ S$; solving algebraic equations, i. e. if $f_i ∈ S$, $i=1,...,n$, and $f^n+f_1f^{n−1}+···+f_n=0$, then $f ∈ S$; solving linear differential equations, i.e. if $f_i ∈ S$, $i=1,...,n$, and $f^{(n)}+f_1f^{(n−1)}+···+f_nf=0$, then $f ∈ S$. Remark. Arithmetic operations and the exponentiation are examples of meromorphic operations, hence the class of $S$-functions is stable under the arithmetic operations and the exponentiation. Corollary 4 (see [2]). If a multivalued function $f$ can be obtained from single valued $S$-functions by integration, differentiation, meromorphic operations, compositions, solutions of algebraic equations and linear differential equations, then the function $f$ has at most countable number of singular points. Corollary 5. A function having uncountably many singular points cannot be represent by generalized quadratures. In particular it cannot be a generalized elementary function and it cannot be represented by k-quadratures or by quadratures." "Theorem 6 (see [2]). The class of all S-functions, having a solvable monodromy group, is stable under composition, meromorphic operations, integration and differentiation." Kanel-Belov, A.; Malistov, A.; Zaytsev, R.: Solvability of equations in elementary functions. 2019: "Suppose we are trying to solve an equation $f(x) = a$ in the complex plane. If $f(x)$ is holomorphic and non-constant, then, by the well-known uniquenuess principle, for any $a$, the set of roots is discrete. ... Corollary 2.2. For any N, the group induced by elementary functions and their compositions of depth not exceeding N is solvable ... The last corollary means that if we can choose some curves, such that the group generated by their induced permutations is unsolvable, we will have proved the unsolvability in elementary functions. Remark. Calculating the group induced by a given set of curves is rather difficult. This could be automated and done by a computer. Also, if the number of critical points is finite, there is only a finite number of curves, and therefore the whole proof could be done by a machine." Zelenko, L.: Generic monodromy group of Riemann surfaces for inverses to entire functions of finite order. 2021: "We consider the vector space $E_{\rho,p}$ of entire functions of finite order $\rho\in\mathbb{N}$, whose types are not more than $p > 0$, endowed with Frechet topology, which is generated by a sequence of weighted norms. We call a function $f\in E_{\rho,p}$ typical if it is surjective and has an infinite number critical points such that each of them is non-degenerate and all the values of $f$ at these points are pairwise different. ... Furthermore, we show that inverse to any typical function has Riemann surface whose monodromy group coincides with finitary symmetric group $FSym(\mathbb{N})$ of permutations of naturals, which is unsolvable in the following strong sense: it does not have a normal tower of subgroups, whose factor groups are or abelian or finite. As a consequence from these facts and Topological Galois Theory, we obtain that generically (in the above sense) for $f\in E_{\rho,p}$ the solution of equation $f(w) = z$ cannot be represented via $z$ and complex constants by a finite number of the following actions: algebraic operations (i.e., rational ones and solutions of polynomial equations) and quadratures (in particular, superpositions with elementary functions)."<|endoftext|> TITLE: Monads on Set with trivial algebras QUESTION [8 upvotes]: In an earlier post, What is known about the category of monads on Set? the following observation was made: What's more, all but two monads on Set have the property that there exists an algebra with more than one element. One of the exceptions is the monad M with M(A)=1 for all sets A; it's the theory generated by a single constant e and the equation x=e. The other is the monad M with M(A)=1 for all nonempty sets A and M(0)=0; that's the theory generated by no operations and the equation x=y. Does anyone know a proof (or a reference to a proof) of this fact--that these are the only two monads on sets having trivial algebras? REPLY [13 votes]: I don't know where this observation was first made, but the proof is short. Let $M$ be a monad on $Set$ such that every $M$-algebra has at most one element. For every set $A$, the set $M(A)$ has the structure of an $M$-algebra (a free one), so $M(A)$ has at most one element. On the other hand, the unit of the monad gives us a map $A \to M(A)$. Since there is no map from a nonempty set to the empty set, this implies that $M(A) = 1$ whenever $A$ is nonempty. So, $M(\emptyset)$ is either $\emptyset$ or $1$, and $M(A) = 1$ for all nonempty $A$. In either case, $M$ becomes a monad in a unique way. This gives the two monads mentioned.<|endoftext|> TITLE: Tensor power of the natural representation of Sn QUESTION [13 upvotes]: The symmetric group $S_n$ acts over $V=\mathbb{R}^n$ by permuting the canonical basis. So it acts over $V^{\otimes p}$ with a diagonal action (acts the same over each element of the tensor product). I'd like to find the decomposition into irreps of it. As I'm a physicist, I'm first interested in the simple cases $p=2$ (and if this is still difficult, $n=3,4$), even if something general would be nice! I also need something constructive: I guess the proof will be constructive, but I'd prefer a reference where the basis vectors are easily tractable. Thanks! REPLY [15 votes]: This question was recently completely solved in this paper. As explained on page 15 of the paper, letting $v_1,...,v_n$ be a basis for $V$, the standard basis vectors $v_{i_1} \otimes \cdots \otimes v_{i_p}$ for $V^{\otimes p}$ naturally correspond to partitions of the set $\{1,...,p\}$ into at most $n$ blocks. The submodule corresponding to a particular partition into $t$ blocks is isomorphic to the permutation module $H^{(n-t,1^t)}=Ind_{S_{n-t}}^{S_n} 1$. Multiplicities of irreducibles in $H^{\lambda}$ are well-known to be Kostka numbers $K_{\lambda, \mu}$, so we get that the multiplicity of the irreducible $S_{\lambda}$ in $V^{\otimes p}$ is $$\sum_{t=0}^n S(p,t) K_{\lambda, (n-t,1^t)}$$ Where $S(p,t)$ denotes the Stirling number. The paper also gives a bijective proof of this equality using paths in the relevant Bratteli diagram. The fact that you are working over $\mathbb{R}$ rather than $\mathbb{C}$ does not matter since all complex irreducible representations of $S_n$ can in fact be realized over $\mathbb{Z}$.<|endoftext|> TITLE: The geometry of the solution set of a symmetric equation in four symmetric matrices QUESTION [15 upvotes]: Let $n$ be a natural number. We can view the space of invertible symmetric matrices over a field as an open subset of$\mathbb A^{(n^2+n)/2}$. Inside the fourth power of this space, we have the closed subscheme consisting of tuples satisfying $A_1 + A_2 = A_3 + A_4$ and $A_1^{-1} + A_2^{-1} = A_3^{-1}+ A_4^{-1}$. Is this subscheme a complete intersection of dimension $n^2+n$? How many irreducible components does this subscheme have? The motivation is that this would evaluate the fourth moment of symplectic Kloosterman sums, in the same way that Kloosterman's classical argument evaluates the fourth moment of the usual Kloosterman sums. However, I don't expect techniques from number theory to be helpful here. The $n=1$ case has three irreducible components of dimension $2$, given by equations as follows $(x_3=x_1,x_4=x_2),(x_3=x_2,x_4=x_1),(x_2=-x_1,x_4=-x_3)$. Using these, we can make $3^n$ $2n$-dimensional families of diagonal examples. All of these are contained in an irreducible component of dimension at least $n^2+n$, as the scheme is defined by only $n^2+n$ equations in $2n^2 +2n$-space. However, only for a few obvious ones can I locate an $n^2+n$-dimensional family containing them. Let me express what I think remains to be done after the two answers already given. Using Alex Gavrilov's algebraic formulation, I think we can classify all the irreducible components that arise when $A_1+A_2$ and $A_1^{-1}+A_2^{-1}$ are invertible. One might conjecture that all irreducible components arise on invertible irreducible component on one subspace and an irreducible component where $A_1+A_2=0$ on another subspace, as in David Speyer's answer. So the main thing to do that I don't know how to do is verify or disprove this conjecture. REPLY [3 votes]: This non-answer grew too long for a comment. Let's change coordinates, $B_i = \check\rho(t)A_i$, where $\check\rho(z)$ is the diagonal matrix $diag(t,t^2,t^3,\ldots,t^n)$. Then the original equations $A_i = A_i^T, A_1+A_2 = A_3+A_4, A_1^{-1}+A_2^{-1} = A_3^{-1}+A_4^{-1}$ become $B_i = \check\rho(t) B_i^T \check\rho(t^{-1})$ and the other two unchanged. If you take $t\to\infty$ then instead of considering symmetric matrices you're considering upper triangular. There's a serious issue that these may not be enough equations to define the limit (may not be "Gröbner enough"), but if they define a complete intersection then (they are Gröbner enough and) the original was a complete intersection. The natural action of $GL(n)$ on the original four matrices, $g\cdot A_i = g A_i g^T$, degenerates/extends to an action of $B\times B$, left and right independent multiplication by upper triangular matrices. Using that we can assume $B_1 = \bf 1$, and still have a conjugation action on the other three $B_i$. So we've reduced to the question, is the space of quadruples $(1,B_2,B_3,B_4)$ of upper triangular matrices, satisfying your original two matrix equations, only $n+1\choose 2$-dimensional. Or even stronger, is (this space)/(the residual $B$-action) only $1$-dimensional. For $n=2$, the resulting space is indeed still a complete intersection, now with 12 components, of degrees {3,3,3,3,3,3, 2,2,2, 1,1,1}. I didn't look into how the original components break into these.<|endoftext|> TITLE: An explicit representation for polynomials generated by a power of $x/\sin(x)$ QUESTION [9 upvotes]: The coefficients $d_{k}(n)$ given by the power series $$\left(\frac{x}{\sin x}\right)^{n}=\sum_{k=0}^{\infty}d_{k}(n)\frac{x^{2k}}{(2k)!}$$ are polynomials in $n$ of degree $k$. First few examples: $$d_{0}(n)=1,\quad d_{1}(n)=\frac{n}{3}, \quad d_{2}(n)=\frac{2 n}{15}+\frac{n^2}{3}, \quad d_{3}(n)=\frac{16n}{63}+\frac{2 n^{2}}{3}+\frac{5n^3}{9}.$$ Question: Is there an explicit formula for the coefficients of polynomials $d_{k}(n)$? Remark: I am aware of their connection with the Bernoulli polynomials of higher order $B_{n}^{(a)}(x)$. Namely, one has $d_{k}(n)=(-4)^{k}B_{2k}^{(n)}(n/2)$. This formula and several other alternative expressions can be found in the book of N. E. Norlund (Springer, 1924) but none of them seems to be very helpful. REPLY [2 votes]: This is really a comment rather than an answer, but it's too long for a comment. Here's a simple proof that the coefficients of the polynomials $d_k(n)$ are positive. We have $$\left(\frac{x}{\sin x}\right)^n = \exp\left(n\log\left(\frac{x}{\sin x}\right)\right)$$ so the coefficient of $n^l$ in $d_k(n)$ is the coefficient of $x^{2k}/(2k)!$ in $\bigl(\log(x/\sin x)\bigr)^l/l!$. So it suffices to show that $\log(x/\sin x)$ has positive coefficients. But $$\frac{d\ }{dx} \log \left(\frac{x}{\sin x}\right) = \frac{1}{x} -\cot x$$ in which the coefficients are easily seen to be positive by expressing the coefficients of $\cot x$ in terms of Bernoulli numbers: \begin{align*}\cot x&= \sum_{n=0}^\infty (-1)^n 2^{2n} B_{2n} \frac{x^{2n-1}}{(2n)!}\\ &=\frac{1}{x}- \sum_{n=1}^\infty 2^{2n} |B_{2n}| \frac{x^{2n-1}}{(2n)!}. \end{align*}<|endoftext|> TITLE: Rational generating function and recursion QUESTION [6 upvotes]: Let $\lambda$ denote a hook of size $d$ and $c(\Box)$ the content of $ \Box \in \lambda $. Let $ \text{Hooks}(d) $ be the set of hooks with $d$ boxes. Define \begin{align} B(d)&= \frac{1}{d!} \sum_{\lambda\in \text{Hooks}(d)} (-1)^{ht(\lambda)-1} \, \dim \lambda \,\color{red}{\prod_{\Box \in \lambda}(1-c(\Box)h)^2}\\ &= \frac{1}{dd!}\sum_{\ell=1}^{d} (-1)^{(\ell+1)}\binom{d-1}{\ell-1}\prod_{i=1}^{d}(1-(\ell-i)h)^2 \end{align} So $B(d)$ is a polynomial in $h$ I have noticed the following recursion for $B(d)$. \begin{align} d(d+1)B(d)=(d-1)(4d-2)B(d-1)+h^2(d-1)^2(d-2)^2B(d-2) \end{align} I am now trying to prove the recursion by comparing the coefficient of $h^k$ on the R.H.S and L.H.S. I have a combinatorial interpretation for the leading coefficient for a fixed $d$ and can prove that they are equal. I still do not have a combinatorial interpretation for other powers of $h^k$. Also proving the recursion in this way cannot be generalised. For example if I replace the $\color{red} {red \, part}$ of my equation with $\prod_{i=1}^{d} (1-(\ell-1)h)^m$ then it would be a different with differential combinatorial interpreation to prove it. I have seen theorems involving rational generating function in chapter 4 volume 1 of Enumerative combinatorics. Theorem 4.1.1 states that if the generating function is rational, then, there is a recurrence among the coefficient. Note that my $B(d)$ are not complex numbers but polynomial in $h$. Showing that $\sum_{d \geqslant 0} B(d) X^d $ is a rational function in $X$ would thus prove the existence of a recurrence, which is what I am looking for. Also, I have noticed that my recursion is a three-term recursion between polynomials in $h$, so I was wondering if there is a known technique from the theory of orthogonal polynomial that would prove such things? Of course, this is not a typical three-term relationship since it does not involve any multiplication by $h$. Any ideas would be very helpful. REPLY [2 votes]: $\textit{Lemma :}$ We have \begin{align*}%$ \frac{1}{d!} \sum_{\lambda\in \text{Hooks}(d)} (-1)^{\ell(\lambda)-1} \, \dim \lambda \prod_{\Box \in \lambda}(x + c(\Box) )(y + c(\Box) ) = [c_d]\prod_{k = 1}^d (x + J_k) (y + J_k) \end{align*} where \begin{align*}%$ J_1 = 0 \, ; J_2 = (1, 2) \, ; \qquad J_k = (1, k) + \cdots + (k - 1, k) \end{align*} are the Jucys-Murphy elements of the symmetric group $ \mathfrak{S}_d $, i.e. particular elements of the group algebra $ \mathbb{C}\mathfrak{S}_d $ that commute with each other and that generate the centre $ Z(\mathbb{C}\mathfrak{S}_d) $ in a certain way (symmetric functions in the JM elements generate it). Here we have noted $ [\sigma]f $ the coefficient of $\sigma$ in $ f \in \mathbb{C}\mathfrak{S}_d $, namely, if $ f = \sum_{ \sigma \in \mathfrak{S}_d } f_\sigma \sigma $, then $ [\sigma]f = f_\sigma $. Last, $ c_d $ is any $d$-cycle, for instance $ (1, 2, \dots, d) $. $ $ $\textit{Proof.}$ We prove this last formula using the dual orthogonality of the characters of $ \mathfrak{S}_d $ : denote by $ \chi^\lambda $ such a character. Then, for all $ \sigma, \tau \in \mathfrak{S}_d $ \begin{align*}%$ \frac{1}{d!} \sum_{ \lambda \vdash d } \chi^\lambda(\sigma) \chi^\lambda(\tau) = \delta_{\sigma, \tau} \end{align*} We have used the fact that $ \chi^\lambda(\sigma) \in \mathbb{Z} $ to avoid writing a conjugation on one of the characters. The next step is the formula (that one can prove with the Murnaghan-Nakayama formula) \begin{align*}%$ \chi^\lambda(c_d) =: \chi^\lambda_{(d)} = (-1)^{\ell(\lambda) + 1} \mathbb{1}_{ \{ \lambda \in \mathrm{Hooks}(d) \} } \end{align*} We thus have \begin{align*}%$ \delta_{\sigma, c_d} & = \frac{1}{d!} \sum_{ \lambda \vdash d } \chi^\lambda(\sigma) \chi^\lambda(c_d) \\ & = \frac{1}{d!} \sum_{ \lambda \in \mathrm{Hooks}(d) } (-1)^{\ell(\lambda) + 1} \dim(\lambda) \frac{\chi^\lambda(\sigma) }{\dim(\lambda)} \end{align*} Defining the normalised character by \begin{align*}%$ \widehat{\chi}^\lambda := \frac{\chi^\lambda }{\dim(\lambda)} = \frac{\chi^\lambda }{ \chi^\lambda(id_d) } \end{align*} we then use the following (fundamental) formula that links content alphabets with Jucys-Murphy elements : for all symmetric function $ F $ in $d$ variables, for all $ \lambda \vdash d $ \begin{align*}%$ \widehat{\chi}^\lambda( F(J_1, \dots, J_d) ) = F(\mathcal{A}^\lambda) \end{align*} where $ \mathcal{A}^\lambda $ is the $\textit{content alphabet}$ defined by \begin{align*}%$ \mathcal{A}^\lambda = \{ c(\square), \square \in \lambda \} \end{align*} My personal reference for this machinery is some articles of Biane (a priori probability theory...), but the people on this forum certainly know some better (more recent) exposition. Using this and the fact that we have extended characters by linearity in the group algebra, i.e. $ \chi^\lambda(f) = \sum_{\sigma \in \mathfrak{S}_d} f_\sigma \chi^\lambda(\sigma) $ for $ f $ previously defined, one gets \begin{align*}%$ [c_d]f & = \sum_{\sigma \in \mathfrak{S}_d} f_\sigma \delta_{\sigma, c_d} \\ & = \sum_{\sigma \in \mathfrak{S}_d} f_\sigma \frac{1}{d!} \sum_{ \lambda \vdash d } \chi^\lambda(\sigma) \chi^\lambda(c_d) \\ & = \frac{1}{d!} \sum_{ \lambda \vdash d } \chi^\lambda(f) \chi^\lambda(c_d) \\ & = \frac{1}{d!} \sum_{ \lambda \in \mathrm{Hooks}(d) } (-1)^{\ell(\lambda) + 1} \dim(\lambda) \widehat{\chi}^\lambda(f) \end{align*} The lemma is thus proven for the following particular element of the (centre of the) symmetric group algebra \begin{align*}%$ F_{x, y}(J_1, \dots, J_d) := \prod_{k = 1}^d (x + J_k)(y + J_k) \end{align*} $ \square $ $ $ Now, to come back to your initial problem, we use the following Jucys-Murphy identity (or chinese restaurant theorem for the Ewens measure in probability theory) \begin{align*}%$ \prod_{k = 1}^d (x + J_k) = \sum_{\sigma \in \mathfrak{S}_d} x^{C(\sigma)}\sigma \end{align*} with $ C(\sigma) $ the total number of cycles of $ \sigma $. Taking the product in the group algebra, namely, the convolution product, one gets \begin{align*}%$ \prod_{k = 1}^d (x + J_k)(y + J_k) & = \sum_{\sigma \in \mathfrak{S}_d} x^{C(\sigma)}\sigma \sum_{\tau \in \mathfrak{S}_d} y^{C(\tau)}\tau \\ & = \sum_{\sigma \in \mathfrak{S}_d} \left( \sum_{ \tau \in \mathfrak{S}_d } x^{ C(\tau) } y^{C(\sigma\tau^{-1})} \right) \sigma \end{align*} Taking the coefficient $ [c_d] $, we thus have \begin{align*}%$ [c_d]\prod_{k = 1}^d (x + J_k)(y + J_k) = \sum_{ \tau \in \mathfrak{S}_d } x^{ C(\tau) } y^{C(c_d\tau^{-1})} \end{align*} You are concerned with the case $ x = y $, which gives \begin{align*}%$ \frac{1}{d!} \sum_{\lambda\in \text{Hooks}(d)} (-1)^{\ell(\lambda)-1} \, \dim \lambda \prod_{\Box \in \lambda}(x + c(\Box) )^2 = \sum_{ \sigma \in \mathfrak{S}_d } x^{ C(\sigma) + C(c_d^{-1}\sigma) } \end{align*} If you want the coefficients of your polynomial, you get \begin{align*}%$ \sum_{ \sigma \in \mathfrak{S}_d } x^{ C(\sigma) + C(c_d^{-1}\sigma) } = \sum_{k = 2}^{2d} a_{k, d} x^k, \qquad a_{k, d} := \#\{ \sigma \in \mathfrak{S}_d \, / \, C(\sigma) + C(c_d^{-1}\sigma) = k \} \end{align*} I personally doubt that this is simple to express. Maybe with some Hurwitz numbers or some free probability (the set defining $ a_{d + 1, d} $ defines the non-crossing permutations). You can try your recurrence relation with this new expression.<|endoftext|> TITLE: Maximum average Euclidean distance between $n$ points in $[-1,1]^n$ QUESTION [7 upvotes]: For my research I have designed a metric that is based on the average Euclidean distance between $n$ points in the $n$-dimensional hypercube $[-1,1]^n$. However, I have a hard time finding the maximal average Euclidean distance between $n$ points in $[-1,1]^n$. I need this maximum to normalize my metric between $0$ and $1$ . Please help! REPLY [4 votes]: Not what you are asking for, but I wanted to point out that the problem is very easy if you work with squared distances. Formulate the problem as the following optimization problem: \begin{align} \max \quad & \sum_{i\neq j} \| x_i- x_j\|_2^2 \\ s.t. \quad & x_i(k) \in \{-1,1\}, \end{align} where $x_i\in \mathbb{R}^n$. If the solution is at the corners, the set of maximizers is non convex (consider permutations). Observe that $ \| x_i- x_j\|_2^2=\sum_{k=1}^n (x_i(k)-x_j(k))^2= \sum_{k=1}^n x_i(k)^2+x_j(k)^2 -2 x_i(k) x_j(k)$. For the corner points, we have $x_i(k)^2=1$, thus the problem becomes \begin{align} z=\min \quad & \sum_k\sum_{i\neq j} x_i(k) x_j(k) \\ s.t. \quad & x_i(k) \in \{-1,1\}, \end{align} where the optimal objective of the former problem is $n^2(n-1)-2z$. The latter problem decouples over $k$. For each $k$, we solve \begin{align} \min \quad & \sum_{i\neq j} x_i(k) x_j(k) \\ s.t. \quad & x_i(k) \in \{-1,1\}, \end{align} This basically is a max-cut problem for an n-node complete graph. Suppose n is even, the optimal solution is achieved by setting half of $x_i(k)$ to $+1$ other half to $-1$. Corresponding objective value is $2\frac{(n/2) ((n/2)-1)}{2}-(n/2)^2=-n/2$. Thus, $z=-n^2/2$. And the optimal objective of the original problem is $n^3$. Average of these squared distances is $n^2$.<|endoftext|> TITLE: Simple question on Kirby move QUESTION [7 upvotes]: From hyperbolic volume computation, I found that the following two 3-manifolds are (possibly orientation-reversal) homeomorphic: surgery on figure-eight knot $4_1$, with slope $-5$, and surgery on $5_2$ knot with slope $5$. Is there simple way to show it using Kirby calculus? It must be easy but I am not that familiar with Kirby calculus. REPLY [16 votes]: Yes, there is a simple way. Below is a sequence of pictures illustrating the procedure (created using Kirby calculator). $5_2$: Blowup at the clasp: Isotopy: Blowdown the purple unknot:<|endoftext|> TITLE: Inequality between nuclear norm and operator norm for positive definite matrices QUESTION [6 upvotes]: I will use $\|\|_*$ to denote the nuclear norm (sum of singular values) and $\|\|_2$ to denote the operator norm / matrix 2-norm (largest singular value). Consider two positive definite $n \times n$ matrices $A$ and $B$ such that $\text{trace}(A)=\text{trace}(B)\equiv k$. I am interested in finding the largest constant $c \geq 0$ such that the following inequality holds: $$ \|\sqrt{B}^{-1} A \sqrt{B}^{-1}\|_2 \geq c \|A - B\|_*.$$ Unable to obtain a proof, I have done some numerical investigation and I have noticed that it's easy to get violations of the inequality if the trace of the matrices $k$ is chosen to be much larger than $n$ (with fixed $c$). However, as the trace decreases, so do the violations, and for smaller values of $k$ I was not able to find numerical counterexamples unless the value of $c$ gets very large. I am curious about whether, given $n$ and $k$, there is a value of $c$ such that the inequality always holds. Or perhaps whether the inequality always holds for some $n$-dependent $c$ when one simply sets $k=1$. I would appreciate any thoughts and ideas. REPLY [4 votes]: There is essentially no $n$ dependence here. It suffices to consider the case $k=1$ (by rescaling), and then $c=1/2$ works for all $n$. This follows because $$ \|B^{-1/2}AB^{-1/2} \| \ge 1 ,\quad\quad\quad\quad (1) $$ and clearly $\|A-B\|_1\le 2$. (So there is no interaction between the terms, the inequality just says that $\min (LHS)\ge\max (RHS)$.) To prove (1), let's write $\lambda_j$ and $\mu_j$ for the eigenvalues of $A$ and $B$, respectively, arranged in increasing order. By our assumptions, $0<\lambda_j,\mu_j\le 1$. I'll compute the norm of $M=B^{-1/2}AB^{-1/2}$ by maximizing the quadratic form. I start out by testing $M$ on $B^{1/2}v_n$, where $Av_n=\lambda_n v_n$; it follows that $\|M\|\ge \lambda_n/\mu_n$. Next, I test $M$ on the space spanned by $B^{1/2}v_j$, $j=n-1,n$. Since this space is two-dimensional, its intersection with the orthogonal complement of $w_n$, $Bw_n=\mu_n w_n$, is still at least one-dimensional, so we now see that also $\|M\|\ge \lambda_{n-1}/\mu_{n-1}$. We can continue in this style, and we in fact have that $\|M\|\ge \lambda_j/\mu_j$ for all $j$. Since $\sum\mu_j=1$ and $\sum (\lambda_j/\mu_j)\mu_j=1$ also, there is a $j$ with $\lambda_j/\mu_j\ge 1$. This gives (1), and it's also clear that (1) is optimal, since we can take $A=B$.<|endoftext|> TITLE: Handle decompositions using only 1-handles QUESTION [7 upvotes]: Let $\Sigma$ be an oriented, compact, connected 2-manifold with boundary. Assume that its boundary is equipped with a disjoint union decomposition into two non-empty parts: $$\partial\Sigma=\partial_{in}\Sigma\cup\partial_{out}\Sigma$$ (both $\partial_{in}\Sigma$ and $\partial_{out}\Sigma$ are disjoint unions of circles). The following should be well known, and yet... $\bullet\quad$ $\Sigma$ admits a relative handle decomposition that only uses $1$-handles (no need to use $0$-handles or $2$-handles). [Ok, that's easy] $\bullet\quad$ One can go from any handle decomposition of the above type to any other one by using: (1) isotopies (2) handle slides (no need to ever introduce $0$-handles or $2$-handles). [That's the harder question] If someone could provide a short and elegant proof of the above result, that would make me very happy. Here, by a relative handle decomposition that only uses $1$-handles, I mean an identification between $\Sigma$ and something of the form $$ \Sigma':=\text{pushout}\left((\partial_{in}\Sigma\times I) \stackrel\varphi\longleftarrow \bigsqcup_{i=1}^n(\partial I\times I) \hookrightarrow \bigsqcup_{i=1}^n(I\times I)\right), $$ where $\varphi$ is an embedding $\bigsqcup_{i=1}^n(\partial I\times I)\stackrel\varphi\to \partial_{in}\Sigma\cong \partial_{in}\Sigma\times\{1\}\hookrightarrow \partial_{in}\Sigma\times[0,1]$. Moreover, the identification $\Sigma\cong\Sigma'$ should commute with the obvious embeddings of $\partial_{in}\Sigma$ into $\Sigma$ and into $\Sigma'$. REPLY [10 votes]: The second statement ought to be in the literature somewhere but I don't know a reference so I'll give an argument. The result can be rephrased in terms of graphs. Let $S$ be a compact connected surface with non-empty boundary and let $P$ be a non-empty finite set of points in the interior of $S$. Consider finite connected graphs $X$ in $S$ with $P$ as vertex set and such that the components of $S-X$ are half-open annuli each having its boundary circle a component of $\partial S$. From $X$ we obtain a handle decomposition of $S$ with no 2-handles, where the 0-handles are disks about the vertices of $X$ and the 1-handles are neighborhoods of the edges of $X$ minus these disks, the rest of $S$ being just a collar on $\partial S$. Let $\cal G$ be the set of all isotopy classes of such graphs $X$, where isotopies fix $P$. The assertion to be proved is that any two graphs in $\cal G$ are related by a finite sequence of handle slides, moving one end of an edge across an adjacent edge. To show this we first enlarge the set $\cal G$ by allowing extra edges to be added to graphs in $\cal G$, keeping the same vertex set. This creates new complementary disks, and we require that none of these is a monogon or bigon. (This restriction may not be necessary for the argument but is usually made to avoid getting graphs that are unnecessarily large.) The resulting enlargement $\overline{\cal G}$ of $\cal G$ is a poset under inclusion, whose geometric realization we also call $\overline{\cal G}$. From a graph $X$ in $\overline{\cal G}-{\cal G}$ we can obtain a graph $X'$ in $\cal G$ by deleting some of its edges. There may be different ways to do this, but we claim that the resulting graphs $X'$ are all related by handle slides (and isotopy). To see this, consider the dual graph $X^*$ whose vertices correspond to the components of $S-X$ and whose edges correspond to the edges of $X$. Then a choice of edges of $X$ to delete to obtain $X'$ corresponds to a choice of maximal tree $T$ in the quotient graph $X^*_{\partial}$ of $X^*$ obtained by identifying all vertices corresponding to components of $S-X$ meeting $\partial S$. It is easy to see that any two choices of maximal tree in a finite connected graph are related by a sequence of elementary moves in which one edge is omitted from a maximal tree and replaced by another edge not in the tree. Changing a maximal tree $T$ in $X^*_{\partial}$ by an elementary move corresponds to adding one edge to $X'$ in an annulus component of $S-X'$ and deleting another edge of $X'$. It is evident that this operation can also be achieved by a sequence of handle slides. This verifies the claim that different graphs $X'$ obtained from $X$ by deleting edges are all related by handle slides. It is a well-known fact that $\overline{\cal G}$ is connected, and in fact contractible. A simple proof of connectedness will be sketched below. Assuming this, let us finish the argument. Given two graphs defining vertices of $\cal G$, there is a sequence of vertices $X_1, X_2,\cdots, X_n$ of $\overline{\cal G}$ with $X_1$ the first given graph and $X_n$ the second one, such that each $X_{i+1}$ is obtained from $X_i$ by adding or deleting a set of edges. (We could refine this sequence so that only a single edge is added or deleted at a time.) By what we have shown in the preceding paragraph, any two graphs $X'_i$ and $X'_{i+1}$ in $\cal G$ obtained from $X_i$ and $X_{i+1}$ by deleting edges are then related by handle slides since if $X_i$, say, is obtained from $X_{i+1}$ by deleting edges, we may choose $X'_{i+1}=X'_i$. Thus the two given graphs $X_1=X'_1$ and $X_n=X'_n$ are related by a finite sequence of handle slides and we are done. We can show $\overline{\cal G}$ is connected by a surgery argument. Let $X$ and $Y$ be two graphs in $\overline{\cal G}$. We can isotope one of them to be transverse to the other and intersect it in the minimum number of points within its isotopy class. Let $x$ be a point of $(X\cap Y)-P$ closest to an endpoint of the edge of $X$ containing it, so $x$ cuts off an arc $\alpha$ in this edge whose interior is disjoint from $Y$, with the endpoints of $\alpha$ being $x$ and a point $p\in P$. We can then cut the edge of $Y$ containing $x$ into two arcs with $x$ as their common endpoint and drag the ends of these arcs at $x$ along $\alpha$ to $p$. After a small isotopy these two arcs can be made disjoint from $Y$ except at their endpoints. We can then enlarge $Y$ by adding these two arcs to get a new graph $Y'$ in $\overline{\cal G}$. If one of the two arcs happens to be isotopic to an existing edge of $Y$, so that a component of $S-Y'$ is a bigon, we can just delete this arc to avoid the duplication. The hypothesis that $X$ and $Y$ intersect minimally guarantees that no complementary monogons are created. There is then an edge in $\overline{\cal G}$ from $Y$ to $Y'$. If we delete from $Y'$ the edge of $Y$ that we cut, we obtain a new graph $Y''$ in $\overline{\cal G}$ joined to $Y'$ by an edge of $\overline{\cal G}$. Since $Y''$ meets $X$ in fewer non-vertex points than $Y$ did, this surgery process can be iterated until we obtain a new graph $Y$ meeting $X$ only in its vertices. Then there are edges of $\overline{\cal G}$ joining $X$ to $X\cup Y$ and $X\cup Y$ to $X$ (after deleting parallel edges of $X\cup Y$ as before).<|endoftext|> TITLE: A Rokhlin lemma with a prescribed height function? QUESTION [15 upvotes]: Let $T$ be a ergodic automorphism of a non-atomic Lebesgue probability space $(X, \mathcal{A}, \mu)$. The celebrated Rokhlin tower lemma says that given an integer $n>0$ and $0 < \epsilon < 1$, there exists $B \in \mathcal{A}$ such that the sets $B$, $TB$, ..., $T^{n-1} B$ are disjoint and their union (called a tower of height $n$) has measure $>1-\epsilon$. Here's an equivalent formulation of the conclusion of Rokhlin's lemma: the return time function $R_B \colon B \to \{1,2,\dots, \infty\}$ satisfies: $R_B \ge n$; $\int_B (R_B - n) d\mu < \epsilon$. (To see that the integral is the measure of the complement of the Rokhlin tower, thing about the Kakutani skyscraper with base $B$.) My question is: Can we replace the constant $n$ above by a function $N$? More precisely: Given an arbitrary measurable function $N : X \to \{1,2,\dots\}$, and $0 < \epsilon < 1$, is there a set $B \in\mathcal{A}$ such that: $R_B(x) \ge N(x)$ for all $x\in B$; $\int_B (R_B - N) d\mu < \epsilon$ ? Note that the integral is the measure of the error set $\{T^i(x) \mid x\in B, N(x) \le i < R_B(x) \}$. If $N$ is assumed to be bounded then the answer is yes. Here's the proof, an easy adaptation of the usual proof of the Rokhlin lemma. Suppose $N$ is bounded by a constant $n$. Take $m > n/\epsilon$. Since the set of periodic points has zero measure, we can take a positive measure set $E$ such that the return time $R_E$ is $\ge m$. Consider the Kakutani skyscraper with base $E$, which by ergodicity covers a full measure set. For each point $x_0$ in the base $E$, color blue each of the points $x_0$, $x_1:=T^{N(x_0)}(x_0)$ (which is at height $N(x_0)$), $x_2:=T^{N(x_1)}(x_1)$ (which is at height $N(x_0)+N(x_1)$), etc., stopping when the height becomes $\ge r_E(x_0) - n$ (i.e., when we first reach the floors of the skyscraper $n$-away from the top). Let $B$ be the set of blue points, which is measurable and satisfies $R_B \ge N$. The error set is contained in the region within height $n$ from the top of the skyscraper, and therefore has measure $0$ be given and let $M$ be such that $\mu({x:N(x)>M})<\epsilon$. Now build a Rokhlin tower with height $M/\epsilon$ and error set of size at most $\epsilon/M$. Let $A$ denote the base of the tower (so that $\mu(A)<\epsilon/M$). For each $x\in A$, let $n_1(x)$ be the least integer (up to $M/\epsilon-M$) such that $N(T^{n_1(x)}x)\le M$ (or $n_1(x)=\infty$ if there is no such finite integer). Then let $n_2(x)$ be the least integer greater than $n_1(x)+N(T^{n_1(x)}x)$ (up to $M/\epsilon-M$) such that $N(T^{n_2(x)})\le M$ etc. For each $x\in A$, let there be $k(x)\ge 0$ finite values of $n_j(x)$. Let $B_x=\{T^{n_j(x)}x:j\le k(x)\}$ and let $B=\bigcup_{x\in A} B_x$. This is a measurable set. I claim it has the properties that you want. Let $F=\{x\colon N(x)>M\}\cup E\cup \bigcup_{j=1}^{M-1}T^{-j}(E\cup A)$. This set has measure at most $4\epsilon$. Now for $x\in B$, notice that $$ R_B(x)-N(x)\le \sum_{j=0}^{R_B(x)-1}\mathbf 1_F(T^jx). $$ In particular, it follows that $$ \int_B (R_B-N)\,d\mu\le \mu(F)<4\epsilon, $$ as required. COMMENT: It was pointed out to me by Jairo Bochi, the poser of the question, that it is not necessary to use Rokhlin's lemma in the proof: it suffices to take $A$ to be any set such that $\mu(A\cap T^{-j}A)=0$ for all $1\le j TITLE: Outer automorphism action on representations of $S_6$ QUESTION [13 upvotes]: Let $S_6$ be the symmetric group on 6 letters and let $\alpha \colon S_6 \to S_6$ be an outer automorphism (note that $S_6$ is the only permutation group that has an outer automorphism and that $\mathrm{Out}(S_6) \cong \mathbb{Z}/2\mathbb{Z}$). For any irreducible representation $\rho \colon S_6 \to \mathrm{GL}(V)$ of $S_6$, the composition $\rho \circ \alpha \colon S_6 \to \mathrm{GL}(V)$ is also an irreducible representation. Question: Is there a reference describing the action of this operation ($\rho \mapsto \rho \circ \alpha$) on the set of all irreducible representations of $S_6$ over $\mathbb{C}$ (that is on the set of partitions of 6)? REPLY [4 votes]: First, notice that this operation preserves the dimension of the representation, this already considerably restricts things: the dimensions of the irreps. are 1,1,5,5,5,5,9,9,10,10,16. The trivial rep. is fixed, thus the sign rep. must also be. The 16-dimensional rep. $V_{(3,2,1)}$ is also fixed. Next, since $\alpha$ maps conjugacy classes to conjugacy classes, it preserves the multiset of character values of $\rho$. Checking this against a character table shows that the 9-dimensional reps. are also all fixed. This does allow for the possibility that the 5-dimensional rep. $V_{(5,1)}$ is sent to $V_{(2,2,2)}$ and its conjugate $V_{(2,1,1,1,1)}$ is sent to $V_{(3,3)}$. You can see that one of these pairs is swapped if and only if the other is by tensoring with the sign representation. The 10-dimensional irreps. $V_{(4,1,1)}$ and $V_{(3,1,1,1)}$ may also be swapped. EDIT: After I have written this, Mark Wildon has linked to a question which subsumes this one. His answer there, using an explicit description of $\alpha$'s action on conjugacy classes, is that all of the possible swaps I identified do in fact occur.<|endoftext|> TITLE: Examples of differential towers of groups QUESTION [10 upvotes]: For $r \in \mathbb{Z}_{>0}$, we say that a tower $1=G_0 \subseteq G_1 \subseteq \cdots$ of finite groups is an $r$-differential tower if for all $n$ the branching rules for restriction of irreducible (complex) representations from $G_n$ to $G_{n-1}$ are multiplicity free, and we have that $$ Res^{G_{n+1}}_{G_n} Ind_{G_{n}}^{G_{n+1}} - Ind_{G_{n-1}}^{G_n} Res^{G_{n}}_{G_{n-1}} = r \cdot id $$ viewed as a linear operator on the representation ring $R(G_n)$. For example, the tower of symmetric groups $\mathfrak{S}_0 \subseteq \mathfrak{S}_1 \subseteq \mathfrak{S}_2 \subseteq \cdots$, and more generally the tower $(A \text{ wr } \mathfrak{S}_n)_{n \geq 0}$ of wreath products of a fixed abelian group $A$ and the symmetric groups give $r$-differential towers where $r=|A|$. This is a restatement of the fact that Young's lattice $Y$ and its powers $Y^r$ are $(r$-)differential posets. Are there any other known examples of differential towers of groups? I would be interested in even a small modification of these families (involving, say, alternating groups instead of symmetric groups, or allowing the abelian group $A$ to depend on $n$). REPLY [2 votes]: After thinking about the problem for a while, I recently posted this paper which answers this question when $r$ is one or prime. I show: Theorem: If $r$ is one or prime, then, even without requiring multiplicity-freeness, the towers $(\mathbb{Z}/r \mathbb{Z} \text{ wr } S_n)_{n \geq 0}$ are the only $r$-differential towers of groups. And I conjecture: Conjecture: For general $r$: If we require multiplicity-freeness, then the only examples are $(A \text{ wr } S_n)_{n \geq 0}$ with $A$ an abelian group of order $r$ (this gives the differential poset $Y^r$). If we allow multiplicities, then the only examples are $(H \text{ wr } S_n)_{n \geq 0}$ where $H$ is any group of order $r$ (this gives the dual graded graph $(d_1 Y) \times \cdots (d_k Y)$ where $d_1,...,d_k$ are the dimensions of the irreducible $H$-reps, and where $dY$ is a copy of $Y$ with all edges having multiplcity $d$). I prove the Theorem by inductively showing that the groups must be complex reflection groups, and applying the Shephard-Todd classification. This method cannot work for general $r$ since $A \text{ wr } S_n$ is not a complex reflection group if $A$ is not cyclic.<|endoftext|> TITLE: Question concerning $\text{Spec}(k[[T]]) $ QUESTION [7 upvotes]: Let $k$ be a field. Then consider the rings $k[T] / (T^n)$ with $n \in \mathbb{N}$. The inverse limit of these is given by $k[[T]]$. Passing on to the category of schemes, one concludes that the direct limit of $\text{Spec}(k[T] / (T^n)) $ is given by $\text{Spec}(k[[T]]) $. Now to the question. $\text{Spec}(k[T] / (T^n)) $ can be viewed as one point on the affine line with an infinitisimal neighborhood that remembers some derivatives of functions, right? As the direct limit, $\text{Spec}(k[[T]]) $ is one point on the affine line, that remembers all derivatives of functions. But this scheme has two points, one being a specialisation of the other. Are there any intuitive geometric explanations for this phenomena? REPLY [11 votes]: I'm not sure if you''re looking for something deeper than this (and therefore would have preferred to make this a comment if I could have squeezed it in), but: $Spec(A)$ has to be just rich enough so that every map from $A$ to a field shows up as a function on $Spec(A)$. Now the issue is that, for a field $F$, the functor $Hom_{\bf Rings}(-,F)$ does not preserve inverse limits. So if $A$ is the inverse limit of rings $A_n$, you can't in general understand maps $A\rightarrow F$ as limits of maps $A_n\rightarrow F$, which means you shouldn't expect to understand the points of $Spec(A)$ as limits of points of $Spec(A_n)$. In particular, put $A_n=Spec\Big(k[[t]]/(t^n)\Big)$ and $A=k[[t]]=\lim_\leftarrow A_n$. Put $F=A[t^{-1}]$. Then there is a ring map $A\rightarrow F$ (namely the obvious inclusion) that does not arise from maps $A_n\rightarrow F.$ This necessitates a point in $Spec(A)$ that does not arise from points in the various $Spec(A_n)$.<|endoftext|> TITLE: Gaussian Surface Area of Positive Semidefinite Cone QUESTION [7 upvotes]: Let $\mathbb{R}^n$ be the Euclidean space and $A \subseteq \mathbb{R}^n$ be a sufficiently regular set, e.g., one that has smooth boundary or is convex. We define the $\epsilon$-neighbor of $A$ in the $\ell_2$ sense as \begin{align} A^{\epsilon} = \{ y \in \mathbb{R}^n \colon \text{there exists}~ x \in A ~\text{such that}~ \| x - y \|_{2} \leq \epsilon \}. \end{align} We denote $\gamma$ as the standard $n$-dimensional Gaussian measure $N(0, I_n)$, then the Gaussian surface measure is defined as \begin{align} \tau (A) = \lim_{\epsilon \rightarrow 0} \frac{ \gamma( A^{\epsilon} \setminus A) }{ \epsilon}. \end{align} By results of Keith Ball (the reverse isoperimetric problem for Gaussian measure https://link.springer.com/article/10.1007/BF02573986), there is a universal upper bound of the Gaussian surface area of any convex set. My questions are: 1) when is Keith Ball's upper bound tight? 2) what is a tight upper bound of the Gaussian surface area of the cone of positive semidefinite matrices? REPLY [3 votes]: Regarding the tightness of Ball's bound, which is $O(n^{1/4})$ for a convex set in dimension $n$, this was proven to be sharp (up to a constant factor) by Nazarov, in Nazarov, Fedor, On the maximal perimeter of a convex set in $\mathbb{R}^n$ with respect to a Gaussian measure, Milman, V. D. (ed.) et al., Geometric aspects of functional analysis. Proceedings of the Israel seminar (GAFA) 2001--2002. Berlin: Springer (ISBN 3-540-00485-8/pbk). Lect. Notes Math. 1807, 169-187 (2003). ZBL1036.52014.<|endoftext|> TITLE: Isometric imbedding of ellipsoidal projective plane QUESTION [9 upvotes]: Identifying antipodal points of an ellipsoid (with axes of different length) defines a Riemannian metric on the real projective plane $\mathbb RP^2$. Is there an explicit global isometric imbedding of this metric into Euclidean space $\mathbb R^N$? By explicit I mean using special functions, ellipsoidal harmonics, etc. The ambient dimension $N$ need not be 5 but perhaps not too large. Of course, the same question can be asked for any $\mathbb RP^n$ with an ellipsoidal metric. REPLY [6 votes]: The Veronese embedding provides an isometric embedding $\mathbb{R}\mathrm{P}^n$ into $\mathbb{S}^N\subset \mathbb{R}^{N+1}$. Taking the cone over this map produce an isometric embedding of the quotient space $\mathbb{R}^{n+1}/\iota$ into $\mathbb{R}^{N+1}$, where $\iota$ is the central symmetry $\iota\colon x\mapsto -x$. (If $n=2$, then the needed length-preserving map $\mathbb{S}^2\to \mathbb{R}^6$ is defined by $$(x,y,z)\mapsto \alpha\cdot(\beta+x^2,\beta+y^2,\beta+z^2,\sqrt{2}{\cdot}x{\cdot}y,\sqrt{2}{\cdot}y{\cdot}z,\sqrt{2}{\cdot}z{\cdot}x)$$ for approprately chosen $\alpha$ and $\beta$.) So, "yes", for any centrally symmetric surface $\Sigma$ in $\mathbb{R}^n$, the quotient $\Sigma/\iota$ admits an explicit embedding in $\mathbb{R}^{N+1}$.<|endoftext|> TITLE: Stability for the Gödel and Jensen hierarchies QUESTION [9 upvotes]: Notations: Let $L_\alpha$ stand for the Gödel constructible hierarchy ($L_0=\varnothing$ and $L_{\alpha+1} = \mathrm{def}(L_\alpha)$ is the set of definable subsets of $L_\alpha$ and $L_\delta = \bigcup_{\beta<\delta} L_\beta$ for limit $\delta$), and $J_\alpha$ for the Jensen hierarchy ($J_0=\varnothing$ and $J_{\alpha+1} = \mathrm{rud}(J_\alpha)$ is the rud-closure of $J_\alpha \cup \{J_\alpha\}$ and $J_\delta = \bigcup_{\beta<\delta} J_\beta$ for limit $\delta$). Let $M \mathrel{\preceq_1} N$ (if $M$ and $N$ are transitive sets mean “$(M,{\in})$ is a $\Sigma_1$-elementary submodel of $(N,{\in})$”. Consider the following two relations between two ordinals $\sigma<\gamma$: say that “$\sigma$ is $\gamma$-stable” when $L_\sigma \mathrel{\preceq_1} L_\gamma$, say that “$\sigma$ is $\gamma$-J-stable” when $J_\sigma \mathrel{\preceq_1} J_\gamma$. Question: Are the above relations equivalent? If not, is there still a way to define “$\sigma$ is $\gamma$-stable” in terms of the Jensen hierarchy and/or “$\sigma$ is $\gamma$-J-stable” in terms of the Gödel hierarchy? Clearly, any one of the above relations implies that $L_\sigma = J_\sigma$ (because $\sigma$ is, at the very least, admissible $>\omega$), so the problem concerns the right-hand side, as $\gamma$ is not required to be admissible, or even a limit ordinal. Certainly $J_\sigma \mathrel{\preceq_1} J_\gamma$ implies $L_\sigma \mathrel{\preceq_1} L_\gamma$, so the question is whether the converse holds, or, if not, whether we can still find a way to express stability for one hierarchy in terms of the other. I know that $\sigma$ is $(\sigma+1)$-stable iff it is $\Pi_m$-reflecting for each $m$, for example, but I don't see whether this also applies to being $(\sigma+1)$-J-stable (or, if not, what the corresponding criterion would be). Bonus question: What if we replace $1$ by $n$ (i.e., $\Sigma_1$ by $\Sigma_n$) throughout? REPLY [11 votes]: As remarked the non-trivial direction is to show $L_\sigma\prec_{\Sigma_1}L_\gamma$ implies $J_\sigma\prec_{\sigma_1}J_\gamma$. Let's take the extreme case that $\gamma=\sigma+1$. Suppose $J_{\sigma+1}\models \exists u \varphi(u,x)$ where $\varphi\in\Sigma_0$ and $x\in J_\sigma = L_\sigma$. The short answer is to copy everything onto subsets of $\sigma$ and use $\mathcal{P}(L_\sigma)\cap J_{\sigma+1}= L_{\sigma+1}$. But in more detail: as $J_{\sigma+1}=\bigcup_{n\in\omega} S_{\omega\sigma+n}$ (see Jensen's original Fine Structure paper, or Devlin's "Constructibility" for the definition of the $S_\nu$-hierarchy, and these and other assertions) we have for some $n$ that $S_{\omega\sigma+n}\models \exists u \varphi(u,x)$. (Here we have $\omega\sigma=\sigma$ as well as the fact that the Goedel pairing function $Q$ is $\Delta_1^{L_\sigma}$). By adding some extra `Basic Functions' to the usual list, we can assume each $S_{\omega\sigma+n}$ is transitive. Now use the fact that there is an onto function $F:\omega\sigma \rightarrow S_{\omega\sigma+n}$ in $J_{\sigma+1}$. Use this to get a subset $f\subseteq \omega\sigma$ with $$\langle \sigma, Q \mbox{''}f \rangle \cong \langle S_{\omega\sigma+n},\in \rangle $$ and $f\in J_{\sigma+1} $ and so also in $L_{\sigma+1} $. We now have an existential statement of the form $L_{\sigma+1} \models \mbox{''}\exists \sigma\exists f \subseteq \sigma \ldots $''. So reflect this to some $\tau<\sigma$ and obtain: $$\langle \tau, Q \mbox{''} \bar f \rangle \models \exists u \varphi(u,Q(\xi)_1) $$ where we ensure for some $\xi \in \bar f$, $ Q(\xi)_1 $ represents $x$ in the structure say. Taking the transitive collapse of $\langle \tau, Q \mbox{''}\bar f\rangle $ then yields a transitive model $U\in J_\sigma$ of $\exists u \varphi(u,x)$ as needed. [Edit: added to address comment. Now use this argument repeatedly to prove by induction on $\delta$ for $\sigma+1\leq\delta<\gamma$ that $L_\sigma\prec_{\Sigma_1}L_\gamma$ implies $J_\sigma\prec_{\sigma_1}J_\gamma$. The successor case of a $\delta+1$ is just a variant of the one given and the case of a limit $\delta$ is of course trivial.] For the Bonus Q just note that for $n>1$, if $L_\sigma\prec_{\Sigma_n}L_\gamma$, then both $L_\sigma$ and $L_\gamma$ are limits of admissible ordinals, so then $L_\sigma=J_\sigma$ and $L_\gamma=J_\gamma$ and there is nothing to do!<|endoftext|> TITLE: $(M,\omega)$ not symplectomorphic to $(M,-\omega)$ QUESTION [22 upvotes]: Looking for an example of a symplectic manifold $(M,\omega)$ that is not symplectomorphic to $(M,-\omega)$. In particular this means that $M$ must be chiral (i.e. doesn't admit an orientation-reversing diffeomorphism). For a topological obstruction, I think it would be enough to find $(M,\omega)$ such that $\mathrm{Diff}(M)$ acts trivially on $H^2(M)\neq0$. A complex projective variety defined by real equations won't work, because the complex conjugation map is antisymplectic. REPLY [5 votes]: The following answer was a suggestion of Ivan Smith's. It seems like a very nice argument, although the proof is quite high-tech. Suppose $X$ is a compact symplectic manifold. By adding a small generic 2-form, we can ensure that the coefficients of $\omega$ with respect to a basis of $H^2(X;\mathbb{Z})$ are linearly independent over $\mathbb{Q}$. Any diffeomorphism that reverses $\omega$ then has to act as $-1$ on $H^2$. The question now is to find an $X$ such that no element of $\mathrm{Diff}$ acts as $-1$ on $H^2$, and Ivan said that a K3 surface would work. $H^2$ has rank 22, and the signature of the intersection form is $(3, 19)$. Apparently Donaldson theory means that $\mathrm{Diff}$ preserves the orientation of a positive-definite 3-dimensional subspace of $H^2$, so in particular cannot act as $-1$. This is explained in Donaldson-Kronheimer (Corollary 9.1.4), but the basic idea is that a choice of such an orientation allows one to orient some gauge-theoretic moduli space. Reversing the orientation reverses the sign of a corresponding Donaldson invariant, but this invariant is non-zero so it's not equal to itself with sign reversed.<|endoftext|> TITLE: Automorphisms of the Lie algebras $\mathfrak{sl}(2,R)$ and $\mathfrak{su}(2)$ QUESTION [7 upvotes]: I would like to know about the literature concerning the group of outer automorphisms of the Lie algebra $\mathfrak{sl}(2,R)$. This question is addressed in different places in a contradictory way. In certain works, e.g. M.A. Farinati and A.P Jancsa, Three dimensional real Lie bialgebras, Revista de la union matematica argentina Vol. 56, No. 1, 2015, Pages 27–62, 2015, it is implicitly claimed that the aforesaid group is trivial and all automorphisms of $\mathfrak{sl}(2,R)$ are inner. By googling the question, I also found several sources claiming that the group of inner automorphisms of $\mathfrak{sl}(2,R)$ is PSL(2,R) (see https://groupprops.subwiki.org/wiki/Special_linear_group:SL(2,R)), and the outer automorphisms are given by PGL(2,R) (see Outer automorphisms of simple Lie Algebras) which are obviously different. Similarly, I would like to know about the structure of the group of outer automorphisms of the Lie algebra $\mathfrak{su}(2)$ Thank you in advance for your comments. REPLY [13 votes]: It is clear that $PGL(2,\mathbb{R})$ acts as automorphisms. It is easy to check that the reflections act in a manner unlike any positive determinant matrices. Hence the automorphism group is larger than $PSL(2,\mathbb{R})$. Since we know the answer over $\mathbb{C}$ (as in Fulton and Harris, Representation Theory, p. 498), by complexification, we know that all automorphisms arise from conjugation by some matrices. We can easily check that complex matrices give us real automorphisms only when they are real up to a constant scaling. So we see that the automorphism group of $\mathfrak{sl}(2,\mathbb{R})$ is $PGL(2,\mathbb{R})$. For $\mathfrak{su}(2)$, any automorphism complexifies to an automorphism of $\mathfrak{sl}(2,\mathbb{C})$, so arises by conjugation from a complex matrix $g$, again from Fulton and Harris. To get conjugation by $g$ to preserve the real subspace $\mathfrak{su}(2)$, we need $gAg^{-1}$ to be special unitary for any special unitary $A$. Plug this in and check that this forces $g^*g$ to commute with all such $A$, so by Schur's lemma, $g^* g=\lambda I$ for some complex number $\lambda$. Take determinant to find that $\lambda=\pm 1$. Replace $g$ by $ig$ if needed to get $\lambda=1$, so $g \in SU(2)$. Check that $g$ acts trivially if and only if $g=-I$ to see that the automorphism group of $\mathfrak{su}(2)$ is $SO(3)=SU(2)/\pm 1$.<|endoftext|> TITLE: Find all integer solutions to $y^2=x^5+4.$ Is it true that the only solutions are $(x, y) = \{( 2,-6), ( 2,6),(0,2), (0,-2) \}.$ QUESTION [7 upvotes]: I think the following problem is difficult, any ideas for solution are welcome. Find all integer solutions to $y^2=x^5+4.$ Is it true that the only solutions are $(x, y) = \{( 2,-6), ( 2,6),(0,2), (0,-2) \}.$? REPLY [15 votes]: Yes, these are the only examples. Either $x$ is zero (your $(0, \pm 2)$ example) or not; assume the latter. Rewrite your equation as $(y - 2)(y + 2) = x^5$, and appeal to unique factorisation. Then $y \pm 2$ are either both fifth powers, or one is of the form $4a^5$ and the other is of the form $8b^5$. The former case is impossible, since no pair of fifth powers differ by 4. Hence let $a, b$ be integers such that $a^5 \pm 1 = 2b^5$. Either $b$ is zero (your $(2, \pm 6)$ example) or not; assume the latter. Factorise the left-hand side over the integer ring of the cyclotomic field generated by a fifth root of unity, and again appeal to unique factorisation. You can show that of the five numbers $a \pm \omega^k$, where $\omega$ is a principal fifth root of unity, that four of them must be fifth powers and the other one (necessarily $a \pm 1$) is of the form $2c^5$. Either way, this involves finding a pair of fifth powers such that $d^5 - e^5 = \omega - \omega^4$, whence you can exhaustively determine there are no further solutions (again by unique factorisation).<|endoftext|> TITLE: What was Smith's proof of Smith's theorem on Hamilton cycles in cubic graphs? QUESTION [13 upvotes]: In a short 1946 paper "On Hamiltonian Circuits", Tutte proved the famous result that an edge in a cubic graph lies in an even number of Hamilton circuits. He attributed the result to his friend CAB Smith, but explicitly mentioned that the proof was not Smith's proof. What was Smith's proof? REPLY [2 votes]: I am fairly sure that I once asked Adrian Bondy about it, who replied something along the lines of Bill Tutte having told him that the original proof by Smith was quite messy.<|endoftext|> TITLE: The category theory of Span-enriched categories / 2-Segal spaces QUESTION [12 upvotes]: The category $\mathsf{Span}$ of spans of sets is symmetric monoidal closed under $\times$ (the cartesian product from $\mathsf{Set}$, which is not the categorical product in $\mathsf{Span}$), complete and cocomplete. So it should be a nice category to enrich in! A $\mathsf{Span}$-enriched category is "a category with multivalued composition". Caveat: Actually, one issue is that $\mathsf{Span}$ is naturally a (2,1)-category, but I think this complication can be dealt with. Motivation: In fact, if we $\infty$-ify and look at $\infty$-categories enriched in spans of spaces, we should get something like the 2-Segal spaces of Dyckerhoff and Kapranov i.e. the decomposition spaces of Galvez-Carillo and Kock and Tonks, which are surprisingly common, arising e.g. from the $S_\bullet$ construction of algebraic $K$-theory. Without $\infty$-fying, a $\mathsf{Span}$-enriched category is essentially a unital 2-Segal set, which is a simplicial set satisfying certain Segal conditions. For the purposes of this question I'm happy to stick to the non - $\infty$-context, but that's where I'd ultimately like to go with this. The problem: However, it seems that the usual notions of enriched category theory shouldn't just be applied blindly to $\mathsf{Span}$-enriched categories. For example: A $\mathsf{Span}$-enriched functor $F: C \to D$ will have spans $F_{c,c'}$ from $C(c,c')$ to $D(Fc,Fc')$, but for many purposes, one will be interested in enriched functors $F$ such that $F_{c,c'}$ is an actual function. Call such an enriched functor map-like. For example, the simplicial maps between 2-Segal sets correspond to the map-like enriched functors. As one illustration of this, note that modulo size issues, the inclusion $I$ of ordinary categories into $\mathsf{Span}$-enriched categories has a right adjoint, given by the usual "underlying category" functor, which in this case sends $C$ to what Dyckerhoff and Kapranov call the "Hall category" $HC$ of $C$, where $HC(c,c') = \mathsf{Set}^{C(c,c')}$ and composition given by an indexed coproduct. The inclusion $I$ also has a left adjoint $F$, which identifies all the different composites of any two morphisms -- but $F$ is only functorial with respect to map-like enriched functors! (Simplicially, this is the Bousfield localization turning a 2-Segal set into a 1-Segal set, i.e. a category.) In fact, if the morphisms of $\mathsf{Span}$-enriched categories are taken to be all enriched functors, then $I$ fails to preserve products and so has no left adjoint. Enriched (co)limits are probably not the correct notion of (co)limits for $\mathsf{Span}$-enriched categories. For example, if $C$ is an ordinary category with (co)products, then the corresponding $\mathsf{Span}$-enriched category $IC$ will typically not have (co)products. This comes down to the fact that (co)limits in a $\mathcal{V}$-enriched category are defined in terms of limits in $\mathcal{V}$, and limits in $\mathsf{Span}$ are not at all related to limts in $\mathsf{Set}$. My question: Has the category theory of $\mathsf{Span}$-enriched categories -- notions of functor, (co)limit, etc -- been developed somewhere in the literature? If it hasn't been developed specifically, is there some existing formal-category-theoretic framework that it should fit into? Equipments, $\mathcal{F}$-categories,... the ones I know of don't seem to quite fit the bill. REPLY [2 votes]: You might like to look at the paper "Algebraic theories, span diagrams and commutative monoids in homotopy theory" (https://arxiv.org/abs/1109.1598) by James Cranch. I think that it does not directly answer any of your questions, but it involves the same circle of ideas.<|endoftext|> TITLE: Are separable F-spaces (completely metrizable topological vector space) homeomorphic to $l_2$? QUESTION [7 upvotes]: An F-space is a completely metrizable topological vector space, i.e. the vector topology is induced by a complete metric. A Fréchet space is, by definition, a locally convex F-space. It is known that all (infinite dimensional) separable Fréchet spaces are homeomorphic to $l_2$, the space of square summable sequences of real numbers. (See e.g. Anderson, R. D.; Bing, R. H., A complete elementary proof that Hilbert space is homeomorphic to the countable infinite product of lines. Bull. Amer. Math. Soc. 74, 1968, 771–792.) More precisely any such space, including $l_2$ is homeomorphic to a countable infinite product of copies of real lines. Can this result be extended to (any) non-locally convex F-spaces? I am interested in any reference or recent literature on the topology (homeomorphism class) of F-spaces. More precisely, I am interested in the following examples: $L_p([0,1])$ or $l_p$ with $01$. Nevertheless $d(f,g)=||f-g||_p^p$ is a complete metric inducing the vector topology. This question arose when wondering if $l_p$ has a structure of a Hilbert manifold. REPLY [6 votes]: There is a famous linear metric space constructed by R. Cauty [Un espace métrique linéaire qui n'est pas un rétracte absolu, Fund. Math. 146 (1994)] whose completion is a separable $F$-space which is not an AR. I do not have Cauty's paper handy but the latter fact is stated on the first page of Cauty's space enhanced in [Topology Appl. 159 (2012), no. 1, 28–33]. Since $\ell_2$ is an AR, the above $F$-space is not homeomorphic to $\ell_2$. Incidentally, a separable $F$-space is homeomorphic to $\ell_2$ if and only if it is a non-locally-compact AR, see Corollary 5.2.2 in [Absorbing sets in Infinite-Dimensional Manifolds, T. Banakh, T. Radul, M. Zarichnyi].<|endoftext|> TITLE: Is $n$ uniformly computable from an oracle for the $n^{\rm th}$ jump $0^{(n)}$? QUESTION [5 upvotes]: This is a little curiosity that came up in a project I am working on, and I thought someone might have a nice way to see the answer. Question. Can we uniformly compute $n$ from an oracle for the $n^{\rm th}$ jump $0^{(n)}$? Of course, if we hard-code $n$ into our representation of $0^{(n)}$, then the answer will be yes. Similarly, if we use a $\Sigma_n$-truth predicate instead of $0^{(n)}$, which is of course Turing equivalent, then we can compute $n$ by looking at the syntactic complexity of the assertions in the oracle. We could also get negative answers by making finite changes to each $0^{(n)}$, which would preserve Turing equivalence at each level but destroy the uniform algorithm. My question, instead, is about using the usual representation of $0^{(n)}$ as sets of halting Turing machine programs. Specifically, using some standard Turing machine architecture and encoding of Turing machine programs, let $0'$ or $0^{(1)}$ be the collection of Turing machine programs (construed as a set of natural numbers) that halt on empty input. For $n\geq 1$, let $0^{(n+1)}$ be the collection of oracle Turing machine programs (construed as a set of natural numbers) that halt on empty input using oracle $0^{(n)}$. The question is whether there is a program $e$ such that on empty input, program $e$ with oracle $0^{(n)}$ outputs $n$. In other words, is there a computer program such that if you give it some $0^{(n)}$ as an oracle, it can recover $n$? I believe that the answer will be yes, by designing certain programs that halt or don't halt in a such a way that one can determine what $n$ must be. If the answer is affirmative, then we can also answer affirmatively the following question: Question. Is $0^{(k)}$ uniformly computable from $0^{(n)}$ for any $n\geq k$? That is, using the specific representations of $0^{(n)}$ I defined above, is there an oracle Turing machine program $e$ such that on input $k$, given oracle $0^{(n)}$ for any $n\geq k$, will write out the contents of $0^{(k)}$? I expect that the answer is yes. REPLY [4 votes]: Well, I am a little embarrassed, but after having written out my question and thinking about it for a time, I now see how to answer it. Somehow, it wasn't as clear to me before. (Should I instead simply delete? If so, please comment.) We can compute any oracle $A$ uniformly in $A'$, by the following process: for any number $k$, there is a program $p_k$ that on any input checks if $k$ is in the oracle, halting if it is and not halting if it isn't. Thus, $k\in A\iff p_k\in A'$, and the map $k\mapsto p_k$ is computable. For any program $e$ to be used with oracle $A$, we can uniformly create a program $q_e$ to be used with $A'$, but computing the same function. Namely, $q_e$ on any input uses the method of the previous paragraph to simulate $e$ with oracle $A$, by systematically generating larger and larger fragments of the oracle $A$ from $A'$ in order to simulate $e$ with oracle $A$. Now we simply iterate this process, so that from $0^{(n)}$ we can uniformly compute the preceding jumps $0^{(n-k)}$ for any given $k< n$. That is, we compute the jumps from the top down. Finally, notice that the way that I defined $0^{(1)}$, it consisted of non-oracle programs only, and so we can determine with oracle $0^{(n)}$ whether $n=1$ or $n>1$ by looking at one of the programs in the oracle, and checking if it is an oracle program or a non-oracle program. So, we simply systematically undo the jumps until we see that we have $0^{(1)}$ or not, counting the number of times, and thereby compute $n$.<|endoftext|> TITLE: Logical complexity of hard functions conjectures QUESTION [6 upvotes]: Let $\phi_1$ and $\phi_2$ be the following statements: $\phi_1:$ There is a function $f:\{0,1\}^*\to\{0,1\}$ computable in $E$ that has circuit complexity $2^{\Omega(n)}$. $\phi_2:$ There is a function $f:\{0,1\}^*\to\{0,1\}$ computable in $NE \cap CoNE$ that has $2^{\Omega(n)}$ hardness on average. Q1. Is there any $\Pi_2$ sentence $\psi$ in the language of arithmetic such that we can prove $\mathbb{N}\models \psi \leftrightarrow \phi_1$? Q2. Is there any $\Pi_2$ sentence $\psi$ in the language of arithmetic such that such that we can prove $\mathbb{N}\models \psi \leftrightarrow \phi_2$? Actually, I want to know if these conjectures are $\Pi_2$ expressible like $P\not = NP$. REPLY [4 votes]: As given, $\phi_1$ and $\phi_2$ are $\Sigma_2$. They cannot be shown equivalent to $\Pi_2$ statements by any proof that relativizes. This follows by the same argument as in Examples of $G_\delta$ sets or https://cstheory.stackexchange.com/a/16644. We only need to know that the relativized statements are unaffected by a finite change of the oracle (which is obvious), and that either statement can be made true or false by relativization with suitable oracles: There are oracles $A$ such that $\mathrm{NEXP}^A=\mathrm{BPP}^A$, whence $\mathrm{NEXP}^A\subseteq \mathrm P^A/\mathrm{poly}$, which implies $\neg\phi_i^A$ for $i=1,2$. See e.g. http://blog.computationalcomplexity.org/2005/08/extreme-oracles.html. There are oracles $A$ such that some $\mathrm{NP}^A$-language needs relativized circuit size at least $2^{\alpha n}$, for some constant $\alpha>0$: see http://www.sciencedirect.com/science/article/pii/030439759390256S. This implies $\phi_1^A$ and (using the Impagliazzo–Wigderson argument that worst-case and average-case complexity is equivalent for $\phi_1$) also $\phi_2^A$.<|endoftext|> TITLE: Cohomology theories as colimits QUESTION [8 upvotes]: I am looking for examples of cohomology theories that can be written as (filtered, or another nice class of) colimits of "simpler" functors, i.e. which $\{h^n : {\bf Top}^2 \to {\bf Ab}\}_n$ are such that $$ h^n(X) \cong \text{colim}_j\; h_j^n(X) $$ for a suitable diagram ${\cal J}\to [{\bf Top}^2,{\bf Ab}]$. Of course this is a really vague question: A "cohomology theory" (and the category thereof) is what (for example) Rudyak I.3.8 defines as such. I'm not asking that the $h^n_j$ are cohomology theories themselves, but you can assume this additional requirement. You're quite free to interpret the word "simple" in the way to like more. I'm in fact explicitly asking for which meaning of "simple" this question has a good answer. The question remains a bit vague: whatever $h^n(X)$ is, you can take a presentation for this abelian group and say that it is a colimit. Nevertheless I think that asking for a colimit of functors is a bit more restrictive and avoids trivial cases. My feeling is that the answer is always "quasi-affirmative": a cohomology theory, i.e. a spectrum, belongs to a presentable quasicategory. But spectra and cohomology theories aren't really the same thing. REPLY [3 votes]: Suppose you just look at colimits indexed by $\mathbb{N}$, and fix the cohomological degree $n$. I claim that there must exist $j$ such that the map $h_j^n(X)\to h^n(X)$ is surjective for all $X$. Indeed, if not, we can choose spaces $X_j$ and classes $a_j\in h^n(X_j)$ that do not come from $h^n_j(X_j)$. Put $X=\coprod_jX_j$. As $h$ is assumed to be a cohomology theory, there is a unique element $a\in h^n(X)$ with $a|_{X_j}=a_j$ for all $j$. If this comes from some $a'\in h^n_j(X)$ then we see that $a'|_{X_j}$ maps to $a_j$, contrary to assumption. (We do not need to assume that $h_j^n$ converts coproducts to products for this, we are just using functoriality.) I suspect that under mild assumptions one can prove something stronger: there is a cofinal sequence $j_1 TITLE: Smallest singular value of $X\mapsto AX^{T}+XA^{T}$ QUESTION [7 upvotes]: Question: Given the long and skinny matrix $A\in\mathbb{R}^{m\times n}$ with $m\ge n$, define the matrix valued operator $$\mathcal{A}:X\mapsto AX^{T}+XA^{T}.$$ What is the tightest nontrivial lower-bound on the singular value $$\sigma_{\min}(\mathcal{A})\triangleq\min_{X}\{\|\mathcal{A}(X)\|_{F}:\|X\|_{F}=1\},$$ where $\|X\|_{F}^{2}=\mathrm{trace}(X^{T}X)$ is the usual matrix Euclidean norm (i.e. Frobenius norm)? Remark 1. In the case that $A=a$ is a vector (i.e. with $n=1$), it is easy to show that $\sigma_{\min}(\mathcal{A})=\sqrt{2}\|a\|$, since $\|ax^{T}+xa^{T}\|^{2}=2\|a\|^{2}\|x\|^{2}+2(a^{T}x)^{2}$, and the term $(a^{T}x)^{2}$ is obviously nonnegative. However in the general matrix case, we have $\|AX^{T}+XA^{T}\|^{2}=2\|AX^{T}\|_{F}^{2}+2\mathrm{trace}[(A^{T}X)^{2}]$, and it appears possible for $\mathrm{trace}[(A^{T}X)^{2}]$ to be negative. Remark 2. In the case that $A$ is square and that $X$ is forced to be symmetric, this problem is closely related to the smallest singular value of the Lyapunov operator $A\otimes I+I\otimes A$, which is known to be related (in a fairly complicated way) to the singular values of $P$, where $P$ solves $AP+PA^{T}=I$. Edit 1. Federico Poloni remarked that the rectangular case should not be easier than the square one. Agreed. In this case, consider where $n$ is very small, say 1 or 2, or at least $n\ll m/2,$ so that $\mathcal{A}(X)$ can be considered low-rank. As mentioned above, the $n=1$ case has an exact solution. Can tight bounds be derived for $n$ small? REPLY [5 votes]: There is always a zero singular value as soon as $n \geq 2$. Write $A = UD V$ with $D$ diagonal and $U, V$ orthogonal. Then we can write $X \mapsto AX^T + X A^T$ as $$X \mapsto UDV X^T + X V^T D^T U^T = U(D (U^T X V^T)^T + (U^T X V^T) D^T ) U^T$$ i.e. the composition of the operation $X \mapsto D X^T + X D^T$ with two orthogonal operatorions. So we may assume $A=D$ is diagonal By performing singular value decomposition on $A$, we may assume that $A$ is diagonal. Say the first two entries are $\lambda_1,\lambda_2$. Then $\begin{pmatrix} 0 & -\lambda_1 \\ \lambda_2 & 0 \end{pmatrix}$, padded with zero entries, is sent to zero. Of course if $\lambda_1 = \lambda_2=0$, then any $2 \times 2$ matrix is sent to $0$. The remaining singular values are either $\sqrt{2}$ times a singular value of $A$, the square root of twice the sum of the squares of two different singular values of $X$, or twice a singular value of $A$.<|endoftext|> TITLE: Twisted-arrow construction for 2-categories QUESTION [6 upvotes]: I've been looking over Lurie's DAG X, and he introduces a combinatorial construction called the twisted arrow construction for simplicial sets that generalizes the following ordinary categorical notion: The Yoneda embedding $\mathcal{C}\to \widehat{\mathcal{C}}$ is adjunct under the hom-tensor adjunction to the functor $$\operatorname{Hom}:\mathcal{C}^{\operatorname{op}}\times\mathcal{C}\to \operatorname{Set}$$. Composing $\operatorname{Hom}$ with the inclusion of $\iota:\operatorname{Set}\hookrightarrow\operatorname{Cat}$ gives us a functor to $\operatorname{Cat}$, and to this functor we can apply the Grothendieck construction to produce a discrete fibration $$\pi:\int^{\mathcal{C}^{\operatorname{op}}\times\mathcal{C}} \operatorname{Hom} \to \mathcal{C}^{\operatorname{op}}\times\mathcal{C}$$ We call the total space of this fibration $\operatorname{Tw}(\mathcal{C})$, the twisted arrow category. In chapter 4 of DAG X, Lurie gives a concrete and combinatorial description of how to generalize this to any simplicial set $X$ by constructing it as the pullback $\sigma^\ast X$ where $$\sigma: \Delta \to \Delta$$ sends $$[n] \mapsto [n]^{\operatorname{op}} \boxplus [n]$$ where $\boxplus$ is the ordinal sum/join (the faces and degeneracies map to the arrows in a canonical way that is induced by the join and the $\operatorname{op}$, so I am omitting them for the sake of brevity). In order to understand things better in the case of a 2-category and maybe come up with a combinatorial description to extend to higher categories, I'm wondering if anyone has seen or could describe that $2$-grothendieck construction assodciated with the $\operatorname{Hom}$ functor in the case of strict $2$-categories. I'm curious if maybe something like just the (horizontal) join could possibly work to generalize it to the case of $\Theta_2$, which plays the role in $2$-categories that $\Delta$ plays in $1$-categories. Thanks! REPLY [3 votes]: While it is more explicit and combinatorial, this probably isn't precisely what you're looking for. It does seem relevant to your question, though. A year or so ago I worked out a 2-categorical analogue of the $(\infty,1)$ twisted arrow construction (I'm working with $(\infty,1)$-categories arising as Joyal fibrant replacements of the Duskin nerves of 2-categories). I initially tried the same approach Mike Schulman suggested in the comments, i.e., using Buckley's 2-categorical Grothendieck construction, but I found that a simpler construction was sufficient for my purposes. It's a quite straightforward generalization of the commutative (up to natural isomorphism) diagram $$ \require{AMScd} \begin{CD} \operatorname{Cat} @>{N}>> \operatorname{Set}_\Delta \\ @V{Tw}VV @VV{Tw}V \\ \operatorname{Cat} @>>{N}> \operatorname{Set}_\Delta \end{CD} $$ to 2-categories. The construction is a little ad-hoc, but the basic idea is the following: From a 2-category $\mathscr{C}$, construct a new 2-category $Tw_2(\mathscr{C})$ by letting Objects of $Tw_2(\mathscr{C})$ are morphisms of $\mathscr{C}$ 1-morphisms in $Tw_2(\mathscr{C})$ from $f$ to $f^\prime$ consist of diagrams $$ \begin{array}{c c c } A & \overset{f}{\rightarrow} & B \\ h \downarrow\hspace{6pt} & \searrow & \hspace{6pt}\uparrow k \\ A^\prime & \underset{f^\prime}{\to} & B^\prime\\ \end{array} $$ and $$ \begin{array}{c c c } A & \overset{f}{\rightarrow} & B \\ h \downarrow\hspace{6pt} & \nearrow& \hspace{6pt}\uparrow k \\ A^\prime & \underset{f^\prime}{\to} & B^\prime\\ \end{array} $$ where each triangle commutes up to a (not nec. invertible) 2-morphism, and the 2-morphisms satisfy the obvious commutativity condition. (Note that, by definition, these are the 3-simplices of the Duskin nerve of $\mathscr{C}$.) 2-morphisms are given by (appropriately coherent) natural transformations of diagrams which are the identity on $f$ and $f^\prime$. The resulting category $Tw_2(\mathscr{C})$ is only lax unital. However, with the appropriate nerve constructions, the diagram $$ \require{AMScd} \begin{CD} \operatorname{2Cat} @>{N}>> \operatorname{Set}_\Delta \\ @V{Tw_2}VV @VV{Tw}V \\ \operatorname{Lax2Cat} @>>{N}> \operatorname{Set}_\Delta \end{CD} $$ commutes up to natural isomorphism. There is an obvious functor $$ Tw_2(\mathscr{C}) \to \mathscr{C}\times \mathscr{C}^{\operatorname{op}}, $$ though I haven't looked at its fibrancy properties yet. Disclaimer: This is my first answer here, please let me know if I have ignored some protocol out of inexperience.<|endoftext|> TITLE: Relating three viewpoints on the semidirect product QUESTION [7 upvotes]: It's known that giving a semidirect product $(X,m)\rtimes G$ of a $G$-group $(X,m)$ with $G$ (as defined in wiki) is the same as giving a split pair over $G$, i.e a pair of arrows $H\overset{s}{\underset{f}{\leftrightarrows}}G$ such that $s$ is a section of the lower arrow. The procedure is described e.g in this MSE answer. Another approach is purely categorical, and my question is about unpacking it concretely. The category of groups is semi-abelian, whence its inverse image functor $\alpha _G^\ast :\mathsf{Pt}_G(\mathsf{Grp})\to \mathsf{Grp}$ is monadic. This functor is the kernel of the rightward arrow of a splitting $H\overset{s}\leftrightarrows G$. Its left adjoint takes a group $X$ to diagram below. $$X\amalg G\overset{\iota_2}{\underset{(0_{XG},1_G)}{\leftrightarrows}} G$$ Thus an algebra for the monad $(T,\eta,\mu)$ induced by this adjunction is given by an arrow $$\xi:\operatorname{Ker}(0_{XG},1_G)\to X.$$ The object $\operatorname{Ker}(0_{XG},1_G)$ is comprised of trivial words in $G$ spliced by elements of $X$, e.g $g_1xg_2,g_1x_1g_2x_2g_3x_3\in X\amalg G$ where $g_1g_2=g_1g_2g_3=1\in G.$ Thus, an algebra $\xi$ seems to identify words in $\operatorname{Ker}(0_{XG},1_G)$ with elements of $X$. Note $\iota_1X\leq \operatorname{Ker}(0_{XG},1_G)$, so $\xi$ makes some inner identifications in $\operatorname{Ker}(0_{XG},1_G)$. By monadicity, a splitting $H\leftrightarrows G$ is the same as an algebra $\xi:\operatorname{Ker}(0_{XG},1_G)\to X.$ Given a splitting, the Eilenberg-Moore comparison functor endows $\operatorname{Ker}(H\overset{f}\to G)$ with the algebra map $$\alpha_G^\ast(\varepsilon_{H\leftrightarrows G}):\operatorname{Ker}(0_{\operatorname{Ker}f,G},1_G)\longrightarrow \operatorname{Ker}f.$$ If I understand correctly, $\varepsilon_{H\leftrightarrows G}=(\ker(0_{HG},1_G),0_{HG})$ so the algebra map acts by $$g_1h_1g_2h_2g_3h_3\mapsto h_1h_2h_3,\; h_1,h_2,h_3\in \operatorname{Ker}f$$ and does nothing interesting. Its adjoint inverse takes an algebra $(X,\xi)$ to the coequalizer of the lift of the pair $(T\xi,\mu_X)$. This lift is the parallel pair below along with arrows to $G$ given by $(0,1_G)$ and arrows from $G$ given by coproduct injections $\iota_2$. $$\operatorname{Ker}(0_{XG},1_G)\amalg G\overset{\xi\amalg G}{\underset{\varepsilon_{X\amalg G\leftrightarrows G}}{\rightrightarrows}} X\amalg G$$ If I understand correctly, $\varepsilon_{X\amalg G\leftrightarrows G}=(\ker(0_{XG},1_G),0_{X\amalg G,G})$. Thus the coequalizer of the above pair identifies e.g $$\xi\amalg G\;(g_1(g_2x_1g_3)g_4(g_5x_2g_6)) \sim \varepsilon_{X\amalg G\leftrightarrows G}(g_1(g_2x_1g_3)g_4(g_5x_2g_6)) $$ i.e $$g_1\xi(g_2x_1g_3)g_4\xi(g_5x_2g_6) \sim \not \! g_1(g_2x_1g_3)\not \! g_4(g_5x_2g_6)= g_2x_1g_3g_5x_2g_6.$$ In particular, considering $g(x)g^{-1}\in \operatorname{Ker}(0_{XG},1_G)\amalg G$ with $(x)\in \operatorname{Ker}(0_{XG},1_G)$, we see $$gxg^{-1}=g\xi(x)g^{-1}=\xi\amalg G\;g(x)g^{-1}\sim \varepsilon_{X\amalg G\leftrightarrows G} \;g(x)g^{-1}=x\in X\amalg G.$$ (The leftmost equality is due to the algebra axiom $\xi \circ \eta_X=1_X$.) This identification resembles the usual presentation of a semidirect product, but has no mention of an action of $G\to \mathsf{Aut}_\mathsf{Grp}(X)$... So, given an algebra $(X,\xi)$, the splitting associated to it is $$\mathrm{Coeq}(\xi\amalg G,(\ker(0_{XG},1_G),0_{X\amalg G,G}))\leftrightarrows G$$ with the coequalizer object being called the semi-direct product $(X,\xi)\rtimes G$. My questions. In this case of groups: How can I directly prove that giving a mysterious algebra on $X$ structure is actually the same as giving a $G$-action on $X$? How can I directly prove that the assignments of the adjoint equivalence of the Eilenberg-Moore comparison functor are mutually inverse? REPLY [2 votes]: First of all, let us see what is an algebra for this monad. One can show that the kernel of $(0,1):X\amalg G\to G$ is generated by the conjugates of elements $X$ by elements of $G$; so an action $\xi$ is in some sense a way to interpret conjugation by $G$ in $X$. To be more precise, we can define for all $g\in G$ and $x\in X$ $g\ast x=\xi (gxg^{-1})$; then the unit condition $\xi \circ\eta_X=1_X$ tells us that $e_G\ast x=x$ for all $x$, while the associativity condition $\xi \circ \mu=\xi \circ T(\xi)$ tells you that $(g_1g_2)\ast x=g_1\ast( g_2\ast x)$ for all $x$, so $\ast$ is an action in the usual sense. Now why are the comparison functor and its adjoint mutually inverse? First of all, notice that you have the wrong counit $\varepsilon_{H \leftrightarrows G}$: it should be a morphism $$\left(\operatorname{Ker} f\amalg G\overset{\iota_2}{\underset{(0_{XG},1_G)}{\leftrightarrows}} G\right) \rightarrow \left(H\overset{s}{\underset{f}{\leftrightarrows}}G\right)$$ in the category $\mathsf{Pt}_G(\mathsf{Grp})$, which means in particular that $\varepsilon_{H\leftrightarrows G}\circ \iota_2=s$. So in fact you should have $\varepsilon_{H\leftrightarrows G}=(\ker(0_{HG},1_G),s)$. Now if you take a word of the form $g_1hg_2$ such that $g_1g_2=1$ in $G$. This condition is equivalent to $g_2=g_1^{-1}$, so such a word is really just the conjugate of $x\in \operatorname{Ker}f$ in $\operatorname{Ker}f\amalg G$; and now the action $\xi=\alpha_G^\ast(\varepsilon_{H\leftrightarrows G})$ will map $g_1xg_1^{-1}$ to $s(g_1)x s(g_1^{-1})=s(g_1)xs(g_1)^{-1}$. So the action $\xi=\alpha_G^\ast(\varepsilon_{H\leftrightarrows G})$ is simply the conjugation of $G$ on $\operatorname{Ker}f$ (through $s$), computed in $H$. And for the left adjoint : this time the counit is given by $\varepsilon_{X\amalg G\leftrightarrows G}=(\ker(0_{XG},1_G),\iota_2)$. So what the coequalizer does is identifying words of the form $gxg^{-1}$ with $\xi(gxg^{-1})=g\ast x$. Thus the semi-direct product is generated by $X$ and $G$ with the condition that conjugation of $G$ on $X$ is given by the (classical) action $\ast$.<|endoftext|> TITLE: On a theorem of Zhang Jinwen about models of arithmetic QUESTION [7 upvotes]: In the paper ''A Nonstandard Model of Arithmetic Constructed by means of Forcing Method'', Zhang Jinwen states the following in his abstract: The first nonstandard model of arithmetic was given by Skolem. A. Robinson has introduced the concepts of standard, internal and external objects (sets, relations, functions, etc.) on the compactness theorem and concurrent relations, and has proved that if a set S is infinite, then S contains nonstandard internal objects. It is interesting to ask whether this is a common property of all non-standard modes of arithmetic. The author's answer to this question is in the negative. We have proved the theorem that there exists a nonstandard model of formal arithmetic in which there are infinitely many infinite internal subsets containing no nonstandard elements. This means that these infinite internal subsets are composed exclusively of finite natural numbers. In order to obtain this theorem we have made use of Cohen's forcing method. The paper is in Chinese, and I could not understand it (I have a copy of it). Question. I am wondering if someone can explain the main idea of his proof. Remark. Based on Mathscinet, the paper is translated in English, but I could not find a copy of it. The file can be seen here: A Nonstandard Model of Arithmetic Constructed by means of Forcing Method REPLY [5 votes]: (The final two sections are translated modulo tweaks...) I do not know the answer to your question, but I can try to translate (part of) this paper. There are some terms that are unfamiliar to me (e.g., "infinite internal sets" as remarked in a comment) so I will indicate where I am quite confused by using brackets. I am making this community wiki as I am sure that a logician who speaks native Chinese could complete this much more quickly, and so I hope others (including non-Chinese speakers who are well versed in mathematical logic!) will not hesitate to edit. I will translate the first two sentences immediately after the abstract, and then attempt the sections in reverse order. The paper is pretty short, but remember that Chinese characters can represent entire words; so, it would be a lot to translate the entire manuscript. I think a more reasonable aim is the main ideas of sections 3-6. No guarantee on a timeline, although it'll obviously be faster if others assist. I don't recall MO being used in this way before, so if it's un-welcome I can delete. p. 7: In order to keep the manuscript self-contained, the article at hand briefly delves into the Peano Axioms and arithmetical definability in Section 1, and provides an overview of a few results around non-standard models of arithmetic in Section 2. Sections 3 through 6 concern our main result. p. 11: $$5, \text{Main Lemma and the structure of the model}^{*}M(G)$$ With regard to forcing notions, following the work of Cohen $[2]$ it is not difficult to prove the lemmata below. Lemma 1. For any $Q$ and $A$, $Q$ cannot simultaneously force $A$ and $\neg A$. Lemma 2. If $Q \Vdash A$ and $Q' \supseteq Q$, then $Q' \Vdash A$. Lemma 3. For any $Q$ and $A$, there exists $Q' \supseteq Q$ for which $Q' \Vdash A$ or $Q' \Vdash \neg A$. Lemma 4. $Q \Vdash A$ if and only if for all $Q' \supseteq Q$ it is the case that $Q' \not\Vdash \neg A$. Definition. A sequence of forcing conditions $Q_0, Q_1, Q_2, \ldots, Q_n, \ldots$ is generic if for any statement $A$ in $^{*}L(G)$ there is a standard natural number $m$ for which $Q_m \Vdash A$ or $Q_m \Vdash \neg A$. Lemma 5. There exists a sequence of generic forcing conditions. Proof. Observe that there are countably many formulae in $^{*}L(G)$. Therefore, one can enumerate the formulae of $^{*}L(G)$ as $A_1, A_2, A_3, \ldots, A_n, \ldots$. Pick any $Q_0$ and suppose one already has $Q_{n-1}$; for any $A_n$, there must exist $Q \supseteq Q_{n-1}$ for which $Q \Vdash A_n$ or $Q \Vdash \neg A_n$ holds. Defining $Q_n := Q$ and continuing in this fashion, one obtains a generic forcing sequence $Q_0, Q_1, Q_2, \ldots$. Now, pick a generic forcing sequence $Q_0, Q_1, \ldots$, from which we define generic sets $S_1, S_2, \ldots, S_i, \ldots$ as below, for $i < \delta$: $$S_i = \{k|\text{there exists }m \text{ for which }O_m \Vdash \bf{k} \in F_i\}$$ In this way, we have defined a generic nonstandard arithmetical model $^{*}M(G) =$ $^{*}M(S_0, S_1, \ldots, S_i, \ldots)$, with underlying structure [?] $(^{*}N, \textbf{,}, +, \cdot, S_0, S_1, \ldots, S_i, \ldots)$ where $i < \delta \leq \omega$. For any formula $A$ in $^{*}L(G)$, the truth value of $A$ in the model $^{*}M(G)$ is defined as follows: $$^{*}M \vDash A \iff (\exists m) Q_m \Vdash A$$ By lemmas 1 through 5, the model $^{*}M(G)$ is well-defined. pp. 11-12: $$6, \text{Characteristics of the model }^{*}M(G)$$ The lemma below follows clearly from the definition of forcing. Lemma 6. For any statement $A$ in $^{*}L$, we have $$^{*}M_1 \vDash A \iff ^{*}M(G) \vDash A$$ Given that $M$, $^{*}M$, and $^{*}M_1$ are all models of $P$ (Peano Arithmetic), it follows from Lemma 6 that $^{*}M(G)$ is also a model of $P$, and it is a non-standard model; however, $^{*}M(G)$ is distinct from $^{*}M$, and $^{*}M_1$; $^{*}M(G)$ contains as a subset $S_i$ $(0 \leq i < \delta)$, and the $S_i$ are generic infinite sets, and their elements are all standard natural numbers, which, in turn, proves our main theorem. By the result of Feferman $[3]$, $S_i$ is hyperarithmetic, thus, the following propositions hold: If $S \subset N$ is any arithmetic set, then a necessary and sufficient criterion for $S^{*}$ to contain a nonstandard number is that $S$ be an infinite set. $N$ is an external subset of $^{*}M(G)$. $^{*}N - N$ is an external subset of $^{*}M(G)$. REPLY [4 votes]: The only way I can make sense of the result of the abstract is as follows: Let $M$ be any nonstandard model of PA (Peano arithmetic), then it is well-known (and first demonstrated by Feferman in [this 1964 paper][1], using an arithmetical adaptation of Cohen forcing), one can build an infinite family $\cal{S}$ of subsets of $M$ such that the structure $(M,S)_{S\in\cal{S}} $ obtained by adding each $S \in \cal{S}$ to $M$ as an extra predicate satisfies the axioms of Peano arithmetic in the extended language (i.e., the language of arithmetic augmented with a distinguished predicate symbol for each $S \in \cal{S}$). Note that the aforementioned Feferman paper is listed in the references of the paper by Zhang Jinwen. I should also add that it is also well-known that the family $\mathcal{S}$ above can be arranged to be of power continuum. More information about forcing (of the Cohen and Sacks variety) in arithmetic can be found in Kossak and Schmerl's monograph The Structure of Models of Peano Arithmetic. Postscript. In light of the translation offered by Benjamin Dickman in his answer to the question, along with clarifications made by 喻 良 in his comments on Dickman's answer, it appears that Zhang Jinwen's claim (in his abstract) cannot be interpreted so as to coincide with the well-known result offered in my answer.<|endoftext|> TITLE: Low rank vector bundles on hypersurfaces QUESTION [10 upvotes]: As far as I know, it is still unknown whether there exists a (holomorphic) indecomposable vector bundle of rank $r$ on $\mathbb{P}^n_{\mathbb{C}}$ with $n\geq 6$ and $1 6$) carrying an indecomposable vector bundle of rank $r$ with $16$. REPLY [6 votes]: What about the paper by GIORGIO OTTAVIANI on spinor bundles on quadrics? There is a rank 4 spinor bundle on a dimension 6 quadric, unless I am mistaken?<|endoftext|> TITLE: Concentration compactness. Can this concept be stated in a theorem? QUESTION [9 upvotes]: I recently attended a talk on NLS which is rather not my main field of interest. Yet, I got interested in a concept called concentration compactness during the talk. When I approached the speaker after the talk whether he could state in a general way what this concept says he was very resilient to state something that is universally true. He rather drifted off into examples in $l^1$ where he talked about non-compact symmetry groups and so on. Although I appreciated this at the moment, I am a bit unsatisfied now, because I would like to see a very dense and general statement what concentration compactness is about. To my surprise, also the internet seems to be full of rather vague explanations what this means on a general Banach space. I do not want to give references at this point, because I think many explanations are well written but do not answer my question: Is there a comprehensive theorem stating the concept of concentration compactness in a most general way? Put differently: What is the analogue of Banach-Alaoglu for concentration compactness? REPLY [3 votes]: This is really just a longwinded comment. The earliest instances of concentrated compactness that I know of are for geometric questions such as existence of energy-minimizing harmonic maps (as studied by Sacks and Uhlenbeck), the Yamabe problem (as studied by Trudinger, Aubin, and Schoen), and the existence of self-dual Yang-Mills connections (as studied by Uhlenbeck and Taubes), which all can be cast as nonlinear elliptic PDEs. They can also be formulated as variational problems. The fact that these arise from geometry means that the energy functional has symmetries, including scale invariance, which leads to the observation that the energy functional is critical in the sense that compactness just barely fails when studying a minimizing sequence. The beautiful and deep insight that I believe was due originally to Sacks and Uhlenbeck is that when compactness just barely fails for a minimizing sequence, one can actually understand quite precisely how compactness fails. In the cases cited above, the local energy (the integrand of the energy functional) can concentrate at a discrete set of points (but not on any larger set) and if one rescales around each point, the limit, often called a "bubble", is a symmetric solution to the original problem. This work led to, for example, new theorems in the differential topology of $4$-manifolds, starting with Donaldson's theorem. I believe Pierre Louis Lions was the first to formulate all of this into a general principle that he called concentrated compactness and that could be applied to a broad class of nonlinear elliptic PDEs. I'm not familiar with his work, but I think he did state and prove general theorems. I do not know whether his theorems imply the geometric results mentioned above. In general, when studying nonlinear PDEs, it is very hard to formulate a general theorem that can be applied directly to different interesting applications. Usually, the best you can do is to formulate a general principle like concentrated compactness and adapt it to specific situations. Based on the original examples and formulation, it appeared that concentrated compactness could not be used for non-elliptic PDEs. However, as Tao describes, concentrated compactness, suitably formulated, has indeed proved to be a powerful tool even for nonlinear wave equations. But a precise statement of a more abstract version that encompasses everything is probably even harder to formulate precisely than the original elliptic version.<|endoftext|> TITLE: 3-folds with "simple" Betti numbers and positive Kodaira dimension QUESTION [5 upvotes]: I am interested to know an example of a simply connected smooth projective 3-fold $X$ (over $\mathbb{C}$) satisfying the following two constraints: $X$ has the same Betti numbers as $\mathbb{C}\mathbb{P}^{3}$ i.e. $b_{1}(X) = b_{3}(X) = 0$ and $b_{2}(X) = 1$ and all of its cohomology groups are torsion-free. $\mathrm{Kod}(X) \geq 0$. ($\mathrm{Kod}(X)$ denotes the Kodaira dimension). REPLY [8 votes]: Let me just mention that the non-existence of such a threefold is an immediate consequence of Yau's inequality. First, as explained in the above comment, the conditions $b_2=1$ and $\mathrm{Kod}(X)\geq 0$ imply that $K_X$ is ample. Then Yau gives $c_1^3\geq \frac{8}{3}c_1c_2 $, which is equivalent by Riemann-Roch to $K_X^3\leq 64 \chi (K_X)$. But the conditions on the Betti numbers imply $H^i(X,\mathcal{O}_X)=0$ for $i>0$, hence $\chi (K_X)=-\chi (\mathcal{O}_X)=-1$, a contradiction.<|endoftext|> TITLE: Interpretations and models of permanent QUESTION [25 upvotes]: The standard interpretation of permanent of a $0/1$ matrix if considered as a biadjacency matrix of a bipartite graph is number of perfect matchings of the graph or if considered as a adjacency matrix of a directed graph is number of vertex-disjoint cycle covers and these are also the standard models. Given the importance of permanent as a $\#P$ complete problem that arises in various contexts such as lattices and polyhedra are there any other non-trivial interpretations of the permanent? I would be interested in any interpretations to number theory (I know none and if there is one it would be very interesting) or algebraic geometry (however tenuous it may be) particularly since the latter is contemplated to be useful in studying the Permanent-Determinant problem. I would even consider this as an interpretation for special matrices. Permanent represents natural numbers in a canonical way through special matrices Is every positive integer the permanent of some 0-1 matrix?. However if there is any multiplicative and additive property that comes along with it that would be very great and so far I do not see it but it looks like it should be manageable to concoct interesting classes of matrices which have specific interpretations. REPLY [7 votes]: The permanent appears in various "weighted" versions of Bézout's theorem (on number of solutions of systems of polynomials). Recall that Bézout's theorem says that the number of isolated solutions of $n$ polynomials $f_1, \ldots, f_n$ in $n$ variables is bounded by $\prod_i \deg(f_i)$. The weighted version of Bézout's theorem generalizes this bound to $\prod_i (\omega(f_i)/\omega(x_i)$, where $\omega$ is any "weighted degree" with positive weights. This is further generalized by the "weighted multi-homogeneous" version which lets you partition the variables into a bunch of (disjoint) groups and consider separately the weights on these groups, and this is where permanent appears naturally. Indeed, let $I_1, \ldots, I_s$ be a partition of $[n] := \{1, \ldots n\}$, i.e. $[n] = \bigcup_j I_j$ and $\sum_j |I_j| = n$. For each $j$, fix a weighted degree $\omega_j$ on $(x_k: k \in I_j)$ such that $\omega_{j,k} := \omega_j(x_k)$ is positive for each $j,k$, and the $\gcd$ of the weights $\omega_{j,k}$ is $1$ for each $j$. Then the weighted multi-homogeneous Bézout bound for the number of isolated zeroes of $f_1, \ldots, f_n$ is $$\frac{\text{perm}(D)}{(\prod_j n_j!)(\prod_{j,k} \omega_{j,k})}$$ where $n_j := |I_j|$ and $D$ is the following matrix: $$ \begin{pmatrix} d_{1,1} & \cdots & d_{1,1} & \cdots & \cdots & d_{1,s} & \cdots & d_{1,s} \\ \vdots & & \vdots & & & \vdots & & \vdots \\ d_{n, 1} & \cdots & d_{n,1} & \cdots & \cdots & d_{n,s} & \cdots & d_{n,s} \end{pmatrix} $$ where each $d_{i,j} := \omega_j(f_i)$ is repeated $n_j$ times in the $i$-th row. One can also try to generalize the weighted bound and the weighted multi-homogeneous bound to incorporate non-positive weights, see Theorems X.36 and X.39 of arxiv:1806.05346 for my attempts - permanent appears naturally in both places. The classical versions described above are also treated in that document (see Chapter VIII). As asked by OP in a comment below, all these are related to mixed volumes of polytopes via the Bernstein-Kushnirenko theorem. More precisely: The Bézout bound $\prod(\deg(f_i))$ is the mixed volume of simplices $\{\alpha \in \mathbb{R}_{\geq 0}^n: \alpha_1 + \cdots + \alpha_n \leq \deg(f_i)\}$. More generally, the weighted Bézout bound is the mixed volume of simplices $\{\alpha \in \mathbb{R}_{\geq 0}^n: \langle \omega, \alpha \rangle \leq \omega(f_i)\}$. More generally, the weighted multi-homogeneous Bézout bound is the mixed volume of products of the above simplices. If $\omega = (\omega_1, \ldots, \omega_n)$ has some negative/zero components, then the problem is that the polyhedron $\{\alpha \in \mathbb{R}_{\geq 0}^n: \langle \omega, \alpha \rangle \leq \omega(f_i)\}$ is unbounded, and one has to decide on a polytope that has a facet on the hyperplane $\{\alpha:\langle \omega, \alpha \rangle = \omega(f_i)\}$. One natural choice is to bound the body by hyperplanes parallel to coordinate hyperplanes, i.e. to consider the polytopes $$\{\alpha \in \mathbb{R}_{\geq 0}^n: \langle \omega, \alpha \rangle \leq \omega(f_i),\alpha_j \leq \deg_{x_j}(f_i)\ \text{for all}\ j\ \text{such that}\ \omega_j \leq 0\}$$ The general weighted (respectively, weighted multi-homogeneous) Bézout bound cited above corresponds to the mixed volume of these (respectively, products of these) polytopes.<|endoftext|> TITLE: Are stably equivalent quadratic forms over Z equivalent? QUESTION [5 upvotes]: Let $Q_1, Q_2, R$ be quadratic froms over $\mathbb{Z}$ such that $Q_1 \oplus R \cong Q_2 \oplus R$ as quadratic forms. Is it necessary that $Q_1 \cong Q_2$? I know that by Witt's theorem it is true for fields. REPLY [8 votes]: I would say no; an example is when $R$ is a hyperbolic plane, the relation you demand says merely that $Q_1$ and $Q_2$ are in the same genus. This is in SPLAG, page 378 in the first (1988) edition, see also Clark Jagy. Notice that rational equivalence "without essential denominator" is exactly how Siegel defined the genus. So, $x^2 + 14 y^2$ and $2 x^2 + 7 y^2$ are rationally equivalent, and we can arrange that the denominators are prime to $14,$ but not integrally equivalent. Give me a few minutes, I am going to display integer equivalence for $Q_1 = x^2 + 14 y^2,$ $Q_2 = 2 x^2 + 7 y^2,$ which pair take up a good deal of room in Cox's book, and $R = 2 z w.$ By solving $P^T A_1P = A_2$ in 4 by 4 integer matrices... $$ A_1 = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 14 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) $$ $$ P = \left( \begin{array}{rrrr} 2 & 7 & -2 & 2 \\ 1 & 5 & -2 & 1 \\ -4 & -14 & 6 & -3 \\ 2 & 14 & -5 & 3 \end{array} \right) $$ $$ A_2 = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) $$ $$ \small \left( \begin{array}{rrrr} 2 & 1 & -4 & 2 \\ 7 & 5 & -14 & 14 \\ -2 & -2 & 6 & -5 \\ 2 & 1 & -3 & 3 \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 14 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) \left( \begin{array}{rrrr} 2 & 7 & -2 & 2 \\ 1 & 5 & -2 & 1 \\ -4 & -14 & 6 & -3 \\ 2 & 14 & -5 & 3 \end{array} \right) = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & 7 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right) $$ check with gp-pari ===================== parisize = 4000000, primelimit = 500509 ? a1 = [ 1,0,0,0; 0,14,0,0; 0,0,0,1; 0,0,1,0] %1 = [1 0 0 0] [0 14 0 0] [0 0 0 1] [0 0 1 0] ? a2 = [ 2,0,0,0; 0,7,0,0; 0,0,0,1; 0,0,1,0] %2 = [2 0 0 0] [0 7 0 0] [0 0 0 1] [0 0 1 0] ? p = [ 2,7,-2,2; 1,5,-2,1; -4,-14,6,-3; 2,14,-5,3] %3 = [2 7 -2 2] [1 5 -2 1] [-4 -14 6 -3] [2 14 -5 3] ? matdet(p) %4 = 1 ? pt = mattranspose(p) %5 = [2 1 -4 2] [7 5 -14 14] [-2 -2 6 -5] [2 1 -3 3] ? a1 %6 = [1 0 0 0] [0 14 0 0] [0 0 0 1] [0 0 1 0] ? pt * a1 * p %7 = [2 0 0 0] [0 7 0 0] [0 0 0 1] [0 0 1 0] ? a2 %8 = [2 0 0 0] [0 7 0 0] [0 0 0 1] [0 0 1 0] ? pt * a1 * p - a2 %9 = [0 0 0 0] [0 0 0 0] [0 0 0 0] [0 0 0 0] ? ===================<|endoftext|> TITLE: What surreal numbers are representable by Red-Blue Hackenbush games? QUESTION [10 upvotes]: Every game of Red-Blue Hackenbush represents a surreal number. Is the converse true? Assuming that it is false, what can be said about the class of surreal numbers that are representable by such games? For instance, it is obvious that this class is a group under addition, but I can't visualize why it should even be closed under multiplication. Edit (by request): A game of Red-Blue Hackenbush is a rooted graph G whose edges are either red or blue. It's helpful to think of the surreal value of the game G as the number of moves by which Blue is ahead. The addition operation is simply the disjoint union with roots identified. The additive inverse of a game is found by changing the color of every edge. [Of course, this is only an inverse after passing to equivalence classes, which are surreal numbers.] A finite path graph beginning with a blue edge at the root represents a positive dyadic rational number, which you find as follows: Each blue edge that appears before the first red edge is worth 1. Beginning with the first red edge, every edge is worth half as much as the edge before. The value of the game is the sum of the values of the blue edges minus the sum of the values of the red edges. This algorithm is presented by Tom Davis in pp. 11-13 of this paper, where you can find a full explanation of what's going on. By using graphs with infinitely many edges, we can represent many surreal numbers. If I understand correctly, the game {1|9} (for example) is not literally a Red-Blue Hackenbush game, but it is equivalent to 2={-1,0,1|3}, which is represented by a 3-edge blue path joined at the root to a single red edge. Is there a surreal number that is not equivalent to any Red-Blue Hackenbush game? REPLY [2 votes]: If you only allow connected graphs, and in particular do not allow ordinal numbers, the answer is no, unless you assign surreal numbers to Hackenbush games by a means other than induction. First it's important to realize that, contrary to what's stated in the question, not every rooted graph with edges colored red and blue corresponds to a surreal number. To assign a surreal number to a game, you can only handle a game if every move takes you to a game already handled, which means you can't handle a game if there is some infinite sequence of legal moves (for either player). This means a graph that corresponds to a surreal number cannot have infinitely many leaves, infinitely many loops, or infinitely many topological ends. Otherwise we could move infinitely often by chopping off leaves, by cutting loops, or by chopping off ends. (For the third case, you can start by cutting off loops until the graph is a tree, then observing that if a vertex has infinitely many ends among its descendents, at some point oen of its descendents must split in such a way that at least one side doesn't contain all but finitely many of the ends, and you can cut off that side, then repeat). If a graph does correspond to a surreal number, the graph is a finite graph glued to finitely many infinite paths. This is because after chopping off the finitely many loops, the graph is a tree with finitely many ends, hence finitely many branch points, and there can only be finitely many edges between any two branch points. Thus, the ordinal for when the game finishes is $< n \omega$, where $n$ is the number of infinite paths - i.e. we can bound the time until we are forced to chop one of the infinite paths, then bound the time until we are forced to chop the next path, etc. a finite number of times. So you cannot express this way any surreal number that occurs in the generation $\omega^2$ or after.<|endoftext|> TITLE: Fermat's Method of Removing Radicals QUESTION [8 upvotes]: When searching for the originator of the method of removing radicals from equations I found the following remark by D. Mooney: "In the 97th Section of his Analysis, Doctor Hales shews the method of taking quadratic surds out of an equation, provided the number of terms be not greater than four, but if there be a fifth term, whether rational or surd, he is of opinion that the equation cannot, by that method, be cleared from surds. Were this the case, we would have no other alternative, than to recur to the method of Monsieur Fermat, by seigning the surds equal to an assumed letter, and thence by means of as many simple equations, as there are surds, to take these letters out of the equation; but this is a work of so much labour, that it is sufficient to deter a person slightly acquainted with algebra." and then details his method of removing an arbitrary number of square roots from an equation by repeated squaring (I assume "surds" are square roots). The description of Fermat's method seems to indicate that he is the original source of the technique of reducing the problem of removing square roots (or even general radicals) from equations to solving a linear system of equations, where the monomials of radicals are the unknowns. Question: what kind of problem did Fermat exactly solve and how has that influenced later work of other mathematicians; apparently Moivre solved the general case? REPLY [3 votes]: Fermat developed his method of clearing radicals to solve the problem of finding the maximum and minimum of a polynomial, as an application of differential calculus "avant la lettre". A 19th century source describes the method as follows: In his method of maxima and minima, for instance, in the function which is to become a maximum, he substitutes for the independent variable $x$, the same increased by a certain quantity $e$; and as the condition of the maximum requires, that when $e$ is infinitely diminished, or zero, these two expressions should be equal, he equates them, clears the equation of surds and radicals, and, striking out the common terms, the whole becomes divisible by $e$, after which $e$ is made zero and the equation of the maximum obtained. As explained here, Fermat's method (developed around 1629 but published only in his collected works) was extended by Newton in his development of calculus. This is consistent with the text by Mooney quoted in the OP, which continues on page 226 with "The example proposed by Newton, where he mentions M. Fermat's method". The procedure for treating surd expressions which Fermat developed in Methode de maximis et minimis is explained in a more modern text here (page 59).<|endoftext|> TITLE: When does a space of endomorphisms contain invertibles? QUESTION [8 upvotes]: Let $V$ be a finite-dimensional vector space over a field $K$. Let $U$ be a linear subspace of $\mathrm{End}(V)$. Write $UV$ for the span of all $Av$ where $A\in U$ and $v\in V$. Suppose that $$ \ker(U)=\bigcap_{A\in U}\ker(A) $$ is zero and that $$ \mathrm{coker}(U)=V/UV $$ is zero. Is it true that $U$ must contain an invertible element of $\mathrm{End}(V)\ $? REPLY [9 votes]: Nope: $$ \left\{ \begin{bmatrix} 0 & x & y\\ x & 0 & 0\\ y & 0 & 0 \end{bmatrix}: x,y \in\mathbb{R} \right\}. $$ This is a classical counterexample in numerical linear algebra -- the simplest singular matrix pencil with a nontrivial Kronecker canonical form.<|endoftext|> TITLE: Number of involutions in a finite group modulo $4$ QUESTION [10 upvotes]: It is a theorem of Alperin, Feit and Thompson [Isaacs Character Theory book (4.9)] that if $G$ is a $2$-group in which the number of involutions is congruent to $1$ modulo $4$, then $G$ is cyclic or $|G:G'|=4$. In the latter case $G$ is dihedral, semidihedral or generalized quaternion. Are there any results of a similar nature which hold for all finite groups? I want a result that says that in a finite group the number of involutions, modulo $4$, influences the structure of a Sylow $2$-subgroup of the group. REPLY [9 votes]: EDIT: Actually, there is an older reference which will give the result below. See Herzog, Marcel Counting group elements of order $p$ modulo $p^2$. Proc. Amer. Math. Soc. 66 (1977), no. 2, 247–250. where you will also find references to other related papers. ------ The following paper seems to give what you are looking for. Murai, Masafumi On the number of p-subgroups of a finite group. J. Math. Kyoto Univ. 42 (2002), no. 1, 161–174. link Let $G$ be a finite group and fix a prime $p$. Denote by $s_e(G)$ the number of subgroups of order $p^e$ in $G$. If $p^e$ divides the order of $G$, a classical result of Frobenius states that $s_e(G) \equiv 1 \mod{p}$. The paper of Murai above studies how $s_e(G) \mod{p^2}$ depends on the structure of a Sylow $p$-subgroup of $G$. For $p = 2$, Theorem D in the paper implies the following: Let $G$ be a finite group with a Sylow $2$-subgroup $P$ of order $2^n$ ($n \geq 2$). If $P$ is cyclic, dihedral, generalized quaternion, or semidihedral, then $s_1(G) \equiv 1 \mod{4}$. Otherwise $s_1(G) \equiv 3 \mod{4}$. Note that in this case $s_1(G)$ is the number of involutions in $G$.<|endoftext|> TITLE: Isoperimetric inequality on the plane QUESTION [8 upvotes]: Let $A$ be a connected compact domain with smooth boundary in the Euclidean 2-plane. Assume its diameter is at most $d$. Assume that the second fundamental form of the boundary is at most $-c$ where $c\geq 0$ (equivalently, the curvature of the boundary is at most $-c$). Is there an upper estimate on the length of the boundary in terms of $d,c$? Remark. If $c=0$ then the domain $A$ is convex and hence the length of the boundary is known to be at most $2\pi d$. REPLY [12 votes]: No, there is no bound --- yinyang is our friend. This example works if $d\cdot c>2$; otherwise there should be an upper bound.<|endoftext|> TITLE: Separating closed $SO(p,q)$ orbits by invariant polynomials QUESTION [6 upvotes]: Consider the real Lie group $SO(p,q)$ (I believe that it happens to be a linearly reductive algebraic group over $\mathbb{R}$, if that's relevant). Also, if relevant, I'm mostly interested in the (Lorentzian) $p=1$ case. I'm also interested in the same question when $SO(p,q)$ is replaced by $O(p,q)$, but I suspect that the answers are similar. Let $V$ be a finite dimensional representation of $SO(p,q)$. Is it true that the scalar $SO(p,q)$-invariant polynomials on $V$ separate the closed $SO(p,q)$-orbits on $V$? I know that in general not all orbits (which are not necessarily closed) can be separated by polynomial invariants. But some results from geometric invariant theory do show that the closure of any orbit contains a unique closed orbit [Richardson & Solodowy (1990) JLMS 42 409-429]. So it's a given that any continuous (let alone polynomial) scalar invariant will not separate a non-closed orbit from the closed one in its closure. But there might still be hope to separate the closed orbits themselves. What I already know is that the polynomial invariants do separate the closed orbits of the complexified representation $V\otimes \mathbb{C}$ of $SO(p,q;\mathbb{C}) \cong SO(p+q;\mathbb{C})$ and that the polynomial invariants of the $SO(p,q)$ representation are equivalent to the polynomial invariants of the $SO(p+q;\mathbb{C})$ representation. However, when we restrict to the smaller real group, the number of orbits may increase and the separation by polynomials is no longer obvious. For instance, when $p=1$, the further restriction to the so called orthochronous subgroup $SO^{\uparrow}(1,q) \subset SO(1,q)$ does break polynomial separability already in the fundamental vector representation. In the language of special relativity, polynomial invariants can recognize whether vectors are timelike or spacelike, but cannot distinguish between future- and past-pointing timelike vectors. On the other hand, orthochronous transformations are precisely those that are not allowed to exchange future- and past-pointing timelike vectors. On the other hand, I do hope that problem does not occur for the slightly larger groups $SO(1,q)$ and $O(1,q)$. Update: Friedrich Knop has answered the bulk of my question, based on the following observation: But then there are also the real points of the algebraic group orbit $Gv$. They consist of all vectors of $V$ which can be obtained from $v$ by an element of the complex group $G(\mathbb{C})$. Hence $(Gv)(\mathbb{R})=G(\mathbb{C})v\cap V$. The latter are sometimes called the stable orbits. The point is now that an invariant $f$ can only separate stable orbits. This is because $G(\mathbb{R})$-invariance implies $G$-invariance. Moreover, any two closed stable orbits can be separated by an invariant. I'm now looking for a specific reference that discusses this fact and its proof. REPLY [9 votes]: The group $SO(p,q)$ is not per se an algebraic group. Rather there is an algebraic group $G$ such that $SO(p,q)=G(\mathbb R)$ is its group of real points. The main point is that also $G(\mathbb C)$ is defined which leads to two different concepts of orbits. First, there are the orbits $G(\mathbb R)v$ which you are probably after. But then there are also the real points of the algebraic group orbit $Gv$. They consist of all vectors of $V$ which can be obtained from $v$ by an element of the complex group $G(\mathbb C)$. Hence $(Gv)(\mathbb R)=G(\mathbb C)v\cap V$. The latter are sometimes called the stable orbits. The point is now that an invariant $f$ can only separate stable orbits. This is because $G(\mathbb R)$-invariance implies $G$-invariance. Moreover, any two closed stable orbits can be separated by an invariant. The stable orbit is always a finite union of open $G(\mathbb R)$-orbits and these cannot be separated by invariants anymore. The good news is that an orbit $G(\mathbb R)v$ is (Hausdorff-)closed iff the associated stable orbit is (Hausdorff-)closed iff $Gv$ is (Zariski-)closed. This follows, e.g., form a theorem of Kempf on the existence of optimal $1$-parameter subgroups. Now back to your real question: is it possible to make all closed $G(\mathbb R)$-orbits stable by enlarging the group, like from $SO^{\uparrow}(1,q)$ to $SO(1,q)$? This works in some cases but in general it is doomed to fail. Example: Take the group $G(\mathbb R)=SO(1,q)$, $q\ge1$ acting on $\mathbb R^{1+q}=\mathbb R\mathbf e_0\oplus\mathbb R\mathbf e_1\oplus\ldots\oplus\mathbb R\mathbf e_q$. Put $V=S^2(\mathbb R^{1+q})$. The transformation $(x_0,x_1,x_2,\ldots,x_q)\mapsto(ix_1,ix_0,x_2,\ldots,x_q)$ is in $G(\mathbb C)$ and maps $v_1=\mathbf e_0^2$ to $v_2=-\mathbf e_1^2$. Thus $v_1$ and $v_2$ cannot be separated by an invariant but their $G(\mathbf R)$-orbits are closed and different (one is timelike the other spacelike).<|endoftext|> TITLE: Is there winning strategy in Tetris ? What if Young diagrams are falling? QUESTION [50 upvotes]: Question 1 Is there a winning strategy (algorithm to play infinitely) in Tetris, or is there a sequence of bricks which is impossible to pack without holes? Consider generalized Tetris with Young diagrams (for some $n$) are falling down. Question 2 Is there winning strategy? If not - consider some probability measure on Young diagrams (e.g. uniform). What will be "losing speed"? I.e. how fast the height of uncancelled rows will grow for best possible algorithm? Question 3 Can one relate such Tetris like questions on Young diagrams with some conceptual/conventional theories where Young diagrams appear - representation theory of symmetric group or something else? REPLY [6 votes]: My answer is regarding the first part of your Question 1. There is a paper linked in comments (below your question) which shows that, in general, there is no winning strategy. The paper is based upon essentially assuming that any given sequence of tetrominoes can occur. The rules for what sequences of tetrominoes are or aren't allowed in a game is usually called a "Randomiser" (in terminology of the game). A randomiser that allows any sequence of tetrominoes is called "memoryless". So the paper shows that given a memoryless-randomiser, there is no general winning strategy. However, the analysis might not apply to games(even otherwise assuming exactly same game mechanics as in paper) which implement a different randomiser because certain sequences of tetrominoes are forbidden with probability 1 (at any possible point in the game). It has already been hinted in the other answer that the winning strategy for single player with, what is called a bag randomiser, and some further rules/features has been known for almost a decade (see for example one of the answers: https://math.stackexchange.com/questions/1135388/an-impossible-sequence-of-tetris-pieces). It should be mentioned that some aspects(I think) have been improved upon for this loop in recent years. But, at any rate, what is surprising for me is that there doesn't seem to be much work done beyond that. Considering "just" the single-player gravity based variant of endless tetris there are three main aspects (keeping it as short as possible): (1) Basic Features: These include, among other things, features such as "Hold" and "Number of Next Piece Previews". (2) Gravity Rules: There are two main choices "0G" and "20G". 0G or very close to 0G occurs in most commercial games. 20G (instant drop) occurs in the well-known TGM series. With 0G one can decide whether to include "softdrop" or not. What that means is that when a tetromino falls on the surface it can be moved around on the surface (of the stack), before finally locking in (the proof I linked actually seems to use that in a couple of situations). With 20G there are certain basic "kick-rules" (the full explanation of rules would be little longer but they are well-documented). (3) Randomiser: Randomiser, as has been mentioned, desrcibes rules for which sequences of tetrominoes are or aren't allowed. In actuality, this is done to reduce probability of unfair situations. As far as I know, "7 piece bag" is the most common randomiser for commercial iterations (atleast for some time now). However, there is no shortage of other randomisers (some with very basic and short descriptions) that have been used in fairly noteable fan-made or commercial games. Note that I have abridged the post enormously (leaving out discussion of myriad of modes that might be interesting for analysis) just to emphasise a somewhat more focused question about which it seems not much is known: " Suppose we set certain rules of play regarding the falling tetrominoes -- coming under points (1) and (2) in my description above. Given a certain randomiser, how do we find out that endless play is possible in general or not? " It would be interesting to see some progress on this question regarding reasonably large classes of randomisers. P.S. Randomisers also include probabilistic considerations with them, which I glossed over (which might only be relevant if we are interested in questions beyond just a guaranteed winning strategy). For example, the more strict definition of memoryless-randomiser would be that at "any" point in the game, regardless of what finite sequence of tetrominoes was handed over, the probability for every tetromino(to be the next one) would be 1/7 (uniform across tetrominoes).<|endoftext|> TITLE: chain condition of a product of posets QUESTION [7 upvotes]: Suppose $P$ and $Q$ are ccc partial orders. Is $P \times Q$ $\omega_2$-cc? Note that this true under CH by the Erdos-Rado Theorem. REPLY [10 votes]: First of all, note (as Monroe does in his question) that if $\mathbb P,\mathbb Q$ are ccc, then $\mathbb P\times\mathbb Q$ is $\mathfrak c^+$-cc, as an immediate consequence of the Erdős-Rado theorem $(2^{\aleph_0})^+\to(\aleph_1)^2_2$. (This is to say, if $\mathbb P$ and $\mathbb Q$ do not admit uncountable antichains, then any antichain in their product has size at most continuum.) The question is essentially whether we can do better. Under Martin's axiom ($\mathsf{MA}_{\aleph_1}$ suffices) we can: $\mathsf{MA}_{\aleph_1}$ implies that any ccc poset has property (K), that is, not only it does not have uncountable antichains, but in fact any uncountable set contains an uncountable subset of pairwise compatible conditions. The product of two property (K) posets has property (K) as well (and, more strongly, so does the finite support product of any number of property (K) posets). Thus, under $\mathsf{MA}_{\aleph_1}$, the product of two ccc posets is again ccc. Without such an assumption, the situation can be completely different. For instance, if $\mathbb S$ is a Suslin tree, then it is ccc but its square is not. The failure of the ccc is not too bad in this case: Since $|\mathbb S|=\aleph_1$, $\mathbb S\times\mathbb S$ is $\aleph_2$-cc. From this, the question presents itself, as it is perhaps natural to wonder whether we can prove that a product of two ccc posets is always $\aleph_2$-cc. This cannot be done in general: It is consistent with the continuum as large as desired that there are ccc posets whose product admits an antichain of size continuum (that is, the trivial upper bound is best possible). This was first proved in MR0493930 (58 #12886). Fleissner, William G. Some spaces related to topological inequalities proven by the Erdős-Rado theorem. Proc. Amer. Math. Soc. 71 (1978), no. 2, 313–320. His model is obtained simply by adding $\kappa$ Cohen reals, for $\kappa$ an uncountable cardinal. The argument is elegant, the two posets are defined in terms of a coloring $P:[\kappa]^2\to 2$: $\mathbb P$ is defined by considering the maximal 0-homogeneous subsets of $\kappa$, and $\mathbb Q$ by considering the maximal 1-homogeneous subsets. (Actually, Fleissner describes topologies on $\mathbb P,\mathbb Q$ and shows that the cellularity of the product is at least $\kappa$.) The difficulty is finding $P$ that ensures $\mathbb P,\mathbb Q$ are ccc to begin with. It is here that forcing is used, with $P$ added generically. See section 5 in Fleissner's paper for details. The topic of cellularity of products is quite interesting and there are many additional results. An excellent survey can be found in MR3271280. Rinot, Assaf. Chain conditions of products, and weakly compact cardinals. Bull. Symb. Log. 20 (2014), no. 3, 293–314. (Assaf's paper is also available at his page.) This situation, of having two ccc posets whose product is not $\mathfrak c$-cc, can also be arranged in a variety of other ways. For example, it is a consequence of the continuum being real-valued measurable. For this, see for instance Theorem 7D in MR1234282 (95f:03084). Fremlin, D. H. Real-valued-measurable cardinals. In Set theory of the reals (Ramat Gan, 1991), 151–304, Israel Math. Conf. Proc., 6, Bar-Ilan Univ., Ramat Gan, 1993.<|endoftext|> TITLE: Estimate of area of 2-dimensional surface QUESTION [7 upvotes]: Let $X$ be a compact smooth 2-dimensional Riemannian manifold with boundary. Assume that the Gauss curvature of $X$ is at least $\kappa$, the diameter is at most $D$, and the second fundamental form of the boundary is at least $\lambda$. Does there exist an upper bound on the area of $X$ in terms of $\kappa, D,\lambda$ only? Remarks. (1) In the case of the empty boundary the answer is known to be positive (e.g. it follows from the Bishop inequality). (2) If the boundary is non-empty and $\lambda\geq 0$, then the answer is also positive. In this case $X$ is locally geodesically convex and hence it is an Alexandrov space of curvature at most $\kappa$ in the sense of Alexandrov, and for such spaces there is a generalization of the Bishop inequality due to Burago-Gromov-Perelman. (Although there is a little bit more direct but longer explanation.) REPLY [6 votes]: Yes, there is a bound; your problem can be reduced to the case of convex boundary using the following trick. Without loss of generality, we can assume that $\lambda=-\tfrac1{10}$ and $\kappa=-1$. In this case you can attach a collar to your surface locally isometric to the tubular neighborhood in of line in the Lobachevsky plane which has curvature of boundary $\tfrac1{10}$. In the obtained surface boundary is convex and it has curvature $\ge -1$ in the sense of Alexandrov (by Alexandrov's gluing theorem). The diameter, area and perimeter of the obtained surface can be bounded in terms of the diameter area and perimeter of the original surface and the other way around.<|endoftext|> TITLE: Graph to Bipartite conversion preserving number of perfect matchings QUESTION [7 upvotes]: Given a graph $G$ on $n$ vertices is there a technique to convert to a balanced bipartite graph $B$ with $O(n^c)$ vertices at some fixed $0 TITLE: Estimate of number of boundary components of a compact Riemannian 2-surface QUESTION [7 upvotes]: Let $X$ be a compact smooth 2-dimensional Riemannian manifold with boundary. Assume that the Gauss curvature of $X$ is at least $-1$ and the diameter is at most $D$. Assume that near the boundary the surface is locally geodesically convex. Is it true that the number of connected components of the boundary is bounded above by a constant depending on $D$ only? Remark. I am not a specialist, but if I understand correctly the answer should be positive and follow from very general results in the Alexandrov geometry. Indeed due to local convexity, $X$ is an Alexandrov space of curvature at least $-1$. Each boundary component is an extremal subset in the sense of Perelman-Petrunin. In general the number of extremal subsets of a compact $n$-dimensional Alexandrov space of curvature at least $-1$ and diameter at most $D$ is bounded above by a constant depending on $n,D$ only. Is this argument correct? Anyway I would prefer to have a more direct argument in my case. REPLY [7 votes]: I think it follows from Gauss-Bonnet. Suppose $X$ has genus $g$ and $n$ boundary components. Gauss-Bonnet says that $$\int_X K\;dA+\int_{\partial X}k\;ds=2\pi\chi(X)=2\pi(2-2g-n),$$ where $K$ is sectional curvature, $k$ is the curvature of the boundary. The local convexity implies that the boundary is positively curved, so $\int_{\partial M}k\;ds\ge 0$ and $$-\mathop{\text{area}}X \le \int_X K\;dA\le 2\pi(2-2g-n),$$ i.e., $2\pi(2g+n-2)\le \mathop{\text{area}}X$. The curvature and diameter bound imply that the area of $X$ is at most the area of a ball of radius $D$ in the hyperbolic plane, so $g$ and $n$ are bounded.<|endoftext|> TITLE: Counting with tensor products QUESTION [7 upvotes]: Suppose I've got vectors $v = (1,-1)$ and $w = (1,1)$ and any $m \in \mathbb{N}$. Let $a = v \otimes v \otimes w^{\otimes m}$ and let $\tilde{a}$ be the sum over all $\binom{m}{2}$ unique vectors obtained by permuting the tensor coordinates of $a$. I'm interested in identifying the asymptotics of a function $f(m) = \|\tilde{a}\|_1$ where $\|\cdot\|_1$ is the usual $1$-norm in terms of $m$. For example, when $m=0$, we simply have $\tilde{a} = v \otimes v$ and so $f(0) = 4$. When $m=1$ we have $\tilde{a} = v \otimes v \otimes w + v \otimes w \otimes v + w \otimes v \otimes v$ and one can check that $f(1) = 12$. When $m = 2$ we have $$\tilde{a} = v \otimes v \otimes w \otimes w + v \otimes w \otimes v \otimes w + v \otimes w \otimes w \otimes v\\+ w \otimes v \otimes v \otimes w + w \otimes v \otimes w \otimes v + w \otimes w\otimes v \otimes v$$ and one can check that $f(2) = 24$. The problem seems simple, but I'm having a tough time finding a nice expression approximating $f(m)$. It seems related to some type of signed Chu-Vandermonde identity. Any insights or suggestions are appreciated. Thank you! REPLY [7 votes]: You are looking at the vector of coefficients of the polynomial $$\sum_{\epsilon}\prod_{i=1}^{m+2}(1-\epsilon_i x_i)$$ where $\epsilon$ runs over all choices of signs $\pm$ provided there are exactly two $-$ signs. There are $\binom{m+2}{2}$ terms being summed. After expanding we get a multilinear symmetric polynomial in the $x_i$'s. this tells us that since we only care about the $|\cdot|_1$ norm, we can set all $x_i=t$ and observe the expression becomes $$\binom{m+2}{2}(1-t)^2(1+t)^m.$$ The coefficient of $t^k$ is $$\binom{m+2}{2}\left[\binom{m}{k}-2\binom{m}{k-1}+\binom{m}{k-2}\right]=\frac{1}{2}\binom{m+2}{k}\left[(m-k+1)(m-k+2)+k(k-1)-2k(m-k+2)\right]=\frac{1}{2}\binom{m+2}{k}\left[(m-2k+2)^2-2-m\right]$$ which means these coefficients are positive everywhere except in the interval $$\frac{m+2-\sqrt{m+2}}{2} TITLE: Definable non measurable set after adding one Cohen real to L QUESTION [12 upvotes]: Suppose $x$ is a Cohen real over the constructible universe $L$. Is there a (parameter free) definable non measurable set of reals in $L[x]$? This appeared in Will Brian's answer here where he noted that the analogous question after adding a random real has a postive answer. REPLY [10 votes]: The paper "$\Delta^1_2$-sets of reals" by Ihoda (=Judah) and Shelah [Annals of Pure and Applied Logic 42 (1989) 207-223] should have the result you want. Theorem 3.1(i) tells you that every $\Delta^1_2(r_0)$ ($\Delta^1_2$ using only $r_0$ as a parameter) set is Lebesgue measurable if and only if there is a real $r$ that is random over $L[r_0]$. In particular, all $\Delta^1_2$ (parameter free) sets of reals are measurable if and only if there is a random real over $L$. This latter fails if we add a single Cohen real to $L$, so there is a non-measurable $\Delta^1_2$ (without parameters) set in $L[c]$ if $c$ is Cohen over $L$. The paper addresses regularity properties of $\Delta^1_2$ sets in many different settings, and not just Lebesgue measurability.<|endoftext|> TITLE: Complexity of $L[\mathrm{cf}]$ QUESTION [8 upvotes]: Assuming large cardinal axioms, which real numbers are in $L[\mathrm{cf}]$, where $\mathrm{cf}$ is the cofinality function on ordinals? $L[\mathrm{cf}]$ is the minimal inner model that 'knows' the cofinality of every ordinal in V, or more precisely, $∀α\,\{(β,\mathrm{cf}(β)):β<α\}∈L[\mathrm{cf}]$. Cofinality in $L[\mathrm{cf}]$ need not equal cofinality in $V$. I have a solution to a simpler problem: Theorem: A real number is in $L$[Card] iff it is in the minimal inner model with a proper class of measurable cardinals (assuming the sharp for this model exists). Here, Card is the cardinality function. Corollary (under the same assumption): The theory of $(L[\mathrm{Card}],∈,\mathrm{Card})$ is $Δ^1_3$. Proof sketch of the theorem: Starting with the minimal inner model $M$ with a proper class of measurable cardinals, iterate the first measurable cardinal until it becomes $ω_1^V$, the second until it becomes $ω_2^V$, and so on for every regular successor cardinal in $V$. For the converse, every $ω_{α+ω}$ can be shown to be measurable in HODL[Card] with $\{ω_{α+1}, ω_{α+2}, ω_{α+3}, ...\}$ Prikry generic over HODL[Card]. Additional details on the proof: If $M$ is iterated until we get $M'$ with measurable cardinals matching regular $V$-cardinals except that the next measurable cardinal after $ω_α^V$ is $ω_{α+ω}^V$, then $S = (ω_{α+1},ω_{α+2},...)$ is Prikry generic over $M'$ since if a measurable cardinal is iterated, a cofinal sequence of length $ω$ of its past values is Prikry generic. Now, $\mathrm{HOD}^{L[\mathrm{Card}]}⊂\mathrm{HOD}^{M'[S]}=M'$ and $S∈L[\mathrm{Card}]$, so $S$ is also Prikry generic over $\mathrm{HOD}^{L[\mathrm{Card}]}$. For the cofinality problem, my conjecture is to use a model $M$ (closely related to $L[\mathrm{cf}]$) such that: nonmeasurable regular cardinals in $M$ have $V$-cofinality $ω$, measurable cardinals of order $α$ in $M$ have $V$-cofinality $ω_{α+1}$ (or $ω_α$ if $α$ is weakly inaccessible (in $V$) plus a finite ordinal), and $M$ is obtained by iterating away the least mouse with a measure concentrating on cardinals with $o(κ)=κ$, and iterating the measures until the above correspondences hold. However, I do not know whether this works, or for that matter, whether the theory of $L[\mathrm{cf}]$ is generically absolute. An extension of the problem is to set $\mathrm{cf}(κ)=κ+α$ iff $κ$ is weakly $α$-Mahlo, which is well-defined for $α≤κ^+$. A $κ^+$-Mahlo $κ$ is called greatly Mahlo. For comparison, a limit ordinal $κ$ has cofinality $≥α$ iff for every (infinite) regular $β<α$, ordinals of cofinality $β$ are stationary below $κ$. (Also, if singular $κ$ pose a problem, I would be happy to see a solution for this extension with $\mathrm{cf}(κ)$ restricted to regular $κ$.) It appears plausible that the construction using orders of measurability can be extended to this problem, with weakly greatly Mahlo cardinals corresponding to $M$-cardinals that are strong up to a measurable. If that is the case, then under large cardinal axioms, all reals in $L[\mathrm{cf}]$ are still $Δ^1_3$. REPLY [5 votes]: Дмитро, can you expand a bit on how you prove that every ℵα+ω is measurable in the HOD of L[Card] and why the sequence of the ℵα+ns is Prikry generic, assuming there is a sharp for an inner model with a proper class of measurable cardinals? Thanks, Ralf. Regardless of that, the result is correct: if L[Card] doesn't have an inner model with a proper class of measurable cardinals, then we may compare KL[Card] with the sharp for an inner model with a proper class of measurable cardinals; a half-open interval from Card may then be used to produce a measure on an iterate of KL[Card] which may be pulled back to KL[Card]; this measure is in L[Card], which gives a contradiction. Your question about the complexity of the reals of L[cf] is related to the question: which reals does C* have (where C* is the least inner model which knows which ordinals have countable cofinality)? Magidor showed C* has 0† and more, and with him I showed (assuming a measurable cardinal above a Woodin cardinal in V) that all the reals of C* are in M1, the least inner model with one Wodin cardinal (so that KC* exists and is 1-small).<|endoftext|> TITLE: Access to papers as an unaffiliated individual QUESTION [27 upvotes]: I have graduated and my employer does not provide access to journals. I don't mind paying, but I cannot afford subscribing directly. Are there public mathematics associations that provide members with access to the most common mathematical journals? REPLY [3 votes]: Often the author has a web page on which he or she posts papers, e.g., T. Tao. Another possible source is Archiv.<|endoftext|> TITLE: Permanent of Nakayama algebras QUESTION [6 upvotes]: See https://en.wikipedia.org/wiki/Nakayama_algebra for the definition of Nakayama algebras and define the permanent of such an algebra to be the permanent of its Cartan matrix. (all algebras are assumed to be connected and finite dimensional) Conjecture: Let $X$ be the set of Nakayama algebras of finite global dimension with $n$ simple modules. Then the maximum of the permanents of algebras in X is given by $\sum\limits_{k=0}^{\infty}{\frac{k^n}{2^{k+1}}}$ and it is uniquely attained. See https://oeis.org/A000670 for the conjectured sequence. The algebras with a line as a quiver should always have permanent equal to one. Here the algebras with permanent higher than 1 and finite global dimension with 3 simple modules (given by their Kupisch series as first entry and the permanent as second entry): [ [ [ 2, 2, 3 ] 3 ], [ [ 2, 4, 3 ], 5 ], [ [ 3, 4, 4 ], 11 ], [ [ 3, 5, 4 ], 13 ] ] REPLY [6 votes]: Mare's answer reduces the problem to showing that the permanant of the $n \times n$ matrix $$\begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & 2 & 2 & \cdots & 2 \\ 1 & 1 & 2 & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & 2 \\ \end{bmatrix}$$ is the number of weak orders on $[n]:=\{ 1,2, \ldots, n \}$. This answer will prove that. First of all, permanent is unchanged by reordering the rows, so I will modify the matrix to $$\begin{bmatrix} 1 & 2 & 2 & \cdots & 2 \\ 1 & 1 & 2 & \cdots & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & 2 \\ 1 & 1 & 1 & \cdots & 1 \\ \end{bmatrix}.$$ By the definition of the permanent, the permanent is $$\sum_{\sigma \in S_n} 2^{\# \{ j \ :\ \sigma(j) > j \}}.$$ We can give a similar formula for the number of weak orders. Given a weak order $\succeq$ on $[n]$, define a total order on the same set by break all ties in according to the standard order on $[n]$. For example, $13 \succeq 5 \succeq 24$ turns into $3 > 1 > 5 > 4 > 2$. Total orders on $[n]$ correspond to permutations in $S_n$; given a permutation $\sigma$, the number of weak orders which give rise to it is $2^{\# \{ k : \sigma(k) > \sigma(k+1) \}}$. For example, to choose a weak order giving rise to the total order $3 > 1 > 5 > 4 > 2$, we just have to decide whether or not to group $(3,1)$, $(5,4)$ and $(4,2)$ together, and these choices are independent. So A000670 is $$\sum_{\sigma \in S_n} 2^{\# \{ k \ :\ \sigma(k) > \sigma(k+1) \}}.$$ For a permutation $\sigma \in S_n$, the exponent $\# \{ k : \sigma(k) > k \}$ is called the number of exceedances of $\sigma$, and $\# \{ k \ :\ \sigma(k) > \sigma(k+1) \}$ is called the number of descents of $\sigma$. So the conjecture follows from the following stronger result: Theorem The number of permutations of $n$ elements with $r$ descents is the same as the number of permutations of $n$ elements with $r$ exceedances. This was originally proved by MacMahon in 1915 and has been reproved many times since. The number of permutations of $n$ elements with $r$ descents is called the Eulerian number, denoted $\left\langle \begin{smallmatrix} n\\r \end{smallmatrix} \right\rangle$ and a statistic $f : S_n \to \mathbb{Z}$ which takes the value $r$ a total of $\left\langle \begin{smallmatrix} n\\r \end{smallmatrix} \right\rangle$ times on $S_n$ is called "Eulerian". A quick proof can be found as Proposition 3.14 in these notes. (No originality claimed for this source; just the first one I found without a lot of extraneous material.)<|endoftext|> TITLE: Questions on permanents of quiver algebras QUESTION [6 upvotes]: Define the permanent of an algebra as the permanent of the Cartan matrix. For simplicity assume all algebras are connected quiver algebras. Questions: Let $X_n$ be the set of all algebras with $n$ simple modules having finite global dimension. Is the permanent of algebras in $X_n$ bounded? If yes, what is the bound? Answer is no: See the answer by Pierre-Guy Plamondon. I suggest the following modified question too: 1.* Let $X_n$ be the set of all representation-finite algebras with $n$ simple modules having finite global dimension. Is the permanent of algebras in $X_n$ bounded? If yes, what is the optimal bound? Let $Y(Q)$ be the set of algebras with quiver $Q$ having finite global dimension. Is the permanent of such algebras bounded for given $Q$? If yes, what are good bounds. For example, my conjecture in the thread Permanent of Nakayama algebras I says that for $Q$ a circle with $n \geq 2$ points, the answer is $\sum\limits_{k=0}^{\infty}{\frac{k^n}{2^{k+1}}}$. Of course answers for special cases are welcome. REPLY [6 votes]: The answer to question 1 is no, unless $n=1$, where the bound on the permanents is $1$. For $n=1$, the algebra is either semisimple (so the permanent is $1$) or has infinite global dimension. Thus it suffices to give examples for $n=2$. Let $A$ be the algebra given by the quiver with two vertices $1$ and $2$, with arrows $x_1, \ldots, x_r$ from $1$ to $2$ and arrow $y$ from $2$ to $1$, subject to the relations $x_iy=0$ for all $i\in \{1, \ldots, r\}$. Then $A$ is of global dimension $2$, and its Cartan matrix is $$ \begin{pmatrix} 1 & 1 \\ r & r+1 \end{pmatrix}, $$ so its permanent is $2r+1$. I do not know the answer to question 2.<|endoftext|> TITLE: Are there curves of genus 2 with real multiplication by a non-maximal order? QUESTION [6 upvotes]: Let us work over $\mathbb{C}$ for the moment. Assume we are given a real quadratic field $K$ with ring of integers $\mathcal{O}_K$. $\mathbf{Question:}$ Is there a smooth projective curve $C$ of genus $g=2$ such that $End(Jac(C))$ is a non-maximal order in $K$, that is $End(Jac(C))=\mathcal{O}_{K,f}=\mathbb{Z}+f\mathcal{O}_K$ for some $f>1$, especially $\neq \mathcal{O}_K$? If yes can we find such a curve for every $K$ and every $f>1$? REPLY [3 votes]: Yes, for every quadratic ring of discriminant $f^2 D$ (where $D$ is the discriminant of $K$ in your notation), there's a Humbert surface's worth of such Jacobians. See David Gruenewald's thesis (available at http://echidna.maths.usyd.edu.au/~davidg/thesis.pdf) and the references there for calculations of Humbert surfaces. The double cover, which is the Hilbert modular surface, can also be computed in various ways. See https://arxiv.org/abs/1209.3527 for a method using elliptic K3 surfaces (for fundamental discriminants) and https://arxiv.org/abs/1412.2849 for non-fundamental discriminants if K is replaced by the disc 1 quadratic algebra $\mathbf{Q} \oplus \mathbf{Q}$.<|endoftext|> TITLE: Analogy between product of conjugacy classes and irreps: is there analog of Thompson conjecture ? QUESTION [14 upvotes]: The Thompson conjecture: in a finite simple non-abelian group, there exists a conjugacy class such that every element of the group can be expressed as a product of two elements from that conjugacy class. The conjecture is still open despite much progress (see below). There are plenty analogies between conjugacy classes and irreducible representations of finite simple groups (see below), so it is natural to ask what is known on the following questions: Question: is it true that in any finite simple group there exists an irreducible representation $\chi$ such that $\chi\otimes \chi$ contains all irreducible representations ? (It might be one should consider $ \chi \otimes \bar \chi$.) What about modular representations ? If is not true - how strongly is it violated ? Background 1: Azad, Fisman 1987 An analogy between products of two conjugacy classes and products of two irreducible characters in finite groups mentions the following nice results: A finite group B is isomorphic to the first Janko group J, if and only if C*C = G for EVERY nontrivial conjugacy class C of G. The analogous theorem in terms of characters is that a finite group G is isomorphic to J, if and only if $Irr(\chi^2) = Irr(G)$ for EVERY nontrivial irreducible character of G, where $Irr(\chi^2)$ is the set of all the irreducible constituents of $\chi^2$ [ACH]. If $C$ and $D$ are non-trivial conjugacy classes of a finite group G such that either $CD = mC + nD$ or $CD = mC^{-1} + nD$ where m,n are non-negative integers, then G is NOT a non-abelian simple group. And similar result on irreps. The paper contains references for further analogies between products of conjugacy classes and irreps. Background 2. There is lots of research devoted to Thompson conjecture. It is strictly stronger than Ore conjecture (see MO-Humphreys for history), which says that every element in finite simple groups is commutator. Now it is proved. Significant progress made by Gordeev, Ellers 1998 On the Conjectures of J. Thompson and O. Ore. Ore conjecture finally proved in [LOST]: M. W. Liebeck, E. A. O’Brien, A. Shalev, P. H. Tiep – The Ore conjecture, J. Eur. Math. Soc. 12 (2010), 939–1008. Bourbaki Seminar Mars 2013 by G. Malle gives overview of that works. See also MO, MO, products of conjugacy classes are related to quantum cohomologies of curves - see MO. REPLY [11 votes]: As Taylor's answer shows, the simple group PSU(3,3) is a counterexample. Interestingly, the conjecture fails rather dramatically for this group since the irreducible character of degree 6 is not a constituent of $\chi\overline\chi$ for ANY irreducible character $\chi$ of $G$. This is a Magma computer observation, and I have no idea if a similar phenomenon occurs more generally.<|endoftext|> TITLE: On independence and large cardinal strength of physical statements QUESTION [27 upvotes]: The present post is intended to tackle the possible interactions of two bizarre realms of extremely large and extremely small creatures, namely large cardinals and quantum physics. Maybe after all those paradoxes and uncertainty phenomena among weird tiny particles, which follow their own weird quantum logic, and after all those controversies surrounding the right interpretation of what is going on in the sub-atomic universe, the last straw that would break the camel's back could be the discovery of a series of statements in quantum theory which are independent or have large cardinal strength set theoretically. The fact that will send such physical statements beyond the realm in which the so-called usual mathematical tools can afford us a solution. Not to mention that inspired by Hilbert's sixth problem and Godel's incompleteness theorems, some prominent physicists already brought up discussions concerning the possibility of obtaining independence results or existence of undecidable facts/theories in physics. In this direction see Stephen Hawking's lecture, Godel and the end of universe. [The corresponding post on MSE might be of some interest as well]. Anyway the bad (good?) news is that the intersection of large cardinal theory and quantum physics is non-empty (if not potentially large). For example one may consider the following theorem of Farah and Magidor in the Independence of the existence of Pitowsky spin models which contains an assumption of consistency strength of measurable cardinals. [cf. R. Solovay, Real-valued measurable cardinals, Axiomatic Set Theory, 1971.] Theorem (Farah - Magidor): If the continuum is real-valued measurable then Pitowsky's kind spin function does not exist. The same holds in the model one gets from any universe of $ZFC$ by adding $(2^{\aleph_0})^+$-many random reals. See also some related philosophical discussions regarding this result in: Menachem Magidor, Some set theories are more equal. Jakob Kellner, Pitowsky's Kolmogorovian models and Super-Determinism. [Related: Super-determinism] One also might be interested in taking a look at the following papers which shed some light on the way forcing, Cohen reals, ultrafilters and various set theoretic concepts and tools play role in connection with some problems of quantum physics including hidden variables: Jerzy Krol, Model and Set-Theoretic Aspects of Exotic Smoothness Structures on $\mathbb{R}^4$. William Boos, Mathematical quantum theory I: Random ultrafilters as hidden variables. Robert Van Wesep, Hidden variables in quantum mechanics: Generic models, set-theoretic forcing, and the emergence of probability. Inspired by Farah-Magidor's theorem and the other mentioned papers, the following question arises: Question: What are some other examples of the statements in (quantum) physics which are mathematically independent or have some large cardinal strength (or at least make use of large cardinal assumptions in their formulation)? Please provide references if you are aware of any such result. REPLY [3 votes]: In my opinion one of the best examples of a physical statement depending on a mathematically independent statement is provided by Malament-Hogarth machines (You can find a bunch of other links just by doing a google scholar search). These are machines which it is argued (though others argue they are impossible) are allowed by general relativity and allow the operator to determine the answer to any $\Pi^0_1$ claim in finite time. As, by Godel's theorem, given any computably axiomitizeable theory $T$ the $\Pi^0_1$ statement $Con(T)$ is independent from $T$. So whatever theory one chooses to work in this lets you produce a physical statement (the machine with such and such construction ... will give answer blah) that depends on the truth of an independent question. Note that this doesn't specifically involve large cardinal assumptions themselves but it does give you a physical system whose outcome depends on the consistency of large cardinal statements (e.g. $Con(ZFC+exists measurable)$) However, one should not assume that how these physical systems turn out tells us what is mathematically true. We develop physical theories by accepting those hypothesises that seem to be good descriptions of reality and it is usually convenient to write those theories in terms of the most natural mathematical structures like the natural numbers or the reals. However, it is equally true that the theory which says experiments turn out the way General Relativity formulated in this non-standard model of the reals/integers predicts (almost certainly it can be formalized in the two-sorted first-order theory of second order arithmetic but if not use a non-standard model of ZFC instead) is as compatible with all our evidence as the theory which applies General Relativity formulated in the standard model. Ohh to put it more simply if your Malament-Hogarth machine tells you that a particular computation converges in finite time you can't really be sure that the computation really converges in finite time or if the temporal structure of the universe is non-standard and the computation only converges at some non-standard time. Thought for a perspective which argues that we could use the results of these MH machines to determine mathematical truth see Sharon Berry's Malament–Hogarth Machines and Tait's Axiomatic Conception of Mathematics with preprint here. (Conflict of interest warning: I'm married to Sharon though I disagree with her conclusions).<|endoftext|> TITLE: Minimum covers of complete graphs by $4$-cycles QUESTION [7 upvotes]: I am interested in coverings of the (edge set of the) complete graph $K_n$ by cycles of length $4$. It is clear that such coverings exist for each $n \ge 4$. I need to find the minimum number of $4$-cycles necessary to cover $K_n$. For example, $K_5$ can be covered by following $4$-cycles $(1, 2, 3, 5), (2, 5, 4, 3), (2, 4, 1, 3)$. I am sure this problem has been studied but unfortunately can't find any results. Can you share some results? Since $K_n$ has $\binom{n}{2}$ edges, the minimum number of $4$-cycles is obviously at least $\binom{n}{2} / 4$. REPLY [6 votes]: If $n$ is odd, the answer is $\lceil \binom{n}{2}/4 \rceil$. If $n$ is even, the answer is $\lceil \binom{n}{2}/4+n/8 \rceil$. This follows from two special cases of a more general conjecture by Alspach. For our purposes, we use a theorem of Heinrich, Horák, and Rosa which says that if $n \geq 7$ is odd and $a,b,c$ are such that $3a+4b+6c=\binom{n}{2}$, then $E(K_n)$ can be partitioned into $a$ $3$-cycles, $b$ $4$-cycles, and $c$ $6$-cycles. Huang and Fu proved the same result with $(3,4,6)$ replaced by $(4,5)$. Thus, if $n \geq 7$ is odd, it is always possible to decompose $E(K_n)$ into $4$-cycles and possibly one extra cycle that is a $3$-cycle, a $5$-cycle or a $6$-cycle. The edge set of the extra cycle can obviously be covered with two $4$-cycles of $K_n$, so we are done. If $n$ is even, then each vertex has odd degree. Let $v$ be an arbitrary vertex. Since every $4$-cycle uses $0$ or $2$ edges incident to $v$, there will be at least one edge incident to $v$ that is covered twice. Thus, in total there will be at least $n/2$ edges that are covered twice. Thus, every covering of $E(K_n)$ by $4$-cycles has size at least $\binom{n}{2}/4+n/8$. We prove that this bound can actually be achieved. Namely, for $n$ even, Heinrich, Horák, and Rosa's result holds except with $K_n$ replaced by $K_n$ minus a perfect matching $M$, and $\binom{n}{2}$ replaced with $\frac{n(n-2)}{2}$. For $n$ even, $\frac{n(n-2)}{2}$ is divisible by $4$. It follows that the edges of $K_n-M$ can be decomposed into $4$-cycles. By then covering pairs of edges of $M$ with $4$-cycles we get a covering of size $\lceil \binom{n}{2}/4+n/8 \rceil$.<|endoftext|> TITLE: Paul Cohen on genesis of method of forcing and mathematical similarities QUESTION [9 upvotes]: We have on record Paul Cohen's comments on being inspired by issues of formalizing algorithms in number theory (this needs to be verified as per comment) as well as related remarks on computability. Beyond such "inspiration", are there mathematical similarities between developments in those fields and Cohen's work on forcing? A related question was posed at MSE with limited success. REPLY [3 votes]: Regarding your question about the connection between forcing and computability theory, the notion of a Cohen generic oracle is used in computational complexity; see for example An oracle builder's toolkit by Fenner et al. For a different kind of connection, see Chapter 13 ("Forcing and Category") of S. Barry Cooper's book Computability Theory (unfortunately I don't think that the complete text is available online). As for the connection to arithmetic, see Forcing over set theory versus forcing over arithmetic<|endoftext|> TITLE: Are all splittings of the normal bundle to a submanifold locally isomorphic? QUESTION [5 upvotes]: Let $S$ be a submanifold of a real smooth manifold $M$. By a splitting of the normal bundle of $S$ I mean a sub-bundle $V$ of $TM|_S$ such that $V\oplus TS = TM|_S$. Question: Given such a splitting, can I always find local coordinates $x_1,\dots,x_s,y_1,\ldots, y_r$ on $M$ around each point $p\in S$, such that the submanifold $S$ is given by $y_1=0,\ldots, y_r=0$ and $V$ is given by the kernels of $dx_1,\ldots,dx_s$? REPLY [8 votes]: Yes, one can always do this. Start with a $p$-centered local coordinate system $(x^\sigma,y^\rho)$ on an open $p$-neighborhood $U\subset M$ such that $S\cap U$ is given by $y^\rho=0$ $(1\le\rho\le r)$. Then there will exist functions $F^\sigma_\rho(x)$ such that $V$ along $S\cap U$ is given by the equations $$ \mathrm{d}x^\sigma + F^\sigma_\rho(x)\,\mathrm{d}y^\rho = 0 $$ (summation convention assumed). This is the same as $$ \mathrm{d}(x^\sigma+F^\sigma_\rho(x)y^\rho) - y^\rho\,\mathrm{d}(F^\sigma_\rho(x)) = 0. $$ Since $y^\rho=0$ along $S$, this says that $V$ along $S$ is defined by $$ \mathrm{d}(x^\sigma+F^\sigma_\rho(x)y^\rho) = 0, $$ so take $\bar x^\sigma = x^\sigma+F^\sigma_\rho(x)y^\rho$, and let $(\bar x^\sigma,y^\rho)$ be your new coordinate system. (One can easily check that these are local coordinates in an open neighborhood of $S\cap U\subset U$.) Added Remark: Just a comment on your terminology: Literally, what you are defining is not a splitting of the normal bundle of $S$, but rather a choice of a normal bundle along $S$, and you are asking whether there are any local invariants that distinguish such choices. As the above shows, the answer is 'no'. In fact, there is even a global version: For two normal bundles $V_1$ and $V_2$ along a closed submanifold $S$, there is a diffeomorphism of $M$ that fixes $S$ and carries $V_1$ to $V_2$. Note also that, in the holomorphic category (in contrast to the smooth case), normal bundles need not be unique in this sense, and may not even exist.<|endoftext|> TITLE: Is Zariski closure of finitely generated matrix semigroup computable? QUESTION [10 upvotes]: In general, can the Zariski closure of the semigroup of matrices $\langle M_1, \ldots, M_k \rangle$ be algorithmically computed (at least in theory)? For this purpose I'm happy to assume the matrices have rational entries, or are otherwise "nice". If the answer is "no", are there known classes of matrices for which these Zariski closures are effectively computable? (I know that the answer is "yes" for unitary matrices over algebraic numbers.) REPLY [4 votes]: This problem has been solved (the answer is 'yes') in the paper "Polynomial Invariants for Affine Programs", by Ehud Hrushovski, Joël Ouaknine, Amaury Pouly, and James Worrell, published in the Proceedings of the 33rd Annual ACM/IEEE Symposium on Logic in Computer Science (LICS) 2018. An author copy is available at https://people.mpi-sws.org/~joel/publications/zariski18abs.html . The main result is stated as Theorem 16 (and Corollary 17 for the real Zariski closure) in Section 6 (it applies in fact more generally to constructible -- not merely finite -- sets of generators).<|endoftext|> TITLE: Inverse of special upper triangular matrix QUESTION [9 upvotes]: Consider the following $n \times n$ upper triangular matrix with a particularly nice structure: \begin{equation}\mathbf{P} = \begin{pmatrix} 1 & \beta & \alpha+\beta & \dots & (n-3)\alpha + \beta & (n-2)\alpha + \beta\\ 0 & 1 & \beta & \dots & (n-4)\alpha + \beta & (n-3)\alpha + \beta\\ 0 & 0 & 1 & \dots & (n-5)\alpha + \beta & (n-4)\alpha + \beta\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & \beta & \alpha+\beta\\ 0 & 0 & 0 & \dots & 1 & \beta\\ 0 & 0 & 0 & \dots & 0 & 1 \end{pmatrix} \end{equation} i.e. \begin{equation} p_{i,j}=\begin{cases} 0, &i>j,\\ 1, &i=j,\\ (j-i-1)\alpha+\beta, &ij,\\ 1, &i=j,\\ \frac{\lambda^2(1+\lambda)^{j-i-1}-\mu^2(1+\mu)^{j-i-1}}{\lambda-\mu},&i TITLE: "Gray code" of all permutations QUESTION [8 upvotes]: Informally asking, can we step through all permutations of the set $\{1,\ldots,n\}$ by just using transpositions? More formally: For any $n\in\mathbb{N}$ let $[n] = \{1,\ldots,n\}$ and let $S_n$ be the set of all bijections (permutations) $\pi:[n]\to [n]$. For any set $X$ let $[X]^2 = \big\{\{x,y\}: x\neq y\in X\big\}$. We let $\pi,\psi\in S_n$ be connected by an edge if "they are one transposition away from each other", or more formally, set $$E_n = \big\{\{\pi,\psi\}\in [S_n]^2:\exists a TITLE: A question about the abelianization of the Johnson kernel QUESTION [5 upvotes]: Let $K_g \le \mbox{Mod}_g$ denote the Johnson kernel subgroup of the mapping class group of a closed surface of genus $g$. Dimca-Hain-Papadima find an explicit presentation for $H_1(K_g; \mathbb{C})$ as a $\mbox{Mod}_g/K_g$-module. According to Theorem B of their paper, $H_1(K_g;\mathbb{C})$ contains a trivial summand. Passing to $H^1$, this implies that there is some homomorphism $f: K_g \to \mathbb{C}$ that is invariant under the conjugation action of $\mbox{Mod}_g$ on $K_g$. Does anyone know an explicit description of this homomorphism? It is not the Casson invariant (which Morita shows determines a (non-canonical) homomorphism $\lambda: K_g \to \mathbb{Z}$). In a remark in his first paper on the Casson invariant, Morita writes down a formula describing how $\lambda$ is affected by conjugation, and this action is nontrivial (this is ultimately a consequence of the fact that the definition of $\lambda$ requires a non-canonical choice of Heegaard splitting on $S^3$). REPLY [2 votes]: I have found an explicit construction in terms of the (higher) Johnson homomorphism. Briefly put, the third Johnson homomorphism is an $\mbox{Sp}_{2g}(\mathbb{Z})$-equivariant map $$ \tau_3: K_g \to M $$ for some $\mbox{Sp}_{2g}(\mathbb{Z})$-module $M$ which can be shown to be isomorphic to $\mbox{Sym}^2(V(2))$ (here we adopt the standard notation for representations of $\mbox{Sp}$). There is a contraction mapping $\mbox{Sym}^2(V(2)) \to \mathbb{C}$, furnishing the required invariant homomorphism. In case anyone is interested, I will sketch the construction. In the setting where the surface has a boundary component, there's more known about the image of the third Johnson homomorphism. Indeed, Morita shows that $$ \tau_3:H_1(K_{g,1}; \mathbb{Q}) \to \mbox{Sym}^2(\wedge^2H) $$ is a surjection. Here $H:= H_1(\Sigma_{g,1};\mathbb{Q})$. We'll work for now over $\mathbb{Q}$. In light of the decomposition $\wedge^2 H = V(2) \oplus V(0)$ into irreps of $\mbox{Sp}_{2g}(\mathbb{Z})$, the target $\mbox{Sym}^2(\wedge^2 H)$ decomposes as $$ \mbox{Sym}^2(\wedge^2 H) = \mbox{Sym}^2(V(2)) \oplus V(2) \oplus V(0). $$ Explicitly, $V(0)$ is spanned by $\omega^2$, where $\omega \in \wedge^2 H$ is the symplectic form. There is a short exact sequence $$ 1 \to N \to K_{g,1} \to K_g \to 1, $$ and hence the Johnson homomorphism on $K_g$ is valued in $\mbox{Sym}^2(\wedge^2 H) / (\tau_3(N))$. The goal now is to identify the image of $N$. This will be done in two steps, reflecting the fact that there is a short exact sequence $$ 1 \to \mathbb{Z} \to N \to \pi_1(\Sigma_g)^{[2]} \to 1, $$ where the notation $G^{[k]}$ indicates the $k^{th}$ term in the lower central series of a group $G$, taking $G^{[1]} = G$. The kernel $\mathbb{Z}$ is generated by the twist about the boundary component $T_\Delta$, and $\pi_1(\Sigma_g)^{[2]}$ is viewed as a subgroup of $K_{g,*}$ via the Birman exact sequence (here $K_{g,*}$ is the Johnson kernel for a surface with one puncture). In the paper mentioned above, Morita shows that $$ \tau_3(T_\Delta) = \omega^2. $$ This implies that there is a well-defined homomorphism $$ \tau_3: H_1(K_{g,*}; \mathbb{Q}) \to \mbox{Sym}^2(V(2)) \oplus V(2). $$ The next step is to identify the image of $\pi_1(\Sigma_g)^{[2]}$ under $\tau_3$. In fact, the image is $V(2)$. This is because the restriction of $\tau_3$ to $\pi_1(\Sigma_g)^{[2]}$ factors through $\pi_1(\Sigma_g)^{[2]} / \pi_1(\Sigma_g)^{[3]}$, and $$ \pi_1(\Sigma_g)^{[2]} / \pi_1(\Sigma_g)^{[3]} \otimes \mathbb{Q} \cong V(2). $$ In summary, we have identified the (rational) image of the third Johnson homomorphism on the closed Johnson kernel: $$ \tau_3: K_g \to \mbox{Sym}^2(V(2)). $$ Now we simply observe that there is the contraction $C: \mbox{Sym}^2(V(2)) \to \mathbb{Q}$. Thus $C \circ \tau_3$ is the required invariant homomorphism. Morita, Shigeyuki, Casson’s invariant for homology 3-spheres and characteristic classes of surface bundles. I, Topology 28, No.3, 305-323 (1989). ZBL0684.57008.<|endoftext|> TITLE: Is every positive integer the permanent of some 0-1 matrix? QUESTION [24 upvotes]: In the course of discussing another MO question we realized that we did not know the answer to a more basic question, namely: Is it true that for every positive integer $k$ there exists a balanced bipartite graph with exactly $k$ perfect matchings? Equivalently, as stated in the title, is every positive integer the permanent of some 0-1 matrix? The answer is surely yes, but it is not clear to me how to prove it. Entry A089479 of the OEIS reports the number $T(n,k)$ of times the permanent of a real $n\times n$ zero-one matrix takes the value $k$ but does not address the question of whether, for every $k$, there exists $n$ such that $T(n,k)\ne 0$. Assuming the answer is yes, the followup question is, what else can we say about the values of $n$ for which $T(n,k)\ne 0$ (e.g., upper and lower bounds)? REPLY [2 votes]: If an order n (0,1) matrix has r rows with all ones, its permanent is a multiple of r! by an easy argument. It follows that the largest odd number which is a permanent of an order n matrix is q = der(n) + der(n-1), which is a sum of numbers of derangements, and is a little larger than $n!/e$. I suspect that the smallest positive number not expressible as a permanent of this size matrix is an odd integer which is not much smaller, perhaps about half, of q. If so, then we can shave a little bit off of n=log(k) in the construction in Timothy Chow's post. For some motivation for this problem, look at Will Orrick's talk mentioned (for determinants) at https://mathoverflow.net/a/271273 . Gerhard "Also A Potential Polymath Project" Paseman, 2017.08.29.<|endoftext|> TITLE: A game with boldface strength QUESTION [5 upvotes]: This is a problem which has been bothering me for a while now; it doesn't seem inherently too hard, but I haven't been able to make any real headway, so I'm putting it out in the open since at this point I just want to know the answer. I don't think it has any deep value, but it's a natural question (at least to me) which I haven't seen addressed in the literature. In brief: is there a single game $G$, defined by a formula in the language of second-order arithmetic, such that the statement "$G$ is determined" isn't true in any Turing ideal with a maximal element (viewed as an $\omega$-model of RCA$_0$)? Say in this case that $G$ has boldface strength (as opposed to the merely lightface strength that we're used to individual instances of principles having, in reverse mathematics). (Note,of course,that because a model $M$ thinks $G$ is determined doesn't mean $M$ contains an actual winning strategy, or even that $M$ is correct about who wins.) Let me say a bit about my thoughts so far: Why the answer might be yes: the statement "$G$ is determined" is $\Sigma^1_2$ on top of $G$; so it easily has enough quantifier complexity to have boldface strength, even if $G$ is low in the projective hierarchy. Now of course $G$ can't be too low in the projective hierarchy - after all, we need the Turing ideal generated by a winning strategy for $G$ to not satisfy "$G$ is determined," so membership in $G$ has to be sufficiently non-absolute. But once $G$ gets complicated enough, all bets seem off. That is: if $G$ is sufficiently complicated, then merely having a winning strategy doesn't matter, the model needs enough structure to be able to verify that the strategy always wins, and there's no obvious upper bound to how complicated this extra structure could be. An obstacle: A natural guess for a game with boldface strength is the following: Player $1$ plays a second-order sentence $\varphi$ of some bounded complexity (say, $\Pi^1_7$), and player $2$ plays "true" or "false" then player $2$ wins iff $\varphi$'s truth value is player $2$'s play. (The bound on the complexity of $\varphi$ is needed for this game to be definable.) However, this doesn't work, since by Martin's cone theorem the models corresponding to all sufficiently large degrees will yield the same theory, so once our topped models get big enough they'll see how to win this game. A response: More generally, the obstacle above suggests that the game needs to involve playing a specific element of the top degree of the model (if there is one). This suggests a game like the following: player $2$ builds a real of maximal Turing degree, and also responds to player $1$'s queries about that real. To give player $2$ a chance, we should demand that player $1$ only gets to ask finitely many questions. Meanwhile, we have a real synergy between this game and the existence of a maximally complicated real: on the one hand, in a model with a maximally complicated real player $2$'s strategy must build something as complicated as itself, while in a model with no maximally complicated real player $1$ wins vacuously (e.g. the classical analysis has no relevance to the case we're interested in). In particular, a game of the following form seems promising: Player $2$ builds a real $r$, while player $1$ plays finitely many second-order sentences of some fixed complexity (say, $\Pi^1_7$), and player $2$ responds to player $1$'s queries with "True" or "False" as above; player $2$ then wins iff $r$ has maximal Turing degree and they never guessed incorrectly about the theory of $(\omega\cup\mathcal{P}(\omega); +,\times, r)$. (Note that all of this is defined internally.) Another obstacle: This suggests some recursion theorem trickery, where player $1$ manages to ask player $2$ about their own strategy in such a way that player $2$ is trapped. In particular, the real that player $2$ builds is guaranteed to compute all of their strategy, if player $2$ is playing according to a winning strategy. However, I haven't managed to get this line of attack to pan out, the main problem being that player $2$'s real (and hence the way player $2$'s real computes player $2$'s strategy) depends on player $1$'s queries. So I'm asking now: Is there a single game with boldface strength? REPLY [6 votes]: The proof of the Kechris-Solovay theorem produces a real $x_0$ such that whenever $x_0 \leq_T x$, every game lightface definable in $M_x = (\omega, \{y : y\leq_T x\},\in)$ is determined in $M_x$. This shows there are many principal Turing ideals in which lightface PD holds. We basically copy the argument given in Woodin-Koellner Theorem 6.6. If there does not exist such an $x_0$, by $\Delta^1_1$ Turing Determinacy, there must be a real $x_0$ such that whenever $x_0\leq_T x$, there is some game lightface definable in $M_x$ that is not determined in $M_x$. For any $x$, let $\varphi_x$ be the least formula defining in $M_x$ a nondetermined set of reals, if there is one, and let $\varphi_x$ be undefined otherwise. If $\varphi_x$ is defined, let $A^x$ be the set of reals defined in $M_x$ by $\varphi_x$. Consider the game $G$ (defined in $V$) where I plays $a,b$ and II plays $c,d$ (alternating in the usual way), and I wins if $\varphi_{\langle a,b,c,d\rangle}$ is defined and $a*d\in A^{\langle a,b,c,d\rangle}$. By $\Delta^1_1$-Determinacy, one player has a winning strategy. We aim to get a contradiction, no matter who wins. Suppose I wins by the strategy $\sigma_0$. Take $x\geq_T\sigma_0,x_0$. We claim there is a strategy $\sigma \leq_T x$ such that for any $d\leq_T x$, $\sigma * d\in A^x$, which contradicts that $A^x$ is not determined in $M_x$. Against $d\in \omega^\omega$, $\sigma$ will respond (well, $\sigma$ goes first but you know what I mean) with the real $a$ such that $\sigma_0$ plays $a,b$ against $x,d$. Then $\sigma$ is recursive in $x$. We must show $\sigma*d\in A^x$ when $d\leq_T x$. But note that with $a,b$ defined as above from $d$, $x\equiv_T \langle a,b,x,d\rangle$, so the fact that $\sigma_0$ is winning for I in $G$ implies $a*d\in A^{\langle a,b,x,d\rangle} = A^x$, and hence $\sigma*d\in A^x$. The contradiction in the case that II wins $G$ by $\tau_0$ is reached by a symmetric argument, with the minor modification that we use the fact that $x\geq_T x_0$ to show that II wins the relevant plays by $\tau_0$ because $a*d\notin A^{\langle a,x,c,d\rangle}$, and not because $\varphi_{\langle a,x,c,d\rangle}$ was undefined.<|endoftext|> TITLE: Why aren't functions used predominantly as a model for mathematics instead of set theory etc.? QUESTION [26 upvotes]: If definitions themselves are informally just maps from words to collections of other words. Then in order for one to define anything, they must inherently already have a notion of a function. I mean of course one could ask what the exact "things" our functions are mapping to and from, and then revert back to set theory or something else to represent those "things" but why not just define everything to be a function so we are mapping functions to functions? (perhaps with some axiom guaranteeing the existence of at least one function which we can use to recursively build up a bunch of other functions allowing us to express most of contemporary mathematics in the same way I've seen this done iteratively with the emptyset in ZFC) I mean how can one read any language, much less a mathematics textbook unless they are capable of mapping words to other words i.e. using functions (albeit even if one doesn't mentally recognize it, their brain is still implementing something akin to a function). Even in formulations of ZFC it seems functions are implicitly being used, for example the Axiom of intersection says given any two sets $A$ and $B$ there exists a set containing all the elements they have in common that one denotes by $A\cap B$. However we essentially have just defined a function $\cap$ of two variables, taking any two sets to another set - their intersection. Similarly in propositional logic when one defines the logical conjunction $\land$ we are again defining a two variable function from truth values to other truth values. Pre-facing a statement about mathematical objects that are being mapped to something with the words "we represent this by" or "we denote it by" merely conceals the fact one has just defined a function by hiding it as a statement in the English language. So re-iterating I don't see how anyone can learn anything without having some mental grasp of what a map/function/arrow etc. is. For example a dictionary can be crudely expressed as a functional relation $f$ where we might have: $$f(\texttt{apple})≡\texttt{a round fruit that grows on trees}$$ So how is it one can accept any mathematical axioms without first taking functions as a primitive, when reading and interpreting words/symbols in of itself requires their usage. If to formally define a function you must use functions then isn't that circular reasoning? How can a person understand a simple map or visual diagram when interpreting objects, color-intensity etc. requires mentally creating a bijective map between the object being labeled and what it represents. I mean the use of any variables for that matter requires creating what amounts to a bijection, at least mentally between the letters/symbols on paper and the objects/ideas they represent. If understanding everything from written language to cave drawings requires some mental notion of a functional relation, then shouldn't it be used as a starting point? Also from an aesthetic point of view, it seems a lot simpler to just accept some mathematical variant of a map/arrow/morphism/function etc. then to define a large number of other objects or "syntactic abbreviations" (I'm not familiar with the proper term but this is what Peter Heinig calls it) which appear to me as objects that hold almost all the same characteristics of functions. Lastly if one gets slightly looser with this idea, couldn't you argue understanding any cause and effect relationship essentially requires some mental abstraction that when modeled symbolically is functional? This could be argued as much simpler as then even non-humans would be capable of grasping similar notions, e.g. gorillas capable of rudimentary sign language can understand that by "inputting" a configuration of their hands they can "output" a banana from their owner. In any case though before I get off topic and venture into philosophy, I want to add I'm not too well versed in mathematical logic, so I'm guessing this exact thing has been covered elsewhere. In which case I would appreciate any references. REPLY [4 votes]: Actually, Jan Kuper has taken your notion seriously. In the abstract to his his paper, "An Axiomatic Theory for Partial Functions" (Information and Computation 107, 104-150 (1993)) he writes: We describe an axiomatic theory for the concept of one-place, partial function, where function is taken in its extensional sense. The theory is rather general; i.e. concepts such as natural number and set are definable, and topics such as nonstrictness and self application can be handled. It contains a model of the (extensional) lambda calculus, and commonly applied mechanisms (such as currying and inductive definability) are possible. Furthermore, the theory is equi-consistent with and equally as powerful as $ZF$ Set Theory. The theory (called Axiomatic Function Theory, $AFT$) is described in the language of classical first order predicate logic with equality and one nonlogical predicat symbol for function application. By means of some notational conventions, we describe a method within this logic to handle undefinedness in a natural way. Note also what Kuper says as regards the concept of function, found in the third paragraph of the Introduction, pg. 104: The concept of function seems to be equally as fundamental as the concept of set; e.g.. notions such as natural number, set, and ordered $n$-tuple can be easily defined, using only the concepts of partial function and function application. Furthermore, a naive way of defining functions and operating on them leads to the Russell Paradox. To avoid the Russell Paradox, we develop a Zermelo-Fraenkel-like system of axioms, which gives us an intuitively very reasonable axiomatization of the concept of partial function. We want the theory to be rather general, so we allow self application and related topics (though not for all functions), we indicate a natural way to handle non-strict functions, and we add a functional form of the Axiom of Choice [see Axiom 8, pg. 139--my comment]. (This paper is a free download from Science Direct. Just google "An Axiomatic Theory for Partial Functions", and click on the appropriate website.) I hope this helps.<|endoftext|> TITLE: Who conjectured the Cartan determinant conjecture QUESTION [15 upvotes]: The Cartan determinant conjecture states that every finite dimensional algebra of finite global dimension has the property that the determinant of its Cartan matrix is equal to one. Who stated this conjecture first and how old is it? The Cartan matrix of a finite dimensional algebra is defined as the matrix having entries $c_{i,j}:=$dimension of $(Hom_{A}(P_i,P_j))$ over the divison ring $D_i$, where $D_i:=End_A(S_i)$. Here $P_i$ are the indecomposable projective modules and $S_i$ the simple modules which are the top of $P_i$. REPLY [12 votes]: This seems to have first been explicitly stated as an open problem by Dan Zacharia in 1983, in On the Cartan matrix of an Artin algebra of global dimension two<|endoftext|> TITLE: Multidimensional gluing theorem for Riemannian manifolds QUESTION [7 upvotes]: I would like to understand whether the following multidimensional (partial) generalization of the A.D. Alexandrov gluing theorem is true and, if yes, whether there is a reference. (The original Alexandrov's gluing theorem was proven in dimension 2 only but under much weaker assumptions on the regularity of the metrics. It is actually much deeper than the situation I consider right now.) Let $M_1, M_2$ be two smooth Riemannian manifolds with boundary with sectional curvatures at least $\kappa$. Let $f\colon \partial M_1\tilde\to \partial M_2$ be a Riemannian isometry between their boundaries. Define a new smooth manifold without boundary $M=M_1\cup_f M_2$ by identifying the boundaries of $M_1$ and $M_2$ via $f$. One defines on $M$ an intrinsic metric: for two points $x,y\in M$ the distance between them is equal to the infimum of the lengths of all paths connecting them when the length of the parts of the path contained in $M_1$ (resp. $M_2$) is computed with respect to the Riemannian metric of $M_1$ (resp. $M_2$). Remark. It is not hard to see that $M$ has a structure of $C^\infty$-smooth manifold, and the above metric is induced by a $C^0$-regular Riemannian metric. Question. Whether the following is known to be true: $M$ has a bounded below curvature in the sense of Alexandrov if and only if the sum of the second fundamental forms of the boundaries of $M_1$ and $M_2$ (after the identification via $f$) is non-negative. In that case the curvature of $M$ is at least the above $\kappa$. A reference would be helpful. REPLY [4 votes]: Just learned that the answer is positive, at least its main part saying that if the sum of second fundamental forms is non-negative then the curvature of $M$ is at least $\kappa$. The answer is published here: N. N. Kosovski˘ı, “Gluing of Riemannian manifolds of curvature ≥ κ”, Algebra i Analiz 14:3 (2002), 140–157.<|endoftext|> TITLE: A search for a sequence of $6$-manifolds QUESTION [18 upvotes]: How to construct closed, orientable, smooth, simply-connected $6$-manifolds such that $H^{*}(M,\mathbb{Z}) \cong \mathbb{Z}[a]/(a^{4})$ (Where $a$ is a generator of degree 2) satisfying $p_{1}(M) = n a^{2}$? ($p_{1}(M) \in H^{4}(M,\mathbb{Z})$ denotes the first Pontryagin class). By Wall's classification of $6$-manifolds for each $n \in \mathbb{Z}$ exists a unique such manifold up to diffeomorphism. The case $n = 4$ is furnished by $\mathbb{C}\mathbb{P}^{3}$, are there at other familiar examples - perhaps the $n=0$ case? How about constructing these manifolds with surgery techniques? REPLY [18 votes]: I looked at Wall's paper Classification problems in differential topology. V On certain 6-manifolds. In theorem 3 of that paper Wall describes some invariants of 6-mainfolds, and the relation between them. These invariants, in the case you are concerned with, are given by: The abelian group $H = \mathbb{Z} = H^2(M;\mathbb{Z})$, and the zero group $G = 0 = H^3(M; \mathbb{Z})$. The symmetric trilinear form $\mu: \mathbb{Z}^3 \to \mathbb{Z}$ given by $\mu(x,y,z) = \langle xyz, [M]\rangle$, i.e. take cup product and pair with the fundamental class. For the cohomology ring you specify this is just the 3-fold multiplication map "xyz". $p_1: H = \mathbb{Z} \to \mathbb{Z}$, which is your integer $n$, corresponding to the 1st Pontryagin class. An element $w_2 \in H / 2H \cong \mathbb{Z}/2\mathbb{Z}$ (2nd Steifel-Whitney class), which has an integral lift $W_2 \in \mathbb{Z}$. These invariants necessarily satisfy three relations for arbitrary $x, y \in H$: $$ \mu(x,y,x+y + W_2) \equiv 0 \; \; \; \textrm{mod 2} \\ p_1(x) \equiv \mu(x,W_2, W_2) \; \; \; \textrm{mod 4} \\ p_1(x) \equiv \mu(x,x,x) \; \; \; \textrm{mod 3}$$ The first equation, with the given $\mu$, implies that $W_2$ is even. So $w_2=0$ and the manifold is spin (rather "spin-able"), which is what Wall calls "condition (H)". The other equations then impose further conditions on the allowed values of $p_1 \leftrightarrow n$. In particular we see that $n$ has to be a multiple of 4 and is congruent to 1 mod 3. All of this comes from some observations about necessarly relations between these invariants. We haven't touched at all on the realization problem, i.e. which of the possible $n$ actually come from 6-manifolds? Wall helps us there too. He gives a more careful analysis in the case of manifolds satisfying "condition (H)". In that case (his thm 5) diffeomorphism classes of manifolds correspond bijectively to systems of invariants $(H,G, p_1, \mu)$ as before subject to two relations $$ \mu(x,x,y) \equiv \mu(x,y,y) \; \; \; \textrm{mod 2} \\ p_1(x) = 4 \mu(x,x,x) \; \; \; \textrm{mod 24}$$ Note that the final condition is stronger than the previous conditions. Some values of $n$ do not occur. In the case at hand where $H = \mathbb{Z}$, $G=0$, and $\mu$ is the standard "3-fold product" trilinear map, the first equation is satisfied, and the second condition tells us that $$ n = 4 + 24k $$ for some integer $k$. Finally, Wall also tells us how to construct this manifold. It can be realized as surgery on a framed knot $S^3 \times D^3 \hookrightarrow S^6$. Specifically it is surgery on the framed knot with invariants $\varphi = -k$ and $\beta' = 1 + 6k$. These knot invariants are discussed at length in sections 4 and 5 of Wall's paper, which has other references.<|endoftext|> TITLE: centre and automorphism groups of finite group schemes QUESTION [5 upvotes]: Let $G$ be a group scheme over a scheme $X$ with centre $Z(G)$, automorphism group $\mathrm{Aut}(G)$ and outer automorphism group $\mathrm{Out}(G)$ (viewed as group schemes on $X$). If $G$ is finite flat over $X$, then are $Z(G), \mathrm{Aut}(G)$ and $\mathrm{Out}(G)$ also finite flat over $X$? If $G$ is finite étale over $X$, then are $Z(G), \mathrm{Aut}(G)$ and $\mathrm{Out}(G)$ also finite étale over $X$? REPLY [2 votes]: The answer to 2. is yes. A sketch of a proof is as follows: $G$, being finite étale, is étale locally on $X$ isomorphic to a constant finite group scheme. Therefore, by a standard descent argument, it suffices to prove the result for constant finite group schemes. However here $Z(G), \mathrm{Aut}(G)$ and $\mathrm{Out}(G)$ are all clearly finite étale, as required.<|endoftext|> TITLE: Stable homotopy theory and physics QUESTION [31 upvotes]: At various points in my life, I have held the following beliefs: 1) Stable homotopy theory is "easy" rationally, and "interesting" integrally. 2) The spectrum of topological modular forms (TMF) is an object that stable homotopy theorists are trying hard to understand integrally. 3) TMF has many connections to physics. 4) The mathematics relevant to physics is a "rational" story, and does not care much about integral or torsion aspects. Taken together, this set of beliefs is evidently inconsistent. But I do not possess the knowledge, especially in physics, to know which one is incorrect (I would suspect the last one). I would be grateful if someone can clarify the situation. Thank you. EDIT (8/31/17): I am grateful to the comments and answer. It seems that the problem indeed lies with (4). But I would love an example explaining a connection between physics and the integral aspects of the study of TMF. REPLY [28 votes]: There's an interesting application of stable homotopy theory to condensed-matter physics, and it makes heavy use of integral and torsion information, contradicting your 4th assumption. Within the general program of understanding topological phases of matter, condensed-matter theorists are interested in symmetry-protected topological phases (SPT phases). Approximately speaking, these are systems which have interesting topological behavior in the presence of a symmetry, but become trivial when that symmetry is broken. Their classification has gradually gotten more homotopical: Kitaev uses real and complex $K$-theory to classify topological insulators and superconductors. Freed-Moore later generalized this to twisted equivariant $K$-theory. Kapustin and Kapustin-Thorngren-Turzillo-Wang use $\mathit{MSpin}\wedge\mathit{BG}$ to classify fermionic SPT phases with symmetry group $G$. Freed and Freed-Hopkins classify SPTs as homotopy classes of maps $$[\mathit{MTH}, \Sigma^{n+1}I_{\mathbb Z}],$$ where $\mathit{MTH}$ is the Madsen-Tillmann spectrum (a kind of Thom spectrum) for the symmetry type $H$ and $I_{\mathbb Z}$ is the Anderson dual of the sphere. The derivation uses some equivariant stable homotopy theory, and the calculations use the Adams spectral sequence. In all of these examples, the torsion information is essential: a $\mathbb Z/2$ classification means there's a phase which is nontrivial, but such that two copies of it stacked together can be continuously deformed to a trivial phase. Examples of such phases have come up in condensed-matter theory and are expected to display this behavior in experiments. There are other places integral information is, well, integral: for example, if an electron moves in a loop $\ell$ around a magnetic monopole, the value of the action depends on the winding number of $\ell$, producing discrete invariants.<|endoftext|> TITLE: Basic Algebraic Applications of Stationary Sets? QUESTION [9 upvotes]: Background: I've been working my way through Thomas Jech's "Set Theory" because I'm working on some problems that have the potential to be logically independent of the usual axioms, or at least involve some hard set-theory about infinite sets, and I want to (eventually) better understand some of the set-theoretical techniques and methods that can be used to show such independence. The first seven chapters of Jech's book have clear applications in my field of expertise, ring theory. I was already quite familiar with much of the material because of this fact. Chapter 1: The basic axioms of set theory are used all the time in the language of ring theory, allowing for the formation of simple objects like unions, products, sequences, etc. Chapter 2: The ordinal numbers have applications in infinitary constructions, and anywhere induction can be pushed further. For instance, one can define the "higher Wedderburn radicals" as a transfinite sequence of ideals which eventually stabilize at the prime radical. Chapter 3: Everyone is familiar with how cardinal numbers give a very easy way to determine whether two objects are the same size, which helps prevent the existence of isomorphisms. Thus, this is a sort of first check one does to prove non-isomorphisms. There are other uses. Chapter 4: Real numbers are a ring, with lots of interesting properties. What more needs to be said? Chapter 5: The axiom of choice turns into Zorn's lemma. Using it you can prove all sorts of neat facts about existence of maximal ideals, algebraic closures, etc. Chapter 6: Well-foundedness of relations comes up when trying to force processes to stop. One application I'm familiar with is in generalizing power series rings to allow for more products, but still requiring a "bottom" so that multiplication is well-defined. Chapter 7: Boolean algebras, not much needs to be said here. However, I'm just not familiar with any immediate uses of stationary sets in ring theory. This bothers me because (1) they seem quite useful in set theory, and (2) Jensen's Diamond, which settled the Whitehead problem (and was one reason I jumped into this current project) is stated in terms of stationary sets. So, here is my formal question. Question: What basic applications do stationary sets (in and of themselves, without appealing to axioms independent of ZFC) have in algebra? If applications to ring theory specifically could be provided, that would be even better. REPLY [4 votes]: In his old paper on the Whitehead problem, Shelah proves that if $p$ is a prime number, $\kappa$ is uncountable regular, then there are $2^\kappa$ Abelian $p$-groups of cardinality $\kappa$, with no element of infinite height, none of them isomorphic to a subgroup to the other. This was a problem of Fuchs.<|endoftext|> TITLE: $SL_2$-action on the free lie algebra on a 2-dimensional vector space QUESTION [5 upvotes]: Let $V$ be a 2-dimensional vector space (over, say, $\mathbb{Q}$). Let $FL$ be the free lie algebra on $V$, then there is a natural action of the group $SL(V)$ on $FL$, such that the action of $-I$ is by multiplication by $(-1)^d$ on the $d$th graded component. An article I'm reading seems to imply that the representation of $SL(V)$ on each graded piece $FL^d$ of $FL$ decomposes into a direct sum of "$[n]$" for various $n\ge 1$ where $[n]$ is the representation of $SL(V)$ on the symmetric power $Sym^n(V)$. Is this true? (or could I be misinterpreting something?) To me, (from my understanding of Serre's "Lie Algebras and Lie Groups") I think of $FL$ as the Lie subalgebra of the tensor algebra $T(V)$ generated in degree 1, and certainly $SL(V)$ acts on the graded pieces of the tensor algebra $T^d(V)$, and hence on $Sym^d(V)$, but $FL^d$ is a subspace of $T^d(V)$, and I don't see why the $SL(V)$ representation on $FL^d$ needs to decompose into representations on the quotient $Sym^d(V)$. I'm a novice in this area, so I apologize if this question is naive. REPLY [3 votes]: (This was meant to be a comment; at the OP's request I copy it as an answer. I leave it CW) Every (finite-dim) irreducible rep of $\mathrm{SL}_2$ is isomorphic to $[n]$ for some $n\ge 1$. Hence every finite-dim rep of $\mathrm{SL}_2$ is a direct sum of such $[n]$. This is just what they say, and is not related to your specific representation (the $d$-th graded component in the free Lie algebra on 2 generators). A reference, among others, for reps of $\mathrm{SL}_2$, is Fulton-Harris' book.<|endoftext|> TITLE: Steps in Geometric Complexity Theory QUESTION [8 upvotes]: GCT purports to provide a program to show that $NP \not \subset P/poly$. At the high level what are the steps involved in the program and what stage is each step in? What difficulties currently are known or envisioned to be roadblocks and what are the easy targets? REPLY [11 votes]: First off, GCT is usually stated as a way of showing that $NP \not \subset P/poly$, which would imply that $P \neq NP$. This strategy was proposed by Mulmuley and several collaborators in a series of papers Geometric complexity theory I-VIII. These papers and several survey articles are available on Mulmuley's website. The first step in the program would be to separate the arithmetic circuit classes $VP$ and $VNP$ (here we think of families of circuits with arithmetic gates which formally compute a family of polynomials in their input variables). This is the arithmetic analogue of the weaker $\#P \not \subset NC$ conjecture. The determinant $\Delta_m$ and permanent $\Omega_n$ polynomials are complete for $VP$ and $VNP$ respectively. Therefore, to separate these classes, it suffices to show that the determinental complexity (the smallest $m$ such that a padded version of $\Omega_n(y)$ can be expressed as the determinant of an $m \times m$ matrix whose entries are affine linear combinations of the entries of the variables $y$) of the permanent is superpolynomial. The main idea is to compare the coordinate rings $A:=\mathbb{C}[\overline{GL_m \cdot \Delta_m}]$ and $B:=\mathbb{C}[\overline{GL_m \cdot z^{m-n}\Omega_n}]$ as $GL_m$-reps. In order to show that $\overline{GL_m \cdot \Delta_m}$ does not contain the (padded) permanent, it suffices to show that there is no surjective map of $GL_m$-reps. from $A \to B$. The idea here is to give labels $\lambda$ of irreps. appearing in $B$ but not in $A$, precluding such a map. To get from here to the boolean result $\#P \not \subset NC$, the idea is the transfer this result (done over $\mathbb{C}$) to finite fields. Mulmuley mentions in several places that the details of this plan are to be sketched in a future paper (GCT IX?), which, as far as I know, is not yet available. This strategy is really only a simpler version of the the strategy for the main problem (although the weaker result would still be one of the most impressive in the history of complexity theory). Mulmuley has identified more complicated polynomials to take the place of the determinant and permanent in the $NP \not \subset P/poly$ version, and the strategy there is analogous. The hoped-for "easy" target was that there would be some irrep. $V_{\lambda}$ appearing in $B$, but not appearing at all in $A$. This has been ruled out in this paper by Burgisser, Ikenmeyer, and Panova. Thus we are left with the more difficult route of being able to estimate (nonzero) multiplicities of irreps. in $A$ and $B$ well enough to preclude the map. This gets into several famous open problems in algebraic combinatorics: how to give positive formulas for Kronecker and plethysm coefficients. Even if these difficulties are surmounted, we are a long way from $P \neq NP$, but separating $VP$ and $VNP$ would itself be an extremely impressive result.<|endoftext|> TITLE: Is the Chow ring of a wonderful model for a hyperplane arrangement isomorphic to the singular cohomology ring? QUESTION [16 upvotes]: In the article "Hodge theory for combinatorial geometries" by Adiprasito, Huh and Katz, it it claimed in the proof of theorem 5.12 that there is a Chow equivalence between the de Concini-Processi wonderful model $Y$ of an arrangement, and a certain toric variety $X$. However, the source they cite only proves that $$H^*(Y)\cong Ch(X)$$ It seems that to draw the conclusion that $Ch(Y)\cong Ch(X)$ it is used that $Ch(Y)\cong H^*(Y)$, but I do not know why that is the case? Here is what I know: $Y\subset X$ as a closed subset, and the relevant maps are (should be) induced by this inclusion. We have a diagram $$ \require{AMScd} \begin{CD} Ch(X) @>{ch}>> H^*(X)\\ @VVV @VVV \\ Ch(Y) @>{ch}>> H^*(Y) \end{CD} $$ and we know that the diagonal arrow is an isomorphism. To establish that $Ch(X)\cong Ch(Y)$ it is enough to show that $Ch(Y)\cong H^*(Y)$. Since the diagonal is an isomorphism, the map $Ch(Y)\to H^*(Y)$ is surjective. I do not see how to prove injectiveness though. REPLY [6 votes]: The isomorphism $H^\cdot(Y)\cong Ch^\cdot(X)$ is shown by Eva Feichtner and Sergey Yuzvinsky in Feichtner, E. & Yuzvinsky, S. Invent. Math. (2004) 155: 515. https://doi.org/10.1007/s00222-003-0327-2 It's worth emphasizing that the inclusion $Y\subset X$ is not a homotopy equivalence, and generally $H^\cdot(X)\not\cong H^\cdot(Y)$. The property that $H^\cdot(Y)\cong Ch^\cdot(Y)$ for iterated blowups of projective spaces I think is due to Sean Keel, Trans. Amer. Math. Soc. 330 (1992), 545-574, https://doi.org/10.1090/S0002-9947-1992-1034665-0, in line with Libli's comment above.<|endoftext|> TITLE: Global dimension of quiver algebra QUESTION [6 upvotes]: Given a representation-finite (finite dimensional over a field) quiver algebra of finite global dimension. Is $eAe$ isomorphic to the field for at least one primitive idempotent $e$? This is true for Nakayama algebras (and trivially for acyclic algebras). REPLY [4 votes]: Edit: When writing this, I missed the condition that the algebra should be representation-finite. The algebra below is not. Not necessarily. Consider the path algebra of the quiver with two vertices and three arrows $$ 1 \substack{\xrightarrow{a} \\ \xleftarrow[b,c]{\xleftarrow{}} } 2 $$ modulo the relations $ab=0$ and $ca=0$ (composing arrows from left to right). This algebra $A$ has dimension $8$ over the base field, and its global dimension $3$, since the projective resolutions of the simple modules have the form $$ 0\to P_1\to P_2\to P_1\to S_1 \to 0 $$ and $$ 0\to P_1 \to P_2 \to P_1\oplus P_1 \to P_2 \to S_2 \to 0, $$ where $P_i$ is the projective cover of the simple module $S_i$. However, we see that $e_1Ae_1 \cong k[x]/(x^2) \cong e_2Ae_2$.<|endoftext|> TITLE: Hilbert 16th problem via hyperbolic geometry QUESTION [7 upvotes]: More than 16 years ago, I heard from someone that he thinks that there is a possible relation between Hilbert's 16th problem(for $n=2$) and Hyperbolic geometry. He says that a possible strategy is that a quadratic vector field $$\begin{cases} x'=P(x,y)\\ y'=Q(x,y)\end{cases}$$ can be rewritten in the form $z'=f(z,\bar{z}) $ with substitution $x=\frac{z+\bar{z}}{2}$ and $y=\frac{z-\bar{z}}{2i}$. He said that this possible strategy does not work for $n>2$. However, he thinks that this strategy leads to finitness of $H(2)$. My immediate reaction at that time was the following: since hyperbolic geometry concerns the upper half plane, are we implicitly assuming that the upper half plane is invariant under flow? So are we assuming that we have an invariant line? If this is the case then the following fact is an obstruction for continuation: Fact: every quadratic vector field with an invariant line has at most one limit cycle. But I think that the story is more complicated. I guess that he was not assuming that the upper half plane is flow invariant. So I guess that there are some thing non trivial in this possible strategy. I did not understand at all what is his strategy.I frequently asked him for more explanation. But I did not get any answer. He allowed me to talk with others about his idea. How does hyperbolic geometry can involve the Hilbert 16th problem? and how does this involvement work only for $n=2$ but not $n>2$. REPLY [10 votes]: This story is somewhat reminiscent of a (probably apocryphal) story of somebody writing a telegram to a mathematical research institute (Steklov Institute in Moscow, if I remember it correctly): "Solved Fermat's Last Theorem. Key idea - move $z^n$ to the left hand side. Details later." No, hyperbolic geometry does not "concern the upper half plane". The upper half plane is just one of many models of the hyperbolic plane and the latter can appear in many different forms. (Besides, maybe hyperbolics space that the speaker had in mind is even the hyperbolic 3-space, or a Kobayashi-hyperbolic manifold, or Anosov-type dynamics, or whatever...) There are some parallels of Hilbert-XVI and the Ahlfors Finiteness Theorem (and its generalizations) in the theory of Kleinian subgroups of $PSL(2,{\mathbb C})$ (so, maybe it is even the hyperbolic 3-space after all!). The parallels between the latter and holomorphic dynamics where exploited, for instance, by Dennis Sullivan - was he the speaker? - (dynamics of rational functions of one variable: proof of Fatou's Wandering Domain conjecture) and Xavier Gomez-Mont (a finiteness theorem for codimension 1 holomorphic foliations; see this 1980 JDG paper). The relation to hyperbolic geometry is only tangential; the idea is to prove a finiteness statement by arguing that otherwise, for some analytical reasons, a certain deformation space (say, a certain cohomology group) would have to be infinite-dimensional, while for some algebraic reasons such a space has to be finite-dimensional. For instance, in Sullivan's proof, the space of rational functions of the given degree is clearly finite-dimensional, while the existence of a wandering domain would create an infinite-dimensional space of quasiconformal deformations of such a function. The common feature of such proofs is that a certain PDE problem in one complex variable (maybe $z, \bar{z}$, to be more precise; for instance, the Beltrami equation) is well-posed, and these proofs break down in higher dimensions since the "right" PDE system turns out to be overdetermined. The rest are details for somebody to figure out and collect his/hers Fields medal.<|endoftext|> TITLE: Tensor product of fields QUESTION [6 upvotes]: I have to say that I have nearly no experience with this (as I always assume that my algebras are quiver algebras) so I do not know if this is apropriate for mathoverflow, but I give it a try. Given a field $k$ and a finite field extension $K$, then $K \otimes_k K$ is a selfinjective algebra. What selfinjective algebras can occur this way? Can they be classified depending on the field $k$? The most extreme case I saw was that the non-semisimple algebra $K[T]/(T^p)$ can occur. Can even a non-representation-finite algebra appear as $K \otimes_k K$? Here a "extreme case": Let $k=F_p(X)$ and $K:=k[T]/(T^p-X)$, then $K \otimes_k K \cong K[T]/(T^p)$. I got this from Example 1.7.17 in the representaton theory book of Zimmermann, where he gives more details and reference. Can we give a concrete example of a symmetric local algebra that is not isomorphic to $K \otimes_k K$ for some field extension $K$ of $k$. For example are the selfinjective algebras $F[x,y]/(x^p-y^q,xy,yx)$ of this form for some natural number $p, q \geq 2$ and some field $F$. Of course we have to make some assumption, like that the field has characteristic p since in characteristic 0 such a thing cant happen. REPLY [4 votes]: If you restrict attention to separable extensions then the picture is as follows. Let $\overline{k}$ be a separable closure of $k$ with Galois group $G$. If $X$ is a finite $G$-set we have a $k$-algebra $A(X)=\text{Map}_G(X,\overline{k})$. This construction gives an equivalence from the category of finite $G$-sets to the category of finite etale $k$-algebras. The finite separable extensions of $k$ just correspond to the $G$-sets on $X$ which $G$ acts transitively (so $X\simeq G/H$ for some $H$, and $A(X)=\overline{k}^H$). We can decompose any $X$ as a disjoint union of transitive $G$-sets, and this writes $A(X)$ as a product of fields. We also have $A(X)\otimes_kA(Y)=A(X\times Y)$. Thus, if $K=\overline{k}^H=A(G/H)$ then we can decompose $K\otimes_kK$ as a product of fields by decomposing $(G/H)\times(G/H)$ as a disjoint union of orbits. On the other hand, you can consider $k=(\mathbb{Z}/p)(x_1,\dotsc,x_r)$ and define an inseparable field extension $K$ by adjoining $y_i$ with $y_i^p=x_i$ for all $i$. Then $K\otimes_kK$ is generated over $K$ by classes $z_i$ satisfying $z_i^p=x_i$, so the classes $u_i=z_i-y_i$ have $u_i^p=0$. It is not hard to see that $K\otimes_kK$ is just the truncated polynomial algebra generated by $u_1,\dotsc,u_r$ subject to $u_i^p=0$. My guess is that you can't get anything much more complicated than that. [ADDED LATER] A commutative finite-dimensional $k$-algebra $A$ is self-injective iff Gorenstein iff $\text{Hom}_k(A,k)$ is isomorphic to $A$ as an $A$-module. There are very many of these. If $(r_1,\dotsc,r_n)$ is any regular sequence in $k[x_1,\dotsc,x_n]$ then the quotient ring $Q=k[x_1,\dotsc,x_n]/(r_1,\dotsc,r_n)$ has this property. The cohomology ring of any closed manifold (with coefficients in $k$) has this property. One small example is $k[x,y]/(x^3,y^2+xy+x^2)$.<|endoftext|> TITLE: Is any dual metrizable locally convex space a Frechet space? QUESTION [7 upvotes]: [I have posted this question on MSE some time ago, but received no answer.] The title basically says all of it. If a normed space $F$ is a dual of a normed space $E$, then $F$ is a Banach space. I wonder if the same holds for Frechet spaces. The strong dual $F$ of a locally convex space $E$ is complete, once $E$ is bornological, but I am not sure if this is the case here. Perhaps the completion of $E$ is though. REPLY [9 votes]: I am not 100% clear what you are asking, but I will answer according to two interpretations: a) Suppose that $F$, a metrizable TVS, is the strong dual of $E$, a locally convex TVS. Need $F$ be complete? The answer to this question is no, by the following counterexample, where we obtain an incomplete normed space as the strong dual of a locally convex space. Take $E$ to be $B(\mathcal{H})$, the bounded operators on an an infinite dimensional Hilbert space $\mathcal{H}$. The topology on $B(\mathcal{H})$ we will use is the strong operator topology (SOT, i.e. pointwise convergence of Hilbert space norm). It is not hard to show that, as a vector space, $F = E'$ is $\mathrm{FinRank}(\mathcal{H})$, the finite-rank operators (see, for example, Takesaki's Theory of Operator Algebras I, Theorem II.2.6). To work out what the strong dual topology on $F$ is, we need to know what the bounded sets are for the SOT on $B(\mathcal{H})$. To this end, we can use the completeness of $\mathcal{H}$ and apply the uniform boundedness principle (see Schaefer's Topological Vector Spaces, III.4.2), showing SOT-bounded sets are exactly equicontinuous sets, a.k.a. norm-bounded sets in $B(\mathcal{H})$. Therefore the strong dual topology on $F$ is that defined by the dual norm, which is the trace-class norm on $\mathrm{FinRank}(\mathcal{H})$. As $\mathcal{H}$ is infinite-dimensional, $F$ is not complete in this norm, as there are trace-class operators that are not of finite rank. (Added in edit: Here, the definition of strong dual topology on $F$ that I used was the unique locally convex topology having the polars of the bounded sets of $E$ as basic 0-neighbourhoods. It can also mean (as it does in Schaefer) the topology obtained by taking the polars of weakly (i.e. $\sigma(E,F)$) bounded sets instead. In this case, as $\sigma(E,F) = \sigma(B(\mathcal{H}),\mathrm{FinRank}(\mathcal{H}))$ is the weak operator topology, these are the WOT-bounded sets. Fortunately, WOT-bounded sets can be shown to be SOT-bounded by applying the uniform boundedness principle pointwise, so the two definitions of strong topology on $F$ coincide.) b) Suppose that $F$, a metrizable TVS, is the strong dual of $E$, a metrizable TVS. Need $F$ be complete? The answer to this question is yes, and also that $F$ is a Banach space (or "Banachable" space, without choosing any norm in particular). This is a consequence of this answer, plus the fact that the dual of a normed space is a Banach space.<|endoftext|> TITLE: Height growth for randomly falling Tetris like blocks ? What if Young diagrams are falling down? QUESTION [5 upvotes]: Question: How the maximal height grows for random Tetris like blocks falling down ? Numeric simulation (see below) shows leading term is linear with some constant depending on shapes of blocks allowed. Can the constant be calculated ? Can the subleading term like $h(T) = a T + b T^p + ...$ be found ( similar to "sticky disks" = "ballistic deposition" model where $p=1/3$) ? Background: Considerable research interest attracted models like "Ballistic Deposition" where blocks randomly fall down and glue to neighbors. See MO Tetris-like falling sticky disks, youtube video from 36 second for illustration, Borodin's lectures, I. Corwin Notices 2016 Kardar-Parisi-Zhang Universality etc... Numeric simulation 1. (Horizontal blocks) Consider horizontal blocks of fixed length $n$ falling down into the "box" (playing field in Tetris) of horizontal size $L$. The model has only two parameters (n,L) and n<=L. One uniformly generates random position (horizontal position x = 1... (L-n+1)) of the block and it falls down straight vertically. The heap of blocks grows - like in Tetris and we are interested in maximal height. In contrast to Tetris to simplify things let us forget about burning out the fully packed raws. We are interested in height growth depending on number $T$ of fallen blocks: $$ h(T) = a T + ... $$ where constant $a$ depends on $(n,L)$. It is clear that for $n>L/2$, $a(n,L)=1$. Numeric estimation for $a(n,L)$ are given in table 1 below. It seems plausible that for $k=1$, $a(1,L) = 1/L$; and $a(n,L) = 2n/L + ...$. It would be interesting to find subleading terms $h(T)$, however from numeric simulation it seems there is no reasonable term. Numeric simulation 2. Young diagrams Consider Young diagrams for some fixed $n$ falling down into the "box" (playing field in Tetris) of horizontal size $L$. Let us use English notation for diagrams. The model has only two parameters (n,L) and n<=L. One uniformly generates random position of the block in horizontal direction and it falls down straight vertically (horizontal position is UNchanged). Again we are interested how the height grows : $$ h(T) = a T + ... $$ where constant $a$ depends on $(n,L)$. Numeric similation results given in the table 2 below. Table 1: Constant $a(n,L)$ for falling horizonal blocks of size $n$ and $L$ is horizontal size of "playing field". Simulation number T= 200000 n\L 10.0000 20.0000 50.0000 80.0000 100.0000 1.0000 0.1014 0.0508 0.0207 0.0129 0.0106 2.0000 0.3808 0.1975 0.0803 0.0508 0.0408 3.0000 0.6100 0.3276 0.1349 0.0843 0.0679 4.0000 0.7999 0.4486 0.1875 0.1188 0.0952 5.0000 0.9510 0.5651 0.2404 0.1521 0.1227 10.0000 1.0000 0.9847 0.4923 0.3167 0.2562 25.0000 1.0000 1.0000 0.9972 0.7424 0.6181 40.0000 1.0000 1.0000 1.0000 0.9989 0.8956 50.0000 1.0000 1.0000 1.0000 1.0000 0.9993 Table 2: Constant $a(n,L)$ for falling Young diagrams (English notation), $n$ is Young diagrams's $n$ and $L$ is horizontal size of "playing field". Simulation number T= 500000 n\L 10.0000 20.0000 50.0000 80.0000 100.0000 1.0000 0.1005 0.0508 0.0203 0.0130 0.0104 2.0000 0.3643 0.1876 0.0763 0.0478 0.0381 3.0000 0.6663 0.3491 0.1427 0.0895 0.0719 4.0000 0.8899 0.4728 0.1942 0.1226 0.0977 5.0000 1.3263 0.7220 0.2998 0.1891 0.1512 6.0000 1.6307 0.9046 0.3772 0.2387 0.1913 7.0000 2.0022 1.1360 0.4767 0.3015 0.2420 8.0000 2.3123 1.3349 0.5675 0.3591 0.2872 There are several other questions on Tetirs here and on MSE: MO Is there winning strategy in Tetris ? What if Young diagrams are falling?, MSE The Mathematics of Tetris, MSE Mathematics of Tetris 2.0, MSE An impossible sequence of Tetris pieces, but they seems to discuss different sides of the game. It would be natural to ask what will happen with the height growth if Tetris rule: burning out the fully packed raws is introduced... It might be next question ... REPLY [5 votes]: This model is known as "Random Walk in Heap Monoids", see Chapter 5 of Vincent Jugé's PhD thesis and the references therein. The law of large numbers, as usually, follows from Kingman's subadditive theorem. A calculation of the rate of growth amounts to finding the limit distribution on infinite heaps. However, I am not sure to what extent it can be made explicit (cf. Abbes-Mairesse). As for the sublinear term, I would still expect the central limit theorem to hold.<|endoftext|> TITLE: Is $\mathscr{M}_{1,1,\mathbb{Z}}$ isomorphic to a quotient stack by a finite group? QUESTION [11 upvotes]: Let $\mathscr{M}_{1,1,\mathbb{Z}}$ denote the moduli stack of elliptic curves. Does there exist a scheme $X$ and a finite group $G$ acting on $X$ such that $\mathscr{M}_{1,1,\mathbb{Z}}$ is isomorphic to the quotient stack $[X/G]$? Remarks/thoughts: For any scheme $S$, set $\mathscr{M}_{1,1,S} := \mathscr{M}_{1,1,\mathbb{Z}} \times_{\operatorname{Spec}\mathbb{Z}} S$. We have $\mathscr{M}_{1,1,\mathbb{Z}} \simeq [W/H]$ where $W = \operatorname{Spec} \mathbb{Z}[a_{1},a_{2},a_{3},a_{4},a_{6},\Delta^{-1}]$ where $\Delta \in \mathbb{Z}[a_{1},a_{2},a_{3},a_{4},a_{6}]$ is the discriminant and $H$ is a subgroup scheme of $\mathrm{GL}_{3,\mathbb{Z}}$ of relative dimension 4 over $\mathbb{Z}$, see [1, Tag 072S] and [6, Section 3]. By [3, 4.7.2], for $N \ge 3$, the restrictions $\mathscr{M}_{1,1,\mathbb{Z}[\frac{1}{N}]}$ are isomorphic to $[Y(N)/\mathrm{GL}_{2}(\mathbb{Z}/N)]$ where $Y(N) \to \mathbb{Z}[\frac{1}{N}]$ is a smooth affine morphism of relative dimension 1. (Thus there is an affine open covering of $\mathbb{Z}$ on which the restriction of $\mathscr{M}_{1,1,\mathbb{Z}}$ is a quotient stack by a finite group.) For $N=2$, see Remark 2.8 and the following paragraph of [4] and also Section 4 of [5]. Such scheme $X$ would have to be affine and smooth over $\mathbb{Z}$ of relative dimension 1. (Reason why $X$ is affine: Fix $N \ge 3$ and set $T_{N}' := Y(N) \times_{\mathscr{M}_{1,1,\mathbb{Z}[\frac{1}{N}]}} X[\frac{1}{N}]$; then $T_{N}' \to Y(N)$ is a $G$-torsor, hence $T_{N}'$ is affine; moreover $T_{N}' \to X[\frac{1}{N}]$ is a $\mathrm{GL}_{2}(\mathbb{Z}/N)$-torsor so $X[\frac{1}{N}]$ is affine; thus $X \to \mathbb{Z}$ is Zariski-locally on the target an affine morphism; thus $X$ is affine.) Such scheme $X$ cannot have a $\mathbb{Z}$-point (otherwise $\mathscr{M}_{1,1,\mathbb{Z}}$ itself has a $\mathbb{Z}$-point, but there are no elliptic curves over $\mathbb{Z}$). References: [1] Stacks Project [2] Olsson, "Algebraic Spaces and Stacks", Colloquium Publications 62, AMS (2016) [3] Katz, Mazur, "Arithmetic Moduli of Elliptic Curves", volume 108 of Annals of Mathematics Studies, Princeton University Press, Princeton, NJ, 1985 [4] Conrad, "Isogenies and level structures", Notes for Stanford Math 248B, link [5] Antieau, Meier, "The Brauer group of the moduli stack of elliptic curves", arxiv [6] Fulton, Olsson, "The Picard group of $\mathscr{M}_{1,1}$", Algebra & Number Theory, vol. 4, no. 1 (2010) link REPLY [11 votes]: I guess I'll post my comments as an answer. The definition of the quotient stack makes $p : X\rightarrow [X/G]$ into a $G$-torsor (in whatever topology $\mathcal{T}$ one chooses). Since here we're working with a finite abstract group, $X\rightarrow[X/G] = \mathcal{M}_{1,1,\mathbb{Z}}$ is $\mathcal{T}$-locally isomorphic to a disjoint union of $|G|$ copies of $\mathcal{M}_{1,1,\mathbb{Z}}$, and hence since being finite etale is local on the target for pretty much any choice of topology $\mathcal{T}$, we find that $p$ is finite etale. However, since the fundamental group of $\mathcal{M}_{1,1,\mathbb{Z}}$ is trivial, $X$ is itself a disjoint union of copies of $\mathcal{M}_{1,1,\mathbb{Z}}$, and hence cannot be a scheme. EDIT: Actually, it seems all this also follows from Chapter 6 of LMB's book (again using the triviality of $\pi_1(\mathcal{M}_{1,1,\mathbb{Z}})$).<|endoftext|> TITLE: What's with equivariant homotopy theory over a compact Lie group? QUESTION [22 upvotes]: For some reason -- I'm not quite sure why -- I've developed the impression that I'm supposed to "tiptoe" around equivariant homotopy theory over groups that are not finite. Should I? Let me explain. Please correct me / tell me what I'm missing. Elmendorff's theorem holds for an arbitrary compact group. So I'm pretty sure at least that there's no disagreement over what the $G$-equivariant categories -- stable and unstable -- should be for an arbitary compact group. Please correct me if I'm wrong here. I sometimes get the impression that there are real problems in setting up $G$-equivariant homotopy theory when $G$ is not finite -- for example, all the foundational $\infty$-categorical work of late seems to be done in the finite or profinite case. But I don't know what these problems could possibly be! By (1), it's perfectly clear what the orbit category should be. When stabilizing, the Haar measure should allow the kind of "sums" required to construct transfers. Where's the problem? If it's possible to do motivic equivariant homotopy theory over non-finite algebraic groups, then surely it's possible to work over compact Lie groups! It's true that non-discrete Lie groups are inconvenient to model as simplicial groups. Is this really a stumbling block? Perhaps people just focus on the groups they actually intend to work with. I don't know many existing or potential applications of equivariant homotopy theory over infinite groups. This is probably just my ignorance, since for that matter I don't know many existing or potential applications even over finite groups. For example in this overflow question I see Kervaire invariant one, the Segal conjecture, Galois descent for $\mathbb{C}$ over $\mathbb{R}$, and the study of spaces with $G$-action. These are all for $G$ finite, except maybe the last one, but it's also the vaguest. The other application that comes to mind is cyclotomic spectra and THH, where $G = S^1$ (but we use a universe with only the proper closed subgroups), which brings me to another confusion: THH should admit a genuine $S^1$-equivariant structure, but people tend to just use the cyclotomic part, using only the proper closed subgroups (which are all finite). Is this because (a) we'd like to use the $S^1$ part, but there's no model of it that's reasonable to work with, (b) we wouldn't have any use for the $S^1$ part even if we could get our hands on it (seems unlikely -- shouldn't this correspond to information at the infinite place?), or (c) even if we could get our hands on the $S^1$ part, thinking about it in equivariant terms would be the "wrong" approach, or require additional corrections, or (d) some other reason? REPLY [12 votes]: Regarding 2, there is no difficulty in defining $G$-spectra in the setting of $\infty$-categories. The only complications I can think of are that (1) the orbit category $\mathrm{Orb}^G$ is now an $\infty$-category and (2) the stabilization process necessarily involves infinitely many representation spheres. Nevertheless, the $\infty$-category of $G$-spaces $\mathrm{Spc}^G$ can still be defined either as presheaves on $\mathrm{Orb}^G$ or by formally inverting the homotopy equivalences in the category of $G$-CW-complexes, and $\mathrm{Spc}^G_*\to\mathrm{Spt}^G$ is still the universal symmetric monoidal colimit-preserving functor that sends representation spheres to invertible objects. However, there is a difficulty in defining $G$-spectra as spectral Mackey functors, because the transfers are dimension-shifting and so cannot be encoded in a simple $2$-category of spans. Regarding 3, it is not clear to me that the orbit $\infty$-category $\mathrm{Orb}^G$ can be reconstructed from the simplicial group $\mathrm{Sing}(G)$ (one would also need the orbits $\mathrm{Sing}(G/H)$). Regarding 5, I believe the answer is (d): THH should not be considered as a genuine $S^1$-spectrum (even though it can be in many ways). This becomes clear when the cyclotomic structure on THH is understood in terms of factorization homology: there is nothing in factorization homology that would induce a genuine $S^1$-equivariant structure on THH.<|endoftext|> TITLE: Hypersurfaces whose equation is not known QUESTION [6 upvotes]: I would like to find some well-known/interesting hypersurfaces which arise as parametrizations where implicitization is computationally too difficult. I have software which computes the Newton polytope of such hypersurfaces and would like to use it on an interesting example! What are some great parametrized hypersurfaces that would be of interest? I am currently working on the Luroth invariant, but would love more examples. For what it's worth, the Luroth invariant is degree 54 in 15 variables and it is proving to be just within reach of the software. Something around this size or slightly smaller would be perfect for me. REPLY [4 votes]: The projective dual of a variety is usually a hypersurface. So, take your favorite variety and try to compute its projective dual.<|endoftext|> TITLE: The sum of the reciprocals of a sequence that increases by its logarithmic terms QUESTION [6 upvotes]: Let $\{a(n)\}$ be a sequence satisfies $a(1)=1$, $a(2)=2$, and $a(n)=a(n-1)+a(\lfloor\ln(n)\rfloor)$ for $n\geq 3$. According to the definition, it seems that $a(n)=\Omega(n\ln n\ln\ln n\ln \ln \ln n ...\ln^{(k)}n)$ for any constant $k$. Does $\sum_{i=1}^{\infty}\frac{1}{a(i)}$ still diverge? If yes, how fast it diverges? REPLY [13 votes]: Consider the function $f(x)=1/x$ on $[1,e]$ and extend it on $[1,\infty)$ by equality $e^xf(e^x)=f(x)$ for $x\geqslant 1$. Then both functions $f(x),xf(x)$ decrease on $[1,\infty)$. At first, I claim that $1/a_n\geqslant f(n)$. This is true for $n=1,2$, and we induct in $n$. Assume that the claim is proved for $1,\dots,n-1$; denote $k=\lfloor \log n\rfloor$. We have $$a_n=a_{n-1}+a_k\leqslant \frac1{f(n-1)}+\frac1{f(k)}\leqslant \frac1{f(n-1)}+\frac1{f(\log n)}=\frac1{f(n-1)}+\frac1{nf(n)}\leqslant \frac1{f(n)},$$ since $xf(x)$ decreases. Next, the integral of $f$ diverges. Indeed, from the change of variables formula $$\int_{e^N}^{e^{e^N}}f(x)dx=\int_{N}^{e^N}f(e^t)e^tdt=\int_N^{e^N}f(t)dt,$$ we get a sequence of disjoint segments over which $f$ has the same integral. Hence $\sum f(n)$ also diverges by the integral test, and so does $\sum a_n$. The growth of divergence for $\int f$ is seen from the proof, and for $\sum 1/a_n$ the growth is essentially the same, since we have suitable reverse estimates aswell.<|endoftext|> TITLE: Is there any published summary of Erdos's published problems in the American Mathematical Monthly journal? QUESTION [16 upvotes]: As we know Erdos has proposed a considerable number of problems in the "American Mathematical Monthly" journal. Is there any published summary of Erdos's published problems in the American Mathematical Monthly journal? Thank you! REPLY [4 votes]: I do not think that anyone ever made an effort to collect ALL problems of Erdos. In the later part of his life he liked to give talks on various conferences under the title "My favorite unsolved problems" (I recomend typing this on Google, there are many versions). But usually these collections were selected according to the topic of the conference. EDIT. In general, his problems are widely scattered. In the last years of his life he was mostly interested in graphs and combinatorics. But I know, for example, a collection of Erdos's problems in complex analysis dated 1964, and included in the book of W. Hayman "Research Problems in Function theory". This example shows that it would be very difficult to collect them all. A person searching for "Erdos problems" is unlikely to find a book like this.<|endoftext|> TITLE: What is the probability that these sets intersect? QUESTION [8 upvotes]: Let $A$ be the subset of $\mathbb{R}^n$ defined by $A=\{x\in\mathbb{R}^{n}:|x_{1}-x_{n}|+\sum_{i=1}^{n-1}|x_{i+1}-x_{i}|\leq d\}$ for a given $d$. Next, sample a point $p$ uniformly in the unit cube, and let $B$ be the $\ell_1$ ball of fixed radius $r$ about $p$. Is there a good upper bound for the probability that $A\cap B$ is nonempty, in terms of $n$, $d$, and $r$? I am most interested in limiting behavior as $n\to\infty$ (in which case, obviously, $d$ and $r$ would have to depend on $n$). REPLY [4 votes]: For $n=3$ this is easy. A triple is in $A$ iff its maximum and minimum are within $d/2$. So $A \cap B_p$ is non-empty iff the maximum and minimum coordinates of $p$ are within $c=\min(1,d/2+2r)$. This has probability $$6\left(\int_{x=0}^{1-c} \int_{y=x}^{x+c} \int_{z=y}^{x+c} dz\, dy\, dx + \int_{x=1-c}^1 \int_{y=x}^1 \int_{z=y}^1 dz\, dy\, dx \right) $$ $$=6\left(\frac{c^2-c^3}{2}+\frac{c^3}{6}\right)=3c^2-2c^3.$$ We'd need other descriptions of $A$ in higher dimensions to make this work for larger $n$.<|endoftext|> TITLE: $G_n$ 's mutually non-isomorphic QUESTION [6 upvotes]: This question was answered by @Jim Belk And he defined $G_n$ as follows: $$ G_n \;=\; \langle a,b \mid [a^{-1}ba,b] = \cdots = [a^{-n}ba^n,b]=1\rangle $$ My question is: Are $G_n$'s mutually non-isomorphic? (i.e., for every two distinct natural numbers $n$ and $m$, $G_n$ and $G_m$ are not isomorphic). Could you please help me to find my answer. Thanks, REPLY [9 votes]: As a complement to YCor's elegant answer, I would like to present three additional ways to prove YCor's Statement 1. The groups $G_m$ and $G_n$ are isomorphic if and only if $m = n$. The first method mimics YCor's but uses the theory of right-angled Artin groups (aka partially commutative groups, or graph groups) instead of Bass-Serre theory. The second is prompted by YCor's proof and consists in the computation of the cohomological dimension of $G_n$ which turns out to be $n + 1$. These two methods enable us to retrieve YCor's stronger statement YCor's Statement 2. The maximal rank $a(G_n)$ of a free Abelian subgroup of $G_n$ is $n + 1$. The last method is the computation of the second integral homology group of $G_n$, also called Schur multiplier. We will establish $\text{H}_2(G_n, \mathbb{Z}) \simeq \mathbb{Z}^n$, which yields, as we will see, some other benefit. Claim 1. Let $\Gamma_n$ be the simplicial graph with vertex group $\mathbb{Z}$ and where an edge binds $i$ to $j$ if and only if $\vert i - j \vert \le n$. Then $G_n \simeq S(\Gamma_n) \rtimes \mathbb{Z}$ where $$S(\Gamma_n) \Doteq \langle s_i, \, i \in \mathbb{Z} \,\vert\, [s_i, s_j] = 1 \text{ for every } i, j \in \mathbb{Z} \text{ such that } \vert i - j \vert \le n \rangle$$ is the right-angled Artin group associated to $\Gamma_n$ and the canonical generator $a$ of $\mathbb{Z}$ acts on $S(\Gamma_n)$ as the right shift operator, i.e., $a^{-1}s_ia = s_{i + 1}$. In addition, any Abelian free subgroup of rank $ > 1$ of $G_n$ is a subgroup of $S(\Gamma_n)$. In particular, $$a(G_n) = a(S(\Gamma_n)) = n + 1.$$ Note that the group $S(\Gamma_n)$ is just the kernel $K_n$ of Jim Beck's answer to the initial MSE question. Proof of Claim 1. Setting as YCor, $s_k \Doteq a^{-k}ba^{k}$ for all $k \in \mathbb{Z}$, we obtain in a similar way $$G_n = \langle a, s_i, \, i \in \mathbb{Z} \,\vert\, a^{-1}s_ka = s_{k + 1},\,[s_i, s_j] = 1 \text{ for all } k, i, j \in \mathbb{Z} \text{ with } \vert i - j \vert \le n \rangle.$$ Hence $G_n \simeq S(\Gamma_n) \rtimes \mathbb{Z}$. To prove the second part of the claim, we consider the basis $t_1 = \sigma_1a^{i_1}, \dots, t_k = \sigma_k a^{i_k}$, with $\sigma_i \in S(\Gamma_n) $, of a free Abelian subgroup of $G_n$ of rank $k > 1$. Replacing, if needed, $(t_1, \dots, t_k)$ by a Nielsen-equivalent $k$-tuple, we can assume that $i_j = 0$ for every $j > 1$ (use the projection $G_n \twoheadrightarrow \mathbb{Z}$ and the Euclidean algorithm in $\mathbb{Z}$). It follows from the Normal Form Theorem of right-angled Artin group [2] that $t_1$ commutes with $t_2$ only if $i_1 = 0$. Thus $\langle t_1, \dots, t_k \rangle \subset S(\Gamma_n)$. It is well-known that for $S(\Gamma)$, the right-angled Artin group associated to a graph $\Gamma$, the number $a(S(\Gamma))$ is the clique number of $\Gamma$, that is the number of vertices in a maximal complete subgraph of $\Gamma$. Obviously, this number is $n + 1$ for $\Gamma_n$. The cohomological dimension of a group $G$ is $\text{cd}(G) = \sup \left\{ q \, \vert \, \text{H}^q(G, \mathbb{Z}) \neq 0 \right\}$ [Section VIII.2, 3] Here is our second proof of YCor's statements. Claim 2. $\text{cd}(G_n) = n + 1$. Proof. Given a group $G$, a subgroup $S \subset G$ and an injective homomorphism $\tau: S \rightarrow G$, let $T = \tau(S)$ and let $\tilde{G}$ denote the corresponding HNN extension, i.e., $\tilde{G} = \langle G, a \, \vert \, a^{-1}sa = \tau(s), \text{ for all } s \in S \rangle$. By [Theorem 2.12, 1], there is a Mayer-Vietoris sequence $$ \cdots \rightarrow \text{H}^{q - 1}(S, \mathbb{Z}) \mathop{\rightarrow}^{\delta} \text{H}^q(\tilde{G}, \mathbb{Z}) \mathop{\rightarrow}^{\text{res}} \text{H}^q(G,\mathbb{Z}) \mathop{\rightarrow}^{\text{res}_S - \tau^{\ast} \circ \text{res}_T} \text{H}^q(S, \mathbb{Z}) \rightarrow \cdots $$ Set $G = \langle s_0, \dots, s_n \rangle$, the free Abelian group generated by the $s_i$ for $0 \le i \le n$. Let $S = \langle s_0, \dots, s_{n - 1} \rangle$ and let $\tau: S \rightarrow G$ is the homomorphism induced by the right shift of indices. As shown by YCor, we have then $\tilde{G} = G_n$. Since $\text{cd}(\mathbb{Z}^q) = q$, inspecting the above exact sequence in dimensions $q \ge n$ yields the result. Since $\text{cd}(H) \le \text{cd}(G)$ whenever $H$ is a subgroup of $G$ [Proposition VIII.2.4, 3], it follows from Claim 2 that $a(G_n) = n + 1$. Let $F(a, b)$ the free group over the alphabet $\{a,b\}$ and let $r_i = [a^{-i}ba^i, b] \in F(a,b)$. We denote by $R_n$ the normal subgroup of $F(a, b)$ generated by $r_1, \dots, r_n$. By Hopf's theorem [Theorem II.5.3, 3], we have $$\text{H}_2(G_n, \mathbb{Z}) \simeq \frac{ R_n}{[F(a,b), R_n]}$$ Claim 3. The Abelian group $\text{H}_2(G_n, \mathbb{Z})$ is freely generated by the images of $r_1, \dots, r_n$. In particular, $\text{H}_2(G_n, \mathbb{Z}) \simeq \mathbb{Z}^n$. Proof. We reuse the notation of the proof of Claim 2, in particular $\tilde{G} = G_n$. By [Theorem 2.12, 1], there is a Mayer-Vietoris sequence $$ \cdots \rightarrow \text{H}_{q}(S, \mathbb{Z}) \mathop{\rightarrow}^{\text{cores}_S - \text{cores}_T \circ \tau_{\ast}} \text{H}_q(G, \mathbb{Z}) \mathop{\rightarrow}^{\text{cores}} \text{H}_q(\tilde{G},\mathbb{Z}) \mathop{\rightarrow}^{\partial} \text{H}_{q - 1}(S, \mathbb{Z}) \rightarrow \cdots $$ Since $$S \simeq \text{H}_{1}(S, \mathbb{Z}) \mathop{\rightarrow}^{\text{cores}_S - \text{cores}_T \circ \tau_{\ast}} \text{H}_1(G, \mathbb{Z}) \simeq G$$ is given by $s_i \mapsto s_i - s_{i + 1}$ for $0 \le i \le n - 1$, this homomorphism is injective. Therefore $\text{H}_2(\tilde{G}, \mathbb{Z})$ is isomorphic to the quotient of $\text{H}_2(G, \mathbb{Z})$ by the image of $\text{H}_2(S, \mathbb{Z})$ via $\text{cores}_S - \text{cores}_T \circ \tau_{\ast}$. By [Theorem V.6.4, 3], the homomorphism $$\Lambda^2(S, \mathbb{Z}) \simeq \text{H}_{2}(S, \mathbb{Z}) \mathop{\rightarrow}^{\text{cores}_S - \text{cores}_T \circ \tau_{\ast}} \text{H}_2(G, \mathbb{Z}) \simeq \Lambda^2(G, \mathbb{Z})$$ is given by $s_i \wedge s_j \mapsto s_i \wedge s_j - s_{i + 1} \wedge s_{j + 1}$ for $0 \le i, j \le n - 1$. Thus $\text{H}_2(\tilde{G}, \mathbb{Z}) \simeq \mathbb{Z}^n$. As the images of $r_1, \dots, r_n$ generates $\text{H}_2(\tilde{G}, \mathbb{Z})$ by Hopf's formula, these elements are necessarily independent over $\mathbb{Z}$. It follows from Claim 3 that the presentation $\langle a,b \, \vert r_i = 1,\, i = 1,\dots, n \rangle$ is minimal in the sense that removing any defining relations yields a non-isomorphic extension. Moreover, any presentation of $G_n$ must involve at least $n$ relators. Claim 4. The cohomological dimension of $\mathbb{Z} \wr \mathbb{Z} = \langle a,b \, \vert r_i = 1,\, i \ge 1 \rangle$ is infinite and $\text{H}_2(\mathbb{Z} \wr \mathbb{Z}, \mathbb{Z})$ is a free Abelian group of infinite countable rank. Proof. The subgroup $S$ of $\mathbb{Z} \wr \mathbb{Z}$ generated by $s_0, s_1, \dots$ is a free Abelian group of infinite countable rank, therefore $\text{cd}(\mathbb{Z} \wr \mathbb{Z}) = \infty$. The last part of the claim is obtained as in the proof of Claim 3, observing that $\mathbb{Z} \wr \mathbb{Z} \simeq S \rtimes \mathbb{Z}$ where the canonical generator of $\mathbb{Z}$ acts on $S$ by shifting indices on the right. Eventually, we note that the answer to OP's question cannot be too simple as we have $$G_n/G_n' \simeq \mathbb{Z}^2,\quad G_n/G_n'' \simeq \mathbb{Z} \wr \mathbb{Z}$$ for every $n$. To get the second isomorphism, observe that $r_n \equiv r_1 \cdot r_1^a \cdots r_1^{a^{n - 1}} \text{mod } (F(a, b))''$. [1] R. Bieri. "Homological dimension of discrete groups", 1981. [2] A. Baudisch. "Subgroups of semifree groups", 1981. [3] K. Brown. "Cohomology of groups", 1982.<|endoftext|> TITLE: Gaussian distribution, maximum entropy and the heat equation QUESTION [7 upvotes]: I have asked this question on MathSE, but I got no replies, so I thought of trying here. Consider the Gaussian distribution on $\mathbb{R}$ with mean $m$ and variance $t=\sigma^2$. This has the explicit density $$ f(x,m,t) = \frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-m)^2}{2t}} $$ The Gaussian distribution has many characterizations, and two of them are the following: The Gaussian distribution maximizes entropy amongst all the distributions on $\mathbb{R}$ with mean $m$ and variance $t$. The density $f(x,m,t)$ provides a solution of the heat equation. More precisely, consider the differential equation $$ \begin{cases} \frac{\partial u}{\partial t} &= \frac{\partial^2 u }{\partial x^2} \quad \text{ for } x\in \mathbb{R},t>0 \\ u(x,0) &= \delta_m(x) \end{cases} $$ where $\delta_m$ is the Dirac concentrated at $m$. Then the function $u(x,t)=f(x,m,2t)$ is a solution of this differential equation. I would like to know whether there is a reason (intuitive or more formal) why the solution of the two problems (maximum entropy distribution and heat equation) is the same. Of course, an explanation is that we can solve explicitly both problems and the solution happens to be the same, but I wonder whether there is a more conceptual reason for this. REPLY [2 votes]: Here is yet another late answer, but I hope it is relevant. Let me make first make clear that I use the mathematical "minus entropy" convention (as is common in my field, which is optimal transport). In other words, for me the entropy of a probability distribution $u$ is $$ \mathcal H(u)=+\int_{\mathbb R}u \log u. $$ A beautiful result due to R. Jordan, D. Kinderlehrer, and F. Otto tells us that one can interpret the heat equation $\partial_t u=\Delta u$ as the gradient-flow of the entropy, $$ \partial_t u=-\operatorname{grad}_{W_2} \mathcal H(u) $$ with respect to the quadratic wasserstein metric $W_2$ over $\mathbb R$ (or $\mathbb R^d$, for that matters). Regardless of the interpretation of this Wasserstein gradient (which would deserve a lot more than this simple post, see e.g. C. Villani's books), this tells us that the entropy (or, again, minus the entropy depending on one's conveniton) decreases (increases) as fast as possible and tends to be minimized (maximized) along the time-evolution. Now, from the purely PDE perspective it is easy to check that the evolution preserves the mass and the average, and that the variance grows linearly in time, which means here that $$ \int u(t,x)dx=1, \qquad \int xu(t,x) dx=m, \qquad \int x^2 u(t,x)=t $$ since we started from $u(0,.)=\delta_m(.)$. (Simply differentiate w.r.t time under the integral signs, use the PDE to substitute $\Delta u$ for $\partial_t u$, and finally integrate by parts.) Owing to the very gradient flow structure (entropy tends to diminish as fast as possible) it becomes then very likely (and true indeed!) that the solution $u(t,.)$ of the heat flow at time $t$ is minimizing the entropy among all probability distributions satisfying the constraints, i-e average $m$ and variance $t$. OF course that's not a proof, but still a damn good hint I believe.<|endoftext|> TITLE: Is there a differentiable random walk? QUESTION [11 upvotes]: Is there a random walk which is differentiable or smooth? Like brownian motion except smoothed out on small distances. I was wondering if there is a "natural" or "canonical" analogue of brownian motion for differentiable curves, it will obviously have some scale parameter associated to it. REPLY [8 votes]: Actually if all you are concerned with is the smoothness of the sample path, the smoothness of a Gaussian process is completely characterized by its covariance function. The following result provides an insight into this issue. In this aspect we can discuss smoothness with probability one, and the sample path smoothness in this sense is relatively clear in stationary case. Theorem 3.4.1[Adler] Let $X(\boldsymbol{t}),\boldsymbol{t}\in\mathbb{R}^N$ be a real-valued zero mean Gaussian random field (in your case take $N=1$ as a special case of this result) with a continuous covariance function. Then if for some $00$ $$\boldsymbol{E}|X(\boldsymbol{t}-X(\boldsymbol{s}))|\leq\frac{C}{|\log\|\boldsymbol{t-s}\||^{1+\epsilon}}$$ for all $\boldsymbol{t},\boldsymbol{s}\in I_0\subset\mathbb{R}^N$. Then $X$ has a continuous sample path over $I_0$ with probability one. In this spirit of "being continuous with probability one", you can actually control the smoothness by choosing different covariance kernels. For example the Matérn covariance function will allow you to have a control over the sample path smoothness by varying its degree of freedom $\nu$; i.e. if you choose a stationary Matérn covariance realization of GP, then the sample path will be in $\mathcal{C}^{[\nu]-1}$ with probability one. Reference [Adler] Adler, Robert J. The geometry of random fields. Society for Industrial and Applied Mathematics, 2010. [Rasmussen] http://ml.dcs.shef.ac.uk/gpip/slides/rasmussen.pdf<|endoftext|> TITLE: Almost Complex manifolds of constant curvature QUESTION [9 upvotes]: Edited (after R. Bryant comment) Let $(M,\cal J,g)$ be a almost Hermitian manifold (not necessary integrable). i.e., ${\cal J}^2=-I$ and $g({\cal J} X,{\cal J} Y)=g(X,Y)$. Suppose that $\{X_i,{\cal J}X_i\}$ be any local orthonormal ${\cal J}$-frame and the following relation hold for $i\neq j$ $$g(Q{\cal J}X_i,{\cal J}X_i)=g(QX_j,X_j),\quad K(X_i,X_j)=K(X_i,{\cal J}X_i);$$ where $Q$ and $K$ are Ricci operator and sectional curvature respectively. Then Can be deduce that $(M,\cal J,g)$ is of constant curvature? Your advice or suggestions will be much appreciated and welcomed. REPLY [8 votes]: Well, first of all, your conditions are vacuous if the dimension of $M$ is $2$, and the conclusion in that case is false. Thus, you must also assume, in order to get the conclusion, that the dimension of $M$ is $2n>2$. It turns out that the answer is 'yes' if $M$ has dimension $4$, but my proof is not particularly simple. I suspect that the answer is 'yes' for all dimensions greater $2n>4$, but I have not checked this yet. Meanwhile, I think it is worth pointing out that, when $M$ has dimension $2n>2$, your condition $$ K(X_i,X_j) = K(X_i,\mathcal{J}X_i)\tag 1 $$ for all local orthonormal $\mathcal{J}$-frames implies that $(M,g)$ is Einstein, so, in particular, the relation $$ g(Q\mathcal{J}X_i,\mathcal{J}X_i) = g(QX_j,X_j) \tag 2 $$ is an automatic consequence. To see this, fix a point $x\in M$ and let $V = T_xM$, endowed with its inner product $g_x$ and complex structure $\mathcal{J}_x$. Let $B_x\subset V\times V$ denote the set of pairs $(v,w)$ such that $v$ and $w$ are unit vectors that satisfy $g_x(v,w) = g_x(\mathcal{J}_xv,w) = 0$. This $B_x$ is a smooth, connected manifold of dimension $4n{-}4$. It has a projection $\ell:B_x\to\mathbb{P}(V)\simeq \mathbb{CP}^{n-1}$ defined by $\ell(v,w) = \mathbb{C}v$, and another projection $p:B_x\to\mathrm{Gr}^+_2(V)$, the space of oriented unit $2$-planes in $V$, defined by $p(v,w) = v\wedge w$ whose image is the smooth hypersurface $L_x\subset\mathrm{Gr}^+_2(V)$ of dimension $4n{-}5$ consisting of the $2$-planes on which the canonical $(1,1)$-form $\omega$ vanishes, where $\omega(v, w) = g_x(\mathcal{J}_xv,w)$. Consider the function $K:B_x\to\mathbb{R}$ defined by the sectional curvature $K(v\wedge w)$. This function is constant on the fibers of $p$ by definition, but it is also constant on the fibers of $\ell$ by (1). Now, it is easy to see that these two constancies imply that $K$ is constant on all of $B_x$. (One way to see this is to note that any two points of $B_x$ can be connected by a sequentially connected chain of $p$-fibers and $\ell$-fibers.) Thus, there is a smooth function $\sigma:M\to\mathbb{R}$ such that $K(v\wedge w) = \sigma(x)$ for all $(v,w)\in B_x$, and, moreover, $K(v\wedge \mathcal{J}_xv) = \sigma(x)$ for all $v\in V = T_xM$. Now, if $v$ is any unit vector in $T_xM$, then there exists a orthonormal $\mathcal{J}$-frame $(X_i,\mathcal{J}X_i)$ on a neighborhood of $x$ such that $X_1(x) = v$. Consequently, $$ \mathrm{Ric}(v,v) = K\bigl(v\wedge \mathcal{J}_xv\bigr) + \sum_{i=2}^nK\bigl(v\wedge X_i(x)\bigr)+ K\bigl(v\wedge \mathcal{J}_xX_i(x)\bigr) = (2n{-}1)\,\sigma(x). $$ In particular, it follows, since $v$ is any unit vector in $T_xM$, that $\mathrm{Ric} = (2n{-}1)\sigma\, g$. Since the dimension of $M$ is greater than $2$, Schur's Lemma implies that $\sigma$ must be constant on $M$, i.e., that $g$ is Einstein, which, of course implies (2). Now, it is well known that if the sectional curvature is constant, i.e, if $K(v\wedge w) = \sigma$ for all $v\wedge w\in\mathrm{Gr}^+_2(TM)$, then $(M,g)$ has constant curvature. However, we do not yet know this property. We only know that $K$ is fiberwise constant on the hypersurface bundle $L\subset \mathrm{Gr}^+_2(TM)$ unioned with the smaller subbundle $\mathbb{P}(TM)\subset \mathrm{Gr}^+_2(TM)$ consisting of the $2$-planes that are complex lines (in either orientation) and that the scalar curvature function is constant. This, by itself, is not sufficient to conclude that the Riemann curvature tensor has constant sectional curvature. For example, when $2n=4$ (the first nontrivial case), when given a $(M,g,\mathcal{J})$, i.e., a $\mathrm{U}(2)$-structure on $M$, the space of potential Riemann curvature tensors for which the sectional curvature $K$ is a given constant $\sigma$ on the union of $L$ and $\mathbb{P}(TM)$ are the sections of an affine bundle of rank $2$ over $M$. One can set up the structure equations for such a $\mathrm{U}(2)$-structure over $M$ and, by applying the Bianchi identities a few times and differentiating and combining results, conclude that any such structure (even a local one) must, in fact, have constant sectional curvature. Thus, when $2n=4$, the answer to your question is 'yes'. However, when $2n>4$, things are not so clear. The first task is to describe the bundle of possible Riemann curvature tensors for $\mathrm{U}(n)$-structures on $M^{2n}$ with the property that the sectional curvature $K$ is constant on $L\cup \mathbb{P}(TM)\subset \mathrm{Gr}^+_2(TM)$. Again, it's an affine bundle, but I don't yet know its rank. It may be that, when $2n>4$, the constant sectional curvature tensor is the only one that satisfies (1), but I have not shown this yet. If that is the case, then the answer will be 'yes' in those cases as well, with a much easier proof than the case $2n=4$. I am interested to know what led you to consider the conditions (1) and (2).<|endoftext|> TITLE: Bound on queries to a tree with unusual probabilities QUESTION [7 upvotes]: Consider a tree $\mathcal{T}(r) = (V,E)$ rooted at $r \in V$. Let $\kappa_r: V \longrightarrow [0,1]$ such that $\sum_{v \in V} \kappa_r(v)^2 = 1$. Furthermore, for any given vertex $v \neq r$, $\kappa_r(v) = \sum_{c\leftarrow v}\kappa_r(c)$ where $c\leftarrow v$ means that $c$ is a child of $v$. Also, assume that $\kappa_r(r)=0$ and that $\kappa_r(c) = \kappa_r(c')$ for $c,c' \leftarrow r$. We want to determine a bound on the expected number of queries required to traverse from $r$ to some leaf $l$ as follows: Sample a random vertex $v$ with probability $\kappa_r(v)^2$ Update the root to the outcome of (1), and repeat the procedure on the subtree $\mathcal{T}(v)$, with probabilities assigned by the updated function $\kappa_{v}$. In the event that the walk is on a path, this clearly scales like $\log(|V|)$ since the probability of sampling anywhere is distributed uniformly. (This is trivial to prove by simply solving the recurrence that results from the probabilities in terms of expected change in depth.) In the event that the tree has more than one leaf, however, the problem IS a bit of a challenge. (Maybe not for someone on here, I hope.) The probabilities become top-heavy due to the square amplitude probabilities, but I'd still anticipate the worst case scaling to occur when a tree with $k$ leaves has $O(\log(k))$ branch points on the path to each leaf or, in other words, when the tree has as many branch points as possible. I have some ideas on how to prove this, but nothing has worked out yet. Thoughts were: Create a new statistic that scales like what I anticipate the expected number of steps to scale like Attempt to prove that for any distribution of vertices, the problem with evenly distributed branch points becomes harder (using some set of moves) Try to explicitly use something like a Jensen inequality. This works out for star-type (sub)trees, but I still get stuck once I go back an ancestor or so. Any thoughts, references, suggestions appreciated. UPDATE: Because of the answer below, I have followed up with a more nuanced question Bound on queries to a tree with unusual probabilties -- follow-up. REPLY [4 votes]: Alas, under the conditions you imposed, your conjecture is way too optimistic. Consider a path of length $mn$ and plant a bush with $m$ leaves at every $m$-th vertex on that path (so $|V|=2nm$). Now, if $w$ is not in a bush, we can define $k_w$ proportional to $1$ on the path until we meet the second bush after $w$ (the leaves of the first bush are assigned the value $0$), then, at the second bush we split $k$ equally between leaves, so we get $m$ leaves with $k\sim \frac{1}{m}$ and assign $0$ to the rest of the path. The result is that during each step we have only at most $2/m$ chance to land in a bush and we never move beyond the second bush. So, if $n\le m/4$, the probability to just move along the path without landing in any bush and covering not more than $2m$ vertices each time during the first $n$ steps is at least $1/2$. However, in such regime, we cannot finish in fewer than $n/2$ steps. Thus, the best upper bound you can hope for in general is something like $\sqrt{|V|}$. Graph for $m=6,n=4$: The function $k_w$ (not normalized, the root $w$ is shown in green)<|endoftext|> TITLE: Is there a continuous function $f:\mathbb R^\omega\to\mathbb R$ with injective restriction $f|\mathbb Q^\omega$? QUESTION [17 upvotes]: Question. Is there a continuous function $f:\mathbb R^\omega\to\mathbb R$ whose restriction $f|\mathbb Q^\omega$ is injective? REPLY [16 votes]: It looks like no. Assume the contrary. We may start with two distinct rationals $q_1,p_1$ such that the sets $f(q_1\times \mathbb{R}^{\omega-1})=f(\{(q_1,\cdot,\cdot,\dots)\})$ and $f(p_1\times \mathbb{R}^{\omega-1})$ intersect. Indeed, for any $p_1$ the set $f(p_1\times \mathbb{R}^{\omega-1})$ has non-empty interior, choose reals such that $f(p_1,x_2,x_3,\dots)$ is interior point of this set. After that any $q_1$ close enough to $p_1$ works, since $f(q_1,x_2,\dots)$ is close to $f(p_1,x_2,\dots)$. After that choose small intervals $\Delta_2,\Delta_3,\dots$ and $\delta_2,\delta_3,\dots$ such that the sets $f(q_1\times \Delta_2\times \Delta_3\times \dots)$ and $f(p_1\times \delta_2\times \delta_3\times \dots)$ intersect and have diameter less than 1. After that choose $q_2\in \Delta_2$ and $p_2\in \delta_2$ such that $f(q_1\times q_2\times \Delta_3\times \dots)$ and $f(p_1\times p_2\times \delta_3\times \dots)$ intersect. Proceeding this way, we get two rational sequences with the same value of $f$.<|endoftext|> TITLE: Fourier expansion of the Saito-Kurokawa lift QUESTION [7 upvotes]: As is well known, the Saito-Kurokawa lifts maps (classical) cusp forms $f$ to Siegel (genus 2) cusp forms $SK(f)$. Is there an explicit formula for the Fourier expansion of a Saito-Kurokawa lift? By explicit I mean something expressing the coefficients of $SK(f)$ in terms of the ones of $f$. Also, I am interested in the level $1$ case (the whole modular groups). I know how the lift is constructed (passing via Jacobi forms and half-integral modular forms, as is well explained in Van der Geer, The 1-2-3 of modular forms - https://link.springer.com/book/10.1007%2F978-3-540-74119-0) but I could not manage to derive an explicit formula nor could I find anything online. I know about the relation between eigenvalues of $f$ and $SK(f)$ - when they are eigenforms - but as far as I remember knowing the eigenvalues of Siegel modular forms is not sufficient to compute the Fourier coefficients (while the converse is true). I would be grateful to be directed to any reference that could potentially help me. REPLY [3 votes]: Let's say $f$ is of weight $2k-2$ and $g$ is the associated form of weight $k - 1/2$. One can relate the Fourier coefficients of $g$ to those of the associated Jacobi form $J$, the Fourier coefficients of $J$ to those of the Saito-Kurokawa lift $F$ of $f$. See Agarwal, Mahesh; Brown, Jim. Saito-Kurokawa lifts of square-free level. Kyoto J. Math. 55 (2015), no. 3, 641–662. and references therein for details. Hence one can relate Fourier coefficients of $g$ to those of $F$, though I do not know the details enough to know if you can get a formula for individual Fourier coefficients of $F$ in terms of those of $g$. From another point of view, one has an explicit relation between square sums of Fourier coefficients of $F$ and squares of Fourier coefficients of $g$. Namely, Bocherer conjectured that twisted central spinor $L$-values of $F$ are essentially squares of Bessel periods, and he proved this for Saito-Kurokawa lifts. Here the $L$-values $L(1/2,F,\chi)$ can be rewritten in terms of twisted $L$-values $L(1/2,f,\chi)$ of $f$, which are essentially squares of Fourier coefficients of $g$ by a formula of Waldspurger. (Here $\chi$ is a quadratic character.) The Bessel periods are sums of Fourier coefficients indexed by the same discriminant. Bocherer's work was unpublished, but you can see Dickson, Pitale, Saha and Schmidt for more details. Moreover, this connection between Fourier coefficients of $g$ and $L$-values $L(1/2,f,\chi)$ means there is no simple connection between Fourier coefficients of $f$ and those of $g$ or $F$. Finally, regarding your penultimate paragraph: it is true that the Hecke eigenvalues (though not just for $T_p$'s) of a holomorphic Siegel modular newform determine the form. Thus the eigenvalues determine (abstractly) the Fourier coefficients of such forms, but indeed this is different from being able to compute the Fourier coefficients given the eigenvalues.<|endoftext|> TITLE: Why are simplicial objects monadic over split (contractible) simplicial objects? QUESTION [6 upvotes]: Given an augmented simplicial object $d_\bullet:X_\bullet \to \Delta X_{-1}$, suppose there's a simplicial map $s_\bullet :\Delta X_{-1}\to X_\bullet$ making $d_\bullet$ a deformation retract, i.e such that $d_\bullet$ is both a retract and a homotopy-section of $s_\bullet$. This is equivalent to providing the augmented simplicial object with an "extra degeneracy" (just an alternative description of the simplicial homotopy axioms in this case). Such an extra degeneracy of an augmented simplicial object will be called a splitting. Let $\mathrm{S}$-$s\mathsf C$ denote the category of split simplicial objects (with fixed splitting) and simplicial arrows between them respecting the simplicial homotopies. This category admits a forgetful functor to simplicial objects $s\mathsf C$. On page 20 of Duskin's Simplicial Methods and the Interpretation of Triple Cohomology (AMS page), the author remarks this forgetful functor is a left adjoint to a shifting functor defined by deleting the top face map, viewing the top degeneracy as an extra one, and shifting to a lower index. (The simplicial homotopy is defined by $h_i=s_0^{n-i}s_{n+1}d_0^{n-i}$.) Moreover, the author writes this shifting functor $s\mathsf C\to \mathrm{S}$-$s\mathsf C$ is monadic. In other words, simplicial objects are monadic over split simplicial objects. What's the intuition behind the fact the shifting functor actually takes values in split simplicial objects? This seems strange to me - as if saying a simplicial object becomes contractible if you forget the top face map. How can that be? What's the intuition behind monadicity? An algebra over a split simplicial object (which is already a structured simplicial object) is an arrow to it, so how can additional structure on an already structured object yield back the original notion of object and arrow? Added. I am looking for naive geometric intuition for these facts, namely: 1. that a simplicial object becomes contractible upon merely forgetting a face map and reindexing; 2. simplicial objects are monadic over split ones. Ideally, I would like an example of what the contraction deformation retract actually does to the Décalage of a non-contractible simplicial complex, say the boundary of a tetrahedron. REPLY [3 votes]: Coming across this old question and noticing it still unanswered, here’s an attempt at the “geometric” answer OP wanted. First note that a split simplicial complex isn’t in general contractible — it’s just 1-truncated, i.e. each connected component is contractible. Concretely, the augmentation indexes the connected components, and the splitting + homotopy gives a distinguished point + contraction for each component. Write $R : \mathrm{SSet} \to \mathrm{SplSSet}$ for the desired right adjoint. We have $(RX)_{-1} = X_0$, so $X_0$ indexes the components of $RX$, and then for each $x \in X_0$, the $n$-simplices of $RX$ in the component over $x$ are the $n+1$-simplices whose $(n+1)$th vertex is $x$ (or $0$th vertex, depending on your indexing convention). In other words the $x$-component of $RX$ is the “out-star” (or “out-cocone”, or “out-path-space”) of $x$, consisting of all simplices whose source is $x$. But now each out-star is contractible, contracting into the degenerate 1-simplex on $x$; this is easy to check algebraically, but also easy to see geometrically/topologically — the space of “paths in $X$ with source $x$” is contractible to the constant path on $x$. Briefly: Under the right adjoint, a simplicial set falls apart into the collection of out-stars of its vertices; and each out-star is clearly contractible to the reflexivity path on that vertex.<|endoftext|> TITLE: Integrality of iterates of rational functions QUESTION [9 upvotes]: Let $f(x)$ be a rational function which is a ratio of two integral polynomials, and $n \in \mathbb Z$. Then the sequence of iterates $n, f(n), f(f(n)), f(f(f(n)), ...$ will be an infinite sequence of rational numbers, except in the rare cases where some iterate is a pole of $f$. In the special case that $f(x) = \frac{a}{x^m}$ it happens that, when $n \ne 0$ is divisible by $a$ (also a nonzero integer), the iterates of $n$ under $f$ are integers infinitely often and non-integers infinitely often. Are there any other examples where $n, f(n), f(f(n)), f(f(f(n)), ...$ contains both infinitely many integers and infinitely many non-integers? REPLY [10 votes]: As Pasten suggested in the comments, the key tool here is Siegel's theorem, and this was already done by Silverman, see "Theorem A" in Joseph H. Silverman: Integer points, Diophantine approximation, and iteration of rational maps, Duke Math. J. 71 (1993) #3, 793--829. Proposition. Fix a rational function $f \in {\bf Q}(x)$, and some $n_0 \in {\bf Q} \cup \{\infty\}$; for $i=1,2,3,\ldots$, define $n_i$ inductively by $n_i = f(n_{i-1})$. Then if $n_i \in \bf Z$ for infinitely many $i$, then either (i) $\{n_i\}$ is periodic, or (ii) $f$ has the form $f(x) = c + a/(x-c)^m$ for some $a,c \in \bf Q$ (with $a \neq 0$) and $m>1$, or (iii) $f$ is a polynomial. In each case it is also possible to have $n_i \notin \bf Z$ for infinitely many $i$. Note that case (ii) contains Kimball's example, and is in fact equivalent to it under conjugation by $x \mapsto x+c$. The point is that if $\{n_i\}$ is not periodic then for each $j=1,2,3,\ldots$ the $j$-th iterate $f^j$ satisfies $f^j(x) \in \bf Z$ for infinitely many distinct $x \in \bf Q$, whence $f^j$ has at most two distinct poles By Siegel's theorem on integral points. Silverman uses this to show that $f^2$ is polynomial, and thence that $f$ is either polynomial or of the form exhibited in (ii) by citing a result from A. Beardon: Iteration of Rational Functions (GTM 132), New York: Springer 1991 ($\S$4.1), which he describes as "elementary" and "well-known", and also proves as Proposition 1.1 of his paper (pages 798--799). A simple example of a polynomial whose iterates can take both integer and non-integer values is $f(x) = x + a$ for non-integral $a \in \bf Q$, say $f(x) = x+\frac12$. More complicated examples such as $f(x) = 2x^2 + \frac12$ can be obtained as $f(x) = P(cx)/c$ for suitable polynomials $P$ of degree $2$ or greater (here $P(x) = x^2+1$ and $c=2$). One can even construct examples such as $f(x) = x^2 + \frac{x}{2}$ for which it is probably true that the iterates of every integer include both integers and non-integers but this is very hard to prove (this example encodes the behavior of the parity of iterates of $x \mapsto (x^2+x)/2$).<|endoftext|> TITLE: Computation of a minimal polynomial QUESTION [8 upvotes]: It is relatively easy (but sometimes quite cumbersome) to compute the minimal polynomial of an algebraic number $\alpha$ when $\alpha$ is expressible in radicals. For example, the simple query "minimal polynomial 2^(1/5)*(1-exp(2*pi*i/5))" to Wolfram Alpha will compute the minimal polynomial of $\sqrt[5]{2}\left(1-\exp\left(\frac{2\pi i}{5}\right)\right)$. However, I do not know how to compute a minimal polynomial of an algebraic number that is not expressible in radicals. Moreover, I want to compute a minimal polynomial of linear combinations of algebraic numbers. For example, take $f(x) = x^5 - x + 1$. This polynomial is irreducible, has Galois group $S_5$, which is not solvable, so by Abel-Ruffini Theorem the roots of $f(x)$ are not expressible in radicals. I want to take two distinct roots of $f(x)$, say $\alpha_1$ and $\alpha_2$, and compute the minimal polynomial $g(x)$ of $\alpha_1 - \alpha_2$. I know how to do this algebraically, because $$g(x) = \prod\limits_{\substack{1 \leq i, j \leq 5,\\i\neq j}}\left(x - (\alpha_i - \alpha_j)\right),$$ but the computation just seems too painful. Is there a function in Sage, or Mathematica, or Maple, or PARI/GP, that allows to find $g(x)$? REPLY [3 votes]: SymPy code : from sympy import * x,y = symbols('x y') roots = solve(x**5 - x + 1) print(minimal_polynomial(roots[0] - roots[1], y)) gives : y**20 - 10*y**16 - 95*y**12 + 625*y**10 - 40*y**8 + 3750*y**6 + 400*y**4 + 5000*y**2 + 2869<|endoftext|> TITLE: Whence “homomorphism” and “homomorphic”? QUESTION [33 upvotes]: Today homomorphism (resp. isomorphism) means what Jordan (1870) had called isomorphism (resp. holoedric isomorphism). How did the switch happen? “Homomorphic” (and “homomorphism” as “property of being homomorphic”) are e.g. in de Séguier (1904, pp. 65–66) and the last edition of Weber (1912, p. 195). “Homomorphism” as map of groups is e.g. in Schur (1924, p. 192). But none of these sound like a first. I asked this on hsm a week ago, but got no answer there. REPLY [49 votes]: I found this footnote on page 195 of Fricke and Klein's Vorlesungen über die Theorie der automorphen Functionen (1897): Translation: The term "homomorphic" seems more appropriate than the previously$^\ast$ used "isomorphic", because it refers not to "equality" but to "similarity" of two groups. The term "isomorphism" will therefore from now on be used in the sense of "1-to-1 homomorphism". $^*$l.c. = loco citato — this seems to refer to Fricke and Klein's Vorlesungen über die Theorie der elliptischen Modulfunctionen (1890).<|endoftext|> TITLE: Cyclic polygons generalized to higher dimensions QUESTION [9 upvotes]: Many theorems hold for cyclic polygons—convex polygons inscribed in a circle. Perhaps the most basic is this, from the reference cited below: Theorem. There exists a cyclic polygon of $n \ge 3$ sides of lengths $\ell_i > 0$ if and only if each $\ell_k$ is less than the sum of the other lengths. And this polygon is unique.                     (Wikipedia image from article: Circumscribed circle.) Kouřimská, Hana, Lara Skuppin, and Boris Springborn. "A variational principle for cyclic polygons with prescribed edge lengths." Advances in Discrete Differential Geometry. Springer Berlin Heidelberg, 2016. 177-195. My question is: Q. What is the closest higher-dimensional analog of this theorem? E.g., in $\mathbb{R}^3$ the areas would be prescribed. I am familiar with Minkowski's theorem on the existence of a polytope realizing given facet areas/volumes and facet normals. What I am wondering is: If one assumes the polytope is inscribed in a sphere, can we reduce the information needed to justify existence/uniqueness? In other words, can Minkowski's theorem be "strengthened" by presuming the inscribed-in-a-sphere condition? Related: Japanese Theorem” on cyclic polygons: Higher-dimensional generalizations? REPLY [5 votes]: The solution to Minkowski's problem already produces a polytope circumscribed around a sphere. Insisting that it be also inscribed may be asking for a little too much. However, if you are happy with replacing "inscribed" by "circumscribed", Minkowski's theorem is your man. For inscribed polyhedra there are other sorts of results, in particular my results on the dihedral angles of ideal polyhedra in $\mathbb{H}^3,$ so if you use those instead of areas, you are golden. In $\mathbb{E}^3$ there is less information. Rivin, Igor, A characterization of ideal polyhedra in hyperbolic 3-space, Ann. Math. (2) 143, No.1, 51-70 (1996). ZBL0874.52006.<|endoftext|> TITLE: Finding a compatible multiplication for a given group QUESTION [8 upvotes]: If you are given an abelian group $\ (G, +)$, is there some algorithm to find all possible semigroups $\ (G, ×)$, such that $\ (G, +, ×)$ is a ring? If not, can you at least decide, if the ring must have zero divisors? REPLY [2 votes]: If not, can you at least decide, if the ring must have zero divisors? It would be interesting to know if there is a good answer to this question. Here I will just make three remarks. I will call an abelian group $G$ good if it can be equipped with a bi-additive multiplication which makes it a ring with no nontrivial zero divisors. By this answer, there is an abelian group $G$ that is elementarily equivalent to the group of integers $\mathbb Z$ such that the only bi-additive operation on $G$ is the zero function. This means that $\mathbb Z$ is good, $G$ is not, yet $\mathbb Z$ and $G$ satisfy the same first-order sentences. If $G$ is good and has a nonidentity element $g$ of finite additive order, then $G$ must be an elementary abelian $p$-group for some prime $p$. Conversely, any elementary abelian $p$-group is good. [For the first assertion, if $m\in\mathbb Z$, $g, h\in G-\{0\}$, and $mg=0$, then $0=(mg)h=g(mh)$, so by the goodness of $G$ we get $mh=0$. This shows that once one nonidentity element of $G$ has finite order, then all nonidentity elements have the same finite order, which must be prime. For the second assertion, any elementary abelian $p$-group supports a field structure, hence is good.] This remark addresses the first interesting cases where the structure of $G$ is not decided by Remark 2, which are the cases where $G$ is torsion-free of rank one. Claim. If $G$ is a torsion-free abelian group of rank one, then $G$ is good iff $G$ is isomorphic to the additive group of a unital subring of $\mathbb Q$. Besides being a torsion-free abelian group of rank one, the additive group of any unital subring of $\mathbb Q$ has the following additional property: Call an element $g\in G$ $m$-divisible for a given positive integer $m$ if the equation $mx=g$ is solvable in $G$. Call $G$ $m$-divisible if every $g\in G$ is $m$-divisible. The additional property is: there is an element $u\in G$ such that for every prime $p$ we have ($u$ is $p$-divisible implies $G$ is $p$-divisible). Pf. The if part of the claim is clear: the additive subgroup of a unital subring of $\mathbb Q$ is a good, torsion-free, abelian group of rank one. It is easy to see why the additive groups of unital subrings of $\mathbb Q$ satisfy the additional property. If $G$ is the additive group of a unital subring of $\mathbb Q$, choose $u=1\in G$. Then if $px=1$ is solved by $x=x_p$, for any $g\in G$ we get that $py=g$ is solved by $y=x_p\cdot g$. Here is a sketch for why the additional property characterizes the subgroups of unital subrings of $\mathbb Q$ among all torsion-free abelian groups of rank one. If $G$ is torsion-free of rank one, then there is a group embedding $\varphi:G\to\mathbb Q$. If $u\in G$ is an element such that for every prime $p$ we have ($u$ is $p$-divisible implies $G$ is $p$-divisible), then we can scale the embedding to $\frac{1}{u}\varphi:G\to Q$ so that the image of $u$ is $1$. Now the property that for every prime $p$ we have ($u$ is $p$-divisible implies $G$ is $p$-divisible) translates into the property that the set $\frac{1}{u}\varphi(G)$ consists of all rational numbers whose denominators are divisible only by those primes for which $u$ is $p$-divisible. This is a unital subring of $\mathbb Q$. What remains to explain is why if $G$ is torsion-free of rank one, and $G$ contains no element $u$ such that ($u$ is $p$-divisible implies $G$ is $p$-divisible), then $G$ is not good. Argument: This is a proof by contradiction, so assume that $G$ is torsion-free of rank one, and $G$ contains no element $u$ such that ($u$ is $p$-divisible implies $G$ is $p$-divisible), YET $G$ is nevertheless good. Applying an embedding if necessary, assume also that $G$ is an additive subgroup of $\mathbb Q$ that contains $1$. Suppose that $G$ is NOT $p$-divisible for exactly the primes $p_1k$, then $u=\frac{1}{p_1^{e_1}\cdots p_k^{e_k}}\in G$ is an element that is not $p$-divisible for the same primes as $G$ ($p_10$. Let $\ast$ denote a bi-additive multiplication on $G$ that witnesses that $G$ is good. Write $1\ast 1$ as $\frac{r}{s}$ in reduced form, and write $\frac{1}{p_i^{e_i}}\ast \frac{1}{p_i^{e_i}}$ as $\frac{r_i}{s_i}$ in reduced form. Since $1 = p_i^{e_i}\cdot \frac{1}{p_i^{e_i}}$ and $\ast$ is bi-additive, $$\frac{r}{s}=1\ast 1 = p_i^{2e_i}\cdot \left(\frac{1}{p^{e_i}}\ast \frac{1}{p^{e_i}}\right) = p^{2e_i}\frac{r_i}{s_i}.$$ Since no reduced fraction in $G$ has denominator divisible by $p_i^{e_i+1}$, it follows that the numerator $r$ of $\frac{r}{s} = \frac{p^{2e_i}r_i}{s_i}$ is divisible by $p_i^{e_i}$. As $i$ ranges we see that the fixed integer $r$ is divisible by infinitely many primes, a contradiction. \\\<|endoftext|> TITLE: Non-isomorphic graphs with bijective graph homomorphisms in both directions between them QUESTION [12 upvotes]: Are there simple, undirected graphs $G, H$ that are non-isomorphic, but there exist graph homomorphisms $f_1: G\to H$ and $f_2: H\to G$ which are bijective set-maps $V(G)\rightarrow V(H)$ and $V(H)\rightarrow V(G)$? Notes. By the argument in Tobias Fritz's comment below, $G, H$ have to be infinite. As suggested by a commenter, one should make it unambiguously clear that here, 'simple, undirected graph'='irreflexive symmetric binary relation on a set'. REPLY [3 votes]: I got here via this question and I thought it may be worth sharing the following uncountable family of locally finite examples: Pick an arbitrary set $S \subseteq \mathbb Z$. The vertex set of the graph $G_S$ is $\mathbb Z \times \mathbb Z$. For the edge set take all "vertical" edges from $(m,n)$ to $(m,n+1)$, horizontal edges $(m,n)$ to $(m+1,n)$ for $n < 0$, and horizontal edges $(s,0)$ to $(s+1,0)$ for $s \in S$; note that no horizontal edges are attached to $(m,n)$ for $n>0$. As long as $S$ and $S'$ are not shifts/reflections of one another, the graphs $G_S$ and $G_{S'}$ are non-isomorphic. The map $f \colon G_S \to G_{S'}$ can be taken as $(m,n) \mapsto (m,n-1)$.<|endoftext|> TITLE: Does there exist a closed geodesic go through a $\epsilon$-net of a hyperbolic surface? QUESTION [5 upvotes]: An $\epsilon$-net of a closed hyperbolic surface $X$ is a finite set of points $p_i$ such that the family of balls centered at $p_i$ with radius $\epsilon$ is a cover of $X$, and the family of balls centered at $p_i$ with radius $\epsilon/2$ are distinct pair by pair. My question is that if there is a closed geodesic goes through all $B(p_i,\epsilon)$ of some $\epsilon$-net, and if not, under which conditions of X and $\epsilon$-net we can find that geodesic? Thank you so much! REPLY [5 votes]: Yes, such a closed geodesic always exists. See Theorem 1.1 of this paper by Basmajian, Parlier, and Souto (which I found by searching under the term "density of closed geodesics on a hyperbolic surface"). Now, to be honest, what you want is simpler than what is proved in that paper. Namely, it is well known that the geodesic flow has a dense (non-closed) geodesic $\gamma : \mathbb{R} \to X$. Take a subsegment $\gamma | [-M,+M]$, close it off with a uniformly short segment --- namely one of length at most the diameter of $X$ --- and straighten. If $M$ is sufficiently large then the result of straightening is a closed geodesic $\gamma_M$. Furthermore, for each $\epsilon>0$, if $M$ is sufficiently large then $\gamma_M$ is $\epsilon$-dense meaning that it hits every $\epsilon$-ball in $X$.<|endoftext|> TITLE: Explicit formula for elementary symmetric sum QUESTION [10 upvotes]: For $k\ge1$, $j\ge1$, Let $$e_k(j)=\sum_{1\le i_1<... TITLE: Can you have a type theory where there is type of all types? QUESTION [20 upvotes]: Normally in a type theory, you can not have a type of all types, due to Girad's paradox. This is somewhat similar to how in set theory, you cannot have a set of all sets. Therefore, usually you just have a type of types that have types, which itself has no type (Calculus of Construction as defined here), or you just define an infinite tower of types, such that every type is of the type of something in that tower (a.k.a. a tower of universes). These two solutions are analogous to the set-theoretic notions of a class of all sets and the Von Neumann hierarchy. In set theory, there is actually a third solution: the New Foundations (and its relatives). In it, there is in fact a set of all sets. My question is, can you similarly have a type theory in which there is a type of all types, possibly using similar techniques as those of New Foundations. You obviously can not just "transfer over" the axioms of New Foundations, since set theory and type theory are quite different, but maybe you can use similar axioms. One direction I was thinking about it somehow extending universe polymorphism. For example, in Coq "Type : Type" is valid. But that doesn't mean Type is of type Type. Rather, there is actually a type Type$_n$ for each $n$, and Coq automatically tries to assign a $n$ to each occurrence of "Type" in the code. If you tried to formalize Girard's paradox in Coq, this assignment would fail. This is rather similar to the idea of stratified formulas in NF. Perhaps we could say that Type is a member of Type, but in any formula involving Type, there must be a way to "stratify" the formula, or something like that. Another possibility would be to establish to have a hierarchy of universes, but declare them all to be isomorphic (although this would make the theory incompatible with the univalence axiom). REPLY [17 votes]: Of course you can have $\mathsf{Type} : \mathsf{Type}$, the consequence of that is that all types are inhabited (by Girard's paradox). Some people call this an inconsistency, but that only makes sense if you view terms as proofs and types as logical statements. In other contexts, for instance when we think of terms as programs, there is nothing wrong with all types being inhabited. After all, in a general-purpose programming language all types are inhabited, thanks to recursion. In fact, if we lean towards programming, $\mathsf{Type} : \mathsf{Type}$ will be desirable because it will say that there is a datatype of all datatypes. There will be no end to Haskell-style hackage after that. There are very nice domain-theoretic models of $\mathsf{Type} : \mathsf{Type}$. For example, we can take countably-based algebraic lattices and continuous maps. Then every such lattice embeds as a retract into $\mathcal{P}(\omega)$, which means that $\mathcal{P}(\omega)$ is a universal lattice. It plays the role of $\mathsf{Type}$. But what is even cooler is that the (algebraic) retracts of $\mathcal{P}(\omega)$ themselves form a countably based algebraic lattice, and this gives us $\mathsf{Type} : \mathsf{Type}$.<|endoftext|> TITLE: Does second countable and functionally Hausdorff imply submetrizable? QUESTION [5 upvotes]: A topological space $\mathbf{X}$ is functionally Hausdorff, if for any two distinct $x, y \in \mathbf{X}$ there exists a continuous function $f_{xy} : \mathbf{X} \to [0,1]$ with $f(x) = 0$ and $f(y) = 1$. A space $\mathbf{X} = (X,\tau)$ is submetrizable, if there exists a topology $\tau' \subseteq \tau$ such that $(X,\tau')$ is metrizable. Equivalently, if there is a continuous injection $\iota : \mathbf{X} \to \mathbf{X}'$ to some metric space $\mathbf{X}'$. It is rather easy to see that any submetrizable space is functionally Hausdorff. I am wondering whether restricted to second-countable spaces, the converse might hold, too. Failed solution attempts: My naive attempt to prove this was to pick a dense sequence $(a_n)_{n \in \mathbb{N}}$, and to consider the continuous map $F : \mathbf{X} \to [0,1]^\omega$ where $F(x)(\langle n,m\rangle) = f_{a_na_m}(x)$ for some tupling functions for unequal pairs. This map can fail to be injective, though: Take an uncountable space with a dense sequence of isolated points, pick the $f_{xy}$ suitably, and $$F[\mathbf{X}] = \{x \in [0,1]^\omega \mid \exists n \ \forall i \neq n \ x_n = 1 \wedge x_i = 0\} \cup \{0^\omega\}$$ is countable. The initial topology induced by all $f_{xy}$ should be regular, and is nested between two countably-based topologies, but I do not see why it should be countably-based itself. Searching on $\pi$-base for secound countable, functionally Hausdorff (aka Urysohn) but not metrizable spaces yields the following: https://topology.jdabbs.com/spaces?q=Second%20Countable%20%2B%20Urysohn%20%2B%20~metrizable Of these examples most are just defined by adding open sets to a metrizable topology. The other two (irregular lattice topology and Roy's Lattice Subspace) are countable, hence the argument above with a total enumeration shows their submetrizability. REPLY [5 votes]: Fact. Each second-countable functionally Hausdorff space is submetrizable. This fact follows from a more general result: Theorem. Each functionally Hausdorff space $X$ with hereditarily Lindelöf square $X\times X$ is submetrizable. Proof. Denote by $\Delta$ the diagonal of the square $X^2:=X\times X$. For any distinct points $x,y\in X$ choose a continuous function $f_{x,y}:X\to[0,1]$ such that $f_{x,y}(x)=0$ and $f_{x,y}(y)=1$. Then $$U_{x,y}=\{(x',y')\in X\times X:f_{x,y}(x')<\tfrac12 TITLE: Detailed modern references for basic properties of Pfaffians over commutative rings QUESTION [11 upvotes]: Pfaffians are important to algebraic combinatorics, at least. This is to propose the making of a 'wiki' list, more modern, precise and compressed than e.g. the relevant Wikipedia page (nothing against Wikipedia; it simply has different format than MO, and produces results of a different kind; both seem rather complementary), and in the spirit of the nice MO page Properties of functors and their adjoints, featuring compressed proofs or precise references to trustworthy references, of the known and not-overly-specializedexample 0 properties of Pfaffians. A (rough) guideline could be to keep the list basic and context-free. (With 'context-free' roughly meaning: to keep to Pfaffians as an algebraic concept, and e.g. leave out any application to graphs without $K^{3,3}$-minors or even planar graphs). A tentative list. Definitions of the Pfaffian. (0) The classical definition.editorializing 0 (1) The definition in Knuth: Overlapping Pfaffians.note on notation 0 (2) 'The' usual definition via exterior algebra.editorializing 1 (3) The definition of A. W. M. Dress, W. Wenzel: Advances in Mathematics Volume 112, Issue 1, April 1995, Pages 120-134.category-theoretic note 0 Basic properties of the Pfaffian. Ideally, all the basic properties will be treated, with either a reasonably precise reference or a reasonably compressed proof. At the minimum, the properties listed---without satisfactory proof or reference--on the current Wikipedia page on Pfaffians will all be treated. Some of these are the following. (I suggest using indices in the finite ordinal $n$, in particular, the indexation starts at $0$.) (0) For any commutative ring $R$ of characteristic different from two, and for each skew-symmetric $A\in R^{n\times n}$, if $n$ is odd, then $\mathrm{pf}(A)=0$. (1) [to be filled with an appropriate property saying how Pfaffians behave under permutations of the factors of the cartesian-product-that-is-the-index-set; property (1) should be suitable to immediately deduce (2) from [Dress--Wenzel, (1.5), p. 122]; such a property is easy to come by; the touchy point is whether to make (1) into more than a mere instrument to prove (2) ] (2) For any commutative ring $R$, and for any $i\in n$, and with $[\cdot]$ being the Iverson bracket and with $|_{}$ being the usual restriction notation applied to matrices-construed-as-functions $n\times n\to R$, for any skew-symmetric $A\in R^{n\times n}$, $\mathrm{pf}(A)=\sum_{j\in n\setminus\{i\}}\ (-1)^{\bigl[j TITLE: What are the odds of a tie in a random election with k candidates? QUESTION [8 upvotes]: Consider an election with $N$ voters and $k$ candidates, where each voter votes randomly for one of the candidates. What are the odds of a tie? Here "tie" means that multiple candidates get the highest number of votes, not necessarily that all candidates get the same number of votes. For $k = 2$ the answer is given by ${N \choose {N/2}} \frac{1}{2^N}$. I'm not sure how to generalize to $k > 2$. If a closed formula doesn't exist, can we still say something about the asymptotic behavior as $k$ grows? Edit: asking for "asymptotic behavior" was ambiguous. The question I'm most interested in is: for $k \ll N$, what can we say about how the odds change if $k$ increases by $1$ (or, if it's a cleaner answer, if $k$ doubles). Perhaps it's incorrect to call this asymptotic behavior at all! And, unfortunately, I'm not sure what "$\ll$" should mean in this context. I suppose that's part of the question as well. REPLY [7 votes]: In general, the probability is $$ \frac{1}{k^N}\sum_{j=2}^k{k \choose j}\sum_{m=0}^{\lfloor N/j\rfloor}{N \choose mj}\frac{(mj)!}{(m!)^j}\sum_{\substack{(a_1,\dots,a_{k-j})\\a_1+\dots+a_{k-j}=N-mj\\0\leq a_iN$.<|endoftext|> TITLE: Action on cohomology of the power map of $K(Z,n)$ QUESTION [8 upvotes]: Consider the "$m$-th power" map $f:K(\mathbb Z,n)\to K(\mathbb Z,n)$ given by $m\in \mathbb Z\cong H^n(K(\mathbb Z,n),\mathbb Z)\cong [K(\mathbb Z,n), K(\mathbb Z,n)]$. Is it true that in any degree the map $f^*$ on integral cohomology sends any element to a multiple of $m$? It's obviously true in degree $n$ where $f^*$ is just multiplication by $m$ but what about higher degrees? It's enough to consider the case when $m=p$ is prime. REPLY [4 votes]: As you say, we can reduce to the case where $m=p$ is prime. By the universal coefficient theorem we know that $H^*(K(\mathbb{Z},n);\mathbb{Z})/p$ injects in the ring $A^*=H^*(K(\mathbb{Z},n);\mathbb{Z}/p)$, so we just need to show that $f^*$ acts as zero on $A^*$ in positive degrees. The kernel of $f^*$ is an ideal and is closed under Steenrod operations (including the Bockstein). There is a tautological class $u\in A^n$ with $f^*(u)=pu=0$. It is known that $A^*$ is the free unstable algebra over the Steenrod algebra generated by $u$ subject only to the relation $\beta(u)=0$. That shows that $f^*=0$ on $A^{>0}$ as required.<|endoftext|> TITLE: What drawbacks are there to using NF(U) for category theory? QUESTION [6 upvotes]: In category theory, you often run into what is known as "size" issues. That is, you run into the issue that the categories you try to define are too "big" to be sets, and so you need to use classes or even larger collections. There are a couple solutions, but one of them would be to use NF or NFU. Since they have a set of all sets, you do not need to worry about your categories being to big. Having a category of all categories which is an object of itself is no problem, for example. My question is, are there any issues with NF or NFU that would make them problematic as a foundation for category theory. If so, what are these issues? REPLY [10 votes]: One serious sticking point is that the category of sets in NF is not cartesian closed.<|endoftext|> TITLE: How to prove Feller property without using heat kernel estimates QUESTION [8 upvotes]: I have a question about Markov processes. Let $\mathbb{M}=(X_t,P_x)$ be a Markov process on a locally compact separable metric measure space $(E,\mu)$. $\mathbb{M}$ is called Feller process if its semigroup $\{p_{t}\}_{t>0}$ satisfies the following: for all $t>0$, \begin{align*} (0)\quad p_{t}(C_{\infty}(E)) \subset C_{\infty}(E), \end{align*} where $ C_{\infty}(E)$ is the set of continuous functions which vanish at infinity. If we know $\mathbb{M}$ has the following property: \begin{align*} (1)\quad\lim_{r \to \infty}\sup_{x \in E}P_{x}(X_{t} \in E \setminus B(x,r))=0, \end{align*} we can prove $p_{t}f$ vanishes at infinity for all $f \in C_{c}(E)$. Indeed, for all $f \in C_{c}(E)$, we have \begin{align*} &|p_{t}f(x)| \le E_{x}[|f(X_t)|] \\ &=\int_{E \cap B(x,r)}p_{t}(x,y)|f(y)|\,\mu(dy)+\int_{E \setminus B(x,r)}p_{t}(x,y)|f(y)|\,\mu(dy) \\ &\le \sup_{y \in E \cap B(x,r)}|f(y)|+\|f\|_{\infty} \sup_{x \in E}P_{x}(X_{t} \in E \setminus B(x,r))\end{align*} Then, letting $x \to \infty$ and then $r \to \infty$, we obtain the assertion. A sufficient condition for (1) Let us consider the case $\mathbb{M}$ is a diffusion process on a Euclidean domain $D$. If the transition density $p_{t}(x,y)$ of $\mathbb{M}$ has the following estimate: \begin{align*} (2) \quad p_{t}(x,y) \le a_{1}e^{t} t^{-d/2} \exp(-|x-y|^2/a_{2}t)\quad (a_1,a_2 \text{ are some constants indep of $t,x,y$}), \end{align*} we can prove (1). My question I am interested in the property (1) of reflecting Brownian motions on smooth domains. These processes are generated by the following classical Dirichlet form: \begin{align*} (3)\quad\mathcal{E}(f,g)=\frac{1}{2}\int_{D}(\nabla f, \nabla g)\,dx,\quad f,g \in H^{1}(D). \end{align*} When the boundary of $D$ is sufficiently smooth, it is known that (3) is regular on $\bar{D}$ and we can construct a processes $(\{X_t\},\{P_x\})$ whose Dirichlet form is (3). Furthermore, $(\{X_t\},\{P_x\})$ solves the following Skorohod SDE: \begin{align*} X_{t}=x+B_{t}+\int_{0}^{t}n(X_s)dL_s, \end{align*} where $B_t$ is the $d$-dim B.M. and $n$ is the inward unit normal on $\partial D$ and $\{L_t\}$ is boundaly local time. If transition density of $X$ has a estimate like (2) and $D$ is bounded, we can compute expectation of $L_t$. REPLY [3 votes]: The answer depends on what you are working with. If you have a pseudo-differential operator and ask whether it generates a Feller semigroup, the best answer known to date is contained in Walter Hoh's works; see a three-volume book Pseudo-Differential Operators and Markov Processes by Niels Jacob for rigorous statements. If you have ask what differential operators generate Feller semigroups, or what SDEs determine Feller diffusions, then this is closely related to a question about the possibility of starting a process "at infinity", and in some sense dual to the "no explosion in finite time" problem. This can be a delicate question. Feller property is known if the coefficients of the generator grow slowly enough (quadratic growth for the coefficients at the second order derivative and linear growth for the linear term). For a detailed discussion and references, see Properties at infinity of diffusion semigroups and stochastic flows via weak uniform covers by Xue-Mei Li. For general conditions, you may see a Primer on Feller semigroups and Feller processes. For example, Theorem 1.10 gives an equivalent condition expressed in terms of transition kernels.<|endoftext|> TITLE: Exceptional primes QUESTION [9 upvotes]: Let $f$ be a cuspidal Hecke Eigenform of weight $k \geq 2$ and let $\rho_{f, \lambda}:G_{\mathbb{Q}} \rightarrow GL_2(E_{\lambda})$ be the corresponding Galois representation with $2 \mid \lambda$ constructed by Deligne. Assume now that $\pi_f={\otimes}' \pi_p$ be the corresponding automorphic representation. My question is the following: If we assume $\pi_2$ non-monomial (it is not induced by character), can you say that projective image $\tilde{\rho_{f, \lambda}}:G_{\mathbb{Q}} \rightarrow PGL_2(E_{\lambda})$ is an exceptional group (S_4). I am aware the image of the corresponding Weil-Deligne representation is $S_4$. REPLY [9 votes]: No, this does not work: for any modular form of weight $k \ge 2$, the image of the projective representation $\tilde\rho_{f, \lambda}$ is infinite for every prime $\lambda$. Proof: if $\tilde{\rho}_{f, \lambda}$ has finite image, then so does the adjoint representation $\operatorname{Ad}^0 \tilde{\rho}_{f, \lambda}$, since the adjoint representation of $GL_2$ factors through $PGL_2$. In particular, the adjoint representation has all Hodge--Tate weights 0, which is a contradiction since $\rho_{f, \lambda}$ has Hodge--Tate weights $0$ and $1-k$ and thus $\operatorname{Ad}^0 \tilde{\rho}_{f, \lambda}$ has weights $\{0, k-1, 1-k\}$. (Note that this even shows that $\tilde\rho_{f, \lambda}(D_\ell)$ is infinite, where $\lambda \mid \ell$ and $D_\ell$ is a decomposition group at $\ell$.)<|endoftext|> TITLE: Construction of representations of the Mathieu groups? QUESTION [13 upvotes]: The Mathieu groups are beautiful simple finite groups. (They were the first sporadic groups to be discovered in 1861-1870). They are related with many other miraculous constructions in mathematics: Golay error-correcting codes, Steiner systems, K3 surfaces and moonshine, etc... One might expect that construction of irreducible representations over complex numbers of such distinguished groups should also be beautiful, however googling I was unable to find something like that. Question What are constructions (hopefully "nice") of irreducible representations of the Mathieu groups ? Googling suggests: Frobenius found character tables of the Mathieu groups in 1901 (I do not have reference). G. James found all modular irreps in The modular characters of the Mathieu groups 1973, but his methods are quite specific to modular case J.F Humphreys in 1980 considered more general questions of projective characters and character for automorphism group. The projective characters of the Mathieu group M12 and of its automorphism group, but there is no free access to that paper ... Some simple considerations leads to the following observations: for example M12 acts on 12 points, hence one has 12-dim permutation representation, it is natural to expect that 11-dim subrepresentation is irreducible and it is restriction of the one of from symmetric group S12 or Alternating groups A12. Mathieu group M12 also has 55-dim irrep it is natural to guess that it is wedge-square of 11-dimensional. One can also see that dimensions of irreps coinciding in M12 and S12 and A12 are: 1, 11,54,55. So 54-dim irrep of Mathieu probably is restriction of the one for S12/A12. (What is this representation for S12/A12 ? ) Character table for e.g. M12 can be obtained by MAGMA http://magma.maths.usyd.edu.au/calc/ for free: load m12; CharacterTable(G); For A12: AlternatingCharacterTable(12) Further info on e.g. M12 irreps can be found here: https://groupprops.subwiki.org/wiki/Linear_representation_theory_of_Mathieu_group:M12, e.g. dimensions of irreps: 1,11,11,16,16,45,54,55,55,55,66,99,120,144,176 Information of complex/real representations: MO Strongly real elements of odd order in sporadic finite simple groups, some discussion: MO Atlas of finite groups, Character table of automorphism group of sporadic group. A part of motivations comes from: MO Monstrous Langlands-McKay ... , the other part from the discussion with S. Galkin, who found in 2010 G-FANO THREEFOLDS ARE MIRROR-MODULAR that Gromov-Witten invariants of certain Fano 3-folds can be expressed via $\eta$-producs related to the Mathieu group M24 by the construction of G.Mason, extending V.Golyshev's results on Fano 3-folds and Moonshine. REPLY [3 votes]: I would like to draw your attention to an article by Nick Gill and (his then MMath student) Sam Hughes The character table of a sharply 5-transitive subgroup of $A_{12}$, that constructs the character table of the Mathieu group $M_{12}$ assuming only the properties that are mentioned in the title.<|endoftext|> TITLE: Possible groups of K-rational points for elliptic curves over arbitrary fields QUESTION [10 upvotes]: It is known that the group $C(\Bbb R)$ has at most two connected components, and the connected component of the identity is isomorphic to $U(1)$ as a topological group (trivially) and $C(\Bbb Q)$ is isomorphic to $\Bbb Z^r\times E(\Bbb Q)$, where all possible $E(\Bbb Q)$ are known. Are there similar results for other fields? I am especially interested in the case K is a number field, or the p-adic numbers (both $\Bbb Z_p$ and $\Bbb Q_p$) REPLY [16 votes]: The answer to one possible interpretation of the title question -- vary over all elliptic curves over all fields and ask which groups arise -- is given in this paper. With regard to the structure of Mordell-Weil groups of elliptic curves over local fields, I think you will find that $\S$5.1 of this joint paper with Allan Lacy relevant. (The title is "Mordell-Weil Groups of Abelian Varieties Over Local Fields".) The case of characteristic $0$ is a slightly more general and less precise variant of Joe Silverman's answer, but things work a bit differently in positive characteristic.<|endoftext|> TITLE: Deformation of the Plücker coordinates QUESTION [5 upvotes]: Let $M_{2,4}(\mathbb{R})$ be the set of real $2\times4$-matrices of rank $2$. For any $A\in M_{2,4}(\mathbb{R})$ and $1\leq i TITLE: Distribution of good diophantine approximations QUESTION [10 upvotes]: Let $\langle x \rangle: \mathbb{R} \to (-1/2,1/2]$ be the periodic function with period $1$ which is $x$ for $x \in (-1/2,1/2]$. Is there some function $D(a,b)$ of real numbers $a TITLE: Quadrics in the Grothendieck ring QUESTION [10 upvotes]: Let $\mathcal{Q}$ be an irreducible quadric in $\mathbb{P}^n(k)$, with $n \geq 2$ and $k$ a finite field. Let $K_0(V_k)$ be the Grothendieck ring of $k$-varieties. It is well known (it appears) that the class $[\mathcal{Q}]$ in $K_0(V_k)$ is contained in $\mathbb{Z}[\mathbb{L}]$, where $\mathbb{L} = [\mathbb{A}^1(k)]$. My question is: what is an easy (elementary) way to prove this rigorously ? The more proofs the better !! REPLY [9 votes]: Edit. Following Remy van Dobben de Bruyn's excellent suggestion, I clarified the use of "irreducible quadrics of dimension $0$." Daniel Loughran's observation about Chevalley-Warning is the key to one solution of this problem. Let $k$ be a finite field, or any quasi-algebraically closed field, or even just a field with $u$-invariant $\leq 2$. This means that every quadric $\mathcal{Q}_n$ in $\mathbb{P}^n$ with $n\geq 2$ has a $k$-rational point. Proposition. For every quadric hypersurface $\mathcal{Q}_n$ in $\mathbb{P}^n$, $n\geq 1$, there exists a field extension $K/k$ of degree $\leq 2$ (possibly degree $1$, i.e., $K$ equals $k$) such that the class $[\mathcal{Q}_n]$ is in the $\mathbb{Z}[\mathbb{L}]$-module generated by $1$ and the class $[\text{Spec}(K)]$ (this is an Artin motive). Proof. This is proved by induction on $n$. The base case is when $n$ is equal to $1$. Then $\mathcal{Q}_1$ is a finite, affine $k$-scheme. If the $k$-scheme is isomorphic to $\text{Spec}(k\times k)$, then $[\mathcal{Q}_1]$ equals $2$. If $\mathcal{Q}_1$ is isomorphic to $\text{Spec}(k[\epsilon]/\langle \epsilon^2 \rangle)$, then $[\mathcal{Q}_1]$ equals $1$. If $\mathcal{Q}_1$ is neither of these, then $\mathcal{Q}_1$ is isomorphic to $\text{Spec}(K)$ for a quadratic field extension $K/k$. Thus $[\mathcal{Q}_1]$ equals $[\text{Spec}(K)]$. So the proposition holds for $n$ equal to $1$. By way of induction, assume that $n\geq 2$, and assume that the result is proved for all quadric hypersurfaces of dimension smaller than $n-1$. First, consider the case when $\mathcal{Q}_n$ is everywhere nonreduced. If the characteristic is different from $2$, then $\mathcal{Q}_n$ is the zero scheme of the square of a linear equation, i.e., a hyperplane with multiplicity $2$. In this case, $$[\mathcal{Q}_n] = [\mathbb{P}^{n-1}_k] = 1+\mathbb{L}+ \dots + \mathbb{L}^{n-1}.$$ Thus, without loss of generality, assume that $\mathcal{Q}_n$ is not a double plane. If the characteristic equals $2$, since the $u$-invariant is $\leq 2$, there is only one other possibility with $\mathcal{Q}_n$ everywhere nonreduced: $\mathcal{Q}_n$ might be the zero scheme of a linear combination of precisely two squares of linear equations. In this case, the common zero locus $\Lambda_{n-2}$ of the $2$-linear equations is a linear space of dimension $n-2$. Moreover, $\mathcal{Q}_n$ is a cone with vertex $\Lambda_{n-2}$ over an everywhere nonreduced quadric $\mathcal{Q}_1'$ in $\mathbb{P}^1$. Similarly, if $\mathcal{Q}_n$ is reduced yet singular, then the singular locus is a linear space $\Lambda_m$ of some dimension $m$, and $\mathcal{Q}_n$ is a cone with vertex $\Lambda_m$ over a smooth quadric hypersurface $\mathcal{Q}'_{n-m-1}$ in $\mathbb{P}^{n-m-1}$. In both of these cases, the reduced case and the nonreduced case, since $\mathcal{Q}_n$ is a cone, we have an identity in the Grothendieck ring of varieties, $$[\mathcal{Q}_n] = [\Lambda_m]+ [\mathcal{Q}_n\setminus \Lambda_m] = $$ $$[\Lambda_m] + [\mathbb{A}^{m+1}_k]\cdot [\mathcal{Q}'_{n-m-1}] = $$ $$(1+\mathbb{L} + \dots + \mathbb{L}^m) + \mathbb{L}^{m+1}\cdot [\mathcal{Q}'_{n-m-1}].$$ Therefore $[\mathcal{Q}_n]$ is a $\mathbb{Z}[\mathbb{L}]$-linear combination of the class $1$ and the class $[\mathcal{Q}'_{n-m-1}]$. By the induction hypothesis, there exists a field extension $K/k$ of degree $\leq 2$ such that $[\mathcal{Q}'_{n-m-1}]$ is in the $\mathbb{Z}[\mathbb{L}]$-submodule generated by $1$ and $[\text{Spec}(K)].$ Thus, from the identity above, also $[\mathcal{Q}_n]$ is in this submodule. So the result is proved in this case. Next assume that $\mathcal{Q}_n$ in $\mathbb{P}^n$ is smooth with $n\geq 3.$ Since $k$ has $u$-invariant $\leq 2$, there exists a $k$-rational point $p$ in $\mathcal{Q}_n$. Projection away from $p$ is birational, with the tangent hyperplane section $Y = H_p\cap \mathcal{Q}$ being equal to a cone over a quadric $\mathcal{Q}''_{n-2}$ in a hyperplane $\Lambda_{n-2}$ inside $\mathbb{P}^{n-1}$. In fact, after base change to $\overline{k}$, it is straightforward to compute that $\mathcal{Q}''_{n-2}$ is smooth and integral. Thus, again we have an identity in the Grothendieck ring, $$[\mathcal{Q}_n] = [\mathbb{P}^{n-1}\setminus \Lambda_{n-2}] + [Y] = $$ $$(\mathbb{L}^{n-1}) + \left( 1 + [\mathcal{Q}''_{n-2}]\cdot \mathbb{L} \right) = $$ $$ 1 + \mathbb{L}^{n-1} + [\mathcal{Q}''_{n-2}]\cdot \mathbb{L}.$$ By the induction hypothesis, there exists a field extension $K/k$ of degree $\leq 2$ such that $[\mathcal{Q}''_{n-2}]$ is in the $\mathbb{Z}[\mathbb{L}]$-submodule generated by $1$ and $[\text{Spec}(K)]$. Thus, from the identity above, also $[\mathcal{Q}_n]$ is in this submodule. So the result is proved in this case. Finally, for a smooth quadric hypersurface $\mathcal{Q}_2$ in $\mathbb{P}^2$, the argument is almost the same as in the previous paragraph. The only difference is that $Y$ is already a $k$-point with multiplicity $2$. So that gives the identity, $$[\mathcal{Q}_2] = [\mathbb{P}^{1}\setminus \Lambda_{0}] + [Y] = $$ $$\mathbb{L} + 1.$$ For $K/k$ equal to the identity field extension, we again conclude that $[\mathcal{Q}_2]$ is in the $\mathbb{Z}[\mathbb{L}]$-submodule generated by $1$ and $[\text{Spec}(K)].$ Thus, in every case, there exists a field extension $K/k$ of degree $\leq 2$ such that $[\mathcal{Q}_n]$ is in the $\mathbb{Z}[\mathbb{L}]$-submodule generated by $1$ and $[\text{Spec}(K)].$ Therefore the proposition is proved by induction on $n$. QED Note. It seems that this argument proves that for every field $k$ with $u$-invariant $\leq r$, for every $n\geq 1$, for every quadric hypersurface $\mathcal{Q}_n$ in $\mathbb{P}^n$, $n\geq r$, there exists a quadric hypersurface $\mathcal{Q}'_m$ with $m\leq r-1$ such that $[\mathcal{Q}_n]$ is in the $\mathbb{Z}[\mathbb{L}]$-submodule generated by $1$ and $[\mathcal{Q}'_m]$.<|endoftext|> TITLE: What is the essence of the constant factor in the standard definitions of the discriminant? QUESTION [6 upvotes]: Let $f(x) = x^m+\sum_{j=0}^{m-1}f_{m-j}x^j\in P[x]$ be a monic polynomial over a field $P$ and let $f(x) = (x-\alpha_1)\cdot\ldots\cdot(x-\alpha_m)$ be a factorization of $f$ over an extension field $Q$ of $P$. Then it is quite natural to consider a value (called discriminant) $$\prod_{1\leq i TITLE: Is a flattening rank a lower bound for the border rank? QUESTION [5 upvotes]: Suppose $T \in V_1 \otimes \cdots \otimes V_k$ is a tensor, where each $V_i$ is a finite dimensional complex vector space. A $1$-flattening (or a flattening) is a realization of $T$ as a matrix in the space of matrices in $k$ essentially different ways as follows: \begin{equation} V^*_{i} \rightarrow V_{1} \otimes \cdots \otimes V_{k-1}. \end{equation} Is it true that the rank of a flattening of $T$ is always a lower bound for the border rank of $T$? If not in general, is it true when $T$ is a symmetric tensor? REPLY [8 votes]: Yes, the flattening rank is a lower bound for border rank. First note that flattening rank is a lower bound for rank. If $T$ is a decomposable tensor (simple tensor, rank one tensor) then every flattening of $T$ has rank one. If $T$ has rank $r$, then $T$ is a sum of $r$ decomposable tensors, and so every flattening of $T$ is a sum of the corresponding $r$ flattenings of decomposable tensors. Each of those flattenings has rank $1$. A sum of $r$ linear maps of rank $1$ has rank at most $r$, so every flattening of $T$ has rank at most $r$. Now, the set of tensors of border rank less than or equal to $r$ is the Zariski closure of the set of tensors of rank $r$. (A digression: This is the $r$th secant variety of the variety of decomposable tensors, i.e., a Segre variety—at least if you set the $k$th secant variety of a variety $X$ to be the closure of the union of $(k-1)$-planes spanned by $k$ points of $X$. In this numbering the "ordinary" secant variety of lines is the $2$nd secant variety. ... At some point in writing this I have rather glossed over any distinction between affine varieties, such as sets of tensors, and projective varieties. But it doesn't matter much for discussions of rank—all the sets of tensors here are cones, so they might as well be projective varieties.) On the other hand, the set of tensors with flattening rank less than or equal to an integer $k$ is also a Zariski closed set. Indeed it is defined precisely by the vanishing of the $(k+1) \times (k+1)$ minors of the flattenings. (I am being intentionally vague as to which flattenings. You may decide if you want to consider $1$-flattenings only, or all flattenings $V_I^* \to V_{[n]\setminus I}$ for $I \subset [n]$, where $V_I = \bigotimes_{i \in I} V_i$, etc.) Putting it all together, the set of tensors of (ordinary) rank less than or equal to $r$ is contained in the set of tensors of flattening rank less than or equal to $r$. The latter is Zariski closed. Therefore the Zariski closure of the first set, the set of tensors of border rank less than or equal to $r$, is also contained in the set of tensors of flattening rank less than or equal to $r$. This is just to say that if $T$ has border rank $r$ (or less), then the flattening rank of $T$ is also less than or equal to $r$. Hence yes, as you asked, the flattening rank of $T$ is less than or equal to the border rank of $T$. It is also true for symmetric tensors with (symmetric) border rank, either by the same argument, or by an inequality between symmetric border rank and (ordinary, non-symmetric) tensor border rank. Some introductions include Landsberg, Tensors: Geometry and Applications (bonus: has a ridiculous drawing of a pickle) or Carlini, Grieve, Oeding, Four Lectures on Secant Varieties (no pickles).<|endoftext|> TITLE: Pencils on del Pezzo surfaces QUESTION [7 upvotes]: Let $X$ be the blow-up of $\mathbb{P}^2$ at three general points $p_1,p_2,p_3$, that is a del Pezzo surface of degree six, and let $\pi_i:X\rightarrow\mathbb{P}^1$ be the morphism induced by the projection from $p_i$. Does anyone know a reference for the following classical fact? Any morphism $f:X\rightarrow\mathbb{P}^1$ factors through $\pi_i$ for some $i = 1,2,3$. REPLY [3 votes]: Let me sketch a proof. Let $D$ be the divisor class giving the map $f$. Then $D^2 = 0$ (because $D$ is the pullback of a point on $\mathbb{P}^1$). By adjunction formula $\deg(K_D) = K_X \cdot D$ is negative since $-K_X$ is ample. Therefore, general fiber of $f$ is a rational curve, hence $K_X \cdot D = -2$, hence $D$ is a conic (with respect to the anticanonical embedding of $X$). This means that $D$ is linearly equivalent to one of $C_i$, hence $f$ coincides with one of $\pi_i$.<|endoftext|> TITLE: Bounds on polynomial values QUESTION [7 upvotes]: Assume $f(x)\in\Bbb{R}[x]$ is a polynomial of degree $n$. Question. If $\int_{-1}^1f^2(x)\,dx=1$, is it true that $$\vert f(x)\vert\leq \frac1{\sqrt2}(n+1), \qquad \text{for $\vert x\vert\leq1$}\,\,\,?$$ REPLY [3 votes]: Inequalities between different $L^{p}$ norms of the type $\|q_{n}\|_{p}\leq C_{n}\|q_{n}\|_{q}$, $0 TITLE: Which ideals have standard Hilbert series? QUESTION [8 upvotes]: Let $m$ and $d$ be two positive integers. Consider the polynomial ring $R = \mathbb{C}[x_1 , \dots , x_m]$. Let $I$ be an ideal of $R$ generated by a finite family of polynomials of degree $d$, and finally let $H(t)$ be the Hilbert series of $R/I$. I say that the Hilbert series $H(t)$ is standard if there exists an integer $n$ such that $$H(t) = \frac{(1-t^d)^n}{(1-t)^m} \, . $$ My question I am looking for a (ideally, necessary and sufficient) condition on $I$ for its Hilbert series to be standard. An example Consider the case $m=3$ and $d=2$. Then For $I=\langle x_1 x_2 \rangle$, we have $H(t) = \frac{1-t^2}{(1-t)^3}$, so $I$ satisfies the condition. For $I=\langle x_1 x_2 , x_1 x_3 \rangle$ we have $H(t) = \frac{t^3-2 t^2+1}{(1-t)^3}$ and this can not be put in standard form, so $I$ does not satisfy the condition. REPLY [2 votes]: I assume you mean that I is generated by homogeneous polynomials of degree d. If so, a necessary and sufficient condition is that I be generated by a regular sequence of n polynomials of degree d. See Corollary 3.3 in Stanley (Advances in Math. 28 (1978) 57-83).<|endoftext|> TITLE: The role of univalence in the homotopy interpretation of type theory QUESTION [11 upvotes]: In Martin-Löf type theory with identity eliminator $$ J : \prod_{B:\prod_{x,y:A}(x=y)\to\mathcal{U}}\left( \prod_{x:A}B(x,x,\mathrm{refl}_x)\to \prod_{x,y:A}\prod_{p:x=y}B(x,y,p) \right) $$ satisfying $J(B,b,x,x,\mathrm{refl}_x)=b(x)$ we can have terms $p:x=x$ that are not equal to $\mathrm{refl}_x$. We can then interpret the $x:A$ as points of a space, and the terms of $x=y$ as the paths joining $x$ and $y$. The loops $p:x=x$ that are not $\mathrm{refl}_x$ are interpreted as non-contractible loops. Terms of type $p=_{x=y}q$ are homotopies between paths, etc. Now, I have seen that the univalence axiom is of great importance in homotopy type theory, but I don't see how it enters in this discussion. The question is: does univalence have consequences in the homotopy interpretation of type theory? If the answer is yes, what role does it play? REPLY [17 votes]: Whenever you’re looking at a logical system, there’s a tension between two main ways of studying it: axioms/theorems in the system show what the world it describes must look like; models show what the world it describes can look like. The interpretation of types as spaces shows that the types of plain Martin-Löf type theory can look like homotopically non-trivial spaces (unlike e.g. the sets of ZF(C), which are always homotopically discrete). However, in plain Martin-Löf type theory, you can’t prove that there must exist homotopically non-trivial types. It’s consistent that every type “is a set”, in the sense of being homotopically discrete. This follows from the fact that you can also model types as plain old sets, in ZF(C) or any similar theory. Univalence implies that there must exist homotopically non-trivial types. Specifically, once you add both univalence and higher inductive types, then (as shown in the HoTT book, and various papers on synthetic homotopy theory) you can reproduce many standard constructions of spaces from classical homotopy theory, and show that they behave how you’d expect them to in many ways: e.g. that the fundamental group of the circle is $\mathbb{Z}$, just to name the simplest non-trivial such fact. So from the point of view of the homotopy interpretation of type theory, the rôle of univalence is ensuring that enough homotopically non-trivial types exist for the world to look like a reasonable homotopy-theoretic world. One approach to making this precise is to say that univalence should force the universe (or a hierarchy of universes, or something) to be an object classifier (or classiying family, or something) in the sense of Lurie’s ∞-topos theory — though this idea hasn’t been made precise yet. This isn’t the only rôle of univalence: it’s not Voevodsky’s original motivation, for example. That was roughly, as I understand it, to allow more powerful and natural reasoning about how constructions respect equivalence, because pragmatically one often needs to use such reasoning when formalising mathematics in type theory. But it’s the essential rôle that univalence plays from the point of view of the homotopy-theoretic interpretation.<|endoftext|> TITLE: What is the relation between 2-Gerstenhaber, CohFT, and Gerstenhaber geometrically? QUESTION [11 upvotes]: Background. As we know from Fred Cohen's Thesis, taking homology of the little 2-discs operad $\mathcal{D}_2$ with coefficients in a field of characteristic zero produces the Gerstenhaber operad $\mathcal{Gerst}$. Algebras over that operad have a commutative product, degree $-1$ Lie bracket, and satisfy a compatibility condition analogues to the Poisson identity. As was defined by Kontsevich and Manin, taking homology of the genus zero Deligne-Mumford operad $$\left\{\overline{\mathcal{M}}_{0,d+1}\right\}_d$$ defines a non-unital, genus zero, tree-level reduction of a CohFT (or a Frobenius manifolds equivalently). Finally, consider the real locus of the DM-spaces. This spaces are known to $K(\pi,1)$ and their fundamental group is called the pure cacti group. It plays a rule analogous to that of the braid group in representation theory is has some interesting applications, e.g., for coboundary categories. The sequence $$\left\{\overline{\mathcal{M}}_{0,d+1}(\mathbb{R})\right\}_d$$ for all $d \geq 1$ defines the Mosaic operad of Devadoss, which is made by ''choping and patching " copies of Tamari-Stasheff associahedra. whose homology was famously computed by P. Etingof, A. Henrquies, J. Kamnitzer, and E. Rains in: ''The cohomology ring of the real locus of the moduli space of stable curves". They concludes that algebras over this operad are ''2-Gerstenhaber" which means they have a commutative product and a ''2-Lie" bracket. Question. What is the algebraic relation between 2-Gerstenhaber algebras and Gerstenhaber algebras? What is the relation between the Gerstenhaber operad and the Mosaic operad in geometric terms? (i.e. can I define a map that sends cycles to cycles in such a way that I see the relation in 1?) I believe that there since the mosaic operad can be seen as the (homotopy) fixed points loci of the DM-operad under involution, there should be some kind of a short exact sequence of operads involving orbit spectra and Tate cohomology (like the one that appeared in Westerland) which would explain the relation ... does anybody know if there is any description of such an operad and algebras over it? REPLY [4 votes]: Well, for your question 1 you presumably may ask yourself first about a relationship between (shifted) Lie algebras and Lie 2-algebras. Lie 2-algebras of Hanlon and Wachs can be viewed as $L_\infty$-algebras where all operations except for the ternary one vanish, and this is the only relationship to Lie algebras that I know. I suppose that from this you can extract a result relating 2-Gerstenhaber algebras to a very special kind of homotopy Poisson algebras. I do not think there is more than that.<|endoftext|> TITLE: Is the $E_\infty$-structure on the cochain complex of a $K(G,n)$ readily understandable? QUESTION [10 upvotes]: One way to construct an $E_\infty$-algebra is to consider the cochain complex $C^*(X;M)$ for $X$ a topological space and $M$ a module over some ring $\Lambda$. From what I can recall, the $E_\infty$-algebra structure should contain the homotopy type of $X$. If $X$ is a $K(G,n)$, can the $E_\infty$-structure on $$ C^*(K;\mathbb{Z}) $$ be made explicit? REPLY [9 votes]: There is a very general framework that gives the $E_\infty$ structure on cochains in McClure and Smith's "Multivariable cochains and little $n$-cubes", with formulas comparable to the Alexander-Whitney diagonal. Roughly, the $E_\infty$ structure is given by a collection of operations as follows: (Warning: I am completely ignoring signs here. They are unpleasant, and McClure-Smith do them more carefully.) Every operation takes a number $r$ of input cochains and has a degree $k$. It is multilinear in these $r$ variables. The operations of this type are denoted by $\langle n_1 n_2 \dots n_{k+r}\rangle$, where $1 \leq n_i \leq r$. If two adjacent $n_i$ are equal, or if one of the integers $1 \dots r$ is not among the $n_i$, then the operation is the zero operation. This is a $\Sigma_r$-equivariant set of operations: permuting the inputs has the same effect as permuting $\{1,\dots,r\}$. For example, the nonzero binary operations are: $\langle 12 \rangle, \langle 21 \rangle, \langle 121 \rangle, \langle 212 \rangle, \langle 1212 \rangle, \langle 2121 \rangle, \dots$, whereas $\langle 236154212\rangle$ is a degree-3 operation of six inputs. For any space $X$, these operations act on the singular cochain complex of $X$ roughly by taking a simplex $[v_0, \dots, v_n]$ and summing up over the ways to: break it up into $k+r$ overlapping partitions, reassemble the pieces corresponding to each number $1,\dots,r$ into a simplex, and feed those simplices to the input cochains. For example, the operation $\langle 12\rangle$ applied to $(f,g)$ gives an operation that partition simplices into two pieces, gives the first half to the input number 1 ($f$), and the second half to input number 2 ($g$). This is the ordinary Alexander-Whitney cup product: if $f$ is a $p$-cocycle and $g$ is a $q$-cocycle, $\langle 12\rangle (f,g)$ is the cocycle that evaluates $f$ on simplex $[v_0,\dots,v_p]$ and evaluates $g$ on the simplex $[v_p,\dots,v_{p+q}]$, then multiplies them together. This is called $f \smile g$. The operation $\langle 2 1\rangle(f,g)$ is an operation that partitions into two pieces, gives the first half to $g$, and the second half to $f$. This is the cup product of $g \smile f$. The operation $\langle 1 2 1\rangle(f,g)$ is an operation that sums up the ways to divide into three pieces, give the first and last to $f$, and the middle to $g$. More explicitly, it sums up all the ways to divide $[v_0,\dots,v_{p+q-1}]$ into $[v_0,\dots,v_i], [v_i,\dots,v_{i+q}],[v_{i+q},\dots,v_{p+q-1}]$, evaluate $f$ on the simplex $[v_0,\dots,v_i,v_{i+q},\dots,v_{p+q-1}]$, evaluate $g$ on the simplex $[v_i,\dots,v_{i+q}]$, and multiply them together. This is called the cup-1 product $f \smile_1 g$ and it satisfies $$ \delta(\langle 121 \rangle (f,g)) = \langle 121 \rangle (\delta f, g) + \langle 121 \rangle (f, \delta g) + \langle 21 \rangle (f, g) + \langle 12 \rangle (f,g) $$ (again, up to signs that I'm not writing). There is a systematic formula for what $\delta$ applied to one of these operations gives: you sum up the ways to apply $\delta$ to the inputs, and then sum up the ways to first delete one of the numbers in $\langle n_1 \dots n_{k+r}\rangle$ and then apply it to the $r$ inputs. There is also a systematic formula for how these operations compose, making them into an operad. These operations work for the singular cochains of any space, or more generally for the cochains on a simplicial set. If you like, there are simplicial models for the $K(G,n)$'s, for example by doing symmetric products of simplicial models of Moore spaces, and at least some of these go back to classical work of Eilenberg and Mac Lane. These tend to look very similar to the "bar complex" of $G.$<|endoftext|> TITLE: The action of the mapping class group of a punctured disk on the boundary at infinity of the universal cover QUESTION [5 upvotes]: Let $\mathbb{D}\subset\mathbb{C}$ be the unit disk, and remove $n\geq 2$ of its points $P$. The resulting object will be called the punctured disk $\mathbb{D}_n$ in the following. I am interested in its mapping class group, that is, the isotopy classes of orientation-preserving self-homeomorphisms/self-diffeomorphisms $\phi:\mathbb{D}_n\rightarrow\mathbb{D}_n$ which fix the boundary, and permute the points in $P$, where isotopy is defined as homotopy 'through' such homeomorphisms. I can equip $\mathbb{D}_n$ with a hyperbolic metric so that the boundary is a geodesic in that metric, and the punctures are cusps. The metric can be lifted to the universal cover $\widetilde{\mathbb{D}_n}$, which can then be embedded into the hyperbolic plane $\mathbb{H}^2$. Specifically, I am interested in how the mapping class group acts on geodesic rays in the universal cover. It is apparently 'well-known' that the mapping class group acts on part of the boundary at infinity of $\mathbb{H}^2$, which is given by equivalence classes $\Gamma$ of geodesic rays, where two geodesic rays are considered equivalent iff they stay a bounded distance apart/are asymptotic. This is used in proving that the braid-groups (which are just the above mapping class groups) admit uncountably many left-orderings by an approach due to Thurston, but the action is never given explicitly. See for example page 113 in this paper. (The paper also has a nice picture of the embedding $\widetilde{\mathbb{D}_2}\subset\mathbb{H}_2$ on page 114.) I am trying to construct this action explicitly, rather than working with high-level arguments. What I tried so far: Fix a point $x\in\mathbb{D}_n$, and a point $y\in\widetilde{\mathbb{D}}_n$ with $p(y)=x$, where $p$ is the covering map. Now I want to create the action: Take an element $[\phi]$ of the mapping class group, and a geodesic ray $\Gamma:[0,\infty)\rightarrow\widetilde{\mathbb{D}}_n$ starting at $y$. The first guess is $[\phi].[\Gamma]:=[\widetilde{\phi}\circ\Gamma]$, where $\widetilde{\phi}$ is the unique lift of $\phi\circ p$ to the universal cover such that $\widetilde{\phi}(y)=y$. This doesn't work though, because $\widetilde{\phi}\circ\Gamma$ need not be a geodesic. Next I thought that maybe the lift $\widetilde{\phi}$ as above is a quasi-isometry, which would allow me to conclude that $\widetilde{\phi}\circ\Gamma$ is a quasi-geodesic ray, which I could then associate to a unique geodesic ray since I am operating in hyperbolic space. According to the answer to this question $\widetilde{\phi}$ need not be a quasi-isometry though. But the answer says there is a fix, where I can construct a representative homeomorphism $\psi$ of each isotopy class which has a quasi-isometry $\widetilde{\psi}:\mathbb{D}_n\rightarrow\mathbb{D}_n$ as lift, and this would give me what I need to construct my action. Sadly, the construction is only sketched and I don't have the knowledge to do it myself, so my first question would be if someone could please elaborate on the sketch given there in the comments. It goes like this: 1. Delete the interior $(0,\infty)\times S_1$ of the cusps. 2. Find a representative which permutes the resulting circles by isometries of circles. 3. Apply Milnor Svarc to the universal cover of the deleted object. 4. Extend over cusps by isometries. I have no idea how to make this precise, but assume for the moment that I accept this result. I assume that I can then explicitly state my action as $[\phi].[\Gamma]=[\widetilde{\phi}\circ\Gamma]$ where I assume w.l.o.g. that $\phi$ lifts to a quasi-isometry. I then still have to show well-definedness of $[\widetilde{\phi}\circ\Gamma]$ for different representatives of the same class which lift to quasi-isometries, which brings me to my second question: Is $\pi_1(\mathbb{D}_n)$ quasi-isometric to $\widetilde{\mathbb{D}}_n$? If I knew that this was true, then I came up with a proof (won't include it here because the post is already so long). But I still need to know whether the statement is true for my proof to hold. I know that at least for the deleted object the analogous statement holds, because the deleted object is compact and thus the fundamental group acts geometrically on the universal cover by deck-transformations. This means that I can apply the Milnor-Svarc-Lemma which states that if a group $G$ acts cocompactly, properly discontinuously, and by isometries on a nice-enough metric space, then group and space are quasi-isometric. But $\mathbb{D}_n$ isn't compact (the points are removed), so I can't apply Milnor-Svarc like I'd like to. Now if somebody could explain to me How to construct the representative as in the sketch in a little more detail and/or Tell me if/why $\pi_1(\mathbb{D}_n)$ is quasi-isometric to $\widetilde{\mathbb{D}}_n$. This is apparently wrong. Most importantly: Since 2) isn't the case, how can I assure well-definedness of $[\phi].[\Gamma]=[\widetilde{\phi}\circ\Gamma]$ for different representatives $\phi_1,\phi_2$ from the same isotopy class and which both lift to quasi-isometries? Or if this doesn't work, how can I construct the action instead? I would be very grateful for answers. Edit 2: An idea regarding 3). Let $D_n$ denote the object I obtain by deleting very small circles around the punctures of $\mathbb{D}_n$. Is it true that the mapping class group of $D_n$ is the same as that of $\mathbb{D}_n$? If this were so, maybe then for any homeomorphism $\phi:D_n\rightarrow D_n$ its lift $\widetilde{\phi}:\widetilde{D}_n\rightarrow\widetilde{D}_n$ would be a quasi-isometry (why?), and because $D_n$ is compact I could use my earlier proof with Svarc-Milnor that relies on $\pi_1(D_n)\overset{quasi-isometric}{\simeq}\widetilde{D_n}$ to show well-definedness of $[\phi].[\Gamma]=[\widetilde{\phi}\circ\Gamma]$. Edit 1: I meant to write $\widetilde{\mathbb{D}}_n$ instead of $\mathbb{D}_n$ in 2). REPLY [2 votes]: This is an exercise. Your second step is the part that will require some thought. For a related discussion, see Rolfsen's book where he classifies simple closed curves in the annulus. They are not quasi-isometric. The fundamental group of a hyperbolic surface with punctures (and perhaps with boundary) is never isometric to its universal cover. This is because the cusps introduce distortion: a group element that winds about the cusp can be geometrically shortened (by entering the cusp) but cannot be algebraically shortened. If $\phi_1$ and $\phi_2$ are a pair of representatives, then there is an isotopy $F$ between them. We have to choose $F$ to fix $x$, to respect the system of peripheral circles, and near the punctures be an isometry. That is, every $F_t$ is a homeomorphism of the same quality as the $\phi_i$ are. Building such an $F$ is an exercise similar to, but harder than (1) above. Now, consider $\tilde{F}$, the lift of $F$ that fixes $y$. You ask in the comments if the punctured disk and the disk with disks removed have the same mapping class group. This depends on how you define the mapping class group. Must isotopies fix boundaries pointwise? Or is each boundary equipped with a marked point, and the set of points is fixed setwise? Or perhaps there is no condition? See the book by Farb-Margalit for a discussion.<|endoftext|> TITLE: Homotopy theoretic description of homotopy fixed points (and obstructions) for an action of group $G$ on a groupoid $X$ QUESTION [8 upvotes]: There are several scattered statements about fixed points and obstructions which I'd very much like to see unified in some framework. To state them let $G$ be a group acting on a connected (1-truncated) groupoid $X$. Firstly let's choose a point in $X$ and take $X=BAut(\pi_1(X))=:BA$. For the purpose of this question i'd like to find a unified homotopy theoretic approach for the following statements: There's an obstruction in $H^2(G,A)$ for the existence of homotopy fixed points. When the obstruction vanishes $\pi_0(X^{hG})\cong H^1(G,A)$ I'm hoping both of these statements can be explained using arguments about the following fiber sequence (and perhaps some additional close constructions): $$BA \to E \to BG$$ Where $E$ is the total space of the $A$-gerbe corresponding to the $G$ action on $X$. Said differently we have by assumption a map $G \to Aut(BA)$ which we can $B(-)$ to get $BG \to BAut(BA)$ which we can use to pullback the universal fibration $BA \to BAut_*(BA) \to BAut(BA)$ to get the above fiber sequence. Ideally both $H^2(G,A)$ and $H^1(G,A)$ will appear in the same long exact sequence (/spectral sequence) of homotopy groups for some fiber sequence. REPLY [4 votes]: I'm not sure exactly what kind of answer you're looking for, but I can try to give some context which may make things sound more reasonable. Let us think of of groupoids as $1$-truncated $\infty$-groupoids, or $1$-truncated spaces. Consider first the case where $A$ is abelian. In this case $BA$ will itself be an $\mathbb{E}_\infty$-group object in spaces, and will have its own classifying space $B^2A$ (which is no longer $1$-truncated, it's a $K(A,2)$). Since the formation of classifying spaces naturally lands in pointed spaces it follows that an arbitrary action of $G$ on $BA$ yields a base-point preserving action of $G$ on $B^2A$. Such a data can be geometrically realized as a fibration $p:F \to BG$ together with an identification $p^{-1}(x_0) \simeq B^2A$ of the fiber over the base point with $B^2A$ and equipped with a $0$-section $s_0:BG \to F$. Furthermore, $F$ will actually form an $\mathbb{E}_\infty$-group object in spaces over $BG$. We note that while we constructed $F$ out of the action of $G$ on $BA$, it actually only depends on the induced action of $G$ on $A$ (something that is well-defined when $A$ is abelian): indeed, as an $\mathbb{E}_\infty$-group object over $BG$, $F$ is just the double delooping of the local coefficients system $BG \to \mathcal{Ab}$ corresponding to the action of $G$ on $A$. The space of sections $\mathrm{Map}_{BG}(BG,F)$ of $F$ is then the space which "represents" the first three cohomology groups of $G$ with coefficients in $A$: namely, $\pi_i\mathrm{Map}_{BG}(BG,F) \cong H^{2-i}(G,A)$ for $i=0,1,2$. We now turn to the fibration $E \to BG$ that you mention, which is associated to the action of $G$ on $BA$. What is the relation between $E$ and $F$? if the action of $G$ on $BA$ had a homotopy fixed point then $E \to BG$ would have a section and we could think of $F$ as a fiberwise delooping of $E$, or, alternatively, identify $E$ with the fiberwise looping of $F$, i.e., with the homotopy fiber product $BG \times^h_F BG$. In the general case (where we don't assume that there is a base point) one can show that a similar thing happens, only twistedly. Instead of taking the homotopy fiber product of $BG$ with itself over $F$, where the map from $BG$ to $F$ is the $0$-section in both cases, one can show that there actually exists (an essentially unique) other section $s: BG \to F$ such that $E$ is equivalent to the homotopy fiber product $BG \times^h_{F} BG$, where now one copy of $BG$ maps to $F$ via $s$ and the other copy via the $0$-section $s_0$. Roughly speaking, we may consider $E$ as the space of pairs $(x,\eta)$ where $x$ is a point of $BG$ and $\eta$ is path in the fiber $F_x$ from $s(x)$ to the base point $s_0(x) \in F_x$. The section $s$ then determines an element in $\pi_0(\mathrm{Map}_{BG}(BG,F)) \cong H^2(G,A)$, which vanishes if and only if $E \to BG$ admits a section (i.e., if and only if $BA$ has a homotopy fixed point). A choice of such a section is essentially the same as an identification of $E \to BG$ with the fiberwise looping of $F \to BG$, in which case the full space of sections satisfies $\pi_i(\mathrm{Map}_{BG}(BG,E)) \cong H^{1-i}(G,A)$ for $i=0,1$. This story is quite general, it will work essentially the same if we replace $A$ by any $\mathbb{E}_{\infty}$-group object, or even a spectrum, equipped with an action of $G$ (the cohomology of $G$ with coefficients in such a thing is exactly what you think it should be). Of course, everything works much less well when $A$ is not abelian. First of all, you might ask yourself what do we even mean by $H^i(G,A)$ when $A$ is not abelian. I have to say I don't really have a good answer for this. There exists an explicit definition using cocycles, which for $i=0,1$ requires an honest action of $G$ on $A$ and for $i=2$ only requires an outer action of $G$, namely, a homomorphism $G \to Out(A)$. In this definition $H^0(G,A)$ is a group (the fixed subgroup of $A$ under the action of $G$), $H^1(G,A)$ is a pointed set and $H^2(G,A)$ is a set with a collection of "neutral elements". This definition enjoys some basic properties: it reduces to the usual definition when $A$ is abelian, and admits some useful (not-so-long) long exact sequences. One can also show that, like in the abelian case, the elements of $H^2(G,A)$ are in one-to-one correspondence with extensions of $G$ by $A$ which are compatible with the given outer action, where the neutral elements correspond to the split extensions (note that there can be several split extensions of $G$ by $A$ with the same outer action). In particular, if we are given an action of $G$ on $BA$ then the corresponding fibration $E \to BG$ determines a class in $H^2(G,A)$ via the group extension $ 1 \to A \to \pi_1(E) \to G \to 1$. The corresponding element is neutral if and only if this sequence splits, which is equivalent to $E \to BG$ having a section. A choice of such a section/splitting determines in particular a lift of the outer action of $G$ on $A$ to an honest action, and hence determines a (non-abelian) local coefficient system $BG \to \mathcal{Grp}$. One may then identify $E \to BG$ with the fiberwise delooping of this local coefficient system and show that, as in the abelian case, $\pi_i\mathrm{Map}_{BG}(BG,E) \cong H^{1-i}(G,A)$ for $i=0,1$. It is worth noting that even though the unabelian $H^2(G,A)$ is not an especially handsome invariant (which makes the obstruction theory above sounds a bit tautological at first), it can occasionally give useful and computable information, and more importantly, it can be shown to vanish under various assumptions. For example, if the map $Aut(A) \to Out(A)$ admits a section and the center $Z(A)$ vanishes (like if $A$ is the alternating group on at least 5 elements) then $H^2(G,A)$ always vanishes. One may then deduce that any action of $G$ on $BA$ admits a homotopy fixed point, a statement that is far from obvious at first sight. Edit: My remark that everything works the same for $A$ an arbitrary $\mathbb{E}_{\infty}$-group should be reserved a bit. This is true if we consider actions of $G$ on $BA$ which are affine-linear, i.e., that factor through the group $\mathrm{Aff}(BA) = BA \rtimes \mathrm{Aut}_{\mathbb{E}_{\infty}}(BA)$ of affine-linear transformations. Given such an action $\alpha: G \to \mathrm{Aff}(BA)$ the projected map $\overline{\alpha}:G \to \mathrm{Aut}_{\mathbb{E}_{\infty}}(BA)$ determines the space $F$ by the homotopy fiber product $F = BG \times^h_{B\mathrm{Aut}_{\mathbb{E}_{\infty}}(BA)} B\mathrm{Aff}(BA)$ while the lift $\alpha$ of $\overline{\alpha}$ determines the section $s: BG \to F$ mentioned above. In the case where $A$ is a discrete abelian group we have an equivalence $\mathrm{Aut}(BA) \simeq \mathrm{Aff}(BA)$, and so we can consider arbitrary actions of $G$ on $BA$, but I don't think this is true in general.<|endoftext|> TITLE: 3D Billiards problem inside a torus QUESTION [39 upvotes]: I have been trying to simulate the behavior of a light particle being reflected inside of a torus (essentially a 3D billiards problem). I have found that after a few thousand bounces, it converges on "rolling" behavior; that is, the bounces get shorter and shorter and it essentially sticks to the surface. The simulation basically can't progress beyond this point as a result. Below is the part of the trajectory that illustrates this behavior; the trajectory goes from chaotic to convergent along the surface of the torus: On the left can be seen the discrete bounces, and on the right is where the particle sticks to the surface (until program termination). It seems counter-intuitive that a the trajectory could become trapped like this after demonstrating chaotic behavior for thousands of prior bounces. Closer examination reveals that the angles become more and more grazing with each bounce, and the distances between bounces shorter, until no progress can be made. If I had infinite numerical precision to peek past this event horizon, would I eventually find that the particle breaks free again? Or is it possible that this is also a theoretical limit, and the motion of the particle has transitioned from finite segments to a continuous curve constrained to the surface? Can this happen in 3D billiards problems, or does chaos dictate that this behavior would reverse after some unknown amount of time? REPLY [16 votes]: As to the fact that this improbable scenario arose from chaos, it seems that numerical error really was at fault (my bisection code was faulty). This behavior is more to be expected. EDIT: According to the literature, as Carlo Beenakker pointed out, such "whispering gallery modes" wherein trajectories can stay arbitrarily close to the surface can only exist if they trace out a curve with strictly concave curvature everywhere -- a condition often violated in a torus. This, together with $C^6$ regularity, are a sufficient condition for the existence of such caustics. Quite an improvement over the previous result by Lazutkin demanding $C^{553}$ regularity. Source: https://mat-web.upc.edu/people/rafael.ramirez/res/pdf/oberwolfach10.pdf<|endoftext|> TITLE: polynomial maps from reducible plane curves to conics QUESTION [5 upvotes]: It is classically known that every smooth plane quartic curve $C$ can be represented by an equation $q_1 q_3 = q_2^2,$ with $q_j\in\mathbb{C}[X,Y,Z]$, $1\leq j\leq 3$ quadratic forms, and the same is true for $C$ being the union of two smooth conics, cf. e.g. C.T.C.Wall's "Is every quartic a conic of conics?" How about the union of a smooth conic and a smooth quartic, can it be represented by $q_1 q_3 = q_2^2,$ with $q_j\in\mathbb{C}[X,Y,Z]$, $1\leq j\leq 3$ cubics? Edit. I do not know whether smoothness assumptions on the quartic and/or on the conic can be completely dropped without changing the outcome. E.g. in a similar setting of the union of a line and a cubic, a representation of the union by $q_1 q_3 = q_2^2,$ (with $\deg q_j=2$) need not exist if the cubic is cuspidal, cf. [loc.cit.]. There is also a natural counterpart to this for $\mathbb{R}$ instead of $\mathbb{C}$, where in the context of positive forms one should talk about the conics and the quartic represented by sums of 3 squares: let $a_i, c_i, b_i\in\mathbb{R}[X,Y,Z]$ be forms. Is it true that for any such $a_i$ of degree 1 and $c_i$ of degree 2, with $1\leq i\leq 3$, there always exist $b_i$ of degree 3 so that $$(a_1^2+a_2^2+a_3^2)(c_1^2+c_2^2+c_3^2)=b_1^2+b_2^2+b_3^2$$ holds? Again, this is classically known to hold for $a_i$ and $c_i$ of degree 1 and $b_i$ of degree 2, the case extensively studied recently too, see e.g. C.Scheiderer's "Hilbert's theorem on positive ternary quartics : a refined analysis". REPLY [4 votes]: Edit. Mistake in the analysis at end of answer. When I computed how $F$ extends to the blowing up, I ignored a particular subvariety contained in the exceptional locus that contains the intersection of the two exceptional divisors. That is a mistake. In fact, there is an affine curve in the image of $F$ that limits to the special point, yet whose lift to $U\otimes_{\mathbb{C}} S_3$ limits to the special subvariety, namely the image under $F$ of the following curve, $$(q_1(t),q_2(t),q_3(t)) = $$ $$t^0(X_1^3,X_1^2(X_2-X_1),X_1(X_2-X_1)^2) +$$ $$ t^1(2X_1(3X_2^2-3X_1X_2+X_1^2), (X_2-X_1)(3X_2^2-4X_1X_2+2X_1^2),2(X_1-X_2)^3) + $$ $$t^2(0,0,X_3^3).$$ So the argument below with Zariski tangent spaces is not complete. I need to work out the image under the extension of $F$ of this special subvariety. It appears, at first blush, that the reducible curves in the image all have a singular point (in the intersection of the two components) with analytic type $x^2 + y(y-x)(y-\lambda x).$ This is certainly not the analytic type of a typical intersection point of a plane conic and plane quartic (that analytic type is simply nodal). So hopefully I can complete this argument. First edit. There was an issue with the answer below, but this has been addressed by the "analysis of stability" at the end of the answer. The induced morphism of quasi-projective varieties, $$\mathbb{P}F : \mathbb{P}(U\otimes_{\mathbb{C}}S_3) \setminus \text{Zero}(F) \to \mathbb{P}(S_6),$$ is not projective. Thus, I need to prove that the image of my special point, $F(q_1,q_2,q_3),$ is not also a "limit" of an affine curve in the image of $F$ whose inverse image under $F$ is a subvariety of the quasi-projective domain whose closure contains no point mapping to $F(q_1,q_2,q_3)$. That sounds complicated, but it almost exactly the analysis of the GIT unstable locus for the natural action of $\textbf{SO}_\beta$ on $U\otimes_{\mathbb{C}} S_3$. I will try to add this GIT analysis to the end of this problem soon. Original answer. A computation of Zariski tangent spaces disproves this, as in my comment. Sorry for introducing more indices, but it is easier for me to organize the computation if I change the names of your variables. Denote by $S$ the polynomial ring $\mathbb{C}[X_1,X_2,X_3]$. Denote by $S_d$ the direct summand of homogeneous polynomials of degree $d.$ Denote the vector space $\mathbb{C}^{\oplus 3}$ by $U$. Denote by $\beta$ the quadratic form, $$\beta:U \to \mathbb{C}, \ \ (t_1,t_2,t_3) \mapsto t_1t_3-t_2^2.$$ There are two relevant homogeneous polynomial maps of $\mathbb{C}$-vector spaces. First, denote by $F$ the following homogeneous polynomial map of $\mathbb{C}$-vector spaces, $$F:U\otimes_{\mathbb{C}}S_3 \to S_6, \ \ (q_1,q_2,q_3)\mapsto q_1q_3-q_2^2.$$ For me, it is fastest to record the derivative of the map in the following way, $$dF_{(q_1,q_2,q_3)}:U\otimes_{\mathbb{C}}S_3 \to S_6, \ \ (\dot{q}_1,\dot{q}_2,\dot{q}_3) \mapsto q_3\dot{q}_1 -2q_2\dot{q}_2 + q_1\dot{q}_3.$$ This comes from expanding $F(q_1+\epsilon \dot{q}_1,q_2+\epsilon \dot{q}_2,q_3+\epsilon \dot{q}_3)$, and gathering the coefficient of $\epsilon$. Note that the kernel of $dF$ always contains the $3$-dimensional vector subspace that is a copy of $\mathfrak{so}_\beta \cong \mathfrak{so}_3$ inside $U\otimes_{\mathbb{C}}\text{span}(q_1,q_3,q_3) \subset U\otimes_{\mathbb{C}} S_3,$ because $F$ is composed with the quadratic form $\beta.$ Second, denote by $M$ the following homogeneous polynomial map of $\mathbb{C}$-vector spaces, $$M:S_2\oplus S_4 \to S_6, \ \ (b,h)\mapsto b\cdot h.$$ This is just a multiplication map. The derivation of this map equals, $$dM_{b,h}:S_2\oplus S_4 \to S_6, \ \ (\dot{b},\dot{h}) \mapsto h\dot{b} + b\dot{h}.$$ This comes from expanding $(b+\epsilon\dot{b})(h+\epsilon \dot{h})$, and gathering the coefficient of $\epsilon$. Now consider the special point $(q_1,q_2,q_3) = (X_1^3,X_2^3,X_3^3)$. The image under $F$ is, $$F(X_1^3,X_2^3,X_3^3) = X_1^3X_3^3-X_2^6= b(X_1,X_2,X_3)h(X_1,X_2,X_3), $$ $$b(X_1,X_2,X_3) = X_1X_3-X_2^2, \ \ h(X_1,X_2,X_3) = X_1^2X_3^2 + X_1X_3X_2^2 + X_2^4.$$ The derivative map of $F$ at this special point is, $$ dF_{q_1,q_2,q_3}:S_3^{\oplus 3} \to S_6, $$ $$(\dot{q}_1,\dot{q}_2,\dot{q}_3) \mapsto X_3^3\dot{q}_1 -2X_2^3\dot{q}_2 + X_1^3\dot{q}_3.$$ By considering the exponent vectors of monomials, and in particular which monomials are divisible by at least one of $X_1^3$, $X_2^3$, and $X_3^3$, the kernel of the derivative map is precisely the copy of $\mathfrak{so}_\beta$. Thus, choosing an affine linear subspace of $S_3^{\oplus 3}$ containing $(q_1,q_2,q_3)$ and whose tangent space is complementary to this copy of $\mathfrak{so}_\beta$, the restriction of $F$ to this linear subspace is immersive with image (locally) equal to the image of $F$. Thus, to prove that the image of $F$ does not contain the locus of reducible curves, it suffices to prove that the image of $dF_{q_1,q_2,q_3}$ does not contain the Zariski tangent space to the locus of reducible curves. By the form of $dF_{q_1,q_2,q_3}$, the image is precisely the vector space spanned by monomials that are divisible either by $X_1^3$, by $X_2^3$, or by $X_3^3$. On the other hand, the Zariski tangent space to the locus of reducible plane curves contains the image of $$dM_{b,h}:S_2\oplus S_4 \mapsto S_6, \ \ (\dot{b},\dot{h}) \mapsto (X_1^2X_3^2 + X_1X_3X_2^2 + X_2^4)\dot{b} + (X_1X_3-X_2^2)\dot{h}.$$ For $(\dot{b},\dot{h}) = (0,X_1X_3X_2^2)$, among many other possibilities, the image tangent vector is not a polynomial whose monomials are each divisible by $X_1^3,$ by $X_2^3,$ or by $X_3^3.$ Indeed, the image is, $$dM_{b,h}(0,X_1X_3X_2^2) = X_1^2X_2^2X_3^2 - X_1X_3X_2^4.$$ The second term is divisible by $X_2^3$, however, the first term is divisible by none of $X_1^3,$ $X_2^3,$ nor $X_3^3.$ Analysis of Stability. The zero locus of $F$ inside $U\otimes_{\mathbb{C}} S_3$ consists of two irreducible components. The simpler component consists of triples $(a_1,a_1,a_3)\otimes q$ for $(a_1,a_2,a_3)\in U$ an isotropic vector of $\beta$. These triples are all GIT unstable for the natural action of $\textbf{SO}_\beta$ on $U\otimes_{\mathbb{C}} S_3$. That means that when we consider $\mathbb{P}F$ as a rational map on the GIT quotient, $$X = \mathbb{P}(U\otimes_{\mathbb{C}} S_3)^{\text{ss}}/\textbf{SO}_\beta,$$ then these triples are not even in $X$. So we can ignore these triples. The second component is more tricky. It is the image of the morphism, $$e:\mathbb{P}(S_1)\times \mathbb{P}(S_1\oplus S_1) \to \mathbb{P}(U\otimes_{\mathbb{C}}S_3), $$ $$ ([L_2],[(L_1,L_3)]) \mapsto [(L_2L_1^2, L_2L_1L_3,L_2L_3^2)].$$ The image of this morphism is singular along the locus of triples where $L_2$ is in the span of $(L_1,L_3)$. So we need to blow up this locus. Then the strict transform of the image is smooth, but $F$ is still zero on the strict transform. So we again need to blow up the strict transform. Once we do this, $F$ is zero only on points of the exceptional divisors that map into the unstable locus. Removing this locus, $\mathbb{P}F$ extends to a regular morphism. Equivalently, performing the two blowings up in the GIT quotient $X$, the morphism $\mathbb{P}F$ extends to a regular morphism. The domain of this extended morphism is projective. Finally, the restriction of the extension of $\mathbb{P}F$ to each exceptional divisor maps into the locus in $\mathbb{P}(S_6)$ of polynomials that are divisible by a linear factor. Since $X_1^3X_3^3-X_2^6$ is not divisible by a linear factor (it is a product of three irreducible quadratic factors), the image of the special point is not contained in the image of the exceptional divisors. That settles the issue raised at the top of the post.<|endoftext|> TITLE: A small corner w.r.t. a masa in a von Neumann algebra QUESTION [10 upvotes]: Let $A \cong L^\infty[0,1]$ be a non-atomic maximal abelian *-subalgebra in $M \cong B(L^2[0,1])$ (or any von Neumann algebra $M$). Is the following true? For every $T \in M$ and $\epsilon>0$, there are non-zero projections $p,q \in A$ such that $\| qTp \| \le \epsilon \| T \|$. REPLY [10 votes]: This is not true. Here's an example for which it fails: Let $\{ r_n \}_{n \in \mathbb N}$ be an enumeration of the rationals. Construct an orthogonal set $\left\{ f_{j, k}^{l, m} \right\}_{j, k, l, m \in \mathbb N} \subset L^2 (\mathbb R)$ such that each $f_{j, k}^{l, m}$ is valued in $\{ -1, 0, 1 \}$, has only finitely many discontinuity points, and satisfies $\left| f_{j, k}^{l, m} \right| = 1_{(r_j - \frac{1}{k}, r_j + \frac{1}{k})}$. (This is easy enough to do inductively on an enumeration of the quadruples $(j, k, l, m)$). Define a partial isometry $T \in \mathcal B(L^2(\mathbb R) )$ such that $T\left( \sqrt{ \frac{k}{2} } f_{j, k}^{l, m} \right) = \sqrt{ \frac{m}{2} } f_{l, m}^{j, k}$. If we're given $\varepsilon > 0$ and $E, F \subset \mathbb R$ having positive measure then by Lebesgue's density theorem there exists a quadruple $(j, k, l, m)$ such that \begin{equation} \left| \left(r_j - \frac{1}{k}, r_j + \frac{1}{k} \right) \setminus E \right| <\varepsilon^2 \frac{2}{k} \ \ \ {\rm and} \ \ \ \left| \left(r_l - \frac{1}{m}, r_l + \frac{1}{m} \right) \setminus F \right| < \varepsilon^2 \frac{2}{m}. \end{equation} Therefore, $\sqrt{ \frac{k}{2} } \left\| 1_E f_{j, k}^{l, m} - f_{j, k}^{l, m} \right\|_2 < \varepsilon$, and $\sqrt{ \frac{m}{2} } \left\| 1_F f_{l, m}^{j, k} - f_{l, m}^{j, k} \right\|_2 < \varepsilon$. Hence, \begin{equation} \left\| \sqrt{ \frac{m}{2} } f_{l, m}^{j, k} - 1_F T 1_E \left( \sqrt{ \frac{k}{2} } f_{j, k}^{l, m} \right) \right\|_2 < 2\varepsilon, \end{equation} so that $\left\| 1_F T 1_E \right\| = 1$.<|endoftext|> TITLE: Is there a first-order theory who does not interpret arithmetic yet still does not have a computable consistent completion? QUESTION [17 upvotes]: By not interpreting arithmetic, I mean it does not interprets enough arithmetic for Godel's argument (coding the syntax, finding the fix point) to work through. In other words, is there any other methods to prove that a theory does not have a computable consistent complete extension, or can we prove the converse that every such theory interprets arithmetic? Related, is there a class of structures whose first-order theory is not computable and arithmetic is not interpreted in it? REPLY [6 votes]: Another example is the theory $\text{Th}(\mathcal{P}(\mathbb{R}),\subset,<)$ of subsets of the reals, where $S$ is less than $S'$ iff every element of $S$ is less than every element of $S'$. The theory is consistent and complete by definition. Shelah proved that it is not decidable, which means it also has no decidable extension, and Gurevich and Shelah jointly proved that it does not interpret arithmetic. To be more precise, Gurevich and Shelah work with a variant of the theory where $<$ applies only to singletons, and they prove that it does not interpret even the weak set theory of null set, singleton and union: \begin{align} \exists y \forall z &[z \notin y]\\ \forall x \exists y \forall z &[z \in y\leftrightarrow z=x]\\ \forall w \forall x \exists y \forall z &[z \in y\leftrightarrow z \in w \text{ or } z \in x]\\ \end{align} The two results on undecidability and non-interpretability are both difficult, but these questions of mine have more details.<|endoftext|> TITLE: Monotonicity of canonical ellipsoids QUESTION [7 upvotes]: Let $\mathcal{C}$ be the set of compact convex centrally symmetric sets in $\mathbb{R}^d$, and let $\mathcal{E} \subset \mathcal{C}$ be the set of ellipsoids centered at the origin. I'm looking for a mapping $\pi:\mathcal{C}\to\mathcal{E}$ that satisfies the following properties: continuity (with respect to the Hausdorff topology, say); equivariance under linear isomorphisms of $\mathbb{R}^d$, i.e., $$C\in \mathcal{C},\ L \in GL(d,\mathbb{R}) \ \Rightarrow \ \pi(L(C))=L(\pi(C));$$ mononicity, i.e., $$C \subseteq D \ \Rightarrow \ \pi(C) \subseteq \pi(D).$$ The Löwner ellipsoid (meaning the unique ellipsoid of minimal volume containing the given compact set) is continuous and equivariant, but unfortunately it is not monotone: see this MO question and answer. Question: Is there another kind of ellipsoid that satisfies the three properties? Or, if no such construction is known explicitly, can it proved abstractly (feel free to use axiom of choice) that such a map $\pi$ exists? PS: One would expect $\pi$ to be a projection, but I don't need that property. REPLY [11 votes]: No, there is no ellipsoid satisfying equivariance and monotonicity. Suppose we have such an ellipsoid. Consider the square whose corners are $(\pm 1, \pm 1)$. Since this is invariant under a quarter-rotation, so is its ellipsoid, and the ellipsoid must be a circle of radius $r$. By equivariance, we get the following table of shapes: \begin{matrix} \text{Quadrilateral with Corners at} & \text{Semi-axes of corresponding ellipsoid}\\ (\pm 1, \pm 1) & r,\ r \\ (\pm \cos(t), \pm \sin(t)) & r \cos(t),\ r \sin(t) \\ (\pm \sqrt{2}, 0) \text{ and } (0, \pm \sqrt{2}) & r,\ r \\ (\pm \sec(t), 0) \text{ and } (0, \pm \csc(t)) & r\sec(t)/\sqrt{2},\ r\csc(t)/\sqrt{2} \\ \end{matrix} But the second quadrilateral is contained in the fourth. So by monotonicity, $r \cos(t) < r \sec(t) / \sqrt{2}$, and $\cos^2(t) < 1/\sqrt{2}$. We can choose $t$ for which this is false, showing the impossibility. EDIT (by Jairo): I'll illustrate Matt's solution (or a minor variation of it) with a figure. As Matt explained, $\pi$(square) = circle. Multiplying $\pi(.)$ be a constant if necessary, we can assume that $\pi$(square) = inscribed circle. By equivariance, $\pi$(rectangle) = inscribed ellipse. But then monotonicity is violated: the red rectangle is contained in the blue square, but the red ellipse is not contained in the blue circle.<|endoftext|> TITLE: When is a direct limit of nice models of set theory is again a model of set theory? QUESTION [12 upvotes]: It is often useful to allow taking direct limits in set theory. This happens all the time when taking about ultrapowers. But let's not limit ourselves to ultrapowers. Suppose that we have a sequence of models of $\sf ZFC$, $M_n$ for $n<\omega$, such that all of them are transitive and even countable, with the same ordinals. What reasonable assumptions would guarantee that $\bigcup M_n$ is again a model of $\sf ZFC$? What happens if we consider longer sequences of models (where we don't require all the models to satisfy $\sf ZFC$, or we don't require the sequence to be continuous at all limits)? I am particularly interested in the case where each model is a forcing extension of its predecessor in the sequence; but not necessarily just that. For example, one could think that the requirement "eventual stabilization of the von Neumann hierarchy" is enough. But it is not. Consider the following counterexample: Start with a [countable transitive] model of $\sf GCH$, $M_0$ and choose (externally) a cofinal sequence $\kappa_n$ for $n<\omega$ in the ordinals of $M_0$; without loss of generality, $M_0\models``\kappa_n\text{ is a strong limit cardinal}"$. Next define $M_{n+1}$ to be the forcing extension of $M_n$ by $\operatorname{Add}(\kappa_n^+,\kappa_n^{+3})^{M_n}$. It is easily verified that the $M_n$'s are all countable with the same ordinals, and that $M_n\subseteq M_{n+1}$, as well as that the von Neumann hierarchy there stabilizes. However, in $M=\bigcup M_n$, we can look at the function $\varphi(n,\alpha)$ stating that $n<\omega$ and $\alpha$ is the unique ordinal such that $\sf GCH$ is violated $n-1$ times below it. Easily, $\varphi$ actually defines the sequence $\kappa_n$, which is a contradiction to Replacement. On the other hand, the stabilization is necessary for us to get a model of the Power Set axiom. Otherwise, there is some $\alpha$ such that $V_\alpha^{M_n}$ never stabilizes, and therefore never becomes an element of $V_{\alpha+1}$ in the limit model. REPLY [9 votes]: This doesn't exactly answer your question, but I find it to be in a similar spirit. Namely, one could naturally consider the version of your question where you ask merely that $\bigcup_n M_n$ is contained within a model of ZFC, rather than necessarily being a model of ZFC itself. For this version of the question, particularly applied to forcing extensions, I have a characterization in my paper: J. D. Hamkins, Upward closure and amalgamation in the generic multiverse of a countable model of set theory, RIMS Kyôkyûroku, pp. 17-31, 2016. Follow the link for further links to slides of talks I've given on the topic. One theorem appearing in the paper is: Theorem. An increasing chain of forcing extensions $$W\subset W[G_0]\subset W[G_1]\subset\cdots$$ of a countable model of ZFC has an upper bound $W[H]$ in a forcing extension of $W$ if and only if the forcing extensions $W[G_n]$ had uniformly bounded essential size in $W$. The essential size of a forcing extension is the size of the smallest complete Boolean algebra by which the extension can be realized as a forcing extension. In particular, any countable tower of extensions by adding a Cohen real has an upper bound, and furthermore in this case, you can find the upper bound which is also the extension by a single Cohen real, which is an attractive little argument, of the style of many computability theory constructions.<|endoftext|> TITLE: In what sense is extensivity a minimal requirement on an opfibration to conform with a notion of "family"? QUESTION [8 upvotes]: The following is an excerpt from Lawvere's Some thoughts on the future of category theory. To clarify the above considerations, generalize to distributive categories and seek philosophical guidance. Even though the determination of which maps are epimorphisms is the more profound question studied with Grothendieck topologies, it takes place within a topos of the following kind. Call a small category $\mathsf C$ "extensive" if it has finite coproducts which yield an equivalence $\mathsf C/A+B = \mathsf C/A \times \mathsf C/B$ and $\mathsf C/ \mathbf{0} = \mathbf{1}$ (this seems a minimum requirement on an op-fibration to conform with the notion of "family" and with Grassmann's "combinatorics of continuous magnitudes"); for example. "the" homotopy category or the category of spaces of dimension at most 4. In what sense does extensivity make the codomain opfibration conform with the notion of family? Why is it worthy of being called minimal? What is Grassman's "combinatorics of continuous magnitudes"? Where can I read about it? How and why does extensivity furnish/oil it? REPLY [7 votes]: The only way to know for sure is to ask Lawvere. You could try mailing him. For (1), I guess what he had in mind is the following. A family $(X_i)_{i \in I}$ of sets indexed by a set $I$ is essentially the same thing as a single set $X$ together with a map $\pi: X \to I$. The idea here is that $X = \coprod_{i \in I} X_i$ and that $\pi$ sends an element $x \in X_i$ to $i$. Once you've also done some thinking about maps between families, you conclude that the category of $I$-indexed families of sets is equivalent to the slice category $\mathbf{Set}/I$. Now, for any two sets $I$ and $J$, a family indexed over $I + J$ should consist of a family indexed over $I$ together with a family indexed over $J$. That's a theorem if we're talking about families of sets: $$ \mathbf{Set}/(I + J) \simeq \mathbf{Set}/I \times \mathbf{Set}/J. $$ But if we're attempting to generalize the notion of family to categories $\mathbf{C}$ other than $\mathbf{Set}$ (as Lawvere is doing), then this identity becomes a sensible axiom. Thus, he requires that $$ \mathbf{C}/(A + B) \simeq \mathbf{C}/A \times \mathbf{C}/B $$ for all $A, B \in \mathbf{C}$. Similarly, but more trivially, it's a theorem that there's exactly one family of sets indexed over the empty set: $$ \mathbf{Set}/\emptyset \simeq \mathbf{1}. $$ Hence Lawvere imposes the axiom $$ \mathbf{C}/0 \simeq \mathbf{1} $$ on $\mathbf{C}$. That's an explanation of why those two axioms on $\mathbf{C}$ are reasonable if you want to be able to talk about "families" in $\mathbf{C}$ and have them behave at all like families of sets. And though one can imagine imposing further axioms on $\mathbf{C}$ to make families in $\mathbf{C}$ behave even more like families of sets, you don't have to: hence, "minimal". I have no idea about the Grassmann thing. A short web search suggests that when Lawvere puts the words "combinatorics of continuous magnitudes" in quotation marks, he's not actually quoting anyone, but indicating that he's coining a term. It's a shame he doesn't give a reference.<|endoftext|> TITLE: field of definition of Fricke involution applied to a modular form QUESTION [8 upvotes]: Let $F\in M_k(\Gamma_0(N),\chi)$, not necessarily an eigenform nor cuspidal, but assume that $\Bbb Q(F)$ is a number field $K$. What can one say of $\Bbb Q(F|_kW_N)$, where $W_N$ is the Fricke involution ? Experiments seem to show that the Fourier coefficients of $F|_kW_N$ divided by $\sqrt{Q}$ for some positive or negative divisor of $N$ (probably linked to the conductor of $\chi$) also lie in $K$. More generally same question for a general Atkin--Lehner involution $W_Q$, and also for $1/2$-integral weight (in which case even a fourth root may be necessary). It may be possible to start with a corresponding result for newforms, but I have not seen how to complete the argument. P.S. Since I tested mainly with real characters, $\sqrt{Q}$ may of course be the Gauss sum associated to $\chi$. But the questions remain. REPLY [7 votes]: Your guess that this is related to the Gauss sum is correct. If $f$ has integer weight, level $N$ and coefficients in $K$, then $W_N(f) / \tau(\chi)$ has coefficients in $K$ also. This follows from a lovely theorem of Shimura, which states that $M_k(\Gamma(N), \mathbf{Q}[\zeta_N])$ is preserved by the action of $SL_2(\mathbf{Z} / N)$, and that the actions of $SL_2(\mathbf{Z} / N)$ and $\operatorname{Gal}(\mathbf{Q}[\zeta_N] / \mathbf{Q})$ piece together into an action of $GL_2(\mathbf{Z} / N)$ if you identify the Galois group with the matrices of the form $\begin{pmatrix} * & 0 \\ 0 & 1\end{pmatrix}$ (or it might be $\begin{pmatrix} 1 & 0 \\ 0 & *\end{pmatrix}$, I can't remember). To use this here, consider $f$ as an element of $M_k(\Gamma(N), K)$, and look at the action of $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$; the result is $W_N(f)(z/N)$ up to a power of $N$, which has the same coefficient field as $W_N(f)$ [edit: possibly up to a factor of $\sqrt N$ in odd weights, depending on you conventions], and Shimura's result now tells you that this has coefficients in $K(\zeta_N)$ and you can read off exactly how $Gal(K(\zeta_N) / K)$ acts. The same argument will tell you how partial Atkin--Lehner operators $W_Q$ for $Q \| N$ act, with a little more bookkeeping -- if I remember correctly, you get the Gauss sum of the Q-primary part of $\chi$ coming out. A reference for this is: Ohta, "P-adic Eichler--Shimura isomorphisms" (Crelle #463, 1995), sections 3.5 and 3.6. Lemma 3.5.2 on page 83 is a reciprocity law describing how Galois acts on $W_Q(f)$, which is (I hope!) equivalent to the statement I gave above; and section 3.6 (pages 86-9) gives a detailed proof of the lemma using Katz's algebraic description of modular forms. (I have no idea about the half-integer weight case.)<|endoftext|> TITLE: Integer homology of double loop space of odd-dimensional sphere QUESTION [6 upvotes]: I have checked everything "homology of loop spaces"-like, but was not able to find what is $H_*(\Omega^2S^3, \mathbb{Z})$. Therefore I ask you how to compute that? REPLY [6 votes]: The calculation you want is described in detail in Joe Neisendorfer's book Algebraic Methods in Unstable Homotopy Theory, Cambridge University Press, 2010. In particular, the Eilenberg--Moore spectral sequence collapses to give that $H_*(\Omega^2 S^3;\mathbb Z) = Cotor_*^{\mathbb Z[y]}(\mathbb Z, \mathbb Z)$. By calculating mod p, and then using the very simple Bockstein spectral sequence, he shows in Corollary 10.26.5, that $p$ annihilates the $p$--torsion for all primes $p$. Thus $H_*(\Omega^2 S^3;\mathbb Z)_{(p)}$ is a graded $\mathbb Z/p$ vector space (above dimension 1) whose Poincare series could easily be worked out from $H_*(\Omega^2 S^3;\mathbb Z/p)$.<|endoftext|> TITLE: About Simon Donaldson's book on four dimensional manifold QUESTION [10 upvotes]: Recently I'm reading Donaldson's Geometry of four manifolds. It seems to me that the book requires a lot for background. Additionally, the proof in the book is too sketchy without too much detail. I had a really hard time to digest the content in the book. Do we have other textbook demonstrating the same topic with more detail and background? REPLY [11 votes]: It's been a while but I remember that I found John Morgan's "An introduction to gauge theory" quite helpful when I was first trying to read about 4-manifolds and gauge theory. While it doesn't go nearly as far as Donaldson-Kronheimer, it does give a gentle introduction to gauge theory. Here's the complete reference: Morgan, John W., An introduction to gauge theory, Friedman, Robert (ed.) et al., Gauge theory and the topology of four-manifolds. Lectures of the graduate summer school. Providence, RI: American Mathematical Society. IAS/Park City Math. Ser. 4, 53-143 (1998). ZBL0911.57024.<|endoftext|> TITLE: Motivation for Hua's identity QUESTION [7 upvotes]: I ran into Hua's identity without intending to, meaning that I do not have a concrete reference available, and my background is not in Ring Theory. It is apparent to me that the identity is something of a big deal, but I couldn't find any explanation why. Authors pretty much assume that if you are reading about it, you already have a feeling for how important it is. Can you give me a hint? REPLY [3 votes]: Hua's identity is used to prove that any additive map of a division ring into itself preserving inverses must be an automorphism or antiautomorphism. His identity puts the Jordan triple product $aba$ in terms of additions and inverses, hence showing that those maps are also Jordan automorphisms; but Jordan isomorphisms between simple algebras had been shown by Ancochea and Kaplansky (1947) to be either automorphisms or antiautomorphisms, and the result for general division rings was proved by Hua himself (1949), completing the proof for additive maps preserving inverses. The study of contexts in which Jordan homomorphisms $f:R\rightarrow S$ for $R,S$ associative rings can be described in terms of (associative) homomorphisms and antihomomorphisms has a long and rich history (e.g., Jacobson and Rickart for $R$ a matrix ring, Bresar for $S$ a semiprime ring). Le théorème de von Staudt en géométrie projective quaternionienne (1942). Ancochea. On semi-automorphisms of division algebras (1947). Ancochea. Semi-automorphisms of rings (1947). Kaplansky. On the automorphisms of a sfield (1949). Hua. Jordan homomorphisms of rings (1950). Jacobson, Rickart. Jordan mappings of semiprime rings I, II (1989-91). Bresar.<|endoftext|> TITLE: Automorphism group of a special commuting graph QUESTION [7 upvotes]: Suppose $S_6$ is the symmetric group on six letters and let $X$ denote the conjugacy class containing $(12)(34)$. Define a graph $\Gamma$ with vertex set $X$ and edges precisely the 2-element subsets of $X$ which commute as elements of $S_6$. I would like to know the automorphism group of the graph $\Gamma$. P.S. These graphs are known in scientific texts as 'commuting graphs'. REPLY [9 votes]: It's an exercise to check it coincides with $\mathrm{Aut}(S_6)$ which has $S_6$ as subgroup of index 2. Here are the steps. First, for arbitrary $n\ge 6$, consider the graph of transpositions. So this is the set $X_n$ of the $n(n-1)/2$ unordered pairs, with an edge between two whenever they are disjoint. On $X_n$, consider the set $Y_n$ all unordered pairs of non-joined vertices (hence of the form $\{\{a,b\},\{a,c\}$ for $a,b,c$ pairwise distinct. There's a canonical map $\phi:Y_n\to\{1,\dots,n\}$ given by $\phi(\{\{a,b\},\{a,c\})=a$; we wish to show it's equivariant for the group action. Link two elements of $Y_n$ if they are disjoint and contained no pair of joined vertices. Then any two linked elements of $Y_n$ have the same image by $\phi$ (check! this uses the disjointness assumption). Next consider the equivalence relation on $Y_n$ generated by this equivalence relation. The only 2-element subsets in $Y_n$ with same image by $\phi$ but not linked have the form, up to permutation, $\{\{12,13\},\{12,14\}\}$. But one indeed have $\{12,13\}-\{15,16\}-\{12,14\}$. So for $n\ge 6$, the equivalence relation on $Y_n$ generated by being linked consists of being in the same fibers of $\phi$, and hence the automorphism group of $X_n$ acts on $\{1,\dots,n\}$. This action is easily seen to be faithful. Since the permutation group $S_n$ already acts, this shows that the automorphism group of $X_n$ is $S_n$. Next in $S_n$ for $n\ge 7$, the transpositions are the only elements whose centralizer has order $2(n-2)!$, i.e., form the only conjugacy class of order $n(n-1)/2$, and hence $\mathrm{Aut}(\mathrm{Comm}(S_n))$ stabilizes its subgraph $X_n$. Again a little argument shows that this action is faithful (i.e. an automorphism of $\mathrm{Comm}(S_n)$ fixing pointwise the transpositions is the identity; also for $n=6$. For $n=6$ on the other hand, there are 2 conjugacy classes of order 15: transpositions and triple transpositions. They are switched by non-inner automorphisms. Hence the previous argument applies to the subgroup of index 2 of $\mathrm{Aut}(\mathrm{Comm}(S_6))$ stabilizing $S_6$ and the result follows. Update: my answer above was incomplete since it does not address double transpositions; let me now answer the question fully; this will actually make use of the above answer about the graph of transpositions! Let us consider the graph $Z_n$ of double transpositions. Let me stick to $n=6$. I'll denote the permutation $(ab)(cd)$ as $(ab|cd)$ to avoid too many parentheses/commas. In $Z_6$, there are triangles defined as follows: Triangles of type I: $S_6$-permutes of the triangle $T_1=\{(12|34),(13|24),(14|23)\}$. Triangles of type II: $S_6$-permutes of the triangles $T_2=\{(12|34),(12|56),(34|56)\}$. It is straightforward that every triangle is of type I or type II, that the $S_6$-action preserves types of triangles, and that elements of $\mathrm{Aut}(S_6)\smallsetminus S_6$ exchange the two types. Consider the graph $W_6$ whose vertices are triangles in $Z_6$ and such that there is an edge between two triangles whenever they have a common vertex. For instance, there is an edge between $T_1$ and $T_2$ since they have the common vertex $(12|34)$. It is easy to check that $W_6$ is connected. Vertices of $W_6$ thus have type I or II, and we see that no vertices of the same type are joined. So $W_6$ is bipartite. Thus, for vertices of $W_6$, the relation "to have the same type" (which is not defined intrinsically) is indeed intrinsic, as it is equivalent to be at even distance. In particular, the isometry group $G$ of $W_6$ preserves the partition by types. Let $G'$ be its index 2 subgroup of type-preserving isometries. Define $W'_6$ as the graph consisting of vertices of type I in $W_6$, linked whenever they have distance at most 2 in $W_6$. We see that $W'_6$ is canonically isomorphic to the transposition graph! The isomorphism is given by mapping the triangle $T_1=\{(12|34),(13|24),(14|23)\}$ to $(56)$, etc. By the previous result on $X_6$, we deduce that $G'=S_6$. It follows that $G$ is reduced to $\mathrm{Aut}(S_6)$ (since clearly an isometry of $W_6$ that is identity on vertices of type I has to be the identity). To conclude that $\mathrm{Aut}(Z_6)$ is reduced to $\mathrm{Aut}(S_6)$, it is enough to show that the canonical homomorphism $\mathrm{Aut}(Z_6)\to\mathrm{Aut}(W_6)$ is injective. This is also immediate: for $f$ in the kernel, any double transposition belongs to exactly two triangles, i.e., two unordered triples of linked vertices, each of which is $f$-invariant, and hence is fixed by $f$. So we have proved that $\mathrm{Aut}(Z_6)$ is reduced to $\mathrm{Aut}(S_6)$, answering the original question. REPLY [6 votes]: Easy enough sage: G=SymmetricGroup(6) sage: cc=G.conjugacy_class([2,2,1,1]) sage: gr=Graph([cc, lambda a,b: a*b==b*a and a!=b]) sage: gr.automorphism_group().cardinality() 1440<|endoftext|> TITLE: Axiom(s) of choice and bases of vector spaces QUESTION [11 upvotes]: I'm not sure this question is more suitable for MO or for MSE, so feel free to move it to MSE if necessary. I work here in ZF theory. Consider the following statements: $(C)$ Axiom of choice: for any non empty family $(E_i)_{i\in I}$ of non empty subsets , there exists a function $f:I\to\displaystyle \bigcup_{i\in I}E_i$ such that $f(i)\in E_i$ for all $i\in I$ $(MC)$ Axiom of multiple choice: for any non empty family $(E_i)_{i\in I}$ of non empty subsets , there exists a function $f:I\to\displaystyle \mathscr{P}(\bigcup_{i\in I}E_i)$ such that $f(i)$ is a non empty finite subset of $E_i$ for all $i\in I$ $(B)$ for any field $K$, any $K$-vector space $E$ has a basis $(G)$ For any field $K$ and any $K$-vector space $E$, any generating family of $E$ contains a basis. In ZF, it is known that all these statements are equivalent. I'm interested in writing a self-contained proof of $(B)\Longrightarrow (C)$. I've found two proofs in the literature: one is proving $(B)\Longrightarrow (MC)$, the other one is proving $(G)\Longrightarrow (C)$. So, if I now how to prove $(MC)\Longrightarrow (C)$ or $(B)\Longrightarrow (G)$, I'm done. Now, here comes my questions. In all the references I found, the proofs of $(MC)\Longrightarrow (C)$ are drowned in the looong proof of the equivalence of 5 statements in ZF, but I couldn't find any direct proof. I'm not that familiar with non elementary set theory, and i'm not sure the effort I'll have to make to understand the proof of the equivalence of all these statements will worth it. So: Question 1. Do you know a direct proof of $(MC)\Longrightarrow (C)$ in ZF ? Moreover, I'm a bit puzzled by the fact that I can't find a direct proof of $(B)\Longrightarrow (G)$ in ZF. I thought it would be straightforward, but I can't find any proof for the moment. It's possible I'm missing an easy argument though. So : Question 2. Do you know a direct proof of $(B)\Longrightarrow (G)$ in ZF ? Thanks in advance for your help ! Edit. After reading Asaf comment, I think I should say that I'm happy to include the axiom of regularity in ZF. I reformulate my question: using ZF+regularity, can we find a direct proof of (MC) implies (C), or at least a proof which is more direct thatn the ones I've read in books (which are a chain of 4 implications, see my comment to Asaf answer). Also, using ZF+regularity, can we find a proof of (B) implies (G) ? Of course, I anticipate the the answers will be probably NO, but I prefer to make sure, before I choose to use a black box in my proof. REPLY [12 votes]: There is no fully elementary proof that you are looking for. The reason is that the axiom of regularity is needed in these proofs. Multiple Choice does not imply Choice without it, and the only proofs we know about vector spaces go through Multiple Choice. While regularity is not a "difficult axiom" it does mean the proof makes some non-trivial use of the structure of the universe of set theory. However, you can look in Jech's The Axiom of Choice, in the 9th chapter he covers the implication of $\sf MC\to AC$, and more.<|endoftext|> TITLE: Existence and uniqueness of geodesics in low regularity QUESTION [14 upvotes]: Consider a Riemannian manifold $(M,g)$. How much regularity is required of $g$ so that for any $x\in M$ and $v\in T_xM$ with $|v|=1$ there exists a unique geodesic $\gamma\colon(-\epsilon,\epsilon)\to M$ so that $\gamma(0)=x$ and $\dot\gamma(0)=v$? All regularity is considered with respect to a fixed smooth (or somewhat less regular) atlas or (equivalently) in a fixed local coordinate chart. If $g\in C^{1,1}$, then the Christoffel symbol is Lipschitz, whence existence and uniqueness follows from standard ODE theory. For ODEs in general Lipschitz continuity is sufficient and necessary for existence and uniqueness. It is not clear to me whether this is the case for geodesics, because the geodesic equation has a very specific structure and that may provide uniqueness in even lower regularity. Existence requires less regularity than uniqueness, and this old question is about existence for continuous metrics. Since existence is known in very low regularity, my question concerns uniqueness (although I want to have both existence and uniqueness in as low regularity as possible). If a metric can have a jump discontinuity, then both existence and uniqueness can be made fail, but I am not aware of any smoother counterexamples. A refined (but equivalent) version my question is: What are the best sufficient and necessary regularity conditions we know (whether or not they coincide in our current knowledge) on a Riemannian metric for uniqueness of geodesics? I am not aware of any positive or negative uniqueness results for $C^{0,\alpha}$ or $C^{1,\alpha}$ below $C^{1,1}$. For example, are geodesics unique if $g\in C^{1,\alpha}$ for some $\alpha>0$, or are there perhaps counterexamples for all $\alpha<1$? If you find the question unclear, please ask for details. At least for $g\in C^1$ the geodesic equation is a well-defined classical ODE. I guess that satisfying the geodesic equation and minimizing arc length locally are equivalent in this regularity, but I may be mistaken. REPLY [7 votes]: There is a classical example by Hartman that shows failure of uniqueness for $C^{1,\alpha}$ metrics. (P. Hartman, On the local uniqueness of geodesics, Amer. J. Math. 1950). You could lower the regularity if the metric is smooth off some hypersurface but globally only $C^{0,1}$, see for example our recent review: On geodesics in low regularity Moreover, being minimzing and solving the geodesic equations is not the same below $C^{1,1}$; see also Hartman or our review.<|endoftext|> TITLE: Two definitions of the narrow Mordell-Weil group QUESTION [6 upvotes]: Let: $K = k(C)$, where $C/k$ is a projective non-singular curve, $E/K$ - an elliptic curve, $\mathcal{E} \to C$ - the minimal elliptic surface associated to $E$. Consider the "narrow Mordell-Weil subgroup", as defined in [1.]: $$E(K)_0 := \{ P \in E(K) : \tau_P(\Gamma) = \Gamma \quad \text{ for every fibral curve } \Gamma \subset \mathcal E \}$$ (where $\tau_P : \mathcal E \to \mathcal E$ is the translation-by-$P$-morphism). The usual definition, that may be found in [2.], [3., p.71], [4, Def. 7.7, p. 33] looks like this: $$E(K)_0' := \{ P \in E(K) : (P) \cdot F = (\mathcal O) \cdot F \quad \text{ for every fibral divisor } F \in Div(\mathcal E) \} = \{ P \in E(K) : \text{ $P$ and $\mathcal O$ intersect each fiber at the same irreducible component} \}$$ (we denote by $\mathcal O \in E(K)$ the neutral element. We treat $P$, $\mathcal O$ as divisors on $\mathcal E$, by identifying them with sections of $\mathcal E \to C$). Silverman claims that $E(K)_0 = E(K)_0'$ (Remark III.9.4.1. and Exercise 3.27.). Question: Is this true? If, yes, how can you prove it? One easily proves that $E(K)_0 \subset E(K)_0'$. The second inclusion is not that easy. Note that $$ P \in E(K)_0' \quad \text{ iff } \quad \tau_P(\Gamma_{t0}) = \Gamma_{t0} \quad \text{ for every $t \in C$},$$ where $\Gamma_{t0}$ is the unique irreducible component of the fiber $\mathcal E_t$, which intersects the zero section $(\mathcal O)$. But why the fact that $\tau_P$ carries $\Gamma_{t0}$ into itself should imply that $\tau_P$ preserves other components of the fiber? I started to disbelieve it, but I don't have an idea to construct a counterexample. I noticed that actually the proofs in [1.] would still be OK, if one would replace $E(K)_0$ by $E(K)_0'$. So maybe this groups are actually different? References: Silverman, Advanced Topics in the arithmetic of Elliptic Curves Tate, Variation of the Canonical Height of a Point Depending on a Parameter Miranda, The basic theory of elliptic surfaces Schuett & Shioda - Elliptic Surfaces Edit: corrected the misleading term "trivial action", as suggested in the answer below. REPLY [8 votes]: Where you say "trivial action on $\Gamma_{t0}$" near the end, you mean "carries $\Gamma_{t0}$ into itself" (i.e., doesn't move it), not that the effect on $\Gamma_{t0}$ is the identity. The global base is a red herring: the core issue here completely concerns the object over the local ring at $t$ (recall that the formation of both the minimal regular proper model and the Neron model of an elliptic curve over the function field of a connected Dedekind scheme commutes with passage to local rings on the base, more-or-less by design). So to clarify matters, let's work in that context. So let $R$ be any discrete valuation ring, say with fraction field $K$, and let $E$ be an elliptic curve over $K$ with minimal regular proper model $\mathcal{E}$ over $R$. Let $\mathcal{E}^{\rm{sm}}$ denote the maximal $R$-smooth open subscheme of $\mathcal{E}$; since $\mathcal{E}$ is $R$-flat of finite type with smooth generic fiber (and the generic point of ${\rm{Spec}}(R)$ is open), $\mathcal{E}^{\rm{sm}}$ is the complement in $\mathcal{E}$ of the closed non-smooth locus in the closed fiber. In general if $X$ is an $R$-scheme of finite type that is regular, any section $x: {\rm{Spec}}(R) \rightarrow X$ lands inside the maximal smooth open subscheme $X^{\rm{sm}}$ (this is a nice application of the characterization of smoothness in terms of completed local rings; see 3.1/2 in the highly-recommended book Neron Models for a proof). Consequently, $\mathcal{E}(R) = \mathcal{E}^{\rm{sm}}(R)$; this is very important (it is the only thing which ensures that $\mathcal{E}^{\rm{sm}}$ has non-empty special fiber, which is to say that the special fiber of $\mathcal{E}$ isn't everywhere non-smooth, since there is at least the "identity section" = the $R$-point extending the identity section of $E$). Now finally come to the real substance of the matter: $\mathcal{E}^{\rm{sm}}$ is the Neron model (and in particular has an $R$-group structure!), or more precisely the unique $R$-map $f:\mathcal{E}^{\rm{sm}} \to N(E)$ to the Neron model extending the identification of $K$-fibers is an isomorphism. This is not obvious, and is not perceived when one builds the Neron model through the general methods that are used for abelian varieties since there is no analogue of $\mathcal{E}$ for abelian varieties of higher dimension (i.e., the contact between the Neron model and the minimal regular proper model is a special feature of the 1-dimensional context). To prove that $f$ is an isomorphism, it is harmless to extend scalars to the strict henselization $R^{\rm{sh}}$ of $R$ (since a map which becomes an isomorphism after a faithfully flat base change was an isomorphism originally, and the formation of both the Neron model and minimal regular proper model commute with such scalar extension more-or-less by design); that case in turn is proved as 1.5/1 in Neron Models (the proof uses much of the work involved in the actual construction of $N(E)$ by general principles for abelian varieties). And even more is true: the translation action of $\mathcal{E}^{\rm{sm}}$ on itself extends to a left action $$\mathcal{E}^{\rm{sm}} \times \mathcal{E} \rightarrow \mathcal{E}$$ on $\mathcal{E}$ (so at the level of $R$-points, this recovers the action of $\mathcal{E}^{\rm{sm}}(R)$ on $\mathcal{E}$ that could be built via other considerations -- do the comparisong of constructions over $K$). This ultimately comes down to good behavior of the formation of the minimal regular proper model under certain kinds of mild extensions of $R$. See Lemma 2.12(c) in section 10.2 of Qing Liu's book Algebraic Geometry and Arithmetic Curves for a proof (expressed in slightly different terms, but yielding the above by composing with a projection map). This is really the most crucial point, since it allows one to actually harness geometric information: if we pass to the special fibers over the residue field $k$ of $R$ we get an action of the smooth $k$-group $G := \mathcal{E}^{\rm{sm}}_k$ on the typically reducible (and generally not everywhere reduced) $k$-curve $X = \mathcal{E}_k$, and your question is now reduced to showing that the action of the identity component $G^0$ on $X$ preserves each irreducible component of $X$. We can now forget about the original setup with elliptic curves, and just want to show that if $H$ is a smooth connected group scheme over a field $k$ (such as $G^0$ above) and $X$ is a $k$-scheme of finite type equipped with a left $H$-action then $H$ preserves each irreducible component $C$ of $X$. Since $C_{\overline{k}}$ is a union of irreducible components of $X_{\overline{k}}$, it suffices to prove the result over $\overline{k}$, so we can assume $k$ is algebraically closed. We can also assume $C$ is reduced. The condition on $H$ of preserving $C$ is now represented by the closed subgroup scheme $${\rm{Stab}}_H(C) := \bigcap_{c \in C(k)} \{h \in H\,|\,h.c \subset C\},$$ yet this has finite index at the level of $k$-points since $X$ has only finitely many irreducible components. But a smooth connected group $H$ over an algebraically closed field $k$ has no proper closed subgroup of finite index at the level of $k$-points (as $H$ is the disjoint union of finitely many $H(k)$-translates of that closed subgroup, forcing the closed subgroup to be open and hence to coincide with $H$), so $H$ preserves $C$ and we win. QED<|endoftext|> TITLE: Is there good reference for proof complexity? QUESTION [8 upvotes]: I am asking if there are some good or standard references for proof complexity theory? I didn't find references when I search in internet. Thanks! REPLY [11 votes]: These are three books that I know: Logical Foundations of Proof Complexity Bounded Arithmetic, Propositional Logic and Complexity Theory Logical Foundations of Mathematics and Computational Complexity, Proof complexity chapter Also, I suggest you read the following papers: The lengths of proofs Bounded Arithmetic and Propositional Proof Complexity ============================================== Update. Jan Krajíček wrote a new book on this topic recently. You can find the final manuscript of it here.<|endoftext|> TITLE: Does every Coxeter group arise from a BN-Pair? Does $\text{PGL}_2(\Bbb{Z})$? QUESTION [10 upvotes]: The question is in the title. Maybe I should explain my interest in it though. To every Coxeter group $(W,S)$ (and even more general groups) and a system of parameters $(a_s,b_s)_{s \in S}$ one can attach a Hecke algebra $\mathcal{H}(W,S)$ (this is an exercise in Bourbaki's Groupes et algèbres de Lie IV-VI) with generators $T_w$ ($w \in W$) and relations $$ \begin{align*} T_w T_{w'} & = T_{ww'} \text{ if } \ell(w)+\ell(w') = \ell(ww') \\ T_s^2 & = a_s + b_s T_s \end{align*} $$ Sometimes, these abstract Hecke algebras turn out to be isomorphic to `convolution type Hecke algebras' $$ \mathcal{H}(G,B) = \text{End}_G(\text{ind}^G_B \Bbb{1}) $$ for a group $G$ and a subgroup $B$ (I am being deliberately vague here), and one therefore gets a functor $$ \text{Rep}(G) \longrightarrow \text{R-Mod}(\mathcal{H}(W,S)),\quad V \mapsto V^B $$ relating representations of the abstract Hecke algebra and of the group $G$. This is for instance the case if $(W,S)$ comes from a BN-pair (by another Bourbaki exercise). REPLY [7 votes]: $PGL_2(\mathbf{Z})$ is a Coxeter group with presentation $\langle s_1,s_2,s_3 \mid s_i^2=(s_1s_2)^3=(s_1s_3)^2=1 \rangle$, i.e. with labels $\{2,3,\infty\}$. To each Coxeter group $W$ with labels in $\{2,3,4,6,\infty\}$ there exists a Kac--Moody $G$ having a (twin) BN-pair with Weyl Group $W$, as constructed by J. Tits. In particular, $PGL_2(\mathbf{Z})$ is the Weyl group of a Kac-Moody group of rank 3. While the definition of a Kac-Moody group might appear quite technical, there are two ways of handwaving to convey a flavour of what a Kac-Moody group is like: (i) A semisimple complex Lie algebra has a nice presentation which can be written down from its Dynkin diagram. Using this presentation for more general Dynkin diagrams, one gets an infinite-dimensional complex Lie algebra $L$ generated by some generators $\langle e_i, f_i, h_i\rangle$. Then the associated adjoint complex Kac-Moody group is by definition the subgroup of Aut$(L)$ generated by ad$(t\cdot e_i)$, ad $(t\cdot f_i)$ for $t \in \mathbf C$. (ii) One can write down a presentation for $SL_n(\mathbf C)$ with generators the elementary matrices and certain relations which can be read off from the group's associated Dynkin diagram. Then a Kac-Moody group can similiarly be written down in terms of generators and relations for more general Dynkin diagrams. Caprace-Rémy's "Groups with a twin root datum" gives a very nice introduction to these matters. On the other hand, by Tits' famous classification of spherical buildings there do not exist thick buildings of types $H_3$ or $H_4$, hence no $BN$-pairs of this type, see e.g. this paper by R. Weiss.<|endoftext|> TITLE: How to improve writing mathematics? QUESTION [40 upvotes]: My first language is not English. How can I improve my mathematical writing. I feel like the only things I can write down are numbers and equations. Is there any good suggestion for improving writing, especially for mathematical writing (math-philosophy)? REPLY [33 votes]: I want to highlight two tools for learning: imitation and practice. Read a lot of mathematics. You will find that some texts are easier to follow than others. What makes you like a text? What texts do you like most? When you write, try to write as your favorite author would. If you keep on writing mathematics long enough, you will find your own voice, but imitation is a necessary first step. Write a lot. You say you can write numbers and equations. What are they about? Explain. It doesn't have to be perfect, but explain it in your own words. Tell a story about your calculation. What would you say out loud to explain your work to a fellow student? Write that down. Try to make a habit out of making explained calculations that anyone could read. Lastly, if you don't know how to write something, ask for help. Composing good mathematical prose is not trivial, and learning it is an important part of any degree in mathematics. You are entitled for help with it, not only with your calculations. If I understand correctly, your problem is in writing relatively simple and short things. Writing and structuring a thesis, a paper, or other extended piece of work is a story I will exclude here. REPLY [26 votes]: At the early stages of my mathematical career, I found the following (little) book very useful. Trzeciak, Jerzy, Writing mathematical papers in English. A practical guide, Zürich: European Mathematical Society Publishing House (ISBN 3-03719-014-0/pbk). 49 p. (2005). ZBL1077.00008. REPLY [22 votes]: Halmos, Paul R. "How to Write Mathematics." L'Enseignement Mathématique 16.2 (1970): 123-152. (PDF download.) Idiosyncratic, but a classic. I especially enjoyed the section, "Think about the alphabet," and his discussion of "frozen" letters: "many readers would feel offended if $n$ were used for a complex number, $\epsilon$ for a positive integer, and $z$ for a topological space. (A mathematician's nightmare is a sequence $n_\epsilon$ that tends to $0$ as $\epsilon$ becomes infinite.)" REPLY [5 votes]: Read what you have written out loud. This way you will hear most of the things that could go wrong. If it sounds awkward, change it. If it bores you, say something more interesting or cut it down. If it confuses you, look for a clearer presentation. I find that to be good advice for writing in any language. It's also useful to do this as the central step with a machine translator. For instance, to write text in Spanish, I will: a) Write a draft in English; b) Run Google Translate on it from English to Spanish; c) Read the result out loud, and edit so that it sounds good to me in Spanish; d) Run Google Translate on the result from Spanish to English; e) Check if anything in this English version shows something ungrammatical or unintended in Spanish, and if so, go back to c. This works best with substantial revisions in the middle step -- substantial enough for me that I feel a physical strain in my cheeks from enunciating all the Spanish sounds. But it saves me from lots of errors of getting genders wrong, and it comes up with better words than I would think of on my own. Anyway, if you've done enough reading in English to have a good ear for it, you may like the results from this technique.<|endoftext|> TITLE: Solution to at least one ODE in a family of ODE's QUESTION [7 upvotes]: In my research I have stumbled across the following 1st order complex differential equation for smooth functions $\eta:\mathbb{R}/2\pi\mathbb{Z}\to\mathbb{C}-\lbrace0\rbrace$ defined on the circle, $$i\frac{\partial\eta}{\partial t}+(re^{it}+\varepsilon i)\bar\eta=0$$ where $\varepsilon\in\mathbb{R}_+$ is sufficiently small and fixed (so choose $0<\varepsilon<<1$ but maybe not $\varepsilon=\frac{1}{2}$) and $r\in\mathbb{R}_+$ is a nonzero positive real parameter. Motivated by some perturbation theory and functional analysis, I believe there must exist a nontrivial solution $\eta$ to this ODE for at least one choice of $r$. What is such an explicit pair $(r,\eta)$? This problem arises when studying the asymptotics of punctured $J$-holomorphic curves, and I need the solutions to do what I want to do (when perturbing $J$). Ultimately I desire the set of all such distinct $r$ and the dimension of the kernel of the corresponding differential operators. $\underline{\text{Attempt}}$ I originally posted this on StackExchange a month ago with many edits but not much luck. My attempt was to Fourier expand $\eta(t)=\sum_{k\in\mathbb{Z}}a_ke^{ikt}$ and obtain the recurrence relation $$-ka_k+r\bar a_{1-k}+\varepsilon i\bar a_{-k}=0$$ I need a specific collection $\lbrace a_k\rbrace\subset\mathbb{C}$ which solves this (for some $r>0$). What I get at the least is $r=\varepsilon i\frac{a_0}{a_1}$ (and subsequently $\frac{a_0}{a_1}$ must be purely complex and nonzero). But there is still a good chance that the coefficients $a_k$ will "blow up" as $k\to\infty$ if not chosen carefully. I've done some manipulations and I cannot parse whether they are helpful or harmful. Also, this complex ODE is equivalent to two coupled real ODEs, but I don't think it helps. Decompose $\eta=x+iy$ and attempt to solve the equivalent system: $$\dot y(t) -r\cdot\cos(t)\cdot x(t) -[\varepsilon +r\cdot\sin(t)]\cdot y(t) = 0$$ $$\dot x(t) -r\cdot\cos(t)\cdot y(t) +[\varepsilon +r\cdot\sin(t)]\cdot x(t) = 0$$ Perhaps there are numerical methods to find "approximate" periodic solutions, or some software to plot $(x,y)$ for various values of $r\in\mathbb{R}_+$? REPLY [3 votes]: This is an expanded version of my comments, to make them somewhat less cryptic. I'm not answering the question, and, I don't really expect to ever make further progress. Update: After I learned from fedja's answer that an $r\simeq \epsilon^{1/2}$ works (much to my surprise), I can now show that too by a perturbation expansion of $D$ below, and this is now included in this new version. If we write $u=(x,y)^t$, then we can rephrase the question as: Does the Dirac operator $L(\epsilon, r)$, $$ Lu = Ju'+\epsilon A u + r R(t) u, \quad J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},\: A =\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\: R= \begin{pmatrix} \cos t & \sin t \\ \sin t & -\cos t \end{pmatrix} $$ on $L^2(0,2\pi)$ have $E=0$ as a periodic eigenvalue (that is, with boundary conditions $u(0)=u(2\pi)$), for suitable $r$? The transfer matrix $T(t,E)$ is defined as the $2\times 2$ matrix solution of $Lu=Eu$ with initial value $T(0,E)=1$. Since the coefficient matrices $JA$ and $JR$ have trace zero, it follows that $\det T=1$, so the eigenvalues of $T(2\pi,E)$ are determined by its trace $D(E):=\textrm{tr }T(2\pi, E)$. In particular, the periodic eigenvalues of $L$ are the solutions of $D(E)=2$. Now $D(E=0,\epsilon,r=0)=2\cosh 2\pi\epsilon$, so existence of an $r$ as desired could be established by showing that $D(0,\epsilon,r)<2$ for some $r$. Moreover, since we now know from fedja's answer that this happens at $r\simeq \epsilon^{1/2}$, we can obtain this from a Taylor expansion of (the entire function) $D$ about $r=0$. Denote the transfer matrix of $L(\epsilon,0)u=0$ by $$ T_0(t) = \begin{pmatrix} e^{-\epsilon t} & 0 \\ 0 & e^{\epsilon t} \end{pmatrix} . $$ I'll do variation of constants about $r=0$, so write $T(t,0)=T_0 M$. Then $M$ solves $M'= r T_0^{-1}JRT_0 M$, and solving this by iteration delivers the Taylor coefficients one after the next. So $M(t,r)=\sum r^n M_n(t)$, with $M_0=1$, $$ M_{n+1}(t) = \int_0^t \begin{pmatrix} -\sin s & e^{2\epsilon s}\cos s \\ e^{-2\epsilon s}\cos s & \sin s \end{pmatrix} M_n(s)\, ds . $$ Everything can be worked out explicitly, though at some risk to one's sanity. We are ultimately interested in $$ D(0,\epsilon,r)=\textrm{tr }T(2\pi,0) = e^{-2\pi\epsilon}a(2\pi, r) + e^{2\pi\epsilon}d(2\pi, r) , \quad \quad M\equiv \begin{pmatrix} a & b\\ c & d \end{pmatrix} . $$ Expanding everything, this becomes $$ D(0) = 2 + 4\pi^2\epsilon^2 + 2r^2\pi\epsilon (d_2-a_2) + \sum_{n=1}^4 r^n (a_n+d_n) + O(r^5) + O(\epsilon^3) \quad\quad\quad\quad (1) $$ (and everything taken at $t=2\pi$, of course). Brute force calculation will show that $d_2-a_2=0$ and $a_1+d_1=0$ (in fact, $a_1=d_1=0$) and $a_2+d_2=O(\epsilon^2)$. As for $a_3+d_3$, maybe the best way is to check that this vanishes when $\epsilon=0$, so we'll pick up at least one extra $\epsilon$ when we reintroduce this, so the third order term is $O(r^3\epsilon)$. The big surprise now (for me) is that after everything vanished at $\epsilon=0$ or went the wrong way, out of the blue comes $a_4+d_4=-4\pi^2$ at $\epsilon=0$. So the bottom line is that when $r=\epsilon^{1/2}$, then this term cancels the $4\pi^2\epsilon^2$ exactly, and everything else in (1) is smaller. Conclusion: There is a solution $r=r(\epsilon)$ of $D(0,\epsilon,r)=2$, and $\epsilon^{-1/2}r\to 1$ as $\epsilon\to0$. Of course, with this perturbative approach, we cannot get the other half of fedja's result, namely that this is the only solution.<|endoftext|> TITLE: English translation of Hilbert's work QUESTION [7 upvotes]: Does anyone know if there is an English translation of Hilbert's: "Grundzuge einer allgemeinen Theorie der linearen Integralgleichungen, Teubner, Leipzig, 1912". ?? Thanks, Andre REPLY [8 votes]: Yes, there is a translation, but not an official one and only of the "Erste Mitteilung", not the whole book: The pdf FREDHOLM, HILBERT, SCHMIDT Three Fundamental Papers on Integral Equations, Translated with commentary by G. W. Stewart contains "Foundations of a General Theory of Linear Integral Equations" (starting on p.55 (59 in the pdf)). This is a translation of the paper Grundzüge einer allgemeinen Theorie der linearen Integralgleichungen, Erste Mitteilung Nachrichten der Wissenschaftlichen Gesellschaft zu Göttingen, Math.-phys. Kl. (1904),49-91". I haven't seen translation of the five other "Mitteilung" under the same title (but I haven't seen version of the second, third and sixth Mitteilung even in German…). The other two papers in the pdf are Fredholm's "On a Class of Functional Equations" and Schmidt's "On the Theory of Linear and Nonlinear Integral Equations. Part I: The Expansion of Arbitrary Functions by Prescribed Systems."<|endoftext|> TITLE: What's the sufficient or necessary conditions for a manifold to have Lie group structure? QUESTION [9 upvotes]: For example, given a Lie group, its fundamental group must be Abelian. So $\Sigma_g$ ($g>1$) can't have Lie group structure. We also know for $S^n$ only $n=0,1,3$ can have Lie group structures. In general, what's the sufficient or necessary conditions for a manifold to have Lie group structure? REPLY [10 votes]: Compact Lie groups are finite loop spaces, but Andersen, Bauer, Grodal, Pedersen (Invent. Math., 2004) give an example of a finite loop space that is not (even rationally) homotopy equivalent to any compact Lie group.<|endoftext|> TITLE: Are topological fiber bundles on the same base with homeomorphic fibers isomorphic? QUESTION [7 upvotes]: First I apologize because this is not a research question, but I can't get any answer on MathStackExchange... Let $\pi \colon E \to B$ and $\pi' \colon E' \to B$ two topological fiber bundles on the same base $B$. A $B$-morphism $f \colon E \to E'$ is an isomorphism of fiber bundles iff for all $b \in B$ the induced morphism between fibers $f_{E_b} \colon E_b \to E'_b$ is a homeomorphism. Is this true ? If not in general, under which additional conditions on the spaces does it become true ? The direct implication is obvious, but I have huge doubts on the reverse. It is clear that $f$ is a bijection, continuous by definition, so we just have to show that $f^{-1}$ is also continuous. Continuity being a local property we can suppose that both fibrations are trivial, i.e. $f \colon B \times F \to B \times F'$ is of the form $(b,y) \mapsto (b,f_b(y))$ where $f_b \colon F \to F'$ is a homeomorphism for all $b \in B$. Define $f^{-1} \colon B \times F' \to B \times F$ by $(b,y') \mapsto (b,(f_b)^{-1}(y'))$. Since $f$ is continuous, the coordinate map $(b,y) \mapsto f_b(y)$ is also continuous, and if we can conclude from this that the coordinate map $(b,y') \mapsto (f_b)^{-1}(y')$ is also continuous, $f^{-1}$ is continuous and we are done. But why should this last step be correct ? It is correct for vector bundles (F and F' being vector spaces of the same dimension) because in this case $f_b \in L(F,F')$ forall $b \in B$, and $f_b$ can be represented by a matrix whose coefficients are continuous fonctions of $b$. Then the polynomial formula for the inverse matrix shows that the coefficients of $(f_b)^{-1}$ are also continuous. It is also trivially true with principal bundles because there any $B$-morphism is an isomorphism of principal bundles. But quid in the general topological case ? Forall $b \in B$, $f_b \in Homeo(F,F')$. With the compact-open topology on this set, can we say that the application $b \mapsto f_b$ is continuous (I hope so) ? And how can we say anything about the map $b \mapsto (f_b)^{-1}$ and $f^{-1}$ ? REPLY [9 votes]: If $F$ is locally connected, locally compact and Hausdorff (or alternatively compact Hausdorff), then the inverse function from $Homeo(F, F)$ to itself is continuous. Moreover, such a space $F$ is exponentiable. This means, in particular, that a map $B\to Homeo(F, F)$ is continuous if and only if the adjoint map $B\times F\to F$ is continuous. So under this assumption the answer to your question is yes. Presumably other (weaker?) sufficient hypotheses are available. But in general, the inverse function from $Homeo(F, F)$ to itself is not continuous (see here). Moreover, there exist exponentiable spaces with this property, namely locally compact, Hausdorff, but not locally connected spaces (see the example here). So in general the answer is no: Let $F$ be an exponentiable space such that the inverse is not continuous on $Homeo(F, F)$. Then the function from $Homeo(F, F)\times F$ to itself given by $(f, x)\mapsto (f, fx)$ provides a counterexample.<|endoftext|> TITLE: Automorphism group of $UT(n,p)$, the group of unitriangular matrices over the field $\mathbb{F}_p$ QUESTION [6 upvotes]: I am interested in understanding (at least roughly, if no such a description exists) the group of automorphisms for the group $UT(n,p)$, of unitriangular matrices over the field $\mathbb{F}_p$ on $p$-elements. Unfortunately the online searching I've carried out wasn't quite helpful. If no description exists, maybe some information how the automorphisms behave on the set of subgroups of it may be helpful too. For instance when $n=3$, I know we have that any $\phi \in Aut(UT(3,p))$, permutes the pairs of non-commuting elements of $UT(3,p)$ (which in that case happens always to be a generating set) and the subgroups of order $p^2$. So am wondering if something similar exists (of course modified somehow) in the general case too. If the above sounds quite general, the case where $n=4$, is of great importance for me either. Any comment might be useful, and of course references too. Thanks! REPLY [3 votes]: for size = 3 https://groupprops.subwiki.org/wiki/Unitriangular_matrix_group:UT(3,p)#Automorphisms Automorphisms The automorphisms essentially permute the subgroups of order containing the center, while leaving the center itself unmoved. over $F_2$, any size: Maginnis J. S., (1993/11)."Outer Automorphisms of Upper Triangular Matrices." Journal of Algebra 161(2): 267-270. Abstract: The outer automorphism group of the upper triangular matrices over the field of two elements is calculated. A. J. Weir (Proc. Amer. Math. Soc.6 (1955), 454-464) performed a similar calculation for Fields of odd characteristic, and we borrow the term extremal auto .... Any field(may be even ring) International Journal of Algebra, Vol. 7, 2013, no. 15, 723 - 733 HIKARI Ltd, The Automorphism Group of the Group of Unitriangular Matrices over a Field1 Ayan Mahalanobis Abstract. This paper finds a set of generators for the automorphism group of the group of unitriangular matrices over a field. Most of this paper is an exposition of the work of V.M. Lev˘chuk, part of which is in Russian. Some proofs are of my own. From the paper: The automorphism group of the group of unitriangular matrices over a field was studied by many [2–4]. In this direction, the first paper was in Russian, published by Pavlov in 1953. Pavlov studies the automorphism group of unitriangular matrices over a finite field of odd prime order. Weir [4] describes the automorphism group of the group of unitriangular matrices over a finite field of odd characteristic. Maginnis 3 describes it for the field of two elements and finally Lev˘chuk 2 describes the automorphism group of the group of unitriangular matrices over an arbitrary ring. In this expository article, we shall study the automorphism group of the group of unitriangular matrices over an arbitrary field F. There are two most commonly used non-abelian finite p-groups in the literature. One is the group 1Research supported by a NBHM research grant. 724 Ayan Mahalanobis of unitriangular matrices over a field and the other is the extra-special pgroups. Weir [4] and Lev˘chuk 2 worked on the automorphism group of the unitriangular matrices. PS I remember I have seen some papers with quite explicit description of automorphism, and I thought I collected such links in comments to my old MO-question: Representation theory of p-groups in particular upper tringular matrices over F_p, but it cannot find it now again, hm-m... I'll try to find more...<|endoftext|> TITLE: Are the semi-simple elements in a non-connected reductive algebraic group dense? QUESTION [6 upvotes]: Let $k$ be an algebraically closed field of characteristic zero and let $G$ be a non-connected linear algebraic group with reductive connected component $G^0$ over $k$. What is known about the semi-simple elements in $G$ ? Are they dense in every connected component of $G$ ? Do they contain an open dense subset of every connected component of $G$ ? If $G$ is instead connected, then the answer is yes, as the regular semi-simple elements form an open dense subset. Also I am aware of a result of Guralnick and Malle [Lemma 6.9 in "Simple groups admit Beauville structures" J. Lond. Math. Soc. (2) 85 (2012), no. 3, 694-721] which gives the answer YES in a special case. Note that in characteristic zero all unipotent elements lie in $G^0$. REPLY [11 votes]: The $G^0$-action on a coset is the same as a so-called twisted action which is pretty well understood. See, e.g., Mohrdieck, S.: Conjugacy classes of non-connected semisimple algebraic groups, Transformation Groups, 8, (2003) 377-395 (MSN). More precisely, let $C=G^0a$ be a connected component of $G$. Then conjugation by $a$ induces an automorphism $\tau$ on $G^0$. Identifying $C$ with $G^0$ via $g\mapsto ga$ converts conjugation on $C$ to twisted conjugation on $G^0$ $$ u(ga)u^{-1}=(ug\tau(u)^{-1})a $$ It is possible to choose $a$ in such a way that $\tau$ preserves a Borel $B$, a maximal torus $T$ and a pinning of $G^0$, i.e., $\tau$ is induced by an automorphism of the Dynkin diagram of $G^0$. Let $T_0:=(T^\tau)^0$ be the connected component of the $\tau$-fixed points in $T$. Then it is not difficult to see that the map $G^0\times T_0\to G^0:(g,t)\mapsto gt\tau(g)^{-1}$ is dominant. Thus the conjugacy classes of $T_0a$ contain an open subset of $C$ and they are all semisimple. Edit: Urs Hartl pointed out to me that the proof of Prop. 3.8 in loc.cit may contain a gap (it is unclear that $t$ exists such that $t^{\mathrm ord\,\tau}$ is regular semisimple). Therefore, I am adding a direct argument for the claim that $G^0\times T_0\to G^0$ is dominant. This is done in two stages: Let $C:=(G^\tau)^0\subseteq G^0$. First one shows that $$ G^0\times C\to G^0:(g,c)\mapsto gc\tau(g)^{-1} $$ is dominant. For this it suffices to show that the map on tangent spaces in $(1,1)$ is surjective. Because of $\mathrm{Lie}\,C=\ker(1-\tau)$ that follows from $$ (1-\tau)\mathfrak g\oplus \ker(1-\tau)=\mathfrak g $$ (observe that $\tau$ is of finite order). So all elements in an open subset of $G^0$ are twisted conjugate to an element of $C$. For the second step observe that twisted conjugation on $C$ is ordinary conjugation and that $T_0$ is a maximal torus of $C$. So all elements in an open subset of $C$ are (twisted) conjugate to an element of $T_0$. q.e.d<|endoftext|> TITLE: Extending the product measure on $2^\omega$ QUESTION [8 upvotes]: Consider the standard completed product measure $P$ on $\Omega=\{0,1\}^\omega$ corresponding to an i.i.d. sequence of fair coin-flips. Given $n\in\omega$, let $\rho_n$ be the bijection of $\Omega$ to itself given by flipping the value of the $n$th coordinate. Say that a subset $A$ of $\Omega$ has property $H$ iff there is an $n$ such that $\rho_n(A)\cap A=\varnothing$ and $\rho_n(A)\cup A=\Omega$. Assume AC. Question: What can we say about the existence of (countably-additive) extensions of $P$ that assign measure $1/2$ to all subsets with property $H$? Are there any? If so, are there any with nice invariance properties, like invariance under $\rho_n$ and/or under bijections of $\{0,1\}^\omega$ induced by permutations of $\omega$? Remarks: Intuitively, the probability of $A$, if $A$ has property $H$, should be $1/2$. This is true with respect to $P$ if $A$ is $P$-measurable. But given AC, there are $P$-nonmeasurable $A$ having $H$. Say that $\alpha\sim\beta$ iff one can get $\beta$ from $\alpha$ by applying an even number of the $\rho_n$. Given a $\sim$-equivalence class $U$, let $U' = \rho_1 U$. Now take a set $B$ of equivalence classes that contains exactly one member of $\{ U,U' \}$ for every $U$, and let $A$ be the union of the members of $B$. Then $A$ has $H$ with respect to every index $n$, and hence is non-measurable by Lévy's zero–one law. REPLY [2 votes]: There is no such measure. Suppose toward contradiction $\mu$ is such a measure. Partition $\Omega$ by the mod finite equivalence relation $\sim_{\text{fin}},$ let $X \subset \Omega$ be a choice of representatives and $\pi: \Omega \rightarrow X$ send each sequence to its representative. Let $X_n = \{x \in \Omega: x \triangle \pi(x) \subset [n]\}.$ Each $X_n$ can be extended to some $H_n$ satisfying $\Omega \setminus H_n = \rho_{n+1}(H_n),$ and therefore having property $H.$ For each $x \in X,$ there are cofinitely many $n$ such that $x \in X_n \subset H_n.$ Thus, $\Omega = \bigcup_{n \in \mathbb{N}} \bigcap_{m \ge n} H_m.$ This is an increasing union of sets of $\mu$-measure bounded by $\frac{1}{2},$ contradiction.<|endoftext|> TITLE: Central Limit Theorem when the sum of the variances is finite QUESTION [6 upvotes]: Suppose $(x_i)_{i\in\mathbb{N}}$ a set of strictly positive numbers such that $L=\sum_{i\in\mathbb{N}}x_i$ is finite. Suppose that $(X_i)_{i\in\mathbb{N}}$ is a set independant (real-valued) random variables, each uniformly distributed in $[-x_i;x_i]$. I am interested in $S_n=\sum_{i=0}^nX_i$ when $n\rightarrow\infty$. $S_n$ is clearly in $[-L;L]$, so I guess the Central Limit theorem can't apply here ($\sum_{i\in\mathbb{N}}x_i^2$ is finite), so $S_n$ doesn't converge to $\mathcal{N}(0,?)$ since the probability density function of $S_n$ has a finite support ($[-L;L]$). Can someone help me to find the distribution of $S_n$ when $n\rightarrow\infty$ (is it something like $\mathcal{N}(0,?)$ restricted to $[-L;L]$ ?) Thanks REPLY [2 votes]: There is no central limite theorem here. But never mind, this is a wonderful example of $C^\infty$ function nowhere analytics ! https://en.wikipedia.org/wiki/Fabius_function What are your favorite instructional counterexamples?<|endoftext|> TITLE: Subdirect product of perfect groups QUESTION [25 upvotes]: Let $G$ be a subdirect product of finitely many perfect groups $S_1,\ldots,S_n$. That is, $G \le S_1 \times \cdots \times S_n$, and $G$ projects onto each of the direct factors $S_i$. It is not difficult to construct examples in which $G$ is not perfect. For example, there is a subdirect product of ${\rm SL}_2(5)^2 = (2.A_5)^2$ with the structure $2^2.A_5$ that is not perfect. Let $G/N$ be a solvable quotient of $G$. I would really like to know whether $G/N$ is necessarily abelian. It can be proved that $G/N$ is nilpotent of class at most $n-1$ (I can give more details of that on request), so we can assume that $n \ge 3$. If it helps to assume that the $S_i$ are finite, then please do so. This problem arose in a study of the complexity of certain algorithms for finite permutation and matrix groups. The group $G$ in the applications is the kernel of the action of a transitive but imprimitive permutation (or matrix) group on a block system. So in that situation the $S_i$ are all isomorphic, and ${\rm Aut}(G)$ induces a transitive action on the direct factors $S_i$, but I am doubtful whether assuming those properties would help much in trying to answer the question. Here is a proof that $G/N$ is nilpotent of class at most $n-1$. This proof is due to Keith Kearnes. Call a normal subgroup $N$ of $G$ co-${\mathcal P}$, if $G/N$ has property ${\mathcal P}$, which could be solvable, nilpotent, or perfect. We claim that if $G$ has co-perfect normal subgroups $K_1,\ldots,K_n$, with $\cap_{i=1}^n K_i = \{ 1 \}$, and $N$ is a co-solvable normal subroup of $G$, then $G/N$ is nilpotent of class at most $n-1$. We can then apply this claim to the original problem with $K_i$ equal to the subgroups of $G$ that projects trivially onto $S_i$ to deduce the required result. To prove the claim, observe first that, since each $NK_i$ is both c-perfect and co-solvable, we have $NK_i = G$ for all $i$. In general, for normal subgroups $A,B$ of a group $X$, we have $[AB,X] = [A,X][B,X]$. We write $[A,B,C]$ for $[[A,B].C]$, etc. Using this gives $$[G,G,\ldots,G] = [NK_1,NK_2,\ldots,NK_n] = [K_1,K_2,\ldots,K_n]\prod_i [A_{i1},A_2,\ldots,A_{in}],$$ where, in each of the terms $[A_{i1},A_2,\ldots,A_{in}]$, at least one of the $A_{ij}$ is equal to $N$. Now the first term $[K_1,K_2,\ldots,K_n]$ lies in $\cap_{i=1}^n K_i$ and so is trivial, while each of the remaining terms in the product lies in $N$. So $[G,G,\ldots,G] \le N$, and hence $G/N$ is nilpotent of class at most $n-1$, as claimed. REPLY [12 votes]: In this problem one has a finite sequence of perfect groups $P_1,\ldots,P_k$, a subdirect subgroup $G\leq (P_1\times \cdots\times P_k)$, and a normal subgroup $N\lhd G$. The question is: Under the above assumptions, if $G/N$ is solvable, must it be abelian? The answer is: No. This is explained in a manuscript Peter Mayr, Nik Ruskuc and I recently submitted to the arxiv. We show that if the quotient $G/N$ is solvable, then it must be nilpotent, but there is no restriction on the possible nilpotence class. The smallest example in the paper where $G/N$ is solvable but not abelian involves a perfect group $P$ satisfying $|P|=2^{22}\cdot 3\cdot 5$, a subdirect subgroup $G\leq P^4$, and a normal $N\lhd G$ with $G/N$ nilpotent of class $2$. The group $P$, which is constructed as a subgroup of $\text{SL}_6(\mathbb F_4)$, has a nilpotent normal subgroup of order $2^{20}$ and a complementary subgroup isomorphic to the simple group of order $60$.<|endoftext|> TITLE: Theorem 3.5.2, page 177 of Mumford's Red Book QUESTION [6 upvotes]: I am a student whose major interest is not algebraic geometry, but for various reasons I began to re-read Mumford's Red Book, and I found myself stuck in a quite simple theorem. Since I had studied some algebraic geometry before, (based on my remaining intuitions) this theorem seems very simple, but I am having trouble understanding the last part of the proof. The theorem is as follows: (Theorem 2, page 177) Let $\mathcal O$ be a complete local ring, with maximal ideal $\mathfrak m$, residue field $k$. Let $f:X\rightarrow Y = \mathrm {Spec} (\mathcal O )$, $x\mapsto y = [\mathfrak m ]$. Assume $k=\kappa (y) \simeq \kappa (x)$. Then $f$ etale near $x$ $\Rightarrow $ $f$ a local isomorphism near $x$. In this book, Mumford first defines etale morphisms non-intrinsically: he first defines that the morphism $X=\mathrm {Spec} R[x_1 , \cdots , x_n ] / (f_1 , \cdots , f_n ) \rightarrow \mathrm {Spec} R$ induced by the inclusion is etale at a point $x\in X$ when the Jacobian of $f_1 , \cdots , f_n $ does not vanish at $x$. Using this definition, he reduces this theorem to the following situation: $$X=\mathrm {Spec} \mathcal O [x_1 , \cdots , x_n ] / (f_1 , \cdots , f_n )$$ $$ x=\mathrm m + (x_1 -a_1 , \cdots , x_n - a_n )$$ $$\mathrm {det} (\partial f_i / \partial x_j )(x)\neq 0,$$ where $a_1 , \cdots , a_n \in \mathcal O$. Then by Hensel's lemma, he finds elements $\alpha _1 , \cdots , \alpha _n \in \mathcal O$ such that $$Z=\mathrm {Spec} \mathcal O [x_1 , \cdots , x_n ] / (x_1 -\alpha _1 , \cdots , x_n -\alpha _n )$$ is a subscheme of $X$ through $x$, which is clearly isomorphic to $Y$. Now he tries to prove that $X=Z$ near $x$, by first defining $\mathcal Q$ to be the $\mathcal O _X $-ideal defining $Z$ and then as follows, which I cannot understand: If this is shown, then the rest is clear from Nakayama's lemma. I think that I am missing something fundamental and this might be a stupid question, but from the start, I can't see why $\mathcal O_{y,Y}=\mathcal Q$. Can someone please explain this thoroughly? This might be an unappropriate question for mathoverflow; if so, I will migrate this down to MSE. REPLY [2 votes]: To make this question non-active, I will post a simple answer: If one replaces $ \mathcal Q$ by $\mathcal O$ in the diagram, we have $\mathcal O _{x,X} = f^* (\mathcal O ) \oplus \mathcal Q _x $ and $\mathcal m _{x} = f^* (m_y ) \oplus \mathcal Q _x $ (they are $k$-algebras). Now we have $$ f^* (m_y) \oplus \mathcal Q _x = m_x = f^* (m_y ) \mathcal O _{y,Y} =f^* (m_y ) (f^* (\mathcal O ) \oplus \mathcal Q _x ) . $$ The rest is clear.<|endoftext|> TITLE: Why is it difficult to obtain the next differential in a spectral sequence? QUESTION [14 upvotes]: I am following Hatcher's notes on spectral sequences. On page 522 an exact couple $(A, E, i, j, k)$ is defined. You can construct a derived exact couple easily from this to get your desired $E'$ however constructing the new differential from this information is said to be not so easy in practice. I would like to know why this is, to me it seems that $d'$ is simply $j'k'$. Is this a difficult thing to calculate or is it simply tedious? It seems to be the same opinion everywhere that given a term in a spectral sequence $(E^r, d_r)$, calculating $E^{r+1}$ is easy and $d_{r+1}$ just isn't practically possible with the given information. I'd like to think there is some algorithm where you give a pair $(E^r, d_r)$ and it spits out a pair $(E^{r+1},d_{r+1})$. So what is the reason this isn't the case? REPLY [12 votes]: Expanding on Tyler Lawson's comment, the point of a spectral sequence is often that we know what $E$ is concretely, and we want to use this to compute $A$. The issue is that if we want to explicitly figure out what $d'$ is, we need to understand $A$ itself, which becomes quite circular. Indeed, to compute $d'$, the procedure is to apply $k$, compute $i^{-1}$ of that, then apply $j$. But $i$ is a map $A \to A$, and in general if we want to understand this map, we need to know something about $A$. For example, if $h_*$ is a (generalized) homology theory and $X$ is a CW complex, then the Atiyah-Hirzebruch spectral sequence is a spectral sequence converging to $h_*(X)$ whose $E_1$ page is the cellular chain complex of $X$ with coefficients in $h_*(*)$. If we are given a cellular decomposition of $X$, then we have some fairly solid grasp on what $E_1$ is, and the differentials $d_1$ are just the differentials we use to compute cellular homology, which can be calculated using local degrees. These are things we can actually work with. However, if we wanted to compute $d_2$, we need to understand $i$. In this case, $i$ is given by the inclusion map $h_*(X^{(n)}) \to h_*(X^{(n+1)})$, where $X^{(n)}$ is the $n$th skeleton of $X$. But our original goal was to compute what $h_*(X)$ is! If we have some concrete information about $i$, chances are we already know what $h_*(X)$ is, and the whole computation is moot. What the machinery of spectral sequences gives us is that we can often use other means to deduce the nature of the differentials $d_r$, without resorting to taking apart the construction (since that is circular). The most common way is that if $d_r$ maps to or from the zero group, then $d_r$ must be zero. At other times, we use certain multiplicative structures on the spectral sequence, or we use the naturality of the constructions to compare the spectral sequence with others. One might object that even understanding $d_1$ requires knowledge about $A$, since $j$ and $k$ also pass through $A$. This is true, and the hard part of setting up a useful spectral sequence is often to identify what $d_1$ is doing and give a concrete description of the $E_2$ page. In the successful cases, we can get away with sufficiently high-level arguments that don't depend on what the details of the space we are working with.<|endoftext|> TITLE: Is this problem of selecting points NP-hard? QUESTION [6 upvotes]: I have an optimization problem related, in a certain way, to the expression of a set of points with the least number of points and I don't know if it is NP-hard (or not). More formally, I have a ground set $U$ of points with coordinates $(x_i,y_i)$ for $i=1,...,n$. Given an integer $k\leq n$, the decision problem consists in finding a subset $I$ (made of $k$ points) of $U$ such that every point of $U$ can be written as the convex combination of at most two points of $I$. For example, let $U=\{(0,0) ; (1,3) ;(2,6); (3,0);(2,3/2);(4,3);(6,0)\}$. There is a solution with $k=4$ wich is $I=\{(0,0) ; (2,6); (4,3);(6,0)\}$. Indeed, we have $(1,3)=\frac{(0,0)+(2,6)}{2}$; $(2,3/2)=\frac{(0,0)+(4,3)}{2}$ and $(3,0)=\frac{(0,0)+(6,0)}{2}$. IP formulation Let $C_{ij}$ be the indices of the points belonging to $U$ which can be expressed as a convex combination of $(x_i,y_i)$ and $(x_j,y_j)$. I think all these $C_{ij}$'s can be computed in $O(n^2)$ time. Then we can write the following IP with $O(n^2)$ binary variables: $$ \min x_1+...+x_n $$ $$\sum_{(i,j)|p\in C_{ij}}y_{ij}\geq 1 \text{ for all } p=1,...,n,$$ $$y_{ij}\leq x_i, y_{ij}\leq x_j \text{ for all } i,j=1,...,n,$$ $$x_i\in\{0,1\} \text{ and } y_{ij}\in\{0,1\} \text{ for all } i,j=1,...,n.$$ I started looking into Garey & Johnson for some similarities with existing problems, but I don't see how to "express" the condition of the convex combination. Thanks for any advice! (related problems, ...) REPLY [5 votes]: This problem is reducible from VERTEX-COVER. A rough description: let the input graph to VERTEX-COVER be $G = (V, E)$ with $|V| = n$. Choose integer $n << t = O(n^c)$, and create a convex polygon $P$ on $1 + n + 2t$ vertices $C_0$, $C_1$, $\ldots$, $C_n$, $D_1$, $\ldots$, $D_t$, $E_1$, $\ldots$, $E_t$. Choose vertices $V_1$, $\ldots$, $V_n$ on sides $C_0 C_1$, $\ldots$, $C_{n - 1} C_n$. Now, for each edge $ij \in E$ with $i, j \in V$ and $i < j$ add all points on intersections of $V_i D_k$ and $V_j E_l$ with $1 \leq k, l \leq t$. We choose coordinates in such a way that no point of the plane has more than two diagonals of $P$ passing through it, and no three intersection points share non-trivial common lines together with vertices $P$ (other than diagonals of $P$ when intended). Obviously, we have to include all vertices of $P$ into $I$. Vertices $V_i$ are already covered by sides of $P$, so we have freedom of including any subset of them. For an edge $ij \in E$ we have to cover all intersections of $V_i D_k$ and $V_j E_l$ either by including one of $\{V_i, V_j\}$ into $I$, or take some of the intersections into $I$; in the latter case, $\Omega(t)$ intersections need to be included, which effectively prohibits us from doing this. We can now see that the minimal size of $I$ is exactly $1 + n + 2t + VC(G)$, where $VC(G)$ is the size of minimal vertex cover of $G$. The last remark is that we have to make sure that it is possible to choose all coordinates so that length of coordinates of each point is polynomial in $n$, which is technical.<|endoftext|> TITLE: Is every weak $\infty$-bicategory (à la Lurie) an $\infty$-bicategory? QUESTION [12 upvotes]: In Definition 4.1.1 of $(\infty,2)$-Categories and the Goodwillie Calculus I, Lurie defines a weak $\infty$-bicategory to be a scaled simplicial set that has the extension property with respect to every scaled anodyne morphism. In Theorem 4.2.7, he defines a model structure on $\operatorname{Set}_{\Delta}^{\operatorname{sc}}$, the category of scaled simplical sets, and in Definition 4.2.8 he defines an $\infty$-bicategory to be a scaled simplicial set that is a fibrant object in the model category $\operatorname{Set}_{\Delta}^{\operatorname{sc}}$. Every $\infty$-bicategory is a weak $\infty$-bicategory, because every scaled anodyne morphism is a bicategorical equivalence (Proposition 3.1.13). What about the converse? Is every weak $\infty$-bicategory an $\infty$-bicategory? EDIT: I think I now have a proof that, indeed, every weak $\infty$-bicategory is an $\infty$-bicategory. I will post it tomorrow. REPLY [8 votes]: A few years later, it has been shown by Gagna, Harpaz, and Lanari that the answer is yes. Every weak $\infty$-bicategory is an $\infty$-bicategory.<|endoftext|> TITLE: Does the orthogonal group act irreducibly on totally symmetric tensors? QUESTION [6 upvotes]: Consider the action of the orthogonal group $\operatorname{O}(d)$ on $k$-way tensors $(\mathbb{R}^d)^{\otimes k}$ defined by $$Q(x_1\otimes\cdots\otimes x_k)=Qx_1\otimes\cdots \otimes Qx_k$$ and extending linearly. I would like to locate a proof that the invariant subspace of totally symmetric tensors $\operatorname{Sym}^k(\mathbb{R}^d)$ carries an irreducible representation of $\operatorname{O}(d)$. If we replace $\operatorname{O}(d)$ with $\operatorname{GL}(V)$ for some vector space $V$, then Schur-Weyl duality gives that the corresponding representation in $\operatorname{Sym}^k(V)$ is irreducible (see Chapter 9 in Procesi's Lie Groups book). In the case where $V=\mathbb{C}^d$, Theorem 1.9 of this thesis suggests that the restriction of any irreducible representation of $\operatorname{GL}(\mathbb{C}^d)$ to $\operatorname{U}(d)$ is irreducible. The proof is said to be contained in Chapter 12 of Carter, Segals and MacDonald's Lectures on Lie Groups and Lie Algebras, but the closest result I could find (Proposition 12.3) gives that every representation of $\operatorname{U}(d)$ is the restriction of a unique holomorphic representation of $\operatorname{GL}(\mathbb{C}^d)$. It is not clear to me that $\operatorname{U}(d)$ then inherits irreducible representations from $\operatorname{GL}(\mathbb{C}^d)$. Furthermore, this result seems to leverage a relationship between $\operatorname{U}(d)$ and $\operatorname{GL}(\mathbb{C}^d)$ known as complexification that is not exhibited between $\operatorname{O}(d)$ and $\operatorname{GL}(\mathbb{R}^d)$. REPLY [12 votes]: It is classical that, as $O(n)$-representation, $$ \text{Sym}^k(\mathbf R^n)=H^k\oplus qH^{k-2}\oplus q^2H^{k-4}\oplus\ldots $$ Here $q=x_1^2+\ldots+x_n^2$ is the quadratic form defining $O(n)$ and $H^k\subseteq \text{Sym}^k(\mathbf R^n)$ is the space of harmonic polynomials, i.e., polynomials which are killed by the Laplacian $\Delta=\partial_{x_1}^2+\ldots+\partial_{x_n}^2$. Moreover all spaces $q^lH^k$ are irreducible provided $n\ge2$.<|endoftext|> TITLE: Is there a standard name for (non-square) matrices with orthonormal columns? QUESTION [10 upvotes]: One encounters often in numerics non-square matrices with orthonormal columns, i.e., $U\in\mathbb{R}^{m\times n}$, with $m > n$, such that $U^TU=I$ (but, clearly, $UU^T \neq I$). Is there a name for these matrices? REPLY [11 votes]: Orthonormal $\boldsymbol n$-frames :  https://en.wikipedia.org/wiki/Stiefel_manifold. Added: This terminology of Hirzebruch (1966), Steenrod (1951) translates the $\boldsymbol n$-Systeme of Stiefel (1936), $\boldsymbol n$-podes, $\boldsymbol n$-pèdes or multipèdes of Einstein (1931), Waelsch (1907), $\boldsymbol n$-Beine or Vielbeine of Hirzebruch (1956), Einstein (1928), Blaschke (1920), or Waelsch (1906). REPLY [3 votes]: Francois Ziegler has already provided a very relevant answer. Let me point out three further relevant things: 1. While the OP is not expressly interested in matrices with integer entries, a very relevant article nevertheless is W. Plesken: Solving $XX^{\mathrm{tr}}=A$ Over the Integers. Linear Algebra and its Applications. Volumes 226–228, September–October 1995, Pages 331-344. Therein, in is in particular proved (let $A=I$ and $X=U^{\mathrm{t}}$) that in the context of the OP $U^{\mathrm{t}}U = I\quad$ ${}\quad$ if and only if${}\qquad\qquad$ $(U U^{\mathrm{t}})^2 = U U^{\mathrm{t}}$ $\quad$and$\quad$ $\mathrm{tr}( UU^{\mathrm{t}})=n$ $\color{white}{( U^{\mathrm{t}}U = I )}{}\quad$ if and only if${}\qquad\qquad$ $UU^{\mathrm{t}}$ is idempotent and has its trace equal to its rank $\color{white}{( U^{\mathrm{t}}U = I )}{}\quad $ if and only if${}\qquad\qquad$ $UU^{\mathrm{t}}$ is idempotent, where the first equivalence holds by Proposition 2.2 in loc. cit., the penultimate holds by a mere reformulation of the first-mentioned equivalence (note that the OP's hypotheses imply that $U U^{\mathrm{t}}$ has rank $n$), and the last step being a widely-known, not-quite-obvious fact from linear algebra (every idempotent matrix has its trace equal to its rank). So we have shown: The matrices of the OP are precisely those $U\in\mathbb{R}^{m\times n}$ for which $UU^T$ is an idempotent in the matrix ring $\mathrm{Mat}(m\times n;\mathbb{R})$. 2. Moreover, in the context defined by the OP we have: If the OP's hypotheses are satisfied, then the Moore-Penrose pseudoinverse of $U$ equals the transpose of $U$. Conversely, if $U$ is a real matrix whose Moore-Penrose pseudoinverse equals its transpose, then the sum of the squares of the rank-sized minors equals $1$. 1 Proof of 2. By the usual formula, $U^+ = \biggl(\overline{U}^{\mathrm{t}}\cdot U\biggr)^{-1}\cdot \overline{U}^{\mathrm{t}}\qquad\qquad$ (0). Sufficiency: If the OP's hypotheses are satisfied, then $\overline{U}^{\mathrm{t}}\cdot U = \mathrm{Id}$ and $\overline{U}^{\mathrm{t}} = U^{\mathrm{t}}$, so (0) implies $U^+=U^{\mathrm{t}}$. Necessity: Suppose conversely that $U^+=U^{\mathrm{t}}$. Then (0) implies that $U^{\mathrm{t}} = (U^{\mathrm{t}}U) U^{\mathrm{t}}$. This implies $U^{\mathrm{t}}U = ((U^{\mathrm{t}}U) U^{\mathrm{t}})\cdot(U(U^{\mathrm{t}}U))$ $=$ (by associativity) $=$ $( U^{\mathrm{t}}U)^3$, hence, applying the homomorphism $\det\colon \mathbb{R}^{m\times m}\to\mathbb{R}$ it follows that, abbreviating $d:=\mathrm{det}(U^{\mathrm{t}}U)$, we have $d = d^3$. Since the OP's hypotheses imply that $U^{\mathrm{t}}U\in\mathbb{R}^n$ has full rank $n$, we know that $d\neq 0$, and hence it follows that $1=d^2$, and hence $d\in\{-1,+1\}$; for further reference, $\det(U^{\mathrm{t}}U)\in\{-1,1\}\qquad\qquad (1)$. By the Cauchy-Binet-theorem, it follows from (1) that (in particular since sums of squares of real numbers are non-negative) $$1 = \sum_{S:\quad\textsf{$n$-element subsets of $m$}} \det( U|_{S\times n})^2 $$ The completes the proof of 2. 1 This is not a "name", yet may lead the OP to useful relevant literature.<|endoftext|> TITLE: Connective spectra and infinite loop spaces QUESTION [11 upvotes]: It seems to be standard that connective spectra are "the same" as infinite loop space. However, I do not understand the reason why the associated spectrum is connective. For me, an infinite loop space is a space $Y_0$ together with a collection of pointed spaces $Y_1, Y_2, \dots$ and homotopy equivalences (or homeomorphisms) $f_j: Y_{j-1} \rightarrow \Omega Y_j$, $j=1, 2, \dots$. Now there is an obvious way to associate an $\Omega$-spectrum $X$ to this data: Just set $$ X_j = \begin{cases} Y_j & j \geq 0 \\ \Omega^j Y_0 & j < 0 \end{cases}$$ for the underlying spaces with the map $g_j: X_{j-1} \rightarrow \Omega X_j$ being given by $f_j$ for $j \geq 0$ and the identity for $j < 0$. Clearly, there is no reason why $X$ should be connective: Just take $Y_0$ the zero space of a non-connective $\Omega$-spectrum. So what is the precise meaning of this statement that infinite loop spaces are the same as connective spectra? REPLY [4 votes]: A simple way to answer this question is: denote by $Spec$ the homotopy category of spectra, $Spec _0$ the category of (-1)-connected spectra, $InfL$ be the homotopy category of infinite loop spaces. The assertion that "connective spectra are the same as infinite loop space" simply means that the composition of functors $$Spec _0 \stackrel{i}{\to} Spec\stackrel{\Omega ^{\infty}}{\to} InfL$$ is an equivalence, where $i$ denotes the inclusion. You gave an example of infinite loop space in the image of $\Omega ^{\infty}$, which doesn't mean that it is not in the image of $\Omega ^{\infty}i$.<|endoftext|> TITLE: divisibility independence QUESTION [14 upvotes]: The following is a standard combinatorics question: Any set of $n+1$ numbers from $1, \dotsc, 2n$ contains a pair of numbers $a, b$ where $a \left| b \right.$ The argument is by pigeonhole principle: consider $A_i = \{2^k (2i-1), k\in \mathbb{N}\}.$ The sets $A_i$ cover $1, \dotsc, 2n,$ and there are $n$ of them. The question is this: to find the maximal division-free set, the obvious way is to do the following (in Mathematica): divGraph[n_] := With[{pairs = Subsets[Range[n], {2}]}, With[{divis = Select[pairs, Mod[#[[2]], #[[1]]] == 0 &]}, Graph[Apply[Rule, #] & /@ divis]]] FindIndependentSet[divGraph[n]] Now, when one runs this for smallish numbers one sees that the maximal set $S_m(2n)$ always has cardinality $n,$ but also, the minimum of $S_m(2n)$ is equal to 1 once 2 thrice 4 nine times 8 $27$ times 16 $81$ times 32 $243$ times And so on. The question is: what is going on? Is this the true size of the minimal element of $S_m?$ As Vladimir Dotsenko points out, the obvious example has the minimum equal to $n+1.$ REPLY [13 votes]: First, I ran some bruteforce myself (I don't have access to Mathematica at the moment), and I'm fairly sure the mysterious numbers obtained in OP ($1, 2, 2, 2, 4\ldots$) are the minimal possible numbers that are present in any $S_m(2n)$. I'm going to explain the pattern in this assumption. Long story short, the minimal possible number in $S_m(n)$ is $2^{\left\lfloor \frac{\log n}{\log 3} \right\rfloor}$. As follows from the OP's explanation, for any odd number $x \leq n$ there is exactly one representative of form $x 2^k$ in any $S_m(n)$. Let us call the exponent $k = d(x)$ the degree of $x$ in a particular $S_m(n)$. Obviously, for non-equal $x \vert y$ we must $d(x) > d(y)$. One can see that the "critical" values where the OP's minimal numbers change are exactly $\frac{3^k + 1}{2}$, that is, the minimum changes whenever a new power of 3 arrives in the set. One can indeed see that in any $S_m(n)$ we have $d(1) \geq \left\lfloor \frac{\log n}{\log 3} \right\rfloor = k_0$ by simply applying the inequality above for a divisor chain $1 \vert 3 \vert \ldots \vert 3^{k_0}$. The numbers $2^{k_0}$ follow the OP's pattern exactly. But why any number that is not a binary power cannot be less than $2^{k_0}$ in an $S_m(n)$? Let $1 < x \leq n$ be an odd number. Then for any $S_m(n)$ the numbers $\{2^{d(x)}, 3\cdot 2^{d(3x)}, \ldots\}$ must be an $S_m(\left\lfloor \frac{n}{x} \right\rfloor)$. As we have shown above, that implies $d(x) \geq \left\lfloor \frac{\log \lfloor n/x \rfloor}{\log 3} \right \rfloor$. It suffices to show $2^{\left\lfloor \frac{\log n}{\log 3} \right\rfloor} < x \cdot 2^{\left\lfloor \frac{\log \lfloor n/x \rfloor}{\log 3} \right\rfloor}$. But $2^{\left\lfloor \frac{\log n}{\log 3} \right\rfloor - \left\lfloor \frac{\log \lfloor n/x \rfloor}{\log 3} \right\rfloor} \leq 2^{\frac{\log x}{\log 3} + 1} = 2 \cdot x^{\frac{\log 2}{\log 3}} < x$ for all $x > 5$ (for $x = 3, 5$ we may still have the claim by handling the rounding error more carefully). To finish the proof, notice that the set $\{x \cdot 2^{\left\lfloor \frac{\log \lfloor n/x \rfloor}{\log 3} \right\rfloor}\}$ with $x$ ranging over odd numbers at most $n$ is an $S_m(n)$. Note that this set is minimal with respect to any measure, e.g. no proper divisor of any of its elements can be in an $S_m(n)$, hence it's also lexicographically minimal when sorted.<|endoftext|> TITLE: Number fields without abstract algebra QUESTION [11 upvotes]: I have a student who has taken some linear algebra but no abstract algebra, and he wants to learn some interesting math. I explained how, by using companion matrices, one can represent and work with roots of polynomials via matrices. The resulting expanded number system is a finite dimensional vector space over the base field (here just the rationals) and multiplication by elements is actually a linear transformation of that space. He seemed intrigued, and I would like to know how much more of the wonderful world of number fields he could learn just from the linear algebra perspective, avoiding abstract algebra (e.g. groups and rings). That probably means most Galois theory is out. I think a very light amount of abstract algebra might be okay, but I don't want that to be what he is spending most time learning during this semester's project. Some modular arithmetic would also be fine. Does anyone know a good reference (book/chapter/article) that might suit this purpose? Or maybe a nice theorem we could study? Thanks. REPLY [11 votes]: I guess you are looking for the appendix by Olga Taussky in Harvey Cohn's "A Classical Invitation to Algebraic Numbers and Class Fields". This being said, there is a reason why algebraic number theory is called "algebraic".<|endoftext|> TITLE: A mysterious connection between Ramanujan-type formulas for $1/\pi^k$ and hypergeometric motives QUESTION [39 upvotes]: The question below is the follow-up of this question on MathOverflow. Motivation: As is stated in the former question, those identities(formula (35)-(44)) of $1/\pi$ attributed to Ramanujan are related to surfaces with Picard rank 20(see the paper of Elkies and Schuett) in the Dwork family $$x_1^4+x_2^4+x_3^4+x_4^4=4\lambda x_1x_2x_3x_4.$$ Jesus Guillera found a few Ramanujan-type formulas for $1/\pi^2$(which can be found in W. Zudilin's paper), three of which(formula (92)(93)(94) in Zudilin's paper) are related to the Dwork family $$x_1^6+x_2^6+x_3^6+x_4^6+x_5^6+x_6^6=6\lambda x_1x_2x_3x_4x_5x_6$$through Picard-Fuchs equation. It is reasonable to conjecture that the Guillera's formulas are related to "singular" members in Dwork family. The end of this paper suggests that the (Hasse-Weil)L-functions of those "singular" members behave differently from those "ordinary" members. It seems that "hypergeometric motive" package(developed by M. Watkins, based on the work of N. Katz et al.) in MAGMA offers a possible way to investigate those L-function numerically(although the L-functions are different from Hasse-Weil L-function). Experiment: M. Watkins and David Roberts tried to find out imprimitive L-function attached to hypergeometric motives in this document(p.29, Table 15), where the L-function attached to the motive can be factorized to the product of two L-functions. One can immediately recognize the numbers corresponding to Guillera formula(formulas (86)(87)(88) in Zudilin's Paper). It is amazing that one can find out that EVERY L-function associated to the Guillera formula is imprimitive. Example: Guillera conjectured that $$\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_n(\frac{1}{8})_n(\frac{3}{8})_n(\frac{5}{8})_n(\frac{7}{8})_n}{(n!)^5}(1920n^2+304n+15)\frac{1}{7^{4n}}=\frac{56\sqrt{7}}{\pi^2}.$$ The HGM package can evaluate Frobenius trace and Euler factors of the corresponding hypergeometric motive $$H([1/2,1/8,3/8,5/8,7/8],[0,0,0,0,0],\tilde{t}),$$ where $$1/\tilde{t}=t=\frac{1}{7^4}.$$ Almost each Euler factor $L_p(T)$ are quintic polynomials, and the absolute values of roots of Euler factors are $p^2.$ Adapting the primitivity test of M. Watkins(p. 25), if one assumes the truth of Selberg's conjecture for Selberg class, then(for a primitive L-function) the second moment of (normalized) Frobenius trace $a_p$ $$\frac{1}{\pi(X)}\sum_{p=10. None of them can pass the test. I am not sure if one can find more similar formulas for higher reciprocal powers of $\pi$. Update2: A recent preprint of David P. Roberts and Fernando Rodriguez Villegas is provided here. It seems that the splitting of underlying hypergeometric motives does not only provide period relations of (generalized) hypergeometric functions at algebraic arguments, but also supercongruences of truncated hypergeometric series. REPLY [8 votes]: It should follow from Grothendieck's period conjecture that whenever a period of an irreducible motive $V$ is also the period of a motive $W$, then $V$ is a summand of $W$ in the category of motives, and hence the L-function of $V$ is a factor of the L-function of $W$. This predicts that in your cases you should see that the normalized average value of $a_p$ is one, except in the first case where you should average $a_p$ against the primitive quadratic Dirichlet character mod $28$ (= character of the field $\mathbb Q(\sqrt{7})$) Conversely, it should be true that if the L-function splits, the motive splits, and so the period splits. However, there is no reason to believe that the motive splits into motives whose periods are given by some kind of closed-form formula. To make this work in full detail one has to understand how the rational functions of $n$ you're throwing in affect the motive. In particular by figuring out what happens to the weight it should tell you what power of $\pi$ to expect and when to expect $\zeta(3)$.<|endoftext|> TITLE: The "strong" measure number QUESTION [11 upvotes]: Beyond measure zero we have yet another measure-y notion of smallness: strong measure zero. A set $S\subseteq\mathbb{R}$ is strong measure zero if, for any $f:\mathbb{N}\rightarrow\mathbb{R}_{>0}$, there is a sequence $U_i$ of open sets with the diameter of $U_i$ is $ TITLE: Feit-Thompson conjecture QUESTION [60 upvotes]: The Feit-Thompson conjecture states: If $p TITLE: How to construct a small coprime? QUESTION [18 upvotes]: Given an integer $n$, is there a deterministic algorithm to find in poly$(\log n)$ time an integer $q$, $n < q< n^{c}$, such that $gcd(q,n!)=1$? Here $c>1$ is some fixed constant. Obviously, a small prime $n< q < 2n$ works. But those are really hard to find deterministically. One can of course, deterministically test primality of $n+1, n+2, \ldots, n+C(\log n)^2,$ but that's not yet proved to work. Taking $q=n!+1$ works for the $gcd$ condition, but such $q$ is too large for my applications. If $q< n^c$ is open, but say for the case $q< n^{O((\log n)^k)}$ this is known, what are the best known bounds? REPLY [2 votes]: If we could find a number co-prime to $2^n!$ in $n^{O(1)}$ time, or even just a number divisible by at least one prime greater than $2^n$, we could factor it to find such a prime. This would constitute a solution to the strong conjecture with factoring, so it is an open problem. As far as I know, it is open whether or not it is possible to find a prime larger than $2^n$ in time $2^{\frac{n}{2}+o(1)}$ with or without a factoring oracle. The finding primes page doesn't say this explicitly but I don't see any obvious method to beat this using a factoring oracle from the information presented. Perhaps someone can confirm that this weaker problem is open.<|endoftext|> TITLE: Hyperbolic Volume and Chern-Simons QUESTION [17 upvotes]: In the paper ``Analytic Continuation Of Chern-Simons Theory'' (arXiv:1001.2933) Witten postulates that hyperbolic volume of 3-dimensional manifold coincides with the value of the Chern-Simons functional of the hyperbolic connection (see section 5.3.4). Let me state this more precisely. Let $M$ be a three dimensional spin manifold. Consider a Riemannian metric $\rho$ on $M$ with constant negative curvature $-1$. The universal cover ($\tilde{M},\tilde{\rho}$) is isometric to the hyperbolic space (${H^3},\rho^{st}$). The fundamental group of $M$ acts on $H^3$ by isometries. It therefore defines a homomorphism $g:\pi_1(M)\to \text {Isom}(H^3)=\text {PSL}(2,{\mathbb{C}})$. Def The hyperbolic connection $A_{\rho}$ on the trivial $PSL(2,\mathbb{C})$-bundle $E$ on $M$ is a flat connection with monodromy representation $g$. Rem The inclusion $\text{SO}(3)\subset \text{PSL}(2,\mathbb{C})$ is a homotopy equivalence. Since $M$ has a spin structure we can lift $A_{\rho}$ to an $SL(2,\mathbb{C})$-connection. Def The value of the Chern-Simons functional on an $\text{SL}(2,\mathbb{C})$-connection $A$ in the trivial bundle on $M$ is given by \begin{equation} CS(A):=\int_{M}tr[A,dA]+\frac{2}{3}tr[A,A\wedge A]. \end{equation} Here $tr[\cdot,\cdot]$ is defined as follows: \begin{equation} tr[\cdot,\cdot]: \Omega^n(M, g)\otimes\Omega^m(M, g) \xrightarrow{\wedge} \Omega^{m+n} (M, g\otimes g) \xrightarrow{tr} \Omega^{m+n} (M,\mathbb{C}). \end{equation} Here $g$ is a simple Lie algebra and the trace over the last arrow is the standard non-degenerate invariant symmetric bilinear form on $g$. Rem A gauge transformation $s \in \Omega^0(M,E)$ changes CS by an integer: $CS(A)-CS(s^*A)\in 2\pi \mathbb{Z}$. Finally, what I'm seeking for is a reference for the formula (which seems to be well known) \begin{equation} 2\pi \text{ Im } \text{CS}(A_{\rho})=\text{Vol}_{\rho}. \end{equation} REPLY [8 votes]: This is a Theorem of Yoshida, the reference is Yoshida, Tomoyoshi: ''The η-invariant of hyperbolic 3-manifolds.'' Invent. Math. 81, 473-514 (1985). http://mathlab.snu.ac.kr/~top/articles/Yoshida.pdf The proof is by explicit computation and comparison of the Chern-Simons form and the volume form. There is an alternative proof using the Extended Bloch Group, the reference is Neumann, Walter D.: ''Extended Bloch group and the Cheeger-Chern-Simons class.'' Geom. Topol. 8, 413-474 (2004). http://arxiv.org/abs/math/0307092 with some more details in Goette, Sebastian; Zickert, Christian K.: ''The extended Bloch group and the Cheeger-Chern-Simons class.'' Geom. Topol. 11, 1623-1635 (2007). http://www2.math.umd.edu/~zickert/The%20extended%20Bloch%20group%20and%20the%20Cheeger-Chern-Simons%20class%20(Goette,%20Zickert).pdf<|endoftext|> TITLE: Intuition for the infinite cardinals p and t (now that p = t)? QUESTION [11 upvotes]: I wonder if there are natural examples of sets whose cardinality is $\frak p$ and $\frak t$? Examples that could provide intuition for non- set-theory experts? I find the definitions in the Malliaris & Shelah paper that established $\frak p=t$ difficult to grok, so to speak. I am not asking to understand the proof, just to understand the cardinals themselves. Malliaris, Maryanthe, and Saharon Shelah. "Cofinality spectrum theorems in model theory, set theory, and general topology." Journal of the American Mathematical Society 29.1 (2016): 237-297. (Journal link; Link to earlier arXiv version.) Related MO question: Short proof of $\frak p=t$. REPLY [4 votes]: A couple of simple examples; they don't have cardinality $\mathfrak p$ or $\mathfrak t$ but depending on your knowledge they may or may not help. They are families of sets of natural numbers that I'll call $\mathcal F$, $\mathcal L$, $\mathcal M$, and $\mathcal D$. Let $\mathcal F=\{X\subseteq\mathbb N: X\text{ is co-finite}\}$ and $\mathcal L=\{[n,\infty)\cap\mathbb N: n\in\mathbb N\}$. In terms of the definitions on page 2 of M. Malliaris, S. Shelah, Cofinality spectrum theorems in model theory, set theory and general topology, arXiv:1208.5424 (or direct pdf link) $\mathcal L$ is not a tower, because even though it is well-ordered by $\supseteq^*$, it has $\mathbb N$ itself as a pseudo-intersection (the elements of $\mathcal L$ are too big, so to speak). Anyway, $\mathcal L$ is countable, so it's plausible that a tower would have to be uncountable, and indeed that's the cardinality $\mathfrak t$. $\mathcal F$ is a bit bigger than $\mathcal L$, is not well-ordered by $\supseteq^*$, but still has the s.f.i.p. since the intersection of finitely many co-finite sets is still co-finite (since the union of finitely many finite sets is finite). Like $\mathcal L$, the family $\mathcal F$ is countable and unfortunately has $\mathbb N$ as pseudo-intersection. So it's again plausible that the corresponding cardinal $\mathfrak p$ is uncountable. So let's consider $\mathcal M$, the family of sets whose limiting (Banach) density is 1. It has the s.f.i.p. and no pseudo-intersection. Great! However, $\mathcal M$ has cardinality $2^{\aleph_0}$. Now consider $\mathcal D=\{\{k: k\text{ is divisible by }2^m\}: m\in\mathbb N\}$. This is also countable, well-ordered by $\supseteq$, but again unfortunately has an infinite pseudo-intersection, namely $\{1,2,4,8,16,\dots\}$. $\begin{eqnarray*} \text{Family }\quad & \text{Well-ordered?}\quad&\text{S.f.i.p.? }\quad & \text{pseudo-$\cap$? }&\text{Cardinality}\\ \mathcal F & \text{no}& \text{yes} &\text{yes}& \aleph_0\\ \mathcal L & \text{yes}& \text{yes} &\text{yes}& \aleph_0\\ \mathcal M & \text{no}& \text{yes} &\text{no}& 2^{\aleph_0}\\ \mathcal D & \text{yes}& \text{yes} &\text{yes}& \aleph_0\\ \end{eqnarray*}$ REPLY [4 votes]: Each element of p and each element of t is a countable family of infinite sets of natural numbers. Let's start by giving an example of such family. A simple example of a countable family of infinite sets of natural numbers would be $$E = \bigotimes_{j\in \Bbb N} \{k \in \Bbb N : \exists m | k = jm \}$$ so that two elements of $E$ would be $\{ 2,4,6,8\cdots\}$ and $\{ 3,6,9,12\cdots\}$. We can say that a countable family of infinite sets of natural numbers is a subset of $[\Bbb N]^{\aleph_0}$. (We will see that this particular $E$ is not in p.) Now let's define a relationship between sets $A$ and $B$ which I will call "is almost a subset of," written as $A \subseteq^* B$. In words, $A$ is almost a subset of $B$ iff only a finite set of elements of $A$ are not also in $B$. That is, $$A \subseteq^* B \Leftrightarrow |\{ x:x\in A \wedge x \not\in B\}| < \infty $$ For example if $A$ is the set of primes and $B$ is the set of odd natural numbers, then $\{ x:x\in A \wedge x \not\in B\} = \{2\}$ and $A \subseteq^* B$. Note that it is quite possible for $A \subseteq^* B$ and $B \subseteq^* A$. Now let's define p and give an example of one element of p. p is the set of all countable families of infinite sets of natural numbers such that family $f \in \mathbf{p}$ if and only if two conditions are met: Every non-empty finite subfamily of $f$ (every finite collection of sets, all of which are in $f$) has an infinite intersection. But there is no infinite set of integers $A$ such that $A$ is almost a subset of each element of $f$. That is, $\left( \forall A \subseteq \Bbb N: B \in f \implies A \subseteq^* B \right) \implies |A| = \infty$. You can see that these two conditions work against one another in some sense. A simple example of a family of sets $f_1$ meeting the first condition is $\left\{ \{ n+0 :n \in \Bbb N\},\{ n+1 :n \in \Bbb N\}, \cdots \right\}$ since the intersection of any finite number of elements of $f_1$ consists of all numbers $\geq$ the highest starting point of any of that finite set of elements. But that family $f_1$ fails the second condition, because if $A$ is the (infinite) set of even integers, then $A$ is almost a subset of each member of $f_1$. An example of an element of p is the family of sets (indexed by $k\in \Bbb N$) of the form $\{ m^k : k\in \Bbb N\}$. The first condition is met since if $\ell$ is the least common multiple of all the indices in the finite sub-family then the set $\{ m^\ell : m\in \Bbb N\}$ is in every set in the finite subfamily. The second condition is met since there are an infinite number of integers that are cubes but not squares and vice-versa, so any infinite set $A$ of integers is either not almost a subset of the squares, or not almost a subset of the cubes, unless all but a finite number of elements of $A$ are sixth powers. But then $A$ won't be almost a subset of the fifth powers, unless all but a finite number of elements of $A$ are $30$-th powers; and so forth. Now let's define t and give an example of one element of t. t is the set of all countable families of infinite sets of natural numbers that can be well-ordered by the $\subseteq^*$ relation. (Remember, well-ordering will mean that if family $T$ is in t then every (non-equivalent under $A \subseteq^* B$) pair of elements of $T$ can be related by $\subseteq^*$ in one direction or the other, and every subset of $T$ has a least element in that ordering.) A simple example of an element of t is the family $f_1$ described above. This "tower" happens not to be in p. The p = t question, then, is whether p can be mapped onto t, or t can be mapped onto p, or they are in $1:1$ correspondence. In the paper by Malliaris and $(\mbox{Shelah})^2$ there is a minor wording error in definition 1.1 whch might be causing some confusion: The first bullet should read $D$ has an infinite pseudo-intersection if there is an infinite $A\subseteq \Bbb N$ such that $\forall B\in D, A \subseteq^* B$.<|endoftext|> TITLE: Are countable dense subspaces of $\mathbb{R}^n$ homeomorphic to ${\mathbb Q}^n$? QUESTION [12 upvotes]: Let $n\geq 1$ be an integer and suppose $S\subseteq {\mathbb R}^n$ is countable and dense. Do we have $S \cong {\mathbb Q}^n$ where both sets carry the topology inherited from the Euclidean topology on ${\mathbb R}^n$? REPLY [9 votes]: You can learn a bit more about countable dense subsets of separable metric spaces by searching for "countable dense homogeneous space". A separable metric space $X$ is Countable Dense Homogeneous (CDH) if given any two countable dense subsets $D$ and $E$ of $X$ there is a homeomorphism $f : X \rightarrow X$ such that $f(D) = E$. The concept was introduced by R. Bennett in Countable dense homogeneous spaces Fund. Math., 74 (1972), pp. 189-194 Theorem 3 in the paper implies that locally euclidean spaces are CDH. So not only any countable dense subset of ${\bf R}^n$ is homeomorphic to ${\bf Q}^n$, but the homeomorphism can be taken as the restriction of a global homeomorphism of ${\bf R}^n$.<|endoftext|> TITLE: cut-elimination provable in PRA QUESTION [8 upvotes]: What is the standard reference for the provability of the cut-elimination theorem in PRA? Update: Rasmus Blanck has offered a reference for a system other than Gentzen's $\mathfrak L \mathfrak K$. The bounty is for a reference for the provability of the cut-elimination theorem for the sequent calculus $\mathfrak L \mathfrak K$ in PRA. REPLY [2 votes]: I think the chapter "Proof Theory: Some Applications of Cut-Elimination", by Helmut Schwichtenberg, from "Handbook of Mathematical Logic", might be what you are looking for.<|endoftext|> TITLE: Sets of unit fractions with sum $\leq 1$ QUESTION [41 upvotes]: Consider a set of fractions $\left\{1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{n}\right\}$. How many subsets of this set have sum at most 1? I'm interested in the asymptotics of this number. Clearly, any subset of $\left\{\frac{1}{\lceil n/2 \rceil}, \ldots, \frac{1}{n}\right\}$ works, hence the answer is $\Omega(2^{n / 2})$. Can we show $\Omega(2^{\beta n})$ for $\beta > \frac{1}{2}$? Can we determine $\beta$ exactly? From numeric estimates of OEIS sequence $\beta$ seems to at least $0.88$ (link to the sequence and correction of the estimate due to Max Alekseyev). The question arose while I was thinking about upper bounds for this question. Clearly, every divisibility antichain $I$ of $[n]$ must satisfy $\sum_{x \in I} \lfloor\frac{n}{x}\rfloor \leq n$, which is a very similar condition. Post-mortem: while I accepted Lucia's answer (simply because it was the first to contain the correct answer and some reasoning to why it is correct), the whole discussion here is very valuable. Be sure to also check out js21's answer with an approach based on large deviations method, and RaphaelB4's answer for a more off-the-ground explanation of the method. In a comment Jay Pantone shared a link to a paper on series analysis, in particular, the differential approximation method allows to obtain the same answer with high precision and is, without doubt, a great practical tool. Kudos to all of you guys! What a great day to learn. REPLY [18 votes]: The bound proposed by Lucia is correct. I add some detail and I stress that It is an application of "large deviation" theorem which is very very standard. Set the following independent Bernoulli random variable defined as $$X_i=\begin{cases} 0 \text{ with } p=1/2\\ \frac{1}{i} \text{ with } p=1/2\end{cases}$$ Then we have the number of subset is given by $$ 2^n \mathbb{P}(\sum_{i=1}^n X_i \leq 1)$$ And we can then adapte the proof of the well known Cramer theorem. For a lower bound $$\mathbb{P}(\sum_{i=1}^n X_i\leq 1)=\mathbb{P}(e^{-x\sum_{i=1}^n X_i}\geq e^{-x})\leq \frac{\mathbb{E}(e^{-x\sum_{i=1}^n X_i})}{e^{-x}}$$ which give because of the independence of $X_i$ the formula that Lucia have already stated (I didn't know it is called Rankin bound) $$\mathbb{P}(\sum_{i=1}^n {X}_i \leq 1)\leq e^{x} \prod_i^n (e^{-\frac{x}{i}}+1)/2^n$$ There exists $x_0$ which minimise the right part. Then introduce $$\tilde{X}_i=\begin{cases} 0 \text{ with } p=\frac{1}{1+e^\frac{-x_0}{i}}\\ \frac{1}{i} \text{ with } p=\frac{e^\frac{-x_0}{i}}{1+e^\frac{-x_0}{i}} \end{cases}$$ Remark that $$ \mathbb{E}(\sum_i \tilde{X}_i)=\frac{\mathbb{E}(\sum_i X_i e^{-x_0\sum_{i=1}^n X_i})}{\mathbb{E}(e^{-x_0\sum_{i=1}^n X_i})}=\partial_x [ln(\mathbb{E}(e^{-x_0\sum_{i=1}^n X_i}))]_{x=x_0}$$ But because $x_0$ is a minimum, $\partial_x [ln(\mathbb{E}(e^{-x_0\sum_{i=1}^n X_i}))-ln(e^{-x})]_{x=x_0}=0$ and therefore $\mathbb{E}(\sum_i \tilde{X}_i)=1$. But because of convexity for any $\epsilon>0$ changing $\tilde{X}$ by $\tilde{X}^\epsilon$ by replacing $x_0$ by $x=x_0+\epsilon n$ in the definition give $\mathbb{E}(\sum_i \tilde{X}_i)\leq 1-\delta(\epsilon)$ with $\delta(\epsilon)>0$. We have then that $\mathbb{P}(\sum\tilde{X}^\epsilon_i\leq 1)\geq \delta(\epsilon)$ We can then state the lower bound $$ \delta(\epsilon)\leq \mathbb{E}(1_{\sum(\tilde{X}_i^\epsilon)\leq 1})\leq \frac{\mathbb{E}(1_{\sum X_i \leq 1} e^{-x \sum_i X_i})}{\mathbb{E}(e^{-x \sum_i X_i})} \leq \frac{\mathbb{E}(1_{\sum X_i \leq 1} )e^{-x}}{\mathbb{E}(e^{-x \sum_i X_i})}$$ And to conclude $$ \delta(\epsilon)\frac{\mathbb{E}(e^{-x \sum_i X_i})}{e^{-x}}\leq \mathbb{P}(\sum X_i \leq 1)$$ and therefore $$\lim \frac{1}{n}\log(\frac{\mathbb{E}(e^{-x \sum_i X_i})}{e^{-x}})\leq \lim \frac{1}{n}\log(\mathbb{P}(\sum X_i \leq 1))$$ which is true for any $\epsilon>0$. This is the end of the proof that the bound is tight<|endoftext|> TITLE: Variations of Hodge structures over the line QUESTION [7 upvotes]: Let $f\colon X\to \mathbb{A}^1$ be a smooth projective morphism of complex algebraic manifolds, where the target $\mathbb{A}^1$ is the affine line. Are there any restrictions on the Hodge structures on the cohomology groups of fibers of $f$ over different complex points of $\mathbb{A}^1$? (Say are there examples where these Hodge structures are not isomorphic to each other?) I apologize if this question is not of the research level; I am not an algebraic geometer. If there is a reference, it would be helpful. REPLY [7 votes]: See Theorem 11 on page 191 of these notes. A special case is as follows. Theorem of the fixed part. Let $S$ be a smooth quasiprojective variety, and $V$ a variation of $\mathbb{Q}$-Hodge structures on $S$ (for example, $R^i\pi_*\underline{\mathbb{Q}}$, for $\pi: X\to S$ a smooth projective morphism). Then $H^0(S,V)$ naturally admits a Hodge structure, such that the map $H^0(S, V)\to V_s$ is a morphism of Hodge structures for any $s\in S$. In particular, if $S$ is simply connected (as it is in your case, where we take $S=\mathbb{A}^1_{\mathbb{C}}$), then the map $H^0(S,V)\to V_s$ is an isomorphism for any $S$. Hence all the $V_s$ are isomorphic.<|endoftext|> TITLE: Has anything (other than what is in the obituary written by M. Noether) survived of Paul Gordan's defense of infinitesimals? QUESTION [23 upvotes]: Question. Has anything other than what can be guessed from this obituary written by Max Noether survived of the 'defense' of infinitesimals that Paul Gordan gave in his doctoral disputation on March 1, 1862 in Berlin? Remarks. This is an interesting historical morsel that I accidentally stumbled over and that Mikhail Katz suggested to be made an MO question. It should be noted that the philosophical substance of this 'discovery' is not surprising at all, rather expected and boring in fact (the only interesting result of this question would be a mathematical argument): that Noether says that Gordan thought that infinitesimals are good and that Gordan defended 'implicit solutions better than explicit solutions' only because he had to, is squarely in line with the typology found in the usual 'narrative', i.e., Gordan : $\mathsf{explicit}$ , Hilbert : $\mathsf{implicit}$ Someone having less reservations against such titles than the writer of these lines might have entitled the question "The King of Invariants and his Defense of the Infinitesimals: a lost Latin swan song?", or something like that. For readers' convenience, here are relevant passages from the obituary. Source is Mathematische Annalen 75, 1914: My translation: In Breslau Gordan got the topic of his dissertation; it arose from an article entitled "De linea geodetica", with which in August 1861 Gordan won a competition started by a prize question asked by the faculty. This doctoral dissertation that Gordan defended on March 1, 1862 in Berlin treated the geodetic line on the spheroid of the earth. Of the questions which Gordan defended at this disputation---in a Latin, of course, which was deemed not exactly classical by Kronecker---the following two may be mentioned; the first, because it is in line with Gordan's views both then and later in life; the second, for the opposite reason: "The method of the Infinitely-Small is, I claim, no less precise than the method of limits." "It is of greater interest to investigate the implicit properties of a function defined by a differential equation than to investigate in terms of which known functions it can be expressed." Complex analysis attracted Gordan in autumn 1862 to go to Riemann [Riemann was employed at the university of Göttingen at the time]. He [...] Max Noether writes as if he knew more about Gordan's thoughts on infinitesimals than what he tells readers in the obituary. Noether insinuates that Kronecker was present at the disputation. I don't know what the regulations for disputations were in the 1860s at University of Berlin. (This is the 'ancestor' of what is by now one of at least three universities in Berlin.) If notes were taken, then they may have survived, though I think this unlikely. (And even if they have survived: anyone who knows what minutes of exams usually look like will not expect there to be much information in them. The best one could hope for is that Gordan later himself wrote carefully about infinitesimals.) One could even hope that this question has (some sort of) mathematical answer; 155 years are not much by historians' standards. Of course, one should not expect there to be much in exam minutes, even conditioning on the, by itself, unlikely event that such were taken and have survived. Update. I now took the time to read the whole of the obituary written by Noether ('op. cit.'). I here summarize all that seems not irrelevant to the present question (to cut a long story short: nothing strictly relevant to the question will follow; the last item will border on the off-topic; Noether is basically saying that Gordan did not use limits simply because he didn't like them, or just didn't care, and avoided them, not because he had a thought-out positive theory about infinitesimals; it seems more of a rather boring avoidance of second-order definitions and a preference for first-order definitions, than any positive theory about infinitesimals, making the chances of any appreciable answer to this question seem even slimmer than they already seem just from thinking about the question; I don't expect there to ever be much of an answer): The most relevant item in this 'update' is the following passage in op. cit., in which Noether describes Gordan as a teacher: My attempt at a translation: In his own field [Noether has written in the immediately preceding paragraph that Gordan liked to read German literature] it was not so much the perusal of the work of others than a 'big picture' of the inner connections and an instinctive feeling for the paths and goals of mathematical endeavours which enabled him to distinguish the valuable from the inferior, and small hints were sufficient for this. However, Gordan never did justice to foundational conceptual developments: also, in his lectures he entirely avoided any basic definitions of a conceptual kind, even the definition of 'limit'. [emphasis added] His lectures extended only to mathematics of the common kind, occasionally also to the theory of binary forms; the exercises were preferably taken from algebra. He liked to lecture on the work of Jacobi [Gordan's doctoral advisor], e.g. on functional determinants, yet never about complex analysis, higher geometry or mechanics1; he also did not have anyone give seminar presentations. The lectures [of Gordan], essentially, had an effect rather due to his vivacious way of speaking, and due to a vigour which encouraged further independent study, than due to systematizing and rigor. Noether on the first page of op. cit. names his 'informants', which explains the pretense Noether's of being informed about so much in Gordan's life: Translation: Part of the relevant [i.e.: relevant for the biographical parts of the obituary] data was taken from [Gordan's] dissertation (I of the bibliography at the end). For other bits of information I am indebted to R. Sturm, a fellow student of Gordan at Breslau, C.F. Geiser, one of the opponents at the Berlin disputation in 1862, J. Thomae, a fellow student at Göttingen, and A. Brill, a colleague at Gießen; furthermore, I am indebted to L. Schlesinger for information from files at Gießen. Noether also describes what he thinks is the only non-algebraic work Gordan ever did after his dissertation:2 My translation: Only once did Gordan ever work on a non-algebraic topic: in 1893 Hilbert had, based on a new idea, given a simple proof of the transcendence of the numbers $e$ and $\pi$; and immediately afterwards Hurwitz had given a modification of the proof, by way of avoiding the integrals used by Hilbert, essentially by using integration-by-parts, and by the use of the simplest case of the intermediate-value-theorem. In an article (54), which by the way only had the aim of giving an exposition of Hurwitz's deduction, 'played backwards', Gordan took one more step in towards an elementary proof, by replacing also the still remaining differential quotients by use of the Taylor series of $e^x$. [emphasis added; by the way: what is nowadays the 'most' elementary proof that $\pi$ and $e$ are transcendental; the only proof I ever studied in detail used integrals, and, as such, limits] Regarding the $r!=h^r$ symbolism, which has been emphasized by various others after Gordan's paper, it must however be said that it does not play any further role in the proof than being a mere abbreviating notation, and that it moreover has simply been lifted from Hurwitz's note. The estimation of the vanishing remainder term, an estimate which is essential for the proof and was left out of the note (54), has been carried out by Gordan in 1900 in an unpublished notebook in Gordan's Nachlass. 3 The 'MathSci-Net' and 'Zentralblatt' of the 19th century was the 'Jahrbuch über die Fortschritte der Mathematik'; therein, Heinrich Burkhardt reviewed Gordan's 'transcendent note'; for convenience, since it is not entirely off-topic (it is 'Gordan on the non-algebraic'), I reproduce the review here (it is also something of a self-contained summary of Hilbert's proof, and as such, maybe of interest in and of itself): The blue link in the yellow lead to a review which is (displayed to me as) empty. 1To my way of thinking, this is all rather expected and unsurprising and in line with the usual reputation of Gordan: all the subjects Noether says Gordan avoided are second-order theories; in particular, in the mechanics of the time variational principles had an important role, and variational principles involve quantifier over sets of functions, i.e., second-order quantifiers; 19th century mechanics/variational principles/principles of least action, etc. to me seem recognizably non-algebraic. Also, one could now very much overinterpret and say something about whether and how much infinitesimals help in complex analysis, and that there, perhaps, limits are more necessary than in real analysis, and Gordan avoided complex analysis because of this. This, however, is evidently an overinterpretation. 2 And to me, it seems evident that if there is any hope of a mathematical answer, then this hope comes from the possibility that Gordan might later have written about infinitesimals/used them for something in published printed work. However, unfortunately, I think this is unlikely. 3 It might be an interesting separate historical problem to locate this "unpublished" notebook. Please note, though, that I don't think this relevant for the question: what Noether describes sounds like a rather ordinary estimate of a remainder term, presumably by inequalities; it does not sound like a use that Gordan made of infinitesimals, which don't yield effective estimates anyway, as far as I know. One should be clear that there is no sign that Gordan ever recognizably used infinitesimals, except for his thesis defense. REPLY [12 votes]: Paul Gordan's theses were published in De linea geodetica and digitised by Google, from which I reproduce the relevant page: Translation: I. The method of functional division, proposed by the respectable Cournot as an analytical and empirical method, appears unsuitable, since one does not have fully empirical functions. II. The method of the infinitely-small is, I claim, no less precise than the method of limits. III. It is of greater interest to investigate the implicit properties of a function defined by a differential equation than to investigate in terms of which known functions it can be expressed. IV. The principles of Democritus remain to the present day as the foundation of the theory of atoms. One can also learn from this publication that the opponents of Gordan at the March 1, 1862 Disputatio were Dr. Kretschmer, Mr. Geiser, and Mr. Rathke, presumably E.E. Kretschmer and C.F. Geiser. Gordan acknowledges as his teachers in mathematics the professors Kummer, Joachimstal, Richelot, Rosenhain, Schroeter, and Galle --- but it is unclear whether any of these were present at the defense ceremony. Concerning the question "what was actually said at the defense"; at least in my University (Leiden) the practice through the centuries has been that the examination itself is not recorded, the secretary records only the outcome (pass/fail/cum laude etc.).<|endoftext|> TITLE: When are the closed convex subsets countable intersections of halfspaces QUESTION [5 upvotes]: For what kind of topological vector spaces (separable maybe?) are the closed convex subsets countable intersections of halfspaces. I've seen somewhere that it's true for separable Hilbert spaces, but without proof or reference. Is there a reference on this fact (and related questions)? REPLY [3 votes]: Those authors have a book on the topic that gives the following more down-to-earth interpretation: J. M. Borwein and J. D. Vanderwerff, Convex Functions: Constructions, Characterizations and Counterexamples Cambridge University Press, 2010. https://carma.newcastle.edu.au/resources/jon/Preprints/Books/CUP/cup-final.pdf Just before Proposition 7.5.6 it defines: "A (closed convex) set in a Banach space is constructible if it is representable as the intersection of countably many closed half spaces." Proposition 7.5.6 in the book: "A closed convex subset [of a Banach space] containing the origin is constructible if and only if its polar is weak*-separable." The book immediately follows Proposition 7.5.6 with this interpretation: "In particular, all closed convex subsets of a separable [Banach] space are constructible." The bracketed phrases in the above quotes are my own additions. Since $\mathbb{R}^n$ is a separable Banach space, it works for all closed and convex subsets of $\mathbb{R}^n$.<|endoftext|> TITLE: Probabilistic Proofs of Key Number-Theoretic Results QUESTION [6 upvotes]: Given a positive integer $n$, let $p$ be the largest prime less than or equal to $n$. Let $N(n)=2^{C_2}\cdots p^{C_p}$ be uniformly distributed from $1$ to $n$, and $M(n)=2^{Z_2}\cdots p^{Z_p}$ where $Z_p's$ are independent geometric with $P(Z_p\ge k)=\frac{1}{p^k}$. It can be shown that $C_p$ has probability mass $P(C_p=k)=\frac{\big\lfloor \frac{n}{p^k}\big\rfloor}{n}.$ Consider the metric $d(M,N)=\sum_{p\le n}\vert C_p-Z_p\vert$ ($d$ counts the number of prime insertions and deletions needed to convert $N$ to $M$). Using this metric, we have the Wasserstein metric $d_W(M,N)=\inf_{\text{couplings}}\mathbb{E}d(M,N).$ Arratia (page 10 of https://arxiv.org/pdf/1305.0941.pdf) claims $d_W(M,N)=o(\log \log n)$ implies the Hardy-Ramanujan Theorem (which states that for almost every positive integer $n$, $\omega(n) \approx \log \log n$, where $\omega(n)$ is the number of distinct prime divisors of $n$). $d_W(M,N)=o(\sqrt{\log \log n})$ implies the Erdos-Kac Central Limit Theorem (which states that $\frac{\omega(n)-\log \log n}{\sqrt{\log \log n}}$ has the standard normal distribution. Further confirmation of this is stated by Bollobas on page 29 of "Contemporary Combinatorics": How can these results be derived from the asymptotics? REPLY [3 votes]: Obviously, one has $$|\omega(M)-\omega(N)|\leq d(M,N).$$ Taking the expectations, we see that for every $n$, we can construct a coupling of $M(n)$ and $N(n)$ such that $$ \mathbb{E}|\omega(M(n))-\omega(N(n))|\leq d_W(M(n),N(n))+\frac{1}{n}. $$ Under the asymptotic estimates above, Chebyshev's inequality implies, for any $\alpha>0$ $$ \mathbb{P}(|\omega(M(n)-\omega(N(n)))|>\alpha \log \log n)=o(1), $$ or, respectively, $$ \mathbb{P}(|\omega(M(n)-\omega(N(n)))|>\alpha \sqrt{\log \log n})=o(1). $$ Hence, Hardy-Ramanujan and Erdos-Kac boil down to the corresponding statements about $\omega(M(n))$. These are just the LLN and the CLT for the independent random variables $$\mathbf{1}_{Z_2\neq 0},\mathbf{1}_{Z_3\neq 0},\mathbf{1}_{Z_5\neq 0}\dots,$$ which are of course well known and standard (e. g., Lindeberg's CLT applies)<|endoftext|> TITLE: What is the status of a problem about cluster categories? QUESTION [6 upvotes]: Let $H$ be a hereditary algebra of Dynkin type. There is a cluster category $\mathcal{C}_H$ defined by Aslak Bakke Buan, Robert Marsh, Markus Reineke, Idun Reiten, and Gordana Todorov in Tilting theory and cluster combinatorics. In the file, Andrei Zelevinsky proposed the following problem. Problem: find an explicit description of $End_H(T)$ ($T$ is a tilting object in $\mathcal{C}_H$) in terms of quivers and relations. What is current status of this problem? Thank you very much. REPLY [6 votes]: For $H$ the path algebra of a 'star-shaped' quiver having three legs with lengths $r$, $s$, $t$, an answer seems to be implicit in Lamberti's combinatorial model for the cluster category: https://arxiv.org/abs/1403.0549 but she doesn't make it explicit. This includes all Dynkin types, where $(r,s,t)\in\{(1,p,q),(2,2,n),(2,3,3),(2,3,4),(2,3,5)\}$. The answer for Dynkin types $\mathsf{A}$ and $\mathsf{D}$ is more explicit, because the quivers with relations are given by the Jacobian algebras associated to triangulations of a disk and a punctured disk respectively: Caldero–Chapoton–Schiffler, type $\mathsf{A}$: https://arxiv.org/abs/math/0411238 Schiffler, type $\mathsf{D}$: https://arxiv.org/abs/math/0608264 This leaves just type $\mathsf{E}$, where one could in principle compute all of the (finitely-many!) cluster-tilted algebras using Lamberti's construction. Some non-Dynkin hereditary algebras are also covered by triangulations of marked bordered surfaces, such as $\tilde{\mathsf{A}}$ by the annulus, but I am not sure precisely which ones. Most are not, since quivers arising from triangulations are very special; for example, the vertices have valency at most $4$.<|endoftext|> TITLE: T-nilpotency and quasinilpotency of ideals QUESTION [5 upvotes]: Let $R$ be a commutative ring and let $\mathfrak{a}\subseteq R$ be an ideal. The ideal $\mathfrak{a}$ is called T-nilpotent if for every sequence $(r_i)_{i\in\mathbb{N}}$ in $\mathfrak{a}$ there exists $n\in\mathbb{N}$ such that $\prod_{i=0}^nr_i=0$, and quasinilpotent if there exists $n\in\mathbb{N}$ such that for every $r\in\mathfrak{a}$ we have $r^n=0$. If $\mathfrak{a}$ is nilpotent (i.e., there exists $n\in\mathbb{N}$ with $\mathfrak{a}^n=0$), then it is T-nilpotent and quasinilpotent. If $\mathfrak{a}$ is T-nilpotent or quasinilpotent, then it is nil (i.e., for every $r\in\mathfrak{a}$ there exists $n\in\mathbb{N}$ such that $r^n=0$). Conversely, a nil ideal need not be T-nilpotent or quasinilpotent, a T-nilpotent ideal need not be nilpotent, and a quasinilpotent ideal need not be nilpotent. I guess that a T-nilpotent and quasinilpotent ideal need not be nilpotent either. However, I was not able to come up with an example of such an ideal. Thus: What is an example of an ideal in a commutative ring that is T-nilpotent, quasinilpotent, but not nilpotent? REPLY [2 votes]: Based on Yves answer above, let me give another example (or maybe another description of his example - I am not sure about this). Let $K$ be a field of characteristic $2$, let $$R=K[(X_i)_{i\in\Bbb N}]/\langle\{X_i^2\mid i\in\Bbb N\}\cup\{X_iX_j\mid i,j\in\Bbb N,2i2j_0$, and thus we get the contradiction $\prod_{i=0}^kr_i=0$.<|endoftext|> TITLE: Why is the catalecticant invariant under coordinate changes? QUESTION [22 upvotes]: Let $\mathbf{k}$ be a commutative $\mathbb{Q}$-algebra. (We could play the same game over any commutative ring $\mathbf{k}$, but this would be a bit more technical, so let me avoid it.) Fix a nonnegative integer $n$. A binary form shall mean a homogeneous polynomial $f=f\left( x,y\right) \in\mathbf{k}\left[ x,y\right] $. Any binary form $f\left( x,y\right) $ of degree $2n$ can be written uniquely in the form $f\left( x,y\right) =\sum\limits_{i=0}^{2n}\dbinom{2n}{i}a_{i} x^{2n-i}y^{i}$ for some scalars $a_{0},a_{1},\ldots,a_{2n}\in\mathbf{k}$ (which are simply the coefficients of $f$, rescaled by binomial coefficients). Let $\mathcal{F}_{2n}$ denote the $\mathbf{k}$-module of all binary forms of degree $2n$. The matrix ring $\mathbf{k}^{2\times2}$ acts on the $\mathbf{k} $-module $\mathcal{F}_{2n}$ by the rule \begin{equation} \left( \begin{array}[c]{cc} a & b\\ c & d \end{array} \right) \cdot f=f\left( ax+cy,bx+dy\right) . \end{equation} (In other words, a matrix $\left( \begin{array}[c]{cc} a & b\\ c & d \end{array} \right) \in\mathbf{k}^{2\times2}$ acts on a binary form by substituting $ax+cy$ and $bx+dy$ for the variables $x$ and $y$. This is called a "change of variables", at least if the matrix is invertible.) The catalecticant $\operatorname*{Cat}f$ of a binary form $f\left( x,y\right) $ of degree $2n$ is a scalar, defined as follows: Write $f\left( x,y\right) $ in the form $f\left( x,y\right) =\sum\limits_{i=0}^{2n}\dbinom{2n} {i}a_{i}x^{2n-i}y^{i}$, then set \begin{equation} \operatorname*{Cat}f=\det\left( \begin{array}[c]{cccc} a_{0} & a_{1} & \cdots & a_{n}\\ a_{1} & a_{2} & \cdots & a_{n+1}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n} & a_{n+1} & \cdots & a_{2n} \end{array} \right) =\det\left( \left( a_{i+j-2}\right) _{1\leq i\leq n+1,\ 1\leq j\leq n+1}\right) . \end{equation} A classical result, going back to Sylvester (On the Principles of the Calculus of Forms) in 1852 (who, in turn, ascribes it to Cayley), then says that $\operatorname*{Cat}f$ is a $\operatorname*{GL} \nolimits_{2}\mathbf{k}$-invariant (in an appropriate sense of this word, i.e., $\operatorname*{GL}\nolimits_{2}\mathbf{k}$ transforms $\operatorname*{Cat}$ by multiplication with an appropriate power of the determinant). More precisely, any binary form $f\left( x,y\right) $ of degree $2n$ and every matrix $A\in\mathbf{k}^{2\times2}$ satisfy \begin{equation} \operatorname*{Cat}\left( A\cdot f\right) =\left( \det A\right) ^{n\left( n+1\right) }\cdot\operatorname*{Cat}f. \label{catalecticant1} \tag{1} \end{equation} Question. Is there a purely algebraic/combinatorial proof of \eqref{catalecticant1}? These days, \eqref{catalecticant1} is usually seen as a consequence of the fact that if $\mathbf{k}$ is an algebraically closed field of characteristic $0$, then a binary form $f\left( x,y\right) $ of degree $2n$ satisfies $\operatorname*{Cat}f=0$ if and only if $f$ can be written as $\sum_{j=1} ^{n}\left( p_{j}x+q_{j}y\right) ^{2n}$ for some $2n$ elements $p_{1} ,p_{2},\ldots,p_{n},q_{1},q_{2},\ldots,q_{n}\in\mathbf{k}$. This is proven, e.g., in J. M. Selig, Sylvester's Catalecticant. Once this fact is granted, the equality \eqref{catalecticant1} follows using some technical Nullstellensatz arguments (I believe so; the details are somewhat annoying to verify, requiring e.g. to show that $\operatorname*{Cat}f$ is irreducible as a polynomial in the $a_{0},a_{1},\ldots,a_{2n}$ for $n>0$). I dislike this argument for the technicalities involved, including the passage to an algebraically closed field and the use of the Nullstellensatz (at least for principal ideals). It seems to me that \eqref{catalecticant1} should have a purely algebraic proof, using properties of determinants. Sylvester might have had one, but I find his paper impossible to decipher. Does anyone know such a proof? REPLY [11 votes]: Let $d=2n$ be the degree of your binary form $f$. Let me introduce $n+1$ pairs of formal variables $\alpha^{(1)}=(\alpha^{(1)}_{1},\alpha^{(1)}_{2}),\ldots, \alpha^{(n+1)}=(\alpha^{(n+1)}_{1},\alpha^{(n+1)}_{2})$. Let $\mathcal{S}$ be the polynomial $$ \mathcal{S}=\prod_{1\le i TITLE: Why Green functions and not Neron functions? QUESTION [7 upvotes]: Arakelov constructed a nice intersection theory on arithmetic surfaces. A key point is the notion of Green function for a Riemann surface, which will be involved in the ''part at infinity'' of the calculation of the intersection number between two Arakelov divisors. Lately I've been reading the paper "P. Hriljac - Heights and Arakelov's intersection theory" and he defines the intersection theory on arithmetic surfaces by using just the notion of Neron function (on a Riemann surface). A Green function is in particular a Neron function and in the paper you may find the following comment: The existence of Neron functions on curves over $\mathbb C$ (...) may be viewed as stemming from Arakelov, where he uses Green's functions in lieu of Neron functions. The Green's functions can be viewed as explicit realizations of Neron functions (...) Observe that there is no unique such family, which corresponds to choices of metrics as in [Arakelov main paper]. Among all Green's functions, there is one which can be selected to be "better" than all others (...) Basically I don't understand what is the author saying with the above lines. In particular I don't understand why we choose Green functions as the "right" Neron functions involved in intersection theory. This question may be relevant. REPLY [2 votes]: A very short answer I learned from summer school - because they want to construct local-global correspondence. A Neron function correspond exactly to the local part and Riemann-Roch is a local-global theorem.<|endoftext|> TITLE: Restricting extenders to a ground model QUESTION [6 upvotes]: Let V=W[g], where g is P-generic over W for some poset P in W. Let F be a V-extender with critical point κ such that P ∈ VκW. If the support of F is sufficiently closed, say strength(F)=length(F)=λ for some inaccessible cardinal λ>κ, then F ∩W ∈ W. To the best of my knowledge, this is due do Hamkins-Woodin; the argument for it which I have in mind is the one written up in https://ivv5hpp.uni-muenster.de/u/rds/VM2_4.pdf . Can this be proven in more generality? I.e., what if we don't assume ult(V;F) to be Card(P)-closed, is it still true that F ∩W ∈ W? Or is there a counterexample? And can one even drop the hypothesis that F be a total V-extender, i.e., that it acts on some transitive model contained in V rather than on all of V? REPLY [6 votes]: Here is one way to make a kind of counterexample, although I'm not exactly sure if this is what you want. Start in a model $W$ where $\kappa$ has two different normal measures $\mu$ and $\nu$. Now, consider the extension where we have added a Cohen real $V=W[c]$. Since this is small forcing, each measure $\mu$ and $\nu$ generates a measure in the extension $V$. Now, use the real $c$ to define a certain $\omega$-iteration of the (extensions of the) normal measures, where the digits of the real tell you which measure to use at each stage. Let $j:V\to M$ be the resulting iteration. This can be realized as an extender embedding with countably many generators. But the restriction $j\upharpoonright W$ cannot be an embedding in $W$ because from this restriction we can tell which measure we used at each stage and therefore we would be able to define $c$ in $W$, which contradicts that $c$ is $W$-generic. Similar examples can be built for larger kinds of extenders. For example, if you have two different rank-to-rank embeddings $j_0:W_\lambda\to W_\lambda$ and $j_1:W_\lambda\to W_\lambda$, then in the extension $V=W[c]$, you can lift the original embeddings and use $c$ to define a certain iteration of them, and then use this to form an extender in $V$, which cannot lift any ground model embedding.<|endoftext|> TITLE: The minimum-perimeter triangle of three sets of points QUESTION [8 upvotes]: If $X$ and $Y$ are two sets of $n$ independent, uniformly sampled points in the unit square, then standard methods can show that the expected minimum distance between points in $X$ and $Y$ is proportional to $1/n$, that is, $$E(\min_{i,j}\|x_i - y_j \|)\sim 1/n$$as $n\to\infty$. Is there anything similar that can be said when we have three sets of points $X,Y,Z$, and we look for the triangle whose perimeter is shortest? I.e., $$E(\min_{i,j,k}\|x_i-y_j\|+\|y_j-z_k\| + \|z_k - x_i\|)$$? REPLY [9 votes]: Partition the unit square into small squares of area roughly $a$. Your question is equivalent to asking for which $a$ do we typically see about 1 small square with points from each of $X$,$Y$ and $Z$? this probability is roughly $(an)^3$ since the probability of seeing a point from each set is about $an$ and they're independent. There are $1/a$ small squares so the expected number of squares with points from each set is roughly $(an)^3/a=a^2n^3$. So if we want this to be roughly 1 we need to set $a=n^{-3/2}$. The diameter of a small square is therefore $n^{-3/4}$. Generalizing for $k$ different sets and dimension $d$ yields $n^{-\frac{k}{d(k-1)}}$.<|endoftext|> TITLE: 71, the Monster, and c = 24 CFTs QUESTION [22 upvotes]: The largest prime in the order of the Monster group is $71$. This number $71$ shows up at various places: The minimal faithful representation has dimension $196883 = 47.59.71$ The Monster group can be realised as a Galois group $Gal$ $L(71)/{\mathbb{Q}(\sqrt{-71})}$ where $L(71)$ is a suitable field. The appearance of $71$ in the above two cases (and possibly others) is not very surprising and might be reasoned out. But the appearance of $71$ in a different area seems very intriguing: The Monster is intimately connected to a special class of conformal field theories. These are the meromorphic $c = 24$ CFTs. The Monster here arises as the discrete automorphism group of the vertex operator algebra of one of the c =24 CFTs. Schellekens in 1992 enumerated such CFTs and he found $71$ such CFTs! All these CFTs have a partition function of the form $$ Z(\tau) = j(\tau) + \mathcal{N} $$ where $j$ is modular invariant and $\mathcal{N} \geq -744$ is an integer. But any value of $\mathcal{N}$ won't work. Schellekens found $71$ values of $\mathcal{N}$ which will work. Unfortunately, it is still not clear if the enumeration Schellekens made is exhaustive, i.e. if there are only exactly $71$ such theories. Is the appearance of $71$ here just a coincidence? Or is it again connected to the Monster? It is hard to believe that this is just a coincidence. REPLY [16 votes]: Schellekens' enumeration is exhaustive in the following sense: the degree 1 subspace of the meromorphic CFT/vertex algebra is naturally a Lie algebra, and it is known that this Lie algebra must be one of the 71 that Schellekens wrote down. Each of these 71 Lie algebras is realised as the weight 1 piece of some holomorphic c=24 vertex algebra, but it is still an open conjecture that the vertex algebra is unique. (Though it is verified in the large majority of cases.) Regarding possible doubts about machine precision arithmetic and such in the linear programming part of Schellekens' calculations: In https://arxiv.org/abs/1507.08142 one of the things my collaborators and I do is to independently confirm Schellekens' result, this time with exact arithmetic, and perhaps you might say a streamlined proof. But it's still a computer proof. We go on to use the result in our construction of some new holomorphic vertex algebras as orbifolds. I find the appearance of 71 as tantalising as you do. I'd certainly like for it not be a coincidence, but for the moment at least I just don't know.<|endoftext|> TITLE: Do contact and CR structures have corresponding $G$-structures? QUESTION [12 upvotes]: For an $n$-dimensional manifold $M$, almost complex and almost symplectic structures on $M$ correspond to reductions on the structure group of the tangent bundle, introducing a $\operatorname{GL}(n/2,\mathbb{C})$ and $\operatorname{Sp}(n)$ structure on $M$, respectively. With additional integrability (the vanishing of the Nijenhuis tensor or the closedness of the almost symplectic form) one has a complex or symplectic structure on $M$. Since CR and Contact geometry are the odd-dimensional cousins of the above geometries, it seems as though it would make sense for CR and Contact structures to come from a reduction of the structure group of the manifold. However, these structures tend to be defined more in terms of their integrability conditions. I suppose what I am really hoping for is a more unified theory among the different geometric structures. The Riemannian, Almost Symplectic, and Almost Complex structures share an intimacy with their "2 out of 3" property arising from properties of the intersections of the reduced structure groups, and it would be nice if the relationship between Almost Contact and Almost CR structures stemmed from a similar place, and furthermore that their integrability conditions corresponded to some sort of "flatness" on the $G$-structure. So I ask, do these $G$-structure correspondences exist for the above geometries, and do they share an analogous relationship with their even-dimensional counterparts? REPLY [6 votes]: In addition to other excellent answers, I would like to approach from another angle. Almost CR structures A most general almost CR structure (i.e. of arbitrary codimension) on a real manifold $M$ is given by a complex subbundle $$V\subset \mathbb C\otimes TM,$$ of the complexified tangent bundle of $M$, satisfying $V\cap \bar V=\{0\}$, where the conjugation on $V$ is induced by the standard conjugation on $\mathbb C\otimes TM$. This definition is equivalent to its "real version" given by the pair $(H, J)$, where $H\subset TM$ is a real subbundle, and $J\colon H\to H$ is a complex structure on $H$. Indeed, given $(H, J)$ as above, the corresponding $V$ is the so-called $(0,1)$-subbundle given by $$ V= H^{01} := \{ L\in \mathbb C\otimes H: JL = -iL \} = \{ X + iJX: X\in H\}.$$ Vice versa, every $V$ as above yields $$H:= (V +\bar V)\cap TM,$$ and $J\colon H\to H$ is uniquely determined by the identity $$X + iJX\in V, \quad X\in H.$$ (An equivalent description can be given via $(1,0)$ instead of $(0,1)$ vectors, but the former is preferred e.g. in the context of the $\bar\partial$ equation.) CR codimension An important invariant of an almost CR structure is its CR codimension, given by the complex codimension of $V\oplus \bar V$ in $\mathbb C\otimes TM$ or, equivalently, by the real codimension of $H$ in $TM$. The almost CR structures of codimension $0$ are precisely the almost complex structures, and those of codimension $1$ are the ones induced on real hypersurfaces in almost complex manifolds. Important examples of almost CR structures of higher codimension are given by real Lie group orbits, for instance, boundary components of bounded symmetric domains may have arbitrarily high CR codimension. Almost CR structures as G-structures To regard an almost CR structure $V$ as a G-structure, consider the subbundle of all frames in $TM$, consisting of all CR isomoprhisms from the flat model $\mathbb C^n\oplus \mathbb R^m$ onto $TM$. Then the group $P$ of all linear CR isomoprhisms of the flat model acts transitively on these frames at each point, and thus defines an $P$-structure on $M$. Vice versa, any such $P$-structure corresponds to an unique almost CR structure. This construction works, in particular, for almost complex structures, since it is the special case of an almost CR structure corresponding to CR codimension 0. Induced almost CR structures on real submanifolds of almost complex manifolds For any real submanifold $M$ in an almost complex manifold $C$, there is canonical almost complex structure on each complex tangent space $H_p:= T_pM\cap J T_pM$ at a point $p\in M$, where $J$ is the complex structure of $C$. If $H_p$ is of constant dimension, $M$ is called CR submanifold of $C$ with the induced almost CR structure. If $M$ is a real hypersurface, it is always a CR submanifold. Almost CR G-structure as a reduction of an almost complex G-structure One way to see such reduction, is to consider a CR submanifold $M\subset C$ of an almost complex manifold $C$, such that $M$ is generic in $C$, which means $TM+JTM = TC|_M$, i.e. $TM$ needs to span the full tangent space to $C$ over $\mathbb C$. That genericity condition guarantees that any CR frame can be uniquely extended to a complex frame in $C$. Then the bundle of all CR frames on $M$ (defining the corresponding $G$-structure) is canonically identified with a subbundle of all complex frames on $C$ resticted to $M$, and hence can be seen as a reduction of the corresponding almost complex G-structure. Integrability An almost CR structure $V\subset \mathbb C\otimes TM$ is (formally) integrable if the subbundle $V$ is closed under Lie brackets. In contrast to almost complex structures, such formal integrability does not guarantee the existence of sufficiently many local CR functions (implying a local CR embedding into a complex manifold), which is a stronger property sometimes called integrability without the "formal" adjective. Most commonly CR structures are defined as formally integrable almost CR structures, whereas the latter ones are called locally embeddable. Note that integrability played no role in the above discussion and comes as additional property. I suppose that was the confusion mentioned in the question. Also note that, in contrast to complex structures, integrability of almost CR structures does not imply any ``flatness'' such as CR equivalences with their flat models. Nondegeneracy conditions The most well-known nondegeneracy condition is the one for the Levi form, which can be defined for any almost CR structure (or any codimension). See e.g. Dmitri Zaitsev, Normal forms for almost non-integrable CR structures, Amer. J. of Math., 134 (2012), no. 4, 915-947, doi:10.1353/ajm.2012.0027, arXiv:0812.1104. In particular, if $M$ is of CR codimension $1$ (i.e. of hypersurface type), adding a nondegenerate Levi tensor allows to reduce the almost CR G-structure to the one admitting a canonical Cartan connection, as more extensively described by Andreas Cap in his answer. There are also many natural generalizations of the Levi-nondegeneracy condition in various directions, where many details can be found in this book: Baouendi, M. Salah; Ebenfelt, Peter; Rothschild, Linda Preiss Real submanifolds in complex space and their mappings. Princeton Mathematical Series 47. Princeton University Press, Princeton, NJ, 1999. xii+404 pp. doi:10.1515/9781400883967 Contact and almost contact structures In contrast to almost CR structures, contact structures already contain a nondegeneracy condition in their definition. The corresponding almost contact structure is then defined by adding a nondegenerate symplectic form to the hyperplane field, unrelated to the one given by the Lie bracket. That can be equivalently described by appropriate G-structure in a similar way as for almost CR structures described above.<|endoftext|> TITLE: deformation theory in positive characteristic QUESTION [9 upvotes]: The idea "Formal deformation theory in characteristic zero is controlled by a differential graded Lie algebra (dgla)" goes back to Goldman-Millson, Deligne, Drinfeld among others; see Lurie's ICM talk. What is the analogue of this, over positive characteristic? Specifically, what replaces the Maurer-Cartan equation $$dx + \frac{1}{2}[x,x] =0$$ in positive characteristic? Do formal groups enter in an essential way, as a substitute for Lie algebras, in deformation theory in positive characteristic? This point is not answered in the responses to a similar question Extended Deformation Theory (dg-Lie algebra principle in positive characteristic?) . REPLY [7 votes]: Since the philosophy that deformation theory is controlled by DGLAs long predates the abstract characterisation of formal moduli problems, I'll break the answer in two. My answer will also hold in mixed characteristic, where things are complicated by the absence of a basepoints for the deformation functor (objects over $\mathbb{F}_p$ don't tend to have canonical deformations over $\mathbb{Z}_p$). A) Deformation problems arising naturally tend to be governed by the SDCs of P-, Deformations of schemes and other bialgebraic structures, Trans. AMS 2008, which is mostly contained in arXiv:math.AG/0311168. An SDC $E^{\bullet}$ is a diagram of formal schemes with an associative product $$ E^i \times E^j \xrightarrow{*} E^{i+j}, $$ together with a unit $1 \in E^0$ and all the structure of a cosimplicial diagram except the outer coface maps $\partial^0, \partial^{n+1}$ (up to relabelling, this is the same structure as an augmented simplicial diagram); these satisfy various compatibility conditions. Cosimplicial formal groups give rise to SDCs, but do not exhaust them. The Maurer-Cartan space of an SDC is then $$ \{\omega \in E^1 ~:~ \partial^1\omega=\omega*\omega\}, $$ on which $E^0$ acts by conjugation. In characteristic $0$, the homotopy categories of SDCs and of non-negatively graded DGLAs are equivalent, with equivalent Deligne groupids. The constructions of P-, The homotopy theory of strong homotopy algebras and bialgebras, HHA 2010, arXiv:0908.0116 show how SDCs give rise to formal derived moduli stacks. Since SDCs have no negative terms, they can only give rise to derived $1$-stacks, but higher stacks can be obtained by incorporating additional simplicial structure. B) In the proof P-, Unifying derived deformation theories, Advances 2010, arXiv:0705.0344 of the equivalence between DGLAs and formal derived stacks (later dubbed "formal moduli problems" by Lurie), most of the steps hold in positive or mixed characteristic. This gives an equivalence between formal derived stacks and certain simplicial cosimplicial formal schemes up to tangent quasi-isomorphism. The relation with DGLAs is that in characteristic $0$, the simplicial and cosimplicial structures can be replaced with dg structures, and these become equivalent to Hinich's dg coalgebras, and hence to $L_{\infty}$-algebras via the bar construction.<|endoftext|> TITLE: Purely complex eigenvalue of matrix product QUESTION [7 upvotes]: Here is a question which arises from physics. Let $A$, $B$ be two symmetric real-valued matrices. What conditions should the matrices meet to make $AB$ has a pure complex eigenvalue ($Im(\lambda) \neq 0$)? Although I'm interested in complex eigenvalues it might be useful to see when $AB$ has real spectrum. For instance, if $A$ is a positive form, then the spectrum of $AB$ is real-valued. Or if $[A,B] = 0$ then again spectrum of product is real. So there is no way to find a purely complex eigenvalue and we are done. What will happen if $A$ isn't positive definite? I'm searching for useful criteria that doesn't require positivity. Or maybe there exist criteria which ensure that $AB$ must have a purely complex eigenvalue. Has anybody heard about this? In what direction should I look? UPD For instance, if matrices are 2x2, the criteria is the following: $AB$ has only real eigenvalues $\Leftrightarrow$ $\exists$ a linear combination $\lambda A + \mu B^{-1}$ with real ${\lambda, \mu}$, s.t. it is positively defined REPLY [5 votes]: Really a remark on Carlo's answer, but somewhat relevant to the OP as well: if you generate random matrices as random products (that is, pick some matrices $A_1, \dotsc, A_N,$ and look at long random products), then the eigenvalues of the products will be almost surely all real. Why? Because the top eigenvalue will a unique eigenvalue of its modulus (this is a proximality statement, which you can find in the reference), and then a similar argument but with your product acting on the second exterior power shows that the biggest sum of two eigenvalues is real, and so on. What is also interesting is that the Galois group of the characteristic polynomial of this long random product (let's assume that the $A_i$ have integer entries) is $S_d.$ Breuillard, Emmanuel; Gelander, Tsachik, A topological Tits alternative., Ann. Math. (2) 166, No. 2, 427-474 (2007). ZBL1149.20039.<|endoftext|> TITLE: The inverse Torsion Mordell-Weil problem QUESTION [5 upvotes]: Let $G$ be a finite abelian group. Is there a field $K$, and an elliptic curve $E$ over $K$ such that $E(K)_{tor} \cong G$? REPLY [7 votes]: Even better, there exists an elliptic curve $E$ over a number field $K$, such that for any group of the form $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/mn\mathbb{Z}$, there is a finite extension $K'/K$ such that $$E(K')_{\text{tors}}\simeq\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/mn\mathbb{Z}.$$ Indeed, take any $(E,K)$ such that the natural Galois representation on the total Tate module of $E$, $$\rho: \text{Gal}(\bar K/K)\to GL_2(\hat{\mathbb{Z}})$$ is surjective (such curves exist by e.g. this paper of Zywina). Now choose a subgroup $$\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/mn\mathbb{Z}\subset E(\bar K)_{\text{tors}}$$ and let $K'$ be the field fixed by its stabilizer under $\rho$. By the surjectivity of $\rho$, the desired subgroup is all that is fixed, and we're done.<|endoftext|> TITLE: Are regular graphs the hardest instance for graph isomorphism? QUESTION [8 upvotes]: Regular graphs are the graphs in which the degree of each vertex is the same. The Weisfeiler-Lehman algorithm fails to distinguish between the given two non-isomorphic regular graphs. Is there a fastest known algorithm for regular graph isomorphism? Are regular graphs the hardest instance for graph isomorphism? Is there any combinatorial or algebraic technique (group theoretic) to deal with this situation efficiently? REPLY [4 votes]: An algorithm which never distinguished between non-isomorphic regular graphs (of the same order) would be spectacularly weak. Some appropriate canonical form of the adjacency matrix is a (totally unusable) invariant which definitively determines isomorphism. Any invariant which sometimes fails to distinguish non-isomorphic graphs is likely to have some pairs $G,G'$ which are non-isomorphic, both regular, and not distinguished. The article to which you link begins, in part: The classical Weisfeiler-Lehman method WL[2] uses edge colors to produce a powerful graph invariant. ... Many traditionally used combinatorial invariants are determined by WL[k] for small k. We show that WL[2] determines the number of cycles of lengths up to 6. So there are some cases of non-isomorphic graphs that WL[2] does not distinguish and certainly some of them are both regular. However such a pair would need to agree on the number of cycles of length $3,4,5,6$ and other things.<|endoftext|> TITLE: Upper bound for number of subpartitions of a partition QUESTION [5 upvotes]: Let $\alpha = (\alpha_1,\alpha_2,\dots)$ be a partition of the positive integer $n$, that is, a nonincreasing sequence of nonnegative integers $\alpha_j$, with only finitely many nonzero terms, whose sum is $n$. A subpartition of $\alpha$ is a partition $\beta = (\beta_1,\beta_2,\dots)$ such that $\beta_j \le \alpha_j$ for all $j\ge1$. How many distinct subpartitions can $\alpha$ have? I am interested in as good an upper bound as possible, in terms of $n$, that holds for every partition of $n$ (that is, I'm interested in bounds that hold in the "worst case"). Every subpartition of $\alpha$ is a partition of some number $0\le m\le n$, and so a trivial upper bound (using the notation $p(m)$ for the number of partitions of $m$) is $$ \sum_{m=0}^n p(m) \le (n+1)p(n) \ll e^{\pi\sqrt{2n/3}} $$ by the Hardy–Ramanujan asymptotics for $p(n)$. However, I suspect the truth is quite a bit smaller than this. References to existing bounds would be optimal, but arguments that improve this trivial bound substantially (including improvements to the constant in the exponent) are very welcome. [Note that we are literally counting partitions here, not weighting the partitions differently according to their shape. (The motivation for my question comes from trying to estimate the number of isomorphism classes of subgroups of a finite $p$-group.) Asking a similar question but where partitions are weighted by a Plancherel measure, while not relevant to this question, is probably still interesting (and, as far as I know, an open problem); there my suspicion is that this upper bound cannot be substantially improved—the intuition being that "typical" partitions $\alpha$ of $n$ give the worst case for the upper bound, and most typical partitions of $(1-\epsilon)n$ might in fact be subpartitions of $\alpha$.] REPLY [9 votes]: Some data suggests that the partition(s) of $n$ maximizing the number of subpartitions has the same limiting curve as for the the partition(s) of $n$ maximizing the number of standard Young tableaux of that shape. This is the famous curve of Logan-Shepp and Vershik-Kerov given parametrically by $$ x = y + 2\cos\theta $$ $$ y = \frac{2}{\pi}(\sin\theta-\theta\cos\theta), $$ for $0\leq\theta\leq \pi$. See for instance Section 3 of https://arxiv.org/pdf/math/0512035.pdf. For instance, when $n=45$ there are two conjugate partitions $\lambda$ that maximize the number of subpartitions. One is $(11,8,6,5,4,3,2,2,1,1,1,1)$. There are also two conjugate partitions maximizing the number of SYT, one being $(11,8,6,5,4,3,3,2,2,1,1)$.<|endoftext|> TITLE: The kernel of a nef line bundle QUESTION [6 upvotes]: Let $V$ be a complex projective variety and $L$ a nef line bundle on $V$ (i.e., $L$ is non-negative on every curve in $V$). Denote, as usual, $\deg_LX = c_1(L)^{\dim{X}}.[X]$ for $X$ a subvariety of $V$, considered as a prime cycle of dimension $\dim{X}$. Question. For subvarieties $X$ and $Y$ of $V$ with $\deg_LX = \deg_LY = 0$, does it follow that all positive-dimensional components $Z$ of $X \cap Y$ necessarily have $\deg_LZ = 0$? REPLY [7 votes]: Consider $V=\mathbb{P}^1\times \mathbb{P}^2$ with projections $p_1\colon V \rightarrow \mathbb{P}^1 \text{ and } p_2\colon V \rightarrow\mathbb{P}^2.$ Let $L = p_1^*(\mathcal{O}_{\mathbb{P}^1}(1))$, let $X = p_2^{-1}(\ell_1)$ and $Y=p_2^{-1}(\ell_2)$ for two distinct lines $\ell_1$ and $\ell_2\subset \mathbb{P}^2$. Then $Z = X\cap Y \cong \mathbb{P}^1$ is a section of $p_1$. Finally we have $\deg_L(X) = \deg_L(Y) = 0$, but $\deg_L(X\cap Y) = 1$.<|endoftext|> TITLE: Does ZF+AD settle the original Suslin hypothesis? QUESTION [25 upvotes]: Everyone knows that the real line $\langle\mathbb{R},<\rangle$ is the unique endless complete dense linear order with a countable dense set. Suslin's hypothesis is the question whether we can replace separability in this characterization with the assertion that the order has the countable chain condition, that is, that every set of disjoint intervals is countable. In other words, Suslin's hypothesis, in the original formulation, is the assertion that the real line is the unique endless complete dense linear order with the countable chain condition. Question. Does ZF plus the axiom of determinacy AD imply the original Suslin hypothesis? Set theorists proved in ZFC that Suslin's hypothesis is equivalent to the assertion that there is no Suslin tree, which is a tree of height $\omega_1$ with no uncountable chains or antichains. And ZF plus the axiom of determinacy AD settles this version of SH by proving that $\omega_1$ is measurable and hence weakly compact and hence has the tree property and so under AD there is there is no Suslin tree. So under AD there is no $\omega_1$-Suslin tree. But does this mean that there is no Suslin line? The point is that refuting a Suslin tree does not seem directly to refute all Suslin lines in the non-AC context, and therefore it does not seem to settle the original Suslin problem under AD. So the question is: does ZF+AD settle the original Suslin problem? Please feel free to post an answer explaining the precise details of the argument that AD implies there is no $\omega_1$-Suslin tree. I heard this question this evening at a party at a conference in honor of Simon Thomas from a certain prominent set theorist, aged Scotch in hand, who told me that he would rather not be mentioned, but who said he was fine for the question to be posted. Let me update the question (after Asaf's very nice answer) to ask about the situation where we also have the axiom of dependent choice DC. And perhaps one really wants to know about the case in $L(\mathbb{R})$ under AD. Question. Does ZF+AD+DC imply the original Suslin hypothesis? Question. Does $L(\mathbb{R})$ have a Suslin line assuming AD? REPLY [10 votes]: Nice question! But which aged set theorist drinks prominent Scotch? The following does not answer this question either. Let $T \in L$ be your favorite Suslin tree of $L$. Consider the $L({\mathbb R})$ of $L[g]$, where $g$ is generic for adding $\omega_1$ Cohen reals. $T$ is still a Suslin tree in $L[g]$. The set of all maximal branches thru $T$ is then a Suslin line in $L({\mathbb R})^{L[g]}$, and $L({\mathbb R})^{L[g]}$ is a model of ZF plus DC plus "there is no w.o. of the reals." So a Suslin line doesn't give a w.o. of ${\mathbb R}$. As has been poited out, this approach can't be generalized to show that in the presence of large cardinals (or just ${\sf AD}^{L({\mathbb R})}$), $L({\mathbb R})$ has a Suslin line, as then no $T \in M$, $M$ any inner model of $L({\mathbb R})$ satisfying choice, can be an Aronzsajn tree in $L({\mathbb R})$. But, assuming large cardinals, the question if $L({\mathbb R})$ has a Suslin line should have an answer as ${\sf ZFC}$ is $\Omega$-complete concerning truth in $L({\mathbb R})$.<|endoftext|> TITLE: Reference for LIL for fractional Brownian motion QUESTION [6 upvotes]: (Cross-posted to https://math.stackexchange.com/questions/2377810/law-of-iterated-logarithm-for-fractional-brownian-motion.) It seems strange but, even after consulting several books, and hours spent on google, nothing came out about a law of iterated logarithm for the fractional Brownian motion. I just need a precise reference, on where I can find such a law. EDIT: My goal is to prove that the fractional Brownian motion of hurst parameter $0 TITLE: Does the image of the exponential map generate the group? QUESTION [16 upvotes]: Let $G$ be a connected Fréchet-Lie group and let $\mathfrak g$ be its Lie algebra. Does the image $\exp(\mathfrak g) \subset G$ of the exponential map generate $G$? REPLY [19 votes]: I believe that this is open. In fact, even for the very special case where $G = \text{Diff}_0(M)$ for a smooth manifold $M$, the only proof I know that $G$ is generated by the image of the exponential map uses a very deep theorem of Thurston that says that in this case $G$ is a simple group (this implies that $G$ is generated by the image of the exponential map since the subgroup generated by the image of the exponential map is a nontrivial normal subgroup). Even giving a genuinely different proof in this special case would be very interesting.<|endoftext|> TITLE: Computing the etale cohomology of spheres QUESTION [9 upvotes]: $\newcommand\Z{\mathbb{Z}}$ Let $K$ be an algebraically closed field of characteristic $\ne 2$. We have the unit sphere $S^n:~x_0^2 + \ldots + x_n^2 = 1$. What are the $\Z/2\Z$ cohomology groups of the sphere - $H^i_{et}(S^n,\Z/2\Z)$? If $K$ is of characteristic 0, then by comparison with the complex numbers we get $$ H^i_{et}(S^n,\Z/2\Z)= \begin{cases} \Z/2\Z, & \text{if}~~ i=0,n \\ 0, & \text{otherwise} \end{cases} $$ and it should be the same result for characteristic $p$. I am sure that this has been calculated before, but I could not find a reference. So I would like a reference to the calculation of this (or practically equivalent) cohomology groups. I have 4 different directions to the computation, all of which require some technical details to complete or a smarter decomposition of the space. Comparison for positive characteristic. I understand that there are methods to make a comparison between the etale cohomology over the complex numbers even when the field is of positive characteristic, by viewing the variety as the specialization of a scheme defined over a $p$-adic ring, such that this scheme is nice enough so that there is an equivalence between the cohomology groups over the generic fiber and the cohomology groups of the fiber over the closed point. $S^n$ as a homotopy fiber. The fiber of the simplicial schemes $O_n\backslash\backslash S^n \to BO_{n+1}$ is the simplicial scheme associated with $S^n$. The conditions given by Friedlander for the identification of the cohomology of the sphere with that of the homotopy fiber of the etale homotopy types seems to hold. Jardine calculated the etale cohomology groups of $BO_{n+1}$, and the map above should be the same as the map $BO_n \to BO_{n+1}$. Then a simple spectral sequence argument calculates the right cohomology groups. Using a (hyper)cover. For sphere of odd dimension, which we can write as $z_0 z_1 +\ldots+ z_{n-1} z_n =1$, we can find a cover given by $U_k: z_{2k} \ne 0$. By taking colimit of the etale homotopy type, this cover can be seen to exhibit $S^n$ as the smash product of the $U_k$, each homotopy equivalent to $\mathbb{G}_m$. Gysin Sequence. There are closed embeddings between the spheres $S^{n-1} \to S^n$, which gives rise to a long exact sequence of a pair. We can calculate easily the cohomology with closed support, but I did not find a closed subscheme isomorphic to $S^{n-1}$ such that the complement open subset is easily computable. REPLY [12 votes]: This is true over any algebraically closed field $k$ of characteristic different from $2$. More generally, if $\ell$ is invertible in $k$, then $$H^i(X,\mathbb Z_\ell) = \left\{\begin{array}{ll}\mathbb Z_\ell & i = 0, n, \\ 0 & i \neq 0, n. \end{array}\right.$$ See for example [SGA 7$_\text{II}$, Exp. XII, Table 3.7]. The result then follows from the universal coefficient theorem (or you could mimic the proof). The proof passes through the projective quadric $\bar X$ by removing a hyperplane (isomorphic to the same quadric of dimension $n-1$). The long exact sequence of compactly supported cohomology and induction computes the compactly supported cohomology of $X$. Then Poincaré duality computes the cohomology of $X$. For the projective quadric, by the Lefschetz hyperplane theorem it suffices to compute the middle cohomology. If you ignore torsion, this is easy: just compute the Euler characteristic (as top Chern class of the tangent bundle). For the statement that $H^n$ is torsion free, you have to a bit more work. It is a little bit unclear to me what argument SGA uses to prove the torsion-freeness (in the proof of Theorem 3.3, it is not addressed), but it certainly follows from the same result over $\mathbb C$ together with the smooth and proper base change theorems (this is implicitly the strategy of Vérification 3.8).<|endoftext|> TITLE: Ramanujan's Lost Notebook page 1 first equation and OEIS sequence A260195 QUESTION [12 upvotes]: In the 1988 Narosa edition of Ramanujan's The Lost Notebook and Other Unpublished Papers, on the first line of page 1 is the following: $$ \Big(1+\frac1a\Big) \Bigg\{\frac{1}{(1-aq)(1-q/a)}+\frac{q(1+q)(1+q^2)}{(1-aq)(1-aq^3)(1-q/a)(1-q^3/a)}+\frac{q^2(1+q)..(1+q^4)}{..}+..\Bigg\} .$$ The infinite sum in braces with $a=-1$ has a $q$-series expansion$$ A(q) := 1 - q + 3q^2 - 2q^3 + 3q^4 - 3q^5 + 4q^6 - 3q^7 + 6q^8 - 4q^9 +\dots$$ which is the generating function of OEIS sequence A292511 and I conjecture that $A292511(n-1) = -(-1)^n A260195(n)$ where $A260195(n)$ is the number of integer triples $[x,y,z]$ such that $1\le\min(x,z),\max(x,z)\le y$ and $y^2-(x^2-x+z^2-z)/2=n$. Is this conjecture true? It may help that G. E. Andrews and B. C. Berndt in Ramanujan's lost notebook, Part I, page 277, in discussing Ramanujan's equation (Entry 12.4.5), have a result in equation (12.4.23) that leads to $$ q A(q) = \prod_{n>0} \frac{1+q^n}{1-q^n} \sum_{n>0} -(-1)^n q^{n^2}\frac{1-q^{2n-1}}{(1+q^{2n-1})^2} $$ which gives an alternate way of getting the $q$-series expansion. REPLY [12 votes]: This conjecture is equivalent to the following $$\frac{q}{(1-q)^2}\sum_{n=0}^\infty(-q)^n \frac{(q;q^2){}_n(-q^2;q^2){}_n}{(q^3;q^2){}_n^2}=\sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)},\tag{1}$$ and after using the identity pointed out by M. Somos (with $q$ relaced by $-q$) $$ \frac{(q;-q)_\infty}{(-q;-q)_\infty}\sum_{n=-\infty}^\infty\frac{q^{n^2}}{(1-q^{2 n-1})^2}=\sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)}.\tag{1a} $$ Initial proof of $(1a)$ utilized the identity $$ \sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)-1}=\frac13\sum_{r,s,t\ge 0}(2q^{rs+st+tr+r+t+s}+(-q)^{rs+st+tr+r+t+s}),\tag{2} $$ which turned out to be another conjecture due to Michael Somos. So that initial proof was not self sufficient. Self sufficient proofs of $(1a)$ and $(2)$ are given below. Proof of $(1a)$. By shifting the summation of variables so that all summations start from $0$ and by elementary rearrangements of terms we get \begin{align} \sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)-1}&=2\sum_{r,s,t\ge 0}q^{\frac{r^2+3r}{2}+s^2+2s+2rs+t(r+1+2s)}-\sum_{1\le r\le t}q^{t^2-(r^2-r)-1}\\ &=\left(\sum_{r,s,t\ge 0}+\sum_{r,s,t< 0}\right)q^{\frac{r^2+3r}{2}+s^2+2s+2rs+t(r+1+2s)}\\ &=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-q^{r+1+2s}}. \end{align} It turns out that it is easier to work with a more general sum $$ g(z,q)=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z\,q^{r+2s}}. $$ Following the general method outlined in the article by E. Mortenson we derive the functional equation satisfied by $g(z,q)$: \begin{align} 0&=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right){q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}\\ &=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z\,q^{r+1+2s}}({1-z\,q^{r+1+2s}})\\ &=g(qz,q)-qz\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+5r}{2}+s^2+4s+2rs}}{1-z\,q^{r+1+2s}}\\ &=g(qz,q)-\frac{z}{q}\left(\sum_{r\ge 1,s\ge 0}-\sum_{r\le 0,s<0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z\,q^{r+2s}}\\ &=g(qz,q)-\frac{z}{q}g(z,q)+\frac{z}{q^2}j\left(-q,q^2\right) m\left(-{z }/{q},q^2,-q\right),\tag{3} \end{align} where in the last line we used notations $$ J_m=(q^m;q^m)_\infty,\quad j(x,q)=(q;q)_\infty(x;q)_\infty(q/x;q)_\infty,\quad m(x,q,z)=\frac1{j(z,q)}\sum_{n=-\infty}^\infty\frac{(-z)^nq^{n(n-1)/2}}{1-xzq^{n-1}}.\tag{4} $$ Since the Appell-Lerch sum $m(x,q,z)$ satisfies the equation (eq. 2.2c in Mortenson's paper) $$ m(qx,q,z)=1-xm(x,q,z),\tag{5} $$ one can show by elementary calculation that the function $$ g_1(z,q)=-\frac{z}{q^3}j\left(-q,q^2\right) m\left(-{z }/{q^2},q^2,-q\right) m\left(-{z}/{q},q^2,-q\right) $$ satisfies the same functional equation $(3)$ as $g(z,q)$. Now consider the function $$ h(z,q)=g(z,q)-g_1(z,q), $$ for which we have the functional equation $$ h(qz,q)=\frac{z}{q}h(z,q).\tag{6} $$ From the definition of $g(z,q)$ it is easy to see that it has simple poles at $z_0=q^n,~n\in\mathbb{Z}$ with residues being equal to finite sums of powers of $q$, i.e. polynomials in $q$. For example for $n\ge 1$ the residue is $$ \sum_{r+2s=n}q^{\frac{r^2-r}{2}+s^2+2rs}. $$ When $m\left(-{z }/{q^2},q^2,-q\right) $ is singular then $m\left(-{z }/{q},q^2,-q\right) $ is not and vice versa. So from the definition of $m(x,q,z)$ one can see that $g_1(z,q)$ has the same set of simple poles as $g(z,q)$. To find residues one can use the identity $m(-1,q^2,-q)=0$ and the functional equation $(5)$. The result is that $h(z,q)$ is analytic. Analytic function with functional equation $(6)$ is $0$ (for details one can consult Mortenson's paper, section 5). So we proved that $$ \left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z q^{r+2s}}=-\frac{z}{q^3}j\left(-q,q^2\right) m\left(-{z }/{q^2},q^2,-q\right) m\left(-{z}/{q},q^2,-q\right).\tag{7} $$ We want to calculate this expression when $z=q$. It is known that $m\left(-{z }/{q^2},q^2,-q\right)$ is singular at $z=q$ and $m\left(-{z }/{q},q^2,-q\right)=0$ at $z=q$. So one needs to resolve this ambiguity by applying L'Hôpital's rule: \begin{align} &\lim_{z\to q}m\left(-{z }/{q^2},q^2,-q\right)m\left(-{z }/{q},q^2,-q\right)\\ &=\frac{q}{j\left(-q,q^2\right) }\lim_{z\to q}\frac{m\left(-{z }/{q},q^2,-q\right)}{1-z/q}\\ &=\frac{q}{j\left(-q,q^2\right)}\frac{\frac d{dz}m\left(-{z }/{q},q^2,-q\right){\Large|_{\small{z=q}}}}{-1/q}\\ &=-\frac{q^2}{j^2\left(-q,q^2\right)}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}q^{2n-2}}{(1-q^{2n-1})^2}\\ &=-\frac{q}{j^2\left(-q,q^2\right)}\left(\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}-\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{1-q^{2n-1}}\right)\\ &=-\frac{q}{j^2\left(-q,q^2\right)}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}. \end{align} So when $z\to q$ eq. $(7)$ becomes $$ \left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-q^{r+1+2s}}=\frac{1}{qj\left(-q,q^2\right)}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}\\ =\frac{(-q;-q)_\infty}{q(q;-q)_\infty}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}.\Box $$ Proof of $(2)$. Here it is proved that $$ \frac13\sum_{r,s,t\ge 0}(2q^{rs+st+tr+r+t+s}+(-q)^{rs+st+tr+r+t+s})=\frac{(-q;-q)_\infty}{q(q;-q)_\infty}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2},(*) $$ which together with $(1a)$ will imply $(2)$. The sums of the type $\sum_{r,s,t\ge 0}q^{rs+st+tr+r+t+s}$ have been studied by Mortenson, Corollary 1.2, where he proved that \begin{align} \sum_{r,s,t\ge 0}q^{rs+st+tr}x^{r+t+s}+&\sum_{r,s,t< 0}q^{rs+st+tr}x^{r+t+s}=\\ &=3\frac{J_1^3j(x^2,q)}{j(x,q)^2}m(-q/x,q^2,qx^2)-2\frac{J_1^3J_2^3j(x^2,q^2)^3}{j(x,q)^3j(-x,q^2)^3}\tag{8} \end{align} with the notations $(4)$. Since the difference of two Appel-Lerch functions $m(x,q,z)$ with different $z$ is a quotient of theta functions (eq. 2.6 in Mortenson's paper) we can write in $(8)$ $$ \small{m(-q/x,q^2,qx^2)=m\left(-{q}/{x},q^2,z\right)+z\frac{J_2^3 j\left({q x^2}/{z},q^2\right) j\left(-xzq^2,q^2\right)}{j\left(q x^2,q^2\right) j\left(-xq^2,q^2\right) j\left(z,q^2\right) j\left(-{q z}/{x},q^2\right)}.}\tag{8a} $$ We want to calculate the triple sum in $(8),(8a)$ when $x=q$. Note that $$ \sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}=\sum_{r,s,t< 0}q^{rs+st+tr}q^{r+t+s}. $$ However $j(q,q)=0$ and since $j(x,q)$ appears in the denominator at the RHS of $(3)$ one needs to use L'Hôpital's rule to calculate the limit $x\to q$ with the help of formulas $\frac d{dx}j(x/q,q){\Large|_{\small{x=q}}}=-\frac1q J_1^3$ and $j(q^nx,q)=(-1/x)^nq^{-n(n-1)/2}j(x,q)$. First, we put $z=-q$ and take the limit $x\to q$ in $(8),(8a)$ to obtain \begin{align} &\sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}\\ &=\frac{3 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}-\frac{8 J_2^{12}}{j\left(-q,q^2\right)^3 J_1^6}-3\frac{j\left(-q^2,q^2\right) J_2^6}{q^2j\left(-q,q^2\right) j\left(q^3,q^2\right) j\left(-q^3,q^2\right) j\left(q,q^2\right)}. \end{align} One can show that $$ \frac{8 J_2^{12}}{j\left(-q,q^2\right)^3 J_1^6}+3\frac{j\left(-q^2,q^2\right) J_2^6}{q^2j\left(-q,q^2\right) j\left(q^3,q^2\right) j\left(-q^3,q^2\right) j\left(q,q^2\right)}=2\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3}. $$ So $$ \sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}=\frac{3 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}-2\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3}.\tag{9} $$ Similarly, the choice $z=q$ in $(8),(8a)$ leads to $$ \sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}=\frac{3 }{q j\left(q,q^2\right)}\sum _{r=-\infty}^\infty \frac{(-1)^rq^{r^2}}{\left(1+q^{2 r-1}\right)^2}+4\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3},\tag{10} $$ and after changing sign $q\to-q$: $$ \sum_{r,s,t\ge 0}(-q)^{rs+st+tr+r+t+s}=-\frac{3 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}+4\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3},\tag{10a} $$ Combine $(9)$ and $(10a)$ to get $$ \frac13\sum_{r,s,t\ge 0}(2q^{rs+st+tr+r+t+s}+(-q)^{rs+st+tr+r+t+s})=\frac{1 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}. $$ Now observe that $\frac{(q;-q)_{\infty } j\left(-q,q^2\right)}{(-q;-q)_{\infty }}=1$ to complete the proof of $(*)$.$~ \Box$<|endoftext|> TITLE: Do these ordinals exist? QUESTION [6 upvotes]: Given an ordinal $\alpha$, I define $F_{n}(\alpha)$ as follows: $F_0(\alpha)=\alpha$ $F_{n+1}(\alpha)$ is the smallest $\beta$ such that no first-order $\phi$ in the language of $\{\in\}$ has $(\mathrm{V}\models\phi(S,F_0(\alpha),F_1(\alpha)...F_{n}(\alpha)))\Leftrightarrow S=\beta$ $F_\omega(\alpha)=\mathrm{sup}\{F_{n}(\alpha):n<\omega\}$ The existance of Reinhardt cardinals implies that $F_1(\alpha)$ exists for every sufficiently small $\alpha$. In fact, every Reinhardt cardinal is $F_1(\alpha)$ for some ordinal $\alpha$. (Yes, even though Reinhardt cardinals are not consistent with AC, they are all ordinals.) However, this is the only information I could find on this. Could somebody else tell me what this is called if it is already named? Is it inconsistent to say $F_n(\alpha)$ exists for $n\leq\omega$? Which $n$ is it inconsistent for? Information gathered post-question: If there is a first-order $\phi$ such that $(\mathrm{V}\models\phi(S))\Leftrightarrow S=\alpha$, then $F_n(\alpha)=F_n(0)$ for every $n$. Here are some $\alpha$ for which this is true: The smallest Erdős initial ordinal, Mahlo cardinal, Ramsey, Rowbottom, Jonsonn, Inaccessible, Strongly Inaccessible, Limit, Weakly compact, ethereal, subtle (assuming each of these exists) Every other minimum initial ordinal for a first-order large cardinal axiom The second of each of those axioms, the third, the $\alpha$-th for $\alpha$ on this list (assuming they exist) Every $\alpha<\omega_1^{CK}$ (hint for proving this: Kleene's $\mathcal{O}$ is first-order definable) Every $\aleph_{\alpha}<\aleph_{\omega_1^{CK}}$ and $\beth_{\alpha}<\beth_{\omega_1^{CK}}$(a corrolary from above) $F_\alpha(\beta)$ is countable or $\omega_1$ (assuming ZF) Honestly I think this is the first time $\beth_{\omega_1^{CK}}$ has been used in a theorem ever. If I'm wrong, please tell me. REPLY [4 votes]: Let $V$ be a transitive model of $ZFC$. Without special assumptions about the model $V$, it is possible that $F_0(0)$ does not exist (in other words - it is possible that every ordinal is definable without parameters). In fact, every model of $ZFC + V=HOD$ has an elementary submodel such that all its elements are definable without parameters - for example, the Skolem closure of the empty set, using the definable Skolem functions. These models are called "Pointwise definable models". The situation in which $F_0(0)$ exists but $F_n(0)$ does not exist for some $n$ may occur as well. For example, let $V$ be a pointwise definable model of $ZFC + V = HOD$ + there is a measurable cardinal (this is a vast overkill). Let $\kappa$ be a measurable cardinal in $V$ and let $j \colon V \to M$ be the ultrapower embedding by a normal measure on $\kappa$. Let us claim that $\kappa$ is the first undefinable ordinal in $M$. Indeed, if $\varphi$ was a definition for $\kappa$ in $M$ then, by elementarity, $\varphi$ defines an unique ordinal in $V$, $\gamma$. But this implies that $j(\gamma) = \kappa$ which is impossible. Every element in $M$ is of the form $j(f)(\kappa)$ for some $f\in V$. Since $f$ is definable without parameters in $V$, $j(f)$ is definable (with the same definition) in $M$. In particular, every element in $M$ is definable from the parameter $\kappa$. Thus, $F_0(0) = \kappa$ and $F_1(0)$ doesn't exist. One can iterate this process in order to get for every $n < \omega$ a model in which $F_n(0)$ exists while $F_{n+1}(0)$ doesn't exists.<|endoftext|> TITLE: How to handle sums in Tait's reducibility proof of strong normalisation? QUESTION [7 upvotes]: I've been reading Girard et al's 'Proofs and Types', which in Chapter 6 presents a proof of strong normalisation for the simply typed lambda calculus with products and base types. The proof is based on Tait's method of defining a set of 'reducible terms for type $T$' by induction on the type $T$. For example a term $t$ is reducible for type $A\to B$ if for all reducible $u$ for type $A$, $t\,u$ is reducible for type $B$. However I don't see how to extend this proof to (binary) sums, which are covered in a later chapter, because the elimination for sums involves a 'parasitic' type which prevents one from defining reducible terms of type $A+B$ via the elimination by induction on $A+B$. How (if at all) does this proof style extend to sums? I can't find a good reference that presents this. REPLY [5 votes]: Let me add to the above (perfectly correct) answers that there is a more general perspective, which works for all positive connectives, of which sums are a special case. The fundamental insight is that reducibility actually arises from a duality between terms and contexts, which is itself parametric in a set of terms closed under certain operations (such as $\beta$-expansion, but I think something trickier must be used for strong normalization), called a pole. Given a pole $P$, one defines $$\mathsf C\perp t\quad\text{just if}\quad\mathsf C[t]\in P$$ where $\mathsf C$ is a context (a term with a hole denoted by $\bullet$) and $t$ a term. The idea is that $P$ is a set of terms which "behave well" (e.g. they have a normal form), so $\mathsf C\perp t$ means that the context $\mathsf C$ and the term $t$ "interact well" with respect to the property expressed by $P$. Given a set $S$ of contexts, one may define a set of terms $$S^\bot:=\{t\mathrm{|}\forall\mathsf C\in S,\,\mathsf C\perp t\}.$$ A similar definition induces a set of contexts from a set of terms $T$, which we still denote by $T^\bot$. We may then define the interpretation of a type $A$, which is a set of contexts denoted by $|A|$, by induction on $A$: $$ \begin{array}{rcl} |X| &:=& \{\bullet\} \\ |A\to B| &:=& \{\mathsf C[\bullet\,u]\mathrel{|} u\in|A|^\bot,\ \mathsf C\in|B|\} \\ |A\times B| &:=& \{\mathsf C[\pi_1\bullet]\mathrel{|}\mathsf C\in|A|\}\cup\{\mathsf C[\pi_2\bullet]\mathrel{|}\mathsf C\in|B|\}\\ |A+B| &:=& \{\mathsf C\mathrel{|}\mathsf C[\mathsf{inl}\,\bullet]\in |A|^{\bot\bot}\text{ and }\mathsf C[\mathsf{inr}\,\bullet]\in|B|^{\bot\bot}\} \end{array} $$ Finally, we let $$\|A\|:=|A|^\bot,$$ which is a set of terms. This is the "reducibility predicate" on $A$. One then goes on to prove the so-called adequacy theorem: if $$x_1:C_1,\ldots,x_n:C_n\vdash t:A$$ is derivable, then for all $u_1,\ldots,u_n$, $u_i\in\|C_i\|$ implies $t[u_1/x_1,\ldots,u_n/x_n]\in\|A\|$. This is proved by induction on the last rule of the derivation, using the properties of the pole (closure under $\beta$-expansion or something like that). Now, if things are setup correctly, one also has $x\in\|A\|$ as well as $\|A\|\subseteq P$ for every variable $x$ and every type $A$, which implies that every term belongs to the pole by adequacy, i.e., every term "behaves well". I am sure that this works for weak normalization, i.e., one may take $P$ to be the set of weakly normalizing terms (which is closed under $\beta$-expansion) and immediately obtain weak normalization of the simply-typed $\lambda$-calculus with sums. For strong normalization, some kind of hack is needed, because $\beta$-expanding a strongly normalizable term does not necessarily yield a strongly normalizable term, but I'm sure that a suitable definition will make things work. This "reducibility by duality" may be found already in Girard's Linear logic paper (1987), although I'm not sure he is the one who introduced it. It has been used by many authors since then and of course it easily extends to higher order quantification. In particular, the above presentation is adapted from Krivine, who uses it as the basis of his classical realizablity theory (see his many papers on the subject, all available on his web page).<|endoftext|> TITLE: Cancelling a graph join from a graph homomorphism QUESTION [12 upvotes]: Given (finite, simple) graphs $G$, $H$ and $K$ and a homomorphism $$ G+K\to H+K $$ where $+$ denotes the join, does it follow that there also exists a graph homomorphism $G\to H$? If this is known, I'd also appreciate a reference. REPLY [13 votes]: If $|K|=\infty$, then this is false, as a counterexample $G=K_2$, $H=K_1$, $K=K_\infty$ shows. Let us prove that the claim is true if $K$ is finite (with no such assumption for $G$ and $H$). Induction on $|K|$; if $|K|=0$, the claim is trivial. For the inductive step, consider a homomorphism $\psi\colon G+K\to H+K$. Set $G_1=\psi(G)\cap K$, $K_1=\psi(K)\cap K$, $H_1=\psi(K)\cap H$. If $G_1=\varnothing$, then $\psi\big|_G$ is a required homomorphism $G\to H$. So now we assume that $|G_1|>0$. Each vertex of $G_1$ is connected with each of $K_1$ since $\psi$ is a homomorphism (thus in particulat $|K_1|<|K|$). Each vertex of $H_1$ is connected with each of $K_1$ by the definition of join. Thus, the induced subgraphs on $G_1\cup K_1(\subseteq K)$ and $H_1\cup K_1$ are isomorphic to $G_1+K_1$ and $H_1+K_1$, respectively. So $\psi\big|_{G_1\cup K_1}$ provides a homomorphism $G_1+K_1\to H_1+K_1$ which by the induction hypothesis implies the existence of a homomorphism $\varphi\colon G_1\to H_1$. Finally, set $M=\psi(G)\cap H$. Each vertex of $M$ is connected with each of $K_1$, since $\psi$ is a homomorphism. Thus the map $\eta\colon G\to H$, $$ \eta(g)=\begin{cases} \psi(g), &\psi(g)\in M;\\ \varphi(\psi(g)), &\psi(g)\in G_1 \end{cases} $$ is a sought homomorphism. The step is proved.<|endoftext|> TITLE: Is there a closed 5-manifold $M$ with $w_1(M)w_2(M)\ne 0$? QUESTION [10 upvotes]: I'm trying to find generating manifolds for the cobordism group $\mathit{MO}_5(K(\mathbb Z/2, 2))\cong (\mathbb Z/2)^4$, which can be represented as the cobordism group of closed 5-manifolds $M$ together with a class $B\in H^2(M;\mathbb Z/2)$. I've found three of the four generators; the fourth should be a 5-manifold $M$ such that $w_1(M)w_2(M)\in H^3(M;\mathbb Z/2)$ is nonzero, and here I've gotten stuck. Such an $M$ cannot be a product of lower-dimensional manifolds, nor can it be the total space of a fiber bundle over the circle. I'm happy to hear general approaches or ideas as well, such as ways of modifying a manifold to change its cohomology or Stiefel-Whitney classes in useful ways. REPLY [13 votes]: Note that the third Wu class is $\nu_3 = w_1w_2$, so on a closed connected smooth $n$-manifold $M$, $\operatorname{Sq}^3 : H^{n-3}(M; \mathbb{Z}_2) \to H^n(M; \mathbb{Z}_2)$ is given by $\operatorname{Sq}^3(x) = w_1(M)w_2(M)x$. As $\operatorname{Sq}^i(x) = 0$ if $i > \deg x$, when $n = 5$ we see that $\operatorname{Sq}^3 : H^2(M; \mathbb{Z}_2) \to H^5(M; \mathbb{Z}_2)$ is the zero map. By Poincaré duality, it follows that $w_1(M)w_2(M) = 0$.<|endoftext|> TITLE: Are theta functions cuspidal representations? QUESTION [6 upvotes]: After reading many textbooks I still can't get the jargon correct. Given a spherical harmonic $u \in L^2(S^n)$ one could construct a theta function: $$ \theta (z;u) = \sum_{ m \in \mathbb{Z}^3} u (m) e^{2\pi i \, |m|^2\,z}$$ with $z\in \mathfrak H$. The situation of interest is that $u$ is not constant. This thing is a $\Gamma_0(4)$ cusp form. Could these theta functions be automorphic forms? REPLY [9 votes]: "Cusp form" means "cuspidal automorphic form" by definition. So yes, $\theta(z;u)$ is an automorphic form. But it is not a form lying in a cuspidal automorphic representation, because it is not a Hecke eigenform. (Actually, this also depends on the context, see the Added section for clarification.) Cuspidal automorphic representations are irreducible subspaces of the relevant cuspidal automorphic $L^2$-space, so most cusp forms are linear combinations of forms coming from distinct (usually infinitely many) cuspidal automorphic representations. For non-cuspidal forms the picture is even more complicated: linear combination is replaced by an integral with respect to some spectral measure which includes various (Hecke-)Eisenstein series. The general theory is due to Langlands (spectral decomposition of $L^2$-automorphic forms). (Again, see the Added section for clarification. The point is that the notion of irreducibility depends on the group, and one can think of the group over the reals $\mathbb{R}$ but also over the rational adeles $\mathbb{A}_\mathbb{Q}$.) Added. I realize now that a serious confusion can arise from considering (irreducible) cuspidal automorphic representations of $\mathrm{GL}_n(\mathbb{R})$ vs. considering those of $\mathrm{GL}_n(\mathbb{A}_\mathbb{Q})$. I like to consider the adelic group $\mathrm{GL}_n(\mathbb{A}_\mathbb{Q})$, because it has more structure and is more relevant for number theory. For simplicity, let us restrict to automorphic forms spherical at the archimedean place (e.g. a classical Maass form on the upper half-plane). Then, a vector from a cuspidal automorphic representation of $\mathrm{GL}_n(\mathbb{A}_\mathbb{Q})$ corresponds to a cusp form on $\mathrm{GL}_n(\mathbb{R})$, which is an eigenfunction of the local spherical Hecke algebra at all but finitely many non-archimedean places, and vice versa. One can refine this notion to talk about newforms and oldforms, by considering non-spherical Hecke algebras (but Hecke algebras corresponding to a level). In the end, the notion of "automorphic form" and "automorphic representation" varies greatly with the context. A good introduction is Borel-Jacquet: Automorphic forms and automorphic representations (Proc. Symp. Pure Math. 53 (1979), 189-202), which discusses both $\mathrm{GL}_n(\mathbb{R})$ and $\mathrm{GL}_n(\mathbb{A}_\mathbb{Q})$.<|endoftext|> TITLE: 2TQFT and commutative Frobenius algebras QUESTION [5 upvotes]: There is an equivalence between the category of commutative finite dimensional Frobenius algebras and 2 dimensional topological quantum field theories, see for example the book by Joachim Kock, which I read a bit (but I have no idea about the topological part). Some questions: Do module theoretic notions have a topological interpretation using this equivalence? For example an open problem for commutative finite dimensional Frobenius algebras A is wheter $Ext_A^1(M,M) \neq 0$ for any non-projective module $M$. Does this have a topological interpretation? Given a commutative f.d. Frobenius algebra $A$, then its trivial extension algebra $T(A)$ (see for example https://math.stackexchange.com/questions/229412/trivial-extension-of-an-algebra for the definition) is again a commutative Frobenius algebra with twice the dimension of $A$. What is the topological interpretation of the trivial extension construction using the equivalence? For two f.d. commutative Frobenius algebras $A,B$ one has $rad(A \otimes B)=rad(A) \otimes B + A \otimes rad(B)$ (here rad dentotes taking the Jacobson radical of an algebra). This reminds me of the formula for the boundary of the product of two topological spaces. Is this just random, or is there a connection between those two formulas? REPLY [2 votes]: Maybe you are not so interested in this, but there is a nice physical interpretation of modules over a Frobenius algebra. (1+1)d TQFTs are used to describe topological phases of matter in 1d. Think of a sequence of spin 1/2 particles connected up in a line or a circle where the global state cannot be specified by local observables. Unfortunately, without more structure, topological phases of matter in 1d are trivial. This is a manifestation of the fact that any frobenius algebra is isomorphic to a product of matrix algebras. In this setting, bi-modules correspond to domain walls between different topological phases of matter. Since topological phases of matter are boring in 1d, this isn't really saying much. In 2d things are much more interesting. Topological phases of matter are now classified by (2+1)d TQFTs which can be specified by fusion categories and slight generalizations of them. There are a lot of interesting fusion categories and domain walls between the corresponding topological phases of matter are specified by bi-modules over fusion categories.<|endoftext|> TITLE: Every contractible smooth loop has a neighbourhood with $H^2=0$ QUESTION [6 upvotes]: Let $c: S^1 \to M$ be a smooth contractible loop (not necesarily an embedding, or even an immersion) on the connected, compact symplectic manifold $(M,\omega)$ (if this helps somehow, $c$ is a $1$-periodic orbit of a $1$-periodic time-dependent Hamiltonian $H_t$). The book by Audin, Damian claims that there exists a neighbourhood $U$ of $c(S^1)$ that retracts (and by this, they mean "deformation retracts") onto $c(S^1)$. What is needed, however, is the fact that there exists a neighbourhood $U$ of $c(S^1)$ such that $H^2(U;\mathbb{R})=0$. My question is: why is this true? If $\dim(M)=2$, then this follows by considering $p \notin c(S^1)$ and taking $U:=M\backslash\{p\}$. Since $U$ is a non-compact connected $2$-manifold, it follows that $H^2(U;\mathbb{R})=0$. However, I can't see how to proceed in generality, since tubular neighbourhoods (which I think is what the authors had in mind) seem elusive etc. REPLY [4 votes]: Inductively isotope the skeleta of a triangulation of $M$ so that the $k$-skeleton is transverse to $c$. (We may use an isotopy to achieve transversality because the simplices are already embedded, and we don't need to homotope the boundary at all; a small homotopy of only the interior of the simplex doesn't change the property of being an embedding. Demand that a neighborhood of the rest of the triangulation is kept fixed, then apply isotopy extension to move the whole mess around.) Once this is true of all skeleta, $\text{Im } c \cap \sigma$ is empty unless $\sigma$ is top dimensional, or one dimension below that. If we consider the union of the open faces of all $n$ and $(n-1)$-dimensional simplices of $M$, we have an open manifold $M_\tau$ with the homotopy type of a graph (the graph whose vertices correspond to $n$-simplices and edges correspond to $(n-1)$-simplices) with $\text{Im } c \subset M_\tau$. In particular, $\text{Im } c$ has a neighborhood with $H^2(M_\tau) = 0$.<|endoftext|> TITLE: GIT quotient vs. largest Hausdorff quotient QUESTION [5 upvotes]: Let a group $G$ act on a (not necessarily irreducible) algebraic variety over ${\bf C}$. It seems to be well-known that the quotient in the sense of geometric invariant theory (i.e., the categorical quotient in the category of algebraic varieties) agrees with the largest Hausdorff quotient (i.e., the categorical quotient in the category of Hausdorff spaces). The reference which I find cited for this fact in the literature are two papers of Luna from the 70s: paper 1, paper 2. For example, this reference says (on page 6): It can be shown that the categorical quotient in the category of affine varieties (over ${\bf C}$ and with respect to a reductive group action) is also the categorical quotient for Hausdorff spaces or complex analytic varieties [Lun75, Lun 76]. My question is just how the claim about the categorical Hausdorff quotient follows from Luna's papers. I understand that Luna is proving that not only (by definition) the invariant polynomials correspond 1-1 to polynomial functions on the GIT-quotient, but that also the same holds true for smooth or analytic functions instead of polynomials. Does this imply by some known results that the GIT-quotient must be the largest Hausdorff quotient? REPLY [6 votes]: I recommend reading the very nice exposition here (which is where I learned this fact): Schwarz, Gerald W. The topology of algebraic quotients. Topological methods in algebraic transformation groups (New Brunswick, NJ, 1988), 135–151, Progr. Math., 80, Birkhäuser Boston, Boston, MA, 1989. In particular, see the proof starting at the bottom of Page 141.<|endoftext|> TITLE: Mathematical games interesting to both you and a 5+-year-old child QUESTION [251 upvotes]: Background: My daughter is 6 years old now, once I wanted to think on some math (about some Young diagrams), but she wanted to play with me... How to make both of us to do what they want ? I guess for everybody who has children, that question comes up. Okay, I said to her: let's play a game which I called "Young diagram" for her: we took a sheet of paper and I tried to explain to her what a Young diagram is, she was asked to draw all the diagrams of some size n=1,2,3,4,5... Question: Do you have some experience/proposals of "games" which you can play with your children, which would be on the one hand would make some fun for them, on the other would somehow develop their logical/thinking/mathematical skills, and on the other hand would be of at least some interest for adult mathematicians ? Related MO questions: “Mathematics talk” for five year olds it is quite related to the present question, but slightly different - it is about a single presentation to children, while the present question is about your own children with whom you play everyday, you can slightly "push", and so on... How do you approach your child's math education? it is also related, but the present questions has a slightly different focus: games interesting for children and adults. The book by Alexandre Zvonkine, "Math for little ones" (in Russian here), recommended in answer there - is really something related to the present question. Which popular games are the most mathematical? is NOT directly related, but may serve as kind of inspiration for answers... I think Allen Knutson's answer on “Mathematics talk” for five year olds: I've spoken (to 5+ years old) about the "puzzles" that Terry Tao and I developed for Schubert calculus, like the left two here: can be a nice example of an answer to the present question as well: on the one hand there is something to explain to the child and some colorful pictures, and on the other hand that is about research level math ... REPLY [2 votes]: You can play Tic-Tac-Toe on the affine plane of order 5. Check up PentacTow. I have a few further ideas for applications gaming finite geometries and I take this opportunity to invite whoever codes and have interest in such a project to contact me (the TicatacToe was programmed by my brother, Gal Bader, but he is too busy now).<|endoftext|> TITLE: Examples of triality in mathematics QUESTION [15 upvotes]: There are tons of interesting examples of duality in mathematics (Poincaré duality, Verdier duality, Stone duality, s-duality, Tannaka duality, Koszul duality, Spanier-Whitehead duality ... ). What examples are there of triality in mathematics? Note: this is not a duplicate of the question about trichotomies in mathematics. A trichotomy is any sort of classification into three. A triality is a classification into three where the relationship between those three is some sort of equivalence relation (especially one emphasizing that though things appear opposite they are in some sense the same). REPLY [5 votes]: An interesting duality is polarity with respect to a conic in the projective plan. A conic is determined by a bilinear form $q$ (the equation of the conic being $q(u,u)=0$) and one way to describe the polarity is to say that the polar of a point corresponding to a vector $v$ in the underlying $3$-space is the line whose equation is $q(v,\cdot)=0$. Now, this yields an immediate equivalent for trialities: consider a cubic, it is determined by a trilinear form $t(\cdot,\cdot,\cdot)$ and two points (with corresponding vectors $u$, $v$) and a line are in triality if $t(u,v,\cdot)=0$ is an equation of the line.<|endoftext|> TITLE: Diameter of $\mathrm{SU}(2)$ endowed with a left-invariant metric QUESTION [9 upvotes]: Basic question: What is the diameter of $\mathrm{SU}(2)$ endowed with a left-invariant metric? Now, let me give more information. Set $$ X_1= \begin{pmatrix} i &\\ &-i \end{pmatrix},\; X_2= \begin{pmatrix} &1\\ -1& \end{pmatrix},\; X_3= \begin{pmatrix} &i\\ i& \end{pmatrix}. $$ It is sufficient to consider the left-invariant metrics with inner product on $\mathfrak{su}(2)$ satisfying that $\{aX_1,bX_2,cX_3\}$ is an orthonormal basis, where $a,b,c$ are positive real numbers. Indeed, any permutation of these numbers will return an isometric metric. Let us denote by $g_{(a,b,c)}$ the left-invariant metric (and also the inner product on $\mathfrak{su}(2)$) mentioned above. The goal is to write an expression for $\mathrm{diam}(\mathrm{SU}(2), g_{(a,b,c)})$ in terms of $a$, $b$ and $c$. It is well known that $\mathrm{SU}(2)$ is diffeomorphic to the $3$-dimensional sphere. For example, a map is given by $$ \begin{pmatrix} u&-\bar v \\ v&\bar u \end{pmatrix} \mapsto \begin{pmatrix} u\\ v \end{pmatrix}, $$ for $u,v\in\mathbb{C}$ satisfying that $|u|^2+|v|^2=1$. Since the Riemannian manifolds at hand are homogeneous, the diamenter is attained at a pair of points $(I_2,h)$, where $I_2$ denotes the $2\times 2$ identity matrix and $h$ is in $\mathrm{SU}(2)$. Since $I_2$ and $-I_2$ are opposite points in $S^3$, the next question seems to be affirmative. Question 1: $\mathrm{diam}(\mathrm{SU}(2), g_{(a,b,c)}) = \mathrm{dist}_{g_{(a,b,c)}} (I_2,-I_2)$? Any idea to prove it (in case is affirmative)? Let's move on, assuming this is true. The geodesics on $(\mathrm{SU}(2), g_{(a,b,c)})$ starting from $I_2$ are not in general one-parameter subgroups (see Robert Bryant's comment in Tsemo Aristide's answer). Geodesics satisfy the Euler equation (see again Bryant's comment or this Bryant's answer). Let $\dot\gamma (t) = d L_{\gamma(t)}(X_{\gamma}(t))$, and for $X\in\mathfrak{su}(2)$, set $\textrm{ad}_{g_{(a,b,c)}}^*(X):\mathfrak{su}(2) \to\mathfrak{su}(2)$ defined by $$ g_{(a,b,c)}(\textrm{ad}_{g_{(a,b,c)}}^*(X)Y,Z)= g_{(a,b,c)}(Y,[X,Z]). $$ The Euler equation is $$ \dot X_{\gamma}(t) = \textrm{ad}_{g_{(a,b,c)}}^*(X_\gamma(t)) X_{\gamma}(t). $$ For $X_\gamma(t)= a_1(t)X_1+a_2(t)X_2+a_3(t)X_3$, it reduces to $$ \begin{array}{rcl} \dot a_1(t) &=&2a^2\, a_2(t)a_3(t) (\frac{1}{b^2}-\frac{1}{c^2}),\\[1mm] \dot a_2(t) &=&2b^2\, a_1(t)a_3(t) (\frac{1}{c^2}-\frac{1}{a^2}),\\[1mm] \dot a_3(t) &=&2c^2\, a_1(t)a_2(t) (\frac{1}{a^2}-\frac{1}{b^2}). \end{array} $$ Example 1: When $a=b=c$ (round metric), it follows immediately that $a_j(t)\equiv a_j(0)$ is constant for all $j$ and then $\gamma(t)=\exp(tX_\gamma(0))$ is an one-parameter subgroup. Moreover, one can check (as expected) that $\mathrm{dist}_{g_{(a,b,c)}} (I_2,-I_2) = \pi/a$ since every geodesic from $I_2$ to $-I_2$ has this length. We go back to the arbitrary case. Although in general not every geodesic is an one-parameter subgroup, it is for some particular cases (remember Bryant's answer). Take the initial velocity vector as $X_1$, thus $a_1(0)=1$ and $a_2(0)=a_3(0)=0$. Euler equation immediately implies that $\dot a_j(0)=0$ for all $j$, thus again $X_\gamma(t)=X_1$ for all $t$ and $\gamma(t)=\exp(tX_1)$. Consequently, since $\gamma(\pi)=-I_2$, then the length of $\gamma$ on $[0,\pi]$ is $$ \int_0^\pi g_{(a,b,c)}(X_\gamma(t),X_\gamma(t))^{1/2} dt = \pi\, g_{(a,b,c)}(X_1,X_1)^{1/2} = \frac{\pi}{a}. $$ Similarly, taking $X_2$ or $X_3$ as initial velocity vector, we obtain that the corresponding geodesics from $I_2$ to $-I_2$ have length $\pi/b$ and $\pi/c$ respectively. Hence $$ \mathrm{dist}_{g_{(a,b,c)}} (I_2,-I_2) \leq \frac{\pi}{\max\{a,b,c\}}. $$ Question 2: $\mathrm{dist}_{g_{(a,b,c)}} (I_2,-I_2) = {\pi}/{\max\{a,b,c\}}$? In case Questions 1 and 2 are affirmative, then we conclude that $$ \mathrm{diam}(\mathrm{SU}(2), g_{(a,b,c)})=\frac{\pi}{\max\{a,b,c\}}. $$ REPLY [2 votes]: Here I propose the following answer to my question. It uses the same ideas as in the two previous answers by Nate Eldredge and Richard Montgomery. Comments are welcome. The basic question is not known. So, it would be useful to have some bounds for $\mathrm{diam}(\mathrm{SU}(2),g_{(a,b,c)})$ in terms of $a$, $b$ and $c$. From now on we assume $a\geq b\geq c>0$. We will prove (I will avoid writing $\mathrm{SU}(2)$ in the diameter function.) $$\frac{\pi}{2b}\leq \mathrm{diam}(g_{(a,b,c)})\leq \frac{\pi}{b}.$$ Moreover, there will be a more precise lower bound in terms of $a$ and $b$. Nate Eldredge had predicted that $\mathrm{diam}(g_{(a,b,c)})$ and $b$ are comparable. We have that $$ g_{(t,b,b)}(X,X)\leq g_{(a,b,b)}(X,X)\leq g_{(a,b,c)}(X,X)\leq g_{(b,b,c)} (X,X) $$ for all left-invariant vector fields $X$, if $t\geq a$. Hence $$ \mathrm{diam}(g_{(t,b,b)})\leq \mathrm{diam}(g_{(a,b,b)})\leq \mathrm{diam}(g_{(a,b,c)})\leq \mathrm{diam}(g_{(b,b,c)}). $$ It is well known that the diameter is equal to the maximum distance between a fixed point and the points in the corresponding cut locus. The cut locus of any Berger's sphere (i.e.\ $(\mathrm{SU}(2),g_{(a,b,c)})$ with $a=b$ or $b=c$) is computed. Sakai did it for $a=b\geq c$, and recently Podobryaev-Sachkov in PS gave the full answer. Although the diameter of a Berger's sphere is not stated in PS, the following expression can be obtained by the same procedure used in PS to obtain the diameter of $\mathrm{SO}(3)$ endowed with a left-invariant metric (in its notation, the multisets $\{\!\{I_1,I_2,I_3\}\!\}$ and $\{\!\{1/(2a)^2,1/(2b)^2,1/(2c)^2\}\!\}$ coincide): $$ \textrm{diam}(\textrm{SU}(2),g_{(a,b,c)}) = \begin{cases} \pi/b &\quad\text{if } a=b\geq c,\\ \pi/a &\quad\text{if } a>b=c\geq a/\sqrt{2},\\ \dfrac{\pi}{2b\sqrt{1-b^2/a^2}} &\quad\text{if } a>\sqrt{2}b=\sqrt{2}c. \end{cases} $$ This expression was kindly communicated by the first named author of PS. It follows immediately that $\mathrm{diam}(g_{(a,b,c)})\leq {\pi}/{b}$ as Richard Montgomery asserted. Furthermore, $$ \mathrm{diam}(g_{(a,b,c)})\geq \mathrm{diam}(g_{(a,b,b)}) = \begin{cases} \pi/a &\quad\text{if } a\leq \sqrt{2}b,\\ \dfrac{\pi}{2b\sqrt{1-b^2/a^2}} &\quad\text{if } a>\sqrt{2}b. \end{cases} $$ Since the above function is decreasing, then $\mathrm{diam}(g_{(a,b,c)})\geq \lim_{t\to\infty} \mathrm{diam}(g_{(t_1,b,b)})={\pi}/{2b}$.<|endoftext|> TITLE: A question about a statement in the paper of C.T.C. Wall QUESTION [5 upvotes]: ‎‎Suppose that ‎‎$‎‎X$ ‎is ‎a finite ‎2-dimensional CW-complex with free fundamental group ‎and ‎‎$‎‎‎\phi‎ :K ‎\longrightarrow ‎X‎$ is a map which ‎induces ‎an ‎isomorphism ‎of ‎fundamental ‎groups, where $K$ is a finite bouquet of circles with the wedge point $a$. Consider the mapping cylinder $M=X\bigcup_{\phi} (K \times \{1\})$‎. ‎Denote $\pi_n (M_{‎\phi‎},K \times \{ 1\} )$ by $\pi_n (\phi )$‎. ‎Recall ‎that ‎an ‎element ‎of ‎‎$‎‎\pi_n (‎\phi‎ )$ ‎is ‎represented ‎by a‎ ‎pair ‎of ‎maps ‎‎$‎‎‎\beta ‎:‎\mathbb{S}^{n-1}‎\longrightarrow ‎K‎$‎ ‎and ‎‎$‎‎‎\gamma ‎:‎\mathbb{D}^n ‎\longrightarrow ‎X‎$‎ ‎with ‎‎$‎‎‎\gamma‎|_{‎\mathbb{S}^{n-1}‎}=\phi‎ \circ ‎\beta‎$. In the paper ''Finiteness conditions for CW-complexes'' of C.T.C.Wall in Propositon 3.3, Wall has mentioned that since ‎‎$‎‎\pi_2 (‎\phi‎)$ ‎is a‎ ‎free ‎‎$‎‎‎\mathbb{Z}\pi_1 (X)‎$-module‎, then we can attach 2-cells ‎to ‎‎$‎‎K$, ‎necessarily ‎with ‎trivial ‎attaching ‎maps, to make $\phi$ a homotopy equivalence. My question is that: What is Wall's mean concerning ''necessarily ‎with ‎trivial ‎attaching ‎maps''? If his mean is that 2-cells are wedged to the wedge point a, then how can I get to this fact? Thank you for your help. REPLY [4 votes]: The attaching map for a 2-dimensional cell is a map $f: S^1 \to K$, and it determines an element $[f]$ of $\pi_1(K)$ (up to conjugacy). What Wall means by trivial attaching maps is that these elements of $\pi_1(K)$ must be trivial. If they weren't, then the map $\pi_1(K) \to \pi_1(K \cup_f D^2) \to \pi_1(X)$ that attaches the cell would send the element $[f] \in \pi_1(K)$ to the trivial element, which contradicts the fact that the map $\phi$ was an isomorphism on $\pi_1$ in the first place. The homotopy type of the complex you get by gluing 2-dimensional cells only depends on the homotopy classes of the attaching maps (to prove this you typically use multiple applications of the homotopy extension property). Therefore, up to homotopy equivalence it really is the case that you can just attach 2-dimensional cells to the basepoint. The second homotopy group $\pi_2(K \vee \bigvee S^2)$ is a free $\Bbb Z[\pi_1(K)]$-module generated by the copies of $S^2$, and Wall's construction is sending these to a set of generators for $\pi_2(X)$ over $\Bbb Z[\pi_1(X)]$.<|endoftext|> TITLE: A game-theoretical question in a political economy model QUESTION [7 upvotes]: My research question in a dynamic model of political competition boils down to the following conjecture. I am confident that it holds (all simulations work), but I have not been able to prove it yet. Let $c,\eta,\mu,i$ be parameters such that $02$ or $i<0$ the result does not hold, so $1<\eta\leq2$ and $0 TITLE: PFA and saturated ideals QUESTION [9 upvotes]: Does PFA imply that there is an $\omega_2$-saturated ideal on $\omega_1$? All I know is that MM implies that $NS_{\omega_1}$ is saturated. Thanks! REPLY [7 votes]: No. It is a theorem of Shelah that PFA is consistent with the failure of weak Chang's conjecture: there exists $\langle f_i\in \omega_1^{\omega_1}: i < \omega_2+1\rangle$ that is increasing mod $NS_{\omega_1}$. We show the existence of $\omega_2$-saturated ideal on $\omega_1$ is incompatible with this fact. Let $g=f_{\omega_2}$. For each $i TITLE: Existence of distribution for certain order statistics QUESTION [6 upvotes]: This is an open question: given a sequence of $n$ real numbers $x_1 0$.<|endoftext|> TITLE: Consistency of Strong reflection principle with the existence of a Suslin tree QUESTION [5 upvotes]: In Woodin's book "The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal" Remark 2.55 (5), it states SRP by Todorcevic (defined below) is consistent with the existence of a Suslin tree (hence it does not imply MM). Is there any reference about this? My guess is we could probably get a restricted version of MM for a Suslin tree, like the one for PFA. More precisely, $PFA(S)$ is the forcing axioms for the class of forcings that preserve $S$. Then maybe we can derive SRP from $MM(S)$. Definition (projectively stationary sets): Let $\lambda\geq \omega_1$, then $S\subset [H(\lambda)]^\omega$ is projectively stationary if for any stationary $T\subset \omega_1$, $\{X\in S: X\cap \omega_1\in T\}$ is stationary in $[\lambda]^\omega$. Definition (SRP): SRP asserts for any projectively stationary $S\subset [H(\lambda)]^\omega$ for regular $\lambda\geq \omega_2$, there exists a continuous increasing $\in$-chain $\langle N_\alpha: \alpha<\omega_1\rangle$ such that $N_\alpha\in S$ for all $\alpha<\omega_1$. REPLY [3 votes]: This appears to be proven in the following paper of Miyamoto: Miyamoto, Tadatoshi, On iterating semiproper preorders, J. Symb. Log. 67, No. 4, 1431-1468 (2002). ZBL1050.03034. In section 5 of the paper, he introduces a forcing axiom, SPFA(Souslin), which is the forcing axiom for semi-proper posets which preserve every Souslin tree. He shows that SPFA(Souslin) is compatible with the existence of a Souslin tree and that SPFA(Souslin) implies SRP.<|endoftext|> TITLE: Free loop space objects and actions QUESTION [8 upvotes]: The free loop space object of an object $X$ in an $(\infty,1)$-category $\mathcal{C}$ can be defined as the pullback $\mathcal{L}X= X\times_{X\times X} X$. Unlike the based loop space, this is not generally a group object in $\mathcal{C}$: there is no way to compose loops with different basepoints. However, if we regard $\mathcal{L}X$ as an object of the slice $\mathcal{C}/X$ (via either projection; they are equivalent), then as sketched here, it should be a group object in $\mathcal{C}/X$ (which "puts together" all the based loop spaces over all possible basepoints), and moreover it should act canonically on all objects of $\mathcal{C}/X$ by "transport". Where can I find this group structure and action defined formally and coherently? REPLY [8 votes]: Suppose $\mathcal{D}$ is an $\infty$-category with finite limits. Given a pointed object $\ast\rightarrow A$, its Cech nerve is a simplicial object $M_\bullet$. By Higher Topos Theory, Proposition 6.1.2.11, it is actually a groupoid and, moreover, since $M_0\cong \ast$, it is a group. This gives $M_1\cong \Omega A$ the structure of a group. Given a morphism $B\rightarrow A$ consider the pullback diagram $$ \require{AMScd} \begin{CD} B\times_A \ast @>>> B\\ @VVV @VVV \\ \ast @>>> A \end{CD} $$ Taking the Cech nerves horizontally you get a morphism of simplicial objects $M'_\bullet\rightarrow M_\bullet$. Using that $M'_\bullet$ is a groupoid, it is easy to see that $M'_\bullet\rightarrow M_\bullet$ is a left action object in the sense of Higher Algebra, Definition 4.2.2.2. So, the group $M_1\cong \Omega A$ acts on $M'_0\cong B\times_A \ast$. Let's restrict to the case $\mathcal{D} = \mathcal{C}_{/X}$. Consider the object $A=(p_1\colon X\times X\rightarrow X)$. The final object in $\mathcal{C}_{/X}$ is $\mathrm{id}\colon X\rightarrow X$ and $A$ has a pointing by the diagonal map $\Delta\colon X\rightarrow X\times X$. Its based loop space is $\Omega_\Delta A = (\mathcal{L} X\rightarrow X)$. Note that using that $\mathcal{C}$ is cotensored over spaces, you can obtain the Cech nerve of $X\rightarrow X\times X$ in $\mathcal{C}_{/X}$ by applying $\mathrm{Map}(-, X)$ to the Cech conerve of $S^0\rightarrow \ast$ in pointed spaces which gives $\Sigma S^0\cong S^1$ the structure of a cogroup in pointed spaces. Given a morphism $f\colon Y\rightarrow X$ in $\mathcal{C}$, let $B = (p_1\colon X\times Y\rightarrow X)$ with $B\rightarrow A$ given by $(\mathrm{id}\times f)\colon X\times Y\rightarrow X\times X$. Then $B\times_A \ast = (X\times Y)\times_{X\times X} X\cong Y$ which by above gives an action of $\mathcal{L} X\rightarrow X$ on $Y\rightarrow X$. Note that you could instead consider $\tilde{B} = (p_1\circ(f\times \mathrm{id})\colon Y\times X\rightarrow X)$. This will give the trivial action of $\mathcal{L} X\rightarrow X$ on $Y\rightarrow X$. In algebraic geometry you can think of it as follows. Let's say $X$ is a (derived) scheme. $\widehat{X\times X} = [X/\mathcal{L} X]$ is the classifying space of the group scheme $\mathcal{L}X\rightarrow X$ and $\widehat{X\times Y} = [Y/\mathcal{L} X]$.<|endoftext|> TITLE: Equivalence classes of norms on $R^n$ under symmetries QUESTION [6 upvotes]: Let $G \leq {\bf GL}_n$ be a symmetry group on $\mathbb{R}^n$. For simplicity, we can consider the case $G = {\bf GL}_n$. Define two norms $\|\cdot\|_1$ and $\| \cdot\|_2$ to be equivalent under $G$ if there exists $A \in G$ such that $ \ \forall \ x \ \| A x \|_1 = \| x\|_2$ It is trivial to show that this defines equivalence classes on the set of all norms on $\mathbb{R}^n$. Now, for $\mathbb{R}^1$, there is only a single class of norms but for $\mathbb{R}^2$ there are infinitely many classes, specifically all $p$ norms live in different equivalence classes (except for the fact that the $1$-norm and the $\infty$-norm are equivalent). For general $\mathbb{R}^n$ the case is even "worse". Is there some way to classify these equivalence classes? REPLY [6 votes]: When $G={\bf GL}_n$, the set of classes of equivalent norms over ${\mathbb R}^n$ is a compact Hausdorff space when equipped with the Banach-Mazur metric $$d(N_1,N_2)=\log\delta(N_1,N_2)$$ where $$\delta(N_1,N_2)=\inf_{T\in{\bf GL}_n}\|T^{-1}\|_{N_2\rightarrow N_1}\cdot\|T\|_{N_1\rightarrow N_2}\,.$$ See this page. Of course, the smaller the group $G$, the bigger the set of equivalence classes.<|endoftext|> TITLE: Variant of Graph coloring QUESTION [6 upvotes]: This is a problem came from social network analysis. In a vertex colored (need not be proper) graph, an edge is monochromatic, if both endpoints of the edge are colored with the same color. Given a partially colored network, my goal is to extend it to total coloring such that the number of monochromatic edges in the network is maximum. If the network (graph) is complete graph then the problem is easy (just see which color is appearing more and assign it all uncolored vertices). Suppose if my graph is a threshold graph, then I don't have any idea how to solve it in polynomial time. Can someone help me? REPLY [6 votes]: Update 2017.09.26: As I've just learnt from Tamas Kiraly, this notion is well-studied and has several names, such as k-way cut, k-terminal cut, multiway cut. For at least three colors, the problem becomes $NP$-complete for general graphs, see E. Dahlhaus, D. S. Johnson, C. H. Papadimitriou, P. D. Seymour, and M. Yannakakis: The complexity of multiterminal cuts. Earlier stuff: For two colors this can be solved in polynomial time for any graph, as it can be converted to a MinCut problem. Just connect all red vertices into a single vertex $s$, and all blue vertices into another single vertex $t$, obtaining a multigraph (or if you prefer, a graph with edge weights). The Minimum cut in this new graph between $s$ and $t$ gives the partition that gives the coloring maximizing the number of the monochromatic edges.<|endoftext|> TITLE: Concurrent bitangents of a quartic curve QUESTION [6 upvotes]: What is the maximum number of concurrent bitangents, i.e. all intersecting at the same point, of a smooth complex projective quartic curve? Can the number of concurrent bitangents be six? REPLY [9 votes]: I think the maximum number of concurrent bitangents is at most 4. Consider the double cover $S\rightarrow \mathbb{P}^2$ branched along your quartic curve. Fix one point $q$ of $S$ above the point of intersection; each bitangent lift to a "line" in $S$ passing through $q$. The description of these lines is well-known. We can represent $S$ as the blown-up of $\mathbb{P}^2$ along 7 points $p_1,\ldots ,p_7 $, and assume that one of the lines is the exceptional divisor above $p_1$. Since any pair of our "lines" in $S$ must meet, we don't have much choice: we can have the strict transform of the line in $\mathbb{P}^2$ joining $p_1$ to another point, say $p_2$, then the strict transform of the conic through $p_1$ and 4 other points $\neq p_2$, say $p_3,\ldots ,p_6$, and finally the the strict transform of the cubic passing through all the $p_i$ and doubly through $p_7$. So the maximum is $\leq 4$; I do not know if 4 can be actually realized. $3$ is easy, by projecting a cubic surface with 3 concurrent lines. REPLY [8 votes]: (This is really a comment to the good answer of abx, but I don't have the reputation for that.) Indeed 4 concurrent bitangents seems to be realisable, for example by the Klein quartic (what else?). In the book Classical Algebraic Geometry by Dolgachev you will find Exercise 6.22 (in the version I have, at least) which reads: "Show that the set of 28 bitangents of the Klein quartic contains 21 subsets of four concurrent bitangents and each bitangent has 3 concurrency points."<|endoftext|> TITLE: Examples of interesting non orientable closed 3-manifolds QUESTION [16 upvotes]: In dimension 2, there are two remarkable non-orientable closed manifolds, the projective plane (from synthetic geometry; has the fixed point property; algebraic compactification of the plane etc) and the Klein bottle (nowhere vanishing vector field; with immersions sold in your nearest nonorientable store). There is also a classification of all closed non-orientable surfaces, as connected sums of projective planes. I am looking for examples of non-orientable 3 dimensional closed (compact, boundaryless) manifolds. Any with some special properties or arising from interesting geometrical problems? Is there a simple classification for them? REPLY [22 votes]: One gets non-orientable closed 3-manifolds by taking a non-orientable surface, and crossing with $S^1$, such as $P^2\times S^1$. In fact, the geometrization theorem hasn't been proven completely for non-orientable 3-manifolds. Of course, a 2-fold cover has a connect sum and geometric decomposition. But the problem is that no one has shown that this decomposition can be made equivariant with respect to the covering translation. One expects that a careful analysis of the proof using Ricci flow could be made equivariant (at least Ricci flow preserves symmetries). One can't do the initial connect sum decomposition, because 1-sided projective planes must be cut along and coned off, resulting in an orbifold with isolated cone points. In any case, I think that the most interesting non-orientable 3-manifold is the smallest known volume closed manifold, which has volume $2.0298...$, the same as the figure 8 complement, and fibers over the circle. This was discovered by Weeks in his census. I think it is known to be arithmetic. See Table 2 of a paper by Hodgson and Weeks. REPLY [11 votes]: A curiosity is obtained as follows: take a solid cube $[-1,1]^3$ and identify one pair of two opposite faces by a symmetry with respect to a coordinate axis, while identifying the other two pairs of opposite faces by translations. The resulting manifold contains an embedded Klein bottle which has two sides (in the mathematical sense that its normal bundle is trivial, i.e. removing its zero section disconnects it), i.e. you can paint one side of the Klein bottle blue and the other side red without the two color meeting. This shows that it is not an intrinsic property of non-orientable surface to "have only one side", it is a property of some of their codimension $1$ embeddings (including all embeddings in orientable $3$-manifold). In fact, having only one side means not being co-orientable rather than being not orientable. The above example also has an embedded $2$-torus which is not coorientable (and thus has only one face!), of course.<|endoftext|> TITLE: How do the existing ways to do formal CT relate to each other? QUESTION [8 upvotes]: I can see three ways to "do formal category theory": Yoneda structures à la Street-Walters. cosmoi à la Street. proarrow equipments. It seems to me that 2. is older than 1., and a particular case thereof (you impose more requests on $y_A : A \to PA$ asking that there is an adjoint to the pseudofunctor $P : {\cal K}^\text{coop} \to {\cal K}$). Is there, instead, a relation between Yoneda structures, cosmoi, and proarrow equipments? REPLY [5 votes]: Street's "fibrational cosmoi" are indeed a special case of, or rather a particular way to construct, a Yoneda structure. They are substantially less general, since Yoneda structures and equipments include the case of enriched categories, but fibrational cosmoi do not. (Street also used the word "cosmos" later for a bicategory of the form $W \mathrm{Prof}$, which is more like the horizontal part of an equipment, but doesn't contain information about the functors as distinct from the representable profunctors.) Yoneda structures and equipments are similar in that they both enhance a 2-category with information about "profunctors", but they do it in slightly different ways, so that in constructing examples one has to make different choices about how to deal with size. Namely, in an equipment (as originally defined), all proarrows are composable; thus if we include large categories as objects, we also need to include large-set-valued profunctors as proarrows; or we can consider only small categories and small-set-valued profunctors. On the other hand, in a Yoneda structure all profunctors are small-"set"-valued, but we are required to include large categories and even "locally large" categories, since the presheaf category of a non-small category is not generally locally small; thus not every category has an identity proarrow and not all proarrows can be composed. Thus, although morally they contain about the same information, it's hard to prove a precise comparison abstractly. The best I'm aware of is this paper of Koudenburg, which uses a generalization of equipments to "hypervirtual double categories" and shows that under reasonable assumptions, presheaf categories can be given a universal property therein that corresponds exactly to the axioms of a Yoneda structure on a certain subcategory.<|endoftext|> TITLE: How to formally split monomorphisms nicely? QUESTION [8 upvotes]: A split monomorphism is a morphism $m \colon A \to B$ for which there exists a morphism $e \colon B \to A$ such that $e \circ m = \mathrm{id}_A \colon A \to A$. Is there an elegant description of a universal construction that splits monomorphisms in a given category, i.e. that adds a new morphism $e$ for every monomorphism $m$? I'm looking for adding one-sided inverses $e$, not two-sided inverses that make every monomorphism $m$ into an isomorphism. A variation on localization would work, but a concrete construction would be much more elegant. Is a variation on calculi of fractions known that might work? By elegant construction, I'm thinking of something like the Karoubi envelope. An idempotent is a morphism $p \colon B \to B$ such that $p \circ p = p$. It splits when there exist $m$ and $e$ as above with $m \circ e = p$. The Karoubi envelope adds new morphisms $e$ and $m$, and a new object $A$, to every $p$, so in a sense is opposite to the question. REPLY [8 votes]: Here are three possibilities: First: you have the obvious 'universal solution': You start from $C$ any category and $J$ a set of maps, you can consider the category $C'$ freely generated from $C$ by adding a retract $r_i$ to each arrow $i \in J$. $C'$ has the same objects as $C$ and rather complicated arrows, they are described as formal sequence of arrow in $C$ and arrow of the form $r_i$, up to the obvious relation (composing the arrow in $C$ and $r_i \circ i = 1$). this being said, I belive one can prove that the functor $C \rightarrow C'$ is faithfull if and only if all arrow in $J$ are monomorphisms. which might be interesting (I will give a proof below in the case where $C$ is small, but it should be true in general either, eventually using a large cardinal axiom, or by just refining the construction a little). Second construction, less general, but closer to the 'category of fractions' you mentioned. Let $C$ be a category with pullback (in fact, we only need pullback of subobjects). Then one can define a category $C'$ of ''partial map'' in $C$ : An arrow $X \rightarrow Y$ in $C'$ is a sub-object $U \subset X$ together with a map $U \rightarrow Y$ (by subobject, I just mean a monomorphism $U \rightarrow X$, but two such maps are identified if there is an isomorphism $U \rightarrow U'$ wich makes everythings commute, such an isomorphism is unique if it exists). Those maps are composed as follows: if $X \overset{(U,f)}{\rightarrow} Y \overset{(V,g)}{\rightarrow} Z$ then you pullback $V$ to $U$ along $f$ you get a $W \subset U \subset X$, the pullback of $f$ induce a map $W \rightarrow V$ which you can compose with $g$ to get your map $W \rightarrow Y$. On easily check that: 1) this form a category. 2) the 'total arrow', i.e. those such that $U =X$ form a subcategory isomorphic to $C$. 3) for each monomorphism $i:X \rightarrow Y$, the partial arrow $r_i: Y \rightarrow X$ which is defined as the identity on $X \subset Y$ and is a retract of $i$. 4) every arrow in this category if of the form $f \circ r_i$ for $f \in C$ and $r_i$ one of the map above. The thrid construction, only applies to a small category $C$ but with no other assumptions. You look at the category $\widehat{C}$ of presheaves on $C$. Monomorphism in $C$ stay monomorphisms in $\widehat{C}$, so one can look at the category whose objects are objects of $C$ and whose arrow are partial maps in $\widehat{C}$. As $\widehat{C}$ has pullbacks this also gives a solution to your problem. This allows to proves the claim made in the first construction, at least for small category: for any small category $C$ and $J$ a set of monomorphisms in $C$, there exists a faithfull functor to a category $C'$ such that all arrow in $J$ have retraction in $C'$, hence the category obtained by universally adding the retraction is indeed a faithful extention of $C$. Final remark: None of the construction above, will produce a locally small category in full generality: In the first if you start form a proper class of arrow $J$ you will almost certainly end up with a non locally small category. In the second construction, the category of partial arrow will be locally small if and only if you started with a well-powered category. For the third one, you need a small category, although it might works for certain non-small category. Also you can adapt the idea and use another an embeddings into a finitely complete category. For example any accessible category can be embedded in $prsh(C)$ of $C$ small.<|endoftext|> TITLE: Open subsets of Euclidean space in dimension 5 and higher admitting exotic smooth structures QUESTION [20 upvotes]: Up to dimension 3, homeomorphic manifolds are diffeomorphic (in particular, homeomorphic open subsets of $\mathbb{R}^n$ ($n\leq 3$) are diffeomorphic). It is known that there are uncountably many mutually non-diffeomorphic open subsets of $\mathbb{R}^4$ that are homeomorphic to $\mathbb{R}^4$ (Demichelis & Freedman, 1992). For $n\neq 4$, $\mathbb{R}^n$ admits a unique smooth structure (up to diffeomorphism), thus the previously mentioned phenomenon is unique to dimension 4. Questions: In dimension $n\geq 5$, is it possible for a manifold to be homeomorphic to an open subset of $\mathbb{R}^n$ but not diffeomorphic to it? is it possible that two homeomorphic open subsets of $\mathbb{R}^n$ may be non-diffeomorphic? Are there examples of such open sets, in each dimension $n\geq 5$, having a simple description up to homeomorphism? (the smooth structure on each open subset of $\mathbb{R}^n$ being the standard one). Edit: If I understand it well, Misha Kapovich points out in his comment below that Robion Kirby told him that the answer to question (2) above is positive, in every dimension $n\geq 5$ (and thus in every dimension $n\geq 4$, with the difference that for $n\geq 5$ no such sets can be homeomorphic to $\mathbb{R}^n$). As (2) implies (1) this would settle the question. It would be interesting to know if other mathematicians heard about these examples of Kirby. From the topological point of view (i.e. up to homeomorphism), what are these sets? REPLY [2 votes]: Misha's answer provides examples (for certain $n$), of pairs of open subsets of $\mathbb{R}^n$ (both with the standard smooth structure), that are homeomorphic but not diffeomorphic to each other, thus giving a partial answer to the more general question 2). One of the open sets is built from an exotic $(n−2)$-dimensional sphere which can be realized as Brieskorn variety. Smoothly embedding in $\mathbb{R}^n$, it can be viewed as a "small exotic $\mathbb{S}^{n-2}\times\mathbb{R}^2$" Meanwhile, I got the following partial answer from an expert (private communication; as above, dimension $4$ is excluded). It goes in the other sense, providing examples of large exotic structures on open subsets of $\mathbb{R}^n$ (again, not for all $n>4$): "Strictly speaking, $\mathbb{R}^n\setminus 0$ answers your question for most values of $n$, since it is diffeomorphic to $\mathbb{S}^{n-1}\times\mathbb{R}$, and the first factor typically has exotic structures that remain nondiffeomorphic to the original after product with $\mathbb{R}$. However, the resulting exotic structures do not embed in $\mathbb{R}^n$. I am guessing that you want pairs of open subsets of $\mathbb{R}^n$ that are homeomorphic to each other but not diffeomorphic. I don't know if such pairs exist. Clearly, they would have to have trivial tangent bundles, but that does not preclude exoticness: there are exotic spheres $S$ that bound parallelizable manifolds, so $S\times\mathbb{R}$ has a trivial tangent bundle and is not diffeomorphic to a standard sphere $\times\mathbb{R}$. However, such an example cannot embed as an open subset of $\mathbb{R}^n$. (Otherwise, $S$ would bound a ball and hence be standard.)" Two short comments: 1) this construction provides examples of manifolds homeomorphic to simple open subsets of $\mathbb{R}^n$, but which do not smoothly embed in $\mathbb{R}^n$ (summing up, an exotic structure in the $(n-1)$-sphere gives rise to an exotic structure on the punctured $\mathbb{R}^n$, which is "large" in the sense that it does not embed smoothly in $\mathbb{R}^n$). 2) It would be also interesting to know analogue (topologically) simple examples in every dimension $n>4$ for which the $(n-1)$-sphere exhibits no exotic structure (in the case $n=5$, for which no exotic structure is known).<|endoftext|> TITLE: Mori: p-adic and real hemispheres of the mathematical universe? QUESTION [8 upvotes]: I recall having read, some time ago, a beautiful and poetic opening of an article (or was it a book?). From memory, it was by Shigefumi Mori, and talked about the (mathematical) universe consisting of two hemispheres, a real and a p-adic; somehow meeting at the equator. We mere humans should strive to contemplate both sides and their connections. I have tried to locate the precise reference of this text, including some of Mori's papers, but failed. In fact, I'm not anymore sure it was Mori. Does some kind soul know which reference I'm trying to recall? REPLY [11 votes]: As the night sky, mathematics has two hemispheres; the archimedean hemisphere and the non-archimedean hemisphere. For some reasons, the latter hemisphere is usually under the horizon of our world, and the study of it is historically behind the study of the former. [...] The aim of this paper is [...] to discuss that we can see an arm of a big galaxy, the galaxy of $p$-adic zeta elements, in the non-archimedean hemisphere of zeta values, but that the total shape of this galaxy is still under the horizon. Precisely speaking, we expect the following: As all automorphic representations have zeta functions with values in $\mathbb{C}$, all Galois representations of number fields with coefficients in any $p$-adic ring $\Lambda$ have $p$-adic zeta elements which are canonical bases of some invertible $\Lambda$-modules. [...] The harmony between the two worlds should be called the generalized Iwasawa theory or generalized Deligne-Beilinson conjectures on zeta values. For [the] Riemann zeta function and Dirichlet $L$-functions, Iwasawa theory is the best theory at present for the arithmetic of zeta values. How nice it would be if we can construct the Iwasawa theory of Hasse-Weil $L$-functions. // Where is the homeland of zeta values to which the true reason of celestial phenomena of zeta values are attributed? How can we find a galaxy train [Mi*] to approach it, which runs through the galaxy of $p$-adic zeta elements and whose engine is the theory of $p$-adic periods? I imagine that one coach of the train has the name 'the explicit reciprocity law of $p$-adic Galois representations'. This is Kazuya Kato speaking of $\zeta$. The reference is the following: Lectures on the approach to Iwasawa theory for Hasse-Weil L-functions via $B_{ \mathrm{dR}}$, Part I, in: Arithmetic Algebraic Geometry, LNM 1553 (1993). More of this zeta poetry, in a more elementary mathematical level, can be found in the three small volumes Number Theory 1-3 (ed. Kato, Kurihara and Saito), published in the Translations of Mathematical Monographs series. Mori, incidentally, is linked to a term called a Mori dream space in the minimal model program. He did make a wonderful use of positive characteristic (if not $p$-adic methods) in his work in complex geometry, and I thought this and the association to dream spaces could be partly responsible for why you thought of him here. With [Mi*] in the above, Kato quotes a Japanese poet, Miyazawa K. (A night on the galaxy train, written circa 1924.)<|endoftext|> TITLE: KPZ-NLS-Burgers relationship QUESTION [6 upvotes]: The Burger's equation $$y_t (t,x) + y\cdot y_x - y_{xx} =0 \, \, ,$$ can be obtained as a limit of the one-dimensional cubic Nonlinear Schrodinger equation (NLS) $$ i\psi _t (t,x) + \psi _{xx} +|\psi|^2\psi =0 \, \, ,$$ but can also be obtained as an approximation of the Kardar-Parizi-Zhang (KPZ) equation $$h_t (t,x) = h_{xx} + (h_x)^2 +\eta (t,x) \, \, ,$$ where $\eta $ is a Gaussian noise term. Question: I don't know any of the details of both approximations/transformations, nor what kind of results can you deduce from it. Do you know any good references for that? The Burger's equation Wiki page offers some information, but not enough (for me) to get the full picture. There are a lot of papers which are somehow related to it, but none that I found were good as an introductory text. Thanks! Edit: I don't want to change the OP, but as noted, the transformation is to a linear Schrodinger equation with a potential, not the NLS. The NLS itself can be obtained as the PDE for the slowly-varying envelope of small perturbations in the KdV and Klein-Gordon equation, see e.g., (Ablowitz, Mark J. Nonlinear dispersive waves: asymptotic analysis and solitons. Vol. 47. Cambridge University Press, 2011). REPLY [4 votes]: A summary of this set of correspondences is outlined in these lecture notes: The three partial differential equations in $x$ and $t$, $$\text{Burgers:}\;\;\partial u/\partial t+u\partial u/\partial x=\nu\partial^2 u/\partial x^2-\partial U/\partial x$$ $$\text{KPZ:}\;\;\partial h/\partial t-\tfrac{1}{2}(\partial h/\partial x)^2=\nu\partial^2 h/\partial x^2+U$$ $$\text{heat equation:}\;\;\partial\psi/\partial t-\partial^2\psi/\partial x^2=(U/2\nu)\psi$$ are mapped onto each other by the substitutions $u=-\partial h/\partial x$, $\psi=e^{h/2\nu}$. The heat equation is also the Schrödinger equation in imaginary time ($t=i\tau$). The term $U$ plays the role of potential energy. Because it is a linear equation (at least if you take $U$ independent of $\psi$), it can be solved exactly. This is useful, because it allows to test the accuracy of a numerical solution of the nonlinear Burgers and KPZ equations.<|endoftext|> TITLE: A nice result for an sub-area I am not working in QUESTION [6 upvotes]: I have recently come across a (recent) problem in a different sub-area, by a well-established mathematician, that he deems likely to be intractable, but to which I have found a particularly nice proof (at least I believe so) using the techniques well known in my own sub-area. I think the result itself has a particularly illuminating interpretation so I would like to get it published somehow, but since I am not familiar with the sub-area nor particularly interested in working in it (by which I simply mean I am not suitable for the sub-area), would the best recourse be to contact the author and ask him to write a paper together on the result? REPLY [9 votes]: Probably more like: Contact the author, and ask for his reaction. It could be: "Great, let me help you publish it". Or it could be: "You made the following well-nown error."<|endoftext|> TITLE: Complexifications of hyperbolic manifolds QUESTION [20 upvotes]: I'm wondering when a compact hyperbolic $n$-manifold ($n \geq 3$) can embed in a complex hyperbolic $n$-manifold as a real algebraic subvariety so that it is a component of the fixed point set of complex conjugation? I suspect it might to be possible to answer this for arithmetic hyperbolic manifolds by an arithmetic construction. But I'm particularly interested in the 3-dimensional case, where most manifolds are not arithmetic. One could vaguely hope to approach this using algebraic geometry. Make the manifold into a real algebraic variety, and then embed into a non-singular complex projective variety by resolving singularities. If this variety satisfies the Yau-Miyaoka inequality, then it is complex hyperbolic (see Theorem 1.3 of this paper for the statement and references). Obviously I have no idea how to achieve this though. The restriction $n\geq 3$ is necessary, since in dimension 2 there are moduli spaces of hyperbolic structures, but compact complex hyperbolic surfaces are rigid and hence countable (the $1$-dimensional case is trivial). REPLY [9 votes]: There exist hyperbolic 3-manifolds which cannot embed totally geodesically in complex hyperbolic manifolds, answering this question in the negative. Recently it was shown that complex hyperbolic manifolds have integral discrete faithful representations; in particular, the traces of the matrices in $SU(n,1)$ have integral traces. See Theorem 1.3.1 of Baldi-Ullmo or Theorem 1.5(3) of Bader et. al.. This follows from a result of Esnault-Groechenig that cohomologically rigid representations of the fundamental group of smooth projective varieties must be integral. Compact hyperbolic $n$-manifolds are projective varieties, and the discrete faithful representation into $SU(n,1)$ is unique up to conjugation and cohomologically rigid by Mostow rigidity. Hence this representation must have integral traces (this corollary was pointed out to me by David Fisher, but see the above citations for more details). However, there are hyperbolic 3-manifolds such that the discrete faithful representation of the fundamental group into $SO(3,1)$ has non-integral traces, implying that they cannot embed isometrically in a complex hyperbolic 3-manifold. See Theorem 1.8.<|endoftext|> TITLE: How to classify continuous maps from 2-spheres to 2-spheres with n fixed points? QUESTION [12 upvotes]: I am a PhD student in Physics. This problem is motivated by representing a spin-$j$ state by Majorana's stellar representation. There are $2j$ points (Majorana stars) on the two-dimensional sphere (Bloch sphere) where the wave function vanishes. Hence, a spin state corresponds to a configuration of points on the sphere. If the spin is placed in an external magnetic field, the configuration of points undergoes rigid rotation. For a general cyclic evolution, the stars will move around the sphere (like wrapping a plastic bag around a ball), and finally come back to their original position. Therefore, the mathematical question is: how to classify continuous maps from $S^2$ to $S^2$ with $n$ fixed points? Since the homotopy group $π_2(S^2)$ is $Z$, I guess the answer will be a subgroup of $Z$. For a general reference, see e.g. https://physics.aps.org/articles/v5/65 Any explanation or reference will be appreciated. REPLY [18 votes]: Let $x_1, \ldots, x_n$ be your points on the sphere. Choose an $n+1$-th point $z$ on the target sphere, and paths from $z$ to each one of the $x_i$s. By moving along these paths you can construct a homotopy from the space of maps that fix the $x_i$s to the space of maps that send all the points $x_i$ to $z$. This latter space is the same as the space of pointed maps from the quotient space of $S^2$ by $n$ points to $S^2$. Easy exercise: this quotient space is homotopy equivalent to the wedge sum $$S^2\vee \bigvee_{n-1}S^1$$ It follows that the space of maps that you are asking about is homotopy equivalent to $$\Omega^2S^2\times (\Omega S^2)^{n-1}$$ Here $\Omega^k X$ denotes the space of pointed maps from $S^k$ to $X$. The set of path components of this space is the same as $\pi_2(S^2)$, which is $\mathbb Z$. Its higher homotopy groups can be written in terms of those of $S^2$. Perhaps the moral is this: the homotopy class of a map that fixes $n$ points is determined by the degree, as usual. But you have more choices in constructing a homotopy between two maps of the same degree.<|endoftext|> TITLE: Does this sequence of ratios of digit sums have a limit? QUESTION [6 upvotes]: I asked this question a few hours ago on MathStackExchange and there it received some attention but we still do not have a proof so I decided to ask it here also, in an unchanged form, and here it is: Digit sums of numbers $3^m$ in base $10$ for $m=1,2,...,50$ are: $3,9,9,9,9,18,18,18,27,27,27,18,27,45,36,27,27,45,36,45,27,45,54,54,63,63,81,72,72,63,81,63,72,99,81,81,90,90,81,90,99,90,108,90,99,108,126,117,108,144$. Ratios $\dfrac {ds_{10}(3^m)}{ds_{10}(3^{m+1})}$ for $m=1,2,...,49$ to three decimal places are: $0.333,1.000,1.000,1.000,0.500,1.000,1.000,0.666,1.000,1.000,1.500,0.666,0.600,1.250,1.333,1.000,0.600,1.250,0.800,1.666,0.600,0.833,1.000,0.857,1.000,0.777,1.125,1.000,1.142,0.777,1.285,0.875,0.727,1.222,1.000,0.900,1.000,1.111,0.900,0.909,1.100,0.833,1.200,0.909,0.916,0.857,1.076,1.083,0.750$ Does there exist limit of the sequence $a(m)=\dfrac {ds_{10}(3^m)}{ds_{10}(3^{m+1})}$? I cannot resist to note some kind of chebyshevness of this question (if there is one) because we know that Chebyshev proved that if limit in the prime number theorem exists then it must be equal to $1$. It could be that this is also the case here. I also welcome any computational effort and results obtained from such an experimental work if the proof is out of reach. REPLY [3 votes]: Experimentally, the sequence converges to $1,$ at the logarithmic rate suggested in Fedor's answer. Here is the graph for the first 20000 numbers: Now, when we fit the actual $ds_{10}(3^m),$ we get the following suggestive graph: Whose slope is pretty close to Fedor's "probabilistic" value. HOWEVER convergence is slow - empirically, Fedor's $o(1)$ term is actually of order $1/\log m.$<|endoftext|> TITLE: Geometric intuition for Fontaine-Wintenberger? QUESTION [23 upvotes]: I asked my advisor the question in the title. He told me it was a stupid question and that I should focus on my research. Thus we're asking here. The statement of Fontaine-Winterberger, per their groundbreaking 1979 paper Extensions algébrique et corps des normes des extensions APF des corps locaux, is as follows. The (absolute) Galois groups of$$\bigcup_n \mathbb{Q}_p\left(p^{1/p^n}\right)$$and$$\bigcup_n \mathbb{F}_p \left(\left(t^{1/p^n}\right)\right)$$are canonically isomorphic. Can someone unpack this statement for me, and give their geometric intuition for it? I do not work in number theory nor arithmetic geometry, but I'm willing to take some stuff on faith and Google around to find what I need to learn to understand whatever answer is provided. This is a result about Galois theory that has a "classical" flavor. I want my questions to be evaluated with that in mind. REPLY [13 votes]: First, the Fontaine-Winterberger isomorphism can also be recovered from a theorem of Deligne, namely Thm 2.8 here. Deligne showed that if two local fields $K_1$ and $K_2$ (possibly of different characteristic) were such that $O_{K_1}/\mathfrak{m}_{K_1}^N \simeq O_{K_2}/\mathfrak{m}_{K_2}^N$ then the category of finite etale extensions of $K_1$ with ramification bounded by $N$ is isomorphic to the corresponding category for $K_2$. One gets the Fontaine-Winterberger isomorphism by taking $K_1 = \mathbb{Q}_p(p^{\frac{1}{p^N}})$, $K_2 = \mathbb{F}_p((t^{\frac{1}{p^N}}))$ and letting $N \rightarrow +\infty$. Abbes and Saito gave a geometric description of the ramification filtration of local fields in this paper. They were motivated by the case of imperfect residue fields, but a quick survey of their idea in the classical case can be found in Section $2$ of this article by Shin Hattori. Shin Hattori used Abbes-Saito's idea in order to reprove and extend Deligne's result. This yields a geometric proof of Deligne's theorem, using rigid geometry and perfectoid spaces. The basic idea is that the ramification of $K_1$, resp. $K_2$, can be read from the connected components of some rigid spaces $X_1^{\leq j}$, $X_2^{\leq j}$ defined by Abbes-Saito, and if $j \leq N$ one can exhibit explicit pro-etale covers $\widetilde{X}_1^{\leq j}$, $\widetilde{X}_2^{\leq j}$ such that $\widetilde{X}_1^{\leq j}$ is the tilt of $\widetilde{X}_2^{\leq j}$. More details can be found in this survey by Shin Hattori, or in the paper here. Remark: the only thing needed about perfectoid spaces is the fact that tilting preserves connected components. This is even easier than showing that the corresponding adic spaces are homeomorphic, and in any case does not use the equivalence of etale site (so no circular reasoning here). For the sake of completeness: if $A$ is a ring then idempotents of $A$ are in bijection with idempotents of the monoid $A^{\flat} = \lim_{x \mapsto x^p} A$.<|endoftext|> TITLE: What are the differences between Woodin and Sy Friedman regarding set theoretic truths? QUESTION [10 upvotes]: Can someone explain in layman's terms or at least give a summary of the key differences of approaches between Woodin and Sy Friedman regarding set-theoretic truths? After this great debate, are there any progress being made? REPLY [13 votes]: It is only a half-joke to say that for a layman, there is no difference between their approaches. Take the following question for example: Does there exist an uncountable subset of $\mathbb R$ that cannot be put into bijection with $\mathbb R$? Both Friedman and Woodin would say that this is a question that might admit a definite yes-or-no answer, but that the jury is still out and we need to do a whole lot more technical set-theoretic work before we will be in a position to say whether it does admit a definite answer, and if so, whether the answer is yes or no. This already sets them aside from many others, e.g., formalists who say that the question does not admit a definite yes-or-no answer, or platonists who say that the question has a definite answer but who also say that we already know that we will never know the answer. To even state the differences between their approaches requires considerable technical machinery. Friedman's approach is called the "Hyperuniverse Program." Roughly speaking, his "hyperuniverse" is the space of all countable transitive models of ZFC, and the idea is that these universes are not all on an equal footing; e.g., some models are in a sense "maximal" and this clues us to which set-theoretic statements are true. Woodin on the other hand may be thought of as taking the inner model program as a starting point and arguing that the mathematical evidence points towards something called $\Omega$-logic as a natural foundation for set-theoretic truth. I do not think that too many further details can really be accurately described in "layman's terms" because the technical prerequisites are considerable. As for whether there has been "progress," both approaches lead to a plethora of technical mathematical questions, and although I do not follow this area, I expect that progress has been made in the same way that any subfield of mathematics makes progress. But if that's what you mean by progress then again I am not sure that it is possible to summarize it "for a layman." On the other hand, a layman could perhaps follow some of the philosophical debates about truth, but then it is less clear (at least for most mathematicians) what "progress" means in the philosophical realm.<|endoftext|> TITLE: Is a topological fiber-bundle, whose total space admits a retraction onto a fiber, trivial? QUESTION [12 upvotes]: Let $\xi = \pi \colon E \to B$ a topological fiber bundle with connected base $B$, $E_x = \pi^{-1}(x)$ the fiber at $x \in B$, $j \colon E_x \hookrightarrow E$ the canonical injection, and let suppose that there exists a retraction $r \colon E \to E_x$, i.e. $r◦j=Id_{E_x}$. Can we conclude that $\xi$ is trivial ? I found this proposition somewhere in a book or a pdf, without proof, but can't remember where. I thought I had a proof but it is plain wrong. The converse is indeed true (i.e. if $\xi$ is trivial, $E \cong B \times F$, there is a retraction of $E$ on any fiber), and this proposition seems intuitively correct... but is it ? REPLY [11 votes]: It was already pointed out that the statement is not true in the point-set sense. It is, however, true up to homotopy. This is a theorem of Dold and follows from his Partitions of unity in the theory of fibrations. Ann. Math. 78 (1963), 223-255. The following formulations can be found in James "Topology of Stiefel manifolds" (with a couple of added assumptions, purely reformulation): Theorem 4.2: Suppose that $B$ is path-connected, that we have fibrations $p:E\to B$ and $p':E'\to B$ such that $E$ and $E'$ have the homotopy type of CW-complexes. Then a fiber-preserving map $f:E\to E'$ is a fiber homotopy equivalence if it induces a homotopy equivalence on fibers. Corollary 4.3: Suppose that $X$ is path-connected, that $p:E\to X$ is a fibration with fiber $F$ and that $E$ and $X\times F$ have the homotopy type of CW-complexes. If there exists a homotopy-retraction $E\to F$ then $E$ is trivial in the sense of fiber homotopy theory.<|endoftext|> TITLE: What is known about relative adjunctions? QUESTION [10 upvotes]: I couldn't add to the well-written $n$Lab page about relative adjoints, so let me start taking that definition for granted. I have a few questions about how the classical results on adjoints remain true, even in this fairly asymmetric setting. I would like to understand better the meaning of this structure (e.g. when, how and why does it arise "in practice"). I collect here a few questions I have no clue how to attack. Let's assume that there are relative adjunctions $f\dashv_j g$ and $f' \,{}_{j'}\!\!\dashv g'$; we can then arrange a pair of commutative triangles $$ \begin{array}{ccc} \bullet & \overset{f}\to & \bullet\\ \hskip-4mm g\uparrow & \nearrow \\ \bullet && \end{array} \qquad \epsilon : fg \Rightarrow j $$ $$ \begin{array}{ccc} &&\bullet \\ &\nearrow& \hskip5mm \uparrow g'\\ \bullet &\underset{f'}\to& \bullet \end{array} \qquad \eta : j' \Rightarrow g'f' $$ assume, now, that there exists a natural transformation $j\Rightarrow j'$; under which assumptions the resulting pasting square is filled by an invertible 2-cell? Assume that $$ g_1 \dashv_{j_1} f \;{}_{j_2}\!\!\dashv g_2 $$ are relative adjoints; in this particular case it is possible to build a 2-cell $j_2 g_1\Rightarrow g_2 j_1$ pasting the unit of $f \;{}_{j_2}\!\!\dashv g_2$ with the counit of $g_1 \dashv_{j_1} f$. Under which conditions is this 2-cell an iso? In the same situation, the composition $j_1 j_2$ is parallel to $f$; is there a "comparison 2-cell" to/from $f$, and if there is, under which conditions it is invertible? The $n$Lab says that $f \; {}_j\!\!\dashv g$ if $f\cong \text{LIFT}_gj$ (absolute lift); is this an iff? This relation means that $g$ uniquely determines $f$; the converse is not true in general: does it mean that there can be two non-isomorphic $g,g'$ such that $f \; {}_j\!\!\dashv g$ and $f \; {}_j\!\!\dashv g'$? Is there an instructive example of this? What is the structure, if any, of the class $\{g\mid f \; {}_j\!\!\dashv g\}$? If $j$ is an isomorphism and $f \;{}_{j}\!\!\dashv g$, then this means that $f\dashv j^{-1}g$; is this condition any different from the particular case $f\dashv g = f \;{}_1\!\!\dashv g$? Is there a nontrivial, instructive example of this situation? I am confused by the loss of uniqueness outlined in the previous point... One of the reasons why relative adjunctions are useful is that they model weighted co/limits: writing the weight in a suitable form, one has that $j\otimes f \dashv_j \hom(f,1)$, which is, well, quite obvious if you write $$ {\bf hom}(j\otimes f,1) \cong {\bf hom}(j, \hom(f,1)) $$ how far can this analogy be pushed? (write $j\otimes(j'\otimes f) \cong \dots$: what does it mean for the associated relative adjunction?) What is the meaning of relations like $1_A \dashv_j g$ for suitable functors? such a situation gives an isomorphism of profunctors $\hom \cong \hom(j,g)$, so in some sense finding "all $j$-right adjoints of the identity" means finding all factorizations of the identity profunctor $\hom_A$. How far can this equivalence be pushed? What is the formal meaning of this analogy in terms of profunctor theory? What does the invertibilty of the relative unit $\eta : j\Rightarrow gf$ imply for $g,f$ ($f$ is "relatively fully faithful"?!). Dual question for the counit. REPLY [3 votes]: Using the answer to (7), if the natural transformation $J \Rightarrow J'$ is invertible, and $G$ and $F'$ are fully faithful, then the pasting square will commute up to natural isomorphism (assuming $J$ is dense and fully faithful). Again using the answer to (7), if $F$ and $G_1$ are fully faithful, then the composite natural transformation will be invertible (assuming $J$ is dense and fully faithful). As far as I can tell, the second part of the question is ill-formed. Yes, $F$ is a $J$-relative left-adjoint to $G$ if and only if $f \cong \mathrm{lift}_G J$ and this left lifting is absolute (this is an easy exercise). An example of non-unique relative right adjoints is given in Ulmer's Properties of Dense and Relative Adjoint Functors: let $J : \mathrm{AbGrp}_f \to \mathrm{AbGrp}$ be the inclusion of finite Abelian groups in all Abelian groups and let $L : \mathrm{AbGrp}_f \to \mathrm{Vect}_{\mathbb Q}$ be the zero functor. Then $L$ is $J$-relative left adjoint both to $0 : \mathrm{Vect}_{\mathbb Q} \to \mathrm{AbGrp}$ and to the forgetful functor $V : \mathrm{Vect}_{\mathbb Q} \to \mathrm{AbGrp}$. I'm not aware of any nice structure on the set of $J$-relative right adjoints to a given functor. Adjunctions relative to isomorphisms are the same as adjunctions relative to the identity. There is no lack of uniqueness here, since isomorphisms are dense, and $J$-relative right adjoints are unique when $J$ is dense. Weighted colimits as relative adjunctions are treated in §4 of Street–Walter's Yoneda structures on 2-categories, where they are called "weak indexed colimits". See there for various useful lemmas making use of this characterisation. If the identity on $\mathbf A$ is $J$-relative left coadjoint to $G : \mathbf B \to \mathbf A$, then there is a natural isomorphism $\mathbf A(a, Jb) \cong \mathbf A(a, Gb)$ which, by Yoneda, just says that $J \cong G$. Assume $J$ is dense and fully faithful. If $L$ is $J$-relative left adjoint to $R$, then the unit $\eta : J \Rightarrow RL$ is an isomorphism if and only if $L$ is fully faithful (easy exercise). Dually, assume $J$ is codense and fully faithful. If $L$ is $J$-relative left coadjoint to $R$, then the counit $\epsilon : LR \Rightarrow J$ is an isomorphism if and only if $R$ is fully faithful.<|endoftext|> TITLE: The Quotients $SO(n)/SO(n-1)$, $O(n)/O(n-1)$ and $SO(n)/O(n−1)$ QUESTION [8 upvotes]: The branching laws for the restricted representation of $SO(n)$ with respect to the subgroup $SO(n-1)$ are discussed in this Wikipedia article. Am I correct in reading from this that any given representation of $SO(n-1)$ can appear in a representation of $SO(n)$ with multiplicity at most $1$? Also what is the name for the corresponding homogeneous space, and does it have any interesting geometric properties? Also, I am interested in the same questions for $SO(n)/O(n-1)$ and $O(n)/O(n-1)$. REPLY [7 votes]: Yes, the branching is multiplicity free. See e.g. Theorem 8.1.3 and Theorem 8.1.4 in Symmetry, Representations, and Invariants by Nolan Wallach and Roe Goodman. The quotient space is a sphere $S^{n-1}$ which you can see for example by calculating the stabilizer of $e_n.$ For orthogonal groups you basically have extra $\mathbb{Z}_2$ action, so for example $SO(n)/O(n-1)$ comes from considering action of $SO(n)$ on $\mathbb{R}^n / \mathbb{Z}_2$ where the action of the $\mathbb{Z}_2$ is by multiplication by $-1.$ As for the branching rules in these cases ... it gets a bit complicated and it's hard to find any good reference. What happens here is that the branching is really a statement about Lie algebra representations (or equivalently about the simply connected covering groups $Spin(n)$) and for disconnected groups such as $O(n)$ one has to specify the action of all connected components. See e.g. https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group#Space_inversion_and_time_reversal for well studied examples.<|endoftext|> TITLE: An elementary question about a sequence of numbers QUESTION [5 upvotes]: Let $\lambda_n$ be an increasing and unbounded sequence of positive real numbers and $a_n$ be a sequence of real numbers such that $$\sum_{n=1}^\infty a_n \lambda_n^k=0 \ \ \text{ for all }\ \ k\geq 0.$$ Is $a_n=0$ for all $n$? REPLY [12 votes]: With $d\mu = \sum a_n\delta_{\lambda_n}$, your question can be rephrased as: Does $\int t^k\, d\mu(t) = 0$ imply that $\mu=0$? Since there are indeterminate moment problems (that is, collections of moments that do not come from one unique measure), it's now clear that the answer is no. To make this more concrete, you can start out with any limit circle Jacobi matrix; any two of its spectral measures $\nu_1,\nu_2$ will have the same moments, and you can take $\mu=\nu_1-\nu_2$. These measures will automatically have most of the extra properties you require here (pure point, with discrete disjoint spectra) if you take two spectral measures corresponding to two distinct boundary conditions at infinity. What is not automatic is that these measures will be supported by $[0,\infty)$, but you can get that, too, since there are indeterminate Stieltjes moment problems.<|endoftext|> TITLE: Multivariate polynomial is never squarefree at integers QUESTION [6 upvotes]: Let $f(x_1,\dots,x_n)$ be squarefree polynomial with integer coefficients. Assume $f$ at integers is not always divisible by a fixed square $m^2 > 1$. Is it possible $f$ to never be squarefree at integers? I suspect this is impossible. REPLY [9 votes]: The answer has been edited, due to a mistake found by joro and explained by Jose Brox --- thanks to both of them! Now we reduce the $n$-variate case to the $(n-1)$-variate one... The claim that an $n$-variate polynomial will always have a square-free value under our assumptions holds if and only if the same claim holds for an $(n-1)$-variate polynomial (and hence for a univariate polynomial, which still seems to be open). The "only if" part is trivial; let us show the "if" part. We may assume that $f(\mathbf 0)\neq 0$. For any $p^2\mid f(\mathbf 0)$ there exists $\mathbf x_p$ such that $p^2\nmid f(\mathbf x_p)$. Using CRT, we find an $\mathbf x$ such that $p^2\nmid f(\mathbf x)$ for all $p^2\mid f(\mathbf 0)$. [EDIT] Moreover, there are $n$ sets of the form $S_i=P\mathbb Z+a_i$ such that every $\mathbf x\in\prod_{i=1}^n S_i$ works. Consider now the polynomial $g(x_1,x_2,x_3,\dots,x_{n-1},t)=f(x_1,\dots,x_{n-1},tx_{n-1})$. It is also square-free (when passing from $g$ to $f$ by substituting $t=x_n/x_{n-1}$, a square may disappear only if it was $x_{n-1}^2$; this cannot happen since $g(\mathbf 0)=f(\mathbf 0)\neq0$). Next, there are only finitely many values $\alpha$ such that $g(x_1,\dots,x_{n-1},\alpha)$ is not square-free: at these exceptional $\alpha$ the discriminant of $g$ with respect to one of the $x_i$ should vanish, and all these discriminants are not identically zero. Now we choose $\alpha\in S_nS_{n-1}^{-1}$ for which $f_1(x_1,\dots,x_{n-1},\alpha)$ is square-free; let $\alpha=k_n/k_{n-1}$ with $k_i\in S_i$. Then the polynomial (with integer coefficients) $h(x_1,\dots,x_{n-2},x_{n-1})=f(x_1,\dots,x_{n-2},k_{n-1}x_{n-1},k_nx_{n-1})=g(x_1,\dots,k_{n-1}x_{n-1},\alpha)$ is square-free, and for $k_i\in S_i$ the values $h(k_1,\dots,k_{n-2},1)=f(k_1,\dots,k_n)$ and $h(\mathbf 0)=f(\mathbf 0)$ have no common square divisors (other than 1). By the $(n-1)$-variate case, $h$ has a square-free value, therefore so does $f$. REPLY [9 votes]: This is a supplement to Ilya Bogdanov's answer. For univariate polynomials Granville (Int. Math. Res. Not. 1998, 991-1009) deduced from the $abc$-conjecture that there are infinitely many natural numbers $x$ such that $f(x)$ is square-free; in fact these numbers have positive density. The same is known unconditionally when $\deg f\leq 3$, by the work of Hooley (Mathematika 14 (1967), 21-26). In addition, Granville's conditional result was extended to multivariate polynomials by Poonen (Duke Math. J. 118 (2003), 353-373).<|endoftext|> TITLE: integer matrices with non-real spectra QUESTION [5 upvotes]: I am interested in infinite order elements $A\in SL(3, {\mathbb Z})$ whose spectra are not contained in ${\mathbb R}$ (i.e. such $A$ has two distinct complex-conjugate eigenvalues which are not roots of unity); I will refer to them as NRS matrices. Question. Is there a pair of commuting NRS matrices $A, B\in SL(3, {\mathbb Z})$ whose product is again an NRS matrix, such that $A, B$ generate a non-cyclic subgroup of $SL(3, {\mathbb Z})$? As the last resort, one can look for such matrices by computer-search, but I would prefer to avoid doing this. REPLY [8 votes]: There are no such pairs. Let $\lambda_1$, $\lambda_2$, $\lambda_3$ be the eigenvalues of $A$, and $\mu_1,\mu_2,\mu_3$ be those of $B$ (with $\lambda_1,\mu_1\in\mathbb R$). Since the eigenvalues are distinct, $A$ and $B$ are diagonalizable; since they commute, they are simultaneously diagonalizable, i.e., an eigenbasis for $A$ is also that for $B$. We assume that in a common eigenbasis, $\lambda_i$ corresponds to $\mu_i$ (clearly, $\lambda_1$ corresponds to $\mu_1$ --- because only these eigenvectors may be chosen real). Now let $e_2$ be an eigenvector for $A$ corresponding to $\lambda_2$. The linear system defining it has coefficients in $K=\mathbb Q[\lambda_2]$, so we may assume that the elements of $e_2$ are also in $K$. Writing down the condition that $e_2$ is an eigenvector of $B$ (with eigenvalue $\mu_2$) we get that $\mu_2\in K$. Thus $\mathbb Q[\mu_2]=K=\mathbb Q[\lambda_2]$. Moreover, $\lambda_2$ and $\mu_2$ belong to the units group of $K$ --- which, as is known, is cyclic. Thus $\lambda_2^k=\mu_2^\ell$ for some $k$ and $\ell$, which yields also $\lambda_i^k=\mu_i^\ell$, and hence $A^k=B^\ell$. Thus the subgroup generated by $A$ and $B$ is cyclic.<|endoftext|> TITLE: Spanning $k$-trees QUESTION [9 upvotes]: ##k-trees A $k$-tree is a graph defined as follows: (They were defined by Harary and Palmer.) a) A complete graph with $k$ vertices is a $k$-tree. b) A $k$-tree on $n$ vertices $T$ is obtained by a $k$-tree on $n-1$ vertices $S$ by adding a new vertex and connecting it to all the $k$ vertices of a complete subgraph of $S$ with $k$ vertices. We can also regard a $k$-tree instead of a graph, as a $k$-dimensional simplicial complex, or as a $k+1$-uniform hypergraph. (We can also regard it as a $r$-uniform hypergraph for $2 \le r \le k+1$.) #The questions The decision problem 1a) Is there an efficient algorithm to tell if a graph contains a spanning $k$-tree. 1b) Is there an efficient algorithm to tell if a $k$-dimensional simplicial complex contains a $k$-tree. (Regarded, this time, as a $k$-dimensional simplicial complex.) Of course, for $k=1$ a graph contains a tree iff it is connected. The counting problem 2a) Given a graph $G$ is there a matrix-tree type theorem for the number of spanning $k$-trees of $G$? 2b) More generally, given a $k$-dimensional simplicial complex $K$ is there a matrix-type formula for the number of spanning $k$-trees. (Regarded as $k$-dimensional simplicial complexes.) The sampling problem Is there a simple way to sample a uniformly-random spanning $k$-tree of a given graph $G$? For $k=1$ there are various miraculous important ways to sample uniform spanning trees. Minimal spanning $k$-trees For $k=1$ the famous greedy algorithm efficiently find the minimum-weight spanning tree. Given weights on edges is there an algorithm for finding minimum spanning $k$-tree? Bach's threshold problem. (Added in a comment, Nov. 2020) What is the threshold for the appearance of a spanning k-tree in the binomial random graph? The general theme In general, to what extent results about trees extend or fail to extend to $k$-trees. ##Some more background Cayley's formula for the number of trees with $n$ labelled vertices was extended to $k$-trees by Beineke and Pippert. A Prufer type correspondence by C. Renyi and A. Renyi allows efficient sampling (uniformly) all $k$-trees with $n$ labelled vertices. REPLY [7 votes]: Regarding the question 1a, Bern showed that checking existence of a spanning $k$-tree in a graph is NP-complete for any fixed $k \geq 2$ (also see another, more accessible relevant paper by Cai and Maffray). This implies that questions 3 and 4 are at least as hard (indeed, each of these problems contains checking existence as a special case), and question 2a is unlikely to have a formula computable within reasonable time (unlike the polynomial-computable Laplacian cofactor for the number of trees). The answers to questions regarding simplicial complexes may depend on the way we represent $k$-trees as complexes. Probably the most natural way to do that is to define a $k$-tree complex as a set of $k$-simplices obtained sequentially (from an initial set of a single $k$-simplex) by choosing a simplex $K$ and adding new simplex $K'$ to the set, so that $K'$ is different from $K$ in a single vertex not present in any previous simplex. However, we can notice that this problem is not easier than the problem of a spanning $k$-tree existence in a graph. Indeed, for any $k$ we can reduce from the graph problem to the simplicial problem simply by listing all $(k + 1)$-cliques (notice that since $k$ is fixed, the reduction is proper polynomial); the answer for that instance of the simplicial problem could be converted to a spanning $k$-tree of the original graph, and vice versa. This implies that the problem 1b is NP-complete whenever $k \geq 2$, and the problem 2b is not easier. As for the general question about extending results from trees to $k$-trees, the intuition is probably that spanning $k$-trees of $K_n$ can be tamed (in light of results mentioned in OP), while spanning $k$-trees of a general graph may possess a much more complex structure not exhibited by spanning trees.<|endoftext|> TITLE: Important open exposition problems? QUESTION [39 upvotes]: Timothy Chow, in his article A beginner's guide to forcing, defines an open exposition problem as a certain concept or topic in mathematics that has yet to be explained "in a way that renders it totally perspicuous." What are some open exposition problems in your field, particularly ones that you think would help interested mathematicians break into it? For instance, there is no shortage of references on chromatic homotopy theory -- Ravenel's Green and Orange books, Lurie's course notes, and Hopkins' COCTALOS notes. Nonetheless, there does not seem to be a complete, smooth, and carefully-put-together exposition of the chromatic story that assumes neither too much homotopy theory nor algebraic geometry. REPLY [4 votes]: I have two such open exposition questions which I hope to address some day (alas, I have a long list of both mathematical research and pedagogical problems which are all worthy of attention - but this kind of question is fun because it is at the intersection of these areas). One is giving a cochain-level treatment of Poincaré duality through intersection theory and then using that to give geometric, cochain level refinements of much of "intermediate" algebraic topology: wrong-way maps, characteristic classes, Thom isomorphism, Steenrod operations, Eilenberg-Moore spectral sequence. A second is to treat loop spaces and classifying spaces in a more unified way, from scratch (course title: "Loop, de-loop.") This would expand on my little expository paper on Koszul duality.<|endoftext|> TITLE: Application of Abhyankhar's lemma QUESTION [5 upvotes]: I am confused by an application of Abhyankhar's lemma in the proof of Theorem 3.4 of Deligne-Rapoport. Here is the question with only the relevant parts of the text: Let $X$ and $Y$ be two curves over $\mathbb{Z}[1/n]$ (ie relative dimension 1). Let $U\subseteq X$ be an open set and let $C$ be its complement. Assume $X$ and $Y$ are normal and that $X$ and $C$ are both smooth over $\mathbb{Z}[1/n]$. Finally, there is a map $f:Y\to X$ identifying $Y$ as the normalization of $X$ in the function field of $Y$. Let $U'$ be the open $f^{-1}(U)\subseteq Y$ and let $C'$ be its complement. Assume lastly that $f|_{U'}:U'\to U$ is finite etale. How can I conclude that the maps $Y\to \operatorname{Spec}\mathbb{Z}[1/n]$ and $C'\to \operatorname{Spec}\mathbb{Z}[1/n]$ are smooth by an application of Abhyrankhar's lemma? Let $P$ be a component of $C$. Let us replace $X$ by a Zariski open subset $X$ so that $P$ is cut out by one equation, $f$. Let $n$ be the degree of ramification of $P$ in $Y$, and let $X'=X[T]/(T^n-f)$. Abhyankhar's lemma suggests that $U'\to U$ when pulled back to $X'$ extends to an etale cover even over the locus $\{f=0\}$. Thus we get a curve $Z$ which fits into the following commutative diagram: $$\require{AMScd} \begin{CD} Z @>>> Y\\ @VVV @VVV \\ X' @>>> X\end{CD}$$ where we know the left most arrow is etale. Now as $X'$ can be computed to be smooth over $\mathbb{Z}[1/n]$, to conclude that $Y$ is smooth it suffices to show the top map is etale. If $Y$ were known to be regular, then Zariski-Nagata purity would facilitate this check, however I cannot seem to figure out why $Y$ should be regular (which it definitely is if it is to be smooth over $\mathbb{Z}[1/n]$!) Thanks for any help. REPLY [2 votes]: Let's work locally near a point $x \in P$. Consider the composed cover $Z \to X' \to X$. Use the fact that etale-locally, each cover splits into irreducible components, where each irreducible component contains at most one point of the fiber over $x$, and those that contain one point are finite. (See here, (13) to reduce to the finite case and then (9)). So we may assume that whichever irreducible component of $Z$ contains a point over $x$ contains only one such point, and is finite, which because it is etale over $X'$, implies that it is a finite etale cover of $X'$ of degree $1$ and thus is $X'$. So locally on $X$ we get a map $X' \to Y$. Because $n$ is the ramification index of $Y$, over $U$ this is a surjective map of finite etale covers of the same degree, hence an isomorphism, and because $X'$ is normal it is an isomorphism everywhere. Thus because $X'$ is smooth, $Y$ is smooth. The point of this is that we can also view Abyankhar's lemma as an etale-local classification result for etale covers with tame ramification around a smooth normal crossings divisor. This is how I generally think of it and it might be how Deligne and/or Rapoport think of it as well.<|endoftext|> TITLE: A question on representation theory of p-adic groups QUESTION [12 upvotes]: Let $V$ be a complex vector space of infinite dimension and let $(\pi,V)$ be a representation of the $p$-adic group $G:=GL_2(\mathbb{Q}_p)$. From representation theory, we know that if the representation $(\pi,V)$ is both smooth and irreducible, then $(\pi,V)$ is admissible. So now it is natural to think about the inverse question. Does there exist a representation $(\pi,V)$ of $G$ which is smooth and admissible but not irreducible? This question is not difficult. You can find two smooth and irreducible representations $(\pi_1,V_1)$ and $(\pi_2,V_2)$ of $G$ and give the example $(\pi_1 \oplus \pi_2, V_1 \oplus V_2)$. Let $\pi=\pi_1 \oplus \pi_2$ and $V=V_1 \oplus V_2$, then $(\pi,V)$ is the representation we want for the question. But if we change a little on the question, does there exist a representation $(\pi,V)$ of $G$ which is smooth, admissible and indecomposable but not irreducible? I have no idea on this question since the above example does not work in this case. REPLY [13 votes]: It is indeed well-known that the category of smooth admissible representations of $G$ (and other reductive $p$-adic groups) is not semi-simple. The principal series, that is the representations induced from a character of the Borel of $G=GL_2(\mathbb Q_p)$, are always indecomposable, but they may nor be irreducible -- think of the case of the trivial character. Another nice way to construct an example is with trees. You may know that $G$ acts transitively on the regular infinite tree $T$ of arity $p+1$ (i.e. every vertex has exactly $p+1$ neighbors -- this tree is the Bruhat-Tits tree of $G$). The stabilizer of a vertex is $ZK$ where $Z$ is the center and $K$ a compact maximal subgroup (= a conjugate of $GL_2(\mathbb Z_p)$). If you take for $W$ the set of functions with finite support on the set of vertices of the tree $T$, then $W$ is a representation of $G$ (since $G$ acts on $T$) which is smooth (stabilizers are intersection of finitely many $K_iZ$, with $K_i$ compact open), admits a non-trivial $G$-morphism to the trivial representation (the linear form which to a function $f$ on $T$ attaches the sum of its value), but certainly does not contain the trivial representation as sub-representation, because $W$ has no function invariant by $G$ except 0 -- they would be of infinite support. Small problem, $W$ is not admissible. But now let $D: W \rightarrow W$ be the operator which to a function $f$ attaches the function $g(x)=\sum f(y)$ with $y$ running over the $p+1$ neighbors of $x$, and define $V = W/ (D-(p+1))W$. You can easily check that the trivial representation is still a quotient of $V$, and with some little geometric reasoning on the tree that I leave as an exercise, that the trivial representation is not a sub-representation of $V$, and that $V$ is admissible. Hence a second example of smooth-admissible non-semi-simple reducible representation which shows that the answer to your question is yes. (Actually this example is the dual of the first one).<|endoftext|> TITLE: Can a sphere glued into a soft 3d-mattress rotate continuously? (manifolds, SU(2) and the belt trick) QUESTION [16 upvotes]: The question is triggered by the wonderful animations by Jason Hise: https://www.youtube.com/watch?v=LLw3BaliDUQ https://www.youtube.com/watch?v=6Ul_-ABYaYU https://www.youtube.com/watch?v=aYVt1UiERIQ All these animations are based on the well-known belt trick (a way to represent SU(2) as double cover of SO(3)) and suggest that a solid sphere that is glued into a soft mattress can rotate continuously, if the region of mattress close to the sphere performs certain appropriate motions around the sphere. Is this conclusion correct? Are there papers on this very counter-intuitive issue? The belt trick is shown impressively in https://youtu.be/DHFdBWU36eY . It is also called the plate trick, the string strick, the balinese dance trick or the balinese candle trick. The trick shows that an object connected to spatial infinity can rotate indefinitely, coming back to its original position after a rotation by 4 pi = 720 degrees. It is not evident that the belt trick implies that a sphere can rotate when glued in a mattress. Many discrete belts do not make a continuous and smooth mattress. So it could be that the mattress property does not follow from the belt trick - it could be that the belt trick requires discontinuities. So the question is: is the mattress trick a continuous operation on the mattress, or does it require cuts in the mattress? Did anybody ever perform the experiment in real life? For two dimensions, this has been done, as shown in this video: https://www.youtube.com/watch?v=UtdljdoFAwg that shows that a ball glued into a handkerchief can be rotated continuously. But it is really possible also for three dimensions, for a full mattress? The issue is interesting because a ball in a mattress can be seen, if this works, as a model for a spin 1/2 particle. Above all, if we assume that the mattress is a model for space itself, the ball in the mattress would be a way to model a spin 1/2 particle with the help of a smooth 3-dimensional manifold, something which is often assumed to be impossible. It would solve one of the contradictions between general relativity and quantum theory. The newest animation by Jason Hise of "Dirac's handkerchief", made after this question was posted, is here: https://youtu.be/tazjVJcxm50 It realizes the answer of André Henriques. The answer to the original question is thus a clear "yes". If anybody every manages to do this in an experiment, I'd like to see the video -- and to put it into my free physics text! REPLY [13 votes]: The answer is "yes": A sphere glued into a soft 3d-mattress can rotate continuously. Let $R_t\in SO(3)$ be the rotation by angle $t$ around the $z$-axis. Pick a nullhomotopy $R_{t,s}$ ($s\in [0,1]$) of the map $[0,4\pi]\to SO(3):t\mapsto R_t$. So $R_{t,0}=R_t$ and $R_{t,1}=\mathrm{id}$, for all $t\in [0,4\pi]$. Now here's the description of the motion of the mattress: At time $t$, the sphere of radius $1+s$ performs the rotation $R_{t,s}$. (The mattress is immobile outside the sphere of radius 2.)<|endoftext|> TITLE: The minimal number of partitions to cover all $k$ tuples QUESTION [8 upvotes]: The set $N=\{1, 2, \ldots, 2k\}$ can be partitioned into pairs (e.g $(1,2),(3,4),\ldots,(2k-1,2k)$) in $\frac{(2k)!}{k!2^k}$ ways. $k$-tuple is subset of size $k$ in $N$. We say that $k$-tuple is covered by partition $\alpha$ if none of pairs of $\alpha$ are in that tuple. For example $(2,4,\ldots,2k)$ tuple is covered by the partition above but $(1,2,\ldots,k)$ is not. I am interested in the following problem: Find the minimal number of partitions of the set $N$ so that all $k$ tuples are covered. There are $2k\choose k$ tuples and every partition covers $2^k$ tuples so the answer is $\ge \frac{2k\choose k}{2^k}$. I hope to find an upper bound that is $c\frac{2k\choose k}{2^k}$ where $c$ is constant. Can anybody help on this problem? REPLY [7 votes]: Denote $N=\frac{2k\choose k}{2^k}$ and choose, say $m=\lceil 10kN\rceil$ independent random partitions (all partitions have equal probability $1/(2k-1)!!$). For any $k$-set $A$, the probability that it is not covered by a single partition equals $1/N$ (indeed, if we denote this probability by $p$, then it does not depend of $A$, and summing up by all choices of $A$ we get $\binom{2k}kp=2^k$, since any partition covers $2^k$ subsets), thus the probability that it is not covered by any of our partitions equals $(1-1/N)^{m} TITLE: Reals which must, can't or might be added by forcing QUESTION [9 upvotes]: Let $W \subseteq V$ be an inner model of ZFC. There are a variety of theorems that characterize when a real $x \in V$ is the generic of a forcing notion $\mathbb P \in W$, for example, the characterization of random reals as the set of reals in every full measure set coded by $W$. There are also theorems characterizing reals which cannot be added by forcing, for example the well known result that $0^\sharp$ cannot be added by forcing. What I want to know is if there are more general theorems characterizing (combinatorial, measure theoretic etc) properties of reals in $V$ guaranteeing that there is or is not a $\mathbb P \in W$ such that $x \in W^\mathbb P$? I am particularly interested in this question with regards to properties related to cardinal characteristics of continuum. For instance, if there is a real $d \in V$ is dominating over the reals in $W$, then is there a nice way to characterize when $W[d]$ is (or is not) a generic extension of $W$? REPLY [7 votes]: The characterization mentioned by Mohammad in his answer really dates back to Lev Bukovský in the early 70s, and, as Ralf and Fabiana recognize in their note, has nothing to do with $L$ or with reals (in their note, they indicate that after proving their result, they realized they had essentially rediscovered Bukovský's theorem). See MR0332477 (48 #10804). Characterization of generic extensions of models of set theory, Fundamenta Mathematica 83 (1973), pp. 35–46. What Bukovský does is to find a characterization for when $V$ is a $\lambda$-cc generic extension of the given inner model $W$ (for a given regular cardinal $\lambda$). Ralf actually has a nice write-up of Bukovský's theorem (organized differently from the presentation in his note with Fabiana) in his recent paper The long extender algebra, preprint. To appear in a special issue of Archive for Mathematical Logic. In fact, it is enough that $W$ uniformly $\lambda$-covers $V$, meaning that for all functions $f\in V$ whose domain is in $W$ and whose range is contained in $W$ there is some function $g\in W$ with the same domain and such that $f(x) \in g(x)$ and $|g(x)| < \lambda$ for all $x \in \operatorname{dom}(f)$. The generality of the result may be a bit of a drawback for the question at hand, however, since it does not seem to provide us with a widely applicable combinatorial characterization allowing us to identify those reals that are generic over $W$. The same comment applies to the version of the result described in the note by Ralf and Fabiana. Essentially, given $r$, they look at $W=L$ and $V=L[r]$, and describe whether $W$ uniformly $\lambda$-covers $V$ in terms of possible liftings of elementary embeddings $j\!:L_\alpha\to L_\beta$. The connection with Bukovský's characterization is made explicit in page 7 of their note.<|endoftext|> TITLE: Can one embed two division rings in a common one? QUESTION [18 upvotes]: I could not get an answer to this question in MathStackExchange, so I dare ask it here. Given any two fields, $\rm F_1,F_2$ over the same prime subfield $\rm F$, the quotient $\rm \mathbf F=F_1\otimes_F F_2/\mathcal M$ of the tensor product $\rm F_1\otimes_F F_2$ by a maximal proper ideal $\mathcal M$ provides a field with two embeddings $$\rm F_1\rightarrow\mathbf F,\ f\mapsto f\otimes 1 +\mathcal M\quad\text{and}\quad F_2\rightarrow\mathbf F,\ f\mapsto 1\otimes f +\mathcal M.$$ Question. Given two division rings $\rm R_1,R_2$ having the same characteristic, is there a way to find a division ring $\rm R$ with two embedings $\rm R_1⊂R$ and $\rm R_2⊂R$ ? REPLY [18 votes]: Yes, this is possible. PM Cohn first showed that the amalgamated product $R_1 * R_2$ over a common subfield is a "fir" (free ideal ring), and then in Cohn, P.M. The embedding of firs in skewfields, Proc. London Math. Soc. (3) 23 (1971), 193–213. that one can adjoin inverses to get a division ring containing $R_1 * R_2$. On the other hand, the analogue is not true for division algebras (division rings which are finite-dimensional over their center). See Rowen, Louis; Saltman, David. Simultaneous embeddings of finite dimensional division algebras. Proc. Amer. Math. Soc. 141 (2013), no. 3, 737–744.<|endoftext|> TITLE: Integrating a family of vector spaces QUESTION [16 upvotes]: Let $X$ be a measure space, or even a subspace of $\mathbb{R}^n$, and suppose I have a family of finite-dimensional vector spaces $\{V_x\}_{x\in X}$ indexed by $X$. Is there any way to "integrate" this family over $X$ (subject to niceness conditions on it) to obtain some sort of "vector-space-like object" $\int V_x \,dx$ possessing a "dimension" (in general not an integer), such that $$\mathrm{dim} \left(\int V_x \,dx\right) = \int (\mathrm{dim}\, V_x)\,dx \;?$$ Bonus points if there is some category-theoretic way to view $\int V_x\,dx$ as a "coproduct" of the family $\{V_x\}_{x\in X}$. REPLY [14 votes]: I'll be brief and (happily) add more details on demand (Edit: Some more details were added). Some Philosophy Slogan: You can do math fibered over a measured space. Most of us are already used to the idea of doing algebraic geometry over schemes and topology over topological spaces, but are less familiar with doing math over measured spaces. Yet, this concept has a long history. Maybe its first appearance is in the notion of a bundle of Hilbert spaces over a measured space aka as direct integral of Hilbert spaces. Also in the theory of von-Neumann algebras one decomposes a general algebra into a direct integral of factors (similarly to the way in which an Azumaya algebra is decomposed over its center). I find Furstenberg's pov on Ergodic Theory parallel to Grothendieck's pov on Algebraic Geometry in the way spaces are treated relative to a base space, only that Ergodic Theory is somehow more generous in allowing further constructions, due to the flexibility of measurable functions. In recent decades Zimmer developed the theory of convex compact spaces, Gaboriau developed the theory of simplicial complexes, Sauer developed the theory of manifolds, all over a base measured space. This pov is quite common nowadays in Ergodic Theory and there are many more examples. I should probably mention that in all of the above examples, theories were developed for an external sake. Maybe it is about time for approaching these theories as a whole and develop a master theory. I don't know. Vector spaces over $X$ Given a measured spaces $X$ (that is, a standard Borel space endowed with a measure class), a (complex) vector space over $X$ is a Borel space $V$ endowed with a Borel map $\pi:V\to X$ such that the fibers $V_x$ over (a.e) point is endowed with a vector space structure which varies measurably. A precise axiomatic definition could be given by means of the standard vector space axioms reinterpreted by means of fiber-products constructions. For example you have the addition map $V\times_X V \to V$ and the scalar multiplication $\mathbb{C}\times V \to V$ which commute with the obvious maps to $X$ and satisfy the obvious compatibility relations. Whatever is ones definition of "a measurably varying $X$-indexed family of vector spaces" it should be equivalent to a vector space over $X$. Unfortunately, I haven't seen this definition published anywhere, so let's say it is a folklore definition. Note that associated with $X$ we have the algebra of bounded (measurable, defined up to a.e equivalence) $\mathbb{C}$-valued functions $L^\infty(X)$, which is a commutative von-Neumann algebra (aka a W*-algebra), that is a C*-algebra which has a predual ($L^1(X)$). To a vector space over $X$, $\pi:V\to X$, one associates the vector space of all (classes of) measurable sections of $\pi$, to be denoted $L(V)$ (or $L(\pi)$ if there is a danger of misunderstanding). This is a module over the algebra $L^\infty(X)$. Dimension Assume now that $X$ is actually endowed with a finite measure (not merely a measure class). Then integration is a finite trace on the algebra $L^\infty(X)$, and this algebra becomes a finite von-Neumann algebra. For modules over such guys there is a well developed notion of dimension, the von-Neumann dimension. For finitely generated projective modules, this dimension is given by taking the trace of a certain projection in a certain matrix algebra over $L^\infty(X)$ (you can guess which projection: a one associated with a presentation of a the module as a direct summand of a free module, which trace is choice independent). The dimension of a general module is defined as the supremum over the dimensions of its f.g projective submodules. This theory is carried in Lueck's book. For an online survey, see his paper. Finally, it is an exercise to show that for a vector space over $X$, $\pi:V\to X$, as defined above, we have that the von-Neumann dimension of the $L^\infty(X)$-module $L(V)$ equals exactly $\int_X \dim V_x$.<|endoftext|> TITLE: Is "weakly good" series in a finite-dimensional Banach space "good"? QUESTION [8 upvotes]: Let us call a series $\sum_n x_n$ in a Banach space "good" if there exists a permutation $\sigma:\mathbb N\to\mathbb N$ such that the rearranged series $\sum_n x_{\sigma(n)}$ converges. Find a simple proof of the following theorem (which was proved by E.Steinitz in 1913 according to V.Kadets). Theorem. A series $\sum_n x_n$ in a finite-dimensional Banach space $X$ is "good" if and only if for every linear function $f:X\to\mathbb R$ the series $\sum_n f(x_n)$ is "good". (This problem was posed 24.09.2017 by Vaja Tarieladze from Tbilisi on page 72 of Volume 1 of the Lviv Scottish Book). REPLY [2 votes]: A relatively short inductive proof of Steinitz Theorem can be founded in this paper. Here we present a sketch of the proof, which is based on 3 lemmas whose proof is left to the reader. First we recall the formulation of the result. Theorem. A series $\sum_{n=1}^\infty x_n$ in a finite-dimensional Banach space $X$ is "good" if and only if for any linear functional $f:X\to\mathbb R$ the series $\sum_{n=1}^\infty f(x_n)$ is "good". Proof. This theorem will be proved by induction on the dimension of $X$. The theorem is trivially true for Banach spaces of dimension $\le 1$. Assume that for some number $d$ the theorem has been proved for all Banach spaces of dimension $0$, and a finite set $\Omega_0\subset \Omega$ there exists a finite set $F\subset \Omega\setminus \Omega_0$ such that 1) $\|x-\sum_{n\in F}pr_0(x_n)\|<\varepsilon$; 2) $\sum_{n\in F}\|x_n\|\le C\max\{\|x\|,\varepsilon\}$. Since $X_1$ has dimension $ TITLE: In what area of study does one encounter this principle in timetabling? QUESTION [9 upvotes]: A while ago I saw an image like the one below in a lecture, which was supposed to represent a rail network in a (square) city: The circles represent trains that are moving either North/South or East/West along the vertical and horizontal tracks. The purpose of this picture was to show that, even though the grid is unevenly spaced, it is always possible to schedule the trains so that connections were perfectly timed: that is, any time a train arrives at an intersection, there is another train also arriving at that intersection, in a perpendicular direction. The math behind this is very simple, so my question is: what is the area of study in which one would encounter an image like this? REPLY [3 votes]: Not an answer, but two related, almost inverse topics suggesting in which areas of mathematics they fall: (1) The Lonely Runners Conjecture. See, e.g., Terry Tao, "Some remarks on the lonely runner conjecture." Perhaps this could be classified as a mixture of number theory and combinatorics. Guillem Perarnau, Oriol Serra, "Correlation Among Runners and Some Results on the Lonely Runner Conjecture." Electr. J. Comb. 2016. (2) "At Museum of Mathematics, Meet 2 Beavers That’ll Never Meet." This construction uses Truchet Tiling. Perhaps this could be classified as a mixture of topology and combinatorics.<|endoftext|> TITLE: Cover of the positive real numbers by intervals QUESTION [9 upvotes]: For which real numbers $x$ and $y$ does the following hold?: $$ \bigcup_{\frac{a}{b} \in \mathbb{Q}^+} \left[\frac{a}{b},\frac{a}{b}+\frac{1}{a^x b^y}\right] \ = \ \mathbb{R}^+ $$ REPLY [6 votes]: Here is the solution for $x>0$ and arbitrary $y$. If $x+y\geq 2$, then the covering won't work. Indeed, consider $\alpha=n+\sqrt{2}$ for large natural number $n$. An elementary argument shows that $|\alpha-a/b|>1/3b^2$ for all $a,b>0$. If $\alpha$ lies in the union of your intervals, then $1/3b^2<|\alpha-a/b|\leq 1/a^xb^y$. Since in particular $|\alpha-a/b|<1$, $a/b>n,a>nb$, thus giving $$\frac{1}{3b^2}<\frac{1}{a^xb^y}<\frac{1}{n^xb^{x+y}}\leq\frac{1}{n^xb^2}$$ which fails for large $n$. Now consider $x+y<2$. Take any $\alpha\in\mathbb R^+$ which we may assume is irrational. A variation of Dirichlet's theorem on diophantine approximations shows that there are infinitely many fractions $a/b$ such that $a/b<\alpha0$ and $\alpha b>a$. Thus $\alpha\in[a/b,a/b+1/a^xb^y]$.<|endoftext|> TITLE: Can we define an isomorphism invariant to measure "dimension" of an undirected simple graph? QUESTION [5 upvotes]: Can we define a characteristic to measure the "dimension" of a graph? Let's start by some simple example. Intuitively, a circuit graph with $n$ nodes and $n$ edges should have dimension $1$. Likewise, an $(n \times n)$-torus should have dimension $2$, etc. So, can we define a general concept of "dimension" for every graphs? Maybe we could draw inspiration from Hausdorff dimension? This question is inspired by When do 3D random walks return to their origin?. As Mikhail Tikhomirov pointed out, we may meet difficulty in defining a certain "dimension". From my point of view, I think that it is beacuse a finite graph does not have enough details or microstructure/macrostructure. Therefore, could we consider this problem on infinite graph such as the infinite $2$-dimensional grid graph? REPLY [4 votes]: Five answers, 'by vague association' and 'lateral thinking' (which is unavoidable for this vague question, I think). All in all, I think that any definition you will give will have an 'air' of arbitrariness: the most straightforward 'take' on this is to point out that (realizations of) graphs are after all just one-dimensional simplicial complexes, so in that sense every graph has dimension one, and one answer to your answer, taken unimaginatively literally, should be Yes we can: we assign each graph the number 1. There is nothing else to say.${}$(short.answer) Four more 'divergent' remarks now follow. 1. The graph invariant called 'genus'. A traditional, important graph invariant spiritually similar to what you are asking about is the genus of a graph. I am aware that one may argue that this does not capture the intuition of 'dimension'. But then again, you yourself tagged this a 'soft-question', so this answer should be sort-of-acceptable to you. Also worth pointing out: as you will probably known, for any fixed dimension $d\geq 3$, any countable graph can be embedded into $\mathbb{R}^d$, a fact which by the way can be proved working only over the 'signature' of graph-theory, plus a little intuitive geometry, by an inductive argument; you do not need the usual argument via a Vandermonde matrix (not that this would be 'bad', yet the proof via the 'moment curve' wields a 'signature' which besides $\sim$ (adjacency) and $\#$ ('intersects') also uses $\mathbb{Z}$, and '$+$' and '$\cdot$'). 2. The methods around the ideas of 'thickening point clouds' and 'persistent homology'. Your question also reminded me of the very active field of 'persistent homology' (and related ideas). Again, it does not fit your specifications precisely. For this to be relevant to your question, of course, the data that you wish to associated your isomorphism invariant to must be given inside a metric space. A purely combinatorial graph will not be enough to make these methods 'bite'. If so inclined, you can start reading about this in the following two references National Science Foundation Mathematical Sciences Institutes: Topology of Shapes, Persistent Homology and Point Clouds: Where Does it Take Us? H. Edelsbrunner and J. Harer. Persistent homology --- a survey. Surveys on Discrete and Computational Geometry. Twenty Years Later, 257-282, eds. J. E. Goodman, J. Pach and R. Pollack, Contemporary Mathematics 453: Persistent Homology — a Survey 3. 'Neighbourhood complexes' of graphs. You can assign to your graph the **vector of the Betti numbers of the abstract simplicial complex whose faces are precisely the subsets of neighborhoods of vertices. Since Bettin numbers are often perceived to be sort-of-a-dimension, this is another approach relevant to your question. One starting point to read up on this could be Matthew Kahle: The neighborhood complex of a random graph. Journal of Combinatorial Theory, Series A 114 (2007) 380–387 4. Graphs as knots. This is something of a non-example, the relation to the idea of 'dimension' being almost non-existant. One can ask knot-theoretic questions about graphs. However, please note the nice summary of Dror Bar-Natan's in this MO thread, which gives a sense in which there is no 'knot-theory of graph' at all. 5. Zariski dimension of associated commutative rings. You could construct some functor1 $F\colon\mathsf{SomeReasonableCategoryOfGraphs}\to\mathsf{CommutativeRings}$, and then take the Krull dimension of $F(G)$ as your "dimension" of $G$. I do not know of anyone who would think this a motivated thing to do. Let me reiterate: all this except (short.answer) seems arbitrary and context-dependent and would say more about the person doing this than about a hard mathematical reality. There are, of course, many wonderful and useful things to learn along any of these roads, but you should be aware that the choice of what road to take is arbitrary. Please forgive this expression: the question in this OP is more of a mathematical Rorschach test than a mathematical question. 1 There are many possibilities to associate rings to graphs. Note, however, that using the word 'functor' immediately implies that your construction will have to be graph-isomorphism-invariant, since any functor takes isos to isos.<|endoftext|> TITLE: How can I calculate eigenvalues of a tridiagonal matrix? QUESTION [7 upvotes]: Are there special methods to get exact eigenvalues of a tridiagonal matrix? REPLY [8 votes]: Yes. For general tridiagonal matrices, see The Numerical Recipes, Chapter 11, or Golub-Van Loan. For symmetric tridiagonal matrices, you can do better, see Coakley/Rochlin's paper. Coakley, Ed S.; Rokhlin, Vladimir, A fast divide-and-conquer algorithm for computing the spectra of real symmetric tridiagonal matrices, Appl. Comput. Harmon. Anal. 34, No. 3, 379-414 (2013). ZBL1264.65051. Golub, Gene; Van Loan, Charles F., Matrix computations., Baltimore, MD: The Johns Hopkins Univ. Press. xxvii, 694 p. (1996). ZBL0865.65009.<|endoftext|> TITLE: An example in Mumford's “Picard Groups of Moduli Problems” QUESTION [6 upvotes]: I tried asking this at math.stackexchange but I didn't get any responses, so hopefully it's ok to try here. I'm reading Mumford's paper "Picard Groups of Moduli Problems" and am confused about an example in the first section. I'll try to explain the situation here, but if I'm not making sense I'm talking about page 40 of the paper. Let $\pi$ be a group, let $\mathfrak{C}$ be the category such that the objects are sets $S$ with an action of $\pi$ and morphisms $\pi$-linear maps $f:S\rightarrow T$. Maps $\{f_\alpha: S_\alpha\rightarrow T\}$ cover $T$ if $T=\bigcup_{\alpha} f_\alpha(S_\alpha)$, so we have defined a site. Let $\langle\pi\rangle$ be $\pi$ considered as a $\pi$-set (so one forgets the group action but retains the left action of $\pi$). The claim that I'm trying to understand is that a sheaf $\mathcal{F}$ on this site is canonically determined by the $\pi$-set $M:=\mathcal{F}(\langle\pi\rangle)$. (This is a $\pi$-set by functoriality - all the elements of $\pi$ give automorphisms of $\langle\pi\rangle$ so by functoriality they give automorphisms of $M$.) The first step of the proof is what I'm having trouble with: let $S$ be a $\pi$-set on which $\pi$ acts transitively, so we have a $\pi$-linear surjection $$p:\langle \pi\rangle\rightarrow S.$$ This is certainly a cover as we've defined it, so we can apply the sheaf axiom to learn... something about the corresponding map $\mathcal F(S)\rightarrow M$. Mumford claims that the sheaf axiom should imply that $\mathcal F(S)\sim M^{H}$ where $H$ is the stabilizer of some element of $S$, but I don't see why this is. REPLY [8 votes]: Picking the point $s \in S$ which is the image of the identity $e \in \langle \pi \rangle$, we can identify the fiber product $\langle \pi \rangle \times_S \langle \pi \rangle$. Any element is uniquely of the form $(g, gh)$ for some $g \in \pi$ and some element $h$ in the stabilizer $H$ of $s$. This makes the fiber product isomorphic to the disjoint union $\coprod_{h \in H} \langle \pi \rangle$ as $\pi$-sets. Under this identification, we can understand the two projections $\coprod_{h \in H} \langle \pi \rangle \to \langle \pi \rangle$: the first projection is the identity on each copy of $\langle \pi \rangle$, and the second projection, on the copy corresponding to $h$, is the map $\langle \pi \rangle \to \langle \pi \rangle$ given by right-multiplication by $h$. Applying $\cal F$, the sheaf condition says that we get an equalizer diagram $$ {\cal F}(S) \to M \rightrightarrows \prod_{h \in H} M $$ where the first map $M \to \prod M$ is the diagonal, and the second map $M \to \prod M$ is given by the action of $h$ on each factor. Therefore, the equalizer is precisely $M^H$.<|endoftext|> TITLE: When does the constant diagram functor preserve fibrant objects in the injective model structure on diagram categories? QUESTION [7 upvotes]: When does the constant diagram functor preserve fibrant objects in the injective model structure on diagram categories? For example, this is the case when the index category of the diagrams is a cofinite filtered poset (see Edwards and Hastings "Čech and Steenrod Homotopy Theories with Applications to Geometric Topology", LNM volume 542). But I would like some more general contexts. REPLY [2 votes]: This is a long comment, rather than an answer. The poset in Edward and Hastings context LNM 542 is elegant reedy, so the injective and the reedy model structures coincide, but independently of this, the constant diagram functor does not preserve fibrations, contrary to the statement in theorem 3.2.4. Consider the poset $A = \{\bullet \to \bullet \gets \bullet \}$, and a fibrant object $X \to 1$ in $\mathcal{C}$. It follows from 3.2.7 that if the constant diagram were a fibration in $\mathcal{C}^A$, then the diagonal $X \to X \times X$ must be a fibration in $\mathcal{C}$. However EH apply this to posets $P$ which are reindexing of filtering categories, so we add the following to include this situation. It is easy to see that this poset is the category $A = P_{< j}$ for the subdiagram $j = \{\bullet \to \bullet\}$ of two consecutive numbers in $\omega$, where $P$ is the result of the "Mardesic trick" (i.e. Deligne construction Prop. 8.1.6 SGA4) applied to $\omega$. Finally, it is easy to see directly that the constant diagram determined by a fibration $X \to Y$ is a fibration in $\mathcal{C}^P$ provided that for any $j \in P$, the finite poset $P_{< j}$ is either empty or connected, which clearly is not the case in the example above.<|endoftext|> TITLE: Pontriagin reflexivity of the character group QUESTION [6 upvotes]: For an Abelian topological group $G$ by $G^{\wedge}$ we denote the Pontryagin dual of $G$, i.e. the group of continuous homomorphisms $G\to\mathbb T:=\{z\in\mathbb C:|z|=1\}$. The group $G^{\wedge}$ is endowed with the topology of uniform convergence on compact subsets of $G$. A topological group $G$ is called Pontryagin reflexive if the canonical homomorphism $G\to (G^\wedge)^\wedge$ is a topological isomorphism. Problem. Let $G$ be a metrizable Abelian topological group. Is the Pontriagin dual $G^{\wedge}$ of $G$ Pontryagin reflexive? (This problem was posed 21.09.2017 by Lydia Aussenhofer on page 71 of Volume 1 of the Lviv Scottish Book). REPLY [2 votes]: Not an answer to this problem. For a non-reflexive $G$ see example: Exercise (23.32) in Hewitt & Ross, Abstract Harmonic Analysis I (Springer 1963). Consider a topological vector space $L^p(\mathbb R)$, $0 < p < 1$. They show that the (separable metrizable) topological group $G = L^p(\mathbb R)$ with operation of addition has no nonzero continuous characters. So $\widehat{G} = \{0\}$ and $\widehat{\widehat{G}} = \{0\}$, not isomorphic to $G$. Facts used (Hewitt & Ross include outlines of proofs): (Day, 1940) this topological vector space has no nonzero continuous linear functionals (Hewitt & Zuckerman, 1950) every continuous character $\chi$ on $G$ has the form $\chi(x) = \exp(2\pi i f(x))$ for some continuous linear functional $f$.<|endoftext|> TITLE: Equivalence of rational Voevodsky motives: partial Converse to Conjecture of Orlov QUESTION [10 upvotes]: There is a conjecture of Orlov stating that if $X$ and $Y$ are smooth projective complex varieties that are derived equivalent (equivalent bounded derived categories of coherent sheaves), then their rational Voevodsky motives $M(X)_{\mathbb{Q}}$ and $M(Y)_{\mathbb{Q}}$ are equivalent. My question is whether the equivalence of $M(X)_{\mathbb{Q}}$ and $M(Y)_{\mathbb{Q}}$ implies the partial converse that the rational(!) bounded derived categories of coherent sheaves are equivalent? There is a result of Cisinski and Tabuada that implies that the equivalence of the rational Voevodsky motives implies that the noncommutative rational motives of $X$ and $Y$ are equivalent in Kontsevich's category of rational noncommutative motives. REPLY [5 votes]: Definitely not. Take $X$ to be the blowup of $P^2$ at a point and $Y$ to be $P^1 \times P^1$. Then $$ M(X) = 1 + 2L + L^2 = M(Y), $$ but the derived categories are different, since both varieties are Fano and non-isomorphic,<|endoftext|> TITLE: Determinant is to Pfaffian as resultant is to what? QUESTION [12 upvotes]: This is an irresponsible question: I do not have done any thinking on it, or even literature search. I just became curious whether there is some modification of the notion of a common root of two polynomials which would be detected by a Pfaffian of some alternate matrix, rather than a determinant of some general matrix, like it is the case with the resultant. (PS Seems like there is no tag "algebra", so I chose commutative algebra instead) REPLY [11 votes]: Pfaffian resultant formulas are obtained in Resultants and Chow forms via Exterior Syzygies (2001), where the polynomials are represented by coordinates on a Grassmanian manifold.<|endoftext|> TITLE: Problems reducing to a graph-theory algorithm QUESTION [6 upvotes]: This is essentially a question in pedagogy -- the answers could be useful to teach (or rather, motivate) graph theory, and especially the algorithmic side of it. I have been very impressed with this example: Question: What is the maximum size of a set of integers between 1 and 100 such that for any pair (a,b), the difference a-b is not a square ? The first, naive algorithms that spring to mind to solve this (finite) problem are awkward. However, let me quote: " This problem can be defined as a Graph problem : we create a vertex for each integer, and link two integers if their difference is a square. We but have to find a maximal independent set in this graph ! " Using this, a few trivial lines of Sage code give the answer (see the link). Sooo much time is saved for the coder! Are there more examples like this one? To be more precise, I'm looking for problems which: do not immediately involve graph theory, on the surface, reduce, after a little translation, into a common problem in graph theory (find a matching, find a colouring...). there should be a "haha!" effect for the mathematician, and a sigh of relief for the programmer. I heard another nice one in a talk by Tim Gowers. A university wants to organize its exams. Some students take several exams. Some rooms and some time slots are available. Help them. Solution: each exam (maths, physics, etc) is a vertex; place one edge between exams if there is one student taking both; define "colours" to be pairs (room, time). Now you have to colour the graph such that adjacent vertices are in different colours. There is a fine line between these great examples, and the articifial ones which I find disappointing, such as "$n$ couples having each $k$ children all want to shake hands on different days such that...". In fact, I should add that I'm certain to have a preference for problems involving maths (like the first one above) rather than "real life" (like the second). Usual precaution: if this is not appropriate for MO, let me know and I will not be offended! REPLY [4 votes]: I expected a flood of responses to this question, since there are so many lovely examples. That flood does not seem to have materialised yet, so let me add an old classic. I hope it isn’t too famous to be worth mentioning, and that adding it as a separate answer is the correct protocol. I’m thinking of Steve Fisk’s celebrated proof of the Art Gallery theorem, which is in the form of an algorithm. We are given the floor plan of an art gallery as a simple polygon in the plane with $n$ vertices, and we wish to place some (motionless point) guards into the polygon so that the whole of the interior of the polygon is visible to at least one guard. We can see that we may need as many as $\lfloor\frac n3\rfloor$ guards by considering comb-shaped polygons with $n=3k$ vertices: (There must be a guard in each of the grey triangles, and there are $k$ disjoint such triangles, so we need at least $k=\frac n3$ guards.) The problem, then, is to place no more than $\lfloor\frac n3\rfloor$ guards into a simple polygon with $n$ vertices in such a way that one of the guards can see any point in the interior of the polygon. Fisk’s algorithm is as follows: triangulate the polygon to obtain a planar graph whose vertices are the vertices of the original polygon and whose faces are triangles, and then three-colour the vertices so each face has a vertex of each colour. At least one of the three colours occurs $\leq\lfloor\frac n3\rfloor$ times: position guards at the vertices that are painted this colour. Since each triangle has a guard at one of its vertices, and a triangle is entirely visible from any one of its vertices, the whole of the polygon must be visible to at least one guard. Thus a difficult-sounding purely geometric problem is solved by three-colouring the vertices of a planar graph. I should perhaps note that prolific MO contributor Joseph O’Rourke literally wrote the book on art gallery theorems and algorithms.<|endoftext|> TITLE: Is the sum of digits of $3^{1000}$ divisible by $7$? QUESTION [35 upvotes]: Is the sum of digits of $3^{1000}$ a multiple of $7$? The sum of the digits of $3^{1000}$ can be computed using a computer. It is equal to $2142$, so the answer is positive. Is there a short proof that the sum of the digits of $3^{1000}$ is a multiple of $7$ without using a computer? Do you have any advice to solve this type of problem (without programming of course!)? The results below are known: $3^{1000}$ has $478$ digits, and so the sum is at most $4302$ ($9\cdot478$). This sum is a multiple of $9$. The last four digits of $3^{1000}$ are $0001$. Context: We are a group of 3 French people working on it since 2007. It's a little exercise I found in my high school book (printed in 2007) which is pretty complicated. The one who created this exercise doesn't know the answer. This question was previously asked on Math.SE (link). REPLY [11 votes]: Middle digits of the numbers $3^n$ are unpredictable. At least it is too hard for current techniques to say anything about them. It means that the their sum is unpredictable as well. Some good random number generators are based "digital" ideas. If we take binary digits of $3^n$ then we immediately get generalization of $(3/2)^n$-problem which is out of reach today. This picture is taken from a New Kind of Science: The pattern is very similar to "rule 30" picture from the same book: It is expected to have very good pseudorandom properties, see discussion at A New Kind of Science: A 15-Year View.<|endoftext|> TITLE: Sum of the divisor function over integers with restricted prime factors QUESTION [6 upvotes]: Let $a,q$ be co-prime integers and let $P(a,q)$ denote the set of primes congruent to $a$ modulo $q$. Is it known whether one can give an asymptotic formula for the expression $$\displaystyle \sum_{\substack{n \leq x \\ p | n \Rightarrow p \in P(a,q)}} d(n),$$ where $d(n)$ is the number of divisors of $n$? REPLY [11 votes]: Sure. The generating function for the sum you want is the Dirichlet series $$ \sum_{\substack{ n=1\\p|n \implies p\equiv a\pmod q}}^{\infty} \frac{d(n)}{n^s} = \prod_{p\equiv a\pmod q} \Big(1- \frac{1}{p^s}\Big)^{-2}. $$ Using Dirichlet characters to isolate primes in progressions, you can express this as $$ \zeta(s)^{2/\phi(q)} \prod_{\chi \neq \chi_0 \pmod q} L(s,\chi)^{\overline{\chi(a)}/\phi(q)} G(s), $$ for a suitable Euler product $G$ converging absolutely to the right of $1/2$. Now use the Selberg-Delange method (see e.g. Tenenbaum's book). The asymptotic will be of the type $$ \sim C x(\log x)^{-1+2/\phi(q)} $$ for a suitable constant $C$.<|endoftext|> TITLE: Morita equivalence and isomorphisms in cohomology theories QUESTION [6 upvotes]: Let $A,B$ be two unital algebras. We say that $A,B$ are Morita equivalent if there are $A-B$ and $B-A$ bimodules $P,Q$ such that $$P \otimes_{B} Q \cong A, Q \otimes_A P \cong B$$ (as $A-A$ and $B-B$ bimodules). Suppose that $A,B$ are Morita equivalent. Then one can show that $K$-theory and cyclic and Hochschild cohomologies of them are isomorphic. How to describe explicitly an isomorphisms $K(A) \cong K(B)$, $HH^{\bullet}(A) \cong HH^{\bullet}(B)$ and $HC^{\bullet}(A) \cong HC^{\bullet}(B)$? REPLY [6 votes]: The conceptual point is that all of these invariants are Morita invariant because they can be defined directly in terms of the category of modules. Explicitly: Starting from the category of modules $\text{Mod}(A)$ we can isolate the subcategory of tiny or compact projective objects: those modules $M$ such that $\text{Hom}(M, -)$ preserves all colimits. These turn out to be precisely the finitely generated projective modules, from which one can define K-theory as usual. So given a Morita equivalence $Q \otimes_A (-) : \text{Mod}(A) \cong \text{Mod}(B)$ the induced map on f.g. projectives comes from restricting to tiny objects, and then the induced map on K-theory comes from passing to the Grothendieck group. Hochschild cohomology of $A$ is the derived endomorphisms of the identity functor $\text{Mod}(A) \to \text{Mod}(A)$. So given a Morita equivalence $\text{Mod}(A) \cong \text{Mod}(B)$ we get an induced equivalence on (cocontinuous) endofunctor categories $\text{Bimod}(A, A) \cong \text{Bimod}(B, B)$ sending the identity to the identity, which furthermore induces an equivalence on self-Ext. I am less familiar with the details of the cyclic case but morally the only additional thing to do is to incorporate the natural (derived) $S^1$-action on Hochschild stuff.<|endoftext|> TITLE: How close $k$-sums of a random set of numbers are on average? QUESTION [6 upvotes]: Consider a set of random iid variables $x_1, \ldots x_n$ uniformly distributed on $[0, 1]$. For each $S \subset [n]$ with $1 \leq |S| = k < n$ take $\sigma_S = \sum_{i \in S}x_i$. Obviously $\sigma_S$ are distinct with probability 1. What is the expectation of $\Delta = \min_{|S| = |S'| = k, S \neq S'} |\sigma_S - \sigma_{S'}|$, at least asymptotically? Can we say something about concentration of $\Delta$? REPLY [3 votes]: Here is some heuristics, which should be possible to strengthen to a completely rigorous answer. Note that $\sigma_S$ is the sum of the sample of size $k$ taken with replacement from the (random) empirical distribution corresponding to the iid sample $x_1, \ldots x_n$ from the uniform distribution on $[0, 1]$. For large $n$, this empirical distribution is close to the uniform distribution on $[0, 1]$. Also, if $k=o(\sqrt n)$ (which let us assume), then sampling with replacement differs little from sampling without replacement. So, the distribution of $\sigma_S$ (with $S$ considered a random subset of $[n]$ of size $k$) is close to the distribution of the sum of $k$ iid random variables (r.v.'s) uniformly distributed on $[0, 1]$, which, by the central limit theorem, is in turn close to $N(k/2,k/12)$ if $k$ is large, which let us assume as well. So, the distribution of the vector $(\sigma_S\colon S \subset [n],|S| = k)$ is close to that of the vector $(X_1,\dots,X_N)$ of $N:=\binom nk$ iid r.v.'s each with a distribution close to $N(k/2,k/12)$. Let $U_j:=F(X_j)$, where $F$ is the cdf of $N(k/2,k/12)$, so that the $U_j$'s are approximately iid uniformly distributed on $[0, 1]$ and hence the random vector $(U_{(j)}-U_{(j-1)}\colon j=2,\dots,N)$ of the spacings between the consecutive order statistics based on the $U_j$'s approximately equals in distribution $(V_2,\dots,V_N)/(V_1+\dots+V_{N+1})$, where the $V_j$'s are iid r.v.'s each with the standard exponential distribution. By the law of large numbers, $V_1+\dots+V_{N+1}\sim N$. So, $(U_{(j)}-U_{(j-1)}\colon j=2,\dots,N)$ is close in distribution to $(V_2,\dots,V_N)/N$. On the other hand, the minimum of $|X_i-X_j|$ over all $i\ne j$ equals $\min\{X_{(j)}-X_{(j-1)}\colon j=2,\dots,N\}$, where the $X_{(j)}$'s are the order statistics based on $X_1,\dots,X_N$. Now write \begin{equation*} U_{(j)}-U_{(j-1)}=F(X_{(j)})-F(X_{(j-1)}) \sim(X_{(j)}-X_{(j-1)})F'(X_{(j)}) \end{equation*} \begin{equation*} =(X_{(j)}-X_{(j-1)})F'(F^{-1}(U_{(j)})) \sim(X_{(j)}-X_{(j-1)})F'(F^{-1}(j/N)), \end{equation*} whence, with $\tau:=F'\circ F^{-1}$, \begin{equation*} X_{(j)}-X_{(j-1)}\sim\frac{U_{(j)}-U_{(j-1)}}{\tau(j/N)} \sim\frac1{\tau(j/N)}\frac{V_j}{V_1+\dots+V_{N+1}}\sim\frac{V_j}{N\tau(j/N)}. \end{equation*} Next, \begin{equation*} E\min\{\frac{V_2}{\tau(2/N)},\dots,\frac{V_N}{\tau(N/N)}\} =\int_0^\infty \prod_{j=2}^N P(V_2>\tau(j/N)x)dx \end{equation*} \begin{equation*} =\int_0^\infty \exp\Big(-\sum_{j=2}^N \tau(j/N)x\Big)dx =\frac1{\sum_{j=2}^N \tau(j/N)}\sim \frac1{N\int_0^1\tau(u)du}=\frac{\sqrt{\pi k/3}}N, \end{equation*} by using the substitution $u=F(z)$ in the integral $\int_0^1\tau(u)du$. So, it should follow that \begin{equation*} E\min\{X_{(j)}-X_{(j-1)}\colon j=2,\dots,N\} \sim \frac1N\,E\min\{\frac{V_2}{\tau(2/N)},\dots,\frac{V_N}{\tau(N/N)}\} \sim\frac{\sqrt{\pi k/3}}{N^2} \end{equation*} and hence \begin{equation*} E\min_{|S| = |S'| = k, S \neq S'} |\sigma_S - \sigma_{S'}|\}\sim \sqrt{\pi k/3}\Big/\binom nk^2. \end{equation*}<|endoftext|> TITLE: necklace reconstruction in the permutation case QUESTION [10 upvotes]: Suppose I want a necklace with $n$ beads labelled (bijectively) by $\{1, 2, \ldots n\}$, that is I want a cyclic order on $\{1, 2, \ldots, n\}$ (so for example $132$ is the same cyclic order as $321$ but different from $231$). Now suppose I know the cyclic order of some subsets of $\{1, 2, \ldots, n\}$ as they should appear in the necklace. I want to know what conditions on the subsets and their cyclic orders would make the necklace uniquely reconstructible, or given a set of subsets with orders how many compatible necklaces there are. I feel like surely this is well studied (somewhere in the world of reconstructibility or elsewhere), but I'm not managing to find any results on it, so perhaps you can help me find where people have looked at this. Thanks. REPLY [3 votes]: This problem is NP-complete, thus there is no easily verifiable condition that would be necessary and sufficient. In fact, it is enough if only some of the triples are prescribed, see Cyclic ordering is NP-complete by Galil and Megiddo. Another, closely related problem is Betweenness.<|endoftext|> TITLE: Splitting the Resultant, as when the Determinant becomes the square of the Pfaffian QUESTION [17 upvotes]: The Determinant of an $n\times n$ matrix, viewed as a polynomial in the entries, is irreducible. But when it is restricted to the subspace of alternate matrices, it becomes reducible, actually the square of a polynomial known as the Pfaffian. Likewise, the Resultant of a pair $(P,Q)$ of polynomials of respective degrees $\le n,m$, when viewed as a polynomial in the coefficients of $P$ and $Q$, is an irreducible polynomial ($n+m+2$ indeterminates, degree $n+m$). Is there a natural subspace $E$ of $k_n[X]\times k_m[X]$, such that the restriction of the Resultant to $E$, viewed as a polynomial in the coordinates, splits in a non-trivial way ? Low temperature example: Let $n,m$ be even, and $E$ be the space of pairs of even polynomials. Then the restriction of the Resultant over $E$ is a square. If $P(X)=p(X^2)$ and $Q(X)=q(X^2)$, then ${\rm Res}(P,Q)=({\rm Res}(p,q))^2$. This follows from the formula ${\rm Res}(P,Q)=\prod(x_i-y_j)$ in terms of the roots $x_i$ and $y_j$ (to be adapted if $P,Q$ are not monic). Less cold example: Let $n,m$ be even, and $F$ be the space of pairs of reciprocal polynomials. Then the restriction of the Resultant over $F$ is a square. If $P(X)=X^np(X+\frac1X)$ and $Q(X)=X^mq(X+\frac1X)$, then ${\rm Res}(P,Q)=({\rm Res}(p,q))^2$. REPLY [11 votes]: I think it is better to homogenize and view $P$, $Q$ as binary forms in $\mathbb{C}[X_1,X_2]$. An invertible $2\times 2$ matrix $g$ acts on the variables $x_i$ by $(gx)_i=\sum_{j=1,2}g_{ij}x_j$ and on forms by $(gP)(x)=P(g^{-1}x)$. The resultant is invariant: $$ {\rm Res}_{p,q}(gP,gQ)=({\rm det}(g))^{-pq}\times {\rm Res}_{p,q}(P,Q) $$ Both examples you gave are given by eigenspaces of a fixed matrix $g$ such as $$ \left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right) $$ or $$ \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right) $$ Note one can also consider mutlidimensional resultants ${\rm Res}_{p_1,\ldots,p_n}(P_1,\ldots,P_n)$ and they satisfy tons of base change formulas (see Jouanolou's article). Hope this helps for the generalization you are looking for. Edit with a few more details: What I said above shows that the two examples are the same. For the first one this follows from the base change formula in Section 5.12 of the article by Jouanolou, which is a vast nonlinear generalization of the transformation formula I wrote above. One has: $$ {\rm Res}_{dp_1,\ldots,dp_n}(P_1\circ G,\ldots,P_n\circ G)= $$ $$ {\rm Res}_{p_1,\ldots,p_n}(P_1,\ldots,P_n)^{d^{n-1}}\ \times \ {\rm Res}_{d,\ldots,d}(G_1,\ldots,G_n)^{p_1\cdots p_n} $$ where each $P_i$ is a homogenous form of degree $p_i$ in $n$ variables and $G$ is a polynomial map $\mathbb{C}^n\rightarrow\mathbb{C}^n$ with components $G_i$ that are homogeneous polynomials of the same degree $d$. The first example is given by $$ \left\{ \begin{array}{ccc} G_1(x_1,x_2) & = & x_1^2\\ G_2(x_1,x_2) & = & x_2^2 \end{array}\right. $$ Spaces of forms of the form (sounds horrible I know) $P\circ G$ should give you tons of examples of this factorization phenomenon. Another thought: this is vaguely reminiscent of the Frobenius determinant factorization. The two matrices $g$ above are involutions and correspond to the discrete group $\mathbb{Z}_2$. I am not familiar with the Galois theory of covers like the one given by $G$ but I think it may help.<|endoftext|> TITLE: Containment of minimal 2-Ramsey-graphs in minimal 3-Ramsey-graphs QUESTION [6 upvotes]: Let $G$ be a minimal $2$-colour Ramsey-graph for $H$. Must there exist a minimal $3$-colour Ramsey-graph $F$ for $H$ with $G\subset F$? I am wondering if anything is known about this, particularly in the case $H=K_n$, at least for $n=3$. (By G being an $r$-Ramsey-graph for $H$ I mean the property that every colouring of the edges of $G$ with $r$ colours admits a monochromatic copy of $H$. By minimal I mean minimal wrt. the subgraph relation.) REPLY [4 votes]: This is now solved: https://arxiv.org/abs/1809.09232 The answer is Yes in general.<|endoftext|> TITLE: MathSciNet Reference Fixes QUESTION [12 upvotes]: Is there an easy way to get MathSciNet to fix minor mistakes in their references? It would be great if there was some sort of web form where you could enter the proposed fix, which would save time for the people working over there. I mention this for two reasons. First, it has become my habit after downloading the MathSciNet BibTeX reference and put into my paper, to double check it. I've found that many times some of the minor information (such as issue number) is simply missing; but occasionally there is a mistake. Second, and more importantly, I just got proofs back for an accepted paper, and they changed my references to those found at MathSciNet and will not let me fix them in the paper if they are incorrect there! [P.S. I think MathSciNet does a great job already! I'm just not familiar with any reference fixing, so thought I'd ask...] REPLY [18 votes]: Please send an email to mathrev@ams.org, explaining the issue. (This is our all-purpose email address; any mistakes you discover, not just regarding references, you can let us know there.) Give us some time, I promise we'll get to it. However, if it seems as if the request somehow fell through the cracks, you can always contact one of your friendly editors :-) and I'll redirect the request to the relevant person. That said, there are some exceptions, at least at the moment. Papers added to our database through a digitization project from the World Digital Mathematics Library (WDML), such as http://www.ams.org/mathscinet-getitem?mr=1576401 sometimes carry (rather annoying) mistakes in their bibliographical data, and these mistakes come directly from the data given to us by the publishers. (That paper by Ramsey is an example.) We hope to address this problem eventually but this is not happening currently. Other mistakes come from how the journal originally printed the author's name. We make an effort to identify authors appropriately, so the name should still be linked to the right author, but we still indicate the printed error. (For example, see Paul Erdős's name here.) See here for more on this. There is one last kind of error that I do not believe is what you mean, but it seems worth mentioning anyway. For errors that appear in the list of references taken from the paper, make sure the mistake or typo is ours and not just a transcription of an error that comes from the paper. (We do not fix the latter: "This list reflects references listed in the original paper as accurately as possible with no attempt to correct error.")<|endoftext|> TITLE: Can you modify solenoids to be locally connected? QUESTION [5 upvotes]: Solenoids are not locally connected. Intuitively, this is because if you look at a neighborhood around a point, the other "strands" will be in the neighborhood, since there are infinitely many strands arbitrarily close to every point. You could say that regular solenoids are "wrapped" too tightly. My question is can you "unwrap" the solenoid so as to make it locally connected. The relationship between the "unwrapped" solenoid and the regular solenoid would be similar to that of the relationship between the rose with infinite petals and the Hawaiian earring. In particular, the "unwrapped" solenoid would be strictly finer than the regular solenoid, just as the infinite rose is strictly finer than the Hawiian earring. REPLY [6 votes]: Sure. If you want to "disentangle" a small piece of arc (a "strand") from all the other strands around it, just declare it to be open. In other words, you can refine your solenoid by declaring every homeomorphic copy of $(0,1)$ to be an open set. The resulting topological space is just a $\mathfrak c$-sized disjoint union of copies of $\mathbb R$. Each copy corresponds to a composant of the original solenoid. But wait! you say. You've disentangled my solenoid too much -- I wanted a topology coarser than this stupid one you've described. Well, fair enough. But I can argue that no coarser topology than the one I've described will succeed in "unwrapping" your solenoid. In other words, mine is the coarsest possible topology that refines the solenoid's original topology and is locally connected. The reason is fairly simple: a connected and locally path connected space is path connected (here's a proof). Your disentangled solenoid should be locally arcwise connected (this is part of what I understand your word "unwrap" to mean -- furthermore, it's true of any space coarser than the one I've described). But I cannot, by refining the topology on a space, create an arc between two points that did not already have an arc between them (because an arc is compact Hausdorff, so it can't be created in a Hausdorff space by refininement). Thus the connected components of our refinement must be the path components of the solenoid. And these are precisely the composants.<|endoftext|> TITLE: Manifolds of maps QUESTION [5 upvotes]: Suppose $M$ is a compact $n$-dimensional manifold without boundary. Let $H^s(M,M)$ denote the Sobolev space on $M$, defined as all maps from $M$ to $M$ whose distributional derivatives up to order $s$ are square integrable. For $s>n/2+1$, let $$\mathcal{D}^s =\{\eta\in H^s(M, M)| \eta ~\text{is bijective and}~ \eta^{-1}\in H^s(M,M)\}. $$ Define right multiplication \begin{align} R_\eta:\mathcal{D}^s \to\mathcal{D}^s \\ \xi\to\xi\circ\eta. \end{align} Why is $R_\eta ~C^{\infty}$ for each $\eta \in \mathcal{D}^s$? Also, if define left multiplication \begin{align} L_\eta :\mathcal{D}^s \to\mathcal{D}^s \\ \xi \to \eta \circ\xi. \end{align} Why is $L_\eta$ only $C^l$ when $\eta\in \mathcal{D}^{s+l}$? Essentially what is the difference between $R_\eta ~\text{and} ~L_\eta$? REPLY [3 votes]: To go a bit more into the details (elaborating on Thomas comment above): The result on the differentiability order is classical (going back to Ebin's work in the 60s, culminating in the seminal paper by Ebin and Marsden: Groups of diffeomorphisms and the motion of an incompressible fluid, Ann. Math., 92(1970), 102-163). However, recently the result was established again in great detail in the paper On the regularity of the composition of diffeomorphisms. For a closed manifold the space $H^s (M,N)$ ($s$ bigh enough) carries a Hilbert manifold structure via the usual manifolds of mappings construction (where charts are constructed using a suitable local addition, aka. the Riemannian exponential map, cf. Kriegl, Michor: The convenient setting of global analysis for an explanation of the construction for the manifold of smooth mappings (instead of Sobolev type). The upshot of this construction is however, that there is a canonical identification of the tangent manifold of the manifold of $H^s$-mappings: $$TH^s(M,N) \cong H^s (M,TN).$$ Now $\mathcal{D}^s \subseteq H^s (M,M)$ inherits its smooth structure from the ambient space and in particular we have an induced identification $T\mathcal{D}^s \subseteq H^s (M,TM)$. Under this identification it is not hard to deduce the following identifications $$TR_\eta (f) = f\circ \eta , \qquad TL_\eta (f) = T\eta \circ f, \quad f \in H^s (M,TM).$$ Here of course $\eta$ needs to be in $H^{s+1}$ for the second formula to make sense, whereas for the first formula we can savely assume that $\eta \in \mathcal{D}^s$. Thus whenever $s< \infty$ we see that every derivative of the left multiplication looses one order of differentiability of the map $\eta$. Of course the right multiplication is also continuous linear, whence smooth. But to me the above formula makes it explicit how one is loosing orders of differentiability on $\mathcal{D}^s$ (which is thus not a Lie group but only a half Lie group).<|endoftext|> TITLE: independence of $\ell$ of characteristic polynomial of Frobenius on $\ell$-adic Tate module of Abelian varieties over number fields QUESTION [5 upvotes]: I am looking for a reference for the independence of $\ell$ of the characteristic polynomial of the Frobenius $\mathrm{det}(1-|\kappa(v)|^{-s}\mathrm{Frob}_v \mid (V_\ell A)^{I_v})$ acting on the $\ell$-adic Tate module of an Abelian variety $A$ over a number field (you may assume $v \nmid \ell$). REPLY [3 votes]: This follows from Grothendieck's semistable reduction theorem -- the precise reference is SGA 7, Exp. IX, Thm 4.3(b). The idea is to express the characteristic polynomial of Frobenius in terms of the special fiber of the Néron model of $A$ and then to write this special fiber as an extension of an abelian variety $B$ by an algebraic group $G$ which is itself an extension of a unipotent group $U$ by a torus $T$. We are thus reduced to show the independence of $\ell$ for $B$ and $T$, which basically follows from the Weil conjectures.<|endoftext|> TITLE: A diophantine equation in $\mathbb{N}$ QUESTION [12 upvotes]: While I was working on a paper on graph theory, I encountered a problem which I think is a number-theory-problem. I don't know if there are any tools to answer the question. Find all natural numbers $n$, or prove there are infinitely many $n$, such that the equation $ab+bc+ca=n$ has no answer in $\mathbb{N}$. Can help me or introduce some tools to answer this question? Thanks REPLY [20 votes]: This is an elaboration of Emil Jeřábek's important comment, and contains no original contribution. The OP's problem was examined in depth by Borwein-Choi (1999), and their article is available for free here. I will summarize the content of this article below. Let us consider an integer $n\geq 2$ that cannot be written as $ab+bc+ca$ with integers $a,b,c\geq 1$. Using Lemma 1.1 with $k=1$, we see that $n$ is even. Using Theorem 2.6, it follows that either $n\in\{4,18\}$ or $n=2p_1\dots p_r$ with distinct odd primes $p_j$. The proof of these two results are elementary but highly nontrivial. Then, using some results of Andrews and Crandall, the authors deduce that the class number $h(-4n)$ equals $2^r$ (which is the number of genera of discriminant $-4n$). It is classical that $h(-4n)$ is of size $n^{1/2+o(1)}$ while $2^r=n^{o(1)}$, hence the list of exceptional $n$'s is certainly finite. In fact, Weinberger (1973) analyzed the condition $h(-4n)=2^r$ carefully, and this way we know that either $$n\in\{2,4,6,10,18,22,30,42,58,70,78,102,130,190,210,330,462\},$$ or $n$ is possibly a further single number beyond $10^{11}$. The last possibility can only occur if the Generalized Riemann Hypothesis (GRH) fails for the $L$-function of some quadratic Dirichlet character. To summarize, there are $18$ exceptional integers $n\geq 1$ if GRH holds, and possibly one further exceptional $n\geq 1$ if GRH fails. (In the above paragraph I restricted to $n\geq 2$ for convenience.)<|endoftext|> TITLE: Simple recurrence that fails to be integer for the first time at the 44th term QUESTION [16 upvotes]: The sequence defined by $a_0=a_1 =1$ and $$ a_n = \frac{1}{n-1}\sum_{i=0}^{n-1}a_i^2, \quad n > 1 $$ fails to be integer for the first time at $a_{44}$. Why?? You can verify the statement by computing the sequence mod 43 (see more commentary here (day 5, problem 3)). That's not a very satisfying answer though. Is there a good reason for this behavior? REPLY [24 votes]: Copying my explanation from https://mathoverflow.net/a/217894/25028 The recurrence formula can be rewritten as $$a_2=2,\qquad a_{n+1}=\frac{a_n\cdot (a_n+n-1)}n,\quad n\geq 2,$$ which somewhat justifies why $a_n$ remains integer for quite a while. It shows that $a_n$ accumulates most of the factors of the previous terms and gains some new ones. The division by $n$ happens to hit the existing factors up until $n=43$. Another example of this kind is given by $$b_2=2,\qquad b_{n+1}=\frac{b_n\cdot (b_n+n+5)}n,\quad n\geq 2,$$ which remains integer for up to $n=59$. ADDED. OEIS A292996 gives indices of first noninteger terms in similar sequences. Some spin-off questions: Is there any positive integer sequence $c_{n+1}=\frac{c_n(c_n+n+d)}n$? Can these sequences stay integer-valued as many times as we want and then fail?<|endoftext|> TITLE: How to check if a box fits in a box? QUESTION [28 upvotes]: How could I calculate if a rectangular cuboid fits in an other rectangular cuboid, it may rotate or be placed in any way inside the bigger one. For example would, (650,220,55) fit in (590,290,160), they are all mm. REPLY [8 votes]: This could also be solved by quantifier elimination, and perhaps someone with more knowledge of quantifier elimination can see how to do it feasibly. A box of side lengths $(a,b,c)$ fits inside a box of side lengths $(x,y,z)$ iff there are vectors $(\mathbf{t},\mathbf u,\mathbf{v})$ for the sides satisfying: \begin{gather*} \begin{aligned} \mathbf{t}\cdot\mathbf{t}=a^2,&& & \mathbf{u}\cdot\mathbf{u}=b^2,&& \mathbf{v}\cdot\mathbf{v}=c^2, \\ \mathbf{t}\cdot\mathbf{u}=0,&&& \mathbf{u}\cdot\mathbf{v}=0,&& \mathbf{v}\cdot\mathbf{t}=0, \end{aligned} \\ \pm\mathbf{t} \pm \mathbf{u} \pm \mathbf{v} \le (x,y,z) \end{gather*} where the last line represents three coordinate inequalities for each choice of signs. In other words, we are checking a statement of the form $\exists t_i t_j t_k u_i u_j u_k v_i v_j v_k\ \phi$, where $\phi$ is the conjunction of 6 equalities and 24 inequalities in those 9 bound variables and in $a$, $b$, $c$, $x$, $y$, $z$. Quantifier elimination should then provide some list of polynomial inequalities in $a$, $b$, $c$, $x$, $y$, $z$ which together are equivalent to the small box fitting inside the large box. Update: This works in the 2-d case, with interesting results. Assuming that $0 TITLE: Who wins infinite Hex? QUESTION [8 upvotes]: In this game, you start with a square. Alice tries to connect the top side to the bottom side, and Bob tries to connect the left side to the right side, like in Hex. Unlike in Hex, Alice and Bob use points instead of hexagons. Now you might say that neither Alice nor Bob can win, since it is impossible to form a line using only finitely many points. Not so, for Alice and Bob can move infinitely many times! In particular, let $S_\alpha$ be the state of the board, where $n$ is an ordinal. Then: If $\alpha=0$, then $S_n=\emptyset$. If $\alpha$ is a successor ordinal, then Alice adds $(n,p,Alice)$, where $p$ is some point in $[0,1] \times [0,1]$ that has not already been played, to $S_{\alpha-1}$, unless $S_{n-1}=[0,1] \times [0,1]$. She can use $S_{\alpha-1}$ to inform her discussion (in other words, Alice has a strategy function that given $S_{\alpha-1}$, gives her move). Bob does likewise. If $n$ is a limit ordinal, then $S_\alpha = \bigcup_{\beta<\alpha}S_\beta$. We say Alice has won if there for some $\alpha$, there is a path in $[0,1] \times [0,1]$, composed of points corresponding to Alice's moves, connecting the top and bottom. Bob wins likewise, (connecting the left and right side of the boards). Since there are ordinals $\alpha$ such that $|\alpha| > |[0,1]\times[0,1]|$, the board must eventually be filled (in particular, it will happen before such an ordinal). One thing to note is that Alice and Bob can not both win, due to the Jordan curve theorem. On the other hand, it is possible for neither of them to win. In this case, Alice would have no curve connecting the top and bottom, and Bob would have no curve connecting the sides. My question is: Does Alice have a winning strategy? (Bob doesn't, due to a strategy stealing argument.) Does Alice or Bob have a drawing strategy (a strategy that guarantees either a win or a draw)? (If Bob does, so does Alice.) REPLY [11 votes]: Let $\mathfrak{c}$ denote the cardinality of real numbers and let $(C_{\alpha}: \alpha < \mathfrak{c})$ be an enumeration of uncountable closed subsets of the unit square. Let Bob's strategy be playing a point $q_{\alpha} \in C_{\alpha}$ not already chosen at stage $\alpha$ for $\alpha < \mathfrak{c}$. This is possible since, at stage $\alpha < \mathfrak{c}$, the players could have chosen only less than $\mathfrak{c}$ many points of the set $C_{\alpha}$ which has cardinality $\mathfrak{c}$. Bob can do whatever he feels like for other ordinals. Note that the game will proceed at least $\mathfrak{c}$ stages since any curve connecting opposite sides contains $\mathfrak{c}$ many points. Any curve $C$ witnessing the victory of a player would be necessarily of the form $C_{\beta}$ for some $\beta < \mathfrak{c}$ and hence contains the point $q_{\beta}$. Thus Bob guarantees not losing if he plays with this strategy. (Of course, by using this strategy, Alice can guarantee not losing as well.)<|endoftext|> TITLE: Maximum area of the intersection of a parallelogram and a triangle QUESTION [16 upvotes]: How large can the intersection of a parallelogram (or a square, if you prefer) with a triangle be, if each of them is of unit area? It is easy to see that the intersection can be of area 3/4 – is this the maximum? (It is not; see Joe's answer below. I now believe $2(\sqrt2−1)≈0.828427$ given by Joe is the correct answer.) The analogous question can be asked in every dimension greater than 2, replacing parallelogram with parallelepiped (or cube, if you prefer) and triangle with a simplex or an affine square pyramid, each of unit volume. REPLY [3 votes]: (I posted earlier a false solution which I have removed. Sorry for all this). Perhaps the following is safe(?), while I'll prove the critical special case (discussed earlier by participants in this thread): CONJECTURE   Let a triangle and a square, both having area $\ 1,\ $ have one edge of each parallel to the other one. Then the intersection of the square and the triangle has an intersection of maximal area for instance when the said edges are contained in the same straight line and both figures are on the same side of that line and the said edges have the same center and the said triangle has the height equal to $\ \sqrt 2\ $ (the @JosephO'Rourke example). THEOREM If we restrict ourselves to the case of an isosceles triangle such that its base and a square edge are contained in the same straight line, and that the centers of the two edges coincide then O'Rourke's example provides the maximal intersection. Let me start with an elementary lemma: Lemma   Let positive $\ a\ b\in\Bbb R\ $ be such that $\ a\cdot b=\frac 12.\ $ Then $$ (1-a)^2 + (1-b)^2\ \ge\ 3-2\sqrt 2 $$ Proof $$ (1-a)\cdot(1-b)\ =\ \frac 32 - (a+b)\ \le \ \frac 32 - 2\cdot\sqrt{a\cdot b}\ = \ \frac 32-\sqrt 2 $$ hence $$ (1-a)^2 + (1-b)^2\ \ge\ 2\cdot(1-a)\cdot(1-b) \ \ge\ 3-2\cdot\sqrt 2 $$ End of Proof Remark Our lemma features equality $\ \Leftrightarrow\ a=b=\frac 1{\sqrt{2}}.$ Proof of the THEOREM  (I'll keep applying Tales Theorem). Let the length of the base be $\ 2\cdot x.\ $ When $\ x=1\ $ then we get the OP Wlodek's example with the intersection area $\ \frac 34.\ $ (We already know that it's not optimal). When $\ x\ge 1\ $ then the area of the intersection is: $$ 1\ -\ \left(\frac{x-\frac 12}x\right)^2\ =\ \frac 34\ + \ \left(\frac 12\right)^2 -\left(\frac{x-\frac 12}x\right)^2 $$ $$ =\,\ \frac 34\ -\ \frac {(x-1)\cdot(3\cdot x-1) } {(2\cdot x)^2} \,\ \le\,\ \frac 34 $$ for $\ x\ge 1.$ Thus, the situation is not better than for $\ x=1.$ Dually, let's now consider the case of $\ 0 TITLE: What is the fundamental group of $\mathcal O_{\mathbb P^n}(k)$ minus the zero section QUESTION [8 upvotes]: Let $L^*$ be the total space of the line bundle $\mathcal{O}_{\mathbb{P}^n}(k)$ minus its zero section. How can one compute the fundamental group of $L^*$? For k = 0 the space $L^*$ is $\mathbb{P}^n \times \mathbb{C}^*$ hence $\pi_1(L^*) = \mathbb{Z}$. For k=-1 the $L^*$ is $\mathbb{C}^{n+1} \setminus \{0\}$, therefore $\pi_1(L^*) = 0$. What about the other $k$ ? The long exact sequence of homotopy of a Serre fibration $\mathbb{C}^* \rightarrow L^* \rightarrow \mathbb{P}^n$ gives $\pi_2(\mathbb{C}^*) = 0 \rightarrow \pi_2(L^*) \rightarrow \pi_2(\mathbb{P}^n) \simeq \mathbb{Z} \rightarrow \pi_1(\mathbb{C}^*)\simeq \mathbb{Z} \rightarrow \pi_1(L^*) \rightarrow \pi_1(\mathbb{P}^n) = 0$. So one needs to understand the map $\pi_2(\mathbb{P}^n)\rightarrow \pi_1(\mathbb{C}^*)$. REPLY [2 votes]: Another way to see that the fundamental group is $\mathbb{Z}/k$ is using geometry of cyclic quotient singularities and it goes as follows. We consider the negative twists first. In this case the total space $\mathcal{O}_{\mathbb{P}^n}(-k)$ is the resolution of singularities of the quotient space $\mathbb{C}^{n+1} / G$ with $G=\mathbb{Z}/k$ acting diagonally by $k$-th roots of unity. To see this one computes the coordinate algebra of the quotient which is precisely the affine cone over the degree $k$ Veronese embedding of $\mathbb{P}^n$. Now the latter cone is resolved by a single blow up of its vertex and the resulting space is the total space $\mathcal{O}_{\mathbb{P}^n}(-k)$. In particular the complement to the zero section of $\mathcal{O}(-k)$ is isomorphic to the punctured cone $(\mathbb{C}^{n+1} - 0) / G$, and so has fundamental group isomorphic to $G$ as $\mathbb{C}^{n+1} - 0$ is simply-connected. Now if the twist was positive, contracting zero section is impossible, but replacing $k$ by $-k$ makes a homeomorphic space, so it is $\mathbb{Z}/k$ in both cases; and the answer is $\mathbb{Z}$ for $k=0$ as the bundle is trivial then.<|endoftext|> TITLE: The $n$-th derivative has $n$ zeros. Can such a function be unbounded? QUESTION [13 upvotes]: I asked this on Math.SE some days ago, but without any success. For some application I need a formal definition of bell-shaped function. So I had the following idea: Definition. A $C^\infty$-function $f:\Bbb R\to\Bbb R$ should be called bell-shaped if for all $n\geq 1$ the $n$-th derivative has exactly $n$ zeros (counted with multiplicity). E.g. this works for $\exp(-x^2)$. As this is my try to formalize an informal term, I have no way to check if this is correct. However, I tried to prove that such functions must be bounded (for me, intuitively bell-shaped functions are always bounded). But I had no success. Question: Is a bell-shaped function (in the sense above) always bounded? REPLY [16 votes]: As suggested by Mateusz Kwaśnicki, the function $f : x \mapsto (1+x^2)^{s}$ is bell-shaped and unbounded for any $s \in (0,\frac{1}{2})$. It is easy to see that $f^{(n)}(x) = P_n(x) (1+x^2)^{s-n}$ where $P_n$ is a polynomial of degree $\leq n$. Actually $$ P_{n+1}(x) = (1+x^2) P_n'(x) - 2(n-s)xP_n(x). $$ Let $a_n$ be the coefficient of $x^n$ in $P_n$. Then $a_{n+1} = n a_n - 2(n-s)a_n = (2s-n) a_n$. In particular $(-1)^{n-1} a_n > 0$ for $n \geq 1$, so that $P_n$ has degree exactly $n$ (this is where we use $s < \frac{1}{2}$). One first checks that $f'$ has a single zero. We then assume that $f^{(n)}$ has exactly $n$ distinct simple zeroes $x_n < x_{n-1} < \dots < x_1$ for some $n \geq 1$. Then $$ (-1)^{n-1} f^{(n+1)}(x_1) = (-1)^{n-1} P_n'(x_1) (1+x_1^2)^{s-n} > 0 $$ since $(-1)^{n-1} P_n(x) > 0$ for $x > x_1$. But $(-1)^{n-1} f^{(n+1)}(x) < 0$ for large $x$ so $f^{(n+1)}$ has a zero in $(x_1,+\infty)$. Similarly $f^{(n+1)}$ has a zero in $(-\infty ,x_n)$. Together with the zeroes of $f^{(n+1)}$ between the $x_i$'s, we get $\geq n+1$ distinct zeroes (and therefore exactly $n+1$ zeroes).<|endoftext|> TITLE: Is the series $\sum_n|\sin n|^n/n$ convergent? QUESTION [102 upvotes]: Problem. Is the series $$\sum_{n=1}^\infty\frac{|\sin(n)|^n}n$$convergent? (The problem was posed on 22.06.2017 by Ph D students of H.Steinhaus Center of Wroclaw Polytechnica. The promised prize for solution is "butelka miodu pitnego", see page 37 of Volume 1 of the Lviv Scottish Book. To get the prize, write to the e-mail: hsc@pwr.edu.pl). REPLY [10 votes]: Let $D_N$ be the discrepancy: $$ D_N=\sup \left| \frac{ A(J:P)}{N} - |J|\right| $$ where $P=\{k/\pi \ \mathrm{mod} \ 1\}_{k=1,2,\ldots, n}$, $J$ is an interval in $[0,1]$. If the irrationality measure $\mu$ of $\pi$ is finite, then we have $$ D_N\ll N^{-\frac1{\mu-1} + \epsilon}. $$ From this result and Terry Tao's answer, the number of $n\in [2^k, 2^{k+1}]$ for which $|\sin n |$ falls in an interval of length $2^{-k}$, is $$ \ll 2^{\frac k2} + 2^{k\left(1-\frac1{\mu-1} + \epsilon\right)} $$ Thus, if $r>\max\left\{\frac12, 1-\frac1{\mu-1} \right\}$, then the series $$ \sum_{n=1}^{\infty} \frac{|\sin n|^n}{n^r} $$ is convergent. It is conjectured that $\mu=2$. If we prove that $2\leq \mu <3$, then we can also prove that $$ \sum_{n=1}^{\infty} \frac{|\sin n|^n}{\sqrt n} $$ diverges. I am not aware of any unconditional proof of the divergence of this series.<|endoftext|> TITLE: Database of integer edge lengths that can form tetrahedrons QUESTION [11 upvotes]: Is there a collection of lists of six integer edge lengths that form a tetrahedron? Is there a computer program for generating such lists? I need to find approximately thirty such tetrahedral combinations. REPLY [2 votes]: It seems to me to be of considerable interest to explore the space of 6-tuples that realize tetrahedra in Euclidean 3-space with the tuples (a,b,c,d,e,f) when the points have coordinates with real numbers or more specifically positive integers. Because a set of 6 numbers as edge lengths can be realized by up to 30 incongruent tetrahedra it is indeed important to have a system to identify a sextuple with a particular tetrahedron - a "canonical" version that realizes a given 6-tuple. I have raised the question of finding sets of 6 real numbers (integers) that are such that for i taking on values from 0 to 30 can be realized with exactly i incongruent tetrahedra. See here: http://www.math.illinois.edu/~ajh/ugresearch/Hildebrand-Calculus-Spring2015-report.pdf though it would be nice to see actual examples of integer 6-tuples that achieve the various values of i. It is also worth noting that there is a simpler determinant (which I refer to as the McCrea determinant) than the Cayley/Menger determinant to find the volume of a tetrahedron, see page 20, of the book by William McCrea - Analytical Geometry: http://vignette3.wikia.nocookie.net/math/images/0/0d/Analytical_Geometry_of_Three_Dimensions.pdf/revision/latest?cb=20150228044133&path-prefix=ro For teachers it is worth pointing out one has to worry about numerical instability in determinant calculations, so one wants to be certain that the volume calculation for a determinant really yields a positive result.<|endoftext|> TITLE: An inequality for expected value of normally distributed variables QUESTION [13 upvotes]: Question. Let $X_1,\dots,X_n$ be random variables with normal distribution. Is it true that $$\mathbb E \prod_{i=1}^nX_i^{2k}\ge\prod_{i=1}^n\mathbb E X_i^{2k}$$for any $k\in\mathbb N$? (The problem was posed on 22.06.2017 by Ph D students of H.Steinhaus Center of Wroclaw Polytechnica. The promised prize for a solution is "zywa kura", see pages 37,38 in the Volume 1 of the Lviv Scottish Book. To get the prize, write to the e-mail: hsc@pwr.edu.pl). REPLY [8 votes]: The problem posed above is a semi-well-known open problem that I believe is equivalent to the real polarization conjecture. A more general version of the question posed above is offered as Conjecture 4 in this paper of Wenbo Li, who considers arbitrary positive powers of the type $|X_i|^{p_i}$. For the case $k=1$, a proof is known. Have a look at Theorem 2.1 in the linked paper for a proof of $k=1$ case (thanks to Iosif for alerting me that my answer was only about the $k=1$ case). In particular, the variables have to be jointly Gaussian with mean zero. Moreover, equality holds if and only if they are independent or at least one of them is a.s. zero.<|endoftext|> TITLE: Tannakian Formalism for the Quaternions and Dihedral Group QUESTION [8 upvotes]: It is a basic fact in representation theory of finite groups over complex numbers that the character tables of $Q_8$ and $D_8$ are identical. I believe, this implies that the corresponding categories of representations are equivalent (as tensor categories). On the other hand, Tannakian Formalism tells us that we can reconstruct a finite group $G$ from its category of representations $\mathbf{Rep}_G$ together with the natural (foregetfull) fibre functor $F_G: \mathbf{Rep}_G\rightarrow \mathbf{Vect}$. Namely, $G$ is canonically isomorphic to the tensor automorphisms of the tensor functor $F_G$. This implies that the fibre functors $F_{D_8}$ and $F_{Q_8}$ are different. How can one see this explicitly in terms of representations of $D_8$ and $Q_8$? REPLY [16 votes]: Let $V_D$ and $V_Q$ be the two dimensional simple representations of $D_4$ and $Q_8$ respectively. Let $1_D$ and $1_Q$ denote their trivial representations. Suppose that there is a tensor equivalence between $\mathbf{Rep}(D_4)$ and $\mathbf{Rep}(Q_8)$ commuting with the fibre functor to $\mathbf{Vect}_\mathbb{C}$. This equivalence sends $1_D$ to $1_Q$ (as they're the unit object) and sends $V_D$ to $V_Q$ (as they're the unique simple of dimension 2). In particular there is a $\mathbb{C}$-linear isomorphism $g$ from $V_D$ to $V_Q$. Consider $$g\otimes g:V_D\otimes V_D\to V_Q\otimes V_Q.$$ It must send the unique copy of $1_D$ in $V_D\otimes V_D$ to the unique copy of $1_Q$ in $V_Q\otimes V_Q$. It is easy to see that there is no such $g$. The slickest way I can see to prove this is to note that the flip map $v\otimes w\mapsto w\otimes v$ acts by -1 on $1_Q$ and by 1 on $1_D$. REPLY [12 votes]: The categories ${\rm Rep}(Q_8)$ and ${\rm Rep}(D_8)$ are not equivalent as tensor categories. They have the same Grothendieck ring, but they have non equivalent associators. As far as I am aware, it is an open problem to classify all tensor categories which have the same Grothendieck ring as ${\rm Rep}(Q_8)$ (there are at least two).<|endoftext|> TITLE: Does an oriented $S^3$ fiber bundle admit the structure of a principal $SU(2)$-bundle? QUESTION [8 upvotes]: Let $S \to X$ be an $S^3$-fiber bundle over a smooth manifold $X$. If $S$ is an oriented manifold does this fiber bundle admit the structure of an $SU(2)$-principal bundle? There is a similar theorem for the case of circle bundles and is proved in Morita's book on differential forms. Unfortunately, I do not see a way to extend his argument to this case and he does not discuss $SU(2)$-principal bundles in detail. REPLY [15 votes]: No. (The main idea here is present in Dylan Wilson's comment.) Every principal $SU(2)$-bundle over $S^2$ is trivial, because $\pi_1 SU(2)$ is trivial. But there is a nontrivial oriented bundle over $S^2$ with fiber $S^3$, namely the unit sphere bundle of the nontrivial rank $4$ vector bundle. (There are precisely two rank $r$ oriented vector bundles over $S^2$ if $r\ge 3$, because $\pi_1 SO(r)$ has order two.) I should explain why the nontrivial vector bundle has nontrivial unit sphere bundle, that is, why $\pi_1 SO(4)$ injects into $\pi_1 SDiff(S^3)$. You can use the big theorem of Hatcher ("Smale Conjecture"), which says that $\pi_k SO(4)$ maps isomorphically to $\pi_k SDiff(S^3)$ for all $k$. Alternatively, you can use that the bundle is detected by the Stiefel-Whitney class $w_2$, and that Stiefel-Whitney classes of vector bundles are invariant under fiber homotopy equivalence of unit sphere bundle. Or you can use $\pi_3$ instead of $\pi_1$ as suggested by Dylan Wilson, making a bundle over $S^4$; there are elements of $\pi_3 SO(4)\cong \mathbb Z\times \mathbb Z$ not coming from $\pi_3 SU(2)\cong\mathbb Z$. Again, the resulting sphere bundle is nontrivial either by Hatcher's theorem or by using characteristic classes. I'm not sure what the most elementary version of the characteristic-class argument would be. One example of a rank $4$ real vector bundle over $S^4$ that does not admit a complex structure is the tangent bundle!<|endoftext|> TITLE: Zeta function of Abelian variety over finite field QUESTION [14 upvotes]: Let $A/\mathbf{F}_q$ be an Abelian variety of dimension $g$. Suppose one knows $|A(\mathbf{F}_{q^n})|$ for all $1 \leq n \leq g$. Does one know then $\zeta(A,s)$ (equivalently, $|A(\mathbf{F}_{q^n})|$ for all $n$)? It is true for $g = 1$ by an elementary computation. REPLY [4 votes]: SORRY, THIS IS AN ANSWER TO THE WRONG QUESTION (count of points on a curve instead of on the abelian variety itself) WILL DELETE OR CORRECT BEFORE LONG Yes, the counts for $q^n$ ($n \leq g$), together with the value of $q$, are enough. Let the eigenvalues of Frobenius be $\lambda_i$ ($1 \leq i \leq 2g$), and let $P(t) = \prod_{i=1}^{2g} (1-\lambda t)$ which is a scaling of the characteristic polynomial of Frobenius. Then the power sums $\sigma_n := \sum_{i=1}^{2g} \lambda_i^n$ are the Taylor coefficients of $$ F(t) := -t \frac{P'(t)}{P(t)} = \sum_{i=1}^{2j} \frac{\lambda_i t}{1-\lambda_i t} = \sum_{i=1}^{2j} \left((\lambda_i t) + (\lambda_i t)^2 + (\lambda_i t)^3 + \cdots \right) = \sum_{n=1}^\infty \sigma_n t^n. $$ We are given $\left| A({\bf F}_{q^n}) \right|$, and thus also $\sigma_n = q^n + 1 - \left| A({\bf F}_{q^n}) \right|$, for $n \leq g$.Thus we know $F(t)$ up to $O(t^{g+1})$. Thus we know $-F(t)/t$ up to $O(t^g)$. But that's the logarithmic derivative of $P(t)$, so we know the power series for $\log P(t)$ up to $O(t^{g+1})$ (the constant term vanishes because $P(0)=1$). Thus $$ P(t) = \exp \left(-\!\int_{\tau=0}^t F(\tau) \frac{d\tau}{\tau}\right) $$ gives $P(t)$ up to $O(t^{g+1})$, i.e.\ up to and including the $t^g$ coefficient. Now the functional equation $P(t) = (qt^2)^g P(1/qt)$ gives us the rest: for each $j=1,\ldots,g$, the $t^{g+j}$ coefficient is $q^j$ times the $t^{g-j}$ coefficient. So we know the full expansion of $P$, and thus the full list of eigenvalues and the zeta function, QED.<|endoftext|> TITLE: Topological version of K-theory of coherent sheaves QUESTION [9 upvotes]: My question is this: what is the topological analog of the Grothendieck group of coherent sheaves $G(X)$? Background: In Algebra/Algebraic Geometry there are two versions of the Grothendieck group of a variety $X$, one is the Grothendieck group of vector bundles, which we can denote as $K(X)$ and the other is the Grothendieck group of coherent sheaves, which we can denote by $G(X)$. As the category of vector bundles is a subcategory of coherent sheaves there is always a group homomorphism $K(X) \to G(X)$ and if $X$ is smooth then this map is an isomorphism (basically because every coherent sheaf will be resolved by a finite complex of vector bundles as soon as $X$ is regular by Serre's Theorem). Now let us assume that $X$ is a complex variety, so that we can compare algebraic K-theory with the topological one. That is there is a ring homomorphism $K(X) \to K^{top}(X)$, typically non-injective and non-surjective. I am asking for groups $G^{top}(X)$ which are covariantly functorial for proper morphisms, admit natural transformations $G(X) \to G^{top}(X)$ as well as natural maps $K^{top}(X) \to G^{top}(X)$, such that the latter maps are isomorphisms in the smooth case. Some remarks: The question makes sense for higher G-theory as well but I only bring up Grothendieck group to keep things simple and to avoid the clash of notation (e.g. $K_0(X)$ in algebra vs $K^0(X)$ in topology). A natural guess is that what I call $G^{top}(X)$ is simply what is called K-homology, could this be the case? Finally, there is a modern procedure of taking topological K-theory of a triangulated category due to Blanc (I think), and as far as I understand, $K^{top}(Perf(X)) = K^{top}(X)$, no matter whether $X$ is smooth or not, so it may be that what I want to take is $K^{top}(D^b(Coh(X))$, is this something sensible? REPLY [2 votes]: I don't think K-homology will satisfy your requirements. The reason people think of G-theory as dual to K-theory is because for proper $X$, the graded space $Ext^*(F,Q)$ is finite-dimensional for $F$ a flat and $Q$ a coherent sheaf, and this defines a pairing. There is no such finite-dimensionality even for pairs of topological vector bundles over a topological manifold. In particular, there is no reason to expect there to be any sort of interesting map $KU^*(X)\to KU_{-*}(X)$ for non-smooth $X$. Your #3 is interesting and would probably satisfy all your requirements, although I'm not sure how well functoriality of Blanc's construction works with respect to functors without any finiteness conditions (it most likely does). A simpler alternative might be as follows. Take some topological Abelian subcategory $\mathcal{C}$ of topological sheaves of modules over $C^\infty(X)$ which is stable under extensions and tensor products with bundles and which contains all coherent sheaves (extended to $C^\infty(X)$). One possibility is to model the theory of constructible sheaves and take sheaves which are repeated extensions of pushforwards of (analytic) bundles on algebraic subvarieties. Now take its topological Waldhausen K theory, and invert $u\in K_{Wald}^2(\mathbb{R}-mod)$ (which acts on the topological Waldhausen K-theory of any $\mathbb{R}$-linear category in an obvious way). If you had started with just the category of bundles on $X$, you would get $KU^*(X)$. Here you will get some new theory which will satisfy your requirements.<|endoftext|> TITLE: Examples of (Git) open math (texts) projects QUESTION [17 upvotes]: I am an active part of a research project on the positive effects of open math projects on the community. With open math projects I have in mind a particular thing, namely a GIT project on mathematics in which everyone can contribute and the final result is a collaborative text about some part of mathematics. The obvious examples are The stacks project, The HoTT book, Open logic project. I would like to know other examples of this kind of project. I am specially interested in those projects that have already finished (succeeded). REPLY [3 votes]: I have a github page for my lecture notes on exterior differential systems. The final document, to date: Introduction to exterior differential systems.<|endoftext|> TITLE: Remove a disc from a manifold. When is the resulting sphere nullhomotopic? QUESTION [5 upvotes]: Let $M$ be a connected topological $n$-manifold (not assumed to be compact or boundaryless) and let $D$ an embedded closed $n$-disc. In this situation, there is an inclusion map $S^{n-1} = \partial D \hookrightarrow M\setminus D^{\circ}$. For which $M$ is the inclusion nullhomotopic? One obvious example of such a manifold is $S^n$. Are there others? If the inclusion is nullhomotopic, then $M$ is homotopy equivalent to $M\setminus D^{\circ}$ with an $n$-cell attached by a constant map, i.e. $M$ is homotopy equivalent to $(M\setminus D^{\circ})\vee S^n$. Therefore $H^n(M) \cong H^n((M\setminus D^{\circ})\vee S^n) = H^n(M\setminus D^{\circ})\oplus H^n(S^n) = \mathbb{Z}$ so $M$ is necessarily closed and orientable. There is also an isomorphism of rings $H^*(M) \cong H^*(M\setminus D^{\circ})\oplus H^*(S^n)$ from which it follows that $H^i(M)$ has no free part for $0 < i < n$ (otherwise Poincaré duality would not hold). By using $\mathbb{Z}_m$ coefficients, the same argument shows that $H^i(M)$ also has no $\mathbb{Z}_m$ part for $0 < i < n$. Therefore $M$ is an integral homology sphere. Are there any (non-trivial) integral homology spheres with the desired property? REPLY [9 votes]: Let $M$ be such an integral homology sphere, with fundamental group $\pi$. As you have said, $M$ is homotopy equivalent to $(M \setminus \mathring{D}) \vee S^n$, so its universal cover $\widetilde{M}$ is homotopy equivalent to $(\widetilde{M} \setminus \pi \mathring{D}) \vee \bigvee^\pi S^n$, and so $H_n(\widetilde{M};\mathbb{Z}) = \mathbb{Z}[\pi]$. But as $\widetilde{M}$ is a connected $n$-manifold its top-dimensional homology must be at most rank 1, so $\pi$ must be trivial.<|endoftext|> TITLE: Can a finite group action by homeomorphisms of a three-manifold be approximated by a smooth action? QUESTION [20 upvotes]: Let $M^3$ be a smooth three-manifold, and let $\gamma:G\to\operatorname{Homeo}(M)$ be a finite group action on $M$ by homeomorphisms. Can $\gamma$ can be $C^0$-approximated by smooth group actions $\tilde\gamma:G\to\operatorname{Diff}(M)$? Note that Bing and Moise proved (independently) that any homeomorphism $h:M\to M$ can be $C^0$-approximated by diffeomorphisms $\tilde h:M\to M$, however this does not a priori imply a positive answer to the question of approximation of group actions. (I presume the answer to the analogous question in two dimensions is positive, but if this is not the case, that would be interesting as well.) EDIT: Bing in this paper defined a continuous involution $\sigma:S^3\to S^3$ whose fixed set is a wildly embedded $S^2\hookrightarrow S^3$ (so, in particular, $\sigma$ is not topologically conjugate to a smooth involution). Bing also showed (in the very same paper!) that $\sigma$ is a $C^0$-limit of smooth involutions. Indeed, Bing considers a smooth involution $r:S^3\to S^3$ fixing a smooth $S^2\subseteq S^3$ and a small unknot $K\subseteq S^3$ stabilized by $r$ and intersecting the fixed locus transversally in two points. He then considers a sequence of diffeomorphisms $\varphi_n:S^3\to S^3$ which shrink the $n$th iterated Bing doubles $B^n(K)$ of $K$, namely every component of $\varphi_n(B^n(K))$ has diameter at most $\varepsilon_n>0$, where $\varepsilon_n\to 0$ as $n\to\infty$. Bing shows that (for judiciously chosen $\varphi_n$), the limit $\sigma:=\varphi_n\circ r\circ\varphi_n^{-1}$ exists and is the desired wild involution of $S^3$ (although, of course, the conjugating diffeomorphisms $\varphi_n$ do not converge to a homeomorphism, otherwise $\sigma$ would be topologically conjugate to $r$). REPLY [8 votes]: See Corollary 3.1 of the following paper by Robert Francis Craggs for the case of any involution of $S^3$ whose fixed set is homeomorphic to $S^2$: http://www.ams.org/mathscinet-getitem?mr=257966 http://msp.org/pjm/1970/32-2/p03.xhtml<|endoftext|> TITLE: Is ZFC+(negation of a large cardinal axiom) arithmetically sound? QUESTION [7 upvotes]: My knowledge in set theory is very limited, so I apologize if this question is naive or trivial: Let $A$ to be a large cardinal axiom. $T=ZFC+\neg A$ is a consistent theory. My question is: Question : Is the theory T sound according to the standard model of arithmetic? in other words, if $\alpha$ is sentence about natural numbers such that $T\vdash \alpha$, is $\alpha$ (necessarily) true in the standard model of arithmetic? REPLY [13 votes]: Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.) Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so, provably in $\mathsf{ZFC}$, the relevant cardinals are weakly inaccessible). If $\kappa$ is the least ordinal such that $L_\kappa $ is a model of $\mathsf{ZFC}$, then $L_\kappa$ and $V$ agree on all arithmetic statements, since they share the same natural numbers, but there are no such cardinals in $L_\kappa$. Notice that there is such an ordinal $\kappa$, since any weakly inaccessible cardinal $\tau$ of $V$ (such as the large cardinal in question) is strongly inaccessible in $L$ and therefore $L_\tau\models\mathsf{ZFC}$. On the other hand, the minimality of $\kappa$ implies that no ordinal smaller than $\kappa$ is weakly (and therefore strongly) inaccessible in $L_\kappa$. ($\kappa$ itself is "tiny": it is countable in $L$.) Now, if instead of weak inaccessibility you assume that consistency statements or the like qualify as large cardinal axioms, then "of course" the answer is no, as $\lnot A$ would be a false arithmetic statement. [Note that H. Friedman's results, providing examples of combinatorial statements of low-level arithmetic complexity whose validity depends on large cardinals, assume more than the negation of the large cardinal in question in the "bad" case, where the nice combinatorial statement fails. The models of set theory where this happens are in fact not $\omega$-models.]<|endoftext|> TITLE: The maximal number of copies of a graph $T$ in an $H$-free graph QUESTION [7 upvotes]: Problem. Let $T,H$ be fixed graphs with $H$ being a tree, not isomorphic to a subgraph of $T$. Let $ex(n,T,H)$ be the maximal number of copies of $T$ in an $H$-free graph on $n$ vertices. Is it always true that $ex(n,T,H)=\Theta(n^k)$ for $k=k(T,H)\in\mathbb N$? (The problem was posed on 13.10.2016 by Clara Shikhelman. The promised prize for solution is "a bottle of wine in Tel-Aviv", see page 20 of Volume 1 of the Lviv Scottish Book). REPLY [7 votes]: This was an answer to a previous version of the question, when $H$ was not claimed to be a tree. It is well-known that a maximal number of edges in a $C_4$-free graph is $\Theta(n^{3/2})$ (since every pair of vertices is connected by at most one path of length 2, so the average square of a vertex degree is at most $\Theta(n)$; this is reached, e.g., by the incidence graph of a projective plane). Thus the conjecture fails for $H=C_4$, $T=K_2$.<|endoftext|> TITLE: Vladimir Voevodsky's works QUESTION [79 upvotes]: Vladimir Voevodsky has made several contributions in abstract algebraic geometry, focused on the homotopy theory of schemes, algebraic K-theory, and interrelations between algebraic geometry, and algebraic topology. Voevodsky was awarded the Fields Medal in 2002. Sadly, he died September 30, 2017. Can one draw a general picture of his works? Any expository reference will be appreciated. REPLY [29 votes]: Aside from his work on the foundations of mathematics, which others have already elaborated on, earlier in his career Voevodsky also proved the Milnor conjecture in algebraic geometry. The Milnor conjecture relates Milnor K-theory to Galois cohomology (or, equivalently, etale cohomology, since we are dealing with the case of a field). It is related to Hilbert's Theorem 90, which has historical origins in number theory (going all the way back to Ernst Kummer's work in 1855). Namely, in its modern formulation, Hilbert's Theorem 90 says that $H^{1}(G(k_{s}|k);k_{s}^{*})$ is trivial. By the Kummer exact sequence, we then get the result that $k^{*}/2\cong H^{1}(G(k_{s}|k);\mathbb{Z}/2)$. Milnor's conjecture states that $K_{n}^{M}(k)/2\cong H^{n}(G(k_{s}|k);\mathbb{Z}/2)$ for $k$ of characteristic not equal to $2$, where $K_{n}^{M}(k)$ is the Milnor K-theory, defined as the quotient of the tensor algebra of $k^{*}$ by the ideal generated by $a\otimes(1-a)$ for $a\in k^{*}-\{1\}$. For $n=1$ we have $K_{1}^{M}(k)=k^{*}$, and the statement of the Milnor conjecture is the same as the result given above. We can therefore think of the Milnor conjecture as its generalization to higher $n$. As a side comment, I'll mention that Milnor K-theory may be viewed as an "approximation" to algebraic K-theory, and agrees with Quillen's later definition of algebraic K-theory for $n\leq 2$. Algebraic K-theory is very much related to ideas in homotopy theory in algebraic topology. A survey of Voevodsky's proof can be found here: Voevodsky's Proof of Milnor's Conjecture by Fabien Morel Later on Voevodsky also worked on the more general version of the Milnor conjecture, the Bloch-Kato conjecture, also now known as the norm residue isomorphism theorem. Material on the Bloch-Kato conjecture may be found on Charles Weibel's website: Charles Weibel's Home Page In the course of his work on the Milnor conjecture and the Bloch-Kato conjecture, Voevodsky developed motivic homotopy theory. This is a version of homotopy theory for algebraic geometry, and in order to transplant this theory from algebraic topology certain technical obstacles must first be surmounted, such as the unit interval not being an algebraic variety, and so it must be replaced by the affine line $\mathbb{A}^{1}$. A nice introduction to these ideas may be found in the book "Motivic Homotopy Theory" by Dundas, Levine, Rondigs, Ostvaer, and Voevodsky himself. Voevodsky also developed, for the same purpose, a triangulated category of "mixed motives". Mixed motives are a generalization of pure motives, which is related to the search for a "universal cohomology theory". An introduction to the theory of motives, which however stops just short of mixed motives, is here (I'm linking to the top level of the site, as per the author's request, so just look for the article in the sidebar): Motives - Grothendieck's Dream by James S. Milne It is worth noting that a theory of mixed motives does not exist at the moment - the triangulated category of mixed motives is kind of the "next best thing". Other approaches were made by Huber-Klawitter, Hanamura, and Levine. A more thorough discussion of these ideas can be found in the book Une Introduction aux Motifs by Yves Andre, or, for an English reference, there's also a brief discussion in the book Feynman Motives by Matilde Marcolli, or the following book available online: Noncommutative Geometry, Quantum Fields, and Motives by Alain Connes and Matilde Marcolli Finally, this is much more elementary, but I've found the following popular exposition (on video) by Voevodsky to be an abstract, yet intuitive, and ultimately elegant presentation of homotopy theory: An Intuitive Introduction to Motivic Homotopy Theory by Vladimir Voevodsky There's also a pdf transcript of that talk.<|endoftext|> TITLE: Proofs of some combinatorial identities QUESTION [11 upvotes]: Just wondering if anyone knows any references in the literature to bijections corresponding to the following simple generating function identities. Let $B(z)=\dfrac{1}{\sqrt{1-4z}}$ and $C(z)=\dfrac{1-\sqrt{1-4z}}{2z}$, the generating functions of the central binomial coefficients and Catalan numbers, respectively. I'm looking for bijections corresponding to the following identities: $$B(z)B(-z)=B(4z^2),$$ $$\frac{C(z)+C(-z)}{2}=B(z)(1-2zC(4z^2))=B(-z)(1+2zC(4z^2))$$ and, taking the product of the last two expressions in the second identity and using the first identity and the fact that $1-zC^2=C/B$, $$\left(\frac{C(z)+C(-z)}{2}\right)^2=C(4z^2).$$ Thanks. Update: In a similar vein, one can try to prove other such identities. For example, let $E(z)=\dfrac{1}{\sqrt{1-2z-3z^2}}$ be the generating function of the central trinomial coefficients, then $$E(4z)=B(-z)B(3z) \quad \text{and} \quad E(2z)E(-2z)=B(z^2)B(9z^2).$$ REPLY [7 votes]: Partial answer: Your first identity is \begin{equation} \sum\limits_{k=0}^n \left(-1\right)^k \dbinom{2k}{k} \dbinom{2\left(n-k\right)}{n-k} = \left[n \text{ is even}\right] 2^n \dbinom{n}{n/2} , \end{equation} where I am using the Iverson bracket notation. (That is, $[\mathcal{A}]$ denotes the truth value of a statement $\mathcal{A}$.) This identity is proven in: Michael Z. Spivey, A Combinatorial Proof for the Alternating Convolution of the Central Binomial Coefficients. This paper actually arose from an m.se question. Disclaimer: I have not read the proof.<|endoftext|> TITLE: Fenchel-Rockafellar Duality in Villani's Book QUESTION [6 upvotes]: Villiani writes (some notation changed) in Topics in Optimal Mass Transportation: Theorem 1.9. Let $E$ be a normed VS, $E^*$ it topological dual. $\Theta$ and $\Psi$ are two convex functions on $E$ with values in $\mathbb{R}\cup{+\infty}$. Let $\Theta^*$ and $\Psi^*$ be Legendre-Fenchel transforms of $\Theta$ and $\Psi$ (Earlier definited as $R^*(z^*) = \sup_{z\in E}[-R(z)]$). Assume that there exists $z_0\in E$ such that $\Theta(z_0)$ and $\Psi(z_0)$ are finite and $\Theta$ is continuous at $z_0$. Then: $$\inf_E[\Theta + \Psi]=\max_{z^*\in E^*}[-\Theta^*(-z^*)-\Psi^*(z^*)]$$ He begins his proof by saying: We want to prove: $$\sup_{z^*\in E^*} \inf_{x,y\in E} [\Theta(x) + \Psi(y) + ]=\inf_{x\in E}[\Theta + \Psi]$$ Then he says: The choice of $x=y$ shows that the LHS is not larger than the RHS. I'm confused about where his "sup inf" expression is coming from. I mean I get there's a substitution but that's it. Does someone understand where this expression is coming from? Doesn't a choice of $x=y$ establish equality? REPLY [4 votes]: More details based on Steve's comment: We have \begin{align} -\Theta^*(-z^*) &= - \sup_{x \in E} \big[ \langle-z^*,x \rangle - \Theta(x) \big] \\ &=\inf_{x \in E} \big[ \langle z^*,x \rangle + \Theta(x) \big] \end{align} and \begin{align} -\Psi^*(z^*) &= - \sup_{y \in E} \big[ \langle z^*,y \rangle - \Psi(y) \big] \\ &=\inf_{y \in E} \big[ \langle z^*,- y \rangle + \Psi(y) \big] \end{align} Then, \begin{align} -\Theta^*(-z^*) -\Psi^*(z^*) &= \inf_{x,y \in E} \big[ \langle z^*,x -y\rangle + \Theta(x) + \Psi(y)\big] \end{align}<|endoftext|> TITLE: How should I formalize that there’s no canonical isomorphism between a finite Abelian group and its Pontryagin dual? QUESTION [13 upvotes]: The title says it all, but let me repeat. We all learn that the Pontryagin dual of a finite Abelian group is abstractly isomorphic as groups, but there’s no canonical isomorphism. I think I understand it, but I don’t know how to formalize this statement, maybe using category theory. Could someone enlighten me? If many thinks that this is more suitable to Math.SE, please move this there. REPLY [30 votes]: Suppose we have a family of isomorphisms $\alpha_A\colon A\to A^*$ for all finite abelian groups $A^*$. If $\phi\colon A\to B$ is an isomorphism, we have a dual isomorphism $\phi^*\colon B^*\to A^*$ and thus an isomorphism $(\phi^*)^{-1}\colon A^*\to B^*$. This makes $A^*$ a covariant functor of $A$ on the category of finite abelian groups and isomorphisms, and it is only reasonable to call $\alpha$ canonical if it is natural with respect to this structure. In other words, we should have $(\phi^*)^{-1}\circ\alpha_A=\alpha_B\circ\phi$ for all $\phi$, or $\alpha_A=\phi^*\circ\alpha_B\circ\phi$. In particular, this must hold when $B=A$ and $\phi=n.1_A$ for some $n$ that is coprime to the order of $A$. This means that $(n^2-1).\alpha_A=0$, but $\alpha_A$ is assumed to be an isomorphism, so the exponent of $A$ must divide $n^2-1$. This fails when $A=\mathbb{Z}/5$ and $n=2$, for example, so there is no natural map $\alpha$ as described. The same argument also shows that there is no natural isomorphism $V\to V^*$ for finite-dimensional vector spaces $V$ over $\mathbb{Z}/p$ provided that $p>3$. The same conclusion holds for $p=2$ or $p=3$ but one needs to use some different choices of $\phi$ to prove it. On the other hand, if we restrict even further to elementary abelian groups of order $4$, then you can check that there is a natural choice of $\alpha_A$. It sends each nonzero element $a\in A$ to the unique map $\theta\colon A\to\mathbb{Z}/2$ such that $\theta\neq 0$ but $\theta(a)=0$.<|endoftext|> TITLE: What subsets of a set of integers can compute it? QUESTION [10 upvotes]: For $x, y \subseteq \omega$, (a) We write $x \leq_T y$ if $x$ is Turing reducible to $y$. (b) We write $x \leq_L y$ if $x \in L(y)$ where $L(y)$ is the smallest model of ZFC that contains all ordinals and $y$. Let $L = L(\emptyset)$. Note that $x \leq_T y \implies x \in L(y)$. I have two questions. (1) Is there an infinite $x \subseteq \omega$ such that for every $y \subseteq x$, if $x \leq_T y$ then $x \setminus y$ is finite? If the answer to (1) is yes, then (2) Assume $\mathbb{R} \cap L \neq \mathbb{R}$. Does there exist an infinite $x \subseteq \omega$ such that for every $y \subseteq x$, if $x \leq_L y$ then $x \setminus y$ is finite? REPLY [7 votes]: For (2), what happens if we are in $L[r]$ with $r$ a Sacks real? Since this has minimal degree, all non-constructible $x\subseteq\omega$ have the same degree of constructibility, and if we split a non-constructible set into two disjoint pieces, at least one must be non-constructible. This would be a consistent "no" answer for (2), as we may split any infinite subset of $\omega$ into two infinite pieces. Edit: Denis' comment on Mathias forcing got me to thinking, and a look back at Mathias' original "Happy Families" paper has what we need for a consistent "yes" answer to (2). In particular, Theorem 8.2 of the paper tells us exactly what we want: if $F$ is a Ramsey ultrafilter in $L$ and $X$ is Mathias generic over $L$ with respect to $F$, then $Z\subseteq X$ and $X\leq_L Z$ implies $X\setminus Z$ is finite. So (2) is actually independent of ZFC. Note as well that (2) has a positive answer if the set of constructible reals is countable, as any infinite pseudo-intersection for the filter generated by $F$ in $V$ is Mathias generic over $L$ for $F$.<|endoftext|> TITLE: Is the Lyness 5-cycle map birationally conjugate to its own square? QUESTION [8 upvotes]: Let $L(x,y) = (y,(y+1)/x)$. On a dense open subset of the plane, $L$ and all its powers are well-defined invertible maps and $L^5$ is the identity ($L$ is sometimes called the Lyness 5-cycle map). $L$ is birationally conjugate to its own inverse via the map that exchanges $x$ and $y$. Is $L$ furthermore birationally conjugate to $L^2$? That is, is there some birational map $f$ satisfying $f^{-1} \circ L \circ f = L^2$? My past MO post Conjugating the Lyness 5-cycle into a rotation of the plane is relevant, and Brunault's reply would give me affirmative answer to my question if I knew that the rotation map of the projective plane associated with $L$ is birationally conjugate to its own square. (I briefly thought Galois theory would bridge this gap, but it smells too much like the problem of extending automorphisms of finite extensions of ${\bf Q}$ to automorphisms of ${\bf C}/{\bf Q}$.) In their article "On Cremona transformations of prime order" (https://arxiv.org/pdf/math/0402037.pdf), Beauville and Blanc (summarizing the state of knowledge prior to their work) write "The linear transformations of a given order are contained in a unique conjugacy class." And apart from these, among other conjugacy classes: " ... As for conjugacy classes of order 5, there is at least one family, given by automorphisms of a special Del Pezzo surface of degree 1." Then they state their main result (Theorem 1): In the group of birational automorphisms of the projective plane, every conjugacy class of order 5 is either "of the above type" or is the class containing the linear automorphisms. It is not clear to me whether this is asserting that there are exactly two conjugacy classes of order-5 automorphisms (one containing all the Del Pezzo automorphisms of order 5 and one containing all the linear automorphisms of order 5). REPLY [7 votes]: The Del Pezzo connection gives another way to see that the map must be conjugate to its square and to construct an explicit conjugation. Blowing the plane up at four points in general position (i.e. no three collinear) yields the Del Pezzo quintic DP5 in projective 5-space, and DP5 has a group $S_5$ of automorphisms that contain the Lyness map as a 5-cycle. Now observe that a 5-cycle is conjugate to its square in $S_5$, and thus a fortiori conjugate in the group of all birational automorphisms of the plane. P.S. An explicit $f$ such that $L \circ f = f \circ L^2$ is $$ f(x,y) = \left( -\,\frac{x}{x+1}, \, -\,\frac{y+1}{x+y+1} \right). $$ Both $L \circ f$ and $f \circ L^2$ take $(x,y)$ to $-(y+1,x+1)/(x+y+1)$. The map $f$ is birational because $f^4$ is the identity.<|endoftext|> TITLE: Clique and chromatic number of cycle graph of permutations QUESTION [6 upvotes]: Let $n>1$ be an integer and let $[n] = \{1,\ldots,n\}$. Let $S_n$ denote the set of all permutations (bijections) $\pi:[n]\to[n]$. We say that $\psi\neq\pi\in S_n$ are a cycle away from each other if there is an integer $k\in[n]$ and $k$ distinct integers $n_1,\ldots n_k\in[n]$ such that $$\psi = \pi \circ (n_1 \; n_2\;\cdots\; n_k) \text{ or } \psi = (n_1 \; n_2\;\cdots\; n_k) \circ \pi.$$ Let $E = \big\{\{\pi,\psi\}: \pi \neq \psi \in S_n\text{, and} \psi,\pi \text{ are a cycle away from each other}\big\}$. This makes $(S_n,E)$ into a connected finite, simple, undirected graph. What is the maximal size of a clique in $(S_n, E)$, what is $\chi(S_n, E)$, and are these numbers equal? REPLY [5 votes]: It turns out that for the $S_5$ graph the clique number is $7$ but the chromatic number is $30.$ Details at the end. The number of cycles is $c_n=\sum_{k=2}^{n}\binom{n}{k}(k-1)!$ The first few terms $(n-1)!+\frac{n(n-2)!}{2}+\frac{n(n-1)(n-3)!}3$ for cycles of length $n,n-1,n-2$ are the largest. This is actually the degree of every vertex in your graph. The problem can also be reduced to one on this many vertices, but with variable degrees. That would still be an advantage. You need only consider multiplication on one side: If $\pi$ is a permutation and $\sigma$ is a cycle Then $\sigma'=\pi \circ \sigma \circ \pi^{-1}$ is another cycle of the same type and $\sigma' \circ \pi = \pi \circ \sigma.$ If $C$ is a clique so is $\{\alpha\circ\pi \mid \pi \in C\}$ where $\alpha$ is an arbitrary fixed permutation. So we could assume that the identity element $i$ is a member. This in turn means that every element is a cycle. Here is a clique of size $13$ for $S_5$ $$ (12345)\ (12354)\ (12534)\ (13245)\ (13254)\ (14235)\ (15243)$$ $$ (132)\ (142)\ (145)\ (253)\ (345) $$ $$i$$ The cycles creating edges are exactly the $44$ cycles of type $(abc)$ and $(abcde).$ However some are used $7$ times, some $3$ and some just once. Note that we could ask the same questions for alternating groups. This particular example lives entirely in $A_5.$ Perhaps there is a generalization. I find an example of size $6$ in $S_4.$ One is $$i\ (12)\ (13)\ (14)\ (234)\ (243)$$ Since we can restrict to the $c_n$ vertices from cycles, just look at the induced graph on these. Perhaps first look for maximal cliques among the $n$-cycles, then the $n-1$-cycles then the $n-2$-cycles. These three graphs are regular and have a high symmetry. Then see how they might be combined. LATER The graph for $S_3$ is $K_6$ so the clique number and chromatic number are the same $6.$ I gave a clique of size $6$ in the $S_4$ graph. For that graph (regular of degree $21$) to have chromatic number $6$ would require $6$ independent sets of size $4$. This occurs. The independent set of $\pi$ is $\{\pi,\pi(12)(34),\pi(13)(24),\pi(14)(23)\}.$ I leave the chromatic number of the $S_5$ graph to others. Since this graph on $120$ vertices is regular of degree $84,$ It might be easier to look for cliques in the complement which has degree $35$ than independent sets in the graph. It is curious that the clique sizes go $1,2,6,6,7$ for $n=1,2,3,4,5.$ UPDATE I let Maple look for the chromatic number of the $S_5$ graph ($120$ vertices and $5040$ edges) but stopped it after several days. However it found a maximum independent set in the blink of an eye. $$i\ \ (12)(34)\ \ (13)(24)\ \ (14)(23).$$ So at least $30$ colors are needed. Since these four elements constitute a subgroup, the left cosets of this group provide a $30$-coloring.<|endoftext|> TITLE: Average measure of intersection of a convex region with its translate QUESTION [24 upvotes]: Let $\lambda$ denote the Lebesgue-measure on $\mathbb{R}^n$, and let $C\subset\mathbb{R}^n$ be a convex region. My question is about $$f(C):=\int_{C} \lambda(C \cap (x + C) ) \mathrm{d} x.$$ How large can $f(C)$ be? Of course, there is the trivial bound $f(C)\leq \lambda(C)^2$ but I would expect more something like $f(C) = O( \lambda(C) )$ at least. This question has probably been answered somewhere, and I suspect that the following might be very well known: for regions $C$ of constant measure, the value of $f(C)$ is maximal if $C$ is an $n$-ball (or something similar). I couldn't find any useful references regarding this question. If someone could help me out with some reference, or even a simple solution, I would be very grateful. REPLY [5 votes]: How about a short solution to a much general problem, and without convexity assumption (which turns out to be really just a distraction here) ? So, more generally, for nonempty compact subsets $A$, $B$, and $C$ of $\mathbb R^n$, define $F(A,B,C)$ by $$ F(A,B,C) := \int_{A}\lambda(B \cap (x + C))dx. $$ In particular, the OP's question is concerned with $f(C) := F(C,C,C)$. We shall proceed in 3 short steps. Step 1. The first step in analyzing $F$ is to realize that we can rewrite it as $F(A,B,C) = G(1_A,1_B,1_C)$, where \begin{align} G(u,v,w) = \int_{\mathbb R^n}\int_{\mathbb R^n} u(x)w(x-y)v(y)\,dxdy. \end{align} Step 2. Now, by the Riez rearrangement inequality, we know that $$ G(u,v,w) \le G(u^\star,v^\star,w^\star), $$ where $u^\star$ is the Symmetric decreasing rearrangement of the function $u$. On the other hand, it is clear that $$ (1_A)^{\star} = 1_{K(A)}, $$ where $K(A)$ is the centered ball of same volume as $A$. We deduce that $$ F(A,B,C) \le F(K(A),K(B),K(C)), $$ and in particular, $f(C) \le f(K(C))$. The solution to the OP's question then follows from computation for the ball, in the begining of the accepted answer https://mathoverflow.net/a/282941/78539.<|endoftext|> TITLE: Hermann Weyl's work on combinatorial topology and Kirchhoff's current law in Spanish QUESTION [24 upvotes]: Hermann Weyl was one of the pioneers in the use of early algebraic/combinatorial topological methods in the problem of electrical currents on graphs and combinatorial complexes. The main article is "Repartición de Corriente en una red conductora", which appeared, written in Spanish, in 1923 in Revista Matemática Hispano-Americana: together with other less known papers. Was there a particular cooperation around this topic between Hermann Weyl and the Spanish-speaking mathematical community? Did these articles just disappear from the mainstream of mathematical research? I am aware of papers of Beno Eckmann from the 40's (Harmonische Funktionen und Randwertaufgaben auf einem Komplex). REPLY [23 votes]: It turns out that Beno Eckmann actually asked Hermann Weyl why he published this work in Spanish, as described here and here. The answer is remarkable and unexpected (and apparently does not involve his Spanish spouse). I quote from the latter source (RK=Robert Kotiuga, BE=Beno Eckmann): RK: In the hallway outside our offices was a high-tech espresso machine and every morning Beno would take a break to sit and enjoy an espresso outside our offices. The first day, I "coincidentally" joined him and he related wonderful anecdotes from 1950-1955, after Hermann Weyl retired from the IAS, resettled in Zürich, and frequented the department. The next day I resolved to ask Beno a question which I didn't think any living person could answer. RK: Beno, there is something I really don't understand about Hermann Weyl. BE: What is it? RK: Well, in his collected works, there are are two papers about electrical circuit theory and topology dating from 1922/3. They are written in Spanish and published in an obscure Mexican mathematics journal. They are also the only papers he ever wrote in Spanish, the only papers published in a relatively obscure place, and just about the only expository papers he ever wrote on algebraic topology. It would seem that he didn't want his colleagues to read these papers. BE: Exactly! RK: What do you mean? BE: Because topology was not respectable! REPLY [9 votes]: This is a relevant comment on an answer given in this thread, which the comment boxes cannot conveniently accommodate. The summary of the 'current' given after "These days" in an answer in this thread seems technically wrong (to me): if by 'planar graphs' is meant (which is quite usual) 'planar undirected simple finite graph-theoretic graph', then this is not how 'currents' are most usually formalized these days: your definition of the 'conservation condition as ' $\forall v\ \sum_{v'\sim v} f(vv')=0$ ', with $f$ having been defined as a function on a set of unordered 2-sets is precisely the central definition of the field of zero-sum flows, which is a, admittedly deceptively similar-looking-to-the-casual-observer, interesting niche-subject mostly practiced in Central Asia; but this is not what Weyl was investigating. The free-abelian-group-that-is-modelled-by-said-zero-sum-flows is also (as yet) rather irrelevant to topology, as it still 'awaits its categorification' (this might be a gap in my knowledge, but to my knowledge the class function $\mathsf{Graphs}\xrightarrow[]{\mathrm{ZeroSumFlowGroup}}\mathsf{FreeAbelianGroups}$ is is-not/has-not-emphasized-to-be a functor for any noteworthy category of graphs (though I don't doubt it can, rather easily, I am simply trying to substantiate that this definition is rather irrelevant to the OP proper). The main point is that with the definition in the answer, the edges are not oriented, so a natural interpretation of a direction in which the quantity flows is lacking, the possibility of having negative values on an edge non-withstanding. Again: what you defined are 0-sum-flows. To add context and a reference, let me mention that a recommendable, light-on-graph-theory-focused-on-the-linear-algebra relevant article is S. Akbari, M. Kano, S. Zare: A generalization of 0-sum flows in graphs. Linear Algebra and its Applications 438 (2013) 3629–3634, from whose abstract I take the following excerpt: about which one should note that (0) calling an arbitrary function from the (unordered-edge)-set to the group-of-coefficients is a bit unusual, (1) the heavily-yellowed equation is essentially exactly your definition of the 'currents'. This is quite a different theory from classical simplicial homology/currents-à-la-Weyl's-Spanish-paper, and rather irrelevant to the OP's question. Please consider the crucial part of Weyl's 1923 paper: whose first sentence I now translate, for convenience of readers: To address the problem of the distribution of current, we assume that each segment is given a sense of traversal [i.e.: Weyl unambiguously does first orient the lines, only later defines a function thereon; this is exactly good old simplicial homology, not zero-sum-flows] It will be necessary for me to interrupt now, but let me summarize that currently the mathematical part of the answer in this thread which starts with 'The Spanish paper of Weyl' is at least a little misleading, since the crucial step in the formalization, usual since the 1920s, to first orient the line segment, and only then define a function on the oriented segments, is not sufficiently emphasized. I am not sure what the best correction is. Maybe this extended footnote is more informative than a slick corrected version. Finally, to my mind the most usual treatment of simplicial homology is of course via exterior algebra: the boundary operator in question is, needless to say, the usual derivation $\partial_2\colon\bigwedge^2\bigoplus_{v\in V}\mathbb{Z}v\longrightarrow\bigoplus_{v\in V}\mathbb{Z}v$, where $\mathbb{Z}v$ denotes the free $\mathbb{Z}$-module (=abelian group) on the set $\{v\}$. I am aware that Baez does it, and does it correctly, with quivers, and hence has 'orientations built-in'. I am also aware that perhaps you simply intended your $G=(V,E)$ to denote a quiver; my little-point-to-be-made is only that this distinction will be lost on the majority of readers, and perhaps also be lost on the OP. The orientation-issue is underemphasized currently. (Whether it should be emphasized I am not sure, as the OP is asking quite a different thing from being served with a summary of simplicial homology. The OP is focusing on historical questions.) Will not be able to respond until tomorrow.<|endoftext|> TITLE: A curiosity on complete homomorphisms of boolean algebras QUESTION [13 upvotes]: The question may be trivial, but has eluded me, may be it is more appropriate for mathstack-exchange. Let $B$, $C$ be boolean algebras and $i:B\to C$ be an homomorphism. By Stone duality to each such $i$ corresponds a continuos map $\pi:St(C)\to St(B)$ defined by $G\mapsto i^{-1}[G]$ for any $G$ ultrafilter on $B$ (where $St(B)$ is the compact $0$-dimensional space of ultrafilters on $B$ with topology given by the base of clopen sets $N_c$ given by ultrafilters to which $c$ belongs, as $c$ ranges in $B$). It is easy to check that if $\pi$ is open then $i$ is a complete homomorphism (i.e. it preserves arbitrary suprema whenever they exist in $B$). Does the converse always hold? The converse holds in case $\pi$ induces an order preserving map $\pi^*:C\to B$ defined by $N_{\pi^*(c)}=\pi[N_c]$ for all $c\in C$, which occurs for example if $B$ is a complete boolean algebra. What if $B$ is not complete and $i$ is complete? In this case I'm stuck. Here is a sketch of the easy implication and of the partial converse: (1) Assume $i$ is not complete. Let $D$ be a predense subset of $B$ such that $i[D]$ is not predense in $C$. Fix $c\in C$ such that $c\wedge i(d)=0$ for all $d\in D$. Then $A=\bigcup_{d\in D} N_d$ is a dense open subset of $St(B)$ and $\pi[N_c]\cap A=\emptyset$. Since $St(B)\setminus A$ is closed nowhere dense, this gives that $\pi[N_c]$ is closed (since $St(B)$ is compact Hausdorff) and nowhere dense, hence it is not open. (2) It suffices to show that $\pi[N_c]$ is clopen for all $c\in C$. Let $A=\{b\in B: i(b)\wedge c=0\}$. By completeness of $B$, $\bigvee A=d$ exists, and by completeness of $i$, $i(d)\wedge c=0$. It is not hard to check that $\pi[N_c]=N_{\neg d}$ (since $G\not\in \pi[N_c]$ if and only if $G\cap A$ is non-empty). REPLY [5 votes]: By Corollary 6.3 of [1], the morphism $i$ of Boolean algebras is complete if and only if the continuous function $\pi$ is quasi-open, which means that $int(\pi(U))$ is nonempty for every nonempty open subset $U$ of the domain. The example given by Don Monk shows that this is a strictly weaker property than being an open continous map, as one would expect. [1] Bezhanishvili, Guram, Stone duality and Gleason covers through de Vries duality, Topology Appl. 157, No. 6, 1064-1080 (2010). ZBL1190.54015.<|endoftext|> TITLE: Convergence of dynamical system on the sphere QUESTION [6 upvotes]: Let $A(x)$ be a symmetric negative semi-definite matrix which depends continuously on the parameter $x\in\mathbb{R}^{d}$. We consider the differential equation $$\dot{x} = (I-xx^*)A(x)x$$ on the unit sphere. Is there anything known about the convergence behaviour of the trajectories for $t\to\infty$? If the matrix is constant it converges to a stable equilibrium. REPLY [6 votes]: I think you can get any flow on the unit sphere in this way. Take $A(x)$ to be the rank-one matrix $$A(x) = -(x-b)(x^*-b^*)$$ with $b = b(x)$ tangent to the sphere (that is, orthogonal to $x$). Then $x^*b=b^*x=0$ and $x^*x = 1$, so that $$\begin{aligned}(I - xx^*)A(x)x & = xx^*(x-b)(x^*-b^*)x-(x-b)(x^*-b^*)x \\ & = x - (x-b) = b .\end{aligned}$$ Therefore, the differential equation under discussion takes form $$\dot{x} = b(x).$$<|endoftext|> TITLE: Strengthening an implication of the abc conjecture QUESTION [6 upvotes]: Granville gives p.5 an implication of the abc conjecture: Assume the abc conjecture. Let $f(x,y)$ be squarefree homogeneous polynomial with integer coefficients. For coprime integers $m,n$ if $q^2 \mid f(m,n)$ then $q \ll \max(|m|,|n|)^{2+\epsilon}$. Can we strengthen this to $q \ll |mn|^{1+\epsilon}$? For constant $n$ this is consistent with the paper. Are there heuristic arguments for small roots modulo squares? REPLY [7 votes]: Yes. Without loss of generality, $x$ and $y$ divide $f(x,y)$. (If not, then multiply by one or the other, and $q$ will still divide it). Without loss of generality $m \geq n$. Then we know that the product of primes dividing $f(m,n)$ is at least $m^{\deg f - 2 - \epsilon}$ and that $f(m,n)$ is at most a constant times $m^{\deg f -1} n$, so $q$ is at most the ratio, which is $ (mn)^{1+\epsilon}$.<|endoftext|> TITLE: Asymptotic behavior of gradient descent on a smooth, convex, non-negative function with no finite minimum QUESTION [5 upvotes]: Let $f: \mathbb{R}^d\rightarrow\mathbb{R}_{\geq 0}$ be a function which is convex and smooth (i.e., in $C^{\infty}$). If $x^* \in \mathbb{R}^d$ is the (global) minimum of $f$, it is well known that gradient descent with a (sufficiently small) fixed step size $\eta$, $$ x_{n+1} = x_{n} - \eta \nabla f(x_n), $$ converges to the global minimum $x_n \rightarrow x^*$. Question: What is the asymptotic behavior of gradient descent when a finite minimizer $x^*$ does not exist (i.e., when $\forall x \in \mathbb{R}^d : \nabla f(x)\neq 0$)? My guess is that $x_n/||x_n||$ should converge to some finite solution, but I could not find a proof. Or perhaps there is some counter-example to this claim? Thanks in advance! REPLY [3 votes]: I think there are simple counterexamples of the form $f(x,y)=\phi(x)+\psi(y)$ (here $x$ and $y$ denote real variables), where $\phi$ and $\psi$ are smooth, positive, decreasing, strictly convex functions. The idea is that, $-\psi'(x)$ and $-\phi'(x)$ being positive decreasing functions, is not an obstruction for their ratio to oscillate between arbitrarily large and small values, on arbitrarily large intervals. This produces zig-zag orbits $z_n:=(x_n,y_n)$ for which $z_n/\|z_n\|$ may accumulate to both $(1,0)$ and $(0,1)$. Construction. Define some smooth, positive, decreasing functions $u(t)$ and $v(t)$ such that for all $n\in\mathbb{N}$ $$u(x):={1\over (n!)^2}\quad\text{for $n$ even, and}\; n!+1\le x\le (n+2)!$$ and $$v(x):={1\over (n!)^2}\quad\text{for $n$ odd, and}\; n!+1\le x\le (n+2)!$$ (so one has only to extend them smoothly for $x\le2$ and between $n!$ and $n!+1$ to have them defined everywhere). As a consequence we have $${v(x)\over u(x)}=\begin{cases} n^{-2} &\text{for $n$ odd, and }\; n!+1\le x\le (n+1)! \\ n^2 &\text{for $n>0$ even, and }\; n!+1\le x\le (n+1)! \\ \end{cases}$$ while $$n^{-2}\le {v(x)\over u(x)}\le n^2 \quad \text{for any $n>0$ and }\; n! \le x\le n!+1 \ .$$ Note also that $u$ (respectively $v$) has finite integral on any right-unbounded interval, because the contribute to the integral on any interval $[n!+1, (n+2)!+1]$ (for $n$ even, resp. odd) is bounded by $(n+2)!/(n!)^2=(n+2)(n+1)/n!$. Therefore we can consider the smooth, decreasing, strictly convex functions $$\phi(x):=\int_x^\infty u(t)dt\quad\text{and }\quad \psi(x):=\int_x^\infty v(t)dt\; .$$ The negative gradient iteration $u_{n+1}=u_n-\eta\nabla f(u_n)$ writes, with $z_n:=(x_n,y_n)$ $$ \begin{cases} x_{n+1}=x_n+\eta \,u(x_n) \\ y_{n+1}=y_n+\eta \,v(y_n)\\ \end{cases} $$ and it follows from the above identities and inequalities on $v/u$ that it produces sequences $(x_n,y_n)$ with $\liminf_{n\to+\infty}y_n/x_n=0$ and $\limsup_{n\to+\infty}y_n/x_n=+\infty$, corresponding to a sequence $z_n/\|z_n\|$ that clusters both to $(1,0)$ and to $(0,1)$. Sketch of the computation. Both sequences $x_m=x_0+\eta\sum_{k=0}^{m-1}u(x_j)$ and $y_m=x_0+\eta\sum_{k=0}^{m-1}v(y_j)$ are increasing and diverging. Let's define $\mu(t)$ as the largest integer $j$ such that $x_j\le n!$ and $\nu(t)$ as the largest integer $j$ such that $y_j\le n!$. The point is to use the above identities and inequalities on $u$ and $v$ to show that, for even $n$ and $m=\mu(n!)$, the principal part in both the above sum is given by the indices $j$ such that $\mu(n!)\le j\le \mu((n+1)!)$, so that $y_m/x_m=O(1/n^2)$. Analogously, for odd $n$ and $m=\nu(n!)$, one should get $x_m/y_m=O(1/n^2)$.<|endoftext|> TITLE: Hasse principle for quadratic forms over finitely generated fields QUESTION [5 upvotes]: Does the Hasse principle hold for quadratic forms over finitely generated fields (e.g. for the Henselisations/completions at height-$1$-primes or all places)? REPLY [4 votes]: The examples that I wrote in the comments are wrong. Thanks to Felipe Voloch for catching my mistake. One can, indeed, form the symbol algebras that I indicated. However, the Brauer-Hasse-Noether(-Albert) theorem from class field theory states that these symbol algebras are actually matrix algebras. In these terms, the Brauer-Hasse-Noether theorem together with Hasse's Global Structure Theorem (injectivity on the left for the "reciprocity short exact sequence") implies that a Severi-Brauer variety over the function field $\mathbb{F}_q(B)$ of a curve has a point if and only if it has local points in the completions for all places. So my examples in the comments were maximally wrong. Sorry for the slip! Nonetheless, there are examples of dimension $2$ or $3$ that are everywhere unramified at height one primes. For instance, the Artin-Mumford threefolds are smooth, projective $3$-dimensional varieties that admit a conic fibration over a rational surface and such that there exists an everywhere unramified conic bundle over the threefold whose corresponding Brauer class is nonzero. The nontriviality of the Brauer class is "witnessed" by geometric computations on codimension $2$ subvarieties of the threefold, so testing with prime ideals of height $>1$ should detect this failure of the Hasse principle. The original construction of the Artin-Mumford examples was in characteristic $0$ (if memory serves). As usual, the smooth, projective $3$-fold $B$, the everywhere smooth conic bundle $C\to B$, and the "witness" subvarieties of $B$ can all be defined over a finitely generated integral domain $R$ that contains $\mathbb{Z}$. Thus, there exists an integer $p_0$ such that for all primes $p$ that are greater than $p_0$, for every field $k$ of characteristic $p$, for every ring homomorphism $R\to k$, the base change $B_k$ will be smooth, the conic bundle $C_k\to B_k$ will be everywhere smooth, and the witnessing subvarieties behave well and prove that there is no rational section of $C_k\to B_k$. Does this conic bundle $C_k\to B_k$ satisfy the Hasse principle? If the restriction of $C_k$ over every surface in $B_k$ has a rational point, then by Hensel's Lemma, there is a local point in the corresponding completion of $k(B_k)$. In that case $C_k\to B_k$ violates the Hasse principle (for height one primes). On the other hand, assume that there exists a surface $B'_k\subset B_k$ over which $C'_k = B'_k\times_{B_k} C_k$ has no rational point. Replace $B_k$ by $B'_k$, replace $C_k$ by $C'_k$, and start over (if you like, you can use resolution of singularities of surfaces to make $B'_k$ smooth). Now the Hasse principle is definitely violated, since the Brauer-Hasse-Noether theorem tells us that the restriction of $C_k$ over every curve in $B_k$ has a rational section. So, one way or another, we get violations of the Hasse principle for conic bundles over a threefold or over a surface over a finite field $k$.<|endoftext|> TITLE: Non-commutative regular sequences and non-commutative Koszul complex QUESTION [5 upvotes]: I'm trying to understand the non-commutative Koszul complex, as can be found in Anick's nice paper "Non-Commutative Graded Algebras and Their Hilbert Series", J. of Algebra 78, (1982) and I'm stuck at two points, which are just where the paper "jumps" from the commutative case to the non-commutative one. Let me set first some notation and assumptions. For the commutative situation, we have: $\mathbf{k}$ is a field, $R$ a connected commutative graded $\mathbf{k}$-algebra; that is, $$ R = \mathbf{k} \oplus R_1 \oplus R_2 \oplus \dots $$ in which every piece $R_n$ is a finite-dimensional $\mathbf{k}$-vector space. Let $\theta_1, \dots , \theta_r \in R$ be a regular sequence. That is, the ideal generated by $\theta_1, \dots , \theta_r$ is smaller than $R$ and each $\theta_n$ is not a zero divisor in $R/(\theta_1, \dots , \theta_{n-1})$ for all $n$. (Also the $\theta_n$ are homogeneous of positive non-zero degree.) First. The main idea in Anick's paper seems to be replacing the notion of non zero divisors, with the help of the following characterization: $$ \theta_1, \dots , \theta_r \quad \text{is a regular sequence}\quad \Longleftrightarrow \quad R \cong \mathbf{k}[\theta_1, \dots , \theta_r] \otimes \frac{R}{(\theta_1, \dots , \theta_r)} \ . $$ Here: $\mathbf{k}[\theta_1, \dots , \theta_r]$ is the polynomial algebra on $\theta_1, \dots, \theta_r$, the isomorphism is as graded $\mathbf{k}$-vector spaces, the tensor product is over $\mathbf{k}$. My first question is about this isomorphism: Anick says it's "well known" and gives a reference: Stanley, "Hilbert Functions of Graded Algebras", Adv. in Math. 28 (1978). Ok, there the closest thing looking like this result is in page 63, where Stanley says: "This is essentially a well-known property..., though an explicit statement is difficult to find in the literature." And gives in turn references for a particular case. So, ok, I'm trying to provide myself of some proof, with the help of what Anick does for the non-commutative case. You can easily produce a morphism of graded vector spaces $$ \mathbf{k}[\theta_1, \dots , \theta_r] \otimes \frac{R}{(\theta_1, \dots , \theta_r)} \longrightarrow R $$ by sending each $\theta_i$ to itself in $R$. For the quotient part, you choose a $\mathbf{k}$-linear section of the projection $R \longrightarrow R /(\theta_1, \dots , \theta_r)$. Anick proves that you can pick no matter which section and the resulting morphism is an epimorphism. So far, so good. Ok so, this is my first question: how do you prove that this morphism is also a monomorphism? In another part of the paper, Anick uses an argument on dimensions: after all, we are dealing with finite dimensional vector spaces. So I'm trying to prove (first for the case $r=1$) that, degree-wise, what we have on both sides are vector spaces of the same dimension. It's a little bit messy (I hate counting!), but I think I got it. My only doubt is: is there a more simple, direct "well-known" proof? Second. What's the problem with the notion of (two-sided) non zero divisor in the non-commutative case that forces Anick to replace it by that isomorphism? (Then he goes on talking about "strongly free" sets, which are those $\theta_1, \dots , \theta_r \in H$, such that you have an isomorphism $$ H \cong \mathbf{k}\langle \theta_1, \dots , \theta_r \rangle \otimes \frac{H}{H(\theta_1, \dots , \theta_r)H} \ , $$ $H$ a connected non-commutative graded $\mathbf{k}$-algebra, $\mathbf{k}\langle \theta_1, \dots , \theta_r \rangle$ is the free associative algebra...) REPLY [4 votes]: First part: Let $\theta_1,\ldots,\theta_r$ be a regular $R$-sequence. Suppose that the kernel of $\mathbb{K}[\mathrm{x}_1,\ldots,\mathrm{x}_n]\otimes R/(\theta_1,\ldots,\theta_r)\rightarrow R$ is not trivial. Then there is a non trivial polynomial $f$ that is being mapped onto $0$. The image of $f$ in $R$ would satisfy an equation $\sum_\alpha a_\alpha \theta_1^{\alpha_1}\cdots \theta_r^{\alpha_r}=0$, where $a_\alpha\notin (\theta_1,\ldots,\theta_r)$. Thus $-a_\beta \theta_1^{\beta_1}\cdots \theta_r^{\beta_r}=\sum_{\alpha\neq \beta} a_\alpha \theta_1^{\alpha_1}\cdots \theta_r^{\alpha_r}$, where $\beta\neq 0$ is the biggest multiindex with respect to the lexiographical ordering. The equation would now imply that it is possible to increase the degree in the $i$-th component by just summing and multiplying monomials by $a_\alpha\notin (\theta_1,\ldots,\theta_r)$. But that is impossible, because $\theta_1,\ldots,\theta_r$ is a regular sequence. Hence the mapping $\mathbb{K}[\theta_1,\ldots,\theta_r]\otimes R/(\theta_1,\ldots,\theta_r)\rightarrow R$ is injective. Second part: The characterization of regular sequences does not hold in the non-commutative polynomial ring. In other words, there is a sequence $\theta_1,\ldots,\theta_r$ that is regular (by transferring the meaning of regular directly to the non-commutative case), but that fails to be regular in the other sense (Anick's definition aka. strongly free). One can see that by using Theorem 3.1. Anick defines regular (strongly free) by $R\cong \mathbb{K}\langle \theta_1,\ldots,\theta_r\rangle\ast R/R(\theta_1,\ldots,\theta_r)R$ because of the following reasons: First: Suppose that we have a commutative regular local ring $R$, then it is well known that the sequence $\theta_1,\ldots,\theta_r$ is regular if and only if $H_0(K(\theta_1,\ldots,\theta_r)\cong R/(\theta_1,\ldots,\theta_r)$ and $H_n(K(\theta_1,\ldots,\theta_r))=0$ for $n>0$, where $K(\theta_1,\ldots,\theta_r)$ is the Koszul-complex with respect to $\theta_1,\ldots,\theta_r$. We want such results in the non-commutative situation. By defining regular as above and defining an appropriate Koszul-complex in the non-commutative situation we get a similar result in form of Theorem 2.9. Second: Under certain circumstances (see Theorem 3.1) being regular (strongly free) has a combinatorial characterization, which is later used to prove the main result (Theorem 3.8).<|endoftext|> TITLE: Characterisation of bell-shaped functions QUESTION [18 upvotes]: This is an open problem that I learned from Thomas Simon. I will completely understand if the question is judged as non-research level (and it is indeed not related to my research), but I believe a solution would result in some nice, publishable mathematics. The point of this post is to popularise the problem. We need two definitions. Definition 1: We say that a function $f$ is bell-shaped on $\mathbb{R}$ if it is infinitely smooth, converges to zero at $\pm \infty$ and the $n$-th derivative of $f$ has $n$ zeroes in $I$ (counting multiplicity). Definition 2: We say that the derivative is interlacing for $f$ on $\mathbb{R}$ if $f$ is infinitely smooth and for every $n \geqslant 0$ the zeroes of $f^{(n)}$ and $f^{(n+1)}$ interlace (counting multiplicity). When we say that the zeroes of $f$ and $g$ interlace, we mean that each connected component of $\{x \in I : f(x) \ne 0\}$ contains exactly one zero of $g$, except perhaps for unbounded components, which are allowed to contain no zero of $g$ — at least when all zeroes of $f$ and $g$ are simple. The definition in the general case is somewhat more involved, but hopefully clear. Problem: Describe all bell-shaped functions. Describe all functions $f$ such that the derivative is interlacing for $f$. Natural modifications are allowed. For example: $f$ can be assumed to be analytic or entire; the problem can be restricted to a finite or semi-infinite interval. By description we mean essentially arbitrary condition that is easier to check than the definition. Ideally, one could hope for a result similar to Bernstein's characterisation of completely monotone functions as Laplace transforms, which can be re-phrased as follows: if $f^{(n)}$ has no zeroes on $\mathbb{R}$, then $f$ is the Laplace transform of either a measure on $[0, \infty)$ (a completely monotone function), or a measure on $(-\infty, 0]$ (a totally monotone function). However, the answer for bell-shaped functions is likely much more complicated. Easy observations: It is clear that $f$ is bell-shaped if and only if it has no zero, it converges to zero at $\pm \infty$ and the derivative is interlacing for $f$. By Rolle's theorem, if $f$ has no zeroes and $f$ converges to zero at $\pm \infty$, then $f^{(n)}$ has at least $n$ zeroes. The function $\exp(-x^2)$ is bell-shaped: its $n$-th derivative is $P_n(x) \exp(-x^2)$ for a polynomial $P_n$ of degree $n$, so it has no more than $n$ zeroes. Similarly, the function $f(x) = (1 + x^2)^{-p}$ is bell-shaped for $p > 0$ (and the derivative is interlacing for $f$ also when $p \in (-\tfrac12, 0)$): the $n$-th derivative of $f$ is of the form $P_n(x) = (1 + x^2)^{-p-n}$ for a polynomial $P_n$ of degree $n$. In the same vein, $f(x) = x^{-p} \exp(-x^{-1})$ is bell-shaped on $(0, \infty)$ for any $p > 0$: $f^{(n)}(x) = P_n(x) x^{-p - 2n} \exp(-x^{-1})$ for a polynomial $P_n$ of degree $n$. The derivative is interlacing for a polynomial if and only if it only has real roots. More generally, the derivative is interlacing for any locally uniform limit of polynomials with only real roots. This class includes $\sin x$, $\exp(x)$, $\exp(-x^2)$, some Bessel functions and many more; see Chapter 5 in Steven Fisk's book. This topic appeared at least in probability literature, in an erroneous article by W. Gawronski, followed by two articles by T. Simon on stable laws and passage times (with W. Jedidi). EDIT: Having read through Widder and Hirschman's book and other references pointed out in Alexandre Eremenko's answer (which was really helpful!), I was able to extend Thomas Simon's work and prove that certain functions are bell-shaped. The preprint is available at arXiv; all comments welcome. The class of functions that I was able to handle includes density functions of infinitely divisible distributions with Lévy measure of the form $\nu(x) dx$, where $x \nu(x)$ and $x \nu(-x)$ are completely monotone on $(0, \infty)$. Such functions are called extended generalised gamma convolutions (EGGC) by L. Bondesson, and it includes all stable distributions (thus correcting an error in W. Gawronski's article). Of course this does not answer the question about characterisation of all bell-shaped functions: this remains an open problem. However, I am not aware of any bell-shaped function outside this class. Is there any? It allows one to show that certain simple functions are bell-shaped. For example, if I did not make a mistake, $f(x) = \dfrac{1}{(1 + x^2)(4 + x^2)}$ is an EGGC, so it is bell-shaped. Out of curiosity: is there any elementary way to prove that this particular function $f$ is bell-shaped? REPLY [4 votes]: Together with Thomas Simon from Lille we just posted to arXiv a paper Characterisation of the class of bell-shaped functions, which answers this question. To my great surprise, the class of bell-shaped functions studied in the paper mentioned in the edit, A new class of bell-shaped functions, exhausts all bell-shaped functions. For the record, let me copy Corollary 1.5 from the new paper (with minor editing): Theorem: The following conditions are equivalent: (a) $f$ is a weakly bell-shaped function; (b) $f$ is the convolution of a Pólya frequency function $h$ and a locally integrable function $g$ which converges to zero at $\pm \infty$, which is absolutely monotone on $(-\infty, 0)$, which is completely monotone on $(0, \infty)$, and which may contain an additional atom at zero; (c) $f$ is a locally integrable function which converges to zero at $\pm \infty$, which is non-decreasing near $-\infty$ and non-increasing near $\infty$, and which satisfies $$ \hat{f}(\xi) = \exp\biggl(-a \xi^2 - i b \xi + c + \int_{-\infty}^\infty \biggl( \frac{1}{i \xi + s} - \biggl(\frac{1}{s} - \frac{i \xi}{s^2} \biggr) \mathbb{1}_{\mathbb{R} \setminus (-1, 1)}(s) \biggr) \varphi(s) ds \biggr) , $$ with $a \geqslant 0$, $b \in \mathbb{R}$, $c \in \mathbb{R}$, and $\varphi : \mathbb{R} \to \mathbb{R}$ a Borel function such that $\varphi - k$ changes its sign at most once for every integer $k$. Furthermore, $f$ is strictly bell-shaped if and only if $f$ is smooth and weakly bell-shaped. The function $\varphi$ necessarily satisfies an appropriate integrability condition at $\pm\infty$, and has some regularity at $0$; see Theorem 1.1 in the new paper for details. In the theorem: a function $f$ is strictly bell-shaped if it is non-negative, it converges to zero at $\pm \infty$, and $f^{(n)}$ changes its sign $n$ times for $n = 0, 1, 2, \ldots$; a function $f$ is weakly bell-shaped if the convolution of $f$ with every Gaussian is strictly bell-shaped; a function $g$ is absolutely monotone on $(-\infty, 0)$ if $g^{(n)}(x) \geqslant 0$ for every $x < 0$; a function $g$ is completely monotone on $(0, \infty)$ if $(-1)^n g^{(n)}(x) \geqslant 0$ for every $x > 0$; a Pólya frequency function $h$ is a convolution of a (possibly degenerate) Gaussian and a finite or infinite number of exponentials, that is, functions of the form $a e^{-1 - a x} \mathbb{1}_{(0, \infty)}(1 + a x)$ (this is the density function of a centred exponential random variable with variance $1 / |a|$, with a negative sign if $a < 0$).<|endoftext|> TITLE: A determinantal formula QUESTION [14 upvotes]: In my research, I encounter the following formula which I believe is correct (checked for $n\le3$). Is it classical ? If so, what is a reference ? I am given a real symmetric matrix $$S:=\int Y(t)Y(t)^Td\mu(t),$$ where $\mu$ is a probability and $Y(t):\Omega\rightarrow{\mathbb R}^n$. Let $\sigma_k(S)$ be the elementary symmetric polynomial in the eigenvalues of $S$. For instance, $\sigma_1(S)$ is the trace and $\sigma_n(S)$ the determinant. The following formula gives $\sigma_k(S)$ in terms of the Gram matrix $G_k(s_1,\ldots,s_k)$ whose entries are the scalar products $Y(s_i)\cdot Y(s_j)$. $$\sigma_k(S)=\frac1{k!}\int^{\otimes k}\det G_k(s_1,\ldots,s_k)\,d\mu(s_1)\cdots d\mu(s_k).$$ Remark that $S$ is positive semi-definite. The integrand is non-negative, as well as $\sigma_k(S)$. The integrand vanishes identically iff $Y(t)$ takes values in a subspace of dimension $ TITLE: Generalizing The Cardinal Characteristics of the Continuum QUESTION [5 upvotes]: Has any work been done on generalizing any characteristics of the cardinality of the continuum? The bounding number $\mathfrak{b}$ and dominating number $\mathfrak{d}$ could be easily generalized for ordinals $\alpha$, as follows: $$\forall f,g\in\alpha^\alpha(f\leq_{\alpha}g\Leftrightarrow|\{\beta\in\alpha:f(\beta)>g(\beta)\}|<|\alpha|)$$ $$\mathfrak{b}_\alpha=\min\{|F|:F\subseteq\alpha^\alpha\land\forall f\in\alpha^\alpha\exists g\in F(g\not\leq_\alpha f)\}$$ $$\mathfrak{d}_\alpha=\min\{|F|:F\subseteq\alpha^\alpha\land\forall f\in\alpha^\alpha\exists g\in F(f\leq_\alpha g)\}$$ A few points to be made at this point: $\mathfrak{b}_\alpha\leq\max\{|F|:F\subseteq\alpha^{\alpha}\}=|\alpha^\alpha|=2^{|\alpha|}$ For cardinals $\kappa,\mathfrak{b}_\kappa>\kappa$ (this is shown by generalized diagonalization) The last two points combined make $\mathfrak{b}_\kappa=\kappa^+$ if $\kappa^+=2^\kappa$ (or if GCH is assumed) $\mathfrak{b}_\omega$ is clearly, in this case, $\mathfrak{b}$, and the same is true with $\mathfrak{d}_\omega$. REPLY [3 votes]: Don Monk has a paper describing generalized $\mathfrak b$ and $\mathfrak d$ as you describe. In his paper he further generalizes to $\mathfrak b_{\kappa,\lambda,\mu}$ (and analogously for $\mathfrak d$), considering families of functions in ${}^\lambda\mu$ with $\lambda, \mu,$ and $\kappa$ all possibly distinct. I think in his notation the numbers you describe are $\mathfrak b_{\alpha, \alpha, \alpha}$ and $\mathfrak d_{\alpha, \alpha, \alpha}$ Generalized ${\frak b}$ and ${\frak d}$. Notre Dame J. Formal Logic 45 (2004), no. 3, 129--146.<|endoftext|> TITLE: $p$-th root of non-torsion points on elliptic curves QUESTION [6 upvotes]: Let $E$ be an elliptic curve over a number field $F$. Assume that $E$ has a $F$-rational non-torsion point $Z$. For each prime $p$, let $\frac{1}{p}Z$ be the set of $X\in E(\bar{F})$ such that $pX=Z$. Is it possible to have $F(\frac{1}{p}Z)=F(E[p])$ for infinitely many prime $p$? REPLY [5 votes]: No. Lemma 3.7 on page 11 from my paper here implies that if $\mathcal{T}_{1} = {\rm Gal}(K(E[p])/K)$ and there is a normal subgroup $H \unlhd \mathcal{T}_{1}$ with order coprime to $p$ for which $E[p]^{H} = 0$, then $E(K) \cap pE(K(E[p])) = pE(K)$. (Full disclosure - I learned the proof of Lemma 3.7 from a very conscientious referee.) This implies that if there is such a subgroup $H$, and the point $Z \ne pW$ for some $W \in E(K)$, then $F(\frac{1}{p} Z) > F(E[p])$. Now, for any elliptic curve $E$, the subgroup $H = \{ kI : k \in \mathbb{F}_{p}^{\times} \}$ of scalar multiples of the identity will be contained in the image of $\rho_{p} : {\rm Gal}(K(E[p])/K) \to GL_{2}(\mathbb{F}_{p})$ if $p$ is large enough. (If $E$ is non-CM, then this follows from Serre's theorem that $\rho_{p}$ will be surjective if $p$ is large enough. If $E$ has CM, then for large $p$ the image of $\rho_{p}$ will be a Cartan subgroup or its normalizer, depending on whether $F$ contains ${\rm End}(E) \otimes \mathbb{Q}$ or not.)<|endoftext|> TITLE: Monadic interpretation of coalgebras over operads QUESTION [5 upvotes]: The structure of an algebra $A$ over a operad $O$ is encoded by an operad morphisms from $O$ to $\{Hom(A^{\otimes k},\, A)\}_{k}$. The same structure can be stored using the structure $M_OA\to A$ of an algebra over the monad $M_O$ which is defined on an object $A$ by $\bigoplus_j O(j)\otimes_{\Sigma_j} A^{\otimes j}$ and its structure maps $M_O\circ M_O\to M_O$ and $1\to M_O$ are induced from composition and unit respectively. The structure of an coalgebra $C$ over a operad $O$ is encoded by an operad morphisms from $O$ to $\{Hom(C,C^{\otimes k})\}_{k}$. Is there a monadic interpretation of coalgebras over an operad? REPLY [2 votes]: What works is a comonadic interpretation of coalgebras over cooperads. Coalgebras over cooperads are like conilpotent coalgebras over operads.<|endoftext|> TITLE: Climbing up subsets of $\omega_1$ using reals QUESTION [10 upvotes]: This is a bit of an odd question, so I've included the motivation below the fold. Throughout we work in ZFC+"$\omega_1^r$ is countable for all $r\in\mathbb{R}$:" Say that a set $X\subset\omega_1$ is really climbable if $X$ is countable (that's the silly case) or there is some real $r$ and some $f\in L[r]$ such that for all countable $\alpha$, $$f(X\cap\alpha)\in X\setminus \alpha.$$ (Note that we don't demand $f(X\cap \alpha)=\min(X\setminus\alpha)$.) That is, an uncountable $X$ is really climbable if there is some real which lets us build a way to climb up $X$. Obviously sufficiently nice sets are really climbable, as are sets containing sufficiently nice sets; at the same time, under CH it's easy to build a non-really-climbable set. My question is: What can be said about the really climbable sets without CH? In particular, it's not clear to me that "Every set is really climbable" is impossible. Motivation: I'm trying to answer a question I left open in my thesis, which has been bothering me for some time (which was also ultimately, but much directly, the motivation for this other question of mine): For $K\subseteq\omega_1$, consider the game (on $\omega$) $Copy(K)$ where Player I builds a single linear order $A$ Player II builds a list of linear orders $B_i$ ($i\in\omega$) Player II wins iff $(i)$ every $B_i$ is isomorphic to some $\alpha_i\in K$, and $(ii)$ if $A\cong\beta$ with $\beta\in K$, then $\beta=\alpha_i$ for some $i\in\omega$. Here I conflate an ordinal $\gamma$ with the linear order $(\gamma, <)$. That is, player II is trying to "guess" what element of $K$ player I is playing; if player I can trick player II into building something not in $K$, they win, and otherwise they win if they build something not isomorphic to anything built by II. (This is an instance of a class of games defined by Montalban.) By $\Sigma^1_1$-bounding, if $K$ is cofinal in $\omega_1$ then $Copy(K)$ can't be a win for player II, and it's easy to see that $Copy(K)$ is a win for II if $K$ is countable. And assuming CH, there is some $K$ such that $Copy(K)$ is undetermined. However, if we drop CH, then I don't know how to produce an undetermined $Copy(K)$, even using Choice - roughly, the reason is that we have continuum-many strategies to defeat but only $\omega_1$-many choices to make in building $K$ (and of course it's even consistent that there are as many strategies as there are subsets of $\omega_1$). "Obviously" there is still such a $K$, but the question is (so far as I know) open for now. The question above is one which grows naturally out of this problem: intuitively, if it's consistent that every set of countable ordinals is really climbable, then it would be more plausible that "$Copy(K)$ is determined for all $K\subseteq\omega_1$" could be consistent with ZFC. REPLY [5 votes]: "Every set is really climbable": this contradicts AC. By AC construct a set $X\subseteq \omega_1$ such that $L[X]\models $"$\omega_1 =\omega_1^V$". Then for arbitrarily large $\gamma < \omega_1$ we have that $L_\gamma[X\cap \gamma] \models $"Every set is countable". But such a set cannot be climbable, for if $f\in L[r]$ were a 'climbing' function, according to the definition we should have such $X\cap \gamma \in dom(f) \subset L[r]$ contradicting the inaccessibility of $\omega^V_1$ there. On the other hand, assuming $AD$, for example in $L({R})$ assuming large cardinals (where also DC holds - but that is not relevant), by an early result of Solovay, every subset of $\omega_1$ is in some $L[r]$ for a real $r$; clearly then every such subset is really, and trivially, climbable. REPLY [3 votes]: Concerning Copy(K)-determinacy: you mention that every set of countable ordinals being climbable would lend plausibility to this in ZFC (that is with AC). We just saw that climbability fails under AC; of course $\forall K$Copy(K)-determinacy holds under AD; and if you make the game harder for II by insisting (I make it that II has the winning strategy, not I, by Boundedness) that II's wellorders code all of $K$ below their supremum, then one can conclude that $K$ is in $L[\tau]$ where $\tau$ is II's strategy. So for this slightly more rigorous game, its determinacy for all $K$ is inconsistent with choice and $\omega_1$ inaccessible to reals.<|endoftext|> TITLE: Expectation of max of Gaussian multiplied by a functional of Gaussian QUESTION [5 upvotes]: Let $X \in \mathbb{R}^{d}$ follows the standard Gaussian distribution $N(0, I_d)$. Let $Y = \max_{j\in[d] } X_j$. It is not hard to see that \begin{align} \mathbb{E}\left [ Y \cdot X\right] = \sum_{j=1}^d \mathbb{P}\left( j = \arg\max_{i \in [d]} X_i \right) \cdot e_j, \end{align} where $e_i$ is the standard basis in $\mathbb{R}^d$. Now I was wondering how to compute $$\mathbb{E} \left[ Y\cdot (X X^\top - I_d) \right ].$$ Is there a closed form solution? REPLY [2 votes]: Let $n:=d$. Assume $n\ge3$. Let $F$ and $f$ denote the cdf and pdf of $N(0,1)$, and let $I\{\cdot\}$ denote the indicator function. Each of the diagonal entries of the matrix $EYXX^T$ equals \begin{equation*} EX_1^2\max_iX_i=E_1+E_2, \end{equation*} where \begin{equation*} E_1:=EX_1^3\,I\{X_1>\max_2^n X_i\},\quad E_2:=(n-1)EX_1^2 X_2\,I\{X_2>X_1\vee\max_3^n X_i\}. \end{equation*} Note that the cdf of the maximum of $k$ iid $N(0,1)$ random variables (r.v.'s) is $F^k$. So, \begin{multline*} E_1=\int_{-\infty}^\infty d(F(c)^{n-1})\,EX_1^3\,I\{X_1>c\} =\int_{-\infty}^\infty d(F(c)^{n-1})\,\int_c^\infty dx\,x^3f(x) \\ =\int_{-\infty}^\infty dc\, F(c)^{n-1}c^3f(c), \end{multline*} by integration by parts. Next, using the identities $$EX_2\,I\{X_2>u\}=f(u)$$ and $$d_c f(u\vee c)=-cf(c)I\{c>u\}\,dc$$ for real $u$, and integrating by parts (twice), we have \begin{multline*} E_2:=\int_{-\infty}^\infty (n-1)d(F(c)^{n-2})\,EX_1^2 X_2\,I\{X_2>X_1\vee c\} \\ =\int_{-\infty}^\infty (n-1)d(F(c)^{n-2})\,EX_1^2 f(X_1\vee c) \\ =\int_{-\infty}^\infty (n-1)dc\, F(c)^{n-2}\,cf(c) EX_1^2\,I\{X_1X_1\vee X_2\vee \max_4^n X_i\} \\ =\int_{-\infty}^\infty d(F(c)^{n-3})\,EX_1 X_2 X_3\,I\{X_3>X_1\vee X_2\vee c\} \\ =\int_{-\infty}^\infty d(F(c)^{n-3})\,EX_1 X_2 f(X_1\vee X_2\vee c) \\ =\int_{-\infty}^\infty dc\,F(c)^{n-3}f(c)c\,EX_1 X_2 I\{X_1\vee X_2X_1\vee \max_3^n X_i\} \\ =\int_{-\infty}^\infty d(F(c)^{n-2})\,EX_1 I\{X_1c\} \\ =-\int_{-\infty}^\infty d(F(c)^{n-2})\,Ef(X_2)X_2^2\,I\{X_2>c\} \\ =-\int_{-\infty}^\infty d(F(c)^{n-2})\,\int_c^\infty dx\, f(x)^2x^2 \\ =-\int_{-\infty}^\infty dc\,F(c)^{n-2}\, f(c)c^2 \\ =\frac1{n-1}\,\int_{-\infty}^\infty dc\, F(c)^{n-1}\,(2c-c^3)f(c). \end{multline*} So, each of the off-diagonal entries of the matrix $EY(XX^T-I)$ equals \begin{multline*} \text{off-diags}=EX_1X_2\max_iX_i=2s_1+(n-2)s_3= \frac2{n-1}\int_{-\infty}^\infty dc\,F(c)^{n-1}(c^3-3c)\,f(c) \\ =\frac2{(n-1)n}\,E(Y^3-3Y) \tag{3} \end{multline*} -- cf. $(1)$, $(2)$. Mathematica claims that the value of the latter integral is $-\frac{1}{4 \sqrt{\pi }}\approx -0.141$ for $n=2,3$, but it cannot evaluate it in closed form for $n=4$ (an approximate value of that integral for $n=4$ is $-0.0969$). Yet, it is now easy to compute the diagonal and off-diagonal entries of the matrix $EY(XX^T-I)$ numerically, and it is also easy to find their asymptotics for large $n(=d)$, using standard methods of asymptotic analysis. Indeed, let us consider the integral \begin{equation*} J_n:=\int_{-\infty}^\infty dc\,F(c)^{n-1}(c^3-3c)\,f(c) \end{equation*} in $(3)$ for large $n$. Write \begin{multline*} J_n=J_{1n}+J_{2n}+J_{3n},\\ J_{1n}:=\int_{-\infty}^{c_n}\dots,\ J_{2n}:=\int_{c_n}^{d_n}\dots,\ J_{3n}:=\int_{d_n}^\infty\dots, \end{multline*} where \begin{gather*} c_n:=F^{-1}(e^{-w_n/n}),\quad d_n:=F^{-1}(e^{-v_n/n}) \\ v_n:=n^{-t_n},\quad w_n:=n^{t_n},\quad t_n:=1/\sqrt{\ln n}. \end{gather*} Then \begin{gather*} 0c_n\to\infty,\\ 1-F(c_n)\sim-\ln F(c_n)=w_n/n=n^{-1+o(1)},\quad 1-F(c_n)=\exp\{-c_n^2/(2+o(1))\}. \end{gather*} Hence and from similar relations for $d_n$, we find that $c_n\sim\sqrt{2\ln n}\sim d_n$, and so, \begin{equation*} c\sim\sqrt{2\ln n}\text{ uniformly in $c\in[c_n,d_n]$}. \tag{4} \end{equation*} Next, for real $c TITLE: Is $\pi_{n-1}$ of a non-trivial connected sum of $n$-manifolds always non-zero? QUESTION [14 upvotes]: Let $X$ and $Y$ be two path-connected $n$-dimensional manifolds. Then one can construct their connected sum $X\# Y$ which can be visualised as removing a disc from $X$, removing a disc from $Y$, and gluing the boundaries together using a cylinder (a copy of $S^{n-1}\times[0,1]$). There is a natural $(n-1)$-sphere in $X\# Y$ due to the cylinder. If $X$ or $Y$ is a sphere, then this $(n-1)$-sphere will bound. Is this the only case in which it bounds? If $X\# Y$ is a non-trivial connected sum (i.e. neither $X$ nor $Y$ is a sphere), is the map $S^{n-1} \to X\# Y$ essential (i.e. not nullhomotopic)? Oscar Randal-Williams' answer to this question shows that the inclusions of the sphere into $X\setminus D^{\circ}$ and $Y\setminus D^{\circ}$ are essential. If the answer is yes, then $\pi_{n-1}(X\# Y) \neq 0$. I don't even know if this is true. If $X \# Y$ is a non-trivial connected sum of $n$-dimensional manifolds, is $\pi_{n-1}(X\# Y)$ necessarily non-zero? I believe the answer to both of these questions is yes, but I have not been able to completely convince myself. REPLY [15 votes]: The answer to this is no, in a stronger sense than Ryan's comment above. There are two sources, Kahn, Peter J. Codimension-one imbedded spheres (Invent. Math. 10 1970 44–56), and an appendix Null-homotopic, codimension-one embedded spheres, to a paper of Terng-Thorbergsson (Taut Immersions into Complete Riemannian Manifolds in Tight and taut submanifolds. MSRI Publications, 32, 1997) that I wrote, not knowing Kahn's result. The main theorem of the appendix states: Suppose that $\imath:S^{n-1} \to M^n$ is a smooth embedding which is null-homotopic. Then either $S$ bounds a homotopy ball on one side, or {1.} M is a rational homology sphere, and therefore X and Y are as well. {2.} The fundamental groups of both X and Y are finite, and at least one of them is trivial. In the appendix, there are constructions of connected sums of rational homology spheres where the summing sphere is null-homotopic that show that this result is essentially best possible.<|endoftext|> TITLE: Is it known whether or not $\aleph_\alpha=\beth_\alpha$ can be proven by ZFC? QUESTION [9 upvotes]: Given a cardinal number $\aleph_\alpha$, is it known whether or not $\aleph_\alpha=\beth_\alpha$ is independent of ZFC? One could define $\mathrm{CH}(\aleph_\alpha)$ as $\aleph_\alpha=\beth_\alpha$. For which cardinals is it true that $\mathrm{ZFC}\models\mathrm{CH(\kappa)}$? Clearly, $\mathrm{ZFC}\models\mathrm{CH}(\aleph_0)$, and $\mathrm{CH}(\aleph_1)$ is equivalent to $\mathrm{CH}$ (and is thus independent of ZFC). Because $\mathrm{GCH}$ is indepenedent of ZFC, $\neg\mathrm{CH}(\kappa)$ cannot be proven for any cardinal $\kappa$. If $\mathrm{ZFC}\models\mathrm{CH}(\aleph_{\alpha+1})$, then $\aleph_{\alpha+1}=\beth_{\alpha+1}$. Thus, if we assume $\aleph_\alpha<\beth_\alpha$, we get a contradiction, because $\aleph_\alpha<\beth_\alpha$ and then $\aleph_{\alpha+1}<\beth_{\alpha+1}$. So, if $\mathrm{CH}(\kappa^+)$ is provable, then $\mathrm{CH}(\kappa)$ is also provable. Thus, for cardinals $\kappa<\aleph_\omega$, $\mathrm{CH}(\kappa)$ is independent of ZFC. It is known that $\alpha\leq\aleph_\alpha$, and thus if $\beth_\kappa=\kappa$ then $\aleph_\kappa=\kappa$ (i.e. all $\beth$-fixed points are also $\aleph$-fixed points). Therefore, $\beth_\kappa=\aleph_\kappa=\kappa$, and $\beth_\kappa=\aleph_\kappa$. So, for all $\beth$-fixed points $\kappa$, $\mathrm{CH}(\kappa)$ is provable from ZFC. Are there any other known $\kappa$ with $\mathrm{CH}(\kappa)$ independent of ZFC? Is $\exists\kappa\neq\aleph_0(\mathrm{CH}(\kappa))$ independent of ZFC? (If so, $\beth$ fixed points are independent of ZFC as well.) Edit: Of course $\beth$-fixed points are not independent of ZFC, they do exist by $\alpha\mapsto\beth_\alpha$ being a normal function. So, ZFC actually proves the existence of cardinals larger than $\aleph_0$ which fulfill GCH. REPLY [8 votes]: Claim. For any $\alpha >0 \colon$ $\mathrm{ZFC} \not \vdash \mathrm{CH}(\aleph_{\alpha})$ Proof. Starting with $L$, we may enlarge the continuum $\beth_1$ arbitrarily without changing cardinals. In particular, we may fix a generic $g$ such that $L$ and $L[g]$ have the same cardinals (thus $\aleph_{\alpha}^L = \aleph_{\alpha}^{L[g]}$) but $\beth_1^{L[g]} > \aleph_{\alpha}^{L[g]}$.<|endoftext|> TITLE: Upper bound of the kissing number in n dimensions QUESTION [7 upvotes]: In geometry, a kissing number is defined as the number of non-overlapping unit spheres that can be arranged such that they each touch another given unit sphere. Let $\tau_n$ be the kissing number in $n$ dimensions. Kabatiansky and Levenshtein proved the following asymptotic upper bound (ppi1518, mr514023, 1978): $$\tau_n \le 2^{0.401n(1+o(1))} = (1.32\dots)^{n(1+o(1))}$$ Question: What is the smallest $\alpha$ such that $\tau_n \le \alpha^n$, for all $n$? ($\alpha := \min_{n \ge 1} \tau_n^{1/n}$) By using volume, we can prove that $\tau_n \le \frac{Vol(B(3))-Vol(B(1))}{Vol(B(1))}=3^n-1$, so $\alpha \le 3$. Now $\tau_2 = 6$, so $\alpha \ge \sqrt 6 \simeq 2.45$. Moreover, for $n \le 24$, $\tau_n^{1/n} \le \sqrt 6$. Is it true that $\alpha = \sqrt 6$? This post is motivated by arXiv:1710.00285, Section 5. REPLY [6 votes]: It’s almost certainly true, and provable, that $\alpha=\sqrt{6}$, although I haven’t worked out the details rigorously. Kabatiansky and Levenshtein give an exact upper bound (not just an asymptotic expression), which is equation (52) in their paper. Numerical calculations indicate that it improves on $\sqrt{6}$ in dimensions 8 and higher, while other kissing bounds cover dimensions 3 through 7 (see Chapter 1 of Conway and Sloane). So all one has to do is prove this inequality for the Kabatianski-Levenshtein bound in dimensions 8 and up. It suffices to prove an asymptotic bound with explicit constants to handle high dimensions, and then check any remaining dimensions not covered by that bound. I haven’t worked it out, but standard techniques should suffice to do this. (I.e., there’s nothing ineffective about the bounds, so one can obtain explicit constants.) This approach looks ugly and tedious enough that I haven’t worked up the energy to actually do it, but it should work in principle. The main technical input needed is explicit bounds for the largest roots of Gegenbauer polynomials.<|endoftext|> TITLE: What functions have the same persistence diagrams? QUESTION [6 upvotes]: The panels in the figure below show, from left to right: a piecewise affine function with support equal to a bounded interval and an indication of its superlevel filtration; the corresponding persistence diagram; a different function in the same spirit as the first, and with the same persistence diagram. This example shows that there is a many-to-one mapping between functions and persistence diagrams, even after accounting for an obvious parity symmetry. However, not every ordering of the blue line segments indicating lifetimes of components is consistent with a filtration of some function. While it's probably not too hard to sort this all out, it seems likely that somebody already has. So: has the equivalence class of [some class of] functions having the same persistence diagram been considered in the literature? 1D is fine for my considerations. REPLY [5 votes]: Your question is precisely the subject of Justin Curry's recent preprint. Bottom line: if you agree to identify functions $f,g:[0,1] \to \mathbb{R}$ whenever they have the same merge-tree, then there are only finitely many equivalence classes of functions which produce a given barcode $B$.<|endoftext|> TITLE: periodic cyclic homology and tilting in the sense of Scholze QUESTION [10 upvotes]: Suppose $R$ is a perfectoid ring in mixed characteristic, and $R'$ its characteristic-$p$ tilt. Scholze's results on tilting say that the étale theories over $R$ and $R'$ are equivalent in an almost sense (i.e. considering the category of extensions in the sense of "almost mathematics"). I want to know whether this is true of other rigid invariants. The question I'm interested in specifically is this: consider the categories $\operatorname{Mod}^a(R), \operatorname{Mod}^a(R')$ of almost modules over $R, R'$. Is it true that (making suitable definitions) the periodic cyclic homology of these two categories over $\mathbb{Z}$ is isomorphic, or related by a spectral sequence? More generally, I'm looking for interesting tilting statements visible on the level of comparing the categories $\operatorname{Mod}^a(R)$ and $\operatorname{Mod}^a(R')$ (possibly with monoidal structure). REPLY [8 votes]: As a complement to Peter's answer, let me try to address the precise question of whether the periodic cyclic homologies of the almost categories are equivalent. The answer is "no". First, what is the definition of the periodic cyclic homology of the almost category? This is not obvious, but Efimov has shown the way (though perhaps his work post-dates your question?). Let me start by recalling his theory, which he calls "continuous K-theory". For more details of what I'm writing you can see Marc Hoyois' nice exposition at http://www.mathematik.ur.de/hoyois/papers/efimov.pdf . First, any invariant of (idempotent-complete $\mathbb{Z}$-linear, but let me not carry these adjectives around) small stable $\infty$-categories, for example periodic cyclic homology, can equivalently be viewed as an invariant of compactly generated stable $\infty$-categories, by taking compact objects. If we want to take care of functoriality we should posit that we only consider those functors between compactly generated $\infty$-categories which preserve the compact objects; equivalently, and this will be better for what's coming next, those functors whose right adjoints also preserve colimits. Now, there is a substantial broadening of the notion of compactly generated table $\infty$-category due to Lurie, the notion of a dualizable stable $\infty$-category. Every dualizable stable $\infty$-category $\mathcal{C}$ is the kernel of a localization functor between compactly generated stable $\infty$-categories, where again the right adjoint to the localization functor is required to preserve all colimits; and actually this is a precise characterization of the dualizable stable $\infty$-categories. This leads to Efimov's fantastic result (previsaged to some extent by Tamme's work on excision, see https://arxiv.org/abs/1703.03331): if $E$ is a localizing invariant, then there is a unique functorial extension of $E$ to a localizing invariant of dualizable stable $\infty$-categories. This tells you what $E(\mathcal{C})$ has to be, if $\mathcal{C}$ is presented as the kernel of a localization functor between compactly generated guys as above: just the fiber of $E$ applied to that localization functor. The almost category provides a good example, because it more or less explicitly arises as the kernel of such a localization functor, namely $\operatorname{Mod}(\mathcal{O}_C)^a$ is the kernel of the base-change functor $\operatorname{Mod}(\mathcal{O}_C)\rightarrow \operatorname{Mod}(k)$ where $k$ is the residue field. The weird property of the residue field that makes this base-change functor a localization is that the derived tensor product of $k$ with itself over $\mathcal{O}_C$ is $k$ again. This is what makes almost mathematics run. Summing up, what this shows is that the perodic cyclic homology of the almost category is basically the same as the periodic cyclic homology of the usual category of modules. They only differ by the periodic cyclic homology of the residue field, which is anyway the same for $\mathcal{O}_C$ and its tilt. You also ask if something weaker could be true, that these invariants of $R$ and its tilt are "related by a spectral sequence". I guess there's a bit a freedom in interpreting what this means, but I suppose with appropriate definitions and interpretations the answer should be "yes". More precisely, if $\pi$ and $\pi^\flat$ are as in Peter's answer, then the associated gradeds for the $\pi$-adic and $\pi^\flat$ adic filtrations are isomorphic (provided, let's say, that $\pi$ and $\pi^\flat$ are non-zerodivisors). If we agree to replace periodic cyclic homology with its "continuous" variant (now not in the sense of Efimov but in the sense of the inverse limit of the periodic cyclic homologies of the quotients by the stages in the $\pi$-adic filtration... though this is also conjecturally a special case of the Efimov construction, applied to the so-called nuclear modules which Peter and I have defined), then it's easy to imagine that there should be a spectral sequence converging from the invariant for the associated graded to the invariant for the ring, so that indeed the two things may well be related by two spectral sequences with the same $E_2$-page.<|endoftext|> TITLE: Commuting hermitian matrices QUESTION [11 upvotes]: Let $A$, $B$ be $n\times n$ hermitian matrices. Denote by $(\alpha_i)$, $(\beta_i)$, $(\gamma_i)$ the eigenvalues of $A$, $B$, and $A+B$. Assume that there exists permutations $\sigma$, $\tau \in \frak S_n$ so that $$\gamma_i = \alpha_{\sigma(i)} + \beta_{\tau(i)}$$ for all $1\le i \le n$. Does this imply that $A$, $B$ commute? REPLY [8 votes]: This question was the topic of Commutativity and spectra of Hermitian matrices, by Wasin So (1994). The statement is negative in general: Richard Stanley's counterexample has size $n\geq 6$, a counterexample for $n=3$ is $$A=\begin{pmatrix}0&0&0\\ 0&6-\sqrt 2&2\\ 0&2&4+2\sqrt 2 \end{pmatrix},\;\;B=\begin{pmatrix} 4&0&0\\ 0&-4&0\\ 0&0&0 \end{pmatrix},$$ There are special cases when the statement holds: If $\gamma_i = \alpha_{\sigma(i)} + \beta_{\tau(i)}$ for a permutation $\sigma(i)$ and $\tau(i)$ of the eigenvalues $\alpha_i,\beta_i,\gamma_i$ in non-increasing order, then the $n\times n$ Hermitian matrices $A,B$ commute if either $n=2$ rank $A=1$ $\sigma(i)=i$ for all $i$ (identity permutation) $\sigma(i)=n+1-i$ for all $i$ (reverse permutation)<|endoftext|> TITLE: Is every real matrix conjugate to a semi antisymmetric matrix? QUESTION [5 upvotes]: Is it true to say that every matrix $A\in M_n(\mathbb{R})$ is similar (conjugate) to a matrix $B=(b_{ij})$ with $b_{ij}=-b_{ji}$ for all $i\neq j$?(With some abuse of terminology,a matrix $B$ with this property is called "Semi antisymmetric"). REPLY [12 votes]: Yes. Every matrix can be written as the sum of a symmetric plus an antisymmetric one: $A = \frac{A+A^T}{2}+\frac{A-A^T}{2}$. Now change basis such that the symmetric part is diagonal.<|endoftext|> TITLE: Holomorphic Sard's theorem? QUESTION [7 upvotes]: I have originally posted this question on math.SE, but it received little attention, so I repost it here. Let $U\subset \mathbb{C}^{n}$ and $V\subset \mathbb{C}^{m}$ be open and connected. Let $\Phi:U\to V$ be a holomorphic map. Is it true that there is a non-zero holomorphic function $w$ on a neighbourhood of $\Phi(U)$, which vanishes at every critical value of $\Phi$? If this is wrong, is there any other refinement of Sard's theorem for the holomorphic maps between complex domains? REPLY [8 votes]: No. We can enumerate $\mathbb{Q}[i]$ as $\{a_0,a_1,a_2,\dotsc\}$ and then choose a holomorphic function $f\colon\mathbb{C}\to\mathbb{C}$ with $f(n)=a_n$ and $f'(n)=0$ for all $n$. This follows from a well-known interpolation theorem; some references are discussed at Which sequences can be extended to analytic functions? (e. g., Ackermann's function), for example. Now the set of critical values is dense, so you probably cannot do much better than Sard's conclusion that it has measure zero.<|endoftext|> TITLE: Mathematical/Physical uses of $SO(8)$ and Spin(8) triality QUESTION [16 upvotes]: Triality is a relationship among three vector spaces. It describes those special features of the Dynkin diagram D4 and the associated Lie group Spin(8), the double cover of 8-dimensional rotation group SO(8). SO(8) is unique among the simple Lie groups in that its Dynkin diagram (below) (D4 under the Dynkin classification) possesses a three-fold symmetry. This gives rise to a surprising feature of Spin(8) known as triality. Related to this is the fact that the two spinor representations, as well as the fundamental vector representation, of Spin(8) are all eight-dimensional (for all other spin groups the spinor representation is either smaller or larger than the vector representation). The triality automorphism of Spin(8) lives in the outer automorphism group of Spin(8) which is isomorphic to the symmetric group $S_3$ that permutes these three representations. What are physics and math applications of $SO(8)$ and Spin(8) triality? For example, one of physics applications of $SO(8)$ and Spin(8) triality is that, in the classifications of interacting fermionic topological phases protected by global symmetries, the 1+1D BDI Time-Reversal invariant Topological Superconductor and 2+1D $Z_2$-Ising-symmetric Topological Superconductor have $\mathbb{Z}_8$ classifications (see a related post here), that can be deduced from adding non-trivial four-fermion interaction terms respect the $SO(8)$ and Spin(8) triality, see for example the Appendix A of this web version (free access). More precisely, the Spin(8) triality specifies the symmetry allowed interation terms for the following systems: a chiral $p_x+i p_y$ with a anti-chiral $p_x-i p_y$ superconductor. Combine a chiral $p_x+i p_y$ with a anti-chiral $p_x-i p_y$ superconductor, what we obtain is a Topological Superconductor respect to $Z_2$-Ising global symmetry as well as a $Z_2^f$-fermionic parity symmetry. So it is a 2+1D $Z_2 \times Z_2^f$-Topological Superconductor. It turns out that stacking from 1 to 8 layers of such $Z_2 \times Z_2^f$-Topological Superconductor ($p_x+i p_y/p_x-i p_y$), you can get 8 distinct classes (and at most 8, mod 8 classes) of TQFTs. They are labeled by $\nu \in \mathbb{Z}_8$ classes of 2+1D fermionic spin-TQFTs: This $\mathbb{Z}_8$ is related to the 3rd spin cobordism group $\Omega^{3,spin}_{tor}(B\mathbb{Z}_2,U(1))=\mathbb{Z}_8$. There may be other examples, other applications in physics and in mathematics(?). In particular, I have an impression that one can use $SO(8)$ and Spin(8) triality in string theory (but may not count as real-world physics) or possibly in disorder condensed matter system or nuclear energy spectrum. If that is true, could one explain how does the $SO(8)$ and Spin(8) triality come in there? In the previous example, I give, the concept of non-perturbative ('t Hooft) anomaly matching is implicit hidden there in the strongly-coupled higher order Majorana interactions. Do we see something similar or different concepts of the $SO(8)$ and Spin(8) triality applications? [Citations/References are encouraged, but some explanations are required.] p.s. This is the modified ellaborated version of a unfortunately closed question from Phys.SE. REPLY [3 votes]: If you have doubts about string theory ("may not count as real-world physics"), then I imagine M-theory may be a step too far for you, but in the recent Twisted Cohomotopy implies M-Theory anomaly cancellation the triality of Spin(8) is used extensively.<|endoftext|> TITLE: Are there functions which are neither convex nor concave everywhere but are continuous? QUESTION [5 upvotes]: By convex/cave I mean by the definition for an interval $(x,y)$ of $f$ is convex iff $f(\frac{a+b}{2})\geq\frac{f(a)+f(b)}{2}$ and is concave if $f(\frac{a+b}{2})\leq\frac{f(a)+f(b)}{2}$ where $a,b\in(x,y)$. This is merely so that the function does not have to be differentiable. If it's not possible to have a function be continuous but not convex or concave on any nonempty interval, is it possible to construct a function which is not convex or concave on any nonempty interval? Has one already been created? If it is possible to have a function be continuous but not convex or concave on any nonempty interval, is it possible to construct it? Has one already been created? Is it possible to have one of these functions be differentiable once? REPLY [12 votes]: It is well known that almost every path of a Brownian motion is nowhere monotone (i.e., not monotone on any interval). Hence the primitive (antiderivative) of almost every path is continuously differentiable but nowhere convex or concave.<|endoftext|> TITLE: Automorphic quotients for inner forms or $GSp(4)$ QUESTION [12 upvotes]: For a quaternion algebra $D$, introduce the quaternionic similitude unitary groups: \begin{equation} \mathrm{GU}_D = \left\{ g \in \mathrm{GL}(D) \ : \ g^\star \left( \begin{array}{cc} & 1 \\ 1 & \end{array} \right) g = \mu(g) \left( \begin{array}{cc} & 1 \\ 1 & \end{array} \right), \mu(g) \in \mathbf{G}_m \right\} \end{equation} The $\mathrm{GU}_D $ are the inner forms of $\mathrm{GSp(4)}$ when $D$ describes the quaternion algebras. For almost every places $v$ of $F$, more precisely the split ones, this group is $G_v \simeq \mathrm{GSp(4,F_v)}$. When are those $\mathrm{GU}_D$ of compact automorphic quotient? I believe assuming $D$ to be totally definite at infinity (i.e.ramified at archimedean places) is enough, but is it necessary? REPLY [8 votes]: Let me give you the inner forms of $\mathrm{GSp}(4)$ with compact adelic quotient. Every nonsplit inner form of $\mathrm{GSp}(4)$ is obtained in the following way: take $D$ a division quaternion algebra over $F$, let $V$ be a $D$-Hermitian space of $D$-dimension $2$, and construct $G = \mathrm{GU}(V)$. By a theorem of Borel and Harish-Chandra, the adelic quotient of $G$ is compact if and only if $G$ modulo its center is anisotropic (in the algebraic group sense: it contains no nontrivial split torus). This turns out to be equivalent to $V$ being anisotropic (in the quadratic/hermitian form sense: it has no nonzero isotropic vector). Over a number field, by the local-global principle for quadratic forms we can test anisotropy locally: over $p$-adic places, $V$ has $8$ variables as a quadratic form and is therefore isotropic, and over real places, $V$ is anisotropic if and only if it is positive definite or negative definite. In summary, $G = \mathrm{GU}(V)$ has compact adelic quotient if and only if $F$ admits a real place at which $D$ is definite and $V$ is positive definite or negative definite. For instance, the $D$-Hermitian form $x\bar{x}+y\bar{y}$ works over a totally definite quaternion algebra, but the one you wrote down is $x\bar{y}+y\bar{x}$.<|endoftext|> TITLE: Endless controversy about the correctness of significant papers QUESTION [156 upvotes]: In principle, a mathematical paper should be complete and correct. New statements should be supported by appropriate proofs. But this is only theory. Because we often cannot enter into the smallest details, we "prove" wrong statements here and then. I plead guilty, having published myself one or two false (fortunately minor) papers. So far, this is not harmful. The research community is able to point out incorrect statements, at least among those which have some importance in the development of mathematics. In time, the errors are fixed; this is the role of monographs to present a universally accepted state of the art of a topic. But sometimes, hopefully rarely, the technicalities are such that a consensus does not emerge and a controversy raises, between the author and their critics. I have an example in the realm of wave stability in PDE models for fluid dynamics. The controversy has lasted for a decade or two and I don't see how it can be resolved some day; it could just kill the topic. Are there famous endless controversies about the correctness of a significant paper? Are there some significant mathematical questions, that remain unsettled because people disagree on the status of released proofs? What should we do in order to salvage mathematical topics that suffer such tensions? In this question, I am not concerned with other kinds of controversy, about priority or citations. REPLY [40 votes]: Stanley Yao Xiao's comment has been upvoted so highly that it seems worth posting as an answer. There is a currently unresolved controversy over Shinichi Mochizuki's claimed proof of the abc conjecture. In a ten-page note, Peter Scholze and Jakob Stix have stated: We, the authors of this note, came to the conclusion that there is no proof. We are going to explain where, in our opinion, the suggested proof has a problem, a problem so severe that in our opinion small modifications will not rescue the proof strategy. Mochizuki, however, maintains that there are no problems with his proof and that Scholze and Stix suffer from "fundamental misunderstandings." UPDATE: Mochizuki's work has been accepted for publication in PRIMS, even though most number theorists believe that the proof is incomplete. For more details, see this April 2020 post on Peter Woit's blog. FURTHER UPDATE: Mochizuki's papers were published in PRIMS in March 2021. However, the publication of the papers has not ended the controversy; see the March 2021 post on Peter Woit's blog.<|endoftext|> TITLE: Periodic function $f$ for which $f(x^2)$ is periodic too QUESTION [10 upvotes]: There is the following question which was asked multiple times on Math.SE (e.g. here and here) without any final result: Question: Is there a periodic function $f:\Bbb R \to\Bbb R$ of smallest period $1$, for which $g(x):=f(x^2)$ is periodic too (of period $\alpha$)? We know that no such function can be continuous. The best approach so far was to define an equivalence relation $\sim$ on $\Bbb R$ via $$x\sim x\pm 1\qquad\text{and}\qquad x^2\sim (x+\alpha)^2.$$ The function $f$ defined by $$f=\begin{cases} 0 &\text{for } x\sim0 \\ 1 &\text{for } x\not\sim0 \end{cases}$$ is then periodic (of period $1$) and $g(x):=f(x^2)$ is periodic too (of period $\alpha$). But how to prove (if true) that $1$ is the smallest period of $f$? Another observation is that any other possible period $\lambda\in(0,1)$ of $f$ must satisfy $\lambda\sim 0$ since $$f(\lambda)=f(0)=0\quad\Rightarrow\quad \lambda\sim 0.$$ REPLY [7 votes]: Here is a proof that Ilya's example works. Fix a positive real transcendental $\alpha$ and define the functions $f_n:x\mapsto x+n$ and $g_n:x\mapsto(\sqrt{x}+n\alpha)^2$ for integer $n$, which generate the equivalence relation $\sim$. Fix a real $r$ (Ilya's $a$) that is algebraically independent of $\alpha$ over $\mathbb{Q}$. Suppose $r+\beta\sim r$ for some $\beta\in\overline{\mathbb{Q}(\alpha)}$. Then there's a witness sequence of (WLOG nonzero) integers $n_i$ and $m_j$ such that $(f_{n_1}\circ g_{m_1}\circ\cdots f_{n_k}\circ g_{m_k}\circ f_{n_{k+1}})(r)=r+\beta$. By the algebraic independence of $r$, this equation holds identically for all real $r$. Therefore it remains true after taking the derivative with respect to $r$ and setting $r$ to be a rational number. It then becomes $(g'_{m_1}\circ\cdots f_{n_k}\circ g_{m_k}\circ f_{n_{k+1}})(r)\cdot (g'_{m_2}\circ\cdots f_{n_k}\circ g_{m_k}\circ f_{n_{k+1}})(r)\cdots(g'_{m_k}\circ f_{n_{k+1}})(r)=1$, since $f'_n=1$. This can be written as an algebraic equation with rational coefficients solved by $\alpha$, so it is true regardless of $\alpha$'s value. As it turns out, if $k>0$ then the left hand side diverges as $\alpha\to\infty$. To wit, for fixed $r$ and as $\alpha\to\infty$, $f_{n_{k+1}}(r)$ stays constant, while $(f_{n_k}\circ g_{m_k}\circ f_{n_{k+1}})(r)=\Theta(\alpha^2)$, $(f_{n_{k-1}}\circ g_{m_{k-1}}\circ f_{n_k}\circ g_{m_k}\circ f_{n_{k+1}})(r)=\Theta(\alpha^2)$, and so on. Therefore $(g'_{m_k}\circ f_{n_{k+1}})(r)=\Theta(\alpha)$ yet every earlier term in the chain rule product (e.g. $(g'_{m_{k-1}}\circ f_{n_k}\circ g_{m_k}\circ f_{n_{k+1}})(r)$) is $\Theta(1)$. Hence there are no $g$ terms in the witness of $r+\beta\sim r$, which means that $\beta\in\mathbb{Z}$.<|endoftext|> TITLE: Conjecture: Finitely many points where gravitational field due to N masses vanishes QUESTION [13 upvotes]: Given a configuration $C$ of $N$ distinct fixed points of equal mass in the plane (eventually in space), let $f_C(N)$ denote the number of points $P$ for which the gravitational field at $P$ vanishes. The gravitational force is Newtonian, i.e $1/r^2$. For example $f_C(2)=1$ for all $C$ and for a configuration $C$ of $3$ unequally spaced collinear points, $f_C(3)=2$. Conjecture: $f_C(N)$ is always finite and nonzero. Is this problem known? Can we establish an upper bound on $f_C(N)$? REPLY [14 votes]: Since you are talking about gravitation (rather than electrostatics) I assume that all charges are positive. (With charges of different sign it is easy to arrange a whole curve of equilibrium points). It is certainly non-zero if the number of points is at least 2. (From very general consideration it must be at least $N-1$ if you count multiplicitis properly. Without counting multiplicities it can be $1$ for any $N$). But an exact upper estimate is not known. J. C. Maxwell conjectured that is is at most $(N-1)^2$, but this is wide open. Even finiteness is not known. I know only computer-assisted proof for $N=3$ but I have not checked it. If one assumes finiteness, then there are upper estimates but they are much worse than expected. Reference: http://www.math.purdue.edu/~eremenko/dvi/equil2.pdf<|endoftext|> TITLE: Examples of representations of quantum groups QUESTION [15 upvotes]: I am trying to learn some basic theory of quantum groups $U_q(\mathfrak{g})$, where $\mathfrak{g}$ is a simple Lie algebra, say $sl_n(\mathbb{C})$. As far as I heard the finite dimensional representation theory of them is well understood. I would be interested to see examples of representations of $U_q(\mathfrak{g})$ which come not from the problem of classification of representations, but rather are either "natural" or typical from the point of view of quantum groups, or appear in applications or situations unrelated to the classification problem. To illustrate what I mean let me give examples of each kind of represenations of Lie algebras since this situation is more familiar to me. 1) Let $\mathfrak{g}$ be a classical complex Lie algebra such as $sl_n,so(n), sp(2n)$. Then one has the standard representation of it, its dual, and tensor products of arbitrary tensor powers of them. 2) Example of very different nature comes from complex geometry. The complex Lie algebra $sl_2$ acts on the cohomology of any compact Kahler manifold (hard Lefschetz theorem). Analogously $so(5)$ acts on the cohomology of any compact hyperKahler manifold (this was shown by M. Verbitsky). REPLY [3 votes]: The noncommutative complex geometry of the quantum flag manifolds provides a rich family of examples of representations of quantized enveloping algebras. For example, there are a number of papers on $q$-deformations of the Borel-Weil theorem, which realises Lie algebra representations on the holomorphic sections of line bundles. One series of papers starts with Majid's paper on the noncommutative spin geometry of the Podles sphere (quantum $\mathbb{CP}^1$, where Borel-Weil is discussed in the final section. A second paper on Borel-Weil for the Podles sphere was later written by Khalkhali, Landi, and van Suijlekom. Moreover, this result was then extended to all quantum projective space by Khalkhali and Moatadelro here. Finally, the extension to all the quantum Grassmannians is given in this paper. A google search will give other approaches to this question. The representation of $\frak{sl}_2$ on the de Rham complex of a Kahler manifold also has a direct noncommutative generalisation. What one needs is a differential calculus endowed with a noncommutative Kahler structure. (This is basically a noncommutative complex structure on the calculus together with a $(1,1)$-form whose Lefschetz map $L$ behaves like a classical Lefschetz map.) The map $L$ admits an adjoint $\Lambda$, the dual Lefschetz map, which together with a counting operator give a representation of $U_q(\frak{sl}_2)$. There is some freedom in the choice of the adjoint, in fact a $q$-parameter, which corresponds to the $q$-parameter in $U_q(\frak{sl}_2)$. This works in quite some generality, but the motivating examples are again the (cominiscule) quantum flag manifolds endowed with their Heckenberger-Kolb calculus. This representation extends to calculi twisted by noncommutative vector bundles as presented here. In the quantum homogeneous space case the calculi admit a natural $L^2$-completion to a Hilbert space, to which the $U_q(\frak{sl}_2)$ representation can be extended to a representation by bounded operators. The representation of the (2,2)-SUSY algebra, as explained in Huybrechts, does not seem to directly $q$-deform, something more subtle happens which is still not well-understood. An analogous formulation of noncommutative hypercomplex structures is most likely possible, but should probably wait until we actually have some examples :)<|endoftext|> TITLE: If a polynomial ring is finite free over a subring, is the subring polynomial? QUESTION [21 upvotes]: Let $R = k[x_1, \ldots, x_n]$ for $k$ a field of characteristic zero and let $S \subset R$ be a graded sub-$k$-algebra (for the standard grading: $\deg x_i = 1$) such that $R$ is a free $S$-module of finite rank. Does this imply $S \cong k[y_1,\ldots,y_n]$? REPLY [8 votes]: It turns out that this result appears in Bourbaki, Groupes et algebres de Lie IV-VI, Ch. 5, $\S$ 5, Lemme 1, with an elementary proof. (However, this proof uses characteristic zero, which Neil's answer does not.) A similar proof of a different theorem occurs as Section 3.5 in Humphrey's Reflection Groups and Coxeter Groups. Warning for those who read the originals: their $R$ and $S$ are reversed from mine! Notation: Let $R_+$ and $S_+$ be the maximal graded ideals of $R_+$ and $S_+$. We will use the following obvious property of free modules: If $M$ is a free $S$-module, and $x \in M$ does not lie in $S_+ M$, then there is a graded $S$-linear map $\lambda: M \to S$ such that $\lambda(x) \not\in S_+$. If $x$ is assumed homogenous, this means that $\lambda(x)$ is a nonzero scalar. Let $y_1$, ..., $y_p$ be homogenous elements of $S_+$ which are minimal generators for the $R$-ideal $S_+ R$. We first claim that the $y_i$ generate $S$ as a $k$-algebra. To this end, let $\lambda_0 : R \to S$ be a graded $S$-linear map with $\lambda_0(1) = 1$, which exists since $R$ is a free $S$-module. We prove by induction on $i$ that $S_i \subset k[y_1, \ldots, y_p]$. For $i=0$, this holds because $S_0=k$. Let us assume the result proved in all degrees $ TITLE: 6-manifolds admitting SO(3) action with 2 orbit types QUESTION [9 upvotes]: Let $M^6$ be a 6-dimensional smooth manifold, on which the group $G=SO(3)$ acts smoothly with 2 orbit types $SO(3)/SO(2)$ and $SO(3)$, such that the orbit space $X=M/SO(3)$ is a 3-ball $B^3$, whose boundary 2-sphere corresponds to the singular isotropy group (stabilizer) $K=SO(2)$ and interior corresponds to the principal isotropy group $H=id$. The problem is to classify all such 6 manifolds admitting such actions. In Bredon's book "Introduction to Compact Tranformation Groups", he has a way to classify group actions with 2 orbit types, which was initially due to Janich. According to Corollary 6.2 on page 257, Chapter V.6 of the book, the class of such 6-manifolds is in bijection with the set $[S^2,N(H)/(N(H)\cap N(K))]/\pi_0(N(H)/H)=\pi_2(N(H)/(N(H)\cap N(K)))/\pi_0(N(H)/H)=\pi_2(\mathbb{RP^2})=\mathbb{Z}.$ Here $K=SO(2),\ H=id$ are the isotropy groups (stabilizers) given in the first paragraph, and $N(H)=id,\ N(K)=O(2)$ are their normalizers in $G=SO(3)$. Thus such actions are parametrized by integers. But Bredon's construction in the book is rather involved. I'm aiming at simple explicit description of this family of $SO(3)$-actions. Moreover, I'm wondering which of those manifolds are simply connected. I'm mainly interested in simply connected examples. Two more remarks: Remark 1. If we ask the orbit space to be a 2-disk instead of 3-ball, then in Bredon's book he gave explicit classification of such $G$-spaces. More precisely, all $O(n)$-spaces with isotropy groups $O(n-1)$ or $O(n-2)$ and orbit space $D^2$ are given by $\Sigma_k^{2n-1}=S^{n-1}\times D^n \cup_{\varphi^k} S^{n-1}\times D^n$, where $\varphi$ is an $O(n)$-equivariant gluing map. For details see Chapter I.7 and V.6 of Bredon's book "Introduction to Compact Tranformation Groups". Those spaces have alternative description: the Brieskorn varieties $B^{2n-1}_k=\{z_0^k+z_1^2+\cdots z_n^2=0\}\cap S^{2n+1}$, see Chapter V.9 of the book. My goal is to find such explicit description in the case when the orbit space is a 3-ball and the group $G=SO(3)$. Remark 2. I have one example of such actions. Consider the 4-dimensional complex representation $\mathbb{C}^2\oplus \mathbb{C}^2$ of $SU(2)$, take its projectivization, then we get a linear $SU(2)$-action on $\mathbb{CP}^3$, which is ineffective (not faithful) and descends to an effective $SO(3)$-action. This action satisfies the requirement. And my intuition tells me that this action corresponds to the integer parameter 0. But I don't know how to show it. I'd like to point out that this question is related to another question I asked: SO(3) action on (simply connected) 6 manifold with discrete fixed point. They are both about $SO(3)$-actions on 6-manifolds, but under different conditions. Update: I suspect they are $S^2$ bundles over $S^4$, as I construct such actions on $\mathbb{CP}^3$ and $S^2\times S^4$. REPLY [4 votes]: Disclaimer: I have not checked Bredon's book. I am trying to describe $M$ as explicitly as I can, but maybe I am simply reproducing Bredon's description. Let $N$ denote the union of singular orbits. From your description, $N$ is an $S^2$-bundle over $S^2$. Let $\nu\to N$ denote the normal bundle in $M$. The group $SO(3)$ acts simply transitively on the fibres of the unit normal bundle, which therefore forms an $SO(3)$ principal bundle $P\to S^2$. On the other hand, the regular orbits form an $SO(3)$-principal bundle over the open ball, which is necessarily trivial. Because both are glued together, we conclude that $P$ is trivial, too. We thus have to glue two trivial $SO(3)$-principal bundles over $S^2$. Two different gluing maps differ by a map from $S^2$ to $SO(3)$. Because $\pi_2(SO(3))=\pi_2(\mathbb RP^3)=0$, there is only one choice. The only way to produce different manifolds comes from the fact that $SO(3)$ contains an $\mathbb RP^2$ worth of one-parameter subgroups, which are all isomorphic to $SO(2)$. The reason is that each unit vector $X$ in the Lie algebra $\mathfrak{so}(3)$ generates one of these groups, and $\pm X$ generate the same subgroup. Hence let $f\colon S^2\to\{\,H\subset SO(3)\mid H$ is a subgroup, $H\cong SO(2)\,\}\cong\mathbb RP^2$ be a map, then we construct the unit disc bundle in the normal bundle over $N$ by dividing the group $f(p)$ out of $P_p=SO(3)$ for each $p\in S^2$. Because $\pi_2(\mathbb RP^2)\cong\mathbb Z$, we get $\mathbb Z$-many homotopy classes of such quotients. We should still show that different homotopy classes lead to different manifolds. The following construction seems to be equivalent and maybe easier to digest. It should give at least half of the manifolds. There are only two $S^2$-bundles over $S^2$ because $\pi_2(BSO(3))\cong\mathbb Z/2$. So I assume that $N=S^2\times S^2\to S^2$ is the trivial bundle. The normal bundle $\nu\to N$ in $M$ can be regarded as a complex line bundle over $N$. It is determined by $c_1(\nu)\in H^2(N)\cong\mathbb Z^2$. Over each fibre $S^2$, it has degree $2$. On the other hand, its degree $d$ over the base $S^2$ has still to be chosen. Again, this would give rise to $\mathbb Z$-many manifolds. Here, any two of them differ as $SO(3)$-manifolds because $d$ can be read off from the way that $SO(3)$ acts. Let $D\nu\to N$ denote its unit disk bundle. We should check that $\partial D\nu$ is a trivial $SO(3)$-principal bundle, so that we can glue it to $D^3\times SO(3)$. It obviously is a bundle with fibre $\mathbb R P^3$, and therefore classified by a map $S^2\to BSO(4)$. But $\pi_2(BSO(4))=\mathbb Z/2$, so for even $d$ we obtain a trivial bundle. Finally, all of these manifolds are simply connected by Seifert-van Kampen. The normal disc bundle $D\nu$ has an $S^2$-bundle over $S^2$ as a deformation retract, which is simply connected. The fundamental group of $D^3\times SO(3)$ comes from $S^2\times SO(3)$, so it vanishes under gluing.<|endoftext|> TITLE: Is $\Gamma(p) := \text{Ker}(SL_2(\mathbb{Z}_p)\rightarrow SL_2(\mathbb{F}_p)$ a "standard" subgroup? QUESTION [7 upvotes]: Let $\Gamma(p) := \text{ker}(SL_2(\mathbb{Z}_p)\rightarrow SL_2(\mathbb{Z}_p/p))$. Viewing $SL_2(\mathbb{Z}_p)$ as an analytic group, is there a formal group law $F$ in three variables, defined over $\mathbb{Z}_p$, such that $\Gamma(p)$ is isomorphic to the group on the set $(p\mathbb{Z}_p)^3$ whose group operation is given by $F$? Here, "standard" is the terminology used in Serre's lecture notes on Lie Algebras and Lie Groups. REPLY [8 votes]: Yes. Let $g = \begin{pmatrix}1+a & b\\ c & 1+d\end{pmatrix} \in \Gamma(p)$ with $a,b,c,d\in p\mathbb{Z}_p$. Then the condition $\det g = 1$ can be rewritten $(1+a)(1+d) = bc+1$, so that $d = (1+a)^{-1}(bc+1) -1$, which is a power series in $a,b,c$ with zero constant term. Then you can turn the group law and the inverse expressed as polynomials in $a,b,c,d$ into power series in $a,b,c$.<|endoftext|> TITLE: Commutation of homotopy groups with filtered colimits QUESTION [6 upvotes]: Let $\mathcal{C}$ be a model category with a forgetful functor towards simplicial sets, and such that fibrations and trivial cofibrations are those whose underlying map on simplicial sets is a fibration, respectively trivial fibration. Is there a criterion to decide whether formation of simplicial homotopy groups commutes with filtered colimits? REPLY [5 votes]: I'm not aware of any such criterion (and I'm doubtful a very general one exists, if any), so this is not meant to be an answer (but for some reason I'm unable to post it as a comment right now). A few remarks. The question is slightly ill posed. What do you mean by "commutation of simplicial homotopy groups with filtered colimits"? You probably mean formation of simplicial homotopy groups turns, in degree zero, filtered colimits in $\mathcal{C}$ into filtered colimits of pointed sets, and in positive degree it turns filtered colimits in $\mathcal{C}$ into filtered colimits of groups. It seems to me a necessary condition on $\mathcal{C}$ is that it is proper (see Goerss-Jardine, Ch. II, $\S$8). Indeed, if formation of higher simplicial homotopy groups commutes with filtered colimits in the above sense, then in particular pushouts along cofibrations preserve weak equivalences. In fact, a sharper necessary condition is that weak equivalences in $\mathcal{C}$ are closed under filtered colimits. This should be seen as a strong "finiteness condition" on $\mathcal{C}$: it is typically met if generating cofibrations and generating trivial cofibrations are small relative to the whole category $\mathcal{C}$ (which is stronger than requiring $\mathcal{C}$ to be "finitely generated"). You may then ask whether it is the case that in a model category satisfying your stated assumptions, and in which, in addition, filtered colimits preserve weak equivalences, formation of higher simplicial homotopy groups commutes with filtered colimits in the above sense. I don't have a counterexample off the top of my head, but I'm led to believe this is not the case in general.<|endoftext|> TITLE: Continuum-many independent vectors over Q in R as a Q-vector space QUESTION [9 upvotes]: Let's put ourselves in the framework of ZF. Is it true that if we think of the set of real numbers as a rational vector space, there are continuum-many linearly independent vectors? I feel that we could then use this to make an injection from $\mathbb{R}$ to $\mathbb{R}/\mathbb{Q}$, the quotient vector space by the 1-dimensional subspace $\mathbb{Q}$. Does this strategy work? This is to help me think about the result that (in ZF) $|\mathbb{R}| \leq |\mathbb{R}/\mathbb{Q}|$, proved by Mycielski in 1964. I haven't looked at his paper, just trying to imagine how one would prove this result. REPLY [15 votes]: Let $f: 2^{\mathbb{N}} \rightarrow \mathbb{R}/\mathbb{Q}$ be the function given by $$ f((a_i)_{i \in \mathbb{N}}) = \text{the equivalence class of }\sum_{k=0}^{\infty} \frac{b_k}{2^{(k+1)!}}$$ where $(b_i)_{i \in \mathbb{N}}=(a_0,a_0,a_1,a_0,a_1,a_2,\dots)$. Notice that a real number of the form $$\sum_{k=0}^{\infty} \frac{c_k}{2^{(k+1)!}}$$ where $(c_i)_{i \in \mathbb{N}} \in 2^{\mathbb{N}}$ is rational if and only if the sequence $(c_k)$ is eventually zero. Consequently, $f((a_i)_{i \in \mathbb{N}})=f((a'_i)_{i \in \mathbb{N}})$ if and only if the corresponding sequences $(b_i)_{i \in \mathbb{N}}$ and $(b'_i)_{i \in \mathbb{N}}$ are eventually equal if and only if $(a_i)_{i \in \mathbb{N}}=(a'_i)_{i \in \mathbb{N}}$. Finally, compose this function with your favorite explicit injection from $\mathbb{R}$ to $2^{\mathbb{N}}$. This gives you an injection from $\mathbb{R}$ to $\mathbb{R}/\mathbb{Q}$. If you want to find a $\mathbb{Q}$-linearly independent subset of $\mathbb{R}$ of size continuum, consider the image of the map $g: \mathcal{A} \rightarrow \mathbb{R}$ given by $$ g(S)=\displaystyle \sum_{k=0}^{\infty} \frac{\chi_{S}(k)}{2^{(k+1)!}}$$ where $\mathcal{A} \subseteq \mathcal{P}(\mathbb{N})$ is an almost disjoint family of size continuum. For example, enumerate the vertices of the full binary tree of height $\omega$ by $\mathbb{N}$ and let $\mathcal{A}$ be the set of (labels of) branches. To see why this set is linearly independent, see this nice answer of Tim Gowers on another MO question. Note that both of these constructions can be done in ZF.<|endoftext|> TITLE: Universal covering of a 2-sphere without $n$ points QUESTION [12 upvotes]: Let $X$ be the $\mathbb{C}\mathbb{P}^1$ with $n$ points deleted. Let $n\geq 3$. If I understand correctly, the universal covering of $X$ is isomorphic to the upper half plane as a complex analytic space. Q. How one can describe the group of deck transformations of the universal covering as a subgroup of $\mathrm{PGL}(2,\mathbb{R})$? I expect this should be a standard material. A reference would be helpful. REPLY [7 votes]: I think this is part of the classical theory of Fuchsian groups. For instance, in the case $n=3$ the fundamental group of the thrice punctured sphere can be explicitly identified with the congruence subgroup of level two $\Gamma(2) \subset \mathrm{PSL}(2, \, \mathbb{Z})$, see Theorem 2.34 in the book E. Girondo, G. González-Diez: Introduction to compact Riemann surfaces and dessins d’enfants, London Mathematical Society Student Texts 79 (2012). ZBL1253.30001. REPLY [7 votes]: Yes, it is the standard material. The group is a discrete group of fractional-linear transformations of the upper half-plane, freely generated by $n-1$ parabolic transformations. There are many books discussing this, for example, L. R. Ford, Automorphic functions. Finding the group generators from the given points is a difficult problem: it is essentially an existence theorem, and the dependence of the generators on the points is complicated and highly transcendental. There are many detailed studies of small cases, for example the first non-trivial case $n=4$ and all points real. This is called the "problem of accessory parameters" for the Heun equation. You may look at the work of Zograf and Takhtajan on this. EDIT. After having finished my answer I noticed the answer by Francesco Polizzi, and I have to apologize for implying that the case $n=3$ is "trivial". It depends on the point of view. The problem of accessory parameters is trivial (there are none). So we have a single group and a single function in this case. Still this important group and this function are not trivial and they are subject of research.<|endoftext|> TITLE: The universal property of the unseparated derived category QUESTION [9 upvotes]: In Appendix C of his book in progress Spectral Algebraic Geometry, Lurie defines the unseparated derived category $\check{{\cal D}}({\cal A})$ (see Definition C.5.8.2 loc.cit) associated to a Grothendieck abelian category. The unseparated derived category $\check{{\cal D}}({\cal A})$ is a stable presentable $\infty$-category equipped with a natural t-structure which is compatible with filtered colimits and whose heart is identified with ${\cal A}$. It is related to the usual derived $\infty$-category ${\cal D}({\cal A})$ by a t-exact functor $\check{{\cal D}}({\cal A}) \to {\cal D}({\cal A})$ which exhibits ${\cal D}({\cal A})$ as the "left separation" of $\check{{\cal D}}({\cal A})$. In addition, $\check{{\cal D}}({\cal A})$ itself enjoys the following universal property (see Theorem C.5.8.8 and Corollary C.5.8.9 of loc.cit): if ${\cal C}$ is any other stable presentable $\infty$-category equipped with a $t$-structure which is compatible with filtered colimits, then restriction along the inclusion of the heart $A \to \check{{\cal D}}({\cal A})$ induces an equivalence $$ {\rm LFun^{{\rm t-ex}}}(\check{{\cal D}}({\cal A}),{\cal C}) \stackrel{\simeq}{\to} {\rm LFun^{{\rm ex}}}({\cal A},{\cal C}^{\heartsuit}) $$ where on the left hand side we have colimit preserving t-exact functors $\check{{\cal D}}({\cal A}) \to {\cal C}$ and on the right hand side we have colimit preserving exact functors ${\cal A} \to {\cal C}^{\heartsuit}$. This is indeed a very satisfying universal characterization of $\check{{\cal D}}({\cal A})$ together with its t-structure. However, for various reasons it can be useful to have a universal characterization of $\check{{\cal D}}({\cal A})$ without the t-structure. To see how this might go, note that t-exact colimit preserving functors $\check{{\cal D}}({\cal A}) \to {\cal C}$ send short exact sequences in ${\cal A}$ to cofiber sequences in ${\cal C}$ and filtered colimits in ${\cal A}$ to filtered colimits in ${\cal C}$. One may hence consider the possibility that restriction along ${\cal A} \to \check{{\cal D}}({\cal A})$ induces an equivalences between all colimit preserving functors $\check{{\cal D}}({\cal A}) \to {\cal C}$ on the one hand and and all functors ${\cal A} \to {\cal C}$ which preserve filtered colimits and send short exact sequences in ${\cal A}$ to cofiber sequences in ${\cal C}$ on the other. Question 1: Is this true? A positive answer to Question 1 would imply that $\check{{\cal D}}({\cal A})$ admits the following explicit description: we would be able to identify it with the $\infty$-category of presheaves of spectra ${\cal A}^{{\rm op}} \to {\rm Sp}$ which send filtered colimits in ${\cal A}$ to filtered limits and send short exact sequences in ${\cal A}$ to fiber sequences of spectra. As a variant to Question 1, one may hope that the connective part $\check{{\cal D}}({\cal A})_{\geq 0}$ enjoys the same universal characterization when ${\cal C}$ is now replaced with a Grothendieck prestable $\infty$-category, or maybe even any presentable $\infty$-category. Such a characterization would imply that $\check{{\cal D}}({\cal A})_{\geq 0}$ can be identified with the $\infty$-category of presheaces of spaces ${\cal A}^{op} \to {\cal S}$ which send filtered colimits to cofiltered limits and short exact sequences to fiber sequences of spaces. Question 2: Is this true? Remarks: 1) A positive answer to Question 2 would imply a positive answer to Question 1 since $\check{\cal D}({\cal A}) \simeq {\rm Sp}(\check{\cal D}({\cal A})_{\geq 0})$ is the stabilization of $\check{\cal D}({\cal A})_{\geq 0}$. 2) If ${\cal A} = {\rm Ind}(A_0)$ with $A_0$ an abelian category with enough projectives in which every object has finite projective dimension then $\check{\cal D}({\cal A}) \simeq {\cal D}({\cal A})$ (Proposition C.5.8.12 in loc.cit) and can also be described using complexes of projective objects. In this case $\check{\cal D}({\cal A})_{\geq 0} \simeq {\cal P}_{\Sigma}((A_0)_{{\rm proj}})$ is the $\infty$-category obtained from $(A_0)_{{\rm proj}}$ by freely adding sifted colimits. In particular, $\check{\cal D}({\cal A})_{\geq 0}$ admits a universal characterization as a presentable $\infty$-category and $\check{\cal D}({\cal A}) \simeq {\rm Sp}(\check{\cal D}({\cal A})_{\geq 0})$ admits a universal characterizations as a stable presentable $\infty$-category (without the t-structure). However, even in this case, I don't know how to deduce from this particular universal characterization the other universal characterization I'm interested in (the missing part is to show that a coproduct preserving functors $(A_0)_{{\rm proj}} \to {\cal C}$ extends in an essentially unique way to a functor ${\cal A}_0 \to {\cal C}$ which sends short exact sequence to cofiber sequences). REPLY [11 votes]: Yes, both of these statements are true (I thought they were in the book, but I can't seem to find them now). Here is a proof sketch. Let's start with the case described in 2). Let $\mathcal{C}$ be any presentable $\infty$-category. As noted in the question, what you need to identify are functors $F: \mathcal{A}_0 \rightarrow \mathcal{C}$ that preserve finite coproducts and carry short exact sequences to cofiber sequences to functors $F_0: \mathcal{A}^{proj}_0 \rightarrow \mathcal{C}$ that preserve finite coproducts. There's an obvious restriction functor in one direction, and there's a functor of left Kan extension in the other (the universal property of $\mathcal{D}( \mathcal{A} )_{\geq 0}$ guarantees that this left Kan extension yields a functor that behaves well on exact sequences). The thing you need to check is that any $F: \mathcal{A}_0 \rightarrow \mathcal{C}$ as above agrees with the left Kan extension of its restriction to $\mathcal{A}_0^{proj}$; let's denote that functor by $F'$, so there's a natural transformation $F' \rightarrow F$ which is an equivalence on projective objects. Since both $F$ and $F'$ carry short exact sequences to cofiber sequences, the collection of objects $X$ for which $F'(X) \rightarrow F(X)$ is an equivalence is closed under taking cokernels of injective maps, and therefore (by induction) contains all objects of finite projective dimension, which is all objects of $\mathcal{A}_0$ by assumption. To handle the general case, we can use the fact that every Grothendieck abelian category can be obtained as a quotient $\mathcal{A} / \mathcal{B}$, where $\mathcal{A}$ is as above and $\mathcal{B}$ is some localizing subcategory. Let $\mathcal{D}_0$ be the smallest localizing subcategory of $\check{\mathcal{D}}( \mathcal{A})_{\geq 0}$ which contains $\mathcal{B}$, so that we can identify $\check{\mathcal{D}}( \mathcal{A} / \mathcal{B} )_{\geq 0}$ with the quotient $\check{\mathcal{D}}( \mathcal{A} )_{\geq 0} / \mathcal{D}_0$. The first part of the proof shows that the following data are equivalent: (1) Functors $F: \mathcal{A} \rightarrow \mathcal{C}$ which preserve finite coproducts, carry short exact sequences to cofiber sequences, and preserve filtered colimits. (2) Functors $F^{+}: \check{\mathcal{D}}(\mathcal{A})_{\geq 0} \rightarrow \mathcal{C}$ which preserve small colimits. We wish to show that if $F$ and $F^+$ correspond under this equivalence, then the following conditions are equivalent: (a) For every map $u: X \rightarrow Y$ in the abelian category $\mathcal{A}$, if the kernel and cokernel of $u$ belong to $\mathcal{B}$, then $F(u)$ is an equivalence (so that $F$ factors through $\mathcal{A} / \mathcal{B}$). (b) For every map $u: X \rightarrow Y$ in $\check{\mathcal{D}}(\mathcal{A})_{\geq 0}$, if the cofiber belongs to $\mathcal{D}_0$, then $F^{+}(u)$ is an equivalence in $\mathcal{C}$ (so that $F^{+}$ factors through $\check{\mathcal{D}}( \mathcal{A} )_{\geq 0} / \mathcal{D}_0$). The implication $(b) \Rightarrow (a)$ is immediate. To proceed in the reverse direction, let $u: X \rightarrow Y$ be as in $(b)$ and choose an object $Y'$ in $\mathcal{A}$ and a map $Y' \rightarrow Y$ which is surjective on $\pi_0$, and set $X' = Y' \times_{Y} X$. Then $u$ is a pushout of the projection map $u': X' \rightarrow Y'$, so it suffices to show that $F^{+}(u')$ is an equivalence. The map $u'$ factors as a composition $X' \xrightarrow{w} Y'_0 \xrightarrow{v} Y'$ where $v$ is a monomorphism in $\mathcal{A}$ and $w$ is surjective on $\pi_0$. Using the assumption that the cofiber of $u$ belongs to $\mathcal{D}_0$, it is not hard to check that $Y' / Y'_0 \in \mathcal{B}$ and that the fiber of $w$ belongs to $\mathcal{D_0}$. Then $F^{+}(v)$ is an equivalence by virtue of $(a)$, and $w$ is a pushout of the map $fib(w) \rightarrow 0$. We are therefore reduced to proving that the functor $F^{+}$ annihilates every object of $\mathcal{D}_0$, which is not hard to check (using the fact that it annihilates every object of $\mathcal{B}$).<|endoftext|> TITLE: (Fictive) story of a time where people reasoned only up to isomorphism QUESTION [15 upvotes]: I seem to remember reading once a story that some mathematician had written to justify the use of categories, or isomorphisms or equivalences, or something like that. The story goes something like this: Once upon a time, people did not know what equality was. Instead, they only thought about things up to isomorphism. For example, they did not say that two sets had the same number of elements, but that they were in bijection. Today with category theory go back to these roots. Does anyone have an idea of who told this story and what the full story is? REPLY [11 votes]: As Todd has already written an answer for me, maybe I can claim it as an Answer: Exercise 1.1 in my book Practical Foundations of Mathematics (CUP 1999) reads, When Bo Peep got too many sheep to see where each one was throughout the day, she found a stick or a pebble for each individual sheep and moved them from a pile outside the pen to another inside, or vice versa, as the corresponding sheep went in or out. Then one evening there was a storm, and the sheep came home too quickly for her to find the proper objects, so for each sheep coming in she just moved ANY one object. She moved all of the objects, but she was still worried about the wolf. By the next morning she had satisfied herself that the less careful method of reckoning was sufficient. Explain her reasoning without the aid of numbers. My reason for putting it in the book was to try to get some anthropologist to say when and what the original "proof" was, ie the cognitive basis of the long-universal belief that this is valid, which provides the justification of counting with numbers. What I am trying to imagine is how one of our distant ancestors with the cognitive abilities but not the education of a mathematician might approach this. Of course they would not have formulated Peano Induction or Euclidean Infinite Descent. They would have an argument (that we would more or less accept as rigorous) that the result is true for three sheep, then four and five. After that they would use Induction in the naive epistemological sense to convince themselves that it is true for arbitrarily large sets. It seems plausible that anyone who is challenged to come up with a proof would give the following (albeit non-constructive) proof. Suppose that some sheep $s_0$ is missing in the evening. Then the pebble $p_0$ that served as its "name" in the morning was used for some other sheep $s_1$ in the evening. But then $s_1$ must have been named by a different pebble $p_1$ in the morning, which named yet another sheep $s_2$ in the evening. And so on. All of the sheep $s_0$, $s_1$, $s_2$, ... are different individuals, Likewise all of the pebbles $p_0$, $p_1$, $p_2$, ... But, essentially as Euclid says in Book VII, Proposition 31, this is impossible for a set of sheep. By chance, this issue came up recently following an internal seminar by Martin Escardo in Birmingham (where I am now an Honorary Research Fellow). He was developing the foundations of arithmetic (in the setting of Homotopy Type Theory, though this was not essential) in such a way that $3\times 5=5\times 3$ could be seen in a primary-school fashion as transposing a rectangle. He based this on a function $F:{\mathbb{N}}\to{\mathsf{Set}}$ with $F0=\emptyset$ and $F(\mathsf{succ} n)=F(n)\coprod{\mathbf{1}}$. In his treatment the most difficult Proposition is $$ F(n)\cong F(m) \Longrightarrow n=m, $$ which he deduced from the Lemma $$ X\coprod{\mathbf{1}}\cong Y\coprod{\mathbf{1}} \Longrightarrow X \cong Y. $$ This Lemma holds in any lextensive category, i.e. one with finite limits and stable disjoint coproducts. The Proposition follows using Peano induction, since then $$ F(n+1)\cong F(m+1) \Longrightarrow F(n)\cong F(m) \Longrightarrow n=m \Longrightarrow n+1=m+1. $$ I think it is reasonable to suppose that Bo Peep could formulate this Lemma, but I feel it is more plausible that she would use the "infinite descent" argument than the Proposition. I am not sure whether this answers the original question about justifying "the use of categories, or isomorphisms or equivalences", although maybe Martin's treatment of arithmetic does so.<|endoftext|> TITLE: The conceptual difference in notations of Cat QUESTION [5 upvotes]: There have been some places in which a conceptual difference (that of criteria of identification, etc) is avoided in the categorical notation, namely Let $Cat$ be at the same time the 2-category of small categories and its subjacent 1-category i.e. the category of small categories (...) Why one can do this? The 2-category CAT and the 1-category Cat are not remarkably different? What is Cat (generally) in the literature..? REPLY [7 votes]: Here I take terminology fromnLab: As long as 2-category means strict 2-category this usage is exactly the same as using $M$ to name both a manifold and its set of points. If $Cat$ is meant to be some weak 2-category then you need to specify what enrichment you want--or as you put it what criterion of identification. There is not only one possibility. In that case the usage only makes sense if one specified enrichment is being assumed in the context.<|endoftext|> TITLE: Extend space to make polyhedra convex hulls of finite sets QUESTION [6 upvotes]: A (convex) polytope is the convex hull of a finite number of points in Euclidean space (this is the so-called "vertex description"). Alternatively, it can defined to be a bounded polyhedron (this is the "facet description"). Here a (convex) polyhedron is the intersection of a finite number of closed half-spaces. My question is whether we can get a vertex description of polyhedra, and not just polytopes: namely, is there some reasonable way to topologically extend Euclidean space so that all polyhedra are convex hulls of finite sets of points in this larger space? REPLY [5 votes]: I am not sure that I understand the question, but unbounded convex polyhedra in $R^3$ are determined, up to a translation, by their vertices and recession cone; see Section 1.4 in Alexandrov's book on Convex polyhedra. Roughly speaking, the recession cone encodes the behavior of the polyhedron at infinity. See also Theorem 3 on page 391 of that book, where it is discussed how to determine an unbounded convex polyhedron from the projection of its vertices into some plane, the curvatures at these vertices, and a given convex polyhedral cone. The cone, in turn, may be determined by a convex polygon in the plane mentioned above and the apex of the cone. In short there is a way to parametrize the space of unbounded convex polyhedra in terms of a finite number of points, which prescribe its vertices and its recession cone. In this conception, the unbounded polyhedron is just the convex hull of the union of its vertices together with the recession cone, once it is properly positioned relative to the vertices. By the way, the space of unbounded convex hypersurfaces of $R^n$ which are homeomorphic to $R^{n-1}$ and the recession cones of the corresponding convex bodies are extensively studied in Deformations of unbounded convex bodies and hypersurfaces, Amer. J. Math.,134 (2012),1585-1611 where it is shown for instance that the space of these convex hypesurfaces admits a strong deformation retraction into the Grassmannian space of hyperplanes. This is true in the polyhedral category as well.<|endoftext|> TITLE: The enigmatic complexity of number theory QUESTION [81 upvotes]: One of the most salient aspects of the discipline of number theory is that from a very small number of definitions, entities and axioms one is led to an extraordinary wealth and diversity of theorems, relations and problems--some of which can be easily stated yet are so complex that they take centuries of concerted efforts by the best mathematicians to find a proof (Fermat's Last Theorem, ...), or have resisted such efforts (Goldbach's Conjecture, distribution of primes, ...), or lead to mathematical entities of astounding complexity, or required extraordinary collective effort, or have been characterized by Paul Erdös as "Mathematics is not ready for such problems" (Collatz Conjecture). (Of course all branches of mathematics have this property to some extent, but number theory seems the prototypical case because it has attracted some of the most thorough efforts at axiomization (Russell and Whitehead), that Gödel's work is most relevant to the foundations of number theory (more than, say, topology), and that there has been a great deal of work on quantifying complexity for number theory--more so than differential equations, say.) Related questions, such as one exploring the relation between Gödel's Theorem and the complexity of mathematics, have been highly reviewed and somehow avoided any efforts to close. This current problem seems even more focused on specific references, theorems, and such. How do professional mathematicians best understand the foundational source of the complexity in number theory? I don't think answers such as "once relations are non-linear things get complicated" or its ilk are intellectually satisfying. One can refer to Gödel's Theorem to "explain" that number theory is so complicated that no finite axiomitization will capture all its theorems, but should we consider this theorem as in some sense the source of such complexity? This is not an "opinion-based" question (though admittedly it may be more appropriate for metamathematics or philosophy of mathematics): I'm seeking theorems and concepts that professional mathematicians (particularly number theorists) recognize as being central to our understanding the source of the breadth and complexity of number theory. Why isn't number theory trivial? What references, especially books, have been devoted to specifically addressing the source of the deep roots of the diversity and complexity of number theory? By contrast, I think physicists can point to a number of sources of the extraordinary wealth and variety of physical phenomena: It is because certain forces (somehow) act on fundamentally different length and time scales. The rules of quantum mechanics govern the interaction of a small number of subatomic particles. At sufficiently larger scales, though, quantum mechanics effective "shuts off" and classical mechanics dominates, including in most statistical mechanics, where it is the large number of particles is relevant. At yet larger scales (e.g., celestial dynamics), we ignore quantum effects. Yes, physicists are trying to unify the laws so that even the quantum mechanics that describes the interactions of quarks is somehow unified with gravitation, which governs at the largest scales... but the fact that these forces have different natural scales leads to qualitatively different classes of phenomena at the different scales, and hence the complexity and variety of physical phenomena. REPLY [39 votes]: What references, especially books, have been devoted to specifically addressing the source of the deep roots of the diversity and complexity of number theory? To a first approximation, I would say that the answer to this question is, there are none. You have emphasized that you are interested in the point of view of professional number theorists. I would say that for most number theorists, a term such as the "diversity and complexity of number theory" brings to mind central problems in modern number theory, such as the generalized Riemann hypothesis, the parity problem in sieve theory, the Langlands conjectures, the structure of Gal(${\overline{\mathbb{Q}}}/\mathbb{Q})$, etc. These are the sorts of things that professional number theorists might cite as the "source" of the diversity and complexity of number theory. Note in particular that things such as Hilbert's Tenth Problem are interesting to relatively few professional number theorists. The kind of "complexity" that logicians and theoretical computer scientists are interested in is not what interests most number theorists. Roughly speaking, it is because undecidability questions are regarded as signs of chaos whereas number theorists are interested in finding structure. If we want an explanation of why there is so much diversity and complexity in number theory even when we focus on the structures that occupy the attention of number theorists, then I do not think that looking towards undecidability will give us the answer. Generalized chess, for example, exhibits that kind of "complexity" but the deeper one studies chess, the more it seems to exhibit seemingly "random" behavior that defies elegant description (just take a look at the record-holders in the endgame tablebases for example). In generalized chess we find no sign of anything with the beautiful and deep structure of, say, class field theory. Seeking an explanation of what number theorists regard as the diversity and complexity of their subject is instead likely to elicit essays with "unreasonable effectiveness" or some such in the title, and the discussion will likely follow the same path that the discussion of Wigner's article has taken. For example it can be pointed out that there is a natural selection process taking place, with number theorists deliberately gravitating towards the areas of diversity and complexity and abandoning areas that are sterile.<|endoftext|> TITLE: Poincaré duality for Deligne (co)homology QUESTION [5 upvotes]: My question is about the papers Hélène Esnault and Eckart Viehweg, Deligne-Beı̆linson cohomology Uwe Jannsen, Deligne homology, Hodge-D-conjecture, and motives (both from the Beı̆linson's conjectures volume). Let $X$ be a smooth complex variety. We may define Deligne-Beı̆linson cohomology groups $$H^\bullet_\mathcal{D} (X, A (n)),$$ where $\mathbb{Z} \subseteq A \subseteq \mathbb{R}$, and $n \in \mathbb{Z}$ is some twist. (See the paper by Esnault and Viehweg.) There is also Deligne homology defined by Jannsen: $$H_\bullet^\mathcal{D} (X, A (n)).$$ Jannsen establishes the "twisted Poincaré duality" $$\tag{*} H^i_\mathcal{D} (X, A (n)) \cong H^\mathcal{D}_{2d-i} (X, A (d-n)),$$ where $d = \dim_\mathbb{C} X$. Does it give some classic duality result as a particular case? My guess is that it's supposed to be some generalization of the duality between singular cohomology and Borel-Moore homology. According to Lemma 1.11 from the same paper of Jannsen, $$H^\mathcal{D}_\bullet (X, \mathbb{Z} (0)) \cong H^\mathrm{BM}_\bullet (X, \mathbb{Z}).$$ So that the above duality gives us as a particular case $$H^i_\mathcal{D} (X, \mathbb{Z} (d)) \cong H^\mathrm{BM}_{2d-i} (X, \mathbb{Z}).$$ But what is the left hand side? Edit: The above is false, I misread the statement of Lemma 1.11. I hope someone can explain me the motivation behind (*). Thank you. REPLY [4 votes]: Deligne complex $A(p)$ is the complex $$A((2\pi i)^p) \rightarrow \mathcal O \rightarrow \Omega^1 \rightarrow {}...{} \rightarrow \Omega^{p-1}.$$ You can work with similar bigraded complexes, taking into account not only holomorphic, but also antiholomorphic forms. Let us define $A(p,q)$ as the complex of sheaves $$A((2\pi i)^{p+q}) \rightarrow \mathcal O \oplus \overline{\mathcal O} \rightarrow \Omega^1\oplus \overline{\Omega^1} \rightarrow {}...{} \rightarrow \Omega^{p-1} \oplus \overline{\Omega^{p-1}} \rightarrow \overline{\Omega^p} \rightarrow ... \rightarrow \overline{\Omega^{q-1}}.$$ (Here it's supposed that $q>p$). In this situation, we may take $A$ to be a subring of $\Bbb C$, not necessarily of $\Bbb R$. Suppose that $A = \Bbb C$. Then we can write a resolution of this complex of sheaves by differential forms, and it's possible to see that $A(p,q)$ is quasiisomorphic to the complex $$ ... \stackrel{d}{\rightarrow} \mathcal{A}^{p-2, q-1} \oplus \mathcal{A}^{p-1, q-2} \stackrel{d}{\rightarrow} \mathcal{A}^{p-1, q-1} \stackrel{\partial\bar\partial}{\rightarrow} \mathcal{A}^{p, q} \stackrel{d}{\rightarrow} \mathcal{A}^{p, q-1} \oplus \mathcal{A}^{p-1, q} \stackrel{d}{\rightarrow}{} ...$$ where the term $\mathcal{A}^{p-1, q-1}$ has the grading $p+q-2$. (The $\partial\bar\partial$ operator arises because of local $\partial\bar\partial$-lemma. You can find the proof that these complexes are quasiisomorphic in M. Schweitzer's text https://arxiv.org/abs/0709.3528). In particular, $\Bbb H^{p+q-2}(X,{~}A(p,q))$ is known as Aeppli cohomology $H_{Ae}^{p-1,q-1}$, and $\Bbb H^{p+q-1}(X,{~}A(p,q))$ is Bott-Chern cohomology $H_{BC}^{p,q}$. If your $X$ satisfies global $\partial\bar\partial$-lemma (for example, if it is compact Kaehler), then these are just isomorphic to the corresponding Dolbeaut cohomology; on a general complex manifold, they are different. The Deligne complex with real coefficients $\Bbb R(p)$ is isomorphic to the fixed points of complex conjugation on $\Bbb R(p,p)$, so your can take real forms in the Aeppli-Bott-Chern resolution and obtain the resolution for the real Deligne cohomology. Now, from the looks of the resolution and a bit of harmonic theory one can conclude that there's a non-degenerate pairing between $\Bbb H^{k}(X,{~}\Bbb C(p,q))$ and $\Bbb H^{2n+1-k}(X,{~}\Bbb C(n+1-p,n+1-q))$ (at least when $X$ is compact, otherwise you need to take cohomology with compact support on one side). This is complex version of the duality on Deligne cohomology, and when $k=p+q$ it specializes to the well-known duality between Bott-Chern and Aeppli cohomology (and to Poincare duality when $X$ satisfies $\partial\bar\partial$-lemma). I don't know what happens if you work with Deligne-Beilinson complex instead of Deligne complex (with logarithmic forms instead of regular), but I'd suppose that everything works the same and gives you the duality between Bott-Chern cohomology and Aeppli cohomology with compact support on an open part of your variety as a particular case.<|endoftext|> TITLE: Rational points on varieties over local fields QUESTION [7 upvotes]: In his expanded lecture notes Rational points on varieties, Bjorn Poonen writes the following: REMARK 2.5.3: There is an algorithm that, given a local field $k$ of characteristic $0$ and a $k$-variety $X$, decides whether $X(k)$ is nonempty. (...) I would like to see if this algorithm can be applied to determine the existence of rational points on a very specific family of varieties over $\mathbb{Q}_p$. Unfortunately, I could not find anything related on the web so far. Could you point to some literature on this matter, or explain some of the ideas? REPLY [11 votes]: Let us assume that $X$ is smooth and projective for simplicity, given by a number of polynomial equations with coefficients in the ring of integers $\mathcal O$ of $k$. Let $\kappa$ denote the residue class field of $k$. Reduce the equation moduloe the maximal ideal to get the reduced variety $\bar X$. Enumerate the $\kappa$-points of $\bar X$ (this is a finite problem). If $\bar X(\kappa) = \emptyset$, then $X(k)$ is empty. If there is a smooth point $\bar P \in \bar X(\kappa)$ (and $\dim \bar X = \dim X$), then it lifts by Hensel's Lemma to a point on $X(k)$, so $X(k)$ is non-empty. Otherwise, consider the lifts of each point to points modulo the square of the maximal ideal. This comes down to looking at equations $f(\dots, a_i + \pi x_i, \dots)$, where $f$ runs through the equations defining an affine patch of $X$ containing a lift of $\bar P$, $a_i$ are lifts of the coordinates of $\bar P$ and $x_i$ are variables; divide the equations by the highest possible power of $\pi$, which is a generator of the maximal ideal, and reduce mod $\pi$. This gives equations for the reduction mod $\pi$ of a different model of $X$. Smoothness of $X$ guarantees that the process will terminate after finitely many lifting steps. If $X$ is not smooth, consider its singular locus $X'$ first (recursively). If $X'(k)$ is non-empty, so is $X(k)$. Otherwise, run the previous procedure on $X$. Since $X'(k)$ is empty, all points that lift indefinitely will end up on the smooth part of $X$, which again gives termination. Magma has an implementation (IsLocallySoluble). See Nils Bruin, Some ternary Diophantine equations of signature (n, n, 2), In Bosma and Cannon, Discovering Mathematics with Magma, Springer-Verlag, Heidelberg, 2004.<|endoftext|> TITLE: Argument principle for matrices QUESTION [7 upvotes]: Let $f,g$ be entire functions, then the argument principle teaches us that $$\frac{1}{2\pi i}\int_{\mathbb{C}} g(z) \frac{f'(z)}{f(z)} dz$$ is equal to $g$ evaluated at the zeros of $f.$ Now, let us assume that $f: \mathbb{C} \rightarrow \mathbb{C}^{2 \times 2}$ is matrix-valued and holomorphic. If $f$ is diagonal, then it follows from the above theorem that $$\operatorname{tr}\left(\frac{1}{2\pi i}\int_{\mathbb{C}} g(z) Df(z)f(z)^{-1} dz\right)$$ is precisely $g$ evaluated at the singular points of $f$, i.e. the points $z$ for which $\operatorname{det}(f(z))=0$ times $2-\text{rank}(f(z))$ at those points. I ask: If $f$ is now not assumed to be diagonal, but only self-adjoint, does this statement hold true, i.e. $$\operatorname{tr}\left(\frac{1}{2\pi i}\int_{\mathbb{C}} g(z) Df(z)f(z)^{-1} dz\right)=\sum_{z \in \mathbb{C}; \operatorname{det}(f(z))=0} g(z) \text{rank}(2-f(z)).$$ An obvious approach would be to use that $f$ is diagonalizable and the properties of the trace, but in this case the unitary transform also depends on $z$, so a bit more care seems to be needed. REPLY [7 votes]: You can write instead $$\frac1{2i\pi}\int_Cg(z){\rm tr}(f'(z)f(z)^{-1})dz.$$ Now use the formula $${\rm tr}(f'(z)f(z)^{-1})=\frac1{\det f(z)}\,(\det f(z))'.$$ And conclude with the formula of residues. Remark that the formula tells us that what matters is the algebraic multiplicities of the zeroes of $\det f$, rather than the dimension of the kernels of $f(z)$.<|endoftext|> TITLE: Edge probability for connected Erdős–Rényi model QUESTION [7 upvotes]: Consider the Erdős–Rényi model $G_{n,p}$ with corresponding probability measure $\mathbb{P}_{n,p}$. For any two vertices $x,y$, $\mathbb{P}_{n,p}[E_{x,y}]=p$, where $E_{x,y}$ is the event that there exists an edge between $x$ and $y$. I need to estimate (especially bound from above) the following probability for two fixed vertices $x,y$ and $G \in G_{n,p}$: $\begin{equation} \mathbb{P}_{n,p}[E_{x,y}|G \text{ is connected}] \end{equation}$ Supplement: $p=\frac{c}{n}$ for some constant $c >1$ (I forgot this in my first version). REPLY [2 votes]: I agree strongly with Peter in the comments: if you're interested in $p \gg \frac{\log n}{n}$, then don't do any hard work, just show the answer is essentially $p$ since the graph is essentially always connected. In particular, let $E$ be the event that the edge exists and $C$ the event that the graph is connected. By the law of total probability, \begin{align*} P(E|C)P(C) + P(E,\lnot C) &= p \end{align*} so \begin{align*} P(E|C) &= \frac{p - P(E,\lnot C)}{P(C)} \\ &\leq \frac{p}{1-o(1)} \end{align*} where a lot is known about bounds on the $o(1)$ (the probability the graph is disconnected). (By the way, naturally, we immediately have $P(E|C) \geq p$. To prove it, $P(E|C) = p P(C|E)/P(C)$, and $P(C|E) \geq P(C)$ as guaranteed existence of $(x,y)$ can only raise the chance of connectedness.)<|endoftext|> TITLE: Quillen + construction for finite groups QUESTION [29 upvotes]: Is there an example of two non isomorphic finite groups $G$ and $H$ such that $BG^{+}$ is homotopy equivalent to $BH^{+}$ ? REPLY [14 votes]: Edit: I thought I had an example, but it does not quite work. It answers a different question. I will edit this post a bit, make it Community Wiki, and leave it up just in case it is of some interest, or might help to find a complete answer. For some finite groups $G$, $BG^+$ is equivalent to the product of the $p$-completions $BG\hat{_p}$. For example, it is true if $G$ is perfect. More generally, it is true if the final term of the lower central series is a perfect subgroup, and the plus construction is taken with respect to this subgroup. So we may consider the related question whether non-isomorphic finite groups can have equivalent $p$-completed classifying spaces for all primes $p$ (in the same spirit as Matthias Wendt's answer). By a theorem of Oliver, the $p$-completion of $BG$ is determined by the $p$-fusion system of $G$, where the $p$-fusion system of $G$ is the category whose objects are subgroups of a $p$-Sylow subgroup $P$, and where morphisms are all homomorphisms that are induced by conjugation by an element of $G$. So this question is equivalent to whether there can be non-isomorphic finite groups whose $p$-fusion systems are equivalent for all primes $p$. Such examples are known to exist. I found the following one in a paper of Martino and Priddy (who attribute it to Minami): $Q_{4p}\times {\mathbb Z}/2$ and $D_{2p}\times {\mathbb Z}/4$. Here $Q_{4p}$ is the generalised Quaternion group of order $4p$, where $p$ is an odd prime. The point is that they have isomorphic $p$-Sylow subgroups at all primes, namely ${\mathbb Z}/2\times {\mathbb Z}/4$ and ${\mathbb Z}/p$, and moreover the $p$-fusion systems are equivalent: you get the trivial structure at the prime $2$, and the only non-trivial morphism at $p$ is the inverse homomorphism. However, these groups are solvable but not nilpotent. So they do not have non-trivial perfect subgroups, and their lower central series do not terminate at the trivial group. So I don't believe their plus constructions are equivalent. Indeed, the plus construction does nothing to these groups. An example where the groups have the property that the lower central series terminate at a perfect subgroup would answer the original question.<|endoftext|> TITLE: A geometric proof of the strong maximal theorem QUESTION [7 upvotes]: While reading the paper "A geometric proof of the strong maximal theorem", by A. Cordoba and R. Fefferman -Annals of Mathematics Vol 102 no. 1, I got stuck trying to understand a main step in the proof. The goal is to prove a inequality for the strong maximal operator: $$ Mf(x) = \sup_R \frac{1}{R}\int_R |f|, $$ where $R$ ranges over all rectangles with sides parallel to the coordinate axes which contain $x$. In particular they state $$ \bigl|\{x:Mf(x)>\lambda\}\bigr| \leq A\int\frac{|f(x)|}{\lambda}\biggl( 1 + \log^+\frac{|f(x)|}{\lambda} \biggr)\,dx. $$ (for dimension 2). To achieve this they prove that given a collection of rectangles $\{R_i\}$ there is a subcollection $\{\tilde{R}_j\}$ such that i) $|\cup_i R_i| \leq C |\cup_j \tilde{R}_j|$ ii) $\|\exp(\sum \chi_{\tilde{R}_j})\| \leq C |\cup_i R_i|$ Then, they assert that the strong maximal theorem follows from this and their proof of the fact that the strong maximal operator if weak-type $(p,p)$ if and only if you have the $V_q$ property where $q$ is the dual exponent of $p$: Given any collection of rectangles $\{R\}$ there exists a subcollection $\{\tilde{R}\}$ such that $(i)$ $|\cup_i R_i| \leq C |\cup_j \tilde{R}_j|$ $(ii)_q$ $\|\sum \chi_{\tilde{R}_j}\|_{q} \leq C |\cup_i R_i|^{1/q}$ I don't understand how their proof, which is based on Holder's inequality, extends to the case $p=1.$ The main obstacle for me is the fact that the right hand side of $(ii)_\infty$ should be just $1$ ($\text{something}^{1/\infty}$) but the exponent in their main result is instead $1$. I tried using the inequality $$ \int fg \leq C \|f\|_{L\log L} \|g\|_{e^L} $$ but I end up with $|\{x: Mf(x)>\lambda\}|$ multiplied on both sides of the equation... How is this proved? REPLY [11 votes]: I am not sure how Cordoba and Fefferman intended the argument to go, but the final argument you indicate basically works, as long as one adjusts things by an epsilon so that the factor of $|\bigcup_i R_i|$ on the RHS is dominated by that on the LHS. [This is an example of a more general principle, namely that one should not be disheartened by finding a "circularity" in one's argument, such as finding the same term on the RHS as on the LHS of an estimate, so long as the circularity is introduced in a non-trivial fashion, as there can be a chance that one can tweak the argument so that the argument becomes non-circular.] More specifically, one can start with the Young type inequality $$ f g \leq C_\varepsilon f (1 + \log^+ f) + \varepsilon e^g \quad (1) $$ that is valid for any $f, g \geq 0$ and $\varepsilon > 0$, and some $C_\varepsilon$ depending on $\varepsilon$. If now $f \in L \log L$ and $\lambda > 0$, then taking a family $R_i$ of rectangles covering $\{ Mf \geq \lambda\}$ with $\int_{R_i} f \geq \lambda |R_i|$ and using the subcollection $\tilde R_j$ and (i) we have $$ \int \frac{|f|}{\lambda} \sum_j \chi_{\tilde R_j} \geq C^{-1} |\bigcup_i R_i|$$ while from (1) and (ii) one has $$ \int \frac{|f|}{\lambda} \sum_j \chi_{\tilde R_j} \leq C_\varepsilon \int \frac{|f|}{\lambda} (1 + \log_+ \frac{|f|}{\lambda}) + C\varepsilon |\bigcup_i R_i|.$$ Choosing $\varepsilon$ small enough we obtain $$ |\bigcup_i R_i| \leq C' \int \frac{|f|}{\lambda} (1 + \log_+ \frac{|f|}{\lambda})$$ which then gives the required bound on $|\{ Mf \geq \lambda \}|$. (There is a technical issue that $|\bigcup_i R_i|$ may be infinite, so that one cannot immediately cancel the terms on both sides of the estimates, but one can first work for instance with finite subcollections of the $R_i$, then use a limiting argument to recover a bound for all of $|\bigcup_i R_i|$.)<|endoftext|> TITLE: Motives associated to a Number Field QUESTION [10 upvotes]: Suppose $k$ is a number field, i.e. an extension of $\mathbb{Q}$ of finite degree, so we have a natural inclusion $\mathbb{Q} \rightarrow k$, which induces a morphism, \begin{equation} \text{Spec}\,k \rightarrow \text{Spec}\,\mathbb{Q} \end{equation} I have a naive (probably wrong) question to bother the mathoverflow community: is there a (mixed)-motive (over $\mathbb{Q}$ with coefficient ring $\mathbb{Q}$ ) associated to the $\mathbb{Q}$-scheme $\text{Spec}\,k$? If so, what is its Betti ($\ell$-adic) realisation? Please forgive my naiveness. REPLY [13 votes]: Write $X=\mathrm{Spec}\, k$, which is a $0$-dimensional variety. Motives of $0$-dimensional varieties are called Artin motives, and they are pure. The Betti realization is the Betti cohomology of $X$, which is $$ H^0(X(\mathbb{C}),\mathbb{Q})=\mathbb{Q}^{X(\mathbb{C})}=\mathbb{Q}^{\mathrm{Hom}(k,\mathbb{C})}. $$ There is an involution induced by complex conjugation on $\mathbb{C}$. To define the $\ell$-adic realization, we need to choose an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$ (of course the one inside $\mathbb{C}$ is a common choice). The basechange $X_{\overline{\mathbb{Q}}}:=X\times_{\mathrm{Spec}\,\mathbb{Q}}\mathrm{Spec}\,\overline{\mathbb{Q}}$ can be identified with the set $\mathrm{Hom}(k,\overline{\mathbb{Q}})$ (viewed as a constant scheme over $\overline{\mathbb{Q}}$), and the $\ell$-adic realization is then $$ H^0_{et}(X_{\overline{\mathbb{Q}}},\mathbb{Q}_\ell)=\mathbb{Q}_{\ell}^{\mathrm{Hom}(k,\overline{\mathbb{Q}})}. $$ This space comes with an action of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. Artin motives are discussed in Section 9.4 of [1] and Section 6 of [2]. [1] Huber, Annette; Müller-Stach, Stefan, Periods and Nori motives, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge 65. Cham: Springer (ISBN 978-3-319-50925-9/hbk; 978-3-319-50926-6/ebook). xxiii, 372 p. (2017). ZBL1369.14001. [2] Deligne, Pierre; Milne, J.S., Tannakian categories, Hodge cycles, motives, and Shimura varieties, Lect. Notes Math. 900, 101-228 (1982). ZBL0477.14004.<|endoftext|> TITLE: Computing kernels of maps of modules over a finitely presented algebra QUESTION [6 upvotes]: I have the following problem: I have an associative (noncommutative) algebra $A$ defined over a rational function field $k = \mathbb{Q}(\delta, \lambda)$. $A$ is given by a presentation in terms of generators and relations, and is finite-dimensional but typically large. I want to consider $A$ as a bimoodule over itself. I can then define some linear maps $f_i : A \to A$ by $f_i(x) = a_i x b_i$ for fixed elements $a_i, b_i \in A$. I want to compute the kernels of the maps $f_i$, in particular their intersection. I'm having a lot of difficulty finding a computer algebra system (CAS) that can do this. Basically the only CAS that will effectively work with such an $A$ is Magma, because it can compute noncommutative Groebner bases over rational function fields. In particular, it can compute a basis of $A$ as a matrix algebra over $k$. Unfortunately, I can't find a good way of working with the matrix algebra representation of $A$; if Magma has a good way of dealing with bimodules or computing kernels of linear maps of algebras, I can't find it. An alternate method would be to simply compute the matrices of the $f_i$ as linear maps and use standard linear algebra techniques. I would want to do this in another CAS, most likely Sage, because I've found it very difficult to script anything in Magma. Unfortunately I can't figure out a good way of exporting the basis of $A$ (given as matrices with entries in $k$), since Magma prints matrices in an odd format and the Sage-Magma interface runs into an error when trying to convert it itself. I might just print the basis to a text file and write a script to parse it into a format Sage can read, but I'd rather not have to do that. Does anyone have any suggestions for dealing with this problem? REPLY [2 votes]: My eventual solution was as follows: Use Magma to compute a Groebner basis of $A$, then compute a matrix representation of $A$. Export this representation to Sage,* then compute the matrices of the linear maps over the vector space of matrices with coefficients in the symbolic ring (which was suitably vectorized.) Compute the intersections of the kernels in Sage or another CAS. I wound up doing (3) in Mathematica, because Sage was extremely slow at computing kernels of matrices with symbolic entries. I think this was because it doesn't know to try to simplify polynomials at each step, because there was no speedup when adding relations on the symbols (setting $\delta = \sqrt2 $ or the like.) Mathematica was much faster in this case. *For some reason there was no automatic way to do this, or at least I couldn't get the Sage-Magma link working properly. Instead I had to print the matrices to a text file, reformat them using a script, then load them into Sage.<|endoftext|> TITLE: A “formalistic” variant of the Gödel completeness theorem QUESTION [8 upvotes]: A week ago I asked this at MathStackExchange, but without success. I think, the following variant of the Gödel completeness theorem must be true, but I can't find the references. I would be grateful if specialists in logic could give me them (or enlighten me in case that something must be corrected). A question. Joseph Shoenfield in his "Mathematical logic" (section 4.7) gives a definition of an interpretation of a theory in another theory. This notion allows to define a model of a first order theory $T$ as its interpretation in some variant of an axiomatic set theory, say, in MK (or in ZFC, I don't think that this choice is important.). (I understand that this is not the custom in mathematical logic, but I invite readers to look at this way, I will explain my motives later.) Suppose now that we have a first order theory $T$. We define its model in MK in the way I described above, and consider the class ${\mathcal M}_T$ of all models of $T$ in MK. Each model $M\in {\mathcal M}_T$ can be considered as another first order theory, an extension by definitions of the theory MK in the sense of Kenneth Kunen's "Foundations of mathematics" (section II.15). Moreover, we can add the symbol ${\mathcal M}_T$ into the signature of MK and the definition of ${\mathcal M}_T$ into the list of axioms of MK, and we'll get another first order theory, an extension by definition of MK. Let us denote this new first order theory by MK+${\mathcal M}_T$. Now let us take a formula $\varphi$ in $T$. It has an analog $\varphi_M$ in each model $M\in {\mathcal M}_T$, and we can consider a formula $\varphi^*$ in MK+${\mathcal M}_T$ which states that $\forall M\quad M\in {\mathcal M}_T\Rightarrow \varphi_M$ My question is if the following proposition is true (up to some possible specifications): Proposition. A formula $\varphi$ is deducible in $T$ if and only if the corresponding formula $\varphi^*$ is deducible in MK+${\mathcal M}_T$. As far as I understand, this can be considered as a "weakened analog" of the Gödel completeness theorem, and the stronger one must state the same about all the formulas $\varphi$ simultaneously. About my motives. This comes from one of my questions at MathOverflow. I believe, there must be a way to explain mathematical logic such that the references to sets and functions appear after the axiomatic construction of the set theory (and not before, as it is now). I am not a specialist in Logic (my field is Analysis), but I am interested in this because (I teach logic sometimes, and) I am writing a textbook on university mathematics where I am planning to add a chapter about Set theory and mathematical logic. From the discussion at MathOverflow I got an impression that the idea to simplify the exposition is not hopeless, it is possible to explain everything inside the standard principle that a mathematician can't use a term before giving a precise definition. That is why in my text (I wrote already a draft of this chapter) Set theory preceeds mathematical logic, so that I can use the notions of set and function after their formal definition. But my problem is a lack of references. I would appreciate very much if somebody could help me with this. EDIT 01.04.2018. Recently one of my friends showed me an article by K.Smorynski in Handbook of Mathematical Logic (edited by Jon Barwise) where he formulates a statement which he calls "the Hilbert-Bernays theorem" (Theorem 6.1.1 in volume 4) and which as far as I understand is equivalent to the following: If a formal theory $T$ is consistent then it has an interpretation in PA. I believe this is more or less equivalent to what Matt F. suggests in his answer: If a formal theory $T$ is consistent then it has an interpretation in MK. And if somebody could give me a reference to this statement (with a proof), this, I believe, will be a proper solution to what I need. Does anybody know such a reference? REPLY [2 votes]: This proposition is equivalent to $Con(MK)$. I'll make free use of the fact that $MK$ can prove the soundness of first-order logic. I'll also use the abbreviation $M(\varphi)$ to mean the interpretation of $\varphi$ under $M$, e.g. if the language of $T$ is the language of groups, and $M$ is the interpretation of $T$ with the symmetric group $S_3$, and $\varphi = \forall x, xx=1$, then $M(\varphi) = \forall x \in S_3, xx=1$. Suppose we have the proposition. Then the case $T=\emptyset$, $\varphi = \bot$ gives $Con(MK)$. Now suppose we have $Con(MK)$ and want to prove the proposition. The easy direction is that if $\varphi$ is deducible (i.e. $T \vdash \varphi$), then so is $\varphi^*$ (i.e. $MK+T^* \vdash \varphi^*$). The harder direction is that if $\varphi$ is not deducible (i.e. $T \nvdash \varphi$), then neither is $\varphi^*$ (i.e. $MK+T^* \nvdash \varphi^*$). First, if $\varphi$ is not deducible, then there is a model $M$ of $T+\neg \varphi$. We can prove this just by formalizing Henkin's proof of the completeness theorem. In particular $MK \vdash M(\neg\varphi^*)$. Now, suppose $MK+T^*\vdash \varphi^*$. Then $MK\vdash M(\neg\varphi^*)$, and $MK\vdash M(\varphi^*)$. But this is impossible by the consistency of $MK$, so $MK+T^*\nvdash \varphi^*$ as desired.<|endoftext|> TITLE: Derivation of certain sums "the hard way" QUESTION [28 upvotes]: It is a well-known fact, that one can derive some spectacular identities, e. g. $\sum^{n-1}_{m=1}\sigma_3(m)\sigma_3(n-m)=\frac {\sigma_7(n)-\sigma_3(n)}{120}$ $\sum^{n-1}_{m=1}\sigma_3(m)\sigma_9(n-m)=\frac {\sigma_{13}(n)-11\sigma_9(n)+10\sigma_3(n)}{2640} $ just by equating several modular forms together. In the book The 1-2-3 of Modular Forms Don Zagier writes: "It is not easy to obtain any of these identities by direct number-theroretical reasoning (although in fact it can be done)" Does anybody know how to derive these identities "the hard way" or at least point me to some discussion? REPLY [22 votes]: A really different proof, and the one Zagier refers to, is by Nils Skoruppa: "A quick combinatorial proof of Eisenstein series identities", J. Number Theory 43 (1993), 68--73.<|endoftext|> TITLE: Do these polynomials have alternating coefficients? QUESTION [21 upvotes]: In answering another MathOverflow question, I stumbled across the sequence of polynomials $Q_n(p)$ defined by the recurrence $$Q_n(p) = 1-\sum_{k=2}^{n-1} \binom{n-2}{k-2}(1-p)^{k(n-k)}Q_k(p).$$ Thus: $Q_{2}(p) = 1$ $Q_{3}(p) = -p^2 + 2 p$ $Q_{4}(p) = -2 p^5 + 9 p^4 - 14 p^3 + 8 p^2$ $Q_{5}(p) = 6 p^9 - 48 p^8 + 162 p^7 - 298 p^6 + 318 p^5 - 189 p^4 + 50 p^3$ Numerical calculations up to $n=60$ suggest that: The lowest-degree term of $Q_n(p)$ is $2n^{n-3}p^{n-2}$. The coefficients of $Q_n(p)$ alternate in sign. Are these true for all $n$? As the title indicates, I'm especially puzzled about 2. Indeed, the original inspiration for the polynomials $Q_n(p)$ comes from a classic paper of E. N. Gilbert (Random graphs, Ann. Math. Stat. 30, 1141-1144 (1959); ZBL0168.40801) where the author studies the sequence of polynomials $P_n(p)$ given by the similar recurrence $$P_n(p) = 1 - \sum_{k=1}^{n-1} \binom{n-1}{k-1}(1-p)^{k(n-k)}P_k(p),$$ which do not have alternating coefficients. REPLY [12 votes]: To illustrate the suggestion of Richard Stanley about positivity of real parts of zeroes, here are the zeroes of $Q_{20}$. The pattern seems to be the same for all of them. Another empirical observation: seems that $$ \frac{Q_n(1-x)}{(1-x)^{n-2}(1+x)}=1+(n-3)x+\left(\binom{n-2}2+1\right)x^2 +\left(\binom{n-1}3+n-3\right)x^3+\left(\binom n4+\binom{n-2}2+1\right)x^4+...+\left(\binom{n+k-4}k+\binom{n+k-6}{k-2}+\binom{n+k-8}{k-4}+...\right)x^k+O(x^{k+1})$$for $n>k+1$ REPLY [7 votes]: To add to the preceding answer: The absolute values of the coefficients appear normal (in particular, unimodal): Which means that the technology developed, in, eg, Lebowitz, J.L.; Pittel, B.; Ruelle, D.; Speer, E.R., Central limit theorems, Lee-Yang zeros, and graph-counting polynomials, J. Comb. Theory, Ser. A 141, 147-183 (2016). ZBL1334.05065. May be relevant.<|endoftext|> TITLE: Getting the most general form of Mayer-Vietoris from the Eilenberg-Steenrod axioms QUESTION [11 upvotes]: I asked this question a while ago on MSE, got no answer, put a bounty on it, still got no answer, was advised to ask here instead, hesitated, forgot about the question for a while and now remembered it. I'm still not sure if it really qualifies as research level. I let you decide... I'd like to derive the most general form of the Mayer-Vietoris sequence from the Eilenberg-Steenrod axioms for homology (in particular: I do not want to use the definition of $H_\ast(X)$ in terms of simplices). By that I mean the existence of an exact sequence of the form $\cdots \to H_n(X_{12},A_{12})\to H_n(X_1,A_1)\oplus H_n(X_2,A_2)\to H_n(X,A) \to H_{n-1}(X_{12},A_{12})\to\cdots$ whenever $(A,A_1,A_2)\subseteq(X,X_1,X_2)$ are two excisive triads and $A_{12}:=A_1\cap A_2$, $X_{12}:=X_1\cap X_2$. It is easy to do that in the special case $A=A_1=A_2$ by looking at the two long exact sequences for pairs from the inclusions $A\subseteq X_{12}\subseteq X_1$ and $A\subseteq X_2\subseteq X$ respectively. These two sequences form a Barratt-Whitehead ladder and the lemma of Mayer-Vietoris applies. Basically the same approach works in the special case $X=X_1=X_2$. Both of these proofs are well known in the literatur, but so far I was unable to find a proof for the general version either in the books or myself that did not go through the realisation of $H_\ast$ as the homology of some chain-complex generated by simplices. All my previous attempts consisted of doodling one diagram after the other, but may be there is a simpler solution. After realising that $H_\ast(X,A) = \tilde{H}_\ast(C_A^X)$, where $C_A^X$ is the mapping cone of the inclusion $A\to X$, one could also ask whether $(C_A^X, C_{A_1}^{X_1}, C_{A_2}^{X_2})$ is an excisive triad. So I'm asking: Is it? REPLY [6 votes]: If you are willing to work with mapping cones, then this follows from looking at the triple (= threefold iterated) mapping cone for the cube with vertices $A_{12} = A_1 \cap A_2$, $A_1$, $A_2$, $A$, $X_{12} = X_1 \cap X_2$, $X_1$, $X_2$ and $X$ in two different ways. Let us use your notation $C_A^X = X \cup_A CA$, so that there is a natural isomorphism $H_*(X, A) \cong \tilde H_*(C_A^X)$. If $(A, A_1, A_2)$ is excisive, then the double (= twofold iterated) mapping cone for the square with vertices $A_{12}$, $A_1$, $A_2$ and $A$ has the homology of a point. Likewise, if $(X, X_1, X_2)$ is excisive, then the double mapping cone for the square with vertices $X_{12}$, $X_1$, $X_2$ and $X$ has the homology of a point. Thus the triple mapping cone for the cube has the homology of a point. This is homeomorphic to the double mapping cone for the square with vertices $C_{A_{12}}^{X_{12}}$, $C_{A_1}^{X_1}$, $C_{A_2}^{X_2}$ and $C_A^X$. Since $C_{A_{12}}^{X_{12}} = C_{A_1}^{X_1} \cap C_{A_2}^{X_2}$, this shows that $(C_A^X, C_{A_1}^{X_1}, C_{A_2}^{X_2})$ is excisive, and gives you the exact Mayer-Vietoris sequence $$ \dots \overset{\partial}\to H_n(X_{12}, A_{12}) \to H_n(X_1, A_1) \oplus H_n(X_2, A_2) \to H_n(X, A) \overset{\partial}\to \dots $$ by the Barratt-Whitehead lemma. (Note the spelling of Michael Barratt's name.)<|endoftext|> TITLE: What are the implications of a zero of zeta off the critical line QUESTION [12 upvotes]: So what happens if there is a non-trivial zero of the Riemann zeta function off the critical line? Has there been any work in the following direction: We know from Landaus theorem that there is a positive proportion of the zeros $\alpha$ on the critical line. Suppose that $\rho$ is a a non-trivial zero of $\zeta$ off the critical line. Then can we use this to cook up an argument to show the existence of another zero off the line? I am being highly optimistic here but continuing in this direction maybe we can show there will be infinite set of zeros off the critical line and going even further we may try to show that the density of this set is greater than $1-\alpha$, yielding the desired contradiction. REPLY [3 votes]: Perhaps this is a duplicate of this question If the Riemann Hypothesis fails, must it fail infinitely often?. The discussion continues here The Hardy Z-function and failure of the Riemann hypothesis, where a hypothesis is proposed, which implies If the Riemann Hypothesis fails, must it fail infinitely often?. To this hypothesis there are two approximations based on: 1. Zeta function universality 2. GUE hypothesis. Concerning 1. I can say that even, $\zeta (s)$ for $Re(s)=\frac{1}{2}$ is dense in $\mathbb{C}$, is unknown. 2. it seems to me it will be easier. It is enough to make an inversion for the expression $N(t_{n})=n$, where $N(t_{n})$ the number of non-trivial zeros of the zeta function, which is well-known function and $t_{n}$ the nth non-trivial zero, and define the property of the recurrence.<|endoftext|> TITLE: Positive integer combination of non-negative integer vectors QUESTION [10 upvotes]: A vector of positive integer numbers with $n$ coordinates is given $a=(a_1,\ldots,a_n)$. It holds that $a_1+\cdots+a_n$ is divisible by some positive integer number $k$. I have checked many cases and arrived to the conjecture that one can always find at most $n$ vectors with $n$ non-negative integer coordinates such that in all the vectors the sum of the coordinates is exactly equal to $k$ and $a$ is represented as a positive integer combination of these vectors. Example: $n=3$, $k=5$ and $a = (12,7,6)$, then the $3$ vectors satisfying above described property are $(2,2,1)$, $(5,0,0)$ and $(1,1,3)$, because $a = 3\cdot (2,2,1) + 1\cdot (5, 0, 0) + 1\cdot (1,1,3)$. One can manually show that the conjecture holds for $k=2,3,4,5$ or $n=2$. It is also easy to see that only $nn/2$, we replace $w$ to $(1-w_1,\dots,1-w_n)$ and $d$ to $n-d$. So, we may suppose that $d\leqslant n/2$. Assume that $w_1\geqslant w_2\geqslant \dots \geqslant w_n$. Denote $p=(1,\dots,1,0,\dots,0,1)\in \{0,1\}^n\cap \Delta(n,d)$. Then $$w=a_n\cdot p+(1-a_n)\cdot \frac{w-a_n\cdot p}{1-a_n}.$$ The vector $\tilde{w}:=\frac{w-a_n\cdot p}{1-a_n}$ has $n$-th coordiante equal to 0; sum of coordinates equal to $d$.Thus if we manage to prove that all coordinates are between 0 and 1, we may do induction step (since $p$ and any unimodular simplex in $\Delta(n-1,d)$ form a unimodular simplex in $\Delta(n,d)$). Clearly all coordinates of $\tilde{w}$ are non-negative, and if they are also at most 1, unless $a_d>1-a_n$. In that case we would have $$d=a_1+\dots+a_n>(1-a_n)d+a_n(n-d)=d+a_n(n-2d)\geqslant d,$$ a contradiction.<|endoftext|> TITLE: Varying a Kahler metric in a neighborhood of a point QUESTION [5 upvotes]: I would like to know if the following statement (or a more general version of it) is contained in some book or article: Statement. Let $(U,g)$ be a complex manifold with a Kahler metric $g$ and let $x\in U$ be a point. Let $U_x\subset U$ be a neighborhood of $x$ and $h$ be a Kahler metric on $U_x$. Then there is a smaller neighborhood $U_x'\subset U_x$ of $x$, and a metric $g'$ on $U$, such that $g'=h$ in $U_x'$ and $g'=g$ in $U\setminus U_x$. I can prove this statement using regularized maximum, but would be grateful for a reference. REPLY [2 votes]: This is not literally the answer that the OP wanted (a reference to the literature, which I am not aware of), but following the comments above let me write down the simple gluing argument. Let $\widetilde{\max}$ be a regularized maximum function, and fix a small coordinate ball $B$ around $x$ (contained in $U_x$), with local coordinates $z=(z_1,\dots,z_n)$, where the Kähler forms of $g$ and $h$ can be written as $\omega_g=dd^c u$, $\omega_h=dd^c v$ for some smooth functions $u,v$ defined in some neighborhood of $\overline{B}$. Fix a smooth nonnegative cutoff function $\chi$ identically $1$ near $x$ and compactly supported in $B$. We can then find $\delta>0$ small enough so that $$\tilde{u}(z)=u(z)+\delta\chi(z)\log|z|,$$ is strictly plurisubharmonic on $B$, and of course it is smooth on $B$ minus $x$. We then choose $C>0$ large enough so that $v-C TITLE: Role of the UCT problem in classification theory for C*-algebras QUESTION [8 upvotes]: Elliott's program for nuclear C*-algebras deals with the problem of classifying nuclear C*-algebras by K-theoretical invariants. A major open question in this context is the UCT problem. A separable C*-algebra $A$ is said to satisfy the UCT if for every separable C*-algebra $B$ a short exact sequence of the form $0\rightarrow\text{Ext}\left(K_{0}\left(A\right)\text{, }K_{0}\left(B\right)\right)\rightarrow KK\left(A\text{, }B\right)\rightarrow\text{Hom}\left(K_{0}\left(A\right)\text{ }K_{0}\left(B\right)\right)\rightarrow0$ exists, where the right hand map is the natural one and the left hand map is the inverse of a map that is always defined. The UCT problem states that every separable, nuclear C*-algebra satisfies the UCT. I'm not sure yet why the UCT is of such great importance concerning Elliott's program since I don't see how it's correctness provides a classification like for example in the AF-algebraic case. Can someone explain? Further I'm interested in the history of Elliott's program and how it led to the UCT question. Are there any good sources about the history of this topic? REPLY [8 votes]: Remark: The UCT-sequence is not correct as you stated it, it does not only involve the $K_0$-groups. Regarding your second quastion: I highly recommend the book "Classification of nuclear $C^*$-algebras. Entropy in operator algebras" written by Rørdam and Størmer. You can find an explanation of the Elliott-conjecture which led to Elliott's classification program, which is as well outlined in this book. The Eliott-conjecture asserts that all nuclear, separable $C^*$-algebras are classified up to the Elliott-invariant, which is a certain functor consiting of K-theory, traces and a pairing between traces and K-theory. The conjecture is known to be false in this generality, but true for certain subclasses of nuclear, separable $C^*$-algebras. Many of these subclasses have in common, that one assumpions for the $C^*$-algebras in these special subclasses is that they should satisfy the UCT. Thus, if you solve the UCT problem, then you can drop the UCT-assumption for these subclasses where the Elliott-conjecture is known to be true, and if the answer of this UCT-problem is yes, then these subclasses are fully understood. In short: The UCT is such a prominent ingredient for many classification theorems in the Elliott program, therefore one whishes to drop the UCT-assumption. But to drop the UCT-assumption, the UCT-problem should have a positive answer. Regarding the first question (although I can not give a full answer of it right now), one possible situation (for Kirchberg algebras, for instance) is the following: For $C^*$-algebras satisfying the UCT, the sequence gives you a surjective map $$\gamma: \operatorname{KK}_*(A,B)\to \operatorname{Hom}(K_*(A),K_*(B)).$$ I.e. if the UCT is satisfied, for an isomorphism $f_*:K_*(A)\to K_*(B)$ in $\operatorname{Hom}(K_*(A),K_*(B))$ there exists a KK-equivalence $x\in \operatorname{KK}_*(A,B)$ such that $\gamma(x)=f_*$, i.e. $A$ and $B$ are KK-euqivalent. Sometimes it is possible to represent every element in $\operatorname{KK}_*(A,B)$ via a *-homomorphism $A\to B$ (see Phillip's classification theorem, theorem 8.2.1 in the recommended book, for instance) so that you can try to prove the existence of an isomorphsm $A\to B$ from there.<|endoftext|> TITLE: Teichmuller groupoids in Grothendieck's esquisse d'un programme QUESTION [7 upvotes]: Grothendieck in his Esquisse d'un programme mentioned without any precise definition and construction that the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ acts on the whole "tower" of the Teichmuller groupoids $\hat T_{g,\nu}$ (see p.5 in the above reference). I do understand what does that mean in a very special case $g=0,\nu=4$. What does it mean in general? What are these Teichmuller groupoids? Is there a more detailed exposition of the above general remark of Grothendieck? ADD: If I understand correctly, $\hat T_{g,\nu}$ is very close to be a profinite completion of the fundamental group of the moduli space $M_{g,\nu}$ of smooth Riemann surfaces of genus $g$ with $\nu$ marked points (am I wrong?). Grothendieck claims that there are various natural morphisms between $\hat T_{g,\nu}$'s. Frankly I do not see any morphisms except $\hat T_{g,\nu}\to \hat T_{g,\mu}$ for $\mu<\nu$ induced by the map $M_{g,\nu}\to M_{g,\mu}$ which is just forgetting several marked points. Are there any other morphisms? REPLY [13 votes]: There is another type of important morphism between the (orbifold) fundamental groups of the moduli spaces $M_{g,\nu}\rightarrow M_{g',\nu'}$ that is considered in Grothendieck's tower. You can see this morphism in three different ways. One is directly on the surfaces of type $(g,\nu)$ and $(g',\nu')$ (of genus $g$ with $\nu$ boundary components, resp. genus $g'$ with $\nu'$ boundary components). This morphism exists if you can put a set of disjoint simple closed loops on the surface of type $(g',\nu')$ such that when you cut along them, you cut your surface into one piece of type $(g,\nu)$, or else into several pieces of which at least one is of type $(g,\nu)$. You can also think of including the smaller surface of type $(g,\nu)$ into the bigger one by gluing it to other smaller pieces along the edges of their boundary components, to form the bigger one of type $(g',\nu')$ (which is the image Grothendieck had in mind when he talked about Lego). The second way to see this morphism is as a morphism of moduli spaces, where $M_{g,\nu}$ is mapped to a boundary component of the Deligne-Mumford compactification $\overline{M}_{g',\nu}$, in fact precisely the boundary component corresponding to taking the simple closed loops on the surface of type $(g',\nu')$ that "cut out" the one of type $(g,\nu)$ and shrinking them to length zero, so they become nodes. The third way to view this same morphism is on the fundamental groups. This is pretty easy, since the (orbifold) fundamental group of $M_{g,\nu}$ is generated by Dehn twists along simple closed loops on the surface of type $(g,\nu)$, and these just map to the Dehn twists along the same simple closed loops when the $(g,\nu)$ surface is included in the $(g',\nu')$ one as above. The Teichmüller tower can be considered to be the collection of all the fundamental groups of the $M_{g,\nu}$ linked by the point-erasing morphisms and by these. Or, as Grothendieck wanted, instead of fundamental groups, that depend on a certain choice of base point, you can replace the groups by more symmetric fundamental groupoids based at all "tangential base points" on the moduli spaces". The automorphism group of the Teichmüller tower basically then consists of tuples $(\phi_{g,\nu})$ such that each $\phi_{g,\nu}$ is an automorphism of $\pi_1(M_{g,\nu})$ and the different $\phi_{g,\nu}$ in the same tuple commute with the homomorphisms of the tower.<|endoftext|> TITLE: Maximize sum of products of binary variable QUESTION [9 upvotes]: Let $n>k$ be positive integers, $r>1$ a positive real number, and $A=\{1,2,\dots,n\}$. For $1\leq i\neq j\leq n$, let $a_{i,j}\in\{r,1\}$ be such that $a_{i,j}=r\Leftrightarrow a_{j,i}=1$. Consider the sum $$S=\sum_{X\subseteq A, |X|=k}\prod_{i\in X, j\in A\backslash X}a_{i,j}.$$ Is it true that $S$ is maximized when $a_{i,j}=r$ for all $i TITLE: Distribution relation in the Euler system of Heegner points QUESTION [13 upvotes]: I am trying to understand the details behind the so-called "distribution relations" between Heegner points on the modular curve $X_0(N)$, as given (for instance) in Gross's paper Kolyvagin's work on modular elliptic curves, [Proposition 3.7, (i)]. More precisely, the relation between Hecke and Galois actions on CM-points is still not clear for me. First of all let me recall in details the general settings, which the familiar reader can skip and go directly to the question. Settings: Let $N$ be a positive integer, let $K$ be an imaginary quadratic field of discriminant $D<0$ such that every rational prime divisor of $N$ splits in $K$ (the so-called Heegner hypothesis). The Heegner hypothesis implies that one can find a (non-unique) ideal $\mathcal{N}$ of $\mathcal{O}_K$ such that $\mathcal{O}_N/\mathcal{N}\simeq \mathbb{Z}/N\mathbb{Z}$. Edit: As noticed by Olivier I also have to assume $D<-4$, to ensure that $\mathcal{O}_K^{\times}=\{\pm 1\}$. For a general $n$, denote by $\mathcal{O}_n=\mathbb{Z}+n\mathcal{O}_K\subset\mathcal{O}_K$ "the" order of $K$ of conductor $n$, and let $Pic(\mathcal{O}_n)$ be the ideal class group of $\mathcal{O}_n$ (which is isomorphic to $I(n)/P(n)$, with $I(n)$ the group of ideals in $\mathcal{O}_K$ that are prime to $n$; and $P(n)$ the subgroup of principal ideals in $\mathcal{O}_K$ generated by an element congruent mod $n\mathcal{O}_K$ to some $r\in\mathbb{Z}$, $(r,n)=1$). Class-field theory provides us with an abelian extension $K_n$ of $K$ such that $\mathrm{Gal}(K_n/K)\simeq Pic(\mathcal{O}_n)$. Let now $n$ be an integer prime to $ND$. Setting $\mathcal{N}_n:=\mathcal{N}\cap\mathcal{O}_n$, we get that $\mathcal{O}_n/\mathcal{N}_n\simeq \mathcal{O}_K/\mathcal{N}\simeq \mathbb{Z}/N\mathbb{Z}$. One can thus define a point $x_n\in X_0(N)$ by setting $$x_n=[\mathbb{C}/\mathcal{O}_n\rightarrow \mathbb{C}/\mathcal{N}_n^{-1}]$$ (here $[E\rightarrow E']$ denotes the isomorphism class of the pair $(E,E')$ of elliptic curves, with $E\rightarrow E'$ a cyclic $N$-isogeny). If $l$ is a rational prime not dividing $ND$ nor $n$, we can mimic the previous discussion and set $$x_{nl}=[\mathbb{C}/\mathcal{O}_{nl}\rightarrow \mathbb{C}/\mathcal{N}_{nl}^{-1}]$$ The theory of complex multiplication ensures that $x_n\in X_0(N)(K_n)$ (resp. $x_{nl}\in X_0(N)(K_{nl})$ ) Here comes my question: assuming $l$ is inert in $K/\mathbb{Q}$, why do we have $$\mathrm{Tr}_{K_{nl}/K_n}x_{nl}:=\sum_{\sigma\in\mathrm{Gal}(K_{nl}/K_n)}\sigma x_{nl} = T_l x_n$$ as divisors on $X_0(N)$ ? What I "understood" is that: For $\sigma\in \mathrm{Gal}(K_{nl}/K)$, the action of $\sigma$ on $x_{nl}$ is given by $\sigma x_{nl}=[\mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\rightarrow \mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\mathcal{N}_{nl}^{-1}]$ where $\mathfrak{a}_{\sigma}$ is any (proper) ideal of $\mathcal{O}_{nl}$ such that $[\mathfrak{a}_{\sigma}]\in Pic(\mathcal{O}_{nl})$ corresponds to $\sigma$ via the isomorphism $\mathrm{Gal}(K_{nl}/K)\simeq Pic(\mathcal{O}_{nl})$. Thus the sum $\mathrm{Tr}_{K_{nl}/K_n}x_{nl}$ rewrites as $$\sum_{\sigma\in\mathrm{Gal}(K_{nl}/K_n)}[\mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\rightarrow \mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\mathcal{N}_{nl}^{-1}]$$ Here the condition $\sigma\in\mathrm{Gal}(K_{nl}/K_n)$ implies, I think, that $\mathfrak{a}_{\sigma}\mathcal{O}_n$ is a principal $\mathcal{O}_n$-ideal. Here $\mathrm{Gal}(K_{nl}/K_n)\simeq \mathbb{F}_{l^2}^{\times}/\mathbb{F}_{l}^{\times}$, so the sum has $l+1$ terms. The action of the Hecke operator $T_l$ ($l$ not dividing $N$) on divisors of the modular curve $X_0(N)$ can be described (at least in characteristic $0$) as $$T_l [E\xrightarrow{\phi} E']=\sum_{C\subset E[l], \#C=l} [E/C\rightarrow E'/\phi(C)]$$ (there are $l+1$ such $C$) In my situation, this can be rewritten (following Gross, Heegner points on $X_0(N)$, §6) as $$T_l x_n=\sum_{\mathfrak{b}\subset\mathcal{O}_n\text{lattice of index } l}[\mathbb{C}/\mathfrak{b}\rightarrow \mathbb{C}/\mathfrak{b}(\mathcal{N}_n\cap End(\mathfrak{b}))^{-1}] $$ What I don't understand is why the exactly the same terms should appear in both sums. I get that $(id)x_{nl}=x_{nl}$ appears in $T_l x_n$, as $\mathcal{O}_{nl}$ is a sub-lattice of order $l$ in $\mathcal{O}_n$ with $End(\mathcal{O}_{nl})=\mathcal{O}_{nl}$, but I don't see why $\mathfrak{a}_{\sigma}^{-1}$ is a sublattice of order $l$ in $\mathcal{O}_n$ if $\sigma$ fixes $K_n$. I think this involves properties about fractional ideals of orders which I don't quite understand. I thank everyone taking the time to read this question and trying to provide me with any help ! REPLY [3 votes]: Thank you Olivier for this great answer, I wish I could be one of your students ;-) ! Though I still not quite fully understand the adelic setting, here is some "classical" explanation I found, using the theory of ideals in orders of imaginary quadratic fields. I guess this is supposed to be a very well-known result but, since I haven't found any complete proof of it, I'll write down what I hope to be a detailed one in case some people are interested. My main reference is D.A. Cox 's "Primes of the form $x^2+ny^2$", chapters 7 and 8. First, slightly modify the notations and set for any $f\in\mathbb{Z}$: $I_K(f)$: group of ideals of $\mathcal{O}_K$ which are prime to $f$ (i.e., whose norm is prime to $f$) $P_{K,\mathbb{Z}}(f)$: subgroup of $I_K(f)$ made of principal ideals of the form $\alpha\mathcal{O}_K$, with $\alpha$ congruent to some $a$ mod $f\mathcal{O}_K$, with $a\in\mathbb{Z}$ and $(a,f)=1$. Here the ideals are meant to be integral ideals, and the group structure is obtained by identifying $\mathfrak{a}$ and $\mathfrak{b}$ iff there are some $\alpha$, $\beta\in \mathcal{O}_K$ such that $\alpha\mathfrak{a}=\beta\mathfrak{b}$. The quotient $I_K(f)/P_{K,\mathbb{Z}}(f)$ is what we call a generalized ideal class group, isomorphic to $Pic(\mathcal{O}_f)\simeq \mathrm{Gal}(K_f/K)$ ($K_f/K $ is the ring class field of conductor $f$, the isomorphism being defined by sending any prime-to-$f$ prime ideal of $\mathcal{O}_K$ to its Frobenius element). Here we set $Pic(\mathcal{O}_f)$ to be the group formed by the quotient of proper ideals of $\mathcal{O}_f $ (i.e., ideals $\mathfrak{a}\subset \mathcal{O}_f$ such that $End(\mathfrak{a})=\mathcal{O}_f)$, by principal ideals. An important fact is that any (proper or not) $\mathcal{O}_f$-ideal is a free $\mathbb{Z}$-module of rank $2$. The isomorphism between $I_K(f)/P_{K,\mathbb{Z}}(f)$ and $Pic(\mathcal{O}_f)$ is given by sending $$\mathcal{O}_K\supset\mathfrak{a} \mapsto \mathfrak{a}\cap\mathcal{O}_f$$ One checks that elements of $P_{K,\mathbb{Z}}(f)$, of the form $\alpha\mathcal{O}_K$ are sent to $\alpha\mathcal{O}_f $ (this follows from $(a,f)=1$, with $a$ "the" integer congruent to $\alpha$ mod $f\mathcal{O}_K$.) The inverse isomorphism is given by sending $$\mathcal{O}_f\supset\mathfrak{a}\mapsto \mathfrak{a}\mathcal{O}_K$$ (see Cox's 7.18, 7.20). Though it's not strictly necessary, I mention that the norm is preserved by these maps. The case$ f=1$ gives us the Hilbert class field $K_1 $ of $K$, such that $I_K/P_K=Pic(\mathcal{O}_K)\simeq \mathrm{Gal}(K_1/K)$. With these identifications, one obtains that the group $\mathrm{Gal}(K_f/K_1)$ corresponds to $I_K(f)\cap P_K/P_{K,\mathbb{Z}}(f)$. An important fact is that, if $\mathcal{O}_K^{\times}=\{\pm 1\}$, we have the following exact sequence: $$1\rightarrow (\mathbb{Z}/f\mathbb{Z})^{\times}\rightarrow (\mathcal{O}_K/f\mathcal{O}_K)^{\times}\rightarrow I_K(f)\cap P_K/P_{K,\mathbb{Z}}(f) \rightarrow 1$$ (see Cox, 7.27) Applying this to $f=n$, $nl$, one gets that $$\mathrm{Gal}(K_{nl}/K_1)\simeq (\mathcal{O}_K/nl\mathcal{O}_K)^{\times}/(\mathbb{Z}/nl\mathbb{Z})^{\times}$$ $$\mathrm{Gal}(K_{n}/K_1)\simeq (\mathcal{O}_K/n\mathcal{O}_K)^{\times}/(\mathbb{Z}/n\mathbb{Z})^{\times}$$ Therefore $\mathrm{Gal}(K_{nl}/K_n)$ is isomorphic to $$\frac{(\mathcal{O}_K/nl\mathcal{O}_K)^{\times}/(\mathbb{Z}/nl\mathbb{Z})^{\times}}{(\mathcal{O}_K/n\mathcal{O}_K)^{\times}/(\mathbb{Z}/n\mathbb{Z})^{\times}} $$ $$\simeq (\mathcal{O}_K/l\mathcal{O}_K)^{\times}/(\mathbb{Z}/l\mathbb{Z})^{\times}$$ as $(n,l)=1$. This last group is isomorphic to $\mathbb{F}_{l^2}^{\times}/\mathbb{F}_{l}^{\times}$, for $l$ is inert in $K/\mathbb{Q}$, so has cardinal $l+1$. Elements of the group $\mathrm{Gal}(K_{nl}/K_n)$ also corresponds to prime-to-$nl$ principal ideals of the form $\mathfrak{a}=\alpha\mathcal{O}_K$, with $\alpha$ congruent to $a\in\mathbb{Z}$ mod $n\mathcal{O}_K$, $(a,n)=1$. Notice that $\alpha\in\mathcal{O}_{nl}$ iff. it corresponds to the identity of $\mathrm{Gal}(K_{nl}/K_n)$. Denote by $\mathfrak{a}(\sigma)=\alpha(\sigma)\mathcal{O}_K$ such an ideal corresponding to $\sigma\in\mathrm{Gal}(K_{nl}/K_n)$, and set $\mathfrak{a}_{\sigma}:=\mathfrak{a}({\sigma})\cap\mathcal{O}_{nl}$ the corresponding proper ideal of $\mathcal{O}_{nl}$. As a consequence, if $\sigma\in\mathrm{Gal}(K_{nl}/K_n)$, then $\mathfrak{a}(\sigma)\cap\mathcal{O}_n=\alpha(\sigma)\mathcal{O}_n$ is principal, thus $\mathfrak{a}_{\sigma}\subset\mathcal{O}_{nl}$ satisfies $$\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma}\subset \mathcal{O}_n.$$ We have $$\frac{\mathcal{O}_n}{\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma}}\xrightarrow{\sim}\frac{\mathfrak{a}(\sigma)\cap\mathcal{O}_n}{\mathfrak{a}_{\sigma}}\hookrightarrow \frac{\mathcal{O}_n}{\mathcal{O}_{nl}},$$ the first map being multiplication by $\alpha(\sigma)$. As $\frac{\mathcal{O}_n}{\mathcal{O}_{nl}}\simeq \mathbb{Z}/l\mathbb{Z}$ and as $\frac{\mathfrak{a}(\sigma)\cap\mathcal{O}_n}{\mathfrak{a}_{\sigma}}$ is non-trivial (if $\sigma$ is not trivial then $\alpha(\sigma$) belongs to $\mathfrak{a}(\sigma)\cap\mathcal{O}_n$ but not to $\mathcal{O}_{nl})$; if $ \sigma$ is trivial then one directly gets $\frac{\mathfrak{a}(\sigma)\cap\mathcal{O}_n}{\mathfrak{a}_{\sigma}}=\frac{\mathcal{O}_{n}}{\mathcal{O}_{nl}})$, this finally implies that $$\frac{\mathcal{O}_n}{\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma}}\simeq \mathbb{Z}/l\mathbb{Z}$$ Combining this with the formula $$\mathrm{Tr}_{K_{nl}/K_n}(x_{nl})=\sum_{\sigma\in\mathrm{Gal}(K_{nl}/K_n)}[\mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\rightarrow \mathbb{C}/\mathfrak{a}_{\sigma}^{-1}\mathcal{N}_{nl}^{-1}]$$ (see my first post), and the facts that 1- $End(\alpha(\sigma)^{-1}\mathfrak{a}_{\sigma})=End(\mathfrak{a}_{\sigma})=\mathcal{O}_{nl}$, 2- the isomorphism class of an elliptic curve $\mathbb{C}/\Lambda$ is invariant under $\Lambda\mapsto \alpha\Lambda$ and 3- if $\sigma\neq \sigma' $ then $ \mathfrak{a}_{\sigma}$ and $\mathfrak{a}_{\sigma'}$ define different elements in $Pic(\mathcal{O}_{nl})$, one gets that $\mathrm{Tr}_{K_{nl}/K_n}(x_{nl})$ is a sum of $l+1$ distinct elements of the form $$[\mathbb{C}/\mathfrak{b}\rightarrow \mathbb{C}/\mathfrak{b}(\mathcal{N_n}\cap End(\mathfrak{b}))^{-1}]$$ with $\mathfrak{b}\subset \mathcal{O}_n$ a sub-lattice of index $l$. This is precisely the equality $$\mathrm{Tr}_{K_{nl}/K_n}(x_{nl})=T_l(x_n)$$ which I was looking for.<|endoftext|> TITLE: Determining the Mordell-Weil group of a universal elliptic curve QUESTION [15 upvotes]: Let $K$ be a number field and let $K(a,b)$ be the field of rational functions with two indeterminates over $K$. Consider the elliptic curve $E$ over $K(a,b)$ defined by the Weierstrass equation \begin{equation*} E : y^2=x^3+ax+b. \end{equation*} What is the torsion subgroup and the rank of the Mordell-Weil group of $E$ over $K(a,b)$? In the case $K=\mathbf{Q}$, this Mordell-Weil group is trivial because $E$ admits specializations $a,b \in \mathbf{Q}$ such that $E_{a,b}$ has trivial Mordell-Weil group. In the general case, I would like to show similarly the existence of $a,b \in K$ such that $E_{a,b}$ has trivial Mordell-Weil group over $K$, but I don't know how to proceed. Since $K(a,b)$ is a finitely generated field, the Lang-Néron theorem tells us that the Mordell-Weil group of $E$ is finitely generated, but I'm not familiar enough to tell whether the proof actually gives us a way to compute this Mordell-Weil group. REPLY [9 votes]: If you want to do this directly, you could "partially specialize" to, say $y^2 = x^3 + Ax + T$ with $A\in\mathbb C$. Then I don't think it's very hard to show, via a standard descent, that as an elliptic curve over $\mathbb C(T)$, the rank is $0$, and that for most $A$, the torsion is trivial, too. Actually, for the rank $0$ part, maybe it's easier to show that $y^2=x(x-A)(x-T)$ has rank 0 over $\mathbb C(T)$, since you can more easily do a 2-descent. And you also get that there is at most 2-torsion for most $A$. But it's pretty clear that your original curve over $\mathbb C(a,b)$ has no 2-torsion.<|endoftext|> TITLE: Square root of the determinant line QUESTION [17 upvotes]: Let $\Sigma$ be a compact Riemann surface equipped with a spin structure (a square root of $\Omega^1_\Sigma$, denoted $\Omega^{1/2}_\Sigma$). Let $\Gamma(\Omega^1_\Sigma)$ be the space of holomorphic differentials on $\Sigma$, and let $\Gamma(\Omega^{1/2}_\Sigma)$ be the space of holomorphic $\frac12$-forms. I believe that there is a canonical isomorphism $$ \Lambda^{top}\big(\Gamma(\Omega^{1/2}_\Sigma)\big)^{\otimes 2} \cong \Lambda^{top}\big(\Gamma(\Omega^1_\Sigma)\big). $$ Did I get this right?  [I got it wrong! See the answers below.] What is the isomorphism? Where can I read about it? PS: Both $\Lambda^{top}(\Gamma(\Omega^{1/2}_\Sigma))$ and $\Lambda^{top}(\Gamma(\Omega^1_\Sigma))$ are line bundles over the moduli stack of spin Riemann surfaces. REPLY [23 votes]: There is no such isomorphism (at least for $g \geq 9$). In O. Randal-Williams, The Picard group of the moduli space of r-Spin Riemann surfaces. Advances in Mathematics 231 (1) (2012) 482-515. I computed the Picard groups of moduli spaces of Spin Riemann surfaces (for $g \geq 9$). Grothendieck--Riemann--Roch shows that, in the notation of that paper, the right hand side of your formula is the class $\lambda$, and the left-hand side is the class $\lambda^{1/2}=2\mu$. (See page 511 of the published version for the calculation of the latter; the preprint has some mistakes at this point.) But the Picard group has presentation $\langle \lambda, \mu \,\vert\, 4(\lambda + 4\mu)\rangle$ as an abelian group, so these are not equal (even modulo torsion). You should not really need my calculation to see this: you can calculate the rational first Chern class of both sides by GRR, and see that they are distinct multiplies of the Miller--Morita--Mumford class $\kappa_1$; all that remains to know is that $\kappa_1 \neq 0$, which was shown in J. L. Harer, The rational Picard group of the moduli space of Riemann surfaces with spin structure. Mapping class groups and moduli spaces of Riemann surfaces (Göttingen, 1991/Seattle, WA, 1991), 107–136, Contemp. Math., 150, Amer. Math. Soc., Providence, RI, 1993. EDIT: To answer the question in the comments. Yes, I think that (in my notation) the relation $4(\lambda + 4\mu)=0$ holds for $g \geq 3$ (for $g < 3$ it can probably be checked by hand). To see this, let me shorten notation by writing $\mathcal{M}_g = \mathcal{M}_g^{1/2}[\epsilon]$ for the moduli space of Spin Riemann surfaces of Arf invatriant $\epsilon \in \{0,1\}$, $\pi : \mathcal{M}_g^1 \to \mathcal{M}_g$ for the universal family (i.e. $\mathcal{M}_g^1$ is the moduli space of Spin Riemann surfaces with one marked point), and $\mathcal{M}_{g,1}$ for the moduli space of Spin Riemann surfaces with one boundary component. Firstly, the Serre spectral sequence for $\pi$ has $$E_2^{0,1} = H^0(\mathcal{M}_g ; H^1(\Sigma_g ; \mathbb{Z}))$$ and one can show that this is zero: the fundamental group of $\mathcal{M}_g$ acts on $H^1(\Sigma_g ; \mathbb{Z}) = \mathbb{Z}^{2g}$ via a surjection onto a finite-index subgroup of $\mathrm{Sp}_{2g}(\mathbb{Z})$, but this finite-index subgroup will still be Zariski-dense in $\mathrm{Sp}_{2g}(\mathbb{C})$, so the (complexified) invariants will be zero. It follows from the Serre spectral sequence that $$\pi^* : H^2(\mathcal{M}_g; \mathbb{Z}) \to H^2(\mathcal{M}_g^1; \mathbb{Z})$$ is injective, so it is enough to prove the relation when there is a marked point. In fact, it even follows that $$H^2(\mathcal{M}_g; \mathbb{Z}) \oplus \mathbb{Z}\cdot e \to H^2(\mathcal{M}_g^1; \mathbb{Z})$$ is injective, where $e$ denotes the Euler class of the vertical tangent bundle of $\pi$. Now there is a fibration sequence $$\mathcal{M}_{g,1} \to \mathcal{M}_g^1 \overset{\frac{e}{2}}\to BSpin(2)$$ and so, from the Serre spectral sequence, an exact sequence $$0 \to \mathbb{Z}\cdot \frac{e}{2} \to H^2(\mathcal{M}_g^1;\mathbb{Z}) \to H^2(\mathcal{M}_{g,1};\mathbb{Z}) \overset{d^2}\to H^1(\mathcal{M}_{g,1};\mathbb{Z}).$$ Now it follows from A. Putman, A note on the abelianizations of finite-index subgroups of the mapping class group, Proc. Amer. Math. Soc. 138 (2010) 753-758. that for $g \geq 3$ the fundamental group of $\mathcal{M}_{g,1}$ has torsion abelianisation, so its first cohomology is zero. Putting it all together, we get an injection $$H^2(\mathcal{M}_g; \mathbb{Z}) \to H^2(\mathcal{M}_{g,1}; \mathbb{Z}),$$ so it is enough to verify the relation $4(\lambda + 4\mu)=0$ on $\mathcal{M}_{g,1}$. But for $g \leq 9$ there is a map $$\mathcal{M}_{g,1} \to \mathcal{M}_{9,1} \to \mathcal{M}_9,$$ given by gluing on 2-holed tori then a disc, so the relation holds because it holds on $\mathcal{M}_9$. REPLY [12 votes]: I'm not sure what you mean by 'canonical'. For example, when $\Sigma$ has genus $1$, there are 4 distinct spin structures, one representing the trivial line bundle, whose space of holomorphic sections has dimension $1$, and three nontrivial, whose spaces of holomorphic sections have dimension $0$. If the above isomorphism were 'canonical' in the three nontrivial cases, since the left hand side is canonically isomorphic to $\mathbb{C}$, it would produce a 'canonical' choice of holomorphic $1$-form on $\Sigma$, i.e., a 'canonical' volume form. Is there such a thing? I don't think so. You may find the paper arXiv:1201.2557 useful. It's possible that the answer depends on whether the chosen spin structure (aka 'theta characteristic') $\Omega^{1/2}_\Sigma$ is even or odd.<|endoftext|> TITLE: Decribe the $S^2$ fibration over $S^2$ that gives $\mathbf{CP}^2\#\overline{\mathbf{CP}}^2$ QUESTION [6 upvotes]: According to this MO post, there is two possible $S^2$ fibration over $S^2$. One is obviously $S^2\times S^2$, another one is the connected sum of two copies of $\mathbf {CP}^2$ with different orientations. Can someone explicitly describe the $S^2$ fibration over $S^2$ that gives $\mathbf{CP}^2\#\overline{\mathbf{CP}}^2$? REPLY [5 votes]: Cheeger described in "Some examples of manifolds with nonnegative curvature" the connected sum $\mathbb C\mathbb P^n\#\overline{\mathbb C\mathbb P^n}$ as a biquotient: Take $(S^{2n-1}\times S^2)/S^1$, where $S^1$ is acting freely on $S^{2n-1}$ and by rotations on $S^2$. This is the associated bundle to the principal fibration $S^1\to S^{2n-1}\to\mathbb C\mathbb P^n$ with fiber $S^2$. See Example 3 in Cheeger's article, why the above quotient is $\mathbb C\mathbb P^n\#\overline{\mathbb C\mathbb P^n}$.<|endoftext|> TITLE: Fake integers for which the Riemann hypothesis fails? QUESTION [73 upvotes]: This question is partly inspired by David Stork's recent question about the enigmatic complexity of number theory. Are there algebraic systems which are similar enough to the integers that one can formulate a "Riemann hypothesis" but for which the Riemann hypothesis is false? One motivation for constructing such things would be to illustrate the barriers to proving the Riemann hypothesis, and another one would be to illustrate how "delicate" the Riemann hypothesis is (i.e., that it's not something that automatically follows from very general considerations). I've run across various zeta functions over the years, but I seem to recall that either the Riemann hypothesis is probably/provably true, or the zeta function is too unlike the classical zeta function to yield much insight. More generally, what happens if we replace "Riemann hypothesis" with some other famous theorem or conjecture of number theory that seems to be "delicate"? Can we construct interesting systems where the result fails to hold? REPLY [3 votes]: In 1936, Davenport and Heilbronn studied certain Dirichlet series with zeros off the critical line. The Davenport–Heilbronn function $f(s)$ may be defined as follows. Let $$\alpha = {\sqrt{10 -2\sqrt{5}} - 2 \over \sqrt{5}-1}$$ and let $\chi$ be a character modulo 5 such that $\chi(2) = i$. Then let $$f(s) = {1 - i\alpha\over 2}\sum_{n=1}^\infty {\chi(n)\over n^s} + {1 + i\alpha\over 2}\sum_{n=1}^\infty {\bar\chi(n)\over n^s}.$$ It can be shown that $f(s)$ satisfies a functional equation $g(s) = g(1-s)$ where $$g(x) = \biggl({\pi \over 5}\biggr)^{-s/2}\Gamma\biggl({s+1\over 2}\biggr)f(s).$$ There is a catch, though, which is that there is no Euler product, so one might object that $f(s)$ is not the "zeta function of fake integers." For more information, see for example Zeros of the Davenport–Heilbronn counterexample by Eugenio P. Balanzaro and Jorge Sánchez-Ortiz, Math. Comp. 76 (2007), 2045–2049 or On the zeros of the Davenport–Heilbronn function by S. A. Gritsenko, Proc. Steklov Inst. Math. 296 (2017), 65–87.<|endoftext|> TITLE: Orthogonal representations of graphs QUESTION [8 upvotes]: A faithful orthonormal representation of a graph $G=(V,E)$ on $n$ vertices $\{1,2,\dotsc,n\}$ is an assignment of unit vectors $v_1,v_2,...,v_n \in \mathbb{R}^d$ to the vertices of $G$ such that $\langle v_i,v_j \rangle =0 \Leftrightarrow ij \in E(G)$, and in addition $|\langle v_i , v_j \rangle| \neq 1$ if $i \neq j$, i.e., distinct vertices are assigned non-parallel vectors. Note that this definition of orthonormal representation is slightly rarer in the literature and differs (by graph complementation) from the definition in [1] where $ij \in E(\bar{G}) \Rightarrow \langle v_i , v_j \rangle = 0$. The question is: Given a graph $G$ that has a faithful orthonormal representation in dimension $d$, form a new graph $G'$ by deleting an edge $uv$ from $G$. Does $G'$ then also have a faithful orthonormal representation in the same dimension $d$? [1] L. Lovász, On the Shannon Capacity of a Graph, IEEE Trans. Inf. Theory, 25 (1):1-7 (1979). REPLY [4 votes]: The following would be a counterexample if we require only $v_i \ne v_j$ but don't forbid different vertices to become antipodes on the sphere. Let us call this a weakly faithful orthogonal representation. The graph of the octahedron has a weakly faithful representation in (real) dimension 2, given by $\pm e_i$. If you remove one edge, there will be no faithful representation in dimension 2: the remaining edges force the two non-connected vertices to lie at distance $\pi/2$ on the sphere. So, for faithful orthogonal representations the question is: can one force the distance (in the standard spherical metric) between two points in the projective space to be $\pi/2$ by imposing distances $\pi/2$ between some pairs of points? REPLY [3 votes]: Thanks to a commenter for pointing out that the following was hasty: the fallacy is of course in the statement that the edgeless graph had no faithful orthogonal representation. I'll leave it at that, but add a warning. No. Fallacious proof by contradiction; the red is wrong. if this were true, then, since you required the graph to be finite, iterating the (hypothetical) statement $\lvert E\rvert$-many times would lead us to the conclusion that the edgeless graph on $n$ vertices has a faithful orthogonal representation in dimension $d$, $\color{red}{\text{which is absurd}}$. This proves that the answer cannot possibly be yes. Useful addition: the relevant technical term is monotone graph property. And the above gives a reason why for any fixed $n$ and $d$, the property(=isomorphism-invariant class of graphs) of all those graphs which admit a faithful orthogonal representation in dimension $d$ is not a monotone decreasing graph property.<|endoftext|> TITLE: Reference request for Hecke operators for principal congruence subgroup of modular group QUESTION [8 upvotes]: I am looking for references that discuss Hecke operators $T_n$ acting on modular forms for the principal congruence subgroup $\Gamma(N)$ of the modular group $SL(2,Z)$ and am happy to restrict to the case that $(n,N)=1$. Most textbooks (Diamond and Shurman, Koblitz etc.) that discuss Hecke operators for congruence subgroups specialize to $\Gamma_0(N)$ or $\Gamma_1(N)$ which contain $T: \tau \rightarrow \tau+1$, but I am specifically interested in the action of Hecke operators on modular forms which are invariant under $T^N$ but not under $T$ and are thus forms for $\Gamma(N)$ but not for $\Gamma_0(N)$ or $\Gamma_1(N)$. It seems reasonably clear how to obtain the answer using the double coset formulation of Hecke operators, but I have not found a reference which writes this out explicitly in terms of the relation between the coefficients of the Fourier expansion of the form at the cusp at infinity and the coefficients of its image under $T_n$. REPLY [4 votes]: The Hecke operators $T(n)$ and the dual Hecke operators $T'(n)$ acting as correspondences on the modular curve $Y(N)$ are defined by Kato in $p$-adic Hodge theory and values of zeta functions of modular forms, section 2.9 (in Kato's notation $Y(N)=Y(N,N)$). The action of $T(p)$ on Fourier expansions is given in section 4.9, there he also describes the relation between his definition and other definitions in the litterature. Actually, when you conjugate the double coset $\Gamma(N) \begin{pmatrix} n & 0 \\ 0 & 1 \end{pmatrix} \Gamma(N)$ in $\mathrm{GL}_2(\mathbf{Q})$ by the matrix $\begin{pmatrix} N & 0 \\ 0 & 1 \end{pmatrix}$, you get the double coset $\Gamma \begin{pmatrix} n & 0 \\ 0 & 1 \end{pmatrix} \Gamma$, where $\Gamma$ is this subgroup intermediate between $\Gamma_1(N^2)$ and $\Gamma_0(N^2)$. So it should be a simple exercise to check that Kato's definition agrees with David's answer.<|endoftext|> TITLE: An abstract nonsense proof of the Hurewicz theorem QUESTION [27 upvotes]: The ordinary homotopy groups of a space $X$ are the homotopy groups of the corresponding singular simplicial set $Sing(X)$. The ordinary homology groups of $X$ are the homotopy groups of the simplicial set $F(Sing(X))$, where $F$ is the functor that replaces each set $Sing(X)_n$ with the free abelian group on that set. There should obviously be some nice property of $F$ that makes the Hurewicz theorem work, but all the proofs I can find in the literature do things entirely by hand at the level of checking individual maps and homotopies to $X$. Is this nice property known? Is there a satisfying answer? REPLY [6 votes]: One approach is to replace the free abelian group functor with the free commutative monoid functor, also known as the infinite symmetric product. Let $X$ be a pointed CW complex, and let $\mbox{SP}^\infty(X)$ be the infinite symmetric product. By Dold-Thom theorem, if $X$ is connected then the homotopy groups of $\mbox{SP}^\infty(X)$ are the homology groups of $X$. The advantage of $\mbox{SP}^\infty$ over the free abelian group functor is that it has a natural filtration, and a cell structure that can be analyzed. Given Dold-Thom theorem, the Huriewicz theorem is equivalent to saying that if $X$ is $k-1$-connected, then the map $X\to \mbox{SP}^\infty(X)$ is $k+1$ connected (for simplicity let's assume $k>1$). This can be proved by analysing the cell structure on $\mbox{SP}^\infty(X)$. You can find this in chapter 6 of the book of Aguilar, Gitler and Prieto.<|endoftext|> TITLE: Natural explanation for a matrix identity QUESTION [5 upvotes]: I recently came across this curious fact in some calculations with the strain tensor in fluid mechanics: Let $A$ be an antisymmetric 3 by 3 matrix and $S$ be a traceless symmetric 3 by 3 matrix. Then it is easy to see that $AS+SA$ is an antisymmetric 3 by 3 matrix. $A$ corresponds to a 3-dimensional vector $v$ in the following way: for any 3-dimensional vector $w$ we have $Aw=v\times w$. When the same procedure is applied to $AS+SA$, it will produce the vector $-Sv$. I have two ways to prove this. The first is to write everything in terms of indices, and watch the big chunk mysteriously fizzles away after massive cancellation. The other way is representation-theoretic: it boils down to the fact that, in terms of $SO(3)$-representations, $3\otimes 5=3\oplus5\oplus7$, and everything except the constant factor follows from Schur's Lemma. The representation-theoretic proof also says that the projection $3\otimes 5\to 3$ is "natural" in some sense. But now I have two descriptions of this projection, so is there a natural way (without indices, but also more concrete than abstract representation-theoretic nonsense) to see that they agree up to the minus sign? REPLY [10 votes]: Your claim can be restated and generalised slightly as $$S(v\times w)+(Sv)\times w+v\times(Sw)=\text{trace}(S)v\times w$$ for all symmetric $S$ and all $v,w\in\mathbb{R}^3$. Just as with the trace-free case, it is straightforward to check the identity by a large and unilluminating calculation. Another approach is as follows. We have an orthonormal basis of eigenvectors $e_i$ with eigenvalues $s_i$ say and the cross product of any two of them will be the third one up to sign. It is then easy to check the identity when $v=e_i$ and $w=e_j$ (noting that both sides are zero if $i=j$). The general case then follows by bilinearity. My guess is that there is a better, more direct proof but I cannot see one at the moment. I agree that this is a curious fact, and I am surprised that I have never seen it before.<|endoftext|> TITLE: Octonion algebras over $\mathbb{F}_p(t)$ QUESTION [6 upvotes]: In their book Octonions, Jordan Algebras and Exceptional groups Springer and Veldkamp have a subsection called 'Classification over special fields' in which they describe the number of division and split quaternion and octonion algebras (and how to find them) over the following fields (as quoted from the book): Algebraically closed fields, the reals, finite fields, complete, discretely valuated fields with finite residue fields and for algebraic number fields. For finite fields, all quaternion and octonion algebras are split (by Wedderburn's Little Theorem). For complete, discretely valuated fields with finite residue fields, there is exactly one isomorphism class of quaternion division algebra (in addition to the split quaternion algebra that always exists) but all octonion algebras are split again. But there is a class of fields `in between' these two classes that is not on the list: NOT-complete, discretely valuated fields with finite residue fields, or more concretely the fields $\mathbb{F}_p(t)$ for $p > 2$. Does anyone know if division octonion algebras over these fields exist? REPLY [9 votes]: The complete classification for fields of characteristic unequal to 2 is given in Section 8 of Serre's paper J.-P. Serre. Cohomologie galoisienne : progrès et problèmes. Séminaire Bourbaki, Volume 36 (1993-1994) , Talk no. 783 , p. 229-257. Numdam. The main result is that octonion algebras over a field $F$ are classified by decomposable classes in $H^3(F,\mathbb{Z}/2\mathbb{Z})$, or alternatively by Pfister 3-forms over the field $F$. Note also that an octonion algebra is either split or division. In particular, if the 2-cohomological dimension of the field is at most 2, then there are no nonsplit octonion algebras. Function fields in one variable over finite fields have cohomological dimension $2$ (p. 85 of Serre's Galois cohomology book), so there are no nonsplit octonion algebras in that case. For a cohomological dimension 3 case: there is an explicit classification of octonion algebras over $K(T)$ where $K$ is a $p$-adic field in Serre's article, section 8.3.<|endoftext|> TITLE: Equivalence of the definitions of a sheaf in SGA4 and in "Categories and Sheaves" QUESTION [6 upvotes]: I asked this question on Mathematics Stack Exchange, but got no answer. I don't understand why the definition of a sheaf (Definition 17.3.1 (ii)) given in the book [KS] Categories and Sheaves by Kashiwara and Schapira is equivalent to the definition of a sheaf (Definition 2.1) given in [V] Verdier, Exposé II, SGA4, http://www.normalesup.org/~forgogozo/SGA4/02/02.pdf To simplify, let me consider only set-valued presheaves. Here is, in the terminology of [KS], how I understand the two definitions. (Warning: my understanding might be incorrect!) Let $\mathcal U$ be a universe, let $X$ be a small site and let $F$ be a $\mathcal U$-set-valued presheaf over $X$. Then: $\bullet\ F$ is sheaf in the sense of [V] if $F(f)$ is an isomorphism for any $A$ in $(\mathcal C_X)^\wedge$, any $U$ in $\mathcal C_X$, and any local isomorphism $f:A\to U$ which is a monomorphism, $\bullet\ F$ is sheaf in the sense of [KS] if $F(f)$ is an isomorphism for any $A$ in $(\mathcal C_X)^\wedge$, any $U$ in $\mathcal C_X$, and any local isomorphism $f:A\to U$. The difference is that the local isomorphism $f$ is supposed to be a monomorphism in Verdier's definition. A "KS-sheaf" is of course a "V-sheaf", but I'm unable to prove the converse. REPLY [4 votes]: Actually I couldn't quite figure out how to do a Ken Brown sort of argument, but here's an argument that works: Let $L$ denote the usual sheafification functor, a la Grothendieck and Verdier etc. Then I claim it's enough to show $L$ sends local epimorphisms to epimorphisms and local monomorphisms to monomorphisms. Indeed, if this is the case then $L$ takes local isomorphisms to epi-monomorphisms, and such things are isomorphisms in toposes. Since $L$ preserves finite limits, we need only check that $L$ takes local epis to epis (since $A \to B$ is a local mono iff $A \to A\times_BA$ is a local epi). But that's not so bad: If $A \to B$ is a local epimorphism then the map $im(A \to B) \to B$ is a local epimorphism and a monomorphism, and hence a monic, local isomorphism. But $L$ preserves images (since they're computed as colimits and L preserves those) and takes monic, local isomorphisms to isomorphisms. Thus $LA \to LB$ has the property that $im(LA \to LB) \to LB$ is an isomorphism (NB: that image is computed in the category of sheaves), and so the map is epi in the category of sheaves, which is what we wanted.<|endoftext|> TITLE: Matroids of rank two QUESTION [8 upvotes]: I am interested in matroids of rank two and would like to understand how interesting/big this class of matroids is. I know that the 2-uniform matroid on (k+2) elements is not representable over any field with at most k elements. (see Oxley, p203). Does there exist any survey on matroids of rank two? REPLY [6 votes]: Up to simplification (suppressing loops and parallel elements), every rank two matroid is just a rank two uniform matroid. Note that the vectors $(1, a_1), \dots, (1, a_n)$ represent the uniform matroid $U_{2,n}$, provided that $a_1, \dots, a_n$ are all distinct. Thus, the rank two matroids are all representable over each infinite field. REPLY [5 votes]: At first, neglect all loops (elements of rank 0). Now call two elements $a,b$ similar, if $\{a,b\}$ is not a base. It follows that if $\{a,c\}$ is a base, than $\{b,c\}$ is also a base (else the set $\{a,b,c\}$ would have inclusion-maximal independent subsets $\{b\}$ and $\{a,c\}$ of different size.) In other words, similarity is an equivalence relation. So, all elements may be partitioned onto groups so that any two elements of different groups form a base, and from the same group not. Of course such a matroid is representable over sufficiently large field. Namely, if we have $m$ groups, we should have $m$ different directions of the lines on the plane, that is, the field must contain at least $m-1$ elements (and this suffices).<|endoftext|> TITLE: Equivalence of $G$-invariant symplectic forms QUESTION [5 upvotes]: Let $V$ be a finite-dimensional complex vector space with a linear action of a complex reductive group $G$. Suppose that $\omega_0$ and $\omega_1$ are two $G$-invariant complex symplectic bilinear forms on $V$. Is there always a $G$-equivariant isomorphism $\varphi:V\to V$ such that $\varphi^*\omega_1=\omega_0$? Remark. This is not true for real symplectic forms and compact Lie group actions [Dellnitz-Melbourne. The equivariant Darboux theorem. Lect. Appl. Math 29 (1993): 163-169]. REPLY [7 votes]: It is true that any two $G$-invariant symplectic forms on a representation are equivalent. A proof of this fact is provided in F. Knop: "Classification of multiplicity free symplectic representations", J. Algebra 301 (2006) 531–553. See Thm 2.1 (b). Since the question is so natural, I suspect that thismust have been observed earlier, though. The argument is roughly as follows: Let $(V,\omega)$ be a symplectic representation and let $U\subseteq V$ be irreducible. If $U$ carries an invariant symplectic form then one can find a copy of $U$ in $V$ such that $\omega|_U\ne0$. Then $V=U\oplus U^\perp$ and one proceeds by induction. If $U$ is not symplectic then $V$ contains a copy of $U^*$ such that $\omega$ restricted to $\overline U=U\oplus U^*$ is non-degenerate. Then $V=\overline U\oplus \overline U^\perp$ and one proceeds by induction. The reason why the proof breaks down over $\mathbb R$ is that on an an absolutely irreducible symplectic representation $V$ a scalar $a\in\mathbb R^*$ transforms $\omega $ into $a^2\omega$. So $\omega$ and $-\omega$ are not equivalent.<|endoftext|> TITLE: The spectrum of the discrete Laplacian QUESTION [5 upvotes]: Consider a connected (we define connected components by defining the set of vertices where every vertex has one neighbour) sublattice $V$ of the square lattice $V \subset\mathbb{Z}^2.$ On this we define the discrete Laplacian as $T:\ell^2(V) \rightarrow \ell^2(V)$ by $$ (Tf)(x)=\sum_{y \text{ neighbour of } x}(f(y)-f(x)).$$ Now, my question is: Is there an infinite(!) connected sublattice $V \subset\mathbb{Z}^2$ on which this objects has eigenfunctions? Why do I ask: It is easy to see that on $\mathbb{Z}^2$ by using the Fourier transform for example, the spectrum is purely absolutely continuous. I would like to understand whether this is because $\mathbb{Z}^2$ is infinite (and so every eigenfunction would dissolve to infinity) or whether this is because $\mathbb{Z}^2$ is translational invariant. Why the assumptions: If the sublattice is finite, we obtain eigenfunctions because this operator is just a matrix. If the sublattice was not connected, it could have a finite connected component. So to exclude this, I stated the assumptions. REPLY [3 votes]: Yes, I'm sure this is possible, and I think there will be many ways to do it, but I don't have a fully worked out rigorous argument right now. But let me throw out some ideas. First of all, a 1D Laplacian with a random potential (aka Anderson model) $$ (Hf)(n) = f(n+1)+f(n-1)+q(n)f(n) $$ on $\ell^2(\mathbb Z)$ will have pure point spectrum with prob $1$. For example, this is true if the $q(n)$ are iid and take two values. Now you could simulate this by taking your $V$ as $V=\mathbb Z\cup \{(2n,1): n\in A\}$, with $A\subseteq \mathbb Z$ a random set produced in the same way. It's not totally clear to me if there is an easy way to reduce this to the original model or if one would have to go through the whole analysis again to prove the result for $\Delta$ on $\ell^2(V)$, but I have no doubt about the result itself (in the classical case, too, the result is fairly robust and doesn't depend on how exactly you introduce the randomness). What is easier to prove is absence of absolutely continuous spectrum. For example, if you take $A$ above (not random, but) as a sparse set, say $A=\{ \pm 2^n \}$, then $\sigma_{ac}=\emptyset$. This follows from my result on reflectionless limit points. Typically, the spectrum will be purely singular continuous; in particular, this will be true if $A$ is something like $A=\{ \pm 2^{n^2}\}$.<|endoftext|> TITLE: Smallest volume representatives of homology QUESTION [11 upvotes]: Given a Riemannian manifold, I have a notion of volume for each of my chains, so it makes sense to ask for a representative of a homology class with the smallest volume. Are there conditions for when such a representative exists? I know it is not always possible: for example, the generator of $H^1$ of the punctured plane doesn't have a shortest representative. But perhaps if we required our space to be compact? I'm especially interested in the case where the class has a submanifold representative. In that case, I would like to run a generalized mean curvature flow on the representative to get one with (locally) smallest volume. REPLY [15 votes]: Finding volume-minimizing representatives of homology classes is one of the major applications of geometric measure theory. It's a theorem of Federer and Fleming that any nontrivial integral $k$-homology class of a smooth compact $n$-manifold (with $n>k$) can be represented by an integer-multiplicity rectifiable current of least volume. That's a bit far from a submanifold representation but there is a fair amount of work that establishes regularity in various cases; from memory, I think if $n<8$ and $k=n-1$ then there's a smooth volume minimizing hypersurface. See e.g. the introduction to Generic regularity of homologically area minimizing hypersurfaces in eight dimensional manifolds for some references.<|endoftext|> TITLE: Todd genus of symplectic $4$-manifolds a smooth invariant? QUESTION [8 upvotes]: Suppose that $(M_{1},\omega_{1})$ and $(M_{2},\omega_{2})$ are compact symplectic $4$-manifolds, that are (oriented) diffeomorphic. Is it true that the Todd genus ($\frac{1}{12} (c_{1}^{2} + c_{2})(M_{i})$) are equal? I know a reference that the answer is yes for algebraic surfaces (since the Todd genus equals $\chi(\mathcal{O}_{S}) = 1 - \frac{1}{2}b_{1}(S) + P_{1}(S) $ and the plurigenera are known to be an (oriented) smooth invariant by Seiberg-Witten theory). REPLY [5 votes]: In dimension 4, the Todd genus does not depend on the choice of a symplectic structure or even on an almost complex structure. If $M$ is an almost complex 4-manifold, then $\langle c_1(M)^2, [M]\rangle = 2\chi(M) + 3\sigma(M)$ (see here, p. 9), and $\langle c_2(M), [M]\rangle = \chi(M)$; here $\chi(M)$ is the Euler characteristic of $M$ and $\sigma(M)$ is its signature. Thus any orientation-preserving diffeomorphism preserves the Todd genus, and the orientation-preserving assumption is necessary, because the signature depends on the choice of orientation. This is also mentioned at the end of §2 of this paper by Łukasz Bąk.<|endoftext|> TITLE: Number of matchings of even cycles QUESTION [14 upvotes]: By doing some calculations on the generating function of matching polynomials of cycles I made the following interesting observation: For all positive integers $n>1$ and $k TITLE: Consistency of Rado's conjecture with not CH QUESTION [7 upvotes]: Rado's conjecture (one of many equivalent formulations) states: any non-special tree has a non-special subtree of cardinality $\aleph_1$. "Special" means a tree can be decomposed into countably many antichains. Hence in particular, special trees have no cofinal branches of uncountable length. I'm asking for a reference of the consistency of RC with not CH. The usual Levy collapse of a supercompact cardinal is good for RC + CH, and the key observation is that no countably closed forcing can specialize a tree of height $\omega_1$. With reals being added I'm not so sure. REPLY [4 votes]: Rado's conjecture holds in Mitchell's model (of course, start with a strongly compact instead of a weakly compact) granted the following: If $T$ if a non-special tree of height $\omega_1$, then $T$ remains non-special after Cohen forcing $Add(\omega, \kappa)$ for any regular $\kappa$. It suffices to show the case where $T$ has size $\aleph_1$ and $\kappa=\omega_1$ (they are in fact equivalent). We will show this. We might assume $T$ is branchless and $\kappa$ is uncountable as if there is a branch through $T$, $T$ can't be special in any $\omega_1$-preserving extension and if $\kappa=\omega$, then we can cook up a function in the ground as the size of the forcing is countable. Now suppose for the sake of contradiction, $\dot{g}: T\to \omega$ is forced to be a specializing function. For each $t\in T$, let $p_t\in Add(\omega,\kappa)$ be a condition deciding the value of $\dot{g}(t)$ to be $n_t\in \omega$. We might shrink to a non-special subtree $T'$ of $T$ such that there exists $n\in \omega$, for all $t\in T'$, $n_t=n$. We will develop a kind of $\Delta$-system lemma on non-special trees, aka the goal is to find non-special subtree $T''$ of $T'$ such that any $t,t'\in T''$ such that $t TITLE: Similar matrices over $\mathbb Z_p$ QUESTION [14 upvotes]: Let $A$ and $B$ be two $n \times n$ matrices with entries in $\mathbb Z_p$, the $p$-adic integers. Is it true that $A$ and $B$ are conjugate iff they're conjugate over $\mathbb Q_p$ and over $\mathbb F_p$? REPLY [23 votes]: A counterexample is $$ A=\left[\begin{array}{cc}0&2\\8&0\end{array}\right],\hspace{5mm}B=\left[\begin{array}{cc}0&4\\4&0\end{array}\right]\in M_2(\mathbb{Z}_2). $$ The matrices are conjugate in $\mathbb{Q}_2$ because they have the same eigenvalues $\pm 4$, and they are conjugate in $\mathbb{F}_2$ because both are $0$. But they are not conjugate in $\mathbb{Z}_2$ because $A\not\in 4\cdot M_2(\mathbb{Z}_2)$ but $B\in 4\cdot M_2(\mathbb{Z}_2)$.<|endoftext|> TITLE: Complexity of the Mandelbrot set on rationals QUESTION [8 upvotes]: Given two rationals $a,b \in \mathbb{Q}$, call $c = a + ib$, i.e., the complex number represented by these two rationals. A point $c$ is contained within the Mandelbrot set $M$ if the following procedure never halts: $z = c$ $while (abs(z) <= 2)$ $\hspace{0.3in} z = z^2+c$ Normally we pick some $k$ (say, 50) and then if it doesn't halt after that many iterations we assume it is in the Mandelbrot set and stop looping. The ordinary algorithm for this takes $O(n2^k)$ time (simply computing the above loop using something like Java's BigInteger class for the numerators and denominators and using fraction arithmetic). Is there a polynomial time algorithm for determining if a given $c$ breaks out of this loop within $k$ steps, in terms of the magnitude of $k$ and in terms of $n$ bits representing the numerator and denominator of $a$ and $b$? For reference, $abs(c) = \sqrt{a^2 + b^2}$ and $c^2 = a^2 + 2abi +b^2i^2 = a^2 - b^2 + 2abi$ because $i^2 = -1$. (this was cross posted at cstheory stackexchange but in hindsight I thought it actually fit better here) REPLY [4 votes]: Okay so I haven't proved this yet but in practice this seems to be working pretty well for me. The trick is that you use interval arithmetic for the real and imag component of your number, using fractions with biginteger type numbers for their numerator and denominator. The bits still grow very quickly, but what you can do is "bound" them at each iter. For example, 0.000123141532 can become [0.000124, 0.000123], then you use those as your new intervals. If you end up having your upper estimate of the norm get >= 4 but the lower estimate of the norm be < 4 then you want to try a closer interval (so like [0.00012314, 0.00012315]). I find that usually 200 bits is plenty enough precision to run 1000 iters which wouldn't be practical otherwise, but these could grow as much as needed. Of course this code isn't super fast and optimized and using floating point numbers for rendering fractals is much easier, but the important part here is that these results are exact, this is guaranteed to give you whether or not it breaks in k steps. It runs basically in time linear in k since you can just fix precision, but there may be worst case behaviors on the edge of the mandelbrot that require the interval to get so small that you get exponential slowdown. There was some stuff on perturbation theory that might help prove that this doesn't happen but I haven't done that yet. I haven't ran into that issue in all my renders I've done with this so far at least. For now, here is that idea implemented in python. I'd like to clean up this code/comment it but I figured I'd just share it first so others can believe me that this works and maybe help with proving perturbation theory stuff I'm not sure about. Specifically, here is the interval class I made in python: import math import fractions class Interval(object): def __init__(self, lo, hi=None): if type(lo) is Interval and hi is None: self.lo = lo.lo self.hi = lo.hi else: self.lo = fractions.Fraction(lo) if hi is None: hi = lo self.hi = fractions.Fraction(hi) if self.lo > self.hi: raise Exception("lo " + str(self.lo) + " is greater than hi: " + str(self.hi)) def applyOp(self, val, op): val = Interval(val) loop = getattr(self.lo, op) hiop = getattr(self.hi, op) things = [loop(val.lo), loop(val.hi), hiop(val.lo), hiop(val.hi)] return Interval(min(things), max(things)) def __mul__(self, val): return self.applyOp(val, '__mul__') def __rmul__(self, val): return self.applyOp(val, '__rmul__') def __add__(self, val): return self.applyOp(val, '__add__') def __sub__(self, val): return self.applyOp(val, '__sub__') def __div__(self, val): return self.applyOp(val, '__div__') def __neg__(self): return Interval(-self.hi, -self.lo) def __repr__(self): return self.__str__() def __pow__(self, val): if val == 2: if self.lo <= 0 and self.hi >= 0: return Interval(0, max(self.lo*self.lo, self.hi*self.hi)) elif self.lo <= 0 and self.hi <= 0: return Interval(self.hi*self.hi, self.lo*self.lo) elif self.lo >= 0 and self.hi >= 0: return Interval(self.lo*self.lo, self.hi*self.hi) raise Exception("don't know how to do higher pow yet") def reduce(self, maxNDigits): lower = reduceFract(self.lo, maxNDigits, roundDown=True) bigger = reduceFract(self.hi, maxNDigits, roundDown=False) return Interval(lower, bigger) def __str__(self): res = "[" + str(self.lo) + "," + str(self.hi) + ']' return res def approx(self): return [float(self.lo), float(self.hi)] def __contains__(self, other): other = Interval(other) return self.lo <= other.lo and self.hi >= other.hi def reduceFract(res, biggestAllowed, roundDown=True): if res.numerator == 0 or res.denominator == 0: return res mag = int(min(math.log(abs(res.numerator), 2),math.log(abs(res.denominator), 2))) if mag > biggestAllowed: mag = mag - biggestAllowed num = res.numerator>>mag den = res.denominator>>mag resFrac = fractions.Fraction(num, den) if roundDown and resFrac > res: resFrac = fractions.Fraction(num-1, den) elif not roundDown and resFrac < res: resFrac = fractions.Fraction(num+1, den) res = resFrac return res now we can do this: def mandelInterval(cr, ci, n, reduceAmount): cr = Interval(cr, cr) ci = Interval(ci, ci) resr, resi, state = mandelIntervalHelper(cr, ci, n, reduceAmount) mag = resr*resr + resi*resi return resr.approx(), resi.approx(), mag.approx(), state def mandelIntervalHelper(cr, ci, n, reduceAmount): if n <= 0: return cr, ci, None else: i = 1 pr, pi, state = mandelIntervalHelper(cr,ci,n-1, reduceAmount) while True: if state == "need better estimates": i += 1 pr, pi, state = mandelIntervalHelper(cr,ci,n-1, reduceAmount*i) else: break if state in ['outside', 'need better estimates']: return pr, pi, state resr = (pr*pr-pi*pi + cr).reduce(reduceAmount) resi = (2*pr*pi+ci).reduce(reduceAmount) mag = resr**2 + resi**2 if mag.lo >= 4 and mag.hi >= 4: return resr, resi, "outside" elif mag.lo < 4 and mag.hi >= 4: return resr, resi, "need better estimates" return resr, resi, "inside" So for example, saving this to ms.py: >>> import ms >>> # c= -0.397959183673 + -0.673469387755i >>> # 1000 iters, 20 bits of precision (goes to 40, 60, 80, ... if needed) >>> ms.mandelInterval(-0.397959183673, -0.673469387755, 1000, 20) ([-0.05965659022331238, -0.033619076013565063], [-2.383112907409668, -2.357083320617676], [5.556972022606055, 5.682786038219633], 'outside') >>> # [-0.05965659022331238, -0.033619076013565063] is interval for real(c) >>> # [-2.383112907409668, -2.357083320617676] is interval for imag(c) >>> # [5.556972022606055, 5.682786038219633] is interval for distance >>> # because both sides of the interval for distance are >= 4, the last value is 'outside', meaning we are outside the mandelbrot set. This will be 'inside' if we are inside it.<|endoftext|> TITLE: What upper bounds are known on the number of non-isomorphic cycle matroids? QUESTION [5 upvotes]: For $n\in\mathbb{N}^{+}$, let $c_{n}$ denote the number of simple non-isomorphic cycle matroids of graphs on $n$ vertices. That is, let $$A(n)=\{M(G)\;;\;G\text{ is a graph on }n\text{ vertices}\},$$ and let $B(n)$ be a largest subset of $A(n)$ such that no two elements of $B(n)$ are isomorphic (as matroids). Then $$c_{n}\equiv|B(n)|.$$ (Here $M(G)$ denotes the cycle matroid of graph $G$. ) A simple upper bound for $c_{n}$ would be something like $2^{2^{\binom{n}{2}}}$, since the ground set of the cycle matroid of a graph on $n$ vertices is of size at most $\binom{n}{2}$. Are there any better upper bounds known for $c_{n}$? Edit: It looks like I am asking for the number of non-$2$-isomorphic graphs on $n$ vertices. Whitney's 2-isomorphism theorem (version of Oxley's Matroid Theory) states: "Let $G$ and $H$ be graphs having no isolated vertices. Then $M(G)$ and $M(H)$ are isomorphic if and only if $G$ and $H$ are $2$-isomorphic." Intuitively, this number looks like it should be significantly smaller than the number of non-isomorphic graphs on $n$ vertices, since non-isomorphic graphs can be $2$-isomorphic. Do we know of upper bounds on the number of non-$2$-isomorphic graphs on $n$ vertices? (Should I ask this as a separate question?) REPLY [4 votes]: It seems now that the question really boins down to the number of non-isomorphic graphs on $n$ vertices. Denote this number by $f(n)$. Clearly, $$ 2^{n\choose 2}\geq f(n)\geq \frac{2^{n\choose 2}}{n!}=\frac{2^{n\choose 2}}{n^{O(n)}}, $$ these two numbers being sufficiently close on the logarithic scale. (Moreover, one may sharpen the upper bound by applying Burnside's formula and investigating the number of fixed points of any permutation of vertices). As Tony Huynh mentioned, $c_n\leq f(n)$. Let us now construct a large family of graphs with a few isomorphisms between their cycle matroids. Let $n$ be odd. Take any graph $H$ on $n-1$ vertices. Add the $n$th vertex $v$ connected to all of them to obtain a graph $G$; non-isomorphic $H$ clearly provide non-isomorphic $G$. In the cycle matroid $M(G)$ there exists a set $S$ of $n-1$ edges (incident to $v$) such that $(*)$ every edge not from $S$ forms a circuit (=minimal non-independent set) with two edges from $S$. Consider any other graph $G'$ (obtained in the described manner) with $M(G')\cong M(G)$; if $S$ corresponds in $G'$ to $n-1$ edges sharing a vertex $v'$, then $G'\cong G$ (every edge not from $S$ connects the other vertices of the two edges from $S$ forming a circuit with it). Thus, the number of such graphs $G'$ is at most the number of sets $S'$ in $M(G)$ satisfying $(*)$. We are now to bound this number from above. Assume that $S'$ contains edges $e_1,\dots,e_k\in S$, and the other edges are not from $S$. Any $e\in S\setminus S'$ should be complemented in $S'$ to a 3-circuit --- by one of the $e_i$ and some other edge. One of the $e_i$ may be chosen in at most $k$ ways, the other one is then found uniquely, and these `other' edges are distinct for distinct $e$. Thus the number of sets $S'$ satisfying $(*)$ is at most $k^{n-k}=n^{O(n)}$, hence we get at least $f(n-1)/n^{O(n)}$ pairwise non-isomorphic cycle matroids. To summarize, $$ 2^{n\choose 2}\geq f(n)\geq c_n\geq \frac{f(n-1)}{n^{O(n)}}\geq \frac{2^{n\choose 2}}{n^{O(n)}}. $$<|endoftext|> TITLE: Upper bound for chromatic number of graphs with $\omega(G)\leq\lfloor\frac{\Delta(G)+1}{2}\rfloor+1$ QUESTION [7 upvotes]: Let $G$ be a simple graph such that $\omega(G)\leq\lfloor\frac{\Delta(G)+1}{2}\rfloor+1$ where $\Delta(G)$ is the maximal degree of $G$. Is it true that \begin{equation} \chi(G)\leq \lfloor\frac{\Delta(G)+1}{2}\rfloor+2? \end{equation} A similar problem can be found here, where, ignoring the exceptional case of odd cycles, it is conjectured that Conjecture (Reed). For any graph $G$ of maximum degree $\Delta(G)$, $\chi(G)$ is at most $\lceil\frac{\Delta(G)+1+\omega(G)}{2}\rceil$. UPDATE: In the case of graphs that satisfy $\omega(G)\leq\lfloor\frac{\Delta(G)+1}{2}\rfloor+1$, the statement in my problem is stronger than Reed's conjecture. For example suppose $\Delta=100$ and $\omega=50$; then by Reed's conjecture we need $76$ colors, while if my conjecture is true, we need just $52$ colors! REPLY [2 votes]: i can't comment yet, but the answer of Ztas Nellets is correct, this conjecture is false, the probabilistic construction in my linked paper with Cranston is from Reed's discussion of variants and tightness of his conjecture. In regards to the paper of Kostochka, i proved a generalization here https://doi.org/10.1002/jgt.21634 which also implies the slightly stronger result Peter Heinig mentions. The proof is essentially the same as Kostochka's original in the paper nobody can find (i had a hardcopy at some point, but cannot find it).<|endoftext|> TITLE: Changing values of digits of an algebraic irrational number QUESTION [5 upvotes]: Consider the set of irrationals $\mathbb{I} \cap (0,1)$. Define the map $f$ that takes $x= \sum_{n\ge 1} \frac{a_n}{3^n}$ to $\sum_{n\ge 1} \frac{ \phi(a_n)}{3^n}$, where $\phi$ is the permutation of $\{0,1,2\}$, $0\mapsto 1 \mapsto 2\mapsto 0$. Is it possible that both $x$ and $f(x)$ are algebraic for some $x$ ? The common belief is that they are "random". A solution would not run contrary to that. Still my feeling is that this would not be possible. REPLY [9 votes]: Let $x_j = \sum_{n \geq 1} \mathrm{1}_{a_n=j} 3^{-n}$. Then $$ x_0 + x_1 + x_2 = \frac{1}{2} \\ x_1 + 2x_2 = x \\ x_0 + 2x_1 = f(x). $$ If $x$ and $f(x)$ are algebraic then the equations above would yield that $x_j$ is algebraic for each $j$. It is conjectured and generally believed that any irrational algebraic number is normal in all bases. Since $x_j$ is obviously not normal in base $3$, this conjecture would imply that $x_j$ is rational for each $j$, and thus that $x$ is rational. REPLY [7 votes]: Just to augment @js21's answer. The question of whether irrational $x_i$ are necessarily transcendent is equivalent to this question. As the answer suggests, even this specific question is wide open, or at least was such in 2012. As for the other bases $b>2$, the result is similar. Defining $x_i$ as in js21's answer, you get that $$ x_{b-1}=\frac{x+\frac1{b-1}-f(x)}{b} $$ is algebraic though not normal. (Surely, if $x_{b-1}$ is just rational, then $x-(b-1)x_{b-1}$ will be an algebraic irrational though not normal.)<|endoftext|> TITLE: Why is the root poset is graded by height? QUESTION [9 upvotes]: Let $\Phi$ be a finite crytallographic root system. Let $\Phi^+$ be the positive roots and $\alpha_1$, ..., $\alpha_n$ be the simple roots. For $\beta = \sum c_i \alpha_i$ in $\Phi^+$, we define $h(\beta) = \sum c_i$. For $\beta = \sum c_i \alpha_i$ and $\gamma = \sum d_i \alpha_i$, we define $\beta \preceq \gamma$ iff $c_i \leq d_i$ for all $i$. Many sources state that $\Phi^+$ is graded by $h$. The nontrivial part of this statement is that, if $\alpha \leq \gamma$ with $h(\gamma) - h(\alpha) \geq 2$, then there is a root $\beta$ with $\alpha \leq \beta \leq \gamma$. Could someone give me a proof or reference to a proof, other than type by type check? To show that I haven't been completely lazy: Humphreys defines $\Phi^+$ and $h$ but doesn't state that $h$ grades $\Phi^+$, Bjorner and Brenti define a different, unrelated partial order on $\Phi^+$ which they call the root poset. Cuntz and Stump cite Armstrong Section 5.4.1 but it doesn't seem to be in there. REPLY [7 votes]: Let $(\cdot,\cdot)$ be a positve definite Weyl group invariant product on $\mathbb{R}\Phi$. Let $\beta$ and $\gamma$ be positive roots with $\beta\leq \gamma$ and $h(\gamma)-h(\beta)\geq 2$. Let $v=\gamma-\beta$. Let $\alpha_i$ be a simple root which occurs in $v$. Suppose for want of a contradiction that $\beta+\alpha_i$ and $\gamma-\alpha_i$ are not positive roots. Then $(\beta,\alpha_i)\geq 0$ and $(\gamma,\alpha_i)\leq 0$. In particular $$(v,\alpha_i)\leq 0.$$ Since $v$ is a positive sum of the $\alpha_i$, we sum these inequalities to get $(v,v)\leq 0$. This is a contradiction since $(\cdot,\cdot)$ is positive definite. QED.<|endoftext|> TITLE: Current interest in geometric properties of Hilbert fundamental domains QUESTION [6 upvotes]: Harvey Cohn published several articles in the 1960's analyzing geometric properties of fundamental domains for Hilbert modular surfaces. H. Cohn, "On the shape of the fundamental domain of the Hilbert modular group," Theory of Numbers, A. L. Whiteman (Ed.), Proc. Svmpos. Pure Math. Vol. 8, Amer. Math. Soc, Providence, R. I., 1965,pp. 190-202. Cohn, Harvey. "A numerical survey of the floors of various Hilbert fundamental domains." Mathematics of Computation 19.92 (1965): 594-605. Cohn, Harvey. "Note on how Hilbert modular domains become increasingly complicated." Journal of Mathematical Analysis and Applications 15.1 (1966): 55-59. Cohn, Harvey. "Some computer-assisted topological models of Hilbert fundamental domains." Mathematics of Computation 23.107 (1969): 475-487. To clarify, he's studying certain fundamental domains for the quotient $$(\mathcal{H}^2\times\mathcal{H}^2)/\mathrm{SL}_2(\mathbb{Z}_K)$$ where $\mathcal{H}^2$ is the upper half-plane, $\mathbb{Z}_K$ is the ring of integers of a real quadratic field $K$, and the group action is by Möbius transformations with a Galois twist. These articles are concerned with nailing down precise geometric properties of the fundamental domains. Especially, he talks about a $3$-manifold boundary for the domain (as a subset of $\mathcal{H}^2\times\mathcal{H}^2$) that arrises in a natural way, which he calls the floor of the domain. The results mostly talk about the complexity of the regions, some obstacles to describing them in general, and eventually move on to computer generated approximations about specific examples. For instance, the main theorem in the second article above is that $\mathbb{Q}(\sqrt 5)$ is the only UFD for which the floor has just one piece. The fourth article above crunches some data and some (circa 1969) computer generated images of cross-sections of floors for the cases $\mathbb{Q}(\sqrt n)$ with $n=2,3,5,6$. My current research project has lead me to a result that allows me to say a lot more about what's happening here. (This wasn't really the plan, as research goes, but to me it is very interesting especially when I look back at what Cohn did.) I'm wondering if (or how much) these results might interest the community. There is a lot already known about these objects' arithmetic and topological properties, and what I have (so far) does not extend these. It does give a different way of seeing/explaining some known facts in a hands-on way, which is cool, but that's not exactly Theorem-(with a capital T)-worthy. What I do have regards putting precise shapes to the objects, giving algorithms to compute sides, showing models for 3D cross-sections (in circa 2017 computer graphics -- interestingly I think Cohn would have noticed some stuff I noticed if he just had better resolution!) ... things like that. But I don't have a good sense of the relevance of these properties to the current study of Hilbert modular surfaces. So, who (if anyone) might find that kind of thing interesting, and why? Would it make a decent paper on its own? Can you direct me to some other relevant literature? Thanks in advance! REPLY [2 votes]: I make the disclaimer that I am by no means an "expert" on computational aspects of modular forms, although I am aware of some of the issues. First, I'd think you should look at the Sage (sagemath.org) situation, where they already have a great many algorithmic things packaged-up. Many elliptic modular forms things are automated, for example. Graphs of zeta at various perspectives are nearly instantaneous. Some decades ago, there were some delicate results in elliptic modular forms exploiting subtle details about choices of fundamental domains... but, I'm very sorry, I do not remember the authors or the results. I should say that it is actually lucky that most basic results about Hilbert modular forms or other groups' automorphic forms do not depend much on details about fundamental domains, since such details are irregular and expensive. So there'd not be an immediate, general benefit, I think, in knowing something about fundamental domains. For that matter, there have been theorems proven indicating that in higher rank, generic Shimura varieties are "of general type" as algebraic varieties, so nothing "sharp" can be said from that side. Likewise, this might cause one to doubt that details about a fundamental domain could be broadly helpful. But perhaps I am mistaken. (Also, there was a paper by Hammond in the Boulder conference about explicit generation of rings of Hilbert modular forms.)<|endoftext|> TITLE: Length of simple closed curve in half-translation surface QUESTION [7 upvotes]: Let $R$ be a Riemann surface of genus $g\ge 2$ and $q$ an holomorphic quadratic differential on $R$. Together they determine a semi-translation structure: an atlas on $X$ such that its changes of charts are of the form $z\mapsto \pm z+c$ and a singular flat metric $|q|$ on $X$. Suppose that $q$ has at least one singularity of odd order: I can't completely grasp what the $-1$ holonomy implies when considering the lengths of simple closed curves on $(X,q)$. In particular, let $c\subset X$ be a closed simple geodesic for $|q|$ which doesn't meet any singular point of $|q|$. Then, is it true that there is always a cylinder in $(X,q)$ foliated by closed geodesics parallel to $c$ (as in the case of translation surfaces with trivial holonomy)? If no, then is it still always possible to assume the existence of a concatenation of saddle connections of $q$ parallel to $c$ and with the same length? REPLY [2 votes]: Yes, it is, of course, true. Even though foliations defined by $q$ are not orientable, the genus $2$ surface is orientable, so an $\varepsilon$-neighborhood of a closed geodesic $\gamma$ is just the cylinder $\gamma\times [-\varepsilon, \varepsilon]$. This statement holds more generally for orientable surfaces with flat metric and conical singularities. If you find a simple closed geodesic (smooth one and disjoint from conical points) its $\varepsilon$-neighborhood is a cylinder.<|endoftext|> TITLE: Stable homotopy type theory? QUESTION [19 upvotes]: This is a combined question, strictly speaking I am asking three questions concerning, respectively, homotopy type theory, stable homotopy theory and Yetter-Drinfeld modules. But I believe in the present context they can be viewed as a single question. After a lecture on HoTT by Voevodsky (at TACL 2013 in Nashville) I asked him whether Homotopy Type Theory can be useful for stable homotopy theory. Without much hesitation he replied that HoTT is not really well suited to describe stable phenomena. We never talked after that, and now we will never talk at all, so I still do not know where the main difficulty lies. Discussing this with several other people (sorry, don't exactly remember with whom) I gradually developed impression that the main difficulty relates to the fact that there seems to be no known good formalism combining constructive type theory and substructural logics, that would lead to some sort of linear type theory. Among possible models one should have triangulated monoidal categories, like stable homotopy category, and it seems not to be known what is the analog of dependent types there. The first question then is whether what I said above makes any sense, and whether there is any work in this direction. A situation where one has something like "linear dependent types" is in the context of Yetter-Drinfeld modules over Hopf algebras. Heuristically, the starting point is the fact that in any category $\mathscr C$ with (Cartesian) products every object $X$ has a unique comonoid structure, and the category of comodules over this comonoid is equivalent to the slice $\mathscr C/X$. One could then try to imitate this in (non-Cartesian) monoidal categories, considering comodules over comonoids as variable families over that comonoid. When this comonoid is equipped with a Hopf monoid structure, one can get a "linear" analog of Freyd-Yetter's crossed G-sets. The second question is then, whether anybody studied features of categories of Yetter-Drinfeld modules over varying Hopf algebras which would resemble either what happens in HoTT or in stable homotopy category. And the third question comes from the fact that while there are plenty of examples of monoids in stable homotopy categories (ring spectra) I cannot remember any considerations of comonoids or Hopf objects there. Does anybody know some works on this? Or even more simple question (actually fourth, sorry): in unstable homotopy theory one knows that for a, say, topological group $G$ there are certain homotopy categories of $G$-spaces and of spaces over $BG$ which are equivalent. Is there an analog of this fact in stable homotopy theory? Or maybe Voevodsky had in mind that there seems to be no analog of univalence in the stable context? (Well that was fifth, I certainly must stop here.) I found only one related question, Are there connections between Homotopy type theory and Grothendieck's theory of motives?, about possible connections between HoTT and motivic homotopy theory. I discovered my pessimistic comment there, remembering that answer by Voevodsky and arguing that motivic homotopy is in a sense even "more linear" than stable homotopy, so it might be even more difficult to relate these two. Anyway. REPLY [8 votes]: There are multiple possible routes to study stable homotopy theory in a type-theoretic system, some of which are work in progress by various people, but none of which are fully developed. As has been pointed out, one can make use of the fact that the category of parametrized spectra is an $(\infty,1)$-topos and therefore (conjecturally) admits an interpretation of HoTT, then use this model to motivate new structures and axioms for HoTT to produce a "homotopy type theory of parametrized spectra" analogous to what we have done for directed and cohesive homotopy type theories. There is work in progress on deciding what exactly these new features should be. On the other hand, one can try to attack the category of spectra directly. As has been pointed out, since it has finite limits it conjecturally admits a model of dependent type theory but without $\Pi$-types and without a universe. This lack makes it almost impossible to do anything resembling current homotopy type theory purely inside spectra. What seems more useful is to have an ambient unstable homotopy type theory which can be related to the stable one in some way. So that, for instance, we have an unstable homotopy type of maps between two spectra, and can talk about homotopies and apply the theory of h-levels and equivalences (using unstable $\Pi$-types) in that way. There could also be an unstable universe of stable types that is "univalent" in an appropriate sense. When one plays with this a bit, it starts to seem very natural, even necessary, to allow stable homotopy types to depend on unstable ones, so that we are heading back in the direction of parametrized spectra. But now instead of considering each parametrized spectrum as a basic object, we are stratifying them over their base spaces, with two context zones for unstable and stable types. In the non-homotopy context, it's not hard to write down such a type theory that has models in "indexed monoidal categories", and that categorical structure can be generalized to "indexed monoidal $(\infty,1)$-categories", which are mostly what the original nLab page "dependent linear type theory" is about (I've now moved that content to Indexed monoidal (∞,1)-category). But as far as I know, no one has yet proposed a good way to write down a homotopy type theory that has semantics in such indexed monoidal $(\infty,1)$-categories. There is work by type theorists on "dependent linear logic". Some of it, such as "linear LF", I believe involves linear types dependent on nonlinear ones. But I'm not aware of any version of such a theory yet that would, for instance, allow linear types to have non-linear identity types, which seems to me to be what we would want to be able to do stable homotopy theory this way.<|endoftext|> TITLE: Examples of Self-Maps of E8-Manifold QUESTION [12 upvotes]: Let $M$ be a simply connected topological 4-manifold with intersection form given by the E8 lattice. Does anyone know of examples of continuous self-maps of $M$ of degree 2 or 3? Or of degree any other prime for that matter? REPLY [13 votes]: There are plenty of integer matrices $A$ with $A^T E_8 A = d E_8$, which give plenty of maps, as in Oscar's answer. First, for $d=1$, these are the automorphisms of the $E_8$ lattice. There are 696729600 of these. Second, for any odd prime $p$, the E_8 quadratic form splits mod $p$, hence we can find a rank $4$ isotropic subspace. Consider the sublattice of elements that are in that subspace mod $p$. It is an index $p^4$ lattice, hence has discriminant $p^8$, and the quadratic form on it is divisible by $p$. Dividing by $p$, we obtain a unimodular lattice. It remains even and positive definite, so it is the $E_8$ lattice again. Fixing an isomorphism with $E_8$ describes a degree $p$ endomorphism of $E_8$. In fact the number of such subspaces should be $2(p^6+p^5+p^4+2p^3+p^2+p+1)(p^4+p^3+2p^2+p+1)(p^2+2p+1)$, so this gives $696729600\cdot 2(p^6+p^5+p^4+2p^3+p^2+p+1)(p^4+p^3+2p^2+p+1)(p^2+2p+1)$ self-maps of degree $p$. I think something similar can be done as well to produce degree $2$ endomorphisms, just with a little more care. REPLY [12 votes]: Such a map $f : M \to M$ of degree $d >0$ satisfies, with respect to the cup-product pairing $\langle -, - \rangle$, $$\langle f^*(x), f^*(y) \rangle = d \langle x, y \rangle.$$ Conversely, I claim that any integer matrix $A$ satisfying $$A^T E_8 A = d E_8$$ arises as $A = f^*$ for a map $f : M \to M$, necessarily of degree $d$. To see this, build $M$ as a CW-complex from $\bigvee_8 S^2$ by attaching a 4-cell along the map $g : S^3 \to\bigvee_8 S^2$ dictated by the $E_8$ form. The composition $$S^3 \overset{g}\to \bigvee_8 S^2 \overset{A}\to \bigvee_8 S^2$$ is the map dictated by the form $A^T E_8 A = d E_8$, so is $d \cdot g$. In particular it becomes nullhomotopic when composed with $\bigvee_8 S^2 \subset M$, giving a map $$f: M = (\bigvee_8 S^2) \cup_g D^4 \to M$$ which induces $A$ on second cohomology. Taking e.g. $A=\lambda \cdot \mathrm{Id}$ yields a self-map of degree $d=\lambda^2$. I am not sure whether it is possible to produce non-square degrees. REPLY [3 votes]: This is more like a long comment than a real answer. This seems like it's an algebra problem that is probably hard to solve. If you had such a map, of degree $d$, then you would get the following. Choose a basis for the homology, so that the induced map on $H_2(M)$ is written as a matrix $A$. Let J be the matrix for the intersection form with respect to this same basis. Then I think you are asking for $A^TJA = dJ$ (1). I don't have much of a feel for this but somehow it seems unlikely that there is any such matrix. Conversely, if you had such a matrix, then perhaps you could define a map of degree $d$ from M to itself. I would use the approach of the Whitehead-Milnor theorem; namely up to homotopy, you can write $M$ as a wedge of spheres union a $4$-cell, where the attaching map for the $4$-cell is determined by the matrix $J$. The matrix $A$ tells you how to map the $2$-skeleton to the $2$-skeleton, and I would guess that you get an extension over the $4$-cell with degree $d$ if (1) holds.<|endoftext|> TITLE: Query about SDG (Synthetic Differential Geometry) QUESTION [16 upvotes]: (Edited 10/17/17): With the hope of obtaining informed responses on the following intriguing remark of Marta Bunge on the status of Synthetic Differential Geometry, I have added a third question to the original two and expanded Bunge's quote to provide further context. In her "A Personal tribute to Bill Lawvere", presented at Union College in January 2013, Marta Bunge made the following observation about synthetic differential geometry (SDG): The basic idea of Synthetic Differential Geometry, in the form of the Kock-Lawvere axiom, requires, for a topos $\mathcal{E}$ with a ring object $\textrm{R}$ in it, that the subobject $\textrm{D}$ of $\textrm{R}$, consisting of those elements of square zero, be tiny and representing of tangent vectors at $0$ of arrows from $\textrm{R}$ to $\textrm{R}$. During the period 1981-88, I devoted myself almost totally to SDG, involving students and collaborators (Murray Heggie, Patrice Sawyer, Eduardo Dubuc, Felipe Gago) and participating in the workshops organized by Anders Kock at Aarhus, as well as in related special meetings. Lawvere’s intuition of the role of atoms (or “tiny objects”) in developing a simple form of Analysis going back to the ideas of Newton and Leibniz, and in the same spirit as in the work of André Weil, was both simple and attractive. In my work with my student Felipe Gago on a synthetic theory of smooth mappings, we used two additional axioms (Bunge-Dubuc 1987) to SDG, to wit, the representability of germs of smooth mappings by the sub object $\Delta = \neg \neg \{0\}$ of $\textrm{R}$, required to be tiny, and the existence and uniqueness of solutions of ordinary differential equations. However, no well adapted model of SDG is known at present to satisfy both of these axioms. This open problem is, in my view, pivotal for further progress in this fascinating area, which includes a synthetic proof of Mather’s theorem on the equivalence of locally stable and infinitesimally stable germs of smooth mappings (Bunge-Gago 1988), as well as Morse theory, developed synthetically in the thesis work of Felipe Gago at McGill. I have three questions regarding this remark. (i) Is the problem of the existence of a well adapted model of SDG that satisfies the above-stated two additional axioms still open? (ii) Assuming it is, how widely shared is her view on this matter in the SDG research community? That is, how widely is it held by members of the community that further progress in SDG is contingent on the existence of such a well adapted model? (iii) (New Question): Assuming the two additional assumptions are natural, which they appear to be, are there any cogent arguments opposed to her view? REPLY [11 votes]: In a paper by Marta Bunge and Eduardo Dubuc. "Local concepts in SDG and germ representability" (1987) certain axioms were laid down towards a synthetic theory of differential topology based on logical infinitesimal notions given by Jacques Penon in his 1985 Universite Paris VII thesis. One of them was Postulate WAII on $\Delta$-integration of vector fields, where $\Delta = \neg \neg \{0\}$ is a subobject of $R$ (the ring of line type in the Kock-Lawvere axioms) that represents germs of mappings from $R$ to $R$ by one of the basic axioms of what was to become SDT. As shown therein, this postulate is equivalent to the existence and uniqueness of (local) solutions to ODE within SDG. On the basis of Postulate WAII, along with further axioms for a theory now called SDT (Synthetic Differential Topology) an extension of SDG, Marta Bunge and her McGill University student Felipe Gago in "Synthetic aspects of $C^{\infty}$-mappings II : Mather's theorem for infinitesimally represented germs" (1988) proved the theorem of the title. Felipe Gago in "Morse theory for infinitesimally represented germs" (1988) developed Morse theory which was part of his 1988 McGill University thesis. In her 1999 thesis, Ana Maria San Luis, a student of Felipe Gago at Sgo de Compostela, gave an alternative proof of Mather's Theorem (without the so-called Preparation Theorem) still using Postulate WAII. Subsequently, Eduardo Dubuc, in "Germ representability and local integration of vector fields in a well adapted model of SDG" (1980) gave proofs of the validity of the basic axiom of germ representability as well as of Postulate WAII both in the topos $(G, R)$ with $G$ the topos of sheaves on the opposite of the category $B$ of finitely generated $C^{\infty}$-rings determined by a local (or germ determined) ideal, known as the Dubuc topos. Unfortunately, an error in the validity of the uniqueness part of the proof of Postulate WAII in the Dubuc topos $G$ given therein was found by Michael Makkai in discussions with Gonzalo Reyes. This meant therefore that there was at that time no known well adapted model of all of the axioms and postulates used in the work of Bunge-Gago-San Luis on a synthetic theory of smooth mappings and their singularities. Hence the remarks made by myself concerning the need to find a suitable well adapted model of SDT to validate our work. At the time this was indeed an open question but it is no longer one. As shown by myself at the Octoberfest, Ottawa, October 31-November 1, 2015, in a talk "Synthetic Theory of Stable Mappings and their Singularities" (whose slides have been posted in my Research Gate page), the uniqueness part of Postulate WAII (called Postulate VIII therein) is in fact valid in the Dubuc topos $G$, which is then a well adapted model of SDT (as well as of SDG). This result is now included in Chapter 12 ("$G$ as a WAM of SDT") of a forthcoming book by Marta Bunge, Felipe Gago and Ana San Luis, "Synthetic Differential Topology", Cambridge University Press, to appear in 2018. In Chapter 12, proofs of the validity of all (general and special) axioms and postulates used in our work are given with references to their various sources. Further developments of differential topology might need additional axioms and postulates, but it seems reasonable now (as it was so before the error was found!) to expect that those too will be shown valid in $(G, R)$. An interesting characterization of well adapted models of SDG (and so also of SDT) involving the Dedekind reals in a topos was given in a 1980 paper by Marta Bunge and Eduardo Dubuc "Archimedian local $C^{\infty}$-rings and models of SDG".<|endoftext|> TITLE: A representation of $F_{\sigma\delta}$-ideals? QUESTION [8 upvotes]: First some definitions. By $\mathcal P(\mathbb N)$ we denote the family of all subsets of $\mathbb N$ endowed with the metrizable separable topology generated by the countable base consisting of the sets $[A;B]=\{C\subset \mathbb N:C\cap B=A\}$ where $A,B$ run over finite subsets of $\mathbb N$. A subfamily $\mathcal F\subset\mathcal P(\mathbb N)$ is called hereditary if for any sets $F\in\mathcal F$ each subset of $F$ belongs to $\mathcal F$. An ideal on $\mathbb N$ is a hereditary subfamily $\mathcal I\subset \mathcal P(\mathbb N)$, closed under unions. An ideal $\mathcal I$ on $\mathbb N$ is called an $F_{\sigma\delta}$-ideal if $\mathcal I=\bigcap_{n=1}^\infty\bigcup_{m=1}^\infty\mathcal K_{n,m}$ for some closed subsets $\mathcal K_{n,m}$ of the compact topological space $\mathcal P(\mathbb N)$. Problem 1. Can each $F_{\sigma\delta}$-ideal $\mathcal I\subset\mathcal P(\mathbb N)$ be written as $\mathcal I=\bigcap_{n=1}^\infty\bigcup_{m=1}^\infty\mathcal K_{n,m}$ for some hereditary closed subsets $\mathcal K_{n,m}$ of the compact topological space $\mathcal P(\mathbb N)$? We can ask also a more general Problem 2. Can each hereditary $F_{\sigma\delta}$-set $\mathcal I\subset\mathcal P(\mathbb N)$ be written as $\mathcal I=\bigcap_{n=1}^\infty\bigcup_{m=1}^\infty\mathcal K_{n,m}$ for some hereditary closed subsets $\mathcal K_{n,m}$ of the compact topological space $\mathcal P(\mathbb N)$? Problem 2 is a related to another open (?) Problem 3. Does each Borel hereditary subset $\mathcal F\subset\mathcal P(\mathbb N)$ belong to the smallest subfamily of $\mathcal P(\mathbb N)$ containing all open hereditary sets and closed under taking countable unions and intersections? REPLY [4 votes]: In [1], Solecki proved (Theorem 2.1 combined with Lemma 2.4), Theorem: $\mathcal{I} \subset \mathcal{P}(\mathbb{N})$ is an analytic P-ideal, if and only if, there is some $\varphi:\mathcal{P}(\mathbb{N})\rightarrow [0,\infty)$ satisfying the following, $\varphi(\emptyset) = 0$, $\varphi(A) \le \varphi(A \cup B) \le \varphi(A) + \varphi(B)$, and $\varphi(A) = \lim_{n\rightarrow \infty} \varphi(A\cap n)$, such that $\mathcal{I} = \{ A \subset \mathbb{N}: \limsup_{n} \varphi(A \backslash n)=0\}$. Furthermore, every analytic ideal on $\mathbb{N}$ is either a $P$-ideal, or finite-to-one reducible to the ideal generated by finite unions of the sets $\{ n\} \times \mathbb{N}.$ Remark: Any $\varphi$ which satisfies the hypothesis of the theorem is called a lower semi-continuous sub-measure on $\mathbb{N}$. An easy corollary to the previous theorem provides a partial answer to Problem 1. Every $F_{\sigma\delta}$ $P$-ideal on $\mathbb{N}$ can be written as $\bigcap_n \bigcup_m K_{n,m}$ for some closed and hereditary $K_{n,m}$. Proof: Using the previous theorem, it follows that for each $F_{\sigma\delta}$ P-ideal $\mathcal{I}$ there is a lower semi-continuous $\varphi$ such that $\mathcal{I} = \{ A \subset \mathbb{N} : \limsup_{n} \varphi(A\backslash n) = 0\}$. Letting $K_{n,m} = \{ A \subset \mathbb{N}: \varphi(A\backslash m) \le \frac{1}{n} \}$, we have $\mathcal{I} = \bigcap_n \bigcup_m K_{n,m}$; so it only remains to show that each $K_{n,m}$ is closed and hereditary. To this end note that, for each $A \in K_{n,m}$ and $B\subset A$ we have $B\backslash m \subset A\backslash m$, therefore using the properties of $\varphi$, it follows that, $\varphi(B\backslash m) \le \varphi(B\backslash m \cup A\backslash m)=\varphi(A\backslash m) \le \frac{1}{n}$ and so $B \in K_{n,m}.$ Next, assume $A \subset \mathbb{N}$ has the property that for every $k \in \mathbb{N}$ letting $a_k = A \cap k$ and $b_k =k$, we have $[a_k, b_k]\cap K_{n,m} \neq \emptyset$. Then, for every $k \ge 0$, $\varphi(a_k \backslash m) \le \frac{1}{n}$ (since the sequence is increasing.) It immediately follows that $\varphi(A \backslash m) = \lim_{k} \varphi(a_k \backslash m) \le \frac{1}{n}$ and $A \in K_{n,m}$. [1] Solecki, Sławomir, Analytic ideals and their applications, Ann. Pure Appl. Logic 99, No.1-3, 51-72 (1999). ZBL0932.03060.<|endoftext|> TITLE: Strengthening of Connes' embedding conjecture QUESTION [10 upvotes]: If $A$ and $B$ are $C^*$ algebras I will write $A \overline{\otimes} B$ for the maximal tensor product and $A \underline{\otimes} B$ for the minimal tensor product. If $G$ is a countable discrete group I will write $C^*(G)$ for the full group $C^*$ algebra of $G$. Let also $\mathbb{F}_\infty$ be the free group on a countably infinite set of generators. A theorem of Kirchberg asserts that Connes' embedding conjecture is equivalent to the statement $C^*(\mathbb{F}_\infty) \overline{\otimes} C^*(\mathbb{F}_\infty) = C^*(\mathbb{F}_\infty) \underline{\otimes} C^*(\mathbb{F}_\infty)$. I would like to find out what is known about the statement that $C^*(\mathbb{F}_\infty) \overline{\otimes} C^*(G) = C^*(\mathbb{F}_\infty) \underline{\otimes} C^*(G)$ for every countable discrete group $G$. Are there counterexamples to this stronger statement, or alternatively is it equivalent to the original conjecture? REPLY [8 votes]: Your more general conjecture is not true. This property for a discrete group $G$ is equivalent to the fact that $G$ is WEP (its $C^{\ast}$-algebra has weak expectation property). In "Examples of hyperlinear groups without factorization property" Andreas Thom constructs a (hyperlinear) property (T) group that is not residually finite. On the other hand, by a result of Kirchberg ("Discrete groups with Kazhdan's property T and factorization property are residually finite") such groups cannot have WEP.<|endoftext|> TITLE: Is there a geometric interpretation for Reidemeister torsion? QUESTION [19 upvotes]: Given a finite CW or simplicial decomposition of a space $X$ and a ring homomorphism $\varphi:\mathbb{Z}[\pi_1(X)]\to F$ for a field $F$, if the $\varphi$-twisted homology is trivial, then the Reidemeister-Franz torsion $\tau^\varphi(X)\in F$ is an invariant of the twisted chain complex, well-defined up to multiplication by $\pm\varphi(\pi_1(X))$. (The value can be pinned down further by choosing a "homology orientation" and an "Euler structure".) The definition of the torsion is rather opaque to me; it involves products of determinants of basis-change matrices. I vaguely see that it has something to do with how well a splitting of the acyclic chain complex respects the cellular basis. My question is whether there is a geometric interpretation of Reidemeister-Franz torsion. For instance, what is it measuring about the space? Is there an object in $X$ representing the torsion? Feel free to restrict the category if it helps. REPLY [18 votes]: Yes, there is a very nice geometric interpretation over any manifold $X$ satisfying $\chi(X)=0$, for a version of Reidemeister torsion spelled out in Turaev's paper "Euler structures, nonsingular vector fields, and torsions of Reidemeister type". Ian Agol's comment mentioned the Seiberg-Witten invariants, which equals this version of torsion in 3 dimensions. It turns out that both of these are equal to an $S^1$-valued Morse theoretic invariant $I(X)$ defined by Hutchings and Lee in their PhD work: "Circle-valued Morse theory and Reidemeister torsion" (Geom. Top. Vol.3, 1999) Roughly speaking, we equip $X$ with a suitable Morse function $f:X\to S^1$ for which all critical points and closed orbits of $-\nabla f$ are nondegenerate. Then $I(X)$ is defined by suitably counting the Morse-flowlines between critical points as well as the closed periodic orbits (remembering their periods). The invariant is independent of the choice of $f$ and it is identified with the Redeimester torsion (for a suitable homology orientation). When $\dim X=3$ and $b_1(X)>0$, this also recovers the Seiberg-Witten invariants. We can view $I(X)$ as the analog of the Gromov invariant in 4 dimensions which suitably counts $J$-holomorphic curves. Here is a simple example which may make the "involvement of products of determinants of basis-change matrices" less opaque to you. When $f:X\to S^1$ is a fiber bundle with fiber $\Sigma$, the periodic flow of $-\nabla f$ defines a self-map $\varphi:\Sigma\to\Sigma$, and $I(X)$ effectively counts fixed points of $\varphi^k$ weighted by their Lefschetz-sign. The identification with torsion translates into the Lefschetz fixed-point theorem. REPLY [6 votes]: Have you looked at the relation with Whitehead torsion and simple homotopy theory? I forget the details and am no expert, but there is a lot of nice low dimensional geometric topology hidden in that theory if that is any help. Look at Milnor's paper, Bull. Amer. Math. Soc., 72 (3):(1966) 358–426. but also look at the Wikipedia articles on both types of torsion for a quick entry point to the ideas.<|endoftext|> TITLE: Picard group of the product of a local normal scheme with $\mathbb{A}^2-\{0\}$ QUESTION [7 upvotes]: Let $X$ be a noetherian local normal scheme, we may even assume that $X$ is complete if necessary. Consider $X\times (\mathbb{A}^2-\{0\})$, is it true that the Picard group of this scheme vanishes? REPLY [5 votes]: My doubt in the first comment was incorrect. There are, indeed, many formal invertible sheaves without trivializations over $\text{Spf} \ k[[t]].$ Thus, for instance, over $R=\text{Spec}\ k[t]/\langle t^n\rangle$ there are invertible sheaves on $\mathbb{A}^2_R\setminus\{0\}$ that are not isomorphic to the structure sheaf. Nonetheless, for every locally Noetherian scheme $S$ that is normal, every invertible sheaf is the pullback of a unique invertible sheaf from $S$. A bit more generally, let $T$ be a locally Noetherian scheme, let $\mathbb{A}_{T}$ be a finite type, separated $T$-scheme, let $\mathbb{A}^{*}_{T}\subset \mathbb{A}_{T}$ be an open subset with closed complement $C,$ and let $\sigma:T\to \mathbb{A}_{T}$ be a section. Hypothesis 0. The $T$-scheme $\mathbb{A}_T$ is smooth, and for every morphism from a field, $t:\text{Spec}\ \kappa \to T,$ the base change scheme $\mathbb{A}_\kappa$ is integral, $C_\kappa$ has codimension $\geq 2$, every invertible sheaf on $\mathbb{A}_\kappa$ is isomorphic to the structure sheaf, and the following pullback homomorphism is an isomorphism, $$\kappa^\times \to \mathcal{O}_{\mathbb{A}_\kappa}(\mathbb{A}_\kappa)^\times.$$ For every $T$-scheme $S$, denote by $\mathbb{A}_S$ the fiber product $\mathbb{A}_T\times_T S$ with its base change section $\sigma_S:S\to \mathbb{A}_S.$ Denote by $\pi_S:\mathbb{A}_S \to S$ the projection morphism. Definition 1. For every invertible sheaf $\mathcal{L}$ on $\mathbb{A}^{*}_S$, a normalized trivialization is an isomorphism of $\mathcal{L}$ with $\pi_S^*\sigma_S^*\mathcal{L}$ whose pullback by $\sigma_S$ equals the tautological trivialization. The normalized invertible sheaf $\mathcal{N}$ is the invertible sheaf $$\mathcal{L}\otimes_{\mathcal{O}}\pi_S^*\sigma_S^*\mathcal{L}^\vee.$$ By construction, there is a canonical trivializing global section $n:\mathcal{O}_S\to \sigma_S^*\mathcal{N}.$ Lemma 2. There exists a normalized trivialization of $\mathcal{L}$ if and only there exists a normalized trivialization of $\mathcal{N}.$ Proof This is just a matter of unwinding definitions. QED Lemma 3. If $S$ is Noetherian, then there exists a coherent sheaf $\overline{\mathcal{N}}$ on $\mathbb{A}_S$ whose restriction to $\mathbb{A}^{*}_S$ is isomorphic to $\mathcal{N}.$ If $S$ is also locally integral, then there exists such a coherent sheaf that is isomorphic to $\textit{Hom}_{\mathcal{O}}(\mathcal{M},\mathcal{O}_{\mathbb{A}^2_S})$ for a coherent, reflexive sheaf $\mathcal{M}.$ Proof. By an Exercise II.5.15 of Hartshorne's Algebraic geometry, there exists a coherent sheaf $\mathcal{N}^{\text{pre}}$ on $\mathbb{A}_S$ whose restriction to $\mathbb{A}^{*}_S$ equals $\mathcal{N}.$ Then both of the following sheaves are also coherent, $$ \mathcal{M} := (\mathcal{N}^{\text{pre}})^\vee=\textit{Hom}_{\mathcal{O}}(\mathcal{N}^\text{pre},\mathcal{O}_{\mathbb{A}_S}),\ \ \ \overline{\mathcal{N}}=\textit{Hom}_{\mathcal{O}}(\mathcal{M},\mathcal{O}_{\mathbb{A}_S}). $$ QED Lemma 4. Let $X$ be a scheme that is locally Noetherian and normal. Let $\phi:\mathcal{E}\to \mathcal{F}$ be an injective homomorphism of reflexive, coherent $\mathcal{O}_X$-modules whose cokernel has support of codimension $\geq 2$. Then $\phi$ is an isomorphism. Proof. Denote by $U$ the open complement of the support $B$ of $\text{Coker}(\phi)$. The restriction $\phi|_U$ is an isomorphism of $\mathcal{O}_U$-modules. Denote by $\psi_U:\mathcal{F}|_U\to \mathcal{E}|_U$ the inverse of $\phi|_U$. Since $X$ is normal, by Serre's criterion, $X$ is $S2$. By hypothesis, the closed subset $B$ has codimension $\geq 2$ everywhere. Thus, by Exercise III.3.5, p. 282, of Hartshorne's Algebraic geometry, there is a unique homomorphism of coherent $\mathcal{O}_X$-modules, $$\psi:\mathcal{F}\to \mathcal{E},$$ whose restriction to $U$ equals $\psi_U.$ By construction, both $\psi\circ \phi$ and $\phi\circ \psi$ restrict on $U$ to equal the respective identity homomorphisms. Thus, by the uniqueness from Exercise III.3.5, these compositions equal the respective identity homomorphisms on all of $X$. QED Proposition 5. For every Noetherian, normal $T$-scheme $S,$ every invertible sheaf on $\mathbb{A}^{*}_S$ has a unique normalized trivialization. Proof. Algebraic geometry. -> Assume that $S=\text{Spec}\ R$ is affine and connected: the uniqueness guarantees the glueing hypothesis relative to an open affine cover of $S.$ By hypothesis, $R$ is an integral domain. Denote by $K$ the fraction field of $R.$ Since Hom for finitely presented modules commutes with localization, the base change $\overline{\mathcal{N}}_K$ is the reflexive extension of $\mathcal{N}_K.$ By hypothesis, there is a unique normalized trivialization of $\mathcal{N}_K.$ Clearing denominators, there exists a unique normalized trivialization $s_r$ after base change to $\text{Spec}\ R[r^{-1}]$ for some nonzero $r\in R.$ By Krull's Hauptidealsatz, for every minimal prime $\mathfrak{p}$ over the principal ideal $\langle r \rangle$, the height of $\mathfrak{p}$ equals one. Since $R$ is normal, it is $R1$. Thus, the local ring $R_{\mathfrak{p}}$ is a DVR, and the scheme $$\mathbb{A}_{\mathfrak{p}}:=\mathbb{A}_{T}\times_T \text{Spec}\ R_{\mathfrak{p}}$$ is a smooth scheme over the regular ring $R_{\mathfrak{p}}$. So $\mathbb{A}_{\mathfrak{p}}$ is a regular scheme. Hence it is locally factorial by the Auslander-Buchsbaum-Nagata theorem. Thus, the base change of the reflexive, rank one sheaf $\overline{\mathcal{N}}$ is an invertible sheaf, cf. Theorem II.6.11 of Hartshorne's Algebraic geometry. In particular, the unique normalized trivialization over the residue field $\mathbb{A}_{\kappa(\mathfrak{p})}$ lifts to a section over $\mathbb{A}_{\mathfrak{p}}$. Since this section is surjective after base change to $\kappa(\mathfrak{p}),$ the section is surjective on $\mathbb{A}_{\kappa(\mathfrak{p})}$ by Nakayama's Lemma. By Lemma 4, the section is an isomorphism, i.e., it is a trivialization. After adjusting the section by the multiplicative inverse of the pullback by $\sigma,$ the section is a normalized trivialization. Since the normalized trivialization is unique, this agrees with $s_r$. Therefore $s_r$ extends over every minimal prime over $\langle r \rangle$. By the usual limit arguments, the maximal open subset on which $s_r$ extends to an $\mathcal{O}_{\mathbb{A}_R}$-module homomorphism that is surjective is an open subscheme of $\mathbb{A}_R$ that contains $\mathbb{A}_{R[r^{-1}]}$ as well as the generic point of every fiber $\mathbb{A}_{\mathfrak{p}}.$ Thus, the complement in $\mathbb{A}_{R[r^{-1}]}$ of this open is a closed subset $B$ of codimension $\geq 2.$ By Lemma 4 once more, $s_r$ extends to a normalized trivialization over all of $\mathbb{A}_R.$ Since the normalized trivialization over $K$ is unique, this normalized trivialization over $R$ is unique. QED<|endoftext|> TITLE: Tubular neighborhoods of embedded manifolds QUESTION [14 upvotes]: Recently I started working on a problem in Differential Geometry (where I'm not a specialist, so I apologize if this question turns out to have a trivial answer) and I had to consider the following situation. Let $M \subset \mathbb{R}^n$ be a smooth submanifold of dimension $n-k$. By a normal tube of radius $\varepsilon$ centered at $M$, I mean a tubular neighborhood of $M$ given by a disjoint union $$\mathscr{B}(M, \, \varepsilon):=\bigsqcup_{p \in M} B(p, \, \varepsilon),$$ where $B(p, \, \varepsilon)$ is a $k$-dimensional ball of radius $\varepsilon$ centred at $p \in M$ and contained in the (embedded) normal subspace $N_pM \subset \mathbb{R}^n$. Q. Under which conditions on $M$ does a normal tube $\mathscr{B}(M, \, \varepsilon)$ exist (for $\varepsilon$ sufficiently small)? It seems to me that this is always the case when $M$ is compact, because then we can identify $\mathscr{B}(M, \, \varepsilon)$ with the open neighborhood $B_M(\varepsilon)$ of $M$ in $\mathbb{R}^n$ given by $$B_M(\varepsilon) := \{x \in \mathbb{R}^n \, | \, d(x, \, M) < \varepsilon\}.$$ On the other hand, there are also examples where $\mathscr{B}(M, \, \varepsilon)$ exists even if $M$ is not compact, for instance in the case where $M$ is a linear subspace (in this case, in fact, the normal spaces to $M$ are pairwise disjoint). My feeling is that $\mathscr{B}(M, \, \varepsilon)$ should always exist when the curvature of $M$ is bounded in some suitable sense, but I'm not able to specify it better. Any answer, example/counterexample and reference to the relevant literature will be greatly appreciated. REPLY [12 votes]: I suggest to look at the very clear proof of the tubular neighborhood theorem in Lee, John M., Introduction to smooth manifolds, Graduate Texts in Mathematics. 218. New York, NY: Springer. xvii, 628 p. (2002). ZBL1030.53001., pag 253, I think it's exactly what you need. He proves the existence of a tubular neighborhood for any embedded submanifold of $\mathbb{R}^n$ (but indeed the proof is the same for any embedded submanifold of a smooth manifold, it just makes use of an auxiliary Riemannian structure to define "normal lines"). Notice that his construction works also in the non-compact case, but in this case the tubular neighborhood has non-constant radius. If you want $\varepsilon$ to be constant, then looking into the proof you just need a lower bound to the injectivity radius from the submanifold, which you always have if the submanifold is compact (and vice-versa, if you have a tubular neighborhood with constant $\epsilon$, then the injectivity radius is $\geq \varepsilon$. In any way, curvature has no relation whatsoever with existence of tubular neighborhoods (more precisely, you cannot give sufficient conditions for existence of a uniform tubular neighborhood in terms of curvature bounds, I'm general). Your problem is deeply related with the regularity of the distance from the submanifold. In fact, you can easily build the tubular neighborhoods using the gradient flow of the distance function from the submanifold, which is as smooth as the submanifold in a neighborhood of the latter. To this regard, I can also suggest you to give a look at this paper : Foote, Robert L., Regularity of the distance function, Proc. Am. Math. Soc. 92, 153-155 (1984). ZBL0528.53005. Ah, and indeed, the magnificent Gray, Alfred, Tubes, Redwood City, CA etc.: Addison-Wesley Publishing Company. xii, 283 p. (1990). ZBL0692.53001.<|endoftext|> TITLE: Cohomology of braid groups with coefficients in the group ring QUESTION [17 upvotes]: Let $\mathbf B_n$ be the braid group on $n$ strings. What is known about the cohomology of $\mathbf B_n$ with coefficients in its integral group ring: $H^*(\mathbf B_n;\mathbb Z \mathbf B_n)$? REPLY [16 votes]: The braid groups $B_n$ are Bieri-Eckmann duality groups of dimension $n-1$. It follows (either by definition or by a standard result depending on how you set things up) that $H^k(B_n;\mathbb{Z}[B_n])$ is $0$ for $k \neq n-1$ and is torsion-free for $k=n-1$. Here is a brief description of how to see this. First, the group $B_n$ are of type F (i.e. they have compact Eilenberg-MacLane spaces). It follows that it is enough to find a finite-index subgroup of $B_n$ that is a duality group of dimension $n-1$. The pure braid group $PB_n$ do the job. To see that $PB_n$ is a duality group of dimension $n-1$, we use induction on $n$. The base case $n=1$ is trivial since $PB_1$ is the trivial group. For $n>1$, we will use the standard short exact sequence $$1 \longrightarrow F_{n-1} \longrightarrow PB_n \longrightarrow PB_{n-1} \longrightarrow 1,$$ where $F_{n-1}$ is the free group on $(n-1)$ generators and the map $PB_n \rightarrow PB_{n-1}$ comes from deleting the final strand. By induction $PB_{n-1}$ is a duality group of dimension $(n-2)$, and it is standard that $F_{n-1}$ is a duality group of dimension $1$. It follows that $PB_n$ is a duality group of dimension $(n-2)+1 = n-1$. There is a more general theorem of Harer that says that all mapping class groups are virtual duality groups (and are actually duality groups if they are torsion-free), but the above argument is much easier than what Harer did.<|endoftext|> TITLE: Converse to Euclid's fifth postulate QUESTION [79 upvotes]: There is a fascinating open problem in Riemannian Geometry which I would like to advertise here because I do not think that it is as well-known as it deserves to be. Euclid's famous fifth postulate, or more precisely Playfair's version of it, states that, in the Euclidean plane, through every point outside a line $\ell$ there passes one and only one line which does not intersect $\ell$. Question: Is the Euclidean plane the only complete Riemannian manifold homeomorphic to $R^2$ which satisfies the fifth postulate, i.e., through any point outside a complete geodesic $\gamma$ there passes one and only one complete geodesic which does not intersect $\gamma$? In other words, does the fifth postulate force the curvature to be zero? The reason this is interesting, or historically significant, is that it was the attempts to reduce the fifth postulate to the other axioms of Euclid, throughout the middle ages, which eventually led to the discovery of non Euclidean geometries and the notion of curvature by Gauss and Riemann. This problem appears to be originally due to Burns and Knieper in 1991: see the survey paper by Burns and Matveev, which also includes other nice problems on geometry of geodesics. The problem is also mentioned in papers of Croke, and Bangert and Emmerich, and has been studied most recently by Ge, Guijarro, and Solorzano. REPLY [5 votes]: It is a very nice question. I am under the impression that a Riemannian metric on $\mathbb R^2$ which has conjugate points cannot satisfy the fifth postulate. Now, there are various results showing that, under some conditions, metrics without conjugate points must be flat. For a $\mathbb Z^2$-periodic metric, for instance, this is a theorem of Hopf. For asymptotically Euclidean metrics, it seems to be a very recent result of Guillarmou, Mazzucchelli and Zuo: https://arxiv.org/pdf/1909.01488.pdf<|endoftext|> TITLE: Push-forward of nef divisors via finite morphisms QUESTION [6 upvotes]: Let $f:X\rightarrow Y$ be a finite morphism between smooth projective varieties, and let $D$ be an effective nef but not ample divisor on $X$. Consider the divisor $f_{*}D$ on $Y$. Is $f_{*}D$ nef but not ample as well ? REPLY [7 votes]: I suppose you want $f$ to be surjective, otherwise $f_*D$ is not defined. Then $f_*D$ is nef: for any curve $C\subset Y$, $\ (f_*D\cdot C)=(D\cdot f^*C)\geq 0$. But it might very well be ample. Consider a smooth quadric $Q\subset \mathbb{P}^3$, and let $f:Q\rightarrow \mathbb{P}^2$ be the projection from a point outside $Q$. Let $D$ be a line contained in $Q$. Then $D$ is nef, not ample, but $f_*D$ is a line in $\mathbb{P}^2$, hence ample.<|endoftext|> TITLE: Question regarding the paper by Atiyah, Bott and Shapiro: alternative description of K-theory QUESTION [8 upvotes]: In Atiyah, Bott, and Shapiro - Clifford modules (journal, MSN), the authors discuss the alternative description of K-theory in terms of sequences of vector bundles. I would like to understand the details of this alternative description and I came across the following question while reading this paper: In the proof of Lemma 7.3 authors claim that (using the notation from the paper) if we take two monomorphic extensions $\sigma_{n+1}',\sigma_{n+1}''$ of $\sigma_{n+1}$ then $E_n' \cong E_n''$ where $E_n'$ and $E_n''$ are cokernels of $\sigma_{n+1}'$ and $\sigma_{n+1}''$. It is true that these two extensions are homotopic rel $Y$ (this follows from the previous lemma). How does it follow that $E_n' \cong E_n''$? . REPLY [5 votes]: The image of the homotopy defines a subbundle of the bundle $E_n\times I$ over $X\times I$. The quotient bundle restricts to $E'_n$ over $X\times 0$ and $E''_n$ over $X\times 1$ and, in general, if you have a vector bundle over $X\times I$ then the restrictions to the two ends are isomorphic.<|endoftext|> TITLE: Models for Higher Inductive Types in Homotopy Type Theory QUESTION [5 upvotes]: Ordinary inductive types is initial algebras for free monads. However, HITs are not initial algebras for endofunctors but presented monads. From nLab, initial algebra of a presentable (infinity,1)-monad and blog comment, Peter and Mike constructed models for HITs in their paper. Where is the reference or link for this paper? REPLY [6 votes]: https://arxiv.org/abs/1705.07088 . I've updated the nLab page. There are also some slides available at my web page.<|endoftext|> TITLE: Gluing Riemann surfaces QUESTION [10 upvotes]: Let $X$ be a compact Riemann surface with boundary $\partial X$. Assume (for simplicity only) that $\partial X$ has a single connected component. Let us fix an orientation preserving diffeomorphism $\phi\colon S^1\tilde\to \partial X$ with the standard unit circle $S^1$ which is the boundary of the standard unit disk $D$. Consider the closed topological surface $Y:=X\cup_\phi D$ obtained from $X$ by gluing the disk $D$ along $\phi$. Q1. Is it true that $Y$ has a unique complex structure such that the interiors of $X$ and of $D$ are complex subspaces of $Y$ with their original complex structures? Q2. If the answer to Q1 is 'yes', is it true that if one chooses another orientation preserving diffeomorphism $\phi'\colon S^1\tilde\to \partial X$ then the new Riemann surface $Y'$ (as in Q1) is isomorphic (as a complex manifold) to $Y$? Q3. Let $X_0\subset X$ be another Riemann surface which is obtained from $X$ by a small deformation of the boundary of $X$ (thus topologically $X_0$ is a deformation retract of $X$). Let $Y_0$ be obtained from $X_0$ as in Q1. Are $Y$ and $Y_0$ isomorphic? REPLY [12 votes]: The answer to the first question is "yes". This is called conformal gluing, and the proof is based on the following lemma due to Lavrentiev: Let $\phi$ be an increasing diffeomorphism of $[-1,1]$ onto itself. Then there is a simple curve in the unit disk connecting $-1$ and $1$ breaking the disk into two domains $D^+$ (above) and $D^-$ (below), so that there exist conformal maps $f_1$ from the upper half-disk to $D_+$ and $f_2$ from the lower half-disk to $D^-$ which satisfy $f_1\circ\phi(z)=f_2(z)$ for $z\in [-1,1]$. For a beautiful proof, see Goluzin, Geometric theory of functions of a complex variable, (English translation: AMS 1969), Chapter 11. The condition that $\phi$ is a diffeomorphism can be substantially relaxed (See Ahlfors, Lectures on quasiconformal mappings, Chap IV, where this is called $M$-condition, modern term is "quasisymmetric"), but for arbitrary homeomorphisms this is not true. The answer to the other two questions is no. The procedure that you describe is called the Schiffer variation of conformal structure, and it indeed changes the conformal structure (of the resulting compact surface). See for example, S. Nag, Schiffer variation of complex structure and coordinates on Teichmuller spaces, Proc. Indian Acad. Sci., 94 (1985) 2-3, p. 111-122. The original source is M. Shiffer and D. Spencer, Functionals of finite Riemann surfaces. Princeton University Press, Princeton, N. J., 1954.<|endoftext|> TITLE: Combinatorial identity: $\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2=\frac{1}{2} \binom{(2a+1)+(2b+1)}{2a+1}$ QUESTION [30 upvotes]: In my research, I found this identity and as I experienced, it's surely right. But I can't give a proof for it. Could someone help me? This is the identity: let $a$ and $b$ be two positive integers; then: $\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2=\frac{1}{2} \binom{(2a+1)+(2b+1)}{2a+1}$. REPLY [17 votes]: When this identity was posted, it struck me as something that ought to have a combinatorial explanation. I have now found one, using a decomposition of NSEW lattice paths: paths in $\mathbb{Z}^2$ consisting of unit steps in the direction N, S, E or W. Many of the ideas here may be found in [GKS], though not the decomposition itself. The expression $\frac12{2a+1\ +\ 2b+1\choose2a+1}$ counts paths of $(a+b+1)$ steps that start at $(0,0)$ and end on the half-line $(a-b,\geq0)$. To see this, decompose each path step as two half-steps $±\left[\begin{smallmatrix}½\\½\end{smallmatrix}\right]$ and $±\left[\begin{smallmatrix}½\\-½\end{smallmatrix}\right]$. If the $+$ option is chosen for $(2a+1)$ of the $2(a+b+1)$ half-steps, and the $-$ option for the other $(2b+1)$, then the $x$-coordinate of the endpoint is $\frac12((2a+1)-(2b+1))=a-b$. Thus there are ${2a+1\ +\ 2b+1\choose2a+1}$ paths of $(a+b+1)$ steps from $(0,0)$ to $x=a-b$. By parity, the end position must have an odd-numbered $y$-coordinate. Reflection in the $x$-axis is therefore a fixpoint-free involution, so half of these paths end on the half-line $(a-b,\geq0)$. Such a path may be split into a pair of paths with $(a+b)$ steps in total. The endpoint of the path is $(a-b, 2k+1)$ for some $k\in\mathbb N$. At least one step of the path must therefore be an N step from $(c,2k)$ to $(c,2k+1)$ for some $c$. Remove the first such step, to give a pair of paths with $a+b$ steps altogether: A path of $n$ steps from $(0,0)$ to $(c,2k)$ that does not cross the line $y=2k$, which we can think of as a 180° rotation of a path from $(0,0)$ to $(c,2k)$ that does not cross the $x$-axis; A path of $a+b-n$ steps from $(c,2k+1)$ to $(a-b,2k+1)$, which we can think of as a translation of a path from $(0,0)$ to $(a-b-c,0)$. This is clearly a bijection. There are ${i+j\choose i}^2$ paths of $(i+j)$ steps from $(0,0)$ to $(i-j,0)$. The four directions N,S,E,W may be obtained by starting with $\left[\begin{smallmatrix}-1\\0\end{smallmatrix}\right]$ and adding neither, one, or both of $\left[\begin{smallmatrix}1\\1 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}1\\-1\end{smallmatrix}\right]$. Build a path of $i+j$ steps, initially all $\left[\begin{smallmatrix}-1\\0\end{smallmatrix}\right]$. Add $\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]$ to $i$ of the steps and, independently, add $\left[\begin{smallmatrix}1\\-1\end{smallmatrix}\right]$ to $i$ of the steps. There are also ${i+j\choose i}^2$ paths of $(i+j)$ steps from $(0,0)$ to $(i-j,\geq0)$ that do not cross the $x$-axis. There is a bijection between these paths and the paths of the previous section using a raising/lowering transformation [GKS]. Suppose we have a path from $(0,0)$ to $(i-j,0)$ that may cross the $x$-axis. While the path crosses the $x$-axis, do the following: Take the initial segment of the path up to the first time it touches the line $y=-1$, and reflect this initial segment about that line. Then translate the entire path up by two units, so it starts at $(0,0)$ again and ends two units higher than before on $x=i-j$. I hope it is clear that this process is reversible. (In reverse: while the endpoint is above the $x$-axis, translate the path two units down, then take the initial segment from $(0,-2)$ to the first intersection with $y=-1$ and reflect this initial segment about that line.) Putting it together Now we have all the ingredients we need. Let us count the pairs of paths as described above. Since $n$ and $c$ have the same parity, we may write $n=i+j$ and $c=i-j$ for $i\in[0,a]$, $j\in[0,b]$. There are ${i+j\choose i}^2$ paths of $(i+j)$ steps from $(0,0)$ to $(i-j,\geq 0)$ that do not cross the $x$-axis. There are ${a-i\ +\ b-j\choose a-i}^2$ paths of $(a+b)-(i+j)$ steps from $(0,0)$ to $(a-b-(i-j),0)$. So in total there are $$\sum_{i=0}^a\sum_{j=0}^b{i+j\choose i}^2{a-i\ +\ b-j\choose a-i}^2$$ such pairs, as required. [GKS] Richard K. Guy, C. Krattenthaler and Bruce E. Sagan (1992). Lattice paths, reflections, & dimension-changing bijections, Ars Combinatoria, 34, 3–15.<|endoftext|> TITLE: Two groups of order 108 QUESTION [5 upvotes]: The Galois group of the polynomial $x^9+x^3+1$ is a semidirect product of the Heisenberg group $He_3$ and the Klein 4-group V. But, in the list of groups of order 108 https://people.maths.bris.ac.uk/~matyd/GroupNames/index.html there are only two such groups: one is the semidirect product of $He_3$ with the cyclic group of order 4, and the other is the semidirect product of $He_3$ with $C_4$ acting as $C_4/C_2=C_2$. Which one is my Galois group? REPLY [6 votes]: In Maple you can enter with(GroupTheory): f := x^9 + x^3 + 1; G := GaloisGroup(f,x); Z := Center(G); Elements(Z); and you will get the response {()}, indicating that the centre is trivial. But the second of the two groups that you mention seems to have a central involution, so your Galois group must instead be the first one. A central involution would correspond to a Galois extension of degree $54$ contained in the full splitting field of $f$. There is probably a direct field-theoretic argument to show that no such extension exists, without assistance from Maple, but I cannot immediately see one. UPDATE: As Jeremy Rouse points out, it seems that the Galois group is not in fact either of the candidates mentioned by the OP. Instead, we can put $g(y)=y^3+y+1$ so $f(x)=g(x^3)$, and let $K$ and $L$ be the splitting fields of $g$ and $f$. Each root of $g$ acquires a full set of cube roots in $L$, and the ratio between two such cube roots is a primitive cube root of $1$. It follows that $L$ contains the field $M=\mathbb{Q}(e^{2\pi i/3})$, which is quadratic over $\mathbb{Q}$. One can check that $g$ has precisely one real root so complex conjugation gives an involution in $G(K/\mathbb{Q})$ so $G(K/\mathbb{Q})\simeq\Sigma_3$. It works out that $KM$ is Galois over $\mathbb{Q}$ with $G(KM/\mathbb{Q})\simeq\Sigma_3\times C_2$ (and one can check that this is also the dihedral group of order $12$). Now let $\alpha$, $\beta$ and $\gamma$ be the roots of $g$. We can obtain $L$ from $KM$ by adjoining a cube root of $\alpha$ and a cube root of $\beta$ (we then have a cube root of $\gamma$ as well because $\alpha\beta\gamma=-1$, and we have a full set of cube roots because $e^{2\pi i/3}\in M$). Using this we see that $G(L/KM)=C_3^2$. We now have an extension $$ C_3^2 \to G(L/\mathbb{Q}) \to \Sigma_3\times C_2 $$ Jeremy Rouse's comment suggests that this must be a semidirect product, and Maple agrees with that, but I do not see a direct argument at the moment.<|endoftext|> TITLE: Intrinsic vs Extrinsic geometry of convex surfaces QUESTION [12 upvotes]: By Alexandrov's isometric embedding theorem, any locally convex metric prescribed on the sphere admits a realization as a convex surface in Euclidean 3-space, which, by Pogorelov's rigidity result, is unique up to an isometry of the ambient space. Thus, the intrinsic geometry of a convex surface completely determines its extrinsic geometry, and in principle it should be possible to describe all geometric quantities in terms of the intrinsic metric. On the other hand, Alexandrov's proof is not constructive, and does not provide any hints as to how one might be able to do this. Hence my first question is: Question 1: Is there any way to compute, recognize, or characterize intrinsically any of the geometric quantities or features of a convex surface which are not invariant under local isometric deformations, e.g., mean curvature, principal directions, principal curvatures, or umbilic points? I recognize this might be too much to ask, because the procedures which the above question asks for would somehow need to take into account the whole metric, not just a neighborhood of a point. So let me try to make things a bit more concrete and maybe a little more accessible. It is well known that any (smooth) convex surface must have at least one umbilic, i.e., a point where the principal curvatures are equal (if not then the direction of the larger principal curvature yields a nonvanishing line field on the surface which violates the Poincare-Hopf index theorem). Can one prove this without reference to the ambient space, or quantities that we only know how to compute extrinsically: Question 2: Is it possible to prove the existence of an umbilic point of a smooth convex surface in a purely intrinsic way? If so, can one also find or approximate the location of the umbilics? One motivation behind this questions is the famous conjecture of Caratheodory, which states that each convex surface must have at least two umbilics. The extrinsic or local approaches to this problem have always been problematic. Adendum: As Robert Bryant correctly points out below, the topological argument for the existence of at least one umbilic described above is really intrinsic and independent of the notion of principal directions. So the main point of Question 2 is its second part, i.e., to somehow get a handle on the location of the umbilic. See also this related question. REPLY [3 votes]: "Umbilic points" have no obvious meaning in Alexandrov's framework (since his theory is not restricted to smooth surfaces), but there has been a fair amount of progress on making the theory more effective (it was always constructive: you could approximate the metric by a polyhedral metric. A polyhedral metric could be embedded in finite time [slowly, true], and then results would converge to the embedding of the surface you started with. The speed of convergence was already addressed in the '50s by A. Volkov). The last word on effectivizing Alexandrov for polyhedra is: Bobenko, Alexander I.; Izmestiev, Ivan, Alexandrov’s theorem, weighted Delaunay triangulations, and mixed volumes, Ann. Inst. Fourier 58, No. 2, 447-505 (2008). ZBL1154.52005.<|endoftext|> TITLE: String diagrams for bimonoidal categories (a.k.a. rig categories)? QUESTION [22 upvotes]: I'm having some fun playing around with string diagrams for monoidal categories, expressing familiar constructions from Riemannian geometry and linear algebra in terms of elegant string diagrams. I've been wondering if there is a nice extension of this diagram calculus to bimonoidal categories (also known as rig categories)? This would allow us to work with direct sums (of bundles, or representations etc.) diagrammatically as well. REPLY [13 votes]: This question is answered in the affirmative in the following preprint: Cole Comfort, Antonin Delpeuch, Jules Hedges, Sheet diagrams for bimonoidal categories, arXiv:2010.13361 The morphisms are represented by so-called proof sheets. Here is an example of such a proof sheet, representing the controlled-not gate using the program sheetshow which Antonin created to typeset the diagrams in the preprint:<|endoftext|> TITLE: How Composite can $2^n-1$ be, infinitely often? QUESTION [8 upvotes]: It seems that as $n$ increases, the ratio $$\frac{\varphi(2^n-1)}{2^n-1},$$ where $\varphi$ denotes the Euler totient function, takes on values reasonably often in the interval $(.3,.4)$. Is there anything known about $$\lim \inf_{n \rightarrow \infty}\frac{\varphi(2^n-1)}{2^n-1}?$$ REPLY [32 votes]: Let $n \in \mathbf N^+$ and $a \in \mathbf N_{\ge 2}$. Every prime $\le n+1$ that doesn't divide $a$, is a divisor of $a^{n!} - 1$ (by Fermat's little theorem). So we have $$\frac{\varphi(a^{n!} - 1)}{a^{n!} - 1} = \prod_{p \,\mid\, a^{n!} - 1} \left(1 - \frac{1}{p}\right) \le \prod_{a < p \le n+1} \left(1 - \frac{1}{p}\right)\! \stackrel{n \to \infty}{\longrightarrow} 0,$$ where $p$ is always a prime and for the first equality we have used Euler's product formula. In particular, this shows that the limit inferior in the OP is $0$.<|endoftext|> TITLE: Models with few types in infinitary logics QUESTION [12 upvotes]: Let $\mathcal{L}_{\kappa \lambda}$ denote the infinitary logic that allows conjunction of less than $\kappa$-many formulas and simultaneous quantification of less than $\lambda$-many variables. It is well-known that if a sentence $\varphi \in \mathcal{L}_{\omega_1 \omega}$ has arbitrarily large models, then for all cardinals $\lambda \geq \omega$ there exists a model of $\varphi$ of cardinality $\lambda$ which realizes only $\omega$-many types. (For example, see Theorem 5.11 of Marker's "Lectures in Infinitary Model Theory".) I would like to have a similar result for $\mathcal{L}_{\kappa^+ \kappa}$ with uncountable $\kappa$ that guarantees the existence of sufficiently large models with $\leq \kappa$-many types. (Indeed, having $\leq \kappa$-many quantifier free 1-types suffices for my purposes.) The problem I am having is that the usual proof of this fact does not generalize to uncountable $\kappa$: The proof proceeds by taking the Skolem hull of a large set of indiscernibles, whose existence is guaranteed by an argument that uses the Erdös-Rado theorem. If one tries to imitate the same proof, in order to guarantee that the Skolem hull is an elementary substructure, one has to introduce Skolem functions of infinite arity, which in turn requires one to consider indiscernibles with respect to formulas possibly having infinitely many variables; and this requires a partition relation where one colors $\geq \omega$-tuples. However, such partition relations are known to fail in general. Is there any way to get around this problem or is it the case that the generalization of this fact fails? If it helps, you may assume that $\kappa$ has various properties such as being strong limit, inaccessible etc. REPLY [6 votes]: Here is a partial answer: consistently, the generalization can fail for all uncountable $\kappa$. Namely: Suppose $\mathbb{V} = \mathbb{L}$ and let $\kappa$ be any uncountable cardinal. Let $\mathcal{L}$ be the language consisting of the binary relation $\epsilon$ and constant symbols $(c_\alpha: \alpha < \kappa)$. Let $\phi$ be a sentence of $\mathcal{L}_{\kappa^+ \omega_1}$ such that $(M, \epsilon, a_\alpha: \alpha < \kappa) \models \phi$ iff $(M, \epsilon)$ is a well-founded model of $ZFC^- + \mathbb{V}=\mathbb{L}$, and $a_\alpha$ is the $\alpha$'th element of $\mbox{ON}^M$ (here $ZFC^-$ is $ZFC$ without powerset, or any other strong enough fragment of $ZFC$ for condensation to work). For each ordinal $\gamma \geq \kappa$ let $M_\gamma$ be the $\mathcal{L}$-structure with universe $\mathbb{L}_\gamma$, where $\epsilon$ is interpreted as $\in$ and where $c_\alpha$ is interpreted as $\alpha$. Then every model of $\phi$ is isomorphic to some $M_\gamma$. But if $|M_\gamma| \geq \kappa^+$ then $\mathcal{P}(\kappa) \subseteq M_\gamma$, and so $M_\gamma$ realizes $2^\kappa$ distinct quantifier-free $1$-types.<|endoftext|> TITLE: A pencil with exactly one multiple fibre QUESTION [10 upvotes]: Does there exist a smooth projective surface $S / \mathbb{C}$ equipped with a dominant morphism $\pi: S \to \mathbb{P}^1$ which has the following properties: The fibre at infinity is "multiple", i.e. as divisors on $S$ one has $\pi^{-1}(\infty) = mD$ for some $m > 1$ and some divisor $D$. All other fibres are integral. REPLY [5 votes]: The answer is yes. In fact, it is possible to construct a rational elliptic fibration $f \colon S \to \mathbb{P}^1$ with exactly one multiple fibre of multiplicity $m \geq 2$, by starting from the blow-up of $\mathbb{P}^2$ at nine points that are the base locus of a pencil $\mathscr{P}$ of elliptic curves and then performing a logarithmic transformation centered at one point of $\mathbb{P}^1$. Since the logarithmic transformation does not change the fibres outside the center, choosing a sufficiently general pencil $\mathscr{P}$ the reduced fibres of $f$ will be all irreducible. For more details and examples, see Y. Fujimoto, On rational elliptic surfaces with multiple fibers, Publ. Res. Inst. Math. Sci. 26, No.1, 1-13 (1990). ZBL0729.14027, in particular Proposition 1.1.<|endoftext|> TITLE: Maximum principle for heat equation, low regularity case QUESTION [6 upvotes]: I meant to assign to my class the following homework problem: If $u\in C^2((0,T)\times \Omega) \cap C^0([0,T]\times\bar{\Omega})$ where $\Omega$ is an open, bounded domain, is such that $\partial_t u - \triangle u \leq - \epsilon < 0$ for some constant $\epsilon > 0$, then $u$ cannot have a local maximum on the set $(0,T)\times \Omega$. This follows from simple second derivative considerations. I made a typo, however, and asked my students to prove that ... $u$ cannot have a local maximum on the set $(0,T\color{red}{]} \times \Omega$. Question: Is the version with the typo still true? Or is there a counterexample? Remark: If we assume that $\nabla u$ and $\nabla^2 u$ (the spatial gradient and Hessian) both extend continuously to $\{T\} \times \Omega$, then the second derivative test argument also goes through. So any potential counterexample must be non-regular at time $T$. Remark 2: the potential lack of regularity also means we cannot directly apply the mean value integral (e.g. that which is given in Evans' textbook). REPLY [5 votes]: This version is still true: if $u$ had a local maximum at $(x,\,T)$, say with $u(x,\,T) = 0$, then $u \leq 0$ in a small parabolic cylinder centered at $(x,\,T)$. After rescaling we can assume that $u \leq 0$ in $\overline{B_1} \times [0,\,1]$, with $u(0,\,1) = 0$. Replacing $u$ with $u - \frac{\epsilon}{4n}t(|x|^2-1) - \frac{\epsilon}{8n}$ we may assume that $(\partial_t - \Delta)u < 0$, with $u \leq -\epsilon / 8n$ on $(B_1 \times \{0\}) \cup (\partial B_1 \times [0,1])$ and $u(0,1) = \epsilon / 8n$. However, by the maximum principle, on the cylinders $\overline{B_1} \times [0,1-\delta]$, $u$ achieves its maximum on the sides or bottom, so $u \leq 0$ on all such cylinders. Taking $\delta \rightarrow 0$ we get a contradiction.<|endoftext|> TITLE: Green's function for fourth order equation QUESTION [5 upvotes]: I know the D'Alembert operator ${\frac {1}{c^{2}}}\partial _{t}^{2}-\Delta _{\text{3D}}$ has a well-known Green's function $\frac{\delta(t-\frac{r}{c})}{4 \pi r}$. This is very useful for studying 3D wave equation / fluids. How about the Green's function of the following operator? $${\frac {1}{\mu^{4}}}\partial _{t}^{2}+\Delta _{\text{3D}}\Delta _{\text{3D}}$$ Is it known? How do I find it? This would be useful for understanding superfluids. REPLY [3 votes]: The time-independent equation is the biharmonic equation, $$-\Delta^2f({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}'),$$ with 3D solution $$f({\bf r},{\bf r}')=\frac{1}{8\pi}|{\bf r}-{\bf r}'|,$$ see Fourier expansions for a logarithmic fundamental solution of the polyharmonic equation (2012). More generally, the Green function of the $k$-the power of the Laplacian, $(-\Delta)^k$ is $\propto |{\bf r}-{\bf r}'|^{2k-d}$ in odd dimensions $d$. The time-dependent case has been studied in 1D in On the biharmonic wave equation (1967) and in higher dimensions in A uniqueness condition for the polyharmonic equation in free space (1990).[The first paper gives the 1D Green function; I don't have access to the second paper, so I don't know if it contains the 3D Green function.]<|endoftext|> TITLE: Volume of convex lattice polytopes with one interior lattice point QUESTION [17 upvotes]: Let $P$ be a convex polytope in $\mathbb{R}^3$ whose every vertex lies in the $\mathbb{Z}^3$ lattice. Question: If $P$ contains exactly one lattice point in its interior, what is the maximum possible volume of $P$? Notice that a convex lattice polytope with no interior lattice point can have arbitrarily large volume, but if the number of interior lattice points in it is a given finite number, then its volume cannot be arbitrarily large. In two dimensions the answer is known: A convex lattice polygon with exactly one interior point has volume at most $9/2$, which follows from a more general inequality by P.R. Scott (see On Convex Lattice Polygons, Bull. Austral. Math. Soc., vol 15; 1976, 395-399). The best example I know in that respect is the tetrahedron with vertices $(0,0,0),\ (4,0,0),\ (0,4,0)$ and $(0,0,4)$, with volume ${32}\over3$. Is this perhaps the maximum volume? Of course, the same question can be asked in higher dimensions as well, with an analogous example to consider as a possible candidate for an answer. In view of several examples presented in answers and comments below, it seems that the optimal polytope should be a simplex, in every dimension. Has this been conjectured already? REPLY [9 votes]: This is addressed in the 2013 paper (appeared in Advances in 2015) by Averkov, Krumpelmann, Nill. The give a sharp bound for the volume of a lattice simplex with one interior lattice point (Theorem 2.2 in the paper), and an improved bound for a general lattice polyhedron with the same property (theorem 2.7) (the two bounds are not the same, indicating that Wlodek's conjecture is either still open or false). The results are stated in terms of the Sylvester sequence: $$s_1 = 2; s_i = 1 + \prod_{j=1}^{i-1} s_j.$$ With that, the volume of tbe biggest simplex in $d$ dimension with one lattice point is bounded (with equality achieved) by $$ \frac2{d!}(s_d-1)^2,$$ while for arbitrary polytopes, the bound is $$(s_{d+1}-1)^d,$$ so a lot worse.<|endoftext|> TITLE: Integral involving Gamma function: density of Kendall-Ressel family of distributions QUESTION [5 upvotes]: I came across the following function when reading the famous paper of Letac and Mora "Natural exponential families with cubic variance functions", i.e., $$f(x) = \frac{x^x e^{-x}}{\Gamma(x+2)}$$ for $x \ge 0$. Proposition 5.5 there showed via a transform of a Levy process that $f$ is a density on $(0,\infty)$ without the need to compute the integral $$\int_{0}^{\infty} f(x)dx$$ In fact, $f$ is called Kendall-Ressel density. Tonight, I used Mathematica to compute the above integral but Mathematica did not give the answer $1$. Instead, it was stuck and returned the orginal integral. So, I guess a direct computation is not trivial!? I am curious on how to directly show that $$\int_{0}^{\infty} f(x)dx = 1$$ Any suggestions? Thank you. Update on June 21, 2018: to follow up on Nemo's solution: is there a general formula or recursive formula for $$I(b,\alpha) = \int_{C} \frac{t^{-1}dt}{(t + \ln(-t)+b)^\alpha},$$ where $\alpha$ is a positive constant? Say, $\alpha$ is a natural number. REPLY [5 votes]: According to Hankel's formula $$ \frac{1}{\Gamma(z)}=\frac{i}{2\pi}\int_C(-t)^{-z}e^{-t}dt, $$ where $C$ is Hankel contour. So $$ \frac{x^x}{\Gamma(x+1)}=\frac{i}{2\pi}\int_C(-t)^{-x-1}e^{-xt}dt,\quad x>0. $$ Consider the integral $$ I(b)=\int_0^\infty \frac{x^xe^{-bx}}{\Gamma(x+1)}dx=\int_0^\infty e^{-bx}dx\frac{i}{2\pi}\int_C(-t)^{-x-1}e^{-xt}dt. $$ Changing the order of integration and calculating the integral over $x$ we get $$ I(b)=-\frac{i}{2\pi}\int_C\frac{dt}{t(t+\ln(-t)+b)}. $$ This integral can be easily calculated using residue theory $$ I(b)=-\frac{1}{1+W_{-1}(-e^{-b})}, $$ where $W_{-1}(z)$ is the Lambert W-function, satisfying $W_{-1}(z)e^{W_{-1}(z)}=z$, which has the derivative $$ W_{-1}'(z)=\frac{e^{-W_{-1}(z)}}{1+W_{-1}(z)}.\tag{*} $$ Now we write the initial integral as follows \begin{align} \int_0^\infty \frac{x^x e^{-x}}{\Gamma(x+2)}dx&=e\int_1^\infty e^{-b}I(b)db\\ &=-e\int_1^\infty e^{-b}\frac{1}{1+W_{-1}(-e^{-b})}db\\ &=-e\int_0^{1/e} \frac{ds}{1+W_{-1}(-s)}\\ &=e\int_0^{-1/e} \frac{ds}{1+W_{-1}(s)}\\ &=e\int_{-\infty}^{-1} e^{W}d{W}=e\cdot 1/e=1. \end{align} Here we used the equation $(*)$ to write $\frac{1}{1+W_{-1}(s)}=e^{W_{-1}(s)}W_{-1}'(s)$ and also the facts $W_{-1}(-1/e)=-1~$, $W_{-1}(0)=-\infty$.<|endoftext|> TITLE: Eigenvalue perturbation theory via Feynman diagrams QUESTION [12 upvotes]: Suppose I have a matrix given by a sum $$A=D+\epsilon B$$ where $D$ is diagonal and $\epsilon$ is small, and I want the eigenvalues of $A$ as a power series in $\epsilon$. The first two orders in perturbation theory are well known. Third and higher orders are briefly discussed here. However, the equations become horrible. I hear that Feynman diagrams are an efficient way to formulate perturbation theory, but I can't find an accessible exposition of this approach. Note that I have in mind the simple matrix setting. I don't want vacuum states, quantum field theory, Dirac equation, etc. Can someone help? REPLY [13 votes]: I think the simplest way is to use the very simple and very useful resolvant formula $$ (A+B)^{-1}=A^{-1}-A^{-1}B(A+B)^{-1}.$$ For perturbative theory, we just iterate this formula $$ (A+B)^{-1}=A^{-1}-A^{-1}BA^{-1}+A^{-1}BA^{-1}B(A+B)^{-1}$$ and obtain a power series in $B$. $$ (A+B)^{-1}=A^{-1}-A^{-1}BA^{-1}+A^{-1}BA^{-1}BA^{-1}-A^{-1}BA^{-1}BA^{-1}BA^{-1} +\cdots$$ With your notation it gives $$ A^{-1}=D^{-1}-\epsilon D^{-1}BD^{-1}+\epsilon^2 D^{-1}BD^{-1}BD^{-1}-\epsilon^3 D^{-1}BD^{-1}BD^{-1}BD^{-1} +\cdots$$ For the eigenvalues there is then the Cauchy formula $$ z_0 = \frac{1}{2i\pi}\oint \frac{z}{z-z_0} dz$$ which for matrices gives $$\lambda_0 = \frac{1}{2i\pi}\oint \lambda Tr(A-\lambda)^{-1}d\lambda$$ where the integral is a closed loop around an eigenvalue of $D$. And we obtain the perturbative expansion: $$ \lambda_0 =\frac{1}{2i\pi} \oint \lambda Tr(D-\lambda)^{-1} d\lambda \\-\epsilon \frac{1}{2i\pi} \oint \lambda Tr (D-\lambda)^{-1}B(D-\lambda)^{-1} \\+\epsilon^2 \frac{1}{2i\pi} \oint \lambda Tr (D-\lambda)^{-1}B(D-\lambda)^{-1}B(D-\lambda)^{-1}\\-\epsilon^3 \cdots$$ Now take a loop sufficiently close to the initial eigenvalue of $D$ and it will give what you wanted.<|endoftext|> TITLE: Is this generalization of Borsuk Ulam true? Roots of unity QUESTION [15 upvotes]: Consider a continous map from $S^2$ to $C$. Is it true that there exists 3 points equially spaced on a great circle, $x_1,x_2,x_3$, such that if $w$ is the third root of unity, $f(x_1)+wf(x_2)+w^2f(x_3)=0$? More generally I'm asking this if we take nth unity roots. Maybe I should add slight motivation: In my topology course we were shown a proof of borsuk ulam that goes through defining $g(x)=(f(x)-f(-x))/(|f(x)-f(-x)|)$, then by looking at it on the great circle, we can lift it to a function to $R$ satisfying (here we view $g$ as a function from $[0,1]$ instead of from the circle, and taking $x+1/2$ mod 1 in the next expression) $g(x+1/2)=n+1/2+g(x)$ for some natural $n$, but since it is continous, this is the same $n$ for all $x$. In particular $g(1)=2n+1+g(0)$ and thus this is a nontrivial path on the circle, but it homotopic to the constant one via returning to the sphere and wrapping the circle around to a constant function. Then notice the argument after the lift works the same when we know $g(x+1/3)=n+1/3+g(x)$. However there is no direct analong we can do to reach the part of lifting, because that would involve choosing for each point a great circle it is contained in a way that partions the sphere into great circles which is clearly impossible. If this turns out to be false, is there a space we can do this trick on to get this cute result? REPLY [10 votes]: $\require{AMScd}$ First, suppose that $G$ acts freely on connected spaces $X$ and $Y$, and that $p\colon X\to Y$ is $G$-equivariant. We then have a diagram of fibrations \begin{CD} X @>>> X/G @>>> BG \\ @VpVV @VVV @VV1V \\ Y @>>> Y/G @>>> BG \end{CD} This gives a diagram of fundamental groups \begin{CD} \pi_1(X) @>>> \pi_1(X/G) @>>> G \\ @Vp_*VV @VVV @VV1V \\ \pi_1(Y) @>>> \pi_1(Y/G) @>>> G \end{CD} and it is not hard to see that the rows are short exact sequences. Now specialise to the case where $Y=\mathbb{C}^\times$ and $G=C_3$ acting by rotation, and $\pi_1(X)$ is finite. Using the top row we see that $\pi_1(X/G)$ is also finite, so the only homomorphism to $\mathbb{Z}$ is trivial. On the other hand, bottom row is then $\mathbb{Z}\xrightarrow{3}\mathbb{Z}\to\mathbb{Z}/3$, and this quickly leads to a contradiction. Now suppose that $f\colon S^2\to\mathbb{C}$ is a counterexample to the OP's question. Fix three points $x_1$, $x_2$ and $x_3$ equally spaced on a great circle. Let $R$ be the rotation that sends $x_1$, $x_2$ and $x_3$ to $x_2$, $x_3$ and $x_1$. Define $p\colon SO(3)\to \mathbb{C}^\times$ by $$ p(T) = f(Tx_1) + \omega f(Tx_2) + \omega^2 f(Tx_3), $$ and note that this satisfies $g(TR)=\omega^{-1}g(T)$. In other words, if we let $C_3$ act on $SO(3)$ using $R$ and on $\mathbb{C}^\times$ by $\omega^{-1}$, we see that $p$ is equivariant. This is impossible by our initial lemma, because $\pi_1(SO(3))=\mathbb{Z}/2$. All the above still works if we replace $3$ by another integer $n>1$.<|endoftext|> TITLE: Skeletal monoidal categories with strict units QUESTION [10 upvotes]: I believe that every skeletal monoidal category is monoidally equivalent to a skeletal monoidal category with strict units. Does anybody know a reference for this fact in the literature? REPLY [5 votes]: See Theorem 3.2 in Turning Monoidal Categories into Strict Ones. Thus, any monoidal category $(\mathcal{C},\otimes, I,\alpha, \lambda,\rho)$ is monoidally equivalent to a monoidal category $(\mathcal{C},\otimes', I,\alpha')$ with strict unit (note that $\mathcal{C}$ is the same underlaing category). The case in that $\mathcal{C}$ is skeletal, answers your question.<|endoftext|> TITLE: Transcendence degree of the fraction field of $k[G]$ for torsion free abelian group $G$ QUESTION [6 upvotes]: Let $k$ be a field of characteristic $p$ and $G$ be a torsion free abelian group . Then the group ring $k[G]$ is an integral domain , let $k(G)$ denote its field of fractions . Then can we say anything about the transcendence degree of $k(G)$ over $F_p$ in terms of $k$ and/or $G$ ? What about the same question for field $k$ of characteristic $0$ ? An answer to this might help in solving https://math.stackexchange.com/questions/2338911/fraction-field-of-group-ring-of-field-over-torsion-free-abelian-group Also asked on https://math.stackexchange.com/questions/2440013/transcendence-degree-of-the-fraction-field-of-kg-for-torsion-free-abelian-gr REPLY [7 votes]: It seems that the transcendence degree of $k(G)$ over $k$ should be the dimension of the $\mathbb Q$-vector space $G \otimes_\mathbb Z \mathbb Q$. Indeed, passing from $G$ to $G \otimes \mathbb Q$ corresponds to adding roots of existing elements, so it does not alter the transcendence degree. Thus we are reduced to $\mathbb Q$-vector spaces. Choosing a basis $\{e_i\ |\ i \in I\}$ gives algebraically independent elements $\{e_i\ |\ i \in I\}$ of $k(G)$ over $k$ such that $k(G)$ is algebraic over $k(\{e_i\ |\ i \in I\})$; therefore the $e_i$ form a transcendence basis.<|endoftext|> TITLE: In a large sparse matrix, how many eigenvalues/eigenvectors are “spurious”? QUESTION [5 upvotes]: In a large (possibly above $5000\times 5000$) matrix, the problem of finding all the eigenvalues and eigenvectors can be solved using iterative methods (Arnoldi, Lanczos etc.). However, there seems to be a convergence happening in these methods suggesting that one can quantify how many eigenvalues/eigenvectors are needed before we have sufficient information and can stop finding more (I apologize for not being able to make this more precise). This is also indirectly suggested in some posts here for example. To take a concrete example, if I have a large antisymmetric matrix made up of only $\pm 1,0$ then there are several eigenvalues that are close to zero but not exactly zero. Their corresponding eigenvectors are almost parallel but not quite. In this example I would be interested to know how many of these are actually numerical residues. Can someone suggest how this distinction between important and spurious eigenvalues can be made? Or even in terms of the spectrum (eigenvectors)? In other words given an $n\times n$ matrix (in my case it is also hermitian, but a general answer, if it exists will be nice) can one quantify how many eigenvalues are spurious? REPLY [5 votes]: Antisymmetric matrices are normal, hence they can be diagonalized with an orthogonal matrix. So $\|A\|_F^2=\sum |\lambda_i|^2$. If you keep only the $k$ largest eigenvalues (ordered in modulus: $|\lambda_1| \geq |\lambda_2| \geq \dots \geq |\lambda_k| \geq |\lambda_{k+1}| \geq \dots \geq |\lambda_n|)$ and replace the other ones with zeros, you get the best possible rank-$k$ approximation $A_k$ to $A$ in Frobenius norm (because of the Eckart-Young theorem: the eigendecomposition of a normal matrix is, up to signs and absolute values, an SVD). Moreover, you can compute exactly the approximation error in Frobenius norm: $\|A-A_k\|_F^2 = \sum_{i=k+1}^n |\lambda_i|^2 = \|A\|_F^2 - \sum_{i=1}^k |\lambda_i|^2$. When this difference is sufficiently small, you can stop your computation (whatever this means).<|endoftext|> TITLE: Is every eigenvector sequence for the Hecke operators a eigenform? QUESTION [5 upvotes]: Let $f(n) = a_n$ be a sequence taking values in $\mathbb C$ for $n=1,2,...$. Let $T_m$ be the Hecke operators (of a fixed weight $k$) defined as usual in terms of the $a_n$. That is: $$T_m(f)(n) = \sum_{r>0, r|(m,n)}r^{k-1}a_{mn/r^2}.$$ Suppose $f$ is such that it is an eigenvalue for all the $T_m$, then is it true that $f$ is actually a cusp form for $SL_2(\mathbb Z)$? That is, is $\sum_{n\geq 1}a_nq^n$ the q-expansion of some modular form? Also, is it true that such sequences always have an analytic (meromorphic?) continuation to the entire plane? This is certainly true if the answer to the first question is yes. One can also generalize a little and consider $f$ that is an eigenvalue for all $T_m$ for $(m,N) = 1$ for some fixed number $N$. In this case, do they necessarily come from some congruence subgroup of level $n$? REPLY [5 votes]: The answer is no. Your form $f$ would have to be of weight $k$, and of level $SL_2(\mathbb Z)$, so should be in a finite dimensional vector space of dimension $d$. That would mean that the Hecke operators $T_n$ acting on the space of all sequences would have at most $d$ different systems of eigenvalues. But this is not true. It is certainly possible to see this by hand. But to avoid any computation: If you take $N$ large enough, the number of different systems of eigenvalues, for the Hecke operators $T_\ell$, $\ell \nmid N$ of forms of weight $k$ and level $N$ goes to infinity. Choose $N$ such that this number is $>d$. Let $(a_\ell)_{\ell \nmid Np}$ be such a system. Since the $T_\ell$'s commute, there exists eigenforms (in your big space of sequences) for all the $T_\ell$ ($\ell$ prime, with no condition) with eigenvalues $a_\ell$ when $\ell \nmid N$. Hence there are strictly more that $d$ systems of eigenvalues appearing in the space of sequence, a contradiction. You could ask a slightly more difficult question by asking if $f$ is always a modular form of some level $N$ instead of $1$. But the answer would still be no: there are sequences which are eigenvectors for all the $T_n$ with system of eigenvalues different from any classical modular forms of weight $k$ and some level $N$. You can get some for instance using $p$-adic modular form of weight $k$.<|endoftext|> TITLE: The space of triangles that fit inside a given triangle, parametrized by edge lengths QUESTION [14 upvotes]: Given a triangle T with sides a, b, and c, describe its "fitting set," the set of all points (x,y,z) in 3-dimensions for which a triangle with sides x, y, z exists that fits in T. Such a set lies in the positive octant, is star-shaped with respect to the origin, and probably is a polyhedron; but it seems difficult to describe. REPLY [17 votes]: Here's the abstract of K.A. Post, "Triangle in a triangle: On a problem of Steinhaus", Geom Dedicata (1993) 45: 115; this paper was cited in the one given in the comment by Nemo. A necessary and sufficient condition on the sides $p, q, r$ of a triangle $PQR$ and the sides $a, b, c$ of a triangle $ABC$ in order that $ABC$ contains a congruent copy of $PQR$ is the following: At least one of the 18 inequalities obtained by cyclic permutation of $\{a, b, c\}$ and arbitrary permutation of $\{p, q, r\}$ in the formula: \begin{array}{l} Max\{ F(q^2 + r^2 - p^2 ), F'(b^2 + c^2 - a^2 )\} \\ + Max\{ F(p^2 + r^2 - q^2 ), F'(a^2 + c^2 - b^2 )\} \le 2Fcr \\ \end{array} is satisfied. In this formula $F$ and $F'$ denote the surface areas of the triangles, i.e. \begin{array}{l} F = {\textstyle{1 \over 4}}(2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 - a^4 - b^4 - c^4 )^{1/2} \\ F' = {\textstyle{1 \over 4}}(2p^2 q^2 + 2q^2 r^2 + 2r^2 p^2 - p^4 - q^4 - r^4 )^{1/2} . \\ \end{array} Here's a picture for the case $a=b=c=1$, which I plotted in this SageMath Jupyter notebook. Note that in this case we need only consider 3 of the above inequalities (plus the triangle inequalities):<|endoftext|> TITLE: Explicit generalizations of Mobius transformations? QUESTION [5 upvotes]: Mobius transformations map circles to circles. Wiki says 'Möbius transformations can be more generally defined in spaces of dimension n>2 as the bijective conformal orientation-preserving maps from the n-sphere to the n-sphere'. Is there an explicit way to given such transformations in $n$-dimenstions? REPLY [6 votes]: Yes, and, indeed, these examples special cases of the situation that $G$ is a simple real Lie group, $K$ is a maximal compact subgroup, and we look at the action of $G$ on the left on the Riemannian symmetric space $G/K$. For the so-called classical groups $G$, the spaces $G/K$ are model-able in terms of matrices, vectors, quadratic forms, and other geometric algebra. For example, the $n$-dimensional real hyperbolic space is isomorphic to $X=O(n,1)/O(n)$, with orthogonal groups of the indicated signatures. Just as the upper half-plane is isomorphic to the unit disc, with a slightly different group acting by linear fractional transformations, there is the real $n$-ball model of $X$, as well as an upper half-space model, using, respectively, different models of $O(n,1)$. Several examples are carried out in considerable detail in some of the notes linked-to from my page http://www.math.umn.edu/~garrett/m/lie/ In particular, the essays "Classical homogeneous spaces" and "Classical groups, domains, cones" treat these and other examples very explicitly. EDIT: Perhaps the easiest example is to give the action of $G=\{g\in GL_{n+1}(\mathbb R): g^TQg=Q\}$, where $Q=\pmatrix{-1_n & 0 \cr 0 & 1}$, on the unit ball in $\mathbb R^n$, by $\pmatrix{a & b \cr c & d}(x)=(ax+b)(cx+d)^{-1}$, where $a$ is $n$-by-$n$, $b$ is $n$-by-$1$, etc., and $x$ is $n$-by-$1$. This is a model of real hyperbolic $n$-space. The upper half-space model is similar.<|endoftext|> TITLE: What is the matter with Hecke operators? QUESTION [22 upvotes]: This question is inspired by some others on MathOverflow. Hecke operators are standardly defined by double cosets acting on automorphic forms, in an explicit way. However, what bother me is that Hecke operators are also mentioned in the automorphic representations language. Those remain very mysterious to me: what are they exactly? Is there a way to define them more explicitly than using the coefficients appearing in the Dirichlet series of the L-function associated to the representation? (What is, even if it appears explicit, unpractical). My main aim is to grasp the Hecke eigenvalues of a representation, for instance using trace formulas. In the "function" language, since Hecke operators are defined as the action of double classes, we can take the Hecke operators themselves as test-functions and obtain some "Eichler-Selberg"-type trace formulas. But on the "representation" language, is there something doing the similar work? More precisely my question could be: is there a function $f_p$ in the Hecke algebra of a local/global group $G(F)$ such that $\mathrm{tr} \ \pi(f_p) = \lambda_\pi(p)$ where $\pi$ is an admissible/automorphic representation of $G(F)$ and $\lambda_p(\pi)$ the Hecke eigenvalue of $\pi$? Any clue or reference is welcome! REPLY [24 votes]: To keep things simple, let $G$ be a finite group and $K$ a subgroup of it. The simplest definition of the Hecke algebra associated to this pair $(G, K)$ is that it is the algebra of $G$-endomorphisms $\text{End}_G(\mathbb{C}[G/K])$ of the permutation representation of $G$ on $G/K$. The significance of this algebra is that $\mathbb{C}[G/K]$ represents the functor sending a $G$-representation $V$ to its $K$-fixed points $V^K$, and hence the Hecke algebra is the endomorphism ring of this functor: it naturally acts on $V^K$ for each $V$ and is the largest thing to do so. Say that the pair $(G, K)$ is a Gelfand pair if any of the following equivalent conditions hold: whenever $V$ is irreducible, $\dim V^K \le 1$; the representation $\mathbb{C}[G/K]$ is multiplicity-free; the Hecke algebra is commutative. Then we can diagonalize the action of the Hecke algebra on $V^K$ and hence associate to any $K$-fixed vector its Hecke eigenvalues, which together give a character of the Hecke algebra. If $V$ is irreducible and $\dim V^K = 1$ (which, by Frobenius reciprocity, is equivalent to $V$ being a component of $\mathbb{C}[G/K]$) then we only have one $K$-fixed vector to choose from up to scale and hence only one collection of Hecke eigenvalues. One thing that is really nice about the Gelfand pair condition is that it can sometimes be verified "geometrically," by thinking explicitly about multiplication in the Hecke algebra in terms of double cosets and "relative positions" as described here and here. For example, $(S_n, S_{n-1})$ is a Gelfand pair because double cosets in this case correspond to relative positions of two elements of $\{ 1, 2, ... \dots n \}$ under the action of $S_n$, and there are two such relative positions: "equal" and "not equal." "Equal" is the identity so the Hecke algebra is generated by "not equal," and in particular must be commutative. This simple argument already implies that branching for the symmetric groups is multiplicity-free, allowing us to construct Gelfand-Tsetlin bases etc. More generally, relative positions have a natural involution which switches the order of the two cosets involved; this gives an involution on the Hecke algebra, and the Hecke algebra is commutative iff every element is fixed by this involution. To check this condition we just need to check whether each relative position is the same when the two cosets involved are switched. A more explicitly geometric example involving Lie groups is $(SO(3), SO(2))$: here the Hecke algebra involves relative positions of two points on the sphere $S^2$, which are labeled by the length of the shortest path between them. All of these relative positions are invariant under switching the two points involved, and so the Hecke algebra, whatever exactly that means, is also commutative in this case.<|endoftext|> TITLE: Stiefel-Whitney class of an orthogonal representation QUESTION [7 upvotes]: Let $BG$ denote the classifying space of a finite group $G$. For which group cohomology classes $c\in H^2(G;\mathbb{Z}/2)$ does there exist a real vector bundle $E$ over $BG$ such that $w_2(E)=c$? REPLY [5 votes]: You can set this up as an obstruction theory problem. Your cohomology class is represented by a map $c: BG\to K(\mathbb{Z}/2,2)$, and the question is whether this map lifts through the universal Stiefel-Whitney class $w_2:BO\to K(\mathbb{Z}/2,2)$. The primary obstruction to such a lift turns out to be the class $$\beta Sq^2(c) = \beta(c\cup c)\in H^5(BG;\mathbb{Z}),$$ where $\beta$ is the Bockstein associated to the coefficient sequence $\mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/2$. This and several other relevant facts can be found in Teichner, Peter, 6-dimensional manifolds without totally algebraic homology, Proc. Am. Math. Soc. 123, No.9, 2909-2914 (1995). ZBL0858.57033. As Neil Strickland mentions, there are higher obstructions which can possibly be defined in terms of secondary cohomology operations, provided you know enough about the Postnikov tower of $BO$.<|endoftext|> TITLE: Are Minkowski sums of upward closed "convex" sets in $\mathbb{N}^k$ still "convex"? (WAS: Comparing mana costs in Magic: The Gathering) QUESTION [28 upvotes]: This was originally a question about comparing mana costs in Magic: The Gathering, but it's turned into a question about Minkowski sums of upward-closed convex sets in $\mathbb{N}^k$. The original question is preserved below, if you want to see the original motivation (or have an answer to the original question that doesn't bear on this new one). The question is this -- say we have two convex polyhedra $A$ and $B$ in $\mathbb{R}_{\ge0}^k$ whose vertices lie in the lattice $\mathbb{Z}^k$ (they're "lattice polyhedra"); and say moreover that both $A$ and $B$ are upward closed in the obvious partial order (i.e., for each of them, the recession cone is equal to $\mathbb{R}_{\ge0}^k$). Then is $(A+B)\cap \mathbb{Z}^k=(A\cap\mathbb{Z}^k)+(B\cap\mathbb{Z}^k)$? This question seems to be more commonly asked in the case that $A$ and $B$ are polytopes, that is bounded, that is both have recession cone equal to $\{0\}$ rather than to $\mathbb{R}_{\ge0}^k$. In this case the answer is no, and just proving conditions under which it holds is as best I can tell an open problem in dimensions greater than 2. But I'm wondering if in this alternate context the problem is easier. That's the question; but if you want to know what this has to do with mana costs in Magic: The Gathering, or whether it might be easier to answer the original question rather than this, read on... Yes, this is a combinatorics question, dealing with something like matchings where sometimes one vertex is allowed to match two other vertices instead of just one like usual! But it arises from Magic and so will need some setup to state. (I would certainly be interested in generalizations, too.) Background: In Magic: The Gathering, there are 6 types of "mana", denoted {W}, {U}, {B}, {R}, {G} (the 5 colors of mana), and {C} (colorless mana). (We can also imagine a 7th type, {P}, representing 2 life; this is not an actual type of mana.) Mana comes in whole number amounts; by an "amount of mana" I mean a specified whole number amount of each type (and some amount of life, I suppose). A mana cost describes a set of amounts of mana sufficient to pay the cost. A mana cost consists of some number of the following symbols: {W}, {U}, {B}, {R}, {G} -- payable only by 1 mana of that color {C} -- payable only by 1 colorless mana {W/U}, {U/B}, {B/R}, {R/G}, {G/W}, {W/B}, {U/R}, {B/G}, {R/W}, {G/U} -- payable by 1 mana of either appropriate color (all 10 color pairs are listed here) {1} -- payable by 1 mana of any type, colored or colorless {W/P}, {U/P}, {B/P}, {R/P}, {G/P} -- payable by the appropriate color or by 2 life (which we can think of as a fictitious type of mana, {P}) {2/W}, {2/U}, {2/B}, {2/R}, {2/G} -- payable by the appropriate color of mana or by any 2 mana (colored or colorless) (Yes, for those familiar with Magic, I am ignoring {X} and I am ignoring the question of snow, since those don't seriously affect the problem. Indeed we didn't actually need all the above, but I figured I'd be thorough...) We can put a partial ordering on the set of mana costs as follows: A≤B if any amount of mana sufficient to pay for B is sufficient to pay for A. Then, the question is, given two mana costs, how can we determine whether one is less than equal to another or not? If we ignore the existence of the symbols of the form {2/M}, then this problem is not hard; using Hall's marriage theorem, we can see that one just has to check a number of inequalities; specifically, that for each set S of types of mana (including the fictitious {P}), the number of mana symbols in A corresponding to a subset of S must be no more than the number of mana symbols in B corresponding to a subset of S. This gives us at most 2^7 inequalities to check (in fact given the symbols that actually exist one only needs to check 65 of them). But the {2/M} symbols are more of a problem. Because they can correspond to 1 or 2 mana, the marriage theorem doesn't apply. One way to handle this is disambiguation -- each {2/M} symbol can be disambiguated to either {2} or {M}. Then A≤B iff for every disambiguation B' of B there's some disambiguation A' of A such that A'≤B'. However, this is slow in the worst case, because it requires checking lots of disambiguations. My question is: Can we do better? I hypothesize the following two statements that will help reduce the number of disambiguations needed when symbols of the form {2/M} are involved: Cancellation applies. That is: If we define addition of mana costs in the obvious way, it's clear that $a\le b$ implies $a+c\le b+c$. If we don't allow symbols of the form {2/M}, the converse also holds, by the reasoning above. Question: Does it also hold when such symbols are allowed? Edit: It seems a positive answer to #3 implies a positive answer to this (see comments). I'm retitling the question to focus on #3. Suppose we have mana costs A and B and suppose M is a color such that the symbol {2/M} does not occur in A. Is it true then that A≤B if and only if A≤B', where B' consists of B but with all instances of {2/M} replaced by {1}? (EDIT: Disproved in the comments by Pace Nielsen. It does however hold if we require A have no symbols of the form {2/M} for all colors M (again, sketch in the comments), but unfortunately that weaker form doesn't form such a useful part of computing the order relation.) (Added after Will Sawin asked it in the comments) -- is the set of amounts of mana sufficient to pay for a given mana cost (including the fictitious {P}) equal to a convex polyhedron intersected with the integer lattice? (Or integer amounts of mana and even amounts of life, if you prefer to think of it that way.) If so this might give us a faster way to compare than disambiguation, by comparing the convex sets (assuming they can be determined quickly; I'm not sure that they can). (This is definitely true if we ignore symbols of the form {2/M}, by the above reasoning.) Note that it's important here that we allow overpaying; it's not true if we don't (see comments). (I don't think there's any way to get rid of the necessity of doing at least some disambiguations; if A has more of some {2/M} symbol than B, I think you are just going to have to handle those manually. But both the above statements at least would bring it down to only doing that.) So far I haven't been able to prove these work, nor find any counterexample. Can anyone prove or disprove these? In addition like I said I'd be interested to see generalizations. As long as each symbol corresponds to 1 mana the whole thing just comes down to the marriage theorem like I said above. But when symbols can correspond to varying amounts I wouldn't expect it to work so nicely in general. It does seem to still work nicely for the set of symbols that actually exist, as listed above (although I might be wrong!). Is there some abstract property of this set of symbols that causes this, so that we can say when this sort of thing happens? It would be interesting to see. Thank you all! REPLY [5 votes]: It seems to me that all of the standard counter-examples to the polytope question easily adapt to be counter examples to this question: Take any polytopes in $\mathbb{Z}^{k-1}$ with $(A_0+B_0) \cap \mathbb{Z}^{k-1} \neq (A_0 \cap \mathbb{Z}^{k-1}) + (B_0 \cap \mathbb{Z}^{k-1})$. Embed $\mathbb{Z}^{k-1}$ into $\mathbb{Z}^k$ as $\{ (x_1, \ldots, x_k) : \sum x_i=0 \}$. Let $A$ and $B$ be the Minkowski sums $A_0+\mathbb{Z}_{\geq 0}^k$ and $B_0 + \mathbb{Z}_{\geq 0}^k$. Then $(A+B) \cap \mathbb{Z}^{k-1} = (A_0+B_0) \cap \mathbb{Z}^{k-1}$, $A \cap \mathbb{Z}^{k-1}=A_0\cap \mathbb{Z}^{k-1}$ and $B \cap \mathbb{Z}^{k-1}=B_0\cap \mathbb{Z}^{k-1}$, so $A$ and $B$ have the same problem. As a concrete example, take $A_0 = \mathrm{Hull}( (0,0,0),\ (1,1,-2))$ and $B_0 = \mathrm{Hull}( (1,0,-1),\ (0,1,-1))$. Then $(1,1,-2) \in A_0 + B_0$, but is not in $(A_0 \cap \mathbb{Z}^3) + (B_0 \cap \mathbb{Z}^3)$. Take $A = A_0 + \mathbb{Z}_{\geq 0}^3$ and $B = B_0 + \mathbb{Z}_{\geq 0}^3$ and you see exactly the same problem occurring with the recession cones your requested.<|endoftext|> TITLE: Does there exist a graph with maximum degree 8, chromatic number 8, clique number 6? QUESTION [18 upvotes]: Question. Does there exist a graph $G$ with $(\Delta(G),\chi(G),\omega(G))=(8,8,6)$? Remarks. Here, 'graph'='undirected simple graph'='irreflexive symmetric relation on a set' any number of vertices is permitted in the question (though, trivially, at least 8 vertices (and quite a bit more) vertices are necessary finiteness of the graph is not required, but finite graphs are the main focus $\Delta(G)$ $=$ maximum vertex degree of $G$ $\chi(G)$ $=$ chromatic number of $G$ $\omega(G)$ $=$ clique number of $G$ Needless to say, what I am asking for is not ruled out by the trivial bounds $\omega\leq\chi\leq\Delta+1$. One should also note that what I am asking for can be regarded as a 'critical instance' for Brooks' theorem (since $(\chi(G),\omega(G))=(8,6)$ by itself implies that $G$ is neither a complete graph nor a circuit, Brooks theorem applies and guarantees that $\chi(G)\leq 8$, hence what is being asked is an explicit example of a graph achieving Brooks' bound; what seems to make it difficult is that the clique number is required to be six.) Needless to say, the condition that $\Delta(G)=8$ rules out most of the 'usual' 'named graphs (e.g., all strongly regular graph that the English Wikipedia currently lists as a named example are not 8-regular, so do not satisfy $\Delta(G)=8$ While I won't go into details to try to 'prove' that I did, I think I did try quite a few things to find an example. This question is motivated both by research on triangle-free graphs (of course, there, $\omega=2)$ and in particular by my writing a comprehensive answer to this interesting research question of 'C.F.G.'. I seem to need a graph as in the question, to make my answer 'more complete', if that's grammatically possible. The particular instance in the question isn't really necessary to my answer, but it would be very nice to have a construction, or proof of non-existence of, such a graph. Let me also mention that constructions of, or proofs of impossibility of, graphs $G$ with (axiom.0) $\quad\Delta(G)\geq 8$ (axiom.1) $\quad\omega(G)\leq\lceil\frac12\Delta(G)\rceil+2$ (axiom.2) $\begin{cases} 3+\frac12\Delta(G)\hspace{25pt} < \quad\chi(G)\quad \leq 2+\frac34(\Delta(G)+0) & \text{if $\Delta(G)\ \mathrm{mod}\ 4\quad = 0$ } \\ 2+\frac12(\Delta(G)+3) < \quad\chi(G)\quad \leq 2+\frac34(\Delta(G)+3) & \text{if $\Delta(G)\ \mathrm{mod}\ 4\quad = 1$ } \\ 3+\frac12\Delta(G)\hspace{25pt} < \quad\chi(G)\quad \leq 0+\frac34(\Delta(G)+2) & \text{if $\Delta(G)\ \mathrm{mod}\ 4\quad = 2$ } \\ 3+\frac12(\Delta(G)+1) < \quad\chi(G)\quad \leq 1+\frac34(\Delta(G)+1) & \text{if $\Delta(G)\ \mathrm{mod}\ 4\quad = 3$ } \end{cases}$ would be helpful for writing the answer to C.F.G.'s question, too, though I chose to make the actual question simpler, by picking out the instance $\Delta(G)=8$, whereupon the first line of the cases above becomes $7<\chi(G)\leq8$, equivalently, $\chi(G)=8$. Note also that then $\lceil\frac12\Delta(G)\rceil+2=6$, explaining the $\omega$-value in the question. Again, an answer to the present answer does not simply imply an answer to the cited thread; I am not asking someone else's question be answered again here; the present question arises as a relevant technical sub-issue in my answer. For the above more general specification of graphs, planar graph evidently are no solution (since they have $\chi(G)\leq4$), and dense random graphs are not a solution either because, very roughly speaking, for them asymptotically almost surely (axiom.1) is satisfied but (axiom.2) is not (because, very very roughly, $G(n,\frac12)$ has $\Delta\in\Theta(\frac12 n)$ but $\chi\in\Theta(\frac{n}{\log n})$, making it impossible to satisfy the lower bound in (axiom.2). Results of McDiarmid, Müller and Penrose make it possible to 'try out' random geometric graphs, but for those, sadly, again not all axioms seem to be satisfied (though getting comprehensible results on the maximum degree in random geometric graph seems to be difficult). Random regular graphs also do not provide an answer, e.g. because [Coja-Oghlan--Efthymiou--Hetterich: On the chromatic number of random regular graphs Journal of Combinatorial Theory, Series B, 116:367-439] implies that for this it is necessary that $8 \geq (2\cdot 8-1)\log(8) - 1$, which is false. REPLY [4 votes]: The example given by Fedor Petrov is well-known and so is the generalization mentioned by Peter Heinig. Another way to view it is as the line graph of a 5-cycle with all tripled edges. The first appearance i know of was in the 1970s, used by Paul Catlin in his counterexample to the Hajos conjecture https://doi.org/10.1016/0095-8956(79)90062-5 The example with $\chi=\Delta=8$ and $\omega=6$ is also the only known (connected) counterexample to the Borodin-Kostochka conjecture for $\Delta=8$. The more general examples prove tightness of a conjecture on the chromatic number of vertex transitive graphs Cranston and i made: http://www.combinatorics.org/ojs/index.php/eljc/article/view/v22i2p1 That conjecture is that vertex transitive graphs with $\chi > \omega$ have $\chi \le (5\Delta + 8) / 6$. This is proved fractionally and the full conjecture follows from Reed's conjecture combined with the strong coloring conjecture. The conjecture is open even for Cayley graphs. Another related conjecture that the example shows tightness for given by Cranston and i in http://epubs.siam.org/doi/abs/10.1137/130929515 is that graphs with $\omega < \Delta - 3$ have $\chi < \Delta$. We prove this for $\Delta \ge 13$. The remaining open cases are $\Delta = 6,8,9,11,12$. Another way to say that is that the OP's question is open in the cases: (6,6,3), (8,8,5), (9,9,6), (11,11,8), (12,12,9).<|endoftext|> TITLE: The limit superior of $\varphi(2^n-1)/(2^n-1)$ QUESTION [13 upvotes]: Let $\varphi$ denote Euler's totient function. It's widely believed that $2^n-1$ is prime for infinitely many (prime) $n$, which, in turn, implies $$\limsup_{n \to \infty} \frac{\varphi(2^n-1)}{2^n-1} = 1. $$ But what about an unconditional proof? This is sort of a follow-up of a question asked by @kodlu yesterday. REPLY [29 votes]: Let $n$ be a large prime. Then $2^n-1=p_1\dots p_k$ (possibly with equal factors), where $n\mid p_i-1$ for all $i$. Therefore, $k\leq \log_n(2^n-1)<\frac{n}{\log_2n}$. Thus, $$ \frac{\varphi(2^n-1)}{2^n-1}\geq\prod_{i=1}^k\left(1-\frac1{p_i}\right) \geq \left(1-\frac1n\right)^{n/\log_2n}\to1, \quad n\to\infty, $$ as required.<|endoftext|> TITLE: Obstructions to lift of representations QUESTION [5 upvotes]: Let $G$ be a profinite group. Let $\Lambda$ be a discrete valuation ring over $\mathbb{Z}_p$. Let $A_0,A_1$ be artinian ring over $\Lambda$. Let $\alpha : A_1 \rightarrow A_0$ is a small map that is $\mathfrak{m}_{A_1}.ker(\alpha) = 0$. Assume that a representation $\rho : G \rightarrow GL_n(A_0)$ is given. I would like to know when can a representation be lifted to $GL_n(A_1)$. I vaguely(and perhaps incorrect) know that the obstruction lies in $H^2(G, Ad\rho)$. What is the cocycle class associated to $\rho$ which must vanish so that $\rho$ can be lifted? Any references/hint/solution are welcome. Thanks! REPLY [3 votes]: Something like what you want is true, though it needs to be formulated slightly differently. Let $k$ be the residue field of $A_0$, and let $\overline{\rho}$ denote the ``residual'' representation $\rho \otimes_{A_0} k$. Let $I$ denote the kernel of the map from $A_1$ to $A_0$; then $I$ is a $k$-vector space. We can associate to $\rho$ a natural ``obstruction class'' that lives in $H^2(G, Ad \overline{\rho}) \otimes_k I$. This class vanishes if, and only if, $\rho$ lifts to $GL_n(A_1)$. (Moreover, if this is the case, then the set of such lifts is a torsor over $H^1(G, Ad \overline{\rho}) \otimes_k I$. Here's a brief sketch of the construction of this obstruction class (ignoring some issues with continuity that are routine but slightly annoying): For each $g \in G$, choose a lift $x_g$ of $\rho(g)$ to $GL_n(A_1)$. Of course the map sending $g$ to $x_g$ need not be a group homomorphism, but for any $g,h$ in $G$, we have: $$x_{gh}^{-1} x_g x_h = 1 + y_{g,h},$$ where $y_{g,h}$ lies in $M_n(I)$. The group $GL_n(A_1)$ acts on $M_n(I)$ by conjugation; this action factors through the quotient $GL_n(k)$. We can thus regard $G$ as acting on $M_n(I)$ via $\overline{\rho}$; this identifies $M_n(I)$ with $(Ad \overline{\rho}) \otimes_k I$. With these identifications it is easy to see that $y_{g,h}$ is a $2$-cocycle with values in $(Ad \overline{\rho}) \otimes_k I$. This cocycle vanishes if, and only if, our chosen set of lifts $x_g$ define a homomorphism of $G$ into $GL_n(A_1)$ lifting $\rho$. Moreover if we replace our chosen lifts $x_g$ with another set of lifts $x'_g$, then this changes $y_{g,h}$ by a coboundary. Thus there exists some lift of $\rho$ if, and only if, $y_{g,h}$ is a coboundary, completing the argument.<|endoftext|> TITLE: Conjecture: The number of points modulo $p$ of certain elliptic curve is $p$ or $p+2$ for $p$ of form $p=27a^2+27a+7$ QUESTION [18 upvotes]: Numerical evidence suggests a conjecture that the number of points of certain elliptic curve over $\mathbb{F}_p$ is either $p$ or $p+2$ for $p$ of certain form. Let $p$ be prime of the form $p=27a^2+27a+7$ and $(a/b)$ denote the Kronecker symbol. For integer $k$ nonzero modulo $p$ define $E_k / \mathbb{F}_p : y^2=x^3+2k^3$. $E_k$ is the quadratic twist of $y^2=x^3+2$. Conjecture 1: $\#E(\mathbb{F}_p)=p+1 + (2k^3 / p)$. In other words for $p$ of the given form $\#E(\mathbb{F}_p)$ is either $p$ or $p+2$. Is Conjecture 1 true? Example sage session with 200 bit $p$ sage: a=2^100+8;p=27*a^2+27*a+7;k=3;Kp=GF(p);E=EllipticCurve([Kp(0),Kp(2*k^3)]) sage: o=E.order();o2=p+1+kronecker(2*k^3,p);o==o2 True REPLY [22 votes]: The conjecture is true. To prove it, we can restrict to $k=1$ as explained in the comments. Let $\chi$ denote a cubic Dirichlet character modulo $p$; this exists as $p\equiv 1\pmod{3}$, and it is unique up to complex conjugation. Let $\psi$ denote the (unique) quadratic Dirichlet character modulo $p$. The number of points of the affine elliptic curve $E$ modulo $p$ equals $$\sum_{y\bmod p}\left(1+\chi(y^2-2)+\overline{\chi}(y^2-2)\right)=p+\sum_{y\bmod p}\chi(y^2-2)+\overline{\sum_{y\bmod p}\chi(y^2-2)}.$$ The $y$-sum on the right hand side equals $$\sum_{y\bmod p}\chi(y^2-2)=\sum_{z\bmod p}\chi(z-2)\bigl(1+\psi(z)\bigr)=\sum_{z\bmod p}\chi(z-2)\psi(z).$$ Here $\chi(z-2)=\chi(2-z)$, because $\chi(-1)=\chi((-1)^3)=\chi^3(-1)=1$. Therefore, $$\sum_{y\bmod p}\chi(y^2-2)=\sum_{z\bmod p}\chi(2-z)\psi(z)=\chi(2)\psi(2)J(\chi,\psi),$$ where $J(\chi,\psi)$ is the corresponding Jacobi sum. To evaluate the Jacobi sum $J(\chi,\psi)$, we rely on Chapter 6 of Rose: A course in number theory (2nd ed., Oxford University Press, 1994). By (the last part of) Exercise 10 at the end of this chapter, $$J(\chi,\psi)=\chi(4)J(\chi,\chi).$$ Now we observe that $4p=1+27(2a+1)^2$, and this is the only way to write $4p$ as $u^2+27v^2$ with $u\equiv 1\pmod{3}$ and $v$ positive, cf. Theorem 2.5 and the subsequent comments. Therefore, by Lemma 2.6, $$J(\chi,\chi)=(3a+2)+(6a+3)e^{\pm 2\pi i/3}=\frac{1\pm(6a+3)i\sqrt{3}}{2},$$ the choice of the $\pm$ sign depending on which cubic character we denoted by $\chi$. Putting everything together, $$\sum_{y\bmod p}\chi(y^2-2)=\chi(2)\psi(2)J(\chi,\psi)=\chi(8)\psi(2)\frac{1\pm(6a+3)i\sqrt{3}}{2}.$$ Here, $\chi(8)=\chi(2^3)=\chi^3(2)=1$, and so $$\sum_{y\bmod p}\chi(y^2-2)+\overline{\sum_{y\bmod p}\chi(y^2-2)}=\psi(2)\sum_{\pm}\frac{1\pm(6a+3)i\sqrt{3}}{2}=\psi(2).$$ To summarize, the number of points of the affine elliptic curve $E$ modulo $p$ equals $p+\psi(2)$.<|endoftext|> TITLE: Module with indecomposable and decomposable reductions mod $p$ QUESTION [5 upvotes]: Let $G$ be a finite group and $p$ a prime dividing the order of $G$. Let $V$ be an irreducible $\mathbb{C}[G]$-module. Let $F$ be the finite field $\mathbb{Z} / p \mathbb{Z}$. Suppose that there exists a $\mathbb{Z}$-lattice $L$ in $V$ which is $G$-invariant, which gives a $F[G]$-module $L/pL$. Take some other $G$-invariant $\mathbb{Z}$-lattice $M$ in $V$. Brauer-Nesbitt tells us that the $F[G]$-modules $L/pL$ and $M/pM$ have the same set of composition factors. Is it possible that $L/pL$ is indecomposable and $M/pM$ is decomposable? If yes, are there any general results which tell us that in some cases this cannot happen? REPLY [5 votes]: [EDIT] Since my various edits got quite long, maybe I can answer the question more directly and refer to previous versions for elaboration. A key elementary result can be found in Feit's 1982 monograph The Representation Theory of Finite Groups (North-Holland Mathematical Library). His Chapters I to XII have no titles but are broken into numerous sections. Already in Chapter I he has a section 17, "Algebras over complete local domains", with an explicit result Corollary 17.13 which applies to a semisimple group algebra. (His conventions on $R, K, A$ are formulated at the beginning of the section, and apply to the group algebra of a finite group when $K$ has characteristic 0 and is the fraction field of a ring $R$ of $p$-adic integers, with finite residue field of characteristic $p>0$.) This result shows that any simple $A$-module has an indecomposable reduction mod $p$. For $A=KG$, it just remains to notice that when such a reduction has two or more composition factors, there is also an $RG$-lattice admitting a direct sum as a module over the residue field. When $G= S_3$ and $p=3$, for example, the 2-dimensional standard module is irreducible in characteristic 0 but has two composition factors mod 3. Since all representations in this case can be realized over $\mathbb{Z}$, the extension to $p$-adic integers in $\mathbb{Q}_p$ is harmless. More generally, as I indicated originally, the Lie family $\mathrm{SL}_2(\mathbb{F}_p)$ provides many more examples if one considers Brauer characters along with ordinary irreducible characters. P.S. I've looked further into the standard textbook literature, which is fairly limited in terms of giving detailed accounts of Brauer theory. In Serre's lectures (translated into English as Springer GTM 42), Exercise 15.2 is close to Ben's answer here, but similarly doesn't start with a simple module in characteristic 0. But the treatise by Curtis-Reiner Methods of Representation Theory in two volumes does have an exercise early in Chapter 2 (which introduces modular representations of finite groups in a modern style). This occurs on p. 416 at the end of $\S16$ (before the introduction of Brauer characters in $\S17$), as Exercise 3: "Find an example of a $p$-modular system $(K,R,k)$, a finite group $G$, and two full $RG$-lattices $M_1$ and $M_2$ in a simple $KG$-module $V$, such that $\overline{M}_1$ and $\overline{M}_2$ are not $kG$-isomorphic." (Here the bar indicates reduction modulo $p$ and the notation differs from that used by spin.) [Alex Zalesskii has reminded me of the Feit result, which I had long ago bookmarked but then forgot about.]<|endoftext|> TITLE: Conjecturally unsafe RSA primes $p=27a^2+27a+7$ QUESTION [44 upvotes]: We got strong numerical evidence that primes of the form $p=27a^2+27a+7$ are unsafe for cryptographic purposes since they can be found in the factorization. Consider the following generic factoring algorithm for factoring $n$ which is divisible by $p$. Suppose you known an elliptic curve in Weierstrass form with known multiple $m$ of the order of $E$ modulo $p$. Let $\psi_n$ denote the $n$-th division polynomial of $E$. Choose random integer $X$. If $X$ is the $x$ coordinate on $E$ modulo $p$ then $\psi_m(X) \equiv 0 \pmod{p}$. If it is not, $\psi_m(X)$ need not vanish modulo $p$. By selecting few random $X$ we probabilistically find $p$ from $\gcd(n,\psi_m(X))$. This question conjectures that if $p=27a^2+27a+7$ and $\textrm{kronecker}(2k^3,p)= -1$ then the order of $y^2=x^3+2k^3$ is $p$. In this case $n$ is multiple of the order, so we can use the above algorithm by trying in addition random $k$. We found $200$ bit factor with pari/gp in just few seconds. Are these conjecturally unsafe RSA primes known? REPLY [7 votes]: This appears special case of "A New Special-Purpose Factorization Algorithm": https://pdfs.semanticscholar.org/1843/73605e846f90b0a9d7252931bab4c47a1ec7.pdf From the abstract: a new factorization algorithm is presented, which finds a prime factor $p$ of an integer $n$ in time $(D \log{n})^{O(1)}$, if $4p - 1 = Db^2$ If $p=27a^2+27a+7$ we have $4p-1=3\square$ which is covered by the paper.<|endoftext|> TITLE: Completely bounded norm for unital maps with completely positive sections QUESTION [6 upvotes]: Consider a completely bounded unital map $\Phi: \mathbf M_h(\mathbb C) \to \mathbf M_k(\mathbb C)$. Suppose that $\Phi$ has right-inverse $\Psi$ which is completely positive. Is the operator norm of $\Phi$ stable under tensor products with other spaces — i.e., is $$\lVert \Phi \rVert \stackrel?= \lVert \Phi \rVert_{\mathrm{cb}} = \bigl\lVert \Phi \otimes \mathrm{id}_{\mathbf A} \bigr\rVert$$ for $\mathbf A = \mathbf M_h(\mathbb C)$? REPLY [7 votes]: Unfortunately, the answer is no. Suppose $\Phi : M_4 \rightarrow M_2$ and $\Psi : M_2 \rightarrow M_4$ are given by $$\Phi\left(\left[\begin{array}{cc} A & B\\ C& D\end{array}\right]\right) = A + 10B^T \ \ \textrm{and} \ \ \Psi(A) = \left[\begin{array}{cc} A&0\\ 0&A\end{array}\right] $$ where $A,B,C,D\in M_2$. Then $\Phi$ is unital, completely bounded, $\Psi$ is unital, completely positive and $\Phi\circ\Psi = I_{M_2}$ but $$\|\Phi\| \leq 11 \ \ \textrm{and} \ \ \|\Phi\|_{\rm cb} \geq 20$$ since the transpose map is cb-norm 2. There is also a negative answer if you look for $\Phi$ being left-invertible by a ucp map: In this case a counterexample is given by $\Phi : M_2 \rightarrow M_4$ and $\Psi : M_4 \rightarrow M_2$ defined by $$ \Phi(A) = \left[\begin{array}{cc}A & 0 \\ 0 & A^T\end{array}\right]\ \ \textrm{and} \ \ \Psi\left(\left[\begin{array}{cc} A & B \\ C & D\end{array}\right]\right) = A $$ where $A,B,C,D \in M_2$. Then $\Phi$ is ucb, $\Psi$ is ucp and $\Psi\circ\Phi = I_{M_2}$ but $$ \|\Phi\| = 1 < 2 = \|\Phi\|_{\rm cb}. $$<|endoftext|> TITLE: Examples of Morse functions on unit tangent bundle of the sphere $T^1(S^2)$ QUESTION [5 upvotes]: In a sense, calculus is all about the study of critical points of functions on flat space $\mathbb{R}^N$ (e.g. here). Let's try a different venue, the unit tangent bundle of the sphere. $$ T^1(S^2) = \big\{ (x,\vec{v}): x \in S^2,\, \vec{v} \in T_x(S^2),\, \big|\big|\,\vec{v}\,\big|\big|=1 \big\} $$ And there are other example of circle bundles over the sphere, such as tensor products of this bundle with itself. I was able to obtain on Math.SE a rather general result: cohomology of circle bundles can be found via the Gysin exact sequence. Can obtain this result via Morse theory? $H^0\big( T^1(S^2) \big) \simeq \mathbb{Z} $ $H^1\big( T^1(S^2) \big) \simeq 0 $ $H^2\big( T^1(S^2) \big) \simeq \mathbb{Z}/2\mathbb{Z} $ $H^3\big( T^1(S^2) \big) \simeq \mathbb{Z} $ I post this question here because I know explicit Morse functions always exist but can be difficult to find. And I'm trying to work out the critical points and visualize the Morse flow. And perhaps I should clarify what I mean by "explicit". I believe the unit tangent bundle could be given the structure of a variety. Looking at our construction, we could try to embed $$ T^1(S^2) \subseteq \mathbb{R}^3 \times \mathbb{R}^3$$ I don't think his is quite the "universal bundle" construction, but it's something. And we could write down the constraints: $x_1^2 + x_2^2 + x_3^2 = 1$ $v_1^2 + v_2^2 + v_3^2 = 1$ $\vec{v} \in T_x(S^2) \subseteq \mathbb{R}^3$ which could be a hyperplane in 3-space. $\langle x, v \rangle = x_1 v_1 + x_2 v_2 + x_3 v_3 = 0$ this shows that $\vec{x} \perp \vec{v} $ and that $\vec{v}$ is tangent to $S^2$. And if we pin down all the relations we obtain the structure of an algebraic variety. Therefore, could it be possible to write down polynomial morse functions of this kind of space? And work out the critical points? How many polynomial Morse functions on the sphere? cohomology module of unit tangent vector bundles over spheres https://en.wikipedia.org/wiki/Gysin_homomorphism Even a derivation of the Gysian homomorphism via Morse theory could be interesting. Certainly I've never seen it. We could try to build Morse functions of out of polynomials in $x$ and $v$. Does the set Morse functions of given degree form a vector space? REPLY [5 votes]: To expand the comment: the function $f(x,v):=x_1+2v_2$ is a Morse function on $M:=T^1 S^2$, in your coordinates $(x,v)$. Denoting $(e_j)_{1\le j\le3}$ the standard basis of $\mathbb{R}^3$, it is easy to see that the only critical points $ p:=( x,v)\in M$ of $f$ are $p_0:=(-e_1,-e_2)$, $p_1:=(e_1,-e_2)$, $p_2:=(-e_1,e_2)$, and $p_3:=(e_1,e_2)$, and that $\text{ind}(p_k)=k$. This is more or less evident from geometrical considerations. To make a formal computation, define for $q:=(q_1,q_2,q_3)$ in a nbd of $0\in\mathbb{R}^3$ the skew simmetric matrix $$Q=Q_q:=\left[ \begin {array}{ccc} 0&-q_{{3}}&q_{{2}}\\ q_{ {3}}&0&-q_{{1}}\\ -q_{{2}}&q_{{1}}&0\end {array} \right], $$ and consider a local chart at $p$ of the form $(q_1,q_2,q_3)\mapsto (e^Qx,e^Qv)$. In this chart the function $f$ reads $\tilde f(q)=[e^Qx]_1+2[e^Qv]_2$; since $e^Q=I+Q+Q^2/2+o(Q^2)$ at $Q=0$, by easy computations this gives the second order expansion at $q=(0,0,0)$; precisely $$\nabla \tilde f(0)= (-2v_3 ,\ x_3 ,\ 2v_1-x_2) $$ $$\text{Hess }\tilde f(0):=\ \left[ \begin {array}{ccc} -4\,v_{{2}}&2\,v_{{1}}+x_{{2}}&x_{{3}} \\ 2\,v_{{1}}+x_{{2}}&-2\,x_{{1}}&2\,v_{{3}} \\ x_{{3}}&2\,v_{{3}}&-4\,v_{{2}}-2\,x_{{1}} \end {array} \right] \ .$$ So if $\nabla\tilde f(0)=0$ then $x_3=v_3=0$ and $x_2=2v_1$; since $x\cdot v=0$ also $|v_1|=|x_2|$, so that $v_1=x_2=0$ and since $\|x\|=\|v\|=1$ we also get $x_1=\pm1$, and $v_2=\pm1$, that is $p$ is one of $p_0,\dots, p_3$. For each of these values $\text{Hess }\tilde f(0)$ is a diagonal matrix with respectively $0,\dots,3$ negative elements, ending the computation.<|endoftext|> TITLE: Fundamental groups of non-orientable closed four-manifolds QUESTION [8 upvotes]: The fundamental group of a closed orientable manifold is finitely presented, and every finitely presented group arises as the fundamental group of a closed orientable four-manifold; see this question. Not every finitely presented group arises as the fundamental group of a closed non-orientable four-manifold. One necessary condition is that it contain an index two subgroup which corresponds to the orientation double cover. Is this the only restriction? Let $G$ be a finitely presented group with an index two subgroup. Is there a closed non-orientable four-manifold $M$ with $\pi_1(M) \cong G$? REPLY [9 votes]: Let $G$ be a finitely presented group $$ G = \langle g_1, g_2, \cdots, g_n | R_1, R_2, \cdots, R_m \rangle$$ One standard way to realize this as the fundamental group of a compact $4$-manifold is to construct a $5$-dimensional manifold. 1) Start with $D^5$ and attach a $1$-handle for every generator $g_i$. Here the attaching map you use does not matter all that much but to get the process moving let's choose the attaching maps so the resulting handlebody will be orientable. We will want to change this to answer your question. . . but this works for now. 2) You then attached a $2$-handle for every relator $R_j$. The $2$-handles are attached via a circle mapping to the handlebody constructed in step (1). A map from a circle to the previous handlebody is determined (up to homotopy) by a conjugacy-class in the fundamental group of the handlebody, i.e. a word in the free group, which is exactly what the relator $R_j$ is. To make this a handle attachment you need to ensure the circle is in the boundary of the handlebody, and it is embedded, and has a trivial normal bundle. This can all be achieved due to our choice of making the handlebody orientable in step (1) and a little transversality. So this is a 5-manifold with the appropriate fundamental group. But the manifold has boundary, so we check to see if the boundary (which is a closed manifold) also has the same fundamental group. Here we use the fact that our 5-manifold deformation-retracts to a $2$-skeleton. So by a little transversality, this $5$-manifold does have the same fundamental group as its boundary. This is where we see we could have made this argument for $4$-manifolds, $5$-manifolds, etc, but this argument falls flat for $3$-manifold fundamental groups. In the case you have an epi-morphism $G \to \mathbb Z_2$, you modify step (1) so that your $1$-handle attachment is orientation-reversing provided the loop maps to something non-trivial via the homomorphism $G \to \mathbb Z_2$. Step (2) is unchanged. The boundary $4$-manifold has the same fundamental group as the $5$-manifold constructed, again by dimension counting (transversality).<|endoftext|> TITLE: The isomorphism class of $\mathrm{Ext}^1_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})$ QUESTION [18 upvotes]: In a recent conversation with a colleague, the following question arose: What is the isomorphism class of $\mathrm{Ext}^1_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})$? That is to say, what is $\mathrm{Ext}^1_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})$ as an Abelian group? I did not have any immediate response other than “That’s interesting; I’ll have to think about it.” Afterwards, I spent some time recalling some standard information about $\mathbb{R}/\mathbb{Z}$ and the Ext functor, and have arrived at the following: $$\mathrm{Ext}^1_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})\simeq {\left(\prod_{x\in\mathbb{R}}{\mathrm{Ext}^1_\mathbb{Z}{\left(\mathbb{Q}_x,\mathbb{Z}\right)}}\right)}\times{\left(\prod_{p\in\mathbb{P}}{\mathrm{Ext}^1_\mathbb{Z}{\left(\mathbb{Z}{\left(p^\infty\right)},\mathbb{Z}\right)}}\right)}\ .$$ Here, $\mathbb{Q}_x=\mathbb{Q}$ for all $x\in\mathbb{R}$, $\mathbb{P}$ denotes the set of prime numbers, and for a prime $p$, $\mathbb{Z}{\left(p^\infty\right)}$ is the Prüfer $p$-group. This decomposition reduces the original problem to the following two questions: What is the isomorphism class of $\mathrm{Ext}^1_\mathbb{Z}(\mathbb{Q},\mathbb{Z})$? Given a prime number $p$, what is the isomorphism class of $\mathrm{Ext}^1_\mathbb{Z}(\mathbb{Z}{\left(p^\infty\right)},\mathbb{Z})$? I was curious to know if anyone happens to know the answer to any of questions (1), (2), or (3), or can point me to such in the literature. REPLY [7 votes]: For context: the group $LA=\text{Ext}(\mathbb{Z}/p^\infty,A)$ is called the derived $p$-completion of $A$. It has a natural map to the ordinary completion $CA=\lim_n A/p^nA$ which is often an isomorphism. In particular it is an isomorphism if $A$ is a free abelian group, or if it is finitely generated, or if there exists $n$ such that $\text{ann}(p^n,A)=\text{ann}(p^{n+1},A)$. In the cases where $CA\neq LA$ it typically works out that $LA$ has better behaviour and is more relevant for applications, especially in algebraic topology and homological algebra. One reference is the book "Homotopy Limits, Completion and Localization" by Bousfield and Kan.<|endoftext|> TITLE: Example of ''annihilation'' of Seiberg-Witten Equation solutions QUESTION [13 upvotes]: The proof that the Seiberg-Witten invariants of a 4-manifold $X$ with fixed Spin$^c$ structure really are invariant wrt the metric used to define them goes roughly as follows (for simplicity let $b_2^+(X) > 1$ and assume the expected dimension of the SW moduli space is 0): If $g_0$ and $g_1$ are metrics on $X$ then they may be joined by a path $(g_t)_{0 \le t\le 1}$ of metrics where each $g_t$ is such that the corresponding SW moduli space avoids reducible solutions. Now the collection of moduli spaces $(M_t)_{0\le t\le1}$ of the SW equations for each metric $g_t$ can be viewed as a cobordism between the 0-dimensional oriented manifolds $M_0$ and $M_1$. Hence the number of solutions with orientation accounted for is the same in each. This allows for an "annihilation" phenomenon to occur among solutions: for example, {+,+,-} is cobordant to {-} via a cobordism consisting of one line from + to - and a U-shaped line along which + and - annihilate. My question is whether anyone can illustrate an explicit example of two solutions to the SW equations wrt a given metric $g_0$ that have opposite orientation and "annihilate" along a path $(g_t)_{0 \le t \le 1}$ to some final metric $g_1$. I would also be interested in any qualitative discussion of this phenomenon. REPLY [3 votes]: I can get you very close to explicit, while giving a qualitative discussion, though you may find what follows as cheating. For starters, I won't be writing down local coordinates. Anyway, Taubes gave a fairly explicit construction for how SW solutions relate to $J$-holomorphic curves in symplectic manifolds (and the signs attached to curves agree with the signs attached to SW solutions). Here, deforming the metric is equivalent to deforming $J$. The relation between curves and SW solutions: Given a curve, the corresponding SW solution will be a pair $(A,\psi)$ of connection and spinor, where $A$ is flat away from the curve while its curvature builds up over the curve (the energy of the curve is related to the norm of the curvature), and $\psi$ decomposes into a pair of sections $(\alpha,\beta)$ for which $\alpha$ vanishes along the curve (roughly speaking). So we can look for annihilations and bifurcations of $J$-holomorphic curves, which you can then build corresponding SW solutions. In one of Taubes' papers is an explicit example with holomorphic tori. But let's make this simpler by taking the explicit example $X=S^1\times T_\phi$, where $T_\phi$ is the mapping torus of a symplectomorphism $\phi:\mathbb{T}^2\to\mathbb{T}^2$ of a torus. Now isotoping $\phi$ ultimately corresponds to deforming the metric (and deforming $J$), and periodic orbits of $\phi$ correspond to SW solutions (and $J$-holomorphic tori). And you can write down a bifurcation in which an elliptic orbit cancels a positive hyperbolic orbit of the same period; the orbits have opposite "Lefschetz sign". So if you trace back through all of that: By writing down an explicit dynamical system where orbits bifurcate, then you can write down explicit connections and spinors which almost solve the SW equations (Taubes' construction is ultimately a perturbative approach, he gets these configurations from a curve and then uses the Implicit Function Theorem to get a nearby honest SW solution).<|endoftext|> TITLE: Values of Artin L-functions at negative integers QUESTION [8 upvotes]: Let $F$ be a number field and $\chi$ a one dimensional Artin character. That is, it is a map $\chi: Gal(\overline F/F) \to \mathbb C^\times$ and let $L(s,\chi)$ be it's L-series. What is known about the values of $L(s,\chi)$ at negative integers? Are they always algebraic integers contained in $F(\chi)$? When are they zero? Is there a (conjectured) explicit representation? What about the case when $\chi$ is trivial? What about higher dimensional characters? Are there any conjectures dealing with these values? I only know the case $F = \mathbb Q$ and $\chi$ a Dirichlet character where you can find an explicit representation in terms of (generalized) Bernoulli numbers. REPLY [7 votes]: The question of order of vanishing is quite elementary: the L-function of a Hecke character (i.e. 1-dimensional Artin representation) over any number field has an Euler product, which is convergent and non-vanishing for $s > 1$, and it satisfies a functional equation relating $L(\chi, s)$ to $L(\bar\chi, 1-s)$; hence the orders of vanishing for negative integer $s$ are completely determined by the Gamma-factors in the functional equation. After some unravelling, one finds that for $s < 0$, or for $s = 0$ and $\chi$ non-trivial, the order of vanishing is $$(\text{# of complex places of $F$}) + \left(\begin{array}{l}\text{# of real places of $F$}\\\text{where $\chi$ has sign $(-1)^s$}\end{array}\right). $$ So there will only be non-vanishing values to study if $F$ is totally real, $\chi$ is either totally even (sign $+1$ at every real place) or totally odd (sign $-1$ at every real place), and $s$ satisfies a parity condition depending on the sign of $\chi$. These cases have been studied in great detail by many mathematicians, notably C.L. Siegel. It's known, for instance, that (for $s$ of the appropriate parity) $L(\chi, s)$ is an algebraic number and depends Galois-equivariantly on $\chi$; that is, we have $L(\chi, s)^\sigma = L(\chi^\sigma, s)$ for $\sigma \in \operatorname{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$. It is not always integral, but its denominator is very well understood (analogously to the von Staudt-Clausen theorem on Bernoulli numbers). Moreover, the crowning theorem of the subject, the Mazur--Wiles theorem (formerly Iwasawa's "main conjecture"), relates the numerator of $L(\chi, s)$ to the order of an ideal class group. For higher-dimensional Artin representations, much less is known (although there are plenty of very precise conjectures). We don't even know if the L-function has analytic continuation, only meromorphic continuation, so asking about its values at negative integers might not even make sense.<|endoftext|> TITLE: Fourier transform of $f$ and $|f|$? QUESTION [5 upvotes]: What is the relationship between the Fourier transform of an $L^1$ function $f: \mathbb{R}^d \to \mathbb{C}$ and the Fourier transform of $|f|$? In other words, what is the relationship between $$ \widehat{f}(\xi) = \int e^{-2\pi i x \cdot \xi} f(x) dx $$ and $$ \widehat{\, | f |\, }(\xi) = \int e^{-2\pi i x \cdot \xi} |f(x)| dx? $$ Writing $f = g|f|$, we have $\widehat{f} = \widehat{g} \ast \widehat{\, | f |\, }$. But can we get something more explicit? More generally, what is the relationship between the Fourier transform of a complex-valued finite measure $\mu$ and the Fourier transform of $|\mu|$, where $|\mu|$ is the the variation of $\mu$. In other words, what is the relationship between $$ \widehat{\mu}(\xi) = \int e^{-2\pi i x \cdot \xi} d\mu(x) $$ and $$ \widehat{\, | \mu |\, }(\xi) = \int e^{-2\pi i x \cdot \xi} d|\mu|(x)? $$ Remark: The title of the following question is similar, but the content is different. Connection between the Fourier transform of f and |f| REPLY [4 votes]: Not sure about $|f|,$ but for $g = |f|^2,$ the fourier transform of $g$ is given by $$\widehat{g(y)} = \frac{1}{2\pi}\int_{-\infty}^\infty \overline{F(t)} F(t + y) d t,$$ where $F$ is the fourier transform of $f.$ The above is apparently known as the Wiener-Khinchin(-Einstein) theorem.<|endoftext|> TITLE: Copies of topological fundamental groups inside etale fundamental groups given by different embeddings of your field into $\mathbb{C}$ QUESTION [10 upvotes]: Let $X$ be a smooth curve over a number field $K$ (not necessarily proper). Fix an algebraic closure $\overline{K}$ of $K$. Let $i,i' : \overline{K}\hookrightarrow\mathbb{C}$ be two abstract embeddings (ie, as $\mathbb{Q}$-algebras). Let $X_\mathbb{C},X_{\mathbb{C}}'$ be the base changes of $X_{\overline{K}}$ to $\mathbb{C}$ via $i$ and $i'$. Then, $X_\mathbb{C}(\mathbb{C})$ has the structure of a Riemann surface. Let $x\in X_\mathbb{C}(\mathbb{C})$ come from a $\overline{K}$-rational point. We may consider its topological fundamental group $\pi_1^{top}(X_\mathbb{C}(\mathbb{C}),x)$. Since every finite cover of the Riemann surface $X_\mathbb{C}(\mathbb{C})$ is algebraic, for every loop in $\pi_1^{top}(X_\mathbb{C}(\mathbb{C}),x)$, its monodromy action on the fibers at $x$ of its finite covers determines an automorphism of the fiber functor at $x$, and hence we obtain a homomorphism $$\pi_1^{top}(X_\mathbb{C}(\mathbb{C}),x)\rightarrow \pi_1^{et}(X_\mathbb{C},x)$$ which is known to be the embedding of the first group into its profinite completion. The map $X_\mathbb{C}\rightarrow X_{\overline{K}}$ given by base change induces an isomorphism on etale fundamental groups, and composing these maps we get $$\pi_1^{top}(X_\mathbb{C}(\mathbb{C}))\longrightarrow \pi_1^{et}(X_\mathbb{C})\stackrel{\sim}{\longrightarrow}\pi_1^{et}(X_\overline{K})$$ where I've omitted the base points because I only care about these maps up to conjugacy (say, inside $\pi_1^{et}(X_{\overline{K}})$). Similarly, with $X_\mathbb{C}'$, we get a map $$\pi_1^{top}(X'_\mathbb{C}(\mathbb{C}))\longrightarrow \pi_1^{et}(X'_\mathbb{C})\stackrel{\sim}{\longrightarrow}\pi_1^{et}(X_\overline{K})$$ Both of these maps give embeddings of the topological fundamental groups inside $\pi_1^{et}(X_{\overline{K}})$, canonical up to conjugation. My question is: When are the images the same (up to conjugation)? Are there examples when the images are not the same? I'm particularly interested in the case when $X_\mathbb{C}(\mathbb{C})$ is hyperbolic. References would also be appreciated. REPLY [2 votes]: For affine hyperbolic curves, when $i$ and $i'$ agree on $K$, this happens only when $i$ and $i'$ are equal or complex conjugate of each other. Let $f: \pi_1^{top}(X) \to \pi_1^{et}(X_{\mathbb C})$ be the natural dense inclusion, and $e_i,e_{i'} \pi_1^{et}(X_{\mathbb C}) \to \pi_1^{et}(X_{\overline{K}})$ be the natural isomorphisms defined by $i$ and $i'$. Let $\sigma$ be the element of the Galois group of $\overline{K}$ over $K$ that sends $i$ to $i'$, then $e_{i'}$ is $e_i$ composed with the action of $\sigma$ by outer automorphism of $\pi_1^{et}(X_{\overline{K}})$. If the image of $e_i \circ f$ is conjugate to $e_{i'} \circ f$, then there must be some automorphism $\alpha$ of $\pi_1^{top}(X)$ such that $e_i \circ f \circ \alpha$ is conjugate to $e_{i'} \circ f = \sigma \circ e_i \circ f$ as a homomorphism. Now because $f$ is dense, we can extend $\alpha$ to an outer automorphism of $\pi_1^{et}(X_{\mathbb C})$ and thus to an outer automorphism of $\pi_1^{et}(X_{\overline{K}})$, and then the condition that these two maps are conjugate becomes the condition that these two outer automorphisms are equal in the outer automorphism group. It is a result of Matsumoto and Tamagawa that this can only happen for the identity and comple conjugation (Mapping-Class-Group Action versus Galois Action on Profinite Fundamental Groups, Remark 2.1). Furthermore, it looks to me that following their argument, the only case that the affineness assumption is used can be replaced with Theorem C(i) of Hoshi and Mochizuki<|endoftext|> TITLE: Why is there no symplectic version of spectral geometry? QUESTION [28 upvotes]: First, recall that on a Riemannian manifold $(M,g)$ the Laplace-Beltrami operator $\Delta_g:C^\infty(M)\to C^\infty(M)$ is defined as $$ \Delta_g=\mathrm{div}_g\circ\mathrm{grad}_g, $$ where the gradient of a function $f\in C^\infty(M)$ is defined by $$\iota_{\mathrm{grad}_g(f)}g=df,$$ $\iota_\bullet$ being the contraction with a vector field, and the divergence of a vector field $X\in \Gamma^\infty(TM)$ is defined by $$ \mathcal{L}_X\,\mathrm{vol}_g=\mathrm{div}_g(X)\,\mathrm{vol}_g, $$ where $\mathcal{L}_\bullet$ is the Lie derivative and $\mathrm{vol}_g$ the Riemannian volume density on $M$. On a symplectic manifold $(M,\omega)$ we can proceed by complete analogy and define the symplectic Laplacian $\Delta_\omega:C^\infty(M)\to C^\infty(M)$ as $$ \Delta_\omega=\mathrm{div}_\omega\circ\mathrm{grad}_\omega, $$ where the symplectic gradient of a function $f\in C^\infty(M)$ is defined by $$\iota_{\mathrm{grad}_\omega(f)}\omega=df,$$ and the symplectic divergence of a vector field $X\in \Gamma^\infty(TM)$ is defined by $$ \mathcal{L}_X\,\mathrm{vol}_\omega=\mathrm{div}_\omega(X)\,\mathrm{vol}_\omega, $$ where $\mathrm{vol}_\omega=\frac{1}{n!}\omega^n$ is the symplectic volume form on $M$ (here $2n=\mathrm{dim}\,M$). Now, a fundamental difference between Riemannian manifolds and symplectic manifolds is that on the latter we have the Darboux theorem, and a simple computation shows that in Darboux coordinates $q_1,\ldots,q_n,p_1,\ldots,p_n$, the symplectic Laplacian defined above is given by the formula $$ \Delta_\omega = \sum_{i=1}^n\Big(\frac{\partial^2}{\partial p_i\partial q_i}-\frac{\partial^2}{\partial q_i\partial p_i}\Big). $$ By the theorem of Schwarz, this means that $$ \Delta_\omega=0. $$ This seems to be a first hint why we don't have a symplectic version of spectral geometry. I would like to know: Can we prove or refer to a theorem of the form: The Darboux theorem is equivalent to the statement that there is no second order differential operator on $M$ that is invariant under all symplectomorphisms ? A broader question is: Is there a deeper reason behind the non-existence of a symplectic version of spectral geometry? Indeed, recall that on a Riemannian manifold the Laplace-Beltrami is not the only natural differential operator that is invariant under isometries, see Laplace-Beltrami and the isometry group https://math.stackexchange.com/questions/833517/is-every-scalar-differential-operator-on-m-g-that-commutes-with-isometries-a/833672#833672 and the related question https://math.stackexchange.com/questions/1348788/why-is-a-diffeomorphism-an-isometry-if-and-only-if-it-commutes-with-the-laplacia It would be nice to have an argument why in fact there cannot be symplectic analogues of any of the natural differential operators commuting with isometries of a Riemannian manifold. Some comments on the answers received so far: I have accepted Ben McKay's answer because it comes closest to what I hoped for. It indeed seems to settle the question for scalar differential operators. As Igor Rivin pointed out, there is a non-trivial study of operators acting on symplectic spinors. It would be interesting to know whether there are more extra structures apart from symplectic spinor bundles that do not correspond to introducing a Riemannian structure but lead to natural differential operators whose study one could call "spectral symplectic geometry". I am not sure if I really agree with Dan Fox' point of view that a smooth setting in which there are no local invariants should be regarded as part of differential topology rather than differential geometry. Maybe it boils down to the question if global analysis is to be considered part of differential geometry or rather of differential topology. I would say it is part of both. REPLY [7 votes]: We talk about spectral geometry because the length spectrum (i.e. the set of lengths of all closed geodesics) on the Riemannian manifold is intimately related to the spectrum of the corresponding Laplace operator (which is an elliptic operator and on the compact manifolds its spectrum is discrete). However, there is no length on the symplectic manifolds, and 'symplectic Laplacian' not only is not elliptic, but it is most likely its spectrum coincides with $\mathbb{R}$. It seems to me that 'spectral geometry without spectra' is a bit of a stretch<|endoftext|> TITLE: Ozsváth-Szabó's contact invariant on the Brieskorn sphere $\Sigma(2,3,6m+1)$ QUESTION [5 upvotes]: According to Theorem 1.7 of Mark-Tosun's paper, the Brieskorn sphere $\Sigma(2,3,6m+1)$ admits two tight contact structure $\xi_{i}\ (i=0,1)$. They are both Stein fillable and they are contactomorphic (but not isotopic). My question: Consider the Ozsváth-Szabó contact invariant $c(\xi_{i})\in HF^{+}(-\Sigma(2,3,6m+1))$. We know it is non-zero by Stein fillability, but do we know whether its image under the natural map $$ HF^{+}(-\Sigma(2,3,6m+1))\rightarrow HF^{\text{red}}(-\Sigma(2,3,6m+1)) $$ is non-zero? Such result can be shown, e.g., if $\xi_{i}$ bounds a non negative definite Stein domain. REPLY [3 votes]: Brieskorn-Pham calculated the signature of the smoothing of the Milnor fiber of that Brieskorn singularity (in this case the smoothing of the complex singularity $x^2+y^3+z^{6m+1}=0$ or the symplectic branched double cover over the smoothing of the algebraic surface bounding $T(3,7)$ ). In the case of $M(2,3,6m+1)$, this is non-definite. See remark 4.6 of http://www.maths.ed.ac.uk/~aar/papers/nemethi1.pdf for Brieskorn's formula. In the case of $\Sigma(2,3,6m+1)$, the contact structure has homotopy type $\theta=-2$. As $c_1=0$ for $M(2,3,6m+1)$ (this can be seen by the double cover representation), Gompf's formula for $\theta$, $c_1^2-3\sigma(X)-2\chi(X)=\theta$ gives $$\sigma(M(2,3,6m+1))=-8m$$<|endoftext|> TITLE: Automorphism of the transfinite rooted binary tree QUESTION [6 upvotes]: I was studying combinatorical group theory recently, and I came across the infinite regular rooted binary tree and its automorphism group $Aut(T^{(2)})$with the Grigorchuk subgroup. Let me now elaborate more on a different topic, the surreal numbers. They are constructed similarly to the binary tree at first, but in the $\omega$-th generation, infinitesimal and transfinite numbers start to appear. I wonder - if we remove, from the surreal number "tree", all numbers and order relations, thus focusing solely on the "transfinite binary tree" (in a graph-theoretic sense), what would be the automorphism group (I allow proper classes to be groups) of that tree? For a better vsiualization of what I mean by such a tree: https://upload.wikimedia.org/wikipedia/commons/thumb/4/49/Surreal_number_tree.svg/532px-Surreal_number_tree.svg.png REPLY [5 votes]: Although this question already has an accepted answer, which is correct for the question as stated, I posit that the surreal number tree is best viewed as a tree in the order-theoretic rather than the graph-theoretic sense. In that case, the automorphism group is isomorphic to the direct product of a class of two-element groups indexed by the surreal numbers: $$ G := \prod_{\alpha \in \textbf{No}} C_2 $$ To formalise this, consider an automorphism $\phi : \textbf{No} \rightarrow \textbf{No}$. We describe the surreal numbers inductively, where element is either: The zero (root) number $0$; The left successor (child) $l(\alpha)$ of an existing number $\alpha \in \textbf{No}$; The right successor (child) $r(\alpha)$ of an existing number $\alpha \in \textbf{No}$; A limit of an ascending chain (with respect to the partial order induced by the tree, not numerical order) of numbers $\sup_{i \in I} \alpha_i$. We shall construct such a $\phi$ naturally from an arbitrary $\psi \in G$ viewed for convenience as a function $\psi : \textbf{No} \rightarrow C_2$. Let $C_2$ act on the two-element set $\{ l, r \}$ in the obvious way, and define: $ \phi(0) := 0 $ $ \phi(l(\alpha)) := (\psi(\alpha)(l))(\phi(\alpha)) $ $ \phi(r(\alpha)) := (\psi(\alpha)(r))(\phi(\alpha)) $ $ \phi(\sup_{i \in I} \alpha_i) := \sup_{i \in I} \phi(\alpha_i) $ We have described a map $\Theta : G \rightarrow \textrm{Aut}(\textbf{No})$ which sends $\psi$ to $\phi$, and it is straightforward to check that this is an isomorphism. EDIT: I should probably clarify that in order to manipulate the surreals set-theoretically without pain, we can think of them as being a subset of a Grothendieck universe within our ambient set theory. It is henceforth safe to treat the surreals of that Grothendieck universe as being a set within the larger ambient set theory, and perform usual set-theoretic constructions such as powersets.<|endoftext|> TITLE: Forcing extensions where meagre sets are covered QUESTION [6 upvotes]: This question fits the Generalised Baire space area. I am interested in the meagre ideal on ${}^\kappa \kappa$, with the bounded topology (or box topology), when, say, $\kappa$ is inaccessible. To be more precise, basic open sets have the form $[s]=\{t\in {}^\kappa \kappa \mid t\supseteq s\}$, where $s\in {}^{<\kappa}\kappa$. A set $X$ is nowhere dense if every open set has an open subset that does not meet $X$. Finally, $X$ is meagre if it is the $\kappa$-union of meagre sets. Are there examples or any criterion for a forcing $\mathbb{P}$ to satisfy the following property: "every meagre set in the generic extension is a subset of a meagre set in the sense of the ground model". REPLY [8 votes]: This is just a comment, not an answer, but the system does not allow me to comment: (1) The proof of Claim 5.3(3) in Shelah's 1004 ( https://arxiv.org/abs/1202.5799 ) may as usual give that "old meager sets are cofinal" is the same as "old non-meager sets are non-meager and old reals are dominating" (2) The A-bounding forcings of Roslanowski/Shelah 860 ( https://arxiv.org/abs/math/0508272 ) and more should fit the bill. (3) Concerning "preserving non-meagerness" (not related here, but...) I would like to point out the iterable "manageable" condition from section 5 of Matet/Roslanowski/Shelah 799 (https://arxiv.org/abs/math/0210087 ) Andrzej<|endoftext|> TITLE: Smooth trivialization of smooth Hilbert bundles QUESTION [8 upvotes]: In page 67 of Topology and Analysis by Booss and Bleecker, it is claimed that any Hilbert bundle is topologically trivial. Clearly, any smooth Hilbert bundle over a smooth manifold is topologically trivial, but it appears to be no reason to believe that this trivialization is smooth.My questions are: Are there known conditions on the manifold or on the Hilbert space to guarantee that such topological trivialization is actually smooth? Are there known counterexamples showing that such smooth trivialization is impossible in the general case? EDIT: The definition of smooth Hilbert bundle is the one defined in 2.1 of: László Lempert and Róbert Szőke, Direct images, fields of Hilbert spaces, and geometric quantization, (arXiv:1004.4863) namely A smooth (always complex) Hilbert bundle is a smooth map $p\colon H\to S$ of Banach manifolds, each fiber $p^{-1}(s)$, $s\in S$, is endowed with the structure of a complex vector space; for each $s\in S$ there should exist a neighborhood $U\subset S$, a complex Hilbert space $X$, and a smooth map (local trivialization) $F\colon p^{-1}U \to X$, whose restriction to each fiber $p^{-1}(t)$, $t\in U$, is linear, and such that $p\times F\colon p^{-1} U \to U \times X$ is a diffeomorphism. REPLY [2 votes]: I believe, the answer is (essentially) contained in the main theorem of the paper Equivalences of smooth and continuous principal bundles with infinite-dimensional structure group by Christoph Müller and Christoph Wockel: Let $K$ be a Lie group, modeled on a locally convex space, and $M$ a finite-dimensional paracompact manifold with corners. Then each continuous principal $K$-bundle over $M$ is equivalent to a smooth principal $K$-bundle. Moreover, two smooth principal $K$-bundles are continuously equivalent if and only if they are smoothly equivalent. Now, every smooth Hilbert bundle in the sense of the post gives rise to a smooth $\mathrm{GL}(\mathcal H)$-principal bundle (defined, for instance, by the transition 1-cocycle) which is topologically trivial, hence by the above theorem smoothly trivial, and so therefore is the original vector bundle.<|endoftext|> TITLE: Chart for Deligne-Mumford Compactification QUESTION [6 upvotes]: Suppose $g\ge 0$ and $n\ge 0$ are integers. We have the space $\overline{\mathcal M}_{g,n}$ of stable curves of arithmetic genus $g$ with $n$ marked points. The topology on this space is the one described in the paper "Compactness Results in SFT" (by Bourgeois et al) starting from page 811. The notes "Holomorphic Curves in Symplectic and Contact Geometry" by Wendl, specifically Theorem 4.26, describe the orbifold structure of $\mathcal M_{g,n}$ (which is the space of non-singular curves of genus $g$ with $n$ marked points). This is done by means of a Banach manifold setup and the proof is an application of the implicit/inverse function theorem. I have seen the following description (which I write here in the simplest case) of a neighborhood of a nodal curve in $\overline{\mathcal M}_{g,n}$ in https://arxiv.org/pdf/dg-ga/9608005.pdf. Assume that $(C_i,x_i)$ are stable curves of genus $g_i$ for $i=1,2$ with only one marked point. Then, the curve $C$ formed by joining $C_1$ and $C_2$ at $x_1$ and $x_2$ is stable. Choose holomorphic local isomorphisms $\varphi_i:U_i\to \mathbb D$ for $i=1,2$ such that $\varphi_i(x_i) = 0$ where $\mathbb D = \{z\in\mathbb C\;|\;|z|<1\}$. Now, fix (non-empty) open subsets $K_i\subset C_i\setminus\overline U_i$, and choose charts for $\mathcal M_{g_i,1}$ around $(C_i,x_i)$ given by a family of almost complex structures $(J_i(z))_{z\in N_i}$ on the underlying smooth manifold $C_i$ parametrized by an open neighborhood $N_i$ of the origin in $\mathbb C^{(3g_i-3)+1}$ with the property that (i) at $0$, the almost complex structure is the original one, say $j_i$, and (ii) even at other points $z\in N_i$ the complex structure $J_i(z)$ coincides with the $j_i$ on the open set $C_i\setminus\overline K_i$. That this can be done is easy to see from the implicit function theorem argument (we have to use the principle of analytic continuation to show that "deformation can be localized"). Now, define the map $\psi:N_1\times N_2\times\mathbb D\to\overline{\mathcal M}_{g_1+g_2}$ as follows. $\psi(z_1,z_2,t) = \Sigma_{z_1,z_2,t}$, where $\Sigma_{z_1,z_2,t}$ is defined as follows. First take $C_i$ with the almost complex structures $J_i(z_i)$, and remove the closed set $\varphi_i^{-1}(\{|z|\le t\})$ from $C_i$. Now, we get $\Sigma_{z_1,z_2,t}$ be identifying the remaining annular parts of the $U_i$ via the holomorphic equivalence relation $p_1\in U_1\sim p_2\in U_2$ iff $\varphi_1(p_1)\varphi_2(p_2) = t$. When $t=0$, we interpret this as simply joining the two curves at the marked points. The claim is that $\psi$ (when restricted to a suitable neighborhood of $(0,0,0)$) maps in a finite to one manner (related to the action of the automorphism group of $C$) onto an open neighborhood of the nodal curve $C$. The proof of this result in https://arxiv.org/pdf/dg-ga/9608005.pdf uses a lot of algebraic formalism from deformation theory which I'm not familiar with. I am interested in seeing a proof of this using some analytical methods (maybe have a suitable Fredholm analysis setup and combine with some gluing theorem). I need to understand the analytic picture as I'm interested in learning about the gluing theorem for pseudoholomorphic curves in symplectic manifolds. I'd appreciate it very much if someone could outline an argument along these lines. I'm also ok with a reference or a sketch with the main ideas and in that case I'll try and work out the details myself. REPLY [6 votes]: For a careful discussion of Deligne-Mumford compactification of the moduli space of curves from a complex-analytic point of view, see the following paper (which also has a good bibliography of earlier work from this perspective): J. Hubbard, S. Koch, An analytic construction of the Deligne-Mumford compactification of the moduli space of curves. J. Differential Geom. 98 (2014), no. 2, 261–313.<|endoftext|> TITLE: Equilaterally triangulated surfaces with prescribed boundary QUESTION [16 upvotes]: There is a problem in Richard Kenyon's list which I would like to post here, because although I have thought about it from time to time, I have not been able to make the slightest progress on it: Question: "Given a closed polygonal path $p$ in $R^3$ composed of unit segments, is there an immersed polygonal surface whose faces are equilateral triangles of edge length $1$, spanning $p$?" The condition that the configuration is an "immersed" surface means that the whole configuration may be parametrized by a continuous locally one-to-one mapping from a compact connected 2-dimensional manifold with boundary into $R^3$. In other words at most two triangles may share an edge, and triangles sharing a vertex form an embedded topological disk, but two triangles which do not share an edge or a vertex may intersect. I cannot answer the question even for the case where $p$ is a quadrilateral. Has there been any partial results for this problem? REPLY [7 votes]: We resolve Kenyon's problem in this paper. We discuss in Section 5 a number of further conjectures and open problems.<|endoftext|> TITLE: Optimization problem with determinant as objective QUESTION [8 upvotes]: Let $A$ be a given symmetric positive definite $N\times N$ matrix. I need to find a symmetric positive semi-definite matrix $S$ which is the solution to the following optimization problem \begin{align} \max_{S}~&\det(A+S) \\s.t.~&\sum_{i}^{N}\sigma_i(S)\,=\,c \\&S\geq0 \end{align}where $\sigma_i(S)$ are the singular values of $S$ and $c$ is a given positive constant. Note that I already know that this can be converted into a standard convex problem by replacing the constraint on sum of singular values by trace and also considering log-determinant. However, I am interested in a different approach. My hunch is that following is the solution. Let $A=U_a\Sigma_aU_a^H$ be its Eigen-decomposition (EVD). Since both $A$ and $S$ are symmetric positive semi-definite, their EVD and SVD are same. Let $S= U_s\Sigma_sU_s^H$ be its EVD which we are seeking. Then first part of the solution is $$U_s = U_a$$ It only remains find $\Sigma_s$ which contains the singular values of $S$. Substituting this in the original optimization problem, it breaks down into the simple form of finding the variables $\sigma_i(S)$ from the optimization problem \begin{align} \max_{\sigma_1(S),\dots,\sigma_N(S)}~&\prod_{i=1}^{N}(\sigma_i(A)+\sigma_i(S)) \\s.t.&\sum_{i=1}^N\sigma_i(S)=c \end{align} This is also equivalent to maximizing the $\log$ of the objective since the objective is a increasing function of each variable. Let us substitute $$x_i=\frac{\sigma_i(S)}{\sigma_i(A)}$$ Also, we can use the fact that determinant is the product of singular values for a symmetric positive definite matrix. Combining all this, we have the optimization problem \begin{align}\max_{x_i}\sum_{i=1}^{N}~&\log(1+x_i)+\log\det(A)\\s.t.~&\sum_{i=1}^{N}\sigma_i(A)x_i\,=\,c\end{align} The constant term in the objective can be neglected. This is an instance of a very famous problem in wireless communications where one tries to maximize the capacity of set of $N$ wireless channels subject to power constraints in each. Its solution is very well known and is typically referred to as the water filling solution. So I know the process after reaching this point. Note that all this was facilitated by the assumption that $U_s = U_a$. Is that true? How do I prove this?. If not, what are other approaches I can try? REPLY [6 votes]: One can verify that $U_{A}=U_{S}$ as follows. Note that $$\det(U_{A}\Sigma_{A}V_{A}^{T}+U_{S}\Sigma_{S}V_{S}^{T})=\det(\Sigma_{A}+U_{A}^{-1}U_{S}\Sigma_{S}V_{S}^{T}V_{A}^{-T})$$ Let $Y=U_{A}^{-1}U_{S}\Sigma_{S}V_{S}^{T}V_{A}^{-T}$ so that the problem can be rephrased as: $$\max\limits_{Y} \det(\Sigma_{A}+Y) \text{ subject to } Tr(Y)=c,\ Y\succ 0$$ Now, let $Z=\Sigma_{A}+Y$ so that the problem is reformulated: $$\max\limits_{Z} \det(Z) \text{ subject to } Tr(Z)=c+\sum\limits_{i=1}^{n} \sigma_{A}(i),\ Z\succ 0,\ Z_{i,i}\geq \sigma_{A}(i)$$ Since $Z$ can be written $Z=UT$ where $U$ is orthogonal and $T$ upper triangular, $\det(Z)=\det(T)$ and the off-diagonal terms of $T$ do not contribute so that $T$ may be assumed diagonal. Thus, finally one arrives at: $$\max\limits_{T} \det(T) \text{ subject to } Tr(T)=c+\sum\limits_{i=1}^{n} \sigma_{A}(i),\ T\succ 0,\ T_{i,i}\geq \sigma_{A}(i),\ \text{T is diagonal}$$ This problem's solution follows from solving the water filling problem above where $t_{k}=T_{k,k}$: $$\max\limits_{t_{k}} \sum\limits_{k=1}^{n} \log(t_{k}+\sigma_{k}) \text{ subject to } t_{k}\geq 0 \text{ and }\sum\limits_{k=1}^{n} t_{k}=c$$<|endoftext|> TITLE: Is there a book on differential geometry that doesn't mention the notion of charts? QUESTION [8 upvotes]: What are some books/texts that use chart free coordinate free language for things otherwise written in a coordinate based formulation? I would like to learn about covariant differentiation, curvature, bundles, characteristic forms etc., but without any charts or local coordinates. The most important areas would be those with applications in physics like gauge theory or Hamiltonian mechanics, where I so often hear all mathematical literature uses coordinate free language, but I never seem to find any such text. They say that they are doing calculations using intrinsic methods; it makes you wonder where all the tedious coordinate manipulations went. Milnor's monograph "Morse Theory for example is a horrible book written in a really bad prosaic style , baez's gauge fields knots and gravity and Mallios's modern differential geometry in gauge theories are the kind of material im interested in. Baez is awesome up untill the point he decides something is too abstract and breaks it down in a chosen basis.. REPLY [15 votes]: For Riemannian geometry (and therefore no gauge theory or Hamiltonian mechanics), I recall two beautiful coordinate-free expositions: 1) Milnor's monograph "Morse Theory" has all of the essentials presented elegantly in one very short chapter. 2) Cheeger and Ebin's book, "Comparison Theorems in Riemannian Geometry" also does everything without coordinates. For $U(1)$ gauge theory (i.e., Maxwell's equations), see Maxwell's equations and differential forms I would add that it was for me very difficult to learn well coordinate-free differential geometry without also grinding through a lot of messy calculations in coordinates. Learning coordinate-free differential geometry is like learning linear algebra using abstract vector spaces. As beautiful as that is, it is hard to appreciate without first learning linear algebra on $\mathbb{R}^n$.<|endoftext|> TITLE: How composite $a^n+b$ is? QUESTION [8 upvotes]: In connection with this question and its follow-up. Suppose that $a\ge 2$ and $b\ne 0$ are integers, and $f$ is a monotonically increasing function such that $f(2)>1$, $f(p)\to\infty$, and the series $\sum_p 1/f(p)$ (extended onto all rational primes) diverges. Suppose further that for any integer $K>0$, there exist infinitely many integers $n\ge 1$ with $\gcd(a^n+b,K)=1$. Is it true that, under the stated assumptions, every point of the interval $[0,1]$ is a limit point of the sequence with the $n$th term $$ \prod_{p\mid a^n+b}\Big(1-\frac1{f(p)}\Big)? $$ In particular, is it true that every point of $[0,1]$ is a limit point of the sequence $\varphi(2^n-1)/(2^n-1)$? REPLY [5 votes]: This is an answer to the part of the question where the OP is asking if the image of the function $$R_{2,\varphi}: \mathbf N^+ \to \mathbf R: n \mapsto \frac{\varphi(2^n - 1)}{2^n-1}$$ is dense in the interval $[0,1]$. The short answer is yes, some more details follow. This week, I met Carlo Sanna in Turin during a meeting of number theory and mentioned the problem to him. He remembered to have read about the same question in a paper of Florian Luca. Today, I got an e-mail from Carlo with the full reference: F. Luca, On the sum of divisors of the Mersenne numbers, Math. Slovaca 53, No. 5 (2003), 457-466 (EuDML link). More precisely, see point ii) of the theorem on the bottom of p. 458. By the way, Luca points out on p. 459 that: his method works for any multiplicative function $f: \mathbf N^+ \to \mathbf R$ for which "there exist [...] $c > 0$ and $\lambda > 1$ so that $f(p^a) = 1 + \frac{c}{p} + O(p^{-\lambda})$ holds for all prime numbers $p$ and all positive integers $a$", on condition that $[0,1]$ is replaced with the interval $[\liminf_n f(n), \limsup_n f(n)]$ (the question in the OP corresponds to the case when $f = R_{2,\varphi}$); the conclusion remains true if the sequence of Mersenne numbers is replaced with a Lucas sequence subjected to some technical conditions. Among other things, Luca's proof makes use of the Siegel-Walfisz theorem.<|endoftext|> TITLE: If $f:[0,1]\to\mathbb{R}$ has continuous approximate deriv., is it $C^1[0,1]$? QUESTION [5 upvotes]: For $f:[0,1]\to\mathbb{R}$, let $f'_{app}(x)$ denote the approximate derivative (that is, the derivative calculated along some set with density $1$ at $x$, if such a thing exists). Assume that $f'_{app}$ exists and is continuous on $[0,1]$. Does this imply that $f'$ exists on $[0,1]$ (of course immediately $f'$ would also be continuous, since $f'_{app}=f'$ when $f'$ exists). Put succinctly: Does $C^1_{app}[0,1]=C^1[0,1]$? REPLY [6 votes]: Yes. Let $g$ be an antiderivative of $f'_{app}$, then $(f-g)'_{app}=0$ , this implies that $h:=f-g=\text{const}$. Sketch of the proof of this fact. It suffices to prove that $h(x)=h(0)$ a.e. Indeed, if $h(a)\ne 0$, than this would imply $h'_{app}(a)=\infty$. We may fix $c>0$ and $\rho\in (0,1)$ and prove that the measure of $x\in [0,1]: h(x)-h(0) TITLE: Attribution of theorem saying that inducing isomorphism on homology implies homotopy equivalence between H spaces that are CW complexes QUESTION [5 upvotes]: Who was the first to prove this theorem and is there an "official" name for it? Let $\phi:X\rightarrow Y$ be a map of H-spaces that are also CW-complexes. Assume $\phi$ induces isomorphisms on homology groups $H_{*}(-,\mathbb{Z})$. Then $\phi$ is a homotopy equivalence. I learned about this theorem from my undergraduate topology professor who called it "Whitehead's theorem for H-spaces". The only source I have for it is V. Srinivas, Algebraic K-theory, Thm. A.53. It does not appear in any of J.H.C. Whitehead's writings, and I guess my professor called it so because of the analogy with Whitehead's theorem on weak homotopy equivalences. Does anybody know where the first proof appeared and if that name is official? REPLY [6 votes]: The result for simply-connected spaces is usually attributed to J.H.C. Whitehead, in particular Theorem 14 in J.H.C Whitehead. Combinatorial homotopy II. Bull. Amer. Math. Soc. 55, (1949). 453–496. The result was generalized to nilpotent spaces in E. Dror. A generalization of the Whitehead theorem. In: Lecture Notes in Mathematics 249 (1971), pp. 13-22. To quote from the corresponding section on simple spaces: If both $X$ and $Y$ are simple, then [the conditions of the Theorem] are trivially satisfied and thus $\pi_\ast f$ is an isomorphism if $H_\ast f$ is. This example includes any map between $H$-spaces. Note that in this case obstruction theory arguments or a careful relative Hurewicz argument would do. Note that e.g. in G.W. Whitehead's book "Elements of homotopy theory", the Whitehead theorem above (for simply connected spaces) is proved as application of the relative Hurewicz theorem. So maybe there is an earlier reference, and maybe people knew before Dror's paper. But it's the earliest explicit statement I could find.<|endoftext|> TITLE: Prefix sums of Pascal triangle = powers of two QUESTION [6 upvotes]: The circle division problem asks for the number of (bounded) regions obtained after choosing $n$ points in general position on a circle and then cutting along all segments connecting the points (cut along the circle too). The first few answers are $1, 2, 4, 8, 16, \mathbf{31}, \ldots$, and the general formula is $f(n) = {n \choose 4} + {n \choose 2} + 1$. The reason why the sequence starts with binary powers is that we can expand the answer as $\sum_{k = 0}^4 {n - 1 \choose k}$. We also have $f(10) = 256$ since the sum above is a half of a Pascal triangle row. Denote $S_{n, m} = \sum_{k = 0}^m {n \choose k}$ the prefix sum of $n$-th row of the Pascal triangle. Obviously, $S_{n, 0}$, $S_{n, n}$, $S_{2n + 1, n}$, and $S_{2^t - 1, 1}$ are binary powers. Are there other non-trivial families of $S_{n, m}$ that are always binary powers? Can we characterize all such entries? REPLY [6 votes]: I think that is it. But the question turns out to be open. Your first example is the case $m=4$ of $S_{n,m}=2^m$ for $n \leq m$ while $S_{m+1,m}=2^{m+1}-1$ and also $S_{2m+1,m}=2^m.$ These could be considered to correspond to “perfect” codes with $1$ and $2$ code words. The celebrated $(23,12,7)-$Golay code would be impossible without the fact that $S_{23,3}=2^{11}.$ The Golay code and the $(2^{n-1},2^n-n-1,3)$-Hamming codes (thanks in part to $S_{2^n-1,1}$) are the only perfect binary codes. I (incorrectly) thought that perhaps this follows from a proof that there are no other non-trivial cases of $S_{n,m}$ a power of $2.$ For a perfect $k$-ary code it would be necessary to have a case of $\sum_0^m\binom{n}{i}(k-1)^i=k^j.$ You were asking about $k=2.$ There is a perfect $(11,6,5)$ $3$-ary code which would not be possible were it not the case that $\binom{11}0+2\binom{11}1+4\binom{11}2=3^5.$ There are no other perfect linear $k$-ary codes. I'm not sure if the proof stems from having no other coincidences as above, at least for $k$ a prime power.<|endoftext|> TITLE: Base change for Borel-Moore homology QUESTION [5 upvotes]: For a seperated scheme of finite type $X$ over $\mathbf{C}$, let $H_*(X)$ denote its Borel-Moore homology, which is defined by $$ H_k(X) = R^{-k}\Gamma(X, \omega_X) $$ where $\omega\in D_c(X, \mathbf{C})$ is a dualising object in the derived category of constructible $\mathbf{C}$-sheaves on $X$. It is immediate from the definition and the six operations for constructible sheaves that $H_*$ is covariant with respect to proper morphisms and contravariant with respect to smooth morphisms. Let $f_*$ and $f^*$ denote respectively the direct and inverse image maps. Question If we are given a cartesian diagramme of schemes over $\mathbf{C}$ $\require{AMScd}$ \begin{CD} Y' @>\tilde g>> Y \\ @VVf'V @VVfV \\ X' @>g>> X \end{CD} where $f$ is proper and $g$ is smooth, do we have an equality of the maps between homology groups $f'_*\tilde g^* = g^*f_*: H_*(Y)\to H_*(X')$? In the book [N. Chriss & V. Ginzburg, Representation theory and complex geometry, 8.3.34] , this is shown in the case where $g$ is locally a trivial fibration, by reducing it to the commutativity of the following diagramme of derived functors \begin{CD} f_*f^! @>>> \mathrm{id}_X @>>> g_*g^* \\ @VVV @. @AAA \\ f_*\tilde g_*\tilde g^*f^! @. = @. g_* f'_*{f'}^! g^* \end{CD} in which all the morphisms are adjunction morphisms except for the base change morphism $\tilde g^*f^! \cong {f'}^! g^*$ in the bottom line, whose definition can be found in [SGA4, Exposé 18, 3.1.14.2]. One gets the result by apply the diagramme to $\omega_X$ and then take global sections. The commutativity of the diagramme above can be easily checked in the case where $g: X'\to X$ is locally a trivial fibration. While I don't know how to do it if $g$ is only assumed to be smooth, I still suspect the commutativity to remain true. Let me also add that there is such an equality $f'_*\tilde g^* = g^*f_*: A_*(Y)\to A_*(X')$ for Chow groups and for $f$ proper and $g$ flat, see for example [Fulton, Intersection theory, Prop. 1.7]. REPLY [3 votes]: $\require{AMScd}$ After the struggle of a whole day due to my unfamiliarity with the category theory, I found the answer. The commutativity is due essentially to the following facts: Given any morphism of schemes $f: X \to Y$ we have the adjunction $(f^*, f_*, \epsilon_f, \eta_f)$: $$ f^*: D_c(Y, \mathbf{C})\to D_c(X, \mathbf{C}),\; f_*: D_c(X, \mathbf{C})\to D_c(Y, \mathbf{C}) $$ $$ \epsilon_f : \mathrm{id} \to f_*f^*, \; \eta_f : f^*f_* \to \mathrm{id} $$ Then, given any two morphisms $X\xrightarrow{f}Y\xrightarrow{g} Z$, we then have $(g_*\epsilon_fg^*)\epsilon_g = \epsilon_{gf}$, as indicated below \begin{CD} \mathrm{id}@>\epsilon_{gf}>> (gf)_*(gf)^* \\ @V\epsilon_{g}VV @| \\ g_*g^*@>g_*\epsilon_fg^*>> g_*f_*f^*g^* \end{CD} Given any morphism of schemes $f: X \to Y$ we have the adjunction $(f_!, f^!, \sigma_f, \tau_f)$: $$ f_!: D_c(X, \mathbf{C})\to D_c(Y, \mathbf{C}),\; f^!: D_c(Y, \mathbf{C})\to D_c(X, \mathbf{C}) $$ $$ \sigma_f : \mathrm{id} \to f^!f_!, \; \tau_f : f_!f^! \to \mathrm{id} $$ Then, given any two morphisms $X\xrightarrow{f}Y\xrightarrow{g} Z$, we then have $\tau_g(g_!\tau_fg^!) = \tau_{gf}$, as indicated below \begin{CD} g_!f_!f^!g^!@>g_!\tau_{f}g^!>> f_!f^! \\ @| @V\tau_{f}VV \\ (gf)_!(gf)^!@>\tau_{gf}>> \mathrm{id} \end{CD} The first one is obvious, while the second one is a sheaf-theoretic version of the Fubini theorem. Now we turn to the diagramme in question \begin{CD} Y' @>\tilde g>> Y \\ @Vf'VV @VfVV\\ X' @>g>> X \end{CD} where $f$ is proper and $g$ is smooth and equidimensional. Suppose we are given two objects $A, B\in D_c(X, \mathbf{C})$ and a morphism $\varphi\in \mathrm{Hom}(f^*A, f^!B)\cong \mathrm{Hom}(A, f_*f^!B)$. Form the following diagramme \begin{CD} g^* A @>g^*\epsilon_f>> g^*f_*f^* A @>g^*f_*\varphi>> g^*f_*f^! B @>g^*\tau_f>> g^* B \\ @V\epsilon_{f'}g^*VV @V\mathrm{BC}VV @V\mathrm{BC}VV @A\tau_{f'}g^*AA \\ f'_*{f'}^*g^*A @= f'_*\tilde g^*f^*A @>f'_*\tilde g^*\varphi>> f'_*\tilde g^*f^!B @= f'_*{f'}^!g^*B \end{CD} This diagramme is commutative: The square in the middle is commutative, by applying the base change isomorphism (BC) $g^*f_*\cong f'_*\tilde g^*$ to $\varphi$. The square in the left is equivalent to the following adjoint \begin{CD} A @>\epsilon_f>> f_*f^* A \\ @V(g_*\epsilon_{f'}g^*)\epsilon_gVV @Vf_*\epsilon_{\tilde g}f^*VV \\ g_*f'_*{f'}^*g^*A @= f_*\tilde g_*\tilde g^*f^*A \end{CD} This is commutative, as we have mentionned in the above fact 1. that $(g_*\epsilon_{f'}g^*)\epsilon_g = \epsilon_{gf'}= \epsilon_{f\tilde g} = (f_*\epsilon_{\tilde g}f^*)\epsilon_f$. Similarly, the square in the right is commutative in regard of the fact 2 and the canonical isomorphisms $g^* \cong g^![2d], \; \tilde g^* \cong \tilde g^![2d]$, where $d = \dim X'/X$. Now, if we set $A = \mathbf{C}$ and $B = \omega_X[-k]$ in $D_c(X, \mathbf{C})$ for $k\in \mathbf{Z}$, then every homology class $\alpha\in H_k(Y)\cong \mathrm{Hom}_{D_c(Y)}(f^*A, f^! B)$ corresponds to a morphism $\varphi_{\alpha}: f^*A \to f^! B$. Then the composite $$(g^*\tau_f)(g^*f_*\varphi_{\alpha})(g^*\epsilon_f): g^*A \to g^*B$$ corresponds exactly to the class $g^*f_*\alpha\in H_{k+2d}(X')$, whereas the composite $$ (\tau_{f'}g^*)(f'_*\tilde g^*\varphi_{\alpha})(\epsilon_{f'}g^*): g^*A \to g^*B $$ corresponds to the class $f'_*\tilde g^*\alpha\in H_{k+2d}(X')$. The commutativity of the above 8-term diagramme applying to $\varphi = \varphi_{\alpha}$ then tells us that $$ (g^*\tau_f)(g^*f_*\varphi_{\alpha})(g^*\epsilon_f) = (\tau_{f'}g^*)(f'_*\tilde g^*\varphi_{\alpha})(\epsilon_{f'}g^*) $$ and consequently $g^*f_*\alpha = f'_*\tilde g^*\alpha$. Thus $g^*f_* = f'_*\tilde g^*:H_*(Y)\to H_*(X')$.<|endoftext|> TITLE: Teaching Steenrod Operations QUESTION [16 upvotes]: I am teaching a class on topology and want to introduce Steenrod Operations. I have talked about simplicial sets and classifying spaces of groups but have not talked about Eilenberg–MacLane spaces. What is the easiest/your favorite way of showing Steenrod operations exist? Is there a good way of constructing an explicit chain map $C_*(E S_2) \otimes_{\mathbb Z[S_2]} (C^*(X)\otimes C^*(X)) \to C^*(X)$ for some model of $E S_2$ that gives Steenrod operations? REPLY [11 votes]: Its nice to look also at Bott's early paper "On symmetric products and the Steenrod squares. " Ann. of Math. (2) 57, (1953). 579–590. He uses an early version of Smith theory. Depending on how you do Smith Theory the homotopy quotient construction lurking somewhere. Anyway worth look.<|endoftext|> TITLE: An identity in Lie algebras over fields of positive characteristic QUESTION [10 upvotes]: Let $L$ be a Lie algebra over a field of characteristic $p>0$ and $D$ a derivation of $L$. For every $x\in L$ denote by $\mathrm{ad} x$ the adjoint map $\mathrm{ad}x: L \rightarrow L, a\mapsto [x,a]$. Is the following relation true? $$D^{p-1}((\mathrm{ad} x)^{p-1}(D(x)))=(\mathrm{ad} x)^{p-1}(D^p(x))$$ I proved it for small values of $p$, but I am not able to find a general argument. In the case the result is well-known, a reference will be very welcome. REPLY [5 votes]: Yes, this identity holds, and both sides are equal to $-D^p(x^p)$. We need the following facts, all of which are straightforward to verify, where $D$ is an arbitrary derivation, still working over characteristic $p>0$: $$(ad_x)^k(y)=\sum_{m=0}^{k}(-1)^{m+1} {k\choose m}x^m y x^{k-m}$$ $$D(x^k)=\sum_{m=0}^{k-1}x^m D(x)x^{k-1-m}$$ $$D^p(xy)=D^p(x)y+xD^p(y)$$ $$(-1)^m {p-1\choose m}=1 \text{mod p}$$ Then we have: \begin{align*} D^{p-1}((ad_x)^{p-1}(D(x))&=D^{p-1}\bigg(\sum_{m=0}^{p-1}(-1)^{m+1}{p-1\choose m}x^mD(x)x^{p-1-m}\bigg)\\ &=-D^{p-1}\bigg(\sum_{m=0}^{p-1}x^mD(x)x^{p-1-m}\bigg)\\ &=-D^{p}(x^p)\\ &=-\sum_{m=0}^{p-1}x^mD^p(x)x^{p-1-m}\\ &=\sum_{m=0}^{p-1}(-1)^{m+1}{p-1\choose m}x^mD^p(x)x^{p-1-m}\\ &=(ad_x)^{p-1}(D^p(x)) \end{align*}<|endoftext|> TITLE: Two embedded symplectic spheres with zero square in a symplectic $4$-manifold QUESTION [7 upvotes]: I am aware that the following result is a classical one (by now). But I am not able to understand who proved it. What should be a proper reference to this statement? Theorem. Let $M^4$ be a compact symplectic manifold with $\pi_1\ne 0$ and let $S_1$ and $S_2$ be two symplectic spheres embedded in it with $S_1^2=S_2^2=0$. Then $S_1$ and $S_2$ are symplectically isotopic in $M^4$. In other words, is there an article/book (say pre 2000) claiming that any two symplectic spheres with zero self-intersection in an irrational ruled (non-minimal) surface are symplectically isotopic? REPLY [2 votes]: It looks indeed that this question is not as classical is it sounds, so let me provide a 2010 reference to a more general statement, at least to show that there is a reference. This is Proposition 3.2 in the following paper: https://arxiv.org/abs/1012.4146<|endoftext|> TITLE: alternating and symmetric powers of the standard representation of the symmetric group QUESTION [9 upvotes]: Let $n \geq 7$ and $V = \mathbb{C}^n$ be the standard representation for $S_{n+1}$, the symmetric group of cardinal $(n+1)!$ Let $k$ be an integer such that $2 \leq k \leq n$. Is it true or false that $\bigwedge^k V$ does not appear as an irreducible sub-representation of $\mathrm{Sym}^k V$? I am looking for a reference which would either prove or give counter-examples to this. Thanks in advance! EDIT : Is there an online software (analogous to LIE) that would compute the decompositions into irreducible representations of the tensor product of two ireps for the symmetric group? REPLY [6 votes]: Here is another attempt. It follows from Enumerative Combinatorics, vol. 2, Exercise 7.73, that the Frobenius characteristic of the representation $\mathrm{Sym}^k V$ is the coefficient of $q^k$ in the power series $$ (1-q)\sum_{\lambda\vdash n+1}s_\lambda(1,q,q^2,\dots)s_\lambda. $$ (The factor $1-q$ appears because Exercise 7.73 deals with the symmetric algebra of the defining representation of degree $n$. The defining representation is the sum of $V$ and the trivial representation.) By Corollary 7.21.3, $$ s_\lambda(1,q,q^2,\dots) = q^{b(\lambda)} +\ \mbox{higher order terms}, $$ where $b(\lambda) = \sum (i-1)\lambda_i$. Hence the least $k$ for which an irrep indexed by $\lambda$ appears in $\mathrm{Sym}^k V$ is $b(\lambda)$. For $\bigwedge^k V$ we have $b(\lambda) =b(n-k+1,1^k) = {k+1\choose 2}>k$ for $k>1$.<|endoftext|> TITLE: Can we approximate any open set by sub-domains with smooth boundary? QUESTION [10 upvotes]: In some books, mainly about PDEs, I read that any open set can be approximated by sub-domain with smooth boundary (not just piecewise smooth). In 2 dimensional case, this seemly to be quite trivial: for any subdomain, use small open balls to cover its boundary and then mollify the connection parts. But in the higher dimensional case, I think this is not that obvious. So the first question is: how can we approximate any open set by sub-domain with smooth boundary? And the second question is: In what meaning the approximation is? Pointwise, i.e., we can find subdomain $D_n$ with smooth boundary such that $D_n\uparrow A$? uniformly pointwise? Or in the Lebesgue measure sense? etc. Here "$D_n\uparrow A$" uniformly pointwise" means means that $\partial D_n\subset A\cap A_n^c$, where $A_n:=\{ x \in A:d(x,\partial A)\geq \frac1n \}$,. REPLY [8 votes]: By a well-known theorem of Whitney, any closed subset of $R^n$ coincides with the zero set of a $C^\infty$ function: Whitney, Hassler: Analytic extensions of differentiable functions defined in closed sets, Trans. AMS 36 (1934), 63--89. Let $f$ be such a function for the boundary of the domain. By Sard's theorem $f$ has a regular value $x$ arbitrarily close to $0$. Then a component of $f^{-1}(x)$ yields the desired approximation. To get a subdomain, make sure that $x$ coincides with a value of $f$ inside the domain. There is an even easier way of doing this when the domain is bounded: cover the domain by balls of radius $\epsilon$, or take the union of a finite collection of balls of radius $\epsilon$ contained in the domain and covering a compact subset. For each ball let $f\colon R^n\to R$ be the function whose zero set coincides with the boundary of the ball (i.e., the distance function from the center of the ball minus the radius). Now multiply all these functions to obtain a function $F$, and take a level set of $F$ close to $0$. The second method yields a subdomain which is not only smooth but is algebraic. In both cases here, the approximation is with regard to the Hausdorff distance.<|endoftext|> TITLE: Waring rank vs tensor rank of symmetric tensors? QUESTION [6 upvotes]: Suppose we work in an algebraically closed field. Then, do the Waring rank (symmetric tensor rank) and tensor rank of a symmetric tensor coincide in general? Recall that tensor rank is rank with respect to the Segre variety and Waring rank is rank with respect to the Veronese variety. REPLY [10 votes]: A counterexample in $\mathbb{C}$ is given in A counterexample to Comon's conjecture: We present an example of a symmetric tensor of size 800×800×800 which can be written a sum of 903 simple tensors with complex entries but not as a sum of 903 symmetric simple tensors. It's a very recent result by Yaroslav Shitov, not yet published but it has survived some scrutiny by the tensor community.<|endoftext|> TITLE: Constant Gaussian curvature surfaces in 3-space containing lines QUESTION [10 upvotes]: How can one construct surfaces in $\mathbb R^3$ of constant negative Gaussian curvature containing a line in $\mathbb R^3$? (this question is inspired by this MSE post). REPLY [17 votes]: Given any point $p$ on a surface $S$ of Gauss curvature -1, there exists an open neighborhood $U\subset S$ and $p$-centered coordinates $(x,y):U\to\mathbb{R}^2$, whose image is a domain $R = (x,y)(U)\subset\mathbb{R}^2$ and a function $u:R\to (0,\pi/2)$ such that the first and second fundamental forms of the surface are given by $$ \mathrm{I} = \cos^2\!u\,\mathrm{d}x^2 + \sin^2\!u\,\mathrm{d}y^2\tag1 $$ and $$ \mathrm{I\!I} = \cos u\,\sin u\,(\mathrm{d}x^2-\mathrm{d}y^2) = \cos u\,\sin u\,(\mathrm{d}x-\mathrm{d}y)(\mathrm{d}x+\mathrm{d}y).\tag2 $$ The function $u$ satisfies the sine-Gordon equation $$ u_{xx}-u_{yy} = \cos u\,\sin u.\tag3 $$ Conversely, given a function $u:R\to (0,\tfrac12\pi)$ that satisfies (3) on a simply connected domain $R\subset\mathbb{R}^2$, there exists an immersion $X:R\to\mathbb{E}^3$ whose first and second fundamental forms are given by $\mathrm{I}$ and $\mathrm{I\!I}$ above; this immersion is unique up to rigid motion and has Gauss curvature $-1$. The asymptotic curves of the immersion $X$ are given by holding $x+y$ or $x-y$ constant. The angle between the asymptotic curves (appropriately oriented) is $2u$, and the ambient curvatures in $\mathbb{E}^3$ of the asymptotic curves described by holding $x\pm y$ constant are $$ \kappa_{\pm} = 2 (u_x \mp u_y).\tag 4 $$ (Here, $\kappa_{\pm}$, which are the curvatures of the asymptotic curves, are not to be mistaken for the principal curvatures of the surface $X(R)$, which are $\tan u$ and $-\cot u$.) Thus, constructing a surface in $\mathbb{E}^3$ with Gauss curvature $-1$ that contains a straight line (which is necessarily asymptotic) is equivalent to finding a (local) solution $u(x,y)$ of the sine-Gordon equation that satisfies the 'characteristic' initial condition $u_x(x,x)+u_y(x,x) = 0$. (It's called 'characteristic' because the curves $x\pm y = c$ are the characteristic curves of the PDE (3).) Unfortunately, characteristic initial value problems for nonlinear PDE can be delicate to study; standard techniques are generally not adequate to prove existence or uniqueness. However, this problem can be partially avoided by looking for special solutions that will satisfy the desired initial condition by virtue of other special properties being assumed. If one seeks a solution to (3) of the form $u(x,y) = \tfrac12\,f(x^2{-}y^2)$, one finds that $f$ must satisfy the (singular, nonlinear) ordinary differential equation $$ 4t\,f''(t) + 4f'(t) = \sin\bigl(f(t)\bigr).\tag5 $$ Moreover, since, for such a solution, one has $(u_x \mp u_y) = f'(x^2{-}y^2)(x \pm y)$, it will follow that the two lines $x{\pm}y = 0$ in $R$ will be mapped by $X$ into straight lines in $\mathbb{E}^3$. To see that there are nontrivial solutions to the singular ODE (5), let $\theta\in(0,\pi)$ be chosen, and write $f(t) = \theta + \tfrac14\,\sin(\theta)\,t + g(t)$. Then (5) becomes the equation $$ 4t\,g''(t)+4g'(t) = \sin\bigl(\theta+\tfrac14\sin\theta\,t+g(t)\bigr) -\sin\theta = G\bigl(\theta,t,g(t)\bigr) \tag6 $$ The left hand side of (6) is a singular linear differential operator at $t=0$ with no resonances, while $G(\theta,t,g)$ is an analytic function on $\mathbb{R}^3$ satisfying $G(\theta,0,0) = 0$. Thus, there is a unique convergent power series solution to (6) satisfying $g(0)=0$, which implies the convergence of the corresponding formal power series for $f$: $$ \begin{aligned} f(t) &= \theta + \frac{\sin\theta}{4}\,t + \frac{\cos\theta\sin\theta}{64}\,t^2 + \frac{(3\cos^2\theta-2)\sin\theta}{2304}\,t^3\\ &\quad\quad + \frac{\cos\theta\,(18\cos^2\theta-17)\sin\theta}{147456}\,t^4 + \cdots, \end{aligned} \tag7 $$ yielding a solution to (5) satisfying $f(0) = \theta$ on an open interval around $t=0$. It is easy to see from (5) that this local solution to (5) satisfying the initial condition $f(0)=\theta$ extends uniquely to a solution to (5) defined on the entire real line $\mathbb{R}$. The range of $f$ will not be contained in the interval $(0,\pi)$, of course, but, there will be an interval $\bigl(a(\theta),b(\theta)\bigr)\subset\mathbb{R}$ containing $0$ that $f$ maps into $(0,\pi)$. Then the corresponding $X$ will immerse the region between the two hyperbolae $x^2-y^2 = a(\theta)<0$ and $x^2-y^2= b(\theta)>0$ into $\mathbb{E}^3$ as a surface of constant Gauss curvature $-1$ and will map the two lines $x\pm y = 0$ into two straight lines that intersect at an angle of $\theta$ in the image surface. Added remark: (23 October 2017) After doing a literature search on examples of surfaces of constant curvature $-1$ in $\mathbb{E}^3$, I have found out that this surface was discovered by M. H. Amsler, Des surfaces à courbure négative constante dans l'espace à trois dimensions et de leurs singularités, Math. Ann. 130 (1955), 234—256.<|endoftext|> TITLE: Pullback and homology QUESTION [14 upvotes]: Suppose I have two maps of topological spaces, $f:X\rightarrow B$ and $g:Y\rightarrow B$, such that $f$ induces a homology isomorphism and $g$ is a fibration and $B$ is connected. Is it true that the natural map $X\times_{B}Y\rightarrow Y$ induces an isomorphism in homology? REPLY [11 votes]: Here is a positive answer to a slightly different question. Call a map $X\to B$ "acyclic" if it induces an isomorphism in homology for every coefficient system on $B$. (If $B$ is simply connected then this is equivalent to your hypothesis.) Note that a map $F\to \ast$ is acyclic if and only if the space $F$ is acyclic in the sense that it has trivial integral homology. Using Serre spectral sequences one can prove the following: If $X\to B$ is acyclic then for any fibration $Y\to B$ the map $X\times_BY\to Y$ is acyclic. A fibration $X\to B$ is acyclic if and only if for every point $b\in B$ the fiber over $B$ is acyclic.<|endoftext|> TITLE: A random walk on an infinite graph is recurrent iff ...? QUESTION [14 upvotes]: Q. Is there a master theorem that can be used to determine whether or not a simple random walk (choose a random neighboring vertex as the next step) on a given infinite graph leads to recurrence (almost surely returns to the start vertex)? Added. Let's restrict the graphs to be locally finite, in the sense that every vertex has finite degree. For example, the infinite path is recurrent, as it is equivalent to $\mathbb{Z}^1$, but I believe the infinite binary tree is not recurrent, i.e., it is transient. I am seeking structural properties of a graph that determines its recurrence/transience behavior. REPLY [5 votes]: In my opinion, the closest to a "master theorem" is the criterion due to Terry Lyons, according to which a reversible Markov chain on a countable state space (in particular, the simple random walk on a locally finite graph) is transient if and only if there exists a flow of finite energy on the state space. PS Contrary to the opinion appearing in the comments, in this criterion no conditions other than reversibility (e.g., uniform bounds on vertex degrees) are imposed. Actually, there are quite instructive examples of reversible random walks with unbounded weights (for instance, any nearest neighbour random walk on a tree).         Theorem from Terry Lyons paper (added by J.O'Rourke).<|endoftext|> TITLE: What is the optimal speed to approach a red light? QUESTION [21 upvotes]: Suppose from distance $d$, while driving at speed $v_0$, I notice that there's a red traffic light in front of me. Suppose that there are no other vehicles, my vehicle has perfect brakes, my maximum acceleration is $a$ and the red light will turn green according to some $\mu$ distribution. My goal is to get to my final destination as soon as possible, i.e., to minimize my estimated time of arrival (ETA), and suppose that the rest of the road has some speed limit $L$. With what speed should I approach the red light to minimize my ETA? Of course, the answer will depend on the parameters; I'm interested in any non-trivial results. As essentially the same problem has been raised earlier here and here (thanks for the links!), but apparently without any solution, let me be more specific. Is there any non-trivial set of parameters for which we know the solution? ps. I'm not interested in reducing milage, but feel free to add comments in this direction, if you wish. REPLY [7 votes]: Similar to the linked variants on this problem, the optimal strategy takes the form of a function $v(t)$, which corresponds to the strategy in which you travel at velocity $v(t)$ until the light turns green, then you slam on the accelerator and accelerate at $a$ until you reach the speed limit $L$. If $T$ is the time when the light turns green, we will be at $D=\int_0^Tdt\;v(t)$ traveling at $V=v(T)$, and the velocity function for $t\ge T$ is $v'(t)=\min(V+a(t-T),L)$. We want to maximize, for $t\gg T$: $$K:=\int_0^tdt\; v'(t)=D-\frac{(L-V)^2}{2a}-LT+Lt$$ The leading term $Lt$ only depends on the chosen time $t$ to measure our distance, so we can ignore it, and similarly with $LT$ which depends only on $T$ which is not under our control. For the rest, we see a quadratic penalty to slowing below top speed. The constraint of not running the red light is expressed as $P[\int_0^Tdt\;v(t)\le d]=1$, which amounts to just $\int_0^xdt\;v(t)\le d$ for all $x$ such that $\mu([x,\infty))>0$. In fact we can just go ahead and assume $\int_0^xdt\;v(t)\le d$ for all $x$ because the strategy is irrelevant once the light is green. So we are looking at: $$E[K]=E\left[\int_0^Tdt\;v(t)-\frac{(L-v(T))^2}{2a}\right]$$ where $T$ is drawn from the distribution $\mu$, and we wish to choose $v$ maximizing $E[K]$. I don't think much more can be done in this generality, so let me focus on the case $\mu$ uniform on $[0,\alpha]$. In this case the expected value becomes (up to a factor $1/\alpha$ which doesn't affect the result) $$\int_0^\alpha dT\left[\int_0^Tdt\;v(t)-\frac{(L-v(T))^2}{2a}\right]=$$ $$\int_0^\alpha dt\left[(\alpha-t)\;v(t)-\frac1{2a}(L-v(t))^2\right].$$ We can add in a Lagrange multiplier to account for the constraint (which will almost certainly turn out to be extremal) $\int_0^\alpha dt\; v(t)=d$, and solve this using calculus of variations: $$\mathcal{L}(t,v,\lambda)=(\alpha-t)\;v-\frac1{2a}(L-v)^2+\lambda v$$ $$\frac{\partial\mathcal{L}}{\partial v}=0\implies v(t)=L+a(\alpha+\lambda)-at$$ Here $\lambda$ is a free constant which should be chosen to maintain the constraint $\int_0^\alpha dt\; v(t)\le d$, which I'll leave to you since the story is more interesting: This says you should decelerate at $a$ starting from an appropriate speed. (Here the appropriate speed is so that you come to a stop at the light, unless this puts the initial speed above $L$, in which case obviously you should drive at the speed limit until you get close enough.) Let's try another simple and reasonable distribution: the exponential distribution with mean $\beta=\alpha^{-1}$. In this case the expectation works out to: $$\int_0^\infty dT\;e^{-\alpha T}\left[\int_0^Tdt\;v(t)-\frac{(L-v(T))^2}{2a}\right]=$$ $$\int_0^\infty dt\;e^{-\alpha t}\left[\beta v(t)-\frac1{2a}(L-v(t))^2\right].$$ As with the uniform distribution case, we add a Lagrange multiplier and solve the Euler-Lagrange equation: $$\mathcal{L}(t,v,\lambda)=e^{-\alpha t}\left[\beta v-\frac1{2a}(L-v)^2\right]+\lambda v$$ $$\frac{\partial\mathcal{L}}{\partial v}=0\implies v(t)=L+a(\beta+\lambda e^{\alpha t})$$ Unfortunately, this solution is formally problematic, since the integral of the velocity doesn't converge for large $t$. This solution is still correct before we hit the discontinuity, but we need to add the constraint in differently. Let's explicitly consider a velocity curve that is zero after a certain fixed time, i.e. we approach the light and come to a stop. For a fixed chosen time $t^*$ to stop at, the functional is the same, and so the solution is the same: exponential deceleration away from the unreachable "steady state" velocity $L+a\beta$. But when we hit the light, we use our magic brakes to stop, and continue after the light from a full stop. $$E[K]=\int_0^{t^*} dT\;e^{-\alpha T}\left[\int_0^Tdt\;v(t)-\frac{(L-v(T))^2}{2a}\right]+\underset{c(t^*)}{\underbrace{\int_{t^*}^\infty dT\;e^{-\alpha T}\left[d-\frac{L^2}{2a}\right]}}$$ $$=\int_0^{t^*} dt\left[\int_t^{t^*}dT\;e^{-\alpha T}\;v(t)-\frac{e^{-\alpha t}}{2a}(L-v(t))^2\right]+c(t^*).$$ $$=\int_0^{t^*} dt\left[\beta(e^{-\alpha t}-e^{-\alpha t^*})\;v(t)-\frac{e^{-\alpha t}}{2a}(L-v(t))^2\right]+c(t^*).$$ The parts that depend on $t^*$ don't affect the variational analysis: $$\mathcal{L}(t,v,\lambda)=\beta(e^{-\alpha t}-e^{-\alpha t^*})\;v-\frac{e^{-\alpha t}}{2a}(L-v)^2+\lambda v$$ $$\frac{\partial\mathcal{L}}{\partial v}=0\implies v(t)=L+a(\beta(1-e^{\alpha(t-t^*)})+\lambda e^{\alpha t})$$ This time we should really solve for $\lambda$ since we need to ensure that $\int_0^{t^*}dt\;v(t)=d$. I'll spare the details as the algebra gets worse, but after putting $\lambda$ back in the equation, we have a one-dimensional optimization for $E[K]$ as a function of $t^*$, which can't be solved with elementary functions, but from numerical simulations it looks like the best option is still the extremal one: Wait until the last moment at the top speed, then execute the exponential deceleration maneuver $v(t)=L+a\beta(1-e^{\alpha t})$ and stop when $v(t)=0$, at $t=\beta\log(1+\frac L{a\beta})$. PS: I don't recommend these maneuvers. They strike me as mildly suicidal.<|endoftext|> TITLE: What are some interesting examples of non-classical dynamical systems? (Group action other than $\mathbb{Z}$ or $\mathbb{R}$ ) QUESTION [15 upvotes]: By classical dynamical system, I mean a measure space together with a measurable action of the integers or the reals. Of course, this action is often interpreted as evolution with respect to discrete or continuous time, respectively. But there is a large theory out there of dynamics/ergodic theory where the action is taken to be either an arbitrary group, or a group from some large class generalizing $\mathbb{Z}$ and $\mathbb{R}$. For example, Chapter 8 of Ergodic Theory with a View Towards Number Theory by Einsiedler and Ward begins with: The facet of ergodic theory coming from abstract mathematical models of dynamical systems evolving in time involves a single, iterated, measure- preserving transformation (action of $\mathbb{N}$ or of $\mathbb{Z}$) or a flow (action of the reals). For many reasons—including geometry, number theory, and the origins of ergodic theory in statistical mechanics—it is useful to study actions of groups more general than the integers or the reals. The rest of the chapter then develops the ergodic theory of amenable group actions. Likewise, there is the classic paper Ergodic theory of amenable groups actions by Ornstein and Weiss, where the authors say in the introduction that in applications they kept encountering groups other than the integers or lattices of the integers, and so it was worth it to develop the theory in the full generality of the amenable setting. But I haven't actually seen many examples where we care about, say, an ergodic action of a group other than $\mathbb{Z}$ or $\mathbb{R}$! Or $\mathbb{Z}^d$ or $\mathbb{R}^d$, I suppose. (A notable exception would be homogeneous dynamics, where one is concerned with the action of a Lie group on a quotient of itself.) What are some other examples? Either in other areas of pure mathematics, or in applied areas like Einsiedler and Ward allude to? (This is a cross-post from Math Stack Exchange, where this question has been sitting unanswered for a while.) REPLY [4 votes]: The simplest example of a non-trivial group action is that of the action of a group on itself. Now suppose that the group is bordified (particular case: compactified) by attaching a certain boundary to it, i.e., the group is embedded into a bigger space. There are numerous constructions of boundaries, which are natural in the sense that the action of the group on itself extends to the boundary (Stone-Cech compactification, end compactification, Martin compactification, hyperbolic compactification, horospheric compactification, etc., etc.). By endowing the boundary with various quasi-invariant measures one can further study ergodic properties of these actions. A notable example is the Patterson (or Patterson-Sullivan) measure on the hyperbolic boundary (in various degrees of generality, the simplest case being the boundary circle of the hyperbolic plane, which was the original setup of Patterson). However, there is also a construction of a boundary action directly in the measure category (without using any topological bordifications). This is the Poisson boundary. Its definition requires an additional piece of data, namely, a probability measure on the group (akin to specifying a Riemannian structure on a smooth manifold).<|endoftext|> TITLE: The Chu-construction and the Int-construction QUESTION [14 upvotes]: The Chu construction is a way of building a star-autonomous category $\mathrm{Chu}(C,\bot)$ from any closed monoidal category $C$ with pullbacks and a choice of an object $\bot\in C$ to become the dualizing object. If $\bot$ is the terminal object, then the underlying category of the Chu construction is just $C\times C^{\mathrm{op}}$. In their paper Traced monoidal categories, Joyal, Street, and Verity described a construction of a compact closed category $\mathrm{Int}(C)$ from any traced monoidal category $C$. The objects of $\mathrm{Int}(C)$ are pairs $(X,U)$ of objects of $C$, and its morphisms $(X,U) \to (Y,V)$ are morphisms $X\otimes V \to Y\otimes U$ in $C$, with composition defined using the trace. There seems to me to be some kind of family resemblance between these constructions. In both cases we are "formally" or "freely" adding some kind of "duals" to a category $C$ by constructing a category whose objects are pairs of objects of $C$, one treated covariantly and one contravariantly. However, the differences are substantial, and I haven't been able to make this similarity precise in any way. Is there any formal relationship between the Chu construction and the Int construction? REPLY [6 votes]: It seems to me that, if $C$ is a traced monoidal category (with unit $I$) that is also closed, there is a comparison functor $Chu(C,I)\to Int(C)$, which is perhaps what you had in mind. However, this functor does not seem particularly well-behaved: in general, it is neither full nor faithful, and only lax monoidal. (It also isn't essentially surjective on objects, for whatever that's worth.) [EDIT: You're right, it isn't even lax monoidal in general, but it might be oplax monoidal.] Consider the discrete case: if (the underlying category of) a traced monoidal category $C$ is discrete, then it is just a cancellative commutative monoid, and $Int(C)$ is the free abelian group on it; however, in this case, $Chu(C,1)$ (where $1$ denotes the unit of the monoid) is the cofree abelian group on it. In fact, a quarter-century ago, Dusko Pavlovic proved that the Chu construction is always (in a suitable sense) the cofree way of turning a pointed symmetric monoidal closed category into a symmetric $*$-autonomous one. Which stands in stark contrast to $Int$ being the free way of turning a traced monoidal category into a compact closed category. One way this shows up is that there is always a forgetful functor $Chu(C,\bot)\to C$---which is the counit of the adjunction---but no obvious embedding $C \to Chu(C,\bot)$, whereas there is always an embedding $C \to Int(C)$---the unit of the adjunction---but no obvious forgetful map $Int(C)\to C$. In conclusion, I don't think that $Int$ and $Chu$ are really all that similar after all. ===== Since you ask, here are some more details. As you said in your question, objects of $Chu(C,\bot)$ and $Int(C)$ comprise two $C$-objects: one treated covariantly (the "positive space"), the other, contravariantly (the "negative space"); moreover, the duality in each case is given by swapping the two. So it seemed to me that if you want to highlight these similarities, one should build a comparison functor that preserves these very features. It turns out---again, you were right about this---that there's no real reason to restrict to the case $\bot=I$; I did so only because I was hoping to build a lax monoidal functor, which would require at least a map $\bot\to I$, and was too lazy to check if the usual coherence conditions on such a map were enough to make this work. But since I've lost hope of making the comparison functor lax monoidal, there's no point for this restriction at all. (To make the functor oplax monoidal would require a suitably coherent map $I \to \bot$ instead.) Anyways, the naïve thing to way to build a functor $Chu(C,\bot)\to Int(C)$ would be as a kind of forgetful functor, where you forget the "inner product" of a Chu space, and remember only the positive space and the negative space. Now a map of Chu spaces comprises two $C$-arrows: one between the positive spaces, and one between the negative spaces; tensoring these two maps together produces a $C$-arrow of the right type to define a map in $Int(C)$. So one can ask if this defines a functor. Amazingly, it does in the symmetric case; moreover, one simply needs to add a $-2\pi$ twist to the negative component to make it work in the balanced case. I'll append a photograph of my calculation, but first, let's settle notation. A Chu space $A$ comprises a positive space $a_+$, a negative space $a_-$, and an inner product $\alpha:a_-\otimes a_+\to\bot$; let $F(A)=(a_+,a_-)$ denote the corresponding object of $Int(C)$. A morphism of Chu spaces $\omega: A=(a_+,a_-,\alpha)\to B=(b_+,b_-,\beta)$ comprises a map $\omega_+:a_+\to b_+$ and a map $\omega_-:b_-\to a_-$ satisfying a diagram which is irrelevant to the purpose at hand; $\omega_+\otimes(\omega_-\circ\xi_{b_-}^{-1}):a_+\otimes b_- \to b_+\otimes a_-$ can be construed as a map $F(\omega):F(A)\to F(B)$ in $Int(C)$, where $\xi$ denotes the balance. I claim that $F$ preserves identities and composition. Identities are easy: the identity Chu morphism is given by a pair of identities, and the identity map $(a_+,a_-)\to(a_+,a_-)$ in $Int(C)$ is given (not by the identity on $a_+\otimes a_-$, but) $\mathrm{id}_{a_+}\otimes\xi_{a_-}^{-1}$ ["Traced monoidal categories", section 4]. As for composition, see attached:<|endoftext|> TITLE: Some Mathematical Questions on Gravitational Waves and Numerical Relativity QUESTION [8 upvotes]: Due to the recent spate of detections of gravitational waves by LIGO, my amateurish interest in the mathematics of general relativity has been revived. The wave-forms of the detected gravitational waves are said to have been compared with predicted wave-forms produced by a combination of post-Newtonian approximations and revamped methods in numerical relativity, such as the puncture method and the excision technique. Here then are my questions. Question 1. In a discussion on the evolution of binary black holes, I suppose that one assumes the ADM formalism, in which space-time is divided into time slices. This reminds me of a famous result by Robert Geroch that a space-time is globally hyperbolic if and only if it admits a foliation by Cauchy hyper-surfaces. Hence, when performing computer simulations of binary black holes, does one assume that space-time is globally hyperbolic? Question 2. I’ve read that the existence of gravitational waves involves analyzing what happens at ‘null infinity’, the definition of which presumably depends on some kind of asymptotic structure on space-time. Hence, when performing computer simulations of gravitational waves emanating from a black-hole merger, what asymptotic structure does one assume on space-time? Besides, I’ve heard that a definition of black holes has been satisfactorily given only for asymptotically flat space-times. Question 3 (Not very mathematically precise). Can the movement of space-time singularities be shown to be consistent with the notion/axiom of general relativity that the world-lines of point particles are geodesics (after perhaps having performed some de-singularization process)? Question 4. Does anyone here know of papers written to establish rigorous error bounds for the approximation methods used in numerical relativity? Question 5. One of the reasons mentioned for justifying the use of the excision technique is that signals originating in a black hole cannot propagate out of it. Now, I understand that light signals cannot propagate out of a black hole because the world-lines of photons are null geodesics, and a black hole is defined as a portion of space-time not contained in the causal past of null infinity. However, in the context of full non-linear general relativity, a gravitational wave is not something that propagates with respect to a background metric but is intimately tied to the metric itself. The Cauchy problem in general relativity says that, due to the nature of hyperbolic PDE’s, perturbing metrical data on a portion of a Cauchy hyper-surface (subject to particular constraint conditions) affects the metric only within the future light-cone of that portion. However, if global hyperbolicity is not assumed (supposing that the answer to Question 1 is negative), then one may not have a Cauchy hyper-surface. Is there, then, an analog of finite propagation speed in any black hole? Thank you for your help! REPLY [2 votes]: Question 3: You should not think of the singularity (corresponding to a black hole) as moving in space-time. It is not. So you are asking the wrong question if your motivation is gravitational waves. The answer to the question you did ask however is "yes", see the work of Einstein-Infeld-Hoffman. Question 5: First, finite propagation speed always holds, by virtue of Einstein's equations being essentially hyperbolic. (This is a purely local property, whereas global hyperbolicity, as its name suggests, is a global property.) Second, you are correct that the definition of a black hole is teleological: you only know what a black hole is if you know what the null infinity looks like. However, for numerical computations a much more acceptable, local substitute is used. Instead of the event horizon (which is defined as the boundary of the past of future null infinity), it is much more common to use the apparent horizon as a proxy for the boundary of the black hole. The apparent horizon will always sit within the black hole, and captures a local notion of "no escape". And in particular global geometry does not come into play in the excision process. (For more about apparent horizons, Wikipedia has a fairly readable lay discussion.)<|endoftext|> TITLE: A different ordering on ${\cal P}(\omega)$ QUESTION [5 upvotes]: For $A, B \subseteq \omega$ we set $A \leq_{\text{inj}} B$ if there is an injective and order-preserving map $f:\omega\to \omega$ , such that $f(A)$ is a down-set of $B$. It is easy to see that $\leq_{\text{inj}}$ is an ordering relation on ${\cal P}(\omega)$. (This is different from the lexicographic ordering discussed in another post.) Let us compare the two posets $({\cal P}(\omega), \subseteq)$ and $({\cal P}(\omega), \leq_{\text{inj}})$: Are there surjective order-preserving maps between them, in either direction? EDITED: Where down-set is printed above, I had the weaker (and confusing) condition $f(A)\subseteq B$ which doesn't make sense, as Andreas Blass pointed out. REPLY [5 votes]: Yes there is a surjective order-preserving map from $\leq_{inj}$ to $\subseteq$, no in the reverse direction. If you restrict the $\leq_{inj}$ order to the family of all infinite subsets of $\omega$ the result is isomorphic to the set of all functions from $\omega$ to $\omega$ with the coordinatewise order ($f \leq g$ if $f(n) \leq g(n)$ for all $n$). Namely, if $(a_n)$ is an infinite subset of $\omega$ given in increasing order, map it to the sequence $(a_{n+1} - a_n - 1)$. I.e. map an infinite set to the sequence of gaps. It's easy to see that there is an order-preserving surjection from this space to $\mathcal{P}(\omega)$. (Identify the latter with the set of functions from $\omega$ to ${0,1}$.) Any finite subset corresponds to a function from an initial segment of $\omega$ to $\omega$; fill it out with a sequence of zeros and map it to the image of that sequence. That gives you one direction. For the nonexistence of a map with the desired properties in the other direction, just note that the standard order had a greatest element (namely, $\omega$) but the inj order does not. So any map has to take $\omega$ somewhere and then nothing could consistently map to any greater element.<|endoftext|> TITLE: Symplectic Lie groups QUESTION [10 upvotes]: Assume that $G$ is a Lie group and at the same time it admits a symplectic structure. Does $G$ necessarily admit a symplectic structure such that the right multiplication preserves the symplectic structure? REPLY [12 votes]: Let $H$ be the universal covering of $\mathrm{SL}_2(\mathbf{R})\times\mathbf{R}$. Then $H$ is diffeomorphic to $\mathbf{R}^4$ and hence has a symplectic structure (as a manifold). However, every Lie group with a right-invariant symplectic structure is solvable (see Baues-Cortès (arXiv link) for references), so $H$ has no such structure.<|endoftext|> TITLE: Finite generation of module of modular forms QUESTION [12 upvotes]: Given a commutative $\mathbb{Z}[\frac1n]$-algebra $R$, we can consider the ring of modular forms $M_*(\Gamma_1(n), R)$. If $R$ is a subring of $\mathbb{C}$, these can be defined as those (holomorphic) modular forms for the congruence subgroup $\Gamma_1(n)$ that have $q$-expansion with coefficients in $R$. More generally, we can define this ring as $H^0(\overline{\mathcal{M}}_1(n)_R; \underline{\omega}^{\otimes *})$, where $\overline{\mathcal{M}}_1(n)_R$ is the compactified moduli stack of elliptic curves with $\Gamma_1(n)$-level structure over $R$ and $\underline{\omega}$ is the pushforward of the sheaf of differentials on the universal elliptic curve. Recently, I have investigated the structure of $M_*(\Gamma_1(n),R)$ as a graded module over $M_*(SL_2(\mathbb{Z}), R)$, the ring of modular forms without level. If $R = k$ is a field of characteristic bigger than $3$, it is quite easy to show that this graded module is free (by using that the compactified moduli stack $\overline{\mathcal{M}}_{ell}$ is in this case a weighted projective stacks line and vector bundles split into line bundles here); the same is actually true for field of characteristic $2$ or $3$, but significantly harder to show. From these results it follows that $M_*(\Gamma_1(n),R)$ is finitely generated (edit: as a graded module) over $M_*(SL_2(\mathbb{Z}), R)$ first in the case that $R$ is a field and one can deduce it also in the general case. This seems to be a rather overkill way to show just this finite generation (and I know another overkill argument using topological modular forms). My question is the following: Is it either an already known result that $M_*(\Gamma_1(n),R)$ is finitely generated (edit: as a graded module) over $M_*(SL_2(\mathbb{Z}), R)$ for all commutative $\mathbb{Z}[\frac1n]$-algebras $R$ or does it admit at least a simple/direct proof? (at least for subrings of $\mathbb{C}$)? REPLY [3 votes]: Edit: This is an answer to the wrong question. This explains the finite generation of $M_{k}(\Gamma_{1}(n), R)$ as an algebra, not as a module over the graded ring of level $1$ modular forms. This result is already known. One good source for this is the paper of Nadim Rustom (Generators of graded rings of modular forms, published in the Journal of Number Theory in 2014, the arXiv version is here). This paper shows more, that the graded ring of modular forms for $\Gamma_{1}(n)$ is generated in weight at most $3$ for $n \geq 5$. This is fairly simple over $\mathbb{C}$ and follows from a lemma (dating back to Mumford, essentially) about surjectivity of the map $H^{0}(X,\mathcal{L}_{1}) \otimes H^{0}(X,\mathcal{L}_{2}) \to H^{0}(X, \mathcal{L}_{1} \otimes \mathcal{L}_{2})$ where $\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ are two line bundles on a curve $X$. However, it takes a bit of work to translate this result into the desired statement for an arbitrary commutative $\mathbb{Z}[1/n]$ algebra. I would expect that there should be a much simpler way to see the finite generation of $M_{k}(\Gamma_{1}(n), R)$, but I don't know one.<|endoftext|> TITLE: Groups for which the $n$-power map is a homomorphism QUESTION [21 upvotes]: Let $\mathcal{V}_n$ be the collection of all groups satisfying $(ab)^n=a^nb^n$ for all $a,b$. In particular $\mathcal{V}_1$ consists of all groups and $\mathcal{V}_2$ consists of all abelian groups, and is contained in $\mathcal{V}_n$ for all $n$. For which coprime integers $n,m$ is $\mathcal{V}_n\cap\mathcal{V}_m$ reduced to abelian groups? REPLY [28 votes]: I assume $m,n>1$. (Edited for readability) A group is said to be "$n$-abelian" if $(xy)^n=x^ny^n$ for all $x,y\in G$. Following the question, let $\mathcal{V}_n$ denote the variety of all $n$-abelian groups, and let $\mathcal{A}b$ denote the variety of all abelian groups. The structure of $n$-abelian groups was determined by Alperin: they are the homomorphic images of subgroups of direct products of an abelian group, a group of exponent $n$, and a group of exponent $n-1$. On the other hand, given a group $G$ we define the "exponent semigroup of $G$" to be $$\mathcal{E}(G) = \{n\in\mathbb{Z}\mid (xy)^n=x^ny^n\text{ for all }x,y\in G\}.$$ This is a multiplicative semigroup of the semigroup of integers under multiplication. Its structure was determined by F.W. Levi, and also by Luise-Charlotte Kappe. The latter also explored the related concepts of $n$-Levi groups (groups for which $[x^n,y]=[x,y]^n$ for all $x,y\in G$) and $n$-Bell groups (groups for which $[x^n,y]=[x,y^n]$ for all $x,y\in G$). The structure of the semigroup of $n$s for which a given $G$ is an $n$-Levi (resp. an $n$-Bell) group is the same as for the exponent semigroup. Kappe defines a Levi system to be a subset of the integers that satisfies the following five conditions: $n,m\in W$ implies $nm\in W$. $n\in W$ implies $1-n\in W$. $0\in W$. There exists $w\in W$, $w\gt 0$, such that for all $n\in W$, $n^2\equiv n\pmod{w}$ and every integer congruent to $n$ modulo $w$ is in $W$. If the congruence classes of both $n$ and $n+1$ modulo $w$ lie in $W$, then $n\equiv 0\pmod{w}$. Kappe proves that a subset of the integers is $\mathcal{E}(G)$ for some $G$ if and only if it is $\{0,1\}$, or is a Levi system. We can drop $\gcd(m,n)=1$ from the hypotheses. The answer to the question is: Theorem. Let $m,n>1$. Then $\mathcal{V}_n\cap\mathcal{V}_m=\mathcal{A}b$ if and only if $\gcd(n^2-n,m^2-m)=2$. Note that $2$ divides $n(n-1)$ and $m(m-1)$, so the gcd is at least two. Proof. If $p>2$ is an odd prime that divides $\gcd(n^2-n,m^2-m)$, then a nonabelian group of exponent $p$ is both $n$-abelian and $m$-abelian. Similarly, if $4$ divides $\gcd(n^2-n,m^2-m)$, then a nonabelian group of exponent $4$ is both $n$-abelian and $m$-abelian (since $2$ will divide either $n$ or $n-1$, but not both, and likewise $m$ or $m-1$, but not both). So the condition is necessary. Conversely, assume $\gcd(n^2-n,m^2-m)=2$, and let $G\in\mathcal{V}_n\cap\mathcal{V}_m$. Then $n,m\in\mathcal{E}(G)$, so the latter is not just $\{0,1\}$, and hence must be a Levi system. Let $w$ be the positive integer guaranteed by property 4. Then $n^2\equiv n\pmod{w}$ and $m^2\equiv m\pmod{w}$, hence $w=1$ or $w=2$. If $w=1$, then because $n\equiv k\pmod{1}$ for all $k$, it follows that $\mathcal{E}(G)=\mathbb{Z}$, so $G$ is abelian. If $w=2$, then $2\in\mathcal{E}(G)$, so $G$ satisfies $(xy)^2=x^2y^2$, a condition well known to imply that $G$ is abelian. Thus, $G\in\mathcal{A}b$ in either case. $\Box$ I note that Theorem 2 in Kappe's paper also gives this result. Given a subset $T$ of integers that includes $0$ and $1$, and such that if $m,n\in T$ then $k(n^2-n)+m\in T$ for all integers $k$, the theorem provides several conditions equivalent to $T=\mathbb{Z}$. One of them is the existence of a subset $U$ of $T$ such that $\gcd(n^2-n\mid n\in U)=2$. Since $\mathcal{E}(G)$ always satisfies these conditions, it follows that $m,n\in\mathcal{E}(G)$ and $\gcd(m^2-m,n^2-n)=2$ implies $\mathcal{E}(G)=\mathbb{Z}$, and hence $G$ is abelian. References. Alperin, J.L. A classification of n-abelian groups. Canad. J. Math. 21 1969 1238–1244. MR0248204 (40 #1458) Kappe , L.C. On n-Levi groups. Arch. Math. (Basel) 47 (1986), no. 3, 198–210. MR0861866 (88a:20048) Levi, F.W. Notes on group theory. VII. The idempotent residue classes and the mappings $\{m\}$. J. Indian Math. Soc. (N. S.) 9, (1945). 37–42. MR0016414 (8,13d)<|endoftext|> TITLE: Galois theory, topos vs fundamental groups QUESTION [8 upvotes]: Classical Galois theory states that the etale topos X of a field k is equivalent to the classifying topos of the absolute Galois group of k. (Marc Hoyois, Higher Galois theory, $\S$3, arXiv:1506.07155, doi:j.jpaa.2017.08.010) I would like to know how is passing from Classical Galois theory (with classical I mean the Galois theory obtained with Grothendieck fundamental group, I know this is not very classic...) to the classical Galois theory referenced (an equivalence of topos, I know this doesn't looks classic also...)? And, I Would like to know if this equivalence has been studied for schemes and not only for fields... REPLY [11 votes]: I'm not sure I understand what you precisely want to know, so I'll try to clarify a few things from the topos theoretic point of view, which I hope will answer your questions: Relation to Galois theory: Classical Galois theory tells you that the category of étale extention of your fields $k$ is equivalent (contravariantly) to the category of finite sets endowed with a continuous action of the galois group (well technically it tells you that transitive action corresponds to fields extention, but étale extention are finite product of fields extention and general action are finite coproduct of transitive action). When you look at the category of sheaves on the category of finite action with the natural topology (covering are surjection of finite action) you gets rather trivially the topos of all continuous action of the galois group (basically because they are justs non-finite coproducts of the finite action). And conversely, if you know that the étale topos is the topos of sets with a continuous $G$ action, as étale extention corresponds to representable sheaves one easily deduce (with more or less work) the classical version of Galois theory. Topos theoretic Galois theory: If what you had in mind is the formalism of Grothendieck of looking at the automorphism of a fiber functor, this is also something we do a lot in topos theory and it is indeed closely related to this representation theorem for the étale topos. In fact a relatively direct extension of Grothendieck Galois theory can be phrased as follow: if you have a connected atomic topos $P$ with a point $p$ then taking the fiber at $p$ induce an equivalence between $T$ and the category of sets endowed with a continuous action of the group of automorphism of $p$. If one wants the theorem to be true in full generality one need to use a 'localic group' but for the toposes which comes from algebraic geometry it is not a problem as those groups are always compact, in fact the original formulation of Grothendieck of his galois theory corresponds precisely to the case where the group is compact. See for example this paper of E.Dubuc for a presentation of these ideas. The étale topos and the étale fundamental group: For the étale topos of a scheme, things are a little bit more complicated, there is still a close relationship between the étale topos and the classifying topos of the étale fundamental group (I mean the topos of set with a continuous action of the étale fundamental group), but they are not isomorphic at all. The classifying topos of the étale fundamental group is closely related to the category of locally constant objects of the étale topos. This is completely similar to the fact that when you have a topological space $X$ the category of locally constant sheaves over $X$ is equivalent to the category of set endowed with an action of the $\pi_1$ (and if you look to space that are not semi-locally simply connected then the $\pi_1$ gets a topology and you have to look at continuous action as for the étale fundamental group and you do not get exactly the locally constant sheaves but something related). In fact there is a general theory of the $\pi_1$ of toposes, where you essentially have that for a general (grothendieck) topos you can prove an equivalence between "covering projection" (which are the same as locally constant object in good cases or if you are only interested in finite covering projection) and sets with continuous action of the $\pi_1$. (also in the most general version, you will need a groupoids and the groups have to be localic). I believe there are several variant of this which are not all fully general and not all clearly equivalent, but this paper also by Dubuc is probably one of the best place to look. the terminology of "covering projection" that I used above is introduced in this paper. It is important to keep in mind that for a general scheme this only corresponds to a small part of the étale topos: the ``locally constant objects'' (at least for the finite ones), it justs happen that for a fields "because it is only a point" (informally) everythings is always locally constant. In general the étale fundamental groups is just the $\pi_1$ of the étal topos, it does not see the full space. General representation theorem for the étale topos: But this is not the end of our geometric understanding of the étale topos. In a hugely important (and sadly hard to found) paper called "an extention of the galois theory of Grothendieck", A.Joyal and M.Tierney proved that this idea of Grothendieck Galois theory can be extended (and coupled with the notion of descent theory in algebra) to obtain a very general representation theorem for toposes: any Grothendieck topos is the category of equivariant sheaves of sets on a localic groupoids. They moreover have a rather explicit construction of the groupoids (but it needs a fair amount of expertise on the subject to be used). There is a variant of this construction due to C.Butz and I.Moerdijk (Representing topoi by topological groupoids) which applies to topos with enough points (which is the case of the étale topos by a famous theorem of Deligne) and produce a topological groupoid instead of a localic groupoid, but most importantly is a lot more explicit, at least if you have a good understanding of the points of our topos and more generally of what it classifies (which in the case of the étale topos is well known: for affine scheme they are the strict localization of your ring). If you go through the construction of Butz & Moerdijk for the étale topos you get that it can be represented by a topological groupoid whose objects corresponds to a large enough set of strict localization of your ring with a certain topology a bit similar to a Zariski topology, with morphism being essentially all the local galois action on those strict localization. The full description of the groupoid is a little bit involved, and I do not know if it has been worked out in details for the étale topos of a scheme (I'm not very familiar with the literature on the subject, maybe someone else can add a reference here, or say if that does not exists ?)<|endoftext|> TITLE: Domination Number equals Independence Number? QUESTION [5 upvotes]: Let $\gamma(G)$ and $\alpha(G)$ be the domination number and independence number for an Graph $G$. Further, let $i(G)$ be the minimum-size Independent Dominating Set. Then, it is known that $\forall G, \gamma(G) \leq i(G) \leq \alpha(G)$. According to this paper, it is also known that $\gamma(G) = i(G)$ for Claw-free Graphs and $i(G) = \alpha(G)$ for Block Graphs. What is the most general family of graphs for which $\gamma(G)=i(G)=\alpha(G)$? Further, what is the most general family of graphs for which $\frac{\alpha(G)}{c}\leq\gamma(G)\leq \alpha(G)$ for a constant c >1? REPLY [5 votes]: Since $\gamma(G)\leq i(G)\leq \alpha(G)$ for all graphs $G$, the problem of characterizing all graphs $G$ for which $\gamma(G)=i(G)=\alpha(G)$ is equivalent to the characterizing all graphs $G$ for which $\gamma(G)=\alpha(G)$. The theorem "$\gamma(G)=i(G)$ holds for all claw-free graphs $G$" was proved by Bollobas and Cockayne in "B. Bollobas and E.J. Cockayne, Graph-theoretic parameters concerning domination, independence, and irredundance, J. Graph Theory" 3 (1979), 241-249. However, the problem of finding the family of all graphs $G$ for which $\gamma(G)=\alpha(G)$ is still open.<|endoftext|> TITLE: How to solve a recursion relation on tensors including derivatives and traces? QUESTION [5 upvotes]: NOTATION I'm a physicist studying higher-spin theory. In my research, we work with fully symmetric tensors using a notation which is implicit both in the dimension of space and the order of the tensor, i.e. an order-$s$ fully symmetric tensor is simply written as $$ \phi_{\mu_1 \cdots \mu_s} \equiv \phi. $$ The $n$-th gradient of $\phi$ is written as $\partial^n \phi$, the $n$-th divergence, as $\partial^n \cdot \phi$ and the $n$-th trace as $\phi^{[n]}$. Lower traces are simply written with a prime, e.g. $\phi''$ for the second trace. All indices are implicitly symmetrized, without weight factors, using the minimal number of terms. For example, if $s=2$, $$ \partial^2 \phi \equiv \partial_\mu \partial_\nu \phi_{\sigma \rho} + \partial_\mu \partial_\sigma \phi_{\nu \rho} + \partial_\mu \partial_\rho \phi_{\sigma \nu} + \partial_\nu \partial_\sigma \phi_{\mu \rho} + \partial_\nu \partial_\rho \phi_{\sigma \mu} + \partial_\sigma \partial_\rho \phi_{\mu \nu} $$ $$ \partial (\partial \cdot \phi) \equiv \partial_\nu (\partial^\lambda \phi_{\mu \lambda}) + \partial_\mu (\partial^\lambda \phi_{\lambda \nu})$$ $$ \eta \partial^2 \cdot \phi \equiv \eta_{\mu\nu} \partial^\rho \partial^\sigma \phi_{\rho \sigma} $$ The formalism implies the following set of rules $$ ( \partial^p \phi )' = \Box \partial^{p-2} \phi + 2 \partial^{p-1} \left( \partial \cdot \phi \right) + \partial^p \phi' $$ $$ \partial \cdot (\partial^p \phi) = \Box \partial^{p-1} \phi + \partial^p \left( \partial \cdot \phi \right) $$ $$ \partial^p \partial^q = {{p+q}\choose{q}} \partial^{p+q} $$ which you can simply take as axioms of the game. By the way, $\Box = \partial \cdot \partial$, which we are allowed to invert to $\frac{1}{\Box}$, but don't worry about what this means, you can just use the three rules written above. PROBLEM So, the thing I'm interested in is the recursion relation $$\mathcal{F}_{n+1} = \mathcal{F}_{n} - \frac{1}{n+1} \frac{\partial}{\Box} \left( \partial \cdot \mathcal{F}_{n} \right) + \frac{1}{(n+1)(2n+1)} \frac{\partial^2}{\Box} \mathcal{F}_{n}{}'$$ with the initial condition $$ \mathcal{F}_0 = \Box \phi.$$ I would like solve the recursion fully, i.e. to express $\mathcal{F}_{n+1}$ only in terms of $\phi$'s and not $\mathcal{F}_{k}'s$ with $k TITLE: integration for fractional laplacian QUESTION [5 upvotes]: Is it possible to integrate by parts the fractional laplacian $(-\Delta)^su+ u=f(u)$ in $\mathbb{R}^N$, or is it true that $\int_{\mathbb{R}^N}u= \int_{\mathbb{R}^N} f(u) $ with $s\in (0, 1)$, $u\in H^s(\mathbb{R}^N)$. The identity holds for the Laplacian operator. REPLY [3 votes]: Yes, this follows from the identity $$(-\Delta^{s}u)(x)=c_{n,s}\int_{\mathbb{R}^n}\frac{u(x)-u(y)}{||x-y||^{n+2s}}\,dy,$$ $$c_{n,s}=\frac{s\,4^{s}\Gamma(s+n/2)}{\pi^{n/2}\Gamma(1-s)},$$ where the principal value of the integral should be taken. (See The Pohozaev identity for the fractional Laplacian.) Integration over $x\in\mathbb{R}^n$ gives zero because of the antisymmetry of the integrand upon interchange $x\leftrightarrow y$, so $$\int_{\mathbb{R}^n} [(-\Delta^{s}u)(x)+u(x)]\,dx=\int_{\mathbb{R}^n} u(x)\,dx=\int_{\mathbb{R}^n} f(u(x))\,dx,$$ as requested.<|endoftext|> TITLE: Duality between Banach spaces and compact convex spaces QUESTION [7 upvotes]: I always had the impression that there was a duality (i.e. a contravariant equivalence of categories) between Banach spaces and certain notion of pointed compact convex set (something like algebras for a certain Giri monad acting on the category of compact topological spaces, maybe with additional conditions) given by: To a Banach space one associate the unit ball of its dual in the weak* topology. And to a convex compact space one associate the space of linear function on it, with the uniform norm. I can't remember if this is something I read somewhere or just something I imagined due to other weaker results in this direction (like the fact that every banach spaces is isomorphic to the space of Weak* continuous linear form on its dual). So, do we indeed have a duality of this kind ? Has this been worked out somewhere, or is it some sort of folk results ? does it have a name ? PS: I had too chose one answer to accept, but all three answers were all equally good and very interesting. REPLY [3 votes]: Yes, see Proposition 8.1.3 of the book Mathematical Quantization. This describes a duality between Banach spaces and "dual unit balls", which are defined as compact, convex, balanced subsets of locally convex TVSs.<|endoftext|> TITLE: Does there exist any non-contractible manifold with fixed point property? QUESTION [30 upvotes]: Does there exist any non-trivial space (i.e not deformation retract onto a point) in $\mathbb R^n$ such that any continuous map from the space onto itself has a fixed point. I highly suspect that the quasi circle on $\mathbb R^2$ is an example. Yet I've not written down the (dirty) proof. But in this case all its homotpy groups are trivial. So if I assume my space as a manifold, then (QUESTION:) does this fixed point property force it to become a contractible manifold? I read somewhere that there exists a contractible compact manifold which does not satisfy this fixed point property. So does there exist any non-contractible manifold (compact) where this property follows? Or otherwise can anyone please provide an outline of how to prove that such a manifold is contractible? REPLY [5 votes]: Another nice example is $\mathbb RP^{2n}$. Here is a simple proof... If $f:\mathbb RP^{2n} \to \mathbb RP^{2n}$ is a continuous function then look at the lift of the map $\bar{f} :S^{2n}\to S^{2n}$. A fixed point of $f$ is equivalent to a point $x\in S^{2n}$ s.t. $\bar{f}(x)=x \ or \ -x$. If such a point doesn't exist, then we can have a homotopy between Identitity and Antipodal map via $H(x,t)=cos(t)x+sin(t)\bar{f}(x)$.[This is well defined map from sphere to sphere]. But for even dimensional sphere antipodal map has degree $-1$. That's a contradiction.<|endoftext|> TITLE: Jordan curves in $\mathbb R^n$ and inscribed equilateral triangles QUESTION [8 upvotes]: Inscribed square problem wants that we know "Does every Jordan curve admit an inscribed square?" From my amateur viewpoint it seems that the concept of Jordan curve can be straightforwardly generalized to $\mathbb R^n$ so as to define those curves so that they are images of continuous functions $f$ from $[0,1]$ to $\mathbb R^n$ in such a way that those functions are injective on $[0,1)$ and $f(0)=f(1)$ On Wikipedia, it is written: " It is known that for any triangle T and Jordan curve C, there is a triangle similar to T and inscribed in C.[9][10] Moreover, the set of the vertices of such triangles is dense in C.[11] In particular, there is always an inscribed equilateral triangle" But, I do not have an access to articles [9], [10], [11] so do not know do they only discuss Jordan curves in the plane. I would just like to know is it known: Do all Jordan curves in all spaces $\mathbb R^n$ admit an inscribed equilateral triangle? Of course, avoid the non-interesting case $n=1$ If i had the access to [9] then I could know are there discussed equilateral triangles only for Jordan curves in $\mathbb R^2$ or in all $\mathbb R^n$ spaces. What is known about this question of mine? The "Inscribed square problem" page is linked because the quote from Wikipedia is on that page and also the links to those articles are on that page. Now I am thinking that I am asking something very trivial here and am in a state of not knowing should I even post this question, but, I will post the question and face the consequences. REPLY [5 votes]: Theorem 5 in [9] proves that any metrizable simple closed curve contains vertices of an equilateral triangle. In particular, this applies to any simple closed curve in $\mathbb{R}^n$.<|endoftext|> TITLE: Prove this conjecture inequality $x\cdot \frac{(1-x)^{k-1}}{(k+1)^{k-2}}+\frac{(1-2x)^k}{k^k}\le \frac{1}{(k+2)^{k-1}}$ QUESTION [5 upvotes]: let $x\in (0,1)$, and $k$ be postive intgers,such $k\ge 2$, I conjecture following inequality maybe hold? $$x\cdot \dfrac{(1-x)^{k-1}}{(k+1)^{k-2}}+\dfrac{(1-2x)^k}{k^k}\le \dfrac{1}{(k+2)^{k-1}}$$ creat by wang yong xi This is my attempt when $k=2$,then inequality can be written as $$x\cdot (1-x)+\dfrac{(1-2x)^2}{4}\le\dfrac{1}{4}$$ it is obviously true. when $k=3$then inequality can be written as $$\dfrac{x(1-x)^2}{4}+\dfrac{(1-2x)^3}{27}\le\dfrac{1}{25}$$ or $$-\dfrac{(5x-1)^2(5x+8)}{2700}\le 0$$ it is clearly true. when $k=4$ it's equivalent $$\dfrac{x(1-x)^3}{25}+\dfrac{(1-2x)^4}{256}-\dfrac{1}{216}\le 0$$ or $$\dfrac{(6x-1)^2(108x^2+12x-125)}{172800}\le 0$$ it is clearly when $k=5$, it's equivalent $$x\cdot\dfrac{(1-x)^4}{216}+\dfrac{(1-2x)^5}{5^5}-\dfrac{1}{7^4}=-\dfrac{(7x-1)^2(185563x^3-181202x^2-127589x+156384)}{1620675000}<0$$ But I can't prove for any postive intgers $k$.and I have found $$LHS-RHS=[(k+2)x-1]^2\cdot h(x,k)$$.so we must prove $h(x,k)\le 0$ Thanks REPLY [9 votes]: Consider only $k>2$. Denote $$f(x)=x\cdot \frac{(1-x)^{k-1}}{(k+1)^{k-2}}+\frac{(1-2x)^k}{k^k},$$ we need to prove that $f(x)\leqslant f(\frac1{k+2})$ for $x\in [0,1]$. We have $f'(\frac1{k+2})=0$ and $$f'(0)=(k+1)^{2-k}-2k^{1-k}=k^{2-k}\left(\left(1+\frac1k\right)^{2-k}-\frac2k\right)>0$$ by Bernoulli inequality $(1+x)^a>1+ax$ for $a=2-k$, $x=1/k$. Also $f(1)$ equals $\pm f(0)$. It means that the maximal value of $f$ on $[0,1]$ is attained at an interior point $a\in (0,1)$, and thus $f'(a)=0$. I claim that $f'$ has unique root (multiplicity counted) on $(0,1/2]$, and this root is $\frac1{k+2}$, and at most one root on $[1/2,1)$. In both cases the only possible maximum point is $\frac{1}{k+2}$ (the second extremal point would be a local minimum) We have $$f'(x)=(1-x)^{k-2}(1-kx)(k+1)^{2-k}-2(1-2x)^{k-1}k^{1-k}=0.$$ Denote $y=\frac{1-x}{1-2x}$, then $x=\frac{1-y}{1-2y}$, $1-kx=\frac{(k-2)y-(k-1)}{1-2y}$. Our equation $f'(x)=0$ in terms of $y$ rewrites as $$y^{k-2}((k-2)y-(k-1))(k+1)^{2-k}+2k^{1-k}=0.$$ LHS is monotone in $y$ for $y\in (1,\infty)$ (corresponds to $x\in (0,1/2)$) and for $y<0$ (corr. $x\in (1/2,1)$), that implies the above claim.<|endoftext|> TITLE: Gauss and primitive roots QUESTION [11 upvotes]: In https://www.math.dartmouth.edu/~carlp/ordertalkunder.pdf Carl Pomerance writes: "... Over two centuries ago, Gauss asked if this deal with the decimal for $1/p$ occurred for infinitely many primes $p$. I.e., do we have $l_{10}(p) = p − 1$ for infinitely many primes $p$?" Meanwhile in http://guests.mpim-bonn.mpg.de/moree/surva.pdf Pieter Moree writes: "... Hence we expect that there are infinitely many primes $p$ having 10 as a primitive root mod $p$. This conjecture is commonly attributed to Gauss, however, to the author’s knowledge there is no written evidence for it." What evidence is there that Gauss raised the question, and what evidence is there that he hazarded a guess? REPLY [5 votes]: (Not really an answer, but too long for a comment.) I found many references writing that Gauss conjectured that there exst infinitely many primes, with $10$ a primitive root, some referring to Disquitiones, others without any reference. Gauss studied the period of $1/p$ from his early days. A link to a manuscript (in German) by Siebeneicher, including handwritten tables from the Göttingen archive. https://www.math.uni-bielefeld.de/~sieben/fermat.pdf In Disquisitiones, section 316, Gauss lists the relevant primes $7, 17, 19, 23, 29, 47, 59, 61, 97$ below 100 and writes he has the decimal periods of prime powers up to 1000 and might publish that list on some other occasion. It seems to me that he did not explictly write in Disquitiones (and possibly elsewhere) that there are infinitely many such primes. This makes this question by James an interesting one, and it should clearly be clarified why it eventually became a "conjecture" of Gauss in the modern literature! However, it seems very very likely to me that Gauss observed from these small primes that this type of primes is not rare: Gauss famously counted primes in very young age, and conjectured a good approximation formula to $\pi(x)$, and the list above "looks like" a positive proportion of all primes below 100.<|endoftext|> TITLE: Translates of measure zero set QUESTION [11 upvotes]: Suppose $X \subseteq \mathbb{R}$ has measure zero. Can we find an uncountable $A \subseteq \mathbb{R}$ such that $X + A = \bigcup \{a + x: a \in A, x \in X\}$ has measure zero? Clearly, the answer is yes under Martin's axiom and $\mathfrak{c} > \aleph_1$. But can we do this without additional assumptions. REPLY [10 votes]: Claim: For every null G-delta set $X$, there exists a perfect set $P$ such that $X + P$ is null. Proof sketch: First assume MA plus $2^{\aleph_0} > \aleph_2$. Choose $A \subseteq \mathbb{R}$ of size $\aleph_2$ and a null G-delta set $G$ such that $X + A \subseteq G$. It follows that $\{y: X + y \subseteq G\}$ is a coanalytic set of size $> \aleph_1$ and hence contains a perfect set $P$. The statement "there exist a perfect set P and a null G-delta set G such that $X + P \subseteq G$" is $\Sigma^1_2(X)$. Apply Shoenfield's absoluteness. Note that the proof is identical to Martin's proof of Claim 1 here.<|endoftext|> TITLE: Do almost all systems of quadratic equations have solutions? QUESTION [11 upvotes]: If I have a system of linear equations, $A x = c$, with $A$ an $n\times n$ complex matrix, it is relatively easy to see that the set of matrices $A$ for which there is no (complex) solution has measure zero, as this is the set of matrices such that $\det(A) = 0$. Can something similar be said for systems of quadratic equations? More precisely, consider a system of $n$ quadratic equations in $n$ variables, which I can always write as $$ \boldsymbol x^\dagger A_i \boldsymbol x + \boldsymbol b_i \cdot \boldsymbol x + c_i = 0, \quad i=1,..., n, $$ where $A_i$ are $n\times n$ complex matrices, $\boldsymbol b_i\in\mathbb C^n$ and $c_i\in\mathbb C$. Does this system have a solution for almost all values of the parameters? In other words, if a given choice of parameters corresponds to no solutions, is it always true that an infinitesimal change of parameters will give me a system which has solutions? REPLY [15 votes]: Yes. The magic words are "elimination theory" and "resultant". In essence, the system has a solution unless some determinant (the iterated resultant) vanishes.<|endoftext|> TITLE: Testing whether two elements of $\text{SL}(2, \mathbb{F}_{2^n})$ generate the entire group QUESTION [12 upvotes]: Given two elements $A,B \in \text{SL}(2, \mathbb{F}_{2^n})$, is there a (computationally inexpensive) test one could perform to check whether together they generate the entire group? REPLY [13 votes]: The answer is yes, it is easy to check whether the ordered pair $(A, B)$ generates $\text{SL}_2(\mathbb{F}_q)$ for $q$ the power of a prime number. Indeed, there exist simple criteria according to Daryl McCullough and Marcus Wanderley, see [1, Section 11]. (Italic means that I am quoting the authors). Claim. The pair $(A, B)$ generates $\text{SL}_2(\mathbb{F}_q)$ if and only if $$\text{Tr}(A, B) \Doteq (\text{Tr}(A), \text{Tr}(B), \text{Tr}(AB))$$ is an essential triple, i.e., doesn't satisfy any of the conditions $(1) - (5)$ of [1, Section 11]. For instance, the condition $(1)$ for a triple $(\alpha, \beta, \gamma) \in \mathbb{F}_q^3$ holds if at least two of $\alpha, \beta$ and $\gamma$ are zero. The image in $\text{PSL}_2(\mathbb{F}_q)$ of a pair $(A, B) \in \text{SL}_2(\mathbb{F}_q)^2$ such that $\text{Tr}(A, B) = (\alpha, \beta, \gamma)$ generates a dihedral group in this case. Condition $(2)$ holds if $\alpha^2 + \beta^2 + \gamma^2 - \alpha \beta \gamma - 2 = 2$, the left-hand side being the Fricke polynomial. This corresponds to the case of an affine subgroup of $\text{PSL}_2(\mathbb{F}_q)$, that is a subgroup which is conjugate to a subgroup of the image in $\text{PSL}_2(\mathbb{F}_q)$ of the subgroup of upper triangular matrices in $\text{SL}_2(\mathbb{F}_q)$. Condition $(4)$ holds if $\alpha, \beta$ and $\gamma$ lie in a proper subfield of $\mathbb{F}_q$. The remaining conditions are presented in the Addendum below. It follows from the claim that OP's decision problem boils down to: deciding whether two elements of $\mathbb{F}_q$ are equal. deciding whether $k$ elements of $\mathbb{F}_q$ ($k \le 4$) generate a proper subfield. deciding whether an element of $\mathbb{F}_q$ belongs to a subfield of $\mathbb{F}_q$ generated by $k$ given elements ($k \le 4$). The above claim follows from a theorem of Macbeath [1, Theorem 8.2] while the criteria are based on Dickson's Theorem [1, Theorem 6.1] and another theorem of Macbeath [1, Theorem 8.1]. Side note. Let $G$ be a finite group generated by $k$ elements. Let $\varphi_k(G) = \frac{\vert \text{Epi}(F_k, G) \vert}{\vert \text{Hom}(F_k, G) \vert}$, where $F_k$ denotes the free group on $k$ generators. The ratio $\varphi_k(G)$ is the probability that two uniformly chosen random elements generate $G$. By a theorem of Liebeck and Shalev, we have $\lim_n \varphi_2(G_n) = 1$ for any sequence of non-isomorphic finite simple groups, see [3, Theorem 1.1.1]. For a simple group $G_n(q)$ of Lie type of untwisted rank $n$, we have $\varphi_2(G_n(q)) = 1 - O(\frac{n^3 \log^2(q)}{q^n})$ by results of Kantor, Lubotzky, Liebeck and Shalev, see [3, Theorem 1.1.2]. Addendum. Let $Q(\alpha, \beta, \gamma) \Doteq \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta \gamma - 2$ be the Fricke polynomial. This name was coined because of Fricke's trace identity $$\text{Tr}([A, B]) = Q(\text{Tr}(A), \text{Tr}(B), \text{Tr}(AB)).$$ Here are the five conditions derived by Daryl McCullough and Marcus Wanderley from Dickson's Subgroup Classification Theorem and Macbeath's Fricke polynomial criterion: $(1)$ Dihedral case. At least two of $\alpha, \beta$ and $\gamma$ are zero. $(2)$ Affine case. $Q(\alpha, \beta, \gamma) = 2$. $(3.A_4)$ The elements $\alpha, \beta$ and $\gamma$ lie in $\{-1, 0, 1\}$ and $Q(\alpha, \beta, \gamma) = 0$. $(3.S_4)$ The elements $\alpha, \beta$ and $\gamma$ lie in $\{\pm 1, \pm \sqrt{2}\}$ and $Q(\alpha, \beta, \gamma) = 1$, where $\sqrt{2}$ denotes one root of $X^2 - 2$. $(3.A_5)$ It's complicated! The authors redirect us to [2, Main Theorem (3)]. $(4)$ Affine or $SL_2(q')$. The elements $\alpha, \beta$ and $\gamma$ lie in a proper subfield of $\mathbb{F}_q$. $(5)$ Affine or $SL_2(q')$ or $PGL_2(q')$. The integer $q$ is odd, $\alpha^2,\beta^2,\gamma^2$ and $\alpha \beta \gamma$ lie in a proper subfield $\mathbb{F}_{q'}$ of $\mathbb{F}_q$ and at least one of $\alpha, \beta$ and $\gamma$ does not lie in $\mathbb{F}_{q'}$. The cited article (a very nice paper!) contains more details, and of course, the corresponding proofs. [1] D. McCullough, M. Wanderley, "Nielsen equivalence of generating pairs of $\text{SL}(2, q)$", 2013. [2] D. McCullough, "Exceptional subgroups of $\text{SL}(2, F )$". Preprint available here, accessed 4 January 2013. [3] I. Pak, "What do we know about the product replacement algorithm?", 2000.<|endoftext|> TITLE: Optimality of the Plünnecke-Ruzsa Inequality QUESTION [14 upvotes]: The Plünnecke–Ruzsa Inequality states that for a finite subset $A$ of an abelian group $G$ with small doubling $|A+A|\le K|A|$, the iterated sum and difference sets are also small: $|tA-sA| \le K^{t+s} |A|$. It seems natural to expect that the optimal exponent on $K$ should actually be $K^{t+s-1}$, since we're thinking of each additional sum or difference with $A$ as magnifying the set by a factor of at most $K$. Also, the choice of $A$ as a basis for $\mathbb{F}_2^n$ leads me to think there could be improvements on the constant in Plünnecke–Ruzsa, so that I would conjecture (roughly) $|tA - sA| \le \frac{K^{t+s-1}}{t!s!} |A|$. Is this possible? I don't see a way to get improvements of either kind directly from the Plünnecke/Ruzsa/Petridis graph approach (see Petridis). REPLY [18 votes]: No. See for instance Exercise 2.3.5 of Tao, Terence; Vu, Van H., Additive combinatorics, Cambridge Studies in Advanced Mathematics 105. Cambridge: Cambridge University Press (ISBN 978-0-521-13656-3/pbk). xviii, 512 p. (2010). ZBL1179.11002. See also a number of papers of Ruzsa constructing various counterexamples, e.g. Ruzsa, I.Z., On the number of sums and differences, Acta Math. Hung. 59, No.3-4, 439-447 (1992). ZBL0773.11010.<|endoftext|> TITLE: Sum of Gaussian pdfs QUESTION [25 upvotes]: I learned from a colleague that if one sums translates of the Gaussian density $f(x)=(2\pi)^{-1/2}e^{-x^2/2}$ translated by the integers (i.e. one considers $F(x)=\sum_{n\in\mathbb Z}f(x+n)$), the resulting function is remarkably constant: that is, the function differs from its average value by less than one part in $10^8$. The significance of translating by multiples of 1 is that the inflection point of the Gaussian occurs at $\pm1$. I attempted to repeat this for $g(x)=e^{-3x^4/4}$ which also has an inflection point at $\pm 1$, and found that the resulting sum is constant to one part in 200. Can anyone offer any conceptual explanation for the remarkable degree of constancy of $F(x)$, or is this just a fluke? REPLY [8 votes]: Interesting! Put another way, the fractional part of the standard Gaussian variable closely approximates the standard uniform variable. You can also closely approximate standard Gaussian from standard uniform: Add 12 independent uniform variables before subtracting 6. This approximation is also shockingly good; see John D. Cook's comment here. (12 isn't "large," but rather uses var(U)=1/12 to get unit variance.) Notice that the fractional part of this approximate Gaussian is exactly uniform. You might think of this as a dual explanation for the phenomenon you observed.<|endoftext|> TITLE: Density of integers with many prime factors QUESTION [7 upvotes]: For a positive integer $n$ put $\omega(n)$ for the number of distinct prime divisors of $n$. It is a well-known theorem of Erdős and Kac that the probability distribution for the quantity $\displaystyle \frac{\omega(n) - \log \log n}{\sqrt{\log \log n}}$ is the standard normal distribution. In other words, we have $$\displaystyle \lim_{X \rightarrow \infty} \left(\frac{1}{X} \# \left\{n \leq X : a \leq \frac{\omega(n) - \log \log n}{\sqrt{\log \log n}} \leq b \right\}\right) = \frac{1}{\sqrt{2\pi}} \int_a^b e^{-t^2/2}dt.$$ My question is, can one give a good estimate for the density of integers which deviates from the mean significantly? The above limit is only sensitive to positive density, where I am expecting a 0-density result. More precisely and concretely, how does one estimate the density of the set $$\displaystyle \{n \leq X : \omega(n) > (\log \log n)^2\}$$ say? REPLY [2 votes]: As a survey in book form I would recommend Tenenbaum's book (Introduction to analytic and probabilistic number theory), chapter II. 6.1 (Integers having $k$ prime factors). Also the notes of the end of the chapter give very useful references (such as the Hildebrand-Tenenbaum paper mentioned by Lucia, the Selberg-Delange method etc.). I would doubt that extending the large deviation techniques from $\omega(n)$ about $\log \log n$ to say $(\log \log n)^2$ is of great use. These end of chapter notes rather direct to Hildebrand-Tenenbaum.<|endoftext|> TITLE: Is there an explicit description of the corestriction map $H^1(H, M) \rightarrow H^1(G, M)$? QUESTION [7 upvotes]: Let $G$ be a group and $M$ a $G$-module. The basic definitions: $H^0(G, M)$ will be the set of $G$-fixed points in $M$. $Z^1(G, M)$ is the group of $1$-cocycles, i.e. the maps $f: G \rightarrow M$ such that $f(gg') = f(g) + g f(g')$ for all $g, g' \in M$. $B^1(G, M)$ is the group of $1$-coboundaries, ie. maps $c_m : G \rightarrow M$ defined by $c_m(g) = gm - m$. $H^1(G, M)$ is the quotient group $Z^1(G, M) / B^1(G, M)$. Let $H < G$ be a subgroup of finite index. We have a map $$tr: H^0(H, M) \rightarrow H^0(G, M)$$ defined by $m \mapsto \sum_{g \in G/H} gm$. This can be extended to a map $H^*(H, M) \rightarrow H^*(G, M)$ of cohomological functors. My question is: Is there an explicit description of the corestriction map $H^1(H, M) \rightarrow H^1(G, M)$, e.g. in terms of $1$-cocycles? For one idea, we have an isomorphism $\psi: \operatorname{Hom}_{\mathbb{Z}H}(\mathbb{Z}G, M) \rightarrow M$ defined by $$\psi(f) = \sum_{g \in G/H} gf(g^{-1}).$$ Hence there is an explicit map $H^1(G, \operatorname{Hom}_{\mathbb{Z}H}(\mathbb{Z}G, M)) \rightarrow H^1(G, M)$. By Shapiro's lemma we have $H^1(H, M) \cong H^1(G, \operatorname{Hom}_{\mathbb{Z}H}(\mathbb{Z}G, M))$, but I don't know if there is a way to make this isomorphism explicit. REPLY [5 votes]: I believe it is the following. Let $f$ be a cocyle for $H$. Take a set of representatives $X$ of $G/H$ in $G$. Then $\operatorname{cor}(f)(g) = \sum_{x \in X} y\cdot f(y^{-1}gx)$ where $y\in X$ is the unique representative such that $gxH=yH$. Then $\operatorname{cor}(f)$ is a cocycle whose class is the well-defined corestriction of the class of $f$. Cohomology of number fields, section I.5 has the formula for right cosets.<|endoftext|> TITLE: Does this geometry theorem have a name? QUESTION [60 upvotes]: Start with a circle and draw two tangent circles inside. The (black) inner tangent lines to the smaller circles intersect the large circle. The (red) lines through these intersection points are parallel to the (green) outer tangents to the small circles. A long time ago I worked on this theorem, but I never knew the name. Without a name it's difficult to find more information. Does anyone know if this theorem has a name and where I can find more information about it? REPLY [11 votes]: This is theorem 2 (the Parallel tangent theorem) in "Two Applications of the Generalized Ptolemy Theorem" by Shay Gueron.<|endoftext|> TITLE: Exponential derivative of delta distribution? QUESTION [5 upvotes]: This question is from here. I'm asking it here as well to increase the number of people who see it and might be able to help. The question is, what is the result of the following integral for integer $n$ and real $x$? $$\int_{-\infty}^\infty dy\, e^{ny}e^{iyx}$$ Does it diverge and give infinity, or is it actually just equal to the following? $$=\sum_{m=0}^\infty\frac{n^m}{m!}\int_{-\infty}^\infty dy\, y^m e^{iyx}=2\pi\sum_{m=0}^\infty\frac{1}{m!}\left(-in\frac{\partial}{\partial x}\right)^m\delta(x)=2\pi \left(e^{-in\frac{\partial}{\partial x}}\delta(x)\right)$$ More importantly, is there some book or scientific article on distributions that discusses this? Thanks for any suggestion! REPLY [3 votes]: I think Schwartz and Gelfand-Shilov, et al, had already considered the possibility of having "Fourier transform" map from all distributions to the dual of the Paley-Wiener space (the latter being the image of test functions under literal Fourier transform). So although the literal integral you write does not converge (nor do many for tempered distributions, either), it does have a sense compatible with other things.<|endoftext|> TITLE: Picking a new set of primes QUESTION [12 upvotes]: If $S$ is a subset of the set of the positive integers $\mathbb N$, we may consider the set $S^*$ of all products of elements of $S$, allowing for repeated factors —this is a multiset, really, in general. For example, if $S$ is the set of all prime numbers, then $S^*$ is simply $\mathbb N$, with all elements of multiplicity one —this is the fundamental theorem of arithmetic— and in fact this set $S$ is the only one with that property, so that this characterizes the set of prime numbers. If $S$ is not the set of prime numbers, $S^*$ will have holes and there will probably be regions of $\mathbb N$ with extra density. Are there standard useful ways to quantify how far from being the set of primes a set $S$ is in this sense, that is, how irregularly distributed is the set $S^*$ in $\mathbb N$? There are sets $S$ with extremely irregular $S^*$. An example is the set of powers of a number. Once one has a qualitative measure of irregularity, one can make sense of the question: If we pick a random set $S$ (for some sense of «random»), what is the expected irregularity? From $S$ one can construct an "Euler product" $\zeta_S(s)=\prod\limits_{p\in S}\frac{1}{1-\frac{1}{p^s}}$ which is the Riemann zeta function if $S$ is the set of prime numbers. The function $\zeta_S$ is the Dirichlet enumerating function for the multiset $S^*$. Are there analytic properties of the function $\zeta_S$ that characterize the set of primes among the other candidates for $S$? Of course, $\zeta_S=\zeta$ iff $S$ is the set of primes, but I imagine there are much coarser conditions that one can put on $S$ that already imply the same conclusion. The sort of thing one could imagine involves the distributions of values, of zeros, growth conditions, and so on. For example (tweaked after the comments of Steve and Greg below): Does the Riemann Hypothesis hold for $\zeta_S$ iff $S$ is the set of primes up to a finite set? REPLY [5 votes]: As Greg Martin has pointed out this topic of Beurling numbers has been extensively investigated. Usually the Beurling numbers are studied as semigroups contained in ${\Bbb R}$. In the particular context of subsets of the natural numbers, you may wish to look at the work of Lagarias who considers such sets satisfying the "Delone property" that the gaps between consecutive elements in $S^*$ are bounded above and below. One of his results then states that the set $S$ consists of all but finitely many primes plus finitely many composite numbers. Lagarias's paper will also give you other references. Since you mentioned irregularities and the Riemann hypothesis, you may also look at my answer to Why does the Riemann zeta function have non-trivial zeros?<|endoftext|> TITLE: Why Calabi-Yau manifolds should be complex? QUESTION [8 upvotes]: I'm aware that mathematically speaking, Calabi-Yau manifolds are complex manifolds with vanishing first Chern number. However from a physics point of view, Calabi-Yau manifolds are related to the solution of Einstein's field equation in vacuum environment (i.e., with vanishing stress–energy tensor). Since Einstein's field equation is on a 4-dimensional real manifold, why Calabi-Yau manifolds are complex? Is there a "real version" of Calabi-Yau manifold? REPLY [5 votes]: I think that one possible answer is that a Calabi-Yau manifold is a Riemannian manifold $M$ with $SU(n)$ Riemannian holonomy, where $2n=\dim_\mathbb R M$. Such a manifold is then necessarily complex, and the Riemannian metric is the real part of a Kähler metric which has zero Ricci curvature. Since the Ricci form in complex geometry is always a representative of the first Chern class of the manifold, what you ask follows. REPLY [4 votes]: You might consider other real forms of $SU(n)$; a Riemannian manifold with holonomy in such a real form will then arise as extra dimensions in string theories with reduced holonomy and a parallel spinor. The Riemannian manifold need not be complex. But the limit as you make the extra dimensions small will not be Lorentzian, unless the extra dimensions are Riemannian with a parallel spinor. To have a reduced dimension of 4 (for standard general relativity) and a total space dimension of 10 (for type A or type B string theory), you need a 6 dimensional Riemannian manifold of extra dimensions, with a parallel spinor. These conditions give you a Calabi-Yau manifold. If you ask for 7 extra dimensions, you get a G2 holonomy manifold, so extra dimensions are not always complex. There are compact Riemannian Einstein 6-manifolds beside Calabi-Yaus, but not with a parallel spinor.