TITLE: The Three Gap Theorem: why should it be true? QUESTION [14 upvotes]: Fix $\alpha\in\mathbb R$ and $N\in \mathbb N$, consider the set $S(\alpha,N)$ of $\{k\alpha\},k=1,\dots,N$, where $\{x\}$ denotes the fractional part of $x$. Let $a_1,\ldots a_N$ be the elements of $S(\alpha, N)$ arranged in increasing order, and consider the gaps sizes $$ |a_{i+1}-a_i|, i=1,\dots,N-1. $$ The well-known Three Gap Theorem tells us that the number of distinct gap sizes is at most three for any $\alpha$ and $N$. There are many proofs and generalisations of this theorem, but I haven't gained any intuition regarding the statement. In particular, is there any heuristic/intuitive reason as to why the theorem should be true? REPLY [2 votes]: Its all about modular arithmetic. Draw $n$ points $P_0 = 0, P_1, \ldots, P_{n-1}$, with the largest $P_{max}$, and smallest (non-zero) $P_{min}$, with $min, max \in \{1, \dots, n-1\}$. Now, the initial gap, $a = P_{min} - P_0$, is repeated for every point with at least $min$ successors, i.e. the first $n - min$ points. Similarly, the final gap, $b = P_0 - P_{max}$, is repeated for every point with at least $max$ predecessors, i.e. the final $n - max$ points. There is also a middle group of $min + max - n$ points/gaps that get bisected into an $a$ and $b$ every time $n$ increments until $n = min + max$, and no more mid-points remain. The next time that $n$ increments, the new point $P_{min + max} = a - b$ bisects the larger of the initial $a$ and final $b$, and it all starts over. There are usually three gaps, $a$, $b$ and $a + b$, that reduce to two, $a$ and $b$, every time $n = min + max$.<|endoftext|> TITLE: Why $S$ cannot be homeomorphic to the $1$-sphere of $\ell^2$? QUESTION [8 upvotes]: Consider the $\ell^2$ complex Hilbert space. Let $m\in \mathbb{N}^*$ be a fixed number, and set $$ S=\left\{ x=(x_n)_n\subset \ell^2\ :\ \sum_{n=1}^m \frac{|x_n|^2}{n^2}=1\right\}.$$ I want to show that $S$ is not homeomorphic to $$ S(0,1)=\left\{ x=(x_n)_n\subset \ell^2\ :\ \sum_{n=1}^\infty |x_n|^2=1\right\}.$$ REPLY [8 votes]: $S$ contains the ellipsoid $\{x\in \mathbb C^m: \sum \frac{|x_n|^2}{n^2}=1\}$ (which is homeomorphic to $S^{2m-1}$) as a strong deformation retract via $A_t(x_1,\dots,x_m,x_{m+1},\dots) = (x_1,x_2,\dots,x_m,(1-t)x_{m+1},(1-t)x_{m+2},\dots)$ for $t\in[0,1]$, thus the homotopy group $\pi_{2m-1}(S) = \mathbb Z$. $S(0,1)$, however, has $\pi_{2m-1}(S(0,1)) = 0$ since it is contractible: shift first all coordinates one step to the right and then contract in the stereographic chart opposite $(1,0,0,\dots)$. Or use the contraction described on page 513 of here.<|endoftext|> TITLE: Rigidity of doubled convex caps QUESTION [12 upvotes]: Suppose that we have a convex cap, i.e., a convex surface in $R^3$ homeomorphic to a disk whose boundary lies in a plane. Reflect the cap through the plane of its boundary and glue it back to the original cap along the common boundary curve. If the closed surface we obtain is convex, then it is rigid by a theorem of Pogorelov, i.e., any convex surface which is isometric to it is congruent to it. My question is: do we still have rigidity when the doubled surface is not convex? I do not know the answer even in the case where each cap is spherical. Here we want to look at isometric surfaces which are smooth except along the curve where the two pieces meet, so that we do not get trivial examples by reflecting a portion of the surface. In general very little is known about rigidity of non-convex surfaces, but this situation is so simple that perhaps someone might have an idea about it. It seems to me that this is the simplest non-convex case. Addendum: There is a theorem of Alexandrov and Sen'kin, see p. 181 in Pogorelov, which seems to be relevant to the above problem. This theorem states that if a pair of isometric convex surfaces lie in the upper half-space, are star-shaped and concave with respect to the origin, and their corresponding boundary points are equidistant from the origin, then they are congruent. REPLY [3 votes]: Here is one possible approach for convex polyhedra. This is only a sketch, not a proof. Truncate a convex polyhedron below its "equator," which I leave undefined. For a regular octahedron, clip off the lowest vertex:           Prove that this convex cap with boundary is rigid as it stands. I believe Alexandrov's results in Section 5.2 of his book,1 "Flexes of Convex Polyhedra," can establish that this particular cap is rigid. Now join two copies along their common boundary, and argue that this closed polyhedron is rigid.           Perhaps this could establish rigidity of your double spherical cap via polyhedral approximations. 1Alexandrov, Aleksandr D. Convex Polyhedra. Springer Monographs in Mathematics, 2005. Translation of Russian edition, 1950. Section 5.2: p.246ff.<|endoftext|> TITLE: Hilbert series of graded Cohen-Macaulay domains, 28 years later? QUESTION [14 upvotes]: I am reading through Richard Stanley's 1990 paper "On the Hilbert Function of a Graded Cohen-Macaulay Domain" to present in a seminar. I am trying to provide a reasonable conclusion for this talk, and would prefer to mention some open problems. But I am not very well-versed in the literature— and since the paper is relatively old, I am not sure if any of the problems are still open. Questions: In section 1, Stanley mentions that some partial characterization of Hilbert series is known for the class of standard connected Gorenstein rings. Have we done any better since then? The main Theorem 2.1 gives some necessary conditions on the $h$-vector of a semistandard graded Cohen-Macaulay domain. Is this still the state of the art? How about for rings coming from polytopes as in Section 4? In the discussion after Proposition 3.4, he suggests that there are no counterexamples known in positive characteristic. Has this been proven, or counterexamples found? I've skimmed through one level mathscinet citations, to no avail. REPLY [5 votes]: Perhaps this is too late for your seminar, but there has been a huge literature on these problems and related ones. The key words to search are: "h-vector of...". For example, here is a paper that deals with Gorenstein domains of codimension three. Here is a paper on h-vector of Gorenstein toric ring, which Section 4 of Stanley's paper addressed. In general, even for Artinian Gorenstein algebras, a complete chracterization of h-vector seems to be out of reach, but there are many surveys and works are being posted at this very moment.<|endoftext|> TITLE: Can an enriched functor be expressed as a colimit of representable functors? QUESTION [11 upvotes]: Suppose that $\mathcal C$ is an ordinary category and $F:\mathcal C^{op}\longrightarrow Set$ a functor. Then, we can form the category $\mathcal C/F$ as follows : each object is a morphism of functors from $h_X\longrightarrow F$, where $X\in \mathcal C$ and $h_X=Hom(\_\_,X):\mathcal C^{op}\longrightarrow Set$ is the representable functor. In other words, we are looking at the category of pairs $(X,x)$, where $X\in \mathcal C$ and $x\in F(X)$. This is clear from Yoneda lemma. Then, we can write down the functor $F$ as a colimit : $$F=\underset{(X,x)\in \mathcal C/F}{colim}\textrm{ }h_X $$ My question is whether we have a version of this for enriched categories. In other words, suppose that $\mathfrak D$ is a category enriched over a closed symmetric monoidal category $(\mathcal V,\otimes,1)$. Let $\mathcal F:\mathfrak D^{op}\longrightarrow \mathcal V$ be an enriched functor. For each $X\in \mathfrak D$, we already have a representable functor: $$H_X: \mathfrak D^{op}\longrightarrow \mathcal V\qquad H_X(Y):=\mathfrak D(Y,X)$$ for any $Y\in \mathfrak D$. Can we construct a category $\mathfrak D/\mathcal F$ in this case and express $\mathcal F$ itself as a colimit of the form: $$\mathcal F = \underset{??}{colim}\textrm{ }H_X$$ Of course the key must be an enriched form of Yoneda lemma. In the case of enriched categories, there are 2 forms of Yoneda lemma, the weak form and the strong form. I would prefer if the answer can be given with the help of the weak form. Of course it would be great if there is a reference where this formula is clearly explained. Thanks! REPLY [9 votes]: The answer is that every enriched presheaf can be expressed as a weighted colimit of representables. The general definition of a weighted colimit goes as follows. Let $\mathcal{A}$ and $\mathcal{C}$ be $\mathcal{V}$-categories, let $F \colon \mathcal{A} \to \mathcal{C}$ be a $\mathcal{V}$-functor (the "diagram"), and let $W \colon \mathcal{A}^\mathrm{op} \to \mathcal{V}$ be a $\mathcal{V}$-functor (the "weight"). A colimit of $F$ weighted by $W$ is an object $W \ast F$ of $\mathcal{C}$ together with a $\mathcal{V}$-natural isomorphism $$\mathcal{C}(W\ast F,C) \cong [\mathcal{A}^\mathrm{op},\mathcal{V}]\left(W,\mathcal{C}(F-,C)\right),$$ where $[\mathcal{A}^\mathrm{op},\mathcal{V}]$ denotes the $\mathcal{V}$-category of $\mathcal{V}$-enriched presheaves on $\mathcal{A}$. To turn to your question, let $\mathcal{A}$ be a small $\mathcal{V}$-category and let $W \colon \mathcal{A}^\mathrm{op} \to \mathcal{V}$ be a $\mathcal{V}$-functor (a "$\mathcal{V}$-enriched presheaf"). It follows from the (strong) enriched Yoneda lemma that $W$ is the colimit of the Yoneda functor $Y \colon \mathcal{A} \to [\mathcal{A}^\mathrm{op},\mathcal{V}]$ weighted by $W$; i.e. $W \cong W\ast Y$. When $\mathcal{V}=Set$, there is a general formula for expressing a weighted colimit as an ordinary "conical" colimit over the category of elements of the weight, which in the above case reproduces the formula you give in your question. For a reference, see around display (3.17) of Kelly's Basic concepts of enriched category theory (that's page 41 of the TAC reprint) or Theorem 6.6.18 of volume 2 of Borceux's Handbook of categorical algebra. You may also wish to read Chapter 7 of Riehl's Categorical Homotopy Theory for a treatment of weighted colimits.<|endoftext|> TITLE: Can infiniteness of finitely generated groups be read by a "paradoxical" decomposition? QUESTION [5 upvotes]: (Edit) Let $G$ be a group. Two subsets $A,B$ of $G$ are said to be equidecomposable if there exists a finite partition $A=\bigsqcup_{i=1}^nA_i$ and $a_i\in G$ such that $B=\bigsqcup_{i=1}^na_iA_i$. Say that a group has Property (X) if it has a subset equidecomposable to a proper subset of itself. Clearly this implies being infinite. The negation of (X) passes to subgroups. The group $\mathbb{Z}$ has (X) (for $n=1$, ans hence all non-torsion (=non-periodic) groups have (X). Using a paradoxical decomposition, all non-amenable groups have (X). Does every infinite finitely generated amenable group have (X)? The only remaining cases are periodic and amenable. REPLY [8 votes]: The answer is yes: every infinite finitely generated group has a subset which is equidecomposable to a proper subset of itself. This follows from a theorem of Brandon Seward: Every finitely generated infinite group $G$ admits a translation-like action by the group $\mathbb{Z}$ of integers. https://arxiv.org/abs/1104.1231 What this means is that there exists a free action of $\mathbb{Z}$ on $G$ such that each element of $\mathbb{Z}$ acts as an element of the wobbling group of $G$. That is, there exists a bijection $f:G\rightarrow G$ having no finite orbits, along with a finite partition $G=\bigsqcup_{i=1}^nX_i$ and finitely many group elements $a_1,\dots , a_n\in G$ such that $f(x)=a_ix$ for all $x\in X_i$. If we take $T$ to be a subset of $G$ containing exactly one point from each orbit, and define $A$ to be the set $A = \bigsqcup _{n\geq 0} f^n(T)$, then $f(A)$ is a proper subset of $A$, and the partition $A=\bigsqcup _{i=1}^n A_i$, where $A_i := A\cap X_i$, along with the group elements $a_1,\dots , a_n$, witness that $A$ is equidecomposable with $f(A)$. Edit: I realized there is also a simple direct argument which also shows that $n$ is bounded above by the minimum size of a finite subset $S$ of $G$ which generates an infinite subsemigroup of $G$. Consider the (left) Cayley graph of this subsemigroup $H$ with respect to its generating set $S$, i.e., with vertex set $H$ and with a directed edge from $h$ to $sh$ for each $s\in S$ and $h\in H$. This graph is infinite and locally finite, so by Kőnig's lemma it contains an infinite geodesic ray, say with vertices $h_0,h_1,\dots$. Let $A := \{ h_0,h_1,h_2,\dots \}$ and let $B=A \setminus \{ h_0 \}$. For each $n\geq 0$ there is some $s_n\in S$ such that $h_{n+1}=s_nh_n$, so $A$ is partitioned into finitely many sets $A_s$, $s\in S$, where $A_s$ consists of those $h_n$ for which $s_n=s$. Then $B$ is partitioned by $sA_s$, $s\in S$, so $A$ and $B$ are equidecomposable.<|endoftext|> TITLE: Is there an "analytical" version of Tao's uncertainty principle? QUESTION [17 upvotes]: Let $p$ be a prime. For $f: \mathbb{Z}/p \mathbb{Z} \rightarrow \mathbb{C}$ let its Fourier transform be: $$\hat f(n) = \frac{1}{\sqrt{p}}\sum_{x \in \mathbb{Z}/p \mathbb{Z}} f(x)\, e\left(\frac{-xn}{p}\right)$$ Terence Tao proves in "An uncertainty principle for cyclic groups of prime order" that for $f: \mathbb{Z}/p \mathbb{Z} \rightarrow \mathbb{C}$ one has: $$|S_1| + |S_2|\ge p+1$$ If $f$ is supported in $S_1$ and $\hat f$ is supported in $S_2$. This is much stronger than the traditional bound: $$|S_1||S_2| \ge p$$ Can we prove an approximate version of this inequality, say, in which we lower bound $$|S_1| + |S_2|$$ assuming only that $S_1$ and $S_2$ contain most of the mass of $f$ and $\hat f$ (rather than all of the mass)? The proof of the lower bound to $|S_1||S_2|$ is "only analytical", so it caries over under this weaker assumption, but Tao's lower bound for $|S_1|+|S_2|$ relies strongly on algebra, in that it is important that $f$ and $\hat f$ are exactly $0$ outside the support, so the proof doesn't immediately carry over. Is there such an approximate version of Tao's uncertainty principle? For a concrete case, if the Fourier transform is normalized so that $\sum |f(x)|^2 = \sum |\hat f(x)|^2 = 1$, can we rule out the possibility that $\sum_{x \in S_1} |f(x)|^2 > 1-\epsilon$, and $\sum_{x\in S_2} |\hat f(x)|^2 > 1-\epsilon$ with $|S_1|, |S_2| < \epsilon p$? For the applications that I am interested I need something quite weaker, but I couldn't really find a counter-example to the above possibility. Thanks in advance REPLY [17 votes]: One does not need Gaussians in the finite case, just take $f$ to be the indicator function of the interval $[-(n-1),n-1]\subset\mathbb F_p$. A simple computation gives $$ |\hat f(x)| = \frac1{\sqrt p} \frac{|\sin\pi(2n-1)x/p|}{|\sin\pi(x/p)|} < \frac1{2\sqrt p\|x/p\|}, $$ where $\|x/p\|$ is the distance to $x/p$ from the nearest integer. As a result, $$ \sum_{x=m}^{p-m} |\hat f(x)|^2 \le \frac p{2m} = \frac p{2mn} \sum_{x=0}^{p-1} |\hat f(x)|^2. $$ Taking, say, $m,n\approx\sqrt{\varepsilon^{-1}p}$, you have both $f$ and $\hat f$ concentrated on intervals of length about $2\sqrt{\varepsilon^{-1}p}$. A historical point worth clarification. The inequality $|{\rm supp}\, f|+|{\rm supp}\,\hat f|\ge p+1$ was in fact proved first by Andras Biro, who has contributed it as a problem to the 1998 Schweitzer competition (Problem 3). Bearing in mind that Tao's paper appeared in 2005, I think it would be most reasonable to call it the Biro-Tao inequality.<|endoftext|> TITLE: Why is subcoalgebra structure unique? QUESTION [10 upvotes]: Sorry for breaking the harmony of MO with a easy and silly questions. I have stuck on elementary category-theoretic reasoning about subcoalgebras, namely: Let $F:\mathcal{Set}\to \mathcal{Set}$ be a functor. An $F$-coalgebra is a pair $\mathcal{A}=(A,\alpha)$ where $\alpha:A\to F(A)$ is an arbitrary map. Given $F$-coalgebras $\mathcal{A}=(A,\alpha)$ and $\mathcal{B}=(B,\beta)$, a homomorphism $\varphi:\mathcal{A}\to \mathcal{B}$ is a map $\varphi:A\to B$ which make the obvious corresponding square commute. A subset $U\subseteq A$ is called closed, if an $F$-coalgebra structure $\mathcal{U}=(U,\delta)$ can be defined on $U$ so that the natural inclusion $\subseteq :U\to A$ is a homomorphism. In this case $\mathcal{U}$ is called subcoalgebra of $\mathcal{A}$. A coalgebra literature indicates that coalgebra structure map $\delta$ is easily seen to be unique. I have stuck on uniqueness, what is the reason, how it can be seen? Is it category-theoretic reason or rather set-theoretic? REPLY [10 votes]: Let us say that a monomorphism $U\subseteq A$ is nontrivial if $U$ is nonempty. Since every nontrivial monomorphism in $Set$ is split, every $F : Set\to Set$ takes nontrivial monomorphisms to monomorphisms. In particular, your $F(\subseteq)$ is again a monomorphism in this case. So if you have two coalgebra structures on $U\subseteq A$ that make the square commute, then they must be equal, since you can cancel $F(\subseteq)$ from the left. The only case left to treat is the trivial case $U = \emptyset$. But then the coalgebra structure is unique by the initiality. The uniqueness of the subcoalgebra structure is false in some categories other than $Set$. For example on $Ab$, consider the functor $-\otimes\mathbb{Z}_2$. The quotient map $\mathbb{Z}\to\mathbb{Z}_2$ makes $\mathbb{Z}$ into a coalgebra. Applying the functor to the inclusion of the even integers $\mathbb{Z}\subseteq\mathbb{Z}$ results in the zero map. Therefore the even integers $\mathbb{Z}$ are a subcoalgebra in two different ways: either via the projection or via the zero map $\mathbb{Z}\to\mathbb{Z}_2$.<|endoftext|> TITLE: Is the category of symmetric bimodules over a commutative ring closed under extensions? QUESTION [6 upvotes]: Let $A$ be a commutative ring, and consider the category of bimodules over $A$. An $A$-bimodule $M$ is called symmetric if $a\cdot m = m \cdot a$ for all $a \in A$, $m \in M$. Is the category of symmetric bimodules over $A$ closed under extensions? Namely, given an exact sequence of $A$-bimodules $0 \to M \to N \to K \to 0$ where $M,K$ are symmetric, must $N$ also be a symmetric $A$-module? REPLY [2 votes]: Here a general abstract argument that shows that the subcategory of symmetric bimodules is never extension closed in case $A$ is a commutative finite dimensional Frobenius algebra that is not semisimple: Let $A$ be such an algebra and $B=A \otimes_K A$ its enveloping algebra and assume that the subcategory of symmetric bimodules is closed under extensions. The simple module $S$ is symmetric and thus the subcategory of symmetric finite dimensional bimodules equals the module category of $B$. It thus also contains $B$, but $Hom_B(A,B) \cong D(A) \cong A$ has dimension less than $B$ and thus $B$ is never symmetric. This is a contradiction and thus the subcategory of symmetric bimodules is never closed under extensions.<|endoftext|> TITLE: Simultaneous failure of weak diamond QUESTION [5 upvotes]: Let $\lambda$ be an infinite cardinal. Recall that Weak diamond $\Phi_S$ on $S\subseteq\lambda^+$ is the following principle: For every function $F:2^{<\lambda^+}\rightarrow 2$, there exists $g\in 2^{\lambda^+}$ such that for all functions $f:\lambda^+\rightarrow 2$, the set $\{\alpha\in S:~F(f\restriction\alpha)=g(\alpha)\}$ is stationary. The following theorem is well known: Theorem: (Devlin-Shelah) For every infinite $\lambda$, $\Phi_{\lambda^+}\Leftrightarrow 2^\lambda<2^{\lambda^+}$. Unlike the above situation, failure of the weak diamond at a stationary set is not enough to get $2^\lambda=2^{\lambda^+}$. In fact, we have the following. Theorem: (Shelah) The theory ${\rm ZFC}+\neg\Phi_{S^{\omega_2}_\omega}+2^{\omega_1}<2^{\omega_2}$ is consistent. where $S^{\omega_2}_{\omega}=\{\alpha<\omega_2: {\rm cof}(\alpha)=\omega\}$, and similarly for $S^{\omega_2}_{\omega_1}=\{\alpha<\omega_2:{\rm cof}(\alpha)=\omega_1\}$ we have: Theorem:(Shelah) The theory ${\rm ZFC}+\neg\Phi_{S^{\omega_2}_{\omega_1}}+{\rm GCH}$ is consistent. Now in the light of above theorems, I would like to know what is the answer to the following question. Question: Does simultaneous failure of $\Phi_{S^{\omega_2}_{\omega }}$ and $\Phi_{S^{\omega_2}_{\omega_1}}$ imply $2^{\aleph_1}=2^{\aleph_2}$? Edit: For a proof of 2nd theorem look at Shelah's paper "More on the weak diamond"-1985. A reference for the 3rd one was given in the comments by Golshani. REPLY [3 votes]: Yes, it does. The collection $I^{\omega_2}_{WD}= \{ S \subset \omega_2: \neg \Phi^{\omega_2}_S \}$ of subsets of $\omega_2$ is an ideal (a proof this fact is provided in the proposition below; moreover I think this result is originally due to Shelah, however a reference escapes me at the moment.) Moreover, $\Phi_{\omega_2}$ fails, precisely when $I^{\omega_2}_{WD}$ is not a proper ideal, which in this case is equivalent to some club subset of $\omega_2$ being a member of $I^{\omega_2}_{WD}$. So if $\Phi^{\omega_2}_S$ fails for both $E^{\omega_2}_\omega$ and $E^{\omega_2}_{\omega_1}$. Then $E^{\omega_2}_{\omega}\cup E^{\omega_2}_{\omega_1} =\lim(\omega_2) \in I^{\omega_2}_{WD}$ and this implies the failure of $\Phi_{\omega_2}$, which is equivalent to $2^{\aleph_1} = 2^{\aleph_2}$. Proposition: For any cardinal $\lambda$, the set $I^{\lambda^{+}}_{WD} = \{S \subset \lambda^{+}: \neg \Phi^{\lambda^{+}}_S \}$ is a $\lambda^{+}$-complete ideal extending the non-stationary ideal on $\lambda^{+}$. Proof: That $I_{WD}^{\lambda^{+}}$ contains the non-stationary ideal on $\lambda^{+}$ follows by definition. As such, all that remains is to show $I^{\lambda^{+}}_{WD}$ is an appropriately complete ideal. To this end suppose $S \in I^{\lambda^{+}}_{WD}$ as witnessed by the sequence $\langle F_\alpha: \alpha \in S \rangle$ where for each $\alpha \in S$, $F_{\alpha}:\,^{\alpha}2\rightarrow 2$. Then, for every $g:\lambda^{+} \rightarrow 2$, there is some $f:\lambda^{+} \rightarrow 2$ and club $C_g \subset \lambda^{+}$ such that $$ \{ \alpha \in S \cap C_g: F_\alpha(f\vert_\alpha) = g(\alpha) \} = \emptyset $$ as such, for every $S_0 \subset S$ we have $ \{ \alpha \in S_0 \cap C_g: F_\alpha(f\vert_\alpha) = g(\alpha) \} = \emptyset $, hence the sub-sequence $\langle F_\alpha : \alpha \in S_0\rangle$ witnesses $S_0 \in I^{\lambda^{+}}_{WD}$. Next, fix a bijection $\varphi:\lambda^{+}\times \lambda^{+}\rightarrow \lambda^{+}$ and let $C = \{ \alpha \in \lim(\lambda^{+}): \varphi[\alpha \times \alpha] = \alpha \}$ be the club subset of $\lambda^{+}$ consisting of closure points of $\varphi$. Now, assume $\{ S_\gamma: \gamma \in \delta \} \subset I_{WD}^{\lambda^{+}}$ (with $\delta \in \lambda^{+}$) are stationary, pairwise disjoint, and that for each $\gamma\in \delta$, the sequence $\langle F^{\gamma}_\alpha : \alpha \in S_\gamma \rangle$ witnesses the failure of $\Phi^{\lambda^{+}}_{S_\gamma}$. Letting $E = \bigcup \{ S_\gamma : \gamma \in \delta \}$, define the sequence $\langle F^{\ast}_\alpha : \alpha \in E\rangle$ of functions $F^{\ast}_\alpha : \,^{\alpha}2\rightarrow 2$ as follows, given $\alpha \in E$ and $h \in \,^{\alpha}2$: if $\alpha \in C\cap S_\gamma$, set $F^{\ast}_{\alpha}(h) = F^{\gamma}_{\alpha}([h]_\gamma)$, where $[h]_\gamma$ denotes the function defined by $[h]_\gamma(\xi) = h(\varphi(\gamma,\xi))$, and if $\alpha \not\in C$, simply take $F^{\ast}_{\alpha}(h) = 0$. Now, for every $g:\lambda^{+} \rightarrow 2$, there are functions $\{ h_\gamma : \gamma \in \delta \} \subset \,^{\lambda^{+}}2$ and clubs $\{ C_\gamma: \gamma \in \delta \} \subset \mathcal{P}(\lambda^{+})$ such that, for each $\gamma \in \delta$, $$ \{ \alpha \in S_\gamma \cap C_\gamma : F^{\gamma}_{\alpha}(h_\gamma\vert_\alpha) = g(\alpha) \} = \emptyset. $$ So, letting $D=\cap \{ C_\gamma : \gamma \in \delta \}$ and defining $h:\lambda^{+}\rightarrow 2$ by $h(\varphi(\gamma,\xi))=h_\gamma(\xi)$ for $\gamma \in \delta$, and $h(\varphi(\gamma, \xi)) = 0$ otherwise; we have $\alpha \in S_\gamma \cap C \cap D\implies $ $$ F^{\ast}_{\alpha}(h\vert_\alpha) = F^{\gamma}_{\alpha}([h]_{\gamma}\vert_\alpha)=F^{\gamma}_{\alpha}(h_{\gamma}\vert_\alpha)=1 - g(\alpha).$$ Therefore, $ \{ \alpha \in E\cap C \cap D: F^{\ast}_{\alpha}(h\vert_\alpha) = g(\alpha) \} = \emptyset $, and it immediately follows that $$E=\cup \{ S_\gamma: \gamma \in \delta \} \in I^{\lambda^{+}}_{WD}$$ $\square$.<|endoftext|> TITLE: expected length of a largest cycle in regular graph QUESTION [5 upvotes]: Consider a simple random regular graph $G_d(n)$ with $d=2$ (that is, I select a $2-$regular graph from the set of all $2-$regular graphs on $n$ vertices, uniformly at random). It is clear that this graph consists of (not-intersecting) cycles only; and asymptotically a.s., it does not contain a cycle of length $n$. My question is: Can we characterize the length of its largest cycle? REPLY [3 votes]: I'm not going to do this calculation, but here is how it can be done. Any interval which contains the length of the longest cycle with probability tending to 1 will also contain the length of the longest cycle of a random pairing (configuration) with probability tending to 1. And vice-versa. With a random pairing you can easily calculate the expectation of the number of sequences of cycles with given lengths. From this the distribution of the number of cycles longer than a given length will follow. I'm not sure how messy the details will be, but it's easier than usual calculations in the pairing model because cycles cannot partially overlap.<|endoftext|> TITLE: indecomposable semi local Bezout ring QUESTION [5 upvotes]: A commutative Bezout ring with $1$ is a ring in which every finitely generated ideal is principal. Is there any charactrization for indecomposable semi local Bezout ring? For example , for a commutative (indecomposable) local rign is Bezout ring if and only if it is a chain ring. REPLY [2 votes]: Some partial ideas: If semilocal is changed to semiperfect, then you can use the fact that commutative semiperfect rings are just products of local rings to extend the fact you mentioned in a different way: A commutative, semiperfect ring is Bézout iff it is a serial ring (Serial ring of course means it is a finite direct product of uniserial submodules, which in the case of commutative rings amounts to a finite product of chain rings.) Things don't seem so nice for semilocal rings that aren't semiperfect. It's easy to create semilocal PIDs which have complex lattices of ideals. For example, you can take a finite collection of maximal primes in $\mathbb Z$ and semilocalize. Commutative Bézout rings are distributive, though, and I don't immediately know if a semilocal, indecomposable, distributive ring has to be Bézout or not. That might be something to explore.<|endoftext|> TITLE: Confusion about a proof from Goresky and MacPherson's "Intersection Homology II" QUESTION [6 upvotes]: Context My question is about the "proof of claim" on page 84 of Goresky and MacPherson's "Intersection Homology II". For ease of reading, here's the claim: Claim: Suppose $X$ is a topological pseudomanifold$^*$ and $\mathbf A^\bullet$ is a topologically constructible$^*$ complex of sheaves on $X$. For any $x\in X$ there is a neighborhood basis $U_1\supset U_2\supset U_3\supset\ldots$ such that for each $i$ and $m$, the restriction map $$H^i(U_m; \mathbf{A}^\bullet) \to H^i(U_{m+1}; \mathbf{A}^\bullet)$$ is an isomorphism. $^*$ See [$\S$1.1 and $\S$1.4, loc. cit.] for the definitions of topological pseudomanifold and topologically constructible. My Question The only part of the proof I'm having issues with is the very beginning, where the authors construct the neighborhood basis itself. They start by choosing a distinguished neighborhood $N\cong \Bbb R^i\times \operatorname{cone}^\circ(L)$ of the point $x$. They then consider the join $Y=S^{i-1}*L$ and assign it a stratification. I'm OK so far. However, they then say, "Choose a stratum preserving homeomorphism $\psi\colon \operatorname{cone}^\circ(Y)\to N$ with $\psi(\text{vertex})=x$." Question: (1) Why are $\operatorname{cone}^\circ(Y)$ and $N$ homeomorphic, and (2) why does there exist a stratum preserving homeomorphism (with $\psi(\text{vertex})=x$) between them? REPLY [5 votes]: As noted by Chris Gerig in the comments, letting $cX$ denote the open cone on the compact space $X$ then $(cX)\times (cY)\cong c(X*Y)$, where $X*Y$ is the join. In the case at hand, Goresky and MacPherson are treating $\mathbb{R}^i$ as $cS^{i-1}$. When $X$ and $Y$ are stratified, there is a natural stratification of the join. I discuss this in Section 2.11 of my book - draft currently available here: http://faculty.tcu.edu/gfriedman/IHbook.pdf This said, the typical stratification of the cone has the vertex as a $0$-dimensional stratum, which can't be the case here thinking of the distinguished neighborhood $N$ as a cone (unless $i=0$). So I would say there isn't quite a stratum preserving homeomorphism here in the way you would expect from the usual definitions. They must be thinking of $cS^{i-1}$ as a single stratum. All of this said, if you're more interested in the theorem in whose proof this statement appears, you can find a much more thorough treatment of the constructibilty issues in Borel's book "Intersection Cohomology," particularly Section V.3. In particular, the Claim you cite is contained in Borel's Proposition V.3.10 on page 77. Note that Goresky and MacPherson make constructibility assumptions, but Borel shows that these assumptions all follow from the definition of the Deligne sheaf and so don't need to be assumed.<|endoftext|> TITLE: Which compact (finite dimensional) Lie groups have $H^1_{DR}(G)\neq 0$ QUESTION [5 upvotes]: In particular, I am wondering if $H^1_{DR}(G)\neq 0$ implies that the group can written as a semidirect product of $\mathbb{S^1}$ and something else, with the $\mathbb{S^1}$ factor being responsible for the first cohomology group. I have no idea if this is true. In case it is not clear by $H^1_{DR}(G)$, I mean the De Rham cohomology of the underlying manifold associated to a group $G$. (I understand this is isomorphic to some kind of Lie algebra cohomology, but as of now know nothing about Lie algebra cohomology). Note that I originally asked this on mathstackexchange and received no answers so I hope this was because perhaps it was more appropriate for here... I have deleted the question on stackexchange. REPLY [10 votes]: The de Rham cohomology is non-trivial if and only if the connected component of $G$ contains a non-trivial central torus. We can assume that $G$ is connected. Then $H^1_{dR}(G)$ is isomorphic with $H_1(G,{\mathbb R})^*$ by Poincare duality. By Hurwitz's Theorem, $H_1(G)$ is the maximal abelian quotient of $\Gamma$, so $H^1_{dR}(G)\cong Hom(\Gamma,{\mathbb R})$, where $\Gamma$ is the fundamental group. Now by, say, Theorem 7.1 of the book by Broecker and tom Dieck, the fundamental group is a quotient of two lattices, so is of the form $F\times {\mathbb Z}^r$, where $F$ is a finite abelian group and $r$ is the dimension of a maximal central torus in $G$. It follows that $H_{dR}^1(G)\cong{\mathbb R}^r$.<|endoftext|> TITLE: Epimorphisms from the genus $2$ surface braid group to finite groups QUESTION [6 upvotes]: This question is somehow related to my previous MO question Explicit description of a subgroup of the braid group $\mathsf{B}_2(C_2)$; for the reader convenience, let me write down again the relevant set-up. Let $C_2$ be a smooth curve of genus $2$ and $X:=\mathrm{Sym}^2(C_2)$ its second symmetric product. If $\delta \subset X$ is the diagonal, then the topological fundamental group $\pi_1(X-\delta)$ is isomorphic to the braid group $\mathsf{B}_2(C_2)$ on two strands on $C_2$. Such a group is generated by five elements $a_1, \, a_2, \, b_1, \, b_2, \, \sigma$ subject to the following set of relations: \begin{equation*} \begin{split} (R2) \quad & \sigma^{-1} a_1 \sigma^{-1} a_1= a_1 \sigma^{-1} a_1 \sigma^{-1} \\ & \sigma^{-1} a_2 \sigma^{-1} a_2= a_2 \sigma^{-1} a_2 \sigma^{-1} \\ & \sigma^{-1} b_1 \sigma^{-1} b_1 = b_1 \sigma^{-1} b_1 \sigma^{-1} \\ & \sigma^{-1} b_2 \sigma^{-1} b_2 = b_2 \sigma^{-1} b_2 \sigma^{-1}\\ & \\ (R3) \quad & \sigma^{-1} a_1 \sigma a_2 = a_2 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} b_1 \sigma b_2 = b_2 \sigma^{-1} b_1 \sigma \\ & \sigma^{-1} a_1 \sigma b_2 = b_2 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} b_1 \sigma a_2 = a_2 \sigma^{-1} b_1 \sigma \\ & \\ (R4) \quad & \sigma^{-1} a_1 \sigma^{-1} b_1 = b_1 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} a_2 \sigma^{-1} b_2 = b_2 \sigma^{-1} a_2 \sigma \\ & \\ (TR) \quad & [a_1, \, b_1^{-1}] [a_2, \, b_2^{-1}]= \sigma^2. \end{split} \end{equation*} The geometric interpretation for the above generators of $\mathsf {B}_2(C_2)$ is the following. The $a_i$ and the $b_i$ are the braids coming from the representation of the topological surface associated with $C_2$ as a polygon of $8$ sides with the standard identification of the edges, whereas $\sigma$ is the classical braid generator on the disk. In terms of the isomorphism with $\pi_1(X-\delta)$, the element $\sigma$ corresponds to the homotopy class in $\textrm{Sym}^2(C_2)-\delta$ of a topological loop that "winds once around $\delta$". For more details see P. Bellingeri's paper On presentations of surface braid groups, Journal of Algebra 274 (2004), 543-563. For some research problems related to algebraic surfaces, I would like to construct some group epimorphism $$\varphi \colon \mathsf {B}_2(C_2) \longrightarrow G, \quad (\ast)$$ where $G$ is a finite group. I also want that the element $s :=\varphi(\sigma)$ is not the identity of $G$. Making some experiments with GAP4, I discovered that, up to order $|G|=64$, if $\varphi$ exists than the order of $s$ is at most $2$. This was a surprise, since I do not see anything in the presentation of $\mathsf {B}_2(C_2)$ forcing this behaviour. So I wonder if this happens just because $64$ is too small or if, instead, I am missing some conceptual point here, maybe some basic result of combinatorial group theory I am not aware of. Of course, the time requested for the machine computations grows rapidly with the order of $G$, hence it is not possible to systematically check all cases when such a order becomes too large (already, $64$ takes a lot of time). So, let me ask the following Q. Is it possible to construct a group epimorphism of type $(*)$, such that the order of $s:=\varphi(\sigma)$ is at least $3$? If yes, what is the minimum order of $G$ such that this happens? If not, why? Note. Here is the Gap4 script I used to define $\mathsf {B}_2(C_2)$: F:=FreeGroup("a1", "b1", "a2", "b2", "s"); a1:=F.1; b1:=F.2; a2:=F.3; b2:=F.4; s:=F.5; R1 := s^(-1)*a1*s^(-1)*a1*(a1*s^(-1)*a1*s^(-1))^(-1); R2 := s^(-1)*a2*s^(-1)*a2*(a2*s^(-1)*a2*s^(-1))^(-1); R3 := s^(-1)*b1*s^(-1)*b1*(b1*s^(-1)*b1*s^(-1))^(-1); R4 := s^(-1)*b2*s^(-1)*b2*(b2*s^(-1)*b2*s^(-1))^(-1); R5 := s^(-1)*a1*s*a2*(a2*s^(-1)*a1*s)^(-1); R6 := s^(-1)*b1*s*b2*(b2*s^(-1)*b1*s)^(-1); R7 := s^(-1)*a1*s*b2*(b2*s^(-1)*a1*s)^(-1); R8 := s^(-1)*b1*s*a2*(a2*s^(-1)*b1*s)^(-1); R9 := s^(-1)*a1*s^(-1)*b1*(b1*s^(-1)*a1*s)^(-1); R10 := s^(-1)*a2*s^(-1)*b2*(b2*s^(-1)*a2*s)^(-1); R11 := a1*b1^(-1)*a1^(-1)*b1*a2*b2^(-1)*a2^(-1)*b2*s^(-2); Br:=F/[R1, R2, R3, R4, R5, R6, R7, R8, R9, R10, R11]; REPLY [7 votes]: I have found a finite image of order $3^9$ in which $\sigma$ has nontrivial image, by using the $\mathtt{pQuotient}$ function which, for a given prime $p$ and class $n$, computes the largest quotient of the group which is a $p$-group of exponent $p$-class at most $n$. gap> pq := PQuotient( Br, 3,2); <3-quotient system of 3-class 2 with 9 generators> gap> phi := EpimorphismQuotientSystem(pq); [ a1, b1, a2, b2, s ] -> [ a1, a2, a3, a4, a5 ] gap> Image(phi, Br.5); a5 gap> Order(Image(phi, Br.5)); 3 In fact this quotient has elementary abelian centre of order $3^5$, so there are certainly solutions to your problem of size $3^5$, and perhaps smaller. But factoring out a normal subgroup of order $3^4$ that lies in the centre and does not contain the image of $\sigma$ gives an extraspecial group of order $3^5$. REPLY [6 votes]: Consider $C_2 \times C_2$ minus the diagonal, this is a double cover of your space. Given a symplectic form on $H_1( C_2 \times C_2 - \Delta, \mathbb Z/p)$, we can form an associated extension $1 \to G \to H_1( C_2 \times C_2, \mathbb Z/p)$ as a Heisenberg group, which will be a quotient of $\pi_1$ if and only if the image of the obstruction class in $H^2 ( \pi_1(C_2 \times C_2 - \Delta), \mathbb Z/p) = H^2 ( C_2 \times C_2 - \Delta, \mathbb Z/p)$ vanishes. The obstruction class should be the class induced by viewing symplectic forms on on $H_1$ as elements of $\wedge^2 H^1$ and applying the cup product. To ensure that $\sigma$ has order $p$ on $G$, it is sufficient to show that the obstruction class does not vanish in $H^2( C_2 \times C_2 - \Delta, \mathbb Z/p)$. In other words, we want the obstruction class to be a multiple of the class of $\Delta$. This is possible as every class in $H^2$ is a cup product. Then because $C_2 \times C_2 - \Delta$ is a double cover of your $X - \delta$, it's fundamental group will map to the wreath product of $\mathbb Z/2$ with $G$, and probably something simpler like a semidirect product if we can choose the form to be invariant. The minimal group size for this method to work is something like $2 \cdot 3^9$, which explains why it wasn't seen by your brute-force search.<|endoftext|> TITLE: Basic questions about formal schemes QUESTION [10 upvotes]: I have some questions related to formal schemes. Essentially I would like to understand how much formal schemes are different from usual schemes. First of all, let me specify the definition I prefere of formal schemes. I call a locally ringed space $(\mathfrak{X},\mathcal{O}_{\mathfrak{X}})$ a locally Noetherian formal scheme if it admits a covering given by affinoid formal schemes ${\text{Spf}(A_i)}$, where each $A_{i}$ is a Noetherian adic ring with ring of definition $I_i$. 1) Is there a functorial interpretation of formal schemes? I mean, is it possible to say, like for schemes, that a formal scheme is the representing object of a functor from the opposite category of Noetherian adic rings to set? I guess it is possible, since we know that $\text{Hom}_{\text{FSchemes}}(\mathfrak{X},\text{Spf}(A))\simeq\text{Hom}_{\text{ cont }}(A,\Gamma(\mathfrak{X},\mathcal{O}_{\mathfrak{X}}))$. I know it is possible to introduce quasi-coherent sheaves in the context of formal schemes. For affinoid formal schemes, quasi-coherent sheaves are given by completions of usual modules, and it is in general possible to glue these sheaves over the whole formal scheme. I guess that also the definition of a quasi-coherent sheaf of adic algebras goes in the same way. 2) Is there a notion of relative affinoid spectrum? I mean, consider a formal scheme $(\mathfrak{X},\mathcal{O}_{\mathfrak{X}})$ and a quasi coherent sheaf of adic $\mathcal{O}_{\mathfrak{X}}$-algebras $\mathcal{E}$. Is it possible to define $\underline{\text{Spf}}_{\mathfrak{X}}(\mathcal{E})$ simply glueing along the intersections of affine pieces things like $\text{Spf}(\Gamma(\mathfrak{X},\mathcal{O}_{\mathfrak{X}}))$? Again I cannnot see any problem in this construction. I am particularly interested in this construction expecially in the case when I construct line bundles. In fact, once we have a notion of coherent sheaf, we also have a notion of invertible sheaf, as coherent modules locally free of rank $1$. So, given an invertible sheaf $\mathfrak{L}$, we can take $\text{Sym}(\mathfrak{L})$. Now, since the construction of symmetric algebra is not left exact, maybe we have to complete before defining $\underline{\text{Spf}}_{\mathfrak{X}}(\widehat{\text{Sym}(\mathfrak{L})})$, where hat means completion. Does this construction, in this particular case, give the total space of a line bundle, i.e. satisfy all the formal properties of the equivalent construction in scheme theory? 3) I know that one example of formal scheme is given by the formal completion of a scheme along a closed subscheme. Is the reverse true? I.e. is it true that every formal scheme comes as the completion of a usual scheme along a closed subscheme? 4) I know that fiber product of formal schemes is locally constructed via completed tensor product. Which properties of usual tensor product does completed tensor product satisfy? In particular, I am interested in the following situation: once I have an invertible sheaf $\mathfrak{L}$ over a formal scheme, may I define the inverse as $\text{Hom}_{\mathcal{O}_{\mathfrak{X},\text{ cont}}}(\mathfrak{L},\mathcal{O}_{\mathfrak{X}})$, where I mean continuous homomorphisms of $\mathcal{O}_{\mathfrak{X}}$-modules? Thanks for any suggestion, I know there are a lot of points here, but also if you can suggest me good references, I would appreciate it! REPLY [6 votes]: For 1) The functorial interpretation is developed by Strickland in Formal schemes and formal groups. Homotopy invariant algebraic structures (Baltimore, MD, 1998), 263–352, Contemp. Math., 239, Amer. Math. Soc., Providence, RI, 1999. an expanded version is on his webpage https://neil-strickland.staff.shef.ac.uk/research/subgp.pdf On quasi-coherent sheaves, the issue is certainly delicate, I would recommend you the initial sections in "Duality and flat base change on formal schemes", which is the first paper in: Alonso Tarrío, Leovigildo; Jeremías López, Ana; Lipman, Joseph: Studies in duality on Noetherian formal schemes and non-Noetherian ordinary schemes. Contemporary Mathematics, 244. American Mathematical Society, Providence, RI, 1999. For a readily available pdf (with some corrections incorporated) http://www.math.purdue.edu/~lipman/papers/formal-duality.pdf For 2) The short answer is no, as far as I know. For 3) there are famous counterexamples, see Hironaka, Heisuke; Matsumura, Hideyuki: Formal functions and formal embeddings. J. Math. Soc. Japan 20 1968, 52–82. and Hartshorne, Robin: Ample subvarieties of algebraic varieties. Lecture Notes in Mathematics, Vol. 156 Springer-Verlag, Berlin-New York 1970 but also Non-algebraizable Formal Scheme? For 4) As far as you stick to coherent sheaves, morphisms are automatically continuous, so invertible sheaves behave much as on ordinary schemes. Another completely different issue is ampleness, where counterexamples in hartshornes "algebraic Geometry" book show that the issue is subtle, and somehow connected to your second question. Generalizing, you would like to seek for a nice system of generators of a substitute of the ill-behaved category of quasi-coherent sheaves. There are several possible candidates, but this line has not been pursued in these terms.<|endoftext|> TITLE: Modular tensor category associated to an even integral lattice and the lattice automorphism QUESTION [5 upvotes]: Let $(L,\langle -,-\rangle)$ be an even integral lattice, and let $(A,q)$ be the associated discriminant form: $$ A=L^*/L, \quad q(a)=e^{\pi i \langle a,a\rangle}. $$ We let $\hat L$ to be the extension of $L$ by $\{\pm1\}$ such that $\hat a\hat b=(-1)^{\langle{a,b}\rangle} \hat b\hat a$. Then we have three natural "isometry groups", $O(L)$, $O(A)$ and $O(\hat L)$; the first two are subgroups of automorphism groups preserving $\langle,\rangle$ or $q$; the last is a natural extension of $O(L)$ by $\mathrm{Hom}(L,\{\pm1\})$. Let us now consider the modular tensor category $\mathcal{C}(A,q)$ associated to $(A,q)$. This is also the category of modules of the lattice VOA $V_L$ constructed from $L$. Let $\mathrm{Aut}(\mathcal{C})$ be the group of braided auto-equivalences of a modular category $\mathcal{C}$ up to natural transformations. $\mathrm{Aut}(\mathcal{C}(A,q))$ is known to be equal to $O(A)$. So we have a natural sequence of group homomorphisms $$ O(\hat L)\to O(L)\to O(A)=\mathrm{Aut}(\mathcal{C}(A,q)) . $$ In [ENO, arXiv:0909.3140], extendsion of a modular tensor category $\mathcal{C}$ by any group $G$ with a homomorphism to $\mathrm{Aut}(\mathcal{C})$ was studied, and two obstructions are identified. The first obstruction takes values in $H^3(G,\mathrm{Inv}(\mathcal{C}))$ where $\mathrm{Inv}(\mathcal{C})$ is the group of invertible objects of $\mathcal{C}$. So it seems natural to study the obstructions to extending $\mathcal{C}(A,q)$ by any subgroup of $O(A)$, $O(L)$ or $O(\hat L)$. The first obstruction takes values in $H^3(G,A)$. (I believe the first obstruction vanishes for (subgroups of) $O(\hat L)$.) Does anybody know of any references concerning this point? I would be happy with any partial results, or any starting point for me to explore the literature. (Also, does the extension theory of ENO appear somewhere in the theory of lattice VOA?) REPLY [4 votes]: Edit: I've thought about this question again, and I think the answer is more positive than what I said in an earlier version. I will assume $L$ is positive-definite, since we need that to make $V_L$ into an honest VOA. I think what you say is still true using vertex algebras of indefinite lattices, but one has to be very careful with definitions to make fusion work well, and no one has done so in that setting yet. Here is what we expect to be true: For subgroups of $\operatorname{Aut}(\mathcal{C}(A,q))$ in the image of the map from $O(\hat{L})$, we expect an extension to exist and to be given by the category of twisted $V_L$-modules (where the twisting ranges over automorphisms of $V_L$). The two obstructions $O_3$ and $O_4$ in the ENO paper concern associativity of tensor product, and are expected to vanish in all cases of interest to us. I have no idea what happens for elements not in the image of $O(\hat{L})$. Here is what we know: If the twisting ranges over a solvable group $G$, then the existence of an extension essentially follows from the regularity (i.e., complete reducibility) properties of the fixed-point subVOA $(V_L)^G$ established in C-Miyamoto, together with Huang's work on modular tensor structure. There is some technical work involving compatibility of twisted intertwining operators, but I don't think anything important is missing. In other words, if the determinant of $L$ is small, you get an extension unconditionally. If the determinant is large, then the existence of an extension given by twisted $V_L$-modules is conditional on the conjecture that taking fixed points under any finite group action preserves complete reducibility. The technical work with intertwining operators comes from the fact that a full definition was only given in Huang's preprint from last year. On the other hand, you can simplify the definition significantly under the assumption that your twisted modules are just finite direct sums of irreducible $(V_L)^G$-modules, and various conjectured properties do not need new proofs.<|endoftext|> TITLE: How does this definition define a größencharakter? QUESTION [9 upvotes]: In Elliptic Modular Forms and Their Applications, p.89, Zagier defines a "größencharakter" $\psi_N$ on the field $K = \mathbb{Q}(i)$. This is a character on ideals. Since $\mathcal{O}_K = \mathbb{Z}[i]$ is a PID we can write every integral ideal $I\subseteq \mathcal{O}_K$ as $I = (\lambda)$ for a choice of $4$ different generators $\lambda\in\mathcal{O}_K$ (since $w_K = \lvert\mathcal{O}_K^\times\rvert=4$). Specifically, define $$\psi_N (I) = \overline{\lambda}^N, \quad \text{ if } w_K\vert N \text{ and } I = (\lambda),$$ and set it to be $0$ for $N$ not divisible by $w_K$. More generally, assume that $K$ is an imaginary quadratic number field with class number $h_K = 1$, and let $w_K$ denote the size of the unit group $\mathcal{O}_K^\times$. Define $$\psi_N(I)=\overline{\lambda}^N, \quad \text{ if } w_K\vert N \text{ and } I = (\lambda).$$ Question: How can $\psi_N$ be interpreted as a größencharakter on $K$? A größencharakter $\chi$ has 2 equivalent formulations (I believe): either as a continuous character on the idele group $\chi: \mathbb{A}_K^\times\to\mathbb{C}^\times$ with $K^\times$ in its kernel, or (modulo some ideal $\mathfrak{m}\subseteq \mathcal{O}_K$) as a character defined on the group $J^\mathfrak{m}$ of ideals coprime to $\mathfrak{m}$ such that there exist characters $$\chi_f: (\mathcal{O}_K/\mathfrak{m})^\times\to\mathbb{C}^\times, \quad \chi_\infty:\mathbf{R}^\times\to\mathbb{C}^\times$$ which determine $\chi$ on principal integral ideals: $\chi((a)) = \chi_f(a)\chi_\infty(a)$ when $a\in\mathcal{O}_K$ is coprime to $\mathfrak{m}$. Here, $\mathbf{R}$ is the $\mathbb{R}$-algebra $$\mathbf{R} = \left\{(z,\overline{z}): z\in\mathbb{C}\right\}$$ and $K\hookrightarrow\mathbf{R}$ via $a\mapsto (a, \overline{a})$. (This definition of a größencharakter can be found, for example, in Neukirch's Algebraic Number Theory, p.470.) Ideally I would like to see how either of these could be written down explicitly for this "größencharakter" $\psi_N$, and to understand how to extend this ad-hoc definition to a größencharakter in either sense. REPLY [4 votes]: $\psi_N$ is a Größencharakter modulo $\mathfrak{m}:=\mathcal{O}_K$ in the following way. Let $\chi_f$ be the trivial character on the $1$-element group $(\mathcal{O}_K/\mathfrak{m})^\times$, and let $\chi_\infty(z,\overline{z}):=\overline{z}^N$. Then, for any nonzero $a\in\mathcal{O}_K$, we have $\psi_N((a))=\overline{a}^N=\chi_f(a)\chi_\infty(a)$, and we are done. Note that the condition $w_K\mid N$ ensures that $\psi_N$ is well-defined: if $a,b\in\mathcal{O}_K$ are nonzero, and $(a)=(b)$, then $a=bu$ for some unit $u\in\mathcal{O}_K^\times$, and hence $a^N=(bu)^N=b^N$.<|endoftext|> TITLE: Comonadicity of spaces over spectra? QUESTION [12 upvotes]: As connective spectra are equivalent to group-like $E_{\infty}$ algebras in spaces, the $\infty$-category of connective spectra is monadic over the $\infty$-category of spaces though the usual $\Sigma^{\infty} \dashv \Omega^{\infty}$ adjunction. Is it also the case that the category of spaces is comonadic over the category of spectra (or maybe connective spectra) ? If so do we have a nice description of "what is a (unstable) space" in terms of stable homotopy theory, i.e. can we say something about what are the $\Sigma^{\infty} \Omega^{\infty}$-coalgebra in spectra in terms of more familiar structure (in the same way that a $\Omega^{\infty} \Sigma^{\infty}$-algebra is the same as a group-like $E_{\infty}$-algebra). For example $\Sigma^{\infty} X$ naturally has the structure of a cocomutative (in the $E_{\infty}$-sense) co-algebras (coming from the diagonal map of $X$). Can we characterize $\Sigma^{\infty} \Omega^{\infty}$-coalgebra, as such cocommutative coalgebra satisfying some additional conditions ? I'm interested in how to deduce some properties of a higher toposes from properties of its category of spectra, and more generally, how much of a higher topos can be understood from its category spectra. so this sort of explicit description could be useful and any description of the co-monad only involving more classical structure on the category of spectra (like the smash product and the subcategory of connective spectra) might be useful. REPLY [5 votes]: Let me record this as an answer: Blomquist and Harper show that $\Sigma^\infty$ is comonadic on simply-connected spaces. It occurs to me that if your ultimate goal is to understand unstable homotopy theory in terms of stable homotopy theory, you may also be interested in the following monadicity result in chromatic homotopy theory: For each $n$, the $v_n$-periodic localization of pointed, $d$-connected spaces (where $d = d(n)$ is some fixed number which grows with $n$) is monadic over the category of $v_n$-periodic spectra, via the Bousfield-Kuhn functor. Moreover, the monad in question is precisely the monad for Lie algebras. That is, the $\infty$-category of pointed, $d$-connected, $v_n$-periodic spaces is equivalent to the category of Lie algebras in the category of $v_n$-periodic spectra. This is almost enough to understand spaces entirely in terms of stable homotopy theory, in that $v_n$-periodic homotopy groups detect weak equivalences (as $n$ varies) between simply-connected finite spaces. However, there's the gap in that the category of Lie algebras corresponds to a category of pointed, $d(n)$-connected spaces. So even understanding all of this Lie algebra data won't tell you anything about a space which depends only on finitely many homotopy groups. And since it's a result about pointed spaces it won't be straightforward to interpret topos-theoretically anyway.<|endoftext|> TITLE: Measurable functions with non measurable image QUESTION [6 upvotes]: I am just curious about examples of measurable functions $f:[0,1]\to[0,1]$ such that $f[0,1]$ is not measurable. This is motivated by the question Is measure preserving function almost surjective?, that asks whether such maps have image of inner measure one. The solution is trivial if $f[0,1]$ is measurable, however, I have not succeeded understanding ``how bad'' it could be assuming that $f[0,1]$ is measurable. I think it is not possible to construct any such $f$ avoiding the use of the axiom of choice (because if not, it would be possible to construct a non measurable set $A=f[0,1]$ without it). On the other hand, if we start from a non-measurable set $V,$ and we try to find $f$ such that $f[0,1]=V,$ then I do not know how to make such $f$ to be measurable. What about examples examples of measurable functions $f:[0,1]\to[0,1]$ such that $f[0,1]=A\cup B$ where $A$ is not Lebesgue, $B$ has Lebesgue measure zero and $f^{-1}A$ has strictly positive Lebesgue measure? REPLY [8 votes]: A measurable function $f:[0,1]\to\mathbb{R}$ maps Lebesgue measurable sets to measurable sets if and only if it has a Lusin property: the image of a set of measure zero has measure zero. Here is an example when the Lusin property is violated. Take a Cantor set of positive Lebesgue measure in $[0,1]$. This set contains a non-measurable subset $E$. Let $F$ be a subset of the ternary Cantor set that is homeomorphic to $E$. Since the ternary Cantor set has measure zero, $F$ is measurable (and of measure zero). Let $f:[0,1]\to [0,1]$ be defined as a homeomorphism of $F$ onto $E$ and a constant map (with the value in $E$) on the complement of $F$. The mapping $f$ is measurable and $f([0,1])=E$ is not measurable. EDIT: I have just realized that my answer is very similar to the comment of Michael Greinecker.<|endoftext|> TITLE: Can There be a 1 dimensional Banach-Tarski paradox in the absence of choice QUESTION [18 upvotes]: Let $\mathbb{R}$ act on itself by translation. Then there is no finite decomposition of a unit interval into pieces which, when translated, yields two distinct unit intervals. More formally does there exist a partition of the unit interval $[0,1]$ into finitely many sets $S_1,...,S_n$ and a collection of finitely many constants ${r_1,...r_n} \in \mathbb{R}$ such that the the union of $S_a+r_a$ is the set $[0,1]\cup [2,3]$. Where $S_a+r_a$ is the set $S_a$ shifted to the right by $r_a$. The only proofs I know of this crucially use the invariant mean-characterization of amenability for $\mathbb{R}$. The standard proofs of this characterization use the ultrafilter lemma and are therefore not in ZF+DC. Is it possible in ZF or (more strongly) ZF+DC for there to be finite decomposition of the unit interval into two pieces. Can we put a bound on how many pieces you would need? (i.e. could you do it with 3 or 4 pieces)? REPLY [14 votes]: No. Let $A$ be the free abelian group generated by the $r_a$s. We can view this as a lattice in the real vector space $A \otimes \mathbb R$. Let $n$ be the rank of this group / the dimension of this vector space. There is a natural evaluation map $f: A \otimes \mathbb R \to \mathbb R$ from this vector space to $\mathbb R$. Applying the assumed bijection between $[2,3] \cup [0,1] $ and $[0,1]$, which is obtained by at each point by subtracting one of the $r_a$s and hence preserved the property of lying in $A$, we obtain a bijection between $A \cap f^{-1} ( [2,3] \cup [0,1])$ and $ A \cap f^{-1} ([0,1])$. In the case $n=1$, this is already impossible as soon as the lattice intersects $[2,3]$, which we can guarantee if necessary by adding additional $r_a$s. Otherwise, this bijection involves moving in the lattice a distance at most $1$. Hence for $B_R$ a ball of radius $R$, we obtain an injection $B_R \cap A \cap f^{-1} ([2,3] \cup [0,1]) \to B_{R+1} \cap A \cap f^{-1} ([0,1])$. We now simply apply lattice point counting to check that the number of lattice points in these sets are asymptotic to the volumes of $B_R \cap f^{-1} ([2,3] \cup [0,1]) $ and $ B_{R+1} \cap f^{-1} ([0,1])$ respectively, which are asymptotic to $2 C R^{n-1}$ and $C R^{n-1}$ respectively, contradicting the claimed existence of an injection. The only subtlety in the lattice point counting is whether the images of the lattice points under $f$ are equidistributed. However, by Weyl equidistribution, it suffices that $A$ is a dense subset of $\mathbb R$, which is automatic as $n>1$. Here is a more general version of my and YCor's argument: Let $\Gamma$ be a group acting by measurable, volume-preserving transformations on a space $X$. Suppose that $\Gamma$ is amenable, in the sense that for each finite set $F$ and each $\epsilon>0$, there is a subset $S$ of $\Gamma$ such that $| FS| < (1+ \epsilon) |S|$ (Følner subsets). Let $I$ and $J$ be two measurable subsets of $X$ such that can each be decomposed into finitely many pieces such that each piece of $I$ is a translate under an element of $\Gamma$ of a corresponding piece of $J$. Then the measure of $I$ equals the measure of $J$. Proof: Let $F$ be the set of elements of $\Gamma$ that appear in the bijection between $I$ and $J$. Let $f_I(x)$ be the number of $\gamma \in F S$ such that $\gamma(f) \in I$, and let $f_J(x)$ be the number of $\gamma \in S$ such that $\gamma(f) \in J$. Then $f_J(x) \leq f_I(x)$ because for each $\gamma$ with $\gamma(f) \in J$, we have chosen a $g\in F$ with $g \gamma(f) \in I$, and for distinct $\gamma(f)$, these give distinct elements of $I$ and hence distinct pairs $g \gamma$. (For distinct $\gamma$ that give the same $\gamma(f)$, we have the same $g$, and so $g \gamma$ remains distinct.) So $\int f_J \leq \int f_I$. But exchanging the order of summation, $\int f_I = |FS| \mu(I)$ and $\int f_J = |S| \mu(J)$. Because so $(1+\epsilon) \mu(I) \leq \mu(J)$. Taking $\epsilon$ to $0$, $\mu(I) \leq \mu(J)$. By symmetry, $\mu(I)=\mu(J)$. REPLY [11 votes]: No choice is needed. First, assuming such a paradoxical decomposition, we get another one in a large circle $C=\mathbf{R}/k\mathbf{Z}$. Now we have, on the subgroup $\Gamma$ generated by the $r_i$ (and by $x\mapsto x+1$), an explicit sequence of Følner subsets, and corresponding normalized $\ell^1$-functions $f_m$. Start from any Dirac measure on $C$, average it by $f_m$ to have a sequence of (finitely supported) measures $\mu_m$. Then $\mu_m$ is approximately invariant, in the sense that for every $g\in\ell^\infty(C)$ and $s\in\Gamma$, we have $\mu_m(g-sg)\to 0$. Indeed, $$\mu_m(g-sg)=(\mu_m-s\mu_m)g,$$ which tends to 0. Split $C$ into $k$ intervals $C_1,\dots,C_k$ of length 1 (say, half-open). Then $\mu_m(C_i)-\mu_m(C_j)\to 0$. In particular, we do not have $\mu_m(C_1)\to 0$, since this would force $\mu_m(C_i)\to 0$ for each fixed $i$ and would entail the contradiction $1=\mu_m(C)=\sum_i\mu_m(C_i)\to 0$. The proof is finished, since the paradoxical decomposition yields a decomposition of the form $\sum_{i=1}^nu_i-r_iu_i=1_{[2,3]}$, and applying $\mu_m$, the left-hand term tends to 0 and not the second one.<|endoftext|> TITLE: Large subgroups of $S_n$ without large symmetric or alternating subgroups QUESTION [5 upvotes]: I'm interested in determining the existence of a permutation group $G\subseteq S_n$ of the following form. $G$ is large. Meaning that $G$ have at least $n!/2^{o(n)}$ elements. Equivalently, their index is at most $2^{o(n)}$. For some fixed $d$, no subgroup $H$ of $G$ is isomorphic to $S_d$ or to $A_d$. Question 1: How small can $d$ be? For instance, is there a subgroup $G$ of $S_n$ of index $2^{o(n)}$ such that no subgroup of $G$ is isomorphic to $S_{\sqrt{n}}$ or $A_{\sqrt{n}}$? The second question is whether there are Ramsey-theoretic results relating symmetric groups of size $d$ with large subgroups of $S_n$. Question 2: Are there functions $f,g:\mathbb{N}\rightarrow \mathbb{N}$ such that for every $d TITLE: What, mathematically speaking, does it mean to say that the continuation monad can simulate all monads? QUESTION [23 upvotes]: In various places it is stated that the continuation monad can simulate all monads in some sense (see for example http://lambda1.jimpryor.net/manipulating_trees_with_monads/)) In particular, in http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.43.8213&rep=rep1&type=pdf it is claimed that any monad whose unit and extension operations are expressible as purely functional terms can be embedded in a call-by-value language with “composable continuations”. I was wondering what (Category-theoretic) mathematical content these claims of simulation have, and what precisely they show us about monads and categories in mathematical terms. In what ways is the continuation monad special, mathematically, compared to other monads, if at all? (I seem to remember some connection between the Yoneda embedding and continuations which might be relevant https://golem.ph.utexas.edu/category/2008/01/the_continuation_passing_trans.html, although I don't know) One other relevant fact might be that the continuation monad is the monad which takes individuals $\alpha$ to the principal ultrafilters containing them (that is it provides the map $\alpha \mapsto (\alpha \rightarrow \omega) \rightarrow \omega$) ) Edit: I have been asked to explain what I mean by the continuation monad. Suppose we have a monad mapping types of the simply typed lambda calculus to types of the simply typed lambda calculus (the relevant types of this calculus are of two kinds: (1) the basic types, (I.e, the type $e$, consisting of individuals and belonging to domain $D_e$; and the type of truth values $\{ \top, \bot\}$ belonging to domain $D_t$) and, (2) for all basic types $\alpha, \beta$, the type of functions between objects of type $\alpha$ and objects of type $\beta$ belonging to domain $D_{\beta}^{D_{\alpha}}$). Let $\alpha, \beta$ denote types and $\rightarrow$ a mapping between types. Let $a : \alpha \hspace{0.2cm}$ (or $b: \beta)$ indicate that $a\hspace{0.2cm}$ (or $b$) is an expression of type $\alpha \hspace{0.2cm}$ (or $\beta)$. Let $\lambda x. t$ denote a function from objects of the type of the variable $x$, to objects of the type of $t$, as in the simply typed lambda calculus. Then a continuation monad is a structure $\thinspace(\mathbb{M}, \eta, ⋆)\thinspace$, with $\mathbb{M}$ an endofunctor on the category of types of the simply typed lambda calculus, $\eta$ the unit (a natural transformation) and ⋆ the binary operation of the monoid) such that: $$\mathbb{M} \thinspace α = (α → ω) → ω, \hspace{1cm} ∀α$$ $$η(a) = λc. c(a) : \mathbb{M} \thinspace α \hspace{1cm} ∀a : α $$ $$m ⋆ k = λc. m (λa. k(a)(c)): \mathbb{M}\thinspace β \hspace{1cm} ∀m : \mathbb{M}\thinspace α, k : α → \mathbb{M}\thinspace β . $$ The continuation of an expression $a$ is $\eta(a) = (a \rightarrow \omega) \rightarrow \omega$. Edit 2: let $ \omega$ denote some fixed type, such as the type of truth values (i.e, $\{ \top, \bot\}$) REPLY [4 votes]: If I understand this paper correctly, the construction in full generality might not really involve the continuation passing monad as such. It is easy to see that if $M$ is an internal monad and $\omega$ is any $M$-algebra, then there is a natural transformation from $M\alpha$ to $ (\alpha → \omega) → \omega$. Indeed, given an element of $(\alpha\to\omega)$, apply $M$ to obtain an element of $(M\alpha\to M\omega)$, plug in $M\alpha$, and compose with the $M\omega\to \omega$ from the algebra structure. (An internal monad is needed for this to make sense on objects, not just algebras). However, there is no universal choice of an algebra. Let $U$ be the functor from the category of $M$-algebras to the base category. Hence for any $\alpha$ in the base category and $z$ a monad algebra, $\operatorname{Hom}(\alpha, Uz)$ is a set. Assume that the base category has arbitrary products, so that we can internally raise objects to the power of sets. The first step is to replace $ (\alpha → Uz) → Uz$ with $Uz^{ \operatorname{Hom}(\alpha,Uz)}$. This is the same in the category of sets, and regardless is formally similar. Next we want to consider elements of this that depend consistently on $z$. You might think this is impossible because one dependence on $z$ is covariant and the other is contravariant, but apparently this is the perfect situation to apply the categorical notion of an end, which gives the correct consistency condition. The end of $Uz^{ \operatorname{Hom}(\alpha,Uz)}$ is naturally isomorphic to $M\alpha$. So in the category of sets the statement is that any monad is isomorphic to the subset of a product of instances of the continuation passing monad consisting of those elements satisfying a certain compatibility condition.<|endoftext|> TITLE: Can one "hear" the shape of a polygon via external reflections? QUESTION [42 upvotes]: This question is a rough analog of Kac's "Can One Hear the Shape of a Drum?" A closer analog is the recent "Bounce Theorem" that says, roughly, the shape of a polygon is determined by its billiard-bounce spectrum.1 Suppose there is a polygon $P$ hidden inside a disk $D$. All its edges are mirrors. You shoot in a light ray, and are able to observe the trajectory of the ray's emergence from $D$. If the ray hits a vertex, it dies; otherwise it reflects across edge perpendiculars.                     Left: Polygon $P$ hidden by disk $D$. Right: The ray reflects from $P$'s edges. Q1. Are there two incongruent polygons $P_1$ and $P_2$ that cannot be distinguished from the bounce behavior of external rays? Here I want to ignore rigid motions of the polygons. By "bounce behavior" I mean comparing the geometry of the incoming and outgoing trajectories of the ray; what happens inside the disk is not known to you. Imagine all possible incoming rays. Can two incongruent polygons have the same bounce behavior for every possible ray, i.e., be equireflective? One can think of several variants. Perhaps a bit more information might help prove a negative result: Q2. Suppose you not only observe the in- and out-trajectories, but also the time it takes for the ray to emerge, effectively yielding the length of the ray path. Perhaps it is easier to construct equireflective shapes if one could make use of sections of parabolas and ellipses and their special reflection properties: Q3. Are there two incongruent piecewise-smooth Jordan curves $C_1$ and $C_2$ that cannot be distinguished from the bounce behavior of external rays? 1 Moon Duchin, Viveka Erlandsson, Christopher J. Leininger, Chandrika Sadanand. "You can hear the shape of a billiard table: Symbolic dynamics and rigidity for flat surfaces." 2018. arXiv abs. REPLY [3 votes]: my answer for Q1: "If the ray hits a vertex, it dies". I take this property as the starting point of my solution Let two neighboring disks D1 and D2 contain two such incongruent equireflective polygons. Let us take a vertex V1 of polygon P1; all rays hitting V1 will die; same behavior for the corresponding point V1' in disk D2, no matter if V1' is a vertex of P2 or not. All rays hitting V1' die, meaning that at some point they hit a vertex of P2, vertex which doesn't necessarily have be in V1' position. But is has to be somewhere on the ray support line, maybe before or after V1'; So this line contains for sure a certain vertex of P2. This property holds for an arbitrary ray hitting V1'. We can have infinitely many rays hitting V1', each containing a vertex of P2. But P2 has a finite number of vertexes. So we have an infinity of lines - all convergent in V1' , each containing a point from a finite set of points. This leads to V1' itself being a vertex of P2. Repeating the same procedure for each vertex of P1, we get that the corresponding point in D2 is also a vertex of P2. So the set of P1 's vertexes is included in the set of P2 's vertexes. Now we start from disk D2 with the same procedure, to obtain that the set of P2 's vertexes is included in the set of P1 's vertexes. The two sets are equal. The two polygons have their vertexes in the exact same (geometric) location on their disks - so they are congruent.<|endoftext|> TITLE: What are Lie groupoids intuitively? QUESTION [7 upvotes]: I am trying to understand about Lie groupoids but not able to get feeling for what it actually is. So, question here is, What are Lie groupoids? How similar are they to Lie groups, Groupoids and what can one expect to do on a Lie groupoid? Any reference is appreciated. REPLY [3 votes]: Here is a somewhat trivial observation, but one that helped me to get an initial 'visual' feeling of the thing. I tried to make it a comment, but since I have just signed up, I didn't have enough reputation for doing so. If someone consider it more appropriate to move it somewhere else, please do it. =] When dealing with purely algebraic groupoids, it is common to call'em 'connected' when two objects are connected by at least one morphism. Also, it is usually OK to deal only with connected groupoids, since the disconnected ones are merely, well, 'disjoint unions'; unless, of course, the groupoid in question had emerged from some problem for which the existence of connected components, and the variability of its isotropy groups, might be relevant. In the case of Lie groupoids, however, this concept of connectedness doesn't make much sense. Indeed, the strictly analogous condition, i.e., each pair of objects being connected by at least one morphism, is in that case called 'transitiveness' of a groupoid (see, for instance, Mackenzie's book 'General theory of Lie groupoids and Lie algebroids'). 'Connectedness', therefore, is in the context of Lie groupoids thought in relation to the topological structure of the groupoid, which is absent in the purely algebraic context. Thus, it seems that an important part of the richness of the Lie groupoid theory resides in the fact that many Lie groupoids are not transitive, but still are connected. This means: the 'connected components', in the purely algebraic sense, of a Lie groupoid are somehow 'glued together' in the topological structure of the Lie groupoid.<|endoftext|> TITLE: Gurski's Definition of a strict functor of tricategories QUESTION [8 upvotes]: In Gurskis definition (page 32 of his thesis) of a strict functor $F$ of tricategories he requires that $F$ maps the adjoint equivalences $a,l,r$ in the source tricategory to the same adjoint equivalences in the target tricategory. This means for example $F (a_{fgh}) = a_{FfFgFh}$. (Note that these morphisms are indeed parallel, because of the strictness of $F$.) However it is not required, that the modifications in the definition of a tricateory $(\pi,\lambda,\mu,\rho)$ are preserved as well. For example $ F (\pi _{fghk}) $ is not required to be equal to $ \pi_{FfFgFhFk} $. So my question is: Why isn't it required that $(\pi,\lambda,\mu,\rho)$ are preserved? Is this because this is already implied by the other requirements? Or are there strict functors which doesn't preserve the modifications $(\pi,\lambda,\mu,\rho)?$ EDIT Let me draw some parallel to the definition of strict functors between bicategories: ($*$ denotes composition along objects) There it is only required that the constraint-2-cells $$ F(f)*F(g) \to F(f*g) $$ and $$ 1_{Fa} \to F(1_a) $$ are identities. That $a,l,r$ are preserved is not demanded by definition, but it follows easily from the functor axioms. Therefore one might hope that in the case of a strict functor of tricategories it follows somewhat analoge that the "highest" constraint data $\pi,\lambda,\mu,\rho$ are preserved, although it is not demanded by definition. REPLY [2 votes]: The axioms in the definition of a non-strict trihomomorphism (Definition 3.3.1, page 31-32) already require that the modifications $\pi,\lambda,\mu,\rho$ are preserved "up to" the coherences $\pmb{\chi}, \pmb{\iota}, \omega,\gamma,\delta$. The definition of a strict functor (3.3.3, p33) then requires that $\pmb{\chi}$ and $\pmb{\iota}$ are identities while $\omega,\gamma,\delta$ are "as close to identities as possible" (they can't literally be identities because their domain is not equal to their codomain), which therefore implies that $\pi,\lambda,\mu,\rho$ are "preserved as much as possible".<|endoftext|> TITLE: Sheaf cohomology with support vanishes QUESTION [5 upvotes]: I am trying to solve the exercise 2.4 chapter III in Hartshorne's "Algebraic Geometry". For this I would like to prove for a sheaf $F$ of Abelian groups on a topological space $X$ and $U$ open subset of $X$ with complement $Z$ the following $$ H^{i}_{Z}(X,j_{.}(F\mid_{U})) = 0$$ where $j:U \longrightarrow X$ the inclusion and $j_{.}$ meaning the extension of a sheaf on U to a sheaf on X by zero. I have proven exercise 2.3 chapter III, so I can use this. Does anyone know a nice proof for this? Or maybe a counterexample if this doesn't hold? I was able to reduce/adapt this case to proving $H^{i}(X,j_{.}(F\mid_{U})) \cong H^{i}(U,F\mid_{U})$ but I also haven't a rigorous proof for this. Also for it is quite clear because $H^{0}_{Z}(X,j_{.}(F\mid_{U})) =0$, it is just for the higher order groups I have't an argument. (Or maybe this all doesn't hold?). Thanks in advance! (Not sure if this is supposed to be on mathstack, but there I have not yet received an answer on this). REPLY [3 votes]: This is not true. Example. Let $X$ be an irreducible topological space with $|X| \geq 2$ that has a closed point $x \in X$. For example, we may take $X = \operatorname{Spec} R$ for $R$ a domain that is not a field, and $x = \mathfrak m$ for $\mathfrak m \subseteq R$ a maximal ideal. Let $Z = \{x\}$, and let $U = X \setminus Z$. Then the constant sheaf $\mathbb Z_X$ is flabby on $X$ [Hart, Exc. II.1.16(a)]. We have a short exact sequence [Hart, Exc. II.1.19(c)] $$0 \to j_!\mathbb Z_U \to \mathbb Z_X \to i_*\mathbb Z_Z \to 0,$$ since $\mathbb Z_X|_U = \mathbb Z_U$ and $\mathbb Z_X|_Z = \mathbb Z_Z$ are the constant sheaves on $U$ and $Z$ respectively. Taking sections with support on $Z$ gives a long exact sequence $$0 \to \Gamma_Z(X,j_!\mathbb Z_U) \to \Gamma_Z(X,\mathbb Z_X) \to \Gamma_Z(X,i_*\mathbb Z_Z) \to H^1_Z(X,j_!\mathbb Z_U) \to \ldots.$$ But $\Gamma_Z(X,\mathbb Z_X) = 0$, since all nonzero sections of $\mathbb Z_X$ are supported everywhere. On the other hand, $\Gamma_Z(X,i_*\mathbb Z_Z) = \mathbb Z$, since all sections of $i_*\mathbb Z_Z$ are supported on $Z$. Thus, the map $$\Gamma_Z(X,\mathbb Z_X) \to \Gamma_Z(X,i_*\mathbb Z_Z)$$ is not surjective, so $H^1_Z(X,j_!\mathbb Z_U)$ cannot be zero. Remark. Your statement that $H^i(X,j_!(\mathscr F|_U)) = H^i(U,\mathscr F|_U)$ is also false. This already fails for $H^0$, because $\Gamma(X,j_!(\mathscr F|_U)) = 0$ if $U \subsetneq X$, by the definition of extension by $0$. References. [Hart] Hartshorne, Robin, Algebraic geometry, Graduate Texts in Mathematics 52. Springer-Verlag, New York-Heidelberg-Berlin (1977). ZBL0367.14001.<|endoftext|> TITLE: Estimating volumes in the trace formula QUESTION [10 upvotes]: Let $G$ be a reductive group. In many instances of the trace formula, elliptic terms corresponding to the $G(k)$-conjugacy class of $\gamma \in G(k)$ are weighted by the following volumes $$v_\gamma = \mathrm{vol}(G_\gamma(k) \backslash G_\gamma(\mathbb{A}))$$ I am interested in knowing how to handle $v_\gamma$ in practice. In very few cases I am able to explicitly compute the quotient (namely, $GL(2)$ or $SL(2)$) for diagonal element, what is possible up to conjugation by semisimplicity). Is there any general method to compute these volumes, or results concerning their values or bounds (for instance depending on the eigenvalues of $\gamma$, on its determinant, and on quantities attached to $k$)? I am interested in inner forms of general linear groups, unitary groups and symplectic $GSp(4)$. REPLY [13 votes]: Up to a normalisation factor for the measure you fixed on $G_\gamma(\mathbb A)$, this is the Tamagawa number of $G_\gamma$. There is indeed a formula for this. It is $$ \frac{ \left| \pi_0\left( Z(\hat{G})^\Gamma\right) \right|}{ \left|\operatorname{ker}^1\left(F, Z(\hat{G}) \right)\right|}$$ where $\hat{G}$ is the Langlands dual group, $Z$ is its centre, $\Gamma$ is the Galois group acting on it, and $\operatorname{ker}^1$ refers to the kernel of the map from Galois cohomology to local Galois cohomology at each place. For a proof see Stable trace formula, cuspidate tempered terms, section 5 where Kottwitz calculates the ratio of the Tamagawa number to the Tamagawa number of the universal cover of the derived subgroup, and Tamagawa numbers, where he shows that the Tamagawa number of the universal cover of the derived subgroup is $1$ (with a conditionality that has since been removed).<|endoftext|> TITLE: Counting primitive lattice points QUESTION [15 upvotes]: In Lemma 2 of [1], Heath-Brown proves the following (I state a simplified version of a more general result): Let $\Lambda \subset \mathbb{Z}^2$ be a lattice of determinant $d(\Lambda)$. Then $$\# \{ (x_1,x_2) \in \Lambda: \max_i |x_i| \leq B, \gcd(x_1,x_2) = 1\} \leq 16\left (\frac{B^2}{d(\Lambda)} + 1\right).$$ My question is whether this generalises to arbitrary dimensions. Does an analogous result hold for lattices in $\Lambda \subset \mathbb{Z}^n$? Namely, is $$\# \{ (x_1,\dots, x_n) \in \Lambda: \max_i |x_i| \leq B, \gcd(x_1,\dots,x_n) = 1\} \leq C_n\left (\frac{B^n}{d(\Lambda)} + 1 \right)$$ for some constant $C_n$? If it helps, I'm primarily interested in the case $n=3$. Obviously I'm aware of standard lattice point counting techniques, but these usually give an error term of the shape $O(\text{boundary of the region/first successive minima})$, and I don't know how to control this in my case. So I'm just looking for uniform upper bounds where this term doesn't appear. [1] Heath-Brown - Diophantine approximation with Square-free numbers REPLY [16 votes]: No, there is no result in this form because in dimension 3 or higher it is allowed to have some non-first minima relatively small even when the first minimum is very small. For example, for any $N>0 $ consider the lattice $$ \Lambda_N = \frac 1 N \mathbb Z \times \mathbb Z \times \mathbb Z^{n-3}\times N\mathbb Z$$ with unit determinant. Then the set $$ S_N = \left\{\left(\frac k N,1 , 0,\ldots, 0\right):\ -N\leq k\leq N\right\}$$ is a set of $2N+1$ primitive vectors of $\Lambda_N$ contained in the unit hypercube. Now, to answer your question, take $B=N$ prime and the lattices $N\Lambda_N\subset \mathbb Z^n$ to see that there is no such constant $C_n$.<|endoftext|> TITLE: Ehresmann's approach to differential geometry QUESTION [11 upvotes]: I have come accross this brief description of Charles Ehresmann's life given by his wife: http://www.cs.le.ac.uk/people/ah83/cat-myths/myth0002.html I quote the part from the text relevant to my question: Indeed, groupoids intervene in fibre bundle theory in two different ways: A. Actions of a topological groupoid. Denote by E a fibre bundle. The isomorphisms from fibre to fibre form a groupoid, which is equipped with a topology compatible with the maps domain, codomain, composition and inversion. This gives a topological groupoid (in the sense: internal groupoid in the category Top of topological spaces), which acts continuously on the topological space E. This topological groupoid G satisfies the axiom: (LT) For each object, say x, of G (identified with a point of the base B of E ), there exists a local section s: U -> G of the codomain map on a neighborhood U of x such that s(y): x -> y for each y in U. Conversely, to a topological groupoid satisfying (LT) (called a locally trivial groupoid) naturally corresponds a principal fibre bundle, and to its actions, the associated fibre bundles. This defines an equivalence from the category of fibre bundles to the category of actions of locally trivial groupoids. In this setting, connections, prolongations of manifolds,... are very easily defined. More generally, the jets between all germs of manifolds form a (big) differentiable category (i.e., internal category in the category Diff of differentiable maps) and Charles described Differential Geometry as the study of this category and of the actions of its subcategories. This 'categorical' point of view is indicated in a series of very concise papers from 1958 to 1969 (CE I). Charles always thought of writing a book on this subject, and he regretted to have spent so much time in Bourbaki's team in the forties instead of developing his own ideas. I find this viewpoint of Differential Geometry as expressed in the quote above very interesting. As stated above, the approach to DG, starting from the category equivalence between actions of locally trivial groupoids and fiber bundles, has never been written up in textbook form by Ehresmann himself and is only to be found in "concise" papers. I wonder if this very natural, transparent and easy approach to differential geometry has been developed in the meanwhile by other authors. I think it would be especially interesting for students as a first introduction to Differential Geometry. Are there papers or textbooks in which connections and other differential geometric structures are developed just starting from the notion of locally trivial categories / groupoids and other structured categories internal to the category of smooth maps? The category equivalence itself is easy to prove, but I find the remark in the quote that the other structures are "very easily defined" - within this framework - not to be really entirely true, at least judging by my own first attempts. I have not yet tried to read Ehresmann's voluminous collected works (which are available online: http://ehres.pagesperso-orange.fr/C.E.WORKS_fichiers/C.E_Works.htm). REPLY [3 votes]: This may not be quite what you are looking for, but a nice reference is Kock's paper: Kock, Anders, Principal bundles, groupoids, and connections, Kubarski, Jan (ed.) et al., Geometry and topology of manifolds. The mathematical legacy of Charles Ehresmann. On the occasion of the hundredth anniversary of his birthday. Proceedings of the conference, Bȩdlewo, Poland, May 8--15, 2005. Warsaw: Polish Academy of Sciences, Institute of Mathematics. Banach Center Publications 76, 185-200 (2007). ZBL1121.51012.<|endoftext|> TITLE: Is it possible to cite a page in n-Lab in a research paper? QUESTION [32 upvotes]: One of the concepts I need in my paper is best summarized on the n-Lab page. I tried my best to find a similar description of the same concept in a more "classical" reference, such as a paper or a book, but I can't. I have two questions: (a) Is it "allowed" to cite n-Lab in a research paper? I mean, are people doing this? Would it make a referee frown? (b) What is the format of citing a page from n-Lab? REPLY [22 votes]: From https://ncatlab.org/nlab/show/FAQ#Citing: How do I cite a page on the nLab? You should, of course, include a link or URL to the page or pages that you wish to cite. You should also give the information about which version of the page you are citing, since pages change over time. The version number can be found as follows: look at the bottom of the page where it says “Back in time ($N$ revisions)”; the current version number is $N+1$. You can link directly to the version you want to cite by using a url such as http://ncatlab.org/nlab/revision/PageName/VERSIONNUMBER We recommend that if you only include one URL, it be of the form show/PageName which will point to whatever version of the page is current when it is accessed. This is because pages generally improve over time, and whoever is following your reference ought to be taken to the best, up-to-date version of the page. Anyone who cares about finding the exact version of the page that you cited can figure out how to find it in the history. On the other hand, if you can give two URLs (this would be cumbersome in a printed paper, but is possible with links on a web page), then it may be helpful to give the appropriate revision link as well as the show one. Here's one possible BibTeX template, which is similar to what I've used in published papers: @Misc{nlab:pagename, author = {{nLab authors}}, title = {Page Name}, howpublished = {\url{http://ncatlab.org/nlab/show/page+name}}, note = {\href{http://ncatlab.org/nlab/revision/page+name/N}{revision N}}, month = {Month}, year = XXXX} The double-brackets around {{nLab authors}} cause the alpha citation style (which I prefer to the numbered one) to cite it as [nLaXX], which is not great but better than [autXX] if you put in only single brackets. (Suggestions for improvements here are welcome.) In some cases, if the page as it existed at the time of citation was largely the work of one person (perhaps with minor things like typo fixes from other contributors --- this requires looking through the page history version-by-version to find out), I have instead attributed it to that person "and others" in the BibTeX (producing "et. al." in the citation --- this is one situation in mathematics where I feel "et. al." is actually warranted). Edit: All nLab pages now have a "Cite" link at the bottom that produces BibTeX according to my template! REPLY [10 votes]: You can cite an nLab entry much as many papers cite an online resource. I recommend searching ArXiv and other published sources for examples of citation formats. Since the Web is not static, I recommend not only that you specify a date retrieved, but that you make a personal copy of the (relevant portion of the) web page and include it in your archive of your paper. In particular, making a hard copy or a JPEG picture format might save you or some researcher in the future. Gerhard "Preparing For The Future Onslaught" Paseman, 2018.04.20.<|endoftext|> TITLE: Is there a Feynman-Kac formula for vector-valued Schrödinger operators? QUESTION [5 upvotes]: Given a vector function $$f=(f_1,\ldots,f_n)\in L^2(\mathbb R,\mathbb R^n)$$ (for some $n\in\mathbb N$), let us define $$\Delta f:=(\Delta f_1,\ldots,\Delta f_n),$$ where $\Delta$ is the Laplacian operator, and let $Q:\mathbb R\to\mathbb R^{n\times n}$ be a potential taking values in symmetric $n\times n$ matrices. I'm interested in vector Schrödinger operators of the form $$Hf=-\Delta f+Qf,\qquad f\in L^2(\mathbb R,\mathbb R^n).$$ Question. Is there a Feynman-Kac type formula known for $H$'s semigroup in the case where $Q(x)$ is not necessarily diagonal? (Note: I specify abote that I'm interested in the case where $Q$ is not diagonal; if $Q=\mathrm{diag}(Q_1,\ldots,Q_n)$, then $$Hf=(-\Delta f_1+Q_1 f_1,\ldots,-\Delta f_1+Q_1 f_1),$$ in which case we can simply apply the one-dimensional Feynman-Kac formula to each component.) REPLY [2 votes]: It's the usual formula with the exponential of $Q$ replaced by a time-ordered exponential --- needed in the integral over $t$ since $Q[x(t)]$ and $Q[x(t')]$ do not commute for $t\neq t'$. One reference where this "chronological'" integral is worked out is Equivalence of Two Definitions of a Chronological Integral.<|endoftext|> TITLE: Surjective group homomorphism from $\text{Sym}(X)$ onto $\mathbb{Z}$ QUESTION [7 upvotes]: For any non-empty set $X$ let $\text{Sym}(X)$ denote the group of bijections $f:X\to X$ with composition. Is there an infinite set $X$ and a surjective group homomorphism $\pi: \text{Sym}(X)\to \mathbb{Z}$? REPLY [12 votes]: The answer here is negative. In fact, any non-trivial quotient group of the symmetric group $\mathrm{Sym}(X)$ contains a copy of $\mathrm{Sym}(X)$. Indeed, by the Baer-Schreier-Ulam Theorem, any normal subgroup $N\ne \mathrm{Sym}(X)$ is contained in the subgroup $\mathrm{Sym}_<(X)$ of permutations having support of cardinality $<\kappa:=|X|$. Let $q:\mathrm{Sym}(X)\to \mathrm{Sym}(X)/N$ be the quotient homomorphism. Since $X$ is infinite, we can choose a family of pairwise distinct $\{x_{p}\}_{p\in \kappa\times\kappa}$ in $X$. For every permutation $\pi\in \mathrm{Sym}(\kappa)$ of $\kappa$ define the permutation $\bar\pi\in \mathrm{Sym}(X)$ letting $\bar\pi(x_{\alpha,\beta})=x_{(\pi(\alpha),\beta)}$ for $(\alpha,\beta)\in \kappa\times \kappa$ and $\bar\pi(x)=x$ for any $x\in X\setminus\{x_{p}:p\in \kappa^2\}$. It is clear that $e:\mathrm{Sym}(\kappa)\to \mathrm{Sym}(X)$, $e:\pi\mapsto\bar\pi$ is a group homomorphism whose image $e(\mathrm{Sym}(\kappa))$ in $\mathrm{Sym}(X)$ is disjoint with the subgroup $\mathrm{Sym}_{<}(X)\supset N$ and hence the composition $q\circ e:\mathrm{Sym}(\kappa)\to \mathrm{Sym}(X)/N$ is injective.<|endoftext|> TITLE: Growth order of numbers whose prime factors are all congruent to +1 or -1 modulo 8 QUESTION [5 upvotes]: Here are the numbers whose prime factors all congruent to $\pm 1\pmod 8$: http://oeis.org/A058529 My questions are: (1) What is the order of growth of these numbers? That is, what is the order of magnitude of the $n$th smallest among them? (2) For the numbers whose prime factors are all congruent to $\pm 3\pmod 8$, is the order of growth the same as in (1)? REPLY [7 votes]: This is an old question (and according to this MO question, the result you seek was proven by Landau). In particular, it follows from this that if $S$ is a set of arithmetic progressions containing a density $1/2$ set of the primes, then number of $n \leq x$ all of whose prime factors in $S$ is asymptotic to some constant times $C \frac{x}{\sqrt{\log(x)}}$. This implies that the $n$th positive integer $q_{n}$ all of whose prime factors are $\pm 1 \pmod{8}$ satisfies $q_{n} \sim C_{1} n \sqrt{\log(n)}$. Similarly, if $r_{n}$ is the $n$th positive integer all of whose prime factors are $\pm 3 \pmod{8}$, then $r_{n} \sim C_{2} n \sqrt{\log(n)}$. If you are looking for a modern treatment of this, which will enable you to compute the constants $C_{1}$ and $C_{2}$, this result is stated in Cojocaru and Murty's book "An introduction to sieve methods and their applications", but I don't have my copy at home right now, so I can't tell you exactly where in the book to look.<|endoftext|> TITLE: Polynomials dense with primes QUESTION [12 upvotes]: Let $p(n)$ be a polynomial with integer coefficients. Define $\Delta( p(n) )$, the prime density of $p(n)$, to be the limit of the ratio with respect to $n$ of the number of primes $p(k)$ generated when the polynomial is evaluated at the natural numbers $k=1,2,\ldots,n$: $$ \Delta( p(n) ) \;=\; \lim_{n \to \infty} \frac{ \textrm{number of } p(k), k \le n, \textrm{that are prime}} {n} $$ For example, Euler's polynomial $p(n)=n^2+n+41$ starts out with ratio $1$, but then diminishes beyond $n=39$:           And it continues to diminish ...           ... and by $n=10^7$ has reached $\Delta=0.22$. Q. What is the largest known $\Delta( p(n) )$ over all polynomials $p(n)$? In particular, are there any polynomials known to have $\Delta > 0$? Maybe these questions can be answered assuming one or more conjectures? REPLY [7 votes]: This has been asked multiple times before (with three variations by yours truly), so is a mega-duplicate, if you will: Bateman-Horn, continued even further Bateman-Horn conjecture, continued Unexpectedly prime rich cubic polynomial And even resulted in a preprint: Some experiments on Bateman-Horn I Rivin - arXiv preprint arXiv:1508.07821, 2015 - arxiv.org<|endoftext|> TITLE: Tate modules of commutative group schemes over finite field QUESTION [8 upvotes]: Let $G$ be a finite type commutative group scheme over a finite field $k=\Bbb F_q$ with $\Gamma=Gal(\bar k/k)$ and $l$ be a prime such that $(l,q)=1$, we can also define the Tate module $T_lG =\varprojlim_nG[l^n](\bar k)$. Then we can also consider the natural map $Hom_k(G_1,G_2) \otimes \Bbb Z_l \rightarrow Hom_\Gamma(T_lG_1,T_lG_2)$ for any two commutative group scheme over $k$, or more generally the map $\text{Ext}^i_k(G_1,G_2) \otimes \Bbb Z_l \rightarrow \text{Ext}^i_\Gamma(T_lG_1,T_lG_2)$ for every natural number $i$ at least when the extension groups are well-defined (This question may be related). My first question is: when is this map surjective, injective or an isomorphism? If $G_i$ are abelian varieties and $i=0$, it's well-known that this map is an isomorphism. If $G_2=\Bbb G_m$ and $G_1$ is an abelian variety, this is also true for $i=0$ as both sides are zero, and is true for $i=1$ and $\text{dim}G_1=1$ as $\text{Ext}^1(A,\Bbb G_m)$ is the dual of $A$ and one can compute both sides explicitly (see this master thesis). However, this map can fail to be injective for trivial reasons (For instance, right side could be zero for some finite abelian groups). So my second question is: how to describe extension groups between commutative group schemes using linear algebra datas in a systematic way? For instance, how to describe extension groups of finite group schemes like $\alpha_p$, $\mu_p$? The last question is about growth of $\#G[l^n]$ for commutative group schemes (in order to study the Tate module). Let $k$ be any field, $G$ be a commutative group scheme over a field $k$ (not necessarily finite type) and $p$ be a prime. Assume $G[p^n]$ is a finite group scheme for every $n$, does there always exist $C >0,h \geq 0$ such that $\#G[p^n]=Cp^{nh}$ for $n >> 0$? Here the order of a finite group scheme means the dimension of its global section ring. I know the result holds for etale group schemes and abelian varieties. For example, if $G$ is constant i.e an abstract abelian group and $\#G[p^n]$ is finite for every $n$, then we know $G[p^\infty]\cong (\Bbb Q_p/\Bbb Z_p)^h\oplus T$ for some $h \geq 0$ and finite $p$-group $T$ (this result is used when studying Tate-Shafarevich group of elliptic curves). Edit: there are good theory for commutative linear algebraic groups over complex number, for example every connected one is a product of $\Bbb G_a$ and $\Bbb G_m$, see here. Here is some discussion for higher extension groups. I am more interested in higher extension groups or non-reduced case. REPLY [5 votes]: I'll try to answer your last question on the order of $\#G[\ell^n]$ for $(\ell,p) = 1$. Assumption: $G$ a commutative connected group scheme locally of finite type over $k$. Since you're interested in the order of the group scheme $G[\ell^n]$, it is harmless to pass to $k = \overline{k}$ (a connected group scheme locally of finite type over $k$ is geometrically connected). Then $G_{\text{red}}$ is a closed reduced subgroup scheme of $G$ with $G_{\text{red}}[\ell^n] = G[\ell^n]$ as the latter is \'{e}tale over $k$. Therefore we may assume that $G$ is reduced, a fortiori smooth. By Chevalley's Theorem, we have an exact sequence $$0 \to H \to G \to A \to 0$$ where $H$ is a closed linear algebraic subgroup and $A$ an abelian variety. Lemma: For every $n > 0$, the sequence $$0 \to H[\ell^n] \to G[\ell^n] \to A[\ell^n] \to 0$$ is still exact. Proof: We have to show that $H/\ell^n H = 0$. Since $(\ell,p) = 1$ the multiplication by $\ell^n$ map is \'{e}tale and therefore open. However a homomorphism of group schemes always has closed image, and so by connectedness of $H$ we conclude the result. Since you know the answer to your last question in the case of an abelian variety $A$, I'll just treat the case of $H$. Now $k$ is algebraically closed, in particular perfect so we know that $$ H \cong D \times U$$ where $D$ is a diagonalizable group and $U$ is unipotent. Any unipotent $U$ group over a perfect field has a filtration with successive quotients isomorphic to $\mathbf{G}_a$. Therefore if $(\ell,p) = 1$ (as is the case here), $U[\ell^n] = 0$ and so $$H[\ell^n] \cong D[\ell^n].$$ In particular, we may assume that $H$ is connected diagonalizable and therefore isomorphic to $\mathbf{G}_m^q$ for some $q \in \mathbf{N}$ (there are no forms of $\mathbf{G}_m^q$ over an algebraically closed field). Hence $$\mathbf{G}_m^q[\ell^n] \cong (\mu_{\ell^n})^q$$ which has order $\ell^{nq}$.<|endoftext|> TITLE: Who wins the Rubik's cube game? QUESTION [20 upvotes]: This game has two players, Spoiler and Solver. We start with a solved 3x3x3 rubik's cube (to make the problem easier). Solver and Spoiler take turns making 90 degree twists (starting with Solver). The cube is forbidden from ever repeating a position (besides the start position). This guarantees the game is finite. If at any point (besides the beginning), the rubik's cube is in a solved state, Solver wins. If the game ends before that (because a position is entered with no valid moves), Spoiler wins. An example game would be F,F;F,F (using basic rubik's cube notation). Solver wins this game. If a game goes through each position that is one move away from the solved state, and afterwards goes to some unsolved state, Spoiler will win (since it is now impossible to get to the solved state). So, which player has the winning strategy? EDIT: It may be simpler to consider the same problem with the 15 puzzle first. REPLY [4 votes]: Not an answer, but just a thought or two about the problem. With the quarter turn metric the Cayley graph is a 12-regular graph. Every element in the Rubik's group can be assigned a "rank", the value of the smallest number of moves to get back to the origin. By making a parity argument, we can see that a quarter turn (move on the Cayley graph) always changes this number by 1 or -1. So we could arrange the Cayley graph like a Hasse diagram. I think understanding the size of the various levels (not sure if that's the term, but the collection of elements having the same minimum solve number) might be key to understanding any strategy for either player. For example on the first turn the state of the cube is at level 1. The Solver can't undo this move and will be forced to move to level 2. The Spoiler clearly wouldn't move the state back to level 1, so moves up to 3. I think a naive strategy for the Spoiler might be to try to move the puzzle up in level all the time. There are more level 3 states than level 2 states, so (being generous with the symmetry and structure of the Cayley graph) I suspect he can move to enough level 3 states any time the puzzle return to a level 2 state and exhaust the level 2 possibilities for the Solver, thus shutting the Solver off from victory. This is just my initial thoughts. I will update if I think of something more, or if the structure of the Cayley graph turns out to be structured in such a way as to stop this strategy. Let me know if anyone manages to build off this idea. EDIT: According to https://www.cube20.org/qtm/, the largest sized level is at 21. Being odd, these are states the Spoiler moves the puzzle to. I think that if the Spoiler tries to keep the puzzle at 21, then he might be able to exhaust the 20 level or the 22 level, before the level 21 states are exhausted.<|endoftext|> TITLE: Idempotent ring spectrum QUESTION [17 upvotes]: Is there a lot of ring spectrum which are idempotent in the sense that the multiplication map $R \wedge R \rightarrow R$ is an equivalence ? The sphere spectrum $\mathbb{S}$ and the $0$ spectrum are clearly examples. I believe that $\mathbb{S}[n^{-1}]$ i.e. the ring spectrum obtained by starting from $\mathbb{S}$ and formally inverting multiplication by a sets of integer, by taking the inductive limits of $\mathbb{S} \overset{\times n}{\rightarrow} \mathbb{S} \overset{\times n}{\rightarrow} \mathbb{S} \dots$ gives us an other example. And one can generalize by localizing at a set of integer. Is there other type of such example ? Can we classify them ? In case there is indeed more examples and they cannot be classified, I have an additional question: Is there any example other than $\mathbb{S}$ and $0$ which are connective and such that the fiber of the unit map $\mathbb{S} \rightarrow R$ is also connective (which if I'm correct amount to saying that $R$ is connective and the map $\mathbb{Z} = \pi_0(\mathbb{S}) \rightarrow \pi_0(R)$ is surjective, and it rules out all the localization mentioned above). REPLY [25 votes]: There are actually quite a number of other examples. In particular, this is necessary and sufficient for $R$ to be a so-called smashing localization of the sphere $\Bbb S$, and there are several prominent examples called the $E(n)$-local spheres as we range over primes $p$ and natural numbers $n > 0$. However, there are not many examples when $R$ is connective: all of them are obtained from the sphere spectrum $\Bbb S$ by inverting some set of primes. Here is a proof. The left unit map $R = R \wedge \Bbb S \to R \wedge R$ is an equivalence, and so is the right unit map. Suppose $E$ is a homology theory with a perfect Kunneth formula: $$E_*(X \wedge Y) \cong E_*(X) \otimes_{E_*} E_*(Y).$$ Then the maps $$ E_* (R) \to E_*(R) \otimes_{E_*} E_*(R) \to E_*(R) $$ induced by the left unit and the multiplication are isomorphisms. In particular, this is true for homology with coefficients in any field $\Bbb F$. Therefore, $H_*(R; \Bbb F)$ is an $\Bbb F$-algebra $A$ such that the left unit $A \to A \otimes_{\Bbb F} A$ is an isomorphism. By choosing a basis, we can conclude that either the unit $\Bbb F \to A$ is an isomorphism or $A = 0$. This gives us a set $S$ of primes such that the mod-$p$ homology of $R$ is trivial, and in the other cases the map $\Bbb S \to R$ is a mod-$p$ homology isomorphism. We can then use the universal coefficient theorem relating integral homology with mod-p homology. For any prime in $S$, both $H_*(R;\Bbb Z) \otimes \Bbb Z/p$ and $Tor(H_*(R;\Bbb Z), \Bbb Z/p)$ are trivial, so multiplication-by-$p$ is an isomorphism on $H_*(R;\Bbb Z)$. Similarly, for primes not in $S$, the map $\Bbb S \to R$ being a mod-$p$ homology isomorphism and naturality of the universal coefficient theorem imply that $Tor(H_*(R;\Bbb Z), \Bbb Z/p) = 0$. Thus $H_*(R; \Bbb Z)$ is torsion-free, hence flat. This means $H_*(R; \Bbb Z)$ is a subring of the rational homology. Now applied to the rational homology, we find that $H_*(R; \Bbb Q)$ is either trivial or $\Bbb Q$, so $H_*(R; \Bbb Z)$ is either $0$ or a subring of $\Bbb Q$. In the latter case, this forces $H_*(R;\Bbb Z)$ to be the localization of $\Bbb Z$ at the set of primes in $S$. If we finally assume that $R$ is connective, then $\pi_0(R) \cong H_0(R;\Bbb Z) = \Bbb Z[S^{-1}]$ by the Hurewicz theorem. This means that we get a factorization $\Bbb S \to \Bbb S[S^{-1}] \to R$ which is an isomorphism on integral homology. Since both sides are connective, by the Whitehead theorem we find that $\Bbb S[S^{-1}] \to R$ is a weak equivalence.<|endoftext|> TITLE: Positive cones in K-groups QUESTION [9 upvotes]: Let $X$ be a topological space or a scheme, and let $K^0(X)$ be $K$-group of vector bundles of $X$. One may ask when an element $x$ of $K^0(X)$ is represented by an actual vector bundle, and not just a formal sum. The set of such classes form a cone in $K^0(X)$. Are there any known statements about the structure of this cone? For example, consider the case $X = \mathbb{P^1}$. Since any vector bundle (at least when we are over a field) is isomorphic to a direct sum of bundles of the form $\mathcal O(k)$, and Euler exact sequence gives us the relation $[\mathcal O(-2)]=2[\mathcal O(-1)]-[\mathcal O]$, we have that $K^0(\mathbb P^1)=\mathbb Z[t]/t^2$, where $t$ is the class of $[\mathcal O] - [\mathcal O(-1)]$. So $[\mathcal O(k)]=1+kt$ in this presentation, and we see that the positive cone is the cone of elements such that free term is bigger than zero. In particular, it is not finitely generated (as a semigroup). Do we have a description of a positive cone in other nontrivial ($K^0(X) \ne \mathbb Z$) cases? Can we say that sometimes it is finitely generated? Can we at least describe this cone for $K^0(\mathbb P^1) = \mathbb Z[t]/t^{n+1}$? We can obtain one necessary condition for a class $x$ to lie in this cone by employing lambda operations on $K^0$: If $x$ is positive, then $\lambda^n(x) = 0$ for $n$ big enough. However, I doubt that this is a sufficient condition --- is there an example of not a positive class satisfying this? Of course, this question makes sense for $K^0$ of any exact category, so it would be interesting to know whether something can be said in these another contexts. For example, if we are considering $K^0$ of $C^*$-algebras, then a huge number of explicit examples arises for approximately finite algebras (which are direct limits of direct sums of matrix algebras). Thank you! REPLY [6 votes]: This is a very complicated problem. However, there is an approach to this type of question which might be helpful. Let us put it in a more general context. Let $R$ be any unital ring (not necessarily commutative), including possibly $C(X)$ for compact $X$ or $C(\tilde X)$ ($\tilde X$ is the one-point compactification of $X$ if $X$ happens to be locally compact). Then $K_0(R),K_0(R)^+$ is a pre-ordered abelian group with positive pre-cone, $K_0(R)^+$, the image of the fg projective modules. If $R$ is stably finite (which trivially holds when $R$ is commutative), then $K_0(R)$ is partially ordered, and the question revolves around determining $K_0(R)^+$. One approach is via traces, specifically, nonzero group homomorphisms $\tau: K_0(R),K_0(R)^+ \to {\bf R},{\bf R}^+$. Obviously, if $q \in K_0(R)^+$, then $\tau(q)\geq 0$, yielding a necessary condition. In the case of compact $X$, each $x \in X$ yields such a trace---but if $X$ is connected, they all yield the same one, corresponding to the rank, and in that case $\tau(q) > 0$ if $q \in K_0(R)^+ \setminus \{0\}$. If $R$ satisfies a somewhat awkward condition (equivalent to unperforation for partially ordered abelian groups---other names are sometimes used), for finitely generated projectives $P,Q$, if there exists an onto module map $P^n \to Q^n$ for some integer $n$ entails that there exists an onto module map $P \to Q$ (this happens often enough that it applies to large classes of rings), then sufficient for $q \in K_0(R)^+$ is that $\tau(q) > 0$ for all (pure) traces $\tau$ (this characterizes the positive elements in the interior of the positive cone). But this isn't the end of the story, because there are going to be lots of elements on the boundary of the positive cone (and still in the positive cone), and sometimes these are much more interesting (and more difficult to characterize). In this case, we look at order ideals of $K_0(R)$, and try to find the traces thereon. Determining the pure traces is itself a generally difficult problem. But if $K_0(R)$ admits a multiplication making it into a partially ordered ring (that is, $K_0(R)^+ \cdot K_0(R)^+ \subseteq K_0(R)^+$---as happens if $R$ is commutative, and occasionally for nice noncommutative $R$---the pure traces are precisely the multiplicative traces, which therefore should be relatively easy to determine. Unfortunately, if $R = C(X)$ and $X$ is compact connected, this yields only one trace ... There is a somewhat limited literature on this sort of thing (for general $R$), mostly for AF C*-algebras, but the techniques apply much more generally.<|endoftext|> TITLE: Alice and Bob playing on a circle QUESTION [27 upvotes]: I want to solve this problem: Let there be $n \ge 2$ points around a circle. Alice and Bob play a game on the circle. They take moves in turn with Alice beginning. At each move: Alice takes one point that has not been colored before and colors it red. Bob takes one point that has not been colored before and colors it blue. When all $n$ points have been colored: Alice finds the maximum number of consecutive red points on the circle and call this $R$. Bob finds the maximum number of consecutive blue points on the circle and call this $B$. If $R \gt B$, Alice wins. If $B \gt R$, Bob wins. If $R = B$, no one wins. Does Alice have a winning strategy for odd $n$? We still seem not to know for which odd $n$ Alice has a winning strategy. She does for $n=3$, and it seems also for $n=5$. But in general, I'm not sure. Could someone help? REPLY [8 votes]: Here is the Java code I used to simulate the game. Feel free to reuse it or translate it into another programming language. The value of $n$ is taken from the program arguments. On my 2013 MacBook Pro, it can prove the first player wins for odd $n$ up to 19 (that takes 2-3 minutes). The runtime increases exponentially for higher $n$ and it's not really optimized. It's programmed in the same way as endgame tablebases for chess; you start with (final) positions you know are winning for the first player. You retract the last move; for the first player, you're looking at moves which will reach a winning position (those are losing positions for the second player); for the second player, you're looking at positions in which all moves will reach a losing position (those are winning positions for the first player). You continue moving backwards until you reach the starting position, or no 'new' positions are left to consider. package net.mathoverflow.test; import java.util.ArrayList; import java.util.Arrays; import java.util.HashMap; import java.util.List; import java.util.Map; // https://mathoverflow.net/q/298443/70594 public class CircleGame { private static int n; // A 'board' is represented by an array of n numbers having one of the values below. // The first player is 'white', the second player 'black'. private static final short WHITE = 1, EMPTY = 0, BLACK = -1; public static void main(String[] args) throws Exception { // Determine the board size from the program argument n = Integer.parseInt(args[0]); if (n > 39) { // 3^40 > 2^63 (the size of a long) so the unique code generation might not be unique anymore. throw new Exception("Maximum board size is 39."); } // First, we need to determine which final positions are won for white and which aren't. // We are going to enumerate all final positions by first concentrating on the positions // of the white nodes. // We can assume WLOG White starts at 0; this is move 0. short[] initialBoard = new short[n]; initialBoard[0] = WHITE; populateFullBoards(initialBoard, 0, (n - 1) / 2); List newBoardsToConsider = new ArrayList<>(); int move = n / 2; do { if (move == n / 2 && n % 2 == 0) { // Black is making the last move, skip the white part in the first loop } else { System.out.println(currentBoardsToConsider.size() + " positions losing for Black on move " + move); if (currentBoardsToConsider.isEmpty()) { break; } // White to move; currentBoardsToConsider contains positions which are losing for Black with Black to // move. Therefore, we need to look for boards where White can reach one of currentBoardsToConsider // with a single move. for (short[] board : currentBoardsToConsider) { for (int i = 1; i < n; i++) { if (board[i] != WHITE) continue; // White can win by moving to i as last move short[] newBoard = Arrays.copyOf(board, n); newBoard[i] = EMPTY; // Already known? long code = getUniqueCode(newBoard); if (results.containsKey(code)) continue; results.put(getUniqueCode(newBoard), WHITE); newBoardsToConsider.add(newBoard); } } // Prepare Black's move currentBoardsToConsider = newBoardsToConsider; newBoardsToConsider = new ArrayList<>(); } System.out.println(currentBoardsToConsider.size() + " positions winning for White on move " + move); if (currentBoardsToConsider.isEmpty()) { break; } // Black to move; currentBoardsToConsider contains positions which are losing for Black with White to // move. Therefore, we need to look for boards where Black can reach only these losing positions, // whatever they move. for (short[] board : currentBoardsToConsider) { for (int i = 1; i < n; i++) { if (board[i] != BLACK) continue; // Undo last move short[] newBoard = Arrays.copyOf(board, n); newBoard[i] = EMPTY; // Already known? long code = getUniqueCode(newBoard); if (results.containsKey(code)) continue; // Try all black moves (except i which we already know loses) boolean isLosing = true; for (int j = 1; j < n; j++) { if (board[j] != EMPTY || j == i) continue; // Move to j short[] newNewBoard = Arrays.copyOf(newBoard, n); newNewBoard[j] = BLACK; Short result = results.get(getUniqueCode(newNewBoard)); if (result == null || result != WHITE) { isLosing = false; break; } } // Losing? if (!isLosing) continue; results.put(code, WHITE); newBoardsToConsider.add(newBoard); } } // Prepare White's move currentBoardsToConsider = newBoardsToConsider; newBoardsToConsider = new ArrayList<>(); move--; } while (move > 0); if (move == 0) { System.out.println(currentBoardsToConsider.size() + " positions losing for Black on move " + move); } } private static List currentBoardsToConsider = new ArrayList<>(); private static Map results = new HashMap<>(); private static long getUniqueCode(short[] board) { // For the unique code of the board, we add 1 to all board values and interpret the result as a ternary number // (node 0 is the lowest (i.e. '1') digit, node 1 is the '3' digit etc.). long power = 1, code = 0; for (int i = 0; i < n; i++) { code += power * (board[i] + 1); power *= 3; } return code; } private static void populateFullBoards(short[] board, int lastWhiteMove, int movesLeft) { if (movesLeft == 0) { // We've positioned all white nodes; the remaining are black nodes for (int i = 1; i < n; i++) { if (board[i] == EMPTY) board[i] = BLACK; } // Determine the result of the game short result = determineResult(board); if (result == WHITE) { currentBoardsToConsider.add(board); } results.put(getUniqueCode(board), result); return; } // Recursive part for (int i = lastWhiteMove + 1; i <= n - movesLeft; i++) { short[] newBoard = Arrays.copyOf(board, n); newBoard[i] = WHITE; populateFullBoards(newBoard, i, movesLeft - 1); } } private static short determineResult(short[] board) { int longestWhiteRun = 0; int longestBlackRun = 0; int currentRun = 1; short currentSide = WHITE; for (int i = 1; i < n; i++) { if (board[i] == currentSide) { currentRun++; } else { // Check if this run is the longest run (so far) for this side if (currentSide == WHITE && currentRun > longestWhiteRun) { longestWhiteRun = currentRun; } if (currentSide == BLACK && currentRun > longestBlackRun) { longestBlackRun = currentRun; } // Reset current run currentRun = 1; currentSide = board[i]; } } // If the last node is black, the current run ends there if (currentSide == BLACK && currentRun > longestBlackRun) { longestBlackRun = currentRun; } // If not, the longest white run might stretch over the end of the array if (longestWhiteRun > longestBlackRun) return WHITE; if (currentSide == WHITE) { for (int i = 0; i < n; i++) { if (board[i] != WHITE) break; currentRun++; if (currentRun > longestBlackRun) return WHITE; } if (currentRun > longestWhiteRun) { longestWhiteRun = currentRun; } } return (longestBlackRun > longestWhiteRun) ? BLACK : EMPTY; } }<|endoftext|> TITLE: Few questions regarding Heath-Brown's identity QUESTION [5 upvotes]: Heath-Brown's identity states: Let $K \geq 1, z \geq 1.$ Then for any $n < 2 z^K$ we have $$ \Lambda(n) = - \sum_{1 \leq k \leq K} (-1)^k {{K}\choose{k}} \sum_{ \substack{ m_1 \cdots m_k n_1 \cdots n_k = n \\ m_1, \ldots, m_k \leq z }} \mu(m_1) \cdots \mu(m_k) \log n_k. $$ I am trying to better understand this well used identity, and I have a few questions regarding this which I list below. I would greatly appreciate any comments on any of them. Thank you. 1) When is it more advantageous to use Heath-Brown's identity over Vaughan's identity? 2) When applying Heath-Brown's identity, does the terms with $1 \leq k \leq K-1$ are they all type I estimates and the term $k=K$ become type II estimate? (I would appreciate comments on how the identity is used normally...) 3) I have seen in few places that states Heath-Brown's identity is 'more flexible' than Vaughan's identity. What is meant by this? REPLY [7 votes]: Heath-Brown's identity is more flexible (than Vaughan's) in the sense that the participating convolutions have several factors (not just two). This allows greater freedom in choosing the supports of the factors, in particular, the factors can be supported on much shorter intervals. I am not sure what you mean by "normal use" of Heath-Brown's identity, but you can have a look at the Polymath8a paper, where the best result is achieved via Heath-Brown's identity and Deligne's work on the Riemann hypothesis for varieties over finite fields, while a weaker (but still very good) result is achieved via Vaughan's identity and Weil's work on the Riemann hypothesis for curves over finite fields.<|endoftext|> TITLE: What is the smallest Lipschitz constant of a Lipschitz retraction of $\ell_\infty([0,1])$ onto $C[0,1]$? QUESTION [7 upvotes]: By Theorem 1.6 in the book "Geometric Nonlinear Functional Analysis" by Benyamini and Lindenstrauss, the Banach space $C[0,1]$ is a Lipschitz retract of the Banach space $\ell_\infty[0,1]$. Unfortunately, the proof does not give any upper bounds on possible Lipschitz constants of the retraction. So, the Problem. Give some upper bounds on the smallest Lipschitz constant $L$ of a retraction $r:\ell_\infty[0,1]\to C[0,1]$. Is $L\le 20$? REPLY [11 votes]: Yes, as we have this theorem of Nigel Kalton: Let $K$ be a compact metric space. Then $C(K)$ is an absolute 2-Lipschitz retract. Please see [1] for details. [1] Kalton, Nigel J. "Extending Lipschitz maps into C (K)-spaces." Israel Journal of Mathematics 162.1 (2007): 275-315.<|endoftext|> TITLE: How is the Eichler-Shimura congruence related to L-functions? QUESTION [7 upvotes]: My understanding is that the Eichler-Shimura relation expresses the Hecke operator $T_p$ in terms of the geometric Frobenius map. Specifically, $T_p = Frob + Ver$ for Frobenius map $Frob$ and it's transpose $Ver$. However, Wikipedia states that "the Eichler–Shimura congruence relation expresses the local L-function of a modular curve at a prime p in terms of the eigenvalues of Hecke operators". I have no idea how that statement comes from the above definition of the congruence. How does such a decomposition of Hecke operators have anything to with its eigenvalues or L-functions of a modular curve? EDIT: Will's amazing answer gives the connection between L-functions and eigenvalues, but how is the local L-function of a modular curve restated explicitly using Eichler-Shimura? REPLY [7 votes]: Recall that the local $L$-function of the modular curve $X$ is $$\frac{1}{\det \left(1 - p^{-s} \operatorname{Frob}_p, H^1(X, \mathbb Q_\ell)\right)}.$$ The denominator is simply a variant of the characteristic polynomial of Frobenius, o it is sufficient to relate the eigenvalues of Frobenius to the Hecke eigenvalues. Now because $FV = p$, $F$ and $V$ commute, and so for each $\lambda$ an eigenvalue of $F$, $\lambda+ p/\lambda$ is an eigenvalue of $F+V$. Using this, one can give formulas for the characteristic polynomial of the Hecke operator in terms of the characteristic polynomial of Frobenius, or vice versa.<|endoftext|> TITLE: Can all contours of a function on a disk be made arbitrarily small? QUESTION [9 upvotes]: Denote $D=\{x^2+y^2\le1\}\subset\mathbb R^2$ a disk. Let $f:D\to\mathbb R$ be a continuous function on it. I am interested in restrictions of simple Morse functions on $\mathbb R^2$, but I suspect the answer is the same for any continuous or any smooth function. A contour of $f$ is a connected component of the preimage of a point, $f^{-1}(\mathrm{const})$. The diameter of a set is the maximum distance between its points. Can the maximum diameter of a contour be made arbitrarily small by a suitable choice of $f$? I suspect that no; namely, I suspect that at least one contour of $f$ will have diameter greater than, say, the radius of the disk, but I can't prove it. The diameter of the disk is not the bound: Consider a Y-shaped figure formed by three radii, and $f$ being the distance from this figure. Then the largest contour is the Y figure, with the diameter greater than the radius of the disk but smaller than the diameter of the disk. REPLY [9 votes]: Every continuous function on a unit disk $D^2$ has a level set containing a connected component of diameter at least $\sqrt{3}$; this constant cannot be increased. Generally, in case of $D^n$, $n>2$, there is a level set with a connected component of diameter at least $2$. See the paper "Level Sets on Disks" by A. Maliszewski and M. Szyszkowski, 2014, https://doi.org/10.4169/amer.math.monthly.121.03.222<|endoftext|> TITLE: Is each compact metric space a subset of a compact absolute 1-Lipschitz retract? QUESTION [6 upvotes]: A metric space $X$ is called an absolute $L$-Lipschitz retract if for any metric space $Y$ containing $X$ there exists a Lipschitz retraction $r:Y\to X$ with Lipschitz constant $Lip(r)\le L$. Question. Is each compact metric space isometric to a subset of a compact absolute 1-Lipschitz retract? Remark 1. Using almost isometric embeddings into the Banach space $c_0$, it can be shown that each compact metric space $X$ is a subset of a compact absolute $(1+\varepsilon)$-Lipschitz retract $Y$, where $\varepsilon$ is any positive real number. The space $Y$ is a suitable cube $\prod_{n\in\omega}[-a_n,a_n]$ in $c_0$ with a bit distorted metric. Remark 2. There exists also a functorial construction of an embedding of compact metric space $X$ into a compact absolute 8-Lipschitz retract $A(X)$. Given a compact metric space $X$, consider the isometric embedding $X\subset\ell_\infty$ identifying each point $x\in X$ with the distance function $d_X(x,\cdot)$. Next, take the closed convex hull $conv(X)$ of $X$ in $\ell_\infty$. Finally, consider the hyperspace $A(X)$ of non-empty convex compact subsets of $conv(X)$, endowed with the Hausdorff metric. By Theorem 1.7 in the book "Geometric Nonlinear Functional Analysis" by Benyamini and Lindenstrauss, the compact metric space $A(X)$ is an absolute 8-Lipschitz retract. I do not know if the constant 8 can be replaced by a smallest constant (say 1). Added at Edit. Thanks to the comment of @Wlod AA, I have found an answer to my question on page 32 of the book of Benyamini and Lindenstrauss. They write that Isbell in 1964 suggested the construction of the injective envelope of a metric space, which is the smallest 1-Lipschitz AR containing a given metric space. For a compact metric space its injective envelope is compact, too. REPLY [5 votes]: The classical paper on the given theme is by Aronszajn and Panitchkpakdi. I am pretty sure that it contains the required result about embedding metric spaces into metric absolute retracts, i.e. in Lip$_1$ category, and perhaps about embeddings compact metric spaces in compact metric absolute retracts. Indeed, these authors have introduced the notion of the hyper-convex metric spaces which is equivalent to metrically injective spaces (i.e. in Lip$_1$ category, which I call simply metric category or Met for short). This and further results are contained in later papers by John Isbell and (a bit later and independently) by wh. Let me provide perhaps the simplest embedding of arbitrary compact metric space $\ \mathbf X:=(X\ d)\ $ into a compact metric absolute retract. Let $\ Y:=\mbox{Met}_\delta\ (\mathbf X)\ $ be the set of all metric maps $\ f:X\rightarrow[0;\delta],\ $ where $\ \delta\ $ is the diameter of $\ \mathbf X\ $ (metric maps means Lip$_1).\ $ Then $\ Y\ $ in its uniform distance function is compact, it's hyper-convex i.e. it's an injective metric space (metric absolute retract), and the embedding $\ i:X\rightarrow Y\ $ is given by Kuratowski-Wojdysławski formula: $$ \forall_{s\ t\in X}\ \ (i(s))(t)\ :=\ d(s\ t) $$ (You need to be archeological to hear about this stuff). Theorem Let $\ \mathbf X:=(X\ d)\ $ be an arbitrary non-empty metric space of an arbitrary finite diameter. Let $\ -\infty\le a\le b\le\infty.\ $ Then space $\ Y\ \:=\ \mbox{Met}(\mathbf X\,\ \mathbb R\!\cap\![a;b])\ $ of all metric functions $\ f:X\rightarrow\mathbb R\cap[a;b]\ $ is hyper-convex. Proof   Let $\ \emptyset\ne F\subseteq Y $ and radia $\ r:F\rightarrow[0;\infty)\ $ be such that $$ \forall_{f\ g\in F}\ r_f+r_g\ge |f-g| $$ Define $\ c : X\rightarrow \mathbb R\!\cap\![a;b]\ $ as follows: $$ \forall_{x\in X}\ \ c(x)\ :=\ \max(a\ \ sup_{_{x\in X}}\ (f(x)-r_f)) $$ Then, by routine applications of the triangle inequality, the function $\ c\ $ has what it takes: $$ c\ \in Y\cap\bigcap_{f\in F} B(f\ r_f) $$ where $\ B(f\ r_f)\ $ is the ball centered in $\ f,\ $ of radius $\ r_f.$<|endoftext|> TITLE: Regarding extenders QUESTION [6 upvotes]: I asked this at math.stackexchange, but got no answers: Let $j\colon M\rightarrow N$ be an elementary embedding (between inner models) with $\operatorname{crit}(j)=\kappa$. Let $\kappa<\lambda$. Let $\mu$ be the minimal $\alpha$ with $\lambda\le j(\alpha)$ and let $E=\langle E_a\mid a\in[\lambda]^{<\omega}\rangle$ be the $(\kappa,\lambda)$-extender derived from $j$. Then, for every $\langle a_n\mid n<\omega\rangle$ (with $a_n\in[\lambda]^{<\omega}$) and $\langle X_n\mid n<\omega\rangle$ (with $X_n\in E_{a_n}$), there is a function $f\colon\bigcup\{a_n\mid n<\omega\}\rightarrow\mu$ such that for each $n<\omega$, we have $f"a_n\in X_n$. I'm interested in a proof/hint for this claim. Any help would be appreciated, thanks. REPLY [3 votes]: Fix $(a_n \mid n < \omega)$, $(x_n \mid n < \omega)$ such that $x_n \in E_{a_n}$ for all $n < \omega$. Without loss of generality we may assume $\{\xi\} \in \{a_n \mid n < \omega \}$ for all $\xi \in \bigcup \{a_n \mid n < \omega\}$ and $a,b \in \{a_n \mid n < \omega\} \implies a \cup b \in \{a_n \mid n < \omega \}$. (Otherwise close the sequence $(a_n \mid n < \omega)$ under these operations and add dummy values for the corresponding $x$'s.) Consider the set $T$ of all functions $$ t \colon \{a_0, \ldots, a_{n-1} \} \to [\kappa]^{< \omega} $$ such that $\forall i < n \colon t(a_i) \in x_i$ (in particular $\mathrm{card}(t(a_i)) = \mathrm{card}(a_i)$), $\forall i < j < n \colon a_i \cup a_j \in \{ a_0, \ldots, a_{n-1} \} \implies t(a_i \cup a_j) || a_i = t(a_i)$. The operation $||$ is defined as follows: Let $a \subseteq b$, $B$ be finite sets of ordinals such that $\mathrm{card}(b) = \mathrm{card}(B)$. Write $b = \{b_1 < \ldots < b_k\}$ and $B = \{B_1 < \ldots < B_k \}$. Let $$ \pi \colon \mathrm{card}(a) \to b $$ be the unique $<$-preserving function such that $a = \pi " \mathrm{card}(a)$. Then $$ B ||a := \{ B_{\pi(0)} < \ldots < B_{\pi(\mathrm{card}(a)-1)} \}. $$ $B || a \subseteq B$ is the subset of $B$ of those elements that correspond to the indexes of $a$'s elements in the increasing enumeration of $b$. Now consider the tree $(T; \subset)$. Claim. $(T; \subset)$ is ill-founded. Proof. Consider $j((T; \subset))$. Since $T$ is countable, elementarity yields that $$ j((T; \subset)) = (j " T; \subset) $$ and $$ j " T = \{ j(t) \colon \{ j(a_0), \ldots, j(a_{n-1}) \} \to [j(\kappa)]^{< \omega} \mid j(t)(j(a_i)) \in j(x_i) \wedge \ldots \}. $$ In $V$ consider $$ b \colon \{ j(a_n) \mid n < \omega \} \to [j(\kappa)]^{< \omega} $$ given by $b(j(a_i)) := a_i$. I'll leave it to you to check that $b \restriction \{j(a_0), \ldots, j(a_{n-1}) \} \in j " T$ for all $n < \omega$ so that $$ V \models (j " T; \subset) \text{ is ill-founded}. $$ By absoluteness of wellfoundedness we have that $$ N \models j((T; \subset)) = (j"T; \subset) \text{ is ill-founded} $$ and hence, by elementarity, that $$ M \models (T; \subset) \text{ is ill-founded}. $$ For the rest of this answer, work in $M$. Let $$ b \colon \{a_n \mid n < \omega \} \to [\kappa]^{< \omega} $$ be a branch through $(T; \subset)$. We define $$ f \colon \bigcup \{a_n \mid n < \omega \} \to \kappa, \xi \mapsto \bigcup b(\{\xi \}) = \text{ the unique element of } b(\{ \xi \}). $$ (This is possible by our assumption 1. on the sequence $(a_n \mid n < \omega)$.) Claim. $f$ is as desired. Proof. Let $a_n = \{\xi_0 < \ldots < \xi_k \}$. Then $$ \begin{align*} f " a_n &= \{ f(\xi_0), \ldots f(\xi_k) \} \\ &= \{ \bigcup b(\{\xi_0\}) , \ldots, \bigcup b (\{\xi_k \}) \} \\ & = \{ \bigcup b(a_n) || \{\xi_0 \}, \ldots, \bigcup b(a_n) || \{\xi_k\} \} \\ &= b(a_n) \in x_n. \end{align*} $$ Q.E.D.<|endoftext|> TITLE: A probabilistic angle inequality QUESTION [12 upvotes]: Conjecture: There is a universal constant $c$ such that for any fixed nonzero real vector $q$ of any dimension $n$ and any random vector $p$ of the same dimension $n$ with independent components uniformly distributed in $[-1,1]$, we have $$(p^Tp)(q^Tq)\le cn(p^Tq)^2$$ with probability $\ge 1/2$. Simulation suggests that the best constant is approximately $c\approx 16/7$. I'd be interested in techniques for proving this and related statements. In particular, can one determine the best constant $c$? An application of the inequality to numerical optimization is reported in the paper M. Kimiaei and Arnold Neumaier, Efficient global unconstrained black box optimization, http://www.optimization-online.org/DB_HTML/2018/08/6783.html REPLY [6 votes]: $\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\X}{\mathcal{X}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$ Let us show more: For each real $\ep\in(0,1)$ there is some real $c>0$ such that for any nonzero $q=(q_1,\dots,q_n)\in\R^n$ we have \begin{equation*} \PP((p^Tp)(q^Tq)\le cn(p^Tq)^2)\ge1-\ep, \tag{0} \end{equation*} where $p:=(U_1,\dots,U_n)$ and $U_1,\dots,U_n$ are iid random variables (r.v.'s) uniformly distributed in $[-1,1]$. Here, instead of $1/2$ in the OP's question, we have $1-\ep$. By rescaling, without loss of generality (wlog) $q^Tq=1$. Also, $p^Tp\le n$, so that the inequality $(p^Tp)(q^Tq)\le cn(p^Tq)^2$ is implied by $(p^Tq)^2\ge1/c$. So, letting $$\de=2/\sqrt c,$$ we see that it is enough to show that $\forall\ep>0$ $\exists\de>0$ $\forall q\in\Si_{n-1}$ $\PP(|S_q|\le\de/2)\le\ep$, where $\Si_{n-1}$ is the unit sphere in $\R^n$ and \begin{equation*} S_q:=p^Tq=\sum_1^n q_i U_i. \end{equation*} In turn, it is enough to show that \begin{equation*} Q_{S_q}(\de)\underset{\de\downarrow0}\longrightarrow0 \tag{1} \end{equation*} uniformly in $q\in\Si_{n-1}$, where \begin{equation*} Q_X(\de):=\sup_{x\in\R}\PP(|X-x|\le\de/2), \end{equation*} so that $Q_X$ is the so-called concentration function of a r.v. $X$. Let us now prove (1). Because the r.v.'s $U_1,\dots,U_n$ are iid and symmetrically distributed, wlog \begin{equation*} q_1\ge\cdots\ge q_n\ge0. \end{equation*} Take any $q_*\in(0,1)$. One of the following cases must occur. Case 1: $q_1\ge q_*$. Then \begin{equation*} Q_{S_q}(\de)\le Q_{q_1 U_1}(\de)\le\frac12\frac\de{q_1}\le\frac\de{2q_*}; \tag{1.5} \end{equation*} the first inequality in the above display is due to the fact that for any independent r.v.'s $X$ and $Y$ \begin{align*} Q_{X+Y}(\de)&=\sup_{x\in\R}\PP(|X+Y-x|\le\de/2) \\ &=\sup_{x\in\R}\int_\R \PP(Y\in dy)\PP(|X+y-x|\le\de/2) \\ &\le\int_\R \PP(Y\in dy)\sup_{x\in\R}\PP(|X+y-x|\le\de/2) =Q_X(\de). \end{align*} Case 2: $q_1\le q_*$. Then, by the Berry--Esseen (BE) inequality, \begin{align*} Q_{S_q}(\de)&\le Q_Z(\de)+2A\frac{\sum_1^n \E|q_iU_i|^3}{(\sum_1^n \E(q_iU_i)^2)^{3/2}} \\ &\le\frac\de{\sqrt{2\pi}}+2A\sum_1^n q_i^3\,\frac{3\sqrt3}4 \le \frac\de{\sqrt{2\pi}}+2Aq_*\,\frac{3\sqrt3}4, \end{align*} where $Z\sim N(0,1)$ and $A$ is the universal positive real constant factor in the BE bound for non-iid independent r.v.'s; the best currently known value for $A$ is $0.56$ (I. G. Shevtsova. Refinement of estimates for the rate of convergence in Lyapunov's theorem. Dokl. Akad. Nauk, 435(1):26--28, 2010); a slightly worse value, $0.5606$, was a bit earlier established by Tyurin. Thus, for all $q\in\Si_{n-1}$ \begin{equation*} Q_{S_q}(\de)\le\max\Big(\frac\de{2q_*},\frac\de{\sqrt{2\pi}}+2Aq_*\,\frac{3\sqrt3}4\Big). \tag{2} \end{equation*} So (1) follows by choosing $q_*=\sqrt\de$, say. The OP later requested an explicit expression for $c$ providing for (0). Let us get this as well. The $\max$ in (2) is minimized in $q_*\in(0,1)$ when the two arguments of the $\max$ are equal to each other, that is, when \begin{equation*} q_*=q_*(\de):=\frac{\sqrt{6 \sqrt{3} \pi A \de +\de ^2}-\de}{3 \sqrt{6 \pi } A}. \end{equation*} Equating now $\frac\de{2q_*(\de)}$ (which equals the $\max$ minimized in $q_*$) with $\ep$ and solving for $\de$, we get \begin{equation*} \de=\de(\ep):=\frac{4 \sqrt{2 \pi } \ep ^2}{3 \sqrt{6 \pi } A+4 \ep }, \end{equation*} whence \begin{equation*} c=c(\ep):=\frac4{\de(\ep)^2}=\frac{\left(3 \sqrt{6 \pi } A+4 \ep\right)^2}{8 \pi \ep^4} \tag{3} \end{equation*} is enough for condition (0). In particular, using the mentioned best known value $0.56$ for $A$, we get $c(1/2)=54.98\dots$. The OP has now also requested a lower bound $c$ providing for (0). Let us get this as well. Suppose that for some real $\ep\in(0,1)$, some real $c>0$, all natural $n$, and all nonzero $q=(q_1,\dots,q_n)\in\R^n$ we have (0), where, as before, $p=(U_1,\dots,U_n)$ and $U_1,\dots,U_n$ are iid r.v.'s uniformly distributed in $[-1,1]$. Then, choosing $q=(1,\dots,1)/\sqrt n$, we have \begin{equation} \PP(|T_n|\ge1/\sqrt c)\ge1-\ep, \end{equation} where \begin{equation} T_n:=\frac{\sum_1^n U_i}{\sqrt{\sum_1^n U_i^2}}. \end{equation} By the central limit theorem and the law of large numbers, $\sum_1^n U_i/\sqrt n\to Z\sim N(0,1)$ in distribution and $\sum_1^n U_i^2/n\to\E U_1^2=1/3$ in probability (say), so that $T_n\to Z\sqrt3\sim N(0,3)$ in distribution; the convergence here is for $n\to\infty$. So, \begin{equation} 2(1-\Phi(1/\sqrt{3c}))\ge1-\ep, \end{equation} where $\Phi$ is the standard normal cumulative distribution function; that is, \begin{equation} c\ge c_\ep:=\tfrac13\,/\Phi^{-1}(\tfrac12+\tfrac\ep2)^2\sim\tfrac2{3\pi\ep^2} \end{equation} as $\ep\downarrow0$, because $\Phi(x)-1/2\sim x/\sqrt{2\pi}$ as $x\to0$ and hence $\Phi^{-1}(\tfrac12+u)\sim u\sqrt{2\pi}$ as $\to0$. This lower bound, $c_\ep$, on $c$ in (0) is quite a bit lower than the upper bound $c(\ep)$ in (3), especially for small $\ep$. For $\ep=1/2$, we have the lower bound $c_{1/2}=0.7327\dots$ on $c$ vs. the mentioned upper bound $c(1/2)=54.98\dots$. The gap is pretty large. One can follow the lines of the above proof to see the causes of this gap. One such cause is the first inequality in (1.5). The other one is due to the use of the Berry--Esseen (BE) bound -- which is true for all distributions (with the worst case being discrete distributions, whereas the distributions of $q_i U_i$ are absolutely continuous with bounded densities) and uniformly for all deviations from the mean. One may try to improve the bounds on $c$, but I believe that would be quite outside the usual scope for MathOverflow answers. Here one may also note that in the possibly simpler case of the so-called nonuniform BE bounds, the gap between the constant factors in the best known upper bound ($25.80$) and in the best known lower bound ($1.0135$) is similarly very large; see e.g. the paper On the Nonuniform Berry--Esseen Bound or its arXiv version.<|endoftext|> TITLE: Embedding the $\ 2^n+1$-point metric 1-space QUESTION [6 upvotes]: A metric space $\ (S\ d)\ $ is said to be a 1-space $\ \Leftarrow:\Rightarrow\ \forall_{x\ y\in S}\ (x\ne y\ \Rightarrow\ d(x\ y)=1).$ Question:   Do there exist a non-negative integer $n,\ $ and an $n$-dimensional Banach space $\ X,\ $ and an isometric embedding of the $\ (2^n+1)$-point metric 1-space $\ (S\ d)\ $ into $\ X$. My obvious conjecture is NO. Added AFTER @Fedor Perov's Answer: The power of the surprisingly simple Fedor's solution is in involving the measure theory. It worked regardless of any choice of any Banach norm. In fact, now Fedor may follow with writing down a solution of my old conjecture from 1972/3 which a priori was harder: CONJECTURE:   Let Banach norm $\ ||.||\ $ in $\ \mathbb R^n\ $ be such that $\ B^n:=(\mathbb R^n\,\ ||.||)\ $ contains a $2^n$-point 1-space $\ S.\ $ Then $\ B^n\ $ is a metrically injective space, i.e. it admits affine coordinates such that the norm is given by $\ \max.\ $ Then--furthermore--under such coordinates, there exists $\ (a_1\ \ldots\ a_n)\in \mathbb R^n\ $ such that $$ S\ =\ \prod_{k=1}^n\ \{a_k\,\ a_k\!+\!1\} $$ Indeed, Fedor's construction provides a decomposition of the convex hull $\ Cv(S)\ $ into $\ 2^n\ $ copies of the $\frac 12$-scaled of $\ Cv(S)\ $ where these copies have disjoint interiors. REPLY [11 votes]: Indeed no. If it exists, for each of our points $p_i$ consider the set $A_i$ homothetic to their convex hull with center $p_i$ and coefficient bit less than $1 /2$. By volume argument some two such sets $A_i, A_j$ must have a common point $q$, but this contradicts to triangle inequality as $p_iq, p_jq$ are less than $1/2$.<|endoftext|> TITLE: PhD dissertations that solve an established open problem QUESTION [61 upvotes]: I search for a big list of open problems which have been solved in a PhD thesis by the Author of the thesis (or with collaboration of her/his supervisor). In my question I search for every possible open problem but I prefer (but not limited) to receive answers about those open problems which had been unsolved for at least (about) 25 years and before the appearance of the ultimate solution, there had been significant attentions and efforts for solving it. I mean that the problem was not a forgotten problem. If the Gauss proof of the fundamental theorem of algebra did not had a gap, then his proof could be an important example of such dissertations. I ask the moderators to consider this question as a wiki question. REPLY [13 votes]: Does Serre's (Jean-Pierre) thesis qualifies ? He computed there a lot of homotopy groups of spheres. But I don't know how old was this problem in 1951.<|endoftext|> TITLE: If the center of a finite group is trivial, are there two elements whose centralizers intersect trivially? QUESTION [16 upvotes]: Let $G$ be a (discrete) group. Define $k^*(G)$ as the minimal cardinality of a set $S \subset G$ such that $C_G(S) = Z(G)$. Define $k(G) = k^*(G)$ if $G$ has trivial center (i.e. $|Z(G)| = 1$), and $k(G) = \bot$ otherwise. If $k(G) = \bot$, then the convention is that neither $k(G) \leq n$ nor $k(G) \geq n$ holds, for any cardinal $n$. Question: Does there exist a finite group $G$ such that $k(G) \geq 3$, and more generally does every natural number $k(G) \geq 3$ occur for some finite group $G$? I do not even know such examples for $G$ infinite, and would also be interested in such, though I do not have an immediate application for this. I am not an expert on group theory (especially finite group theory), so I do not know very effective search terms for this, and would also be interested in pointers to the literature. What I have tried so far (though don't take my word on these): No abelian group or a p-group or a nilpotent group is an example, since they have nontrivial centers (in the finite case), thus $k(G) = \bot$. No finite simple group is an example, since they are all 2-generated (by CFSG), thus satisfy $k(G) \leq 2$ or $k(G) = \bot$. $k(G \times H) = \max(k(G), k(H))$ for any groups $G, H$ (by a simple proof). I did a quick search in GAP and seems that there are no finite groups of size up to $1151$ with this property (this is the first time I used GAP, so not sure how much proof value this has). $k(G) = 0$ for precisely the trivial group, and $k(G) = 1$ is impossible (since any $g$ commutes with itself). For infinite cardinal $\kappa$, $k(G) = \kappa$ where $G$ is the group of finite-support permutations on a set of cardinality $\kappa$, but of course $k(G)$ is finite (or $\bot$) for finite groups. Arbitrarily large $k^*(G)$ are provided by wreath products $\mathbb{Z}_2 \wr \mathbb{Z}_2^d$, where $\mathbb{Z}_2 = \mathbb{Z}/2\mathbb{Z}$, but I wrote a quick proof sketch that $k(G \wr H) = 2$ whenever $|H| \geq 2$ and $k(G) = 2$ so it seems one cannot use wreath products to get examples. For a single permutation $\pi$ on a set of size $n$, The centralizer of $\pi$ in $S_n$ never has $k(G) \geq 3$ (I prove this as Proposition 7.9 in this paper of mine, by a rather ad hoc case analysis). In terms of the commuting graph $\Gamma(G)$ with vertices $G \setminus Z(G)$ and edges $\{(g, h) \;|\; gh = hg\}$, the question of whether $k(G) \geq 3$ is possible is equivalent to whether there exists a finite group $G$ with trivial center such that $\mathrm{diam}(\Gamma(G)) = 2$, where $\mathrm{diam}$ is the diameter, i.e. maximal minimal distance between a pair of vertices. For a finite minimal nonsolvable group the diameter is always at least $3$ according to this paper which implies $k(G) = 2$ for a minimal nonsolvable group (a different definition is used in that paper, but it should be equivalent to mine for minimal nonsolvable groups, as they have trivial center). Most literature I know about this graph and its diameter are about finding upper bounds, but I do not know if that has a relation to $k(G)$. Larger values of $k(G)$ also correspond to statements about this graph, but not about its diameter. For context, the question arose from the study of automorphism groups of one-dimensional subshifts. I am interested in quantitative versions (or lack thereof) of the so-called Ryan's theorem, which states that the center of the automorphism group of a mixing subshift of finite type consists of only the shift maps. I ask the above question about finite groups after Lemma 7.7 here and Lemma 7.7 is my application for it. The paper of Boyle, Lind and Rudolph is a standard reference for these groups. REPLY [14 votes]: Here's another family of examples with arbitrary large $k$, more based on linear algebra. Fix any odd prime $p$. Consider the group $G_{p,s}$ of square matrices of size $s+2$ over the field $F=\mathbf{Z}/p\mathbf{Z}$ of the form $$m^\pm(u,v,z)=\begin{pmatrix}\pm 1 & ^tu & z\\ 0 & I_s & v\\ 0 & 0 & 1\end{pmatrix};$$ with $u,v\in F^s$ and $z\in F$. Its order is $2p^{2s+1}$. Its center is trivial (the center of the subgroup of index 2 is reduced to the cyclic group of elements $m^+(0,0,*)$. For any $s-1$ elements in $G_{p,s}$. Then they are contained, for some hyperplane $H$ of $F^s$, in the subgroup $\Gamma$ consisting of those $m^\pm(u,v,z)$ with $u\in H$. Then there exists $w\in F^p\smallsetminus\{0\}$ such that $^tuw=0$ for all $u\in H$. Then $m^+(0,w,0)$ belongs to the centralizer of $\Gamma$. Hence $k(G_{p,s})\ge s$ (actually $\le s+1$).<|endoftext|> TITLE: Status of the three-body problem QUESTION [9 upvotes]: I find many numerical results on the three-body problem, but what is rigorously proved? Especially I would be interested in the parameter domains for which we have rigorous lower bounds on the topological entropy or Lyapunov exponent of the system. REPLY [9 votes]: Although there is no solution for the general three body problem there are many related results, especially for the restricted case. Here is a study of LE's in restricted 3BP. And here is a classification scheme for the mean motion resonance of the restricted 3BP including using the maximal LE for characterizing the resonances. These will point you to many related results. As a historical aside, Gutzwiller, in his book[1] says that the ancient Greeks knew there were 3 different lunar months (depending on how you measure them in they sky): the Sidereal month (27.32166 days), the anomalistic month (27.55455 days) and the nodical month (27.21222 days). These result from 3 body interactions (earth, moon, sun). The Greeks were apparently able to observe these to an accuracy of 1 second per month. Gutzwiller calls these the first precision scientific measurements in history. Every time I think about this I am astonished; it must have taken immense effort over centuries to compile the needed data. [1] Martin C. Gutzwiller Chaos in Classical and Quantum systems.<|endoftext|> TITLE: Minimal refinements of open covers of $T_2$-spaces QUESTION [9 upvotes]: Let $(X,\tau)$ be a topological space. We say ${\cal U}\subseteq \tau$ is an open cover if $\bigcup {\cal U} = X$, and $X\notin {\cal U}$. ${\cal U}$ is minimal if for all $U_0\in {\cal U}$ we have $\bigcup \big({\cal U}\setminus \{U_0\}\big) \neq X$. Clearly, every $T_1$-space on more than $1$ point possesses a minimal cover: pick $x\neq y\in X$ and let ${\cal U} = \big\{X\setminus\{x\}, X\setminus\{y\}\big\}$. Question. Given any open cover of a Hausdorff space $(X,\tau)$ with $|X|>1$, does it have a refinement that is a minimal cover? Note. I did not ask for subcovers in the question because of the following example: Let $X = \mathbb{R}$ with the Euclidean topology, then ${\cal U} = \big\{ \{x\in\mathbb{R}: x < n\} : n\in\mathbb{N}, n\geq 1 \big\}$ does not have a minimal subcover. REPLY [5 votes]: I believe that $\omega_1$ with the order topology is a counter example. Call a subset of $\omega_1$ bounded if it is contained in some initial interval $[0,\alpha]$ and unbounded otherwise. Lemma: Any cover of $\omega_1$ by bounded open subsets is not minimal. I am sure that this is standard, but here is a proof. It contains way too many notations and is a bit clumsy, I am afraid, but I hope it is readable. We can assume that such a cover is is of cardinality $\omega_1$. Let thus $\mathcal{V} = \{V_\alpha\,:\,\alpha\in\omega_1\}$ be a cover of $\omega_1$ by bounded open subsets. By a standard argument the sets below are closed and unbounded in $\omega_1$: \begin{align*} C_1 &= \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta\subset[0,\alpha)\}\\ C_2 &= \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta\supset[0,\alpha)\} \end{align*} Hence their intersection $C = \{\alpha\,:\, \cup_{\beta<\alpha} V_\beta =[0,\alpha)\}$ is closed and unbounded as well. We may assume that each member of $C$ is a limit ordinal. When $\alpha\in \omega_1$, let $\alpha^*$ denote the smallest element of $C$ which is $>\alpha$. For each $\alpha\in C$ choose $V_{s(\alpha)}\ni\alpha$ with $s(\alpha)\in[\alpha,\alpha^*)$. Since $V_{s(\alpha)}$ is open and $\alpha$ is limit, $V_{s(\alpha)}$ contains some interval $[\gamma(\alpha),\alpha]$. By Fodor's Lemma, there is some $\gamma\in\omega_1$ such that the set $S$ of $\alpha\in C$ such that $\gamma(\alpha) = \gamma$ is stationary and in particular unbounded. For any $\alpha\in S$ with $\alpha>\gamma^*$, $V_{s(\alpha)}$ contains in particular the interval $[\gamma^*,\alpha]$. Fix now $\eta\in[\gamma^*,\gamma^{**})$. Choose $\alpha \ge \gamma^{**}$ in $S$. Then $$ V_\eta \subset V_{s(\alpha)} \cup \left(\cup_{\beta<\gamma^*}V_\beta\right). $$ Notice that $\gamma^*\le\eta<\alpha\le s(\alpha)$, hence the indices in the union at the righthand side do not contain $\eta$. It follows that $\cup (\mathcal{V} \backslash \{V_\eta\}) = \omega_1$. Corollary: The cover of $\omega_1$ by the intervals $[0,\alpha]$, $\alpha\in\omega_1$, has no refinement that is a minimal cover.<|endoftext|> TITLE: What is the minimal volume of the intersection of a self-dual cone and the unit ball? QUESTION [24 upvotes]: When thinking of some other problem, I stumbled upon the following innocently looking question that is natural enough to have been considered (and, possibly, solved) many years ago. However my attempts to search the literature for an answer resulted in next to nothing. Let $K\subset\mathbb R^n$ be a convex cone and let $K^*=\{y:\langle x,y\rangle\ge 0\text{ for every }x\in K\}$ be its dual cone. Suppose that $K\supset K^*$ (or, if you prefer, even that $K=K^*$). What is the minimal possible ratio $\frac{|K\cap B|}{|B|}$ where $B$ is the unit ball in $\mathbb R^n$ when $n$ is large? The answer should, probably, be of order $2^{-n}$ (positive orthant) but the best clean lower bound I can prove myself with my "homemade tools" is $(\sqrt 2+1)^{-n}$ (it can be improved a bit further to something like $2.317^{-n}$ but the argument gets somewhat messy and it is clear that this way won't lead to the optimal estimate). Any help would be appreciated. REPLY [5 votes]: $\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\X}{\mathcal{X}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$ I thought the following simple restatement of a bit narrower problem might help. Suppose that the convex cone $K$ is polyhedral: \begin{equation} K=\{x\in\R^n\colon a_i\cdot x\ge0\ \;\forall i\in\{1,\dots,N\}\}, \end{equation} where $a_1,\dots,a_N$ are nonzero vectors in $\R^n$ and $\cdot$ denotes the dot product (the case of a general convex cone $K$ can hopefully be done by approximation). Then, by Farkas' lemma (say), the dual cone $K^*$ is the conical span of the $a_i$'s. So, the condition $K\supseteq K^*$ means that $a_i\cdot a_j\ge0$ for all $i,j$, that is, all the angles between the vectors $a_i$ are $\le\pi/2$. If now $N\le n$, then it should be comparatively easy to move the $a_i$'s so that they become pairwise orthogonal and in the process the cone $K^*$ (which is the conical span of the $a_i$'s) only increases, so that $K$ only decreases, and then so does its "spherical angle" measure $\mu(K):=\frac{|K\cap B|}{|B|}$. Thus, in the case when $N\le n$, we will have $\mu(K)\ge1/2^n$, as desired. Then it will "only" remain to consider the case when $N>n$. Then one may try to find movements of $a_i$'s which will bring all of them into the conical hull of $n$ of the $a_i$'s without increasing $\mu(K)$; this certainly seems much more difficult to do than the above. Another thing to try here may be to embed the $a_i$'s into, or approximate them by vectors in, or otherwise map them into, $\R^N$.<|endoftext|> TITLE: Is there something like internal language of an abelian category? QUESTION [10 upvotes]: While studying topos theory I was wondering if there is something like internal logic of an abelian category. Aparently the answer is yes (by 7º slide in https://www.mimuw.edu.pl/~gael/xxi/files/slides_posters/Blechschmidt.pdf) but I can't see what this weaker variant could be. Besides that, is there relations between this internal logic and the internal logic of a topos? This relations would be interesting for topos theory? Maybe for homological algebra? Is there more relations between homological algebra and topos theory beyond Chapther 8 (Cohomology) of "Topos Theory" Johnstone's book? I know that SGA4 is a reference in this topic to. It's my first time making a question here, sorry for any mistake and thanks in advance REPLY [9 votes]: I don't know exactly what Blechschmidt had in mind, but an abelian category is in particular a regular category (indeed, a Barr-exact category), and hence it has an internal logic that is a regular logic, having $\exists,\wedge,\top$ but no other connectives (except "$\forall$ and $\Rightarrow$ at top-level" in the sense of an entailment). This regular logic doesn't include the "additiveness" of an abelian category, but you could augment it by axioms making every object an abelian group object and every morphism an abelian group homomorphism.<|endoftext|> TITLE: Limit of a Combinatorial Function QUESTION [6 upvotes]: I need help with the following problem, proposed by Iurie Boreico: Two players, $A$ and $B$, play the following game: $A$ divides an $n \times n $ square into strips of unit width (and various integer lengths). After that, player $B$ picks an integer $k$, $1 \leq k \leq n$, and removes all strips of length $k$. Let $l(n)$ be the largest area that $B$ can remove, regardless how $A$ divides the square into strips. Evaluate $$ \lim_{n \to \infty} \frac{l(n)}{n}. $$ Some progress that I made on this problem: Observe that there is at most $l(n)/k$ strips of length $k$. So, $l(n)n - \sum_{k = 1}^n l(n) \pmod k = \sum_{k = 1}^n k \lfloor l(n)/k \rfloor \geq n^2$. Now I'm stumped on how to asymptotically bound the left hand side, and I cannot find this problem posted anywhere online. Any solutions, observations, progress is appreciated. REPLY [2 votes]: Here is another way to get the bound obtained by Fedor Petrov. One observation is that from the covering of the board using strips of size $1 \times i$ and $1 \times n-i$ to cover row $r$ for rows $1 \leq r \leq n$, then $l(n) \leq (n-1) + (n-1) = 2n-2$. From the Pigeonhole principle, we also have that $n \leq l(n)$. So, $n \leq l(n) \leq 2n-2$. As stated in the OP, we have that $$ l(n)n - \sum_{k = 1}^n l(n) \pmod k \geq n^2. $$ We will try to get a bound the second sum. To make notation more simple, we let $l(n) = L$. Splitting the sum from $1 \leq k \leq L/2$ and $L/2 < k \leq n$, we have that \begin{align*} \sum_{n \geq k > L/2} l(n) \pmod k & = \sum_{n \geq k > L/2} L - k \\ & = \left ( n - \frac{L}{2} \right ) \left ( \frac{3L}{4} - \frac{n}{2} \right) + O(L) \\ & = nL - \frac{n^2}{2} - \frac{3L^2}{8} + O(L). \end{align*} For the second sum, we have that \begin{align*} \sum_{k \leq L/2} L \pmod k & = \sum_{i = 2}^L \sum_{\frac{L}{i+1} < k \leq \frac{L}{i}} L \pmod k\\ & = \sum_{i = 2}^L \sum_{\frac{L}{i+1} < k \leq \frac{L}{i}} L - ik \\ & = \sum_{i = 2}^L \frac{L^2}{2i(i+1)^2} + L \cdot O\left (\frac{1}{i+1} \right) \\ & = O(L \log L) + \frac{L^2}{2} \sum_{i = 2}^L \frac{1}{i(i+1)^2}. \end{align*} Since $$\sum_{i = 2}^L \frac{1}{i(i+1)^2} = \sum_{i = 2}^\infty \frac{1}{i(i+1)^2} - \sum_{i \geq L+1} \frac{1}{i(i+1)^2} = \frac{7}{4} - \frac{\pi^2}{6} + O \left (\frac{1}{L^2}\right ).$$ So, $$\sum_{k \leq L/2} L \pmod k = O(L \log L) + \left ( \frac{7}{8} - \frac{\pi^2}{12} \right ) L^2.$$ Adding our two expressions together, we have that $$\sum_{k = 1}^n L \pmod k = nL - \frac{n^2}{2} + \left ( \frac{1}{2} - \frac{\pi^2}{12} \right ) L^2 + O(L \log L).$$ Hence, \begin{align*} nL - \sum_{k = 1}^n l(n) \pmod k & \geq n^2 \\ \left ( \frac{\pi^2}{12} - \frac{1}{2} \right ) L^2 + O(L \log L) & \geq \frac{n^2}{2} \\ \left ( \frac{L}{n} \right )^2 & \geq \frac{6}{\pi^2 - 6} + o(1) \\ \frac{L}{n} & \geq \sqrt{ \frac{6}{\pi^2 - 6} } + o(1). \end{align*} Taking the limit as $n \to \infty$, we find that the limit is $\boxed{\geq \sqrt{6/(\pi^2 - 6)}}$. The only upper bound I've arrived at is $2$ which can be achieved by the first observation.<|endoftext|> TITLE: Why does the class equation have rational coefficients? QUESTION [7 upvotes]: Let $j$ be the Klein modular invariant and let $\Gamma=SL_2(\mathbb Z)$ be the modular group. Consider the set of primitive quadratic forms $ax^2+bxy+cy^2$ of discriminant $d<0$. The root of a quadratic form $f(x,y)$ is the unique complex number $\omega$ from the upper half-plane such that $f(\omega, 1)=0$. It follows, that two quadratic forms are equivalent, if and only if their respective roots are in the same orbit of the action of $\Gamma$ on the upper half plane. Therefore we may associate to each equivalence class of quadratic forms of the discriminant $d$ a unique value of the $j$-invariant. Now choose a root from every equivalence class, say $\omega_1,...,\omega_h$, where $h=h(d)$ is the class number and consider the class equation $$H(x)=(x-j(\omega_1))\cdots(x-j(\omega_h)).$$ Why should this polynomial have integral coefficients? I am seeking some simple reason why $j(\omega)$, where $\omega$ is a root of a quadratic form of discriminant $d$, should be an algebraic integer of degree at most $h(d)$. The modular equation shows, that $j(\omega)$ is an algebraic integer, see my question Singular values of the j-invariant. I am aware that more advanced theory shows that the degree of $j(\omega)$ is actually $h(d)$. Also there is an elementary proof of the weaker statement in the Cox's book Primes of the form $x^2+ny^2$, but this requires a lot of work. On the other hand Weber in his Lehrbuch der Algebra proves that $H(x)$ has rational coefficients by some sort of induction. However, this is scarcely readable. My ultimate goal is to understand the proof of Heegner-Stark theorem as described in Stark's article. I asked a question about it before: Modular functions of the type $\mathfrak f(\cdot)^{k}\mathfrak f(\cdot)^{23nk}$, but got no response. Fortunately I had resolved that problem and now this is the next question to be answered. REPLY [8 votes]: I am not sure there is a simple reason you are looking for, because any proof will relate ideal classes to Galois automorphisms, and this is what class field theory is about. At any rate, Shimura in his book "Introduction to the arithmetic theory of automorphic functions" answers your questions nicely. Theorem 4.14 states that $j(\omega)$ is always an algebraic integer, and the proof is rather short and elementary. Then, Theorem 5.7 shows that $H(x)$ has rational coefficients (and it is irreducible over $\mathbb{Q}$), whence the coefficients are in fact integers (because they are algebraic integers).<|endoftext|> TITLE: Minimum euclidean spanning tree in n dimensional space QUESTION [5 upvotes]: I need to compute the minimum euclidean spanning tree in $R^d$ and do it with some algorithm that can do it with complexity near to $\Omega(nlogn)$ where $n$ is the size of the point set. Right now I'm thinking about usage of delaunay triangulation for my set of points and after that use some MST algorithms like Prim\Kruskal, but as far as I see this won't give me needed complexity bounds. Could someone point me to the right references about this problem from which I could: Learn theory behind Write exact algorithm to solve this problem REPLY [5 votes]: The best you can hope for is $O(n \log n)$ plus a term dependent upon accuracy, for finding an approximate Euclidean MST. The fastest known exact algorithm is just a hair better than quadratic, $O(n^2)$. Below is a central paper on the topic. Note especially: "The algorithm is deterministic and very simple." Arya, Sunil, and David M. Mount. "A fast and simple algorithm for computing approximate Euclidean minimum spanning trees." Proceedings 27th Annual ACM-SIAM Symposium Discrete Algorithms. SIAM, 2016. (PDF download.)                                                             Fig.6a.<|endoftext|> TITLE: continuity points of elementary embeddings from $0^\sharp$ QUESTION [7 upvotes]: Suppose $0^\sharp$ exists, and let $\langle \alpha_i : i \in \text{Ord}\rangle$ be the Silver indiscernibles for $L$. Let $j : L \to L$ be the embedding generated by mapping $\alpha_n$ to $\alpha_{n+1}$ for finite $n$ and fixing the rest of the indiscernibles. Suppose $\alpha_0 < \delta <\alpha_\omega$ and $\delta$ is regular in $L$. Is it the case that $\sup j[\delta] < j(\delta)$? More generally, pick any order preserving map from the indiscernibles to itself and consider the generated embedding. How does one compute at what points the embedding is continuous? REPLY [9 votes]: Claim: Suppose $i : L \to L$ is an elementary embedding. If $\kappa$ is an $L$-regular cardinal and $\sup i[\kappa] < i(\kappa)$, then $\kappa$ is a Silver indiscernible. The claim answers your more general question: the discontinuity points of $i$ are then precisely the ordinals with $L$-cofinality equal to some Silver indiscernible at which $i$ is discontinuous (since if $i$ is discontinuous at $\alpha$ then $i$ is discontinuous at the $L$-regular cardinal $\text{cf}^L(\alpha)$, which must therefore be a Silver indiscernible). Proof of claim. Let $\kappa' = \sup i[\kappa]$. Consider the hull $H = H^L(i[L]\cup \kappa')\prec L$. Let $k : L\to L$ be the inverse of the transitive collapse of $H$ and let $j = k^{-1}\circ i$. (If you're comfortable with long extenders, it is easier to see $j$ as the extender ultrapower by the $L$-extender of length $\kappa'$ derived from $i$, and $k$ as the factor embedding.) We claim that $j(\kappa) = \kappa'$. Since $\kappa'\subseteq H$, we have $\text{crit}(k)\geq \kappa'$ and therefore $\sup j[\kappa] = \sup i[\kappa] = \kappa'$. Hence $\kappa' = \sup j[\kappa] \leq j(\kappa)$. Conversely, we must show $j(\kappa) \leq \kappa'$. Suppose $\alpha < j(\kappa)$; we'll show $\alpha < \kappa'$. Then $\alpha = j(f)(\xi)$ for some ordinal $\xi < \kappa'$ and some function $f\in L$. Fix $\bar \xi < \kappa$ be such that $j(\bar \xi) > \xi$ and let $\bar \alpha = \sup (f\restriction \bar \xi)$. Then $\bar \alpha < \kappa$ since $\kappa$ is $L$-regular, and $j(\bar \alpha) \geq \sup j(f\restriction \bar \xi) \geq j(f\restriction \bar \xi)(\xi) = \alpha$. Hence $\alpha < \kappa'$. So $j(\kappa) = \kappa'$. (This is the proof of a well-known extender fact: if $\kappa$ is $M$-regular then any $M$-extender whose constituent $M$-ultrafilters lie on ordinals less than $\kappa$ gives rise to an embedding of $M$ that is continuous at $\kappa$.) But now $\kappa' = \text{crit}(k)$: we have seen that $\text{crit}(k)\geq \kappa'$. But $k(\kappa') = k(j(\kappa)) = i(\kappa) > \kappa'$, so $\text{crit}(k) = \kappa'$. It follows that $\kappa'$ is a Silver indiscernible: it is the critical point of an elementary embedding of $L$. But then since $j(\kappa) = \kappa'$, $\kappa$ is also an indiscernible. (Otherwise fix a finite set of ordinals $a\subseteq \kappa$ and a finite set of indiscernibles $b$ above $\kappa$ such that $\kappa$ is definable in $L$ from $a\cup b$. Then $j(\kappa) = \kappa'$ is definable in $L$ from $j(a)\cup j(b)$. But $j(a)\subseteq \kappa'$ and $j(b)$ is a finite set of indiscernibles above $\kappa'$. This contradicts that $\kappa'$ is an indiscernible, since an indiscernible is never definable from ordinals below it and indiscernibles above it.)<|endoftext|> TITLE: Let $f \in \mathbb{Z}[x]$. Does $\bar{f}$ have as many roots in $\mathbb{F}_p$ as $f$ has in $\mathbb{C}$ for infinitely many primes $p$? QUESTION [8 upvotes]: Let $f \in \mathbb{Z}[x]$ be a nonconstant polynomial. Consider $\bar{f} \in \mathbb{F}_p[x].$ Let $\rho_p$ be the number of distinct roots of $\bar{f}$ in $\mathbb{F}_p$, and let $\rho$ be the number of distinct roots of $f$ in $\mathbb{C}$. Do you have a reference for showing that $\rho_p = \rho$ for infinitely many primes $p$? My background to this question: I actually use this result to calculate the precise maximal subgroup growth of groups of the type $\mathbb{Z}^k \rtimes \mathbb{Z}$ (and of similar metabelian groups). REPLY [13 votes]: Yes. I will focus on the case that $f$ is irreducible, thus also separable, over $\mathbf Q$. (The general case can be deduced from this, the main case of interest.) For large primes $p$ the reduced polynomial mod $p$ is separable. There is a theorem in algebraic number theory that "in each number field, infinitely many primes split completely". This follows from the zeta-function of each number field having a simple pole at $s=1$. In the application of this to the field $K = \mathbf Q(a)$, where $f(a) = 0$, a large prime $p$ splitting completely in $K$ will be a prime $p$ for which $f \bmod p$ splits into distinct linear factors. REPLY [11 votes]: Yes. Chebotareff (really Frobenius here) density theorem says that each cycle type in the Galois group of $f$ occurs as the splitting type of $f$ modulo $p$ infinitely often. Your case corresponds to the cycle type of the identity element. The Chebotareff theorem says that this happens asymptotically once in $|Gal(f)|$ primes (the statement about infinite number of primes is actually much easier, and follows from the Chinese remainder theorem, but what the heck).<|endoftext|> TITLE: Is totalization (of a cosimplicial category) a part of some adjunction? QUESTION [8 upvotes]: For a diagram category $\Gamma$ and and a cocomplete category $\mathcal{C}$, we have an equivalence $$\mathrm{Fun}(\Gamma,\mathcal{C}) \simeq \mathrm{Adj}(Set^{\Gamma^{Op}},C)$$ where for $F: \Gamma \to C$ we have a pair of adjoint functors $\mathcal{L}_F: Set^{\Gamma^{Op}} \to \mathcal{C}: S \mapsto \mathrm{colim}_{h_\gamma \to S}F(\gamma)$ (left adjoint) and $\mathcal{N}_F: \mathcal{C} \to Set^{\Gamma^{Op}}: c \mapsto \mathcal{C}(F(-),c)$ (right adjoint). For $\Gamma = \Delta$, this states that a cosimplicial object in $\mathcal{C}$ defines an adjunction between simplicial sets and $\mathcal{C}$. For example, when $\mathcal{C}=Top$ and the cosimpicial object consists of standard toplogical simplices, $\mathcal{L}_F$ is geometric realization and $\mathcal{N}_F$ is singular set. Generally for $\Gamma=\Delta$, the explicit formula for $\mathcal{L}_F$ is $$\mathcal{L}_F(S_\bullet) = Coeq(\amalg_{\phi:[m]\to[n]}F([m])\times S_n \rightrightarrows \amalg_{[n]}F([n])\times S_n).$$ When instead of a simplicial set $S_\bullet$ we have a cosimplicial category $A^\bullet$, there exists a (somewhat dual) notion of totalization: $$\mathrm{Tot}(A^\bullet):=Eq(\Pi_{[n]}\mathrm{Fun}(Iso(n),A^n) \rightrightarrows \Pi_{[n] \to [m]}\mathrm{Fun}(Iso(n),A^m))$$ where $Iso(n)$ stands for the category "string of $n$ isomorphisms". This totalization is a correct notion (e.g. it works well for $A^\bullet$ being sheaves on a Čech nerve of a cover). I would expect it to be a part of some general picture similar to $\mathrm{Fun}(\Gamma,\mathcal{C}) \simeq \mathrm{Adj}(Set^{\Gamma^{Op}},C)$, only in enriched setting (sets are replaced by categories) and for $\Gamma=\Delta^{Op}$. However, it is not exactly the picture above, as $Iso(n)$ is a functor $\Delta \to Cat$ not $\Delta^{Op} \to Cat$. What is this general picture for totalizations? Is there an adjoint functor? REPLY [9 votes]: There are a number of ways of expressing this duality. The colimit in the question may be written as the coend $\mathcal L_F(S_\bullet) = \int^{[n] \in \Delta} F([n]) \times S_n$, while the limit may be written as $\mathrm{Tot}(A^\bullet) = \int_{[n] \in \Delta} \mathrm{Hom}(Iso([n]), A^n))$. The colimit in the question is the weighted colimit $\mathcal L_F(S_\bullet) = F \otimes S$ (the colimit "of" the functor $S$, "weighted" by the presheaf $F$). The limit in the question is the weighted limit $\mathrm{Tot}(A^\bullet) = \{Iso,A\}$ (the limit "of" the functor $A$, "weighted" by the copresheaf $A$). The colimit in the question computes the the left Kan extension $\mathcal L_{(-)}(S_\bullet)$ of the functor $S_\bullet$ along the Yoneda embedding $\Delta \to Fun(\Delta^{op},\mathsf{Set})$. The limit in the question (generalized from the case of the functor $Iso$ in the obvious way) computes the right Kan extension $\mathrm{Tot}^{(-)}(A^\bullet)$ of the functor $A^\bullet$ along the coYoneda embedding $\Delta \to Fun(\Delta,\mathsf{Set})^{op}$. The category of presheaves $Fun(\Delta^{op},\mathsf{Set})$ is the free cocompletion of $\Delta$, so that $Fun(\Delta \simeq Fun^{cocts}(Set^{\Delta^{op}},\mathcal C)$ for cocomplete $\mathcal C$ (where $Fun^{cocts}$ means cocontinuous functors). Dually, the opposite of the category copresheaves $Fun(\Delta,\mathsf{Set})^{op}$ is the free completion of $\Delta$, i.e. there is an equivalence $Fun(\Delta,\mathcal{C}) \simeq Fun^{cts}((Set^{\Delta})^{op},\mathcal{C})$ (where $Fun^{cts}$ means continuous functors). Applying the adjoint functor theorem to the previous point, we obtain the equivalence mentioned in the question $Fun(\Delta,\mathcal C) \simeq LAdj(Set^{\Delta^{op}},\mathcal C)$ (the cateogry of left adjoint functors). Dually, we have $Fun(\Delta,\mathcal{C}) \simeq RAdj((Set^{\Delta})^{op},\mathcal{C})$ (the category of right adjoint functors. And all of this works for any small $\Gamma$ in place of $\Delta$, and any complete and cocomplete $\mathcal{C}$.<|endoftext|> TITLE: Compact manifolds which do not admit a diffeomorphism with a dense orbit QUESTION [6 upvotes]: What is an example of a compact manifold which does not admit a diffeomorphism with at least one dense orbit? Moreover, is it true to say that every isometry of $\mathbb{C}P^n$ with the Fubini-Study metric do not possess any dense orbit? REPLY [9 votes]: Dolgopyat and Pesin proved (ETDS 2002) that "every compact manifold of dimension $\geq 2$ admits a Bernoulli diffeomorphism with non-zero Lyapunov exponents". That is, on any such manifold $M$ there is a $C^\infty$ diffeomorphism $f$ that preserves volume $m$ and has the property that $(M,m,f)$ is a measure-preserving transformation that is measure-theoretically isomorphic to a Bernoulli shift. In particular, by the Birkhoff ergodic theorem, $m$-a.e. point $x\in M$ has the property that its orbit equidistributes with respect to $m$, meaning that for every continuous $\phi\colon M\to \mathbb{R}$, we have $\frac 1n \sum_{k=0}^{n-1} \phi(f^kx) \to \int \phi\,dm$ as $n\to\infty$. In particular, any such $x$ has a dense orbit.<|endoftext|> TITLE: Natural bijection between Dyck paths and tilting modules QUESTION [6 upvotes]: Let $A_n=kQ_n$ be the path algebra of linear oriented Dynkin graph $Q_n$ of Dynkin type $\mathcal{A_n}$ (so $A_n$ is the unique hereditary Nakayama algebra given by quiver and relations). The number of tilting $A_n$-modules equals the Catalan numbers $C_n$ when $n$ is the number of simple $A_n$-modules. Question: Is there a natural bijection between Dyck paths of length $2n$ to the tilting modules of $A_n$? I would think that this might be noted somewhere in the literature, but I did not find it. REPLY [3 votes]: Start with a tilting module for the quiver $0\to 1\to 2\to ...\to n$. Throw away the components with support at 0. You get a support tilting module $T_1,...,T_k$ for $1\to 2\to ...\to n$. Take, for each $T_i$, the part $\lambda_i$ equal to the top of $T_i$. For example, every injective module gives $\lambda_i=1$. Take the Young diagram of that partition. The shadow of the Young diagram is the corresponding Dyck path of length $2n+2$. Conversely, take a Dyck path of length $2n+2$. Then each down-step $D_i$ corresponds to a component $T_i$ of the corresponding tilting module. The top of $T_i$ is at vertex $k$ where $k$ is the number of upsteps after $D_i$. If $D_i$ goes from level $\ell+1$ to level $\ell$, the length of $T_i$ is equal to the number of upsteps since the last time the Dyck path went up to level $\ell$ (if none exists, $T_i$ is projective). For example, $T_{n+1}$ will be the projective injective $P_0$.<|endoftext|> TITLE: Contemporary philosophy of mathematics QUESTION [39 upvotes]: Starting to write an introduction to the philosophy of mathematics, I find tons of positions that are of historical interest. Which philosophical positions are explicitly considered these days, say in the last ten years? REPLY [2 votes]: Scott Aaronson's article "Why philosophers should care about computational complexity" might be of interest: https://www.scottaaronson.com/papers/philos.pdf<|endoftext|> TITLE: Is the strong operator topology metrizable? QUESTION [10 upvotes]: Let $X$ be a separable Banach space. Is the strong operator topology metrizable on $B(X)$, the space of all bounded operators on $X$? SOT-$\lim T_i=0~$ if and only if $~\lim \|T_ix\|=0$ for every $x\in X$. REPLY [2 votes]: Not sure whether this is equivalent to former answers. For maps between metric spaces, continuous is equivalent to sequentially continuous (see this wiki page). Consider the composition of operators in $B(X)$, it is a map from $B(X) \times B(X) \to B(X)$. It can be shown that this map is sequentially continuous (by playing with the definitions), but it is not continuous (See a counterexample here). So $B(X)$ is not metrizable.<|endoftext|> TITLE: List Ramsey numbers? QUESTION [23 upvotes]: The diagonal Ramsey number $R(n,n)$ is the least number $m$ for which the following holds: in any edge-colouring of the complete graph $K_m$ in which each edge is coloured blue or red, there is a monochromatic $K_n.$ The list Ramsey number (my name for it) $R_\ell(n)$ is the least number $m$ for which the following holds: we can assign to each edge of the complete graph $K_m$ a list of two distinct colours so that, in any edge-colouring of $K_m$ where each edge receives a colour from its assigned list, there is a monochromatic $K_n.$ This definition is what we get if we regard the diagonal cases of Ramsey's theorem as assertions about the chromatic numbers of certain hypergraphs, and then replace chromatic numbers with list chromatic numbers. Plainly $R_\ell(n)\le R(n,n),$ since we can simply assign the list $\{\text{blue, red}\}$ to each edge. On the other hand, the probabilistic lower bound works just the same for $R_\ell(n)$ as for $R(n,n),$ so we have $R_\ell(n)\ge2^{n/2}.$ Questions. Have such things been studied? Does it ever happen that $R_\ell(n)\lt R(n,n)$? Of course we can define analogous variants of the (diagonal) multicolour and hypergraph Ramsey numbers, but maybe the above is enough insanity for today. P.S. In a comment Thomas Bloom brought up the list colouring conjecture, which says that the list edge chromatic number of any graph is equal to the ordinary edge chromatic number. Here is a common generalization of the list colouring conjecture (for simple graphs) and the conjecture that $R_\ell(n)=R(n,n)$: General Conjecture. For any simple finite graphs $G$ and $H$ and any $c\in\mathbb N$ the following statements are equivalent: (a) any edge-colouring of $G$ with at most $c$ colours contains a monochromatic copy of $H$; (b) we can assign to each edge of $G$ a list of $c$ distinct colours so that, in any edge-colouring of $G$ where each edge receives a colour from its assigned list, there is a monochromatic copy of $H.$ This reduces to the list colouring conjecture (for simple graphs) when $H=K_{1,2},$ and it reduces to my question about $R_\ell(n)$ versus $R(n,n)$ when $G,H$ are complete graphs and $c=2.$ I suppose such questions are much too general to be worth thinking about. Maybe I should just ask: Question 0. Is $R_\ell(3)=6$? [Oops, see below.] I guess that's true but the proof might be a bit tedious. P.P.S. Actually it's quite easy to see that $R_\ell(3)=6=R(3,3),$ i.e., if each edge of $K_5$ is assigned a list of two colours, we can give each edge a colour from its list without creating a monochromatic triangle. Since $R(4,4)=18,$ I guess the following is the smallest nontrivial case: Question 1. Is $R_\ell(4)=18$? P.P.P.S. Although it doesn't fit in with my original question (which asked only about lists of size $2$), perhaps a better starting point would be to ask about the list version of the tricolour Ramsey number $R(3,3,3)=17.$ This amounts to asking whether the General Conjecture holds when $c=3$ and $H=K_3$ and $G=K_{16}:$ Question 2. If a list of $3$ different colours is assigned arbitrarily to each edge of the complete graph $K_{16},$ can we colour each edge with a colour from its assigned list, without creating a monochromatic triangle? REPLY [11 votes]: we stumbled upon the same concept in a different way and discovered this topic when looking if someone studied it already. We have figured out some interesting things, such as: disproving your general conjecture (it is false for matchings), proving it is true for stars (except for very small stars), proving behaviour is very different between the Ramsey numbers even for 3-uniform hypergraphs, obtaining several other results giving various bounds for general (hyper)graphs. Your original question of whether it is equal for cliques stays open, even for multiple colours. Here is the link to the paper: https://arxiv.org/abs/1902.07018<|endoftext|> TITLE: "This category obviously leads to paradoxes of set theory." What is the paradox? QUESTION [12 upvotes]: Eilenberg and Mac Lane formally defined categories in their 1945 paper General Theory of Natural Equivalences. Their definition of a category starts as follows: "A category {A,a} is an aggregate of abstract elements A (for example, groups), called the objects of the category and etc." When they consider their first example of categories on P. 239, namely the category of all sets, they immediately remark: "This category obviously leads to paradoxes of set theory." Obviously, all sets collected together don't form a set. However, their definition does not require the objects of a category to form a set. They use the term "aggregate" in their definition, perhaps to avoid this particular issue. That is, a set can be an aggregate, a class (in the sense of NBG) can also be an aggregate. Since the "aggregate" of all sets is a class, I do not see the "paradox" that they say will arise from considering the category of all sets. Unless by "aggregate" they really meant set. So my question is: what is the paradox that they were referring to? Please note, my question is specific to Eilenberg and Mac Lane's comment about the category of all sets. Obviously, there are other paradoxes caused by their definition of categories, and allowing the notion of a class doesn't eliminate all set-theoretic paradoxes from category theory. But I am not asking about these topics. On the other hand, as pointed out on page 245 of R. Kromer's book (Tool and Object): "Eilenberg and Mac Lane use the term 'set' in the combination the set of all objects of [a] category, on page 238 of their paper." Is this an evidence that by "aggregate" they really meant "set"? REPLY [24 votes]: I interpret the question as not being about the problems of size in category theory in general and how to deal with them (which are fairly well-understood and the subject of other questions on this site), but about what Eilenberg and MacLane actually meant in their original paper. The phrasing of that particular footnote is sloppy, but I think section 6 of their paper ("Foundations") suggests that what they meant is that "this category would lead to paradoxes if we required the objects of a category to form a set rather than something like a proper class". My guess is that they used the word "aggregate" in the definition in section 2 as a nod to the fact that to be formal, one may want to take these to be proper classes (or something related), but assumed that the average mathematician reading the paper would interpret "aggregate" as "set" at least until they got to section 6. So they added a footnote pointing out that they were aware of the issue, but deferred a fuller discussion of it (and an explanation of what "aggregate" can formally be defined to mean, or other ways one can deal with the problem while still interpreting "aggregate" as "set") to the later section. For instance, in section 6 they wrote "we have chosen to adopt the intuitive standpoint, leaving the reader free to insert whatever type of logical foundation (or absence thereof) he may prefer".<|endoftext|> TITLE: Free graded Lie algebras QUESTION [11 upvotes]: Let $R$ be a graded commutative unital ring and $M$ a graded $R$-module (all gradings are over $\mathbb{Z}$). I'm looking for a reference for the following statement: If $M$ is $R$-free, then the free graded Lie algebra over $R$ generated by $M$ is also $R$-free. For the purposes of this question, a graded Lie algebra over $R$ is a graded $R$-module $L$ with an $R$-linear map $[-, - ] : L\otimes_RL\to L$ satisfying the following identities for all homogeneous elements $a,b,c\in L$: (1) $[a,b]+(-1)^{|a||b|}[b,a]=0$ (2) $(-1)^{|a||c|} [a,[b,c]] + (-1)^{|b||a|}[b,[c,a]] + (-1)^{|c| |b|} [c,[a,b]] = 0 $ (3) $[a,a]=0$ if $a$ is of even degree (4) $[a,[a,a]]=0$. I should also say that I'm aware of Reutenauer's book, and, as far as I can tell, it does not deal with the graded case. REPLY [4 votes]: I will answer under the assumption that $2$ is invertible in $R$, the general case of which follows after extension of scalars from the case $R=\mathbb{Z}[1/2]$, which is a particular case of Proposition 8.5.1 in Neisendorfer’s “Algebraic Methods in Unstable Homotopy Theory.” The assumption that $2$ is invertible is necessary to ensure that Definition 8.1.1 coincides with the definition stated above (see the following remark on p. 262). I do not know if this assumption is necessary.<|endoftext|> TITLE: 0-1 matrix corresponding to an abstract simplicial complex QUESTION [6 upvotes]: Let $A$ be a 0-1 matrix whose columns are maximal. We can associate its rows with vertices and columns with simplices in an abstract simplicial complex. Conversely, given an ASC, we can encode it in a 0-1 matrix. What is the name of this matrix? Surely it must be used in the literature, but I can't seem to find an instance. REPLY [2 votes]: This looks like the incidence matrix of a clutter or hypergraph (whose edges are precisely the facets of your complex). It shows up quite a bit in optimization and combinatorial commutative algebra. For example see this question for some interesting conjectures. If you are interested in this area, the key words are: max-flow min-cut, packing problem, facet ideals, etc.<|endoftext|> TITLE: Computational complexity of finding the class number QUESTION [8 upvotes]: Let $f(x)$ be an integral irreducible monic polynomial and $\alpha$ be its root. What is the computational complexity of finding representatives of ideal classes of the integral ring $\mathbb{Q}[\alpha]$? What is a natural bound for "sizes" of the representatives in terms of the coefficients of $f$? As is explained in Keith Conrad's notes, this is related to the conjugacy problem for integral matrices. REPLY [5 votes]: I'm not an expert, but this is what I learned from Lenstra's Algorithms in Algebraic Number Theory (Bull AMS, 1992) and Kirschmer and Voight's paper Algorithmic enumeration of ideal classes for quaternion orders: For a number field $F$ of degree $n$ and absolute discriminant $d_F$, it seems that the best general results about computing the ideal class group comes from Buchmann's 1990 papers A subexponential algorithm for the determination of class groups and regulators of algebraic number fields and Complexity of algorithms in algebraic number theory: One can determine the class group with a deterministic algorithm which runs in time $d_F^{3/4}(2+\log d_F)^{O(n)}$, or in expected run time $d_F^{1/2}(2+\log d_F)^{O(n)})$ with a probabilistic algorithm. (Under GRH, one can improve the latter bound for fixed $n$.) Edit: Aurel points out in a comment below that an improvement of Schoof gives a deterministic algorithm with runtime $d_F^{1/2}(2+\log d_F)^{O(n)})$, and heuristically it should be computable in subexponential (wrt $\log d_F$) time. While this does not quite answer your question, it suggests that a natural bound for the "size" is the pair of the degree and discriminant of $F=\mathbb Q(\alpha)$. Your question of course involves more general orders than the full ring of integers of $F$, and I have not thought about the details, but because one can reduce the class number calculation for $\mathbb Q[\alpha]$ to that for $F$ and the conductor of this suborder, I would expect your problem to be reducible to the class group problem for $F$ in time that is essentially the size of the ring of integers of $F$ modulo the size of your suborder. (This seems similar to a reduction Kirschmer and Voight do for Eichler orders in quaternion algebras.)<|endoftext|> TITLE: Odds on rolling a rhombicosidodecahedron QUESTION [10 upvotes]: This is more of a curiosity to me, but I'm sure I don't have the mathematical skills to answer it. That said... I took a look at several other posts with questions that relate to this one, but I haven't seen this specifically addressed... Given a uniform solid rhombicosidodecahedron, what are the odds of rolling each of the faces: triangle, square and pentagon? I'm interested in a theoretical answer here, I'm not concerned with reality... like a roller trying to "cheat" the roll for a specific outcome. And I'm obviously not concerned with the die being "fair". So let's assume that the roll is truly random. REPLY [8 votes]: As a first pass, we can approximate the odds of landing on a face by projecting the polyhedron on a sphere, and taking the fraction of the sphere which it covers. Then the odds of a triangular, square or pentagonal roll overall are $$14.4\%,\ 50.4\%,\ 35.1\%$$ and the odds for an individual triangular, square or pentagonal face are $$0.72\%,\ 1.68\%,\ 2.93\%.$$ This is Simpson's method. It also represents the result of throwing the die high above an adhesive surface, so that the die is well-randomized in the air, and then after touching the surface falls onto the nearest side. The angles have exact formulas which are easy enough to calculate in Mathematica: data = PolyhedronData["SmallRhombicosidodecahedron"]; faces = Map[data[[1, 1]][[#]] &, data[[1, 2, 1]]]; angle[a_, b_, c_] := 2 ArcTan[Abs[a.Cross[b, c]]/ (Norm[a] Norm[b] Norm[c] + a.b Norm[c] + b.c Norm[a] + c.a Norm[b])]; Map[Length[#] angle[# // Mean // Simplify, #[[1]], #[[2]]] &, faces] // FullSimplify // Union This gives the following solid angle measures for each triangular, square or pentagonal face: $$6 \cot^{-1}\left(2 \sqrt{3} u+\sqrt{124 u-61}\right),\\ 8 \cot^{-1}\left(2u+\sqrt{40 u-21}\right),\\ 10 \cot^{-1}\left(2 \sqrt{5u}+3 \sqrt{2u+1}\right)$$ where $u=5+2\sqrt{5}$.<|endoftext|> TITLE: Is the strict henselization isomorphic to the filtered colimit of finite etale algebras? QUESTION [7 upvotes]: Let $(A,\mathfrak{m})$ be a local ring, and let $A^{\mathrm{sh}}$ be the strict henselization of $A$ at $\mathfrak{m}$. Let me denote $A^{\mathrm{sh},\mathrm{fin}}$ for the filtered colimit of finite etale $A$-algebras (with a fixed map to the separable closure of $A/\mathfrak{m}$). There is a canonical map \begin{align} \varphi : A^{\mathrm{sh},\mathrm{fin}} \to A^{\mathrm{sh}} \end{align} of $A$-algebras. When is $\varphi$ an isomorphism? Remarks/thoughts: This is true if $A$ is a field. We can consider using Zariski's Main Theorem (say this version) in some way. If $A^{\mathrm{sh},\mathrm{fin}} \to B$ is an etale ring map, there exists a factorization $A^{\mathrm{sh},\mathrm{fin}} \to C \to B$ where $A^{\mathrm{sh},\mathrm{fin}} \to C$ is finite and $C \to B$ induces an open immersion $\operatorname{Spec} B \to \operatorname{Spec} C$, but I don't know whether $A^{\mathrm{sh},\mathrm{fin}} \to C$ is etale (or whether it can be made etale after a refinement of $B$). It seems we can make $A^{\mathrm{sh},\mathrm{fin}} \to C$ finite flat, namely using the structure theorem for etale ring maps (e.g. 00UE) which says we can take $C = A^{\mathrm{sh},\mathrm{fin}}[t]/(f(t))$ for some monic polynomial $f(t) \in A^{\mathrm{sh},\mathrm{fin}}[t]$ and $B$ to be a principal localization of $C$. REPLY [4 votes]: If your proposed description is correct, then the strict henselization of $A$ would be integral over $A$, and hence the same holds true for any subring of the strict henselization. But this is essentially never true (unless $A$ is henselian). For an explicit example, take $A = \mathbf{C}[x]_{(x)}$. Choose a map $f:X \to \mathbf{A}^1$ of smooth affine curves of degree $3$ such that $f^{-1}(0) = \{x,y\}$ with $x$ unramified and $y$ ramified. Take $B = \mathcal{O}_{X,x}$, so $\mathrm{Spec}(B)$ is the affine open subscheme of $X \times_{\mathbf{A}^1} \mathrm{Spec}(A)$ obtained by removing $y$. Then $A \to B$ is a local \'etale map of local domains with the same residue field, so $B$ occurs as a subring of the strict henselization of $A$. But $A \to B$ is not integral as it fails the valuative criterion of properness: there is a valuation on $\mathrm{Frac}(B) = K(X)$ corresponding to the point $y \in X$ that has a center on $\mathrm{Spec}(A)$ but not one on $\mathrm{Spec}(B)$.<|endoftext|> TITLE: Why Use Hypercohomology When Defining the de Rham Cohomology of a Smooth Scheme over $k$? QUESTION [12 upvotes]: Hopefully this question is of an appropriate level for this site: I'm reading some notes by Claire Voisin titled Géométrie Algébrique et Géométrie Complexe. Let $X$ be a smooth $k-$scheme. In these notes, one constructs the (algebraic) de Rham complex $$ 0\xrightarrow{}\mathscr{O}_X\xrightarrow{d}\Omega_{X/k}\xrightarrow{d}\Omega_{X/k}^2\xrightarrow{d}\cdots \xrightarrow{d}\Omega_{X/k}^n\xrightarrow{}0$$ where $n=\dim(X)$. Now, one constructs the algebraic de Rham cohomology by way of using the hypercohomology of the complex, $H^l_{dR}(X/k):=\mathbb{H}^l(\Omega_{X/k}^{\cdot})$. My question is not regarding the construction of the hypercohomology. I am just wondering why one should think to not use the ordinary cohomology of the complex in this case. REPLY [16 votes]: This is pretty much explained in the comments, but let me put it into an answer. One wants algebraic de Rham cohomology to be isomorphic to the usual de Rham cohomology (using $C^\infty$ forms) when $k=\mathbb{C}$, and have similar properties when $k$ is an arbitrary field of characteristic zero. From this point of view, as Grothendieck observed in the 1960's, the definition using hypercohomology is the correct one to use. To understand why, one should observe that a quasi-isomorphism of (bounded below) complexes induces an isomorphism of hypercohomologies. The holomorphic Poincar\'e lemma, says that on the complex manifold $X^{an}$, $\mathbb{C}_{X^{an}}$ and $\Omega_{X^{an}}^\bullet$ are quasi-isomorphic and so have the same hypercohomologies. We are halfway there. Now a (not completely straightforward) application of GAGA shows that $$\mathbb{H}^i(X,\Omega_X^\bullet)\cong H^i(X^{an},\mathbb{C})$$<|endoftext|> TITLE: What computer program for automorphic forms QUESTION [19 upvotes]: This question has its origins in this entertaining discussion on MO. There are many programs (CAS) and libraries that are able to handle algebraic expressions. These are both a verification tool for (sometimes nightmarish) computations and a way to explore objects for which a good intuition is not available. Among them, I have in mind Mathematica, Maple, Magma, and also the Python-interface Sage (including lots of packages like NumPy, SciPy, Maxima, GAP, etc.). Sage has the appeal to be free and open source, however I wonder if it is at the same level of the others. To narrow the question to a more specific field, for every language has its advantages somewhere, I am interested in automorphic forms and number theory. Thus the basic uses will be to manipulate automorphic forms for different groups and congruence subgroups, to locate zeros of the associated $L$-functions, compute Fourier coefficients, etc. What are the pros and cons of these programs to work with automorphic forms? Every direction of answer is welcome, in particular taking into account ease of use available literature (not on the program itself, but related to automorphic forms) regular updating of packages and functions community size for support and discussion REPLY [21 votes]: The only CAS's that have built-in support for modular and automorphic forms, as far as I know, are Sage and Magma. [Edit: I had forgotten Pari/GP, which will introduce substantial modular forms functionality as of version 2.10 which is currently in alpha testing; see Aurel's comment below]. Both Sage and Magma offer roughly comparable functionality. In Sage you can do something like this: ┌────────────────────────────────────────────────────────────────────┐ │ SageMath version 8.1, Release Date: 2017-12-07 │ │ Type "notebook()" for the browser-based notebook interface. │ │ Type "help()" for help. │ └────────────────────────────────────────────────────────────────────┘ sage: S = Newforms(Gamma1(7), 4, names='a'); S [q - q^2 - 2*q^3 - 7*q^4 + 16*q^5 + O(q^6), q + a1*q^2 + (-7/2*a1 - 7)*q^3 + (2*a1 + 4)*q^4 + 7/2*a1*q^5 + O(q^6)] sage: f = S[0] sage: f[next_prime(1000)] -8930 sage: L = S[0].lseries() sage: L(0.5 + 4.0*I) 6.68198475901674 + 1.21937727142056*I William Stein (original author of most of this code) has written a wonderful book "Modular Forms -- A Computational Approach", which describes all the theory and algorithms, with copious Sage code examples. You ask: "Among them, I have in mind Mathematica, Maple, Magma, and also the Python-interface Sage. [...] Sage has the appeal to be free and open source, however I wonder if it is at the same level of the others." Where modular forms are concerned, it's certainly not the case that Magma has a clear lead over Sage -- some functionality is better implemented in one or the other, but there's not an obvious winner overall. Both have flourishing user communities, regular updates, etc. On the other hand, Mathematica and Maple are both completely useless as tools for number theory; they're great at symbolic manipulation of algebraic expressions, but that's more or less all they do. (I work on modular forms, and I can count on my fingers the number of times I've found Maple or Mathematica useful, whereas I use Sage and/or Magma every couple of weeks at least.)<|endoftext|> TITLE: Diplomacy when reporting errors QUESTION [31 upvotes]: I am a young researcher and sometimes I face an uncomfortable situation : I find an error in a research paper! Of course, most of the time, it is all just my misunderstanding but it happens that after checking tens of times, there is indeed an error and therefore I am not sure what I should do. "Am I the first one that notices this mistake or someone has already reported it? and then where?" "Hass this error been corrected in a later version? How can I know? ", "Should I send an email to the authors ? But still how could I be diplomatic enough not to bother them (and maybe the mistake is still mine, I am not an expert of the field)? " The situation gets worse when -The paper is already old. Personally I would find annoying if someone asked me questions on a project finished years ago. And for very old papers the authors can just be retired already. -It comes from a different field like physics or chemistry. Where they use a different language. How could I explain myself correctly and without creating a "diplomatic crisis"? So I am here to ask advice to the experimented researchers how to deal with this awkward situation. REPLY [29 votes]: Of course, reporting an error benefits the author, so I advise to write to him/her. A polite form is: Dear X, I am reading your interesting paper Y, and I have difficulty in understanding why A implies B on p. C. I will appreciate it very much if you explain this to me (help me to understand). I received such letters several times. In most cases I was able to explain. In one case I published a correction. And of course I am thankful to all who wrote these letters. When I was a student I wrote to one famous mathematician: Dear X. Let me bring to your attention that conjecture W which you state in your paper Y seems to be proved by Picard in 1880 in his paper Z. Do you think his proof is correct?<|endoftext|> TITLE: Is there any work on the Gauss circle problem over function fields? QUESTION [5 upvotes]: I would be thankful if someone had references to provide... REPLY [5 votes]: In function fields, the Gauss Circle Problem, at least in its usual formulation, is much simpler than in number fields. In a number field $K$, one studies the difference $\sum_{n \le x} r_{K}(n) - C_{K} x$, where: $r_K$ is the coefficient of $n^{-s}$ in $\zeta_{K}$, the Dedekind zeta function for $K$, and $C_{K}$ is the residue of $\zeta_K$ at $s=1$. The 'Circle' corresponds to the geometric interpretation of $r_K$ when $K=\mathbb{Q}(i)$. One may also define $r_K$ as $$r_K(n) = \# \{ I \text{ ideal in }\mathcal{O}_K : \mathrm{Nm}(I)=(n) \}.$$ Now consider a function field $K$ which is a finite field extension of $\mathbb{F}_q(T)$ , the field of rational functions over the finite field with $q$ elements. Let $\mathcal{O}_K$ be the integral closure of $\mathbb{F}_q[T]$ in $K$. We define for any monic $f \in \mathbb{F}_q[T]$: $$r_{K/\mathbb{F}_q(T)}(f) = \# \{ I \text{ ideal in }\mathcal{O}_K : \mathrm{Nm}_{K/\mathbb{F}_q(T)}(I)=(f)\}.$$ It turns out that in fact we have a closed form formula for $\sum_{f \in \mathbb{F}_q[T]: \deg(f) =n, \text{ monic}}r_{K/\mathbb{F}_q(T)}(f)$. Indeed, consider the (incomplete) Dedekind zeta function $$\mathcal{Z}_{\mathcal{O}_K}(u) = \prod_{\mathcal{P} \text{ prime in }\mathcal{O}_K} (1-u^{\deg \mathrm{Nm}_{K/\mathbb{F}_q(T)} \mathcal{P}})^{-1} = \sum_{f \in \mathbb{F}_q[T]: \text{ monic}} r_{K/\mathbb{F}_q(T)}(f) u^{\deg f}.$$ Then $\mathcal{Z}_{\mathcal{O}_K}(u)$ is a rational function, so the $n$-th coefficient has a "finite formula", as explained in Vesselin's comment. For instance, if $K/\mathbb{F}_q(T)$ is a geometric extension, than $$\mathcal{Z}_{\mathcal{O}_K}(u) = \frac{P(u)}{1-qu}$$ for some polynomial $P$. Thus, for $n \ge \deg P$, we have $$\sum_{f \in \mathbb{F}_q[T] : \deg(f)=n, \text{ monic}} r_{K/\mathbb{F}_q(T)}(f) =q^n P(\frac{1}{q}), $$ and there is no error term in this formula. Although the Gauss Circle Problem itself was not studied in the function field case AFAIK, norms (and sums of two squares) were indeed studied. There are two common analogues for $\mathbb{Q}(i)$ in the function field case: $\mathbb{F}_q(\sqrt{-T})$ (if $q$ is odd) or $\mathbb{F}_{q^2}(T)$ (works for any $q$). See for instance this work and this one.<|endoftext|> TITLE: What are the maximal ideals in $C(0,1)$ apart from $M_c = \{f\in C(0,1) | f(c)=0\}$? QUESTION [6 upvotes]: I have studied about maximal ideals in $C[0,1]$ . They are precisely of the form $$\{f\in C[0,1] | f(c)=0\} \text{ for } c\in [0,1].$$ If we replace $[0,1]$ by $(0,1)$ and then look at $C(0,1)$, then obviously $M_c = \{f\in C(0,1) | f(c)=0\}$ are maximal ideals in $C(0,1)$ , $\forall c \in (0,1)$ , but since we do not have the compactness of $[0,1]$ anymore, I guess there are some other maximal ideals as well. I am trying to solve the problem by first giving an existential argument and then by showing an explicit maximal ideal other that of the form $M_c$ for some $c \in (0,1)$ . REPLY [5 votes]: Since $\mathbb{R}$ and $(0,1)$ are homeomorphic, you can look at maximal ideal in $C(\mathbb R)$. There is the idea form Dummit and Foote Abstract Algebra, page 259 Ex 34. Let $I\subset C(\mathbb R)$ be the set of all continuous functions with compact support. It can be shown that $I$ is an ideal. Let $M$ be a maximal ideal containing $I$. Then $M\neq M_{c}$ for any $c\in \mathbb R$. For suppose $M=M_{c}$, and let $r=|c|+1$. There is a function $f\in C(\mathbb R)$, such that $f(x)>0$ when $x\in (-|c|,|c|)$ and $f(x)=0$ when $x>r$ or $x<-r$. Then $f\in I$ but $f\notin M_{c}$.<|endoftext|> TITLE: An inequality related to Lagrange's identity and $L_p$ norm QUESTION [5 upvotes]: Let $a_1, a_2, \cdots , a_n$, $b_1, b_2, \cdots, b_n$ be real numbers, $p \in [1, +\infty)$, prove that $$\sum_{1\leq i < j \leq n} |a_ib_j - a_jb_i|^p \leq c_p \sum_{i=1}^n |a_i|^p \sum_{i=1}^n |b_i|^p$$ where $c_p = \max(1, 2^{p-2})$. Remark: When $p=1$, the proof is straightforward since $|a_ib_j-a_jb_i| \leq |a_i||b_j| + |a_j||b_i|$, and summing up all these inequalities is enough. When $p=2$, the inequality is a direct consequence of Lagrange's identity: $RHS - LHS = (\sum_{i=1}^n a_i b_i)^2 \geq 0$ I have generated millions of sets of random $(a_i, b_i)$, for various values of $p$, and didn't find a counterexample. Thanks! REPLY [6 votes]: Note: the answer here stems from the comment of user fedja above. The following Lemma is another version of the Riesz–Thorin interpolation theorem (See Lemma 8.5 in this book). Lemma. Let $(X_i , \mathfrak M_i , μ_i )$, $i = 0,1, 2, . . . , n$ be measure spaces. Let $V_i$ represent the complex vector space of simple functions on $X_i$. Suppose that $$\Lambda : V_1\times V_2 \times \cdots \times V_n \to V_0$$ is a multilinear operator of types $p_0$ and $p_1$ where $p_0,p_1 \in [1,\infty]$, with constants $C_0$ and $C_1$, respectively. i.e., $$ \|\Lambda(f_1 , f_2 , . . . , f_n) \|_{p_i} \leq C_i \|f_1\|_{p_i} \|f_2 \|_{p_i} \cdots \|f_n\|_{p_i} \tag{1} $$ for $i=0,1$. Let $t \in [0, 1]$ and define $$\frac 1{p_t} := \frac{1-t}{p_0}+\frac t{p_1}$$ Then, $\Lambda$ is of type $p_t$ with constant $C_t :=C_0^{t-1}C_1^t$, that is, (1) holds true for $i=t$. Now, Let $(X_i , \mathfrak M_i , μ_i )$ be a uniform measure on $[n] := \{1,\ldots,n\}$ for $i=1,2$, and be a uniform measure on $[n^2]$ for $i=0$. In this case, we have $V_1= V_2=\mathbb{C}^n$, and $V_0 = \mathbb{C}^{n^2}$. Define $\Lambda : V_1\times V_2 \to V_0$, by $[\Lambda(a,b)]_{i,j} := a_ib_j -a_jb_i$ for $a,b\in \mathbb{C}^n$ and $1\leq i,j\leq n$. Note that, in this setting, we have $$ \|\Lambda(a,b)\|_p^p = \frac 2{n^2} \sum_{1\leq i TITLE: Tangled random triangles: One giant component? QUESTION [10 upvotes]: Suppose you have $n$ triangles whose corners are random points on a sphere $S$ in $\mathbb{R}^3$. Viewing the triangles as built from rigid bars as edges, two triangles are linked if they cannot be separated without two edges passing through one another. A triangle that is not topologically linked with any other is loose. In the example below of $n=15$ triangles, $11$ are linked to at least one other triangle, and $4$ are loose.                     $n=15$. Magenta triangles $\{1,5,10,11\}$ are loose. It is easy to surmise that the proportion of linked triangles approaches $1$ as $n \to \infty$:                     Fraction of triangles linked to at least one other triangle. But I wonder about the largest linked component, the collection of all triangles linked into one "giant" component, in the sense that if you picked up one triangle all the others would follow. I wonder that when $n \to \infty$, what is the probability that this linked component includes all the triangles. Q. As $n \to \infty$, what is the probability that every triangle is linked into one giant component? My sense is that this probability is zero: Even though the probability that each triangle is linked to another approaches $1$, the probability that all triangles are linked to one another approaches $0$. This would contrast with an earlier related question, Random rings linked into one component?, whose answer was the opposite: The rings form one component as $n \to \infty$. REPLY [6 votes]: Here is @fedja's clever example: The magenta triangle is topologically loose but metrically "stuck":                     fedja: "(link two large ... triangles [blue & green] near vertices           and surround the linkage by a small ... triangle [magenta])"<|endoftext|> TITLE: Name for $\omega_1$-DCC / Noetherian condition? QUESTION [6 upvotes]: I recently asked (and then answered) this question: https://math.stackexchange.com/questions/2756777/decreasing-sequence-of-closed-sets-in-a-separable-metric-space. In a separable metric space there is no strictly decreasing sequence of closed sets $(X_\alpha)_{\alpha<\omega_1}$, where $X_\beta\supsetneq X_\alpha$ iff $\beta<\alpha$. This sounds a lot like the Descending Chain Condition in Algebra. I assume it comes up frequently in topology (maybe not?). In that case, is there a name for it? REPLY [7 votes]: For any topological space $X$, the following statements are easily seen to be equivalent: (1) there is no strictly decreasing $\omega_1$-sequence of closed sets in $X$; (2) there is no strictly increasing $\omega_1$-sequence of open sets in $X$; (3) every open subspace of $X$ is Lindelöf; (4) every subspace of $X$ is Lindelöf. Spaces satisfying those equivalent conditions are called hereditarily Lindelöf spaces by some, strongly Lindelöf spaces by others.<|endoftext|> TITLE: Arithmetic motivations for modularity in higher rank QUESTION [16 upvotes]: The classical setting of modularity is that one can associate elliptic modular forms (or automorphic representations of GL(2)/$\mathbb Q$) to elliptic curves over $\mathbb Q$. This has far-reaching consequences to elementary problems in arithmetic: Fermat's last theorem, the congruence number problem, sums of 2 cubes, ... Conjectures and some results are also known over other number fields. Now there are various conjectural generalizations of modularity to higher dimensions that are part of Langlands' philosphy. For instance the paramodular conjecture, asserting abelian surfaces are associated to certain kinds of Siegel modular forms (cf. Modularity of higher dimensional abelian varieties, Langlands in dimension 2: the Yoshida conjecture). Certainly knowing modularity in higher rank would be phenomenal, and is inherently interesting. But... Question: Are there appealing "elementary" applications, or suspected applications (analogous to connections with elliptic curves) of knowing modularity (automorphy) in higher rank, i.e., that certain motives (beyond GL(2) type) are associated to automorphic representations? The issue, to my naive mind, is that elliptic curves are given by rather simple equations that naturally arise in classical number theory, whereas abelian surfaces are not described such simple equations. However, philosophically, I feel that there is a lot of arithmetic in these conjectures, which should manifest itself in concrete (and hopefully easily to explain) ways---the question is how? (Note Langlands also brings up this issues at the end of his introduction to his 2010 Notre Dame talk Is there beauty in mathematical theories?, writing "[a] very hard question is what serious, concrete number-theoretical results does the theory suggest or, if not suggest, at least entail. A few mathematicians have begun, implicitly or explicitly, to reflect on this. I have not.") REPLY [16 votes]: A quick fix for the lack of nice equations defining abelian surfaces is to write down equations for the corresponding genus $2$ curves. Something similar works in higher genus, but there most abelian varieties won't be Jacobians of curves. However, this is no problem as the modularity conjecture is hard enough for curves already. I think many would say that the most important concrete consequences of modularity are statistical statements that follow from the existence of $L$-functions. For instance, the following statements should follow from the $GL_n$ modularity of curves: $$ \sum_{ p < X} \frac{ p + 1 - |C(\mathbb F_p)|}{\sqrt{p}} = o (\pi(X))$$ $$\sum_{ p < X} \frac{ (p + 1 - |C_1(\mathbb F_p)|) (p + 1 - |C_2(\mathbb F_p)|) }{p} = \operatorname{rank} (\operatorname{NS}( C_1 \times C_2) )\pi(X) - 2\pi(X) + o (\pi(X)) $$ where $\operatorname{NS}$ is the Neron-Severi group of divisors defined over $\mathbb Q$ modulo numerical equivalence. The first from the ordinary $L$-function and the second from the Rankin-Selberg $L$-function plus the Tate conjecture for abelian varieties due to Faltings. These will all fit into the framework of the generalized Sato-Tate conjecture, which would follow from super-general modularity statements.<|endoftext|> TITLE: The actual Satake diagram EIV QUESTION [8 upvotes]: In table 9 of "Lie groups and algebraic groups" (1990) [OV], Onishchik and Vinberg present the Satake diagrams. The diagram corresponding to EIV has the "orthogonal" node blackened. In table 4 of "Lie groups and Lie algebras III" (1993), the same authors together with Gorbatsevich present the Satake diagrams again. The diagram corresponding to EIV has the "orthogonal" node colored white (as can be seen here, page 230: Lie groups and Lie algebras III). I want to know which option is the right one. Could be that the 1993 version is a correction of the 1990 one, but could also be a plain mistake from 1993. Other sources paint the node black, but I cannot discard that they have taken their diagrams from either [OV] or Helgason's "Differential geometry, Lie groups, and symmetric spaces" (1978), which is the first place the diagrams were drawn (and the original source for [OV]), which paints it black. REPLY [6 votes]: One simple way to tell which is correct is to remember that deleting an orbit of white nodes from a valid Satake-Tits diagram should give another valid Satake-Tits diagram. If the node you are asking about (which is node number 2 in Bourbaki's labeling) were white, then removing one of the other white nodes would give a supposed Satake-Tits diagram for $D_5$ which is not possible, whereas if it is black we get the diagram for the real form of $D_5$ for a real quadratic form with signature $(1,9)$. For more explanations on the subject, I recommend Tits's 1966 survey paper, "Classification of algebraic semi-simple groups" (in Algebraic groups and discontinuous subgroups, Boulder 1965, Proc. Symp. Pure Math. 9 33–62, p.372ff in volume 2 of the EMS edition of the complete works of Tits), see in particular §3.2.2 (keep in mind that what Tits called a "distinguished orbit" is an orbit of white nodes, and what Tits calls a "Witt index" is what is now generally called a Satake-Tits diagram). The possible diagrams over any field are given in the appendix tables of the paper in question, with a specific mention of which are realized over $\mathbb{R}$. Another independent confirmation is Araki's 1962 paper "On root systems and an infinitesimal classification of irreducible symmetric spaces" J. Math. Osaka City Univ. 13 1962 (1–34), which proceeds to derive the classification of simple real lie Groups directly from the Satake diagrams without using Cartan involutions. The diagram you seek is on page 29. (I was about to also refer to Tits's 1967 book Tabellen zu den einfachen Lie Gruppen und ihren Darstellungen, which contains fewer typos than Onishchik and Vinberg, but I'm surprised to find that he does not give Satake diagrams nor any information from which the answer could be obviously derived, such as the real rank of the real form.)<|endoftext|> TITLE: Convolution of two Brownian motions QUESTION [5 upvotes]: Suppose $B_1(t)$ and $B_2(t)$ are two independent, standard Brownian motions. What is the distribution of \begin{align*} G(t) = \int_0^t B_1(\tau)B_2(t-\tau)d\tau \qquad \end{align*} Or, at least an approximation to $\text{Var}(G(t))$? Generalizations (1) How about $B_1, B_2$ being any Gaussian process with covariance functions $c_1(t, s)$ and $c_2(t, s)$? (2) How about \begin{align*} G(t) = \int_0^t B_2(t-\tau)dB_1(\tau) \qquad \end{align*} For either $B_1, B_2$ being Brownian motions or, more generally, Gaussian processes? REPLY [4 votes]: use $E[B_n(t)]=0$, $E[B_n(t)B_n(t')]=\min(t,t')$: $${\rm Var}\,G(t)=E[G(t)^2]=\int_0^t \int_0^t E[B_1(\tau)B_1(\tau')]E[B_2(t-\tau)B_2(t-\tau')]\,d\tau d\tau'$$ $$=\int_0^t \int_0^t \min(\tau,\tau')\min(t-\tau,t-\tau')\,d\tau d\tau'=t^4/12$$<|endoftext|> TITLE: Positive polynomial equations mod integers QUESTION [5 upvotes]: Let $f(x)$ and $g(x)$ be polynomials with integer coefficients such that $f(x)>0$ and $g(x)>0$ for all real values of $x$. Suppose that for every integer $n$, if $f(x)=0$ has a solution mod $n$, then $g(x)=0$ has a solution mod $n$. What can generally be said about $f(x)$ and $g(x)$? My guess is that $g(x)$ must have a factor $h(x) \in \mathbb{Z}[x]$ such that $f(x)=h\circ k(x)$ for some $k(x) \in \mathbb{Z}[x]$. REPLY [4 votes]: The question becomes more tractable if we assume that $f$ and $g$ are monic irreducible polynomials, and we restrict to $n$'s coprime to the discriminants of $f$ and $g$. In this case, by Hensel's lemma and standard facts on discriminants, the modified question remains unchanged if we restrict to primes $n$ coprime to the discriminants of $f$ and $g$. Let us consider the corresponding number fields $K=\mathbb{Q}[x]/(f(x))$ and $L=\mathbb{Q}[x]/(g(x))$ whose discrimants divide the discriminants of $f$ and $g$, respectively. Then, in the special case when $L$ is Galois (i.e. $L$ contains all the roots of $g$), the modified condition holds for $f$ and $g$ if and only if $K$ contains $L$. This follows from Prop. 15 in Ch. VIII-5 of Weil: Basic number theory. Example. Let $p$ be a prime congruent to $1$ mod $4$. Let $f(x)=x^{p-1}+\dots+1$ be the $p$-th cyclotomic polynomial, and let $g(x)=x^2-p$. Consider an arbitrary prime $n\nmid 2p$. Then, $f(x)$ has a root modulo $n$ if and only if $n$ is congruent to $1$ mod $p$, and $g(x)$ has a root modulo $n$ if and only if $n$ is a quadratic residue mod $p$. Indeed, in this case, $K=\mathbb{Q}(e^{2\pi i/p})$ is the $p$-th cyclotomic field and $L=\mathbb{Q}(\sqrt{p})$ is its unique quadratic subfield, and we just recorded the splitting primes in $K$ and $L$.<|endoftext|> TITLE: What is a model category from an $\infty$ point of view? QUESTION [14 upvotes]: A number of different models for $\infty$ categories can seen to have analogs in $\infty$-category theory. For example: Quasicategories: $\Delta \subseteq \mathrm{Cat}_{(\infty, 1)}$ is (?) a dense generator, so we can see $\infty$-categories as presheaves on $\Delta$ Simplicially enriched categories: Simplicial enrichment is a proxy for enrichment in $\mathrm{Cat}_{(\infty, 0)}$ Simplicial categories: These can be seen as the objects in an intermediate step towards adjoining simplicially-indexed colimits to $\mathrm{Cat}_{(1,1)}$ Complete segal spaces: These look to be inspired by models in $\mathrm{Cat}_{(\infty, 0)}$ of the finite limit sketch defining categories Relative categories: A pair $(C,W)$ is a proxy for the localization $C[W^{-1}]$. However, I don't know of a suitable interpretation of a model structure. Every model category is in particular a relative category, so that's how they correspond to $\infty$-categories. But that doesn't involve the fibrations and cofibrations. My question is if there is some notion in $\infty$-category theory that is analogous to equipping a relative category with a choice of classes of fibrations and cofibrations. REPLY [8 votes]: This is just a long comment, which does not really address the main question. Instead, I give few examples where fibrations are an additional rigidity structure allowing "homotopical commutativity". In all of the simplicial models for $(\infty, 1)$-categories that you listed, only the fibrant objects of the respective model category structure behave as "categories weakly enriched in $\infty$-groupoids". Inside this categorical framework (if you want, formal 2-categorical), there is a notion of localisation $C[W^{-1}]$ of an $\infty$-category $C$ with respect to a sub-$\infty$-category $W$. Given a relative category $(C, W)$, it is a theorem that the $\infty$-localisation $N(C)[N(W)^{-1}]$ models the $\infty$-category associated to $(C, W)$ (see for instance the appendix of Brown categories and bicategories of Geoffroy Horel). However if $(C, W)$ is not fibrant, nor will be its Rezk's relative nerve, i.e. $N^R(C, W)$ is not really an $\infty$-category. Suppose now your relative category $(C, W)$ has an additional structure of fibrantions that interacts gently with the weak equivalences (for instance, a model category), let's call such a structure a fibration category. Lennart Meier showed in Fibration categories are fibrant relative categories that any such relative category is fibrant in the Barwick-Kan model category structure and thus $N^R(C, W)$ is an $\infty$-category modelling the $\infty$-categorical localisation $N(C)[N(W)^{-1}]$. This is just the reflection of something happening entirely in the $\infty$-categorical world. Indeed, in Chapter 7 of Higher categories and homotopical algebra Denis-Charles Cisinski studies the $\infty$-localisation theory of fibration $\infty$-categories. Within this framework, we have the beautiful feature that the $\infty$-localisation of a fibration $\infty$-category has finite $\infty$-limits (see Section 7.5). Moreover the identification of homotopy limits and $\infty$-limits can be stated as an elegant theorem about an equivalence of $\infty$-categories (see Theorem 7.9.8 and Remark 7.9.10). The longer story tells that fibration $\infty$-categories are closed by exponentiation, that is if $(C, W)$ is a fibration $\infty$-category and $I$ is an $\infty$-category, then $\underline{\text{Hom}}(I, C)$ is still a fibration $\infty$-category, with fiber-wise weak equivalences $W_I$. If $(C, W)$ is just (the nerve of) a regular fibration category and $I$ (the nerve of) a small category, then the $\infty$-localisation $\text{Fun}(I, C)[W_I^{-1}]$ is equivalent to the $\infty$-functor category $\underline{\text{Hom}}(I, C[W^{-1}])$. (Notice that $(\text{Fun}(I, C), W_I)$ is a fibrant relative category!) In the case where the fibration $\infty$-category is actually a model $\infty$-category, the theory has been studied by Aaron Mazel-Gee, as Dmitri Pavlov says in his nice answer.<|endoftext|> TITLE: $\mathbb{E}[X^4]=1$, $X,Y$ iid, what's the best upper bound of $\mathbb{E}[(X-Y)^4]$? QUESTION [31 upvotes]: Let $X,Y$ be i.i.d. random variables, $\mathbb{E}[X^4]=1$, what's the best upper bound for $\mathbb{E}[(X-Y)^4]$ ? A trivial upper bound is $16$, since $(X-Y)^4 \leq 8 (X^4+Y^4)$ then take expectation on both sides. However, equality cannot be achieved. My guess of the best upper bound will be $8$, achieved when $X$ is uniform at random from $\{-1, +1\}$. REPLY [35 votes]: $\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$ Your guess is correct. Indeed, it is well known (see e.g. Bertsimas--Popesku, page 781) that real numbers $m_0=1,m_1,\dots,m_{2\ell}$ are the moments of orders $0,1,\dots,2\ell$ of a real-valued random variable $X$ iff the matrix $M:=(m_{i+j})_{i,j=0}^\ell$ is nonnegative-definite, that is, iff all the principal minors of $M$ are $\ge0$; here $\ell$ is a natural number; in our case, $\ell=2$. Also, given $\E X^4=1$, we have \begin{equation} \E(X-Y)^4=2-8m_3m_1+6m_2^2. \end{equation} Thus, the problem is a simple problem of real algebraic geometry, which can be solved algorithmically. Using the Mathematica command Maximize[], we get the result: Added: Here is an elementary solution, without using Mathematica: Since $m_4=1$, the condition $\det M\ge0$ implies \begin{equation} 2m_3m_1\ge m_2^2-m_4+(m_3^2+m_4m_1^4)/m_2\ge m_2^2-1, \end{equation} whence \begin{equation} \E(X-Y)^4=2-8m_3m_1+6m_2^2\le2-4(m_2^2-1)+6m_2^2=2m_2^2+6\le8, \end{equation} since $m_2^2\le m_4=1$. The equality in the inequality in question is attained only if $m_2^2=m_4=1$ and $2m_3m_1=m_2^2-1=0$, that is, only if $\PP(X=1)=\PP(X=-1)=1/2$.<|endoftext|> TITLE: Is there a purely inseparable covering $\mathbb{A}^2 \to K$ of a Kummer surface $K$ over $\mathbb{F}_{p^2}$? QUESTION [7 upvotes]: Let $E_i\!: y_i^2 = x_i^3 + a_4x_i + a_6$ be two copies ($i = 1$, $2$) of a supersingular elliptic curve over a finite field $\mathbb{F}_{p^2}$, for odd prime $p > 3$. Consider the Kummer surface $K = \mathrm{Kum}(A)$ for the superspecial abelian surface $A = E_1\times E_2$. Katsura and Schütt proved in the articles "Generalized Kummer surfaces and their unirationality in characteristic p (1987)" and "Zariski K3 surfaces (2017)" that the surface $K$ is a Zariski surface, i.e., there is a purely inseparable covering $\mathbb{A}^2 \to K$ over $\overline{\mathbb{F}_p}$, where $p \not\equiv 1$ (mod $12$). Is it still true over $\mathbb{F}_{p^2}$ at least for some supersingular elliptic curve and some $p$? A positive answer to this question may have value for cryptography :) If $K$ is a Zariski surface over $\mathbb{F}_{p^2}$, then we have a method to compress a pair $(P_1, P_2) \in A(\mathbb{F}_{p^2})$ by computation of a map from $A$ to the affine plane $\mathbb{A}^2$ (through $K$) of separable degree 2. This is very compact and efficient method, because for decompression we need to solve only one quadratic equation. Computation of a preimage for a purely inseparable map is very fast. People usually take two projections on x-coordinates for $P_1$ and $P_2$ independently, hence they should solve two quadratic equations. The state of the art for this part of cryptography is represented, for example, in the article https://eprint.iacr.org/2017/1143. REPLY [5 votes]: See Proposition 4.5 of my paper with D. Abramovich, "Lang’s Conjectures, Fibered Powers, and Uniformity", New York J. Math. 2 (1996) 20–34. A supersingular elliptic curve in characteristic $p>2$ will always be covered by the hyperelliptic curve $y^2=x^p-x$ as the Jacobian of the latter is isogenous to a product of supersingular elliptic curves. When $p \equiv 2 \pmod{3}$ this can be made very explicit for $y^2=x^3-1$, which is covered by $y^2=x^{p+1}-1$ in the obvious way and $y^2=x^{p+1}-1$ is isomorphic to $y^2=x^p-x$ by sending a Weierstrass point to infinity.<|endoftext|> TITLE: Equations in free groups satisfying all elements QUESTION [8 upvotes]: please help me to solve the following problem. Let $F$ be a non-abelian free group and $w(x)=1$ be an equation in one variable $x$ ($w(x)$ may contain elements of $F$ as constants). Clearly, one can consider $w(x)$ as an element of free product $F\ast \langle x\rangle$. Suppose $w(a)=1$ for all $a\in F$. Is it true that $w(x)$ equals $1$ in $F\ast \langle x\rangle$? Do you know a simple proof? Probably, you can remember the papers which can be useful for this propblem? REPLY [2 votes]: If $F=\langle a_1,\ldots,a_n\rangle=\langle\underline{a}\rangle$ is finitely generated, we can replace $w$ with a coefficient-free word $w'(x,y_1,\ldots,y_n)=w'(x,\underline y)$ such that $w(x)=w(x,\underline{a})$. Now consider the first-order sentence $\phi$: $$\exists y_1,\ldots,y_n\forall x\ (w'(x,\underline{y})=1)$$ Then the result follows from Sela's proof of Tarski's problem: Theorem [Sela, Theorem 3]: All finitely generated non-abelian free groups have equivalent first-order theories. In particular, $F$ satisfies $\phi$ (where $\underline{y}=\underline{a}$ works to satisfiy the existential part of $\phi$) if and only if $F*\langle a_{n+1}\rangle$ satisfies $\phi$. I'm not sure about the case where $F$ is not finitely generated, but because at most finitely many of the generators of $F$ could appear as coefficients in $w(x)$, you might be able to reduce to this case, depending on what you need this result for.<|endoftext|> TITLE: Surreal Numbers, Proving $x1=x$ QUESTION [6 upvotes]: I am trying to learn the theory of the Surreal numbers and I am therefore going over all the theorems and trying to prove them for myself. I am struggling to complete the proof of $x1 = x$. I have the following. Assume x is a surreal number. Then $x1 = \{X_L1 + x0 - X_L0, \ X_R1 + xØ - X_RØ | X_L1 + xØ - X_LØ, \ X_R1 + x0 - X_R0 \} = \{X_L, X_R | X_R, X_L\}$ Why is this the same as x?. I would argue that this is the same as $\{X_R | X_L\}$, but this is not a surreal number (per definition). Could someone explain to me how you go from $\{X_L, X_R | X_R, X_L\} \equiv \{X_L | X_R\}=x$ REPLY [12 votes]: If you look at the Wikipedia entry for surreal multiplication, you find The recursive formula for multiplication contains arithmetic expressions involving the operands and their left and right sets, such as the expression $$X_{R}y+xY_{R}-X_{R}Y_{R}$$ that appears in the left set of the product of x and y. This is to be understood as the set of surreal numbers resulting from choosing one number from each set that appears in the expression and evaluating the expression on these numbers. (In each individual evaluation of the expression, only one number is chosen from each set, and is substituted in each place where that set appears in the expression.) In your case, since $1=\{0\mid\ \}$, you have that $1_R=\emptyset$, the empty set. Thus, any part of the term involving $1_R$ will have no elements and therefore contribute no elements to that side of the final surreal number value for the product. For example, the term $X_R1+x\emptyset−X_R\emptyset$ in your expression adds no elements to the final value. But in your calculation, you seem to have indicated that you think it adds $X_R1$; this is wrong and this is exactly where your calculation makes a mistake. Perhaps you thought that in the expression $X_R1+x\emptyset-X_R\emptyset$, the latter two terms just evaluate to $0$ and therefore cancel out, giving $X_R1$. But that is not right. The correct meaning of this term is: for every element $r\in X_R$ and every $y\in\emptyset$, we contribute the surreal numbers $r1+xy-ry$ to that part of the product surreal value. But since there are no such $y$, we actually end up contributing nothing to the product from this term, even though the first part of the expression $X_R1$, if it had appeared alone, would have contributed $r1$. The situation is something like the Cartesian product $X\times\emptyset$, which is empty even when $X$ is not empty, simply because elements $x\in X$ have no partner $y\in\emptyset$ with which to form a pair $(x,y)$.<|endoftext|> TITLE: Is the category of left exact functors abelian? QUESTION [17 upvotes]: Recently I asked on Math Stack Exchange here, if the category $\mathbf{Lex(\mathcal{A,B})}$ of left exact functors between two abelian categories $\mathcal{A,B}$ is abelian? In his thesis Des catégories abéliennes, Gabriel proved that under stronger conditions the category $\mathbf{Lex(\mathcal{A,B})}$ is abelian. In fact, if the categories $\mathcal{A,B}$ are abelian and $\mathcal{B}$ has generators and exact inductive limits, then the category $\mathbf{Lex(\mathcal{A,B})}$ is abelian. One particular example which is often used in this thesis (and other works) is the category $\mathbf{Lex(\mathcal{A},Ab)}$. Thus if we want a counterexample, we know that $\mathcal{B}$ should not be a Grothendieck category. Jeremy Rickard gave a counterexample but it is not trivial, and till now it is the only one I know. So my question is : Is there any simpler counterexample or explanation? REPLY [23 votes]: The following pair of examples follows the idea of Jeremy Rickard suggested in a comment on Math Stack Exchange under the link. Inverting the arrows, it suffices to construct an example of abelian category $\mathcal A$ such that the category of right exact functors $\mathcal A \to \mathcal Ab$ or $\mathcal A \to k{-}\mathcal{V}ect$ is not abelian (where $k$ is a field). Let $p$ be a prime number, and let $\mathcal A_p$ be the abelian category of finite abelian $p$-groups. Let me start with a very brief introductory discussion of the category of left exact functors $\mathcal A_p\to\mathcal Ab$. The category of finite abelian $p$-groups is self-dual (by Pontryagin duality, taking a finite abelian $p$-group $A$ to the group $\operatorname{Hom}_{\mathbb Z}(A,\mathbb Q/\mathbb Z)$). So the category of left exact functors $\mathcal A_p\to\mathcal Ab$ is equivalent to the category of left exact functors $\mathcal A_p^{op}\to\mathcal Ab$. The category of left exact functors $E\colon\mathcal A_p^{op}\to\mathcal Ab$ is equivalent to the category of $p$-primary torsion abelian groups $T$. To a $p$-primary torsion abelian group $T$ one assigns the contravariant functor $E_T$ taking a finite abelian $p$-group $A$ to the abelian group $\operatorname{Hom}_{\mathbb Z}(A,T)$. To construct the inverse functor, consider the projective system of finite abelian $p$-groups $\mathbb Z/p\mathbb Z \leftarrow \mathbb Z/p^2\mathbb Z\leftarrow \mathbb Z/p^3\mathbb Z\leftarrow\dotsb$ and apply the contravariant functor $E$ to it. The $p$-primary torsion abelian group $T_E$ corresponding to $E$ is the inductive limit of the sequence $E(\mathbb Z/p\mathbb Z)\to E(\mathbb Z/p^2\mathbb Z)\to E(\mathbb Z/p^3\mathbb Z)\to\dotsb$. From the left exactness property of the functor $E$ one can see that the group $E_m=E(\mathbb Z/p^m\mathbb Z$) is identified with the subgroup of elements annihilated by $p^m$ in the group $E_n$, for every $n\ge m$. The category of $p$-primary torsion abelian groups is abelian. Hence so is the category of left exact functors $\mathcal A_p\to \mathcal Ab$ (as we know it should be by the additive sheaf theory). Now let us turn to right exact functors $\mathcal A_p\to \mathcal Ab$, which we are really interested in. Let $F\colon\mathcal A_p\to\mathcal Ab$ be a right exact functor. Then $F_n=F(\mathbb Z/p^n\mathbb Z)$ is an abelian group annihilated by the multiplication with $p^n$. For any $n\ge m$, from the right exact sequence $\mathbb Z/p^n\mathbb Z\to \mathbb Z/p^n\mathbb Z \to \mathbb Z/p^m\mathbb Z\to 0$ we get an isomorphism $F_n/p^mF_n\cong F_m$. The functor $F$ is uniquely determined by the sequence of abelian groups $F_n$ together with these isomorphisms. Indeed, for every abelian group $A$ annihilated by $p^n$ one has $F(A)=F_n\otimes_{\mathbb Z/p^n\mathbb Z} A$. To any right exact functor $F\colon\mathcal A_p\to\mathcal Ab$, one assigns the abelian group $C_F=\varprojlim_n F_n$. This assignment is an equivalence between the category of right exact functors $\mathcal A_p\to\mathcal Ab$ and the full subcategory in abelian groups consisting of all the $p$-separated $p$-complete abelian groups, i.e., abelian groups $C$ for which the natural map $C\to\varprojlim_n C/p^nC$ is an isomorphism. The inverse functor assigns to an abelian group $C$ the functor $F_C$ taking a finite abelian $p$-group $A$ to the abelian group $C\otimes_{\mathbb Z}A$. The corresponding groups $F_n$ are $F_n=C/p^nC$. The category of all $p$-separated $p$-complete abelian groups is not abelian. There is a closely related (locally $\aleph_1$-presentable) abelian category of "weakly $p$-complete" or "Ext-$p$-complete" abelian groups, but it is a different (bigger) category: all $p$-separated $p$-complete abelian groups are Ext-$p$-complete, and all Ext-$p$-complete abelian groups are $p$-complete, but they need not be $p$-separated. In other words, the map $C\to \varprojlim_n C/p^nC$ for an Ext-$p$-complete abelian group $C$ is always surjective, but it need not be injective. Similarly, let $k$ be a field, and let $\mathcal A_k$ be the abelian category of finite-dimensional $k$-vector spaces $V$ endowed with a nilpotent $k$-linear operator $x\colon V\to V$. Let $G\colon\mathcal A_k\to k{-}\mathcal{V}ect$ be a $k$-linear right exact functor. Then $G_n=G(k[x]/x^nk[x])$ is a $k[x]/x^nk[x]$-module. Just as in the first example above, one constructs a natural isomorphism $G_n/x^mG_n\cong G_m$ for every $n\ge m$. The functor $G$ is uniquely determined by the sequence of modules $G_n$ together with these isomorphisms. To any $k$-linear right exact functor $G\colon\mathcal A_k\to k{-}\mathcal{V}ect$, one assigns the $k[x]$-module (or, if one wishes, $k[[x]]$-module) $D_G=\varprojlim_n G_n$. This assignment is an equivalence between the category of $k$-linear right exact functors $G\colon\mathcal A_k\to k{-}\mathcal{V}ect$ and the full subcategory in $k[x]$-modules (or, equivalently, in $k[[x]]$-modules) consisting of all the $x$-separated and $x$-complete modules, i.e., $k[x]$-modules or $k[[x]]$-modules $D$ such that the natural map $D\to\varprojlim_n D/x^nD$ is an isomorphism. The inverse functor assigns to a $k[x]$-module $D$ the functor $G_D$ taking a $k$-finite-dimensional $k[x]$-module $B$ with $x$ acting by a nilpotent operator to the $k$-vector space $D\otimes_{k[x]}B$. The corresponding modules $G_n$ are $G_n=D/x^nD$. Taking $k=\mathbb Z/p\mathbb Z$ or $k=\mathbb Q$, one can drop the adjective "$k$-linear" before the words "right exact functor" (as any abelian group homomorphism between $k$-vector spaces is a $k$-linear map, so any additive functor between $k$-linear categories is $k$-linear, for such fields $k$). As in the first example, the category of all $x$-separated $x$-complete $k[x]$-modules is not abelian. There is a closely related (locally $\aleph_1$-presentable) abelian category of "weakly $x$-complete" or "Ext-$x$-complete" $k[x]$-modules, but it is a different (bigger) full subcategory in $k[x]{-}\mathcal{M}od$. Any Ext-$x$-complete $k[x]$-module is $x$-complete, but it need not be $x$-separated. The categories of $p$-separated $p$-complete abelian groups and $x$-separated $x$-complete $k[x]$-modules are complete and cocomplete. In fact, they are locally $\aleph_1$-presentable, being reflective and closed under $\aleph_1$-filtered colimits as full subcategories in the locally $\aleph_1$-presentable abelian categories of Ext-$p$-complete abelian groups and Ext-$x$-complete $k[x]$-modules (respectively). So, in particular, all morphisms in these two categories (of right exact functors) have kernels and cokernels. Still, these two categories are not abelian. The counterexample showing that the categories of $p$-separated $p$-complete abelian groups and $x$-separated $x$-complete $k[x]$-modules are not abelian is now well-known. It has been rediscovered and discussed by many authors, including Example 2.5 in A.-M. Simon, "Approximations of complete modules by complete big Cohen-Macaulay modules over a Cohen-Macaulay local ring", Algebras and Represent. Theory 12, 2009, and Example 3.20 in A. Yekutieli, "On flatness and completion for infinitely generated modules over noetherian rings", Communic. in Algebra 39, 2010, https://arxiv.org/abs/0902.4378 . The specific assertion that these two categories (or, rather, the first one, but there is no difference), viewed as abstract categories, are not abelian, can be found in Example 2.7(1) in my paper L. Positselski, "Contraadjusted modules, contramodules, and reduced cotorsion modules", Moscow Math. Journ. 17 #3, 2017, https://arxiv.org/abs/1605.03934 . The above descriptions of two right exact functor categories as categories of separated and complete modules (= "separated contramodules") are the simplest examples to a much more general theory developed in Section 6 of our paper L. Positselski, J. Rosicky, "Covers, envelopes, and cotorsion theories in locally presentable abelian categories and contramodule categories", Journ. Algebra 483, 2017, https://arxiv.org/abs/1512.08119 . Specifically, the discussion in Remark 6.5 in this paper is relevant.<|endoftext|> TITLE: Consecutive polynomial non-residues modulo a prime QUESTION [8 upvotes]: Given a polynomial $f(x)\in \mathbb{Z}[x]$ of degree at least 2 and a positive integer $t$, does there always exist infinitely many primes $p$ such that the range of $f(x)$ modulo $p$ does not contain the $t$ consecutive numbers $s+1,\ldots, s+t$ for some integer $s$ (where $s$ can depend on $p$)? This is known to be true if $f(x)=x^n$, $n\geq 2$. I am wondering if anything is known beyond this example. REPLY [3 votes]: Let $d = \deg f$. Consider the covering of $\mathbb A^1_\mathbb Q$, with coordinate $s$, defined by $f(x_1) =s+1 ,\dots, f(x_t) =s +t$. The monodromy representation of this covering gives a map from the etale fundamental group of $\mathbb A^1$ minus the finite set of critical values to $\prod_{i=1}^t S_d$. The image of this representation has a normal subgroup which is the image of the geometric fundamental group, and a quotient corresponding to some Galois extension of $\mathbb Q$. Then any sufficiently large prime $p$ which splits in this quotient has a sequence of $t$ consecutive non-residues. First, because $p$ splits, the arithmetic monodromy group of the covering over $\mathbb F_p$ is contained in the geometric monodromy group in characteristic zero. Then because $p$ is sufficiently large, the geometric monodromy group in characteristic $p$ matches the one in characteristic zero, so these are equal. Then we may apply the function field Chebotarev theorem, which says that as long as $p$ is sufficiently large (with regards to the degree and genus of this covering, say) then for each conjugacy class $\sigma$ in the image of $\pi_1$, we can find a prime where $\operatorname{Frob}_q$ acts on the fiber by $\sigma$. We take $\sigma$ to be a generator of the local monodromy at $\infty$, which for all $i$ acts on the roots of $f(x_i)= s+i$ by a $d$-cycle, i.e. it is a tuple of $t$ $d$-cycles in $\prod_{i=1}^t S_d$. In particular, because $d>1$, it has no fixed points. Hence the Frobenius element has no fixed points when acting on the roots of $f(x_i)=s+i$ for any $i$ from $1$ to $t$, and thus none of the roots lie in $\mathbb F_p$, as desired.<|endoftext|> TITLE: Is it true that every mapping class in $\mathrm{Mod}(\Sigma_3)$ commutes with some hyperelliptic involution? QUESTION [5 upvotes]: Two questions. First, let $\Sigma_3$ be the closed genus 3 surface and let $\rm Mod(\Sigma_3)$ be its mapping class group. Is it true that for any mapping class $g\in\rm Mod(\Sigma_3)$ there is some hyperelliptic involution of $\Sigma_3$ which it commutes with? By hyperelliptic involution, I mean a mapping class with an order two representative $f$ with $8$ fixed points. Also, another question, are there any results on choosing representatives for each coset of $\rm HMod(\Sigma_3)$, the hyperelliptic mapping class group? Thank you for any help. REPLY [3 votes]: V.I. Arnold proved that the image of the hyperelliptic group in the mod 2 symplectic group (ie, the image of $HMod_g\to Mod_g\to Sp_{2g}(\mathbb Z)\to Sp_{2g}(\mathbb F_2)$) is the symmetric group on the branch points: $HMod_g\twoheadrightarrow Mod_{0,2g+2}\twoheadrightarrow S_{2g+2}\hookrightarrow Sp_{2g}(\mathbb F_2)$. For $g=1,2$, this inclusion of finite groups is an isomorphism, but not for $g\geqslant 3$. Moreover, for $g=3$, not every element is conjugate into the symmetric group. In particular, there are elements of order 9 in $Sp_6(\mathbb F_2)$ but not in $S_8$. This is an exercise in linear algebra which I messed up when preparing this answer, so me rely on the symmetry Dan mentioned, which is thus a finite element of the mapping class group, and thus retains its order in the coprime symplectic group. Thus not only does this finite element of the mapping class group not commute with any hyperelliptic element, no element which reduces to it in $Sp_6(\mathbb F_2)$ commutes with any hyperelliptic element. (Linear algebra: given a Galois extensions $K\subset M$ which is a composite of two Galois extensions $M=L_1L_2$ of degrees $g$ ($L_1/K$) and $2$ ($L_2/K$); put a symplectic form on $M/K$ so that $U_1(M/L_1)\subset U_g(L_2/K)\subset Sp_{2g}(K)$. Ultimately $K=\mathbb F_2$, $L_1=\mathbb F_8$, and $L_2=\mathbb F_4$; so $U_1(F_{64}/F_{8})=\mu_9\subset Sp_6(\mathbb F_2)$.)<|endoftext|> TITLE: Compactifications of SL(2)-character varieties of surfaces QUESTION [6 upvotes]: Thurston compactified the Teichmüller space ${\cal T}(F)$ of a closed, oriented surface $F$ with a piecewise-linear sphere. Furthermore, as far as I understand, its linear pieces have natural symplectic structure. I am looking for an analogous statement for $SL(2,\mathbb C)$-character varieties $X(F)$ of surfaces $F$. Q1: Is there a natural extension of $X(F)$ whose points have some geometric structure analogous to that of Thurtson's boundary? That may be a compactification or perhaps there is something composed of piecewise complex linear pieces which is not a compactification? Q2: In particular, what can be said about the boundary points of the Morgan-Shalen compactification (in $\mathbb P^S$, where $S$ is the set of all closed loops of $F$)? Q3: There are infinitely many finite subsets $S'\subset S$ whose corresponding trace functions $\tau_s,$ $s\in S',$ embed $X(F)$ into $\mathbb C^{S'}$ and, consequently, compactify $X(F)$ by its algebraic closure in $\mathbb C\mathbb P^{S'}$. These compactifications have geometric structure but are not canonical. But I expect a relation between such compactifications taken for all suitable sets $S'$ and the Morgan-Shalen compactification. Did someone work it out? REPLY [3 votes]: I don't think this actually answers any of your questions explicitly, but it might help. In my recent paper Wonderful Compactification of Character Varieties (co-authored with I. Biswas and D. Ramras) we construct a compactification of any $G$-character variety (including surface groups) where $G$ is of adjoint-type. In the free group case (open surface group), we also show the compactification is simply-connected, admits a natural Poisson structure, and identify the boundaries as "parabolic character varieties". For the closed surface group case you seem to be interested in, you can try to "cut-out" the surface group boundaries from the free group case. Regardless, the introduction of the above paper has many references that might be useful to you (including work by Manon, and also Parreau). In particular, the work of Manon shows that for each quiver-theoretic avatar of a free group character variety discussed here there is a natural compactification (this might speak to your Question 3). The work of Parreau might speak to your interest in geometrically significant boundary divisors.<|endoftext|> TITLE: Is there a ''simple'' formula for the inverse of the Drinfeld associator? QUESTION [6 upvotes]: The Drinfeld associator $\Phi(x_0, x_1)$ encodes the parallel transport of the Knizhnik-Zamolodchikov (KZ) connection $\nabla$ on the bundle $\mathbb{C}\langle\langle x_0, x_1\rangle\rangle$ of formal power series in noncommutating variables $x_0, x_1$ over $X:=\mathbb{P}^1(\mathbb{C})\backslash\left\{0,1,\infty\right\}$. Writing $$\omega_0=\frac{dz}{z}, \quad \omega_1 = \frac{dz}{1-z},$$ we can express the KZ connection as $$\nabla = d-\omega_{KZ}$$ where $\omega_{KZ}\in\Omega^1(X)\otimes\mathbb{C}\langle x_0, x_1\rangle$ is the ''connection form'' $$\omega_{KZ} = x_0\omega_0 + x_1\omega_1.$$ The Drinfeld associator $\Phi(x_0,x_1)\in\mathbb{C}\langle\langle x_0, x_1\rangle\rangle$ (sometimes called KZ associator) is the result of analytically continuing a global horizontal section of $\nabla$ taking the asymptotic value $1$ at $z=0$ along the ''straight line path'' $\text{dch}(t)=t$, $0 < t < 1$. In other words, it is the parallel transport along $\text{dch}$. It can be expressed using iterated integrals as $$\Phi(x_0,x_1) = 1 + \int_{\text{dch}}\omega_{KZ} + \int_{\text{dch}}\omega_{KZ}\omega_{KZ}+\dots$$ It is well-known (e.g. see Lemma 4.2 of Brown's Iterated integrals in quantum field theory) that the coefficients of the Drinfeld associator are multiple zeta values (MZVs). Specifically, we can write $$\Phi(x_0,x_1) = \sum_{w \text{ word in } x_0, x_1}\zeta(w)w$$ where $\zeta(w)$ is the ''shuffle-regularised'' MZV such that $$\zeta(x_0^{k_1-1}x_1\dots x_0^{k_n-1}x_1) = \zeta(k_1, \dots,k_n), \quad \zeta(x_0) = \zeta(x_1)=0.$$ (This depends on your conventions for writing iterated integrals etc.) The inverse of the Drinfeld associator (as power series in noncommuting variables) is $$\Phi(x_1,x_0) = \Phi(x_0,x_1)^{-1}.$$ I take this to mean $$\Phi(x_1,x_0) = \sum_{w \text{ word in } x_0, x_1}\zeta(\overline{w})w$$ where $\overline{w}$ denotes $w$ with $x_0$ and $x_1$ swapped, and the zeta value requires shuffle-regularising. Is there a simple formula for $\Phi(x_1,x_0) = \Phi(x_0,x_1)^{-1}$ in terms of MZVs without having to apply a combinatorial action like swapping to the words? The desire to not deal with things like $\zeta(\overline{w})$ is that the shuffle-regularistion procedure is very combinatorially complicated. As an example of what I mean by ''simple formula'', I originally thought perhaps $$\Phi(x_1,x_0)=\sum_{w} (-1)^{\text{weight}(w)}\zeta(w)w, \quad \text{weight}(w) := \text{length of } w.$$ However this is not correct, because we know that $\zeta(2)=\zeta(x_0 x_1) = -\zeta(x_1 x_0)$ (using shuffle-regularisation). I expect that some formula would use the path-reversal formula for iterated integrals (because we are now integrating along $\text{dch}^{-1}$) and perhaps the change of variables $z\mapsto 1-z$ on $X$. REPLY [2 votes]: Try \begin{equation} \Phi(x_1,x_2)=\sum_w \left(-1\right)^{\textrm{weight}(w)} \zeta(w) \tilde{w} \end{equation} where $\tilde{w}$ is the word with reversed order. Because the $\zeta(w)$ obey the shuffle product, that should give you a proper inverse. Try the first few orders!<|endoftext|> TITLE: Continuity of disintegrations QUESTION [6 upvotes]: Suppose that $\pi:Y\to X$ is a continuous surjection from one compact metric space to another. Given a regular probability measure $\mu$ on $Y$ with pushforward measure $\nu:=\pi^*\mu$, it is known that there is a family $(\mu_x)_{x\in X}$ of probability measures on $Y$ such that $x\mapsto \mu_x(B)$ is Borel measurable for each Borel subset $B$ of $Y$, such that almost every $\mu_x$ lives on the fiber above $x$, and such that $\int_Y f(y)d\mu(y)=\int_{x\in X}\int_{\pi^{-1}(x)}f(y)d\mu_x(y)d\nu(x)$. My question is: suppose in advance that $\mathcal{O}$ is some fixed nonempty open subset of $Y$. Is it true that there is a probability measure $\mu$ as above for which $\mu(\mathcal{O})>0$ and such that there is a disintegration as in the previous paragraph for which the map $x\mapsto \mu_x(\mathcal{O})$ is upper semi-continuous? (A previous version of this question asked for continuity.) This is motivated by the fact that any Radon measure on a compact metric space is upper semi-continuous with respect to the Hausdorff metric. REPLY [3 votes]: Let me start by addressing the question hinted on in the title, which in my view is more interesting than the question eventually asked in the body. Given $\pi:X\to Y$ as above, we may endow the space $\text{Prob}(Y)$ with the weak* topology and consider its closed subspace $Z=\cup_x \text{Prob}(\pi^{-1}(x))$ consisting of all probability measures which images in $X$ are delta measures. Identifying $\delta_x$ with $x$ we consider the map $\pi_*:Z\to X$. The disintegration of a probability measure $\mu\in\text{prob}(Y)$ is a $\pi_*(\mu)$-a.e defined measurable map $X\to Z$, $x\mapsto \mu_x$, as described in the question. One may ask whether such a map could be represented as a continuous map (and that was my initial interpenetration of the question above). The answer to this question is that in case $Y=X=[0,1]$ and $\pi:Y\to X$ is the Cantor-Lebesgue function (see https://en.wikipedia.org/wiki/Cantor_function) there are no local continuous sections $X\to Z$. Indeed, any such section is easily seen to be discontinuous at every diadic point of $X=[0,1]$. In particular, the disintegration function of any probability measure on $Y$ could not be represented by a continuous function. Below is an edit answering the new version in the question's body. Consider $X=Y=[-1,1]$ and $\mathcal{O}=(0,1]\subset Y$. Let $\pi_1:Y\to X$ be the identity map. Then for any $\mu\in \text{Prob}(Y)$ we have that $\mu_x=\delta_x$, and the function $x\mapsto \mu_x(\mathcal{O})$ equals $\chi_{\mathcal{O}}$ which is lower semicontinuous, but not upper semi-continuous. Let $\pi_2:Y\to X$ be given by defined by $\pi_2(x)=\min\{x,0\}$. Then for every $\mu\in \text{Prob}(Y)$ the function $x\mapsto \mu_x(\mathcal{O})$ equals $\mu(\mathcal{O})\cdot\delta_0$ and if $\mu(\mathcal{O})>0$ this function is upper semi-continuous but not lower semi-continuous. We conclude that no semi-continuity is guaranteed in general.<|endoftext|> TITLE: Left Kan extension along Yoneda of pullback-preserving functor preserving pullbacks QUESTION [9 upvotes]: Let $F\colon C \to D$ be a functor. The Kan Extension of $y_D \circ F$ along $y_C$ yields a functor $F_!: Fun(C^{op},Set) \to Fun(D^{op},Set)$. Here, $y_C$ and $y_D$ denotes the respective Yoneda embeddings. It is well-known that if $C$ and $D$ have finite limits and $F$ preserves them, then so does $F_!$. Question: Suppose that we only assume that $C$ has pullbacks and that $F$ preserves them (so some pullbacks exist in $D$, too). Is it true that $F_!$ preserves pullbacks as well? (I'm willing to assume that both $C$ and $D$ have finite products, but in my case, the functor $F$ does certainly not preserve these.) I went through Lurie's ∞-categorical version of the "well-known" part (6.1.5.2 in HTT), but I was not able to verify that the last step works without assuming that $F$ preserves products because this step involves some heavy machinery. I could give a different proof for this last step if $F_!$ preserved monomorphisms, but I believe that this is not the case in general. REPLY [2 votes]: My previous answer left open the following: Proposition: Let $C$ be a small $\infty$-category with all fiber products, let $\mathcal{T}$ be an $\infty$-topos and let $F : C \rightarrow \mathcal{T}$ be a functor preserving fiber products. Then the left Kan extention: $$ \widehat{F} : \text{Prsh}(C) \rightarrow \mathcal{T} $$ also preserves fiber product. Indeed in the previous answer I showed that the statement was true as soon as $C$ has a terminal object and I have given an example showing that the $1$-categorical version of the proposition is false. But the example I have given somehow suggested that the proposition might be true in the $\infty$-categorical settings. I recently needed an answer to this question so I thought about it, and I think I have a proof of the proposition above. Proof: As $C$ has all fiber product, and $F$ preserves them, one has that for each object $c\in C$, the slice category $C/c$ has all finite limits, and the functor induced by $F$: $$ F/c : C/c \rightarrow \mathcal{T}/F(c) $$ preserves them. In particular the left Kan extention: $$ \widehat{F/c} : \text{Prsh}(C/c) \rightarrow \mathcal{T}/F(c) $$ preserve all finite limit by the results of Lurie. This functor is isomorphic to: $$ \widehat{F}/c : \text{Prsh}(C)/c \rightarrow \mathcal{T}/F(c) $$ It follows in particular that $\widehat{F}$ preserves all fiber product of the form $X \times_c Y$ where $c$ is a representable object. Note that up to this point, everything also works in the $1$-categorical case, and this is exactly how we concluded in the special case where $C$ has a terminal object. The idea is now to use the descent property to extend this to more general pullback. Let $V \rightarrow X$ be any morphism in $\text{Prsh}(C)$, and write $X$ as the canonical colimits: $$ X = \underset{c \in Elt(X)}{\text{colim }} c $$ (where $Elt(X)$ is the category of elements of $X$ and I have identified objects of $C$ with their image by the Yoneda embedding) For each elements $c \rightarrow X$, let $V_c$ be the fiber of $V \rightarrow X$ over $c$. By descent, one has: $$ V = \underset{c \in Elt(X)}{\text{colim }} V_c $$ Now, as $\widehat{F}$ preserves all pullbacks whose bottom corner is representable, it follows that the the natural transforation: $\widehat{F}(V_c) \rightarrow F(c)$ is cartesian (in $c$), and hence, by descent in $\mathcal{T}$ it follows that all the maps $\widehat{F}(V_c) \rightarrow F(c)$ are pullback of the maps between the colimits: $$ \widehat{F}(V) \simeq \underset{c \in Elt(X)}{\text{colim }} \widehat{F}(V_c) \rightarrow \underset{c \in Elt(X)}{\text{colim }} F(c) \simeq \widehat{F}(X)$$ Which shows that $\widehat{F}$ also preserves all pullback of the form $V \times_X c$ as soon as $c$ is representable. To conclude, one either use a again a similar (but simpler) colimit/descent argument in the variable $c$, or one simply write (using descent) that the functor $\text{Prsh}(C)/X \rightarrow \mathcal{T}/\widehat{F}(X)$ can be decomposed as a limits: $$ \text{Prsh}(C)/X \simeq \lim_{c \in Elt(X)} \text{Prsh}(C)/c \rightarrow \lim_{c \in Elt(X)} \mathcal{T}/F(c) \simeq \mathcal{T}/\widehat{F}(X) $$ (Note: one needs the fiber product preservation proved above to show this) But as all the functors $\text{Prsh}(C)/c \rightarrow \mathcal{T}/F(c)$ preserves finite limits, their limits also preserve finite limits and hence for all $X \in \text{Prsh}(C)$ the functor $\text{Prsh}(C)/X \rightarrow \mathcal{T}/\widehat{F}(X)$ preserve all finite limits, which concludes the proof.<|endoftext|> TITLE: Are classes still "larger" than sets without the axiom of choice? QUESTION [22 upvotes]: Classes are often informally thought of as being "larger" than sets. Usually, the notion of "larger" is formalized via an injection: $B$ is "at least as large" as $A$ iff there is an injection from $A$ to $B$, and strictly "larger" if there is no injection going the other way. Even though ZFC does not formalize the notion of proper classes, we can still speak sensibly of the notion of an injection from a set to a class, which is simply a function such that its image is within the class. So we can still say a set injects into some proper class if the relevant function exists. It is known that without AC, we can have sets that are "incomparable" in size, because the universe is now only partially ordered rather than well-ordered. But it seems that we must now also have the situation where sets can be incomparable in size to classes! Otherwise, if every set injected into every class, they would all inject into the ordinals, and hence be well-ordered. (Right?) So without AC, we cannot say that classes are necessarily "larger" than sets -- just "different"! My questions: Is this the right understanding? Is the axiom "every set injects into every class" equivalent to the axiom of choice? Is there some sensible way to interpret the notion of "class" other than as an entity too "large" to be a set, since this makes no sense without AC? I would be happy to discuss within a theory like NBG as well that explicitly formalizes classes, although I tried to word this in such a way that it wasn't necessary. REPLY [26 votes]: Yes, your remarks about incomparability of sets and classes without the axiom of choice are correct. Yes, in ZF (or in GB), the axiom of choice is equivalent to the assertion that every set injects into every proper class (one must say proper class, since in the usual terminology, every set also counts as a class). The forward direction holds, since every proper class $A$ must have elements of unboundedly many ranks, and so it has arbitrarily large subclasses. Conversely, if a set $a$ injects into the ordinals, then it is well-orderable. In ZF or GB, that is, without the axiom of choice, one can interpret the notion of proper class as those classes that do not contain all their elements at some stage of the cumulative hierarchy. That is, they must contain elements of ranks unbounded in the ordinals. So the essence of what makes something a proper class is not its size, as such, but rather the fact that at no stage of the cumulative hierarchy did one finish completing the class in the sense of already having all of the elements of the class by that stage. Meanwhile, one can recover the idea that proper classes are large, even in ZF or GB, by considering quotients of the class, for a class $A$ is a proper class if and only if there is an equivalence relation $\sim$ on $A$ (and one may freely assume that all equivalence classes are sets) such that the class $\text{Ord}$ injects into the quotient $A/\sim$. Indeed, the same equivalence relation $\sim$ works for all classes $A$: let $a\sim b$ if and only if they have the same rank. A class $A$ is a proper class if and only if it has elements from $\text{Ord}$ many distinct ranks.<|endoftext|> TITLE: Ring structures on algebraic K-theory spectrum, and its non-connective counterpart QUESTION [6 upvotes]: I have a few naive questions on the algebraic K-theory spectrum construction, but whose answers I couldn't figure out using the internet. I'm mostly interested in the case of a commutative ring, but I guess most of these questions apply somewhat more generally as well - e.g. $E_{\infty}$ ring spectra, or maybe more abstractly for something like exact monoidal categories. 1) I've seen several ways to construct the cup product in algebraic K-theory, which generally result in something on the level of the K-space, i.e. a map $K\wedge K\to K$ which is associative, commutative up to homotopy and distributive over the H-space structure up to homotopy. Am I correct that this also implies the corresponding connective spectrum is a ring spectrum? 2) Under what circumstances do the above structures lift to $\mathbb{E}_\infty$-ring structures? I would guess if you had similar coherence structures already on the symmetric monoidal category (so in particular all $R$-module categories since they're strict), but I am hopeless with all of that. Also, I don't understand the issues surrounding $E_\infty$-structures; it seems like this sometimes depends in what category of spectra you're working in. 3) In Weibel's K-book pg 79, a non-connective version of the K-theory spectrum due to Bass is given, basically by homotopizing the fundamental exact sequence. Do the structures (ring, $E_{\infty}$) from the previous questions extend here? (Also, as a digression, this construction seemingly only applies for rings - is there a version of this construct whose input is more categorical?) 4) (a bit more farfetched) For the kinds of rings you find in arithmetic geometry (esp. rings of integers in global fields), is there any relationship between the Bass non-connective spectrum and $K(1)$-localization? cf. the Lichtenbaum-Quillen conjectures implying that such rings are essentially $K(1)$-local. REPLY [6 votes]: Proposition 5.9 in [Blumberg, Gepner, Tabuada. Uniqueness of the multiplicative cyclotomic trace. arXiv:1103.3923] shows that both connective and nonconnective K-theory define lax symmetric monoidal functors (on the infinity-category of stable idempotent-complete infinity-categories). It follows that if you apply either variant of K-theory to a symmetric monoidal stable idempotent-complete infinity-category, you end up with an $E_\infty$-ring spectrum. Of course this applies to the infinity-category of perfect complexes on a commutative ring or $E_\infty$-ring spectrum.<|endoftext|> TITLE: Relations of axioms of choice QUESTION [7 upvotes]: We start with $ZF$. The axiom of countable choice, $AC_\omega$, says that any set product of nonempty sets with a countable index set is nonempty. For any $ZF$-definable set $A$, we should be able to define $AC_A$ in an analogous manner: any product of nonempty sets with index set $A$ is nonempty. What is the implication structure of these axioms? There is the obvious fact that if $A$ injects into $B$ (or rather, if that can be proven by $ZF$), then $AC_B$ implies $AC_A$ (Proof: choose $1$-element sets for $B\backslash A$, and restrict the choice function); is there anything else beyond the trivial? Also, the axiom of choice implies that all of these axioms are true; is the converse true? REPLY [12 votes]: Yes, the axiom of choice is equivalent to the assertion that $\text{AC}_X$ holds for every definable $X$. One usually has to take a little care with foundational matters when definability is involved, since one cannot ordinarily express assertions of the form "every definable $X$ has a certain property", as the class of definable $X$ is not itself necessarily definable. Nevertheless, in your case, it turns out that we can get around this obstacle. Specifically, you may be surprised to hear that I claim that there is a single definable set $X$, such that $\text{ZF}$ proves that $\text{AC}_X$ is equivalent to the axiom of choice. That is, we need only to consider one suitably defined set $X$, and $\text{AC}$ is provably equivalent to this special single instance of your axiom $\text{AC}_X$ for this specific definable set $X$. My set $X$ is: the smallest set $V_\theta$, a rank initial segment of the cumulative hierarchy, such that there is an $X$-indexed family of nonempty sets with no choice function, if there is such a family, and $X=\emptyset$, otherwise. This is a perfectly good definition of a set $X$ in the theory $\text{ZF}$. If the axiom of choice holds, then $\text{AC}_X$ holds for every set $X$, including the one I just defined. Suppose conversely that the axiom of choice fails. So there is some index set $I$ and indexed family $\langle A_i\mid i\in I\rangle$ of nonempty sets $A_i$ having no choice function. The set $I$ must be contained in some $V_\theta$, since every set is a subset of some rank initial segment of the universe. Let $A_i=\{0\}$ for all $i\in V_\theta\setminus I$. So now we have a family $\langle A_i\mid i\in V_\theta\rangle$ of nonempty sets, and still there can be no choice function. So there is such a $\theta$ as in the definition of my set $X$. Thus, by design, it cannot be that $\text{AC}_X$ holds for this set $X$. Of course, this is a kind of trick. It is like saying that every natural number has a certain property if a certain single number $n$ has the property, where $n$ is defined to be the smallest counterexample, if there is one, and $0$ otherwise. If that number $n$ has the property, then every number does. Many years ago I was at a party in Berkeley at the home of Ken Ribet and a mischevious Hendrik Lenstra mentioned to me and a small group of my fellow graduate students that although it was not known whether there are infinitely many prime pairs, nevertheless there is a number $N$ such that if there is any prime pair above $N$, then there are infinitely many prime pairs. "If we could only find that $N$!," he said, with a wink in his eye. The other students nodded at this profound-seeming statement, but I thought a bit and said: case 1 is that there are only finitely many pairs, and we can take any $N$ above them; case 2 is that there are infinitely many pairs, in which case we can take $N=17$ or anything. Lenstra shook my hand and said, yes, it is really trivial, isn't it?<|endoftext|> TITLE: Some strange multinomial averaging QUESTION [6 upvotes]: How do I prove : $\sum_{j=2}^{n} (-1)^j {\frac {M(n+j,j;2)}{j!}} = (-1)^n n! + 1$? where $M(n+j,j;2)$ is the multinomial sum $M(n+j,j;2) = \sum_{t_1 + t_2 + \dotsc + t_j = n+j, t_k \geq 2} {n+j \choose t_1 \dotsc t_k}$ which denotes the number of surjective functions from $n+j$ points to $j$ points with at least two elements in each pre-image. I have tried using the recursion formulas of counting doubly surjective functions which didn't help much. My thought was the alternate signs appear for some Moebius maps on some partial order on the disjoint set of surjective maps from $n+j$ points to $j$ points up to permuting the image points (which explains division by $j!$). My definition of reshuffling images which give a partial order which give $0$ to distant points. REPLY [10 votes]: Denote by $f(n,m)=M(n+m,m;2)/m!$ the number of partitions of $\{1,2,\dots,n+m\}$ onto $m$ subsets of size at least 2 (observation of Darij Grinberg). Your identity rewrites as $A(n):=\sum_{j\geqslant 1} (-1)^jf(n,j)=(-1)^nn!$. We have $f(n,m)=mf(n-1,m)+(n+m-1)f(n-1,m-1)$: the first summand corresponds to the partitions in which $n+m$ belongs to a subset of size at least 3; the second to the partitions in which $n+m$ belongs to a subset of size 2. Substitute this to $A(n)$, we get $A(n)=-nA(n-1)$ and your identity follows by induction (with base $n=1$, say).<|endoftext|> TITLE: Summing the infinite series $\sum_{k=0}^{\infty} \frac{x^k}{(k!)^2}$ QUESTION [6 upvotes]: Is there a closed form sum of $\sum_{k=0}^{\infty} \frac{x^k}{(k!)^2}$ It is trivial to show that it is less than $e^x$ but is there a tighter bound? Thanks REPLY [2 votes]: $\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \renewcommand{\th}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$ This answer is based on ideas quite different from those used in my previous answer to this question, and the result is much better. As in that answer, let \begin{equation*} S(x):=\sum_{k=0}^{\infty} \frac{x^k}{(k!)^2}. \end{equation*} Term-wise differentiation shows that \begin{equation*} (xS'(x))'=S(x). \end{equation*} This differential equation can be rewritten as \begin{equation*} 16x^2a''(x)+8x(1+4\sqrt x\,\tanh(2\sqrt x))a'(x)+a(x)=0, \tag{1} \end{equation*} where \begin{equation*} a(x):=S(x)/S_*(x), \end{equation*} \begin{equation*} S_*(x):=\cosh\sqrt{4x}\big/\sqrt{\pi\sqrt x}\sim e^{\sqrt{4x}}\big/\sqrt{4\pi\sqrt x}\sim S(x) \end{equation*} for large $x$, as was noted by Carlo Beenaker. Note that $a(x)>0$ for $x>0$. So, by (1), if $a'(x)=0$ for some $x>0$, then $a''(x)<0$. So, the only local extrema of $a$ are local maxima; therefore and because between any two local maxima there is a local minimum, we see that there is at most one local maximum of $a$ on $(0,\infty)$. Since $a(0+)=0$, $a(1)>1$, and $a(\infty-)=1$, we conclude that $a$ has precisely one local (and hence global) maximum on $(0,\infty)$. In fact, this maximum occurs at $x=x_*=0.7277\dots$, and $a(x_*)=1.0769\ldots<1.08$. Moreover, $a>1$ on $[0.2,\infty)$. Thus, \begin{equation*} S_* TITLE: A Non-quasi-affine scheme with quasi-affine irreducible components QUESTION [7 upvotes]: Let $X$ be a scheme finite type over a field $k$ such that each of its irreducible components $X_i$ is quasi-affine. Is it true that $X$ is quasi-affine? REPLY [6 votes]: I am just writing my comment as an answer. Probably this example has appeared before on MO. Let $\mathbb{P}^3_k$ denote the projective space $\text{Proj}\ k[s_1,s_2,t_1,t_2]$. Let $\overline{X}$ denote the closed, reduced subscheme $$\overline{X} =\text{Zero}(s_1s_2) = \overline{X}_1 \cup \overline{X}_2, \ \ \overline{X}_1 = \text{Zero}(s_2), \ \ \overline{X}_2 = \text{Zero}(s_1).$$ This is a union of two hyperplanes, $\overline{X}_1$ and $\overline{X}_2,$ whose intersection is a line, $$\overline{L} = \overline{X}_1\cap \overline{X}_2 = \text{Zero}(s_1,s_2) = \text{Proj} \ k[t_1,t_2].$$ Inside $\overline{X}_1$, define $C_1$ to be the closed subscheme that is a line, $$C_1 = \text{Zero}(s_2,t_1).$$ Inside $\overline{X}_2$, define $C_2$ to be the closed subscheme that is a line, $$C_2=\text{Zero}(s_1,t_2).$$ Note that $C_1\cap \overline{X}_2$ is the singleton set of $p_2=[0,0,0,1]$, and $C_2\cap \overline{X}_1$ is the singleton set of $p_1=[0,0,1,0].$ Define $C$ to be the closed union of $C_1$ and $C_2$ in $\overline{X}$. Define $X$ to be the open complement in $\overline{X}$ of the closed subset $C$. By construction, $X$ is quasi-projective. Moreover, $X$ is the union of the two irreducible components, $$X_1 = \overline{X}_1\setminus(C_1\cup \{p_1\}), \ \ X_2 = \overline{X}_2 \setminus (C_2\cup \{p_2\}).$$ The intersection of the two irreducible components is $$L=\overline{L}\setminus\{p_1,p_2\}.$$ Each of the two irreducible components is isomorphic to a punctured affine plane. In particular, $\mathcal{O}_{X_1}(X_1)$ equals $k[s_1/t_1,t_2/t_1]$. Similarly, $\mathcal{O}_{X_2}(X_2)$ equals $k[s_2/t_2,t_1/t_2]$. Finally, the intersection $L$ is a twice-punctured projective line with coordinate ring $\mathcal{O}_L(L)$ equal to $k[t_2/t_1,t_1/t_2]$. The restriction map $\rho_1$, resp. $\rho_2$, to $\mathcal{O}_L(L)$ from $\mathcal{O}_{X_1}(X_1)$, resp. from $\mathcal{O}_{X_2}(X_2)$, is the $k$-algebra map that sends $s_1/t_1$ to $0$, resp. that sends $s_2/t_2$ to $0$, and that sends $t_2/t_1$ to itself, resp. that sends $t_1/t_2$ to itself. The fibre product of these restriction maps is the $k$-algebra $$\mathcal{O}_X(X) = \{(f_1,f_2)\in \mathcal{O}_{X_1}(X_1)\times \mathcal{O}_{X_2}(X_2) : \rho_1(f_1) = \rho_2(f_2)\}.$$ This is infinitely generated as a $k$-algebra by the following monomials, $$\frac{s_1}{t_1}, \ \frac{s_2}{t_2}, \ \left(\frac{s_1^mt_2}{t_1^{m+1}}\right)_{m\geq 1}, \left(\frac{s_2^nt_1}{t_2^{n+1}}\right)_{n\geq 1}. $$ In particular, the image in $\mathcal{O}_L(L)$ of $\mathcal{O}_X(X)$ equals the constant subfield $k$. Thus, $X$ is not quasi-affine.<|endoftext|> TITLE: Intuition behind the canonical projective resolution of a quiver representation QUESTION [8 upvotes]: Let $Q$ be a finite acyclic quiver, and $X$ some representation of $Q$. For $i \in Q_0$ define the $kQ$-modules $P_i = kQe_i$, and $X(i) = e_i X$. The representation $X$ has a canonical projective resolution, called the Ringel resolution, given by $$ 0 \longrightarrow \bigoplus_{a \in Q_1} P_{h(a)} \otimes X(t(a)) \xrightarrow{\quad d \quad} \bigoplus_{i \in Q_0} P_i \otimes X(i) \xrightarrow{\quad f \quad} X \longrightarrow 0 $$ where in each direct summand the maps are given by $$ d(p \otimes x) = p \otimes (a \cdot x) - pa \otimes x \qquad\quad f(p \otimes x) = p \cdot x $$ Does anyone have some intuition for this resolution that they could share? Can the modules and the maps in this resolution be nicely interpreted in terms of the paths in $Q$? And can this interpretation be extended to get us projective resolutions for quivers with relations? Or for non-acylic quivers? REPLY [3 votes]: There's some exposition along these lines in Theorem 2.15 in Schiffler's "Quiver Representations". Using your notation, let $Q = (Q_0,Q_1)$ be a finite acyclic quiver and $X$ be some representation of $Q$. Let $d_i$ denote the dimension of $X_i$. Then the standard projective resolution of $X$ is $$0 \rightarrow P_1 \rightarrow P_0 \rightarrow X \rightarrow 0$$ where $$P_1 = \bigoplus_{\alpha \in Q_1} d_{s(\alpha)} P(t(\alpha)), \qquad P_0 = \bigoplus_{i \in Q_0} d_i P(i).$$ Looking at $P_0$, it contains $d_i$ copies of $P(i)$ for each vertex $i$. We want $P_0 \rightarrow X$ to be surjective; this choice of $P_0$ is definitely "big enough" to make that possible, since $(P_0)_i$ has at least dimension $d_i$. Basically what we've done is the "cheapest" way to find a projective representation that can surject onto $X$. Then the natural thing to do for $P_0 \rightarrow X$ is to have the map act like the identity on $d_i$ copies of $(P_0)_i$ and like the zero map on any "extra" copies. So then we need to understand what $P_1 = \textrm{ker}(P_0 \rightarrow X)$ needs to look like (intuitively, $P_1$ is all the "extra" stuff in $P_0$ that's not in $X$). Essentially we want to take $P_0$ and reduce the dimension of each $(P_0)_i$ by $d_i$. This is equivalent to replacing $d_i P(i)$ with $\oplus_{\alpha: i \rightarrow j} d_i P(j)$ (i.e., you replace each copy of $P(i)$ with the direct sum of the projective representations at each vertex $j$ such that there's an arrow $\alpha: i \rightarrow j$ in $Q$). The same reasoning should carry through for bound quivers, with the relations showing up in the way $P(i)$ is defined for a bound quiver. I hope this was somewhat useful and not just a repetition of information in the previous answers and/or something that you already knew.<|endoftext|> TITLE: How many rich directions does a set in $\mathbb F_p^2$ determine? QUESTION [10 upvotes]: $\newcommand{\F}{\mathbb F}$ A subset $P$ of the affine plane $\F_p^2$ is said to determine a direction if there is a line in this direction containing at least two points of $P$. A set of size $|P|>p$ determines all $p+1$ directions, a set of size $|P|\le p$ not contained in a single line determines at least $(|P|+3)/2$ directions; the former is an immediate consequence of the pigeonhole principle, the latter is a highly non-trivial result of Szonyi that builds on works of Rédei. Let's say that a direction in $\F_p^2$ is rich if there is a line in this direction containing at least three points of $P$. If $P$ is trapped in a line, or a union of two lines, then there are just one or two rich directions. What is the smallest possible number of rich directions for a set $P\subset\F_p^2$ of size $\frac53\,p<|P|\le 2p$ given that $P$ is not contained in a union of two lines? It is easy to construct sets $P$ with about $|P|-p$ rich directions, but it is not clear to me how much better can one do. Added May 04, 2018 The largest possible number of points in general position in $\F_p^2$ is $p+1$, which is quite easy to prove; thus, we are guaranteed to have at least one rich direction. Indeed, I can show that if $\frac32(1+\delta)p\le|P|\le 2p$, then there are at least $\Omega_\delta(\sqrt p)$ rich directions, but I suspect that this can be far from being sharp. REPLY [3 votes]: A cool generalization of Rédei's method furnishes the following lower bound: Theorem-lower bound. Let $U\subset \mathbb F_p$ have cardinality $2p$. Then $U$ is contained in the union of two lines or it determines at least $\lfloor\frac{p+4}{3}\rfloor$ rich directions. The key idea is to consider the y-derivative of the Rédei polynomial $$ H_{U}(x,y):=\prod_{(a,b)\in U} (x-ay+b) = x^{2p}+h_1(y)x^{2p-1}+\dots+h_{2p}(y)$$ and the following proposition on lacunary fully reducible polynomials: Proposition Let $p\neq 2$ be prime and let $f(x)\in\mathbb F_p[x]$ be a (monic) fully reducible polynomial of degree $2p$. Write $f(x)=x^{2p}+g(x) x^p + h(x)$ with $deg(g)=m$, $deg(h)=n$ and $m,np$$ or $f(x)$ is one of the following: $f(x)=(x^p-x)^2$ or $f(x)=(x-t)^p(x-s)^p$ or $f(x)=(x^p-x)(x-t)^p$ for some $s,t\in\mathbb F_p$. Fully reducible means that it factors completely, as a product linear factors, over $\mathbb F_p$. The Theorem should be compared to the following Example. Example-upper bound. For every odd prime $p$ there is a set $P\subseteq \mathbb F_p^2$, with cardinality $|P|=2p$ that determines exactly $(p+3)/2$ rich directions, and which is contained in 3 lines but not in 2 lines. Remarks The theorem is of course optimal up to multiplicative constant, but I do not know if it is actually close to be sharp. To see if it is, I suggest to think first about this MO problem. Following Szonyi, it is not difficult to get a result valid for $|U|<2p$. Essentially, all is needed is to prove the version of the Proposition corresponding to almost-fully-reducible $f(x)$. The theorem seems to generalize to counting $n$-rich directions, while the proposition suggests interesting generalizations to lacunary almost-fully reducible polynomials of degree $np$. It would be cool to apply them to blocking-set problems. Question: can the Proposition be proved more directly from the corresponding classical result for lacunary polynomials of degree $p$? Last thing: Rédei-s theorem has a proof that avoids lacunary polynomials due to Lovász, L., and A. Schrijver, but I have been unable to generalize their method. Proofs follow Proof of the Proposition Let $F(x) = gcd(f(x),x^p-x)$, then $$F(x)\ |\ xg(x)+h(x)+x^2$$ because $xg(x)+h(x)+x^2=f(x)-(x^p-x)(x^p+x+g(x))$ and $$\frac{f(x)}{F(x)}\ |\ f'(x) = g'(x)x^p + h'(x).$$ Therefore $f(x) | (g'(x)x^p + h'(x))(xg(x)+h(x)+x^2)$. Case1: $(g'(x)x^p + h'(x))(xg(x)+h(x)+x^2)\neq 0$. Then $2p\leq m-1+p+N$ with $N:=max(n,m+1,2)$. Case2: $g'(x)x^p + h'(x)+0$. Then $f(x)=(x^2+g(0)x+h(0))^p=(x-t)^p(x-s)^p$. Case3: $h(x)=-xg(x)-x^2$. Then $f(x)=(x^p+g(x)+x)(x^p-x)$. By Rédei's proposition on lacunary fully reducible polynomials of degree $p$, we have either $x^p+g(x)+x=(x-t)^p$, or $x^p+g(x)+x=x^p-x$, or $2m\geq p+1$. Here $n=m+1$. Proof of the Theorem Let $D$ be the set of rich directions determined by $U$. Without loss of generality we may assume that $\infty\in D$ (i.e. that there is a vertical line $X=c$ containing $\geq 3$ points of $U$. See for example Seva's addendum, May 4, 2018). Notice that for fixed slope $y\in \mathbb F_p$ we have that $(x-ay+b)=(x-a'y+b')$ if and only if $y=\tfrac{b-b'}{a-a'}$. Thus $y\not \in D$ if and only if $$H_U(x,y)=\prod_{k\in\mathbb F_p}(x-k)^2=x^{2p}-2x^{p+1}-x^2.$$ In particular for $j\neq p-1,2p-2$ the polynomial $h_j(y)$ vanishes at least at $p+1-|D|$ values of $y$ (i.e. $y\not\in D$). Since $deg(h_j)\leq j$, we have that $h_j(y)\equiv 0$ identically for $j=1,\ldots,p-|D|$. Now the y-derivative trick to gain vanishing Consider the y-derivative $$ \partial_y H_U(x,y) = h'_1(y)x^{2p-1}+\dots+h'_{2p}(y).$$ By Leibniz formula we see that for $y\not \in D$ we have that $H_U(x,y)$ is divisible by $(x-k)$ for all $k\in\mathbb F_p$. In other words $$ H_U(x,y) = (x^p-x)S_y(x) = S_y(x)x^p-S_y(x)x$$ for some polynomial with $deg(S_y)\leq p-1$. We deduce for $j=2,\ldots,p-1$ and $y\not \in D$ the equality $h'_j(y)=h'_{j+p-1}(y)$. Since $h_j(y)\equiv 0$ identically for $j=1,\ldots,p-|D|$, we get that $h_{p+j}$ vanishes at $y\not\in D$ with order 2 for $j=1,\ldots,p-|D|-1$. Therefore $h_{p+j}(y)\equiv 0$ identically as soon as $2(p+1-|D|)\geq p+j+1$. As a consequence, for all $y\in \mathbb F_p$ we have $$ H_U(x,y) = x^{2p} + g(x)x^p+h(x)$$ with $deg(g)\leq |D|-1$ and $deg(h)\leq 2(|D|-1)$. Take $y\in\mathbb F_p$. $H_U(x,y)=(x^p-x)^2$ when $y\not\in D$; $H_U(x,y)=(x-s)^p(x-t)^p$ when $U$ is the support of the two lines $Y=yX+s$, $Y=yX+t$; $H_U(x,y)=(x^p-x)(x-t)^p$ means in particular that $p$ points of $U$ are on the line $Y=yX+t$. It is not difficult to show that, if $U$ is not supported on two lines only, then here is a line containing at least 3 points of $U$ but less than $p$ points. Such line corresponds to $y=y_0\in D$ for which 1-2-3 above do not hold. By the Proposition applied to $f(x)=H_U(x,y_0)$ we get $3(|D|-1)\geq p+1$, so the Theorem. Proof of the Example. Let $P'=\{(k,k^{\frac{p+1}2}): k\in\mathbb F_p\}$, which is a set contained in the two lines $y=\pm x$ with cardinality $|P'|=p$. It is shown in [1] that $P'$ attains the equality in the theorem of Rédei. In other words, $P'$ determines exactly $(|P'|+3)/2=(p+3)/2$ directions. Now let $P''=\{(k,k-2): k\in\mathbb F_p\}$, i.e the line parallel to $y=x$ passing through $(1,-1)$. We observe that $P'\cap P''=\emptyset$ and that the direction of $P''$ is one of the directions determined by $P'$. Finally we let $P=P'\cup P''$ and we observe immediately that the rich directions of $P$ are exactly the directions determined by $P'$. [1] Lovász, L., and A. Schrijver. "Remarks on a theorem of Rédei." Studia Scient. Math. Hungar 16.449-454 (1981): 15-32.<|endoftext|> TITLE: Do all closed connected subgroups of $SO(2n+1)$ embed into $SO(2n)$? QUESTION [8 upvotes]: Is every closed connected proper subgroup of $SO(2n+1)$ isomorphic (as a Lie group) to a subgroup of $SO(2n)$? The answer is yes for abelian closed connected subgroups. REPLY [10 votes]: As an answer to your weaker version: No. For example, $\mathrm{G}_2$ is a closed proper subgroup of $\mathrm{SO}(7)$, but it is not isomorphic to any subgroup of $\mathrm{SO}(6)$. Of course, this follows from the fact that ${\frak{g}}_2$ has no nontrivial representation of dimension less than $7$.<|endoftext|> TITLE: Odd primary dual Steenrod algebra QUESTION [20 upvotes]: My question is related to this, this, and this older questions. Let $\mathcal A_*$ be the dual Steenrod algebra. This is a super-commutative Hopf algebra, and so its $Spec$ is an algebraic super-group. At  p = 2: ... the algebraic group $Spec(\mathcal A_*)$ can be naturally identified with the group of automorphisms of $Spf(\mathbb F_2[[x]])$ (with $x$ in degree $1$) that preserve: • the group law $x_1,x_2\mapsto x_1+x_2$, • the tangent space at the identity. When p is an odd prime: ... the algebraic super-group $Spec(\mathcal A_*)$ can be naturally identified with the super-group of automorphisms of $Spf(\mathbb F_p[[x,\theta]])$ (with $\theta$ in degree $1$, and $x$ in degree $2$ — this is an exterior algebra on $\theta$ tensor a polynomial algebra on $x$) that preserve: • the group law $(x_1,\theta_1),(x_2,\theta_2)\mapsto (x_1+x_2+\theta_1\theta_2,\theta_1+\theta_2)$ ???? • the tangent space at the identity. ???? Are these statements correct? If yes, where can I read about them? If not, how does one fix them to make them correct? REPLY [8 votes]: This is not a full answer but it was too long for a comment so I decided to write it as detailed answer (EDIT: I added what I believe to be a full answer at the end resolving both points (1) and (2) in my previous answer). I think it helps approaching this story from the other side. Instead of trying to find the correct formal group by staring at the formulas lets try to see if we can squeeze the formal group out from the topology. Homology is represented by $H_*(-) = \pi_*(- \otimes H\mathbb{F}_p)$ from which we see (by virtue of the fact that $\pi_*$ is symmetric monoidal over a field) that the hopf algebra $\pi_* (H\mathbb{F}_p \otimes H\mathbb{F}_p) := \mathcal{A}_*$ must coact on $H_*(-)$. (i will write $\otimes$ for smash product of spectra throughout). So homology determines a symmetric monoidal functor from spectra to graded $\mathcal{A}_*$ comodules. And in fact even before taking $\pi_*$ we still have (in some vague sense which might be difficult to make precise and I won't need) that $- \otimes H\mathbb{F}_p$ lands in the category of comodules over the "derived" hopf algebra $H\mathbb{F}_p \otimes H\mathbb{F}_p$. Interpreting this algebro-geometrically we may think of $\pi_*(H\mathbb{F}_p \otimes H\mathbb{F}_p)$ as the ring of functions on some sort of "super" (which just means "in a graded sense") algebraic group $\mathbb{G} = Spec \mathcal{A}_*$ and of comodules over it as algebraic representations of $\mathbb{G}$. In this language homology determines a symmetric monoidal functor $\mathcal{Sp} \to Rep(\mathbb{G})$. We may now precompose this functor with $\Sigma^{\infty}_+$ to get a symmetric monoidal functor (with respect to the product on spaces) $\mathcal{S} \to Rep(\mathbb{G})$ (where $\mathcal{S}$ stands for the category of spaces). Since every object in $\mathcal{S}$ is canonically a cocommutative coalgebra via the diagonal we immediately conclude that the functor actually lands in $CoAlg(Rep(\mathbb{G}))$. The algebro-geometric interpretation of coalgebras is as formal schemes by taking $Spf$ of the linear dual (there's a caveat here in that the filtration on the dual should come from the filtration on the homology coming from expressing a space as a filtered colimit of finite spaces). Moreover in this form the functor sends products to products since the categorical product in coalgebras is given by the tensor product on the underlying vector space. Summarizing we now have homology as a functor $\mathcal{S} \to CoAlg(Rep(\mathbb{G}))$ which preserves products and filtered colimits. We can also compose with the fogetful $CoAlg(Rep(\mathbb{G})) \to CoAlg(grVect)$ which only remembers homology with the coalgebra structure. We may now use this to conjecture about the structure of $\mathcal{A}_*$ as follows. An eilenberg maclane space $K(A,n)$ is a commutative group object in $\mathcal{S}$. Its homology is therefore a commutative group object in $CoAlg(Rep(\mathbb{G}))$. Interpreted algebro-geometrically the homology of $K(A,n)$ determines a commutative formal super group $\mathbb{G}_{A,n}$ together with a compatible action of the algebraic super group $\mathbb{G}$. In other words we get a map which we will hence forth call the comparison map: $$\mathbb{G} \to Aut(\mathbb{G}_{A,n})$$ Where by $Aut$ I mean the full algebraic group of automorphisms of $\mathbb{G}_{A,n}$. We may now try to plug in values for $A$ and $n$ and check whether we get something valuable. The most obvious first choice is $A = \mathbb{Z}/p\mathbb{Z}$ and $n=1$. It turns out that this brings us surprisingly close to the correct answer. Let $V$ be an $\mathbb{F}_p$ vector space (equivalently a finite $p$-torsion abelian group) then there's the following computation: $$H^{\bullet}(K(V,1),\mathbb{F}_p) = \begin{cases} Sym^{\bullet}(V^{\ast}[-1]) & \text{if $p = 2$} \\ Sym^{\bullet}(V^{\ast}[-2]) \otimes \bigwedge^{\bullet}(V^{\ast}[-1]) & \text{if $p \gt 2$} \end{cases}$$ Algebro geometrically this can be interpreted as saying that $Spf H^{\bullet}(K(V,1),\mathbb{F}_p)$ is the formal super scheme corresponding to the super vector space $V[1] \oplus V[2]$ completed at the origin in the case where $p$ is odd and the completion of $V[1]$ when $p$ is even. It hence inherits the additive formal group structure from the addition on the corresponding vector space. Taking $V = \mathbb{Z}/p\mathbb{Z}$ we get something which one might call an additive formal super group. Now comes the part i'm least sure about. Its straightforward to compute the automorphism hopf algebra of $\mathbb{G}_{\mathbb{Z}/p\mathbb{Z},1}$. Then using the fact that we already know $\mathcal{A}_*$ and staring at it for a bit its easy to become convinced $\mathbb{G}$ is a sub algebraic-group of $Aut(\mathbb{G}_{\mathbb{Z}/p\mathbb{Z},1})$. There are therefore 2 questions left: Find precisely the a subgroup of $Aut(\mathbb{G}_{\mathbb{Z}/p\mathbb{Z},1})$ isomorphic to $\mathbb{G}$. In other words what kind of extra structure on $\mathbb{G}_{\mathbb{Z}/p\mathbb{Z},1}$ do homology co-operation preserve? Analyze the comparison map and determine whether it induces the inclusion from (1) I think the for the even case we know that (1) is solved by requiring the automorphism be strict and that (2) is an isomorphism (I'm less sure about this point - so i'd love to be corrected about this if i'm wrong). For the case of $p$ odd I think the answer to (1) is that one should take automorphisms preserving the even coordinate to first order and the odd coordinate to first order modulo the even coordinate so it preserves the descending degree filtration on the tangent space. I'm less sure about this because I may have made a mistake in the calculation. Regarding (2) it's straightforward that the map must be injective since if a stable cohomology operation in mod p cohqomology is trivial on $B \mathbb{Z}/p\mathbb{Z}$ it must be the identity. And finally by computing the action on the cohomology using the axiomatic definition of the steenrod algebra we see that it's the same action. EDIT: Recently I discovered the story is actually not terribly complicated. Let $CAlg_{\mathbb{F}_p}^{\mathbb{Z}/2}$ the category of graded commutative $\mathbb{F}_p$-algebras. Importing the definition from the ordinary case we we will define the additive formal "super" group as the formal group $\hat{\mathbb{G}}^{1|1}_{a}$ whose functor of points is: $$R \mapsto (Nil(R),+) \in Ab$$ Where $Nil(R)$ is the nilpotent radical of $R$. This is of course the functor of points of $Spec H^{\ast}(B \mathbb{Z}/p)$ by the above discussion. Similarly to the previous case we can define the functor of points of the automorphisms of $\hat{\mathbb{G}}^{1|1}_{a}$ as $$R \mapsto Aut(Nil(R))$$ At this point in the $p=2$ all we needed to do was restrict to the subgroup of automorphisms which preserve the tangent space. Here however the tangent space is graded and the requirement that the derivative preserves the grading on the tangent space turns out not give the full dual steenrod algebra. The correct condition in the odd $p$ case will be that the automorphism must preserve the 2-step filtration on the tangent space (as observed by André Henriques in the comments). This means that the even part of the tangent space will be preserved and the odd will be preserved modulo the even. We now describe this in coordinates: In coordinates a general automorphism $f$ (with no condition on the tangent space) of $(Nil(R),+)$ can be written in coordinates as $$(f_+(x,\theta), f_-(x,\theta)) : Nil(R)^{\text{even}}_x \oplus Nil(R)^{\text{odd}}_{\theta} \to Nil(R)^{\text{even}} \oplus Nil(R)^{\text{odd}} $$ Where $f_+(x,\theta) = \Sigma_{n \ge 0} a_n x^{p^n} + \alpha \theta, f_-(x,\theta) = \Sigma_{n \ge 0} b_n x^{p^n} + \beta \theta$. We claimed that the correct "strictness" condition is $gr(df) = Id$ where $gr$ is the associated graded w.r.t. the $\mathbb{Z}/2$-filtration. In coordinates this condition translates to: $$\frac{\partial f_+}{\partial x} = a_0 = 1, \frac{\partial f_+}{\partial \theta} = \alpha = 0, \frac{\partial f_-}{\partial \theta} = \beta = 1 $$ So the general form of a strict automorphism $f \in Aut^{\text{strict}}(\hat{\mathbb{G}}^{1|1}_{a})$ is $f_+(x, \theta) = x + \Sigma_{n \ge 1} a_n x^{p^n} , f_-(x,\theta) = \theta + \Sigma_{n \ge 0} b_n x^{p^n}$. Writing everything explicitly we see that the functor of points we defined is given by $$R \mapsto \{ (f_+(x,\theta),f_-(x,\theta)) \in R[ [x] ][\theta]^{\times 2} : f_+(x, \theta) = x + \Sigma_{n \ge 1} a_n x^{p^n} ,$$ $$ f_-(x,\theta) = \theta + \Sigma_{n \ge 0} b_n x^{p^n} \}$$ This is a group for composition. Abstractly this functor can be defined as the functor of strict automorphisms of the $\mathbb{Z}/2$-filtered formal additive super group (where strict now means the same thing as before but we are now only enforcing it on the associated graded): $$CAlg^{\mathbb{Z}/2} \to Fil_{\mathbb{Z}/2}(Ab), R \mapsto (Nil(R)^{\text{even}} \subset Nil(R),+)$$ The elements from the dual steenrod algebra define functionals on this group as follows $$\xi_n : f \mapsto a_n , \tau_n: f \mapsto b_n $$ We can now check that the composition of power series induces a comultiplication on the dual steenrod algebra coinciding with the one we know. Therefore this map extends to an isomorphism $\mathcal{A}_{\ast} \to \mathcal{O}(Aut^{\text{strict}}(\hat{\mathbb{G}}^{1|1}_{a}))$. This resolves point (1) above. We can now resolve (2) as well by noting that the map above induces a comodule structure on $H^{\ast}(B \mathbb{Z}/p)$ which when dualized corresponds to the usual action of the steenrod algebra on $H^{\ast}(B \mathbb{Z}/p)$ via cohomology operations.<|endoftext|> TITLE: Commutative algebra counterexample QUESTION [8 upvotes]: Let $M$ be an $R[x]$-module, such that $M$ is finitely generated as an $R$-module. Does there exist one such $M$, such that $M\otimes_{R[x]}R[x,x^{-1}]$ is not finitely generated as an $R$-module? REPLY [16 votes]: Let $R=\mathbb{Z}$ and let $M=\mathbb{Z}$ with $x$ acting by $2$. Then $M\otimes_{R[x]}R[x,x^{-1}]\cong \mathbb{Z}[1/2]$ is not finitely generated over $R$.<|endoftext|> TITLE: An equivariant map from sphere to a Lie group of lower dimension which is not null homotopic? QUESTION [15 upvotes]: Is there a natural number $n$, a compact Lie group $G$ of dimension less than $n$ and a continuous map $f:S^n \to G$ with $f(-x)=f(x)^{-1}$, such that $f$ is not a null homotopic map? This question was included in this MSE post but I did not receive any answer. REPLY [31 votes]: The Blakers-Massey element in $\pi_6(S^3)\cong\mathbb{Z}_{12}$ can be represented by such a map. This is done explicitly on page 3 of the paper https://arxiv.org/abs/math/0501091, published as Abresch, U.; Durán, C.; Püttmann, T.; Rigas, A., Wiedersehen metrics and exotic involutions of Euclidean spheres, J. Reine Angew. Math. 605, 1-21 (2007). ZBL1125.57017. Let $\mathbb{H}$ denote the quaternions, and represent the $6$-sphere as $$ S^6 = \{(p,w)\in \mathbb{H}\times\mathbb{H} \mid \mathfrak{Re}(p)=0\mbox{ and } |p|^2+|w|^2=1\}. $$ The map $b:S^6\to S^3\subseteq \mathbb{H}$ is given by $$ b(p,w) = \left\{\begin{array}{ll} \frac{w}{|w|} e^{\pi p} \frac{\overline w}{|w|}, & w\neq 0 \\ -1, & w=0, \end{array}\right. $$ where $e^{\pi p} = \cos(\pi |p|) + \sin(\pi|p|) \dfrac{p}{|p|}$ is the quaternionic exponential. The fact that $b(-p,-w)=\overline{b(p,w)}$ is easily checked (and is noted in the proof of Theorem 1 in the linked paper).<|endoftext|> TITLE: Pointwise convergence in functional calculus QUESTION [5 upvotes]: Let $A_n$ be a family of (bounded) self-adjoint operator converging pointwise to some (unbounded) self-adjoint operator $A,$ i.e. for all $x$ in the domain of $A$ $$\left\lVert A_n x-Ax \right\rVert \rightarrow 0.$$ Does this imply that $e^{it A_n}$ converges pointwise to $e^{itA}$? I know it holds, if the $A_n$ commute with each other as in this case $$\left\lVert T_n(t)x-T_m(t)x \right\rVert \le \int_0^t \left\lVert \frac{d}{ds} T_m(t-s)T_n(s) x \right\rVert \ ds \le t \left\lVert A_n x- A_m x \right\rVert$$ where $T_n =e^{itA_n}.$ Searching the literature myself I noticed that this might be related (especially the frist paragraph of the question): click me. REPLY [4 votes]: There are easyer and more direct ways to prove it, but this follows immediatelly as a special case from the Trotter-Kato approximation theorem, see Theorem III.4.8 in 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.<|endoftext|> TITLE: Obstruction Theory for Vector Bundles and Connections QUESTION [10 upvotes]: I know that the property that a vector bundle on a manifold is flat is equivalent to it being a pull-back from the fundamental groupoid. Namely, there is a map $X \to P_{\le 1} X$ from $X$ to its first Postnikov truncation, and giving a flat connection on a bundle is essentially the same as writing it as a pull-back along this map. Are there any known "higher equations" on a connection that guarantee that $V$ is a pull-back from a higher Postnikov truncation, say, that it is a pull-back from $P_{\le 2} X$? Intuitively that would mean that the vector bundle "depends only on the first 3 homotopies, including $\pi_0$", in some sense. Edit: As mentioned in John Klein's answer and comments, the condition of being flat is stronger than being a pullback from the 1-st Postnikov filtra, since it is actually equivlent to lifting the map $X \to BG$ through the classifying space of the discrete group $G^\delta$. Thus, one can at most ask for a $\textbf{sufficient}$ differential geometric condition on a connection to guarantee that the vector bundle factor through $P_{\le 2}X$, say. Thus, the remaining question is: Can we find such a sufficient condition on a connection, which is weaker than flatness? REPLY [5 votes]: Correction: the definition below is wrong. It isn't true that a 1-flat reduction is the same as a flat reduction. One also needs to require that the map $Z \to BG$ factors through $BG^\delta$, where $G^\delta$ is $G$ with the discrete topology. (Also, one may as well replace $Z$ by the $k$-th Postnikov section in the definition of a $k$-structure). The following is thoroughly useless general nonsense. Its main problem is that it lacks geometry. Even so, it kind of does the job and it is probably worth mentioning. Let $G$ be the structure group of your vector bundle. For simplicity, I will assume that $G$ is connected (for example, $G$ could be $U(n)$ or $SO(n)$). Let our vector bundle be classified by a map $f: X\to BG$. Definition: Let $k \ge 1$ be an integer. A $k$-structure is a pair $(Z,g)$ such that $\bullet$ $Z$ is a path connected space. $\bullet$ $Z$ has vanishing homotopy groups above dimension $k$, and $\bullet$ $g: Z\to BG\, $ is a map. [To such pairs $(Z_i,g_i)$ for $i=0,1$ and are equivalent there is a (finite chain of) weak homotopy equivalences over $BG$ from $Z_0$ to $Z_1$.] Example: A 1-structure amounts to a discrete group $H$ and a map $BH\to BG$. Definition: A $k$-flat reduction of $f$ consists of a $k$-structure $(Z,g)$ and a factorization of $f$ up to homotopy as $\require{AMScd}$ $$ \begin{CD} X \to Z @>g>> BG\, . \\ \end{CD} $$ Examples: (1). A 1-flat reduction of $f$ is the same thing as a flat reduction of the associated vector bundle. (2). An oriented line bundle over $X$ automatically admits a $2$-flat reduction (this is a tautology, since $BSO(2) = K(\Bbb Z,2)$. In particular, the canonical line bundle over $\Bbb CP^1$ admits a 2-flat reduction but not a 1-flat reduction (since it isn't flat). More generally, a finite Whitney sum of oriented line bundles admits a $2$-flat reduction, since $BSO(2)^{\times j} = K(\Bbb Z^j,2)$. (3). I don't know of examples in degrees $\ge 2$. However, if one is willing to work instead in the rational homotopy category, there are examples in higher degrees. Here's one: let $X= S^4$. This includes into $BS^3 = BSU(2)$. Rationally, this is a $K(\Bbb Q,4)$. Taking $G = SU(2)$ and $f: X\to BG$ to be the inclusion (which represents the quaternionic line bundle), we see that $f$ is rationally $4$-flat but not rationally $3$-flat.<|endoftext|> TITLE: Perfect numbers and perfect powers QUESTION [14 upvotes]: This was asked earlier at MSE. The observation that 28 = 27 + 1 shows that it is possible to have consecutive perfect numbers and perfect powers. However, this must be extremely rare. Is it unique? Questions: (1) Is there another example known of such a consecutive pair? (2) Is there a rigorous or heuristic proof showing that there are only finitely many such pairs? Thanks REPLY [2 votes]: Actually a stronger result (for cube powers) is proved here: Luis H. Gallardo, "On a remark of Makowski about perfect numbers" Elem. Math. 65 (2010) 121 – 126 The only even perfect number that is also a sum of two cubes is 28. The paper is easy (and free) to read. I believe you will enjoy it!<|endoftext|> TITLE: name for monoids inducing bimonoids in Rel? QUESTION [7 upvotes]: Let Rel be the category of sets and relations, which is a (compact closed) symmetric monoidal category under the cartesian product of sets. We write $A \nrightarrow B$ to indicate a relation from $A$ to $B$, and given a function $f : A \to B$, we write $f^+ : A \nrightarrow B$ and $f^- : B \nrightarrow A$ for the corresponding functional/anti-functional relations. Now, any ordinary monoid $(X,m,e)$ (i.e., monoid in Set, where $m : X \times X \to X$ is the multiplication and $e : 1 \to X$ is the unit) induces a pair of a "relational monoid" $(X,m^+,e^+)$ and a "relational comonoid" $(X,m^-,e^-)$, that is, an object $X$ which is both a monoid and a comonoid in Rel. I am interested in when $(X,m^+,e^+,m^-,e^-)$ forms a bimonoid in Rel. The condition that $(X,m^+,e^+,m^-,e^-)$ is a relational bimonoid reduces to the following conditions on the original set-theoretic monoid (writing $m(x,y)$ as $x \cdot y$): $x\cdot y = x'\cdot y'$ if and only if there exist $t,u,v,w$ such that $x = t\cdot u$ and $y = v\cdot w$ and $x' = t\cdot v$ and $y' = u\cdot w$. If $x\cdot y = e$ then $x = e$ and $y = e$. Question: Do monoids satisfying conditions (1) and (2) have a name? REPLY [3 votes]: I think describing all monoids satisfying these conditions is an essentially impossible task. But for a finite band (monoid where each element is idempotent) this condition is equivalent to being a distributive lattice with meet as the binary operation. As observed by Zeilberger in the comments $M$ must be commutative. If all its elements are idempotent, then it is a meet semilattice with top via the order $e\leq f$ iff $ef=e$. The product is the meet. By finiteness any set with a lower bound has a meet. Everything I'm about to write is true for meet semilattice with this property (ie completeness). Actually all I need is a top and binary joins. It was observed in the comments if $M$ is a distributive lattice with top, then it satisfies the above properties (the commenter used the dual convention of join as the operation). If $M$ is a meet semilattice which is either finite or complete then since it has a top it has joins. So it is a lattice. From now on just assume $M$ is a lattice with top. If it is not distributive it contains a sublattice isomorphic to $M_3$ or $N_5$. See https://en.m.wikipedia.org/wiki/Distributive_lattice If it contains $M_3$ then we can find distinct $x,y,x'$ in $M$ so that any two of these elements have the same join $m$ and same meet. From $xy=x'y$ we get $x=tu$, $x'=tv$ and $y=uw=vw$. Thus $t,u,v$ are each greater than the join $m$. This contradicts the given equalities. Similarly, if $M$ has a sublattice isomorphic to $N_5$ we can find $x,x',y$ with $x' TITLE: Coefficients of shifted Bernoulli polynomials QUESTION [8 upvotes]: I stumbled across the following curious empirical properties of the Bernoulli polynomials $B_n(x)$. Can anyone provide a reference or proof? Let $k\in\mathbb{Z}$, $k\geq 2$. Then (empirically): The coefficients of $x^{n-3}$, $x^{n-5}$, $x^{n-7}, \dots$ in the polynomial $\frac{1}{n}B_n(x+k)$ are integers. Define a sequence $a_2,a_3,\dots$ by $a_2=20$, and if $b_i=a_{i+1}-a_i$ then the sequence $b_2,b_3,\dots$ is periodic of period four, with terms $16,16,16,20,16,16,16,20,\dots$. Let $n\equiv j\,(\mathrm{mod}\,4)$, $0\leq j\leq 3$. Then the coefficient of $x^i$ in $B_n(x+k)$ is negative if and only if $n\geq a_k$ and $i=j,j+4,j+8,\dots,j+4\lfloor \frac{n-a_k}{4}\rfloor$. I have checked this for $n\leq 324$ and $2\leq k\leq 8$. REPLY [2 votes]: I think I see how to prove (2), or whatever version of it is true, but I haven't checked the details. Namely, use the identity $$ B_n(x+d) = \sum_{k=0}^n {n\choose k}B_k(d)x^{n-k}, $$ together with known results on the real zeros of Bernoulli polynomials such as https://ac.els-cdn.com/0021904589900166/1-s2.0-0021904589900166-main.pdf?_tid=147e6852-c89e-44e9-a805-69dd4bc54a5b&acdnat=1525395947_25f4432d47ebd7c61d3d59b2cff877a9.<|endoftext|> TITLE: Weak complicial sets: Are the morphisms too strict? QUESTION [13 upvotes]: In Verity's first paper on weak complicial sets, he shows that every strict complicial set is a weak complicial set. He also showed in an earlier paper that the full subcategory of stratified simplicial sets spanned by the strict complicial sets is equivalent to the category of globular strict $\omega$-categories. A troubling corollary of these two facts together is that upon passing to the homotopy category, since all strict complicial sets are both fibrant and cofibrant, that all pseudofunctors between strict $\omega$-categories are representable by strict $\omega$-functors. Is there a general coherence result for strict $n$-categories showing that indeed every pseudofunctor can be strictified up to a sufficiently weak notion of pseudonatural homotopy? (I am defining a pseudofunctor $A\to B$ of strict $n$-categories following Garner to be a strict $n$-functor $C(A)\to B$ where $C$ is the cofibrant replacement comonad arising from the Free-Forgetful adjunction to $n$-computads). If not, has anyone conjectured a class of monomorphisms of stratified simplicial sets at which we could localize in order to obtain the correct model category? REPLY [11 votes]: Indeed there is no such coherence result: it is false already for $2$-categories (see for instance Lemma 2 of this paper of Steve Lack). The solution to your troubling corollary is that the "correct" models for weak $\omega$-categories are not the weak complicial sets but the saturated weak complicial sets, i.e. those weak complicial sets in which all the equivalences are marked. In the Street nerve of a strict $\omega$-category only the identities are marked, and it is thus not saturated in general (unless the $\omega$-category has no non-identity equivalences, in which case all pseudofunctors into it are strict). The saturated weak complicial sets are indeed the fibrant objects of a model structure on the category of stratified simplicial sets, which is a localisation of the model structure whose fibrant objects are the weak complicial sets. See Emily Riehl's lecture notes Complicial sets, an overture, in particular Example 3.3.5.<|endoftext|> TITLE: higher order analogues of sylvester's law of inertia? QUESTION [5 upvotes]: Sylvester's law of inertia (here I quote wikipedia) If A is the symmetric matrix that defines the quadratic form, and S is any invertible matrix such that D = SAS^{T} is diagonal, then the number of negative entries in the diagonal of D is always the same, for all such S; and the same goes for the number of positive elements. Is there an analgoue of Sylvester's theorem for say cubic forms? My gut feeling is that this should be false for higher order tensors. So maybe the question here is what is the right analogue for higher degree forms? Have such questions been investigated in the literature. Apologies if this is too elementary, but I couldn't find anything related when I searched on the web. REPLY [8 votes]: A generalization of Sylvester's classification of canonical quadratic forms (which is the "law of inertia") to cubic forms has been presented in Canonical forms for symmetric tensors (1984). The matrices $A,D$ are now $2\times 2\times 2$ tensors, and the congruence relation is $D_{ijk}=\sum_{l,m,n}S_{li}S_{mj}S_{nk}A_{lmm}$. The canonical cubic forms are given on page 276 of the cited paper.<|endoftext|> TITLE: A family of polynomials whose zeros all lie on the unit circle QUESTION [35 upvotes]: I had posted the following problem on stack exchange before. Suppose $\lambda$ is a real number in $\left( 0,1\right)$, and let $n$ be a positive integer. Prove that all the roots of the polynomial $$ f\left ( x \right )=\sum_{k=0}^{n}\binom{n}{k}\lambda^{k\left ( n-k \right )}x^{k} $$ have modulus $1$. I do not seem to know how to do it, to show if the roots of the polynomial have modulus one. Putnam 2014 B4 Show that for each positive integer $n,$ all the roots of the polynomial $\sum_{k=0}^n 2^{k(n-k)}x^k$ are real numbers. This problem is very similar to the Putnam problem. REPLY [22 votes]: This follows from Julienne's answer. Fix $\lambda \in (0,1)$ and let $$ f_n(x) = \sum_{k=0}^n {n \choose k} \lambda^{k(n-k)}x ^k. $$ Each $f_n$ satisfies condition (a) for $\mu = 1$. As for (b), we can compute the derivative of $f_n$ to get $$f_n ' (x) = n \lambda^{n-1} f_{n-1} \left(\frac{x}{\lambda}\right).$$ Now use induction on $n$: for $n=1$ we have $f_1(x) = 1+x$, so $x=-1$ is its only root. If $f_{n-1}$ has all roots on the unit circle, then all roots $x_0$ of $f_n'$ satisfy $$n\lambda^{n-1}f_{n-1}\left( \frac{x_0}{\lambda}\right)=0$$ and so $x_0=\lambda z_0$ for some $z_0$ on the unit circle, by the induction hypothesis. Hence, as $\lambda \in (0,1)$, $x_0$ lies inside the unit circle. By induction on $n$, (b) holds for all $f_n$ and so the result follows from Cohn. (Sorry for not leaving this as a comment to Julienne's answer, I do not have enough reputation to comment.) Edit: corrected formula for $f_n'$ thanks to Ian Agol's correction.<|endoftext|> TITLE: Spaces without maximal homogeneous subspaces QUESTION [8 upvotes]: A homogeneous space $(X,\tau)$ is a topological space such that for all $x,y\in X$ there is a homeomorphism $\varphi:X\to X$ such that $\varphi(x)=y$. As a previous question implies, the union of an ascending chain of homogeneous spaces need not be homogeneous (example, see below). What is an example of a Hausdorff space $(X,\tau)$ that does not contain a homogeneous subspace that is maximal with respect to $\subseteq$? Note. The supremum (union) of an ascending chain of homogeneous spaces need not be homogeneous: Endow $\mathbb{N}$ with the discrete topology and consider the disjoint union $X = (\mathbb{N}\times\{0\}) \cup (\mathbb{Q}\times\{1\})$, where $\mathbb{Q}$ carries the Euclidean topology. Since $X$ is countable, it is the the union of an ascending chain of finite sets, all of which carry the discrete topology and are homogeneous, but $X$ is not. REPLY [2 votes]: A comment on Taras Banakh's beautiful answer (with his corrections). If I understand it correctly, the key ingredients of the construction are: 1) $X$ is a regular space, countable union of disjoint open sets $X_k$, for $k\in\omega$; 2) Any nonempty open subset of $X_{k+1}$ contains a copy of $ X_k$ as a subspace. As a consequence of 1) and 2), any open subset of $X$ that meets infinitely many $X_k$, contains a copy of $X$ itself as a subspace (it has room enough to include a topological sum of copies $X_k$, for all $k$); in fact even with its closure. 3) Each subset of $X_{k+1}$ , which is homeomorphic to a subset of $X_{k}$ is not dense in $X_{k+1}$ In this situation if $H\subset X$ is a homogeneous subspace of $X$, then $ X_k\setminus \overline H\neq\emptyset$ for all large $k $. But then $X$ also contains a shifted copy $H'$ of $H$, separated by neighbourhoods, so that it also contains their topological sum, which is a strictly larger homogeneous space.<|endoftext|> TITLE: Constructive proof of a rational version of Perron-Frobenius? QUESTION [16 upvotes]: In the following, we work with vectors and matrices whose entries are rational numbers. Inequalities between such vectors are understood to be coordinatewise: e.g., two vectors $a = \left(a_1,a_2,\ldots,a_n\right)^T$ and $b = \left(b_1,b_2,\ldots,b_n\right)^T$ are said to satisfy $a > b$ if and only if each $i \in \left\{1,2,\ldots,n\right\}$ satisfies $a_i > b_i$. Theorem 1. Let $n$ be a positive integer. Let $A$ be an $n\times n$-matrix whose entries are nonnegative rational numbers. Then, there exists a nonzero vector $v \in \mathbb{Q}^n$ with nonnegative coordinates such that either $Av \geq v$ or $Av < v$. The next theorem concerns a more restricted class of matrices. If $A$ is an $n\times n$-matrix whose entries are nonnegative rational numbers, then we define $D_A$ to be the directed graph whose vertices are the numbers $1,2,\ldots,n$, and which has an arc from vertex $i$ to vertex $j$ if and only if the $\left(i,j\right)$-th entry of $A$ is positive. We say that the matrix $A$ is irreducible if and only if this digraph $D_A$ is strongly connected. (Other equivalent definitions of irreducibility appear on the Wikipedia page for Perron-Frobenius.) Theorem 2. Let $n$ be a positive integer. Let $A$ be an irreducible $n\times n$-matrix whose entries are nonnegative rational numbers. Then, there exists a nonzero vector $v \in \mathbb{Q}^n$ with positive coordinates such that either $Av > v$ or $Av = v$ or $Av < v$. These two theorems can both be derived (with some nontrivial but not too hard work) from the Perron-Frobenius theorem. The latter constructs a Perron-Frobenius eigenvector of $A$ with real coordinates; one just needs to approximate its coordinates by rational numbers (unless it has eigenvalue $1$, in which case we have to make sure the approximation happens in the $1$-eigenspace). Question. How can Theorem 1 and Theorem 2 be proven in constructive logic? (Roughly speaking, this is probably tantamount to not using real numbers in the proof -- or building up the theory of real algebraics in constructive logic, which seems to have been done but I haven't seen it properly written up, and even then I don't know if Perron-Frobenius still applies and how you would prove it. Simon Henry's approximate version of Brouwer's fixed point theorem might help, but I don't know how it is proven.) I'm asking partly because of recent uses of Theorems 1 and 2 in combinatorics, but this question has been in the back of my mind for many years, ever after I solved a restricted version of the $n = 3$ case (IMO Shortlist 2003 problem A1) (official solution) on an exam. REPLY [4 votes]: Yes, both Theorem 1 and Theorem 2 have constructive proofs. In the following, I will work with rational numbers, but the same arguments work for any ordered field. (Note that in constructive logic, fields are automatically discrete -- i.e., any two elements of a field are either equal or not. Therefore, I don't think that $\mathbb{R}$ is a field for any reasonable constructive definition of $\mathbb{R}$. Perhaps $\mathbb{R}$ is some sort of "pro-field" in the same way as Laurent series over a field are; to be frank, I don't care enough about $\mathbb{R}$ to find out.) I set $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $. Nonnegative matrices My main tool will be the following result from linear optimization: Theorem 3. Let $n\in\mathbb{N}$, $n^{\prime}\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$ be an $m\times n$-matrix. Let $A^{\prime}$ be an $m\times n^{\prime}$-matrix. Then, exactly one of the following two assertions holds: Assertion L1: There exist two vectors $x\in\mathbb{Q}^{n}$ and $x^{\prime }\in\mathbb{Q}^{n^{\prime}}$ such that $x>0$, $x^{\prime}\geq0$ and $Ax+A^{\prime}x^{\prime}=0$. Assertion L2: There exists a vector $y\in\mathbb{Q}^{m}$ such that $y^{T}A\geq0$, $y^{T}A^{\prime}\geq0$ and $y^{T}A\neq0$. Theorem 3 is Theorem 2.5l in my Elementary derivations of some results of linear optimization (where I prove it for $\mathbb{R}$ instead of $\mathbb{Q}$, but the proof works over any ordered field). It seems to be due to Motzkin; it is actually a particular case of what is called "solvability of f)" on the EoM page for the Motzkin transposition theorem (note that the general case is easily seen to follow from this particular case). It is one of several results that are roughly equivalent to Farkas's lemma or the duality of linear programming. Its constructive proof (there are probably much better sources than my notes -- perhaps Motzkin's thesis?) is more or less based on Fourier-Motzkin elimination. Recall that if $a=\left( a_{1},a_{2},\ldots,a_{n}\right) ^{T}\in \mathbb{Q}^{n}$ and $b=\left( b_{1},b_{2},\ldots,b_{n}\right) ^{T} \in\mathbb{Q}^{n}$ are two vectors of the same size, then $a$ and $<$. With this out of our way, let's state a few simple lemmas: Lemma 4. Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A\in \mathbb{Q}^{n\times m}$ be a matrix whose entries are nonnegative. Let $v\in\mathbb{Q}^{m}$ be such that $v\geq0$. Then, $Av\geq0$. Proof of Lemma 4. Clear from the definition of $Av$. Lemma 5. Let $n$ be a positive integer. Let $A\in\mathbb{Q}^{n\times n}$ be a matrix whose entries are nonnegative. Let $y\in\mathbb{Q}^{n}$ and $z\in\mathbb{Q}^{n}$ be such that $y\geq0$, $z\geq0$, $Ay\geq y$ and $AzAz\geq0$ (by Lemma 4). Write $y$ as $y=\left( y_{1},y_{2},\ldots,y_{n}\right) ^{T}$, and write $z$ as $z=\left( z_{1},z_{2},\ldots,z_{n}\right) ^{T}$. The coordinates $z_{1},z_{2} ,\ldots,z_{n}$ of $z$ are positive (since $z>0$), while the coordinates $y_{1},y_{2},\ldots,y_{n}$ of $y$ are nonnegative (since $y\geq0$), and at least one of them is positive (since $y\neq0$). Hence, $\max\left\{ y_{i}/z_{i}\ \mid\ i\in\left\{ 1,2,\ldots,n\right\} \right\} $ is a well-defined positive rational number. Denote this number by $\mu$. Thus, for each $j\in\left\{ 1,2,\ldots,n\right\} $, we have $y_{j}/z_{j}\leq\mu$, so that \begin{equation} y_{j}\leq z_{j}\mu. \tag{1} \label{pf.l5.0} \end{equation} Now, write the $n\times n$-matrix $A$ in the form $A=\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq n}$. Then, $a_{i,j}\geq0$ for all $i$ and $j$ (since all entries of $A$ are nonnegative). We have $Ay\geq y$; in other words, \begin{equation} \sum_{j=1}^{n}a_{i,j}y_{j}\geq y_{i}\ \ \ \ \ \ \ \ \ \ \text{for each } i\in\left\{ 1,2,\ldots,n\right\} . \tag{2} \label{pf.l5.1} \end{equation} Also, $Az0$, $x^{\prime}\geq0$ and $I_{n}x+\left( A-I_{n}\right) x^{\prime}=0$. Case 2: There exists a vector $y\in\mathbb{Q}^{n}$ such that $y^{T} I_{n}\geq0$, $y^{T}\left( A-I_{n}\right) \geq0$ and $y^{T}I_{n}\neq0$. Let us first consider Case 1. In this case, there exist two vectors $x\in\mathbb{Q}^{n}$ and $x^{\prime}\in\mathbb{Q}^{n}$ such that $x>0$, $x^{\prime}\geq0$ and $I_{n}x+\left( A-I_{n}\right) x^{\prime}=0$. Consider these $x$ and $x^{\prime}$. The vector $x^{\prime}$ has nonnegative coordinates (since $x^{\prime}\geq0$). From $I_{n}x+\left( A-I_{n}\right) x^{\prime}=0$, we obtain $0=I_{n}x+\left( A-I_{n}\right) x^{\prime }=x+Ax^{\prime}-x^{\prime}$, so that $Ax^{\prime}-x^{\prime}=-x<0$ (since $x>0$). In other words, $Ax^{\prime}0$, $x^{\prime}\geq0$ and $I_{n}x+\left( A^{T}-I_{n}\right) x^{\prime}=0$. Subcase 2.2: There exists a vector $y\in\mathbb{Q}^{n}$ such that $y^{T}I_{n}\geq0$, $y^{T}\left( A^{T}-I_{n}\right) \geq0$ and $y^{T} I_{n}\neq0$. Let us first consider Subcase 2.1. In this subcase, there exist two vectors $x\in\mathbb{Q}^{n}$ and $x^{\prime}\in\mathbb{Q}^{n}$ such that $x>0$, $x^{\prime}\geq0$ and $I_{n}x+\left( A^{T}-I_{n}\right) x^{\prime}=0$. Consider these $x$ and $x^{\prime}$. From $I_{n}x+\left( A^{T}-I_{n}\right) x^{\prime}=0$, we obtain $0=I_{n}x+\left( A^{T}-I_{n}\right) x^{\prime }=x+A^{T}x^{\prime}-x^{\prime}$, so that $A^{T}x^{\prime}-x^{\prime}=-x<0$ (since $x>0$). In other words, $A^{T}x^{\prime}0$. Statement 3: There exists some $m\in\mathbb{N}$ such that all entries of the matrix $A^{0}+A^{1}+\cdots+A^{m}$ are positive. Statement 4: If $\left\{ U,V\right\} $ is a partition of $\left\{ 1,2,\ldots,n\right\} $ into two nonempty subsets $U$ and $V$, then there exist $u\in U$ and $v\in V$ satisfying $A_{u,v}>0$. Theorem 6 is well-known (it is the algebraic analogue of the equivalence of different definitions of "strong connectivity" for a directed graph), and the proof you will find in the literature (e.g., on Markov chains, I believe) is already constructive. All I will need is the implication from Statement 1 to Statement 3. We will also need the following simple lemma: Lemma 7. Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A\in \mathbb{Q}^{n\times m}$ be a matrix whose entries are positive. Let $v\in\mathbb{Q}^{m}$ be such that $v\geq0$ and $v\neq0$. Then, $Av>0$. Proof of Lemma 7. This is similar to Lemma 4, except that each coordinate of $Av$ is a sum of nonnegative addends containing at least one positive addend, and thus is positive. Proof of Theorem 2. The matrix $A$ is irreducible. Thus, by Theorem 6 (more precisely, by the implication $\left( \text{Statement 1}\right) \Longrightarrow\left( \text{Statement 3}\right) $), there exists some $m\in\mathbb{N}$ such that all entries of the matrix $A^{0}+A^{1}+\cdots +A^{m}$ are positive. Consider this $m$, and set $B=A^{0}+A^{1}+\cdots +A^{m}\in\mathbb{Q}^{n\times n}$. Then, all entries of the matrix $B$ are positive. Theorem 1 shows that there exists a nonzero vector $v\in\mathbb{Q}^{n}$ with nonnegative coordinates such that either $Av\geq v$ or $Avv$ or $Av=v$ or $Av0$. Thus, $v=Bw>0$. In other words, $v$ is a vector with positive coordinates. Hence, $v$ is nonzero (since $n>0$). We have $B=A^{0}+A^{1}+\cdots+A^{m}$. Thus, $AB=A^{1}+A^{2}+\cdots+A^{m+1} =BA$. Now, recall that either $Aw\geq w$ or $Aw0$. From $v=Bw$, we obtain \begin{equation} Av-v=\underbrace{AB}_{=BA}w-Bw=BAw-Bw=B\left( Aw-w\right) >0. \end{equation} In other words, $Av>v$. Thus, in Case 1, we have proven on our goal (namely, that either $Av>v$ or $Av=v$ or $Avv$ or $Av=v$ or $Av0$. Therefore, $w-Aw\geq0$ and $w-Aw\neq0$. Hence, Lemma 7 (applied to $n$, $B$ and $w-Aw$ instead of $m$, $A$ and $v$) yields $B\left( w-Aw\right) >0$. From $v=Bw$, we obtain \begin{equation} v-Av=Bw-\underbrace{AB}_{=BA}w=Bw-BAw=B\left( w-Aw\right) >0. \end{equation} In other words, $Avv$ or $Av=v$ or $Avv$ or $Av=v$ or $Av TITLE: Cup products in the Mayer-Vietoris sequence QUESTION [9 upvotes]: Let $(X;U,V)$ be an excisive triad and consider the corresponding part of the Mayer-Vietoris sequence $H^{\bullet-1}(U\cap V)\stackrel{\delta^*}{\to} H^\bullet(X)\to H^\bullet(U)\oplus H^\bullet(V)$. Now let $\alpha,\beta\in H^{\bullet-1}(U\cap V)$. Can we say anything about $\delta^*(\alpha\smile\beta)$? REPLY [6 votes]: Here are some comments. The map $\delta^\ast$ is a composition of the form $\require{AMScd}$ \begin{CD} H^{\ast-1}(U\cap V) @> \sigma >\cong > H^{\ast}(\Sigma (U\cap V)) \to H^\ast X \end{CD} where the first map is given by the suspension isomorphism and the second is induced by the map $X\to \Sigma(U\cap V)$ which cones off $U$ and $V$ in $X$ (here $\Sigma$ means suspension). This second displayed map is cup product preserving so it is sufficient to investigate the first map. Setting $Z = U\cap V$, represent your cohomology classes as maps $a: X \to K(\Bbb Z,k)$, $b: X \to K(\Bbb Z,\ell)$ (where we the Hopf classification theorem to identify $H^k(X)$ with $[X,K(\Bbb Z,k)]$, where $K(\Bbb Z,k)$ is the Eilenberg-Mac Lane space). Then $a\cup b$ is given by $\require{AMScd}$ \begin{CD} X @>\text{diag} >> X \wedge X @>a\wedge b>> K(\Bbb Z,k)\wedge K(\Bbb Z,\ell) @> m >> K(\Bbb Z,k+\ell)\, . \end{CD} where $m$ represents the evident cohomology class in degree $k+\ell$ of the smash product (using the Künneth formula). The class $\sigma(a\cup b)$ is given by \begin{CD} \Sigma X @>\Sigma \text{diag} >> \Sigma X \wedge X @>\Sigma a\wedge b>> \Sigma K(\Bbb Z,k))\wedge K(\Bbb Z,\ell) @>\Sigma m>> \Sigma K(\Bbb Z,k+\ell)\to K(\Bbb Z,k+\ell+1)\, , \end{CD} where the last map represents the generator of $H^{k+\ell+1}(\Sigma K(\Bbb Z,k+\ell))$.<|endoftext|> TITLE: Direct limit of Cantor sets QUESTION [5 upvotes]: Let $C$ be the Cantor set, and $\omega$ the discrete space of integers $\{0,1,2,...\}$. My conjecture: (1) For each $n<\omega$ let $f_n:C\to C$ be a continuous function (possibly not onto). Let $$X=(\omega\times C)/\sim.$$ This is the quotient of $\omega\times C$ under the relation $\sim$: $$\langle n,c\rangle\sim\langle m,d\rangle\iff(\exists l\geq n,m)(f_l\circ ...\circ f_n(c)=f_l\circ ...\circ f_m(d)).$$ Then $\dim(X)=0$. Is it true? I have not found it anywhere, despite doing a Google and looking in Dimension Theory texts. The space $X$ is sometimes called a direct limit https://en.wikipedia.org/wiki/Direct_limit. By $\dim(X)=0$, I mean $X$ has a basis of clopen sets (small inductive dimension is $0$). More generally, I suppose one could ask if the direct limit of zero-dimensional spaces has dimension zero. But letting each factor be equal to the Cantor set potentially makes things easier. What leads me to believe the statement is true, is two extreme cases of $f_n$'s. If each is constant, then $X$ is a singleton. And if each is the identity, $X$ is an increasing union of $\omega$-many Cantor sets. In this regard, the statement seems to be a generalization of the well-known "countable sum theorem" for (separable metric) spaces: (2) If a separable metric space $X$ can be represented as the union of a sequence $F_0 , F_1, ...$ of closed zero-dimensional subspaces, then X is zero-dimensional. I'm unsure if this is question is "research-level", so feel free to move to MSE if appropriate. EDIT: Just to be clear, I am asking for either a reference or a proof of (1). In the meantime I'll try to prove it by mimicing the proof of the (2), which can be found on page 20 (item 1.3.1) of this book: http://www.maths.ed.ac.uk/~v1ranick/papers/engelking.pdf REPLY [4 votes]: It's not true. Here's an easy way to get $[0,1]$ as a limit. Consider $C_0$ as the set of all sequences of $0,1$: each such sequence defines a binary expansion of some element in $[0,1]$. Let $C_n$ be obtained from $C$ by identifying, for each dyadic number of the form $k/2^n\in \mathopen] 0,1\mathclose[$ (possibly not reduced), its two dyadic expansions. Then $C_n$ is still homeomorphic to the Cantor space, and the inductive limit is naturally homeomorphic to $[0,1]$.<|endoftext|> TITLE: Is there a simple argument that shows that two unitary fusion categories are Morita equivalent if their Drinfeld centers are equal? QUESTION [8 upvotes]: By Morita equivalent I mean that there is an invertible bi-module between the two fusion categories. [Feel free to replace the Drinfeld centers being "equal" by an appropriate categorial notion of "equivalent"; for me (braided) fusion categories are just F-tensors (and R-tensors), in which case it's really "equal" (up to a gauge on the fusion vector spaces)] I'm asking because there is a very direct physical translation of that statement: "Two gapped 2+1-dimensional non-chiral topologically ordered systems are in the same phase if their anyon content is the same." By "in the same phase" here I mean equivalence under a short-range local unitary circuit, or in other words an invertible domain wall. The statement seems so accepted in the physics community that hardly anyone talks about it. Still I never heard a simple argument why it should be so. For the Turaev-Viro-Levin-Wen fixedpoint models, the physical systems are constructed from fusion categories, and invertible domain walls can be constructed from invertible bi-modules. REPLY [11 votes]: In the non-unitary setting ENO proved that if $Z(C)$ and $Z(D)$ are equivalent as braided tensor categories, then C and D are Morita equivalent. This is Theorem 3.1 of this paper. Note that they say the result was already known to Kitaev and Müger. This is also an iff, though the other direction is easier. The analogous result also holds in the unitary setting. That is if $Z(C)$ and $Z(D)$ are equivalent as unitary braided tensor categories then C and D are unitarily Morita equivalent. (Unitary Morita equivalence means that the actions on the bimodule are compatible with the star structure.) I don't know whether this has appeared in the literature, but the proof is completely analogous just replacing "algebra" with "$C^*$-algebra" (or Q-System if you prefer) everywhere. Again this is an iff. You do need to be careful that you put in the word "unitary" everywhere. For example, it's not at all clear to me whether two unitary fusion categories being algebraically Morita equivalent is enough to say that they're unitarily Morita equivalent. (Aside: I'm not sure at all what you mean by "really are equal" since they're really not equal, you still may need to gauge.)<|endoftext|> TITLE: Do combinatorial model categories and Quillen adjunctions model presentable $\infty$-categories? QUESTION [16 upvotes]: Let $Q$ be the homotopical category of combinatorial model categories and left Quillen functors, with left Quillen equivalences for weak equivalences. Let $\mathbf Q$ be the corresponding $\infty$-category (in whatever foundations you prefer). Let $Pres^L$ be the $\infty$-category of presentable $\infty$-categories and left adjoint functors. The usual functor from relative categories to $\infty$-categories (modeled however you prefer) descends to a functor $N: \mathbf Q \to Pres^L$. Variation A: I'm happy to work with simplicial combinatorial model categories rather than ordinary ones. Variation B: It would also be interesting to know the answer to the following questions for combinatorial model $\infty$-categories and left $\infty$-Quillen functors. This is probably easier because one has more flexibility in this setting. Question 1: Is $N: \mathbf Q \to Pres^L$ an equivalence of $\infty$-categories? This functor is known to be essentially surjective -- for simplicial model categories this is in HTT, I think. Question 2: Can the homotopical category $Q$ be refined to a model category? There are size issues here; I'm happy with any way of handling them. I might suspect that something like Dugger's universal homotopy theories provide cofibrant resolutions. More specifically, I'd like to know: Question 3: Are there model categories $C, D$ such that every left adjoint functor $NC \to ND$ is modeled by a left Quillen functor? REPLY [5 votes]: As pointed out in the answer to Localizing $\mathrm{CombModCat}$ at the Quillen equivalences, the answer to Question 1 is affirmative and is provided by the paper Combinatorial model categories are equivalent to presentable quasicategories.<|endoftext|> TITLE: Torsion in the integral cohomology of $BPU_{n}$ QUESTION [8 upvotes]: I would like to prove that the integral cohomology of $BPU_{n}$ the classifying space of the projective unitary group of order $n$ has $n-$primary torsion. We have a fiber sequence of the form $BSU_{n}\rightarrow BPU_{n} \rightarrow K(\mathbb{Z}/n,2)$. Then we can consider the Serre spectral sequence associated to this fibration. Let's suppose that the smallest prime dividing $n$ is a very large prime $p$. For general reasons, we know that the integral cohomology of $H^{i}(K(\mathbb{Z}/n,2))$ is $\mathbb{Z}$, 0, 0, $\mathbb{Z}/n$, 0, $\mathbb{Z}/n$, 0, $\mathbb{Z}/n$, ... where it repeats this way in a range $2p+\delta$. We also know that $H^{*}(BSU_{n})\cong \mathbb{Z}[c_{2},\dots,c_{n}]$ where deg$(c_{i})=2i$. Therefore, third page of the spectral sequences will have all its columns (but the zeroth column) conformed by zeros and cyclic groups of order $n$: How can I answer my question using this setup? Thanks. REPLY [10 votes]: You may want to have a look at this paper: X. Gu. On the cohomology of classifying spaces of projective unitary groups. arXiv:1612.00506, (link to arXiv) The spectral sequence involving $BSU_n$ appears in the discussion in Section 5. The paper contains a lot of spectral sequence calculations, computations of differentials and determines cohomology (including ring structure) up to degree 10. Results like "the only torsion is $n$-torsion" and "there exists some nontrivial $n$-torsion" follow from these computations. Further references to other computations can also be found in the paper.<|endoftext|> TITLE: Examples of normal subgroups of a surface group with no embedded elements QUESTION [6 upvotes]: As a consequence of the loop theorem, if $F$ is a closed surface in the boundary of a 3-manifold and if the kernel $N = \ker(\pi_1(F) \to \pi_1(M))$ is nonempty then there is a nontrivial element of $N$ that can be represented by an embedded curve. I read in Hempel's book that there are lots of normal subgroups of $\pi_1(F)$ that contain no nontrivial elements that can be represented by embedded curves. How can I cook up such subgroups and prove that they have this property? REPLY [8 votes]: One source of interesting examples is the lower central series. In my paper J. Malestein, A. Putman, On the self-intersections of curves deep in the lower central series of a surface group, Geom. Dedicata 149 (2010), no. 1, 73–84. my coauthor and I show that the minimal number of self-intersections among nontrivial elements of the kth term of the lower central series of a surface group goes to infinity as k goes to infinity.<|endoftext|> TITLE: Riccati-type recurrence: infinitely many sign changes? QUESTION [9 upvotes]: Suppose $b_1, b_2, b_3, \dots \in \Bbb{R}$ satisfy the Riccati-type recurrence $$b_{k+1}=\frac{1+kb_k}{k-b_k},\quad k\ge 1.$$ Is it true that such a sequence reaches infinitely many positive as well as negative values? It appears to be so. REPLY [8 votes]: Yes, choose $a_1$ such that $\frac{1}{b_1}=\tan(a_1)$, and let $a_{k+1}=a_k-\arctan\frac{1}{k}$, then we have $\frac{1}{b_k}=\tan(a_k)$, since $\sum_k \arctan\frac{1}{k}=\infty$, the conjecture follows.<|endoftext|> TITLE: Congruences Ramanujan-style QUESTION [17 upvotes]: Let $t\in\Bbb{N}$ and consider the sequences $p_t(n)$ defined by $$\sum_{n\geq0}p_t(n)x^n=\prod_{i\geq1}\frac1{(1-x^i)^t}=(x;x)_{\infty}^{-t}.$$ The numbers $p_t(n)$ can be regarded as enumerating partitions of $n$ into parts that come with $t$ colors. Furthermore, $p_t(n)=\sum_{\lambda\vdash n}\prod_{j\geq1}\binom{k_j+t-1}{t-1}$ where $\lambda=1^{k_1}2^{k_2}\cdots$ and each $k_j\geq0$. Note also that $p_1(n)=p(n)$ is the usual number of (unrestricted) integer partitions of $n$. Ramanujan's famous congruences state $$\begin{cases} p(5n+4)\equiv0\mod 5, \\ p(7n+5)\equiv0\mod 7, \\ p(11n+6)\equiv0\mod 11. \end{cases}$$ In the same spirit, the following appear to be true. Are they? $$\begin{cases} p_t(5n+4)\equiv0\mod 5, \qquad t\equiv0,1,2,4\mod 5 \\ \,p_t(7n+5)\equiv0\mod 7, \qquad \,\,t\equiv0,1,4 \,\,\,\, \mod 7\\ p_t(11n+6)\equiv0\mod 11, \qquad t\equiv0,1,10\mod 11. \end{cases}$$ REPLY [14 votes]: More general versions of this have been established: see in particular Theorem 2 of Kiming and Olsson, and for other work see (for example) Locus and Wagner. To answer the question fully, as Ofir Gorodetsky observed the problem is trivial when $t$ is $0$ or $1$ modulo the prime modulus. The other cases correspond to $t \equiv \ell-1$ or $t\equiv \ell-3$ modulo $\ell$ (with $\ell$ being $5$, $7$, or $11$). These cases are the "non-exceptional congruences" covered by Theorem 2 of the paper by Kiming and Olsson (and the argument there is essentially elementary following easily from Euler's pentagonal number theorem in the case of $\ell-1$, and an identity of Jacobi in the case of $\ell-3$). The real interest is in the situation of exceptional congruences, and Theorem 4 of Kiming and Olsson gives some examples of these. For instance, if $t\equiv 3 \mod 11$ then $$ p_{t}(11n+7) \equiv 0 \pmod{11}. $$<|endoftext|> TITLE: Intersection of $\pi_1$-injective surfaces QUESTION [5 upvotes]: Let $M$ be a closed non-Haken hyperbolic $3$-manifold. Are there two, nonhomotopic, $\pi_1$-injective closed surfaces in $M$, and which do not intersect? REPLY [6 votes]: No, this is impossible. Assume that the surfaces $\Sigma_1, \Sigma_2$ are immersed and disjoint realized by immersions $f_i:\Sigma_i\to M$. Take a region $N$ in the complement of these surfaces, and whose boundary intersects both surfaces (such a region must exist if the manifold is connected, which is implicit in your question). Then this submanifold has non-trivial $H_2$, and hence contains an embedded $\pi_1$-injective orientable surface which is homologically non-trivial in $N$ and separating the two immersed surfaces. Any sphere component must bound a ball since hyperbolic manifolds are irreducible, and this ball cannot contain either immersed $\pi_1$-injective surface, hence it will bound a ball in $N$. Hence there must be a component $\Sigma$ of genus at least one which is incompressible in $N$ and still separates the two surfaces. Assume that $\Sigma$ has minimal genus with respect to these properties. Let $D$ be a compressing disk for $\Sigma$, say on the side containing $\Sigma_1$. Then we may homotope $\Sigma_1$ off of $D$. There is an immersion $f_1:\Sigma_1\to M$ such that $f_1$ is transverse to $D$. Then the preimage $f_1^{-1}(D)$ will be curves in $\Sigma_1$ which are homtopically trivial in $D$. Since $f_1$ is $\pi_1$-injective in $M$, these also bound embedded disks in $\Sigma_1$. Take an innermost disk and surger $\Sigma$ along this disk to remove the component of $f_1^{-1}(D)$, which may be accomplished by a homotopy by irreducibility. One may repeat this until $f_1^{-1}(D)=\emptyset$. Then we may compress $\Sigma$ along $D$ to obtain a surface $\Sigma'$ which still separates $f_1(\Sigma_1)$ from $f_2(\Sigma_2)$, a contradiction to the minimality assumption of $\Sigma$. Thus, we may find an incompressible surface in $M$, contradicting the assumption that it was non-Haken. These arguments are fairly standard in classical 3-manifold topology going back at least to Waldhausen, and probably earlier.<|endoftext|> TITLE: Point-wise limit of finite valued functions QUESTION [6 upvotes]: Let $X$ be a second countable topological vector space. Does there exist any sequence of finite valued functions $f_n\colon X\to X$ converging point-wise to the identity mapping on $X$? REPLY [7 votes]: To answer a question raised in comments: if $X$ is merely separable then this can fail. Let $X = \mathbb{R}^\mathbb{R}$ with the product topology, which is a Hausdorff topological vector space. Then $X$ is separable. (Any product of continuum many separable spaces is separable. Alternatively, viewing $X$ as the space of all functions from $\mathbb{R}$ to $\mathbb{R}$, one can show that the set of polynomials with rational coefficients is dense.) Now let $f_n : X \to X$ be a sequence of finite-range functions converging pointwise to some $f$. Let $A = \bigcup_n f_n(X)$ be the union of the images of the $f_n$, so that $A$ is countable. Then the number of sequences of elements of $A$ has cardinality at most $(\aleph_0)^{\aleph_0} = \mathfrak{c}$, so that the set $B$ of possible limits of sequences from $A$ likewise has cardinality at most $\mathfrak{c}$. But the image of $f$ is contained in $B$. Since $X$ has cardinality $\mathfrak{c}^\mathfrak{c} > \mathfrak{c}$, the function $f$ cannot be surjective.<|endoftext|> TITLE: How to compute Dedekind eta function efficiently? QUESTION [8 upvotes]: According to wiki: https://en.wikipedia.org/wiki/Dedekind_eta_function, Dedekind eta function is defined in many equivalent forms. But none of them is an explicit description (say in algorithmic format) on how to computing it. Where to find such one? Thanks! REPLY [4 votes]: Euler's formula $$ \sum\limits_{n \in \mathbb{Z}} {( - 1)^n q^{\frac{{(3n^2 - n)}} {2}} } = \prod\limits_{n = 1}^\infty {(1 - q^n ),} $$ (which can be proven from Jacobi’s triple product identity by using the fact that $\prod\limits_{n = 1}^\infty {(1 - q^{3n} )(1 - q^{3n - 2} )} (1 - q^{3n - 1} ) = \prod\limits_{n = 1}^\infty {(1 - q^n )} $) provides a good way of numerically computing $$ \eta (\tau )=e^{\frac {\pi {\rm {{i}\tau }}}{12}}\prod _{n=1}^{\infty }(1-e^{2n\pi {\rm {{i}\tau }}})=q^{\frac {1}{24}}\prod _{n=1}^{\infty }(1-q^{n}). $$ I hope this answers your question.<|endoftext|> TITLE: Cycles in the hyperoctahedral group (symmetries of the hypercube) QUESTION [6 upvotes]: Let $B_n$ be the hyperoctahedral group (the isometries of the $n$-dimensional hypercube). Let $k TITLE: Bertini's theorem over non-algebraically closed field QUESTION [7 upvotes]: Let $K$ be a non-algebraically closed (infinite) field of characteristic $0$ and $X$ a smooth, projective $K$-variety. Does there exist an ample invertible sheaf $\mathcal{L}$ on $X$ such that a general element of the linear system $|\mathcal{L}|$ is a smooth $K$-variety? If not true in general, is there any condition on $X$ under which this holds true? REPLY [11 votes]: This is true both over finite and infinite fields. For infinite fields, see [Jou, Cor. I.6.11(2)]. It works for a general section of any very ample line bundle $\mathscr L$, using that over an infinite field a nonempty open subset of $|\mathscr L| \cong \mathbb P^N$ contains a rational point. For finite fields, see [Poo, Thm. 1.1]. It requires taking a high power of your given very ample line bundle $\mathscr L$. References. [Jou] Jouanolou, Jean-Pierre, Théorèmes de Bertini et applications. Progress in Mathematics, 42. Birkhäuser, Boston-Basel-Stuttgart (1983). ZBL0519.14002. [Poo] Poonen, Bjorn, Bertini theorems over finite fields. Ann. Math. (2) 160.3 (2005), p. 1099-1127. ZBL1084.14026.<|endoftext|> TITLE: Summing moments and Riemann zeta values QUESTION [9 upvotes]: Let $d\mu_n(x)=\cos^{2n}x\,dx$ and consider the averages of moments $$\alpha_n=\frac{\int_0^{\pi/2}x^4d\mu_n(x)}{\int_0^{\pi/2}d\mu_n(x)}.$$ Then, I have encountered a curious evaluation $$\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^2(\alpha_{n-1}-\alpha_n) =\frac{21}8\zeta(4)-\frac32\zeta(2);$$ where $\zeta(s)$ is the Riemann zet function. Is this true? REPLY [11 votes]: After my first failed attempt I now follow the route suggested by Nemo --- which works smoothly. Starting from Nemo's identities $$F(b)\equiv\int_0^{\pi/2}\cos^{2n}x\cos bx\,dx=\frac{\pi (2n)!}{2^{2n+1}\Gamma(n+1+b/2)\Gamma(n+1-b/2)}$$ $$\int_0^{\pi/2}x^4\cos^{2n}x\,dx=\lim_{b\rightarrow 0}\frac{d^4}{db^4}F(b),$$ I arrive at $$\alpha_n=\tfrac{3}{4}\left[ \psi ^{(1)}(n+1)\right]^2-\tfrac{1}{8}\psi ^{(3)}(n+1),$$ where $\psi^{(m)}(x)$ is the Polygamma function $$\psi^{(m)}(x)=\frac{d^m}{dx^m}\left(\frac{1}{\Gamma(x)}\frac{d}{dx}\Gamma(x)\right),$$ $$\psi^{(m)}(n)=(-1)^{m+1}m!\sum_{k=n}^\infty\frac{1}{k^{m+1}},\;\;m\geq 1,\;\;n\in\mathbb{N}.$$ (The last equation gives the connection to Harmonic numbers mentioned by Nemo.) The recurrence relation $$\psi^{(m)}(x+1)=\psi^{(m)}(x)+\frac{(-1)^m m!}{x^{m+1}}$$ implies that $$\alpha_{n-1}-\alpha_n=\tfrac{3}{2} n^{-2} \psi ^{(1)}(n)-\tfrac{3}{2}n^{-4}.$$ From here we arrive at the result in the OP, $$\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^2(\alpha_{n-1}-\alpha_n)=\frac{3}{2}\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac{1}{k^2(n+1)^2}=\frac{21}8\zeta(4)-\frac32\zeta(2).$$ This simple final expression suggests a generalization to higher powers of $x$, but that does not seem to work. If I replace $x^4\cos^{2n}x$ in the definition of $\alpha_n$ by say $x^6\cos^{2n}x$ or $x^8\cos^{2n}x$, I find that $\alpha_{n-1}-\alpha_n$ contains products of Polygamma functions, which I have not been able to sum up in closed form.<|endoftext|> TITLE: Is there a position in infinite Go for which the life of a particular stone has transfinite game value? QUESTION [32 upvotes]: As follow up to Checkmate in $\omega$ moves?, we can ask the same question about go. Is there a position on a $\mathbb Z \times \mathbb Z$ goban such that either black can kill a white group, but white can stave off the capture for $n$ moves, or white can make a group live, but black can stave off life of the group for $n$ moves (for any $n \in \mathbb N$). This is considerably trickier than chess, since go is more of a local game. My first thought is to use ladders somehow. The other thought is that the superko rule, since it actual has global influence. (Although it does not matter too much, using Tromp-Taylor rules is probably best, since they are the simplest.) REPLY [37 votes]: This is a really great question! Previous attempts to make sense of infinite Go have sometimes had problems because it wasn't clear how to define the winner of a game of Go after transfinite play. The problem was that perhaps a black group was surrounded by a white group, which was surrounded by a black group, and so on forever in such a way that the asymptotic density did not converge. In this case it wouldn't necessarily be clear who had more area. (Another separate issue is to define the state of the board at time $\omega$, in situations where stones had been placed and then killed unboundedly often; it seems most natural to keep only stones that had persisted from some time on.) Your formulation of the problem gets around this difficulty by making the winner of the game depend on the life or death of a single designated stone. This makes the game an open game and therefore subject to the theory of transfinite open game values. The answer is yes. There is a position with a black stone that is part of a group that white can definitely kill in finitely many moves, but black can make this take as long as desired. Consider the following position. There are infinitely many stones, with black to play, and one should imagine that the pattern continues to the right. The designated black stone is part of the main infinite horizontal group of black stones, which continues to the right, with little black knobs poking down on the underside. This group is nearly surrounded and has only three liberties (at top) and no eyes. Because of the knobs, the surrounding white stones form separate groups, although they are nearly connected. For each knob, black has the opportunity to make a hopeless cut, attempting to capture the two isolated white stones touching the end of the knob. For example, black could play at $A$, cutting the two white stones off on one side. This move is hopeless by itself, of course, because white can simply connect the two isolated stones on the other side at $B$, and then aim to kill the main black group shortly thereafter. So the hopeless initial cut is only valuable for black, if white should ignore it. There are similar such hopeless cutting moves at each knob. A key point to observe, however, is that white cannot afford to follow the policy of always answering the hopeless cuts, since this would lead to infinite play, which is what black wants. So eventually, white will indeed have to ignore one of the hopeless cuts, in order to make progess on surrounding the main black group. When white ignores the hopeless cut, then black will immediately play the other cut, with Atari on the two white stones. White cannot allow that black will connect the main black group to the live black group, and so white must extend. This leads to a ladder, which is ultimately broken by the lower isolated white stones. The main thing to observe is that these ladder-breaking stones are further and further away as you consider successive knobs. Thus, black can choose to act on knob $n$, aiming to have a ladder of size at least $n$. After the ladder is played out, then the white stones in the ladder will gain sufficient liberties for white to kill the main black group shortly thereafter. So ultimately, the ladder is also hopeless for black, since white will eventually kill the main group; but the point is that black can prolong play as long as she likes, by choosing to act on a knob whose ladder is that long, even if she does lose after that. So the main line play is as follows. Black makes a hopeless cut on the $n^{th}$ knob for some particular $n$, chosen as large as black likes. White cannot afford to always answer these hopeless cuts, because this would lead to infinite play. And so white will eventually choose to ignore the hopeless cut, playing instead so as to reduce the main black group to only two liberties. At this point, black gets to make the other cut, which leads to a ladder of $n$ forcing moves with black playing Atari on the white group at each step in the ladder. Ultimately, however, the ladder was hopeless, because it is broken by the isolated white stones. When this happens, the white stones in the ladder will get three liberties, which is not enough, since at this point white will make Atari on the main black group and then capture it, as black has no response. Note that after the ladder is broken, further hopeless cuts on other knobs are now truly hopeless, since white can answer them by capturing the main group. In summary, white will eventually kill the main black group in finitely many moves, but black can choose any $n$ and play so as to postpone her loss of that group for at least $n$ steps. So the life-and-death of that group has game value $\omega$. I believe that one could hope to modify the position to achieve much larger game values in Go. One could also modify the example by making the main black group much larger, so that it has asymptotic density 3/4, say, which would find tranfinite game values in the asymptotic density area calculation formulation of the game. Update. Let me explain how to modify the position to achieve game value $\omega\cdot n$ for any finite $n$. What we do is give the main black group $n+2$ liberties, instead of only 3, and then also make the ladder-blocking white stone a live group. In addition, we can add a diagonal of white stones parallel to the ladder, just to ensure that black has no advantage to deviating from the main line. With these revisions, black will be able to make $n$ hopeless cuts in succession, each leading to a ladder of length $n$ before the next begins, giving the game value $\omega\cdot n$ overall. Black may make $n$ announcements in all, with each being the length of play before the next announcement, until he loses his group at the end. I'll try to post an image later. Next challenge: $\omega^2$.<|endoftext|> TITLE: Looking for holomorphic function on a sector with specified boundary behavior QUESTION [7 upvotes]: Fix $h \in (0,\pi/2)$. I am trying to explicitly exhibit a holomorphic function $f\colon \Sigma \to \mathbb{C}$, where $\Sigma$ is the punctured sector $$\Sigma := \{z \in \mathbb{C} \:|\: z\neq 0, 0\leq\arg z\leq h\},$$ and where we require $f$ to satisfy the following conditions on $\partial\Sigma$: $f(x) \in \mathbb{R}$ for $x \in \mathbb{R}_+$. $|f(e^{ih}x)|^2 = x^2 + 1$ for $x \in \mathbb{R}_+$. The only thing I tried so far is to start with $f(z) := z + 1$, which satisfies all the conditions except for having $|f(e^{ih}x)|^2 = x^2 + 2\cos(h) x + 1$, and to tweak that -- but I didn't get anywhere. I'd be happy for any tips, e.g. a proof that such an $f$ exists abstractly or doesn't exist at all. Best of all would be "$f$ exists, and here's a formula: ... ." I did think twice about this being "research level", but, well, it is part of my research! I posted the same question at https://math.stackexchange.com/questions/2766653/looking-for-holomorphic-function-on-a-sector-with-specified-boundary-behavior , but didn't get any answers. REPLY [7 votes]: Such a function can be written with the help of the Poisson integral. First, your function must extend by reflection to $\{ z:|\arg z|\leq h\}$. Let $g(z)=f((-iz)^{2h/\pi})$, then $g$ must be analytic in the upper half-plane, and $$g(t)=\sqrt{1+|t|^{4h/\pi}}.\quad\quad\quad\quad\quad (1)$$ To construct $g$, first define a harmonic function $u$ using the Poisson integral: $$u(x+iy)=\frac{y}{\pi} \int_{-\infty}^\infty\frac{\log g(t)dt}{(t-x)^2+y^2},$$ where $g$ in the RHS is defined by (1), and let $v$ be a conjugate harmonic function, so that $g(z)=\exp(u+iv)$ is analytic in the upper half-plane and has the desired value of $|g|=u$ on the real line. It remains to check that $g$ is real on the positive imaginary axis. This follows because $u(x+iy)=u(-x+iy),$ so we can choose $v$ so that $v$ is $0$ on the positive imaginary axis, therefore $g$ is positive on the positive imaginary axis. This construction gives you $g$, and $f(z)=g(iz^{\pi/2h}).$ EDIT. The conjugate function can be also written as a formula. Combining it with Poisson's formula one can write $$\log g(z)=\frac{1}{\pi i}\int_{-\infty}^\infty\frac{1+z^2}{1+t^2}\frac{\log g(t)dt}{t-z}+\frac{z}{\pi i}\int_{-\infty}^\infty\frac{\log g(t)dt}{1+t^2}.$$ See, for example de Branges, Hilbert spaces of entire functions, Thm. 2 on p. 3. EDIT 2. Your function is not unique: one can multiply $g$ on any bounded function analytic in the upper half-plane, whose absolute value is $1$ on the real line. Such functions can be completely described: they are of the form $$e^{iaz}\prod_{-\infty}^\infty\frac{z-a_k}{z-\overline{a_k}},\quad\quad\Im a_k>0,\quad a\in{\mathbf{R}}, $$ where $(a_k)$ is any sequence in the upper half-plane which satisfies the Blaschke condition (convergence of the product), and which is mapped by $z\mapsto -\overline{z}$ onto itself. Then the construction gives all such functions satisfying a mild growth restriction at $\infty$.<|endoftext|> TITLE: Do a Hausdorff space and its associated completely regular space have the same Borel subsets? QUESTION [5 upvotes]: Let $(X,T)$ be a Hausdorff topological space. Let $C_b(X)$ be its algebra of continuous bounded functions. Let $T'$ be the initial topology on $X$ given by $C_b(X)$. It is known that $T=T'$ if and only if $(X,T)$ is completely regular. If $\mathcal B(X,T)$ and $\mathcal B(X,T')$ are the Borel $\sigma$-algebras generated by the two topologies, we have $\mathcal B(X,T') \subseteq \mathcal B(X,T)$ because $T' \subseteq T$. Is it true that $\mathcal B(X,T') = \mathcal B(X,T)$? (Motivation: I am trying to pull a Borel regular measure from $(X,T')$ back to $(X,T)$. I could use Henry's extension theorem or follow Bourbaki and extend an additive set-function of compact subsets, but my secret hope is that in this very convenient setting the two $\sigma$-algebras coincide, so I don't need to resort to heavy artillery (in particular, Henry's extension theorem requires Zorn's lemma).) REPLY [8 votes]: There are Hausdorff spaces all of whose real-valued functions are constant. A classical example (of a countable space) was given by Uryshon (Über die Mächtigkeit der zusammenhängenden Mengen, Math.Annalen, 1925). This implies that the initial topology on $X$ induced by $C(X)$ is trivial, and so is its Borel sigma-algebra. REPLY [5 votes]: The Borel $\sigma$-algebras may be unequal. Let $\omega_1$ be the least uncountable ordinal. Let $L_0=[0,\omega_1)\times I/\sim$ be the "long line", where $[0,\omega_1)$ is the space of ordinals less than $\omega_1$ with the order topology, $I=[0,1]$ is the unit interval, and the identifications are: $(\alpha,1)\sim(\alpha+1,0)$. Let $L'=L_0\cup\{*\}$ be the one point compactification of $L_0$. Such $L'$ is a connected, linearly ordered compact space. Note that the Stone-Cech compactification of $L_0$ is again $L'$. The subspace $\omega_1=[0,\omega_1)\times\{0\}\subseteq L'$ is not Borel. Also for every closed subset $A\subseteq L_0$ that is not bounded above, every continuous function $f:A\to\mathbb{R}$ is eventually constant. Let $L$ be a refinement of $L'$: a subset $U$ is open in $L$ if it is of the form $U=V\cup (W\setminus\omega_1)$ where $V,W$ are open in $L'$. Here $\omega_1$ is a closed hence a Borel subset of $L$. But if $f:L\to\mathbb{R}$ is continuous then it has to be eventually constant then it is continuous as a function on $L'$. Thus $C_b(L')=C_b(L)$, but $\mathcal{B}(L')\subsetneq\mathcal{B}(L)$.<|endoftext|> TITLE: The anti-symmetrization of a kind of polynomials in $\mathbb{Z}[x_1,x_2,\ldots,x_n]$ QUESTION [7 upvotes]: Let $n$ be a positive integer and $S_n$ be the symmetric group on $\{1,2,\ldots,n\}$. Let $\mathcal{A}_n$ be the anti-symmetrization operator on $\mathbb{Z}[x_1,x_2,\ldots,x_n]$ such that for any $f(x_1,x_2,\ldots,x_n)\in \mathbb{Z}[x_1,x_2,\ldots,x_n]$, $$\mathcal{A}_n(f)=\sum_{w\in S_n}\varepsilon(w)f(x_{w(1)},x_{w(2)},\ldots,x_{w(n)}),$$ where $\varepsilon(w)=(-1)^{l(w)}$ is the sign of $w$. My question: For any integer $n>1$ with $n(n-1)\equiv 0 \pmod 4$, is it true that we can always choose $\displaystyle \frac{n(n-1)}{4}$ different polynomials $$x_{i_k}+x_{j_k},\ 1\leq k\leq \frac{n(n-1)}{4}$$ from the $\displaystyle \frac{n(n-1)}{2}$ polynomials $$x_i+x_j,\ 1\leq i TITLE: index of smooth varieties QUESTION [7 upvotes]: What are the simplest examples of smooth, projective varieties defined over the fraction field of an Henselian DVR of characteristic $0$ which have index $>1$? EDIT: Also assume that the residue field of the DVR is algebraically closed. REPLY [4 votes]: Let $d\geq 3$ be any integer that is relatively prime to the residue characteristic of $R$. Let $s$ be any generator of the maximal ideal $\mathfrak{m}$ of $R$. Let $\widetilde{R}$ be the finite, flat extension $R[\sigma]/\langle \sigma^d -s \rangle$ with its natural action of the group scheme $\mu_d=\text{Spec} \ \mathbb{Z}[\zeta]/\langle \zeta^d-1\rangle$, $$\text{Spec}\ \mathbb{Z}[\zeta]\langle \zeta^d-1 \rangle \times \text{Spec}\ R[\sigma]/\langle \sigma^d-s\rangle \to \text{Spec}\ R[\sigma]/\langle \sigma^d - s\rangle, \ \ \sigma \mapsto \zeta\otimes \sigma.$$ Denote the induced morphism of schemes by $$q:\text{Spec}\ \widetilde{R} \to \text{Spec}\ R.$$ The generic fiber of $q$ is a $\mu_d$-torsor. Let $X_R$ be the smooth, projective $R$-scheme, $$X_R = \text{Proj}\ R[t_0,t_1,\dots,t_{d-1}]/\langle t_0^d + t_1^d + \dots + t_{d-1}^d \rangle.$$ This has a natural action of $\mu_d$ by $$\zeta\bullet [t_0,t_1,\dots,t_{d-1}] = [\zeta^0t_0,\zeta^1 t_1, \dots, \zeta^{d-1}t_{d-1}].$$ Denote by $X_{\widetilde{R}}$ the base change, $$X_{\widetilde{R}} = \text{Spec}\ \widetilde{R} \times_{\text{Spec}\ R} X_R.$$ This has a diagonal action of $\mu_d$ that is a free action. The projection to the first factor is equivariant for this action, $$\text{pr}_1:X_{\widetilde{R}} \to \text{Spec}\ \widetilde{R}.$$ Thus, for the geometric quotient by this free $\mu_d$-action, $$q_X:X_{\widetilde{R}} \to \mathcal{X}_R^s,$$ there is a unique morphism, $$\pi:\widetilde{X}_R\to \text{Spec}\ R,$$ such that $\pi\circ q_X$ equals $q\circ \text{pr}_1$. Since the base change of $\pi$ by the fppf morphism $q$ equals the smooth morphism $\text{pr}_1$, also the proper morphism $\pi$ is smooth. Since the geometric generic fiber of $\text{pr}_1$ is a smooth hypersurface of dimension $d-2$, it is integral. Thus, the geometric generic fiber of $\pi$ is also integral. If $d$ is a prime, then the index of the generic fiber of $\pi$ equals $d$. Indeed, every zero-dimensional, reduced, closed subscheme of the generic fiber pulls back to a zero-dimensional, reduced, closed subscheme of the generic fiber of $\text{pr}_1$ that is $\mu_d$-invariant. Thus, for every irreducible component of this closed subscheme, the closure in $X_{\widetilde{R}}$ is an integral, closed subscheme that is $\mu_d$-invariant and finite over $\text{Spec}\ \widetilde{R}$. By the valuative criterion of properness for $\text{pr}_1$, the intersection of this closed subscheme with the closed fiber of $\text{pr}_1$ is a closed subscheme that is $\mu_d$-invariant. Since the action of $\mu_d$ on the closed fiber is free, the length is divisible by $d$. Since this holds for every irreducible component, the total length is divisible by $d$.<|endoftext|> TITLE: Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$? QUESTION [14 upvotes]: The q-Vandermonde identity reads: $$ \binom{m + n}{k}_{\!\!q} =\sum_{j} \binom{m}{k - j}_{\!\!q} \binom{n}{j}_{\!\!q} q^{j(m-k+j)} $$ The q-binomial coefficients: $$ \binom{ a }{ b}_{\!\!q} $$ are known to count the number of points over the Grassmannians $Gr(a,b)$ over $F_q$, and $q^{j(m-k+j)}$ is the number of points over $A^{j(m-k+j)}$ over $F_q$. So the identity above states that number of points of $Gr(k, m+n)$ can be written as the sum of numbers of points of smaller Grassmannians and $A^l$. That would be naturally explained if there would be some geometric relation between these manifolds, like fiber bundles. Question 1 What could be the geometric interpretation of the identity above, if any? If yes, what about other fields, e.g. $C$? Question 2 Is the identity above true on the level of motives? Question 2b If yes, then is it true for arbitrary fields? For the case of q=1 - the hypothetical "field with one element", everything is "Okay": the $Gr(a,b,F_1)$ are just the sets of all a-combinations out of b and so the usual Vandermonde identity implies that $Gr(m+n,k)$ can be factorized in terms of smaller Grassmannians. That is, however, a kind of cheating, since over $F_1$, geometry disappears and counting is enough to give a "geometric" identity. REPLY [20 votes]: Assume $V$ is a vector space of dimension $m+n$, $M \subset V$ is a subspace of dimension $m$, and $N = V/M$. Let $p:V \to N$ be the projection. Consider the Grassmannian $X = Gr(k,V)$ and its stratification by the dimension of intersection with $M$, i.e., set $$ X_j = \{ U \in Gr(k,V) | \dim(U \cap M) = k - j \}. $$ If $U \in X_k$ then the projection of $U$ to $N$ has dimension $j$. Thus, we have a natural map $$ \pi \colon X_j \to Gr(k-j,M) \times Gr(j,N),\qquad U \mapsto (U \cap M,p(U)). $$ Finally, for any $U_M \in Gr(k-j,M)$ and $U_N \in Gr(j,N)$ we have $$ \pi^{-1}(U_M,U_N) \cong \{U' \in Gr(j,p^{-1}(U_N)/U_M) | U' \cap (U/U_M) = 0\}. $$ This is an open Schubert cell, hence is isomorphic to $\mathbb{A}^{j(m-k-j)}$. Moreover, the map $\pi$ is locally trivial, hence $$ [X_j] = [Gr(k-j,m)][Gr(j,n)][\mathbb{A}^{j(m-k-j)}] $$ in the Grothendieck ring of varieties. Summing up over $j$, we obtain $$ [Gr(k,m+n)] = \sum_j [Gr(k-j,m)][Gr(j,n)][\mathbb{A}^{j(m-k-j)}], $$ a motivic version of the formula..<|endoftext|> TITLE: A topos where every object is internally projective but not every object is projective QUESTION [6 upvotes]: An object $P$ in a topos $\mathcal{E}$ is said to be projective if $Hom_{\mathcal{E}}(p,-)$ preserves epis, internally projective if $(-)^P$ preserves epis. Can anyone give an example of a topos where every object is internally projective but not every object is projective? Can such a topos be a presheaf topos? Generally what properties for $\mathbf{C}$ such that $\mathbf{Sets}^{\mathbf{C}}$ is such a presheaf topos? REPLY [10 votes]: See Johnstone's Sketches of an Elephant, section D.4.5 for the following and more examples. If $G$ is a group, then the topos of $G$-sets satisfies the internal axiom of choice -- every object is internally projective. But the only projective objects are the free $G$-sets. So if $G$ is nontrivial, then this topos does not have every object projective -- it does not satisfy the external axiom of choice.<|endoftext|> TITLE: The Tits alternative for $\operatorname{Out}(F_n)$ QUESTION [9 upvotes]: Not sure if this is the right place to ask this, but the paper I am reading seems to be too specialised for mathstack (if you do not agree, pleas let me know and I will take down this question) I am reading the paper 'The Tits alternative for $\operatorname{Out}(F_n)$ I: dynamics of exponentially growing automorphisms' by Bestvina, Feighn and Handel. In it, the following lemma occurs: There is a homomorphism $PF_{\Lambda^+}: \operatorname{Stab}(\Lambda^+) \to \mathbb{Z}^k$ such that $\psi \in \ker PF_{\Lambda^+}$ if and only if $\Lambda^+ \notin \mathcal{L}(\psi)$ and $\Lambda^+ \notin \mathcal{L}(\psi^{-1})$ ($\psi \in \operatorname{Out}(F_n)$). This lemma can be found on page 546 of the paper. The proof is the following: and proposition 3.3.3. is question I do not understand why each $\mu(\Psi)$ other than $1$ occurs as the Perron-Frobenius eigenvalue of some irreducible matrix and why the set of Perron-Frobenius eigenvalues is discrete. Any chance someone knows this paper/has read this paper or could point me to some extra lecture which could clarify this proof for me? My reasoning My confusion came from the following reasoning, which dr. Bestvina (through mail) told me was false (I can not figure out why): I think I could show from the definition of $\mu$ that $\mu(\operatorname{Id}) = 1$. Let $\psi \in \operatorname{Stab}(\Lambda^+)$ be an outer automorphism with $\mu(\psi) > 1$, then $\mu(\psi)$ is indeed a Perron-Frobenius eigenvalue (proposition 3.3.3.(4)). However, this would imply that $\mu(\psi^{-1})$ is not, by proposition 3.3.3.(2): $mu(\psi^{-1}) < 1$ and Perron-Frobenius eigenvalues $\lambda$ of non-negative, integer irreducible matrices satisfy $\lambda \geq 1$. I can not see where my reasoning is flawed, nor can I see why then every $\mu$ has to be a Perron-Frobenius eigenvalue REPLY [5 votes]: I think the first sentence in the proof of 3.3.1, with "$...PF_\Lambda^*(\psi^{-1}) = -PF_\Lambda^*(\psi)...$" was meant to tell you to just think about when that is positive because, as you mention, $PF^*_\Lambda(\psi)$ could not have come from a Perron-Frobenius matrix if it was negative, but up to taking inverse it will, unless $\mu(\psi)=1$. This also makes sense since the theorem is about whether or not $\Lambda^+ \in \mathcal{L}(\phi^\pm)$, and proposition 3.3.3 implies that $\mu(\psi^{-1})<1$ then $\Lambda^+ \in \mathcal{L}(\psi).$ As for discreteness of the eigenvalues, this takes a bit of an argument and it is in [BH1]=Train tracks and Automorphisms of Free Groups, and is in the proof of theorem 1.7, along with the comment about the more general setting on page 37. The outline of the proof that irreducible integer Perron-Frobenius matrices, of bounded dimensions will have discrete set of Perron-Frobenius eigenvalues goes like this: Bound the eigenvalue $\lambda$, of $M=(m_{ij})$, from below by the minimum row sum of the matrix. Show that there is a uniform $k$ so that $M^k$ will have row sums larger than the largest entry of the original matrix $M$. I would use the graph theoretic definition of irreducible(which is in the paper) and there is a uniform bound on the dimensions to get the "mixing of the largest term" The above shows $\lambda^k \geq $ min row sum of $M^k$ $\geq \max{m_{ij}}$, which bounds the eigenvalues with the terms of a matrix, so only finitely many eigenvalues below any $K$, so the set of the eigenvalues is discrete.<|endoftext|> TITLE: Can operads (or category theoretic structures more generally) be compared? QUESTION [5 upvotes]: I was reading John Baez’s paper on operads and phylogenetics trees where he formalizes a Jukes–Cantor model of phylogenetics. Because biological questions receive different answers depending on the model used, I was wondering if the operad that represents that model could be compared to an operad representing a different model. i.e., with some distance metric or structural similarity, or perhaps by choosing an appropriate topology on them? So, for example, we might instead adopt a model with more parameters like the Kimura one to better treat the patterns in our data. And it would be interesting to quantify, conceiving of the first model as a predecessor and the second as a recently discovered successor, what was retained in the successor model. I've been hoping to measure structural continuity across theory change (e.g., the preservation of Newtonian mechanics in a limit of GR, and the same for classical mechanics to quantum which can be measured by deformation theory). But I'm having trouble doing this with biological models, specifically phylogenetic ones formalized in a category theoretic way. I've talked to Baez about this a little on twitter, so let me know if attaching that would help answer this. REPLY [5 votes]: Operads are the one-object version of a multicategory. This perspective is expanded in the excellent book by Leinster "Higher operads, higher categories" If you want to be not-so-profound but get the idea, a multicategory is like a category, but instead of having a single domain, morphisms have a tuple of objects for domain; so the generic morphism is a pair $(n,f) : \vec X\to Y$ where $\vec X = (X_1,\dots,X_n)$ is a tuple of objects of $\cal C$. Morphisms of this kind can be composed grafting the "root" of the tree-like shape of this morphism You have identities, and this botanical composition is associative. Now try to unravel the definition of what such a structure is, when $|\cal C|=*$! From this, you define a "morphism of operads" as a functor between these gadgets.<|endoftext|> TITLE: Bichromatic pencils QUESTION [6 upvotes]: A pencil is a collection of some lines through a point, called the center of the pencil. If the points of the plane are colored, then call a pencil bichromatic if there is a color that is present on all the lines of the pencil such that this color is different from the color of the center of the pencil. Given any non-monochromatic coloring of the plane with finitely many colors, and $m$ directions, $\alpha_1,\ldots, \alpha_m$, is it true that there is a point $p$ and an angle $\varphi$ such that the pencil determined by the lines of direction $\alpha_1+ \varphi,\ldots, \alpha_m+ \varphi$ through $p$ is bichromatic? This is related to polymath16, see why here. I can only prove the statement for $m=2$. REPLY [5 votes]: Observe that we are done if there is a monochromatic circle $C$, say blue: If a point inside the circle is not blue, then any pencil with that centre is bichromatic, so we may assume that the whole disk bounded by $C$ is blue. If a point outside is close enough and not blue, then we can find a bichromatic pencil with that centre. With this, we can extend blue to the whole plane or find a bichromatic pencil. We now find a monochromatic circle: Let the angles determined by cyclically consecutive lines be $\beta_1,\dots,\beta_{m-1}$, and $S=\{b_1,\dots,b_m\}$ a set of points on a circle $C$ such that for any $x\in C$ the angle of the line through $x$ and $b_1$ and through $x$ and $b_{i+1}$ is $\beta_i$ for any $1\leq i \leq m-1$. If every $b_i$ is blue, and $x\in C$ is not blue, then there is a bichromatic pencil centered at $x$. So it would be enough to find a monochromatic set similar to $S$. But this exists by Gallai's theorem.<|endoftext|> TITLE: New articles by Errett Bishop on constructive type theory? QUESTION [19 upvotes]: Recently two formerly unknown articles by Errett Bishop (1928-1983) were posted online by Martín Escardó. One is entitled "A general language", deals with constructive type theory, and is 28 pages long. The other is entitled "How to compile mathematics into Algol", deals with implementation of his constructive type theory, and is 29 pages long. Both pdfs can be found here. Numerous intriguing questions arise in connection with these articles, such as the following: Question 1. What is the precise date of composition of the articles? Question 2. Why weren't the articles published? Question 3. What led Bishop to develop a constructive type theory, given what is often considered to be his lack of interest in formal logic and formalisation of constructive mathematics? (see Note 1 below). Question 4. Should our perceived views of Bishop's foundational stance be reconsidered in light of these manuscripts? Question 5. Since the manuscript "A general language" and Bishop's 1970 published article "Mathematics as a numerical language" are very different, could somebody summarize the similarities and differences between the two? Note 1. For example, one finds the following comment by a selfdeclared constructivist: the constructive mathematician dismisses classical mathematics as an exercise in formal logic (see Richman, Fred. Interview with a constructive mathematician. Modern Logic 6 (1996), no. 3, 247-271). This is not an indication of great esteem for formal logic on the part of this particular constructivist. Note 2. It must be said that Bishop himself was not averse to making statements of this sort: It is not surprising that some of Brouwer's precepts were then formalized, giving rise to so-called intuitionistic number theory, and that the formal system so obtained turned out not to be of any constructive value. In fairness to Brouwer it should be said that he did not associate himself with these efforts to formalize reality; it is the fault of the logicians that many mathematicians who think they know something of the constructive point of view have in mind a dinky formal system or, just as bad, confuse constructivism with recursive function theory. (page 4 in "A constructivist manifesto", chapter 1 in Bishop, Errett. Foundations of constructive analysis. McGraw-Hill Book Co., New York-Toronto, Ont.-London 1967) REPLY [2 votes]: Regarding Question (2), I have heard the following story from (non-constructivist) people who knew E. Bishop well: Bishop had submitted (some of) these papers for publication, but they were not accepted in their then form. The reasons (as pointed out by either the editor or referees) was that other people had been working on similar topics with better/similar results. Bishop then set about improving his own papers/results, trying to formulate (what we now call) explicit type constructors everywhere. Unfortunately, Bishop died before he could finish this endeavour.<|endoftext|> TITLE: Suspensions are H-cogroup objects QUESTION [5 upvotes]: Do you know any reference where you have a formal justification for the following statement that appears in nLab? https://ncatlab.org/nlab/show/suspensions+are+H-cogroup+objects "Let $\mathcal{C}$ be an $(∞,1)$-category with finite $(∞,1)$-colimits and with a zero object. Write $\Sigma \colon X \mapsto 0 \underset{X}{\coprod} 0$ for the reduced suspension functor. Then the pinch map $$ \Sigma X \simeq 0 \underset{X}{\sqcup} 0 \simeq 0 \underset{X}{\sqcup} X \underset{X}{\sqcup} 0 \longrightarrow 0 \underset{X}{\sqcup} 0 \underset{X}{\sqcup} 0 \simeq \Sigma X \coprod \Sigma X $$ exhibits a cogroup structure on the image of $\Sigma X$ in the homotopy category of an $(∞,1)$-category homotopy category $Ho(\mathcal{C})$". REPLY [7 votes]: Ok, let me try to give you a proof of something that is a lot stronger than what you asked for, but which hopefully is a bit more natural. I am basically going to smother the problem under the abstract nonsense, so I recommend you look at special cases of this proof, e.g. when $X=S^n$ in pointed spaces, in order to get intuition. Recall that an associative monoid in an ∞-category $D$ with all finite products is a functor $$M:\Delta^{op}\to D$$ such that for all $[n]\in \Delta^{op}$ the map $$M([n])\xrightarrow{\prod M(e_i)} \prod_{1\le i\le n} M([1])$$ is an equivalence, where $e_i:[1]\to [n]$ is the map picking out the edge $\{i-1,i\}$ (this is known as the Segal condition). Hence, an associative comonoid in an ∞-category with all finite coproducts $C$ is a functor $$M:\Delta\to C$$ such that $$\coprod_{1\le i\le n}M([1])\xrightarrow{\coprod M(e_i)} M([n])$$ is an equivalence. Moreover, by postcomposing with the map $C\to hC$, we see that an associative comonoid in $C$ induces an associative comonoid in $hC$. What I will prove is that, if $C$ is a pointed ∞-category with finite colimits, $\Sigma X$ is an associative comonoid for all $X$ (the same proof, maybe more recognizably, will prove that $\Omega X$ is a group object for all $X$). Let us consider the category $\Delta_+$ of all finite (possibly empty!) totally ordered sets. This contains two full subcategories we are going to use: $\Delta$ (the category of finite nonempty totally ordered sets) and $\Delta^1$ (the full subcategory spanned by $[0]$ and by the empty set $\varnothing$). We can construct a functor $M_0:\Delta^1\to C$ sending $[0]$ to $0$ and $\varnothing$ to $X$. Let $M_+$ be the left Kan extension of $M_0$ to $\Delta_+$. Lemma The restriction of $M_+$ to $\Delta$ is an associative comonoid $M$ such that $M([1])\cong \Sigma X$. Proof. Unwrapping the standard formula for left Kan extensions we obtain $$ M([n])= \mathrm{colim}_{i\in \Delta^1_{[n]/}} M_0(1) = 0\amalg_X 0\amalg_X \cdots \amalg_X 0$$ where in the formula there are $n+1$ 0s. In particular $M([1])=\Sigma X$. For the sake of clarity of notation let me prove the Segal condition only for $n=2$. Then we need to prove the map $$(0\amalg_X 0)\amalg (0\amalg_X 0) \to 0\amalg_X 0 \amalg_X 0$$ induced by the inclusions of the corresponding summands is an equivalence. But this is simply the associativity of the pushout. $\square$ Finally let me say a few words about how to prove $\Sigma X$ has the structure of a cogroup and not simply a comonoid. If we let $$m:M([1])\to M([2])\cong M([1])\amalg M(1)$$ be the map induced by the edge $\{0,2\}$ (the "comultiplication"), the condition of being a cogroup is equivalent as asking the map $$M([1])\amalg M([1])\xrightarrow{m\amalg i_1} M([1])\amalg M([1])$$ is an equivalence. To check this is the case for the $M$ I constructed above is left as an exercise for the reader.<|endoftext|> TITLE: Comparing two $\sigma$-algebras on $B(\ell^1)$ QUESTION [9 upvotes]: Let us consider $B(\ell^1)$, bounded linear operators on $\ell^1$. We recall the weak operator topology, denoted by $w$, on $B(\ell^1)$ is determined as follow $$w-\lim T_i=T \Longleftrightarrow \lim\langle (T_i-T)x,f\rangle=0$$ for every $x\in \ell^1 , f\in\ell^{\infty}$. Let $M$ be the Borel sigma algebra coming from the weak operator topology on $B(\ell^1)$ and let $M_0$ be the $\sigma$-algebra generated by the $w$-basic neighborhoods in $B(\ell^1)$. Q. Do we have $M=M_0$? Let us give some information concerning this problem which may be helpful. It is proved by Ju Myung Kim that the weak operator topology on the unit ball of $B(X)$ is relatively second countable if $X^*$ is separable. By an interesting argument in enter link description here, Matthew Daws showed the converse is also valid. So we have that: $X^*$ is separable if and only if the weak operator topology on the unit ball of $B(X)$ is relatively second countable. It is clear that, if $X^*$ is separable then $M_0=M$. I feel the converse is not true and $B(\ell^1)$ may serve a counterexample, however I have no clear proof. With taking account to the comments mentioned in this problem, another approach of this question is: Is $(B(\ell^1),w)$ hereditary lindelof? REPLY [2 votes]: Combination of the following two facts implies that $(B(\ell^1),w)$ is hereditary Lindelof. 1- $(B(\ell^1),sot)$ is hereditary Lindelof. 2- (Kuratowski and Sierpinski [1921]) A regular space $X$ is hereditary Lindelof if and only if every uncountable set $A\subseteq X$ has a condensation point which is contained in $A$<|endoftext|> TITLE: Example of a connected finite group scheme which is not solvable QUESTION [7 upvotes]: What would be an example of a connected finite group scheme over a field $k$ that is not solvable? Here $k$ is algebraically closed. Let $\operatorname{GL}_n$ be the general linear group scheme over a field of characteristic $p>0$. Now consider the $H=\operatorname{ker}{F^r:\operatorname{GL}_n \to \operatorname{GL}_n}$, where $F^r$ is the frobenius map which raises each element of the matrix $\operatorname{GL}_n(R)$ to $p^r$. In this case, is $H$ connected and not solvable? REPLY [5 votes]: The connected finite kernel $H$ is not solvable. $\def\eps{\varepsilon} \def\m{\mathfrak{m}}$Suppose by contradiction that $H$ is solvable and $m$-th derived group of $H$ is trivial (where $[H,H]$ is the first derived group). It implies that for any $k$-algebra $R$ the $m$-th derived group of the abstract group $H(R)$ is also trivial(because for any $G$ we have $[G(R), G(R)]\subset D(G)(R)$). Put $l=2^{m+1}$ and consider $R=k[\eps_1,\dots,\eps_l]/(\eps_1,\dots,\eps_l)^{(p)}$ where $(p)$ means taking $p$-th powers of all the elements of the ideal. Put also $\m=(\eps_1,\dots,\eps_l)$ to be the maximal ideal. I claim that $D(D(...D(H(R)))\neq 1$ ($D$ iterated $m$ times). Indeed, $H(R)=1+\m\mathrm{Mat}_n(R)$. Now note that for any two matrices $A, B\in \mathrm{Mat}_n(R)$ we have (sums represented by dots are actually finite) $$[1+\eps_i A,1+\eps_j B]=(1+\eps_i A)(1+\eps_j B)(1-\eps_i A+\eps^{2}_iA^2-\dots)(1-\eps_j B+\eps_j^2 B^2-\dots)=1+\eps_i\eps_j(AB-BA) \,\,(\rm{mod}\,\, \m^3)$$ Now pick a tuple $A_1,\dots A_{2^{m+1}}$ of elements of the Lie algebra $\mathfrak{gl}_n$ over $k$ such that $(m+1)$ times nested commutator of them is $B\neq 0$. The $(m+1)$ times nested commutator of the elements $1+\eps_iA_i\in H(R)$ then equals $1+\eps_1\dots\eps_{2^{m+1}}B$ modulo $\m^{2^{m+1}+1}$ and thus is not equal to $1$.<|endoftext|> TITLE: Second fundamental form blows up at minimal hypersurface singularity QUESTION [5 upvotes]: I have seen the claim that "$A$ bounded on an area minimizing current implies no singular set" in a couple of papers by Lohkamp, but with no reference (see https://arxiv.org/abs/1805.02180 e.g.). Precisely, consider $T$ an i.m.r. $(n-1)$-current in a Riemannian $n$-manifold $M$. Suppose $T$ is locally mass minimizing. If, on the regular set of $T$, the second fundamental form $A$ is uniformly bounded up to the singular set, then the singular set is empty. Alternatively, if $x_j\in \mathrm{reg}(T)$ and $x_j\to x\in \mathrm{sing}(T)$, then $|A(x_j)|\to\infty$. Lohkamp says this a folklore theorem and gives no proof. Is there a reference for this out there or is the proof really simple? REPLY [2 votes]: Well I know that you can rule out higher multiplicity planes; that argument is in Federer somewhere. So we just have to show that every tangent cone is a plane. Here's a sketch that comes to mind: If the curvature is bounded by say $C$ near a singular point $x_0$, then consider taking a tangent cone at $x_0$. You would rescale a ball $B_{\epsilon_j}(x_0)$ to size $1$. Then the curvature on the regular part of the rescaled current would be $C\epsilon_j$. So when you take the tangent cone, the regular part would converge smoothly to flat pieces. The closure of the regular part of the cone has to be the whole cone. But flat pieces can't meet or intersect in any way other than all be part of the same plane.<|endoftext|> TITLE: lower bound for absolute value of a hypergeometric function QUESTION [7 upvotes]: I am working with a certain Gauss hypergeometric function, $_{2}F_{1}(a,a-b;2a;1-z)$, where $a, b \in {\mathbb R}$ with $a, a-b > 0$ and $00$ and $c \geq \max( a,b)$. For example, if $0 \leq z \leq 1$, then this is true since all the coefficients in the expansion of $_{2}F_{1}(a,b;c;z)$ are positive. But I have been unable to prove the more general result for all $z$ with $|z| \leq 1$ myself or find a proof in the literature. Would anyone have any ideas on this, or references I may have missed? REPLY [4 votes]: As I said, the problem is to show that the absolute value of $$ z\mapsto\int_0^\infty t^\alpha(t+1)^{-\beta}(t+z)^{-\gamma}\frac {dt}t $$ attains its minimum in $\Omega=\{z:|z|\le 1, z\notin[-1,0]\}$ at $z=1$. Note that we are free to change the contour of integration to any curve that stays in the open angle bounded by the rays $\{-\tau w:\tau>0\}$ and $\{-\tau:\tau>0\}$ containing the positive semi-axis. So, I'll immediately change it to the ray $\{\tau\sqrt w:\tau>0\}$, which reduces the problem to the following. Let $z,w\in\mathbb C, \Re z,\Re w>0, zw\in\mathbb R_+$. Then $$ \left|\int_0^\infty t^\alpha(t+w)^{-\beta}(t+z)^{-\gamma}\frac {dt}t\right|\ge \left|\int_0^\infty t^\alpha(t+|w|)^{-\beta}(t+|z|)^{-\gamma}\frac {dt}t\right|\,. $$ Note that this inequality is immediate if $\beta=\gamma$ and $z=\bar w$ (which corresponds to $c=2a$, $|z|=1$ in the original formulation) because the integrand on the left is then positive and obviously greater than the one on the right. However, in general the integrand on the left oscillates so we cannot take advantage of its being large in absolute value immediately. Still it suggests the following idea: change the contour (we are now free to choose any curve $\Gamma$ connecting $0$ and $\infty$ in the right half-plane) so that the argument of the product $F(t)=t^\alpha(t+w)^{-\beta}(t+z)^{-\gamma}$ stays fixed (say, $\theta$) on $\Gamma$. If, in addition, we knew that $\Gamma$ crossed each circle centered at the origin just once, we would be able to write $$ \left|\int_\Gamma f(t)\frac{dt}t\right|\ge \Re\left[\int_\Gamma e^{-i\theta}F(t)\frac{dt}t\right]=\int_\Gamma |F(t)|d(\log|t|) \\ \ge\int_\Gamma |t|^\alpha(|t|+|w|)^{-\beta}(|t|+|z|)^{-\gamma}d(\log|t|) =\int_0^\infty t^\alpha(t+|w|)^{-\beta}(t+|z|)^{-\gamma}\frac{dt}{t}\,, $$ making everything obvious again. However we face two problems here: we have to demonstrate the existence of such a curve and the circle crossing property. It turns out that the fixed argument condition determines the curve uniquely and we'll define it shortly. However, as far as proving the circle crossing property is concerned (numerical experiments indicate it should be true), I have lost my competition with the tree stump in my backyard: we both couldn't do it after two days of thinking but the stump was obviously much more concentrated on the problem since it wasn't getting distracted by anything (hey, it was so deep in thought that it even didn't move for 2 days!), so we'll have to circumvent it. The next thing we need is that for $\Re z>0$, the function $$ [-\pi/2,\pi/2]\ni \theta\mapsto |re^{i\theta}+z|^2=r^2+|z|^2+2sr\Im z+2\sqrt{1-s^2}r\Re z $$ is a non-negative strongly (in the sense of non-degenerate second derivative) concave function of $s=\sin\theta\in[-1,1]$. It follows that the function $$ \theta\mapsto -\beta\log|w+re^{i\theta}|-\gamma\log|z+re^{i\theta}| $$ is a strongly convex function of $s$. Moreover, when going slightly away from the endpoints $\theta=\pm\pi/2$ it obviously decreases (both absolute value increase). Thus on each semicircle $S_r=\{re^{i\theta}:\theta\in[-\frac\pi 2,\frac\pi 2]\}\ni t$ the function $L(t)=\alpha\log|t|-\beta\log|w+t|-\gamma\log|z+t|$ has a unique strong minimum $t(r)$, which, thereby, depends on $r$ continuously, and when you move from $t(r)$ towards either endpoint of $S_r$, it increases monotonically. Clearly, $L(t(r))\to -\infty$ for $r\to 0$ and $r\to\infty$ (that's where we use the condition $\alpha<\beta+\gamma$), so there is at least one maximum $r_*$ of $r\mapsto L(t(r))$. Any such maximum corresponds to a non-degenerate saddle point of $L(t)$. However, the critical points of $L(t)$ are just the solutions of $$ D(t)=\frac \alpha t-\frac\beta{t+w}-\frac\gamma{t+z}=0\,, $$ i.e., the roots of the quadratic polynomial $(\alpha-\beta-\gamma)t^2+[\text{some junk}]t+\alpha zw$. Since the leading coefficient is negative and the free term positive, there are two roots whose product is negative none of which lies on the imaginary axis (because $\Re D(t)<0$ for imaginary $t\ne 0$), so the right half-plane contains exactly one root, which means that $r_*$ and the associated saddle point $t_*=t(r_*)$ are unique. Now we are ready to define $\Gamma$. It is just the union of two steepest descents for $L(t)$ originating at $t_*$ one to $\infty$ and one to $0$. Since the boundary of the right half-plane is repelling for the descending gradient flow of $L$ except at the origin (it is the same condition $\Re D(t)<0$ for imaginary $t\ne 0$; the vector field in question is just $-\bar D(t)$), we are guaranteed to stay in the right half-plane all the way. The descents cannot cross $S_{r_*}$ because the values of $L$ on them are smaller than $L(t_*)=\min_{S_{r_*}}L(t)$, and $L$ goes down at fixed rate on any compact subset of $\mathbb C_+\setminus\{0,t_*,\infty\}$, so the only thing the descents can do is to escape to $0$ inside the circle and to $\infty$ outside. Near $0$ and $\infty$ the flow is asymptotically like a straight line (the direction of the field is quite clear there), so we, indeed, get an admissible contour analytic everywhere except, perhaps, $t_*$. Finally, the argument of $F$ stays fixed because the gradient curves of a harmonic function $L=\Re\log F$ are exactly the level curves of its harmonic conjugate $\arg F=\Im\log F$. As I said, if we had the circle crossing property, we would be done by now. Alas, we (or at least I) don't know how to prove that, so we'll have to consider the possibility of multiple crossings. In the multiple crossing case the formula gets a bit more complicated: $$ \Re\left[\int_\Gamma e^{-i\theta}F(t)\frac{dt}t\right]=\int_0^\infty \left[\sum_{t\in\Gamma\cap S_r}\varepsilon(t)|F(t)|\right]d(\log r) $$ where $\varepsilon(t)$ is $+1$ if $\Gamma$ (oriented from $0$ to $\infty$) crosses $S_r$ at $t$ in the outward direction and $-1$ if it does so in the inward direction. What saves the day is that the outward and the inward crossings alternate on $S_r$ (note that it is by no means necessary that the adjacent crossings on $S_r$ are adjacent on $\Gamma$; still the alternation is always there because if we complement $\Gamma$ by a piece of the imaginary line and close the contour near infinity by a circular arc, we'll get a Jourdan curve bounding a domain and the signs at the crossings will correspond to going in or out of that domain when moving along the semi-circle and traversing the boundary of the domain). Now we have the following simple Lemma: Assume that a function $f(s)$ has a unique minimum on $[-1,1]$ and is monotonically increasing if we go from that minimum to any endpoint. Then for any $-1\le s_00$ in $|z| TITLE: Link such that deleting any two components leaves an unlink QUESTION [16 upvotes]: Brunnian links are well known, where deleting any component allows you to isotope the rest to an unlink. It's common to construct them by taking an $n-1$ component unlink and defining the $n$th component as a chain of commutators in the Wirtinger presentation such as $[x_1,[x_2,[x_3,x_4]]]]$ Can you construct $n \ge 4$ component links for any $n$ where deleting any two components leaves an unlink, but the link will not be split after deleting any single component? REPLY [18 votes]: Here is a figure of a (4,2)-Brunnian link (in the terminology of Mark Grant's answer): And here is an image of a (5,3)-Brunnian link: These are taken from G.C. Shephard's 2006 article "Interlinked Loops". He was not aware of the work of Debrunner and Penney at that time as he states essentially this MO question as an open problem. However, the 2009 followup article "More Interlinked Loops" by W. R. Brakes and G. C. Shephard remedies this by giving a nice pictorial explanation of a version of Penney's construction. Here's an image of a (5,2)-Brunnian link from that paper, where I believe the letters $a,b,c,d$ correspond to elements of the Wirtinger presentation of the fundamental group of the link complement:<|endoftext|> TITLE: invertibility of matrix over free associative algebra QUESTION [7 upvotes]: For a commutative ring $R$, a matrix $A \in M_n(R)$ is invertible iff $\det (A)$ is a unit in $R$. Is there a similar criterion to determine invertibility (having two-sided inverse) of a matrix over a noncommutative ring ? I am particularly interested in determining invertibility of matrices in $M_n(F \langle x, y \rangle )$, the free associative algebra in $2$ indeterminates over a field $F$. Passing onto the polynomial ring using the relation $xy - yx = 0$, we may conclude that the `determinant' should be in $F$ which serves as a necessary condition to filter out certain matrices as being non-invertible. Based on this, I have a question as stated below. Let $f \in F \langle x, y \rangle$ and $a \in F - \{ 0 \}$. Clearly matrices of the form $$\begin{bmatrix} 1 & f\\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}, \begin{bmatrix} a & 0\\ 0 & 1 \end{bmatrix} $$ are invertible in $M_2(F \langle x, y \rangle)$. By composing such matrices, one gets other examples of invertible matrices in $M_2(F \langle x, y \rangle)$. Question : Can any invertible matrix in $M_2(F \langle x, y \rangle)$ be decomposed as a product of matrices of the above $3$ types ? REPLY [6 votes]: You are asking whether the following holds. Claim. Let $F$ be a field. Then the free associative algebra $F\langle x,y\rangle$ is a $GE_2$-ring in the sense of P. M. Cohn [2]. The answer is yes. Indeed, free associative algebras over a field are more generally $GE$-rings [Theorem 3.4, 2]. This fact has been used by Bass [1] to prove that a finitely generated projective module over a free algebra with coefficients in a PID is free. By contrast, free commutative algebras over a field, i.e., polynomial rings over a field aren't $GE$-rings if the number of indeterminates exceeds $1$ [Proposition 7.3, 2]. (I have made this remark in another MO post). We recall below the definition of a $GE$-ring in the sense of P. M. Cohn and we give subsequently a proof of the claim. Let $R$ be an associative ring with identity and let $E_n(R)$ be the subgroup of $GL_n(R)$ generated by matrices obtained from the identity matrix by replacing an off-diagonal entry by some $r∈R$. Let $D_n(R)$ be the group of invertible diagonal matrices; note that it normalizes $E_n(R)$. A ring $R$ is termed a $GE_n$-ring if $GL_n(R)=E_n(R)D_n(R)$. The ring $R$ is a $GE$-ring in the sense of P. M. Cohn if it is a $GE_n$-ring for every $n > 1$. Let us turn to the proof of the claim. Let $S$ be the free monoid on $\{x, y\}$ endowed with the shortlex order, i.e., words are ordered according to word length and words of the same length are ordered lexicographically. Then $S$ is a bi-ordered monoid and it is not difficult to show that $F\langle x,y \rangle = F[S]$ is an integral domain with unit group $F^{\times}$. Let $\varphi: S \rightarrow \omega$ be an order-preserving bijection. The following lemma is instrumental. Lemma. Let $R = F\langle x,y \rangle$ and let $a_1, a_2 \in R$. Assume there are $b_1, b_2 \in F\langle x,y \rangle$, not both zero, such that $a_1b_1 + a_2 b_2 = 0$. Then we can find $E \in E_2(R)$ and $a \in R$ such that $(a_1, a_2)E = (a, 0)$. Proof. For $r \in R$, write $r = \sum_{w \in S} r_w w$, where each $r_w \in F$ and $\text{supp}(r) \Doteq \{w \in S \,\vert\, r_w \neq 0\}$ is a finite subset of $S$. Since $R$ is an integral domain, we can assume, without loss of generality, that none of $a_1, a_2, b_1$ and $b_2$ is zero. Setting $n(r) \Doteq \varphi(\max(\text{supp}(r))) \in \omega$ for $r \neq 0$, we argue by induction on $n(a_1) + n(a_2)$. Let $w_i = \max(\text{supp}(a_i)) \in S$, for $i = 1,2$. Since $a_1b_1 + a_2 b_2 = 0$, we can find $v_1, v_2 \in S$ such that $w_1 v_1 = w_2 v_2$. Swapping $a_1$ and $a_2$ if needed, we can assume that $w_1 \le w_2$, so that $w_1$ is an initial subword of $w_2$. Therefore we can find $s \in S$ and $\lambda \in F$ such that either $a_2' = 0$ or $n(a_2') < n(a_2)$ where $a_2' = a_2 - \lambda a_1 s$. We set $a_1' = a_1$, $b_1' = b_1 + \lambda sb_2$ and $b_2' = b_2$. Then we have $a_1' b_1' + a_2'b_2' = 0$ and $(a_1', a_2')$ is obtained from $(a_1, a_2)$ by right-multiplication of a matrix in $E_2(R)$. If $a_2' = 0$, we are done. Otherwise the induction hypothesis applies to $(a_1', a_2')$, which yields the result. We are now in position to prove the claim. Proof of the claim. Let $A = (a_{ij}) \in GL_2(R)$ with $R = F\langle x,y \rangle$. As $A$ has a right inverse, we can find $b_1, b_2 \in R$, not both zero, such that $a_{11} b_1 + a_{12}b_2 = 0$. By the above lemma, we can assume that $a_{12} = 0$. It follows that $a_{11} \in F^{\times}$, so that $A$ can be reduced to an invertible diagonal matrix through an elementary row operation. Therefore $A \in E_2(R)D_2(R)$. [1] H. Bass, "Projective modules over free groups are free", 1964. [2] P. M. Cohn, "On the structure of the $GL_2$ of a ring", 1966.<|endoftext|> TITLE: Do simplicial join and product form a duoidal category structure? QUESTION [5 upvotes]: The join $\ast$ and the product $\times$ are both important monoidal structures on simplicial sets, but the way they interact is not so simple. For instance, neither distributes over the other. However, I believe there is a comparison map $(A\times B) \ast (C \times D) \to (A \ast C) \times (B \ast D)$, suggesting that they fit together into a duoidal category structure on simplicial sets. Is this in fact the case? If so, has this duoidal category been studied before? REPLY [4 votes]: I probably should have read more closely. If $C$ is any monoidal category which is also cartesian monoidal, then the combination of the two monoidal structures is duoidal. This is on the nlab page.<|endoftext|> TITLE: $L^p$ estimates and functions with positive Fourier transform QUESTION [5 upvotes]: For $f\in\mathcal{S}$ a Schwartz function on $\mathbb{R}^n$ and $m$ a bounded function, define $Tf$ by $\widehat{T f}=m\cdot \widehat{f}$. Fix $10$ such that $$ \|Tf\|_p \le C \|f\|_p$$ holds for all $f\in\mathcal{S}$ satisfying $\widehat{f}\ge 0$. Is it still possible that $T$ is not $L^p\to L^p$ bounded? I believe that the answer is yes, but I don't know of an example. I'd also be grateful for any reference where this or similar situations (possibly in the context of other kinds of operators) are discussed. REPLY [2 votes]: I apologize for answering my own question (and for asking a trivial question). The answer is yes. If $\widehat {f}\ge 0$, then we always have by the triangle inequality, $$\|Tf\|_4=\|\widehat{Tf}*\widehat{Tf}\|_2^{1/2}\le C \|\widehat{f}*\widehat{f}\|_2^{1/2}=C\|f\|_4 $$ so any multiplier that is not bounded on $L^4$ is a counterexample.<|endoftext|> TITLE: Can one divide algebraic manifolds ? Make sense: $Gr(2,n)/ Gr(2,n+m) = P^{n-1}/P^{n+m-1} P^{n-2}/P^{n+m-2}$ QUESTION [7 upvotes]: Let's start from a little bit far. Basic probability theory - chain rule reads: $$ P(AB) = P(A)P(B|A)$$ Example: consider n+m balls, where n - white balls, m - black balls, consider A - first chosen ball is white, B - second chosen ball is also white. The formula above gives: $$ \frac{ \binom{ n}{2} } {\binom{ n + m}{2}} = \frac{n} {n+m} \frac{n - 1} {n+m - 1 } $$ q-Example: consider q-analogs of binomial coefficients, obviosly, similar fact is true for them: $$ \frac{ \binom{ n}{2}_{\!q} } {\binom{ n + m}{2}_{\!q} } = \frac{ [n]_{q} } {[n+m]_{q} } \frac{[n - 1]_{q} } {[n+m - 1]_{q} } $$ Now, number of points of the Grassmanian over the finite field $F_q$ is exactly given by the q-binomial coefficient and $[n]_q$ is the one for $P^{n-1}(F_q)$. Optimistically enumaration relation is manifstation of some deeper relation on the geometric level. (See for example: Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over Fq?) So we come up with: Question 1: Can one make sense of: $$Gr(2,n)/ Gr(2,n+m) = P^{n-1}/P^{n+m-1} P^{n-2}/P^{n+m-2} $$ Motivation. Combinatorics of binomial coefficients is very very related to probability theory, on the other hand they might be thought as $F_{q=1}$ of the Grassmanian geometry, since q-binomial coefficients behave pretty much similar to ordinary, it would be tempting to have a lift of probability ideas to generic $q$, and if so, we should have the relation above as a kind of q-chain rule. It might be easier to analyse more simple question: Question 2: (Weak version of 1). Is there any geometric relation between: $$ Gr(2,n) P^{n+m-1} P^{n+m-2} = Gr(2,n+m) P^{n-1} P^{n-2} $$ The idea to divide manyfolds might sound crazy, however, it might have some ground in the past. There are ideas of the so-called "fractional motives". Which might be something like motives of such results of the division. What I vaguely remember from Yuri Manin's talks about ten years ago, is that one the main example comes from considerations by physicts Doron Gepner who discovered that CFT on certain CY can be factorized to product of other CFT (minimal models) which does not have direct geometric interpetation but if the that would exist - it should be a manyfold with some fractional dimension. Manin mentioned some similar constructions for some Frobenius manyfolds... REPLY [14 votes]: Let me begin with a sketch of an answer to your question two, namely giving an interpretation of the cross-multiplied equality $$\text{Gr}(2,n)\mathbb{P}^{n+m-1}\mathbb{P}^{n+m-2}=\text{Gr}(2,n+m)\mathbb{P}^{n-1}\mathbb{P}^{n-2}.$$ I claim this equality holds in the Grothendieck ring of stacks. This is the Grothendieck ring of varieties (the free Abelian group on isomorphism classes of algebraic varieties, subject to the cut-and-paste relation $$[X]=[U]+[X\setminus U]$$ where $U$ is Zariski-open in $X$, with multiplication given by the Cartesian product), and with the classes $[\mathbb{A}^1], [\mathbb{A}^n]-1$ inverted. Indeed, let $$Y=\text{Fl}_{1,2}(U)\times \text{Fl}_{1,2}(W),$$ where $\text{Fl}_{1,2}(V)$ denotes the space of flags of signature $(1,2)$ in a vector space $V$, and where $\dim(U)=n, \dim(W)=n+m$. Now $$[\text{Fl}_{1,2}(k^r)]=[\mathbb{P}^{r-1}][\mathbb{P}^{r-2}],$$ because forgetting the subspace of dimension $2$ witnesses $\text{Fl}_{1,2}(k^r)$ as a Zariski-locally trivial $\mathbb{P}^{r-2}$ bundle over $\mathbb{P}^{r-1}$. On the other hand, $$[\text{Fl}_{1,2}(k^r)]=[\mathbb{P}^1][\text{Gr}(2,r)],$$ because forgetting the subspace of rank $1$ witness $\text{Fl}_{1,2}(k^r)$ as a Zariski-locally trivial $\mathbb{P}^1$-bundle over $\text{Gr}(2,r)$. So we have $$[Y]=[\mathbb{P}^1][\text{Gr}(2,n)][\mathbb{P}^{n+m-1}][\mathbb{P}^{n+m-2}]=[\mathbb{P}^{n-1}][\mathbb{P}^{n-2}][\text{Gr}(2,n+m)][\mathbb{P}^1].$$ This is your desired equality, with both sides multiplied by $\mathbb{P}^1$. But in the Grothendieck ring of stacks, $[\mathbb{P}^1]=1+[\mathbb{A}^1]$ is invertible, because $[\mathbb{P}^1]([\mathbb{A}^1]-1)=[\mathbb{A}^2]-1$, which we inverted by fiat. So the desired equality holds in this ring. In fact, it is an easy exercies to show that $\text{Gr}(a,b)$ and $\mathbb{P}^c$ are invertible for any $a,b,c$, so one may divide out to obtain your equality in Question 1. That said, I'm dubious that there is any real value in thinking about this in terms of "fractional motives" -- as you can see, the underlying geometry is just a linearization of the combinatorial reasoning in your example.<|endoftext|> TITLE: Open projections and Murray-von Neumann equivalence QUESTION [7 upvotes]: Let $\mathcal{A}$ be a $C^*$-algebra and $p\in\mathcal{A}^{**}$ be an open projection, that is, $p=p^*=p^2$ and $p\in\overline{(p\mathcal{A}^{**}p\cap\hat{\mathcal{A}})}^{\operatorname{w}^*}$, where $\hat{\mathcal{A}}$ is the canonical copy of $\mathcal{A}$ in $\mathcal{A}^{**}$ and the closure is taken in the weak$^*$ topology on $\mathcal{A}^{**}$. Question: Is every orthogonal projection in $\mathcal{A}^{**}$ which is Murray-von Neumann equivalent to $p$ open? REPLY [2 votes]: The answer is no. Proof (Thomas Schick). The idea of the proof is due to Thomas Schick. I thank him for allowing me to reproduce it here. Let $\mathcal{A}:=C([0,1])\otimes\mathbb{M}_2$, where $\mathbb{M}_2$ is the $W^{\star}$-algebra of $2\times2$ matrices with entries in $\mathbb{C}$. Since the set of bounded Borel functions $B^{\infty}([0,1])$ is contained in $C([0,1])^{**}$, we canonically identify $B^{\infty}([0,1])\otimes\mathbb{M}_2$ with a $C^*$-subalgebra of $\mathcal{A}^{**}$. Let $F:=\{0\}$ and $F^c:=(0,1]$, then the characteristic functions $\chi_F$ and $\chi_{F^c}$ are Borel measurable. Define a partial isometry $u:=\chi_F\otimes E_{11}+\chi_{F^c}\otimes E_{12}\in\mathcal{A}^{**}$, where $E_{ij}\in\mathbb{M}_2$ is the canonical matrix unit, that is, it has $1$ in its $(i.j)$-entry and $0$ elsewhere. Then the orthogonal projection $p:=uu^{\star}=1\otimes E_{11}$ is open, whereas $q:=u^{\star}u=\chi_F\otimes E_{11}+\chi_{F^c}\otimes E_{22}$ is not. $\square$ Cf. Here is my related question.<|endoftext|> TITLE: Is every symmetric monoidal combinatorial model category symmetric monoidally Quillen equivalent to a simplicial such? QUESTION [10 upvotes]: Is every symmetric monoidal combinatorial model category symmetric monoidally Quillen equivalent (by a zig-zag of symmetric monoidal Quillen equivalences) to a symmetric monoidal combinatorial simplicial model category? Do we have a symmetric monoidal version of Dugger's theorem? Is this true and if yes, where do I find a reference for that? REPLY [6 votes]: Yes, this is proved in https://arxiv.org/abs/1506.01475 by Thomas Nikolaus and Steffen Sagave. If you need a specific monoidal left Quillen equivalence, you can upgrade Dugger's result to the symmetric monoidal setting. One proceeds as in Dugger's proof. Start with the Reedy model structure on simplicial objects. Equip it with the degreewise monoidal product, which turns it into a monoidal model category (see, e.g., Ghazel-Kadhi). Now perform a left Bousfield localization with respect to hocolim-equivalences, as in Dugger's proof. The resulting model category is a monoidal because monoidal products preserve filtered colimits (see, e.g., Gorchinskiy-Guletskii). A symmetric monoidal Quillen equivalence connects it to the original category, like in Dugger's proof.<|endoftext|> TITLE: Is the cohomology ring of a finite group computable? QUESTION [26 upvotes]: Is there an algorithm which halts on all inputs that takes as input a finite group ($p$-group if you like) and outputs a finite presentation of the cohomology ring (with trivial coefficients $\mathbb{F}_p$) in terms of generators and relations. For concreteness, let's say the input is given by a generating set of matrices, or permutations, or by giving its multiplication table - from the computability point of view, these are all equivalent. I've read much of David J. Green's book "Grobner bases and the computation of group cohomology," in which he presents an algorithm that produces "partial presentation" of the cohomology ring degree-by-degree. There is a sufficient criterion due to J. F. Carlson which says when you're done - that is, when this partial presentation is actually a correct presentation of the cohomology ring - but the book seems to indicate that Carlson's criterion either is not necessary or at least not known to be necessary (as of its writing, 2003). Now, the algorithm used in Green's book will eventually get a complete presentation of the cohomology ring, but the issue is whether the algorithm can tell when it's reached a high enough degree to be done. Following this strategy, a related question is: Consider the function $b_p:FinGrp \to \mathbb{N}$ defined by $b_p(G)$ is the least $n \in \mathbb{N}$ such that the cohomology ring of $G$ with coefficients in $\mathbb{F}_p$ is completely determined by the partial presentation one gets by going up to degree $n$. Is $b_p$ bounded by a computable function? REPLY [19 votes]: As I understand it this follows from Benson's Regularity Conjecture, proved by Symonds fairly recently. It says that $b_p = 2(|G|-1)$ will do.<|endoftext|> TITLE: Interpolation space between $L^1\cap L^2$ and $L^1$ QUESTION [6 upvotes]: In the paper of Bourgain, the way equation (3.78) is deduced from (3.69) and (3.76) seems via the following interpolation result. Let $(X,\mu)$ and $(Y,\nu)$ be two measure spaces and let $T$ be a linear operator from $L^1(X,\mu)+L^2(X,\mu)$ to $L^\infty(Y,\nu)+L^2(Y,\nu)$ such that \begin{align} \|Tf\|_{L^\infty}&\leq A\|f\|_{L^1}, \\ \|Tf\|_{L^2}&\leq B\|f\|_{L^2}+D\|f\|_{L^1}. \end{align} For $0<\theta<1$, let $p=\frac{1-\theta}{1}+\frac{\theta}{2}$, $\frac{1}{p}+\frac{1}{q}=1$. Then for some constant $C$ independent of $A,B,D$, we have $$\|Tf\|_{L^{q}}\leq C(A^{1-\theta}B^\theta\|f\|_{L^p}+A^{1-\theta}D^\theta\|f\|_{L^1}).$$ By scaling the measures $\mu$ and $\nu$ as well as the linear operator $T$, one can reduce it to the case when \begin{align} A=B=D=1. \end{align} I tried to prove it by looking for the complex interpolation space between $L^1$ and $L^2\cap L^1$, as well as to directly prove it as to prove the Marcinkiewicz's interpolation inequality. But I could not find the correct proof. Any help is appreciated. Thank you! REPLY [3 votes]: As requested, I post my comment as an answer (although this is not a true answer, just a possibly useful reference; feel free to edit it if this approach works out). In Section 3 of the article Interpolation of sum and intersection spaces of $L^q$-type and applications to the Stokes problem in general unbounded domains, P.F. Riechwald studies interpolation between spaces $L^2 \cap L^q$ for $q \geqslant 2$ (and also $L^2 + L^q$ for $q \leqslant 2$). His main tool is Theorem 4, which I reproduce below. Theorem. Let $\Omega \subseteq \mathbb{R}^n$ be a domain and let $f \in L^1(\Omega) + L^\infty(\Omega)$ be a given and fixed function. Then there exist linear maps $$ S_1 : L^1(\Omega) + L^\infty(\Omega) \to L^1((0, 1)) , \qquad S_2 : L^1(\Omega) + L^\infty(\Omega) \to \ell^\infty $$ and $$ T_1 : L^1((0, 1)) \to L^1(\Omega) + L^\infty(\Omega) , \qquad T_2 : \ell^\infty \to L^1(\Omega) + L^\infty(\Omega) $$ satisfying the equality $$ f = T_1 S_1 f + T_2 S_2 f $$ almost everywhere. Moreover, these maps satisfy the estimates $$ \|S_1 u\|_{L^p((0, 1))} \leqslant \|u\|_{L^p(\Omega)} , \qquad \|S_2 u\|_{\ell^p} \leqslant \|u\|_{L^p(\Omega)} $$ and $$ \|T_1 u\|_{L^p(\Omega)} \leqslant \|u\|_{L^p((0, 1))} , \qquad \|T_2 u\|_{L^p(\Omega)} \leqslant \|u\|_{\ell^p} $$ for all $1 \leqslant p \leqslant \infty$ and all $u$ in the respective $L^p$-spaces. I suppose the argument used in the proof of Theorem 3 implies that the complex interpolation space between two spaces $L^1(\Omega) \cap L^p(\Omega)$ is again a space of this form. This, in turn, should imply the desired bound on $\|T f\|_p$.<|endoftext|> TITLE: Fundamental group of Alexandrov space. QUESTION [5 upvotes]: Is it true that the fundamental group of a compact finite dimensional Alexandrov space with curvature bounded below is finitely generated? REPLY [4 votes]: Under these hypotheses the systole of $X$ is clearly bounded from below, and the usual comparison arguments would give an upper bound on the number of points in an $\epsilon$-net in $X$. Then every loop can be discretized and the finite-generation follows.<|endoftext|> TITLE: A question about the Chern-Weil construction of Euler class QUESTION [5 upvotes]: I'm reading Appendix C of "Characteristic Classes" by Milnor & Stasheff and something confuses me. When proving that the Pfaffian of the curvature form (of an oriented $2n$-plane bundle $\xi$ over a smooth manifold $M$, with a metric and a connection compatible with that metric) represents a multiple of the Euler class of $\xi$, the authors say: (here $K$ stands for curvature tensor) The notation $Pf(K(\tilde{\gamma}))$ suggests that $\tilde{\gamma}$ has a connection on it, thus the base space must be a smooth manifold. On the other hand, $\tilde{\gamma}$ is required to be "universal", so it should be the tautological bundle over the oriented Grassmannian $\tilde{G}_{2n}(\mathbb{R}^{\infty})$, but then the base space is not a manifold, seemingly a contradiction. So what should this bundle $\tilde{\gamma}$ be? REPLY [9 votes]: In the second line of the quoted passage, it says that $\tilde\gamma$ should be `universal in dimensions $\leq 4n$'. This means that it's universal for bundles over complexes of dimension at most $4n$, and so we can take a finite dimensional oriented Grassmannian $\tilde{G}_{2n}(R^N)$ for sufficiently large $N$. This is a manifold, and the restriction of $\tilde\gamma$ has a connection. You can read about $n$-universal bundles in Steenrod's book, Fiber Bundles.<|endoftext|> TITLE: Knot Factorization Homology inputs QUESTION [5 upvotes]: Following the paper by Ayala, Francis, and Tanaka: https://arxiv.org/pdf/1409.0848.pdf If we are talking about knots we are talking about framed 3-manifolds with a framed 1-dimensional sub-manifold and trivialized normal bundle. Lets assume the monoidal category is dg-vector spaces with $\otimes$. Corollary 0.3. says there is an equivalence between framed $1\subset3$ disk-algebras and a triple $(A,B,\alpha)$ where $A$ is a 3-disk algebra ($E_3$-algebra), $B$ is a 1-disk algebra ($E_1$-algebra or $A_\infty$) and $$\alpha:HC_*(A)\to HC^*(B)$$ is a map of 2-disk algebras. My questions are: If $A$ and $B$ are commutative algebras (concentrated in degree 0) over some field $\mathbb{k}$ (smooth, if you wish) then $A$ is in particular also $E_3$ and $B$ is $E_1$. 1) What kind of map should $\alpha$ be in this case? 2) If we take $HC_*$ to mean the hochschild chain complex of $A$ then $HC_*(A)$ is concentrated in (homological) non-negative degree and $HC^*(B)$ is concentrated in non-positive degree. Should our map have to respect degrees of elements or is there some sort of degree shift going on? 3) It seems this structure should make $B$ an $A$-module (in the usual sense). This should be a ring map from $A$ into $End(B)$ or a series of compatible linear maps one such being $A\otimes A\otimes B\to B$. Is there a way to see this from the map $\alpha$? REPLY [6 votes]: Let me start with the second question: 2) The map $\alpha$ will indeed respect the degrees. You should think of it as a map between two $E_2$-algebras in unbounded chain-complexes (whose domain happens to be concentrated in non-negative degrees, while its codomain is concentrated in non-positive degrees). 1) As a result of (2), the data of the map $\alpha$ is actually equivalent to the data of a map of discrete commutative algebras $A \to B$. To see this, observe that the inclusion $\iota_{\leq 0}: {\rm Ch}_{\leq 0}(\mathbb{k}) \to {\rm Ch}(\mathbb{k})$ of non-positively graded chain-complexes into unbounded chain-complexes has a left adjoint $\tau_{\leq 0}: {\rm Ch}(\mathbb{k}) \to {\rm Ch}(\mathbb{k})_{\leq 0}$, the truncation functor. Both $\iota_{\leq 0}$ and $\tau_{\leq 0}$ are symmetric monoidal functors and descend to an adjunction between $E_2$-algebras in ${\rm Ch}(\mathbb{k})$ and $E_2$-algebras ${\rm Ch}(\mathbb{k})_{\leq 0}$. This means that the data of an $E_2$-algebra map $\alpha: HC_*(A) \to HC^*(B)$ is equivalent to the data of an $E_2$-algebra map $A = \tau_{\leq 0}HC_*(A) \to HC^*(B)$. A similar argument using cotruncation instead of truncation shows that the data of such a map is equivalent to the data of an $E_2$-algebra map $A \to \tau^{\geq 0}HC^*(B) = B$. Since $A$ and $B$ are discrete this is the same as a map of commutative algebras. 3) Suppose $A$ was an $E_1$-algebra and $B$ a chain-complex (which we can think of as an $E_0$-algebra object in ${\rm Ch}(\mathbb{k})$). Then the structure of an $A$-module on $B$ is given by a map of $E_1$-algebras $A \to {\rm End}(B)$. This is because the monoidal structure on ${\rm Ch}(\mathbb{k})$ is closed, i.e., admits a compatible system of internal mapping objects. If $A$ is now an $E_2$-algebra and $B$ is an $E_1$-algebra, then we can think of $A$ as an $E_1$-algebra in $E_1$-algebras, and so there is an associated notion of an action of $A$ on $B$ (in $E_1$-algebras) in which case one usually says that $B$ has the structure of an $A$-algebra. However, the induced tensor product on $E_1$-algebras is not a closed monoidal structure: given two $E_1$-algebras $B,B'$, there is no internal mapping object ${\rm Map}(B,B')$ in $E_1$-algebras. Nonetheless, there is still an object which controls $A$-actions on $B$, and that is the center $HC^*(B)$ of $B$, considered as an $E_2$-algebra. In particular, $A$-algebra structures on $B$ are classified by $E_2$-algebra maps $A \to HC^*(B)$. Note that for $A$, as an $E_1$-algebra object ${\rm Alg}_{E_1}({\rm Ch}(\mathbb{k}))$, there is also the associated notion of an $A$-bimodule structure on a given $B \in {\rm Alg}_{E_1}({\rm Ch}(\mathbb{k}))$. This, in turn, can be described by a map of $E_2$-algebras $A^{\rm op} \otimes A \to HC^*(B)$. Now let's consider the case at hand where $A$ is an $E_3$-algebra and $B$ is an $E_1$-algebra. Then we can think of $A$ as an $E_2$-algebra in $E_1$-algebras. Now, just like associative algebras have the associated notion of a bimodule, for $E_2$-algebras there is the analogous notion of an $E_2$-module. Roughly speaking, an $E_2$-action of $A$ on $B$ means a "continuous family" of $A$-actions on $B$ parameterized by the circle (with suitable compatibilities with the $E_2$-structure of $A$). If $A$ is an $E_2$-algebra in ${\rm Alg}_{E_1}({\rm Ch}(\mathbb{k}))$ and $B$ is an object of ${\rm Alg}_{E_1}({\rm Ch}(\mathbb{k}))$ then the data of an $E_2$-action of $A$ on $B$ will be given by an $E_2$-algebra map of the form $\alpha: HC_*(A) \to HC^*(B)$. When $A$ and $B$ are discrete commutative algebras this is the same as just an $A$-algebra structure on $B$, which, in turn, is given by a map of commutative algebras $A \to B$.<|endoftext|> TITLE: The largest topological copy of a Hilbert space contained in $\ell^1$ QUESTION [5 upvotes]: Let us consider $\ell^1$, the space of absolutely summable sequences in the space of complex numbers. Clearly every finite dimensional Hilbert space is topologically embedded into $\ell^1$. Convention. For given Hilbert spaces $H$ and $K$, let us write $H\leq K$ if the Hilbertian dimension of $H$ is less than $K$. Q. What is the Hilbertian dimension of the largest Hilbert space which can be topologically embedded into $\ell^1$? REPLY [4 votes]: No infinite dimensional reflexive space can be embedded into $\ell_1$, because every infinite dimensional closed subspace of $\ell_1$ has a non separable dual.<|endoftext|> TITLE: How many "steps" does one need to obtain a group from a semigroup QUESTION [8 upvotes]: Given a countable group $G$ and a generating subsemigroup $S\subset G$, let us consider the increasing sequence of alternating products of $S$ and $S^{-1}$ beginning with $S$ $$ A_1=S, A_2=S S^{-1}, A_3=S S^{-1}S, A_4=S S^{-1}S S^{-1}, A_5=S S^{-1}S S^{-1}S, \dots \;, $$ and put $$ \kappa(S) = \min \{k: G=A_k \} \;. $$ Obviously, $\kappa(S)=1$ if and only if $S=G$, and $\kappa(S)=\infty$ if $S$ is the semigroup of positive words in a free group $G$. One can show that if $G$ is nilpotent, then $\kappa(S)= 2$ for any generating subsemigroup $S\neq G$, and, more generally, that $\kappa(S)=2$ for any generating subsemigroup iff $G$ does not contain a free subsemigroup. What else is known? REPLY [10 votes]: Marek Kuczma asked in 1980 whether for every positive integer $n$ there exists a subsemigroup $M$ of a group $G$ such that $G$ is equal to the $n$-fold product $MM^{-1}MM^{-1}\cdots M^{(-1)^{n-1}}$, but not to any proper initial subproduct of the product. George Bergman proved that the answer is affirmative for all $n$. The result (and sundry generalizations) appear in Submonoids of groups, and group-representability of restricted relation algebras, which will appear in Algebra Universalis this year. It is also available on his website.<|endoftext|> TITLE: Lower bound on the entries of the Perron vector QUESTION [7 upvotes]: Let $A$ be a matrix that satisfies all the conditions of Perron- Frobenius theorem. From the theorem it is known that the entries of the eigenvector corresponding to the largest eigenvalue will be strictly positive. Are their any good lowerbounds known for these entries ? (Assuming that the eigenvector is normalized in some suitable way) Imagining $A$ as some kind of a Markov chain, I would expect the lowerbound to depend on how the vertex corresponding to the entry in question is connected to other verticies (degree, weight of self-loops etc.). Any results along those lines would be very-helpful REPLY [5 votes]: This seems to be answered in the accepted answer to this question: The height of the Perron-Frobenius eigenvector For convenience, here is the estimate:<|endoftext|> TITLE: Symmetric Powers for Lie Algebras QUESTION [9 upvotes]: Let $V_\lambda$ be the irreducible representation of $sl_{n}(\mathfrak{C})$ with highest weight $\lambda$. There are well known formulas for the decomposition of $V_\lambda^{\otimes^k}= V_\lambda\otimes V_\lambda\otimes \cdots\otimes V_\lambda$ into irreducble representations using Littelwood-Richardson, or Littelmann paths, among others. Is there any similar for the symmetric power $S^k(V_\lambda)$ or for $\wedge^k(V_\lambda)$, even for $sl_2(\mathfrak{C})$ or$k=3, 4$, or just some results in some cases?? I know that small cases can be computed, for instance using LiE http://wwwmathlabo.univ-poitiers.fr/~maavl/LiE/form.html but this formulas seems to be recursive. Is there any closed formula like Littelwood-Richardson, or Littelmann paths? What is the best recent reference? Thanks REPLY [7 votes]: For $sl_2$ one has formulas. If $V_n=S^n(\mathbb{C}^2)$, then the multiplicity of the irreducible $V_k$ inside $S^m(V_n)$ is $$ M_{m,n,k}=\left[m,n,\frac{mn-k}{2}\right]-\left[m,n,\frac{mn-k}{2}-1\right] $$ where $[m,n,w]$ is the number of integer partitions of $w$ with at most $n$ parts of length $\le m$. This is the Cayley-Sylvester formula. It was discovered by Cayley in a letter to Sylvester around 1854/1855 but only proved in 1878 by Sylvester (see this article). For simple cases like $S^3(V_n)\simeq S^n(V_3)$ (by Hermite's reciprocity of 1854) one has even more explicit formulas, see Theorem 1.3 in this article. For $sl_n$, as far as I know the state-of-the-art is summarized in this article by Khale and Michałek.<|endoftext|> TITLE: Conformal boundary and cusp of figure-8 complement QUESTION [5 upvotes]: As we know the figure-8 ($4_1$) complement can be obtained by quotienting $\mathbb{H}^3$ with an arithmetic Kleinian group, which has index 12 inside $PSL(2,\mathcal{O}_3)$. The resulting complete hyperbolic manifold has a cusp corresponding to the torus boundary. On the other hand, the "conformal boundary" (boundary at infinity) of $\mathbb{H}^3$ is the sphere at infinity $S^2_{\infty}$. For a Kleinian group $\Gamma$, the hyperbolic manifold $\mathbb{H}^3/\Gamma$ inherits a conformal boundary $\Lambda/\Gamma$, where $\Lambda$ is the domain of discontinuity ($S^2_{\infty}$ with the limit set of $\Gamma$ excluded). Then how is the empty "conformal boundary" of $4_1$ complement related to its cusp? (I know that the conformal structure of the torus boundary is called "cusp shape"); i.e., should both of them correspond to the $z=0$ "end" in Poincare half-space model? (since the torus boundary results from the truncated vertices on $S^2_{\infty}$ of 2 ideal tetrahedra under developing map upon gluing). Update: the limit set of $\Gamma$ here is the entire $S^2_{\infty}$ (Corollary 11.8 in Bonahon), so domain of discontinuity as well as the conformal boundary is $\emptyset$. Confusion: according to section 5 of Thurston, the entire boundary of a Kleinian manifold $(\mathbb{H}^3\cup\Lambda)/\Gamma$ is $\Lambda/\Gamma$, which is empty for $4_1$ complement. Then what is the role of the torus boundary (cusp)? P.S.: As a physicist, I may not have stated the definition of "conformal boundary" rigorously, but in my mind it is the boundary in the sense of AdS/CFT correspondence (in Euclidean signature, $AdS_3$ is $\mathbb{H}^3$). REPLY [3 votes]: The group of parabolic isometries fixing a point at infinity is isomorphic to ${\mathbb C}$. (Because it acts simply transitively on a horosphere $H$.) The discrete group $\Gamma$ intersects this stabilizer of the parabolic fixed point (i.e., the stabilizer of $H$) in a lattice $\Lambda\subset{\mathbb C}$. The quotient $H/\Lambda={\mathbb C}/\Lambda$ is a torus with a conformal structure.<|endoftext|> TITLE: Where to find the results of Onishchik? QUESTION [6 upvotes]: I would like to have a good reference where the results in "Inclusion relations between transitive compact transformation groups" https://mathscinet.ams.org/mathscinet-getitem?mr=27:3740 can be found in English/French/Spanish. I am particularly interested about decompositions of compact simple Lie groups. I have the translated version of the subsequent paper "Decompositions of Reductive Lie groups" where some results are mentioned, but still I am interested in the previous work. Thanks. REPLY [7 votes]: This paper was translated to English and appears in Fifteen papers on algebra. American Mathematical Society Translations. Series 2. Vol. 50; American Mathematical Society, Providence, R.I. 1966; MR189949. This particular paper is on pages 5 to 58; Zbl 0207.33604 (and Zentralblatt entries for other papers from this book). I did not find a copy that is freely available online, but you can see some parts in Google Books. In case it is useful for somebody, here is also a link to the Russian original: http://mi.mathnet.ru/eng/mmo134 REPLY [7 votes]: Onishchik has a book "Topology of transitive transformation groups", written in English, where this material is presented in Chapter 4. The book is excellent but hard to find. I recall being unable to find it in the US and reading it in the Oberwolfach library. REPLY [4 votes]: if you have the Russian paper, which I do not, Google translate should do a satisfactory job; I tried it on a related paper by Onishchik, for which I do have the source: Semi-simple decompositions of semi-simple Lie algebras. A pass of the first paragraph through Google translate produces a result that seems quite workable: I don't read Russian: no edits on my part, other than formatting. Let $G$ be a Lie algebra, let $G'$ and $G''$ be its subalgebras, we say that the triple ($G, G', G''$) is a decomposition if $G = G' + G''$. A Lie group acting on a manifold $M$ is called locally transitive if at least one of its orbitals on $M$ is open. It is easy to see that studying decompositions of real Lie algebras is equivalent to studying the inclusion relations between locally transitive groups. Lee transformations. If $G$ is a complex Lie algebra, $G'$ and $G''$ are its complex subalgebras, then the decomposition $(G, G', G'')$ is said to be complex. The decomposition is said to be semisimple if $G, G'$, and $G''$ are semisimple.The present paper is devoted to finding all the real and complex semisimple expansions.<|endoftext|> TITLE: Is there a non-atomic finite positive measure in the plane, of which uncountably many projections have atoms? QUESTION [8 upvotes]: I would like to know whether or not there exists a finite probability measure $\mu$ on $\mathbb R^2$ which has no atoms, but such that there exists an uncountable set $A\subset \mathbb S^1$, such that for every $v\in A$ there exists $x_v\in \mathbb R^2$ such that $\mu(x_v+v\mathbb R)>0$. (A weaker requirement on $\mu$ would be that there exists an uncountable set of lines (i.e. affine $1$-dimensional subspaces of $\mathbb R^2$) $L$ such that for each $\ell \in L$ there holds $\mu(\ell)>0$, so proving that no positive finite Radon measure $\mu$ satisfies this property, would imply a negative answer to my question.) REPLY [12 votes]: Such a measure cannot exist. Suppose to the contrary that we have an uncountable family of lines $\ell$ such that $\mu(\ell)>0$. Then there is $\epsilon>0$ and an infinite family of lines $\{\ell_i\}_i$ with $\mu(\ell_i)\geq\epsilon$. These lines intersect at countably many points. Since the measure has no atoms, this set $E$ has measure zero. Removing this set from the lines we obtain a countable family of pairwise disjoint sets $\{\ell_i\setminus E\}_i$ with $\mu(\ell_i\setminus E)\geq\epsilon$. Hence $\infty>\mu(\mathbb{R}^2)\geq\mu(\bigcup_i (\ell_i\setminus E))=\infty$ which is a contradiction.<|endoftext|> TITLE: Differentiability of Eigenvalues - Perturbation Theory QUESTION [12 upvotes]: first, I have a general question. In perturbation theory, I saw perturbations in eigenvalues and eigenvectors of square, non-symmetric matrices and the calculations were all right but no one ever proved the differentiability of the eigenvalues/eigenvectors in the perturbation-parameter! They just assumed it. So, why is that? Second, does someone know one way to prove differentiability (at least twice) of eigenvalues/eigenvectors in the perturbation-parameter for those matrices in the perturbation-parameter? REPLY [18 votes]: The complete reference is Kato's book Perturbation theory .... But perhaps you need only the most basic results. Then see my book Matrices (Springer GTM #216), 2nd edition. This is Section 5.2. Mind however that this is a much more elaborate topic than what you could think at first glance. There are at least four completely distinct aspects: On the one hand, the spectrum is always a continuous function of the entries. But this continuity is false in full generality for individual eigenvalues. Next, if an eigenvalue $\lambda_0$ of a given matrix $M_0$ is algebraically simple, then there exists locally an analytic function $M\longmapsto\lambda(M)$ where $\lambda(M)$ is an eigenvalue of $M$ and $\lambda(M_0)=\lambda_0$. This is the case where you can compute by hand the derivatives of an eigenvalue. Finally, if $t\mapsto H(t)$ is a one-parameter family (a curve) of Hermitian matrices (think to real symmetric matrices), then the eigenvalues of $H(t)$ can be arranged as a list of functions $t\mapsto\lambda_j(t)$ which inherit the regularity ($C^k$, analyticity) of the data. Mind that there are crossings, the list $\vec\lambda(t)$ is not ordered. The statement becomes deadly false for a two-parameter family (a surface). However, the ordered list $\lambda_1(H)\le\cdots\le\lambda_n(H)$ is Lipschitz continuous over the space of Hermitian matrices, with Lipschitz constant one: $$|\lambda_j(K)-\lambda_j(H)|\le\|K-H\|$$ where the norm is the operator norm (equal to the spectral radius for Hermitian matrices). REPLY [9 votes]: Eigenvalues are roots of the characteristic polynomial. In the case the polynomial has simple roots, they smoothly depend on the coefficients. Below is the proof in the real case. The argument below can be easily modified to more general situations. Theorem. For $a=(a_0,a_1,\ldots,a_n)\in\mathbb{R}^{n+1}$, $a_n\neq 0$ let $P_a(x)=a_nx^n+\ldots+a_1x+a_0$. Suppose that for $a^0=(a_0^0,\ldots,a_n^0)$, $a^0_n\neq 0$ the polynomial $P_{a^0}(x)$ has $n$ distinct real roots. Then, there is $\epsilon>0$ and $C^\infty$ smooth functions $$ \lambda_1,\ldots,\lambda_n: B^{n+1}(a^0,\epsilon)\to\mathbb{R} $$ such that for any $a\in B^{n+1}(a^0,\epsilon)$, $\lambda_1(a),\ldots,\lambda_n(a)$ are distinct roots of the polynomial $P_a(x)$. In other words, prove that in a small neighborhood of $a^0$, roots of the polynomial $P_a$ depend smoothly on the coefficients $a_0, a_1,\ldots, a_n$. Proof. Denote the roots of $P_{a^0}$ by $\lambda_1^0,\ldots,\lambda_n^0$. That is $P_{a^0}(\lambda)=a_n^0(\lambda-\lambda_1^0)\cdots(\lambda-\lambda_n^0)$ and clearly $$ \left.\frac{d}{d\lambda}\right|_{\lambda=\lambda_i^0} P_{a^0}(\lambda)\neq 0 \quad \mbox{for all $i=1,2,\ldots,n$.} $$ This is where we employ the assumption that the roots are distinct. Let $F(a,\lambda)=F(a_0,a_1,\ldots,a_n,\lambda)=P_a(\lambda)$. For each $i=1,2,\ldots,n$ we have $$ F(a_0^0,a_1^0,\ldots,a_n^0,\lambda_i^0)=P_{a^0}(\lambda_i^0)=0 $$ and $$ \left.\frac{\partial}{\partial\lambda}\right|_{\lambda=\lambda_i^0} F(a_0^0,a_1^0,\ldots a_n^0,\lambda) = \left.\frac{d}{d\lambda}\right|_{\lambda=\lambda_i^0} P_{a^0}(\lambda)\neq 0. $$ Thus according to the Implicit Function Theorem, there is a unique $C^\infty$ smooth function $\lambda(a)$ defined for $a=(a_0,a_1,\ldots,a_n)$ in a neighborhood of $a^0=(a_0^0,a_1^0,\ldots,a_n^0)$ (neighborhood in $\mathbb{R}^{n+1}$) such that $\lambda(a^0)=\lambda^0_i$ and $P_a(\lambda(a))=F(a,\lambda(a))=0$. Denote this function by $\lambda_i(a)$. Hence $\lambda_i(a)$, $i=1,2,\ldots,n$ are roots of the polynomial $P_a$. Since $\lambda_i(a^0)\neq \lambda_j(a^0)$ for $i\neq j$, we see that these roots are distinct in a neighborhood $B^{n+1}(a^0,\epsilon)$ of $a^0$, provided $\epsilon>0$ is sufficiently small. $\Box$ REPLY [7 votes]: The eigenvalues of a square matrix $A$ are the roots of the characteristic polynomial, and are analytic except where their multiplicities change. Thus if (in a certain open region of parameter space) there is one eigenvalue, of algebraic multiplicity $m$, inside a positively oriented closed contour $\Gamma$ in $\mathbb C$, and no eigenvalues on $\Gamma$, that eigenvalue is $$ \frac{1}{2\pi im}\oint_\Gamma \frac{z P'(z)}{P(z)}\; dz$$ where $P(z) = \det(A - z I)$, and this is analytic in the entries of $A$. The corresponding eigenvectors (with an appropriate normalization) can be found by solving a linear system depending linearly on the parameters and the eigenvalue, and these are also analytic.<|endoftext|> TITLE: Spectrum of the product of operators QUESTION [5 upvotes]: Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$. Let $A,B\in \mathcal{B}(F)^+:=\left\{T\in \mathcal{B}(F);\,\langle Tx, x\rangle\geq 0,\;\forall\;x\in F\;\right\}$, be such that $AB\neq 0$. I want to show that $$\sigma(AB)\neq\{0\}\;.$$ REPLY [5 votes]: Use the fact that $\sigma(ST) \cup \{0\} = \sigma(TS) \cup \{0\}$. So if $A$ and $B$ are positive then, except possibly for the point $0$, $\sigma(AB)$ equals $\sigma(A^{1/2}BA^{1/2})$. If $AB \neq 0$ then the latter is a nonzero positive operator, and hence it has a nonzero element in its spectrum. Therefore so does $AB$.<|endoftext|> TITLE: $S^{2}$-bundles over complex projective varieties QUESTION [5 upvotes]: Is there an example of a smooth complex projective variety and an $S^{2}$-bundle over it which is not diffeomorphic to a complex projective variety? REPLY [3 votes]: Yes. Let $E\to T^2$ be the $S^2$-bundle over the torus $T^2$ obtained by gluing the trivial bundle over $[0,1]\times[0,1]$ by the identity map in the first coordinate and a reflection in the second coordinate. The torus admits the structure of a smooth complex projective variety, but $E$ isn't orientable, so cannot be diffeomorphic to a smooth complex variety.<|endoftext|> TITLE: The Origin(s) of Modular and Moduli QUESTION [41 upvotes]: In mathematics and in physics, people use the terms "modular..." and "moduli space" very often. I was puzzled by the etymology, the origins and the similarity/equivalence/differences for these usages/concepts behind for some time. Could I seek for expert's explanations on the contrast/comparison of the "modular..." and "moduli space" in the following contexts: "Modular" in the modular tensor category. (Also the "pre-modular" category.) "Modular" in the modular form. "Modular" in the topological modular form (tmf). "Modular" in the modular invariance (e.g. in Conformal Field Theory). "Moduli space" in algebraic geometry "Moduli space" in Non-supersymetric Quantum Field Theories or gauge theories "Moduli space" in Supersymetric gauge theories "Module" in the "mod" or modular arithmetic. "Modular" in modular representation theory of (finite) groups/algebraic groups. [As André Henriques suggested] "Module": $G$-module in the group (co)homology and topological $G$-module for the topological group. [Likely related to 9.] "Modulus" of convergence and "Modulus" of continuity in analysis. Thank you in advance for solving/illuminating this rather puzzling issue. REPLY [7 votes]: I was similarly curious when writing the introduction of my PhD thesis, since the context was moduli spaces of Abelian differentials. I felt the need to dig a bit and include a discussion there. Here is an English translation of what I wrote then. The original is here (see page 9 of the introduction). 1. Moduli spaces The term "moduli space" is often preferred to "parameter space" in contexts where one is interested in describing geometric objects up to a certain equivalence. For instance, we consider the modulus of a cylinder (ratio of its height to its circumference) when we are interested in its shape without caring about its size. The moduli space of cylinders is the set of positive reals. A more instructive example is the moduli space of tori. 1.1. Tori. Here again, we seek to describe the shape of a torus (of real dimension 2, endowed with a complex structure) without taking its size into account. A torus can be defined as a quotient $\mathbb{C}/\Lambda$ where $\Lambda$ is a lattice in $\mathbf{C}$. Consider tori equivalent when they correspond to lattices obtained from each other by rotation and dilatation. Given a torus, we can then consider it as a quotient $\mathbb{C}/\Lambda$ where $\Lambda$ is a lattice in $\mathbb{C}$ with basis $(1, \tau)$. Two parameters $\tau$ and $\tau'$ define equivalent tori if their difference is an integer, or if they are opposite or inverse of each other. We can therefore assume that $\lvert \operatorname{Re} \tau \rvert < 1/2$, $\operatorname{Im} \tau > 0$, and $\lvert \tau \rvert > 1$. This draws a domain in the upper half-plane, located between the vertical lines at x-coordinate $-1/2$ and $1/2$, and outside the circle of radius $1$ centered at the origin. Some points in this domain still need to be identified: vertical half-lines at x-coordinate $-1/2$ and $1/2$ by horizontal translation, and the two halves of the arc of circle which bounds the domain below by the map $z \mapsto -1/z$. The identifications of these parts of the boundary of the domain can be described globally: they are obtained by reflection about its (vertical) symmetry axis. We call moduli space of tori the space obtained after performing these identifications. Its topology is that of a sphere minus a point. Its geometry is richer, and is inherited from the hyperbolic metric on the upper half-plane. The points corresponding to $i$ and to $i 2 \pi / 3$ represent the square torus and the hexagonal torus, which respectively admit automorphisms of order 4 and 6. They are cone points of respective angles $\pi$ and $2\pi/3$ in the moduli space of tori. This very classical moduli space is called the modular surface or the modular curve, depending on whether one chooses to adopt the real or complex viewpoint. Because of these cone points, it is not quite a manifold; we call it an orbifold. I stopped the discussion there but wish I had expanded a bit on how Riemann found that it takes "more moduli" to describe the geometry of closed compact Riemann surfaces of higher genus, and concluded it took "$3g-3$ complex moduli" for genus $g > 1$. Or if one prefers, the "modulus" of a genus $g$ closed compact Riemann surface, for $g > 2$, is no longer a real number or a complex number or a point in a quotient of the upper half-plane, but a point in a $(3g-3)$-dimensional complex orbifold...<|endoftext|> TITLE: Quasi-symmetric generalizations of classical symmetric functions QUESTION [5 upvotes]: I am looking for quasi-symmetric versions of the classical $e_\lambda$ and $h_\lambda$ (the elementary and complete homogeneous symmetric functions). Is there some reference for this? I am aware of at least four quasi-symmetric generalizations of Schur polynomials: Extended Schur (Assaf, Searles) Quasi-symmetric Schur (Haglund et al) Young quasi-symmetric Schur Dual Immaculate functions There are also two quasi-symmetric power-sums which I know of. I have defined a non-symmetric extension of the elementary symmetric functions, (as a certain specialization of non-symmetric Macdonald polynomials), and this can perhaps be turned into some quasi-symmetric version. However, I have not yet managed to do this. REPLY [2 votes]: I sent an email to Stephanie van Willigenburg, and she suggested the following reference, which contains such generalizations.<|endoftext|> TITLE: Existence of subset of reals such that any real number is unique sum of exactly two elements of the subset QUESTION [11 upvotes]: It is easy to see (using AC, of course) that there exist two sets $U\subset\mathbb{R}$ and $V\subset\mathbb{R}$ such that any real number $x$ can be represented as unique sum $x=u+v$, where $u\in U$ and $v\in V$. There are $2^{2^\omega}$ such $(U,V)$ pairs. The question given by my son: prove existence of pair $(U,V)$ with $U=V$. In other words, prove the existence of $U\subset\mathbb{R}$ such that for any $x\in\mathbb{R}$ there exists unique pair $\{u,v\}$ with $x=u+v$ REPLY [9 votes]: The usual transfinite construction works. Let $\{r_{\alpha} : \alpha < \mathfrak{c}\}$ list $\mathbb{R}$. Construct $\{U_{\alpha}: \alpha < \mathfrak{c}\}$ by induction on $\alpha$ such that the following hold. (a) For every $a, b, c, d \in U_{\alpha}$, $a + b = c + d \implies \{a, b\} = \{c, d\}$. (b) There are $a, b \in U_{\alpha + 1}$ such that $r_{\alpha} = a + b$. (c) For limit $\alpha$, $U_{\alpha} = \bigcup_{\beta < \alpha} U_{\beta}$. Then $U = \bigcup_{\alpha < \mathfrak{c}} U_{\alpha}$ is as required. Requirements (a), (c) are trivially satisfied. To ensure (b), at stage $\alpha + 1$, if $r_{\alpha} \notin U + U$, choose $x$ outside the $\mathbb{Q}$-linear span of $U_{\alpha} \cup \{r_{\alpha}\}$ and put $U_{\alpha+1} = U_{\alpha} \cup \{x, r_{\alpha} - x\}$ and note that this does not violate (a).<|endoftext|> TITLE: About the Fourier transform of the logarithm function QUESTION [8 upvotes]: I want to calculate / simplify: $$\mathcal{F} (\ln(|x|)\mathcal{F(f)}(x))=\mathcal{F} (\ln(|x|)) \star f$$ where $\mathcal{F}$ is the Fourier transform ($\mathcal[f](\xi)=\int_{\mathbb R}f(x)e^{ix\xi}\,dx$) and where $f$ is an even function. Looking here: wiki, we find that $$\mathcal{F}[\log|x|](\xi)=-2\pi\gamma\delta(\xi)-\frac\pi{|\xi|},$$ so we should have: $$\mathcal{F} (\ln(|x|)) \star f = (-2\pi\gamma\delta(x)-\frac\pi{|x|}) \star f(x) $$ $$ = -2\pi\gamma f(x)- \pi \int_{-\infty}^{\infty} \frac{f(t)}{|x-t|} dt $$ but the integral of the second term does not converge... whereas the term $\mathcal{F} (\ln(x)\mathcal{F(f)}(x))$ is well defined providing the function $f$ is of rapide decrease near zero and infinity. So where is the problem ? and what is finally the "simplified expression" of $\mathcal{F} (\ln(x)\mathcal{F(f)}(x))$ ? We cannot use this distribution in a convolution product with a function? I already post this on Stackexchange but did not receive an answer. REPLY [4 votes]: I thought that it might be instructive to present an approach to deriving the Fourier transform of $\log(|x|)$. The result includes a distributional interpretation of $\frac1{|x|}$. Finally, we show that the distributional interpretation of $\frac1{|x|}$ is non-unique and that it differs from other interpretations by a multiple of the Dirac Delta distribution. With that introduction, we now proceed. PRELIMARIES Let $\psi(x)=\log(|x|)$ and let $\Psi$ denote its Fourier Transform . Then, we write $$\Psi(x)=\mathscr{F}\{\psi\}(x)\tag 1$$ where $(1)$ is interpreted as a Tempered Distribution. That is, for any $\phi \in \mathbb{S}$, we can write $$\langle \mathscr{F}\{\psi\}, \phi\rangle =\langle \psi, \mathscr{F}\{\phi\}\rangle$$ Now, let $\psi_\epsilon(k) =e^{-\varepsilon|k|}\log(|k|)$. Therefore, $\psi(k)=\lim_{\varepsilon\to 0^+}\psi_\varepsilon(k)$ and we see that $$\begin{align} \lim_{\varepsilon\to 0^+}\langle \mathscr{F}\{\psi_\varepsilon\}, \phi\rangle&=\lim_{\varepsilon\to 0^+}\langle \psi_\varepsilon, \mathscr{F}\{\phi\}\rangle \\\\ &=\langle \psi,\mathscr{F}\{\phi\}\rangle\\\\ &=\langle \mathscr{F}\{\psi\}, \phi\rangle \end{align}$$ Next, we evaluate the Fourier transform of $\psi_\varepsilon$. EVALUATING THE FOURIER TRANSFORM OF $\displaystyle \psi_\varepsilon$ Denote by $\Psi_\epsilon$, the Fourier transform of $\psi_\varepsilon$. Then, we have $$\begin{align} \Psi_\varepsilon(x)&=\mathscr{F}\{\psi_\epsilon\}(x)\\\\ &=\int_{-\infty}^\infty e^{-\varepsilon|k|}\log(|k|) e^{ikx}\,dk\\\\ &=2\text{Re}\left(\int_0^\infty e^{-(\varepsilon -ix)k}\log(k) \,dk\right)\\\\ &=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma -\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\\\\ &=\psi^{(1)}_\varepsilon(x)+\psi^{(2)}_\varepsilon(x)+\psi^{(3)}_\varepsilon(x)\tag2 \end{align}$$ where $$\begin{align} \psi^{(1)}_\varepsilon(x)&=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma\\\\ \psi^{(2)}_\varepsilon(x)&=-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)\\\\ \psi^{(3)}_\varepsilon(x)&=-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \end{align}$$ Next, we will find the distributional limits of $\psi^{(1)}_\varepsilon$, $\psi^{(2)}_\varepsilon$, and $\psi^{(3)}_\varepsilon$. DISTRIBUTIONAL LIMITS OF $\displaystyle \psi^{(1)}_\varepsilon$, $\displaystyle \psi^{(2)}_\varepsilon$, and $\displaystyle \psi^{(3)}_\varepsilon$ Again, let $\phi\in \mathbb{S}$. Then, $$\begin{align} \lim_{\varepsilon\to 0^+}\langle \psi^{(1)}_\varepsilon,\phi \rangle &=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \psi^{(1)}_\varepsilon(x)\phi(x)\,dx\\\\ &=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \left(-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma \right)\phi(x)\,dx\\\\ &=-2\gamma\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx\\\\ &=-2\pi \gamma \phi(0)\tag3 \end{align}$$ $$\begin{align} \langle \psi^{(2)}_\varepsilon,\phi \rangle &=\int_{-\infty}^\infty \left(-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2) \right)\phi(x)\,dx\\\\ &=-2\log(\varepsilon)\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx-\int_{-\infty}^\infty \frac{\log(1+x^2)}{1+x^2}\phi(\varepsilon x)\,dx\\\\ &= -2\pi \log(\varepsilon)\phi(0)-2\pi \log(2) \phi(0)+o(\varepsilon) \end{align}\tag4$$ $$\begin{align} \langle \psi^{(3)}_\varepsilon,\phi \rangle &=\int_{-\infty}^\infty \left(-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\right)\phi(x)\,dx\\\\ &-\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx-\int_{|x|\ge 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx\\\\ &=-\phi(0)\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \,dx\\\\ &-\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) (\phi(x)-\phi(0))\,dx-\int_{|x|\ge 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx\\\\ &= \left(2\pi \log(\varepsilon) +2\pi \log(2)\right)\phi(0)+o(\varepsilon)\\\\ &-\pi \int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx-\pi \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx\tag5 \end{align}$$ FINAL RESULTS Substituting $(3)$, $(4)$, and $(5)$ into $(2)$, we find that $$\begin{align} \lim_{\varepsilon\to 0^+}\langle \mathscr{F}\{\psi_\varepsilon\},\phi\rangle =-2\pi \gamma \phi(0)-\pi \int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx-\pi\int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx\\\\ \end{align}$$ from which we assert that in distribution $$\mathscr{F}\{\psi\}(x)=-2\pi \gamma \delta(x)-\pi \text{PV}\left(\frac1{|x|}\right)$$ where we interpret $\text{PV}\left(\frac1{|x|}\right)$ to mean that for any $\phi\in \mathbb{S}$, $$\int_{-\infty}^\infty \phi(x) \text{PV}\left(\frac1{|x|}\right)\,dx=\int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx$$ NOTE: It was arbitrary to split the integration in $(5)$ into inervals $|x|\le 1$ and $|x|\ge 1$. Had we chosen instead the intervals $|x|\le \nu$ and $|x|\ge \nu$ for any $\nu>0$, we would have obtained $$\mathscr{F}\{\psi\}(x)=-2\pi (\gamma+\log(\nu)) \delta(x)-\pi \text{PV}\left(\frac1{|x|}\right)$$ where we interpret $\text{PV}\left(\frac1{|x|}\right)$ to mean that for any $\phi\in \mathbb{S}$, $$\int_{-\infty}^\infty \phi(x) \text{PV}\left(\frac1{|x|}\right)\,dx=\int_{|x|\le \nu}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge \nu}\frac{\phi(x)}{|x|}\,dx$$<|endoftext|> TITLE: Does foundation/regularity have any categorical/structural consequences, in ZF? QUESTION [14 upvotes]: (Prompted by reflection on this old answer, and its suggestion of the “harmlessness” of the axiom of regularity.) In ZFC, one may justify the axiom of foundation (AF, aka the axiom of regularity) as being “convenient, and harmless”, as follows. If $V = (V,\epsilon)$ is a model of ZFC–AF, then its well-founded part $V_{wf}$ models ZFC. Moreover, ZFC–AF suffices for the standard proof of the well-ordering principle, so $V$ believes that every set is isomorphic to some von Neumann ordinal, which will lie in $V_{wf}$. Hence the inclusion $V_{wf} \hookrightarrow V$ underlies an equivalence of their (internal, large) categories of sets; so any “purely structural” statement should hold in $V$ if and only if it holds in $V_{wf}$. “Purely structural” can be made precise in various reasonable ways (e.g. any statement expressible in the language of categories/elementary toposes/higher-order logic/type theory, interpreted into set theory), and includes essentially all of ordinary mathematics, excluding only a few explicitly material-set-theoretic statements such as AF itself. Summing up: given any model of ZFC–AF, one can replace it by its well-founded part, which is now a model of ZFC, believing the same purely structural statements as the original model. So relative to ZFC–AF, AF has no purely structural consequences — i.e. it’s not quite conservative, but pretty close to it. However, this argument relied essentially on the axiom of choice! Relative to ZF–AF, is AF still harmless, or does it have some structural consequences? Precisely: is there some “purely structural” statement $\varphi$ (in one of the senses above, or some similar sense), such that $ZF \vdash \varphi$, but $ZF-AF \not \vdash \varphi$? I’m also interested in the same question with ZF weakened to IZF, CZF, and similar theories. Over these of course AF should be replaced by $\in$-induction (a classically equivalent and constructively better statement), as described in this older question (which considers closely related issues to the present question, but doesn’t as far as I can see directly imply an answer here). REPLY [8 votes]: The following provides an "engine" for generating many structural statements provable in ZF but not in ZF without foundation: Theorem. Let $S$ be any structural equivalent of the axiom of choice within ZF without foundation (e.g., Zorn's Lemma or Tychonoff's compactness theorem). Then the following structural statement $R_{S}$ (in honor of Herman and Jean Rubin) is provable in ZF; but NOT in ZF without foundation: $R_{S}$ := "$S$ holds iff every linearly orderable set is well-orderdable". Explanation. The nontrivial direction ($\leftarrow$) of $R_{S}$ follows from the theorem below, and the observation that the powerset of every well-orderable set $S$ is linearly orderable (by deeming $X$ to be less than $Y$, where $X$ and $Y$ are subsets of $S$, and $S$ well-ordered by $<$, if the $<$-first element in the symmetric difference of $X$ and $Y$ is in $Y$. The proof is readily implementable in Zermelo set theory, i.e., ZF without the foundation axiom and without the replacement cheme, but with the separation scheme). Theorem (H. Rubin & J. Rubin). ZF proves that if the powerset of every well-orderable set is well-orderable, then the axiom of choice holds. The above theorem was first published the 1963 book Equivalents of the Axiom of Choice, by Herman and Jean Rubin. A proof can also be found in these lecture notes by Andrés E. Caicedo (see Theorem 13). Later, in this 1973 paper of Felgner & Jech, a model of ZF with atoms was built in which the powerset of every ordinal is well-orderable, but the axiom of choice fails; this proof also appears in chapter 9 of the book Axiom of Choice by Jech.<|endoftext|> TITLE: Homeomorphism/ homotopy types of non-negatively curved manifolds QUESTION [8 upvotes]: A (special case of a) theorem of Gromov says for any $n\in \mathbb{N}$ there exists a constant $C(n)$ such that for any smooth connected closed $n$-dimensional Riemannian manifold with non-negative sectional curvature the sum of all of its Betti numbers is at most $C(n)$. On the other hand, for $n=2$ only sphere, torus, real projective plane, and (possibly- I do not know) connected sum of 2 real projective planes, can carry a non-negatively curved metric. (This follows from the Gauss-Bonnet theorem and topological classification of surfaces with non-negative Euler characteristic.) What happens when $n>2$? For which $n>2$ it is known that there are infinitely many homeomorphism/ homotopy types of $n$-manifolds which can carry a Riemannian metric with non-negative sectional curvature? REPLY [14 votes]: As mentioned in comments, the first dimension where an infinite family of pairwise non-homeomorphic closed nonnegatively curved manifolds occurs is $3$ (the lens spaces). The question becomes more challenging for simply-connected manifolds. If (as expected) simply-connected nonnegatively curved manifolds are rationally elliptic, then the first infinite family occurs in dimension 6, see Curvature, diameter, and quotient manifolds by B. Totaro. In general, the question is well-studied and there are many examples (easily found by searching online).<|endoftext|> TITLE: Variant of mutual information QUESTION [9 upvotes]: Given a discrete random variable $(X,Y)$, one can consider the smallest entropy of a random variable $Z$ such that $X$ and $Y$ are independent conditioned to $Z$. This quantity is akin to the mutual information, in fact one can see that it is always larger than half the mutual information. Does this number have a name, and has it been studied? REPLY [5 votes]: This is related to Wyner's common information which is always larger than the mutual information. https://ieeexplore.ieee.org/document/1055346/<|endoftext|> TITLE: Linear permutations commuting with $x\rightarrow x^{-1}$ QUESTION [7 upvotes]: Let $F = \operatorname{GF}(2^n)$ be a finite field. Define a permutation $\phi:F \rightarrow F$ by the formula $$ \phi(x) = x^{-1}, \ x\neq 0; \ \phi(0) =0. $$ We say that a permutations $\psi$ of $F$ is linear if $\psi(x+y) = \psi(x)+\psi(y)$. It is not hard to see that all automorphisms $\sigma_k: x\rightarrow x^{2^k}$ are commuting with $\phi$. That is in the group $S(F)$ of permutations of $F$ we have the following equality: $$ \phi\cdot \sigma_k = \sigma_k\cdot\phi. $$ My question. Are there other linear permutations of $F$, commuting with $\phi$? REPLY [16 votes]: The answer is no: a linear transformation of $F$ which commutes with $\phi$ is an automorphism of $F$. This is a seemingly inelegant but simple argument. Any map $\psi: F \rightarrow F$ can be represented uniquely as a polynomial $P_\psi (x) = \sum a_i x^i$ of degree less than $2^n$, with each $a_i \in F$. If the map $\psi$ is linear, then in fact $a_i =0$ for all $i\neq 2^j$, so that the degree of $P_\psi$ is in fact at most $2^{n-1}$. Indeed, there are $2^{n^2}$ linear transformations of $F$ and the same number of polynomials of the form $\sum_{0 \le j \le n - 1}b_jx^{2^j}$, all of which induce linear transformations. Let $Q_\psi(x) = x^{2^{n-1}} P_\psi(\frac{1}{x})$ as polynomials; in other words, $Q_\psi$ is the "reverse" of $P_{\psi}$. Note that the degree of $Q_{\psi}$ is also at most $2^{n-1}-1$, because the coefficient of $1$ in $P_\psi$ is $0$. Your condition is equivalent to the equation $(P_\psi(x))^{-1} = P_\psi(x^{-1})$ in value for any $x \neq 0$. Multiplying it out so everything is a polynomial, it's equivalent to $P_\psi(x)Q_\psi(x) = x^{2^{n-1}}$ in value for all $x$ including $0$. But two polynomials are equal in value iff they are equivalent modulo $x^{2^n}-x$. Both of these polynomials are of smaller degree than $2^n$, so they must in fact be equal. Unique factorization therefore says that $P_{\psi}$ must be a monomial, with coefficient $a$ such that $a^2=1$. As we are in characteristic $2$, this is unique: $a=1$. I wouldn't be surprised if there were a more elegant answer based on automorphisms, but I don't see it immediately. I figured out an answer based on multiplication-preserving endomorphisms that works generally in any characteristic other than $2$, and for any finite field (or any field in which every element is a square) in characteristic $2$. The commutation equation says that for any $x$, we have $\psi(x^{-1}) = \psi(x)^{-1}$. Assume $x, y, x + y \neq 0$. Let's apply it to $x^{-1} + y^{-1}$: $\psi((x^{-1} + y^{-1})^{-1}) = (\psi(x^{-1} + y^{-1}))^{-1}$ $\psi(\frac{xy}{x + y}) = (\psi(x^{-1}) + \psi(y^{-1}))^{-1}$ (using linearity) $\psi(\frac{xy}{x + y}) = (\psi(x)^{-1} + \psi(y)^{-1})^{-1} = \frac{\psi(x) \psi(y)}{\psi(x) + \psi(y)}$ Multiplying it out, we get: $\psi(\frac{xy}{x + y}) (\psi(x + y)) = \psi(x) \psi(y)$ Choose $y = y' - x$. Our conditions become $x, y', y'-x \neq 0$. Then $\psi(\frac{x (y' - x)}{y'}) \psi(y') = \psi(x) \psi(y' - x)$ Using linearity and canceling terms, we get: $\psi(\frac{x^2}{y'})\psi(y') = \psi(x)^2$ From here, this separates into two proofs: one for characteristic $2$ fields, and one for any field not of characteristic $2$. For characteristic $2$ fields: $\psi(z z')^2 \psi(1)^2 = \psi(z^2) \psi(1) \psi(z'^2) \psi(1) = \psi(z)^2 \psi(z')^2$ for any $z, z'$. In fields of characteristic $2$, square roots are unique, if they exist. Therefore: $\psi(z z') \psi(1) = \psi(z) \psi(z')$ For characteristic not $2$: $\psi(x^2)\psi(1) = \psi(x)^2$ for any $x \neq 0, 1$. However, for both of those cases, this is obvious, so this is true for any $x$. Let $x = z + z'$. Then: $\psi((z + z')^2) \psi(1) = \psi(z + z') \psi(z + z')$. We can again cancel terms and divide by $2$ (except in characteristic $2$) to get: $\psi(z z') \psi(1) = \psi(z) \psi(z')$. Here, the proofs reunite. In other words, $\psi$ must be a multiplication-preserving morphism up to scaling by $\psi(1)$. But $\psi(1) = \psi(1^{-1}) = \psi(1)^{-1}$, so $\psi(1) = \pm 1$. In characteristic $2$, these are the same, and the only answers are multiplication-preserving morphisms; more generally, negation (and composing negation with a multiplication-preserving morphism) is also possible.<|endoftext|> TITLE: Well known matrix inequality? QUESTION [9 upvotes]: I suspect that the following matrix inequality is well known, but I can't find a reference or proof: Given $n \times n$ symmetric matrices $A,B$ such that $I_n \leq A,B$, is the following true? $${Tr}\big[\big( \log A^{\frac{1}{2}} B A^{\frac{1}{2}}\big)^p\big] \geq {Tr}\big[\big( \log B \big)^p\big] + {Tr}\big[\big( \log A\big)^p\big].$$ For clarity, for a continuous function $f$ and symmetric matrix $C$, $f(C)$ is simply equal to $U {diag}(f(\lambda_1),...,f(\lambda_n))U^t$, where $U$ is a unitary matrix diagonalizing $C$, i.e., $C = U diag(\lambda_1,...,\lambda_n)U^t$. UPDATE: after the discussion with amakelov below, I edited the statement to be more aesthetic. REPLY [7 votes]: The desired inequality follows from an majorisation argument : According to Bhatia Matrix Analysis Corollary III.4.6 we have : $log \lambda(AB) \succ log \lambda^\downarrow(A) + log \lambda^\uparrow(B)$. Using Corollary II.3.4 from the same book, we conclude that $(log \lambda(AB))^p \succ_w (log \lambda^\downarrow(A) + log \lambda^\uparrow(B))^p \succ_w (log \lambda^\downarrow(A))^p + (log \lambda^\uparrow(B))^p$ , since $x \mapsto x^p$ is convex and monoton increasing for $x \ge 0$ . From this it follows the trace inequality.<|endoftext|> TITLE: Has there been any new development on the Freudenthal Problem? QUESTION [25 upvotes]: Background I have seen a few variants of this Sum-and-Product puzzle in the past. The premise of these puzzles is as follows Sam hears the sum of two numbers, Polly the product. The numbers are known to be between m and M. S: "You don't know the numbers" P: "That was true, but now I do" S: "Now I do too" What are the numbers? A set of papers from 2006 refer to this as the Freudenthal Problem Freudenthal(m,M). I have been specifically interested in classifying solutions when $m=3$, and $M$ is unbounded. Assuming a modified form of Goldbach's conjecture, the authors prove that whether the numbers are known to be distinct or not does not change the solutions when $m=2$ and $m=3$, so I have removed their superscript distinguishing the case. The authors also give a rather naive algorithm that enumerates solutions ordered by sum. They generate solutions for Freudenthal(3,*) up to a sum of 50,000, and find that there are 804 stable solutions, and 288 phantom solutions that rely on the presence of an upper bound. My own findings I wrote a program that very efficiently generates solutions, also ordered by sum, and improved the highest sum by an order of magnitude overnight. I then defined the "Freudenthal Partition Function" $F(s)$. The domain of this function is any sum $s$ which allows the first statement by Sam. The value $F(s)$ is the number of options remaining in Sam's mind after Polly's statement. There is a lot to unpack in this function, so I will go into detail. Domain of $F$ With $m=3$, the products which allow Polly to immediately deduce the numbers are any of the following: Odd semiprimes 4 times an odd prime (numbers must be 4 and p) 2 times the square of an odd prime (numbers must be p and 2p) For Sam to make his first claim, the number cannot be the sum of two odd primes, sum of a prime and 4, or 3p for some p. What remains is the set $S$, which is the domain of $F$. Assume the Goldbach conjecture so that all even numbers are eliminated from $S$ outright. Then $s\in S$ iff $s$ is odd, $s-4$ is composite, and $s\ne 3p$ for some prime $p$. This set can be efficiently generated. Computing $F(s)$ Start with a list of all even integers. Iterate over all pairs $\{(s,a)\mid s \in S, a \in \mathbb{Z}, 3 \le a \le s/2\}$ and place a tally next to $a(s-a)$. For computability's sake, proceed lexicographically by $s$ then $a$. After this procedure passes $s_0$, all tally counts on integers up to $3s_0-9$ are stable. Define the set $P$ to be all the integers with exactly one tally. $p \in P$ has the following properties: $p$ is not uniquely factorable into two divisors in the set $\{z\mid z \in \mathbb{Z}, z\ge 3\}$ (guaranteed since they were reached from $S$). $p$ has a unique divisor pair $\{d,p/d\}$ such that $d + p/d \in S$. Equivalently, if Polly was told the number $p$, then the first two lines of the puzzle are satisfiable. To compute $F(s)$ given $P$, iterate over $\{a \mid 3\le a\le s/2\}$ and count the number of $a$ for which $a(s-a) \in P$. Using $F(s)$ If $F(s)=1$, then Sam, upon hearing Polly's statement, can deduce the numbers. This corresponds to a solution to the puzzle. Other claims I will make about the behavior of $F(s)$ are just conjectures. If $F(s)=0$, then there seems to always exists some upper bound $M$ such that an integer which had two tallies loses one, and $P$ gets a new element which causes the puzzle to be satisfiable. These correspond identically to what the authors of the 2006 paper called "phantom solutions". Indeed, for $s<50,000$, there are 804 instances where $F(s)=1$, and 288 instances where $F(s)=0$. A graph of $F(s)$ is found here for $s$ up to around 260,000. The function has three very distinct branches, corresponding to congruence classes mod 3. The bottom branch, which produces all known solutions, has $s\equiv 2 \pmod{3}$. The solution pairs themselves A list of the 3141 smallest pairs ordered by sum can be found here. I list the even number, then the odd number. The same pairs ordered by even number can be found here, and is more illuminating. A notable oddity is $(4, 137233)$, the only currently-known pair to include the value 4. This was missed entirely by the original papers. The pairs seem to all be of the form $(4^np,q)$, where $n>0$, $p$ is either 1 or prime, and $q$ is either prime or the product of a small number of primes. All prime factors of $p$ and $q$ are congruent to $1 \pmod{3}$. Non-solutions and what we can learn from them For every $s$, there are $F(s)$ pairs that can be considered "candidate solutions." If $F(s)>1$, they are not solutions, but in the lowest branch they share many properties of solutions. For example, $F(53) = 2$, and the contributing pairs are $(4,49)$ and $(16,37)$. This could have allowed one to see that 4 is viable as one of the two numbers, even without finding the first example of $(4, 137233)$. Looking at all candidate solutions for $s\equiv 2 \pmod{3}$, we find a very small set of exceptions to the $(4^np,q)$ trend. The exceptional pairs with sum less than 500,000 are $(32,27)$ $(128,9)$ $(512,3)$ $(2048,3)$ $(32768,3)$ $(32768,27)$ All are $(2^k,3^l)$ for $k$ odd and $l$ small. They all have $F(s)>1$ for their sum, so they are not solutions, but they lie in the bottom branch of the partition function and allow the puzzle to proceed through its first two statements. I checked (inefficiently) all pairs $(2^k,3^l)$ for odd $k\le 31$ and $l\le 10$, and found 5 other pairs which allow the first two statements to be satisfied, but in every case $F(s)>1$. It remains open whether there exists a true solution with this form. Edit: I have a program running to find pairs $(2^k,3^l)$ for which the first two statements can be satisfied, and for each pair, search for a witness to $F(s)>1$. For $k\le 219$ and $m\le 25$, there are 13 pairs with no witness of the form $(4^n,q)$. The much more computationally intensive task of confirming absence of a witness of the form $(4^n p,q)$ is likely intractable if the pair is a solution. Witnesses have been found for the smallest 10 pairs, the first pair with no known witness yet is $(2^{187}, 3)$. Results for $m\ge 4$ I used the same program to examine other values of the lower bound, with what I believe were the appropriate changes and assumptions. The header comment can help others verify the correctness. I did not find any solutions, in agreement with the conjecture put forth in in the papers. Graphs of the corresponding Freudenthal Partition Functions diverge from 0, leaving no branch which could be believed to provide solutions. So, my question to mathoverflow is: Has there been anyone else thinking about this problem, and do they have results that supplement/contradict/dwarf those I have gathered? Asymptotic behavior of $F(s)$ I did some asymptotic analysis of $F(s)$. For a given $s$, I determined all the products with at most 3 odd factor sums such that one of those sums is $s$, and what numbers on the order of $s$ must be prime for $F(s)$ to increase by 1. This gives remarkably good fits for the three congruence classes. To reduce noise, I show here a moving average of $F(s)$, averaging the nearest hundred values in the same branch. Notably, those $s \equiv 2 \pmod{3}$ have no growth at order $s/\log(s)^3$, always requiring an additional number to be prime and resulting in asymptotic growth at order $s/\log(s)^4$. This keeps $F(s)$ very near zero for those values we have seen, but means that $F(s)$ still grows without bound for these $s$. This has potential to imply that there are only finitely many solutions, although such a claim is still far off. I don't know how best to determine $\liminf\limits_{s \to \infty} F(s)$. Arithmetic progressions $s_a$ with $F(s_a)$ exceptionally low If one rephrases the argument which gave the asymptotic formula in terms of the probability $F(s)=1$, they predict that there are only finitely many solutions, and that over 95% have been seen by the point $s=100,000$. We know at least the latter to be false, so there are systematic failures of the argument. The leading order contribution to the lowest branch of $F(s)$ counts the instances where three things hold: $s=4^kp+q$ for primes $p$ and $q$ with $k\ge 2$ $s^\prime=4^k+pq \not\in S$, that is, $4^k+pq-4$ is prime. $s^{\prime\prime}=4^kq+p \not\in S$, that is, $4^kq+p-4$ is prime. Note that $s^{\prime}-s=(4^k-p)(q-1)$ and $s^{\prime\prime}-s=(4^k-1)(q-p)$. The factor $4^k-1$ is independent of $p$ and $q$. When $s-4$ shares a factor with $4^k-1$, $s^{\prime\prime}$ is not prime for any $p,q$. Since $s\equiv 2\pmod{3}$ is necessary, $s-4$ cannot be a multiple of 3. If we define $k_m(s)$ to be the smallest $k\ge 2$ such that $(4^k-1,s-4)=1$, then we can refine the first bullet point above: $s=4^kp+q$ for primes $p$ and $q$ with $k\ge k_m(s)$ We then refine our prediction about the asymptotic behavior of the low branch of $F(s)$: $F(s) \sim s\log(s)^{-4}4^{-k_m(s)}$ We may define $g(k)$ to be the smallest value not divisible by 3 and sharing a factor with $4^n-1$ for all $n\le k$, so $k_m(g(k))=k+1$, then $g(2)=5$, $g(3)=35$, $g(5)=385$, etc. Each defines an arithmetic progression with exceptionally low values of $F(s)$. Specifically, $s_{a,k}=\left\{\begin{array}{11} (6a+1)g(k)+4 & g(k)\equiv 1\pmod{6}\\ (6a+5)g(k)+4 & g(k)\equiv 5\pmod{6} \end{array} \right.$ Determining whether $F(s)$ takes the value 1 often, we must know the growth pattern of $g(k)$. $F(s_{0,k})\sim g(k)\log(g(k))^{-4}4^{-k}$ This appears to grow unbounded, though erratically. It returns to values near 1 at $k=30$ and $k=60$ before jumping upward. I conjecture that $\liminf\limits_{s \to \infty} F(s) = \liminf\limits_{k \to \infty} F(s_{0,k})$. I have checked the value of $F(s_{a,30})$ for $a \le 500$ and found 4 probable zeros and 7 probable solutions, the largest of which is $(4^{31}*459703, s_{34,30}-4^{31}*459703)$. This might be the largest solution. (Update: I did finish the check of the less-probable cases, completing a proof that this pair is a solution. The proof is found here). To determine $g(k)$, we must make a statement about the smallest prime factor of $(4^k-1)/3$. If $k$ is composite, then the smallest prime factor of $(4^k-1)/3$ is the smallest prime factor of $(4^d-1)/3$ for some divisor $d$ of $k$. This factor is present in $g(q)$ for all $q\ge d$, so if $k$ is composite we must have $g(k)=g(k-1)$. If $k$ and $2k+1$ are both prime, then given the recurrence $r_0=0$, $r_{n+1}=4r_n+3 \pmod{2k+1}$, we have $r_k=0$, and so $4^k-1$ is divisible by $2k+1$. If $k$ is prime and there is not a value of the form $ck+1$ for which the recurrence relation above holds, then the smallest prime factor of $(4^k-1)/3$ might be $(2^k+1)/3$. This happens for $k=7,13,17,19,31,61,...$. At the moment I don't have a full understanding of why primality of both $k$ and $6k+1$ is able to predict a factor of $6k+1$ in $4^k-1$ when $k=37$, but not when $k=17$. The jumps upward in $g(k)$ when it must now include a factor of $(2^k+1)/3$ have increasing magnitude and seem to indicate unbounded growth of the expression $g(k)\log(g(k))^{-4}4^{-k}$, although proving this is beyond my ability. Unbounded growth of this expression would imply with high certainty that there are finitely many solutions to the Freudenthal problem Freudenthal(3,*). Is there an approach to proving this? REPLY [2 votes]: I have a new result that might be a step towards proving there are infinitely many solutions. We show that there is an infinite sequence of values for which $F(s)$ is at most 1. Consider $s_k = 2^{2^k} - 5$ for $k \ge 3$. We have stated that a number is in $S$ if it is not $4 + p$ for prime $p$, and is not $3q$ for prime $q$. We also stated that a product is in $P$ iff exactly one divisor pair has sum in $S$. We will show that $s_k \in S$ and at most one of the products $l (s_k - l)$ is in $P$. $s_k$ is not a multiple of 3, so the latter condition for inclusion in $S$ is always satisfied. Subtract 4 to get $2^{2^k} - 9$. This number is not prime. It is divisible by 7 is $k$ is even, and 13 if $k$ is odd. This can be proven by induction. Therefore $s_k \in S$ for all $k \ge 3$. Consider ways to sum to $s_k$. Since $s_k$ is odd, all pairs of summands must consist of one even and one odd number. Let the even number be $a'$, the odd number $b$. Let $v_p(x)$ be the p-adic valuation of $x$. Define $m = v_2(a')$ and $a = 2^{-m} a'$. We know $m \in [1, 2^k)$ and $a$ is odd. We can write $s_k = 2^m a + b$. If $a \ge 3$, write $s'_k = 2^m b + a$. Add $s_k$ and subtract it: $s'_k = 2^m b + a + (2^m a + b) - (2^{2^k} - 5)$ Factor and subtract 4: $s'_k - 4 = (2^m + 1)(a + b) - (2^{2^k} - 1)$ Since $m < 2^k$, we know $2^m + 1$ shares a nontrivial factor with $2^{2^k} - 1$. This factor divides $s'_k - 4$, and thus $s'_k - 4$ is composite. Furthermore, $s'_k$ is congruent mod 3 to $(-1)^m s_k$, so it is not a multiple of 3. $s'_k \in S$, so the product $2^m a b$ with $a \ge 3$ has at least two factorizations with sum in $S$, and cannot be in $P$. If $a = 1$, we have $b = 2^{2^k} - 5 - 2^m$. For $m \ge 3$, we can examine $b$ modulo $y = 2^{2^{v_2(m-2)}} + 1$. Write $2^m$ as $4 * 2^{m-2}$, and then write $2^{m-2}$ as $2^{2^{v_2(m-2)} m'}$ with $m'$ odd. Modulo $y$, $2^m = 4(-1)^{m'} = -4$. Since $2^k > m$, we know $2^{2^k}$ is $2^{2^{v_2(m-2)}}$ raised to some even power. Thus, modulo $y$, $b = 1 - 5 + 4 = 0$. We found a divisor of $b$. When $m$ is even, the alternate factor sum $y + 2^m (b/y)$ is 1 mod 6, and thus is in $S$, so $2^m b \not \in P$. When $m$ is odd, $y = 3$, so we know $b$ is a multiple of 3. Define $r = v_3(b)$, $z = 3^{-r} b$. We know $r$ is at least 1 and $z$ is congruent to either 1 or 5 mod 6. If $z \ge 5$, we can write the factor sums $s'' = z + 3^r * 2^m$ and $s'''= 3^r + 2^m z$. If $z$ is 1 mod 6, $s''$ is 1 mod 6. If $z$ is 5 mod 6, $s'''$ is 1 mod 6. Either way, one of the two values is in $S$, so $2^m 3^r z \not \in P$. $z = 1$ occurs when there is a solution to the diophantine equation $2^{2^k} - 5 = 2^m + 3^n$. I am to believe that there is exactly one solution for $k \ge 3$. When $k = 3$, we have $2^8 - 5 = 251 = 8 + 243$. In this case, we can consider the alternate factor sum $27 + 72 = 99$, which is in $S$. A result told to me here claims there can be no other cases where $z = 1$ can occur. We have eliminated $a \ge 3$ unconditionally and $a = 1$ when $m \ge 3$. We are left with only one possible pair of summands: 4 and $2^{2^k} - 9$. I don't have a proof that $4(2^{2^k} - 9)$ is or is not in $P$. By my counts it certainly can be, if 3 derived numbers are all prime. That is a rare event, and it would be reasonable for it never to occur given how quickly $2^{2^k}$ grows. However, our initial claim was that at most one derived product is in $P$. We have shown that. This completes the proof.<|endoftext|> TITLE: A question on Steenbrink's paper, limit of Hodge structures QUESTION [5 upvotes]: Steenbrink in his paper "Limit of Hodge Structures", (supplemented by the book "Mixed Hodge Structures" by Peters and Steenbrink) discuss the limit mixed Hodge structures for a fibration over the unit disc $\Delta \subset \mathbb{C}$, i.e. a fibration $$\pi: X \rightarrow \Delta$$ whose fibers are projective varieties and the only singular fiber is over 0, which is a reduced divisor with smooth normal crossing components. Question: Does Steenbrink's construction in his paper (or book) work in the general case i.e. for a fibration $$\pi: X \rightarrow \Delta^n$$ I guess this also involves the issues with "simultaneous semi-stable reduction" of this family, which has been discussed in the paper "A tour of stable reduction with applications". REPLY [3 votes]: If $\pi$ is semistable, then Fujisawa, Limits of Hodge structures in several variables. Compositio (1999), does this. You might also look at some later papers by the same author for some refinements.<|endoftext|> TITLE: Can't one walk to infinity on the prime numbers with finitely many distinct affine steps? QUESTION [43 upvotes]: Let $(a_1,b_1), \dots, (a_k,b_k)$ be finitely many pairs of positive integers, and let $\Gamma$ be the graph whose vertices are the prime numbers and in which two vertices $p$ and $q$ are connected by an edge if and only if $a_ip + b_i = q$ or $a_iq + b_i = p$ for some $i \in \{1, \dots, k\}$. Question: Is it true that necessarily all connected components of $\Gamma$ are finite? Update of May 21, 2018: So far this question has gathered two deleted answers. REPLY [2 votes]: (After I finished writing the paragraphs below, I realized that the assumption that the affine step was a partial matching (paths only of length 1) was wrong. I suggest studying the model anyway with a subset of the affine step represented as a matching, and then relaxing this to K steps where J of them are matchings, and see how the estimates change. My guess is that when J is more than K/2, the quality of the estimates does not change much and the answer will still be no, the connected components will not grow significantly.) Short (non-)answer: I don't know. Long (also non-) answer: I suspect not, based on heuristics involving random partial matchings in graphs. In addition to other computer experiments to try, I suggest modelling the problem along the lines below. The affine step for a given relation can be thought of as a partial matching between some vertices in a graph. Instead of walking to infinity, let us consider how large a connected component can be when using a graph with N primes and picking K many partial matchings. I will freely ignore multiplicative constant factors and make reasonable but unverified estimates in this model. Each step thus is represented by about N/(log N) edges in the graph. (If Hardy and Littlewood can guess at the number of twin prime pairs, I can assume 1/log N fraction of edges with likely the same reasoning and different numeric, but don't take my word for it. A reasonable experiment would see how reality contrasts with this estimate.) So a subset of vertices about the 1/log N the size of the whole graph is involved in the matching. If we wanted to connect the whole graph, we would need K to be at least log N. If we were choosing subsets at random, two of these subsets would have an intersection about 1/(log N)^2 the size of the whole set, and for K of these the size of the common intersection would vary widely but would be expected to be about 1/(log N)^K, for K reasonably small. The union would be expected to be about a little less than K*(log N) in size. For K=2, note that if we want a path longer than length 2, one of the edges of the matching has to belong to the intersection of (the subsets of vertices involved in) the two matchings. This is less likely, and I imagine that of the vertices in the intersection, much less than 1/log N match one another. While longer paths are possible, I expect adding an edge decreases the count of such paths of that length by a factor of log N or more. My guess is that for K fixed, increasing N drops the likelihood of a path of length L to zero. Gerhard "Start With Assuming No Iterates" Paseman, 2018.05.21.<|endoftext|> TITLE: Is the intersection of ramification groups in upper numbering of a $p$-adic local field trivial? QUESTION [6 upvotes]: Let $K$ be a $p$-adic local field, for example $\mathbb{Q}_p$. Let $G$ be the absolute Galois group of $K$, and let $G^v$($v\ge -1$) be the ramification groups in upper numbering, then is it true that $\bigcap_{v=0}^\infty G^v=\{0\}$? REPLY [7 votes]: Yes, at least if the upper ramification groups $G^\nu$ are defined as $\varprojlim_L\mathrm{Gal}(L/K)^\nu$ for $L/K$ finite Galois (e.g. as in [1]). This makes sense because the upper-numbering is compatible with quotients. Then it follows from the fact that for each finite $L/K$, $\mathrm{Gal}(L/K)^\nu$ is trivial for $\nu$ large enough ([2], IV.3). [1]: Fontaine, Jean-Marc. Il n'y a pas de variété abélienne sur Z. Invent. 1985 [2]: Serre, Jean-Pierre. Corps Locaux<|endoftext|> TITLE: On the failure of extending a probability measure on uncountable $\Omega$ QUESTION [6 upvotes]: It is a well known fact that if $(\Omega, \mathcal{F}, P)$ is a probability triple and $\{A_i : i < k\}$ is a finite collection subsets of $\Omega$, then there is a $P' \supset P$ and $\mathcal{F'} \supset (\mathcal{F} \cup \{A_i : i < k\})$ such that $(\Omega, \mathcal{F'}, P')$ is a probability triple. In this case, let's say that $P$ can be extended to include $\{A_i : i < k\}$, or that $\{A_i : i < k\}$ can be adjoined to $P$. I would like to know what is the best, and the worst, we can say when we replace the word "finite" in the statement above with "countable". More precisely, I am asking the following: What is the biggest class of probability triples and countable collections (that has been proven in $ZFC$) for which the statement above holds? For example, is it true that if $A \subset \mathscr{P}(\Omega)$ is countable, and the complete subalgebra of $(\mathscr{P}(\Omega), \subset)$ generated by $A$ includes no dense linear order, then any probability measure can be extended to include $A$? Modulo large cardinal assumptions, what is the best case scenario consistent with $ZFC$? For example, is it consistent that we can extend any probability measure to include any countable collection of subsets of the sample space? Modulo large cardinal assumptions, what is the worst case scenario consistent with $ZFC$? For example, is it consistent that for each uncountable cardinal $\kappa$ there is a probability triple $(\kappa, \mathcal{F}, P)$ such that no countable $S \subset \mathscr{P}(\kappa)$ can be adjoined to $P$? REPLY [3 votes]: Claim: Let $\mathcal{F}$ consist of all countable and co-countable subsets of $\omega_1$ and $m: \mathcal{F} \to \{0, 1\}$ be defined by $m(X) = 0$ iff $X$ is countable. There is a countable family $\mathcal{A}$ of subsets of $\omega_1$ such that there is no probability measure defined on the sigma-algebra generated by $\mathcal{F} \cup \mathcal{A}$ that extends $m$. Proof: Recall that the sigma-algebra generated by $\{A \times B: A, B \subseteq \omega_1\}$ is $\mathcal{P}(\omega_1 \times \omega_1)$ (B. V. Rao). Let $W = \{(\alpha, \beta): \alpha \leq \beta < \omega_1\}$. Since the sigma-algebra generated by any family is the union of the sigma-algebras generated by its countable subfamilies, we can choose a countable family $\mathcal{A} \subseteq \mathcal{P}(\omega_1)$ such that $W$ belongs to the sigma-algebra generated by $\{A \times B: A, B \in \mathcal{A}\}$. Let $\mathcal{G}$ be the sigma-algebra generated by $\mathcal{F} \cup \mathcal{A}$ and towards a contradiction, suppose $m':\mathcal{G} \to [0, 1]$ is a probability measure that extends $m$. As $W \in\mathcal{G} \otimes \mathcal{G}$, every horizontal section $W^{\beta} = \{\alpha: (\alpha, \beta) \in W\}$ of $W$ has $m'$-measure zero and every vertical section $W_{\alpha} = \{\beta: (\alpha, \beta) \in W \}$ of $W$ has $m'$-measure one, we get a contradiction by Fubini's theorem. Answer 1: Every infinite sigma algebra $\mathcal{F}$ contains an infinite disjoint family. It follows that $(\mathbb{R}, \leq)$ embeds into $(\mathcal{F}, \subseteq)$. This makes Question 1 trivial. Answer 2. By the first paragraph, the answer is no. We can assume that $\omega_1 \subseteq X$. Choose $(X, \mathcal{F}, m)$ such that $\omega_1 \in \mathcal{F}$, $m(\omega_1) = 1$ and $m \upharpoonright \omega_1$ is the countable cocountable measure. Answer 3. Since any measure can be extended to the sigma-algebra generated by a disjoint family of sets, the answer is no. So to avoid trivialities, we must put restrictions on the base measure. With this in mind, let me mention the following. (1) Carlson showed that in the random real model, the Lebesgue measure on $[0, 1]$ can be extended to any countably generated sigma-algebra containing the Borel algebra. (2) Section 8 in Fremlin's "Real valued measurable cardinals" addresses similar issues when the underlying measure space is Radon.<|endoftext|> TITLE: An interesting triple integral QUESTION [18 upvotes]: What is the value of this triple integral $$\int\limits_0^{2\pi}\int\limits_0^{2\pi}\int\limits_0^{2\pi}|\cos x+\cos y+\cos z|\ dx\ dy\ dz?$$ It has to do with some Schwarz lemma. REPLY [27 votes]: Following Gerald Edgar, $$W_3(1)\equiv\mathbb{E}[|Y|]=\int dx\int dy \,\sqrt{x^2+y^2}\,p(x,y)=2\pi\int_0^\infty r^2p(r)dr$$ with $p(x,y)dxdy=p(r)\,rdrd\phi$ the rotationally invariant distribution of the complex variable $Y=x+iy=r\cos\phi+ir\sin\phi$. From here $$\mathbb{E}[|X|]=\int dx\int dy \,|x|\,p(x,y)=\int_0^\infty \int_0^{2\pi} |\cos\phi|r^2 p(r)\,drd\phi=\frac{2}{\pi}W_3(1).$$ According to equation (12) of Borwein, Straub, and Wan the numbers $W_n(1)$ are given by $$W_1(1)=1,\;\;W_n(1)=n\int_0^\infty J_1(x)J_0(x)^{n-1}\frac{dx}{x},\;\;n\geq 2.$$ We thus arrive at $$\int\limits_0^{2\pi}\int\limits_0^{2\pi}\int\limits_0^{2\pi}|\cos x+\cos y+\cos z|\ dx\ dy\ dz=(2\pi)^3\mathbb{E}[|X|]=16\pi^2 W_3(1)$$ $$\qquad=48\pi^2\int_0^\infty J_1(x)J_0(x)^{2}\frac{dx}{x}=3\;\frac{2^{1/3}}{\pi^2}\Gamma({\textstyle \frac{1}{3}})^6 + 108\;\frac{2^{2/3}}{\pi^2}\Gamma({\textstyle \frac{2}{3}})^6 \approx 248.65$$ comments: since $(2\pi)^3\approx 248.05$, the average of the absolute value of the sum of three cosines equals unity within one-quarter of a percent. the $n$-fold integral $$I_n\equiv\int_0^{2\pi}\cdots\int_0^{2\pi}|\cos x_1+\cdots + \cos x_n|\,dx_1\cdots dx_n=2^{n+1}\pi^{n-1}n\int_0^\infty J_1(x)J_0(x)^{n-1}\frac{dx}{x}$$ equals a power of 2 for $n=1$ and $n=2$, but not for $n=3$ (nor for larger $n$, as far as I have checked). for $n\gg 1$ one can use the diffusion equation to deduce that $$\lim_{n\rightarrow\infty}(\pi/n)^{1/2}(2\pi)^{-n}I_n=1.$$ (I have not been able to derive this asymptotics directly from the Bessel function expression for $I_n$.) REPLY [25 votes]: Comment This is related to some papers (e.g. this) by P.M. Borwein et. al. on short random walks in the plane. Let $X_1, X_2, \dots$ be i.i.d. random variables, uniformly distributed on the unit circle $|z|=1$ in the complex plane. Then $$ X_1+X_2+X_3 $$ is a random variable in the plane, and your integral is $$ (2\pi)^3\;\mathbb{E}\Big[\big|\mathrm{Re}(X_1+X_2+X_3)\big|\Big] $$ Borwein and collaborators have information on moments $$ W_3(s) := \mathbb{E}\Big[\big|X_1+X_2+X_3\big|^s\Big] $$ including "closed form" in terms of hypergeometric functions when $s \in \mathbb N$. In particular $$ W_3(1) = \frac{3}{16}\;\frac{2^{1/3}}{\pi^4}\Gamma({\textstyle \frac{1}{3}})^6 + \frac{27}{4}\;\frac{2^{2/3}}{\pi^4}\Gamma({\textstyle \frac{2}{3}})^6 . $$ Now, if the distribution of $$ Y = X_1+X_2+X_3 $$ is rotationally symmetric in the complex plane, and we have the exact value of $\mathbb E[|Y|]$, can we find $\mathbb E[|\mathrm{Re}\;Y|]$ ??<|endoftext|> TITLE: A compact set in the plane with small sum-set and large projections QUESTION [5 upvotes]: Problem. Let $K$ be a compact subset of the plane such that the projection of $K$ on each line has non-empty interior in the line. Has $K+K$ or $K-K$ non-empty interior in the plane? Remark. The results of this paper imply the affirmative answer to this problem for compact subsets $K$ of positive dimension in the plane. REPLY [3 votes]: The answer is negative. The key point is the following Lemma: For every $L>0$ and $x\in\mathbb R^2, x\ne 0$, there exist $\ell>0$ and a closed origin-symmetric set $F\subset\mathbb R^2$ such that every interval $I$ of length $L$ on the plane contains a subinterval $J\subset I\cap F$ of length $\ell$ while $F\cap(F+x)=\varnothing$. Proof: WLOG, $x=(0,1)$. Now choose a huge integer $N=N(L)$ and consider the sinusoidal strip $S=\{(x,y):|y-\sin x|\le\frac 1{10(2N+1)}\}$. Take $F=\cup_{k\in\mathbb Z}(S+(0,\frac{2k}{2N+1}))$. Now just enumerate some dense set $x_1,x_2,\dots$ on the plane, take $L_1=1$ and use $x_k$ and $L_k$ to construct $F_k$ and $L_{k+1}$. Put $K=\cap_k F_k\cap \bar D(0,10)$, say (note that each finite intersection of $F_k$ intersects every interval of length $1$, so this property will survive for the full intersection $\cap_k F_k$).<|endoftext|> TITLE: Why does $E\otimes_KE\cong EG$ imply that Galois theory works? QUESTION [18 upvotes]: This is a part of statement in the book I do not fully appreciate. Suppose $E/K$ is Galois extension and $G$ the Galois group of $E/K$. $E[G]$ is the group ring formed by finite group $G$. "It is worthwhile remarking that $E\otimes_KE\cong EG$ can be viewed as a deep reason why Galois theory works." Ref: Algebra Vol 1: Fields and Galois Theory by Falko Lorenz. Q: (solved by the comments below) What is the implication above? I thought $E\otimes_KE\cong EG$'s proof has a major ingredient that the trace map is non degenerate (i.e $E/K$ is separable). Is this affording some representation of $G\to \operatorname{Aut}_K(E)$? What is the author trying to express? Does this extend to infinite Galois extension case? This question (Q) has been asked on Stackexchange but there is no answer. Hence, I asked this on MO. Q': As Qiaochu commented that this is related to torsor theory, I do not see why the two are related. It would be nice if someone could elaborate this a bit. REPLY [3 votes]: This is a CW answer to note that I answered this question over at math.SE. If someone upvotes this, it will be removed from the unanswered list.<|endoftext|> TITLE: Am I counting quantifiers correctly? QUESTION [6 upvotes]: I think this is right but I want to check. The theory $\mathsf{WKL}^*_0$ is conservative over EFA for $\Pi^0_2$ sentences. And the first order part of $\mathsf{WKL}^*_0$ is axiomatized by EFA plus the following formula scheme, known as the $\Sigma^0_1$ bounding principle. For every $\Sigma^0_1$ formula $\varphi(i,j)$ ( with $n$ not free): $$(\forall i TITLE: Is there a natural relationship between OEIS A127670 and Cayley's tree formula? QUESTION [7 upvotes]: I apologize in advance that this question must sound highly amateurish, but I am wondering if there is any connection between the formula https://oeis.org/A127670 , which counts the number of fixed $n$-cell polycubes that are proper in $n-1$ dimensions, and Cayley's tree formula. The expression for the former is $$a_n = 2^n(n+1)^{n-2}$$ and the latter is $$b_n = n^{n-2}$$, which means that $$a_n = \frac{2^n}{n+1} b_{n+1}$$. Is there any significance to this at all? The reason I ask is that I'm working on a Cayley-type tree enumeration problem that yields the sequence $a_n$ according to numerical computations, and I am struggling to see the connection. REPLY [14 votes]: Yes there is a connection. While $n^{n-2}$ counts the number of vertex labeled trees on $n$ vertices, the expression $2^n(n+1)^{n-2}$ counts the number of edge labeled trees on $n$ edges. There is a bijection between edge labeled trees on $n$ vertices and proper $(n-1)$-dimensional polycubes of size $n$. See lemma 2 (which combinatorially proves the relation $a_n=\frac{2^n}{n+1}b_{n+1}$) and theorem 1 in Formulae and growth rates of high-dimensional polycubes by R. Barequet, G. Barequet, G Rote.<|endoftext|> TITLE: Is the following result about congruences of lines classical? QUESTION [13 upvotes]: The following result must be classical; I am looking for a reference or an easy proof (I have a rather sophisticated argument). Let $\mathbb{G}$ be the Grassmannian of lines in $\mathbb{P}^3$, and let $S$ be the complete intersection of two general hypersurfaces in $\mathbb{G}$, of fixed degrees $>1$. Then $S$ parametrizes a 2-dimensional family of lines (a congruence in classical language). The claim is that through any point of $\mathbb{P}^3$ pass only finitely many such lines (in classical language again, the fundamental locus is empty). REPLY [5 votes]: Let $V = \mathbb{C}^4$ and let $a \in V$. The set of $\mathbb{C}^2 \subset V$ which passes through $a$ is a $\mathbb{P}^2$ linearly embedded in $G(2,V) \subset \mathbb{P}(\bigwedge^2 V)$. The Grassmannian $G(2,4) \subset \mathbb{P}(\bigwedge^2 V)$ is identified with a quadric and the $\mathbb{P}^2$ described above is a maximal isotropic subspace of $\mathbb{P}(\bigwedge^2 V)$ for the corresponding quadratic form. When $a$ varies in $\mathbb{P}^3$, we get a $3$-dimensional family of isotropic subspaces of $ \mathbb{P}(\bigwedge^2 V)$. This family is identified with the spinor variety of isotropic $\mathbb{C}^3 \subset \bigwedge^2 V$. The spinor variety is in this case isomorphic to $\mathbb{P}^3$, but for reading convenience, we denote it by $\mathrm{Spin}^{3}$. Let $\mathcal{R}$ be the taulogical bundle of $\mathrm{Spin}^{3}$. We consider $ p :X = \mathbb{P}(\mathcal{R}) \longrightarrow \mathrm{Spin}^3$. By the Borel-Bott-Weyl Theorem, we have an identification (to check?) : $$ H^0(X, \mathcal{O}_{X/\mathrm{Spin}^3}(d)) = H^0(\mathrm{Spin^3}, S^d \mathcal{R}^*) = S^d (\bigwedge^2 V)^* = H^0(\mathbb{P}(\bigwedge^2 V), \mathcal{O}(d)).$$ We deduce that a pair of general hypersurfaces of degree $d$ and $e$ in $\mathbb{P}(\bigwedge^2 V)$ can be seen as a generic section of $S^d \mathcal{R}^* \oplus S^e \mathcal{R}^*$ on $\mathrm{Spin}^3$ and can be also seen as a generic section of $\mathcal{O}_{X/\mathrm{Spin}^3}(d) \oplus \mathcal{O}_{X/\mathrm{Spin}^3}(e)$. Your statement can now be reformulated as follows : Statement: Let $d,e \geq 2$ and $f \in H^0(X,\mathcal{O}_{X/\mathrm{Spin}^3}(d) \oplus \mathcal{O}_{X/\mathrm{Spin}^3}(e))$ be a general section. For all $u \in \mathrm{Spin}^{3}$, the scheme $\{f = 0\} \cap p^{-1}(u)$ is of codimension $2$ in $p^{-1}(u)$. This is looks very like a Bertini-type result. In that direction, one can prove the following : Lemma 1 : Let $e \geq 2$ and general $h \in H^0(X,\mathcal{O}_{X/\mathrm{Spinor}_3}(e))$. Then, for all $u \in \mathrm{Spin}^{3}$, the scheme $\{h = 0\} \cap p^{-1}(u)$ is of pure codimension $1$ in $p^{-1}(u)$. proof: the section $h$ induces a section a generic section of $S^e \mathcal{R}^*$. The vanishing locus of this section corresponds to the variety of isotropic $\mathbb{P}^2 \subset \mathbb{P}(\bigwedge^2 V)$ which are included in the hypersurface in $\mathbb{P}(\bigwedge^2 V)$ corresponding to $h$. We only have to check that the vanishing locus of this section is empty. We have $\dim S^e \mathcal{R}^* = \binom{e-1+3}{e}> \dim \mathrm{Spin}^{3}$ when $e \geq 2$. Furthermore, the vector bundle $\mathcal{R}^*$ is globally generated, we deduce that the vanishing locus of $h$ is empty. Lemma 2 [see here, the idea is due to Zak] Let $X \subset \mathbb{P}^N$ be an equidimensional scheme of dimension $n$. Denote by $\mathcal{H}_{N,e}$ the projective space of hypersurfaces of degree $e$ in $\mathbb{P}^N$. Then, the set of points in $\mathcal{H}_e$ which contains an irreducible component of $X$ is of codimension at least $\binom{e+n}{e}$. proof : [see here]. Using Lemma $1$ and $2$, one proves the following: proposition : Let $e \geq 2$, $d \geq 3$ and $(g,h) \in H^0(X,\mathcal{O}_{X/\mathrm{Spin}^3}(d) \oplus \mathcal{O}_{X/\mathrm{Spin}^3}(e))$ be a general section. For all $u \in \mathrm{Spin}^{3}$, the scheme $\{(g,h) = 0\} \cap p^{-1}(u)$ is of codimension $2$ in $p^{-1}(u)$. proof : Denote by $X_h$ the vanishing locus of $h$ in $X$. By Lemma $1$, we know that all the fibers of the map: $$ p_h : X_h \longrightarrow \mathrm{Spin}^3$$ are of pure dimension $1$. Let $\mathcal{H}_{\bigwedge^2V,d}$ be the projective spaces of hypersurfaces of degree $d$ in $\mathbb{P}(\bigwedge^2 V)$. Denote by $J \subset \mathcal{H}_{\bigwedge^2V,d} \times S$ the subvariety defined by: $$ J = \{(H,s) \in \mathcal{H}_{\bigwedge^2V,d} \times S, H \ \textrm{contains an irreducible component of} \ p_h^{-1}(s) \}$$ By Lemma $2$, we know that each fiber of $q : J \longrightarrow S$ has codimension at least $\binom{d+1}{d}$ in $\mathcal{H}_{\bigwedge^2V,d}$. As a consequence, the codimension of $J$ in $S \times \mathcal{H}_{\bigwedge^2V,d}$ is at least $\binom{d+1}{d}$. Since $\dim S = 3$, we get that $\mathrm{codim}(J) - \dim S - 1 \geq 0$ for $d \geq 3$. Hence, the map $p : J \longrightarrow \mathcal{H}_{\bigwedge^2V,d}$ can't be dominant. We deduce that for generic $g \in \mathcal{H}_{\bigwedge^2V,d}$, the scheme $\{(g,h) = 0\} \cap p^{-1}(s)$ has dimension $0$ in $p^{-1}(s)$, for all $s \in S$.<|endoftext|> TITLE: Is $GL_n(\mathbb{Q}_p)=GL_n(\mathbb{Z}_p)GL_n(\mathbb{Q})$? QUESTION [17 upvotes]: Is $GL_n(\mathbb{Q}_p)=GL_n(\mathbb{Z}_p)GL_n(\mathbb{Q})$? Generally, let $R$ be a discrete valuation ring and $K$ its fraction field. Let $\widehat{R}$ be the completion and $\widehat{K}$ the fraction field of $\widehat{R}$. Is $GL_n(\widehat{K})=GL_n(\widehat{R})GL_n(K)$? We know it is true when $n=1$. REPLY [26 votes]: Yes: more generally for every topological group $G$, dense subgroup $H$ and open subgroup $U$, we have $G=UH$. This applies when $F$ is a valued field, $A$ an open subring (typically, elements of non-negative valuation), and $K$ any dense subfield of $F$, $G=\mathrm{GL}_n(F)$, $H=\mathrm{GL}_n(K)$, $U=\mathrm{GL}_n(A)$.<|endoftext|> TITLE: $T_2$-spaces where all non-empty open sets are homeomorphic QUESTION [13 upvotes]: We say that a $T_2$-space $(X,\tau)$ has homeomorphic open sets if every non-empty open set $U\subseteq X$ endowed with the subspace topology is homeomorphic to $(X,\tau)$. The rationals with the Euclidean topology are an example of a space with homemorphic open sets, as well as $\{0,1\}^\lambda$, where $\lambda$ is an infinite cardinal and $\{0,1\}$ carries the discrete topology. Is there for every infinite cardinal $\kappa$ a space of cardinality $\kappa$ with homeomorphic open sets? Edit. Apologies for falsely claiming that $\{0,1\}^\lambda$ is a space with homeomorphic open sets - Joel David Hamkins' comment below gives an argument refuting my claim. REPLY [19 votes]: The Baire space $\omega^\omega$ has homeomorphic open sets: it is an immediate consequence of the fact that it is the unique nonempty zero-dimensional Polish space, up to homeomorphism, every compact subset of whose has empty interior (theorem 7.7 in Kechris). In fact we actually have better: Proposition. For every infinite set $X$ with the discrete topology, $X^\omega$ has homeomorphic open sets. Proof. Let $U \subseteq X$ be nonempty open. I claim that $U$ is a union of exactly $\kappa$ disjoint cones, where $\kappa = |X|$. To see this, it is enough to show that $U$ is a union of $\leqslant \kappa$ disjoint cones, as a cone is itself a union of $\kappa$ disjoint cones. But if $T \subseteq X^{< \omega}$ is a nonempty tree such that $[T] = U^c$, then $U$ is the disjoint union of all cones $N_s$ where $s \notin T$ but $s_{\restriction |s| - 1} \in T$. Now, writing $U$ as a union of $\kappa$ disjoint cones, we can get a homeomorphism between $U$ and $X$ by gluing together homeomorphism between these cones and cones of the form $N_{(x)}$, $x \in X$. So the answer to Dominic's question is positive at least for cardinals of the form $\kappa^\omega$. Edit: Will Brian remarked, in the comments, that the same argument actually enables to show that the subspace of $X^\omega$ whose elements are eventually constant also had homeomorphic open sets. So for every infinite cardinal, there exists a space with homeomorphic open sets.<|endoftext|> TITLE: Definition of Non-commutative de-Rham-Cohomology QUESTION [5 upvotes]: Let $A$ be a (not necessarily commutative, associative) $k$-algebra. The bimodule of non-commutative one-forms $\Omega^1_A$ is the free $A$-bimodule generated by symbols $da$, $a \in A$, subject to the relations $$ d(\lambda a + \mu b) = \lambda da + \mu db, ~~~~~~ d(ab) = a\,db + da \,b$$ for $a, b \in A$, $\lambda, \mu \in k$. Higher degree forms are then defined by $$\Omega^n_A = \Omega^1_A \otimes_A \cdots \otimes_A \Omega^1_A.$$ The direct sum of all forms $\Omega_A = \bigoplus_{n=0}^\infty \Omega_A^n$ becomes a differential graded algebra with the obvious differential. Now Loday's book "cyclic cohomology", he defines the non-commutative de-Rham cohomology of $A$ to be the cohomology of the complex $((\Omega_A)_{\mathrm{ab}}, d)$, i.e. he takes the abelianization of $\Omega_A$. Q: Why is that? What is wrong with defining the non-commutative de-Rham-cohomology just as the cohomology of $\Omega_A$? /edit: To answer the question regarding the definition of $d$: Using the relations of $\Omega^1_A$, every element of $\Omega^n_A$ can be brought into the form $a_0 da_1\cdots da_n$. For these, $$d(a_0 da_1\cdots da_n) = da_0da_1 \cdots da_n.$$ REPLY [2 votes]: I do not really know the reason, but if I would guess I think it might be $\Omega(A)$ is no longer an abelian category. The exact sequences you written down are not just exact sequences of abelian groups, but may be interpreted as exact sequences of algebras (note that we have $d(ab)=da*b+a*db=0*b+a*0=0$ for $a,b\in \ker(d)$). I am not entirely sure if the construction of abelian category carries through for non-commutative algebras. I do not know any counter-example, but it sounds very unlikely to be true as the construction is so general (consider some infinite dimensional non-commutative $\mathbb{C}$-Banach algebras, for example).<|endoftext|> TITLE: Does every model of ZF-foundation have an extension, with no new well-founded sets, where every set is bijective with a well-founded set? QUESTION [15 upvotes]: This question follows up on an issue arising in Peter LeFanu Lumsdaine's nice question: Does foundation/regularity have any categorical/structural consequences, in ZF? Let me mention first that my view of the role of the axiom of foundation in the foundation of set theory differs from Peter's. (But see Peter's remarks about this in the comments below; our views may not actually be so different.) Specifically, Peter explains how to view the axiom of foundation in ZFC as an axiom of convenience (see his answer here). Namely, since every set is well-orderable, all the non-well-founded sets are bijective with an ordinal and therefore with a set in the well-founded part of the universe $\bigcup_\alpha V_\alpha$. In particular, any mathematical structure to be found at all can be found up to isomorphism in the well-founded realm, where the axiom of foundation is true. Peter's follow-up question arose essentially from his observation that this argument uses the axiom of choice, and my answer shows that indeed this is the case, for there are models of ZF-foundation where some sets are not bijective with any well-founded set. My view of the axiom of foundation, in contrast, is that it is not merely an axiom of convenience, but rather expresses a fundamental truth of the intended realm of sets the theory is trying to describe. Specifically, the cumulative universe of all sets arises from the urelements in a well-ordered series of set-building stages; every set comes into existence at the first stage after all of its elements exist. This cumulative universe picture leads immediately to the axiom of foundation. Andreas Blass expresses a similar view in answer to an earlier question. (To my way of thinking, the real step of convenience in ZFC consists instead of the insight that we don't actually need urelements for any purpose; and the resulting ZFC theory without urelements is more uniform and elegant.) My question here is whether one can resurrect the idea of the harmlessness of foundation by showing that every model of ZF-foundation can be extended, without changing the well-founded part, to one where every set is bijective with a well-founded set. Question 1. Does every model of ZF-foundation have an extension, with the same well-founded sets, in which every set is bijective with a well-founded set? A weaker version of the question asks about doing this for just one set at a time: Question 2. In any model of ZF-foundation, if $A$ is any set, then is there an extension of the model, without changing the well-founded part, in which $A$ becomes bijective with a well-founded set? For example, perhaps we might hope to make $A$ well-orderable or even countable. The trouble would be to do so without adding any well-founded sets. Definitely we cannot hope to always make a set $A$ well-orderable, if there are non-well-orderable well-founded sets. If you can answer the question for countable models, or by making further assumptions on the models, I would be interested. REPLY [12 votes]: Well. The question is what do you mean by an extension exactly. There's a theorem of Paul Larson and Shelah (originally the work started with Eric Hall, then continued with Paul Larson a few years later) that you can construct a permutation model that has no extension to models of choice with the same pure sets. And of course moving from atoms to Quine atoms, pure sets become well-founded sets. The idea is roughly to sort of code into the structure of the model a collapse of some cardinals, with the pure sets satisfying $V=L$. Then if you well-order the atoms, then you necessarily had to collapse cardinals and thus add new pure sets. So from a Platonistic point of view, the answer is negative. If your universe has only those pure sets, then there is no possible way to extend this model further. But if you want to allow forcing extensions, then the answer is yes, you can just add the necessary sets and collapse these cardinals. In general, if you are willing to add new well-founded sets, then simply collapsing more and more cardinals should in principle work. But I suspect that one can modify Larson and Shelah's work to a class setting such that for every $\alpha$, there is a set of atoms with structure that codes collapsing $\omega_\alpha$ to $\omega$. And then the answer is negative. But this would require a model where the atoms form a proper class.<|endoftext|> TITLE: An extension of the Izergin-Korepin determinant to the eight-vertex model QUESTION [5 upvotes]: In the six-vertex model, edges in a square lattice are oriented so that the in-degree of each vertex is exactly two. This gives six types of allowable vertices: $$\begin{array}{cccccc} \begin{array}{ccc} & \uparrow & \\ \leftarrow & \bullet & \leftarrow \\ & \uparrow & \\ \end{array} & \begin{array}{ccc} & \downarrow & \\ \rightarrow & \bullet & \rightarrow \\ & \downarrow & \\ \end{array} & \begin{array}{ccc} & \uparrow & \\ \rightarrow & \bullet & \rightarrow \\ & \uparrow & \\ \end{array} & \begin{array}{ccc} & \downarrow & \\ \leftarrow & \bullet & \leftarrow \\ & \downarrow & \\ \end{array} & \begin{array}{ccc} & \downarrow & \\ \leftarrow & \bullet & \rightarrow \\ & \uparrow & \\ \end{array} &\begin{array}{ccc} & \uparrow & \\ \rightarrow & \bullet &\leftarrow \\ & \downarrow & \\ \end{array} \\ [z]&[z]&[az]&[az]&z&z^{-1} \end{array}$$ If a vertex has label $z$ (along with a global parameter $a$), the vertex is assigned the weight listed above, where $[z]$ means $\frac{z-z^{-1}}{a-a^{-1}}$. As usual, the partition function $Z$ of a lattice is defined to be the sum (over all allowable orientations of the edges) of the product of the individual weights of all the vertices. In a $n\times n$ lattice with column parameters $x_1,\dots,x_n$ and row parameters $y_1,\dots,y_n$, the label of the vertex in row $i$ and column $j$ is $z_{ij}=x_i/y_j$. Then, the Izergin-Korepin determinant gives an exact formula for the partition function of a lattice in the six-vertex model $$Z=\frac{\textstyle\prod_{i=1}^n x_i/y_i \prod_{1\le i,j\le n}[z_{ij}][az_{ij}]}{\textstyle \prod_{1\le i TITLE: What is Chemlambda? In which ways could it be interesting for a mathematician? QUESTION [19 upvotes]: I${}^{*}$ have randomly come across a couple of websites (Chemlambda project, chorasimilarity) that seem to be about a certain "thing" (a computer program, I think) called Chemlambda that does "stuff" with some kind of decorated graphs called molecules according to some "rules". ${}^*$ Disclaimer: I don't know anything about computer science, computability theory, rewriting systems, distributed computing, or the like. Chemlambda molecules look more or less like this: and can make up much more complicated shapes: Looks cool, doesn't it? Also, there are links to very cool looking youtube videos which show such decorated graphs in action, such as this one or this one, and a gallery of videos perhaps trying to explain some elementary Chemlambda "operations". Notice that the title of one of the youtube videos contains "factorial of $4$", so it seems to suggest that some kind of computation is being performed... Question 0. Does all this contain some legitimate and/or interesting maths ̶o̶r̶ ̶i̶s̶ ̶i̶t̶ ̶b̶u̶l̶l̶s̶h̶i̶t̶ ? There are some articles on the arXiv authored by M. Buliga, such as this one or this one, that are relevant to the topic. Notice that the second linked article is coauthored with L.H. Kauffman, who is a well known mathematician (for reasons unrelated to this project). Question 1. What is Chemlambda? Provided the definitions on which it is based are mathematically sound/rigorous, I would like to roughly understand what this program does with such decorated graphs and why it would be interesting (from the point of view of mathematical logic, or more generally of mathematics) to consider such kind of operations with decorated graphs. Skimming through the websites and the articles, I wasn't able to precisely understand what the point of Chemlambda is, besides providing cute animations: Question(s) 2. As far as I -kind of- understand, Chemlambda graphically models $\lambda$-calculus, so it's suitable for computation. Why would one want to encode computations graphically? Does it provide any conceptual insights or practical advantages? How are, roughly, inputs encoded and the output read off the resulting molecules? Are the operations on molecules performed deterministically or randomly? If randomly, how? REPLY [16 votes]: UPDATE (2020) There is now chemlambda.github.io which collects all the chemlambda related projects and articles. For the history of chemlambda see arXiv:2007.10288 or the associated html version Graph rewrites, from emergent algebras to chemlambda. Author here. I think the most informative link is the GitHub repository (and links therein). I'll try to answer as short as possible to questions 1 and 2 and then comment on the mathematical relevance. Question 1. What is Chemlambda? Chemlambda is an asynchronous graph rewrite automaton. There are 3 parts: graphs, rewrites and the algorithm of reduction. Source, alternative formalism of sticks and rings which might be interesting for this community. The graphs are called molecules and they are oriented ribbon graphs, with the nodes decorated with colors. Molecules are built from 3valent nodes (A, L, FI, FO, FOE), a 2valent node Arrow and 1valent nodes FRIN (free in) FROUT (free out) T (termination). All the information (oriented ribbon graph, type of nodes) can be encoded as a 3 colors decoration of nodes and half-arrows. As a graph rewrite system is of the same family as interaction nets. List of rewrites here. The algorithm of the application of rewrites is very important as well. There are two of them and they are both the most stupid ones: deterministic, with a priority of moves in case there are conflicts (i.e. overlapping patterns to be rewritten), or the random one. Question 2. Chemlambda, with the random reduction algorithm, is a model of computation which has the properties: is Turing universal, random and local, i.e. there is a small number N, a priori, so that the rewrites or the decision to apply them act (or need) at most N nodes and half-arrows of the graph. On purpose, there are no global notions used: there is no typing, there is no categorical approach considered, there is no equivalence relation obtained from the closure of rewrites application, there is no limitation on the possible graphs, nor there is any other algorithm which has as input the whole molecule (graph) and as output a decoration of particular half-edges (as is the case with the association of a lambda term with its AST and then back from the AST to the term). Chemlambda can model untyped lambda beta (but not eta) calculus. As well, by the introduction of supplementary 2valent nodes, it can as well model Turing machines with multiple heads, working asynchronously. Comments. Chemlambda was not made with the goal to model lambda calculus, nor is lambda calculus the main interest here. I arrived to chemlambda from sub-riemannian geometry, where I used uniform idempotent right quasigroups, aka emergent algebras. That formalism can, almost entirely, be written as a graph rewrite system and the question I had was: is it Turing complete? The exact relation between chemlambda and emergent algebras is still work in progress, but basically emergent algebras resemble very much to a type system for chemlambda (which shows that emergent algebras, as they are presently, are too limited and hide something more interesting).<|endoftext|> TITLE: Join Density in R.E. Degrees: Are there r.e. B, C with all r.e. X below B computable or C join X computes B QUESTION [7 upvotes]: Are there r.e. sets $B >_T 0$ and $C >_T 0$, $C \not\geq_T B$ such that for all r.e. $W \leq_T B$ either $W \leq_T 0$ or $C \oplus W \geq_T B$. The explanation for the title is because one can think of this as saying that there is a gap between $0$ and $B$ in the r.e. degrees join $C$. I've been hitting my head against this problem for awhile. Originally I thought I had a construction of r.e. degrees with the following property but now I think it's impossible and worry I'm missing something dumb so before I waste more time trying to decide the question one way or another I figured I should find out if I'm missing anything dumb. Note that it's easy to see that if $B,C$ satisfy then $B$ and $C$ are a minimal pair. Now the initial temptation is to try and use a proof something along the lines of Sack's density theorem for r.e. degrees but one quickly runs into problems since trying to restrain $W$ to ensure $\phi_i(C \oplus W) \neq B$ might impose unbounded restraint (though with finite lim inf) but since $B$ can't compute the stages at which elements enter $C$ to disrupt $\phi_i(C \oplus W)$ one quickly runs into a conflict between the requirement that $W \leq_T B$ and the attempt to ensure $C \oplus W \not\geq_T B$. Indeed, I can show there is no uniform solution in the following sense. Given a finite lits $i_0\ldots i_n$ of indexes I can uniformly build $C, B$ such that if $W = \phi_{i_j}(B)$ then $W \leq_T 0$ or $C \oplus W \geq_T B$. Anyway, before I waste any more time trying to decide this question I want to make sure I'm not missing anything obvious. REPLY [7 votes]: You could check barmpalias, cai, lempp and slaman. It shows that there is a pair of the sort that you asked about. As far as I know, realizing the fully symmetric condition is open.<|endoftext|> TITLE: Homotopy limits of Simplicial sets QUESTION [5 upvotes]: Consider a small diagram $X_\bullet$ of simplicial sets. Under what sufficient conditions (apart from $X_\bullet$ being injective fibrant), is the map $\text{lim }X_\bullet \to \text{holim } X_\bullet$ a weak equivalence? Is it sufficient that the morphisms of simplicial sets $X_{i} \rightarrow X_{j}$ be fibrations $\forall i,j$? I'm looking for something analogous to what one has in case of homotopy pullbacks of simplicial sets where one map being a fibration is sufficient to ensure that homotopy pullback is same as categorical pullback. EDIT: I'm interested in diagrams indexed by cosimplex category $\Delta$. REPLY [4 votes]: I'll just address the second question. It's not enough for the maps $X_i \to X_j$ to be fibrations, it's not even enough for them to be identity maps on a Kan complex! Take $X$ to be a Kan complex, $G$ to be a group and consider the trivial $G$ action on $X$ as a diagram whose shape is the one-object category $G$. Then the fixed points, $\lim_G X$, are just $X$, but the homotopy fixed points, $\mathrm{holim}_G X$, have the homotopy type of $\mathrm{Map}(BG, |X|)$. So any reasonable answer to the first question must take into account the diagram shape, not just the individual simplicial sets $X_i$ and the maps $X_i \to X_j$.<|endoftext|> TITLE: Closed geodesics on constant positive Gauss curvature surfaces QUESTION [8 upvotes]: Can we show that geodesics with a rational radius $ r_{mid-equator} =a q/p \,(p>q ) $ at mid-equator on a spindle type surface of revolution of constant Gauss curvature $ (K=1/a^2 \, $in $\, \mathbb R^3 ) $ form closed loops around the symmetery axis in terms of $ (p, q)$? Is the arc length of a loop/orbit expressible as being related to $ (2 \pi a, p,q)?$ If a Riemann sphere radius $a$ is isometrically mapped/deformed without twist (roll squeezed) along equator will its geodesics deform this way due to isometry? The figure shown below is drawn for $ q=2,p=3 $ respecting geodesy/constant $K$ in the present investigation. In all such cases numerical work established/ confirmed existence of such closed loops. What existing related literature can be found on the topic? Thanks in advance of your response. EDIT1: At first $k_g=$ for constant geodesic curvature closed loops were expected (by me) but not the entire geodesic line closing in as a single loop line as that would be counter-intuitive. In the picture shown numerical value $a k_g=1.4$ is assumed. EDIT2: The following geometrical results of radius and arc length are now verified in this investigation and reported here: Radius variation with respect to arc length $s$, $ \alpha$ being angle between arc projection and symmetry axis at mid-equator start: $$ \left(\frac{r}{r_{mid-equator}}\right)^2 + \left(\cos \alpha \cdot \sin \frac{s}{a}\right)^2=1 \tag1 $$ which can be also written $ r_o= r_{cuspidal-equator}$ $$ \cos \dfrac{s}{a} =\sqrt{\dfrac{r^2-r_0^2}{r_1^2-r_0^2}} \tag2 $$ Arc length of a single circuit $ L_{circuit}$ from right cusp equator to right cusp equator back via left cusp equator is found to be: $$ L_{circuit}= 2 \pi a \tag 3$$ as can be expected in full isometric mappings. REPLY [10 votes]: The answers to your questions about geodesics on such surfaces can be found in A. L. Besse's book Manifolds all of whose geodesics are closed (1978, Springer Ergebnisse series). Especially, you should have a look at Chapter 4 on surfaces. In particular, it is true that all of the geodesics that stay in the smooth part of the surface are closed, and their length is known in terms of $a$, $p$, and $q$.<|endoftext|> TITLE: Rational homotopy groups of a projective hypersurface QUESTION [15 upvotes]: Let $X$ be a smooth projective hypersurface in $\mathbb{P}^n$. Has anyone computed the rational homotopy groups $\pi_i(X)\otimes \mathbb{Q}$ of $X$? I tried Google, but did not find anything. One could use the formality and Sullivan's minimal model to compute these groups up to some level by hand. And in principle this can be done for all the groups. But is there any nice way of organising these things? For example, give the ranks using some generating functions, or express the generators in a nice way? REPLY [10 votes]: The paper "On real homotopy properties of complete intersections" by Babenko gives an answer to your question. I will adapt Corollary 1 presented therein to the case of hypersurfaces: Let $X$ be a degree $d$ hypersurface in $\mathbb{C}\mathbb{P}^{n+1}$, and denote its topological Euler characteristic by $\chi$ (note that $\chi$ depends only on $d$ and $n$). If $\chi = n+1$, then the only rationally non-zero homotopy groups of $X$ are $\pi_2 (X) \otimes \mathbb{Q} = \pi_{2n+1} (X) \otimes \mathbb{Q} \cong \mathbb{Q}$. If $\chi \neq n+1$, then for any $j\geq 1$ we have $$\dim\pi_{j+1}(X)\otimes \mathbb{Q} = \frac{(-1)^j}{j} \sum_{d|j, \ \ d\geq 1} (-1)^d \mu\bigl(\frac{j}{d}\bigr)\sum_{\alpha=1}^{2n-2} \xi_\alpha^{-d},$$ where $\xi_\alpha$ are the roots of $1-(-1)^n(\chi-n-1)(1+z)z^{n-1} + z^{2n-1}$ that are not the root $-1$, and $\mu$ is the Möbius function. For example, in the case of a $K3$ surface, $n=3$ and $\chi = 24$, and we get $\pi_3(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{252}$, $\pi_4(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{3520}$, $\pi_5(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{57960}$, $\pi_6(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{1020096}$, $\pi_7(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{18664800}$, $\pi_8(K3)\otimes \mathbb{Q} \cong \mathbb{Q}^{351204480}$, $\ldots$. This is consistent with Corollary 4.10 in Samik Basu and Somnath Basu's Homotopy groups and periodic geodesics of closed 4-manifolds and the answers to this question What are the higher homotopy groups of a K3 suface? For convenience, here is a formula for the topological Euler characteristic of a degree $d$ hypersurface in $\mathbb{C}\mathbb{P}^{n+1}$ (simplified thanks to abx's comment); $$\chi = \frac{1}{d}((1-d)^{n+2} + d(n+2) -1).$$<|endoftext|> TITLE: Transporting a model category structure along a left adjoint QUESTION [6 upvotes]: There is a well-known theorem for transporting a model category structure along a left adjoint $F:\mathcal{M}\to \mathcal{N}$ which is explained here and which is due to Sjoerd Crans. The difficult part is to check the third condition. By Axiomatic homotopy theory for operads (2.6), it suffices to check that $\mathcal{N}$ has a functorial fibrant replacement and a functorial path-object for fibrant objects. This paper cites as references: D. G. Quillen, Homotopical algebra, Lect. Notes Math. 43 (1967) Theorem II.4 Every homotopy theory of simplicial algebras admits a proper model, Theorem 7.6 Algebras and modules in monoidal model categories, Theorem A.3 The first reference is about the construction of a model category structure on the category of simplicial objects of a category satisfying some conditions. The second reference is a "useful lemma" Lemma 7.6. As for the third reference, I just can't find where is Theorem A.3. Could someone give a reference for the proof of this fact (or the proof) ? REPLY [2 votes]: It suffices to dualize the proof of Theorem 2.2.1 in Necessary and sufficient conditions for induced model structures. It uses indeed an argument coming from Quillen's book "Homotopical Algebra", II page 4.9 (the diagram in the bottom part of the page). And it is also necessary to use the fact that the class of weak equivalences satisfies the 2-to-6 property.<|endoftext|> TITLE: Classification of compact globally symmetric spaces QUESTION [8 upvotes]: It is known that any connected compact Lie group $G$ is a finite quotient of the product of a compact simply connected semisimple Lie group $\tilde{G}$ and a torus $\mathbb{T}^n$ (see for example Chapter V Theorem 8.1 in Brocker). In other words, the universal covering map $\tilde{G}\times \mathbb{R}^n\to G$ factors through some $\tilde{G}\times\mathbb{T}^n$. More generally, consider a compact globally symmetric space $M$. Then the universal cover of $M$ is the product of a simply connected globally symmetric space $\tilde{M}$ of compact type and a Euclidean space $\mathbb{R}^n$. My question is, is it still true that the universal covering map $\tilde{M}\times \mathbb{R}^n\to M$ factors through some $\tilde{M}\times \mathbb{T}^n$ so that $\tilde{M}\times \mathbb{T}^n\to M$ is a finite Riemannian cover? Any help is appreciated. Thank you! REPLY [2 votes]: See Theorem A here: https://www.ams.org/journals/proc/2001-129-12/S0002-9939-01-06008-7/S0002-9939-01-06008-7.pdf Your $M$ is compact globally symmetric so any geodesic is contained in a compact flat. So you get a covering as you wanted.<|endoftext|> TITLE: Reference request for statement on nlab: Reedy (co)fibrancy of (co)simplicial objects QUESTION [5 upvotes]: On the nlab http://ncatlab.org/nlab/show/Reedy+model+structure#fibrant_and_cofibrant_objects it is claimed that a simplicial object in a model category, in which all monomorphisms are cofibrations, is always Reedy-cofibrant. Does anyone know a reference for this? In the case of simplicial sets the statement is Theorem 15.8.7 in Hirschhorn's book. Follow up question: Is it dually true that any cosimplicial object in a model category, in which all epimorphisms are fibrations, is Reedy-fibrant? Again, is there a reference? Thanks a lot. REPLY [3 votes]: The claim would be true if it were the case that for any simplicial object $X_\bullet$ in some (cocomplete) category $C$, the maps $L_nX\to X_n$ from the latching objects are always monomorphisms. This is the case when $C$ is any topos for instance. It is also true when $C$ is any additive category (because of the Dold-Kan theorem on (co)simplicial objects in an additive category with retracts), which would seem to include Faelivrin's example. But it is not true generally. For instance, if $C$ is the category of augmented commutative $k$-algebras. Pick an augmented commutative $k$-algebra $A$, and form the simplicial object with $X_n= A\times_k\cdots \times_k A$ ($n$ copies of $A$); this is an example of a 1-coskeleton. The map $L_2X\to X_2$ has the form $$ A\otimes_k A \to A\times_k A, $$ and this can't generally be a monomorphism, e.g., when $A=k[x]$. Probably an actually counterexample to the claim can be constructed from an idea like this.<|endoftext|> TITLE: Some examples of vertex algebra modules QUESTION [5 upvotes]: Recently I'm learning the vertex modules. In the paper, there are a lot of abstract theory about the module theory,for instance the $C_{2}-$cofinite conditions and associated variety. I hope to find some fundamental examples to get some intuition for these theory. As we all know, a vertex algebra is a module of itself. But this is too trivial to reveal some ideas behind it. I want to know if there are some nontrivial basic examples of vertex algebra modules for instance over Heisenberg vertex algebra, Affine vertex algebra or Virasora vertex algebra? I would appreciate if you can provide some details or reference. REPLY [3 votes]: I don't know what paper you are reading, but you can find examples in most textbooks. For example, Frenkel and Ben-Zvi's book "Vertex algebras and algebraic curves" has a treatment of modules in chapter 5 that goes into some detail for the Heisenberg case. One thing they don't mention is that there are non-trivial self-extensions of irreducible modules, given by a rather easy Jordan block method. There is a brief note in 5.5.5 about various rational quotients of affine and Virasoro vertex algebras, with references to the literature. Any conformal vertex algebra is a module for the Virasoro vertex algebra of that central charge, so this particular case has an overabundance of examples.<|endoftext|> TITLE: Expected cardinality of a randomly chosen element of the family of subsets of $\{1,\ldots,n\}$ with at most $k$-elements QUESTION [7 upvotes]: Assume that $1\le k \le n$ and let $\mathscr{Z}$ be the family of all subsets of $\{1,\ldots,n\}$ with at most $k$ elements. Pick a random element $X$ of $\mathscr{Z}$ (we consider the probablity distribution on $\mathscr{Z}$ is uniform, that is, each $X$ is chosen with probability $1/|\mathscr{Z}|$). What is the expected value for the random variable $\xi_n^k=|X|$, where $|X|$ stands for the cardinality of $X$? In addition, what can be said about asymptotic behaviour of $\xi_n^{\lfloor n\delta\rfloor}$ for a fixed $\delta>0$? REPLY [8 votes]: $\DeclareMathOperator\E{E}$As already noted by the other answers, $$\E\xi^k_n=\frac{\sum_{i=0}^ki\binom ni}{\sum_{i=0}^k\binom ni}.$$ One can then easily determine the asymptotics of $\E\xi_n^{\lfloor n\delta\rfloor}$ for fixed $0\le\delta\le1$: Case 1: $0\le\delta<1/2$. Then $$k-\frac{\delta}{1-2\delta}\le\E\xi_n^k\le k,$$ where $k=\lfloor n\delta\rfloor$. This can be shown by approximation of $\binom ni$ by a geometric series. That is, we have $$0 TITLE: How is this HA unprovable formula recursive realizable? QUESTION [5 upvotes]: In Realizability: A Historical Essay [Jaap van Oosten, 2002], it is said that recursive realizability and HA provability do not concur, because although every HA provable closed formula is realizable, not every realizable closed formula is HA provable. Van Oosten points out that Propositional calculus and realizability [G.F. Rose] first proved this, but his example is a bit unwieldy and the old syntax a bit terse. Van Oosten also notes, in a footnote, the following simple formula due to G.S. Tseitin, which would be realizable but not HA provable: $$ \big[\neg (A \land B) \land (\neg A \to (C \lor D)) \land (\neg B \to (C \lor D))\big] \to \big[(\neg A \to C)\lor(\neg B \to C)\lor(\neg A \to D)\lor(\neg B \to D)\big] $$ I get how this formula is not HA provable, because it's essentially just De Morgan. But I don't see how this would be recursive realizable? Somehow De Morgen is intertwined twice, but I don't see how this would cleverly sitestep the otherwise unrealizable $\neg (A\land B) \to (\neg A \lor \neg B)$. REPLY [2 votes]: Let us first have a look at the core obstace for $$\neg (A \wedge B) \rightarrow (\neg A \vee \neg B)$$ Negated formulas do not have computational content, so $\neg (A \wedge B)$ is only a promise that either $A$ or $B$ is false, but does not provide any useful information. But we need to explicit compute a bit telling us whether $\neg A$ or $\neg B$ is going to be realized, and we have no input to compute this from. Now in the given formula, we have three premises: As before, $\neg (A \wedge B)$ is merely a promise, but contains no information. $\neg A \rightarrow (C \vee D)$ is realized by a Turing machine $M_1$ that if $\neg A$ holds, will terminate, and tell us either that $C$ holds (and provide a realizer), or that $D$ holds (and provide a realizer). Likewise, $\neg B \rightarrow (C \vee D)$ gives a Turing machine $M_2$, working analogously. The promise that $\neg (A \wedge B)$ guarantees that at least one of $M_1$ and $M_2$ halts. Markov's principle lets us compute the index of one that does. For example, assume that $M_1$ halts and claims that $C$ holds (and gives a claimed realizer of $C$). Classically, this means one of two things: Either $\neg A$ holds, in which case $M_1$ is required to be correct, or $\neg A$ does not hold. In either case, we can now claim that $\neg A \rightarrow C$ (on the righthand side of the big implication), and give the claimed realizer of $C$ as the output here. Any other combination of what machine $M_1$, $M_2$ halts, and what they are claiming then, also gives us one of the cases on the right.<|endoftext|> TITLE: Equivariant bundles invisible in K-theory and Borel cohomology QUESTION [8 upvotes]: For a given topological group $G$ there are natural transformations $$K^* \leftarrow K^*_G \overset a\to H^{**}(EG \times_G -;\mathbb Q)$$ from equivariant K-theory, the first forgetting the $G$-structure of a bundle and $a$ inducing from an equivariant bundle over $X$ a nonequivariant one over $EG \times_G X$ and taking the Chern character. These maps do respectably at detecting equivariant K-theory classes in nice cases, so I'm interested in how they can fail. Let us simplify matters by rationalizing and taking $X$ and $G$ compact. Then the forgetful map factors as $$K^*_G(X;\mathbb Q) \overset a\to H^{**}(EG \times_G X;\mathbb Q) \to H^{**}(X;\mathbb Q) \overset\cong\to K^*(X;\mathbb Q),$$ so we really only want the kernel of $a$. As $a$ can be viewed as completion of the $R(G) \otimes \mathbb Q$-module $K_G^*(X;\mathbb{Q})$ at the augmentation ideal $I(G;\mathbb Q)$, its kernel consists of classes annihilated by some element of $1 + I(G;\mathbb Q)$. Such classes exist when $G$ is discrete (already in $R(G) \otimes \mathbb Q$), but I don't know any where $G$ is connected. Does anyone have an explicit example of this happening when $G$ is connected? REPLY [11 votes]: The simplest example of what you are looking for occurs when $G = S^1$ and $X=S^1/C_6$, where $C_6$ is the group of 6th roots of unity. Then the map $$ K_G(X) \rightarrow K(EG\times_G X)$$ identifies with $$ R(C_6) \rightarrow K(BC_6),$$ which has free abelian kernel of rank 2 (corresponding to the two conjugacy classes in $C_6$ of order not a power of a prime).<|endoftext|> TITLE: Spinozistic partitionings [sic] QUESTION [6 upvotes]: Joel Friedman in 1974 invented these things called Spinozistic partitionings [sic] (of a set) where the pieces are all pairwise isomorphic as binary structures (piece, $\in$). He shows that $V_\omega$ has one. The idea seems to me to hold possibilities for exercises for a set theory course, but beyond that i can't see any real reason why i should care about them. Is there some model theory angle i'm missing..? REPLY [3 votes]: I hope a negative answer isn't inherently rude - that's certainly not my intention - but that does seem to be the situation, as far as I can tell. Friedman's original paper appears in the journal Synthese, which has a strong philosophical focus. In particular, quoting from Friedman's paper: In Friedman (1972), it was presented as a philosophical thesis that, in general, maximization implies decidability in set theory. ... [I]n this paper, we attempt to support our general thesis by offering still another concrete, and hopefully, non-trivial example, showing that Spinoza's metaphysical system generates some settheoretical concepts and theorems. Conversely, we hope to show that Set Theory sheds light on Spinoza's metaphysics. So Friedman's paper was not motivated by a particular mathematical relevance of the combinatorial principles involved, but rather the (claimed) connection between them and specific philosophical issues. Obviously I'm not qualified to judge the validity of this connection, but I think this does explain that the notion of Spinozistic partitionings emerged as an attempt to find a mathematical topic corresponding to a philosophical topic rather than as a concept relevant to a specific problem (or similar). The situation, as far as I can tell, hasn't changed since then. I could only find two follow-up papers (both also published in Synthese, indeed in the same issue): "The universal class has a Spinozistic partitioning" by Friedman and "Spinozistic partitions of classes" by John Lake. The papers address specific mathematical questions appearing in Friedman's original paper. At a quick glance, neither paper appears to give more mathematical motivation for the notion beyond independent interest (or further develop the philosophical angle); between this and the general lack of work on the topic, I think that the answer to your question is no. Of course, it's essentially impossible to prove a negative like this, and I wouldn't be too surprised if there turned out to be a relation between these and some more mainstream notion, but I don't see any at the moment.<|endoftext|> TITLE: Why does this quasi-modular function have integral values? QUESTION [21 upvotes]: It is a well-known result that the modular function $1728J(\tau) := \frac{1728E_4(\tau)^3}{E_4(\tau)^3-E_6(\tau)^2}$ has integral values if $\tau$ has class number 1 - for example at $\tau_{163}:=\frac{1+i\cdot\sqrt{163}}{2}$ you get $1728J(\tau_{163})=-640320^3$. Now define the quasi-modular function $s_2(\tau):=\frac{E_4(\tau)}{E_6(\tau)}\cdot\left(E_2(\tau)-\frac{3}{\pi Im(\tau)}\right)$. Then I have looked at all 13 class 1 discriminants and have verified numerically that $$M(\tau):=s_2(\tau)\cdot(1728J(\tau)-1728)=\frac{1728E_4(\tau)E_6(\tau)}{E_4(\tau)^3-E_6(\tau)^2}\left(E_2(\tau)-\frac{3}{\pi Im(\tau)}\right)$$ also has integral values for all these $\tau_N$. For example $M(\tau_{163})=-2^{13}\cdot3^5\cdot5\cdot7\cdot11\cdot19\cdot23\cdot29\cdot127\cdot181$. My Question: How can I prove that $M(\tau)\in\mathbb Z$ for all $\tau$ with class number 1? Or where can I find a proof for it? Edit: After the answer of @Zavosh (thank you!!!) it remains to prove this question. Who can help? REPLY [6 votes]: Algebraicity is proven in appendix one of [1]. The function considered there is $$ \psi(\tau) = \frac{3E_4(\tau)}{2E_6(\tau)} (E_2(\tau) - \frac{3}{\pi \rm{Im} \tau}) = \frac{3}{2} s_2(\tau).$$ The proof is by establishing, for quadratic irrationals $\tau$, the identity $$ \psi(\tau) = 9j(\tau)\gamma + \frac{3(7j(\tau)-6912)}{2(j(\tau)-1728)}$$ where $\gamma$ is a rational function in the coefficients of the Taylor series of a certain polynomial $\Phi(X,Y)$ around the point $(j(\tau),j(\tau))$ where it vanishes. Then $$M(\tau) = 1728 s_2(\tau) (j(\tau)-1) = 1728(j(\tau)-1) ( 6j(\tau)\gamma + \frac{(7j(\tau) - 4\cdot1728)}{j(\tau)-1728}.$$ There is also a different identity for calculations, for when $\tau$ is not equivalent to $i$ or $\rho$, and satisfies $$ A \tau^2 + B \tau + C = 0$$ with $A,B,C$ coprime positive integers: $$\psi(\tau) = \frac{-g_2 S}{Cg_3 (2A+B\tau)}$$ where $S$ is the sum of $\wp(z)$ as $z$ ranges over $C\tau$-torsion points of $\mathbb{C}/\langle 1,\tau \rangle$. Masser gives a table of $\psi(\tau)$, for a list of generators of rings of integers of imaginary quadratic fields with class number 1. They are all rational numbers, and one can indeed check that $$ M(\frac{-1+\sqrt{-163}}{2}) = 223263987730882560,$$ which agrees with the factorization in the question. Also $$\begin{align}M(\frac{-1+\sqrt{-67}}{2})&=-112852776960=-2^{11}\cdot 3^5 \cdot 5 \cdot 7 \cdot 11 \cdot 19 \cdot 31\\ M(\frac{-1+\sqrt{-43}}{2})&=-627056640=-2^{13}\cdot 3^7 \cdot 5 \cdot 7\\ M(\frac{-1+\sqrt{-27}}{2})&=-7772161=-1181\cdot 6581\\ M(\frac{-1+\sqrt{-19}}{2})&=-497664=-2^{11} \cdot 3^5\\ M(\frac{-1+\sqrt{-11}}{2})&=-14336=-2^{11} \cdot 7\end{align},$$ etc. Probably one can prove $M(\tau)$ is an algebraic integer by going through Masser's calculations and clearing denominators. [1] Masser, David W. Elliptic Functions and Transcendence<|endoftext|> TITLE: 2-natural operations on toposes QUESTION [9 upvotes]: Any pseudonatural endomorphism $\Phi$ of the forgetful 2-functor $U:Topos^{coop}\to Cat$ is essentially determined by its component $\Phi_{Set}$. But which endofunctors of $Set$ induce such a $\Phi$? More generally, one can consider pseudonatural transformations $U^n \Rightarrow U$, which are determined by a functor $\mathsf{Set}^n \to \mathsf{Set}$. Here $\mathsf{Topos}^\mathrm{coop}$ is the 2-category of Grothendieck toposes and left exact left adjoint functors -- doubly dual to the 2-category of toposes and geometric morphisms. The functor $U : \mathsf{Topos}^\mathrm{coop} \to \mathsf{Cat}$ sends a topos to its underyling category and a left exact left adjoint to its underlying functor. For $n \in \mathbb N$, the functor $U^n$ is the composite of $U$ with the $n$th power functor $\mathcal C \mapsto \mathcal C^n$. So it's natural to define a 2-natural operation on toposes to be a pseudonatural transformation $U^n \to U$ for $n \in \mathbb N$. In this language, the question is: Question: What are all the 2-natural operations on toposes? To see that such an operation $\Phi: U^n \Rightarrow U$ is determined by the component $\Phi_\mathsf{Set}: \mathsf{Set}^n \to \mathsf{Set}$, first note that $\Phi$ is determined by its components on presheaf toposes, because left exact localizations are essentially split epimorphisms in $\mathsf{Cat}$. Then the components at presheaf toposes are determined by the components at $\mathsf{Set}$ because $U$ preserves $\mathsf{Cat}$-cotensors. I would also be interested in the analogous question for $\infty$-toposes. REPLY [11 votes]: (For me the category of toposes is the opposite of the category of left exact left adjoint functors and natural transformations, so $Topos^{co}$ in your sense) The functor $U$ is representable by the classifying topos of objects, i.e. the topos $S[\mathbb{O}]$ which as a category is the category of functors from finite sets to sets, i.e. : $$ U( \mathcal{T}) = Hom(\mathcal{T}, S[ \mathbb{O}] ) $$ Similarly, $$ U(\mathcal{T}) ^n = Hom(\mathcal{T}, S[\mathbb{O}]^n)$$ $S[\mathbb{O}]^n$ being the category of functors from (finite set)$^n$ to Set. Now by the Yoneda lemma, and up to $2$-categorical details that I will totally ignore, a natural transformation from $U^n$ to $U$ is the same as a morphism from $S[\mathbb{O}]$ to $S[\mathbb{O}]^n$, i.e. it is given by an object of $S[\mathbb{O}]^n$, i.e. a functor from (finite set)$^n$ to Set. Claim/exercice: given a functor from (finite set)$^n$ to $Set$ its actions $Set^n \rightarrow Set$ corresponds to the unique extension commuting to directed colimits. So in the end those "operations" are exactly the same as the operations $Sets^n \rightarrow Sets$ which commutes to directed colimits. I believe everything works exactly the same for $\infty$-toposes, replacing sets and finite sets by "spaces" and "finitely generated spaces" (I mean finite CW-complexes), (Of course this will relies on a large amount of results from Lurie's books, although I think one can avoid manipulating $(\infty,2)$-categories by just forgeting the non-invertible $2$-cells at least in a first time...) REPLY [10 votes]: The "2-natural operations" $U^n \to U$ correspond to functors $\mathbf{FinSet}^n \to \mathbf{Set}$. (edit: As Simon points out, these correspond to the finitary functors $\mathbf{Set}^n \to \mathbf{Set}$.) The 2-functor $U$ is birepresented by the object classifier $\mathbf{Set}[\mathbb{O}] = [\mathbf{FinSet},\mathbf{Set}]$ (the classifying topos for the theory of objects). Thus by the bicategorical Yoneda lemma, the category of pseudonatural transformations $U^n \to U$ is equivalent to the category $U(n\cdot \mathbf{Set}[\mathbb{O}])$ (where $ n \cdot {}$ denotes $n$-fold (bi)coproduct in $\mathrm{Topos}^\mathrm{coop}$), which is equivalent to the category $[\mathbf{FinSet}^n,\mathbf{Set}]$ (see Cole's paper The bicategory of topoi and spectra).<|endoftext|> TITLE: Smooth algebras always lift QUESTION [6 upvotes]: Let $k$ be a finite field, $A$ a smooth $k$-algebra. Does there exists a smooth algebra $B$ over the Witt vectors $W(k)$, such that $B/p\simeq A$? How is it constructed? REPLY [4 votes]: This follows from a result of Elkik in [R. Elkik Solutions d’équations à coefficients dans un anneau hensélien Annales scientifiques de l’É.N.S. 4e série, tome 6, no 4 (1973), p. 553-603.]. Theorem (Theorem 6 in Section 4 on p. 580, changed notation to fit yours). Let $(R, J)$ be a noetherian henselian couple. Then every smooth $R/J$-algebra $A$ lifts to a smooth $R$-algebra $B$. Of course if you only want to lift formally over $R=W(k)$, that is to find a system $B_n$ over $W_{n+1}(k)$ lifting $B_{n-1}$ with $B_0 = A$, then the result follows from elementary deformation theory: the obstruction to lifting $B_{n-1}$ over $W_{n+1}(k)$ lies in $H^2({\rm Spec}(A), T_{{\rm Spec}(A)})=0$. But why should the inverse limit of the $B_n$ be the $p$-adic completion of a smooth $W(k)$-algebra? Similarly, if $A$ is a smooth complete intersection, then it is easy to lift $A$ over $W(k)$: just lift the equations, and then maybe localize if the lifting is not smooth.<|endoftext|> TITLE: Can a hyperbolic, one ended, one relator group, have a shorter trivial word? QUESTION [21 upvotes]: Let $G= \langle S \mid r \rangle$ be a one-relator presentation for a one-ended hyperbolic group, with $r$ cyclically reduced. Question: Can there be a nontrivial word $w(S)$ which is trivial in the group $G$ but has length shorter than $r$? What if $r$ is the shortest possible word for a one-relator presentation of $G$? Note that it follows from Newman's spelling theorem that in the torsion case there are no shorter words, since you can apply Dehn's algorithm. Similarly if $r$ gives a $C'(1/6)$ presentation there are no shorter words. Generally it is known that subwords of $r$ will not be trivial either. This is proved by Weinbaum in On relators and diagrams for groups with one defining relation. This question grew out of this question on math.se and my answer to it. One thing to note is that without hyperbolicity you can find that some Baumslaug-Solitar groups provide examples with shorter trivial words. REPLY [14 votes]: I think I found an example with shorter trivial words using a handy characterizations in a paper by Ivanov and Schupp called On hyperbolicity of small cancellation groups and one-relator groups. Consider $\langle a,b,c \mid ab^2ac^{12}\rangle$. By checking Whitehead automorphisms this relation is as short as possible in the $Aut(F_3)$ orbit and has every generator in the relation, so does not have infinitely many ends(Thanks to ADL for the correction). One can look at the abelianization to rule out zero or two ends. Theorem 3 in the above paper says that this group is hyperbolic since it has exactly two occurrences of $a$, no $a^{-1}$, and $b^2c^{-12}$ is not a proper power in the free group generated by $a,b,c$. Now $$(ab^2ac^{12})c(c^{-12}a^{-1}b^{-2}a^{-1})c^{-1}=ab^2aca^{-1}b^{-2}{a^{-1}}c^{-1}$$ which is shorter than the defining relation.<|endoftext|> TITLE: $\operatorname{Out}(F_n)$ is not linear for $n > 3$ QUESTION [7 upvotes]: The paper The Tits alternative for $\operatorname{Out}(F_n)$ I by Bestvina, Feighn and Handel and the paper Automorphisms of free groups and Outer space by Vogtmann both state that $\operatorname{Aut}(F_n)$ and $\operatorname{Out}(F_n)$ are not linear groups for $n \geq 3$ respectively $n \geq 4$. They both cite The automorphism group of a free group is not linear by Formanek and Procesi. This text proves the claim about the automorphism group of a free group, but does not mention the claim about the outer automorphism group. I was wondering if some properties about linear groups imply this result, but I couldn't figure out what property would. Any help would be appreciated. Once again, not really sure if this is the right place to post this question. If not, I'll take it down (or if someone knows how to relocate it to mathstack, that would be great too) REPLY [17 votes]: There is an embedding $\text{Aut}(F_{n-1}) \hookrightarrow \text{Out}(F_n)$ for any $n \ge 2$, as follows. First one embeds $\text{Aut}(F_{n-1}) \hookrightarrow \text{Aut}(F_n)$ by extending any automorphism of $F_{n-1}$ to an automorphism of $$F_n \approx F_{n-1} * \mathbb{Z} $$ using the identity automorphism of $\mathbb{Z}$. Next one notices that none of the resulting automorphisms of $F_n$ are inner, hence we get an embedding $\text{Aut}(F_{n-1}) \hookrightarrow \text{Out}(F_n)$. It follows that if $\text{Out}(F_n)$ is linear then $\text{Aut}(F_{n-1})$ is linear.<|endoftext|> TITLE: What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$? QUESTION [5 upvotes]: What is $\mathrm{Hom}(\mathbb{Q},\mathbb{Z}(p^{\infty}))$? I have a reference that says the group in question is $\mathbb{Q}_p,$ the additive group of the quotient field of the $p$-adic integers. Can anyone provide a reasonable derivation of this result? REPLY [7 votes]: YCor's answer is good, but here's a slightly different way of seeing it. Given $x \in \mathbb Q$, let $[f(x)]$ be any lift of $f(x)$ from $\mathbb Z[1/p]/\mathbb Z$ to $ \mathbb Z[1/p]$. Then $p^n [ f(1/p^n)]$ is a $p$-adic Cauchy sequence because $p^n [f(1/p^n)] - p^m [ f(1/p^m)]$ is an integer multiple of $p^{\min (n,m)}$. Changing the lift also causes a difference by an integer multiple of $p^n$ and thus does not affect the limit. This defines for each $f$ an element of $\mathbb Q_p$. To go the other way, given $\alpha \in \mathbb Q_p$, observe that multiplication by $\alpha$ defines a map $\mathbb Q$ to $\mathbb Q_p$. Compose this with the projection to $\mathbb Q_p / \mathbb Z_p \cong \mathbb Z[1/p] / \mathbb Z$ to obtain a map $\mathbb Q \to \mathbb Z[1/p]$. It is straightforward to check that these are inverses. Starting with $\alpha$, $[f(1/p^n)]$ is an element of $\mathbb Z_p$ plus $\alpha /p^n$ so $p^n [f(1/p^n)]$ is an approximation of $\alpha$. Starting with $f$, the limit $\alpha$ differs from $p^n [f(1/p^n)]$ by an element of $p^n \mathbb Z_p$, so $f(1/p^n)$ differs from $\alpha p^{-n}$ by an element of $\mathbb Z_p$, and then multiply these by elements of $\mathbb Z$ inverting every prime other than $p$ to get all rational numbers.<|endoftext|> TITLE: Intuition of the energy of a graph QUESTION [7 upvotes]: The energy of a a simple graph $G$ is defined to be the sum of the absolute values of the eigenvalues of $G$. What is a good intuition of it? REPLY [2 votes]: The energy of a graph equals to the nuclear norm $\|A\|_*$ of the adjacency matrix $A$ of the graph. So, an intuition of the nuclear norm can give an intuition of the energy of a graph. Therefore answers to the following question may be useful: What is the intuition for the trace norm (nuclear norm)? The following intuition may be more interesting which has some "energy" sense $$ \|A\|_{\mathrm{*}} = \inf\big\{ \sum_j \|X_j\| : A = \sum_j X_j, \ \ \mathrm{rank}(X_j) = 1 \ \ \forall j \big\}. $$<|endoftext|> TITLE: Linear independence of genus-one correlation functions QUESTION [11 upvotes]: Let $V$ be a vertex operator algebra with all the good finiteness properties that people usually assume (positively graded, $C_2$-cofinite, $V\cong V'$, etc.) Let $W$ be a module for $V$, not necessarily irreducible. How does one prove, or where can I find a proof of the following fact: Claim: The map $$ \bigoplus_{\tilde W\in\text{irrep}(V)} \operatorname{Hom}_{\mathrm{Rep}(V)}(W\boxtimes\tilde W,\tilde W)\to \operatorname{Hom}(W,\{\text{functions of $q$ and $z$}\}) $$ which sends an intertwining operator $\mathcal Y$ type $\binom {\tilde W} {W \tilde W}$ to the function that sends $w\in W$ to $$ (q,z)\mapsto \operatorname{Tr}_{\tilde W}(\mathcal Y(w,z)q^{L(0)}) $$ is injective. Thank you for your help. REPLY [6 votes]: A proof of this property can be found in Proposition 2.2 of Huang's paper Vertex operator algebras and the Verlinde conjecture. This proof actually uses Zhu's algebras to simplify the discussions (similar to Zhu's proof of the linear independence of characters in Modular invariance of characters of vertex operator algebras) although I think this is not necessary. So let me give an elementary argument without using Zhu's algebras. Indeed I think this argument should also work for proving the injectivity of the sewings of (the vector spaces of) general conformal blocks. Let me change your notations a little bit, and restate your claim as follows: Proposition. Let $M_1,\dots,M_n$ be a list of inequivalent irreducible $V$-modules. For any $k=1,\dots,n$ we choose any intertwining operator $\mathcal Y_k$ of type $M_k\choose WM_k$. Assume that $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)q^{L_0})=0$ for any $w,z,q$. Then $\mathcal Y_k=0$ for all $k$. Proof. It suffices to show that there exits a $\mathcal Y_k$ equaling $0$. Then, by induction, all $\mathcal Y_k$ are $0$. For each $k$ we have a natural injective linear map $A_k:M_k\otimes\overline M_k\rightarrow End(M_k)$, where $\overline M_k$ is the contragredient (conjugate) $V$-module of $M_k$. Set $\mathfrak M=\bigoplus_kM_k\otimes\overline M_k$, and set $A=\bigoplus_k A_k$. We then have $$A:\mathfrak M\hookrightarrow\bigoplus_k End(M_k).$$ Now let $\mathfrak N$ be the subspace of all $\xi\in\mathfrak M$ satifying that $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)A(\xi))=0$ for any $w,z$. (Here we regard the action of $\mathcal Y_{k_1}$ on $M_{k_2}$ as zero if $k_1\neq k_2$.) $$\mathfrak N := \left\{\xi\in\mathfrak M:\sum_k Tr_{M_k}(\mathcal Y_k(w,z)A(\xi))=0\right\}$$ We claim that $\mathfrak N\neq0$. Indeed, since $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)q^{L_0})=0$, if we regard it as a series of $q$: $\sum_{\Delta}\sum_k Tr_{M_k(\Delta)}(\mathcal Y_k(w,z))q^\Delta$, then any coefficient $\sum_k Tr_{M_k(\Delta)}(\mathcal Y_k(w,z))$ must be $0$ 1. Choose a conformal weight $\Delta$ such that there exists $k$ such that $M_k(\Delta)$ is non-zero. Then we immediately have the non-zeroness of $\mathfrak N$. Note that $\mathfrak M$ is indeed a $V\otimes V$-module. We now show that $\mathfrak N$ is a $V\otimes V$-submodule. Choose any $\xi\in\mathfrak N$. Then, for any $w,z$, since $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)A(\xi))=0$, we have $\sum_k Tr_{M_k}(\mathcal Y_k(Y(u,z_0-z)w,z)A(\xi))=0$ for any $u$ and any $z_0$ such that $0<|z_0-z|<|z|$. If moreover $0<|z_0-z|<|z_0|<|z|$, then $\mathcal Y_k(Y(u,z_0-z)w,z)=\mathcal Y_k(w,z)Y(u,z_0)$. Thus $$\sum_k Tr_{M_k}(\mathcal Y_k(w,z)Y(u,z_0)A(\xi))=0$$ for all $z_0$ satisfying $0<|z_0-z|<|z_0|<|z|$. By uniqueness of analytic continuations, we have $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)Y(u,z_0)A(\xi))=0$ for all $z_0$ satisfying only $0<|z_0|<|z|$. Take contour integrals of $z_0$. Then we obtain $\sum_k Tr_{M_k}(\mathcal Y_k(w,z)Y(u)_mA(\xi))=0$ for any $m\in\mathbb Z$, where we write $Y(u,z_0)=\sum_{m}Y(u)_mz_0^{-m-1}$. Since $Y(u)_mA(\xi)=A((Y(u)_m\otimes 1)\xi)$, we've actually shown that $\mathfrak N$ is invariant under the action of $V\otimes \Omega$ (where $\Omega$ is the vacuum vector of $V$). Similarly $\mathfrak N$ is $\Omega\otimes V$-invariant. So $\mathfrak N$ is $V\otimes V$-invariant, i.e., a $V\otimes V$ submodule of $\mathfrak M$. Since $\mathfrak M$ has irreducible decomposition $\mathfrak M=\bigoplus_k M_k\otimes\overline M_k$, and since $\mathfrak N$ is a non-trivial submodule of $\mathfrak M$, there exists $k$ such that $M_k\otimes\overline M_k\subset \mathfrak N$. Therefore, for any $w_1\in M_k,w'_2\in \overline M_k$, we have $\langle\mathcal Y_k(w,z)w_1,w'_2\rangle=Tr_{W_k}(\mathcal Y_k(w,z)A(w_1\otimes w_2'))=0$ for any $w,z$. Thus $\mathcal Y_k=0$. Q.E.D. 1 We know that the coefficients $c_n$ of a convergent power series $f(z)=\sum_{n\in\mathbb N}c_nz^n$ are determined by the values of $f$. This is also true if $f$ can be written as a finite sum $f(z)=\sum_m z^{\mu_m}g_m(z)$ where each $g_m(z)$ is a convergent power series of $z$ and $\mu_m\in\mathbb C$. This is an easy exercise. See for example A theory of tensor products for module categories for a vertex operator algebra, IV Lemma 14.5 and section 15.4.<|endoftext|> TITLE: Is the Riemann zeta function surjective? QUESTION [46 upvotes]: Is the Riemann zeta function surjective or does it miss one value? REPLY [8 votes]: (Just a quick aside: the set of solutions to the equation $\zeta(s) = a$ when $a \neq 0$ are usually called the $a$-points of the Riemann zeta function in the literature, in case you want to look up the state of the art) As Steve Huntsman mentions in a comment, it is indeed possible to use the universality of the zeta function to prove that $\zeta(s)$ is surjective with a simple argument based on Rouché's theorem. In particular, let $U$ be a compact connected subset with a smooth boundary in the critical strip on which universality holds (for simplicity, take $U = \{ z \in \mathbb{C} : |z - 3/4| \leqslant 1/8 \}$), and let $f:U \to \mathbb{C}$ be a function holomorphic on $U$ (i.e. holomorphic in the interior, and continuous on the boundary) with the following properties: $f$ has no zeroes in $U$. There exists a $z \in U$ such that $f(z) = a$. For all $z \in \partial U$, $f(z) \neq a$. As an example, for $U$ as above, one could take $f(z) = a + |a| (z-3/4)$. Now note that 3 implies that there is an $\epsilon$ such that $$0 < \epsilon < \inf_{z \in \partial U} |f(z) - a|.$$ Then, due to 1, universality guarantees that there is a $t \geqslant 0$ such that $$ |\zeta(s+it) - f(s)| < \epsilon$$ for every $s \in U$. In particular, this inequality holds for every $s \in \partial U$. Thus, for every $s \in \partial U$, we have that $$ |\zeta(s+it) - f(s)| < \epsilon < \inf_{z\in\partial U} |f(z) - a| \leqslant |f(s) - a|. $$ Thus, Rouché's theorem implies that $f(s) - a$ and $f(s) - a + \zeta(s+it) - f(s) = \zeta(s+it) - a$ have the same number of roots in the interior of $U$. Since $f$ has such a root, it follows that $\zeta$ has such a root as well, and we are done. This argument is robust in a few ways. For one, it applies to other functions which exhibit universality (which is known for fairly large classes of $L$-functions and other zeta-like families of interest). For another, effective versions of Voronin's universality theorem can be used to give a good lower bound for the number of $a$-points in a rectangle within the critical strip. For a third, this argument actually proves that there are simple $a$-points for every $a \neq 0$. This argument is probably classical, but I first saw it in a paper on simple $a$-points by Gonek, Lester and Milinovich.<|endoftext|> TITLE: On models of $Th_{\Pi_2}(PA)$ QUESTION [7 upvotes]: Let $M$ be a nonstandard model of $PA$. Q1. Is there any way to get a submodel $N\subset M$ such that $N\models Th_{\Pi_2}(PA)$, but $N\not\models PA$? Q2. Especially, what combinatorial principle can be used to construct a model $N\models Th_{\Pi_2}(PA)$ such that $N\not\models PA$? Or $Th_{\Pi_2}(PA)\vdash PA$? REPLY [8 votes]: $\def\pa{\mathit{PA}}\def\rfn{\mathrm{RFN}}\def\pr{\mathrm{Pr}}\def\num#1{\ulcorner#1\urcorner}\def\Th{\mathrm{Th}}$ $\Th_{\Pi_2}(\pa)$ certainly does not prove $\pa$: Theorem: For any constant $k$, there is no consistent $\Pi_k$-axiomatized theory $T$ such that $T\vdash\pa$. More generally, if $S$ is a sequential theory that proves full induction, then $S$ is not derivable from any consistent theory $T$ axiomatized by formulas of restricted quantifier complexity. (So, for example, $\mathit{ZF}$ is not derivable from any consistent theory axiomatized by $\Pi_k$-sentences in the Levy hierarchy.) The reason is that $\pa$ proves the uniform reflection principles $$\rfn_Q(\Pi_k)=\forall x\,(\pr_Q(x)\land x\in\Pi_k\to\mathrm{Tr}_k(x)),$$ where $\mathrm{Tr}_k$ is the truth predicate for $\Pi_k$-sentences, $\mathrm{Pr}_Q$ is the provability predicate for $Q$, and $Q$ could really stand for any fixed finitely axiomatized subtheory of $\pa$ (the choice does not matter much). So, if $T$ is a $\Pi_k$-axiomatized theory that proves $\pa$, then there is a finitely axiomatized subtheory $T_0\subseteq T$ that proves $\rfn_Q(\Pi_{k+1})$, and (say) $I\Sigma_1$. Let $\psi=\bigwedge T_0$. Then $$\pr_Q(\num{\neg\psi})\to\neg\psi$$ is an instance of $\rfn_Q(\Pi_{k+1})$, which implies that $T_0$ proves its own consistency. By Gödel’s theorem, $T_0$ is inconsistent, hence so is $T$. As for constructing submodels of $M\models\pa$ satisfying $\Th_{\Pi_2}(\pa)$ but not $\pa$, there is the following general result due to McAloon [1]: Theorem (McAloon): Let $T$ be a recursively axiomatizable $\Sigma_1$-sound extension of $I\Sigma_1$. Then for any countable nonstandard model $M\models I\Delta_0$, there is a nonstandard cut $N\subseteq M$ such that $N\models T$. This was generalized to $M\models IE_1$ by Wilmers [2]. McAloon’s argument relies on the indicator theory, hence one may consider it a combinatorial construction. In order to get $N\models\Th_{\Pi_2}(\pa)$, $N\not\models\pa$, we can apply this theorem to $$T=I\Sigma_1+\Th_{\Pi_2}(\pa)+\neg\rfn_Q(\Pi_4).$$ We only need to check that $T$ is $\Sigma_1$-sound. In fact, $T$ is $\Pi_4$-conservative over the (sound) theory $I\Sigma_1+\Th_{\Pi_2}(\pa)$. To see this, let $\phi$ be a $\Pi_4$-sentence provable in $T$. Notice that $I\Sigma_1+\Th_{\Pi_2(\pa)}$ is $\Pi_3$-axiomatizable, hence there exists a $\Pi_3$-sentence $\psi$ derivable in $I\Sigma_1+\Th_{\Pi_2(\pa)}$ such that $$\psi\land\neg\phi\vdash\rfn_Q(\Pi_4).$$ As before, an instance of $\rfn$ for the $\Pi_4$-sentence $\neg(\psi\land\neg\phi)$ shows $$\psi\land\neg\phi\vdash\mathrm{Con}_{\psi\land\neg\phi},$$ hence $\psi\land\neg\phi$ is inconsistent by Gödel’s theorem, i.e., $$I\Sigma_1+\Th_{\Pi_2}(\pa)\vdash\psi\vdash\phi.$$ [1] Kenneth McAloon, On the complexity of models of arithmetic, Journal of Symbolic Logic 47 (1982), no. 2, pp. 403–415. [2] George Wilmers, Bounded existential induction, Journal of Symbolic Logic 50 (1985), no. 1, pp. 72–90. REPLY [5 votes]: Emil Jeřábek has already provided a nice answer. Here is another answer to complement his. Answer to Q1: Yes, indeed $2$ can be replaced with any natural number $n>0$. Explanation: Let $M$ be a nonstandard model of $PA$ and let $a$ be a nonstandard element of $M$, and $n$ be a fixed natural number, and let $K^{n}(M,a)$ be the submodel of $M$ consisting of elements that are definable in $(M,a)$ using a $\Sigma_n$ unary formula with parameter $a$. Then we have: Theorem (Kirby & Paris). For all natural numbers $n\geq 1$: (1) $K^{n}(M,a)$ is a $\Sigma_n$-elementary submodel of $(M,a)$. (2) The collection scheme $B\Sigma_n$ fails in $K^{n}(M,a)$. Note that the first condition of the theorem assures us that $K^{n}(M,a)$ satisfies all $\Pi_n$-consequences of $PA$, and the second condition tells us that $K^{n}(M,a)$ is not a model of $PA$. [You can find an exposition of the theorem above in Chapter 10 of Kaye's textbook Models of Peano Arithmetic, or in chapter IV of the Hajek-Pudlak textbook Metamathematics of First Order Arithmetic]. Answer to Q2: The usual method is to arrange a model that supports a definable map whose domain is the predecessors of a "number", and whose range is unbounded in the model. This is equivalent to the failure of the scheme $B\Sigma_n$ mentioned above.<|endoftext|> TITLE: Adjoints of scalar extension and scalar coextension QUESTION [8 upvotes]: Let $h\colon R\rightarrow S$ be a morphism of commutative rings. We consider the following functors (I am aware that the notations may be different in other contexts): $h^*$: Scalar extension by means of $h$, i.e. $h^*(M)=M\otimes_RS$; $h_*$: Scalar restriction by means of $h$; $\widetilde{h}$: Scalar coextension by means of $h$, i.e., $\widetilde{h}(M)={\rm Hom}_R(S,M)$. These functor form the adjoint pairs $(h^*,h_*)$ and $(h_*,\widetilde{h})$. Moreover, using some general nonsense one can show that the following statements are equivalent: (i) $h^*$ has a left adjoint; (ii) $\widetilde{h}$ has a right adjoint; (iii) The $R$-module $h_*(S)$ is projective and of finite type. So, suppose that (i)-(iii) are fulfilled. My question is then: What are the left adjoint of $h^*$ and the right adjoint of $\widetilde{h}$? ADDENDUM: It is a result of Morita (Theorem 4.1 in K. Morita, Adjoint pairs of functors and Frobenius extensions, Sci. Rep. Tokyo Kyoiku Daigaku Sect. A 9 (1965), 40--71) that (still under the above conditions (i)-(iii)) $$h^*\cong\widetilde{h}\quad\Longleftrightarrow\quad\widetilde{h}(R)\cong S.$$ Therefore, it remains to consider situations where $\widetilde{h}(R)$ and $S$ are not isomorphic. REPLY [7 votes]: If $X$ is an $R$-module, there is a natural map $M\otimes_RX\to\text{Hom}_R\left(\text{Hom}_R(X,R),M\right)$ given by $m\otimes x\mapsto[\varphi\mapsto m\varphi(x)]$ that is easily checked to be an isomorphism when $X=R$, and hence (by additivity) when $X$ is a finitely generated projective. So assuming (i)-(iii), $h^*(M)=M\otimes_RS\cong\text{Hom}_R\left(\text{Hom}_R(S,R),M\right)$. This is an isomorphism of $S$-modules, since for $x,s\in S$ and $m\in M$, $m\otimes xs\mapsto[\varphi\mapsto m\varphi(xs)$. Thus $h^*$ has left adjoint $N\mapsto N\otimes_S\text{Hom}_R(S,R)$. Similarly, there is a natural map $M\otimes_R\text{Hom}_R(X,R)\to\text{Hom}_R(X,M)$ given by $m\otimes\vartheta\mapsto[x\mapsto m\vartheta(x)]$ that is an isomorphism for $X=R$ and hence for $X$ a finitely generated projective. So assuming (i)-(iii), $\widetilde{h}(M)=\text{Hom}_R(S,M)\cong M\otimes_R\text{Hom}_R(S,R)$. Again, this is an isomorphism of $S$-modules, since for $m\in M$, $\vartheta\in\text{Hom}_R(S,R)$ and $x,s\in S$, $(\vartheta s)(x)=\vartheta(sx)$, so the isomorphism (in the reverse direction) maps $m\otimes(\vartheta s)\mapsto [x\mapsto m\vartheta(sx)]$. Thus $\widetilde{h}$ has right adjoint $N\mapsto\text{Hom}_S\left(\text{Hom}_R(S,R),N\right)$.<|endoftext|> TITLE: Intuition about $\mathrm{Spec}\mathbb{C}[[t]]$ versus $\mathrm{Spf}\mathbb{C}[[t]]$ versus $\mathrm{Specan}\mathbb{C}[[t]]$ (and similar objects) QUESTION [20 upvotes]: The first one $\mathrm{Spec}\mathbb{C}[[t]]$ is a scheme, the second one $\mathrm{Spf}\mathbb{C}[[t]]$ is a formal scheme. In my mind they both realize an "infinite order infinitesimal neighbourhood of a point in $\mathbb{A}^1_{\mathbb{C}}$ ". The "formal disc" is an increasing limit of finite order infinitesimal neighbourhoods; pretty much like an increasing union of closed subschemes that are thickenings of the reduced point. How should I think of the others? So What's the difference between these objects, intuitively? How are these objects related to the fact that $\mathbb{C}[[t]]$ actually carries an adic topology? What is each of them best suitable for? When to "use" one and when the other(s)? Of course the same questions could be asked replacing $\mathbb{C}[[t]]$ with any complete local ring $A=\hat A$. REPLY [4 votes]: To complement the other answers, I would like to add a word on the analytic spectrum $\mathrm{Specan}(\mathbb{C}[[t]])$. First, let me say that I am not sure what $\mathrm{Specan}$ means and have no idea where the notation comes from. On the other hand, it makes sense to consider the analytic spectrum of $\mathbb{C}[[t]]$ in R. Huber's theory of adic analytic spaces (where it would be called $\mathrm{Spa}(\mathbb{C}[[t]],\mathbb{C}[[t]])$) and in V. Berkovich's theory of analytic spaces (where it would be called $\mathcal{M}(\mathbb{C}[[t]])$ ; one would also need to prescribe the absolute of $t$, say $|t| = r \in (0,1)$, because the theory requires actual absolute values and not merely equivalence classes, i.e. valuations, but this is not really an issue). Let me start with adic spaces. In this case, the spectrum is made of one closed point ($t=0$) and one open point (associated to the $t$-adic valuation). This open point is really the generic point of the space. More generally, you can associate (fully faithfully) an adic space to any sufficiently nice formal scheme and take its generic fiber inside the category of adic spaces. So here you really have the formal scheme again but, in the underlying space, you see its special fiber and its generic fiber (whereas $\mathrm{Spf}(\mathbb{C}[[t]])$ only shows the special fiber). (Everything here looks quite similar to $\mathrm{Spec}(\mathbb{C}[[t]])$, so one may wonder why bother with formal schemes, fancy analytic spaces, etc. This is a only because the chosen situation is very simple (affine in particular) and you will have a hard time representing infinitesimal neighbourhoods by algebraic objects very quickly as soon as you start glueing.) The situation with Berkovich spaces is very similar except that you will see a segment $[0,r]$ instead of two points. The point $0$ corresponds to $t=0$ and the other points all correspond to $t$-adic absolute values with different normalizations (given by $|t| = s$ for $s\in (0,r]$). Here again, you see a special fiber and a generic fiber (and even several equivalent copies of it). Note that the underlying space is Hausdorff and compact. This is a general feature of Berkovich spaces which is quite nice, although it is not clear how useful it would be in this particular situation.<|endoftext|> TITLE: Approximation of the identity by simple functions QUESTION [8 upvotes]: Let $X$ be a topological space. Assume that there exists a sequence of simple functions $\phi_n:X\to X$ (finite range and measurable) with $\lim\phi_n(x)=x$. Can we concluded $X$ may be written by a countable union $X=\cup X_n$ where all $X_n$'s are all relatively second countable? REPLY [7 votes]: The answer is no. The Moore plane is a counterexample. For $z \in \mathbb{R}^2$ and $r \geqslant 0$ denote by $\bar D(z, r)$ the closed Euclidean disc of center $z$ and radius $r$. Recall that the Moore plane is $M = [0, + \infty) \times \mathbb{R}$, where: A basis of neighborhoods of $(x, y) \in (0, +\infty) \times \mathbb{R}$ consists in all discs $\bar D((x, y), r)$ with $0 < r < x$; A basis of neighborhoods of $(0, y)$ consists in all discs $\bar D((r, y), r)$, with $r>0$. This space is a usual example of a separable topological space that is not second-countable. To see that $M$ is separable, remark that the set $D = \{(x, y) \in \mathbb{Q}^2 \mid x> 0\}$ is dense in $M$. However $M$ cannot be written as a countable union of second-countable spaces, otherwise the subspace $\{0\} \times\mathbb{R}$ could also be written as such a union; but the inherited topolgy on $\{0\} \times\mathbb{R}$ is discrete. To show that an approximation of the identity as required exists in $M$, let $(r_k)_{k \in \mathbb{N}}$ be an enumeration of $D$ and define $\phi_n$ as follows: for $z \in M$, consider $r \geqslant 0$ minimal such that the basic neighborhood of $z$ of radius $r$ contains an element of $\{r_0, \ldots r_n\}$, and let $\phi_n(z) = r_k$ be such an element with $k$ minimal. Then $\phi_n(z) \rightarrow z$: indeed, if $(r_{k_i})$ is a sequence of elements of $D$ converging to $z$, then for $U$ a basic neighborhood of $z$, there is $i_0 \in \mathbb{N}$ such that $r_{k_{i_0}} \in U$; and then, for all $n \geqslant k_{i_0}$, we have $\phi_n(z) \in U$. Obviously, $\phi_n$ has finite range, included in $\{r_0, \ldots, r_n\}$. And to show that $\phi_n$ is Borel, remark that for $k \in \{0, \ldots, n\}$, $\phi_n^{-1}(\{r_k\}) = A \sqcup B$, where $A \subseteq \{0\}\times \mathbb{R}$ and $$B = \{z \in (0, \infty) \times \mathbb{R} \mid (\forall 0 \leqslant i < k, \, d(z, r_k) < d(z, r_i)) \wedge (\forall k < i \leqslant n, \, d(z, r_k) \leqslant d(z, r_i))\},$$ where $d$ is the Euclidean distance in $\mathbb{R}^2$. $A$ is relatively open in $\{0\} \times \mathbb{R}$ as the inherited topology on this subset is discrete. As $\{0\} \times \mathbb{R}$ is closed in $M$, then $A$ is Borel in $M$. In the same way, $B$ is Borel in $\mathbb{R}^2$, so since the topology inherited from $M$ on $(0, +\infty) \times \mathbb{R}$ is the same as the topology inherited from $\mathbb{R}^2$, and since $(0, +\infty) \times \mathbb{R}$ is open in $M$, we deduce that $B$ is Borel in $M$. So $\phi_n^{-1}(\{r_k\})$ is Borel in $M$. REPLY [6 votes]: Here is a counterexample where the $\phi_n$'s are continuous. Let $X$ be the Sorgenfrey line, that is $X=\mathbb{R}$ with the topology generated by intervals of the form $[a,b)$ with $a,b \in \mathbb{R}$. Note that $X$ is not a countable union of second countable subspaces since for any $Y \subseteq X$ the size of a base for $Y$ must have cardinality at least $|Y|$ (so that $Y$ is second countable only if $Y$ is countable). Fixing an enumeration of the rationals $\mathbb{Q}=\{q_i:i \in \omega\}$, we can define $$\phi_n(x)=\min \{q_i:i\leq n\}\cap (x,\infty)$$ if the set $\{q_i:i\leq n\}\cap (x,\infty)$ is non-empty and $\phi_n(x)=\max \{q_i:i \leq n\}$ otherwise. It is easy to see that each $\phi_n$ is continuous and $\lim\phi_n(x)=x$ for every $x \in X$.<|endoftext|> TITLE: Expansion of elementary symmetric function in Jack's? QUESTION [6 upvotes]: Consider the expansion $$ e_\mu(x) = \sum_\lambda c_{\mu\lambda}(\alpha) J_\lambda^{(\alpha)}(x) $$ where $J_\lambda^{(\alpha)}(x)$ are the integral-form Jack polynomials (the ones with $n!$ as coefficient of $m_{1^n}(x)$). Is there some result which proves that each $c_{\mu\lambda}(\alpha)$ is of the form $P(\alpha)/Q(\alpha)$, where $P,Q \in \mathbb{N}[\alpha]$? It seems that $Q$ can be chosen as some $Q_\mu(\alpha)$ that factors in a nice way (some product of deformed hook values perhaps?) REPLY [2 votes]: Regarding my comment above, note that by induction we only need to show that $\langle J_\lambda, J_{1^n}J_\nu\rangle\in\mathbb{N}[\alpha]$. This follows from the dual Pieri formula for Jack polynomials (obtained by combining Theorems 3.3 and 6.1 of http://math.mit.edu/~rstan/pubs/pubfiles/73.pdf), so we do get $c_{\mu\lambda}(\alpha)\in\mathbb{N}[\alpha]$, as well as a factorization of $Q_\mu(\alpha)$ into linear factors.<|endoftext|> TITLE: Papers in which the questions were more interesting than the results QUESTION [33 upvotes]: I am looking for examples of recently (last 20 years, say) published math papers such that: the results/examples were fairly trivial (by this I mean anyone with the definitions and standard background in the area of research could have thought of them, but never took the time to do it, or it simply never occurred to them); and yet the questions posed in the papers, which were motivated by the results, lead to future research and solutions which were non-trivial. These should not be foundational papers in the sense that they introduced an entirely new field. Assume those papers published long ago, with books written on the subjects, etc. I guess this question stems from a fear that my papers fit this mold. The questions (often my own) I am unable to answer seem far more intriguing than what's actually in my papers... REPLY [8 votes]: It's older than 20 years but otherwise the paper where Kashiwara and Vergne introduced their conjecture definitely qualify https://link.springer.com/article/10.1007/BF01579213 There is an important theorem of Duflo stating that the PBW isomorphism for a finite dimensional Lie algebra (in characteristic 0) can be twisted so that its restriction to the invariant part actually become an algebra isomorphism $U(\mathfrak g)^{\mathfrak g}\cong S(\mathfrak g)^{\mathfrak g}$. The original proof is complicated and require a lot of case by case study. Kashiwara and Vergne suggest a very natural uniform proof of that result. Roughly speaking they postulate that the pull-back of the product on $U(\mathfrak g)$ to $S(\mathfrak g)$ through this twisted version of the PBW isomorphism can be written as $m\circ F$ where $m$ is the multiplication of $S(\mathfrak g)$ and $F$ is an automorphism of a very special form. Then they "rescale" it by introducting a parameter $t$, and observe that Duflo's theorem would follow from the existence of such an $F$ satisfying a certain differential equation w.r.t the parameter $t$. They also observe this would in fact imply a stronger statement, giving an isomorphism between invariant distribution on $\mathfrak g$ and its group $G$ with small suport. Their conjecture is thus really important in harmonic analysis and representation theory. The main result of their paper is a proof of that conjecture in the solvable case, which can be done in a fairly elementary way. After that, several special cases were proven, but a general proof was given only in 2005 (by Alekseev--Meinrenken) and uses some high level machinery related to Kontsevich's proof of his deformation quantization theorem. Since then this now theorem has been connected to a surprisingly wide range of topics, including Grothenideck-Teichmueller theory, Etingof--Kazhdan theorem, quantum topology and the study of the Atiyah--Bott--Goldman Poisson structure on moduli spaces of flat connections.<|endoftext|> TITLE: A question on the injectivity of a canonical map between galois cohomology groups QUESTION [5 upvotes]: I'm currently reading the book "Galois theory of $p$-extensions" by Helmut Koch. There, we calculate the cohomological dimension of the galois group $G(K/k)$ where $K$ is the maximal (normal) $p$-extension of $k$. (Here $p$ is a prime and $k$ is a local field or global field of finite type, i.e finite extension of $\mathbb{Q}$ of $\mathbb{Q}_p$.) As $G(K/k)$ is a pro-p group, we study $H^2(G(K/k), \mathbb{F}_p)$ where $\mathbb{F}_p$ is the finite field with $p$ elements with trivial group action. Let $k'$ be the field generated by $k$ and the $p$'th roots of unity, and let $K'$ be the maximal $p$-extension of $k'$. Then there is a canonical group homomorphism from $G(K'/k')$ to $G(K/k)$ (restriction map). This induces a homomorphism from $H^2(G(K/k),\mathbb{F}_p)$ to $H^2(G(K'/k'),\mathbb{F}_p)$. The question is, is this map injective? REPLY [2 votes]: I believe this map is always injective. Here is a quick argument: first note that $K'$ is normal over $k$ (because it is invariant under any automorphism of the algebraic closure of $k$ which preserves $k'$, and $k'$ is normal over $k$). This means that we can view the map $G(K'/k') \to G(K/k)$ as a composition of two maps $$ G(K'/k') \stackrel{f}{\to} G(K'/k) \stackrel{g}{\to} G(K/k) .$$ It will hence suffice to show that both $f^*:H^2(G(K'/k),\mathbb{F}_p) \to H^2(G(K'/k'),\mathbb{F}_p)$ and $g^*:H^2(G(K/k),\mathbb{F}_p) \to H^2(G(K'/k),\mathbb{F}_p)$ are injective. Now $f: G(K'/k') \to G(K'/k)$ is an inclusion of a subgroup of finite index $[k':k]$, and hence there exists a corestriction map $f_!:H^2(G(K'/k'),\mathbb{F}_p) \to H^2(G(K'/k),\mathbb{F}_p)$ such that $f_! \circ f^*$ is multiplication by $[k':k]$. Since $[k':k]$ is coprime to $p$ and $H^2(G(K'/k),\mathbb{F}_p)$ is a $p$-torsion group it follows that $f^*$ is injective. To show that $g^*$ is injective, we note that $g: G(K'/k) \to G(K/k)$ is a surjective group homomorphism with kernel $G(K'/K)$, and so the Hochschild-Serre spectral sequence gives an exact sequence $$ H^1(G(K'/K),\mathbb{F}_p)^{G(K/k)} \to H^2(G(K/k),\mathbb{F}_p) \to H^2(G(K'/k),\mathbb{F}_p) .$$ To show that $g^*$ is injective it will hence suffice to show that $H^1(G(K'/K),\mathbb{F}_p)^{G(K/k)} = 0$, i.e., that $K$ does not admit any non-trivial cyclic $p$-extensions inside $K'$ which are normal over $k$. But this is just because $K$ does not admit any non-trivial cyclic $p$-extensions which are normal over $k$: indeed, it is the maximal $p$-extension of $k$.<|endoftext|> TITLE: On the probability of the truth of the continuum hypothesis QUESTION [29 upvotes]: First note that there exists a natural measure $\mu$ on $P(\omega \times \omega)$, inherited from the Lebesgue measure on the reals (by identifying the reals with $P(\omega)$ and $\omega$ with $\omega \times \omega$ in the natural ways) Let us consider models of $ZFC$ of the form $(\omega, E),$ where $E \subseteq \omega \times \omega$ is interpreted as the $\in$-relation. Given a theory $T \supseteq ZFC,$ let us consider the set $M_T=\{E \subseteq \omega \times \omega:(\omega, E)$ is a model of $T\}$ Question 1. Is the set $M_{ZFC}$ measurable? If so, what is its measure? In the case that the answer to the above question is yes, and the measure is positive, then one may ask questions like the following which can measure the truth or falsity of statements like $CH$. Question 2 What is the measure of $M_{ZFC+CH}?$ What about $M_{ZFC+\neg CH}?$ If the answer to question 1 is negative, then one may ask the following: Question 3. What about questions 1 and 2 if we replace the measure with the outer measure of the required sets? Edit: By the answer given by Wojowu, and his suggestion, I would also like to ask the following: Question 4. What happens if we replace $ZFC$ with $ZFC -$foundation? REPLY [9 votes]: This answer is really just an extensive elaboration of the comments of Andreas Blass and fedja. For any relational language $L$, let $X_L$ be the space of all $L$-structures with domain $\omega$. To determine an $L$-structure with domain $\omega$, we just need to decide, for each relation symbol $R\in L$ of arity $\text{ar}(R)$ and each tuple $a_1,\dots,a_{\text{ar}(R)}$ from $\omega$, whether or not $R(a_1,\dots,a_{\text{ar}(R)})$ holds. So $X_L$ is in bijection with the Cantor space $\prod_{R\in L}2^{\omega^{\text{ar}(R)}}$ (and the topology on $X_L$ is defined so that this bijection is a homeomorphism). Let's denote by $\mu$ the measure on $X_L$ inherited from the natural measure on the Cantor space. The space $X_L$ also comes with an action (called the logic action) of $S_\infty$, the permutation group of $\omega$. A permutation $\sigma\in S_\infty$ acts on $X_L$ by permuating the domains of structures. That is, $$\sigma(M)\models R(a_1,\dots,a_{\text{ar}(R)})\text{ iff }M\models R(\sigma^{-1}(a_1),\dots,\sigma^{-1}(a_{\text{ar}(R)})).$$ Then $\sigma$ is an isomorphism $M\cong \sigma(M)$, and the orbit of a point $M\in X_L$ under the action of $S_\infty$ is $$\text{Iso}(M) = \{N\in X_L\mid N\cong M\}.$$ Note that the natural measure on $X_L$ is invariant for the logic action. For more on this setting, see Section 16.C of Kechris's book Classical Descriptive Set Theory. Now the class of all finite $L$-structures is a Fraïssé class with Fraïssé limit $M_L$. It's a fact that $\mu(\text{Iso}(M_L)) = 1$ and $\text{Iso}(M_L)$ is comeager in $X_L$. So the structure $M_L$ is "generic up to isomorphism" from the point of view of both measure and Baire category. In your question, the space on which your measure lives is $X_L$ with $L = \{E\}$, and your measure is the natural measure $\mu$. By the fact above, for any $L$-sentence $\varphi$, writing $[\varphi] = \{N\in X_L\mid N\models \varphi\}$, we have $$\mu([\varphi]) = \begin{cases} 1&\text{if } M_L\models \varphi \\ 0&\text{otherwise}\end{cases}$$ Of course, $M_L$ looks nothing like a model of set theory - it should fail to satisfy almost all the axioms, though it does satisfy extensionality. Ok, so the natural measure $\mu$ gives measure $0$ to the models of ZFC (and the same goes for any first-order theory which is not contained in $\text{Th}(M_L)$. You might ask if there are other measures on $X_L$ which give measure $1$ to the models of ZFC. Of course, you need to impose some restrictions, since you could always pick the Dirac measure concentrating on a single point (well, assuming ZFC is consistent!). A natural idea is to find such a measure which is still invariant for the logic action. But this is impossible, for the reason pointed out by fedja. Explicitly, suppose $\mu$ is an invariant measure on $X_L$ which gives measure $1$ to models of ZFC. For $n\in \omega$, let $Y_n$ be the set of all structures in $X_L$ such that the "set" $n$ has no elements. By invariance, $\mu(Y_n) = \mu(Y_m)$ for all $m$ and $n$, but $\mu(Y_n\cap Y_m) = 0$ for all $m\neq n$. But by countable additivity, we can't have an infinite family of almost-surely disjoint sets a probability space, all with the same positive measure. More generally, if $\mu$ is an invariant measure on $X_L$ which gives positive measure to the set of models of a first-order theory $T$, then $T$ must have a completion with trivial definable closure: That is, there should be a model $M\models T$ such that if $\varphi(x,y)$ is a formula, where $x$ is a tuple of variables and $y$ is a singleton, and $a$ is a tuple from $M$ such that there is a unique $b\in M$ such that $M\models \varphi(a,b)$, then already $b$ is an element of the tuple $a$. Ackerman, Freer, and Patel proved that nontrivial definable closure is actually the only obstruction to the existence of invariant measures on $X_L$ giving positive measure to a given theory. See Theorem 1.2 in this paper. So one thing you can do is take ZFC and replace equality with an equivalence relation with infinitely many infinite classes (i.e. "blow up" each element of a model to an infinite equivalence class). This is convenient to do in the language of sets, by weakening extensionality to the axiom that if $X$ and $Y$ have exactly the same elements, then they are elements of exactly the same sets. The resulting theory ZFC' does admit invariant measures which give its set of models measure $1$ (again, assuming that ZFC is consistent). But so does (ZFC + CH)' and (ZFC + $\lnot$CH)', and I think it would be hard to make an argument to prefer the measures concentrating on (ZFC + CH)' over those concentrating on (ZFC + $\lnot$CH)' or vice versa. So unfortunately I don't think this path leads to a meaningful way to assign a "probability" to CH.<|endoftext|> TITLE: Hodge Laplacian in local coordinates QUESTION [8 upvotes]: On a Riemannian Manifold $M^n$, the Hodge Laplacian is defined on k-forms by $\Delta\omega=\operatorname{d}\operatorname{d}^*\omega+\operatorname{d}^*\operatorname{d}\omega$. For 0-forms, e.i. smooth functions, one can easily see that w.r.t. some local coordinates with inverse metric tensor $g^{ij}$ it holds:$$\Delta\omega=-\partial_j(g^{ij}\partial_i\omega)$$ This can be used to apply results of 1-dimensional elliptic regularity to $\Delta$. My goal is to understand elliptic regularity on all k-forms, $\Delta:\Omega^k(M)\rightarrow\Omega^k(M)$. My question is: How does the term of highest order (2nd order in derivatives) look like in a local trivialisation of k-forms? (e.g. $\omega=\sum_{I=(i_1,...,i_k)\\0≤i_1≤...≤i_k≤n}\omega_I \operatorname{d}x^{i_{1}}\wedge...\wedge\operatorname{d}x^{i_{k}}$ or alternatively using some local orthogonal frames). Is it true or wrong that its highest order is diagonal?$$(\Delta\omega)_I=A^{I,ij}\partial_i\partial_j\omega_I+\sum_JB^{I,J,i}\partial_i\omega_J+\sum_J C^{I,J}\omega_J$$ Is it maybe even true that the $A^{I,ij}=-g^{ij}$ as in the k=0 case? REPLY [4 votes]: Your formula is true. The Weitzenböck formula states that the Hodge Laplacian on $k$-forms satisfies $$ \Delta\omega=(d\delta+\delta d)\omega=\nabla^*\nabla\omega +\operatorname{Ric}(\omega), $$ where $\nabla^*\nabla$ is the Bochner Laplacian. For a proof, see Theorem 9.4.1 in: P. Petersen, Riemannian geometry. Third edition. Graduate Texts in Mathematics, 171. Springer, Cham, 2016. Finally, the Bochner Laplacian in suitable local coordinates can be represented as $$ \nabla^*\nabla=-\sum_{k,j} \big\{\ g^{kj}\nabla_k\nabla_j+\frac{1}{\sqrt{|g|}}\partial_{x^k}\big(\sqrt{|g|}g^{kj}\big)\cdot\nabla_j\ \big\}. $$ This is proved in Example 10.1.32 in L. I. Nicolaescu, Lectures on Geometry of Manifolds. Beautiful Nicolaescu's lectures are freely available on his homepage.<|endoftext|> TITLE: Modular parametrization of a curve of Heegner and Weber QUESTION [5 upvotes]: The curve $$(X-16)^3=XY\tag{1}\label{1}$$ is essential to Heegner's approach to the class number one problem for imaginary quadratic fields. We have the following “modular” parametrization \begin{equation}\tag{2}\label{2}(X,Y)=\left(2^{12}\Phi(\tau),j(\tau)\right),\end{equation} where $\Phi(\tau)=\frac{\Delta(2\tau)}{\Delta(\tau)}$. Note that the function $\Phi(\tau)$ is a Hauptmodul for the group $\Gamma_0(2)$. My questions are: The parametrization \eqref{2} can be deduced, quite laboriously, using Weierstrass elliptic functions and the product expansion for $\Delta$. However, when we clear the denominators, equation \eqref{1} (possibly) becomes an identity between modular forms and such identities should be easy to prove using the fact, that the space of modular forms has a finite dimension. Can this be done? What can we say about $\Delta(2\tau)$? It is a modular form with respect to some group? REPLY [2 votes]: We can also use Ramanujan's functions $Q(q), R(q) $ given by $$Q(q) =1+240\sum_{n=1}^{\infty} \frac{n^3q^n}{1-q^n},R(q)=1-504\sum_{n=1}^{\infty} \frac{n^5q^n}{1-q^n}\tag{1}$$ The expression $\Delta(q) $ is defined as $$\Delta(q) = Q^3(q)-R^2(q)\tag{2}$$ It is well known that $$\Delta(q) =1728\eta^{24}(q)\tag{3}$$ where by definition $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty} (1-q^n)\tag{4}$$ The invariant $j(q) $ is then defined by $$j(q) =\frac{1728 Q^{3}(q)}{Q^{3}(q)-R^{2}(q)}=\frac{Q^3(q)}{\eta^{24}(q)}\tag{5}$$ The relation we seek to verify can be written as $$(256\Delta(q^2)+\Delta(q))^3=\Delta(q^2)\Delta^2(q)j(q)$$ and replacing $q$ by $q^2$ we get $$ (256\Delta(q^4)+\Delta(q^2))^3=\Delta(q^4)\Delta^2(q^2)j(q^2)\tag{6}$$ Using identity $(3)$ and definition of $j(q) $ (via equation $(5)$) the above identity can be reduced to $$(256\eta^{24}(q^4)+\eta^{24}(q^2))^3=\eta^{24}(q^4)\eta^{24}(q^2)Q^{3}(q^2)$$ or $$256\eta^{24}(q^4)+\eta^{24}(q^2)=\eta^8(q^4)\eta^{8}(q^2)Q(q^2)\tag{7}$$ We can now use the link between these functions and elliptic integral with modulus $k$ corresponding to nome $q$. We have the following identities $$\eta(q^2) = 2^{-1/3}\sqrt{\frac{2K}{\pi}}k^{1/6}k'^{1/6}\tag{8a}$$ $$\eta(q^{4}) = 2^{-2/3}\sqrt{\frac{2K}{\pi}}k^{1/3}k'^{1/12}\tag{8b}$$ $$Q(q^2) = \left(\frac{2K}{\pi}\right)^4(1-k^2+k^4)\tag{8c}$$ Using these identities we can see that both the LHS and RHS of equation $(7)$ can be written as $$2^{-8}\left(\frac{2K}{\pi}\right)^{12}k^4k'^2\left(1-k^2+k^4\right)$$ and the proof is thus complete.<|endoftext|> TITLE: The logarithm of Kähler metric is not globally defined QUESTION [5 upvotes]: In reducing the existence of Kähler-Einstein metrics to the complex Monge Ampere equation, the logarithm $$-\log \det (\omega + \partial \overline{\partial} \phi)$$ appears, where $\omega$ is a Kähler metric and $\phi$ is a smooth function. In Tian's book, he writes that while this function is not globally defined, we may form the quotient $$\log \left( \frac{(\omega + \partial \overline{\partial} \phi)^n}{\omega^n} \right)$$ and this is globally defined. I would just like to clarify some things: The reason why the first logarithm is not necessarily globally defined is because $\omega + \partial \overline{\partial} \phi$ is possibly zero? The reason why the second logarithm is globally defined is because $(\omega + \partial \overline{\partial} \phi)^n$ and $\omega^n$ are now volume forms which are positive? Thanks. REPLY [10 votes]: Question 1: Note, $\omega + \partial\bar{\partial}\phi$ cannot be zero. Recall that $\phi$ is chosen so that $\omega + \partial\bar{\partial}\phi$ is another metric (in particular, a Kähler-Einstein one). What does $\log\det(\omega + \partial\bar{\partial}\phi)$ mean? In holomorphic coordinates $(U, (z^1, \dots, z^n))$ we have $$(\omega + \partial\bar{\partial}\phi)|_U = \omega_{ij}dz^i\wedge d\bar{z}^j + \frac{\partial^2\phi}{\partial z^i\partial\bar{z}^j}dz^i\wedge\ d\bar{z}^j = \left(\omega_{ij} + \frac{\partial^2\phi}{\partial z^i\partial\bar{z}^j}\right)dz^i\wedge d\bar{z}^j.$$ By $\log\det(\omega + \partial\bar{\partial}\phi)$ we mean the function $\log\det(a_{ij})$ where $a_{ij}$ is the coefficient of $dz^i\wedge d\bar{z}^j$ in the local expression of $\omega + \partial\bar{\partial}\phi$; this is only defined on $U$ in terms of the coordinates $z^1, \dots, z^n$. Given another coordinate system $(V, (w^1, \dots, w^n))$ with $U\cap V \neq \emptyset$ and $(\omega + \partial\bar{\partial}\phi)|_V = b_{ij}dw^i\wedge d\bar{w}^j$, the functions $\log\det(a_{ij})$ and $\log\det(b_{ij})$ need not agree on $U\cap V$ and hence do not give rise to a well-defined function. Note however that $\partial\bar{\partial}\log\det(a_{ij}) = \partial\bar{\partial}\log\det(b_{ij})$ as they are both local expressions for $-\operatorname{Ric}(\omega + \partial\bar{\partial}\phi)$. Question 2: Yes. Given two volume forms $\eta_1$ and $\eta_2$ on a manifold $M$ which induce the same orientation, then $\eta_2 = g\eta_1$ where $g : M \to (0, \infty)$ is smooth. So $(\omega + \partial\bar{\partial}\phi)^n = g\omega^n$ and hence $$\log\left(\frac{(\omega + \partial\bar{\partial}\phi)^n}{\omega^n}\right) = \log\left(\frac{g\omega^n}{\omega^n}\right) = \log g$$ which is globally defined.<|endoftext|> TITLE: Group completion of topological monoids QUESTION [10 upvotes]: Let $M$ be an abelian monoid. For sake of simplicity we shall assume that in $M$ the cancellation law holds true. With this last assumption we define the group completion $G$ of $M$ as $$G:=M\times M/\sim$$ where $(a,b)\sim (a',b')$ if and only if $a+b'=a'+b$. It has been quite surprizing find out, reading the paper Completions and Fibrations for Topological Monoids - Paulo Lima-Filho, that in general, if $M$ is a topological abelian monoid then its group completion $G$, endowed with the quotient topology, is not a topological group. Can anyone help me to provide an example of this fact? REPLY [9 votes]: A suitable counterexample can be constructed as follows. Let $(e_n)_{n\in\omega}$ be the standard orthonormal basis of the Hilbert space $\ell_2$. For every $n\in\mathbb N$ consider the linear hull $L_n$ of the vectors $e_1,\dots,e_{n}$ in $\ell_2$. On the union $L^\infty:=\bigcup_{n=1}^\infty L_n$ consider the strongest topology that induces the Euclidean topology on each space $L_n$. It is well-known (and easy to see) that the space $L^\infty$ is a non-metrizable $k_\omega$-space (i.e., the direct limit of a sequence of compact Hausdorff spaces). Observe that the vector $e_0$ is orthogonal to the subset $L^\infty$ in $\ell_2$. In the Hilbert space $\ell_2$ consider the submonoid $S:=\bigcup_{n=0}^\infty(n e_0+L_n)$. Now consider the product $S\times S$ and the map $$q_S:S\times S\to S-S=\mathbb Z+\bigcup_{n=1}^\infty L_n,\;\;q_S:(x,t)\mapsto x-y.$$ It is easy to see that the quotient topology on $S-S$ is not metrizable -- it coincides with the topology of the direct limit of the sequence $(\mathbb Z+L_n)_{n=1}^\infty$. Finally consider the topological monoid $M:=\ell_2\times S$. We claim that $M$ is a required counterexample. Indeed, consider the subgroup $M-M\subset \ell_2\times\ell_2$ and the map $q:M\times M\to M-M$, $q:(x,y)\mapsto x-y$. Since the space $M\times M$ is second-countable, the set $M-M$ endowed with the quotient topology is a sequential $\aleph_0$-space (see Theorem 11.3 on p.494 here) and hence $M-M$ has countable $cs^*$-character. Assuming that $M-M$ is a topological group, we can apply Theorem 1 from this paper of Banakh and Zdomskyy and conclude that $M-M$ is either metrizable or contains an open $k_\omega$-subgroup. But none of these two conditions applies to $M-M$: this space contains a topological copy of the $k_\omega$-space $L^\infty$ and hence is not metrizable and contains a copy of the Hilbert space $\ell_2$, so cannot contain an open (and hence closed) $k_\omega$-subgroup.<|endoftext|> TITLE: Are all compact subsets of Banach spaces small in a measure-theoretic sense? QUESTION [10 upvotes]: Definition. A subset $K$ of a topological group $X$ is called measure-continuous if there exists a $\sigma$-additive Borel probability measure $\mu$ on $X$ such that for every compact subset $C\subset X$ the map $f:K\to [0,1]$, $f:x\mapsto \mu(C+x)$ is continuous. Remark 1. Each measure-continuous set $K$ in a Polish group $X$ is contained in a $\sigma$-compact subgroup; so $K$ is small in a sense. What about the converse? Problem. Is each compact subset of a Banach space $X$ measure-continuous? What is the answer for classical Banach spaces $c_0$ or $\ell_p$? Remark 2. It can be shown that each compact subset in the countable product of locally compact topological groups is measure-continuous. REPLY [6 votes]: I was informed by Vladimir Bogachev that the answer to Problem is negative at least for the Hilbert space $\ell_2$ as every measure-continuous compact subset of $\ell_2$ is contained in the image of a Hilbert-Schmidt operator $T:\ell_2\to\ell_2$ (for which there exists an orthonormal basis $(e_n)_{n\in\omega}$ in $\ell_2$ such that $\sum_{n=1}^\infty \|T(e_n)\|^2<+\infty$). According to Bogachev the answer to the Problem also is negative for the Banach space $c_0$ in which the compact subset $K=\{(x_n)_{n=1}^\infty\in c_0:\sup_{n\in\mathbb N}(|x_n|\cdot\ln n)<\infty\}$ is not measure-continuous. On the other hand, in nuclear Frechet spaces all compact subsets are contained in Hilbert-Schmidt ellipsoids and hence are measure-continuous. More information on this topic can be found in Chapter 3 of the book "Differentiable measures and Malliavin Calculus" of V. Bogachev.<|endoftext|> TITLE: how to view homology of affine Grassmannian as a subring of symmetric function QUESTION [10 upvotes]: Let $G=SL_n$, it is proven that $R:=H_*(Gr_G)\cong \mathbb{C}[\sigma_1,...,\sigma_{n-1}]$ where $\sigma_i$ are of degree $2i$ as a polynomial ring generated by $n-1$ variables and the ring structure comes from the fact the $Gr_G$ is homotopic to $\Omega K$ ($K$ is the maximal compact subgroup of $G$) and $\Omega K$ has a group structure. It is known that $R$ can be viewed as a subring of symmetric function $Sym$ by mapping each $\sigma_i$ to $h_i$ ($i$-th complete homogeneous symmetric function of degree $i$ ). My question is: Why should we view $R$ as a subring in $Sym$ in such way? What is the geometric reason behind it? REPLY [4 votes]: You should look at the book of Lam, Lapointe, Morse, Schilling, Shimozono and Zabrocki. More specifically, under k-Schur functions and how/why they constitute a basis of $H_*(Gr_{SL_k})$. They mainly work in K-theory, but one of the main results in loc. cit originally proved by Lam is that there is a natural Hopf algebra isomorphism between the Hopf subalgebra of k-Schur functions $\Lambda_{(k)}:=\mathbb{Q}[h_1,\ldots,h_{k-1}]\subset \Lambda$ and $H_*(Gr_{SL_k})$. See the references therein for the Hopf algebra structures. Geometrically the k-Schurs are just the Schubert basis in the (K-)homology of the affine Grassmannian. Multiplication by $h_i$ is more interesting in this basis, it's given by the k-Pieri rule.<|endoftext|> TITLE: Looking for examples of maps $\Omega^lS^{n+l}\to\Omega^kS^{m+k}$ with $l>k$ QUESTION [8 upvotes]: As the title says I am looking for examples of essential maps $\Omega^lS^{n+l}\to\Omega^kS^{m+k}$, with $l>k$, which may or may not be an iterated loop map. As an example, James fibration $S^n\to \Omega S^{n+1}\to \Omega S^{2n+1}$ provides a map $\Omega^2S^{2n+1}\to S^n$ which we may look at its iterated loops and get maps $\Omega^lS^{2n+1}\to\Omega^{l-2}S^n$. One source which I have thought of is James-Hopf maps, or maps which are somehow constructed in a combinatorial manner, but I don't see how to construct them. Let me add that the usual James-Hopf maps produce maps into infinite loop spaces, but I strictly want $0\leqslant kN$. This involves finding the minimum possible value for the integer $N$. I don't have time to work through this (so I don't know whether or not these resulting maps can possibly satisfy your condition, although I suspect they don't), but Cohen's "The unstable decomposition of $\Omega^2 \Sigma^2 X$ and its applications" has a very readable introduction, and refers to Cohen-Mahowald's "Unstable properties of $\Omega^n S^{n+k}$".<|endoftext|> TITLE: Quantum tunneling on the line with non-symmetric double well potential QUESTION [5 upvotes]: Consider the Schroedinger equation on the line $$i\frac{\partial \Psi(x,t)}{\partial t}=[-\frac{d^2}{dx^2}+V(x)]\Psi(x,t),$$ where one assumes that $V(x)\to +\infty$ as $|x|\to +\infty$, and $V$ has two local minima. The case of symmetric potential, i.e. $V(-x)=V(x)$, seems to be discussed in physics textbooks using the WKB method (see e.g. the 3rd volume of the Landau-Lifshitz textbook on theoretical physics). I am interested in the situation of non-symmetric potential $V$ especially when it takes different values at the two local minima (the global minimum is called the true vacuum, and the other one the fake vacuum). This case seems to be very different from the symmetric one. In physics literature it is often stated that if a wave function $\Psi(x,0)$ is localized near the fake vacuum, then after a long time $\Psi(x,t)$ will be localized near the true vacuum. Is there a mathematically rigorous treatment of this problem? REMARK. It seems that the above statement is not always true and it is very sensitive to the shape of $V$. In the paper "Resonances in quantum mechanical tunneling" by M.M.Nieto et al. in Phys.Lett.B (1985) what they claim is the following: "In asymmetric double-well potentials, it can be tacitly assumed that a wave function in the higher-energy well (false vacuum) will always tunnel to the lower well, given enough time. However, in general this is not true. Whether a state can significantly tunnel to the true vacuum is a very sensitive function of the shape of the potential. We illustrate this with analytic and numerical examples. Thus, if there is not dissipation or coupling to other modes, a wave function may not tunnel." The method of the paper seems to be not mathematically rigorous and to large extend numerical. REPLY [3 votes]: Tunnel splitting of the spectrum and bilocalization of eigenfunctions in an asymmetric double well has a "theorem" on the bilocalization phenomenon (wave function localized in both asymmetric wells); The theorem might satisfy some requirements of rigor...<|endoftext|> TITLE: How to compute Coefficients in Chudnovsky's Formula? QUESTION [6 upvotes]: My aim is to understand all three coefficients arising in the Chudnovsky-Formula (see also Question 300385). Two of them are easily computed, but I failed with the third: It is known that for all $\tau$ with $Im(\tau)>1.25$ we have \begin{align*} \frac{1}{2\pi Im(\tau)}\sqrt{\frac{J(\tau)}{J(\tau)-1}} &= \sum_{n=0}^\infty \left( \frac{1-s_2(\tau)}{6} + n \right)\cdot \frac{(6n)!}{(3n)!(n!)^3}\cdot \frac{1}{\left(1728J(\tau)\right)^n}\\ \text{with }s_2(\tau) &:= \frac{E_4(\tau)}{E_6(\tau)}\left(E_2(\tau)-\frac{3}{\pi Im(\tau)}\right) \end{align*} Then for $\tau=\frac{1+i\sqrt{163}}{2}$ it is known that $1728J(\tau)=-640320^3$. This gives us \begin{align*} \frac{1}{\pi} &= \sum_{n=0}^\infty \frac{(-1)^n(6n)!}{(3n)!(n!)^3}\cdot \frac{A + B\cdot n}{640320^{3n+3/2}} \end{align*} with $B = \sqrt{163\cdot(1728+640320^3)} = 12\cdot 545140134$ $$\text{and } A = 12\cdot 545140134\cdot\left( \frac{1-s_2(\tau)}{6} \right)$$ It can be easily computed that $A$ is approximately $12\cdot13591409$ Question: How can I prove that the value of $A$ is exactly this number? Edit: Thanks to the answer of @HenriCohen, the only thing left to prove is this: a reference is needed, why $\sqrt{D}E_2^*(\tau)/\eta^4(\tau)$ is an algebraic integer. EDIT: Answer: Thanks to the answers of Henri Cohen (see below) and Michael Griffin (see here), an explicit calculation of the coefficients along with a complete proof of their exactness can now be found in chapter 10 and the appendix of this arXiv-preprint. REPLY [8 votes]: Let $\tau$ be any CM point. By basic theorems of complex multiplication, if you choose a suitable period $\omega(\tau)$, $E_4(\tau)/\omega(\tau)^4$, $E_6(\tau)/\omega(\tau)^6$, and $\sqrt{D}E_2^*(\tau)/\omega(\tau)^2$ (with $E_2^*(\tau)=E_2(\tau)-3/(\pi\Im(\tau))$ and $D$ the discriminant of $\tau$) will be algebraic numbers of known degree, and if you choose $\omega(\tau)=\eta(\tau)^2$, they will even be algebraic integers. Incidentally (but this is not needed) the Chowla--Selberg formula gives $\eta(\tau)$ explicitly as a product of gamma values. In the case $\tau=(1+\sqrt{-163})/2$, we easily find that $E_4/\omega^4=-640320\rho$, $E_6/\omega^6=-40133016\sqrt{-163}$, and $\sqrt{-163}E_2^*/\omega^2=-8688\rho^2$ with $\rho=(-1+\sqrt{-3})/2$ a cube root of unity. This gives $s_2(\tau)=77265280/90856689$ which implies the result.<|endoftext|> TITLE: Variance modulo 1 QUESTION [5 upvotes]: The fact that the variance of the sum of independent random variables is the sum of their variances allows one to have a good understanding of how well-concentrated each term $X_i$ in a sum of $n$ independent random variables $S= X_1+X_2+\dots+X_n$ has to be in order for $S$ to be concentraded around the mean. It seems that an analogous fact should be true for the sum taken modulo $1$. More concretely, I believe that the following should be true, but I cannot seem to find a proof or a reference. Help finding either of these would be highly appreciated. Lemma (tentative). Suppose that $X_1, X_2, ..., X_n$ are (jointly) independent real random variables with $\mathbb{E} X_i = 0$ for each $i \leq n$, and that there exists a real number $\gamma$ such that $$ \mathbb{E} [ \lVert X_1 + X_2 + \dots + X_n - \gamma \rVert_{\mathbb{R}/\mathbb{Z}}^2] \leq \varepsilon $$ for some small $\varepsilon > 0$. Then $$ \mathbb{E} [ \lVert X_1\rVert_{\mathbb{R}/\mathbb{Z}}^2 ] + \mathbb{E} [ \lVert X_2\rVert_{\mathbb{R}/\mathbb{Z}} ^2 ] + \dots + \mathbb{E} [ \lVert X_n\rVert_{\mathbb{R}/\mathbb{Z}}^2 ] \leq C \varepsilon,$$ where $C$ is an absolute constant. Here, $\lVert x\rVert_{\mathbb{R}/\mathbb{Z}} $ denotes the distance of $x$ from the nearest integer. Rationale: The lemma would be true if $\lVert x\rVert_{\mathbb{R}/\mathbb{Z}} $ were replaced with $\lvert x\rvert_{\mathbb{R}/\mathbb{Z}} $; indeed this is just the additivity of variance. It is not hard to check that the lamma is true if $X_i$ have normal distribution or if $X_i$ have Bernoulli distribution with vales $0$ and $1/2$. On the intuitive level, if we imagine that the distribution of $X_1+\dots+X_n$ is roughly bell-shaped with a single maximum, the only way for $\mathbb{E} [ \lVert X_1 + X_2 + \dots + X_n - \gamma \rVert_{\mathbb{R}/\mathbb{Z}}^2]$ to be small is if $X_1 + X_2 + \dots + X_n$ is very strongly concentrated around that maximum. In particular the ``wrap-around'' issues should not play much role and the above lemma should be roughly equivalent to the version with absolute value in place of the distance from $\mathbb{Z}$. REPLY [6 votes]: On the one hand, the proof is very cheap. Let $Z_j=e^{2\pi iX_j}$. $X=\sum_j X_j$, $Z=e^{2\pi i X}$. Note that $\operatorname{Var}_{\mathbb R/\mathbb Z}X\approx 1-|EZ|$ and similarly for $X_j$ and $Z_j$. Now just use the identity $EZ=\prod_j EZ_j$ to conclude. On the other hand, finding the reference may be a highly non-trivial task, so I leave it to somebody else :-). P.S. What can be really concluded here is that there exist $\gamma_j$ summing to $\gamma$ with $\sum_jE\|X_j-\gamma_j\|_{\mathbb R/\mathbb Z}^2\le C\varepsilon$. The condition $EX_j=0$ does not allow one to conclude from here that we can take $\gamma_j=0$: take $X_j$ symmetric with values about $\pm \frac 12$ for a counterexample.<|endoftext|> TITLE: Example similar to the Griffiths twin cone but with fundamental group that allows surjection onto $\mathbb Z$ QUESTION [6 upvotes]: The Griffiths twin cone is an example of a wedge sum of two contractible spaces being non-contractible. Namely, it is the wedge sum $\mathbb G=C\mathbb H\vee_p C\mathbb H$ of two coni over the Hawaiian earring by the bad point $p\in\mathbb H$. However, as stated at the bottom of this post, $Hom(\pi_1(\mathbb G),\mathbb Z)=0$. Is there any example of two contractible spaces whose wedge sum's fundamental group would allow a surjection onto $\mathbb Z$? REPLY [4 votes]: No such homomorphism is possible. The prototypical nature of the Griffiths twin cone guarantees this. Let $X,Y$ be contractible with basepoints $x,y$ respectively. We only need to assume $\{x\}$ and $\{y\}$ are closed. Let $X\vee Y$ be the wedge with basepoint $\ast$. Suppose $h:\pi_1(X\vee Y)\to \mathbb{Z}$ is non-trivial. Pick a loop $\alpha$ in $X\vee Y$ based at $\ast$ such that $h([\alpha])\neq 0$. Consider the set $\mathscr{L}$ of components of $[0,1]\backslash \alpha^{-1}(\ast)$ with the natural ordering inherited from $[0,1]$. Let $\mathscr{L}_X$ be the set of components $(a,b)$ of $[0,1]\backslash \alpha^{-1}(\ast)$ such that $\alpha([a,b])\subseteq X$. Let $\mathscr{L}_Y$ be the set of components $(c,d)$ of $[0,1]\backslash \alpha^{-1}(\ast)$ such that $\alpha([c,d])\subseteq Y$. Since $X$ and $Y$ are both simply connected, in order to have $[\alpha]\neq 1$, both $\mathscr{L}_X$ and $\mathscr{L}_Y$ must be infinite. Choose enumerations $\mathscr{L}_X=\{(a_1,b_1),(a_2,b_2),...\}$ and $\mathscr{L}_X=\{(c_1,d_1),(c_2,d_2),...\}$. Let $\mathbb{H}$ be the Hawaiian earring and $\ell_n$ be the standard loop traversing the n-th circle. Let $\mathbb{H}_{odd}$ and $\mathbb{H}_{even}$ be the sub-Hawaiian earrings consisting of the odd and even indexed circles respectively. Construct a map $f:\mathbb{H}\to X\vee Y$ such that $f \circ \ell_{2n-1}= \alpha|_{[a_n,b_n]}$ and $f\circ \ell_{2n}= \alpha|_{[c_n,d_n]}$ after reparameterization. Using the enumerations and ordering of $\mathscr{L}$, it is easy to build a loop $\beta:[0,1]\to \mathbb{H}$ such that $f\circ\beta$ is a reparmeterization of $\alpha$. Since $\mathbb{H}=\mathbb{H}_{odd}\vee \mathbb{H}_{even}$, $f(\mathbb{H}_{odd})\subseteq X$, and $f(\mathbb{H}_{even})\subseteq Y$ where $X$ and $Y$ are contractible, $f$ extends to a map $g:\mathbb{G}=C\mathbb{H}_{odd}\vee C\mathbb{H}_{even}\to X\vee Y$ (where $C$ is the cone and $\mathbb{G}$ is the Griffiths twin cone from the question). Recalling the fact $Hom(\pi_1(\mathbb{G}),\mathbb{Z})=0$ from the question, we see that the composition $h\circ g_{\#}:\pi_1(\mathbb{G})\to\mathbb{Z}$ is trivial; However, if $i:\mathbb{H}\to\mathbb{G}$ is inclusion, then $h\circ g_{\#}(i_{\#}([\beta]))=h([\alpha])\neq 0$; a contradiction.<|endoftext|> TITLE: Hilbert 90 for higher K-groups QUESTION [11 upvotes]: For a field $F$, Let $K_n(F)$ be the Quillen's $n$-th K-group of $F$.Then $K_0(F)\cong \mathbb{Z}$, $K_1(F)\cong F^\times$. For a finite Galois extension $L/K$, $K_n(L)$ are Galois modules. Then $\mathrm{H}^1(L/K,K_0(L))=0$ trivially holds, and $\mathrm{H}^1(L/K,K_1(L))=0$ by Hilbert 90. In Srinivas's book Algebraic K-Theory, Page 161, it is proved there if $L/K$ is a cyclic extension, then $\mathrm{H}^1(L/K,K_2(L))=0$. For general Galois extension $L/K$, do we have $\mathrm{H}^1(L/K,K_2(L))=0$? Do we have more Hilbert 90 style results on higher K-groups? REPLY [2 votes]: Another good reference is Weibel's K-book, see section III.6. Voevodsky has proved that Hilbert 90 indeed generalizes to higher Milnor K-theory, see III.7.8.4 in the K-book.<|endoftext|> TITLE: Twisting bordism classes QUESTION [11 upvotes]: Let $X$ be a reasonable topological space (I'd be happy to assume that $X$ is a smooth closed manifold) and let $f\colon M^n \rightarrow X$ be a continuous map from a smooth oriented $n$-manifold $M^n$ to $X$. Let $\phi\colon M^n \rightarrow M^n$ be an orientation-preserving diffeomorphism. Question: Must it be the case that the maps $f\colon M^n \rightarrow X$ and $f \circ \phi \colon M^n \rightarrow X$ are oriented bordant to each other? The reason I'm asking is that $f$ and $f \circ \phi$ induce the same element of $H_n(X;\mathbb{Z})$, and it is known that oriented bordism agrees with integral homology up to degree $3$. If the answer to the question is negative, I'd also be interested in an explicit proof that this is the case for $n \leq 3$. REPLY [14 votes]: The correct definition of bordism should have this built in. More precisely, two singular manifolds $(M_0^n,f_0)$ and $(M_1^n,f_1)$ in a space $X$ are oriented bordant if there exists a smooth manifold $W^{n+1}$ with a diffeomorphism $\varphi: M_0\sqcup M_1\stackrel{\simeq}{\to}\partial W$ and a map $F:W\to X$ such that $f_i=(\partial F\circ \varphi)|_{M_i}$ for $i=0,1$. It's true that some authors surpress this from the definition, but it's there for example in tom Dieck's recent book "Algebraic Topology".<|endoftext|> TITLE: A Question on Graph Limits QUESTION [7 upvotes]: I have a somewhat technical question about the concept of graph limits: Suppose that $G_n$ is a sequence of labelled, simple, unweighted graphs, and let $W_n$ denote the graphon of $G_n$ (i.e. $W_n(x,y) = 1$ for all $\frac{i-1}{n} TITLE: A Bitwise Xor Problem QUESTION [7 upvotes]: Consider a sequence $a_i$ defined by $$ \begin{align*} a_1&=p,\\ a_2&=q,\\ a_i&=a_{i-1} \oplus a_{i-2}+1, \end{align*}$$ where $\oplus$ is the bitwise xor operation. How can we give an upper bound for $a_n$ as a function of $p,q,n$? There are lots of $p,q$ which satisfy $\mathop {\lim }\limits_{n \to \infty } {a_n} = \infty $, but it seems from experimentation that $a_n=O(n)$. I do not have any idea how to prove it, nor do I know whether it is correct. If this problem is difficult to solve, I only need to prove that, for each $1 \le i \le n$, we have $a_i \le 2^{12}$ when $p,q,n \le 1000$. REPLY [6 votes]: $\def\U#1{\underline{#1}}\def\O#1{\overline{#1}}$The bound $a_n=O(n)$ is true, in fact, we have $a_n\le\max\{8p,8q,\frac{16}3n\}$. The argument is quite elementary, but a bit tedious to write down properly, hence I will only sketch it, and rely on the reader to fill in the details. Rather than estimating $a_n$ directly, we will try to bound the position of the most significant bit of $\max_{m\le n}a_m$. Notice that $\oplus$ cannot raise this quantity, and ${}+1$ can raise it only by one bit, when it results in a power of $2$ larger than all the previously seen numbers. In this case the ${}+1$ operation involves carries all the way from the $0$th bit to the new most significant bit. Thus, the crucial thing is to investigate carries in the sequence. For any $k\ge1$, the operations of $\oplus$ and ${}+1$ are well defined modulo $2^k$, hence we can look at the sequence $a_n\bmod2^k$. Since the sequence is uniquely reversible mod $2^k$ by $a_{n-1}=a_n\oplus(a_{n+1}-1)$, it must be periodic mod $2^k$. When computing $a_n$, there is a carry from $(k-1)$th bit to $k$th bit iff $a_n\equiv0\pmod{2^k}$; for $k>1$, this holds iff there was a carry to the $(k-1)$th bit, and the $(k-1)$th bit of $a_n$ is $0$. Let us denote the $k$th bit of $a_n$ as $a_{n,k}\in\{0,1\}$. Case 1: $a_n\bmod2$ (that is, $a_{n,0}$) is $\dots1111\dots$. Then there are no carries whatsoever, hence $a_n=O(1)$ and we are done. Case 2: $a_n\bmod2$ is $\dots001001001\dots$ of period $3$. Carries from $0$th to $1$st bit happen at the positions of the $0$s in the period. (Here and below, the pattern shown is not necessarily aligned with the start of the sequence, hence strictly speaking the period is one of its cyclic shifts.) This again splits to two subcases: Case 2a: $a_n\bmod4$ is $\dots122320122320\dots$ of period $6$. That is, the sequence of $a_{n,1}$ is $\dots\O{\U0}\U11\U1\U10\dots$, where underlines signify carries from the $0$th bit to the $1$st bit, and overlines carries from the $1$st bit to the $2$nd bit. Case 2b: $a_n\bmod4$ is $\dots120300120300\dots$ of period $6$; that is, the sequence of $a_{n,1}$ is $\dots0\U1\O{\U0}1\O{\U0}\O{\U0}\dots$. I claim that in Case 2a, for all $k\ge2$, the sequence $a_n$ has period $3\cdot2^{k-1}$ modulo $2^k$, and there is carry to the $k$th bit at exactly one position per period. This is proved by induction on $k$. It holds for $k=2$. When going from $k$ to $k+1$, we see that the sequence $a_{n,k}$ is a $\oplus$-linear combination of: A solution of the homogeneous recurrence $b_n=b_{n-2}\oplus b_{n-1}$ modulo $2$. This is either $\dots00000\dots$, or (some shift of) $\dots011011011\dots$. One particular solution of $b_n=b_{n-2}\oplus b_{n-1}\oplus c_n$, where $c_n$ is the sequence of carries to the $k$th bit. By the induction hypothesis, $c_n$ is $3\cdot2^{k-1}$-periodic, and the period has the form $100\dots0$. Thus, one solution $b_n$ is given by the $3\cdot2^k$-periodic sequence consisting of $110110110\dots110$ for $3\cdot2^{k-1}$ positions, followed by $00\dots0$ for $3\cdot2^{k-1}$ positions. Clearly, this makes $a_{n,k}$ (and thus $a_n\bmod2^{k+1}$) $3\cdot2^k$-periodic. Carries to the $(k+1)$th bit may happen only at the two positions in this period that correspond to carries to the $k$th bit; these two positions are at distance $3\cdot2^{k-1}$ apart, and they are occupied by two opposite bits: the sequence from 1. gives them the same value, whereas the sequence from 2. gives them opposite values. Thus, exactly one of them gives rise to a carry to the $(k+1)$th bit. In Case 2b, a similar inductive argument establishes the following claim: for any $k\ge2$, $a_n\bmod2^k$ is $3\cdot2^{k-1}$-periodic, and there are exactly $3$ carries to the $k$th bit in each period. Two of these carries have the same position modulo $3$, whereas the third is at relative position $2$ modulo $3$ with respect to them. This implies a bound on $a_n$ as follows. For a given $n$, let $k$ be the least integer such that $n<3\cdot 2^k$ and $p,q<2^{k+1}$. Thus, $2^k\le\max\{p,q,2n/3\}$. We know from above that up to $a_n$, there are at most three carries to the $(k+1)$th bit. The first of these will attain the value $2^{k+1}$, the other two may at most get to $2^{k+3}$. Thus, the most significant bit of $a_n$ is at position at most $k+3$. Actually, taking into account the relative position of the carries modulo 3, one can check that the most significant bit can only become at most $k+2$. Thus, $a_n<2^{k+3}\le\max\{16n/3,8p,8q\}$.<|endoftext|> TITLE: Number of hypercube unfoldings QUESTION [24 upvotes]: While writing the code for this answer, I noticed that I not only could calculate the number of unfoldings of the $4$-cube, but also the number of the $n$-cube for more values of $n$. Basically, we count pairs $(T, P)$, where $T$ is a tree on $2n$ nodes and $P$ is a perfect matching of the complement of $P$. We consider $(T, P)$ equivalent to $(T', P')$ if there is an automorphism $\phi$ of $T$, such that $T=\phi(T)=T'$ and $\phi(P) = P'$. Here's what I found: $$\begin{array}{c|cccccc}n&3&4&5&6&7&8\\\hline \text{# of unfoldings} &11& 261& 9694& 502110& 33064966& 2642657228\end{array}$$ For $n=7$ the calculation took about $7$ hours on a desktop pc with 8 cores. The calculation can almost trivially be parallelized, since we can consider each tree $T$ at the same time. The case $n=8$ was done on a small cluster. Have these unfoldings been counted before? The sequence is in the oeis: A091159, but only two terms are given (as of today). I searched for the number $33064966$ on the net but couldn't find anything. Update: Improving the algorithm to generate hypercube unfodings, together with Luca Versari, we were able to also calculate the number of unfoldings of the $n$-cube for $n=9$ and $n=10$. The code for that can be found here:  https://github.com/google-research/google-research/tree/master/cube_unfoldings dimension number of unfoldings file 2         1                     2.cnt.txt 3         11                   3.cnt.txt 4         261                   4.cnt.txt 5         9694                 5.cnt.txt 6         502110               6.cnt.txt 7         33064966             7.cnt.txt 8         2642657228           8.cnt.txt 9         248639631948         9.cnt.txt 10         26941775019280       10.cnt.txt Might there be a better approach in getting these numbers? There is a nice generating function for unlabelled unrooted trees, so perhaps a generating approach could help here too? Matt Parker made a video about these hypercube unfoldings. [1, 11, 261, 9694, 502110, 33064966, 2642657228, 248639631948, 26941775019280] REPLY [6 votes]: Following up on the comments about using spanning trees and automorphism groups. It does seem to me that perhaps the best way to do this would be to compute the number of different spanning trees of the hyperoctahedral graph up to symmetries of the hyperoctahedron. As Peter Taylor and Brendan McKay have noted, one can use Burnside's lemma to do this. I think this will be easier, because it seems likely that about half of all symmetries will not fix any spanning trees (see below), and that most spanning trees are not fixed by any symmetry. If this is true, most of the computation would be finding a small number of spanning trees that are fixed by reflections of the hyperoctohedron. On the other hand, it seems like computing the number of trees with an appropriate pairing would require solving several instances of the graph isomorphism problem to make sure you aren't overcounting. The rest of this post is not about computation, but rather about general conjectures. I think (tho I haven't sat down to prove) that rotations will never fix a spanning tree. If this is so, then, after counting the number of spanning trees of the hyperoctohedral graph (which can be done using the matrix-tree theorem), we could use a very simple version of Burnside's lemma to get bounds on the number of nets. Specifically, let $N_d$ be the number of nets of the hypercube in $\mathbf{R}^d$, and let $F(d) = \frac{(2d)^{d-2} (d-1)^d}{d!}$ (this is the same number that Brendan mentions). Then, assuming (as I believe) that no rotation fixes a spanning tree, we get that $F(d) \leq N_d \leq 2F(d).$ For some small $d$, the bounds look like this: $d$ Bounds 2 $\frac{1}{2} \leq 1 \leq 1$ 3 $8 \leq 11 \leq 16$ 4 $216 \leq 261 \leq 432$ 5 $8533\frac{1}{3} \leq 9694 \leq 17066\frac{2}{3}$ The table makes it look like the lower bound is probably closer to the truth. Looking at the ratios $N_d/F(d)$ for the numbers you've computed, Brendan says they appear to be going to 1. I'm a little less sure, since it looks like the rate at which they decrease is going down as $d$ increases, but with such a small dataset it's hard to tell. I'd guess that $N_d/F(d)$ approaches a constant near $1.08$. If it turns out I'm wrong, at least I'll have Legendre to keep me company. Update: So I started being a little more sophisticated and factoring in what to me are the most obvious symmetries that fix some spanning trees – reflections that interchange only two opposite points. Let the number of spanning trees of the hyperoctohedron in $\mathbf{R}^d$ be $T_d$. Then we can choose a pair of opposite points in $d$ ways. For a spanning tree to be fixed by this reflection, these opposite vertices must both connect to the same vertex, and in order for it to be a tree they can only connect to a single vertex, and we can choose that vertex in $(2d-2)$ ways. Finally, we need a spanning tree of the rest of the hyperoctohedron, but that's actually the same as a spanning tree of the $d-1$ octohedron. So the number of trees fixed by one of these reflections is $(2d-2)T_d$, and there are $d$ such reflections. When we factor this into our lower bound we get $N_d \geq F(d) + (d-1)F(d-1)$. This is enough to show that $N_d/F(d)$ does not approach 1, since $\frac{N_d}{F(d)} \geq \frac{F(d) + (d-1)F(d-1)}{F(d)} \xrightarrow{x\to\infty} 1 + \frac{1}{2e^2}.$ I think there's still one more class of reflections contributing a significant number of fixed spanning trees. I'll update as soon as I find it. Right now my prediction for $N_{11}$ is $3.30\times 10^{15}$. Update: I realized this construction can be extended a little bit. After you fix two opposite vertices and remove them, you're looking at a hyperoctohedron in $\mathbf{R}^{d-1}$. Now, instead of just taking any spanning tree, you can take a spanning tree and an automorphism fixing that spanning tree $\sigma$. If you let $\alpha$ be the automorphism reflecting the two opposite vertices we chose, then you can connect up your two chosen vertices to the chosen tree, and then you get that the whole thing is fixed by $\sigma \circ \alpha$. I haven't taken the time to work this into the lower bound. However, it does show that my initial conjecture that no rotation fixes a spanning tree is off, which means there's a more subtle reason why certain automorphisms cannot fix a spanning tree. I think that if you're careful enough, then doing this construction with spanning forests on the $d-1$ hyperoctohedron, and then connecting the forest into a tree with your two chosen vertices, should exhaust all the possibilities when $d$ is odd. When $d$ is even I think there's going to be special spanning trees that don't fit this construction. Update: I think actually there's actually a better repair than the one I mentioned above. The graph we're considering can actually be thought of as two complete graphs $K_d$ connected up in a certain way. So, you have several symmetries that flop two copies of $K_d$. If you fix one of these symmetries, you can get trees which are fixed by this symmetry by the following recipe: Choose a rooted spanning tree for $K^d$. Connect the two copies of $K^d$ by an edge between the two roots. It seems like understanding the exact count of these spanning trees is going to involve understanding automorphisms of rooted trees on $d$ vertices. As far as I can tell, there is no closed form solution for this problem, but perhaps there are good enough asymptotics on the rooted tree problem to give good asymptotics for the problem of nets of cubes.<|endoftext|> TITLE: If $(X,\tau)$ has more than $1$ point and is $T_2$ and connected, do we have $|X| =|\tau|$? QUESTION [6 upvotes]: If $(X,\tau)$ has more than $1$ point and is $T_2$ and connected, do we necessarily have $|X| =|\tau|$? REPLY [4 votes]: Here are: 1) a metrizable example. Namely, consider a complete graph on $\alpha\ge\mathfrak{c}$. Its 1-skeleton, endowed with the geodesic metric with edges of length one, has cardinal $\alpha$ and has $2^{\alpha}$ open subsets (since for any subset of the set of vertices, the open ball of radius $1/2$ around this subset determines this subset. 2) a separable example. Namely, to fix ideas choose $S=\mathbf{R}/\mathbf{Z}$. Start from $X=S^{\mathfrak{c}}$, which is separable. Choose $D$ a countable dense subgroup (e.g., generated by any dense countable subset), and $F=S^{(\mathfrak{c})}$, the finitely supported subgroup. Then the topological group $G=FD$ is a dense subgroup, is connected because it has the dense connected subgroup $F$, and is separable because it has the dense countable subset $D$ (beware that $F$ is not separable), and $|G|=\mathfrak{c}$. Using suitable inverse images of projections, we see that $G$ has $2^{\mathfrak{c}}$ open subsets. (Of course, a separable metrizable connected space of cardinal $\ge 2$ has cardinal $\mathfrak{c}$ and the same number of open subsets.)<|endoftext|> TITLE: Is there a Baldwin-Lachlan style characterization of countable unidimensional theories? QUESTION [7 upvotes]: It's well known that a countable theory is uncountably categorical if and only if it is $\omega$-stable and has no Vaughtian pairs. One of the old definitions of unidimensional theory (all $\omega_1$-saturated models of the same (sufficiently large) size are isomorphic) is a weak categoricity notion, so it's natural to wonder if an analogous characterization can be made. Hrushovski showed that unidimensional theories are supserstable. Although I'm sure it's already known, I couldn't find a proof that unidimensional theories do not have any Vaughtian pairs, so I worked out a silly proof: Let $T$ be a countable, superstable theory with a Vaughtian pair. By Vaught's two-cardinal theorem, since $T$ has a Vaughtian pair, it has a $(\omega_1,\omega)$-model. By a result of Shelah, in a stable theory the existence of any two-cardinal model implies the existence of $(\kappa,\lambda)$-models for any $\kappa\geq\lambda\geq\omega$, so we can find a $(\kappa,\omega)$-model, $\mathfrak{M}$, of $T$, with $\kappa \gg 2^\omega$. Then if we take any non-principal ultrafilter $\mathcal{U}$ on $\omega$, $\mathfrak{M}^\omega/\mathcal{U}$ will still be a two-cardinal, $\omega_1$-saturated model of $T$, contradicting unidimensionality. Again, I'm sure there's a better, published proof somewhere, but I couldn't find it. I would guess it's far too much to hope that a countable theory $T$ is unidimensional if and only if it is superstable and has no Vaughtian pairs, but I also can't quite find a counterexample. So my first question is: What is an example of a countable superstable theory with no Vaughtian pairs that fails to be unidimensional, if such a thing exists? EDIT: I believe theory of the following structure is a counterexample to this: $(2^\omega, P_i, f_i)$ where $P_i$ is a sequence of unary predicates such that $P_i(\alpha)$ is true if and only if $\alpha(i)=1$, and $f_i$ is a sequence of unary functions such that $f_i(\alpha)(j)=\alpha(j)$ if $i\neq j$ and $f_i(\alpha)(i)=1-\alpha(i)$. Another related characterization of uncountably categorical countable theories is that a countable theory $T$ is uncountably categorical if and only if there is a definable strongly minimal set $\varphi$ such that the entire structure is layered by $\varphi$ (where layering is in the sense of Hodges' "Model Theory"). In Pillay's "Geometric Stability Theory," he shows in chapter 6 proposition 1.1 that if $T$ is a countable, superstable, unidimensional theory that fails to be $\omega$-stable, then there is a weakly minimal group definable over $\text{acl}(\varnothing)$. Since we know that $T$ doesn't have any Vaughtian pairs, by the results on layerings in Hodges we know that the entire theory is layered over this weakly minimal group. I suspect that it's also way too much to hope that any weakly minimal group is unidimensional, so my next questions are: What is an example of a weakly minimal group that fails to be unidimensional? Is there a good characterization of which weakly minimal groups are unidimensional? Aside from those parts, I'm pretty confident that if a theory is layered over a unidimensional set, then it will be unidimensional, so if there is a good characterization of which weakly minimal groups are unidimensional, then I'm expecting there is a statement like: "A theory $T$ is unidimensional if and only if it is layered over a strongly minimal set or a weakly minimal group with property X." Where 'property X' is the missing ingredient to ensure that a weakly minimal group is unidimensional. REPLY [3 votes]: I believe the answer to the second question is that all weakly minimal groups are unidimensional. Proof: Let $T$ be the theory of a weakly minimal group. It suffices to show that if $p$ and $q$ are non-algebraic $1$-types over a sufficiently saturated model $G$, then $p$ and $q$ are not orthogonal. So fix such $p$ and $q$. By weak minimality, $p$ and $q$ have $U$-rank $1$, and thus are generic. By stability, there is some $g\in G$ such that $q=gp$. So $p$ and $q$ are not orthogonal. Edit: The fact that stable unidimensional theories have no Vaughtian pairs is in Lemma IX.1.10 of Shelah's book.<|endoftext|> TITLE: Are all generalized Scott sets realized as generalized standard systems? QUESTION [11 upvotes]: Below, I've focused on PA when lots of other theories would do. If replacing PA with a different theory leads to a more answerable question, feel free to do so. The standard system of a nonstandard model $M$ of PA is the set of sets of natural numbers coded by elements of $M$: $$SS(M)=\{X\subseteq\omega: \exists a\in M\forall x\in\mathbb{N}(x\in X\iff p_x\vert a)\}.$$ (Here "$p_i$" denotes the $i$th prime.) An easy overspill argument shows that $SS(M)$ is always a Scott set,$^1$ and Scott proved that every countable Scott set is the standard system of some nonstandard model.$^2$ We can define an analogous notion$^3$ of standard system with $\mathbb{N}$ replaced with more general initial segments: if $M\subsetneq_{end} N$ are models of PA, then we let $SS_M(N)$ be the set of elements of $M$ coded by elements of $N$: $$SS_M(N)=\{X\subseteq M: \exists a\in N\forall x\in M(x\in X\iff p_x\vert a)\}.$$ (Here "$p_i$" denotes the $i$th prime in the sense of $N$, or equivalently in this case of $M$.) The same overspill argument shows that the second-order structure $(M, SS_M(N))$ is a model of WKL. My question is whether the analogue of Scott's theorem holds, at least for countable $M$: Question. Suppose $M$ is a countable model of PA and $\mathcal{X}$ is a countable family of subsets of $M$ such that $(M, \mathcal{X})\models$ WKL. Is there an $N\supsetneq_{end} M$ such that $SS_M(N)=\mathcal{X}$? The problem here is that in the usual case, we don't need to worry about $\subseteq$ versus $\subseteq_{end}$, whereas that poses a real problem here. $^1$A set of sets of natural numbers closed under join and Turing reducibility such that for every infinite binary tree in the set, an infinite path through that tree is also in the set. Equivalently, the second-order part of an $\omega$-model of WKL. $^2$The generalization of Scott's theorem to uncountable Scott sets is wildly open. Knight and Nadel generalized Scott's theorem to Scott systems of cardinality $\aleph_1$, thus solving the problem under the assumption of CH, but their argument breaks down immediately for models of cardinality $\aleph_2$ or greater. Meanwhile, Gitman has recently shown that the question has an affirmative answer for a class of Scott sets characterized in terms of forcing, assuming the set-theoretic hypothesis PFA (which contradicts CH); however, my understanding is that her arguments really only apply to that particular class of Scott sets and that the hypothesis of PFA is currently necessary. $^3$I've had trouble finding much literature on this, especially compared to the usual notion of a standard system, but Kossak and Schmerl's book does contain some information about them (especially in Chapter $7$). REPLY [4 votes]: Your question has a positive answer: there do exist proper end extensions for countable models with the required coded sets. The PA requirement can be weakened. Generalizing results of Friedman[1973], the MacDowell-Specker Theorem extensions of Phillips[1974] and Gaifman[1976], and the results of Kirby, MacAloon, and Murawski[1979], Joseph Quinsey proved in his unpublished (and perhaps unreadable) 1980 thesis Lemma 5.9: Suppose $M$ is a countable model of arithmetic and $\mathcal{X}$ is a countable family of subsets of $M$ such that $$(M, \mathcal{X})\models exp + \Delta_0^0\text{-Induction} + \Delta_0^0\text{-Arith-Collection} + \text{WKL}$$ where Arith-Collection is the schema (with $q$ not free in $\theta$): $$\forall m \lt n\, \exists p\,\theta \rightarrow \exists q\,\forall m \lt n \, \exists p \lt q\,\theta \,.$$ Let $\text{T} \in \mathcal{X}$ be such that in $(M, \mathcal{X})$, $\text{T}$ is is the encoding of a consistent set of sentences of arithmetic. Then there exists a proper end-extension $N$ of $M$ which is a model of the standard sentences of $\text{T}$ and is such that $SS_M(N)=\mathcal{X}$. Moreover, if $(M, \mathcal{X})$ is a model of $\Pi_1^0\text{-overspill}$ (or if $M$ is the standard model), $N$ can be chosen to be recursively saturated. If additional conditions are placed on $(M, \mathcal{X})$, then using the embedding techniques of Friedman[1973], Wilkie[1977], and Wilmers[1977], we may obtain elementary end extensions. Theorem 5.8 of Quinsey's thesis gives: Let $k \ge 1$. Suppose $M$ is a countable model of arithmetic and $\mathcal{X}$ is a countable family of subsets of $M$ such that $$(M, \mathcal{X})\models \Delta_0^0\text{-CA} + \Delta_0^0\text{-Arith-Collection} + \text{WKL} + \text{“the $\bar{k}$th Turing jump of the empty set exists"}$$ Further suppose that $\Pi_1^0\text{-overspill}$ holds in $(M, \mathcal{X})$. Let $c \in M$. Then there exists a structure $(N, \mathcal{Y})$ isomorphic to $(M, \mathcal{X})$ such that $N$ is a $(k - 1)$-elementary initial segment of $M$, $SS_N(M)=\mathcal{Y}$, and the isomorphism fixes $c$. If in addition we have that $\Sigma_1^1\text{-overspill}$ holds in $(M, \mathcal{X})$ and that $$(M, \mathcal{X})\models \text{“the $\bar{k}$th Turing jump of the empty set exists"}$$ for each standard integer $k$, then we may require that $N \prec M$. Since the two structures are isomorphic, we may swap them to get the end-extension. Note in passing that these two results have nothing to do with the main thrust of the thesis, which is to explore applications of Kripke's notion of fulfillability.<|endoftext|> TITLE: Nice partition of $\mathbb{R}$ into uncountably many uncountable sets QUESTION [5 upvotes]: A recent issue of American Math. Monthly has a paper that partitions $\mathbb{R}$ into an arbitrary finite number of uncountable sets such that every real number is a condensation point of all the sets in the partition. In fact, we can find uncountably many uncountable subsets $S_\alpha$ of $\mathbb{R}$ that are (1) pairwise disjoint, (2) have union $\mathbb{R}$, and (3) every real number is a condensation point of all the sets. One way to do this is the following. For any infinite binary sequence $\alpha=(a_1, a_2, \dots)$, let $S_\alpha$ be the set of all real numbers whose fractional part has the binary expansion $0.b_1b_2\cdots$ such that the sequence $(b_2,b_4,b_6,\dots)$ differs from $\alpha$ in only finitely many terms. When $\alpha=(1,1,1,\dots)$ we must remove from $S_\alpha$ all numbers whose binary expansion ends in infinitely many 1's, due to the nonuniqueness of the binary expansion. The set of distinct $S_\alpha$'s then has the desired properties. This partition of $\mathbb{R}$ also has the property that any two of the sets $S_\alpha$ and $S_\beta$ are translates of each other, except for $S_{1,1,1,\dots}$. This suggests the question: does there exist a partition of $\mathbb{R}$ satisfying (1), (2), and (3) such that every pair of sets in the partition are translates? REPLY [8 votes]: I think the following works: Fix an uncountable subgroup $G$ of $(\mathbb{R}, +)$ - note that any such is dense in $\mathbb{R}$, and in fact intersects each nonempty open interval uncountably often - of uncountable index, and consider the set of cosets of $G$. (Such a group can be produced by transfinite recursion using a well-ordering of $\mathbb{R}$: alternate between throwing reals into the group, and barring reals from entering the group.) Note that the argument above relies on the axiom of choice. However, this isn't necessary: provably in ZF, there is a $\mathbb{Q}$-linearly independent set $S$ of reals of size continuum. We can split this into two disjoint sets $S_1, S_2$ of size continuum (pick some appropriate real $a$ and consider the points to the left of $a$ versus the right of $a$). Now consider the $\mathbb{Q}$-vector space generated by $S_1$.<|endoftext|> TITLE: A characterisation of $\mathbb{P}^n$ QUESTION [24 upvotes]: Let $X$ be a projective variety (so, with some (edit: fixed nondegenerate closed) embedding) with the following curious property: for every hyperplane section $H$, we have that $X-H \cong \mathbb{A}^n$. Then is $X$ necessarily isomorphic to $\mathbb{P}^n$? Assuming we are over the complex numbers for simplicity, one can show that $X$ has the same Hodge diamond as $\mathbb{CP}^n$ using purely topological arguments (such as long exact sequence of a pair, duality as in the proof of Lefschetz hyperplane, etc.) but this does not rule out these potential fake projective spaces. This question is motivated by the fact that spheres are characterized by a similar property; i.e., a closed oriented manifold that is contractible upon removal of any point will be homeomorphic to a sphere. REPLY [9 votes]: Here's a quick proof over a finite field $\mathbb F_q$, if we assume that the isomorphisms with $\mathbb A^n$ are defined over $\mathbb F_q$. Suppose $X$ of dimension $n$ embedded in $\mathbb P^N_{\mathbb F_q} $ has this property. Every hyperplane complement in $\mathbb P^N(\mathbb F_q)$ contains $q^N$ points, of which $q^n$ lie in $X(\mathbb F_q)$. So the fraction of points in the hyperplane complement that also lie in $X(\mathbb F_q) $ is $q^{n-N}$. Averaging over all hyperplane complements, as each point in $\mathbb P^{N}(\mathbb F_q)$ is in the same number of hyperplane complements, we see that the fraction of points in $\mathbb P^N(\mathbb F_q)$ that also lie in $X(\mathbb F_q)$ is $q^{n-N}$. Thus $$|X(\mathbb F_q) |= q^{n-N} |\mathbb P^N(\mathbb F_q)| = q^n + q^{n-1} + \dots q^{n-N}$$ which is not an integer unless $N \leq n$, in which case $X = \mathbb P^n$ because it is an $n$-dimensional subscheme of $\mathbb P^N$. If we replace the assumption by the statement that every hyperplane complement defined over $\mathbb F_q$ has an isomorphism with $\mathbb A^n$ defined over $\overline{\mathbb F}_q$, we can do it with a quick etale cohomology lemma. Every variety $Y$ over $\mathbb F_q$ with $Y_{\overline{\mathbb F}_q} \cong \mathbb A^n_{\overline{\mathbb F}_q}$ has $Y(\mathbb F_q) = q^n$. Proof: By the Grothendieck-Lefschetz fixed-point formula and the calculation of the cohomology of affine space we have $$Y(\mathbb F_q) =\sum_{i}(-1)^i \operatorname{tr}(\operatorname{Frob}_q, H^i_c(X, \mathbb Q_{\ell}) = \operatorname{tr} (\operatorname{Frob}_q, H^{2n}_c(\mathbb Q_{\ell})).$$ Furthermore, $H^{2n}_c$ is one-dimensional and the Frobenius eigenvalue on it can be computed by Poincare duality as $q^n$, because the unique Frobenius eigenvalue on $H^0(Y_{\overline{\mathbb F}_q}, \mathbb Q_\ell)$ must be one. With this enhanced form, we can get the statement over an arbitrary field by a spreading out argument. First, define a stratification of the dual variety, and, over a finite etale covering each stratum, an explicit isomorphism from the family of hyperplane complements to $\mathbb A^n$. To do this, choose an isomorphism at the geometric generic point, which automatically extends to a finite etale cover of an open set, and induct. Next, choose a finitely generated ring over which $X$, the projective embedding, this stratification, the finite etale coverings, and all these isomorphisms are defined. Over an open subset, the stratification will remain a stratification and the isomorphisms will remain isomorphisms. Choosing a finite field-valued point in that subset and performing a counting argument, we see $n=N$ and conclude $X = \mathbb P^n= \mathbb P^N$.<|endoftext|> TITLE: Solution of an equation with Jacobi theta function QUESTION [6 upvotes]: I have been struggling with this equation for some time and I do not seem to find any conclusive answer (it's from my research, not a homework). It has to do with the real solutions $x$ to the following equation $$ x + x f(x) = 1 + f(1),$$ where $$ f(x) = 2\sum_{n=1}^\infty \mathrm{e}^{(-ax^2-b) n^2} $$ with $a$ and $b$ strictly positive. I know that $x=1$ solves the equation trivially; from simulations, I cannot find a contradiction to the fact that it should be the only solution. However, I cannot prove nor disprove that $x=1$ is the only solution. I have tried using the upper bound $$ f(x) \leq \frac{ax^2+b+1}{ax^2+b}\mathrm{e}^{-ax^2 -b},$$ and i have tried relating $f(x)$ with the Elliptic theta function $$ f(x) = -1 + \theta_3(0,\mathrm{e}^{-ax^2 -b});$$ I have also tried to prove that $x = -x f(x) + 1 +f(1)$ is a contractive mapping; however, I have only found (quite restrictive) sufficient conditions on $a$ and $b$ for it to be true. If someone manages to solve it or help me find a counterexample, I will gladly acknowledge their contribution in the paper i am writing. EDIT: $x = -x f(x) +1 +f(1)$ is not a contractive mapping in general. I have found counterexamples where it is not (but the equation still only has one solution $x=1$). REPLY [6 votes]: Have you tried Jacobi's identities for theta-functions? At least for $a=1, b=0$ the identity $$xg(x)=\sqrt{\pi}\,g(\pi/x)$$ implies that the function $xg(x):=xf(x)+x$ is monotone increasing. Edit: the identity corrected, thanks to Somos.<|endoftext|> TITLE: Compatibility of Grothendieck construction with pullback QUESTION [7 upvotes]: Suppose $D$ is an $\infty$-category, then we have the equivalence $$ \text{Fib} (D) \substack{ \text{St} \\ \longrightarrow \\ \cong \\ \longleftarrow \\ \text{Un}} [ D^\text{op}, \mathbf{Kan}]$$ between (right) fibrations over $D$ and functors from the opposite of $D$ to Kan complexes ($\infty$-groupoids) via the Grothendieck construction. Given a map $f: C \to D$ of $\infty$-categories, there is a functor $f^* : \text{Fib} (D) \to \text{Fib}(C)$ by forming pullbacks. Since $f$ induces a functor $C^\text{opp} \to D^\text{opp}$, there is also a functor $f^* : [ D^\text{op}, \mathbf{Kan}] \to [ C^\text{op}, \mathbf{Kan}]$. Question: For a fibration $\pi : X \to D$ do we have $$ \text{St} f^* \pi \cong f^* \text{St} \pi \quad ?$$ So is the Grothendieck construction compatible with pullbacks? Thanks for any hints. REPLY [6 votes]: Yes, though it is usually written as the commutativity of unstraightening with pullback (on the $\infty$-categorical level it doesn't matter, since straightening and unstraightening are inverse equivalences). This compatibility even holds on the point-set level if one uses a suitable model categorical presentation of the straightening-unstraightening equivalence, see Proposition 2.2.1.1 of higher topos theory (if you combine (1) and (2) of that proposition then you get the compatibility of straightening with left Kan extensions. The compatibility of unstraightening and pullbacks is obtained by switching to right adjoints).<|endoftext|> TITLE: A map of spaces implementing the Pontryagin Thom collapse map? (collapse maps in families) QUESTION [8 upvotes]: Let $M$ be an $n$ dimensional smooth manifold and let $j: M \to \mathbb{R}^{m}$ be an embedding. Associated to this embedding we can form the "collapse map" which is a pointed map from a sphere to the Thom space of the normal bundle $S^{m}=(\mathbb{R}^m)^{+} \to Th(N_j)$ (which depends on the choice of tubular neighborhood). The homotopy class of the collpase map is however well defined as a function of the isotopy class of the embedding (notice the homotopy type of the Thom space depends only on the isotopy class of the embedding). The "collapse map" is, in laconic terms, nothing more than a well defined function of sets: $$\{{ \text{embeddings } M \hookrightarrow \mathbb{R}^m \}/\sim_{\text{isotopy}}} =: E \longrightarrow \coprod_{[j] \in E} [S^m,Th(N_j)]$$ This might be fine for the most part but it seems desirable (even from the POV of applications) to have a version of this construction which works for families. Question: Is there a "natural" (in the sense that it does what you expect it to do for families of embeddings) map (more accurately a homotopy class of maps) of spaces: $$Emb(M,\mathbb{R}^m) \to \coprod_{[j] \in \pi_0(Emb(M,\mathbb{R}^m))} Map_*(S^m,Th(N_j))$$ Which on connected components induces the collapse map construction above? REPLY [4 votes]: The situation is somewhat easier to describe if one replaces the embeddings of the closed codimension $(m-n)$manifold $M$ with the embeddings of the total space of a disk bundle of a rank $(m-n)$-vector bundle over $M$. If $\xi$ is a smooth vector bundle over $M$ of rank $(m-n)$, with disk bundle $D(\xi)$, then there is a restriction map of embedding spaces $$ E(D(\xi),\Bbb R^m) \to E(M,\Bbb R^m)\, , $$ There is an evident Pontryagin-Thom map $$ E(D(\xi),\Bbb R^m) \to \Omega^mM^\xi $$ where the target it the $m$-fold loop space of the Thom space of $\xi$. The map is given by sending an embedding to the map $S^m \to M^\xi$ given by collapsing the complement of the image of the embedding to a point. The above restriction map sits in a homotopy fiber sequence, $$ E(D(\xi),\Bbb R^m) \to E(M,\Bbb R^m) \to \text{maps}(M,BO_{m-n}) $$ where the base space is the space of maps from $M$ to the Grasmannian of $(m-n)$-planes in $\Bbb R^\infty$. The displayed fiber is the one taken at the basepoint defined by $\xi$. There is another homotopy fiber sequence $$ \Omega^m M^\xi \to D_m(M) \to \text{maps}(M,BO_{m-n}) $$ where $M^\xi$ is the Thom space of $\xi$ and $D_m(M)$ is the space consisting of pairs $(\xi,g)$ in which $\xi$ is as above and $g: S^m \to M^\xi$ is a based map. The first fiber sequence maps to the second one, after thickening up $E(M,\Bbb R^m)$ in the way that Neil describes.<|endoftext|> TITLE: $RO(Q)$-graded homotopy fixed point spectral sequence QUESTION [5 upvotes]: I am trying to understand some part of J. Greenlees's "Four approaches to cohomology theories with reality": https://arxiv.org/abs/1705.09365 I have a problem with understanding $RO(Q)$-graded homotopy fixed point spectral sequence. Namely: 1. In section 2.C, proof of Lemma 2.1 - how filtration on $EQ_+$ helps us in describing $E_1$-page of this spectral sequence? And why it looks as described? Which leads to seceond question: How this spectral sequence arises? Any help or reference would be strongly appreciated. REPLY [6 votes]: For a based $G$-space or $G$-spectrum $X$, the homotopy fixed point object $X^{hG}$ is by definition $F_G(EG_+,X)$. Suppose we write $EG$ as the colimit of a sequence of $G$-subspaces $A_k$. This then gives a tower of objects $F_G((A_k)_+,X)$ whose inverse limit is $X^{hG}$, and the fibres of the maps in the tower are $F_G(A_k/A_{k-1},X)$. There is a standard spectral sequence for the homotopy groups of the inverse limit of any tower, with the $E^1$ term being given by the homotopy groups of the fibres. In particular, we have a spectral sequence for $\pi_*(X^{hG})$ with $E^1$ term given by $[A_k/A_{k-1},X]^G_*$. In the example in question, $G$ has order $2$, so we can take $EG$ to be $S(\mathbb{R}^\infty)$ with the antipodal action, or $S(\infty\sigma)$ in the notation of Greenlees. We can then take $A_k=S((k+1)\sigma)$ and check that $A_k/A_{k-1}\simeq G_+\wedge S^k$ (with trivial action on $S^k$). This gives $[A_k/A_{k-1},X]^G_*=\pi_{k+*}(X)$.<|endoftext|> TITLE: Representing elements of $\pi_2(M)$ by embedded spheres in 3-manifolds QUESTION [23 upvotes]: I am sorry that this question is probably too basic - I could not seem to find the answer though. I know the following - let $S$ be a closed orientable surface, an element of $H_1(S;\mathbb{Z})$ is represented by an embedded curve if and only if it is primitive, and the question of whether or not an element of $\pi_1(S)$ can be represented by an embedded curve is at least algorithmically decidable. Also - the sphere theorem implies that if $\pi_2(M) \neq 0$ then there will be some nonzero element represented by an embedded sphere. Let $M$ be a (closed, orientable if you like) 3-manifold. What can be said about elements of $\pi_2(M)$ being represented by embedded spheres? Are there some simple obstructions? Are there some necessary/sufficient conditions? REPLY [7 votes]: If one has a connect sum of two non-trivial irreducible 3-manifolds $M=M_1\#M_2$, then there is a unique embedded essential 2-sphere $S^2$ up to isotopy (forgetting orientation). Hence the elements of $\pi_2(M)$ realized by an embedded 2-sphere will just be the orbit of this 2-sphere under the action of $\pi_1(M)=\pi_1(M_1)\ast\pi_1(M_2)$. With the identification $\pi_2(M) \cong \pi_2(\tilde{M})\cong H_2(\tilde{M})$ via Hurewicz, where $\tilde{M}$ is the universal cover, we can see that most classes (even primitive) will not be realized by an embedded 2-sphere. From the proof of the geometrization conjecture, one may conclude that $\tilde{M} \cong \mathbb{S^3}-K$, where $K$ is either 2 points (if $M_1\cong M_2\cong \mathbb{RP}^3$, or $K$ is a tamely embedded Cantor set. I have some notes describing how this goes. In the first case, $H_2(\tilde{M})\cong \mathbb{Z}$, and the element is represented by an embedded sphere iff it is primitive. In the second case, one may represent any element of $H_2(\tilde{M})$ by a disjoint union of oriented 2-spheres $\Sigma$. Moreover, we may assume this is minimal, so that if components of $\Sigma$ are adjacent, then the orientations agree (otherwise, we may tube them together to get fewer spheres). If $\Sigma$ is disconnected, the the corresponding class of $\pi_2(\tilde{M})$ cannot be represented by an embedded 2-sphere, and hence the same for $\pi_2(M)$. But $\Sigma$ may be disconnected without being non-primitive, since there may be components of $K$ between each of the components of $\Sigma$ which prevent them from being parallel. I believe figuring out if such a class is connected should be algorithmic (but not practically), by constructing enough of the universal cover, and then performing 3-manifold algorithms to find the embedded spheres. However, embeddedness of spheres in $\pi_2(\tilde{M})$ is not equivalent to embeddedness in $\pi_2(M)$. In the case at hand, where there is only one class of embedded 2-sphere (up to the action of $\pi_1(M)$), I believe that there ought to be an algorithm. Pull back the embedded 2-sphere $S$ from $M$ to $\tilde{M}$ to get a surface $\tilde{S}$ with infinitely many components. Then we want to be able to determine whether an embedded 2-sphere $\Sigma$ in $\pi_2(\tilde{M})$ is isotopic to one of the components of $\tilde{S}$. As described in my notes, $\tilde{M}$ is a union of thrice-punctured spheres. We may assume that the components of $\tilde{S}$ are contained in this union of 3-punctured spheres. Then take enough of these 3-punctured spheres to contain $\Sigma$ and be connected, giving a submanifold $N\subset \tilde{M}$. The by Hurewicz, $\pi_2(N)=H_2(N)$. Hence $\Sigma$ will be parallel to a sphere of $\tilde{S}$ iff it is homologically equivalent to a sphere of $\tilde{S}$ in $N$ (up to orientation). I believe that this approach could be made algorithmic, although not in a practical way. For the general case, I think there ought to be an abstract criterion that implies embeddedness. By the previous consideration, a 2-sphere in $\pi_2(M)$ lifts to a class of $H_2(\tilde{M})$ represented by an embedded sphere. This sphere separates the ends $E=\partial_{\infty}\tilde{M}$ into two pieces $E_+, E_-$. Then one can associate to this partition an action on a CAT(0) cube complex by Sageev's theory. There is a way to do this canonically in this case using a wall space on $E$, so that the sphere is embedded iff the complex is a tree. I won't delve into the theory, but suffice it to say that the sphere will be embedded iff the partition $\{E_+,E_-\}$ has the property that for each element $g\in \pi_1(M)$ acting on $\tilde{M}$ by covering translations and hence on $E$, either $g(E_+)\subset E_+$ or $g(E_+)\subset E_-$. Again, it should be possible to make this criterion algorithmic by building "enough" of $\tilde{M}$ to see if a lift of $S$ crosses any of its covering translates in an essential way. One might even be able to make this computable using PL minimal surface theory.<|endoftext|> TITLE: Homeomorphism-fixing subsets QUESTION [5 upvotes]: If $(X,\tau)$ is a topological space, we say $Y\subseteq X$ is homeomorphism-fixing if the only homeomorphism $\varphi:X\to X$ such that for all $y\in Y$ we have $\varphi(y)=y$ is the identity map. Moreover we say that $Y$ is minimally homeomorphism-fixing if for all $y\in Y$ the set $Y\setminus \{y\}$ is not homemorphism-fixing. What is an example of an infinite Hausdorff space without minimally homeomorphism-fixing subsets? REPLY [11 votes]: The real numbers. Any dense subset, like the rationals is homeomorphism fixing. Conversely if that subset $Y$ is not dense, we can find a point $x\in X$ and a small interval around $x$ that does not contain any element of $Y$. Now it is easy to construct a homeomorphism that is the identity outside of that interval (and thus fixes $Y$) and does something on that interval (reparametrizing). If you remove a single point from a dense subset, that subset is still dense and thus homeomorphism fixing. Thus there are no minimal homeomorphism fixing subsets.<|endoftext|> TITLE: What does the group action of a rough path in a Lie group look like? QUESTION [7 upvotes]: Rough paths can be thought of as taking values in a Lie group embedded in the tensor algebra of $\Bbb R^d$. See page 17/section 2.3. Lie groups represent the continuous symmetries of some object. That is, elements of the Lie group act on some other object in a way that preserves symmetry. I am curious, since rough paths are elements of a Lie group, what exactly are they acting on? What symmetry do rough paths preserve? What do those actions look like? REPLY [14 votes]: Though I like the Arnoldian spirit of the question ("a group is not some set with a forgettable system of axioms but something which acts on a space"), I think the comments given above are already spot on: Rough paths are paths of a certain regularity (often one takes finite $p$-Variation or equivalently $1/p$-Hölder type paths) with values in certain Lie groups, e.g. taking paths with finite $p$-Variation in $\mathbb{R}^d$,Chen's theorem asserts that the signature of such a path takes values (in a suitable sense) in the free $\lfloor p\rfloor$-step nilpotent Lie group $G^{\lfloor p\rfloor}(\mathbb{R}^d)$ (a so called Carnot-Caratheodory group). However, due to the seminal results of Lyons, one can always (uniquely) lift these paths to paths with values in $G^{N}(\mathbb{R}^d)$ for $N \geq \lfloor p \rfloor$ and this process (the so called Lyons lift) is continuous in the $p$-variation topologies. So rough paths (of a prescribed regularity) can be seen to have values in many Lie groups. Indeed one can lift these rough paths even to the projective limit of the nilpotent groups, which happens to be an infinite-dimensional Lie group (and can be identified with the character group of a certain Hopf algebra). Moreover, the paths (of a given regularity) form a group under concatenation of paths. The signature map (taking a path to its signature) identifies the paths of a given regularity with a subgroup of the projective limit $G = \lim_{p \rightarrow \infty} G^{p} (\mathbb{R}^d)$ of the Carnot-Caratheodory groups. However, it is not an isomorphism. One can show that its kernel consists exactly of the tree-like paths (see The Signature of a Rough Path: Uniqueness for the (involved) proof). Before continuing, let me mention that that the projective limit $G$ is an infinite-dimensional Lie group and can be identified with the character group of a Hopf algebra (see Character groups of Hopf algebras as infinite-dimensional Lie groups for an article developing the Lie group structure) So the quotient of the group of paths (the so called reduced path group) is isomorphic to a subgroup of an infinite-dimensional Lie group. Does this imply that the rough paths of a given regularity (modulo tree like equivalences) are a Lie group itself? Unfortunately, this does not seem to be the case: On one hand the projective limit Lie group the subgroup is sitting in has very strong Lie theoretic properties which seem not to be shared by the reduced path subgroups. To illustrate this, let me mention that for a given regularity not every element in a reduced path subgroup $R$ admits a square root, i.e. in general for $g \in R$ one can not find an element $h \in R$ (the root) such that $g = h\cdot h$. (This seems to be a folklore fact in the community.) As a consequence if $R$ is a Lie group, then $g$ can not be contained in the image of its Lie group exponential (if it were, say $g = \exp_R (X)$, then $\exp_R(\frac{X}{2})$ would be a root). Now if $R$ was a (closed) Lie subgroup of $G$, then the Lie group exponentials of $R$ and $G$ would be related (this is explained nicely in Neeb's Towards a Lie theory of locally convex groups in the chapter on locally exponential Lie groups) via the inclusion map $R \rightarrow G$. However, this would imply that the Lie group exponential of $R$ needs to be a bijection as the Lie group exponential $\exp_G$ of the character group is a bijection (see Theorem B of 2). Since this contradicts the existence of group elements without roots we conclude that $R$ can not be a closed Lie subgroup of $G$. So the upshot is that the canonical Lie group structure of $G$ will not be inherited by the reduced path subgroups (though this is not saying that the reduced path subgroups could not be a Lie group in some sense, possibly with a finer topology). This is all a beautiful theory and yields an elegant way of combining algebraic and analytic properties of rough paths in geometric language. However, as the original question is concerned: There is no known canonical Lie group structure on the space of rough paths and to my knowledge there is a priori no distinguished Lie group action (of the finite or infinite-dimensional groups the paths take their values in) which gives a deeper meaning to this. To make it perfectly clear: Though there is no group action satisfying your curiosity, I do not want to imply by any means that it is useless to consider these paths as taking values in the groups. Indeed it is an elegant way to formulate many properties and to use (Carnot-Caratheodory) group techniques in some of the proofs in rough path theory. For example, the $1/p$-Hölder regularity condition required in rough path theory makes sense as an actual Hölder-continuity estimate with respect to the the Carnot-Caratheordory metric on the group.<|endoftext|> TITLE: Egyptian representations of $1$ QUESTION [11 upvotes]: Let $\ F(n)\ (\mbox{where}\ n\in\mathbb N:=\{1\ 2\ \ldots\})\ $ be the least cardinality $\ |A|\ $ of a set $\ A\subseteq\mathbb N $ such that: $\ \min A=n $ $\ \sum_{x\in A}\frac 1x = 1 $ QUESTION   Is set $\ \{n\in\mathbb N:\ \frac{F(n)}n\le 2\}\ $ finite? Background: Roughly speaking, $\ \frac {F(n)}n\ \ge\ e-1,\ $ where $\ e:=exp(1)\ $ is one of the Euler's constants, $\ 2.718\ldots.\ $ This induces a bunch of natural questions, I have restricted myself to just one. Please, feel free to enjoy addressing also related questions. REPLY [14 votes]: Lots of questions along these lines were raised by Erdos and Graham and many have been solved by Croot, Greg Martin and others. In particular, Croot has shown that any rational number $r$ can be represented as a sum of unit fractions with denominators lying in the interval $[N, (e^r+o(1))N]$ for all large $N$ (indeed his result is more precise, quantifying also the $o(1)$ term). In particular your $F(N)$ is $\sim (e-1)N$ for all large $N$. For related work, see also Martin.<|endoftext|> TITLE: Left Bousfield localization without properness, what is known? QUESTION [8 upvotes]: I'm interested in the existence of several example of left Bousfield localization of model categories that are not left proper (nor simplicial). I'm relatively convince that I can construct all those I need by hand, but that got me curious about what is known in general about existence of Left Bousfield localization of combinatorial model category at a set of maps: Is there any known example of a combinatorial model category with a set of maps such that the left Bousfield localization does not exist ? (Edit: this is answered there) Is there other kind of assumption under which we have a general theorem of existence ? For example I have the impression that putting condition on the set of maps we localize at (like localizing at a set of cofibrations between cofibration objects) might remove the need for left properness. In On left and right model categories and left and right Bousfield localizations C.Barwick claims (4.13) that if we only want to construct a left semi-model category then the left Bousfield localization always exists. But he does not prove it. What is the status of this claim ? is it proved somewhere ? REPLY [7 votes]: I have an unpublished note that proves Barwick's claim. Aspects of this story have appeared in some papers of mine with Michael Batanin, including one we published in the proceedings of the 2015 CRM conference in Barcelona on "Interactions between representation theory, algebraic topology, and commutative algebra." I'm working to make the note into a real paper. I can share a draft at some point if you want. Specifically, what I can prove is that, if $M$ is a combinatorial semi-model category (i.e. a locally presentable category with a cofibrantly generated semi-model structure), and $C$ is a set of morphisms, then the left Bousfield localization $L_CM$ with the classes of maps in Hirschhorn's book, has a combinatorial semi-model structure and satisfies the universal property of localization (that any left Quillen functor of semi-model categories $F: M\to D$ taking $C$ to weak equivalences factors through $L_CM$). By the way, this was certainly known to Hirschhorn, as you can see from the comments in Moduli problems for structured ring spectra, by Goerss and Hopkins. Also, I think Cisinski probably knew this. Regarding your other questions...I am pretty certain that it is not true that if you replace the maps by cofibrations between cofibrant objects, then you can get a full model structure without left properness. But I don't know an example offhand. And, I don't know an example for your first question, but of course this would be a phenomenon only visible on the model category level because every combinatorial (semi-)model category is Quillen equivalent to a left proper one, basically by Giraud's theorem.<|endoftext|> TITLE: Are uniformly continuous functions dense in all continuous functions? QUESTION [14 upvotes]: Suppose that $X$ is a metric space. Is the family of all real-valued uniformly continuous functions on $X$ dense in the space of all continuous functions with respect to the topology of uniform convergence on compact sets? REPLY [8 votes]: Let $C(X)$ be the algebra of continuous functions endowed with the compact-open topology. Notice that the functions $d_x : X \to [0, \infty)$ given by $d_x (y) = \min( d(x,y), 1)$ separate the points of $X$: if $d_x (p) = d_x (q)$ for all $x \in X$, taking $x = p$ gives $d_p(q) = 0$, whence $d(p,q) = 0$. The functions $d_x$ are Lipschitz (with Lipschitz constant $1$). There are 3 cases to examine: if $d(x,y) \le 1$ and $d(x,z) \le 1$, then $| d_x(y) - d_x(z) | = |d(x,y) - d(x,z)| \le d(y,z)$ by the triangle inequality; if $d(x,y) \le 1$ and $d(x,z) > 1$ then $$| d_x(y) - d_x(z) | = 1 - d(x,y) \le d(x,z) - d(x,y) \le d(y,z)$$ again by the triangle inequality; similarly for $d(x,y) > 1$ and $d(x,z) \le 1$; if $d(x,y) > 1$ and $d(x,z) > 1$ then $| d_x(y) - d_x(z) | = | 1 - 1 | = 0 \le d(y,z)$. Remember now that the sum and product with scalars of Lipschitz functions are again Lipschitz, and that the product of bounded Lipschitz functions (and the $d_x$ are bounded) is again Lipschitz (of course, the Lipschitz constant changes). Consider the subalgebra $A$ of $C(X)$ generated by the family $(d_x)_{x \in X}$ and the constant function $1$. By the preceding paragraph it will be made only of Lipschitz functions. Using the version of the Stone-Weierstrass theorem for the compact-open topology (Stephen Willard, "General Topology", 1970, p. 293, par. 44B.3), the closure of $A$ is $C(X)$, so you get that not only uniformly continuous, but even Lipschitz functions are dense in $C(X)$<|endoftext|> TITLE: Cocycle superrigidity QUESTION [6 upvotes]: Let $\Gamma$ be a group with a probability measure preserving action on $(X,\mu)$, and $H$ another group. Recall that a cocycle is a map $c:\Gamma\times X\to H$ such that $c(gg',x)=c(g,g'x)c(g',x)$. Two cocycles $c,c'$ are cohomologous if there is $f:X\to H$ such that $c(g,x)=f(gx)^{-1}c'(g,x)f(x)$. Let $H$ be a discrete group. Say that $H$ is superrigid if for every lattice $\Gamma$ in a simple, higher rank Lie group $G$, and for every ergodic probability measure preserving action of $\Gamma$ on $(X,\mu)$, every measurable cocycle $c:\Gamma\times X\to H$ is cohomologous to a cocycle with values in a finite group. For example, if $H$ is a hyperbolic group, then a result of Adams proves that $H$ is superrigid. My question is : Assume that $H$ contains a normal subgroup $H'$ of finite index, and that $H'$ is superrigid. Does it follow that $H$ is superrigid ? REPLY [5 votes]: Yes. Instead of writing down explicitly such a cocycle (which is not hard), let me take this opportunity to explain how to think of such objects in a "cooridnate free" manner. Given a cocycle $c:\Gamma\times X\to H$ you can consider the space $Y=X\times H$ and endow it with the product measure and with the $\Gamma\times H$ measure preserving action given by the right action of $H$ on the second coordinate and the twisted left action of $\Gamma$ $$(*)\quad\gamma\cdot(x,h)=(\gamma x,c(\gamma,x)h).$$ Clearly, $Y$ is $\Gamma\times H$-ergodic iff $X$ is $\Gamma$-ergodic. Conversely, let $Y$ be an ergodic measure preserving $\Gamma\times H$-space on which $H$ acts freely and with a finite measured fundamental domain. Then we may form the space of orbits $X=Y/H$ and endow it with a $\Gamma$-invariant finite measure, and any choice of a fundamental domain (ie a section $X\to Y$) will give us an identification $Y=X\times H$ such that the $\Gamma$-action on $Y$ will be realized as in $(*)$ for some cocycle $c$. Different choices of fundamental domains will give cohomologous realization cocycles. In this cohomology class we could find a cocycle taking values in a subgroup $F TITLE: Character theoretic proof of the Littlewood–Richardson rule? QUESTION [15 upvotes]: The Littlewood–Richardson coefficients are the multiplicities $$ c(\lambda,\mu,\nu)= \dim_{\mathbb{C}}\operatorname{Hom}_{S_n}(S(\nu),S(\lambda/\mu)) $$ and the Littlewood–Richardson rule says that these coefficients are enumerated by the semistandard Young tableaux satisfying the lattice permutation condition. Moreover, using James' approach one can actually construct the homomorphisms from these tableaux (so we can work on a structural level). There are also plenty of proofs using only combinatorics (of Schur functions). Is there a proof on a level between these two? The characters of the symmetric groups have "more structure" than simple combinatorics but have less structure than representations. Is there an entirely character-theoretic proof of this rule? I imagine that any such proof would use the Murnaghan–Nakayama rule for (skew representations of) symmetric groups. P.S. James gave one of the first proofs of the Littlewood–Richardson rule in 1977 (funnily enough, this wasn't proven by Littlewood and Richardson). James' proof calculates these homomorphisms over any field (whereas his contemporaries Schützenberger (1977) and Thomas (1974) only considered the semisimple case using combinatorics) and so his work is far more impressive and benefits from being self-contained. Yet he gets little recognition of this (see Wikipedia for example). His proof can be found here https://doi.org/10.1016/0021-8693(77)90380-5 REPLY [2 votes]: I might have misunderstood the question, or what it asks for, but doesn't the method of crystals do this? See Prop. 2.62. The idea is that there is a notion of a crystal structure on a combinatorial set. The cool thing is that it automatically makes your set into an $S_n$-module (so you get a representation). On the other hand, a crystal structure is very combinatorial, so you have a strong connection to combinatorics.<|endoftext|> TITLE: Handel's Theorem for surfaces with boundary QUESTION [6 upvotes]: Handel's Theorem(Entropy and semi-conjugacy in dimension two, 1987): let $M$ denote a closed surface. Let $\vartheta$ be a pseudo-Anosov (orientation-presrv.) homeomorphism of $M$ and $g$ be an (orientation-presrv.) homeomorphism of $M$ isotopic to $\vartheta$. If the topological entropy of $\vartheta$ equals the topological entropy of $g$, then $\vartheta$ is semi-conjugate to $g$. Is there any similar statement that holds for surfaces with boundary ? REPLY [2 votes]: This question is addressed in detail in Isotopy Stability of Dynamics on Surfaces, Contemp. Math., 246, 17-46, 1999 or arXiv:math/9904160 [math.DS].<|endoftext|> TITLE: How different can the constituents of an Ehrhart quasi-polynomial be? QUESTION [7 upvotes]: Consider a $d$-dimensional convex rational polytope $P\subset\mathbb{Q}^d\subset\mathbb{R}^d$. Then, it's a standard fact that in general the function counting the number of lattice points inside the multiples $t\cdot P$ of $P$ is a quasi-polynomial instead of a polynomial, as in the integral case. This is the so-called Ehrhart quasi-polynomial of $P$. This means that if $L(t,P)=\#\{x\in\mathbb{Z}^d:x\in t\cdot P\}$, then there exists a finite number of polynomial functions $f_1,\dots, f_D$, all of degree $\dim P$, such that $L(t,P)=f_i(t)$ whenever $t\equiv i\mod D$ (n.b. $D$ is the period of the polytope and if the polytope is integral we have $D=1$ and recover the usual Ehrhart polynomial of $P$). What I want to know is how much can these polynomial functions differ from one another. That is, are there known bounds to the order of $f_i-f_j$ for all $i$ and $j$? How sharp are these bounds? The case I'm dealing with is for $\dim P=2$, so if there are sharper results for this case than in general that'd be great. REPLY [2 votes]: I'll complement Christian's answer with an example in the other direction. Consider the polytope of $8\times 8$ symmetric doubly-stochastic matrices with 0 diagonal. The period of the Ehrhart quasipolymonial is 2 and the degree is 20. However the difference between the polynomials for even and odd dilations has degree only 5. I don't know (but would like to know) what happens for larger matrices.<|endoftext|> TITLE: If number of points on a manifold is $q^n ( [n+1]_q )$ does it imply a geometric relation to $A^n (P^n)$? QUESTION [6 upvotes]: Consider an algebraic manifold whose number of points is $q^n ([n+1]_q)$. Is there a geometric relation to $A^n (P^n)$? In particular, is there an equivalence in the Grothendieck ring of varieties or could there be a birational equivalence? If there is no such equivalence in general, might some additional reasonable requirements on a manifold force there to be such an equivalence? Motivation: one can see that some examples of identities on the level of enumerating $F_q$ points can be lifted to geometric relations: Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$? Can one divide algebraic manifolds ? Make sense: $Gr(2,n)/ Gr(2,n+m) = P^{n-1}/P^{n+m-1} P^{n-2}/P^{n+m-2}$ REPLY [6 votes]: The Russell Cubic $R:=V(x + x^2 y + z^2 + t^3)\subset \mathbb{A}^4$ is not isomorphic to $\mathbb{A}^3$ although over $\mathbb{C}$ they are both diffeomorphic to $\mathbb{R}^{6}$ (see this Wikipedia page). I ran a Mathematica program I quickly wrote to compute the number of solutions of $x + x^2 y + z^2 + t^3$ over $\mathbb{F}_p$. For the first twelve primes $p$ (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37) I got $p^3$. Based on this evidence I guessed that although $R$ and $\mathbb{A}^3$ are not isomorphic that they had the same counting polynomial. In the comments below, Vladimir Dotsenko, provides an elementary proof of my guess: "Consider the zero set of $x+x^2y+z^2+t^3$. Note that for $x\not=0$ we have the unique value for $y$, so this gives $(p−1)p^2$ points ($p−1$ choice for $x$, $p$ choices for $z$, $p$ choices for $t$). For $x=0$, the polynomial becomes $z^2+t^3$, so there are $p$ choices for $y$ and a choice of a zero $(z,t)$ of that polynomial. However, it clearly has $p$ zeros by the usual parametrization of a singular cubic curve: for $t=0$ there is just $z=0$, and for $t\not=0$, we have $t=−(z/t)^2$ so denoting $z/t=u$, we have $(p−1)$ solutions $(u^3,−u^2)$. "<|endoftext|> TITLE: How to add essentially new knots to the universe? QUESTION [53 upvotes]: A knot is an embedding of a circle $S^{1}$ in $3$-dimensional Euclidean space, $\mathbb{R}^3$. Knots are considered equivalent under ambient isotopy. There are two different types of knots, tame and wild. A tame knot is any knot equivalent to a polygonal knot, that is a knot whose image in $\mathbb{R}^3$ is the union of a finite set of line segments. A wild knot is a knot that is not tame. In contrary to the tame knots, wild knots are not well-behaved creatures. They often provide counterexamples to the general version of those theorems which hold for the tame knots. There are several invariants associated with the knots. For example, the crossing number of a knot is a knot invariant defined as the smallest number of crossings of any diagram of the knot. Wild knots (such as the following picture) may have infinite crossing numbers anywhere between $\aleph_0$ and $2^{\aleph_0}$. Remark 1. Some cautiousness is needed in the previous statement. According to what Joel has mentioned in his comment, the last statement can be made more accurate. In fact, the set of crossing points of a wild knot seems to be projective which implies that assuming the existence of infinitely many Woodin cardinals, it is either countable or of size continuum because the axiom of Projective Continuum Hypothesis holds under this plausible large cardinal assumption. Furthermore, based on Andreas' comment, one may remove the large cardinal assumption in the last argument too. While there has been an extensive amount of research dedicated to the classification of tame knots (particularly the iconic prime ones), up to ambient isotopy, it turned out that the classification of wild knots is generally much more complicated. (In this direction see Kulikov's paper: A non-classification result for wild knots). My question simply is how to make the chaos caused by wild knots even worse by adding new ones to the universe through forcing. Question 1. Assuming failure of Continuum Hypothesis, let $\aleph_0<\kappa\leq 2^{\aleph_0}$ be an uncountable cardinal. Is there a cardinal preserving forcing notion $\mathbb{P}$ of the universe $V$ such that in $V^{\mathbb{P}}$ there exists a wild knot $K$ of crossing number $\kappa$ which is different from all wild knots of crossing number $\kappa$ in the ground model up to ambient isotopy (definable in $V^{\mathbb{P}}$)? Remark 2. Potentially, one may find some questions, related to the possible number/behavior of wild knots of the certain knot invariant (such as bridge, unknotting, stick numbers, etc.) in a forcing extension, interesting. However, for the sake of this question, I preferred to stick to a very specific one concerning crossing number. Please feel free to share your thoughts about other possibly interesting problems along these lines in the comments below. Update. According to Ian's remark in the comments section, it turned out that the notion of crossing number for wild knots retains some ambiguity, as (in contrary to the case for tame knots) there is no straightforward theorem guaranteeing the existence of a regular projection for wild knots, which is somehow essential for counting the number of crossing points in a wild knot. Roughly speaking, if you have a wild knot, you can't even be sure how much knotted it is! Another indication of why these crazy creatures deserve the name "wild"! Anyway, the soul of the original question remains valid even without any specification to the case of tame/wild knots of any particular characteristics in terms of knot invariants. The main problem is whether it is possible to add a morphologically new knot to the mathematical universe via forcing. Precisely: Question 2. Starting from a model $V$ of $ZFC$, is there a (cardinal preserving) forcing notion $\mathbb{P}$ in $V$ such that in $V^{\mathbb{P}}$ there exists a (possibly wild) knot $K$ which is different from all knots of the ground model up to ambient isotopy (definable in $V^{\mathbb{P}}$)? REPLY [4 votes]: Awesome. In fact the way I define the reduction of a turbulent equivalence relation into wild knots is by taking infinite descending intersection of knotted tori. This can be directly converted into a forcing notion where the conditions are finite descending sequences of such tori. This forcing will directly give new knots and is probably equivalent to Cohen forcing (I am not sure). But in this way one can have greater control of what kind of a knot one wants to add, e.g. if one is interested in some particular invariants for example.<|endoftext|> TITLE: Cogenerator of Categories of Topological Spaces Satisfying Some Separation Axiom QUESTION [6 upvotes]: This question begins with a sort of mysterious comment at the bottom of this Wikipedia page on injective cogenerators. There, it is said, without citation or proof, that as a result of the Tietze Extension Theorem, the interval $I=[0,1]$ is an injective cogenerator for categories of topological spaces satisfying separation axioms (e.g. Hausdorff, Tychonoff, Kolmogorov, or other of the various $T_{i}$ conditions). So there are a few questions here: (0) This question is basically terminological. The Wikipedia page says that an injective cogenerator is simply an object that admits a non-zero (although in a general category I'm not sure what a zero map would be anyway) map from every non-zero object. But the nlab page indicates that we should rather define a cogenerator to be an object whose represented functor is faithful, so that's the definition I'm working with for now. Are these equivalent in the case that "zero map" has a meaning? (1) Is it true that the interval is an injective cogenerator for any of the categories of spaces subject to separation axioms? If so, can you give an idea of how the proof would go? It seems likely that it would work for Tychonoff spaces at least? (2) If not the interval, do the other categories of spaces subject to separation axioms admit any (set of) cogenerators? (3) Does the full category of topological spaces have a cogenerator? I was thinking that it might be the two point space $X=\{a,b\}$ with topology given by $\{\{a,b\},\{b\},\emptyset\}$, but I might be wrong here. REPLY [2 votes]: The answer to your question (0) is "no". The Wikipedia page is in the context of a category with zero object, in which case a zero morphism is one that factors through the zero object. Consider a category with four objects $A,B,C,0$, with $0$ the zero object, and four nonzero nonidentity morphisms $f,g:A\rightrightarrows B$, $h:B\to C$, and $k:A\to C$, where $h\circ f = h\circ g = k$. Then $C$ is a cogenerator in the Wikipedia sense, but $\hom(-,C)$ is not faithful. The Wikipedia page seems to believe that the two definitions are equivalent at least in an abelian category, since the subsequent sections use instead the property (equivalent to the usual "faithful represented functor" definition) that every object injects into a product of copies of the cogenerator. I'd be surprised if this were true; I would only expect defining (co)generators to detect triviality of objects rather than morphisms to work in contexts like a triangulated category, where whether a morphism is an isomorphism can be detected by whether its cone is zero. The most I can prove from the Wikipedia definition in general is that $\hom(-,C)$ reflects epimorphisms.<|endoftext|> TITLE: Examples of function fields Langlands for small genus (<= 2) QUESTION [6 upvotes]: See Edward Frenkel's article "Lectures on the Langlands program and conformal field theory" for an exposition of the function fields Langlands correspondence (now a theorem of Drinfel'd, L.Lafforgue & V.Lafforgue). It's available here: https://arxiv.org/abs/hep-th/0512172 My question is about function fields Langlands for $GL_n$ (see page 32 of the article). Drinfeld-Lafforgue's theorem gives a bijection between $n$-dimensional irreducible representations of $Gal(\overline{F}/F)$ and irreducible cuspidal automorphic representations of $GL_n(\mathbb{A}_F)$. Here $X$ is a projective curve over $k = \mathbb{F}_q$ (a finite field with $q$ elements), and $F = k(X)$ is the field of rational functions; it's algebraic closure is $\overline{F}$. See Section 2.2 (on pg 27) for more details. The ring of adeles, $\mathbb{A}_F$ is the restricted product of the fields $F_x$, where $x$ runs over all closed points of $X$; see Section 2.3 (on pg 29) for more details, and a definition of ``irreducible cuspidal automorphic representations". Let $X$ be a curve with genus either 0 (i.e. $\mathbb{P}^1$), 1 (i.e. an elliptic curve) or 2 (i.e. a hyperelliptic curves). What are some examples of the function fields Langlands correspondence (i.e. an example of a Galois representation, with a computation of the corresponding irreducible cuspidal automorphic representation)? If there are any papers computing this (for instance, using the theory of Drinfel'd shtukas), please let me know. Thanks. REPLY [10 votes]: If you really want an example of a representation, there's something funny you will find. Any irreducible cuspidal automorphic representation of $GL_n(\mathbb A_F)$ factors as a restricted tensor product of representations of the group over local fields $GL_n(F_x)$. So giving an example really means giving an example of a local representation at each place. It follows from Lafforgue's theorem (and prior work) that the local representation in this case is determined by the restriction of the Galois representation to the Galois group of $F_x$. So if you want an example of the global Langlands correspondence, all you really need is a set of examples of the local Langlands correspondence sufficient to contain all the local Galois representations of a global representation. For instance, I could tell you how to construct the representations associated to unramified Galois representations using the Satake isomorphism or unramified principal series, and how to construct the representations associated to unipotent Galois representations using Steinberg representations, and then write down an example of a semistable elliptic curve, whose local representations are all either unramified or unipotent. However, a problem with this approach is that it doesn't provide any intuition for why the representation actually appears in $GL_n(F) \backslash GL_n (\mathbb A_F)$. Thus, another approach is to, rather than explicitly describing the representation, describe a function on $GL_n(F) \backslash GL_n (\mathbb A_F)$ which generates the representation. There are some papers doing this in the literature, in the special case of rigid Galois representations, where this is the easiest (because the function can then usually be characterized by certain symmetry properties). For instance, see Heinloth-Ngo-Yun section 2.1 (though that may be a little too general/abstract for your purposes). Here is what I believe is the simplest example: Let $F= \mathbb F_q(t)$, with $q$ odd. Let $E$ be the elliptic curve over $\mathbb F_q(t)$ given by the equation $y^2 = x(x-1)(x-t)$ . It has singular fibers at $0, 1$, and $\infty$. Let $J$ be the subgroup of $K = \prod_x GL_2 (\mathcal O_{x})$ consisting of tuples of matrices whose elements at the primes $0, 1$, and $\infty$ are congruent modulo the uniformizer to upper-triangular matrices. Let $\chi$ be the character of $J$ which is defined by restricting to the matrix over the prime at infinity and taking a quadratic character of the ratio of the two diagonal entries mod the uniformizer. Then it is not too hard to calculate the space of functions on $GL_2(\mathbb A_F)$, left invariant under $GL_2(F)$, right $\chi$-equivariant under $J$, and invariant under $GL_1(\mathbb A_F)$, is two-dimensional. Indeed, one sees directly that for each double coset of $GL_2(F)$ by $JGL_1(\mathbb A_F)$, generated by an element $g$, the space of such functions on the coset is either zero or one-dimensional, and is one-dimensional if and only if $\chi$ vanishes on $g^{-1} GL_2(F) g \cap J GL_1(\mathbb A_F)$. The cosets are in bijection with rank two vector bundles on $\mathbb P^1$ with a fixed one-dimensional subsace of the fiber at $0,1,$ and $\infty$, up to twisting by line bundles, and that intersection is simply their automorphism group preserving the subspaces. So one is looking to find such vector bundles whose automorphism group does not map surjectively to $\mathbb G_m \times \mathbb G_m$ under its natural action on the one-dimensional space at $\infty$ and its quotient. Classifying, the vector bundles, one sees that there are two such. This two-dimensional space is generated by two different forms, each of which generates an automorphic representation. One corresponds to the Tate module of $E$, and the other to the Tate module of the quadratic twist of $E$. It should be possible to check this by calculating the Hecke eigenvalues by hand.<|endoftext|> TITLE: Homeomorphic open sets and homogeneity QUESTION [6 upvotes]: If $(X,\tau)$ is a $T_2$-space such that all non-empty open sets are homeomorphic (with the subspace topology) to $X$, is $(X,\tau)$ necessarily homogeneous? REPLY [7 votes]: For an infinite Hausdorff space the diversity of a space is the number of homeomorphism types of non-empty open sets, so if all non-empty open sets are homeomorphic, the space is said to be of diversity one. According to this paper, reference [11]: S.P Franklin and M. Rajagopolan, spaces of diversity one, Ramanujan Math. Soc. 5 (1990), 7-31 has an example such that a space of diversity one need not be homogeneous. I have no access to this paper, so I couldn't look to see and describe the example. Maybe someone else can. But it's not hard to prove that a zero-dimensional, first countable space of diversity one is homogeneous (it's also in that paper, but the proof idea is classical); examples of such spaces are $\mathbb{Q}$ and $\mathbb{P}$ (the irrationals). Any compact space of diversity two is homogeneous (reference in the linked paper) and a compact metric space of diversity two (diversity one is impossible for a compact space) is homeomorphic to the Cantor set. (the two types being clopen = Cantor set, and open, non-clopen = Cantor set minus a point).<|endoftext|> TITLE: Kozsul resolution of $\mathcal{O}_X$ QUESTION [7 upvotes]: Let $i: X \hookrightarrow Y$ be a closed embedding of smooth algebraic varieties. In the book D-modules, perverse sheaves and representation theory the authors say that there exists a locally free resolution of the $i^{-1}\mathcal{O}_Y$-module $\mathcal{O}_X$ called the Koszul resolution. Namely, they say it is $$ 0 \rightarrow K_{n-r} \rightarrow \dots \rightarrow K_1 \rightarrow K_0 = i^{-1}\mathcal{O}_Y \rightarrow \mathcal{O}_X, $$ and choosing a local coordinate system such that on the open subset $U \subseteq Y$ we have $X \cap U = \{ y_{r+1} = \dots = y_n = 0 \}$, we have $$ K_{j} = \bigwedge^j \left( \bigoplus_{s = r+1}^n \; i^{-1}\mathcal{O}_Y dy_{r} \right). $$ I don't understand whether this resolution exists globally or only locally. I guess it exists globally, but I don't understand how to define it. Both answers and references are welcome. REPLY [10 votes]: It exists locally, and more generally when $X$ is the zero locus of a global section $s$ of a rank $r$ vector bundle $E$ on $Y$, where $r$ is the codimension of $X$ in $Y$. Then the resolution is given by the celebrated Koszul complex $$0\rightarrow \bigwedge^rE^*\xrightarrow{\ i(s)\ }\bigwedge^{r-1}E^*\rightarrow \cdots \rightarrow E^* \xrightarrow{\ i(s)\ } \mathcal{O}_Y\rightarrow \mathcal{O}_X\rightarrow 0\, ,$$where $i(s)$ denotes the interior product with the section $s$ (googling "Koszul complex" should give you thousands of references). Locally $X$ is given by $r$ equations $f_1=\cdots =f_r=0$, and you can just take $E=\mathcal{O}_Y^r$ and $s=(f_1,\ldots ,f_r)$. Globally such a vector bundle may or may not exist, depending on the geometry of $X$.<|endoftext|> TITLE: Galois descent in motivic cohomology QUESTION [6 upvotes]: Let $X_N$ denote the Fermat curve defined over $\mathbb{Q}$ by the equation $x^N+y^N-z^N=0$ and let $X_{N,\mathbb{Q}(\mu_N)}$ be the base change. Let $G$ be the Galois group of $\mathbb{Q}(\mu_N)/\mathbb{Q}$, hence $G \cong (\mathbb{Z}/N\mathbb{Z})^\times$. Consider the motivic cohomology $H^2_\mathcal{M}(X_N,\mathbb{Q}(2))$. I would like to know whether $$ H^2_\mathcal{M}(X_N,\mathbb{Q}(2)) \cong H^2_\mathcal{M}(X_{N,\mathbb{Q}(\mu_N)},\mathbb{Q}(2))^G.$$ Motivation: Let $W:=(\mathbb{Z}/N\mathbb{Z})^2$ act on $X_{N,\mathbb{Q}(\mu_N)}$ as $$(r,s)(x:y:z) := (\zeta^rx:\zeta^sy:z),$$ where $\zeta$ is a primitive $N-$th root of unity. I am reading On the regulator of Fermat motives and generalized hypergeometric functions by N. Otsubo, where he proves some results of surjectivity of regulators using the splitting of the motive $h^1(X_N)$ into motives $X_N^{[a,b]}=(X_N,p_N^{[a,b]})$, where $p_N^{[a,b]}$ is in $\mathbb{Q}[W]^G$. I am working on something related, but I am not much acquainted with motivic cohomology. My definitions (which are quite ad hoc for this situation) come from Milnor $K-$theory: $$H^2_\mathcal{M}(X_N,\mathbb{Q}(2)) \cong K_2(X_N)_\mathbb{Q} := \ker\left( K_2^M(k(X_N))\otimes \mathbb{Q} \overset{T\otimes\mathbb{Q}}{\longrightarrow} \bigoplus_{x \in X_N(\overline{\mathbb{Q}})} \overline{\mathbb{Q}}^\times \otimes \mathbb{Q} \right),$$ where $T$ denotes the Tame symbol (see for instance page 27 of the above article). My original problem is that I want to show that if $e_N$ is in $K_2(X_N)_\mathbb{Q}$, then the elements $e_N^{[a,b]}:=p_N^{[a,b]}(e_N)$ defined by Otsubo are in $K_2(X_N)_\mathbb{Q}$ as well. But, in principle, I only know that they are in $K_2(X_{N,\mathbb{Q}(\mu_N)})_\mathbb{Q}^G$. So I would like to know whether the two vector spaces are equal. The problem is that, starting from my definition, I think that we have $$k(X_N)^\times \otimes k(X_N)^\times \subsetneqq (k(X_{N,\mathbb{Q}(\mu_N)})^\times \otimes k(X_{N,\mathbb{Q}(\mu_N)})^\times)^G.$$ So if things work out in the end it must be thanks to the Steinberg relation $a\otimes(1-a)$ that still has to be quotiented out in order to define $K_2(k(X_N))$, or thanks to the Tame symbol. But this seems to be difficult to check, hence I was wondering if there is some general argument for the equality $$K_2(X_N)_\mathbb{Q}=K_2(X_{N,\mathbb{Q}(\mu_N)})_\mathbb{Q}^G$$ for example coming from motivic cohomology. REPLY [3 votes]: What you need is the existence of the transfer map $N : K_2(L) \to K_2(K)$ for any finite field extension $L/K$, which is due to Bass and Tate, see Introduction to algebraic $K$-theory by Milnor. I don't know of any definition of the transfer map using Matsumoto's decription of $K_2$, one should rather use Milnor's definition of $K_2$. In any case, if $L/K$ is Galois and $j : K_2(K) \to K_2(L)$ is the canonical map, then $N \circ j = |G| \cdot \mathrm{id}$ and $j \circ N = \sum_{\sigma \in G} \sigma^*$. We deduce in particular that $j$ induces an isomorphism $K_2(K)_{\mathbb{Q}} \cong (K_2(L)_{\mathbb{Q}})^G$. Granted these facts and returning to your situation, we have $K_2(\mathbb{Q}(X_N))_{\mathbb{Q}} \cong (K_2(\mathbb{Q}(\mu_N)(X_N))_{\mathbb{Q}})^G$ for the function fields, and you should get the isomorphism you want by taking the kernel of tame symbols. I should add that transfer maps exist in much greater generality for motivic cohomology with rational coefficients, and for finite Galois covers the obvious formulas for $N \circ j$ and $j \circ N$ are still valid. See e.g. Deninger--Scholl, The Beilinson conjectures, (1.3) for a nice summary of the properties of motivic cohomology.<|endoftext|> TITLE: Is the product of Jacobson rings a Jacobson ring? QUESTION [10 upvotes]: I asked this question on Mathematics Stackexchange, but got no answer. Is the product $$ \prod_{i\in I}A_i $$ of a family $(A_i)_{i\in I}$ of Jacobson rings a Jacobson ring? (Here "ring" means "commutative ring with one".) REPLY [12 votes]: The answer is no in general. Take $R = \prod_{n \in \mathbb{N}_{> 0}} \mathbb{Z}/2^n\mathbb{Z}$. Then the Jacobson radical of $R$ is $\prod_{n \in \mathbb{N}_{> 0}} 2\mathbb{Z}/2^n\mathbb{Z}$, and it contains a non-nilpotent element, namely $(2 + 2^n \mathbb{Z})_n$. Therefore the Jacobson radical of $R$ doesn't coincide with the nilradical of $R$. All credits go to Georges Elencwajg, see this MO post. Side note. Let $J(R)$ denote the Jacobson radical of a commutative unital ring $R$. Knowing that $x \in J(R)$ holds if and only if $1 + rx \in R^{\times}$ holds for every $r \in R$, the identity $J(\prod_{i \in I} A_i) = \prod_{i \in I} J(A_i)$ is immediate.<|endoftext|> TITLE: Are all exotic affine spaces count equivalent to affine space? QUESTION [10 upvotes]: This question is inspired by this MO question; indeed it is a special case on which to focus. An exotic affine space is an affine variety $V$ whose $\mathbb{C}$-points are diffeomorphic to $\mathbb{R}^{2n}$ yet $V$ is not algebraically isomorphic to $\mathbb{A}^n$. Say that two varieties are count equivalent if they are both polynomial count varieties with the same counting polynomial. As shown in the comments here, the Russell Cubic is count equivalent to $\mathbb{A}^3$ although it is not isomorphic to $\mathbb{A}^3$. Question: Are all exotic affine spaces count equivalent to affine space? REPLY [11 votes]: For all but finitely many primes, yes. Any such $V$ has $V_{\mathbb C}$ smooth and has $H^i(V_{\mathbb C}, \mathbb Q_\ell)=0$ for $i\neq 0$ and $=\mathbb Q_\ell$ for $i=0$. Both these properties are known to be constructible, so they hold for $V_{\mathbb F_q}$ for all but finitely many $q$. For any such $q$, the action of $\operatorname{Frob}_q$ on $H^0(V_{\overline{\mathbb F}_q}, \mathbb Q_\ell)$ must be trivial as it is one-dimensional and contains at least one invariant class. So its cohomology, with Frobenius action, matches affine space. Applying Poincare duality (using smoothness), the same is true for its cohomology with compact supports. Applying the Grothendieck-Lefschetz formula, its number of points matches affine space.<|endoftext|> TITLE: Determining if a rational function has a subtraction-free expression QUESTION [28 upvotes]: This question was first asked by Mehtaab Sawhney in Alex Postnikov's combinatorics class. Given a rational function $F=P(x_1,...,x_n)/Q(x_1,...,x_n)$ with (say) integer coefficients, it is often of interest in combinatorics whether $F$ has a subtraction-free expression, that is, whether $F$ is formally equal to some rational function $G=R(x_1,...,x_n)/S(x_1,...,x_n)$ where $R,S$ have nonnegative coefficients. I consider $F$ to be equal to $G$ when $PS=QR$ as polynomials. For example, although $1-x+x^2$ contains a minus sign, it has the subtraction-free expression $\frac{1+x^3}{1+x}$. It is clear that not all rational functions have such an expression, since for example any subtraction-free rational function takes positive values when all inputs are positive. Is the problem of determining whether $F$ has a subtraction-free expression decidable? Is there a reasonably efficient algorithm in terms of $n$, $\deg(P)$, $\deg(Q)$? REPLY [16 votes]: $\def\RR{\mathbb{R}}$I'm pretty sure I know what the answer is, but I found it surprisingly hard to find references for the facts I needed. So, here is what I think the truth is: Let $f(x_1, \ldots, x_n)$ be a polynomial with real coefficients; we write $$f(x_1, \ldots, x_n) = \sum_{a \in A} f_a x^a$$ for some finite subset $A$ of $\mathbb{Z}^n$. Let $N(f)$ be the convex hull of $A$, also known as the Newton polytope of $f$. For each face $F$ of $N(f)$, let $f_F = \sum_{a \in A \cap F} f_a x^a$. Then I claim that $f$ has a subtraction free expression if, for all faces $F$ of $N(f)$, and all $(x_1, \ldots, x_n) \in \RR_{>0}^n$, we have $f_F(x_1, \ldots, x_n)>0$. Furthermore, for any integer polytope $\Delta$, I claim we can find a polynomial $h$ with positive coefficients such that, if $N(f) = \Delta$ and $f$ obeys the above condition, then $f h^M$ has positive coefficients for all sufficiently large $M$. Claim 1: If $f$ has a subtraction free expression, then $f$ obeys this condition. Write $f = \tfrac{p}{q}$ where $p$ and $q$ have positive coefficients. Suppose for the sake of contradiction that $f_F$ is negative somewhere. Choose a linear functional $\lambda$ on $N(f)$ which is maximized on $F$. Let $P$ and $Q$ be the faces of $N(p)$ and $N(q)$ where $\lambda$ is maximized. Then $f_F = \tfrac{p_P}{q_Q}$. Since $p$ and $q$ have positive coefficients, that's a contradiction. $\square$ If $f$ obeys this condition, then $f$ has a subtraction free expression This is the one I don't have a complete proof of. If $N(f)$ is a multiple of the standard simplex, meaning that $f$ is homogenous of degree $d$ and the coefficients of the monomials $x_1^d$, $x_2^d$, ... and $x_n^d$ are nonzero, then this is a result of Polya, taking $h = x_1 + x_2 + \cdots + x_n$. It seems to me that the generalization to other Newton polytopes should be a matter of bookkeeping and saddlepoint approximations. (Specifically, I think we should take $h = \sum_{a \in d N(f)} x^a$ for $d$ chosen large enough.) I thought this would be in the literature -- it seems to me like low hanging fruit on Polya's tree -- but I can't find it. ADDED Sam Hopkins points out that I only addressed the case of polynomials, not rational expressions. I claim that, if the above is right, this also resolves the rational function case. Specifically, I claim the test is the following: Write your rational expression in lowest terms as $f/g$. Choose a point $x \in \mathbb{R}_{>0}^n$ where $f$ and $g$ are both nonzero. If $f(x)/g(x)<0$, your function does not have a subtraction free expression. Otherwise, after replacing $f$ and $g$ by $-f$ and $-g$, we may assume that $f(x)$ and $g(x)>0$. After making this replacement, there is a positive expression for $f/g$ if and only if there is separately for $f$ and for $g$. Clearly, if $f/g$ passes this test, it has a positive expression. What we need to show is that, if there is a positive expression for $f/g$ then, after sign normalization, there is for $f$ and $g$ separately. IF there is such a positive expression, then there is some $h$ such that $fh$ and $gh$ have positive coefficients. We wish that we knew that $h$ also had positive coefficients. In other words we need: Lemma Let $f$ be a polynomial which is positive somewhere in $\mathbb{R}_{>0}^n$ and suppose there exists a polynomial $h$ such that $fh$ has positive coefficients. Then there is a polynomial $h'$ with positive coefficients such that this occurs. Proof The condition that $h$ exists implies that $f$ has constant sign on $\mathbb{R}_{>0}^n$, and this sign is by hypothesis positive. The same is true for each $f_F$. Thus, by the unproven part of the above claims, there is an $h'$ with positive coefficients such that $f h'$ has positive coefficients. $\square$<|endoftext|> TITLE: Hochschild homology of a category of modules over an algebra QUESTION [5 upvotes]: Suppose $A$ is an algebra over some field, say the complex numbers if that helps. Then we can consider the category $\mathbf{C}_A$ of finite-dimensional modules over $A$. This category can be seen as a dg-category and hence has a Hochschild homology $HH_*(\mathbf{C}_A)$. What is known about $HH_*(\mathbf{C}_A)$ for instance in low degree? How does it relate to the traditional Hochschild homology $HH_*(A)$ of the algebra $A$? Note that this question is related to http://mathoverflow.net/questions/295876/hochschild-homology-of-a-hopf-algebra but still different. Thank you for any hints. REPLY [7 votes]: This is merely a couple of examples showing how $HH_*(\mathbf{C}_A)$ may behave. Let first $A$ be the polynomial algebra $\Bbb C[x]$. Then $\mathbf{C}_A$ is the category of coherent sheaves with zero-dimensional support on $\Bbb A^1$. Since sheaves with supports on different points are orthogonal,this category is just the direct sum over all closed points of $\Bbb A^1$ of the corresponding categories, $\mathbf{C}_A = \oplus_{\lambda \in \Bbb C}~ \mathrm{Coh}_{\lambda}(\Bbb A^1).$ $Coh_\lambda(\Bbb A^1)$ is the category of vector spaces endowed with an endomorphism with all eigenvalues equal to $\lambda$. In particular, all these categories are equivalent to each other and, in particular, to $\mathrm{Coh}_{0}(\Bbb A^1)$, the category of vector spaces with nilpotent endomorphism, in other words, to $\mathbf{C}_{\Bbb C[[x]]}$, where $\Bbb C[[x]]$ is the power series algebra. Now, there's a localization sequence of triangulated categories: $$\mathbf{C}_{\Bbb C[[x]]} \longrightarrow \Bbb C[[x]]-\mathrm{mod}_{\mathrm{fin. gen.}} \longrightarrow \Bbb C[x^{-1}, x]] - \mathrm{vect}_{\mathrm{fin. dim.}}$$ where $\Bbb C[x^{-1}, x]] - \mathrm{vect}_{\mathrm{fin. dim.}}$ is the category of finite dimensional vector spaces over the field of Laurent series. This sequence gives you a long exact sequence of Hochschild homology $$\dots \longrightarrow HH_0(\mathbf{C}_{\Bbb C[[x]]}) \longrightarrow HH_0(\Bbb C[[x]]-\mathrm{mod}) \longrightarrow HH_0(\Bbb C[x^{-1}, x]] - \mathrm{vect}) \longrightarrow HH_{-1}(\mathbf{C}_{\Bbb C[[x]]}) \longrightarrow 0.$$ (see Keller, "On the cyclic homology of exact categories", https://www.sciencedirect.com/science/article/pii/S0022404997001527) In particular, you can see that $HH_{-1}(\mathbf{C}_{\Bbb C[[x]]})$ is non-zero and is isomorphic to the polynomials in $x^{-1}$. I don't know for sure, but I'd think that using something like adelic resolutions will show you that the category of sheaves with zero-dimensional support on a smooth variety of dim $n$ will have non-zero (in general) $HH_k$ for $k > -n$. On the other hand, if you take $U{\mathfrak g}$, the universal enveloping algebra of some semisimple Lie algebra, then $\mathbf{C}_{U{\mathfrak g}}$ will be semisimple, so only $HH_0(\mathbf{C}_{U{\mathfrak g}})$ will be non-zero, while $HH_*(U\mathfrak g -\mathrm{mod})$ is isomorphic to the Lie algebra homology of $\mathfrak g$ with coefficients in the ring of invariant polynomials and is non-zero in arbitrary large degrees. And of course if your algebra was finite dimensional from the beginning, you'll get nothing new.<|endoftext|> TITLE: Mixed norm inequality QUESTION [5 upvotes]: Suppose we have a product space $(X_1\times X_2,\mu_1\otimes\mu_2)$, with finite measures $\mu_1,\mu_2$ and $p>1$. Is there a possibility that an inequality of this form holds on the product space? $$\|f\|_{L^pL^p}\leq C_1\|f\|_{L^1L^p} + C_2\|f\|_{L^pL^1},$$ where $\|f\|_{L^pL^q}=\big(\int_X\big(\int_Y |f|^pd\mu_2\big)^{q/p}d\mu_1 \big)^{1/q}$. REPLY [8 votes]: The answer is no. E.g., suppose that $X_1=X_2=[0,1]$, $\mu_1=\mu_2=$ Lebesgue measure, $f(x_1,x_2)=g(x_1)g(x_2)$, $g=1_{[0,u]}$, $u\in(0,1)$. Then your proposed inequality becomes $$u^{2/p}\le(C_1+C_2)u^{1+1/p},$$ which fails to hold for any given real $p>1$, $C_1$, $C_2$ if $u$ is small enough.<|endoftext|> TITLE: Periods of the continued fraction expansions of Galois-conjugate quadratic-irrationals QUESTION [5 upvotes]: Question: Given a quadratic irrational $x = a + b\sqrt{D}$ ($a,b \in \Bbb{Q}$, $D \in \Bbb{N}_{> 0}$ square-free) and its Galois conjugate $x' = a - b\sqrt{D}$, is it true that the continued fraction expansions of $x$ and $x'$ have the same period? Computations of a few random example seems to suggest that is indeed the case, and I think I even saw a statement to this effect somewhere, but can't remember where. Motivation: The continued fraction expansions of two numbers $x,x' \in \Bbb{R}$ conjugate via an element of $\text{PGL}_2(\Bbb{Z})$ are 'almost the same', i.e. $a_{k+n} = a'_{k+m}$ for some $n,m$ and all $k \geq 0$, and I suspect that in the case of two conjugate quadratic-irrationals there always exists a 'generalized reflection' $s_H = wsw^{-1}$, $w \in \text{PGL}_2(\Bbb{Z})$, $s \in \left\{\begin{bmatrix} & 1 \\ 1& \end{bmatrix}, \begin{bmatrix} -1 & 1 \\ & 1\end{bmatrix}, \begin{bmatrix} -1 & \\ & 1\end{bmatrix}\right\}$ such that $s_H(x) = x'$. REPLY [6 votes]: The periodic part of the continued fraction for the Galois conjugate is always the mirror image of the periodic part for the original quadratic irrational. Here we are viewing the periodic part as a cycle arranged in a circle, and the mirror image is with respect to reflecting across a suitable diameter of this circle. The periodic part of every irrational number $\sqrt{p/q}$ (with $p$ and $q$ integers) is always palindromic in the sense that its cyclic arrangement has mirror symmetry. This follows from the classical fact that the numbers $\sqrt{p/q}$ with $p/q>1$ are exactly the numbers whose continued fraction has the form $[a_0;\overline{a_1,\cdots,a_n}]$ with the terms $a_1,\cdots,a_{n-1}$ forming a palindrome and $a_n=2a_0$. Inserting the last term $a_n$ into the palindromic cycle formed by $a_1,\cdots,a_{n-1}$ then produces a cycle with mirror symmetry across the diameter through $a_n$. All this becomes visually clear using Conway's notion of topographs for binary quadratic forms over the integers. One place to read about this viewpoint is in Chapter 4 of my unfinished book "Topology of Numbers", available on my webpage.<|endoftext|> TITLE: Explicit expression of the unstraightening functor QUESTION [8 upvotes]: Hard as I tried, I couldn't find a proof of Remark 2.2.2.11 in Higher Topos Theory, or prove it myself. It seems to need an explicit formulation for the unstraightening functor, so my question is: is an explicit expression known for the unstraightening? Anyway, is it possible to obtain Remark 2.2.2.11 without having one? REPLY [4 votes]: This is an old question, but it seems worthwhile to give the full explicit description of unstraightening, for convenient reference. I'll do this for contravariant unstraightening, and trying to match the notation used in HTT 2.2.1 (i.e., resisting the temptation to replace $\mathcal{C}$ with $\mathcal{C}^{\mathrm{op}}$ everywhere). Fix functors $\phi\colon \mathfrak{C}[S]\to \mathcal{C}^{\mathrm{op}}$ and $F\colon \mathcal{C}\to \mathrm{Set}_\Delta$ of simplicial categories. I will describe the $n$-simplices of $\mathrm{Un}_\phi F$, which is a simplicial set mapping to $S$. Given a map $s\colon \Delta^n\to S$ let $\phi_s= \phi\circ \mathfrak{C}[s]$, a functor $\mathfrak{C}[\Delta^n]\to \mathcal{C}^{\mathrm{op}}$. Then there is a bijective correspondence between $n$-simplices of $\mathrm{Un}_\phi F$, and pairs $(s,g)$, where $s\colon \Delta^n\to S$ is a map of simplicial sets, and $$ g\colon D^n\to F\circ \phi_s^{\mathrm{op}} $$ is a map of simplicial functors $\mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathrm{Set}_\Delta$. Here $D^n$ is a particular functor $\mathfrak{C}[\Delta^n]^{\mathrm{op}} \to \mathrm{Set}_\Delta$, defined as follows. Consider the simplicial category $\mathfrak{C}[(\Delta^n)^\rhd]\approx \mathfrak{C}[\Delta^{n+1}]$, which contains $\mathfrak{C}[\Delta^n]$ as a subcategory. Then $D^n$ is the functor represented by the object corresponding to the cone point $v$ of $(\Delta^n)^\rhd$. When you unwind this, you get that $D^n(x)$ is isomorphic to the nerve of a poset: $$ D^n(x) \approx N\bigl\{ S\subseteq \{x,x+1,\dots, n\} \;\bigm|\; x\in S \bigr\} \approx (\Delta^1)^{n-x}. $$ (In fact, $D^n$ is nothing other than the straightening of $(\mathrm{id}\colon \Delta^n\to \Delta^n)$ along $\phi=\mathrm{id}\colon \mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathfrak{C}[\Delta^n]^{\mathrm{op}}$.) In HTT 2.2.2.11, we are interested in $s\colon \Delta^n\to \{s_0\}\to S$ which factor through a single vertex in $S$, which maps under $\phi$ to some object $C$ in $\mathcal{C}$. In this case $F\circ \phi_s^{\mathrm{op}}$ is a constant simplicial functor $\mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathrm{Set}_\Delta$ with value $F(C)$. So natural transformations $D^n\to F\circ \phi_s^{\mathrm{op}}$ are the same as maps of simplicial sets $Q^n\to F(C)$, where $Q^n$ is the enriched left Kan extension of $D^n$ along $\mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathfrak{C}[\Delta^0]^{\mathrm{op}}=*$. Unwinding this should yield Lurie's description of $Q^n$, which will be as a quotient of $D^n(0)\approx (\Delta^1)^n$.<|endoftext|> TITLE: complicated combinatorial algorithms with good descriptions QUESTION [7 upvotes]: For educational purposes, I am looking for an example of a complicated, elementary, but very well-explained combinatorial algorithm. Such an example might be a bijection between two easily described sets, where the bijection involves many case distinctions. Of course, "very well-explained" is completely subjective, but that is really the focus of the question. REPLY [2 votes]: I have a candidate. Let $\mathcal{A_n} = \{w \in S_n: w_1 = w_n+1, \mbox{Fl}(w_2 \dots w_{n-1}) \in \mathcal{A}_{n-2} \}$ where $\mbox{Fl}$ is the flatten operator mapping a word with distinct entries to the permutation with the same relative order. In one of my papers, we show that $$ \sum_{w \in \mathcal{A}_{2n+1}} F_w = s_{\delta_n}^2 $$ where $F_w$ is the Stanley symmetric function and $\delta_n = (n{-}1,n{-}2,\ldots,1)$ (a similar result holds for $\mathcal{A}_n$ with $n$ even). We have since given two bijective proofs of this identity - one based on an insertion algorithm and the other based on a variant of the Little bijection. What we don't mention is that I have a more or less complete document defining a bijection $$ \mbox{SYT}(\delta_{2n}\setminus (n)^n) \to \mathcal{R}(n{+}1\ n \dots1 \ 2n{+}2\ 2n{+}1 \dots n{+}1) \to \bigcup_{w \in \mathcal{A}_{2n+1}} \mathcal{R}(w) $$ where $\mathcal{R}(w)$ is the set of reduced words of $w$. The first arrow here follows immediately from the literature, but the second requires repeated application of a very precise choice of Little bumps. My implementation of the algorithm in SAGE has a little over 100 lines of code (not counting an implementation of the quite technical Little bijection), mostly to describe which Little bumps to perform. The description of the algorithm in the uncompleted paper takes about 2/3 of a page once all terminology has been established. The three main steps call each other recursively and repeatedly, with various cases determining which step to return to. Converting the description to pseudo-code would make it much longer. One attractive feature of this map for your purposes is that it runs into quite subtle issues. Initial attempts to code the bijection would usually succeed for instances of the problem with $n \sim 50$, while for $n \sim 100$ they tended to get caught in infinite loops. With a few more major pieces of casework, the algorithm could be proved to terminate. In the end, it was relatively easy to see the map is an injection, but I was missing a key technical lemma about Little bumps to show it was invertible. Since better proofs are available, I will probably never write the full details, but I am happy to share a correct implementation or a complete description of the map.<|endoftext|> TITLE: Density of the Klarner-Rado Sequence QUESTION [8 upvotes]: Consider the Klarner-Rado sequence OEIS A005658 defined by the rule: the sequence starts with 1, and if it contains $n$ it also contains $2n$, $3n+2$ and $6n+3$. According to R. Guy's popular article, Klarner conjectures that this set has positive density. Lagarias in The Ultimate Challenge book says the conjecture is open. Question: should we believe or disbelieve the conjecture? Is there a heuristic argument explaining what's going on here? REPLY [2 votes]: I have performed a heuristic calculation of the density of the Klarner-Rado sequence. My calculation suggests that the Klarner's conjecture is false. Instead, I expect the number of terms in the Klarner-Rado sequence between $1$ and $N$ approaches $c N / \log N$ as $N$ approaches infinity, for some positive constant $c$. The Klarner-Rado sequence is the range of the function $f : \{0, 1, 2\}^* \to \mathbb {N}$ whose domain is the set of sequence of the letters $\{0, 1, 2\}$, defined by \begin{align*} f (0x) &= 2 f (x) \\ f (1x) &= 3 f (x) + 2 \\ f (2x) &= 6 f (x) + 3 \\ f (e) &= 1 \end{align*} where $e$ is the empty sequence. I will start with the easier question of counting Klarner-Rado numbers less than $N$ with multiplicity: that is, how many $x$ are there with $f (x) < N$? To simplify this further, I will replace $f$ with a purely multiplicative version of itself which differs from $f$ only by a constant factor: \begin{align*} \tilde {f} (0x) &= 2 \tilde {f} (x) \\ \tilde {f} (1x) &= 3 \tilde {f} (x) \\ \tilde {f} (2x) &= 6 \tilde {f} (x) \\ \tilde {f} (e) &= 1 \end{align*} Then $g (x) = \log \tilde {f} (x)$ has a simple expression $g (c_0 c_1 \dots c_{r-1}) = a_{c_0} + a_{c_1} + \dots + a_{c_{r-1}}$ where $a_0 = \log 2$, $a_1 = \log 3$, $a_2 = \log 6$. To understand the distribution of $g$, and hence $\tilde {f}$, I will consider the function $$ G (s) = \sum _{x \in \{0, 1, 2\}^*} e^{- s g (x)} = \sum _{x \in \{0, 1, 2\}^*} Z^{g (x)} $$ where $Z = e^{-s}$. The second expression displays how $G (s)$ is a sort of generating function, except that arbitrary real-valued exponents are allowed. (In a similar manner the Riemann zeta function can be thought of as the generating function $\zeta (s) = \sum_{n=1}^\infty Z^{\log n}$ where $Z = e^{-s}$.) $G (s)$ can be evaluated as $$ G (s) = \sum_{n=0}^\infty (e^{-s \log 2} + e^{-s \log 3} + e^{-s \log 6})^n = \frac {1} {1 - (2^{-s} + 3^{-s} + 6^{-s})} $$ $2^{-s} + 3^{-s} + 6^{-s}$ is decreasing in $s$, and $2^{-1} + 3^{-1} + 6^{-1} = 1$. It follows that $G (s)$ converges for $s > 1$ and approaches infinity as $s \to 1$. In fact, $G (s) \sim c / (s-1)$ for $s \sim 1$ and some $c$. This suggests that the number of $x$ with $g (x) < k$ is approximately $e^k$, with an averaged multiplicative error that is $O (e^{\epsilon k})$ for all $\epsilon > 0$. Moreover, the similar series $$ \sum_{n=0}^{\infty} e^{-n s} = \frac {1} {1 - e^{-s}} $$ has a similar behavior near the divergence point, it is $\sim \frac {1} {s}$ for $s \sim 0$, and it is distributed uniformly. This suggests the multiplicative error for my estimate is even smaller than subexponential. On the other hand, I believe the analytic continuation of $G (s)$ has infinitely many poles with imaginary parts arbitrarily close to $1$, which suggests a significant oscillatory component to the density. Ignoring these difficulties, I will assume that the number of $x$ with $g (x) < k$ is around $c e^k$ for some $c$. This is the same as the number of $x$ with $\tilde {f} (x) < N = e^k$, and this number is $c N$. Since $f$ and $\tilde {f}$ only differ by a bounded multiplicative factor, I expect that density of the Klarner-Rado sequence with multiplicity is bounded below by a constant. What about the density of the Klarner-Rado sequence without multiplicity? With the above, this can be reformulated as the question: If $x$ is a sequence with $f (x) < N$, what is the probability that $x$ is first in dictionary order with this value for $f (x)$? If the density of the Klarner-Rado sequence is bounded below, so should this probability. On the other hand, this occurs for $x$ if and only if $x$ doesn't have any suffix $1 y$ with $f (1y) = f (0z)$ for some $z$. Heuristically the probability that such a $z$ exists for a given $y$ should be the same as the density of $\mathrm {ran} f$, times a constant to account for the nonuniformity of the sequence at different residues. So this probability is bounded below. But as $N$ goes to infinity, so does the number of suffixes of $x$ of the form $1 y$, and so the probability the $x$ is least in dictionary order goes to zero! We get a (heuristic) contradiction. Therefore I don't expect Klarner's conjecture about the density to hold. Note that these suffixes are distributed in logarithmically uniform distribution on $[1, N]$. Let $h (k)$ be the density of the Klarner-Rado sequence on $[1, e^k]$. Then the probabilistic argument above suggests that $$ \log h (k) \sim - c^{-1} \int_0^k h (t) dt $$ and so $$ \frac {h' (k)} {h (k)} \sim - c^{-1} h (k) $$ $$ h' (k) \sim - c^{-1} h (k)^2 $$ Since the integral actually takes the form of a discrete sum close to $k$, this "differential equation" is more properly a complicated difference equation with the same general behavior as the differential equation I wrote. Ignoring this complication, we get a solution $h (k) \sim c/(k-k_0) \sim c/k$ ($k_0$ is irrelevant for the lowest-order asymptotic estimate). Therefore I expect the density on $[1, N]$ is $h (\log N) = c/\log N$, and there are $c N / \log N$ terms of the sequence on this interval.<|endoftext|> TITLE: (pro)Étale cohomology of adic spaces and inverse limit QUESTION [11 upvotes]: I am studying Peter Scholze's paper $p$-adic Hodge theory for rigid-analytic varieties and I am confused by the following. Let $X$ be a finite type scheme over $\mathbb{C}_p$ (proper and smooth if you want) and let $X_{\mathrm{ét}}$ be the usual étale site. Usually, the (algebraic) étale cohomology with coefficients in $\mathbb{Z}_p$ is defined as $$ H^i(X_{\mathrm{ét}}, \mathbb{Z}_p) := \varprojlim_n H^i(X_{\mathrm{ét}}, \mathbb{Z}/p^n), $$ and we cannot move the inverse limit inside. Indeed one can consider the cohomology of the constant sheaf $\mathbb{Z}_p$, but this is not well behaved. Let $X^{\mathrm{ad}}$ the adic space associated to $X$, with étale site $X_{\mathrm{ét}}^{\mathrm{ad}}$. The étale cohomology is defined as above, i.e. as the inverse limit of the cohomology groups of the sheaves $\mathbb{Z}/p^n$. Now let us consider the pro-étale cohomology. If I understand correctly, the paper, Lemma 3.18 (together with other results) says that cohomology commutes with inverse limit, so $$ H^i (X^{\mathrm{ad}}_{\mathrm{pro-ét}}, \mathbb{Z}_p) \cong \varprojlim_n H^i(X^{\mathrm{ad}}_{\mathrm{pro-ét}}, \mathbb{Z}/p^n), $$ where this time the group on the left is just the cohomology of the constant sheaf $\mathbb{Z}_p$. So far so good, but let $\nu \colon X^{\mathrm{ad}}_{\mathrm{pro-ét}} \to X^{\mathrm{ad}}_{\mathrm{ét}}$ be the natural morphism of sites. Corollary 3.17 says that for any abelian sheaf $\mathcal{F}$ on $X^{\mathrm{ad}}_{\mathrm{ét}}$ we have $$ H^i(X^{\mathrm{ad}}_{\mathrm{ét}}, \mathcal{F}) \cong H^i(X^{\mathrm{ad}}_{\mathrm{pro-ét}}, \nu^\ast\mathcal{F}), $$ and if we apply this to $\mathcal{F} = \mathbb{Z}_p$ (the constant sheaf, no inverse system here) we find $$ H^i(X^{\mathrm{ad}}_{\mathrm{ét}}, \mathbb{Z}_p) \cong H^i(X^{\mathrm{ad}}_{\mathrm{pro-ét}}, \nu^\ast\mathbb{Z}_p) = H^i(X^{\mathrm{ad}}_{\mathrm{pro-ét}}, \mathbb{Z}_p)= \\ \varprojlim_n H^i(X^{\mathrm{ad}}_{\mathrm{pro-ét}}, \mathbb{Z}/p^n) \cong \varprojlim_n H^i(X^{\mathrm{ad}}_{\mathrm{ét}}, \mathbb{Z}/p^n). $$ In particular it seems that, for the étale cohomology of adic spaces, there is no need of taking the inverse limit outside the cohomology. Is this correct/well known or am I missing something? Thank you! REPLY [3 votes]: This problem came up in a reading group yesterday, and the consensus was that the problem is exactly the one that user45878 pointed out in the comments. Thanks to Bogdan Zavyalov for explaining this issue to me and providing the example below. Let's see why these are different with an example: Consider the scheme $X = \mathrm{Spec}(\mathbf C)$ and its pro-étale site. Note that the category $X_{\mathrm{et}}$ is equivalent to the category of finite sets, with covers surjective maps. Via this equivalence, we can see that $X_{\mathrm{pro-et}}$ consists of pro-finite sets, i.e. topological spaces $S$ which are isomorphic to $\varprojlim_{i \in I} S_i$ for some finite sets $S_i$ with surjective transition maps. Covers are jointly surjective families of open maps, at least when the index sets $I$ for the limits are countable. (See the erratum to the paper for a subtle set-theoretic issue coming up here; the definition used in more recent papers requires that $\{S^i \rightarrow S\}$ is a covering if for each quasi-compact open in $S$, there is a quasi-compact open in $\sqcup_{i \in I} S^i$ covering it). Now, we want to consider the constant sheaves $\underline{\mathbf{Z}/\ell^n\mathbf{Z}}$ on $X_{\mathrm{pro-et}}$ and compute their limit. I claim that the sheaf $\widehat{\mathbf{Z}_\ell} := \varprojlim_n \underline{\mathbf{Z}/\ell^n\mathbf{Z}}$ is isomorphic to the sheaf $S \mapsto \mathrm{Hom}_{\mathrm{cont}}(S, \mathbf{Z}_\ell)$, where $\mathbf{Z}_\ell$ has its \emph{$\ell$-adic topology}. On the other hand, we can think of the constant sheaf $\underline{\mathbf{Z}_\ell}$ as the sheaf $S \mapsto \mathrm{Hom}_{\mathrm{loc. cont}}(S, \mathbf{Z}_\ell)$ (note that the topology on $\mathbf{Z}_\ell$ does not effect the notion of locally constant maps to it). Now, the constant sheaves $\underline{\mathbf{Z}/\ell^n}\mathbf{Z}$ are given by $$S \mapsto \mathrm{Hom}_{\mathrm{loc. const.}}(S, \mathbf{Z}/\ell^n \mathbf{Z}) = \mathrm{Hom}_{\mathrm{cont}}(S, \mathbf{Z}/\ell^n\mathbf{Z})$$ with the groups $\mathbf{Z}/\ell^n\mathbf{Z}$ given the discrete topology. Thus, the inverse limit presheaf is given by $$S \mapsto \varprojlim_n \mathrm{Hom}_{\mathrm{cont.}}(S, \mathbf{Z}/\ell^n \mathbf{Z}) =: \mathrm{Hom}_{\mathrm{cont.}}(S, \varprojlim_n \mathbf{Z}/\ell^n \mathbf{Z}) = \mathrm{Hom}_{\mathrm{cont.}}(S, \mathbf{Z}_\ell)$$ Here, we are using the definition of the inverse limit of topological spaces. Now, it turns out that this is actually already a sheaf: the continuity condition is local in the pro-étale topology since pro-étale covers are open. Here's a weird thing that can happen: this argument shows that we have an injective map of sheaves $\underline{\mathbf{Z}_\ell} \rightarrow \widehat{\mathbf{Z}_\ell}$. Call its cokernel $\mathscr{F}$. Note that in the terminology of Scholze's paper, we've shown that $X_{\mathrm{pro-et}}$ is equivalent as a site to the site $G-\mathrm{pfsets}$ with $G$ the trivial group. Here, $\widehat{\mathbf{Z}_\ell}$ is the sheaf $\mathscr{F}_M$ associated to the topological $G$-module (i.e. topological space) $\mathbf{Z}_\ell$ given the $\ell$-adic topology and $\underline{\mathbf{Z}_\ell}$ is $\mathscr{F}_M$ with $M = \mathbf{Z}_\ell$ with the discrete topology. Now, since $G$ acts freely on the profinite set $\{1\}$, corresponding to $\mathrm{Spec}(\mathbf{C})$, Scholze's Proposition 3.7 shows us that the functor $\mathscr{F} \rightarrow \mathscr{F}(\mathrm{Spec} \mathbf{C})$ is exact. Thus, we have an exact sequence: $$ 0 \rightarrow \underline{\mathbf{Z}_\ell(\mathrm{Spec} \mathbf{C})} \rightarrow \widehat{\mathbf{Z}_\ell}(\mathrm{Spec} \mathbf{C})) \rightarrow \mathscr{F}(\mathrm{Spec} \mathbf{C}) \rightarrow 0 $$ But we know that $\underline{\mathbf{Z}_\ell}(\mathrm{Spec}\mathbf{C}) \rightarrow \widehat{\mathbf{Z}_\ell}(\mathrm{Spec}\mathbf{C})$ is an isomorphism (as locally constant functions and continuous functions from the one-point set to $\mathbf{Z}_\ell$ are the same thing), so $\mathscr{F}$ furnishes an example of a non-zero sheaf on the pro-etale site of $\mathrm{Spec}(\mathbf{C})$ which has no sections over the spectrum of any geometric point of $\mathrm{Spec}(\mathbf{C})$. Note that in the erratum, Scholze removed the discussion of points after Proposition 3.13, but this is surely the sort of phenomenon he had in mind: in the notation of that proposition, the functors $i_x^*$ are jointly conservative (i.e. they detect whether a sheaf is $0$) for $x$ ranging over geometric points, but (unlike in the case of the étale topology) the functors $\Gamma \circ i_x^*$ are not!<|endoftext|> TITLE: Extending a weak version of Sullivan's generalized conjecture QUESTION [12 upvotes]: Apologies for the title. Miller's theorem (formerly Sullivan's conjecture) gives that for a finite group $G$ and a finite dimensional connected CW complex $X$, the based mapping space $\operatorname{Map}_*(BG,X)$ is weakly contractible. In particular, by consideration of $\pi_0$ we see that every map $BG\to X$ is null-homotopic. A generalization of Sullivan's conjecture to $G$-CW complexes $X$ with non-trivial action was proved Dwyer-Miller-Neisendorfer, Lannes and Carlsson. It states that if $G$ is a $p$-group and $X$ is a finite dimensional $G$-CW complex, then the natural map $$ X^G \to \operatorname{Map}_G(EG,X) $$ from the fixed points to the homotopy fixed points is a weak equivalence after $p$-completion. In particular, this map induces an isomorphism on mod $p$ homology and consequently induces a bijection $$ \pi_0(X^G) \cong \pi_0(\operatorname{Map}_G(EG,X)) $$ on path components. This could be viewed as a weakened generalized Sullivan's conjecture. Does the weakened Sullivan conjecture hold for all finite groups? That is, let $G$ be a finite group and $X$ a finite dimensional $G$-CW complex. Is it true that $$ \pi_0(X^G)\cong\pi_0(\operatorname{Map}_G(EG,X))? $$ If not, what extra hypothesis could be put on $X$ so that this holds? REPLY [13 votes]: The answer to the first question is negative by a result of Dror Farjoun and Zabrodsky. In Fixed points and homotopy fixed points, they prove that if a finite group $G$ is not a $p$-group, then there exists a finite $G$-CW complex $X$ with empty fixed points $X^G$, but non-empty homotopy fixed points $\mathrm{Map}_G(EG,X)$.<|endoftext|> TITLE: Research in applied algebraic geometry that essentially needs a background of modern algebraic geometry at Hartshorne's level QUESTION [23 upvotes]: By applied algebraic geometry, I don't mean applications of algebraic geometry to pure mathematics or super-pure theoretical physics. Not number theory, representation theory, algebraic topology,differential geometry, string theory etc. However, since the border between pure math and applied math is kind of vague, if you are uncertain whether your answer is really about the applied algebraic geometry (for example, you know a direction that is half-pure and half-applied), please don't hesitate to add you answer or leave a comment. To be specific, I wonder if the modern theory of schemes (and coherent sheaves, if applicable) has any applications outside of pure math while the classical theory of varieties won't be sufficient. I know there is an area called statistical algebraic geometry, but I think so far it still only uses classical algebraic geometry (no need to know schemes and sheaves). I hope to find an applied algebraic geometry area in which a background as strong as finishing most of the Hartshorne's exercises is not wasted. Updates: As per some of the comments below, I want to clarify some points: (1) By "...while the classical theory of varieties won't be sufficient", you don't have to demonstrated that the research work in applied algebraic geometry you have in mind (that involves modern algebraic geometry notions) can't be translated to classical languages. I think as long as the author chooses modern language to write an applied algebraic geometry paper, there should be a reason behind it and we will find out why. (2) As for "applicable" vs "potentially-applicable", I think nowadays it is clear many (if not most) applied math and statistics papers are only "potentially-applicable" for the time being (Just look at SIAM journals and conferences). Hence, I think "potentially-applicable" answers are welcome. If your answer is "too pure to be even potentially-applicable", someone would leave a comment below... REPLY [3 votes]: Maybe research at the intersection of algebraic geometry, representation theory, and deep learning might be of interest to you. Double framed moduli spaces of quiver representationshttps://arxiv.org/abs/2109.14589 This builds on previous work on the representation theory of neural networks, giving category-theoretic formulations too! This one also has the same flavor; authors define a Kähler metric to perform gradient descent on the moduli space of quivers. Kähler Geometry of Quiver Varieties and Machine Learning For an even more abstract/categorical approach. Try this by Yuri Manin and Matilde Marcolli. It's Algebraic topology and category theory, not algebraic geometry though. HOMOTOPY THEORETIC AND CATEGORICAL MODELS OF NEURAL INFORMATION NETWORKS Finally, the paper below cites the work by all of the authors above. But it bridges the abstract approach with the more concrete goal of developing better optimization techniques of neural networks derived from the geometry and symmetries of their parameter spaces. THE QR DECOMPOSITION FOR RADIAL NEURAL NETWORKS The following quote of the authors encapsulates their more pragmatic goal. "Manin and Marcolli advance the study of neural networks using homotopy theory, and the “partly oriented graphs” appearing in their work are generalizations of quivers. In comparison to the two aforementioned works, our approach is inspired by similar algebro-geometric and categorical perspectives, but our emphasis is on practical consequences for optimization techniques at the core of machine learning."<|endoftext|> TITLE: Version of Banach-Steinhaus theorem QUESTION [15 upvotes]: I am wondering about the following version of the Banach-Steinhaus theorem. Let $A$ be a closed convex subset contained in the unit ball of a Banach space $X$ and consider bounded operators $T_n \in \mathcal L(X).$ Assume we know that for every $x \in A$ the sequence $\left\lVert T_n x \right\rVert$ is bounded uniformly in $n.$ Does this imply that $$\sup_{x \in A} \left\lVert T_n x \right\rVert$$ is bounded uniformly in $n$? If $A=X$ then the theorem is undoubtedly true by the folklore Banach-Steinhaus theorem but I was wondering whether this version holds as well? REPLY [3 votes]: I like Jochen Wengenroth's approach, and I think there is a point that it is worth to clarify. If we want to make a norm out of $A$, we need it to be a balanced set, so we'd like to pass to the bounded absolutely convex set $\overline{\operatorname{co}}\left(A\cup(-A)\right)$ or to $A-A$. Any family of linear operators which is point-wise bounded on $A$ is clearly also point-wise bounded on $A-A$. However these sets are in general not closed, so some care is needed, because a bounded absolutely convex not closed set $B$ in general would not produce a Banach disk on its linear span, and in fact in general the statement itself does not hold on such $B$ (see the example in the initial comment). A cheap solution to make the argument work smoothly is to use the notion of $\sigma$-convexity (see e.g. this MO thread) which also generalize slightly the statement); in particular, it covers both the case of a closed and an open bounded convex set $A$. Recall that for a subset $A$ of a Banach space $X$ the following easy facts hold: If $A$ is $\sigma$-convex, it is bounded; If $A$ is $\sigma$-convex, $A-A$ is $\sigma$-convex and symmetric (that is, $\sigma$-absolutely convex); If $A$ is $\sigma$-absolutely convex, it is a Banach disk, that is, its Minkowski functional is a Banach norm on the linear span of $A$. As a conclusion, we can follow Jochen Wengenroth's reduction to the standard Banach-Steinhaus theorem. We thus have: Any family of linear operators on a Banach space, which is point-wise bounded on a $\sigma$-convex set $A$, is also uniformly bounded on $A$.<|endoftext|> TITLE: When is a linear combination of the elementary symmetric polynomials reducible? QUESTION [5 upvotes]: Let $n\ge 2$ and consider the polynomial ring $\mathbb F [X_1,...,X_n]$, where $\mathbb F$ is a field. Let $e_j:=e_j(X_1,...,X_n)$ be the elementary symmetric polynomial of degree $j$ in $X_1,...,X_n$ (https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial). Now is there a way to characterize for which $(c_0,...,c_n)\in \mathbb F^{n+1} \setminus \{0\}$, is the polynomial $f(X_1,...,X_n)=\sum_{j=0}^n c_je_j \in \mathbb F[X_1,...,X_n]$ reducible in $\mathbb F [X_1,...,X_n]$ ? For example, if $n$-many among $c_0,...,c_n$ are zero, i.e. if we have $f=e_k$ for some $k$, then we must have $k=n$, because $e_1,...,e_{n-1}$ are all irreducible as seen here Is an elementary symmetric polynomial an irreducible element in the polynomial ring? . Apart from this, I don't know ... REPLY [4 votes]: Suppose that $f=p_1 p_2\cdots p_k$, where each $p_i$ is irreducible. Since $f$ is a symmetric function, for every $w\in S_n$ and every $i$ we must have $w\cdot p_i = cp_j$ for some $j$, where $c$ is a nonzero constant (depending on $w$ and $i$). In each factor $p_i$ pick out the term of highest degree $d$ that is first in lex order, e.g., for $d=3$ the lex order is $$ X_1^3 TITLE: Which large cardinals have a Matryoshka characterization? QUESTION [7 upvotes]: What on Earth do Russian Matryoshka dolls have in common with large cardinal axioms?! Well, the answer lies in Jónsson algebras! Here is how: As illustrated in the pictures, a Matryoshka set is a self-replicating container in which the mother doll contains a similar copy of itself. This sub-doll also contains a rather smaller similar copy of itself and so on. The dolls in a Matryoshka set could be completely isomorphic in shape (often depicted as a chubby Russian girl wearing Sarafan, the beautiful Persian-named traditional Russian dress) or with some slight differences in spite of general similarity (e.g. Different species of Taiga). As the allegory suggests, the Matryoshka phenomenon is not uncommon in mathematics. There are (infinite) self-replicating mathematical structures which contain some proper isomorphic copies of themselves or at least have a fairly similar (e.g. elementary) proper sub-structures of the same size as the mother structure. Jónsson algebras provide an iconic example of this type. Definition 1. Let $M$ be a model in a countable language. $M$ is called a Jónsson algebra (model) if it does not have any proper (elementary) submodel of the size $|M|$. It is easy to see that for an infinite cardinal $\kappa$, the existence of a Jónsson algebra of size $\kappa$ is equivalent to the existence of a Jónsson model of size $\kappa$. By the way, there is a combinatorial large cardinal of the same name: Definition 2. An uncountable cardinal number $\kappa$ is called Jónsson if for every function $f:[\kappa]^{<\omega}\rightarrow \kappa$ there is a set $H$ of order type $\kappa$ such that for each $n$, $f$ restricted to $n$-element subsets of $H$ omits at least one value in $\kappa$. According to the following fact, it turns out that the demand for every infinite first-order structure in a countable language to have a proper sub-structure of the same size (i.e. non-existence of Jónsson algebras) has the large cardinal strength of a Jónsson cardinal which is a fairly strong large cardinal assumption incompatible with the Axiom of Constructibility. Theorem. For an infinite cardinal $\kappa$, the followings are equivalent: (a) $\kappa$ is a Jónsson cardinal. (b) There are no Jónsson algebras of size $\kappa$. (i.e. Every algebra of size $\kappa$ is a Matryoshka, namely every algebra of size $\kappa$ contains a proper sub-algebra of the same size.) The above fact indicates that the occurrence of a Matryoshka phenomenon (i.e. having self-replicating structures) among certain types of mathematical structures may give rise to some large cardinals. In other words, a structure needs to be so large in order to be able to support a self-replicating Matryoshka-like structure. In this sense, Jónsson cardinals have a characterization in terms of a Matryoshka phenomenon for certain type of mathematical structures, namely algebras. Here, the natural question to ask is whether there are other examples of a Matryoshka characterization for large cardinal axioms. In other words, what other Matryoshka phenomenona for mathematical structures do have some large cardinal strength? Precisely: Question 1. What are some other examples of large cardinal characterizations of the following form? The followings are equivalent/equiconsistent: (a) $\kappa$ is large. (b) Any structure of type X and size $\kappa$ has a proper (isomorphic, elementary, "similar", etc.) sub-structure of size $\kappa$. (i.e. There is a Matryoshka phenomenon among mathematical structures of type X). Note that we are not limiting ourselves to the first order structures or structures in countable language. The question is considering the possible existence of such a characterization for large cardinal axioms anywhere in mathematics where a theory around certain type of "mathematical structures" (whatever it means depending on the context, say functional space or a manifold) evolves. Question 2. How much does the consistency strength of $\kappa$ increase if we expand our demand in the existing Matroshka chracaterization of Jónsson cardinals to contain isomorphic proper sub-algebras? Will we hit the inconsistency barrier? Precisely, what is the consistency strength of the following statement? Is it consistent at all? "Every algebra of size $\kappa$ contains a proper sub-algebra of the same size which is isomorphic to the original structure." REPLY [11 votes]: Adding to Joel's answer, there is a weak form of Berkeley cardinals having almost the property you want and whose existence is consistent with the axiom of choice (in fact with ZFC + $V = L$.) Note: the cardinals defined in Joel's answer are sometimes instead called proto-Berkeley cardinals, but I'll call them Berkeley cardinals here for the sake of consistency. (The difference is whether or not the critical points of such $j$ are unbounded in $\kappa$.) We may call a cardinal $\kappa$ a virtual Berkeley cardinal if for every transitive set $M$ containing $\kappa$ there is a generic elementary embedding $j : M \to M$ with critical point below $\kappa$, where a generic elementary embedding is defined as an elementary embedding that exists in some generic extension of $V$. (This is a slight abuse of terminology: what is really being defined is the phrase "there is a generic elementary embedding.") This is a type of virtual large cardinal property, other examples being remarkability, virtual extendibility, and the generic Vopěnka principle. Note that the range of such a generic elementary embedding is a proper elementary substructure of $(M;\in)$ that is isomorphic to $(M;\in)$. However, the embedding and its associated substructure typically exist only in a generic extension in which $M$ (and therefore $\kappa$) is collapsed to be countable. If $0^\sharp$ exists, then one can show that every Silver indiscernible is a virtual Berkeley cardinal in $L$. This is fairly easy, using the fact that the existence of an elementary embedding $j : M \to M$ with critical point below $\kappa$ is absolute from $V$ to any generic extension of $L$ in which $M$ is countable. It turns out that a cardinal $\kappa$ is a virtual Berkeley cardinal (as defined above, which should perhaps be called a virtual proto-Berkeley cardinal) if and only if $\kappa \to (\omega)^{<\omega}_2$. Therefore there is a virtual Berkeley cardinal if and only if there is an $\omega$-Erdős cardinal. The forward direction is proved using methods of Silver [1] and the reverse direction is proved similarly to Gitman and Schindler [2, Theorem 4.17]. [1] Jack Silver. A large cardinal in the constructible universe. Fund. Math. 69:93–100, 1970. [2] Victoria Gitman and Ralf Schindler. Virtual large cardinals. Preprint available at http://ivv5hpp.uni-muenster.de/u/rds/virtualLargeCardinalsEdited4.pdf<|endoftext|> TITLE: Reference Request: Existence of Ordinal Rank Theory? QUESTION [7 upvotes]: Notations: Recall that $\omega_1$ is the first uncountable ordinal. Let $X$ be a Polish space (completely metrizable and separable) and $F(X)$ be the collection of all real-valued functions on $X.$ A function $\rho:F(X)\to \omega_1$ is called an ordinal rank. In other words, ordinal rank assigns an ordinal to a function, typically measuring complexity of the function. In $1990,$Kechris and Louveau conducted an investigation on three types of ordinal ranks acting on the class of Baire Class $1$ functions on a compact metric space (recall that $f$ is Baire Class $1$ if it is a pointwise limit of a sequence of continuous functions). The three ranks are separation rank (introduced by Bourgain), oscillation rank and convergence rank (introduced by Zalcwasser, Gillespie and Hurwicz). They measure complexity of the respective definition of Baire Class $1$ functions (Baire Class $1$ functions have three equivalent definitions). $26$ years later ($2016$), Elekes, Kiss and Vidnyánszky generalised from compact metric space to Polish space and the three ranks to Baire Class $\xi$ functions where $\xi$ is any countable ordinal (Recall that $f$ is Baire class $\xi$ if it is a pointwise limit of a sequence of Baire Class $\xi_n$ functions where $\xi_n<\xi$ for all natural number $n$). One of the generalisations is through topology refinement (For interested reader, please refer to section $5$ of their paper). Question: Does there exist a literature on ordinal rank? More precisely, where can I find more information about ordinal rank? REPLY [3 votes]: This is a topic I briefly looked at in 1992-1993 (I was mostly interested in ranks for differentiable functions and for nowhere differentiable continuous functions), when I was working on my dissertation, and at that time I collected a few papers on the topic (and some more throughout the 1990s). However, even though I have a “designated notebook binder” for papers on this topic (which I stopped adding papers to around 1999), I've never posted a list of such papers, or even made such a list for my own personal use, so I suppose this is something I can do today since today is a holiday for me. Below is a chronological list of items from my $\leq$ 1990s collection of stuff. The dates at the end of each entry are submission dates or book preface dates, which I included to make it easier for me to chronologically order the items. I had planned to include all relevant items after the 1990s, but I found way too much by googling, so I’ll leave the more recent work for you to find. There does not seem to be any survey/expository paper on ordinal ranks in general (however, I’ve included 3 items from after the 1990s that seem useful in this regard), so this is something you might want to consider writing if you wind up getting very engrossed in this topic. Regarding searching online, besides googling (variations of) the authors’ names, the following phrases, when googled, will lead you to more than I’m willing to look at. For some of these phrases, such as “convergence rank”, you will need to include “Kechris” as an additional search term. Lusin-Sierpinski index [or Luzin-Sierpinski index], Piatetski-Shapiro rank, Zalcwasser rank, Szlenk index, Denjoy rank, Kechris-Woodin rank, oscillation rank, convergence rank, differentiability rank. Also, try googling “coanalytic rank” and (as a separate search) “co-analytic rank”. [1] Miklós Ajtai and Alexander Sotirios Kechris, The set of continuous functions with everywhere convergent Fourier series, Transactions of the American Mathematical Society 302 #1 (July 1987), 207-221. [11 June 1985] [2] Alexander Sotirios Kechris and William Hugh Woodin, Ranks of differentiable functions, Mathematika 33 #2 (December 1986), 252-278. [1 July 1985] [3] Taje Indralall Ramsamujh, Some Topics in Descriptive Set Theory and Analysis, Ph.D. Dissertation (under Alexander Sotirios Kechris), California Institute of Technology, 1986, vi + 147 pages. abstract [5 May 1986] [4] Taje Indralall Ramsamujh, A comparison of the Jordan and Dini tests, Real Analysis Exchange 12 #2 (1986-1987), 510-515. [correction in RAE 14 #1 (1988-1989), pp. 251–252] [11 November 1986] [5] Alexander Sotirios Kechris and Alain Louveau, Descriptive Set Theory and the Structure of Sets of Uniqueness, London Mathematical Society Lecture Notes Series #128, Cambridge University Press, 1987, viii + 367 pages. [June 1987] [6] Alexander Sotirios Kechris and Russell David Lyons, Ordinal rankings on measures annihilating thin sets, Transactions of the American Mathematical Society 310 #2 (December 1988), 747-758. [10 September 1987] [7] Taje Indralall Ramsamujh, The complexity of nowhere differentiable continuous functions, Canadian Journal of Mathematics 41 #1 (1989), 83-105. [22 October 1987] [8] Alexander Sotirios Kechris and Alain Louveau, A classification of Baire class $1$ functions, Transactions of the American Mathematical Society 318 #1 (March 1990), 209-236. [15 December 1987] [9] Alexander Sotirios Kechris, Alain Louveau, and Valérie Tardivel [Tardivel-Nachef], The class of synthesizable pseudomeasures, Illinois Journal of Mathematics 35 #1 (Spring 1991), 107-146. [22 November 1988] [10] Taje Indralall Ramsamujh, Three ordinal ranks for the set of differentiable functions, Journal of Mathematical Analysis and Applications 158 #2 (1 July 1991), 539-555. [25 August 1989] [11] Alexander Sotirios Kechris, Classical Descriptive Set Theory, Graduate Texts in Mathematics #156, Springer-Verlag, 1995, xviii + 402 pages. corrections and updated [September 1994] An introduction to ordinal ranks is given in Chapter V.1 (pp. 140-150), and several ordinal ranks are defined and used in the remainder of the book. [12] Haseo Ki, The Kechris-Woodin rank is finer than the Zalcwasser rank, Transactions of the American Mathematical Society 347 #11 (November 1995), 4471-4484. [14 September 1994] Ordinal ranks are discussed in Sections 34-36 (pp. 267-312). [13] Haseo Ki, Topics in Descriptive Set Theory Related to Number Theory and Analysis, Ph.D. Dissertation (under Alexander Sotirios Kechris), California Institute of Technology, 1995, v + 72 pages. abstract [15 March 1995] [14] Haseo Ki, On the Denjoy rank, the Kechris-Woodin rank and the Zalcwasser rank, Transactions of the American Mathematical Society 349 #7 (July 1997), 2845-2870. [13 April 1995] [15] Taje Indralall Ramsamujh, Simply connected compact subsets of the plane, Journal of Mathematical Analysis and Applications 237 #1 (1 September 1999), 240-252. [19 August 1998] [16] Spiros A. Argyros, Gilles Godefroy, and Haskell Paul Rosenthal, Descriptive set theory and Banach spaces, pp. 1007-1069 in Johnson/Lindenstrauss (editors), Handbook of the Geometry of Banach Spaces, Volume 2, Elsevier Science B.V., 2003, xii + 1007-1866 pages. [17] Yiannis Nicholas Moschovakis, Descriptive Set Theory, 2nd edition, Mathematical Surveys and Monographs #155, American Mathematical Society, 2009, xiv + 502 pages. See Section 2D (pp. 62-64). [18] Zoltán Vidnyánszky, Descriptive Set Theoretical Methods and Their Applications, Ph.D. Dissertation (under Márton Elekes), Eötvös Loránd University, 2014, 112 + 2 pages.<|endoftext|> TITLE: Left split subobject in a $2$-category QUESTION [5 upvotes]: Let $\mathcal{K}$ be a $2$-category. Keeping in mind the Cat-intuition on $\mathcal{K}$ I say that: Def : A $1$-cell $A \stackrel{f}{\to} B$ is a left split subobject if $f$ is a right adjoint and the counit is invertible. In Cat this condition is equivalent to be fully faithful (and a right adjoint) for $f$. In Lemma 2.3 of his 2-categories companion Steve Lack points out that: In a 2-category, when the counit is invertible then the right adjoint is representably fully faithful. Thus one implication remains true. Is it possible to have sort of a converse for this statement? Even adding somme additional structure on $\mathcal{K}$. I was hoping to have something like the following: A $1$-cell is a left split subobject if and only if $f$ is a right adjoint and a monomorphism in $\mathcal{K}$, which - I understand - is a vain expectation. I was hoping to use the left cancellation in the triangle equality somehow. Unfortunately it is not working as I expected. REPLY [5 votes]: This is true if by "monomorphism" you mean "representably fully faithful" as in the result of Lack: just argue representably. If $f:A\to B$ has a left adjoint $\ell:B\to A$ in $\mathcal K$, then $\mathcal K(X,f): \mathcal K(X,A) \to \mathcal K(X,B)$ has a left adjoint $\mathcal K(X,\ell)$ in $\mathrm{Cat}$ for any $X\in \mathcal K$. If $f$ is representably fully faithful, then by definition $\mathcal K(X,f)$ is fully faithful. Thus, the counit of the adjunction $\mathcal K(X,\ell)\dashv \mathcal K(X,f)$ is an isomorphism in $\mathrm{Cat}$. This counit has components that are 2-cells $\ell f g \to g$ in $\mathcal K$ for $g:X\to A$, obtained by whiskering with the counit $\epsilon : \ell f \to 1_A$. So just take $X=A$ and $g = 1_A$ to conclude that $\epsilon$ is an isomorphism.<|endoftext|> TITLE: Subspaces isomorphic with quotients QUESTION [8 upvotes]: Suppose $X$ is a Banach space not isomorphic to a Hilbert space. Can we always find a subspace of $X$ that is not isomorphic to a quotient of $X$? REPLY [11 votes]: Every separable Banach space is a quotient of $\ell_1$, so in particular every subspace of $\ell_1$ is a quotient of $\ell_1$.<|endoftext|> TITLE: How are simplicial sets with Quillen model structure a simplicial model category? QUESTION [5 upvotes]: I got very lost in checking that simplicial sets with Quillen model structure are indeed a simplicial model category. Recall that a model category $\mathcal{M}$ is simplicial if it is enriched in $\mathbf{SSet}$, powered by $\mathbf{SSet}$ and tensored by $\mathbf{SSet}$ so that the natural adjunctions exist and and also the enrichment is compatible with the model structure -- namely, for $i: A \to B$ a cofibration in $\mathcal{M}$ and $p: X \to Y$ a fibration in $M$, the map of simplicial sets $i^*\times p_*: \operatorname{Map}(B,X) \to \operatorname{Map}(A,X)\underset{\operatorname{Map}(A,Y)}{\times}\operatorname{Map}(B,Y)$ is required to be a fibration in $\mathbf{SSet}_{Quillen}$, and it is required to be trivial if either $i$ or $p$ is. Recall that for simplicial sets, the enrichment is given by $\operatorname{Map}(K,M)_n = \mathbf{SSet}(\Delta^n \times K, M).$ $\mathbf{SSet}_{Quillen}$ is a cofibrantly generated model category, so to check the compatibility one must fill the horns: $\require{AMScd}$ \begin{CD} \Lambda^n_i @>h>> \operatorname{Map}(B,X) \\ @V \iota V V @VV i^* \times p_* V\\ \Delta^n @>(\alpha,\beta)>> \operatorname{Map}(A,X)\underset{\operatorname{Map}(A,Y)}{\times}\operatorname{Map}(B,Y) \end{CD} The filling would be the lift in the diagram $\require{AMScd}$ \begin{CD} \Delta^n \times A @>\alpha>> X \\ @V 1\times i V V @VV p V\\ \Delta^n \times B@>\beta>> Y \end{CD} The natural idea is to consider the diagram $\require{AMScd}$ \begin{CD} \Lambda^n_i \times B @>h>> X \\ @V \iota\times 1 V V @VV p V\\ \Delta^n \times B@>\beta>> Y \end{CD} where the left map is a trivial cofibration, the right map is a fibration and thus from the lifting properties I obtain $\sigma: \Delta^n \times B \to X$ satisfying that $p\sigma = \beta$. However, if I want this $\sigma$ to be the answer, I must check $\sigma(1 \times i) = \alpha$... Would that be so? I see that I have not made use of $i$ being an inclusion, but I don't see how it helps. REPLY [6 votes]: The trick is to check that the corner map $$\lambda^n_k\bar{\times}\delta^m:\Lambda^n_k \times \Delta^m \coprod_{\Lambda^n_k\times \partial \Delta^m} \Delta^n \times \partial \Delta^m \hookrightarrow \Delta^n\times \Delta^m$$ is anodyne for all $k, m, n$ appropriate. This is proven in Higher Topos Theory chapter 2, for your reference. The fact that $i$ was an inclusion allows you to build it up as a relative cell complex of boundary inclusions, so it is enough to prove the case where $i$ is a boundary inclusion.<|endoftext|> TITLE: What are the $2 \times 2$ matrix generators of $\text{SL}_2\big(\mathbb{Z}[i]\big)(2+i)$? QUESTION [5 upvotes]: I have been trying to learn about congruence groups. Here is an example: \begin{eqnarray*} \Gamma\big(1+2i\big) &=& \text{SL}_2\big(\mathbb{Z}[i]\big)(1+2i) \\ \\ &=& \left\{ \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) : ad-bc = 1 \text{ and } \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \equiv \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \pmod{1+2i} \right\} \\ \\ &\subseteq & \text{SL}_2\big(\mathbb{Z}[i]\big) \end{eqnarray*} While the proof is uses the theory of algebraic groups, we can prove that is finitely generated. This particular case is look elementary, these are $2 \times 2$ invertible matrices, $\mathbb{Z}[i]$ is a Eucliean domain, and we can solve $ad-bc = 1$ by finding two primes (e.g. $6+i $ or $ 2+3i$) and looking for their greatest common divisor. Since this group is finitely generated, how can I find a generating set? What are the generators? Even computer code would be helpful. In order to specify what I am looking for here is the result for $\text{SL}_2(\mathbb{Z})$: $$ \text{SL}_2(\mathbb{Z}) \simeq \big\{ S,T : S^2 = 1,\; (ST)^3 = 1 \big\} \quad\text{with}\quad S = \left( \begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array} \right) \text{ and } T = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right)$$ At least over $\mathbb{Z}$ there an exact sequence relating the congruence groups to the special linear group over finite fields: $$ 1 \to \Gamma(N) \to \Gamma \to \text{PSL}_2\big(\mathbb{Z}/N\mathbb{Z}\big) \to 1 $$ And so it's likely that congruence groups of $\text{SL}_2(\mathbb{Z})$ should have finite presentations. Even if I do something slightly more inefficient and use the exact sequence. Also Are congruence subgroups of the modular group finitely presented? REPLY [2 votes]: $PSL_2(\mathbb{O}_d)$ acts on the upper half-space model of hyperbolic 3-space in a nice way, namely the quotient can be viewed as a finite volume 3-orbifold. Since all principal congruence subgroups are finite index in $PSL_2(\mathbb{O}_d)$, all principal congruence subgroups are finitely generated and correspond to some (finite-sided/geometrically finite) finite volume orbifold cover of $H^3/PSL_2(\mathbb{O}_d)$. (There are probably more direct ways of showing this, but might help frame this discussion.) As the OP notes in the question, after specifying a $d$, one can draw a direct analogy between group, and $PSL_2(\mathbb{Z})$ acting on $H^2$. There is natural question, "Which principal congruence subgroups of $PSL_2(\mathbb{Z})$ are genus zero (aka fill to a sphere)?" One 3d analog of this question is "Which principal congruence subgroups of $PSL_2(\mathbb{O}_d)$ are homeomorphic to $S^3\setminus L$ for some link $L$ in $S^3$?" Happily, over a series of papers the subsets of collection of Baker, Goerner and Reid provide a complete solution to this question. More relevantly, $PSL_2(\mathbb{Z}[i])(2+i)$ is one of the links they study. Proposition 3.1 and Lemma 3.2 of Baker, Mark D.; Reid, Alan W., Principal congruence link complements, Ann. Fac. Sci. Toulouse, Math. (6) 23, No. 5, 1063-1092 (2014). ZBL1322.57015.http://www.numdam.org/item/AFST_2014_6_23_5_1063_0 discusses this question directly. It gives a six generator presentation of $PSL_2(\mathbb{Z}[i])(2+i)$. The index of this group is confirmed using a magma computation. Also the generating set is minimal as the abelianization of this group has rank 6. Selecting the correct coset representative from each of these elements in $PSL_2(\mathbb{Z}[i])$ should give you group you are looking for. Some of the other papers in there line of attack are here if you are interested: Baker, Mark D., Link complements and the Bianchi modular groups, Trans. Am. Math. Soc. 353, No. 8, 3229-3246 (2001). ZBL0986.20049. Baker, Mark D., Link complements and integer rings of class number greater than one, Topology ’90, Contrib. Res. Semester Low Dimensional Topol., Columbus/OH (USA) 1990, Ohio State Univ. Math. Res. Inst. Publ. 1, 55-59 (1992). ZBL0768.57005. Görner, Matthias, Regular tessellation link complements, Exp. Math. 24, No. 2, 225-246 (2015). ZBL1319.57002. Baker, Mark D., Goerner, Matthias, and Reid, Alan W., All principal congruence link groups arXiv preprint arXiv:1802.01275 (2018). Note that as the title suggests, the last paper gives a complete classification of principal congruence link complements. Of course, the methods of these papers which make a good guess for what the generators of the principal congruence subgroups are can be adjusted, especially if the pair $(d,I)$ is not on their lists. In this case, one would definitely need more than just a trace 2 generator corresponding to each cusp. Nevertheless, the magma code can be adapted to more general searches for generating sets of principal congruence groups of $PSL_2(\mathbb{O}_d)$.<|endoftext|> TITLE: Isotopy of periodic homeomorphisms of a surface along periodic homeomorphisms QUESTION [9 upvotes]: Let $\Sigma$ be an oriented compact surface with non-empty boundary that is not a disk or a cylinder (i.e. negative Euler characteristic). Let $\phi, \psi: \Sigma \to \Sigma$ be two orientation preserving periodic homeomorphisms, that is, $\phi^n = \psi^m = id$ with $n,m$ their periods. Suppose that $\phi$ is isotopic to $\psi$ (so in particular $n=m$). Does there exist an isotopy $H(x,t)$ between these two homeomorphisms such that $H(\cdot, t)$ is a periodic homeomorphism for all $t$? Edit: Just in case this helps to solve the question. Notice that since $\phi$ and $\psi$ are isotopic and periodic, they have the same fixed-point data. This is enough to say that they are conjugate in $Homeo^+(\Sigma)$ by the classical theory of Nielsen. So $\phi = \gamma \circ \psi \circ \gamma^{-1}$. To prove the original question it would be enough to show that $\gamma$ is isotopic to the identity. REPLY [5 votes]: The answer is "yes". It follows as part of F. Bonahon's determination of the bordism group of surface diffeomorphisms. See Bonahon, Francis, Cobordism of automorphisms of surfaces, Ann. Sci. Éc. Norm. Supér. (4) 16, 237-270 (1983). ZBL0535.57016. Bonahon justified the relevant step by appeal to some 3-dimensional topology, including geometrization. Independently about the same time J. Ewing and I proved the same result using 2-dimensional techniques and some number theory related to the G-signature theorem. Edmonds, A. L.; Ewing, J. H., Remarks on the cobordism group of surface diffeomorphisms, Math. Ann. 259, 497-504 (1982). ZBL0468.57023. We wrote a conference proceedings note that has exactly the result you ask for as its title! See Ewing, John; Edmonds, Allan, Periodic surface diffeomorphisms which bound, bound periodically, Publ., Secc. Mat., Univ. Auton. Barc. 26, No. 3, 37-42 (1982). ZBL0545.57009.<|endoftext|> TITLE: Automated geometry theorem provers QUESTION [8 upvotes]: What is the state of the art concerning automated geometry theorem provers (AGTP)? I can see that a few computer algebra softwares and dynamic geometry softwares (e.g. geogebra) have embedded provers but I failed to find a comparison of these solvers nor a thorough list of AGTP. Are there still AGTP using the quantifier elimination method of Tarski and its refinement such as cylindrical algebraic decomposition, or has it been superseded by other methods such as Wu-Ritts or Grobner bases? REPLY [5 votes]: There is the work of Narboux and collaborators on those topics, see e.g. his software page http://dpt-info.u-strasbg.fr/~narboux/software.html and his papers too. I don't know if there's a paper comparing everything, but perhaps his last one with Boutry and Braun fits the bill https://hal.inria.fr/hal-01483457<|endoftext|> TITLE: Are exotic affine spaces motivic/whatever equivalent to affine space? QUESTION [6 upvotes]: This question is inspired by this MO question; in turn by this MO; in turn by these MO, MO. An exotic affine space is an affine variety $V$ whose $\mathbb{C}$-points are diffeomorphic to $\mathbb{R}^{2n}$ yet $V$ is not algebraically isomorphic to $\mathbb{A}^n$. Will Sawin showed in his answer MO that the number of $F_q$ points is the same for fake affine spaces as for non-fake (for generic $q$). Question: Are exotic affine spaces equivalent to affine space in Grothendieck ring of varietes ? Or may be there are some other simple geometric equivalence which is stronger than just point counting and weaker than isomorphism ? PS Evgeny Shinder in comments to MO states that " Among nonsingular projective varieties, a fake projective plane or odd-dimensional quadrics have the same point count as projective planes. In the case of a fake projective plane its class in the Grothendieck ring is not L^2 + L + 1 (and in fact not congruent to 1 (mod. L) because it's not stably rational). The class of a quadric is same as [P^n] (using projection from the point). " REPLY [3 votes]: Here is an argument showing that if $V$ is a smooth complex surface with trivial integral homology groups (note that exotic $\mathbf{A}^2$ do not exist, as explained in the comments), then $[V] = \mathbb{L}^2$ in the Grothendieck ring of varieties. We use the notation $\mathbb{L} = [\mathbb{A}^1]$. It follows from this paper: https://projecteuclid.org/euclid.jmsj/1230128845 (which I learnt about from this MO post Topologically contractible algebraic varieties) that $V$ is a rational surface. Let $X$ be a non-singular compactification of $V$ such that the divisor $D = X \setminus V$ has simple normal crossings with smooth components $C_1, \dots, C_t$. Since $X$ is a non-singular projective rational surface, we have $[X] = 1 + k\mathbb{L} + \mathbb{L}^2$. We have $[D] = [C_1] + \dots + [C_t] + r$, where $r \in \mathbb{Z}$ ($r$ depends on the intersection graph of these curves). Altogether $[V] = 1 + k\mathbb{L} + \mathbb{L}^2 - [C_1] - \dots - [C_t] - r$. Applying Hodge realization to this equality, using the fact that $V$ has a trivial Hodge structure, we deduce that all curves $C_i$ are rational so that $[C_i] = [\mathbb{P}^1] = 1 + \mathbb{L}$. Thus $V$ is a polynomial in $\mathbb{L}$ and the only possibility is $[V] = \mathbb{L}^2$. P.S. If $V$ an exotic affine space of dimension one, then $V$ is isomorphic to $\mathbb{A}^1$, because it has to be a genus zero curve. If $V$ is an exotic affine space of dimension $3$ or higher, rationality of $V$ is an open question.<|endoftext|> TITLE: About fibrations with fibre Eilenberg-MacLane spaces QUESTION [8 upvotes]: Let $f: E\rightarrow B$ be a Kan fibration between pointed connected Kan complexes with fibre the Eilenberg-MacLane space $\mathrm{K}(M, n), n\geq 2, M$ an abelian group. Assume $f$ induces an isomorphism on $\pi_1$ with common value $G$, and so $G$ acts naturally on $\pi_n\mathrm{K}(M, n)=M$ hence on (the simplicial abelian group) $\mathrm{K}(M, n)$. Let $\tilde{B}$ be the universal cover of $B$, on which $\pi_1B=G$ acts. Is there a fibration equivalence $\tilde{B}\times_G\mathrm{K}(M, n)\sim E$ over $B$? REPLY [7 votes]: No. If this were the case then there would be a section $s: B \to E$ to $f$ induced by the $G$-equivariant map $\widetilde{s}:\widetilde{B} \to \widetilde{B} \times {\rm K}(M,n)$ sending $x$ to $(x,0)$ (where $0 \in {\rm K}(M,n)$ is the neutral element, which is fixed by the action of $G$). However, there exists fibrations with fibers ${\rm K}(M,n)$ for $n \geq 2$ which do not admit a section, such as the quaternionic Hopf fibration $S^7 \to S^4$. In general fibrations $E \to B$ with fiber ${\rm K}(M,n)$ with a fixed $\pi_1(B)$-action on $M$ are classified by the cohomology group with local coefficients $H^{n+1}(B,M)$, where the ones of the form you describe correspond to the $0$ element. Edit: As pointed in the remarks below, the quaternionic Hopf fibration is indeed not a counter-example for what the OP is asking since its fibers are $S^3$, which is not an Eilenberg-MacLane space. One way to fix this is to take the relative Postnikov truncation (which still doesn't have a section). Alternatively, there are probably other more natural examples.<|endoftext|> TITLE: Square root of a large sparse symmetric positive definite matrix QUESTION [7 upvotes]: I am trying to calculate $$Y = A^{\frac 12} X$$ where $A$ is a very large and sparse positive definite matrix, say, $10^4 \times 10^4$. Matrix $X$ is known and, say, $10^4 \times 100$. Is there any method that can compute or approximate $A^{\frac{1}{2}}$ efficiently? I noticed that matrix $A$ can be written as $D + B$, where $D=\mbox{diag}(A)$ and $B=A-D$, which is very sparse. But I still do not know how to compute the square root of the sum of matrices, $(D + B)^{\frac{1}{2}}$. REPLY [3 votes]: I completely agree with fedja: there is a nice method here (which, unfortunately, does not always work well). If you know bounds for the spectrum of $A$, say $0 TITLE: Arithmetic symplectic geometry via mirror symmetry? QUESTION [8 upvotes]: Homological mirror symmetry in the classical setting relates the bounded derived category of coherent sheaves on a Calabi-Yau manifold to the split-closure of the derived Fukaya category of the mirror Calabi-Yau. In the paper 'Arithmetic mirror symmetry for the 2-torus', authors construct a $\mathbb{Z}$-linear equivalence between exact Fukaya category of a punctured torus and the category of perfect complexes of coherent sheaves on the central fiber of Tate curve. Intuitively, this statement is only a half of the HMS conjecture since we also would like to have an equivalence between the 'Fukaya category of the central fiber of Tate curve' and the bounded derived category of coherent sheaves on the punctured elliptic curve. The question is: is it possible to somehow define the notion of Fukaya category for the central fiber of the Tate curve (which is a curve in $\mathbb{P}^2(\mathbb{Z})$, not a smooth manifold)? If it is possible, can we construct an equivalence to the derived bounded category of coherent sheaves on the punctured elliptic curve? REPLY [2 votes]: Yes. Recently Auroux (jointly with Efimov and Katzarkov) has proposed a definition of the Fukaya category for trivalent configurations of rational curves. If $\Sigma_g$ is a genus $g$ Riemann surface with $g\geq2$, then its mirror is a trivalent configuration of $3g-3$ rational curves meeting in $2g-2$ triple points. In the case when $g=1$, a nodal curve is obviously not a trivalent configuration. However, one can replace it with a nodal curve with an affine line attached at the node, which then enables one to make sense of its Fukaya category. The objects of this version of Fukaya category are embedded graphs with trivalent vertices at the triple points, and morphisms are linear combinations of intersection points as in usual Floer theory. Note that since the punctured elliptic curve $E^\circ:=E\setminus\{pt\}$ is not compact, there are two versions of derived categories of coherent sheaves, namely one can consider either the usual $D^b\mathit{Coh}(E^\circ)$ or its compactly supported version $D^b\mathit{Coh}_\mathit{cpt}(E^\circ)$. On the mirror side, this determines whether one needs to do wrapping on the affine line component or not.<|endoftext|> TITLE: Is there a notion of 'amenable ring' QUESTION [12 upvotes]: Amenable groups are everywhere these days, as examples of all kinds of lovely phenomena. And there are various ways of defining notions of 'amenable monoid' or possibly 'amenable semigroup'. But for monoids and semigroups, amenability conditions that are known to be equivalent for groups are not always equivalent and there is a more complicated story unfolding. Nevertheless it seems natural to ask if there is a notion for rings. REPLY [15 votes]: Yes, see Elek, Gábor The amenability of affine algebras. J. Algebra 264 (2003), no. 2, 469–478. and Gromov, Misha Entropy and isoperimetry for linear and non-linear group actions. Groups Geom. Dyn. 2 (2008), no. 4, 499-–593.<|endoftext|> TITLE: commutative "weakly" Frobenius algebras and 2d TQFT QUESTION [7 upvotes]: Fix a field $k$. A classic result written up carefully by Abrams in the article "Two-Dimensional Topological Quantum Field Theories and Frobenius Algebras" says that there is a bijective correspondence between finite dimensional commutative Frobenius algebras and 2d TQFTs in the sense of Atiyah. Suppose we do not require that $A$ is finite dimensional but still require that the pairing is nondegenerate i.e. the map $A \to Hom(A,k)$ is injective. I think the center $C_k[G]$ of any group ring $k[G]$ satisfies this but I haven't checked it. Is there a similar description in terms of 2D surfaces? REPLY [9 votes]: Let's first think about 1-dimensional TQFTs. As is well-known, these correspond exactly to finite dimensional vector spaces as follows. The positive point is assigned to some vector space V, the negative point is assigned to some vector space W, and the "cap" gives a pairing between them, the "cup" gives an element $\sum v_i \otimes w_i$. The zig-zag relations tell you that the pairing is non-degenerate, that the cup is the copairing $\sum e_i \otimes e^i$, and that the vector space is finite dimensional. (This is a great exercise to work out if you haven't done so before.) The value of the circle is the dimension of the vector space. Ok, now we want to find a way to "break" this to generalize to infinite dimensional vector spaces. The only good way I can see to do this is to replace the bordism category with the "non-compact" or "punctured" bordism category whose morphisms are 1-dimensional bordisms such that every connected component has non-trivial incoming boundary. This keeps the pairing, but gets rid of the co-pairing. A functor from the punctured 1-dimensional bordism category to vector spaces is a 1-dimensional punctured TQFT. Now an infinite dimensional vector space gives you a perfectly good punctured TQFT, however these aren't the only ones. In fact any pair of vector space V and W with a pairing between them should do, and there's no reason that pairing needs to be non-degenerate in any sense whatsoever. You could also ask about unoriented 1-dimensional TQFTs and those would correspond to a vector space V with a (possibly degenerate) symmetric billinear form. Ok, now let's turn to 2-dimensions. Note that the above discussion is highly relevant because any 2-dimensional TQFT gives you a 1-dimensional one by "dimensional reduction." That is if Z is a 2-dimensional TQFT then we get a 1-dimensional one via $Z(M \times S^1)$. In particular, as pointed out in comments the value of the torus in your 2-dimensional TQFT is $\dim Z(S^1)$ and so you have to break either the pairing or copairing in some way. Again we can look at punctured bordisms and punctured TQFTs. This means we have perfectly good bordisms for pants, copants, downwards macaroni, and cap. However, we no longer get upwards macaroni or cup. Translating across to algebra, we still get a multiplication, comultiplication, and trace, and many compatibility rules like associativity. But we no longer have a unit or a copairing. In particular, infinite dimensional Frobenius algebras give such examples, but they're not the only examples because there's no requirement of being unital or of the pairing being non-degenerate. Punctured TQFTs (more commonly called "non-compact TQFTs") are studied by Costello and others in the context of "Calabi-Yau algebras."<|endoftext|> TITLE: On Mathematical Analysis of MathSciNet & MathOverflow QUESTION [36 upvotes]: This question has two original motivations: mathematical and social. The mathematical motivation is mainly based on what I have seen about Zipf's law here and there. The Zipf's law simply states that a Zipfian distribution (a variant of power-law probability distribution) provides a good approximation of many types of data corresponding to physical or social phenomena. The iconic example is the frequency of the words used in a natural language where the Zipf's law indicates that the most frequent word in the language will occur approximately twice as often as the second most frequent word, three times as often as the third most frequent word, etc. There are plenty of other surprising appearances of Zipf's (and power) law in various real-world situations. For instance, Silagadze in his paper, "Citations and the Zipf-Mandelbrot's law", shows that the number of citations in scientific papers obeys the Zipf's law! These made me think whether Zipf/power-law distributions appear, in other ways, in large mathematical research databases such as Arxiv, MathSciNet, or MathOverflow, say in the number of publications, co-authors, reviews, reputation points, etc. The social motivation, however, comes from occasional general claims by some mathematicians about how the mathematical society was, is or will be. Statements such as the followings: The mathematical paradigm is slowly shifting from pure mathematics to applied as more and more mathematicians are doing research of applied nature. In comparison with other branches of logic, model theory has the strongest ties with other mathematical disciplines. Due to the urgent need and rapid advancement in AI and computer science, computability and complexity are the most rapidly expanding branches of mathematical logic. The most influential work of a mathematician usually happens before age 40. Such controversial claims often provoke intense arguments among the mathematicians. The question is how to settle them once for all. Indeed, a rigorous approach towards verification of any such social claim is to mathematically analyze the large mathematical databases such as Arxiv, MathSciNet, or MathOverflow in order to extract general patterns including the distribution of research topics, possible changes in mathematical research fashion and evaluating the interaction and intersection of various mathematical disciplines with each other. These motivations lead to the following general question: Question. Have large mathematical databases such as Arxiv, MathSciNet, or MathOverflow been the subject of any social network and database analysis so far? What are examples of published mathematical research about possible mathematical patterns that may exist in them as databases? What patterns are found? (I am particularly interested in the case of Zipf's law and other naturally occurring statistical patterns such as Benford's law.) It is likely that some journal ranking organizations have already conducted some research along these lines but I haven't seen any outline so far. It is also interesting if the research has been done in a comparative way which allows one to compare the general characteristics of math community with its other counterparts, say physics or biology. Update. A MathOverflow fellow just emailed me expressing his interest in conducting some statistical research on MathSciNet and MathOverflow databases. He asked whether I know how to get access to the corresponding background data, which I actually don't. I am not even sure if it is publicly accessible or free (particularly in the case of MathSciNet and Arxiv). I also suspect that there must be some non-disclosure rules and restrictions which any corresponding research along these lines should follow. It would be nice if somebody who knows how to get access to the raw material needed for any research concerning these databases, sheds some light on these issues. REPLY [22 votes]: With regards to reputation on Stack Exchange, I did a very short analysis last year on the distribution of reputation on Stack Overflow. Thanks to the Stack Exchange Data Explorer, I can easily run the same analysis for MathOverflow: x-axis: logarithm of reputation; y-axis: logarithm of number of users; logarithms are base-10, so the 2.0 on the x-axis corresponds to $10^2 = 100$ reputation and there are about $10^{4.25} \approx 18000$ users with this much reputation. Some interesting points, caused by 'oddities' in the Stack Exchange reputation system: are the many no-activity users with just 1 reputation the sawtooth until x = 2.0 (±100 reputation) looks strange, but makes sense once you realize how hard it is to get a total reputation of 2 (1 question upvote followed by 2 downvotes). a peak on and just after x = 2.0, corresponding to 101 reputation; these are mainly users from other Stack Exchange sites who have the association bonus on Math Overflow plus some optional minor additional activity. the peak at 300 reputation is also caused by the association bonus. Users with 200-300 reputation either don't have other accounts on the network, or have another site where they have more reputation. Feel free to fork the query to experiment yourselves with the data. One of the things here that more closely follows Zipf's Law is the number of questions with a certain number of answers:<|endoftext|> TITLE: Patterns in roots of integer-coefficient polynomials QUESTION [8 upvotes]: Below are shown two displays of all the roots of polynomials $$c_n x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0 \;=\; 0$$ with each coefficient $c_i$ an integer $|c_i| \le M$ (including $c_i=0$). No doubt this is all well-known, but I would be interested to learn what results explain the patterns in the distributions, especially the holes, both surrounding the real axis—in both shape and location—and the off-axis holes, perhaps more evident in the degree-$3$ plot than in the degree-$5$ plot.                     Roots of polynomials of degree $\le 3$ and integer coefficients $|c_i| \le 6$.                     Roots of polynomials of degree $\le 5$ and integer coefficients $|c_i| \le 5$. Addendum. User j.c. cited the article by John Baez on Dan Christensen's impressively detailed images, one of which I include below: REPLY [7 votes]: Suppose $P(z)$ is a polynomial with integer coefficients $|c_j| \le M$ and $z \ne 0$. If $c_k$ is the first nonzero coefficient, $$|P(z)|/|z|^k \ge |c_k| - \sum_{j=k+1}^d |c_j| |z|^j \ge 1 - \frac{M |z|}{1-|z|} >0 \ \ \text{if}\ \ |z| < \frac{1}{1+M}$$ This explains the hole around $z=0$. For the holes around $z=\pm 1$, first translate $z \to z\pm 1$.<|endoftext|> TITLE: A characterization of $L_1(\mu)$ in $L_\infty(\mu)^*$ QUESTION [13 upvotes]: Let $\mu$ be a finite positive measure on a set $M$: $$ \mu(M)<\infty. $$ As is known, the Banach dual space $L_\infty(\mu)^*$ to the space $L_\infty(\mu)$ contains $L_1(\mu)$, but (excluding some trivial situations) does not coincide with $L_1(\mu)$: $$ L_1(\mu)\subseteq L_\infty(\mu)^*, \qquad L_1(\mu)\ne L_\infty(\mu)^*. $$ Is there a reasonable description of $L_1(\mu)$ as a space of functionals on $L_\infty(\mu)$? For example, is the following true? Conjecture. A functional $f\in L_\infty(\mu)^*$ belongs to $L_1(\mu)$ if and only if $$ f(x_n)\underset{n\to\infty}{\longrightarrow} 0 $$ for each sequence of functions $x_n\in L_\infty(\mu)$ which is bounded in $L_\infty(\mu)$ as in a Banach space $$ ||x_n||_\infty\le 1, $$ and converges to zero almost everywhere $$ x_n\overset{\text{almost everywhere}}{\underset{n\to\infty}{\longrightarrow}}0. $$ This condition is necessary due to the Lebesgue's dominated convergence theorem. Is it sufficient? REPLY [9 votes]: The criterion suggested in the question works fine for $\sigma$-finite spaces, and Michael Greinecker's answer is correct under this assumption. However, the suggested criterion is not (provably) sufficient in general. I can give two examples. The first is quite pathological, and the second is not, but goes beyond ZFC. (1) Let $X$ be an uncountable set, and take $\Sigma$ to be the countable-cocountable $\sigma$-algebra (the one generated by singletons). Take $\kappa$ to be the counting measure, i.e. a set $S \in \Sigma$ has measure $|S|$ if it is finite and $\infty$ if it is of any infinite cardinality. A function is integrable with respect to this measure iff it is an element of $\ell^1(X)$ (the small $\ell$ is intentional). Therefore a functional in $L^\infty(X,\Sigma,\kappa)$ that is in the image of $L^1(X,\Sigma,\kappa)$ must take a nonzero value on some $\chi_{\{x\}}$ for some $x \in X$. Now consider the following measure $\nu : \Sigma \rightarrow [0,1]$. It takes the value $0$ on all countable sets and $1$ on all cocountable sets. A quick check shows that it is countably additive, and it is absolutely continuous to $\kappa$. But for all $x \in X$, $\int_X \chi_{\{x\}} \mathrm{d}\nu = 0$, so integration against it does not come from an element of $L^1(X,\Sigma,\kappa)$, even though it satisfies your criterion by the dominated convergence theorem. (2) The reason why that space is somewhat pathological is that it is not localizable, i.e. $L^\infty(X,\Sigma,\kappa)$ is not a W$^*$-algebra (isomorphic to a von Neumann algebra). The following is a way of getting a localizable counterexample. Suppose that there exists a real-valued measurable cardinal, i.e. that there exists a set $X$ and a countably additive probability measure $\mu : \mathcal{P}(X) \rightarrow [0,1]$ vanishing on singletons. Such a measure is necessarily not completely additive, because it would then be zero (and the set $X$ is necessarily uncountable for the same reason). So there exists a family of sets $(S_i)_{i \in I}$ where $\mu(S_i) = 0$ for all $i \in I$, but $\bigcup_{i \in I}S_i = X$ (for instance, take the family of all singletons). Now recall the following facts. The spaces $\ell^p(X)$ are exactly the spaces $L^p(X,\mathcal{P}(X),\kappa)$, where $\kappa$ is the counting measure (on $\mathcal{P}(X)$ this time), and each element of $\ell^1(X)$ defines a completely additive signed measure $\nu : \mathcal{P}(X) \rightarrow \mathbb{R}$ by integration (because all $\phi \in \ell^1(X)$ have countable support). By the dominated convergence theorem, the element of $L^\infty(X,\mathcal{P},\kappa)^*$ defined by $f \mapsto \int_X f \mathrm{d}\mu$ satisfies your criterion, but cannot be equal to any $f \mapsto \int_X f\phi\mathrm{d}\kappa$ for any $\phi \in \ell^1(X)$ because $$ \lim_{i \to \infty} \int_X \chi_{S_i} \mathrm{d}\mu = \lim_{i \to \infty} 0 = 0 $$ for all $i \in I$, but for all $\phi \in \ell^1(X)$ that are density functions of probability measures: $$ \lim_{i \to \infty} \int_X \chi_{S_i} \phi \mathrm{d}\kappa = \int_X \chi_X \phi \mathrm{d}\kappa = 1 $$ Conclusion If you wish to go beyond the $\sigma$-finite case to the localizable case, your criterion is not far off. Essentially all you need to do is replace sequences by nets. The image of $L^1(X,\Sigma,\mu)$ in $L^\infty(X, \Sigma,\mu)^*$ is the normal linear functionals, the span of those positive linear functionals that preserve suprema of bounded nets. See Sakai's C$^*$-algebras and W$^*$-algebras, Theorem 1.13.2, taking $\mathcal{M} = L^\infty(X,\Sigma,\mu)$ and $\mathcal{M}_* = L^1(X,\Sigma,\mu)$.<|endoftext|> TITLE: Padé multipoint approximants of the exponential function QUESTION [5 upvotes]: One says that a pair of polynomials $(P_m,Q_n)$ over $\mathbb C[z]$, with $\text{deg }P_m=m$, $\text{deg }Q_n=n$, is a "multipoint Padé approximant of the exponential function" if $P_m(z)e^z-Q_n(z)$ vanishes for $z=0,\ldots, n+m$. Is there an explicit formula for the $P_m$ and $Q_n$? REPLY [4 votes]: Yes, indeed there is. Explicit expressions for multipoint Padé approximants to the exponential (and power) function at points $z=0,\ldots,m+n$, were given in A. Zhedanov, Explicit multipoint rational interpolation Padé table for exponential and power functions, Workshop on Group Theory and Numerical Analysis, CRM, Montréal, May 26 - 31, 2003, CRM Proceedings and Lecture Notes 39, 2004, 285--298. With your notations, the formulas read as follows, \begin{align*} Q _ { n } ( z ) & = ( - 1 ) ^ { m } ( 1 - 1 / e ) ^ { - m } ( 1 + n ) _ { m }~ {}_{2}F _ { 1 } \left( \begin{array} { c } { - n , - z } \\ { - m - n } \end{array} ; 1 - e \right),\\ P_ {m } ( z ) & = ( - 1 ) ^ { m } ( 1 - 1 / e ) ^ { - m } ( 1 + n ) _ { m }~ {}_{2}F _ { 1 } \left( \begin{array} { c } { - m , - z } \\ { - m - n } \end{array} ; 1 - 1/e \right), \end{align*} where $P_{m}(z)$ is a monic polynomial, and here is a plot of $e^z-Q_n/P_n(z)$ for $n=2,3,4$,<|endoftext|> TITLE: Obstructions for the wedge of coordinate differentials to be harmonic QUESTION [8 upvotes]: Let $(M,g)$ be a smooth $d$-dimensional Riemannian manifold, $d$ even. Are there obstructions (I guess in terms of curvature) for $g$ to have the following property: For every $p \in M$ there exist a coordinate system around $p$, such that the co-frame associated with it satisfy: $$ \delta(dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_{\frac{d}{2}}})=0 \, \, \text{for every choice of indices } 1 \le i_1 < i_2 < \dots < i_{\frac{d}{2}} \le d.$$ Here $\delta=d^*$ is the adjoint of the exterior derivative. In fact, for my purposes it suffices that $dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_{\frac{d}{2}}}$ would be co-closed only for two complementing sets of indices, but I am not sure this problem is easier to analyse in practice. (though any ideas on that would be welcomed). For the full problem, we have $\binom{d}{\frac{d}{2}} \cdot \binom{d}{\frac{d}{2}-1}$ equations, while the metric has $\frac{d(d+1)}{2}$ degress of freedom, so this problem is probably overdetermined. (Can we prove that a generic metric admits no solutions?). Given a coordinate system, we can write $dx^i=a^i_je^j$ where $e^j$ is some (positive) orthonormal coframe. Writing $A=a^i_j$, we get that $A^TA=G^{-1}$, where $G=g_{ij}$ is the coordinate representation of the metric. This means we can assume that $A= \sqrt{G^{-1}}$. Then $$ dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_{k}}=a^{i_1}_{j_1}\dots a^{i_k}_{j_k} e^{j_1} \wedge \dots \wedge e^{j_k}=$$ $$\sum_{1 \le j_1 < j_2 < \dots < j_k \le d} \sum_{\sigma \in S^k} a^{i_1}_{j_{\sigma(1)}}\dots a^{i_k}_{j_{\sigma(k)}} \text{sgn}(\sigma) e^{j_1} \wedge \dots \wedge e^{j_k}= \sum_J A^I_Je^J,$$ where $A^I_J$ is the $k$-minor of the matrix $A= \sqrt{G^{-1}}$ corresponding to columns $I=(i_1,\dots,i_k)$ and rows $J=(j_1,\dots,j_k)$. Here $e^J:=e^{j_1} \wedge \dots \wedge e^{j_k}$. Taking the Hodge dual, we obtain $$\star dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_{k}}=\sum_J A^I_Je^{J^c},$$ so the final equation is $$d \star dx^{i_1} \wedge dx^{i_2} \dots \wedge dx^{i_{k}}=\sum_J dA^I_J \wedge e^{J^c}+A^I_J \wedge de^{J^c}.$$ I am not sure how to proceed from here. My idea was to expand $dA^I_J ,de^{J^c}$ and get some first order equation on $G$ and its minors. Then, I guess that second differentiation might give us something which is related to the curvature. However, it doesn't seem easy to do so. Finally, we might simplify things a bit by assuming $e^{J^c} $ is closed. This raises the question what are the obstructions for such an orthonormal co-frame $e^j$ to exist. Note that if we want the $e^i$ themselves to be closed (not just their wedge product) then this forces the metric to be flat. I am not sure what is the obstruction when $|J|=k>1$. REPLY [6 votes]: Update (1 June 2018): I have now figured out the 'little linear algebra lemma' in all dimensions $d = 2n$ and can give a complete answer to the OP's question: A metric $(M^{2n},g)$ possesses coordinate charts of the kind that the OP desires if and only if it is 'locally conformally unimodular Hessian', i.e., every point $p\in M$ has an open neighborhood $V$ on which there exist coordinates $x^1,\ldots,x^{2n}$ and a function $u$ with positive definite $x$-Hessian satisfying the Monge-Ampère equation $\det(\mathrm{Hess}_x(u)) = 1$ so that $g$ is a multiple of the Hessian metric $$ h = \frac{\partial^2u}{\partial x^i\partial x^j}\,\mathrm{d}x^i\mathrm{d}x^j\,. \tag0 $$ When $d = 2n > 2$, the generic metric $g$ is not locally conformally unimodular Hessian, so such coordinate charts do not exist. Because the proof is a bit clearer in the case $d=4$, I'm going to leave that in and simply indicate how the fundamental lemma (equation (2) below) changes in higher dimensions after the $d=4$ argument. What follows up until then is what I had written before: I now have a complete answer in the case $d=4$ (the case $d=2$ being trivial). (I suspect that the answer holds for all (even) $d$, but that would require proving a little linear algebra lemma that I don't see an immediate proof of, but maybe it will come to me.) Here is the answer: When $d=2n>2$ there are definitely obstructions (though, what they are explicitly in terms of curvature, I haven't a clue at this point), as my original argument (retained below) shows. Meanwhile, in the case $d=4$, one has the following necessary and sufficient condition: $(M^4,g)$ has the desired local coordinate systems if and only if $(M^4,g)$ is locally conformally unimodular Hessian, i.e., every point $p\in M$ has a neighborhood $V$ on which there exist coordinates $x^1,x^2,x^3,x^4$ and a function $u$ with positive definite $x$-Hessian satisfying the Monge-Ampère equation $\det(\mathrm{Hess}_x(u)) = 1$ so that $g$ is a multiple of the Hessian metric $$ h = \frac{\partial^2u}{\partial x^i\partial x^j}\,\mathrm{d}x^i\mathrm{d}x^j\,. \tag1 $$ In fact, for such coordinates, we have $\mathrm{d}\bigl(\ast_g(\mathrm{d}x^i\wedge\mathrm{d}x^j)\bigr) = 0$ for all $1\le i1$, the generic metric $(M^{2n},g)$ does not admit any local coordinate system satisfying even the single condition $$ \mathrm{d}\bigl(\ast(\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n)\bigr)=0. \tag2 $$ In the OP's formulation, one is asking for the conditions that $g$ admit a coordinate system such that $$ \mathrm{d}\bigl(\ast(\mathrm{d}x^{i_1}\wedge\cdots\wedge\mathrm{d}x^{i_n})\bigr) =0.\tag3 $$ for each choice of $1\le i_1 TITLE: Does a torus action with isolated fixed points imply rational? QUESTION [7 upvotes]: Suppose that $X$ is a smooth projective variety $/\mathbb{C}$ with a $\mathbb{C}^{*}$-action with isolated fixed points. Must $X$ be rational? REPLY [9 votes]: Yes. This follows from the Białynicki-Birula decomposition (see Theorem 4.4 in the original paper).<|endoftext|> TITLE: Riemannian holonomy of generic manifolds QUESTION [7 upvotes]: It is well known, as well as absolutely intuitive, that the Riemannian holonomy of a generic Riemannian manifold is $O(n)$, the Riemannian holonomy of a generic orientable Riemannian manifold is $SO(n)$, and the Riemannian holonomy of a generic Kähler manifold is $U(n/2)$. I guess that in this context by "generic" one means that given any Riemannian metric $g_0$, it suffices to perturb it a little bit in order to get the biggest possible holonomy group. What I would like to ask is if you could provide one or more simple and/or elementary proofs of this general principle, which is often stated without specifying much more. Thanks in advance! REPLY [5 votes]: Here are proofs for the Riemannian and Kähler case which rely on the fact that the curvature can be seen as parallel transport around infinitesimal loops. It uses explicit deformations which are hard to build with the holonomy is further restricted, that's why this proof doesn't work for smaller homotopy groups. Riemannian case Let $(M,g_0)$ be a Riemannian manifold, $p\in M$, and $\mathrm{Hol}_{g_0}(p)$ be the reduced holonomy group of $g$ at $p$. We know that the associated Lie algebra $\mathfrak{hol}_{g_0}(p)$ contains every two forms $R_p^0(x\wedge y)$ where $R^0_p:\Lambda^2T_pM\to\Lambda^2T_pM$ is the Riemann curvature tensor of $g_0$ (seen as a symmetric operator on $\Lambda^2TM\simeq \mathfrak{so}(TM)$). I will exhibit a one parameter family of Riemannian metrics $g_t$ going to $g_0$ such that $R^t_p(\Lambda^2T_pM)=\mathfrak{so}(T_pM)$ and hence $Hol_{g_t}(p)=SO(n)$. I will use the following fact, which follows from the expression of the curvature in term of metric components : Theorem : Let $g$ be a metric on $\mathbb{R}^n$ such that : $$ g=\delta_{ij}dx^i\otimes dx^j+\tfrac{1}{3}\left(A_{ikjl}x^kx^ldx^i\otimes dx^j\right)+O(\|x\|^3)$$ Then the $(4,0)$-curvature tensor of $g$ a the origin is given by $A_{ikjl}dx^i\otimes dx^k\otimes dx^j\otimes dx^l$. Now pick $n$ smooth functions $x^1,\dots,x^n$ supported in a small neighborhood of $p$ and which form a normal coordinate system in a smaller neighborhood of $p$. Set : $$g_t=g_0+\tfrac{t}{3}\left(I_{ikjl}x^kx^ldx^i\otimes dx^j\right)$$ where $I_{ikjl}=\tfrac{1}{2}\left(g(\partial_i,\partial_j)g(\partial_k,\partial_l)-g(\partial_i,\partial_l)g(\partial_k,\partial_j)\right)$ is the $(4,0)$ tensor corresponding at $p$ to the identity $I_p:\Lambda^2T_pM\to\Lambda^2T_pM$. Since we are working in normal coordinates : $$ g_0=\delta_{ij}dx^i\otimes dx^j+\tfrac{1}{3}\left(R^0_{ikjl}x^kx^ldx^i\otimes dx^j\right)+O(\|x\|^3)$$ and : $$ g_t=\delta_{ij}dx^i\otimes dx^j+\tfrac{1}{3}\left((R^0_{ikjl}+tI_{ikjl})x^kx^ldx^i\otimes dx^j\right)+O(\|x\|^3)$$ It follows from the theorem above that the curvature of $g_t$ at $p$ is given by : $R_p^t=R_p^0+t I_p$ which will be invertible as an symmetric operator on $\Lambda^2T_pM$ for all $t$ in some interval $(0,\varepsilon)$. Hence for $t\in (0,\varepsilon)$, we have that $R^t_p(\Lambda^2T_pM)=\mathfrak{so}(T_pM)$. And hence $\mathrm{Hol}_{g_t}(p)=SO(n)$. This should actually work for the non-orientable case too. (The image of any curve which reverse the orientation will give the rest of $O(n)$.) Kähler case I think the proof below works however my knowledge of Kähler geometry is quite limited and I may have missed something. First we need to understand the symmetries of the curvature tensor of a Kähler manifold $(X^m,\omega_0,J)$ with $m=\dim_\mathbb{C}X$. Its symmetries show that $R^0:\Lambda^2TM\to\Lambda^2TM$ vanishes on $(\Lambda^{1,1}TM)^\perp$ and thus can be viewed as a symmetric endomorphism $K^0:\Lambda^{1,1}TM\to \Lambda^{1,1}TM$ which we call the Kähler curvature operator. The holonomy algebra $\mathfrak{hol}_{g_0}(p)$ will contain $K^0(\Lambda^{1,1}TM)$. Note that $\Lambda^{1,1}TM\simeq\mathfrak{u}(m)$. The Kähler curvature operator of $\mathbb{CP}^m$ is given by (Besse) : $$I=(m+1)\mathrm{Id}_{\mathbb{R}\omega_0}+\mathrm{Id}_{(\mathbb{R}\omega_0)^\perp}. $$ The equivalent of the Theorem above should be (I haven't been able to locate a reference) : Theorem : Let $g$ be a Kähler metric on $\mathbb{C}^m$ such that : $$ g=\delta_{i\bar{j}}dz^i\otimes d\bar{z}^{j}+\tfrac{1}{3}\left(A_{i\bar{k}\bar{j}l}\bar{z}^kz^ldz^i\otimes d\bar{z}^{j}\right)+O(\|z\|^3)$$ Then the $(4,0)$-curvature tensor of $g$ a the origin is given by $A_{i\bar{k}\bar{j}l}dz^i\otimes d\bar{z}^k\otimes d\bar{z}^j\otimes dx^l$. (Take my indexed formula with a grain of salt here, they might not be exactly the right one. The main point is that if the metric is of the form euclidean plus something of order 2, then this second order term is exactly the curvature; just as in the Riemannian case.) We can find holomorphic coordinates on $(X,J)$ $(z^1,\dots,z^m)$ such that $g_0$ has an expansion as above around $p\in X$. (There existence is proven by Moroianu in Lectures on Kähler Geometry.) Now consider deformations of $(X,\omega_0,J)$ of the form $\omega_t=\omega_0+ti\partial\bar\partial\varphi$, where $\varphi(z)=\ln(1+|z|^2)$ near $p$ (this is the potential of the Fubini Study metric). We have the following expansion around $p$ for the associated metric : $$ g_t=(1+t)\delta_{ij}dz^i\otimes d\bar{z}^{j}+\tfrac{1}{3}\left((K^0+tI)_{i\bar{k}\bar{j}l}\bar{z}^kz^ldz^i\otimes d\bar{z}^{j}\right)+O(\|z\|^3)$$. So the Kähler curvature operator $K^t$ of $g_t$ at $p$ will be (some multiple of) $K^0+tI$, which will be invertible on $\Lambda^{1,1}$ for $t\in(0,\varepsilon)$. Hence $K^t(\Lambda^{1,1})=\Lambda^{1,1}$ and $\mathfrak{hol}_{g_t}(p)=\mathfrak{u}(m)$ thus $\mathrm{Hol}_{g_t}(p)=U(m)$.<|endoftext|> TITLE: Surjectivity of a map on inverse limits QUESTION [7 upvotes]: (The following is crossposted from Math.SE, where the question did not receive any answers.) I am looking for a proof of the following lemma from P. Gabriel's Des catégories abéliennes (Chap. IV, §3, Lemme 1): Lemma. Let $B$ be a ring, $(I, \leq)$ be a directed poset, and $(M_i, f_{ji}:M_j\rightarrow M_i)_{j\geq i},$ $(N_i, g_{ji}:N_j\rightarrow N_i)_{j\geq i}$ two inverse systems of $B$-modules. Let $(h_i: M_i\rightarrow N_i)_i$ be a morphism of the inverse systems, and assume that all the maps $h_i$ are surjective with an Artinian kernel. Then the limit map $$\varprojlim_i h_i: \varprojlim_i M_i\rightarrow \varprojlim_i N_i$$ is surjective. I am looking either for a proof (I haven't manage to come up with one so far), or for a reference to a proof - Gabriel refers to "Bourbaki, Topologie, I Appendice, $3^{\text{e}}$ éd.", which is a reference I cannot find anywhere. I do not want to assume that e.g. $I$ is countable - the reason is that I need this lemma, similarly as Gabriel in his thesis, to establish some properties of pseudo-compact modules over a pseudo-compact ring (namely that quotient of a pseudo-compact module by a closed submodule is pseudo-compact, and exactness of inverse limits). For this reason, as far as I can tell, I cannot use countable index sets (i.e. countable bases of neighbourhoods) in general. Thank you in advance for any help. REPLY [4 votes]: Your question can be interpreted as the vanishing of the first derived projective limit functor ${\lim\limits_{\leftarrow}}^{(1)} \mathcal M$ for the projective spectrum of the kernels. If these kernels are Artenian even all derivatives vanish. This is shown in Corollary 7.2 of C.U. Jensen's Les Foncteurs Dérivés de lim et leurs Applications en Théorie des Modules, Springer Lecture Notes 254 (1972). He also refers to his article On the vanishing of ${\lim\limits_{\leftarrow}}^{(i)}$, J. Algebra 15 (1970), 151-166 (but I did not have a look at it). It seems to me that Jensen does not claim for originality as he writes "Il est bien connu...". However, he does not give an explicit reference. A main step in his proof is a result in Bourbaki's Topologie Générale.<|endoftext|> TITLE: Hopf invariant and the Freudenthal theorem QUESTION [6 upvotes]: Let $\mathcal{H}(f)$ be the Hopf invariant of a map $f:\mathbb{S}^{4n-1}\to \mathbb{S}^{2n}$. When is the suspension map $$ \sigma:\{f\in \pi_{4n-1}(\mathbb{S}^{2n}):\, \mathcal{H}(f)\neq 0\}\to \pi_{4n}(\mathbb{S}^{2n+1}) $$ a non-zero map? When is it a surjection? Clearly it is a surjection when $n=1$, but I am interested in the higher dimensional cases. In other words, I would like to know if there exist non-trivial elements in $ \pi_{4n}(\mathbb{S}^{2n+1})$, $n\geq 2$, that are suspensions of a map with a non-zero Hopf invariant. REPLY [6 votes]: I'm on shaky ground here, this is what I think happens. There are the Hopf invariant one maps $\nu:S^7\to S^4$ and $\sigma:S^{15}\to S^8$, and these suspend to generators of the stable stems $\pi^S_3\cong \mathbb{Z}/24$ and $\pi^S_7\cong\mathbb{Z}/240$. So you also have surjectivity when $n=2, 4$, also. The Hopf invariant is a homomorphism $H:\pi_{4n-1}(S^{2n})\to \mathbb{Z}$, so it vanishes on torsion elements. It takes the value $\pm 2$ on the Whitehead square $[\iota_{2n},\iota_{2n}]$, which generates the $\mathbb{Z}$ summand in $\pi_{4n-1}(S^{2n})$. The suspension of $[\iota_{2n},\iota_{2n}]$ is trivial. These facts together imply that the answer to your final question is no in the other dimensions. The facts above are stated in Hatcher's book on pp 473-474, and no doubt are proved in Toda's book.<|endoftext|> TITLE: A second Ph.D. in mathematics? QUESTION [46 upvotes]: I have now some problems about my research Career, I would like to tell my stories. I am a Chinese guy, but now a Ph.D. candidate in Germany, in the field of so called 'Geometric Analysis', but I do not feel happy when I work in such a field. Somebody said that a wrong choice of study field or supervisor means several years' Frustration, it describes my situation very well. I choose this geometric analysis only because I want to study geometry with some analytical methods, but gradually I found that the most researcher in this field only have very narrow knowledge about PDE and Riemannian geometry, I think this field is lack of real beautiful idea, but full of papers, I am not judging that this is not good, but this is not my taste. Compare to the normal researchers in this field, I have relative comprehensive mathematical knowledge, I am not only familiar with second order elliptic PDE and differential geometry (Riemannian geometry, geometry of fiber bundles, Chern-Weil theory) but also with algebraic topology (homology, cohomology, characteristic class and spectral sequence), complex geometry(the whole book of Demailly'Complex differential and analytic geometry' and some complex Hodge theory in the first book of C.Voisin). Several years ago, I have also studied abstract algebraic geometry and Index theorem in courses, but now I am not familiar with such kind of material. In China I have already known, for my future, I should not only have one math-tool. At that time, a professor in algebraic geometry has suggested me that as a young guy, algebraic geometry may be a better choice for the future and invited me to be his student, but I hesitated and refused, that was maybe a stupid decision. During my research in Germany, I was always very depressive because what I was working were only some trivial and unnature PDE estimates. Now my career is a little hopeless, somebody suggested that I could try to contact some experts who work in Symplectic geometry with analytic methods, like members of Hofer's school in Germany. Yes, I actually want to study some deep topics in (complex-) algebraic geometry or symplectic geometry, but the problem is, I will soon have a Ph.D. degree, how can I change my research area after my graduation? As a normal young guy the experts in another field do not know me at all, maybe it's very hard to get a postdoc position from them, so should I do another Ph.D. in math? Is that worthy? I am frustrated and asking for help, maybe some guys have similar experience. If I cannot research what I like, I must try to find a Job in Industry. REPLY [33 votes]: Finish this degree, then switch to whatever interests you. Many (most?) mathematicians change fields at some point in their research careers. Bob Solovay told me once that the most important research was the first new thing you did after you got your degree. His thesis was A Functorial Form of the Differentiable Riemann–Roch theorem. He finished it in a hurry so he could move on to mathematical logic, where he's famous. https://en.wikipedia.org/wiki/Robert_M._Solovay<|endoftext|> TITLE: What is the archimedean Hecke algebra? QUESTION [16 upvotes]: Let $\mathbf G$ be a connected, reductive group over $\mathbb Q$. For each nonarchimedean place $v$, let $K_v$ be a maximal compact subgroup of $\mathbf G(\mathbb Q_v)$. The space $\mathscr H(\mathbf G(\mathbb Q_v),K_v)$ of bi $K_v$-invariant compact supported functions $\mathbf G(\mathbb Q_v) \rightarrow \mathbb C$ is a unital algebra over $\mathbb C$ with respect to convolution. This is the non-archimedean Hecke algebra. I'm not familiar with the archimedean version of this. Jayce Getz's notes define the Hecke algebra of $G = \mathbf G(\mathbb R)$ to be the "convolution algebra of distributions" of $G$ supported on $K_{\infty}$. What is this exactly? If it is difficult to describe, I would appreciate any good references on this. Is there a more general notion of a Hecke algebra associated to a real Lie group and a maximal compact subgroup? REPLY [8 votes]: Here is the picture, as I understand it, for $\mathrm{GL}_n$; this is described in chapter 8 of Godement-Jacquet (and see also the archimedean theory in Jacquet-Langlands). Let $F \in \{\mathbb{R},\mathbb{C}\}$ be an archimedean local field. Let $\mathcal{H}_1$ denote the space of smooth compactly supported functions on $\mathrm{GL}_n(F)$ that are bi-$K$-finite, where $K = \mathrm{U}(n)$ if $F = \mathbb{C}$ and $K = \mathrm{O}(n)$ if $F = \mathbb{R}$. These may be regarded as measures on $\mathrm{GL}_n(F)$, in which case $\mathcal{H}_1$ is an algebra under convolution: for $f_1, f_2 \in \mathcal{H}_1$, \[f_1 \ast f_2(g) = \int_{\mathrm{GL}_n(F)} f_1(gh^{-1}) f_2(h) \, dh.\] Every function $\xi$ on $K$ that is a finite sum of matrix coefficients of irreducible representations $\tau$ of $K$ may be identified with a measure on $K$, and hence on $\mathrm{GL}_n(F)$. Under convolution, these measures form an algebra $\mathcal{H}_2$. We let $\mathcal{H}_F = \mathcal{H}_1 \oplus \mathcal{H}_2$. This is an algebra under convolution of measures: for $f \in \mathcal{H}_1$ and $\xi \in \mathcal{H}_2$, \[\xi \ast f(g) = \int_{K} \xi(k) f(k^{-1} g) \, dk\] and \[f \ast \xi(g) = \int_{K} f(gk^{-1}) \xi(k) \, dk.\] This is the Hecke algebra of $\mathrm{GL}_n(F)$. Given a representation $(\pi,V)$ of $\mathrm{GL}_n(F)$, we define the action of $f \in \mathcal{H}_F$ on $v \in V$ by \[\pi(f) \cdot v = \int_{\mathrm{GL}_n(F)} f(g) \pi(g) \cdot v \, dg.\] REPLY [8 votes]: The terminology is a bit misleading, and the analogy with the non-archimedean situation is a bit forced. The goal was/is to have a $\mathfrak g,K$-module be a "Hecke algebra module", for some suitable notion of "Hecke algebra". One wanted/wants all differential operators (identified with the universal enveloping algebra of $\mathfrak g$) and also the action of $K$. Well, as a consequence of some old, standard lemmas, to say "convolution algebra of distributions supported on $K$" is equivalent to that $\mathfrak g,K$-module structure. Some work to be done, though. Also, the relevant vector-valued integral (as in Peter Humphries's answer) needs a bit of shoring-up to be guaranteed to do what we expect. And, indeed, the Gelfand–Pettis "weak" (ironically, "weak" mostly in terms of assumptions, rather than conclusions) integral has been around for a long time and does the job.<|endoftext|> TITLE: "Oriented representation" sphere QUESTION [9 upvotes]: I am trying to understand basic notions from Hill-Hopkins-Ravenel paper: https://arxiv.org/abs/0908.3724 In the Example 3.10 we are considering equviariant cellular chain complex for $n$-dimensional representation $V$ of a group $G$ $$ \ldots\to C^{cell}_n(S^V;\underline{\mathbb{Z}})\to C^{cell}_{n-1}(S^V;\underline{\mathbb{Z}})\to\ldots\to C^{cell}_0(S^V;\underline{\mathbb{Z}}). $$ The underlying homology groups are those of the sphere $S^V$ - in particular kernel of the map $C^{cell}_n(S^V;\underline{\mathbb{Z}})\to C^{cell}_{n-1}(S^V;\underline{\mathbb{Z}})$ is isomorphic as a $G$-module to nonequivariant homology of $S^V$ (denoted as $H^u_n(S^V;\mathbb{Z}))$. Now if $V$ is oriented (so I suppose this means that group action preserves orientation), then $G$ acts trivially on $C^{cell}_n(S^V;\underline{\mathbb{Z}})$ (? I am not sure about this statement) and we obtain that $H^G_*(S^V;\mathbb{Z})\cong H^u_n(S^V;\mathbb{Z}).$ Question: Why if $V$ is oriented we obtain such result? I moved this question from MathStack Exchange. REPLY [9 votes]: First of all, note that right before example 3.9 they prove that $$H^G_*(S^V;\underline{\mathbb{Z}})=H_*(C^{cell}_*(S^V)^G)\,,$$ where $C^{cell}_*(S^V)$ is the cellular complex for some $G$-CW-structure on $S^V$ (and so levelwise is just a sum of permutation modules). In particular, since $S^V$ is $n$-dimensional $$H^G_n(S^V;\underline{\mathbb{Z}})=\ker\left(C^{cell}_n(S^V)^G\to C^{cell}_{n-1}(S^V)^G\right)\,.$$ Hence $$H^G_n(S^V;\underline{\mathbb{Z}})=\ker\left(C^{cell}_n(S^V)\to C^{cell}_{n-1}(S^V)\right)^G=H^u_n(S^V;\mathbb{Z})^G\,,$$ since taking fixed points commute with taking kernels. Now, $V$ is orientable iff the action of $G$ preserves the orientation iff $G$ acts trivially on $H^u_n(S^V;\mathbb{Z})$ (recall that an orientation of $V$ is the same thing as a generator of $H^u_n(S^V;\mathbb{Z})$). So the restriction map $H^G_n(S^V;\underline{\mathbb{Z}})\to H^u_n(S^V;\mathbb{Z})$ is an isomorphism iff $V$ is orientable. WARNING: In general it is not possible to find a $G$-CW structure such that $G$ acts trivially on $C^{cell}_n(S^V)$, even if $V$ is orientable.<|endoftext|> TITLE: An example of an open discontinuous function QUESTION [5 upvotes]: Consider the following simple example of a function $f: \mathbb{R}\to\mathbb{R}$ which is open and discontinuous at all points. If $x\in\mathbb{R}$ is represented as something.$x_1x_2x_3\dots$ in the binary system, then set $$f(x)=\lim_{n\to\infty}\frac{x_1+\cdots+x_n}{n}$$ if the limit exists and belongs to $(0,1)$, and set $f(x)$ to (say) $\frac{1}{2}$ otherwise. Is this example known (I suppose it is), and what's the reference for it? REPLY [5 votes]: The following open question was popular in some places: Q:   does there exist $\ x\in(0;1)\ $ such that $\,\ s(x)=x,\,\ $ where $$ s(x)\ :=\ \lim_{n=\infty} \frac{\sum_{k=1}^n x_k}n $$ and $\ x_n\ $ are binary digits $\ 0\ $ or $\ 1,\ $ and $$ x\ =\ \sum_{n=1}^\infty\frac{x_n}{2^n} $$ Thus, in 1959/60 I've formulated and proved a more general theorem, and it was recognized in the spring of 1961, namely: Theorem   For every function $\ f: (0;1)\rightarrow (0;1)\ $ which is a pointwise limit of a sequence of continuous functions, and for every non-empty $\ (a;b)\subseteq (0;1),\ $ the equation $$ s(x)\ =\ f(x) $$ has $\ 2^{\aleph_0} $ of different solutions $\ x\in (a;b).$ Remark   there is something of a paradox due to the juxtaposition of the equation $\ s(x)=f(x)\ $ and the classical Jacques Hadamard equation $\ s(x)=\frac 12,\ $ which holds for almost all $\ x\in[0;1].$<|endoftext|> TITLE: Is the sheaf associated to a differential structure of a specific type? QUESTION [5 upvotes]: On a set $X$, let us define a set $\mathcal{D}$ of functions from $X$ to $\mathbb{R}$. Consider first the initial topology $\tau_\mathcal{D}$ on $X$ with respect to $\mathcal{D}$, i.e. the coarsest topology that makes all functions in $\mathcal{D}$ continuous (with the usual topology on $\mathbb{R}$). Given a $X$ and a set of functions $\mathcal{D}$, for $W \subseteq X $ we say $f:W \to \mathbb{R}$ verifies the property $P_W$ if $$\forall x \in W, \ \exists (V,g) \in \tau_\mathcal{D} \times \mathcal{D} \text{ such that } V \ni x, \text{ and } f|_V = g|_V. $$ Now, a differential structure on a set $X$ is a set $\mathcal{D}$ of functions from $X$ to $\mathbb{R}$ that verifies the following axioms $\mathcal{D}$ is a $\mathcal{C}^\infty$-ring: if $u_1,\ldots,u_N \in \mathcal{D}$ and $g \in \mathcal{C}^\infty(\mathbb{R}^N,\mathbb{R})$, then $g \circ (u_1,\ldots,u_N) \in \mathcal{D}$ It is locally determined: If $f:X \to \mathbb{R}$ verifies $P_X$ then $f \in \mathcal{D}$. We can consider the functor that associates to an open set $U \in \tau_\mathcal{D}$ the set $$\mathbf{C}_\mathcal{D}(U) := \{ f:U \to \mathbb{R} \mid f \text{ verifies the property } P_U \}.$$ The two axioms ensure this is a sheaf of $\mathcal{C}^\infty$-ring. The set of smooth functions of a paracompact manifold defines a sheaf of $\mathcal{C}^\infty$-ring which is fine, and it is for sure an important feature. Differential structure are meant to mimick these functions, when one does not have a manifold structure. On the other hand, we have by definition $\mathcal{D} = \Gamma(X,\mathbf{C}_\mathcal{D})$, so it seems the associated sheaf is somehow "determined" by its global sections. I understand there is a whole typology of sheaves related to the local-VS-global properties: injective, flabby, soft, acyclic, fine. Is $\mathbf{C}_\mathcal{D}$ of one of these types? Can I actually define a differential structure as a sheaf of $\mathcal{C}^\infty$-ring of some well-known type ? Note that it is ok if you need some topological assumption, like $(X, \tau_\mathcal{D})$ is paracompact. REPLY [2 votes]: I have found some answer, in a context that is large enough to satisfy me. Any other comment in any other context is still welcome. In the book, Differential geometry of singular space and reduction of symmetry 2013, Sniatycki proves in section 2.2 the existence of a partition of unity for any locally compact, second countable Hausdorff differential spaces. So in this context, we have indeed a fine sheaf.<|endoftext|> TITLE: What kind of category is generated by Cubical type theory? QUESTION [15 upvotes]: What kind of ‘category’ is Cubical type theory the internal language of? Its known that Martin-Löf type theories are the internal language of Locally cartesian closed categories, adding higher inductive types you get the internal language of locally cartesian closed $(\infty , 1)$-categories, a.k.a HoTT without Univalence. And as far as I know, it is suspected that plain HoTT is the internal language of whatever an $(\infty , 1)$-topos is. Cubical type theory doesn’t quite fit into this sequence of type theories but it is surprisingly good at modelling HoTT. It is a strange kind of type theory in that it uses De Morgan algebras to reason about cubes but non the less it begs the question: What kind of category does it generate? I suspect the answer to this question isn’t known, however I would be more than happy to see peoples suspicions. REPLY [17 votes]: There are two kinds of answers as to what kind of category a "homotopy type theory" is the internal language of. On the one hand there is a kind of $(\infty,1)$-category that is the semantic object of real interest; but on the other hand there is a 1-categorical presentation of the latter that corresponds more closely to the syntax of the type theory. The latter kind of category (whose variants go by names like "contextual category", "category with families", "category with attributes", "type-theoretic fibration category", "tribe", etc.) is also closely related to the model categories and fibration categories used to present $(\infty,1)$-categories in classical abstract homotopy theory. For instance, HoTT with $\mathrm{Id},\Sigma,\Pi$ and function extensionality, but no universes, corresponds (conjecturally) to locally cartesian closed $(\infty,1)$-categories — presented by means of "$\Pi$-tribes" (or whatever other name you prefer for the latter). Cubical type theory is a syntactic variation which changes the corresponding 1-categorical presentations, but not the desired $(\infty,1)$-categorical semantics. That is, the analogous cubical type theory, with $\mathrm{Id},\Sigma,\Pi$ (and, in the cubical case, provable function extensionality), but no universes, also corresponds (conjecturally) to locally cartesian closed $(\infty,1)$-categories, but now presented in a different way by $\Pi$-tribes containing (or more precisely, fibered over the theory of) some kind of "interval object". There is a sketch of the latter kind of category (for a more general kind of "type theory with shapes") in appendix A of https://arxiv.org/abs/1705.07442.<|endoftext|> TITLE: Why are values of Eisenstein $E_2^*$ algebraic integers? QUESTION [23 upvotes]: I'm looking for a proof that the following term is an algebraic integer whenever $\tau_N=\frac{N+\sqrt{-N}}{2}$ is a quadratic irrationality with class number $1$: $$A_N:=\sqrt{-N}\cdot\frac{E_2(\tau_N)-\frac{3}{\pi\cdot Im(\tau_N)}}{\eta^4(\tau_N)}$$ Here $\eta$ denotes the Dedekind $\eta$-Function and $E_2$ is the Eisenstein series of weight $2$. As @HenriCohen said here: How to compute Coefficients in Chudnovsky's Formula? it follows from theorems of complex multiplication, but I couldn't find such theorems. I calculated the numerical value of $A_N$ for all discriminants with class number 1. The results are: $A_3 = 0$ $A_4 = 0$ $A_{7}=3\cdot e^{i\pi/3}$ $A_{8}=4\cdot e^{i\pi/2}$ $A_{11}=8\cdot e^{i\pi/3}$ $A_{12}=6\cdot4^{1/3}\cdot e^{i\pi/2}$ $A_{16}=12\cdot2^{1/2}\cdot e^{i\pi/2}$ $A_{19}=24\cdot e^{i\pi/3}$ $A_{27}=24\cdot9^{1/3}\cdot e^{i\pi/3}$ $A_{28}=54\cdot e^{i\pi/2}$ $A_{43}=144\cdot e^{i\pi/3}$ $A_{67}=456\cdot e^{i\pi/3}$ $A_{163}=8688\cdot e^{i\pi/3}$ Thus we get numerically, that these $A_N$ are algebraic integers, but I don't see how I can prove it. Does anyone know how to do that? EDIT: Thanks to the answer of Nikos Bagis, only the $A_N$ with odd $N$ remain to be proven. I moved the remaining part of the question here, where a complete answer was given by Michael Griffin. EDIT: Complete Solution: The answer of Michael Griffin (see here) can now be found in more details in the appendix of this arXiv-preprint. Edit: Ramanujan-Sato-Series Tito Piezas III found some Ramanujan-Sato-Series of level 9 which can be expressed with these numbers $A_N$ (see this question). REPLY [4 votes]: This is too long to fit in a comment. Ramanujan established in his monumental paper Modular Equations and Approximations to $\pi$ that the desired expression is an algebraic number if $\tau_n=\sqrt{-n} $ where $n$ is a positive rational number. Let $$P(q) =1-24\sum_{j=1}^{\infty}\frac{jq^{j} }{1-q^{j}}\tag{1}$$ and then Ramanujan proved that $$P(e^{-2\pi\sqrt{n}}) =\left( \frac{2K}{\pi}\right)^2A_n+\frac{3}{\pi\sqrt{n}}\tag{2}$$ where $A_n$ is an algebraic number dependent on $n$ provided that $n$ is a positive rational number. Here $K=K(k) $ is the complete elliptic integral of first kind with modulus $k$ and $k$ corresponds to nome $q$ so that $$k=\frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)},q=e^{-\pi\sqrt{n}}\tag{3}$$ and $k$ is an algebraic number if $n$ is a positive rational number. Ramanujan's proof is presented in one of my blog posts. If $q=\exp(\pi i\tau) $ then we have $P(q^2) =E_2(\tau)$. Further we have $$\eta(\tau) =q^{1/12}\prod_{j=1}^{\infty}(1-q^{2j}),q=e^{\pi i\tau}\tag{4}$$ It is well known that the eta function can be expressed in terms of $k, K$ as $$\eta(\tau)=2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{5}$$ and therefore the equation $(2)$ can be written as $$A_n=\dfrac{E_{2}(\tau_n)-\dfrac{3}{\pi\Im{\tau_n}}}{\eta^4(\tau_n)}$$ which is an algebraic number. Note that if $\tau_n=\dfrac{n+\sqrt{-n} }{2}$ then we have $E_2(\tau_n)=P(q^2)$ if $n$ is even and $q=e^{\pi i\tau_n} $ and if $n$ is odd then $E_2(\tau_n)=P(-q^2)$. Using Ramanujan's technique one can prove that $$P(e^{-\pi\sqrt{n}})=\left(\frac{2K}{\pi}\right) ^2B_n+\frac{6}{\pi\sqrt{n}}\tag{6}$$ (just replace $n$ in $(2)$ by $n/4$ and $B_n=A_{n/4}$) and $$P(-e^{-\pi\sqrt{n}}) =\left(\frac{2K}{\pi}\right)^2C_n+\frac{6}{\pi\sqrt{n}}\tag{7}$$ where $B_n, C_n$ are algebraic numbers and $n$ is a positive rational number and thus the expression mentioned in the question is an algebraic number. Proving that it is an algebraic integer is unfortunately not possible via Ramanujan methods.<|endoftext|> TITLE: Is there a triangle which makes dense set of angles by drawing medians? QUESTION [10 upvotes]: This problem is a restatement of this question, first announced in MathStackExchange. We start with a triangle $T$ in the Euclidean plane and we define $A_n$ as the set of angles of the $6^n$ triangles obtained from the $n$-th barycentric subdivision of $T$. (Angles are geometric angles, i.e., the angles of $T$ and its children lie in $[0, \pi]$). What we can say about set $A=\bigcup_{n=0}^{\infty}A_n$? Is there a specific predefined triangle for which $A$ is dense on $(0,\pi)$? If no, does there exist another simple alternative to this iterative procedure, i.e., barycentric subdivision, using other concurrent lines in triangles, which achieves our goal? REPLY [15 votes]: The answer to the second question is yes, for any non-flat triangle $T$, the set of angles $A$ is dense in $(0, \pi)$. This follows from a stronger result of Barany et al. [Theorem 1, 1]: Theorem. Successive barycentric subdivisions of a non-flat triangle contain triangles which, to within a similarity, approximate arbitrarily closely any given triangle. Here is a nice divulgative article on the dynamics of iterated barycentric subdivisions. It also has a list of further readings on the topic. (The joy of barycentric subdivision, by Bill Casselman)- [1] I. Barany, A. Beardon and T. Carne, "Barycentric subdivision of triangles and semigroups of Möbius maps", 1996. MR1401715.<|endoftext|> TITLE: Groups that satisfy ${ [x,y]^2 \approx 1 }$ QUESTION [16 upvotes]: Lately, I have been constructing finite involution monoids that generate varieties with $2^{\aleph_0}$ subvarieties. One construction requires groups that violate the identity ${ [x,y]^2 \approx 1 }$, where ${ [x,y] = x^{-1} y^{-1} xy }$. Is there a name for groups satisfying the identity ${ [x,y]^2 \approx 1 }$? Has there been any work done on these groups? REPLY [6 votes]: Here are some additional details for the answer of M. Farrokhi. In the paper " On a conjecture of Hanna Neumann", B.H.Neumann constructs a certain group $G$. There are generators $a_1,\dotsc,a_4$, and additional elements defined in terms of these as follows: \begin{align*} b_{12} &= [a_1,a_2] = [a_2,a_1] \\ b_{13} &= [a_1,a_3] = [a_3,a_1] \\ b_{14} &= [a_1,a_4] = [a_4,a_1] \\ b_{23} &= [a_2,a_3] = [a_3,a_2] \\ b_{24} &= [a_2,a_4] = [a_4,a_2] \\ b_{34} &= [a_3,a_4] = [a_4,a_3] \\ c_1 &= [a_2,b_{34}] = [a_4,b_{23}] \\ c_2 &= [a_3,b_{14}] = [a_4,b_{13}] \\ c_3 &= [a_4,b_{12}] = [a_1,b_{24}] \\ d &= [a_1,[a_2,[a_3,a_4]] = [a_2,[a_3,[a_4,a_1]] = [a_3,[a_4,[a_1,a_2]] \end{align*} There are some relations implicit in the above equations. There are also additional relations as follows: $a_i^2=b_{jk}^2=c_l^2=d^2=1$ All commutators $[a_i,b_{jk}]$ that have not already been listed, are trivial. $[a_i,c_j]=1$ whenever $i\neq j$, and $[a_i,d]=1$ We can define a map $\phi\colon\{0,1\}^{14}\to G$ by $$ \phi(u) = a_1^{u_1}\dotsb a_4^{u_4} b_{12}^{u_5} \dotsb b_{34}^{u_{10}} c_1^{u_{11}}c_2^{u_{12}}c_3^{u_{13}}d^{u_{14}} $$ One can check that this is bijective, and one can write formulae for the permutations of $\{0,1\}^{14}$ corresponding to right multiplication by the elements $a_i$, $b_{jk}$, $c_l$ and $d$. In particular, this proves that $|G|=2^{14}$. One can also check that $$ [b_{12},b_{34}] = [b_{13},b_{24}] = [b_{14},b_{23}] = d \neq 1, $$ so $G$ is not metabelian. In the paper "On certain varieties of groups", Macdonald states that it is easy to verify that the above group has $[x,y]^2=1$ for all $x,y\in G$. I don't see how to prove this myself. However, I have checked it by computer for 10000 randomly chosen pairs $(x,y)$, so it must be true.<|endoftext|> TITLE: De-Nesting Absolute Value Function into Linear Combination of Absolute Value Functions QUESTION [9 upvotes]: Context: In formulating problems for secondary school mathematics teachers (and students) about absolute value functions, which we define as functions $\mathbb{R} \rightarrow \mathbb{R}$ that send $x \mapsto a|x-h|+k$ for fixed parameters $a, h, k \in \mathbb{R}$, I was able to rewrite the nested absolute value function $$f(x) = \Big||x|-1\Big|$$ as a linear combination of absolute value functions, $$g(x) = |x+1| + |x-1| - (|x|+1)$$ (You can view the graphs of $f$ and $g$ here; although not delved into in this post, my colleagues enjoyed finding similar relationships even when there is a quadratic $x$ term, for example, in the graphs/functions depicted here.) My question is twofold (although the follow-up question depends on the first answer): 1. Is it true that every nested absolute value function (NAVF) or linear combination of NAVFs can be written as a linear combination of AVFs? 2a. If not, what is a counterexample, and what criteria must be satisfied for de-nesting to be possible? 2b. If so, is there an algorithm for de-nesting, i.e., rewriting an arbitrary NAVF as a linear combination of AVFs? Pointers to related literature/references would be welcome, even if they do not explicitly answer the questions above. Please edit the questions, title, or tags if you believe it will improve clarity. REPLY [7 votes]: $\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}}$ Lemma 3 in Explicit additive decomposition of norms (which is Lemma 1.2 in the arXiv version of that note) states the following: Suppose that $f\colon\R\to\R$ is a convex function such that for some real $k$ there exist finite limits \begin{equation*} d_+:=d_{f,k;+}:=\lim_{u\to\infty}[f(u)-ku]\quad\text{and}\quad d_-:=d_{f,k;-}:=\lim_{u\to-\infty}[f(u)+ku]. \end{equation*} Then for all $x\in\R$ \begin{equation*} f(x)=\frac{d_++d_-}2+\frac12\,\int_\R|x-t|\,d f'(t). \tag{1} \end{equation*} As is clear from the short proof, this lemma holds for any absolutely continuous function $f$ with (possibly infinite) limits $\lim_{x\to\pm\infty}f'(x)$. So, it is easy to see by induction on the nesting depth that the lemma holds for any nested absolute value function. Added details: For a finite nesting depth, the function $f$ is piecewise-affine: \begin{multline*} f(x)=(a_1+b_1 x)\ii{x\le t_1}+(a_2+b_2 x)\ii{t_1t_n} \end{multline*} for some natural $n$, some real "switch points" $t_1<\dots TITLE: Some clarifications on Connes' approach to RH QUESTION [13 upvotes]: How serious/promising is Connes' work on the Riemann Hypothesis? Connes is mostly known these days for his work in non-commutative geometry, having previously earned a Fields medal* for his work on the classification of factors. He is a trusted mathematician with deep insights. On the other hand, his essay on RH doesn't seem to identify one clear idea on how to approach the problem. At least, none that I can see. I'm wondering if anybody looked into it. After all, he seems to have been working on RH, one way or another, for twenty years, now. I am mostly curious, specifically, in conceptual reasons (if any) why this approach is, or may be, regarded as "not promising", or "conceptually incorrect", in contrast to the other questions already on MO, that usually inquire about what such approach may or may not be "still missing", or what it already accomplishes. *See Araki's article for the 1983 ICM Proceedings (requires subscription; the entire proceedings are free via https://www.mathunion.org/icm/proceedings REPLY [40 votes]: Answering to this kind of questions is not an easy task, but one can sympathize with the curiosity. Just some personal thoughts. To my mind, an approach to RH could be regarded as "serious" (by which I mean "with hope to at least yield interesting and genuinely new partial or related results") essentially in the sole case it also yields, as a bi-product of the methods, a "cohomological construction" of Hasse-Weil $L$-functions and zeta functions (completed with their respective archimedean local factors as predicted by Serre) for all smooth projective varieties over number fields, much as for Grothendieck's cohomological construction of $L$- and zeta functions of smooth projective varieties over finite fields using geometric $\ell$-adic cohomology. This would be a first hint at how promising such approach can hope to be. We currently lack insight into where such cohomological interpretation could possibly come from, and such accomplishment will probably entail the introduction of some kind of analytic geometry "over the integers", imbuing both complex, real, and $p$-adic analytic geometry, on equal or comparable footing. The literature does contain evidence, at times even explicit and quite strikingly strong, for the existence of such cohomological interpretation to (completed) $L$- and zeta functions, or for why one should expect it. Connes' work seems to suggest non-commutative geometry should somehow provide the (or, maybe more appropriately, "an") answer, although I honestly cannot fathom how (but, probably, that's just me). It seems to me Connes does humbly acknowledge that he doesn't either. I'm not sure to what extent he himself puts trust into his more recent preprints on the matter ("arithmetic and scaling sites" etc. I think). I personally do not believe for a moment, that the datum of a site (essentially a piece of algebra) and as simple minded as those put forth there, can possibly encode deep analytic information such as some arithmetic analog of Hodge-Weil positivity, to which RH should amount. There's also, spread over some literature, a line of thought that seems to suggest that a cohomological interpretation to $L$- and/or zeta functions of smooth projective varieties over number fields could come out of a version of Topological Hochschild Homology, but this really doesn't sound promising either. Even if this were the case, remember that even over finite fields, the crux of the matter was not, in the end, (just) coming up with a robust enough cohomological formalism able to provide spectral interpretations to zeta functions of smooth projective varieties over finite fields. Grothendieck did so, but he, remarkably enough, did not succeed in proving RH anyway. So, betting everything (or much) on the fruitfulness of a cohomology theory in its own right, amounts to me to burying the problem under a pile of formal nonsense in the hope it solves itself on its own, without actually gaining a deeper understanding of it. Back to Connes' work, I believe one thing he's trying to do, by keeping working on it to some extent to this day (which he can do fearlessly, since he's already had a life rather full of remarkable accomplishments) is keeping the interest and fascination into the problem "alive", the risk being, otherwise, that nobody will ever try to ever even touch it, given the mythology surrounding it. Of course, at the end of the day, what we lack of is not a cohomology theory, technical prowess, motivation, or inspiration, but, simply, a real idea.<|endoftext|> TITLE: Picard group and reduced schemes QUESTION [9 upvotes]: $\DeclareMathOperator\Pic{Pic}$If $A$ is a ring, then we know that $\Pic(A)=\Pic(A_\text{red})$, but for a scheme $X$ it is false in general. On the other hand, we have that $\Pic(X)=H^{1}_{et}(X,\mathbb{G}_m)$ and étale cohomology doesn't see the nilpotents, so there should be no difference of the right hand side if we replace $X$ by $X_\text{red}$. How can we accomodate the two? REPLY [6 votes]: Let $f : X \to Y$ be a universal homeomorphism of schemes. Then as you noted above, the pullback functor on (small) étale sites $$\begin{eqnarray} Y_{\text{ét}} &\to& X_{\text{ét}}\\ U &\mapsto& U \times_X Y \end{eqnarray} $$ is an equivalence of categories. In this case, the natural transformations $\text{id} \to f_\ast f^\ast$ and $f^\ast f_\ast \to \text{id}$ are isomorphisms. So now let $X = Y_{\text{red}}$. If $f^\ast \mathbf{G}_{m,Y} = \mathbf{G}_{m, Y_{\text{red}}}$, then it would follow for any etale open $U_{\text{red}}$ over $Y_{\text{red}}$ that $$\mathbf{G}_{m,{\text{red}}}(U_{\text{red}}) = \mathbf{G}_m(U), $$ where $U$ is the unique scheme étale over $Y$ that pulls back to $U_{\text{red}}$ over $Y_{\text{red}}$. The following example shows this is already false for rings: Take $U= Y = \operatorname{Spec} k[\epsilon]/(\epsilon^2)$. Then $\mathbf{G}_m(Y) = k[\epsilon]^\times$ which is evidently not equal to $k^\times$. For instance, $1+\epsilon$ is a unit because $(1+ \epsilon)(1-\epsilon) = 1 - \epsilon^2 = 1$.<|endoftext|> TITLE: Classification of fibrations for classifying spaces $B^2\mathbb{Z}_2$ and $BSO(3)$ or $BO(3)$ QUESTION [5 upvotes]: I am interested in knowing what can we say about the classification of fibrations for classifying spaces $B^2M \equiv B^2\mathbb{Z}_2$ and $BG \equiv BSO(3)$ or $BO(3)$. Here we can take either: $B^2M=B^2\mathbb{Z}_2$ as fibers, and $BG=BSO(3)$ (or $BO(3)$) as base manifold. or $BG=BSO(3)$ (or $BO(3)$) as fibers, and $B^2M=B^2\mathbb{Z}_2$ as base manifold. For example, (A) We consider the Postinikov classes $\omega \in H^3(BG,M)$. This classifies the fibrations of $$ B^2M \hookrightarrow BG_\text{{new}} \to BG. $$ If I understand correctly, the distinct $BG_\text{{new}}$ are determined by the classes of cocycles $\omega \in H^3(BG,M)$. (B) On the other hand, we have $1\to \mathbb{Z}_2 \to O(3) \to SO(3) \to 1$, this gives rise $B\mathbb{Z}_2 \to BO(3) \to BSO(3) \to B^2\mathbb{Z}_2,$ $$ BO(3) \to BSO(3) \to B^2\mathbb{Z}_2, $$ which seems to suggest a fibration of $BO(3)$ over $B^2\mathbb{Z}_2$ (C) On the other hand, we further have $1\to \mathbb{Z}_2 \to SU(2) \to SO(3) \to 1$, this gives rise $B\mathbb{Z}_2 \to BSU(2) \to BSO(3) \to B^2\mathbb{Z}_2,$ $$ BSU(2)\to BSO(3) \to B^2\mathbb{Z}_2, $$ which seems to suggest a fibration of $BSU(2)$ over $B^2\mathbb{Z}_2$ My questions: (a) Do (B) and (C) above really suggest a valid fibration of classifying space fibered over another classifying spaces? How do we classify them? Are there similar classifications like Postinikov classes? (b) The first example (A) and the later (B) and (C) look very different from each other. Are there any relations or generalizations to relate each of them? (c) In general, given $B^2M \equiv B^2\mathbb{Z}_2$ and $BG \equiv BSO(3)$ or $BO(3)$, what are other possible fibrations between them (these classifying spaces)? REPLY [8 votes]: Firstly, yes, your examples are all correct. However, in example~(B) we just have $O(3)=\{\pm I\}\times SO(3)$ as groups, so your fibration is just the product of $$BSO(3)\xrightarrow{1}BSO(3)\to 1$$ with the path-loop fibration $$ B\mathbb{Z}/2\to 1\to B^2\mathbb{Z}/2 . $$ The same happens with $BSO(m)$ and $BO(m)$ whenever $m$ is odd. When $m$ is even, there is no normal subgroup of order two in $O(m)$, so we cannot do an analogous thing. However, for any $m$ we have an extension $SO(m)\to O(m)\to \mathbb{Z}/2$ giving a fibration $BSO(m)\to BO(m)\to B\mathbb{Z}/2$, the other way around from your example. As Dylan mentioned, fibrations with fibre $F$ are classified by maps to $B\text{hAut}(F)$. If $F$ has a group structure then it acts on itself by translation, giving a map $F\to\text{hAut}(F)$. In this context we have a homeomorphism $\text{Map}(F,F)=F\times\text{Map}_*(F,F)$ and similarly $\text{hAut}(F)=F\times\text{hAut}_*(F)$, where $\text{hAut}_*(F)$ is the space of based weak equivalences from $F$ to itself. If $F=K(\mathbb{Z}/2,d)=B^d(\mathbb{Z}/2)$ then for $k>0$ we have $\pi_k(\text{Map}_*(F,F))=[\Sigma^kF,F]=H^d(\Sigma^kF,\mathbb{Z}/2)=0$ (because $\Sigma^kF$ is $(d+k-1)$-connected). Moreover, in this case we have $\pi_0(\text{Map}_*(F,F))=H^d(F;\mathbb{Z}/2)=\{0,1\}$, with the $0$-component consisting of nullhomotopic maps, and the $1$-component consisting of weak equivalences. From this we find that the translation map $F\to\text{hAut}(F)$ is a weak equivalence in this case (as well as being a homomorphism of topological groups). This gives $B\text{hAut}(F)\simeq K(\mathbb{Z}/2,d+1)$, so fibrations with fibre $F$ and base $B$ are classified by maps $B\to K(\mathbb{Z}/2,d+1)$, or in other words by Postnikov invariants in $H^{d+1}(B;\mathbb{Z}/2)$. None of this relies on $B$ being the classifying space of a Lie group, which is a bit of a distraction from the real issues here. Whenever you have a fibration as above, with total space $E$ say, you get a longer sequence $$ \dotsb \to \Omega^2B\to K(\mathbb{Z}/2,d-1)\to \Omega E \to \Omega B \to K(\mathbb{Z}/2,d-1) \to E \to B \to K(\mathbb{Z}/2,d)$$ in which any three adjacent terms give a fibration. In particular, we have some fibrations with Eilenberg-MacLane spaces as the base, and others with Eilenberg-MacLane spaces as the fibre. All of the examples that we have discussed arise in this way, for suitable choices of $B$ and $d$. Given a compact Lie group $G$, we could in principle classify fibrations with fibre $BG$ using $B\text{hAut}(BG)$, which is similar to $\text{Map}(BG,BG)$. If $G$ is an $n$-torus then $BG=K(\mathbb{Z}^n,2)$ and$\text{hAut}(BG)\simeq BG\times GL_n(\mathbb{Z})$ by an argument similar to that given above, so this is not too hard to understand. However, even in the simplest nonabelian case of $G=SU(2)$, the set $[BG,BG]$ is uncountable and has no tractable structure, so this approach is not very useful. We can instead try to produce extensions $G\to P\to Q$ of topological groups, for various $Q$, and then take classifying spaces to produce fibrations $BG\to BP\to BQ$. This will not produce all $BG$-fibrations over $BQ$, but it will produce a class of such fibrations that one can reasonably hope to analyse.<|endoftext|> TITLE: A variational problem - some guidance QUESTION [7 upvotes]: This is a problem I'm thinking about, to learn some more advanced calculus of variations on my own. I would appreciate some help, or a solution, just to have a sample to compare in the future. Let $\Omega$ be a bounded open subset of $\mathbf{R}^N$, $N\geq3$; $a(x)$ an almost everywhere bounded measurable function on $\Omega$ ($0<\alpha\leq a(x)\leq\beta$, with $\alpha,\beta\in\mathbf{R}_{>0}$, for almost all $x\in\Omega$); $1 \frac{p^*}{p^*-1}=p_*$. Finally, if $\text{div}(F)\in L^s(\Omega)$, then $F\cdot\nabla v\in L^1(\Omega)$, whence, since $\nabla v\in (L^{s'}(\Omega))^N$, we have $F\in (L^s(\Omega))^N$, that is, $s=m$. It follows the question is indeed well-posed for $m> p_*$. 2. Existence by minimization In order to determine existence of solutions to the Euler equation: $$\forall v\in W_0^{1,p}(\Omega)\ \ \ J'(u)(v)=0$$ by minimizing the functional $J:W_0^{1,p}(\Omega)\to\mathbf{R}$, we'll use a well known minimization Theorem of Weierstrass asserting existence of a minimum for coercive weakly lower semi-continuous operators on reflexive Banach spaces. Since $W_0^{1,p}(\Omega)$ is Banach and reflexive, we need only prove: (a) $J$ is weakly lower semi-continuous; (b) $J$ is coercive. Proof of (a). We call $g(x,s,\xi):=\frac{1}{p}a(x)|\xi|^p$, a measurable function of $x\in\Omega$, costante (hence continuous) in $s$, continuous, and in fact differentiable, in $\xi$, for almost all $x\in\Omega$. A Theorem of De Giorgi would already yield the contention, but we give a direct argument below anyway. Let $v_n\to v$ weakly in $W_0^{1,p}(\Omega)$. We now use the assumptions on $a$, hence on $g$. Since $g$ is differentiable as a function of $\xi$, we have: $$g(x,s,\xi)=g(x,s,\eta)+\partial_{\xi}g(x,s,\eta)\cdot(\xi-\eta)+O(|\xi-\eta|^2)\geq g(x,s,\eta)+\partial_{\xi}g(x,s,\eta)\cdot(\xi-\eta)$$ Upon integrating both sides, we have: $$\frac{1}{p}\int_{\Omega}a(x)|\nabla v_n|^p\geq\frac{1}{p}\int_{\Omega}a(x)|\nabla v|^p+\int_{\Omega}\partial_{\xi}g(x,s,\nabla v)\cdot(\nabla v_n-\nabla v)$$ $\nabla v_n\to\nabla v$ weakly in $(L^p(\Omega))^N$ by assumption, and: $$\partial_{\xi}a(x)|\nabla v|^p\leq p\beta|\nabla v|^{p-1}\in L^{p'}(\Omega),$$ whence, by the Bounded Convergence Theorem: $$\int_{\Omega}\partial_{\xi}\{a(x)|\nabla v|^p\}\cdot(\nabla v_n-\nabla v)\to 0.$$ We deduce: $$\liminf_n\int_{\Omega}a(x)|\nabla v_n|^p\geq\int_{\Omega}a(x)|\nabla v|^p.$$ Since the term $-\int_{\Omega}F(x)\cdot\nabla u$ is continuous in $u$, the contention follows. QED Proof of (b). This is immediate by the assumptions on $a$: $$a(x,s)\geq\alpha>0,$$ which yields: $$g(x,s,\xi)\geq \alpha|\xi|^p.$$ By the Hölder inequality, we have: $$\int_{\Omega}|F(x)\cdot\nabla u|\leq\left(\int_{\Omega}|F(x)|^s\right)^{1/s}\cdot\left(\int_{\Omega}|\nabla u|^r\right)^{1/r},$$ and upon choosing $s=m$ and $r=m'$, we have: $$\int_{\Omega}|F(x)\cdot\nabla u|\leq\||F|\|_m\cdot\left(\int_{\Omega}|\nabla u|^{m'}\right)^{1/m'}\leq\||F|\|_m\cdot \left(\int_{\Omega}|\nabla u|^{rt}\right)^{1/rt}\cdot|\Omega|^{\frac{t-1}{t}}$$ which is: $$\int_{\Omega}|F(x)\cdot\nabla u|\leq\||F|\|_m\cdot\||\nabla|\|_p\cdot|\Omega|^{\frac{pm-p-m}{pm-p}}.$$ Note that the steps in the foregoing are indeed legitimate, as $m>p^*$. It follows: $$|J(u)|\geq \frac{\alpha}{C_{p,\Omega}}\cdot \|u\|_{W_0^{1,p}(\Omega)}^p-c_p\||f|\|_m\cdot\|u\|_{W_0^{1,p}(\Omega)},$$ by the Poincaré inequality, whence $\alpha/C_{p,\Omega}$ and $c_p$ are constants, depending only on $p$ and $\Omega$. Coerciveness of $J$ follows. QED A weak solution $u\in W_0^{1,p}(\Omega)$ to the problem $(P)$, therefore, exists, in the form of a critical point (a minimum, in fact) of the functional $J$. 3. Uniqueness Assume $u_1$ and $u_2$ in $W_0^{1,p}(\Omega)$ are two solutions. We have: $$J'(u_1)(v)=\int_{\Omega} a(x)|\nabla u_1|^{p-2}\cdot\nabla u_1\cdot\nabla v=\int_{\Omega}F\cdot\nabla v$$ $$J'(u_2)(v)=\int_{\Omega} a(x)|\nabla u_2|^{p-2}\cdot\nabla u_2\cdot\nabla v=\int_{\Omega}F\cdot\nabla v.$$ We choose $u_1-u_2$ as $v$. Upon subtracting on both sides, we get: $$I_p:=\int_{\Omega}a(x)\left\{|\nabla u_1|^{p-2}\nabla u_1-|\nabla u_2|^{p-2}\nabla u_2\right\}\cdot\nabla(u_1-u_2)=0.$$ We treat the case $p\geq 2$, using the inequality: $$(|t|^{p-2}t-|s|^{p-2}s)\geq c_p\cdot(t-s)^{p-1}$$ for a positive constant $c_p$ and all $s,t\in\mathbf{R}$, and using the assumption $a(x)\geq\alpha>0$ (for almost all $x\in\Omega$). We get: $$0=I_p\geq\alpha\cdot c_p\int_{\Omega}|\nabla u_1-\nabla u_2|^{p}\geq\alpha\cdot c_p\cdot C_{p,\Omega}\|u_1-u_2\|_{W_0^{1,p}(\Omega)}$$ from the Poincaré inequality again. It follows $u_1=u_2$ almost everywhere on $\Omega$. The case $1\frac{Np}{Np-2N+p}.$ Upon defining $g(x,\xi)=a(x)|\xi|^{p-2}\xi$, we use $G_k(u)$, with $u$ a solution to $(P)$, as test function. We have, for $J'(u)(G_k(u))$, $$\int_{\Omega_k}g(x,\nabla G_k(u))\cdot\nabla G_k(u)=\int_{\Omega}g(x,\nabla u)\cdot \nabla G_k(u)=\int_{\Omega}F\cdot\nabla G_k(u).$$ We have, on the left side, by ellipticity of $g$, a consequence of the assumptions on $a$, and on the right side, by the Hölder inequality with exponents $(r,s)$: $$\alpha\int_{\Omega}|\nabla G_k(u)|^p\leq\left(\int_{\Omega}\{|F|\cdot \chi_{\Omega_k}\}^r\right)^{1/r}\cdot\left(\int_{\Omega}|\nabla G_k(u)|^s\right)^{1/s}$$ For the right side to make sense, we choose $r=m$, whence $s=m'$, upon observing that if we have $u\in W_0^{1,p}(\Omega)$, then also $G_k(u)\in W_0^{1,p}(\Omega)$ (check!). We now estimate the last integral, again using the Hölder inequality with exponents $(h,k)$: $$\left(\int_{\Omega_k}|\nabla G_k(u)|^{m'h}\right)^{1/m'h}\cdot|\Omega_k|^{\frac{1}{m'k}}.$$ For this to make sense, we choose $h=p/m'$. It follows $$m'k=\frac{m'p/m'}{h-1}=\frac{mp}{mp-p-m},$$ We obtain: $$\alpha\||\nabla G_k(u)|\|_p\leq\||F|\|_m\cdot|\Omega_k|^{\frac{mp-p-m}{mp}}.$$ By Sobolev-Gagliardo-Niremberg, we get: $$\alpha\cdot S_{p,\Omega}\left(\int_{\Omega}|G_k(u)|^{p^*}\right)^{1/p^*}\leq\||F|\|_{m}\cdot|\Omega_k|^{\frac{1}{m'}-\frac{1}{p}}.$$ We again estimate $\left(\int_{\Omega}|G_k(u)|^{p^*}\right)^{1/p^*}$ using the Holder ''backwards'': $$\int_{\Omega}|G_k(u)|\leq\left(\int_{\Omega}|G_k(u)|^{p^*}\right)^{1/p^*}\cdot|\Omega_k|^{\frac{1}{p_*}}.$$ Tying everything together, we get: $$\int_{\Omega}|G_k(u)|\leq C_{p,\Omega}(\alpha)\cdot\||F|\|_m\cdot|\Omega_k|^{\frac{m-1}{m}-\frac{1}{p}+\frac{Np-N+p}{Np}}$$ where $C_{p,\Omega}(\alpha):=(\alpha\cdot S_{p,\Omega})^{-1}$. $C_{p,\Omega}(\alpha)>0$, whence we are left to check: $$\frac{m-1}{m}-\frac{1}{p}+\frac{Np-N+p}{Np}>1,$$ that is: $$\frac{m-1}{m}>\frac{2N-p}{Np}.$$ We have: $$m>\frac{Np}{Np-2N+p}.$$ We observe that for $p=2$ we have $m>N$, as expected. 5. Summarizing (P) is well posed for $m> p_*$, with unique solution $u\in W_0^{1,p}(\Omega)$, and with $u\in W_0^{1,p}(\Omega)\cap L^{\infty}(\Omega)$ for $$m>\frac{Np}{Np-2N+p}>p_*.$$ Remark Since the operator: $$A:W_0^{1,p}(\Omega)\longrightarrow W_0^{-1,p'}(\Omega)$$ assigned by $u\mapsto -\text{div}(a(x)|\nabla u|^{p-2}\nabla u)$ is well posed, and since for $m>p_*$, we have $-\text{div}(F)\in W_0^{-1,p'}(\Omega)$ (by Sobolev-Gagliardo-Niremberg). existence of solutions also follows from the Leray-Lions Theorem. I think another way to see existence is by the Ekeland principle, since we have $J\in C^1(W_0^{1,p}(\Omega),\mathbf{R})$, bounded below, and easily seen to satisfy the Palais-Smale condition. Some references for the Thms used I'll track them back when I have some time. Spare time is over now<|endoftext|> TITLE: Obstructions to simultaneous resolution QUESTION [7 upvotes]: The following question on simultaneous resolutions is a follow-up to earlier questions posed here (e.g. Resolution of singularities for flat families.). What I'm interested in is an "obstruction theory" for simultaneous resolutions. More precisely: Let $f:X\to B$ be a flat projective (for simplicity) morphism of smooth varieties. What are obstructions that prevent the existence of a simultaneous resolution (*) of $f$? If we further simplify and assume that $B\cong \mathbb{A}^n$ for some $n$, then an obvious obstruction is the monodromy (**) of $f^{sm}:X^{sm}\to B^{sm}$, the restriction of $f$ to its smooth locus. In some cases, this obstruction can be "killed" by a base change along a finite branched covering $\tilde{B}\to B$. For example, this is the case for the adjoint quotient $\chi:\mathfrak{g}\to \mathfrak{t}/W$ of a semisimple complex Lie algebra $\mathfrak{g}$ where the base change along the quotient $\mathfrak{t}\to \mathfrak{t}/W$ is sufficient (Grothendieck's simultaneous resolution). Is this the only obstruction to the existence of simultaneous resolutions or are there more refined ones? Does it matter if we work in the algebraic or analytic category? Edit: For simplicity, let us assume that we work over $\mathbb{C}$ but feel free to make comments/give answers for characteristic $p$. To further clarify my question: (*) A simultaneous resolution of $f:X\to B$ is a pair $(\tilde{f},g)$ of a morphisms $\tilde{f}:\tilde{X}\to B$ and a $B$-morphism $g:\tilde{X}\to X$ such that the induced morphism $g_b:\tilde{X}_b\to X_b$ between the (scheme-theoretic) fibers $\tilde{X}_b=\tilde{f}^{-1}(b)$ and $X_b=f^{-1}(b)$ is a resolution of singularities ($b\in B$). (**) More precisely, what I had in mind: The monodromy of the local systems $R^kf^{sm}_*\mathbb{Z}$. REPLY [5 votes]: I am writing my comments as a solution. Assume that $k$ has characteristic $0$, for simplicity. First I will make precise one interpretation of "vanishing monodromy around the discriminant". Denote by $\Delta$ the complement of $B^{\text{sm}}$ in $B$. For simplicity, assume that $\Delta$ has pure codimension $1$ in $B$ (components of $\Delta$ that have codimension $\geq 2$ have no "local monodromy" by the Purity Theorem for fundamental groups). For every irreducible component $\Delta_i$ of $\Delta$, for the generic point $\eta_i$ of $\Delta_i$, for the strict Henselization of the local ring $$R_i:=\mathcal{O}^{\text{sh}}_{B,\eta_i},$$ the (profinite) Galois group of the fraction field of this local ring, $$\widehat{\pi}^{\text{inertia}}_1(B^\text{sm};\overline{\eta}_i):= \text{Aut}(\overline{\text{Frac}(R_i)}/\text{Frac}(R_i)),$$ equals the profinite completion of $\mathbb{Z}$ (this is essentially the Newton-Puiseux Theorem), i.e., it is procyclic. If $k$ equals $\mathbb{C}$, for a general $\mathbb{C}$-point $p_i$ of $\Delta_i$, for a transverse slice $N$ to $\Delta_i$ at $p_i$, for the punctured neighborhood $N^*=N\setminus\{p\}$, this procyclic group corresponds to the profinite completion of $\pi_1(N^*)\cong \mathbb{Z}$. Denote by $X_{\overline{\eta}_i}$ the fiber product of $X\to B$ and $\text{Spec}(\overline{\text{Frac}(R_i)}) \to B.$ Then the profinite Galois group $\text{Aut}(\overline{\text{Frac}(R_i)}/\text{Frac}(R_i))$ acts continuously on the etale cohomology of the $\overline{\text{Frac}(R_i)}$-scheme $X_{\overline{\eta}_i}$. If there is a modification of $X$ that makes $f$ smooth over $\eta_i$, then this action is trivial. This is what is usually meant by vanishing monodromy on cohomology of the fiber. If $k$ equals $\mathbb{C}$, then (via the usual comparison theorems) this is equivalent to saying that the linear representation of $\pi_1(N^*,q)$ on the singular cohomology of the fiber $X_q$ over $q\in N^*$ is a trivial representation. Vanishing of monodromy on cohomology of the fiber is not sufficient for the existence of a modification. Here is one example. Let $B$ be the projective plane $\text{Proj}\ k[r,s,t]$. Let $\mathbb{P}^2_k$ be the projective plane $\text{Proj}\ k[u,v,w]$. Let $X$ be the closed subscheme of $B\times \mathbb{P}^2_k$ with defining equation, $$f=u^2r+v^2s+w^2t.$$ Then $X$ is smooth, and the projection $f$ to $B$ is flat. Moreover, the fiber $X_{\overline{\eta}_i}$ is just $\mathbb{P}^1_{\overline{\text{Frac}(R_i)}}$. The étale cohomology of this is straightforward: $H^q(X_{\overline{\eta}_i},\widehat{\mathbb{Z}}_\ell)$ equals $\widehat{\mathbb{Z}}_\ell$ for $q=0$, it is zero for $q=1$, and it equals $\widehat{\mathbb{Z}}_\ell(-1)$ for $q=2$. Of course the action of monodromy on $H^0$ and $H^1$ is trivial. For $q=2$, observe that the first Chern class of the relative dualizing sheaf is a nonzero element that is preserved under monodromy. Since $H^2$ is procyclic and the action is continuous, it follows that the monodromy also acts trivially on $H^2$. However, there is no modification of $X$ that makes $f$ smooth over the generic points $\eta_i$ of the three coordinate axes in $B$. If there were such a modification, then by Zariski glueing, there exists a modification of $X$ that is smooth over a dense Zariski open $B^o$ in $B$ containing both $B^{\text{sm}}$ and containing the generic point $\eta_i$ of each irreducible component $\Delta_i$. Thus, the complement of $B^o$ is a closed subset of $B$ that has codimension $\geq 2$. By the Purity Theorem for Brauer groups, Théorème 6.1 of the following, the restriction map from the Brauer group of $B$ to the Brauer group of $B^o$ is an isomorphism. MR0244271 (39 #5586c) Grothendieck, Alexander Le groupe de Brauer. III. Exemples et compléments. (French) Dix exposés sur la cohomologie des schémas, 88–188, Adv. Stud. Pure Math., 3, North-Holland, Amsterdam, 1968. The Brauer group of $B$ is zero. This implies that the Brauer class of $X^{\text{sm}}/B^{\text{sm}}$ is zero, and thus there exists an element of the Picard group of $X^{\text{sm}}$ whose base change to $X_{\overline{\eta}_i} \cong \mathbb{P}^1_{\overline{\text{Frac}(R_i)}}$ is a generator of the Picard group. Since $X$ is smooth, every divisor class on $X^{\text{sm}}$ is the restriction of a divisor class on $X$. The projection from $X$ to $\mathbb{P}^2_k$ is the projective bundle associated to a locally free sheaf of rank $2$, from which it is straightforward to compute the Picard group of $X$. There is no divisor class on $X$ whose base change to $X_{\overline{\eta}_i}$ is a generator of the Picard group. This contradiction proves that there is no modification of $X$ that is smooth over a generic point $\eta_i$ of $D_i$.<|endoftext|> TITLE: Is a Sobolev map with smooth minors smooth on the whole domain? QUESTION [7 upvotes]: Let $d\ge 3$ and $2 \le k \le d-1$ be integers, where at least one of $k,d$ is odd. Let $\Omega \subseteq \mathbb{R}^d$ be open, and let $f \in W^{1,p}(\Omega,\mathbb{R}^d)$, for some $p \ge 1$. Question: Suppose that $\det df>0$ a.e. and that $\bigwedge^k df $ is smooth. Is $f$ smooth? $f$ must be smooth on $\Omega_0=\{ x \in \Omega \, | \, \bigwedge^k df_x \in \text{GL}(\bigwedge^{k}\mathbb{R}^d) \}$, which is an open subset of full measure on $\Omega$. I ask if $f$ is smooth on all $\Omega$. A proof sketch for regularity on $\Omega_0$: (for the full details see theorems 1.1 and 1.2 here). On $\Omega_0$, we can smoothly reconstruct $df$ from $\, \,\bigwedge^k df$: If $A,B \in \text{GL}^+(\mathbb{R}^d)$ and $\bigwedge^k A=\bigwedge^k B$, then $A=B$. The exterior power map $\psi: A \to \bigwedge^k A$ is a smooth embedding when considered as a map $\text{GL}^+(\mathbb{R}^d) \to \text{GL}(\bigwedge^{k}\mathbb{R}^d)$; $\text{Image} (\psi)$ is a closed embedded submanifold of $\text{GL}(\bigwedge^{k}\mathbb{R}^d)$, which makes $\psi:\text{GL}^+(\mathbb{R}^d) \to \text{Image} (\psi)$ a diffeomorphism. Composing $x \to \bigwedge^k df_x$ with the smooth inverse of $\psi$ finishes the job. Without the invertibility assumption $\bigwedge^k df \in \text{GL}(\bigwedge^{k}\mathbb{R}^d)$ everywhere on $\Omega$, we cannot inverse the map $df \to \bigwedge^k df$ on all $\Omega$ (as the rank can fall on a subset of measure zero). Thus, I don't see how to advance beyond the "good" set $\Omega_0$. Edit: If $k$ is odd, I can say a bit more under additional assumptions: Suppose that $f \in W^{1,\infty}(\Omega,\mathbb{R}^d)$, and that $\text{rank}(\bigwedge^k df)>k$ everywhere on $\Omega$ and that $\bigwedge^k df$ is smooth. Then $f$ is smooth. Proof sketch: Since $\text{rank}(\bigwedge^k df)=\binom{\text{rank}(df)}{k}$, $\text{rank}(\bigwedge^k df)>k$ implies $\text{rank}(df) > k$ a.e., and in this regime the map $df \to \bigwedge^k df$ is smoothly invertible. (This relies upon the pointwise algebraic fact that the exterior power map $A \to \bigwedge^k A$ is invertible when restricting the domain to the set of matrices of rank larger than $k$). For full details, see theorem 5.10 here. Comment: I don't think that (naive) quantitative estimates are possible here: Small $\|\bigwedge^k df\|$ does not imply small $\|df\|$; if one singular value of $df$ is $n$, and all the rest are $\frac{1}{n}$, then all the singular values $\bigwedge^k df$ are bounded below by $1$ (If $k>1$). REPLY [2 votes]: The theorem quoted below is from: Z. Liu, J. Malý, A strictly convex Sobolev function with null Hessian minors. Calc. Var. Partial Differential Equations 55 (2016), no. 3, Art. 58, 19 pp. This is not really a counterexample to the question above since we relax the assumption about the determinant. The determinant of the mapping is $\det df= 0$ a.e. On the other hand the mapping is a homeomorphism so intuitively it is very close to a mapping with Jacobian positive a.e. Theorem. Given $1\leq p TITLE: A one-dimensional integral minimization problem QUESTION [6 upvotes]: Let $\mathscr F$ be the collection of smooth functions $f \colon \mathbb R \to \mathbb R$ such that $f \in C^\infty_c(\mathbb R)$, with $\text{supp } f \subset [-1,1]$; $\int_0^1 x f(x) dx = \frac{1}{2\pi}$. I would like to compute $$ \inf_{\mathscr F} \int_0^1 \vert f^\prime (x) \vert x^2 \, dx. $$ My (last) hope is that this infimum is $0$ but I do not manage neither to prove it nor to disprove it. The point is that the functional I am trying to minimize is invariant under rescaling, in the sense that given $f \in \mathscr F$ the functional computed on $f$ and on $f_\varepsilon(x) = \varepsilon^{-3} f(\frac{x}{\varepsilon})$ (for every $\varepsilon>0$) has the same value. What is this telling? REPLY [8 votes]: $\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}}$ This infimum is $\frac1\pi$. Indeed, integrating by parts, for any $f\in\F$ we have \begin{equation} \frac1{2\pi}=I(f):=\int_0^1 x f(x) dx =-J(f)/2,\quad J(f):=\int_0^1 x^2 f'(x) dx, \end{equation} whence \begin{equation} H(f):=\int_0^1 x^2 |f'(x)| dx\ge|J(f)|=\frac1\pi, \end{equation} so that the infimum in question is $\ge\frac1\pi$. On the other hand, for any $\ep\in(0,1)$ let \begin{equation} f_\ep:=c_\ep g_\ep,\quad g_\ep(x):=\frac x\ep\,1_{(0,\ep]}(x)+\frac{1-x}{1-\ep}\,1_{(\ep,1)}(x), \end{equation} where $c_\ep$ is such that $I(f_\ep)=\frac1{2\pi}$, so that $c_\ep\to3/\pi$; the convergence everywhere here is for $\ep\downarrow0$. Then $H(f_\ep)\to\frac1\pi$. It remains to approximate in $L^1[0,1]$, however closely, both $f_\ep$ and $f'_\ep$ by $f$ and $f'$ for some $f\in\F$; this can be done by using, say, the convolution of $f_\ep$ (or, rather, of a close version of $f_\ep$ with support on $(0,1)$) with a smooth kernel and the multiplication by a constant factor close to $1$. So, the infimum in question is indeed $\frac1\pi$.<|endoftext|> TITLE: Erroneous proof of recursion theorem examples QUESTION [10 upvotes]: In his book Elements of Set Theory, Herbert Enderton defines (p. 70) a Peano system as a triple $(N, S, e)$ where $N$ is a set, $S$ is an $N$-valued function defined on $N$ and $e$ is a member of $N$ such that $e \notin S(N)$; $S$ is one-to-one; Any subset $A \subset N$ that contains $e$ with $S(x)\in A$ whenever $x \in A$ equals $N$. Then, on page 76 he has the following cryptic remark: There is a classic erroneous proof of the recursion theorem that people have sometimes tried (even in print!) to give. Of course, he gives some explanation of why these "proofs" are wrong (they do not use all the three conditions given above) but anyway: can someone give some concrete examples that Enderton could have in his mind of such an flawed proof? REPLY [3 votes]: An example, citing Saunders Mac Lane's 'Mathematics Form and Function', is 'Natural Numbers, Integers, and Rational Numbers (Following MacLane)' by Alexander Nita, http://math.colorado.edu/~nita/Numbers.pdf, page 12. I did not check what Mac Lane said, though. Addendum 1 (2021-03-17): Mac Lane (1986 edition, pages 43 - 46) omits the use of several of the Peano axioms in his proof, hence furnishing an erroneous proof. Another more subtle example: 'An Innocent Investigation' by D. Joyce, http://homepages.math.uic.edu/~kauffman/EvenOddJoyce.pdf . Here, for the definition of even/odd it is shown that every number besides 1 has a predecessor. It would be necessary, however, to show that it has exactly one predecessor (injectivity of the successor map). In a model with a loop, consisting of 1,2,3,... where the successor of 7 is 3, it is not possible to define the even/odd property recursively for 3 because this number has the predecessors 2 and 7. Addendum 2 (2021-03-17): An erroneous proof, as of pre-1960, can be found in the brillant article 'On mathematical induction' by Leon Henkin (page 327, in the paragraph starting with 'Clearly (the argument goes), ...'). The (valid) proof methods explained in the article go back to Kalmar (construction by partial functions), Lorenzen (intersections of relations) and Landau (definition of addition and multiplication, with credit to Kalmar).<|endoftext|> TITLE: Consequences of foundation/regularity in ordinary mathematics (over ZF–AF)? QUESTION [11 upvotes]: This is a followup question to Does foundation/regularity have any categorical/structural consequences, in ZF? As shown in answers to that question, the axiom of foundation (AF, aka regularity) has consequences over ZF–AF that can be formulated purely in structural terms (i.e. using sets only in ways that respect isomorphism). In particular, it implies the statement “Every set can be embedded into a set equipped with a well-founded extensional relation”, which Mike Shulman has called the axiom of well-founded materialisation; and (by the arguments in that question) this axiom should have the same structural consequences as foundation. This statement is a very mathematically reasonable one. However, I don’t think I have ever seen this or anything similar appearing in practice in ordinary mathematics — only in logical investigations like this one, where it occurs specifically because of its relation to AF. So my question here is: Does the axiom of foundation have any consequences (over ZF–AF) that have appeared in ordinary mathematics? Slightly more precisely: By “consequence of foundation over ZF–AF”, I of course mean principally a statement that’s provable in ZF, but not in ZF–AF. But I’d also be interested in e.g. a proof in ZF that doesn’t go through in ZF–AF and can’t be trivially modified to do so. By “ordinary mathematics”, I mean… this of course is subjective, and mostly I think “we know it when we see it”, and hope that most of us will find our judgement on it mostly agrees. But to articulate my own intent here a little: a sufficient (but not necessary) condition to would be that the statement has been considered in a field of mathematics other than logic; necessary conditions would be that it’s purely structural, and that it’s been considered outside work specifically investigating foundation (or related axioms) or modelling set theories that include foundation. To illustrate the non-necessity of the “outside logic” criterion — I would certainly be happy to consider consequences in e.g. general classical model theory. See e.g. Where in ordinary math do we need unbounded separation and replacement? for a similar question, about different axioms, and relevant discussion of what should be considered “ordinary mathematics”. REPLY [7 votes]: What about Scott's trick? In more detail: here, I claim, is a proof-schema in ZF that belongs to "ordinary mathematics" but cannot be trivially modified to go through in ZF–AF. Let $C$ be a locally-large category, by which I mean a proper class of objects, a proper class of morphisms, class-functions for domain, codomain, composition, and identities, satisfying the axioms of a category. (Since a proper class in ZF is specified by a logical formula, this is why the proof is a schema rather than a single one.) Let $W$ be a subclass of the morphisms of $C$. Then we can construct the localization $C[W^{-1}]$ as another locally-large category, as described here: we consider the (large) directed graph whose edges are the morphisms of $C$ and the morphisms of $W^{\rm op}$, generate the free (large) category on it, and then quotient by an appropriate equivalence relation. The free large category on a large directed graph is unproblematic even in ZF–AF: its morphisms are finite lists of composable edges, and we can define a formula that specifies the proper class of finite sequences of elements of some other proper class. But there is no obvious way to form the quotient of a generic proper class by a proper-class equivalence relation in ZF–AF: the usual construction of the quotient of a set by an equivalence relation takes the elements of the quotient to be equivalence classes, but if the "equivalence classes" are proper classes then they cannot be elements of some other class. Scott's trick is to instead define the elements of the quotient to be the sub-sets of the equivalence proper-classes consisting of all their elements of minimal rank, which are sets since each $V_\kappa$ is a set. But without the axiom of foundation, this doesn't work since we can't assert that each equivalence proper-class contains any well-founded elements.<|endoftext|> TITLE: Covering number of Lipschitz functions QUESTION [6 upvotes]: What do we know about the covering number of $L$-Lipschitz functions mapping say, $\mathbb{R}^n \rightarrow \mathbb{R}$ for some $L >0$? Only 2 results I have found so far are, That the $\infty$-norm covering number for $L$-Lipschitz functions constrained to map $[0,1]^d \rightarrow [0,1]$ is $\exp\left(\Theta\left( L/\epsilon\right)^d\right)$. And for this I could not find a reference for the proof. Another such $\infty$-norm covering number count for $1$-Lipschitz functions mapping an unit diameter metric space to $[-1,1]$ was given in this previously unanswered question here, metric entropy for Lipschitz functions Even here I can't find the proof anywhere and the link in the MO question seems to go to something else (and the original reference in the Uspekhi Mat. Nauk, 1959, Volume 14, Issue 2(86), Pages 3–86, is in Russian and hence I can't read it). Is there some reason why we don't have (is provably impossible?) such counts when either the range or the domain of the function space is unbounded? It would be great if someone can maybe reference the proofs for the two results quoted here! REPLY [4 votes]: Here is a reference to a more general result: Lipschitz functions over a doubling metric space (rather than $[0,1]^d$). The $||\cdot||_\infty$ $\epsilon$-metric entropy of such functions is, disregarding log factors, of order $(D/\epsilon)^{ddim}$, where $D$ and $ddim$ are the diameter and doubling dimension of the metric space, respectively. A modern proof may be found in Lemma 4.2 here: https://www.sciencedirect.com/science/article/pii/S0304397515009469 Edit: As pointed out in the comments, the result above is stated without proof. However, a proof is given in Lemma 6 here: https://ieeexplore.ieee.org/document/7944658/<|endoftext|> TITLE: Do all finite-dimensional division algebras appear as Wedderburn factors of rational group rings? QUESTION [5 upvotes]: Suppose that $D$ is a division algebra that is finite-dimensional over $\Bbb Q$, does there exist a finite group $G$ such that one of the factors in the Wedderburn decomposition of $\Bbb Q[G]$ is a matrix ring over $D$? (Note that the answer with general base field is no, as there are countably many finite groups and each group algebra has only finitely factors in the Wedderburn decomposition, so the division algebras which appear as Wedderburn factors are countable, thus any field with uncountably many finite extensions is a counterexample) REPLY [10 votes]: The subgroup of the Brauer group generated by (and in fact, consisting of) such division algebras is called "the Schur group". Brauer-Witt theorem asserts that it is given by cyclotomic algebras, so the answer is negative. I learned this here.<|endoftext|> TITLE: UFD containing element with finite quotient QUESTION [13 upvotes]: Since this question could not be answered at math.stackexchange, I would like to try my luck here now: Does anyone know an example of a unique factorization domain $R$ that is (i) not a Dedekind domain (or equivalently, not a principal ideal domain) and (ii) contains some irreducible element $r \in R$ such that the quotient $R/rR$ is finite? I am grateful for any suggestions. REPLY [4 votes]: This is an attempt to give a more specific and concrete version of Neil Epstein's answer. The initial version was incorrect, and the current version is incomplete. Let $k$ be a finite field, and fix a sequence of polynomials $u_i(x)\in k[x]$. For $1\leq n<\infty$ put $$ R_n = k[x,y_0,\dotsc,y_n]/(y_i = x(y_{i+1}+u_i(x))+1), $$ then let $R_\infty$ be the colimit of the rings $R_n$. Because the relations give $y_i$ in terms of $y_{i+1}$ we just have $R_n=k[x,y_n]$ for $n<\infty$, and this is a UFD. We have $y_i=1\pmod{x}$ for all $i$ so $R_\infty/x=k$. On the other hand, we have $y_i=y_0-i\pmod{x-1}$ for all $i$, so $R_\infty/(x-1)=k[y_0]$. From this it follows that the ideal $(x-1,y_0)$ cannot be principal. Thus, if we can prove that $R_\infty$ is a UFD, then we are done. One can check that if $f$ is irreducible in $R_n$ and does not lie in $R_n.(x,y_n-1)$ then $f$ remains irreducible in $R_{n+1}$. Initially I had hoped to take $u_i(x)=0$ for all $i$. However, in this case we find that the elements $p_n=(1-x)y_n-1\in R_n$ satisfy $p_n=x\,p_{n+1}$ in $R_{n+1}$, and it follows that $p_0$ cannot be factored as a product of irreducibles in $R_\infty$. I still think (by comparison with the details of the example mentioned by Neil Epstein) that it should be possible to produce an example by choosing the polynomials $u_k(x)$ appropriately, possibly as $u_k(x)=x^{m_k}$ for some rapidly increasing sequence $m_k$, perhaps $m_k=k!$. The point is that a certain power series defined in terms of the numbers $m_k$ should be transcendental over $k(x)$. However, I have not understood all the details yet.<|endoftext|> TITLE: Examples of Baire Class $\xi+1$ but not $\xi$ functions for each countable ordinal $\xi.$ QUESTION [8 upvotes]: We say that $f:\mathbb{R}\to\mathbb{R}$ is of Baire Class $1$ if it is a pointwise limit of a sequence of continuous functions. One can generalize the definition above by taking pointwise limit of each 'previous' level(s) to obtain 'next' level. More precisely, For any countable ordinal $\xi\geq 1,$ we say that $f:\mathbb{R}\to\mathbb{R}$ is of Baire Class $\xi$ if it is a pointwise limit of a sequence of Baire Class $\zeta$ functions where $\zeta<\xi.$ The generalization covers the case $\xi=1$ as every continuous function is of Baire Class $0.$ When $\xi = 2,$ it is well-known that $\chi_\mathbb{Q}$ is of Baire Class $2$ as it is a pointwise limit of $(g_n)_{n=1}^\infty$ where $g_n(x) = \max\{0,1-n d(x,K)\}$ and $K$ is a finite collection of rationals. (extracted from Wiki) Since $\chi_\mathbb{Q}$ is discontinuous everywhere, so it is not of Baire Class $1.$ This MSE post also contains other Baire Class $2$ functions. However, I fail to obtain any Baire Class $3$ function and above. Question: For each countable ordinal $\xi\geq 3,$ what are some examples of Baire Class $\xi+1$ but not $\xi$ function by using the pointwise limit definition? REPLY [5 votes]: A somewhat concrete example of function which is Baire class $\zeta$ but not Baire class $\gamma$ for any $\gamma < \zeta$ is the $\zeta$-th Turing jump. This is essentially the iterated version of Shoenfield's limit lemma. The (iterated) Turing jump naturally gives us a function $J : \{0,1\}^\omega \to \{0,1\}^\omega$, but of course we can embed $\{0,1\}^\omega$ into $\mathbb{R}$ as the Cantor middle set and extend the function, if we prefer to work with $\mathbb{R}$. Another approach is to piggy-back on the non-collapse of the Borel hierarchy: By the Banach-Lebesgue-Hausdorff theorem, Baire class $\zeta$ and the $\Sigma^0_{1+\zeta}$-measurable functions coincide. So the characteristic function of a $\Sigma^0_{1+\zeta}$-complete set is Baire class $\zeta+1$, but not Baire class $\zeta$. The example given in the question is of course a special case of the second procedure. A general construction of $\Sigma^0_{1+\zeta}$-complete sets is found eg in this answer<|endoftext|> TITLE: What is the dimension of the mathematical universe? QUESTION [29 upvotes]: Forcing construction in set theory leads to a new understanding of the mathematical (multi)universe by providing a machinery through which one can construct new models of the universe from the existing ones in a fairly controlled and comprehensible way and connect them to one another through forcing extensions. Since the very beginning, the true nature of this rather bizarre construction method has been a matter of debate. A widely accepted (?) analogy emphasizes on a degree of similarity between forcing method in set theory and field extensions in Galois theory where one constructs new fields using the existing ones by adding new elements. Besides some papers and textbooks, a good description of this analogy could be found in this MathOverflow good oldie: Forcing as a new chapter of Galois Theory?. However, as Dorais mentioned in his answer, despite the undeniable similarity, this analogy is not as perfect as what it seems at the first glance. So, forcing is not a one hundred percent field extension-like construction anyway. This fact causes some confusion concerning the possible translation of Galois theory machinery into the language of forcing and set theory. It is often not immediately clear what the forcing analogy of a given notion in the field extensions would be. Also, sometimes there is more than one approach towards defining a corresponding Galois theoretic notion in forcing so that one needs to decide which the natural one is. One such central concept in Galois theory is the notion of the degree of a field extension, that is the dimension of the extended field while viewed as a vector space over the ground field. The question that arises here is whether there is any corresponding well-defined, well-behaved and natural similar notion in set-theoretic forcing. One may think if a forcing extension could be viewed as a kind of vector space-like structure over the ground model. In this sense, there might be a kind of basis associated with any such pair of models which itself may satisfy some uniqueness properties that give rise to the existence of a well-defined notion of (relative) dimension of a generic forcing extension with respect to its ground model. Our axiomatic expectations of the possible behavior of any such notion of basis/dimension is also an interesting topic to explore even before defining any such notion. Of course, we need them to behave as natural as possible and resemble their Galois theoretic counterparts fairly closely. Question. What are examples of defining a notion of dimension or basis for forcing generic extensions in the set-theoretic literature (possibly close to the same fashion that exists for the field extensions and vector spaces)? I couldn't find much along these lines in the literature except a short unpublished note of Golshani in which he takes a Galois theoretic approach towards dimension in forcing by dealing with mutually generic sequences and the chain of forcing extensions. It also shares many features with Hamkins' set-theoretic geology project. However, this notion of forcing dimension seems not to be complete enough to cover all aspects of a vector space view towards generic extensions but could be a really good starting point anyway. Remark. As Peter stated in his comment, absoluteness would be an issue while defining a notion of (relative) dimension in the set-theoretic multiverse. One may ask, from whose perspective are you trying to calculate the relative dimension of two set-theoretic universes and why? In fact, due to the highly contradictory views of different models of $\sf ZFC$ towards anything beyond the narrow scope of universally absolute properties, there is not more than a little hope to obtain an absolute notion of forcing dimension after all. However, one may argue that absoluteness is not a crucial condition in the long list of our expectations from a possible natural definition of dimension for generic models. The same situation happens in the Hamkins' Well-foundedness Mirage Principle stating that "Every universe $V$ is non-well-founded from the perspective of another universe" and so there is no standard model of $\sf ZFC$. Update. According to the Joel and Mohammad's answers, it turned out that there is more than one approach towards developing a dimension theory for forcing extensions; each with their own characteristics. While Joel's definition requires all forcing dimensions to be infinite, Mohammad's approach allows finite dimensions as well as infinite ones. Also, as Monroe pointed out both definitions fail to satisfy the so-called downwards closure property. In general, Joel's approach sounds a little bit vector space-like to me while Mohammad's reminds me the Krull's dimension in commutative algebra. There is also a chance that these different dimensions be related or coincide with each other under certain circumstances. The point is that in the absence of a general common sense about the expected behavior of a nice forcing dimension, it is not immediately clear whether the mentioned features are flaws or advantages of the presented definitions. Maybe one needs to fix some abstract list of required properties for any such dimension operator and then search for its existence in the forcing extensions. Anyway, if there is any suggestion for such an axiomatic approach towards forcing dimension, I will be so happy to hear about it. REPLY [12 votes]: Let me first give the motivations of the definitions given in my paper. So suppose $F \subseteq K$ be fields. Note that $[K:F],$ the dimension of $K$ over $F$ considered as a vector space is equal to $\kappa$ iff $\kappa$ is maximal length of a chain $F=F_0 \subset F_1 \dots F_\kappa=K$, where each $F_i$ is a vector space over $F$. My definition of dimension is essentially taken from this fact. The note is that based on Hamkins definition, we don't have finite dimension, while in my case for example a one dimenional extension of a model of ZFC is just a minimal generic extension, which in my opinion seems reasonable. Also note that if $[K:F]=\kappa$  and if $\lambda < \kappa,$ then there exists a vector space $G$ over $F$ with $F \subset G \subset K$ and $[G:F]=\lambda.$ However this is not true in general nor for the Hamkins definition nor for my definition. Also my definition allows to consider pairs $V \subset W$, where $W$ is not necessarily a set forcing extension of $V$, and for this reason I have defined the upward generic dimension and the downward generic dimension, the latter being related to set theoretic geology. Another important notion is the notion of independence.Note that $x, y \in K$ are independent over $F$ iff $F[x] \cap F[y]=F$ (where for $X \subseteq K,$ $F[X]$ is the vector space generated by $F \cup X$). Note that a subset $X$ of $K$ is independent over $F$ iff for any finite disjoint subsets $X_1, X_2$ of $X$ we have $F[X_1] \cap F[X_2]=F$. On the other hand, using a theorem of Solovay, it is natural  for $V \subset W$ and for $x, y \in W$ to say that $x$ and $y$ are independet over $V$ if they are mutually generic. This is  my motivation for $\aleph_0$-transendence degree between models of set theory, of course in this case we can define $\kappa$-transendence degree for any $\kappa \geq \aleph_0.$   Note that as it is stated in the comments, in general we don't need for example $x$ and $y$ to be mutually generic over a model $V$ to have  $V[x] \cap V[y]=V$, however the assumption of they being mutually generic has some advantages, for example we can form $V[x, y]$ and it is again a model of ZFC, while in general this may not be the case. Regarding your update, I don't see any essential similarities between these two concepts: 1) Suppose $\mathbb{P}$ is a forcing notion of size $\kappa$ which gives a minimal generic extension of the universe and let $\mathbb{Q}=Add(\omega, \kappa).$ Then: (1-a) In the sense of Hamkins definition $[V^{\mathbb{P}}, V]=[V^{\mathbb{Q}}, V]=\kappa$. (1-b) In the sense of my definition $[V^{\mathbb{P}}, V]=1$ while $[V^{\mathbb{Q}}, V]=\kappa^+$. 2) Suppose $\mathbb{P}$ and $\mathbb{Q}$ gives minimal generic extensions of the universe and are of size $\kappa$ and $\lambda$ respectively where $\kappa < \lambda$. Then (2-a) In the sense of Hamkins definition $[V^{\mathbb{P}}, V]=\kappa$ and $[V^{\mathbb{Q}}, V]=\lambda$. (2-b) In the sense of my definition $[V^{\mathbb{P}}, V]=[V^{\mathbb{Q}}, V]=1$.<|endoftext|> TITLE: Is this inequality on sums of powers of two sequences correct? QUESTION [27 upvotes]: Let $a_1,\dots,a_n$ and $b_1,\dots,b_n$ be two sequences of non negative numbers such that for every positive integer $k$, $$ a_1^k+\cdots+a_n^k \leq b_1^k+\cdots+b_n^k,$$ and $$a_1+\cdots+a_n = b_1+\cdots+b_n.$$ Can we conclude $$\sqrt{a_1}+\cdots+\sqrt{a_n}\geq \sqrt{b_1}+\cdots+\sqrt{b_n}$$ REPLY [22 votes]: As a counter-example with $n=3$ $$a_1=6, a_2=42,a_3=52$$ $$b_1=12, b_2=22, b_3=66$$ Then $a_1+a_2+a_3 = 100 = b_1+b_2+b_3$ $a_1^p+a_2^p+a_3^p \lt b_1^p+b_2^p+b_3^p$ for $p \gt 1$ $\sqrt{a_1}+\sqrt{a_2}+\sqrt{a_3} \lt 16.2 \lt \sqrt{b_1}+\sqrt{b_2}+\sqrt{b_3}$ though notice that $a_1^p+a_2^p+a_3^p \gt b_1^p+b_2^p+b_3^p$ for $0.851 \le p \lt 1$, and I suspect all such counterexamples with $n=3$ reverse the inequality in a small interval below $1$ Added Perhaps a more interesting counter-example is $$a_1=1, a_2=4,a_3\approx 5.3931524748543$$ $$b_1=2, b_2=2, b_3\approx 6.3931524748543$$ where $ 6.3931524748543$ is an approximation to the solution of $x^x=16 (x-1)^{x-1}$, so $\sum a_i = \sum b_i$ and $\prod a_i^{a_i} = \prod {b_i}^{b_i}$ This has $$a_1^p+a_2^p+a_3^p \le b_1^p+b_2^p+b_3^p$$ for all non-negative real $p$ (integer or not), and equality only when $p=0$ or $1$<|endoftext|> TITLE: Do $\mathbb{HP}^2\#\overline{\mathbb{HP}^2}$ and $\mathbb{OP}^2\#\overline{\mathbb{OP}^2}$ arise as sphere bundles over spheres? QUESTION [11 upvotes]: Recall that $\mathbb{RP}^2\#\mathbb{RP}^2$ is the Klein bottle and can be seen as a non-trivial $S^1$-bundle over $S^1$. In particular, it is the total space of the sphere bundle of $\gamma\oplus\varepsilon^1$ over $S^1$ where $\gamma$ is the Möbius line bundle and $\varepsilon^1$ is the trivial line bundle; note, $\gamma\oplus \varepsilon^1$ is the unique non-trivial rank two vector bundle on $S^1$ (because $\pi_0(O(2)) = \mathbb{Z}_2$) and it is not orientable (i.e. $w_1(\gamma\oplus\varepsilon^1) \neq 0$) as is $K$. Also $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$ can be seen as a non-trivial $S^2$-bundle over $S^2$. One way to do this is to view it as the first Hirzebruch surface $\mathbb{P}(\mathcal{O}(-1)\oplus\mathcal{O})$; here $\mathcal{O}(-1)$ and $\mathcal{O}$ are holomorphic line bundles over $\mathbb{CP}^1$. Alternatively, it is the total space of the sphere bundle of $\mathcal{O}(-1)\oplus\varepsilon^1$ over $S^2$. Note that $\mathcal{O}(-1)\oplus\varepsilon^1$ is the unique non-trivial rank three vector bundle on $S^2$ (because $\pi_1(SO(3)) = \mathbb{Z}_2$) and it is not spin (i.e. $w_2(\mathcal{O}(1)\oplus\varepsilon^1)) \neq 0$) as is $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. This naturally leads one to ask: Can $\mathbb{HP}^2\#\overline{\mathbb{HP}^2}$ and $\mathbb{OP}^2\#\overline{\mathbb{OP}^2}$ be realised as (non-trivial) sphere bundles over $S^4$ and $S^8$ respectively? Not all sphere bundles in these dimensions necessarily arise as sphere bundles of vector bundles. However, given the constructions above, I would expect them to in this case. Unlike in the real and complex cases, there is not a unique non-trivial vector bundle to take the sphere bundle of because $\pi_3(SO(5)) = \mathbb{Z}$ and $\pi_7(SO(9)) = \mathbb{Z}$. All of these homotopy groups are in the stable range, so it is no coincidence that there we have two $\mathbb{Z}_2$'s and two $\mathbb{Z}$'s appearing. As $w_4(\mathbb{HP}^2\#\overline{\mathbb{HP}^2}) \neq 0$ and $w_8(\mathbb{OP}^2\#\overline{\mathbb{OP}^2}) \neq 0$, the desired vector bundles should have non-zero $w_4$ and $w_8$ respectively (in direct analogy with the cases above). Note that both $\gamma$ and $\mathcal{O}(-1)$ can be viewed as the tautological (real/complex) line bundles over $\mathbb{RP}^1$ and $\mathbb{CP}^1$ respectively. This suggests that we should try the direct sum of the tautological (quaternionic/octionic) line bundles over $\mathbb{HP}^2$ and $\mathbb{OP}^2$ with $\varepsilon^1$. This is perfectly reasonable in the quaternionic case, but $\mathbb{OP}^2$ does not arise as the projectivisation of $\mathbb{O}^3$, so there is no such bundle; maybe there is a natural replacelement though (probably associated to one of the generators of $\pi_7(SO(8)) \cong \mathbb{Z}\oplus\mathbb{Z}$). REPLY [5 votes]: Another way of describing the content of the previous answers is as follows. Suppose one starts with one of the Hopf bundles $p:S^{2n-1}\to S^n$ for $n=2,4,8$ coming from $\mathbb C$, $\mathbb H$, or $\mathbb O$, with fibers $S^{n-1}$. The mapping cone of $p$ is then the associated projective plane $\mathbb {CP}^2$, $\mathbb {HP}^2$, or $\mathbb {OP}^2$. The mapping cylinder is obtained from the mapping cone by deleting an open cone on the domain $S^{2n-1}$ of $p$ so this amounts to deleting the interior of a ball $B^{2n}$. The mapping cylinder of a fiber bundle with fibers $S^{n-1}$ is a fiber bundle with fibers $B^n$ over the same base space, $S^n$ in this case. The double of this mapping cylinder is then a bundle over the same base with fibers $S^n$, the double of $B^n$. On the other hand, the double is also the connected sum of two copies of the projective plane since we are taking taking two copies of the projective plane, deleting the interior of a ball $B^{2n}$ from each, then identifying the two resulting boundary spheres $S^{2n-1}$.<|endoftext|> TITLE: Graphs with only disjoint perfect matchings, with coloring QUESTION [11 upvotes]: The following purely graph-theoretic question is motivated by quantum mechanics. Definitions: A bi-colored graph $G$ is an undirected graph where every edge is colored. An edge can either be monochromatic or bi-chromatic. The later means that the two endpoints of an edge are allowed to have different colors. We will be interested in perfect matchings of bi-colored graphs. If $G$ is a bi-colored graph and PM is a perfect matching in $G$ then we can associate a coloring of the vertices of $G$ to the matching PM in the natural way: every vertex gets the color of the edge of PM that is incident to said edge. (Note that a red-blue bi-chromatic edge of PM results in one endpoint of this edge being colored red but the other one blue.) We call these associated (inherited) colorings the inherited vertex coloring (IVC) of the perfect matching PM. Question: Is there a bi-colored graph with $|V|>4$ with the following properties: For all three colors (red, green, blue) there is at least one perfect matching that has a monochromatic IVC. For every non-monochromatic vertex coloring $c$, there is either no perfect matching of $G$ that has its IVC equal to $c$, or there are at least two such perfect matchings. The special case for graphs with no non-monochromatic IVC has been solved by Ilya Bogdanov. The connection to quantum mechanics is described in this article (where I also cite Ilya's solution at MO). Example of Inherited Vertex Coloring: Here I show an example Graph with 8 vertices, with monochromatic and bichromatic edges. It has 7 perfect matchings. For each of the three colors red, blue and green, there is exactly one perfect matching with a monochromatic IVC (PM1-3). For the coloring ($g$=green, $b$=blue, $r$=red) $c=ggggbbbb$, there are two PMs with that IVC (PM4 and PM6) - this is property 2. However, for the non-monochromatic vertex coloring $c=bbbbgggg$ and $c=rgggbrrr$, there is exactly one such perfect matching with this IVC - this the graph $G(V,E)$ does not have the required property. REPLY [4 votes]: Look at the strong product of an even complete graph with an even cycle, $G=K_{2n}\boxtimes C_{2m}$. We can give an edge coloring to this graph such that any vertex coloring is either not inherited from a perfect matching, or it is inherited by exponentially many matchings, and all monochromatic vertex colorings are induced from some perfect matching. We can denote the vertices by pairs $(x,y)$ with $x\in \{1,2,\dots,2n\}$ and $y\in \mathbb Z/2m\mathbb Z$, so that we have edges $(x_1,y_1)\sim(x_2,y_2)$ iff $y_1-y_2\in \{-1,0,1\}$. We will pick the coloring of the edges as follows: the blue edges are all edges of the form $\left((x_1,y),(x_2,y)\right)$, the green edges are all edges of the form $\left((x_1,2y)(x_2,2y+1)\right)$, and the red edges are all edges of the form $\left((x_1,2y)(x_2,2y-1)\right)$. For a given coloring, $C$, of the vertices of $G$ we let $C(b,y), C(g,y), C(r,y)$ denote the number of values of $x$ for which $(x,y)$ is colored blue, green, or red, respectively. It is easy to prove that $C$ is the inherited coloring of a perfect matching in $G$ iff $C(g,2y)=C(g,2y+1)$ for all $y$ $C(r,2y)=C(r,2y-1)$ for all $y$ $C(b,y)$ is even for all $y$ in particular this is satisfied for all monochromatic colorings. Given such a coloring $C$, the number of perfect matchings that induce it is given by $$\prod_{y=1}^mC(g,2y)!C(r,2y)!\prod_{y=1}^{2m}(C(b,y)-1)!!$$ which is not only greater than 1, but grows like $e^{O(nm\log n)}$.<|endoftext|> TITLE: English translation of Borel-Serre, Le théorème de Riemann-Roch? QUESTION [5 upvotes]: Would be happy to receive a translation in English of Borel and Serre's Le théorème de Riemann-Roch, Bulletin de la Société Mathématique de France, Tome 86 (1958) pp. 97-136, doi:10.24033/bsmf.1500, EuDML. REPLY [19 votes]: I have just finished this translation! Here is the TeX source, and here is a (live) PDF. Hopefully there shouldn't be too many mistakes :-) Edit: due to moving links, it's probably best if I just leave the following, where all my translations can be found: thosgood.com/translations.<|endoftext|> TITLE: What is the consistency strength of this theory? QUESTION [6 upvotes]: Language: first-order logic Primitives: $=, S, \in $ (the first denotes identity, the second denotes “is a successor of”, and the third denotes membership relation). Axioms: those of identity theory + Existence: $\exists y, x (y \ S \ x)$ Define: $y \text{ is a successor } \iff \exists x (y \ S \ x)$ Succession: $\forall x,y (y \ S \ x \to \exists z (z \ S \ y))$ Uniqueness: $\forall a,b,c,d (a \ S \ b \wedge c \ S \ d \to[a=c \leftrightarrow b=d])$ Extensionality: $\forall x,y (\forall z (z \in x \leftrightarrow z \in y) \to x=y) $ Comprehension: If $\phi$ is a formula in which $x$ is not free, then all closures of $(\exists x \forall y (y \in x \leftrightarrow [y \text{ is a successor} \lor \forall z \in y ( z \text{ is a successor})] \wedge \phi))$ are axioms. Well-foundedness: $\forall x [\exists m \in x \to \ \exists y (y \in x \wedge \forall z \in x (\neg [y \ S \ z])) ]$ This is, I think, equivalent to third-order arithmetic ($\text{Z}_3)$. Now if we weaken Comprehension to the following: Comprehension': If $\phi$ is a formula in which $x$ is not free, then $ \forall w_1,..,w_n \exists x \forall y (y \in x \leftrightarrow [y=w_1 \lor ..\lor y=w_n] \wedge \phi)$ is an axiom. Question: Could that weakened theory be stronger than bounded arithmetic? Or is it actually weaker than that? REPLY [11 votes]: $\let\itp\vartriangleright\def\mr{\mathrm}\DeclareMathOperator\dom{dom}\let\bez\smallsetminus\let\sset\subseteq$The weakened theory is much weaker than bounded arithmetic. Let me denote the theory as $\mr{WT}$. Notice that the weakened comperehension schema is simply equivalent to the schema $$\tag{$V_n$}\forall w_1,\dots,w_n\,\exists x\,\forall y\,\Bigl(y\in x\leftrightarrow\bigvee_{i=1}^ny=w_i\Bigr)$$ for $n\in\omega$ (here $V_0$ asserts the existence of the empty set). Indeed, by excluded middle, $\{y:(y=w_1\lor\dots\lor y=w_n)\land\phi\}$ is one of the $2^n$ classes $\varnothing$, $\{w_1\}$, $\{w_2\}$, $\{w_1,w_2\}$, $\{w_3\}$, ..., $\{w_1,w_2,\dots,w_n\}$, each of which is a set by $V_n$. Let $\mr{WT}_k$ denote the finite fragment of $\mr{WT}$ with axioms $V_n$ only for $n\le k$ (it is enough to take $V_0\land V_k$). First, $\mr{WT}$ is at most as strong as bounded arithmetic. Note that all reasonable fragments of bounded arithmetic are mutually interpretable with Robinson’s arithmetic $Q$, hence I will use that as a gauge. Theorem 1: $\mr{WT}$ is interpretable in $Q$ (which I will henceforth write as $Q\itp \mr{WT}$). Proof sketch: Let us start by interpreting in $Q$ a decent fragment of arithmetic such as $S^1_2$. Inside this theory, we can define a well-behaved coding of sequences e.g. as follows: we first represent a sequence $\langle x_0,x_1,\dots,x_n\rangle$ by a string over the alphabet $\{0,1,2\}$ as $2_\smile x_0{}_\smile2_\smile x_1{}_\smile\dots_\smile x_n$ where each $x_i$ is written in binary (with no leading $0$s), and then we reinterpret the string as an integer written in ternary. Using this, we represent finite sets of numbers by strictly increasing sequences. We define $x\in y$ in accordance with this encoding scheme; if $y$ is not a valid representation of a set, we put $x\in y\iff x\ne y$ in order to maintain extensionality. We define $y\mathrel Sx$ as the usual successor relation $y=x+1$. Then it is routine to check that all axioms of $WT$ are valid under this interpretation. QED We aim to show that $\mr{WT}$ is in fact strictly weaker than $Q$ in terms of strength, i.e., $\mr{WT}\not\itp Q$. To this end, we will relate $\mr{WT}$ with other theories. Vaught’s set theory $\mr{VS}$ is a theory in the language of set theory $\langle\in\rangle$, axiomatized just by the schema $V_n$ for $n\in\omega$. Let $\mr{VSE}$ denote $\mr{VS}$ + the extensionality axiom. I will write $\mr{VS}_k$ and $\mr{VSE}_k$ for the finite fragments of $\mr{VS}$ and $\mr{VSE}$ (resp.) with $V_n$ axioms only for $n\le k$. As mentioned in Visser [1,§3], $\mr{VS}$ interprets Robinson’s theory $R$, and as such it is essentially undecidable, however all finite fragments $\mr{VS}_k$ are interpretable in $\mr{VS}_2$, which is just a theory of unordered pairs (denoted $\mr{PAIR_{uno,ns}}$ in [1]); in particular, since there are decidable theories with pairing, it follows that each finite fragment of $\mr{VS}$ has a decidable extension. Consequently, $\mr{VS}\not\itp Q$, and more generally, $\mr{VS}$ does not interpret any finitely axiomatizable essentially undecidable theory. Notice that $\mr{VSE}$ is a subtheory of $\mr{WT}$. We will show that these two theories are close enough to each other, and generalize to them the results above: finite fragments of $\mr{WT}$ are interpretable in finite fragments of $\mr{VSE}$, and both have decidable extensions. Theorem 2: $\mr{WT}$ is locally interpretable in $\mr{VSE}$. Specifically, $\mr{VSE}_k\itp\mr{WT}_k$. Proof: In $\mr{VSE}_k$, we can use the standard Kuratowski pairing function $\langle x,y\rangle=\{\{x\},\{x,y\}\}$ to define $(k+1)$-tuples as $\langle x_0,\dots,x_k\rangle=\langle x_0,\langle x_1,\langle\dots\rangle\rangle\rangle$. Let us put $$\begin{align} x\in^*y&\iff\begin{cases}x\in y&\text{if }|y|\le k,\\x\ne y&\text{otherwise,}\end{cases}\\ y\mathrel{S^*}x&\iff\exists\text{ distinct }x_0,\dots,x_k\:(x=\langle x_0,\dots,x_k\rangle\land y=\langle x_1,\dots,x_k,x_0\rangle). \end{align}$$ It is easy to see that under this interpretation, axioms of $\mr{VSE}_k$ are valid. Since $S^*$ consists of a nonempty collection of $(k+1)$-cycles, the axioms of Existence, Succession, and Uniqueness also hold. Finally, Well-foundedness also holds: any counterexample would have to include at least one full $S^*$-cycle, but none of the infinitely many elements that are not part of any cycle. However, all sets under the interpretation have either at most $k$ elements, or include all but one elements of the universe. QED Note: it is perfectly possible that $\mr{WT}$ is in fact globally interpretable in $\mr{VSE}$, but I didn’t try. Theorem 3: Each finite fragment $\mr{VSE}_k$ has a decidable consistent extension. Corollary: Each finite fragment $\mr{WT}_k$ has a decidable consistent extension. Neither $\mr{VSE}$ nor $\mr{WT}$ interprets $Q$, or any finite essentially undecidable theory for that matter. I will assume $k\ge2$. In order to prove Theorem 3, working in a usual set theory (e.g., ZFC), let $H_k$ denote the collection of all hereditarily $\le k$-element sets: the least family of sets such that $$x\sset H_k\land|x|\le k\implies x\in H_k.$$ Note that $H_k\sset V_\omega$. Clearly, $$(H_k,\in)\models\mr{VSE}_k,$$ thus Theorem 3 follows from Theorem 4: $\mathrm{Th}(H_k,\in)$ is decidable. This is proved in detail in [2]; let me sketch the basic argument. Since any r.e. complete theory is decidable, it is enough to show that $\mathrm{Th}(H_k,\in)$ is recursively axiomatizable. In fact, I claim that it coincides with the theory $S_k$ axiomatized as follows: $\mr{VSE}_k$ (i.e., extensionality, $V_0$ and $V_k$), every set has at most $k$ elements, there are no finite $\in$-cycles: i.e., the schema $$\forall x_0,\dots,x_n\,\neg(x_0\in x_1\in\dots\in x_n\in x_0)$$ for $n\in\omega$. Obviously $H_k\models S_k$, it thus suffices to show that $S_k$ is complete, i.e., that any two models of $S_k$ are elementarily equivalent. We will prove more generally a characterization of elementary equivalence of finite tuples in models of $S_k$. Let $A,B$ be models of $S_k$, and $\bar a\in A$, $\bar b\in B$ finite tuples of the same length. We write $A,\bar a\equiv B,\bar b$ if $\bar a$ and $\bar b$ are elementarily equivalent. We define levels of the transitive closure of $\bar a$ (as subsets of $A$): $$\begin{align} t^A_0(\bar a)&=\{\bar a\},\\ t^A_{n+1}(\bar a)&=t^A_n(\bar a)\cup\bigcup_{u\in t^A_n(\bar a)}u^A,\\ t^A_\infty(\bar a)&=\bigcup_{n\in\omega}t^A_n(\bar a), \end{align}$$ where $u^A$ denotes the extension of $u\in A$ in $A$: $u^A=\{v\in A:v\in^A u\}$. Notice that each $t^A_n(\bar a)$ is finite: $|t^A_n(\bar a)|\le|\bar a|k^{n+1}$ or so. Also, each $t^A_n(\bar a)$ is definable (as a class) in $A$ from $\bar a$; the definition is uniform if $n$ and $|\bar a|$ are fixed. Let us define $$\begin{align} A,\bar a\sim_n B,\bar b&\iff(t^A_n(\bar a),\in,\bar a)\simeq(t^B_n(\bar b),\in,\bar b),\\ A,\bar a\sim B,\bar b&\iff(t^A_\infty(\bar a),\in,\bar a)\simeq(t^B_\infty(\bar b),\in,\bar b). \end{align}$$ Recall that elementarily equivalent finite structures are isomorphic. Thus, the finiteness and definability of $t^A_n(\bar a)$ implies Claim 5: $A,\bar a\equiv B,\bar b\implies A,\bar a\sim_n B,\bar b$ for all $n\in\omega$. Notice that $\bar a\sim_m\bar b$ implies $\bar a\sim_n\bar b$ for $n\le m$, as an isomorphism $t^A_m(\bar a)\simeq t^B_m(\bar b)$ must map $t^A_n(\bar a)$ to $t^B_n(\bar b)$. Thus, if we look at the set of all isomorphisms $f\colon t^A_n(\bar a)\simeq t^B_n(\bar b)$ for all $n\in\omega$, it forms a tree ordered by inclusion, and finiteness of the structures implies that the tree is finitely branching. Thus, if it is infinite, it has an infinite branch by König’s lemma, which is then an isomorphism $t^A_\infty(\bar a)\simeq t^B_\infty(\bar b)$. Thus, Claim 6: $A,\bar a\sim B,\bar b\iff\forall n\in\omega\:A,\bar a\sim_n B,\bar b$. Corollary 7: $A,\bar a\equiv B,\bar b\implies A,\bar a\sim B,\bar b$. We intend to prove the converse Claim 8: $A,\bar a\sim B,\bar b\implies A,\bar a\equiv B,\bar b$. This finishes the proofs of Theorems 3 and 4, as it implies $A,\varnothing\equiv B,\varnothing$ for any $A,B\models S_k$. By definition, it’s clear that $A,\bar a\sim_0 B,\bar b$ implies $\bar a$ and $\bar b$ have the same atomic diagram. Thus, by standard model-theoretic machinery, Claim 8 follows from the following back-and-forth property: Claim 9: For every $n$, there exists $m$ such that whenever $A,\bar a\sim_m B,\bar b$ and $c\in A$, there exists $d\in B$ such that $A,\bar a,c\sim_n B,\bar b,d$. Finally, using the fact that $\sim_n$ can be characterized by satisfaction of suitable formulas and the compactness theorem, it is enough to show Claim 9’: If $A,\bar a\sim B,\bar b$, $c\in A$, and $n\in\omega$, there exists $d\in B$ such that $A,\bar a,c\sim_n B,\bar b,d$. Proof: Let us fix an isomorphism $f\colon t^A_\infty(\bar a)\simeq t^B_\infty(\bar b)$. We need to extend $f$ to an isomorphism $g$ whose domain includes $t^A_n(c)$. For convenience, we will actually define $g$ as a map on $t^A_{n+1}(c)$, but it will not necessarily behave as an isomorphism on the extra points. Notice that $t^A_{n+1}(c)$ is a finite set, and $\in^A$ is acyclic, hence well founded, on it. If there is $u\in t^A_{n+1}(c)\bez\dom(f)$ such that $u\sset\dom(f)$, we may extend $f$ right away so that $f(u)=\{f(x):x\in^A u\}^B$ (this exists in $B$, as it has at most $k$ elements). This maintains the salient property that $f$ is an isomorphism between two transitive closures of finite tuples. By repeating this construction as many times as necessary, we may assume without loss of generality that every $u\in t^A_{n+1}(c)\bez\dom(f)$ contains an element outside $\dom(f)$. Now, let $\{u_i:i TITLE: Tying knots via gravity-assisted spaceship trajectories QUESTION [22 upvotes]: Q. Can every knot be realized as the trajectory of a spaceship weaving among a finite number of fixed planets, subject to gravity alone?           To make this more specific, let $S$ be a large sphere, containing a finite number of planet-points $P=\{p_1, \ldots, p_n\}$. The planet-points have (in general) different masses, and are fixed in $\mathbb{R}^3$. A spaceship $x$ approaches $S$ from $\infty$, interacts via gravity with the point masses, and eventually exits $S$ to $\infty$. Define the knot $K$ realized by the ship's trajectory as the path of $x$ plus a connection between the two ends at $\infty$. (Assume those two $\infty$-ends are distinct.) Q'. For any given knot $K$, can one arrange point masses in $P \subset S$ and a line and speed of approach to $S$ so that $x$'s path weaves $K$ by interacting with the planets via gravity alone, i.e., without the use of rocket fuel? One approach might be to design a "gadget" that mimics a vertex $v$ of a stick knot and $v$'s two incident segments. But preventing the vertex gadgets from interfering with one another might not be straightforward.                     Cassini gravity-assist trajectory. Image from NASA/JPL. REPLY [8 votes]: Yes, with some caveats. For an authoritative source, please see the monograph Koon et al. "Dynamical systems, the three-body problem and space mission design" The short story is that with 2 or more large bodies, trajectory of a spacecraft is "chaotic", and hence under some conditions, it can be shown that horseshoe-type dynamics exist. In other words, if you label the regions around each large body with an alphabet, any arbitrary string of alphabets can be achieved" Also see: http://www2.esm.vt.edu/~sdross/papers/AmericanScientist2006.pdf<|endoftext|> TITLE: A generalization of Witt's theorem for quaternion algebra isomorphism QUESTION [5 upvotes]: Let $Q$ be a quaternion $k$-algebra (namely, a dimension 4 $k$-central simple algebra). Then it is possible to (canonically) attach a smooth projective conic $C_Q\subseteq \mathbf{P}_k^2$ to $Q$: if $\mathrm{char}\,k \neq 2$, then $Q\simeq \langle x,y : x^2 = a, y^2 = b, xy+yx=0\rangle$ for certain $a\in k,b\in k^\times$ so it suffices to set $C_Q\simeq\{ aX_0^2+bX_1^2 - X_2^2 =0\}\subseteq \mathbb{P}^2_k$. If instead $\mathrm{char}\,k =2$ one has $Q=\langle x,y : x^2+x=a,y^2=b, xy+yx=yy\rangle$ for some $a\in k,b\in k^\times$ and assign $C_Q=\{aX_0^2 +bX_1^2 +X_2^2 + X_0X_2 =0\}$. There exists an interesting relationship between splitting of quaternion algberas (i.e. $Q\simeq M(2,k)$) and existence of $k$-rational points in the associated conic $C_Q$. This works for every characteristic too. But there is an even more elegant result, attributed to Witt, which relates isomorphism of two quaternion algebras with the birational equivalence of their associated conics. More precisely: Theorem. Let $Q,R$ be two quaternion $k$-algebras. Then $Q\simeq R$ if and only if $C_Q$ and $C_R$ are birational (namely $k(C_Q)\simeq k(C_R)$). Such result can be found, for instance, in the great book by Gille and Szamuely, Central simple algebras and Galois cohomology where, much to my dismay, almost all the results are shown for a field of characteristic $\neq 2$. In particular, the above Theorem is proved using the definitions for quaternion algebras and conics in characteristic not 2. I have searched in some more specific book (like Vigneras' lectures and Voight's draft) but this result does not seem to be covered. In the Vigneras' book there is a chapter called "Geometry" in which something similar is done, but the whole chapter works in the non-dyadic case. I was wondering if such result held for a field of characteristic 2 and, if yes, is there a reference for this? REPLY [3 votes]: Norms of quaternion algebras are particular cases of n-fold Pfister forms for n=2 (in any characteristic), and the result holds true for any Pfister forms. See, for example, Elman, Karpenko, Merkurjev "The algebraic and geometric theory of quadratic forms": https://sites.ualberta.ca/~karpenko/publ/Kniga.pdf, Corollary 23.5.<|endoftext|> TITLE: Probability of a large random integer Matrix to have zero determinant QUESTION [6 upvotes]: Suppose we have a matrix $A \in \{0,1\}^{n \times n}$ where $$A_{ij} = \begin{cases} 1 & \text{with probability} \quad p\\ 0 &\text{with probability} \quad 1-p\end{cases}$$ I would like to know the probability $$\mathbb P( \det (A) =0) \ \text{ for large }n,$$ i.e., the asymptotic probability of the determinant being zero as $n$ becomes large. I know that the probability $$\lim_{n \rightarrow \infty} \mathbb P \left( \det (A) =0 \right) = 0$$ if $p \neq 0,1$ but could not find results for finite $n$/the asymptotic behavior. Edit: I changed the symbol $\mathbb{E}$ to $\mathbb{P}$ since I meant the probability and not the expectation value as some comments assumed correctly. REPLY [2 votes]: NOTE: This answer was posted in response to an earlier version of the question that was about the expected value and not the probability. So it has now become irrelevant. I decided not to delete it because in a comment there is some useful general result provided. If moderators think the answer should be deleted, no problem. It appears I am missing something here, but as long as I cannot find what I am missing: "$\det$" is an operator on all the elements of a matrix that results in an expression that combines them by using addition, subtraction and multiplication. Then, due to the linearity of the expected value we have $$\mathbb E[A_{ij} + A_{k\ell}] = \mathbb E[A_{ij}] + \mathbb E[A_{k\ell}] \tag{1}$$ Moreover, in the OP's case as explained in a comment, the elements of the matrix are i.i.d. random variables. Due to the statistical independence assumption we have $$\mathbb E[A_{ij} A_{k\ell}] = \mathbb E[A_{ij}] \mathbb E[A_{k\ell}] \tag{2}$$ But in such a case we have $$\mathbb E[\det(A)] = \det(\mathbb E[A]) \tag{3}$$ $\mathbb E(A)$ is a matrix with all elements identical and equal to $p$ (due to the "identically distributed" assumption). Then $\det(\mathbb E[A]) = 0 = \mathbb E[\det(A)]$ for all $n$, and so $$\Pr\{\mathbb E[\det(A)]=0\} = 1, n\geq 2$$<|endoftext|> TITLE: Entering to the K-theory realm QUESTION [13 upvotes]: I am looking for a guidance in $K$-theory. My master thesis was in the field of Algebraic K-theory and its relation and interaction with the field of Algebraic Topology. I mainly had concentrated on the study of the third K-group of an infinite field. I studied the Anderie Suslin paper, which was titled as the "$K_3$ of a field, and the Bloch group". Entering to this field required my hard work on algebraic topology. By now I am a Ph.D. student, and I do not have my master supervisor here. I always see $K$-theory as a giant which has many things in his heart. I am much interested in it, but I do not how should I walk into this realm. I may have a question from the experts: I know some preliminaries, and I want to fulfill my dreams about $K$-theory, what is the main plan? What kind of requirements I need? I have studied several books in this branch, Hatcher for first, Sirinivas, Atiyah, and many lectures and some papers. I have good background in Homology and Algebraic geometry too. Your comments and suggestions would be greatly appreciated. REPLY [19 votes]: I think that doing algebraic K-theory properly certainly requires a good background on stable homotopy theory, that is to say the homotopy theory of spectra. Unfortunately there are not many textbooks in the subject. Let me mention two of them: Stable homotopy and generalized cohomology by J. Frank Adams is an old classic. Its treatment of some topics is far from modern though, and in particular the development of localizations is flawed and should be complemented by reading Bousfield's original papers. He also doesn't talk about commutative (i.e. $E_∞$) ring spectra, which you're going to need. Symmetric spectra by S. Schwede. A much more modern approach, covering a variety of topics you're going to need. Just don't get too hung on the model categorical subtleties of the model he chose (I'm thinking mainly semistability here), 'cause they won't come up in practice. Categories and cohomology theories by G. Segal. This is a short paper, but if you want to learn about group completion and its relation to spectra, reading this is probably the quickest thing you can do. Despite its age it is surprising modern in its approach. Moreover it is a pleasure to read. When you have a sufficiently good background in homotopy theory that the words spectra and group completion don't make you scream in terror, it's time to start with actual algebraic K-theory. Here are some useful starting points The K-book by Charles Weibel has a lot of classical material and it is a useful bridge from the low dimensional, hand-defined groups to the more modern algebraic K-theory spectrum. It is a bit long though, and I'd treat it more as a reference than a book to be read from top to bottom. Higher Algebraic K-theory of Schemes and of Derived Categories by Thomason and Trobaugh. One of the fundamental papers on algebraic K-theory. It also has a decent introduction to Waldhausen's S-construction and it is worth reading in full. Highly highly recommended. On the Lichtenbaum-Quillen Conjectures from a Stable Homotopy-Theoretic Viewpoint by Stephen A. Mitchell. This is a general survey of algebraic K-theory. It is extremely useful and will acquaint you with most classical theorems that you might find used in more specialized papers. Absolutely on a to-read list. Algebraic K-theory and traces by I. Madsen. Filling a gap in the previous survey, this talks about trace methods, the best way we have to actually compute the K-theory groups. Algebraic K-theory of spaces Waldhausen's original paper on his approach to algebraic K-theory. It's worth taking a look at some of the proofs, and the motivation is explained clearly and naturally. I also think some familiarity with ∞-category theory might be useful from a technical standpoint, but you shouldn't enter into a full dive into the foundations. This is something best coordinated with your advisor, who will know how much of it is actually useful for you. As a rule of thumb, if you're spending more than a week on it (at first, of course), you're doing too much. Feel free to jump into the homotopy theory chatroom to hang around and ask questions if you feel like it.<|endoftext|> TITLE: Examples of extensions of non-solvable groups by one another QUESTION [5 upvotes]: Apart from the direct products, what are some "interesting" or "naturally occurring" examples of extensions $$ 1 \to N \to G \to Q \to 1 $$ of finite groups such that neither $N$ nor $Q$ is solvable? I feel a bit stupid for asking the question, but I don't think I know a single example that isn't split. Just to set the bar pretty low: essentially the only example I can think of is to take the wreath product $\mathfrak{S}_n\wr\mathfrak{S}_m$ (group of permutations of $mn$ elements preserving a partition into $m$ sets of $n$), for $m,n\geq 5$, which is a semidirect product $(\mathfrak{S}_n)^m \rtimes \mathfrak{S}_m$. Anything which is not a trivial variation on this construction interests me. REPLY [9 votes]: Here is a way you can construct nonsplit examples in which all composition factors are nonabelian. Many finite nonabelian simple groups do not split over their automorphism groups. The smallest such example is the group often known as $M_{10}$, which is the point stabilizer in the Mathieu group $M_{11}$. It has order $720$ and is the extension of ${\rm PSL}(2,9)$ by a product of a field and a diagonal automorphism. So it has a normal subgroup $H$ of index $2$ with $H \cong A_6 \cong {\rm PSL}(2,9)$. Now $M_{10}/H$ is of course solvable so we don't have an example yet, but we can use a wreath product type construction to get what we want. Let $K$ be any finite nonabelian simple group, and let $S$ be any subgroup of order $2$ in $K$. Let $P$ denote the image of the permutation action of $K$ on the cosets of $S$. So, for example, if $K=A_5$ then $P$ is a subgroup of $S_{30}$. Now the permutation wreath product $W$ of $H$ by $P$ is of course a split extension. $1 \to H^{|K:S|} \to W \to P \to 1$. But there are in general other extensions of this type, and there is a theorem that says that the equivalence classes of such extensions are in one-one correspondence with the extensions of $H$ by $S$. So, if there is a nonsplit extension of $H$ by $S$, then there is also one of $H^{|K:S|}$ by $P$. (I will look for a reference for that result later.) For example, if we take $H={\rm PSL}(2,9) \cong A_6$ as above, and $K=A_5$, then we get a nonsplit extension with normal subgroup $A_6^{30}$ and quotient $A_5$. I found a reference for the result about wreath product type extensions in an old paper of my own. It is proved as Theorem 1 of D.F. Holt, Embeddings of group extensions into Wreath products, Quar. J. Math. (Oxford) 29 (1978), 463--468. But I suspect it was known earlier. The main result of that paper is a generalization of the Krasner Kaloujnine theorem about embedding group extensions into standard wreath products.<|endoftext|> TITLE: When does completion preserve injectivity? QUESTION [9 upvotes]: Let $ f:A\to B$ be an injective, local homorphism between two Noetherian local rings. Consider the completions $\hat A$ and $\hat B$ with respect the maximal ideals. We have an induced homomorphism $\hat f: \hat A \to\hat B$. What assumptions do we need in order to ensure that also $\hat f$ is injective? My main interest is geometric, so for example the local map induced by a surjective morphism between Noetherian schemes. Many thanks in advance REPLY [4 votes]: This may be useful: Proposition (Zariski) (see EGA I, (3.9.8) in Springer edition) Let $f: (A,\mathfrak{m})\to (B,\mathfrak{n})$ be a local homomorphism of noetherian local rings. Assume that: $f$ is injective. $\hat{A}$ is a domain. $f$ is essentially of finite type. Then the $\mathfrak{m}$-adic topology on $A$ is induced by the $\mathfrak{n}$-adic topology on $B$. Of course this implies that $\hat{f}$ is injective. If we don't assume that $\hat{A}$ is a domain (but $A$ is) it is easy to construct counterexamples.<|endoftext|> TITLE: Countable (?) dependent choice QUESTION [9 upvotes]: In some circumstances I've been using a form of choice over the first uncountable ordinal knowing a priori that only a countable number of choices were going to be made (without any a priori upper bound). I would like to know whether I was using the classical dependent choice or something stronger and in case how much stronger. I think the argument that I have in mind is better explained with an example and the best that I know is a proof of Ekeland's variation principle. Such principle establishes the existence of `almost mininizers' of a function on a non-compact, but complete, metric space. Let me give a - somehow suboptimal - formulation of the result: Theorem Let $(X,d)$ be a complete metric space and $f:X\to[0,+\infty)$ lower semicontinuous (one can assume $f$ to be continuous if he wishes). Let $\epsilon>0$. Then there exists a point $x\in X$ such that $f(x)\leq\inf f+\epsilon$ and satisfying $$ f(y)\geq f(x)-\epsilon d(x,y)\qquad\forall y\in X. $$ Thus if we could choose $\epsilon=0$ the theorem would provide the existence of a minimum for $f$, so that the requirement $\epsilon>0$ might be seen as a sort of almost minimization (also known as quasi-minimization in some contexts). Here is a possible proof of the theorem: Let $\Omega$ be the first uncountable ordinal and let's recursively define a map $\Omega\ni \alpha\to x_\alpha\in X$ as follows. $x_0\in X$ is taken arbitrarily such that $f(x_0)<\inf f+\epsilon$. If $x_\alpha$ has already been defined, we define $x_{\alpha+1}$ as follows: if $x_{\alpha}$ satisfies the conclusion of the theorem, we put $x_{\alpha+1}:=x_{\alpha}$. Otherwise the set of $y$'s such that $$ f(y)< f(x_{\alpha})-\epsilon d(x_{\alpha},y) $$ is not empty: let $x_{\alpha+1}$ be any of these $y$'s. Thus in particular we have $$ \epsilon d(x_\alpha,x_{\alpha+1}) f(x(\alpha))$ when $\beta < \alpha < \omega_1$. This is impossible since there is no uncountable decreasing sequence of real numbers. From this contradiction, we conclude that our hypothesis from step 1 is false: it is not true that $\mathsf{DC}_{\omega_1}$ holds and Ekeland's theorem is false. In other words, $\mathsf{DC}_{\omega_1}$ implies Ekeland's theorem. This does not mean that Ekeland's theorem requires $\mathsf{DC}_{\omega_1}$, but this argument makes essential use of $\mathsf{DC}_{\omega_1}$. Unfortunately, $\mathsf{DC}_{\omega_1}$ also implies that there is a nonmeasurable set of reals. Indeed, it is trivial to use $\mathsf{DC}_{\omega_1}$ to construct an injection $\omega_1 \to \mathbb{R}$. A celebrated result of Raisonnier says that this implies that there is a nonmeasurable set of reals. Raisonnier, Jean, A mathematical proof of S. Shelah’s theorem on the measure problem and related results, Isr. J. Math. 48, 48-56 (1984). ZBL0596.03056.<|endoftext|> TITLE: A strengthening of the Cauchy-Schwarz inequality QUESTION [47 upvotes]: Suppose $\mathbf{v},\mathbf{w} \in \mathbb{R}^n$ (and if it helps, you can assume they each have non-negative entries), and let $\mathbf{v}^2,\mathbf{w}^2$ denote the vectors whose entries are the squares of the entries of $\mathbf{v}$ and $\mathbf{w}$. My question is how to prove that \begin{align*} \|\mathbf{v}^2\|\|\mathbf{w}^2\| - \langle \mathbf{v}^2,\mathbf{w}^2\rangle \leq \|\mathbf{v}\|^2\|\mathbf{w}\|^2 - \langle \mathbf{v},\mathbf{w}\rangle^2. \end{align*} Some notes are in order: The Cauchy-Schwarz inequality tells us that both sides of this inequality are non-negative. Thus the proposed inequality is a strengthening of Cauchy-Schwarz that gives a non-zero bound on the RHS. I know that this inequality is true, but my method of proving it is extremely long and roundabout. It seems like it should have a straightforward-ish proof, or should follow from another well-known inequality, and that's what I'm looking for. REPLY [41 votes]: Here is a proof for every $n$. Using the notation $\mathbf{v}=(v_1,\dots,v_n)$ and $\mathbf{w}=(w_1,\dots,w_n)$, the inequality reads $$\left(\sum_i v_i^4\right)^{1/2}\left(\sum_i w_i^4\right)^{1/2}-\sum_i v_i^2 w_i^2\leq \left(\sum_i v_i^2\right)\left(\sum_i w_i^2\right)-\left(\sum_i v_i w_i\right)^2.$$ Rewriting the right hand side in a familiar way, and then rearranging and squaring, we obtain the equivalent form $$\left(\sum_i v_i^4\right)\left(\sum_i w_i^4\right)\leq\left(\sum_i v_i^2 w_i^2+\sum_{i TITLE: Sporadic subgroup of E7 QUESTION [19 upvotes]: The dimensions of some representations of the Janko group J1 coinside with dimensions of smallest representations of the Lie algebra of type E7 (56, 133). It seems to be natural that there is a subgroup of the Lie group of type E7 isomorphic to J1. Is it true? REPLY [7 votes]: The Frobenius–Schur indicators of both $56$-dimensional representations are $+1$, meaning that in both cases the representation supports an invariant symmetric bilinear form. Since these are irreps, they do not support any other invariant bilinear forms. On the other hand, the $56$-dimensional irrep of $\mathrm{E}_7$ supports an invariant antisymmetric bilinear form. So $\mathrm{J}_1 \not \subset \mathrm{E}_7$. According to the ATLAS, the finite group of Lie type $\mathrm{E}_7(2)$ has a $132$-dimensional "adjoint" representation over $\mathbf{F}_2$. The ATLAS does not list such a representation for $\mathrm{J}_1$, but I don't know how to confirm that it does not exist, so I cannot rule out an inclusion $\mathrm{J}_1 \overset?\subset \mathrm{E}_7(2)$. I remark that of the three $133$-dimensional representations of $\mathrm{J}_1$, the first supports one invariant antisymmetric 3-form, which could in principle be a Lie bracket, and the other two support five each.<|endoftext|> TITLE: Detecting the Brown-Comenetz dualizing spectrum QUESTION [18 upvotes]: The Brown-Comenetz dualizing spectrum $I_{\mathbf{Q/Z}}$ is not detected by very many spectra: it is $BP, \mathbf{Z}, \mathbf{F}_2, X(n)$ for $n\geq 2$, and even $I_{\mathbf{Q/Z}}$-acyclic. However, if $X$ is any nontrivial finite spectrum, then $X\wedge I_{\mathbf{Q/Z}}$ is not contractible. This motivates a natural question: let $E$ be any spectrum such that $E\wedge I_{\mathbf{Q/Z}}$ is not contractible. Then, is it true that $\langle E\rangle \geq \langle X\rangle$ for some nontrivial finite spectrum $X$? REPLY [12 votes]: Your question appears to be equivalent to the 'dichotomy conjecture' of Hovey, which I believe is still open. First, note that any finite spectrum has a type, and all finite spectrum of type $n$ have the same Bousfield class, usually denoted $F(n)$. In Hovey and Strickland's memoir (Appendix B) they conjecture that if $E \wedge I \ne 0$, then $\langle E \rangle \ge \langle F(n) \rangle$ for some $n$. This is precisely your question. In his paper on the chromatic splitting conjecture, Hovey conjectured that every spectrum has either a finite acyclic or a finite local (the dichotomy conjecture). With Palmieri (The structure of the Bousfield lattice) he proved that the following conjectures are equivalent: (1) If $E \wedge I \ne 0$, then $\langle E \rangle \ge \langle F(n) \rangle$ for some $n$. (2) The dichotomy conjecture. (3) If $E$ has no finite acyclics, then $\langle E \rangle \ge \langle I \rangle$. The analog of the dichotomy conjecture is known to hold in other categories with a good theory of support satisfying the tensor product property - see 'The Bousfield lattice of a triangulated category and stratification' by Iyengar and Krause.<|endoftext|> TITLE: Sheaf-theoretically characterize a Riemannian structure? QUESTION [14 upvotes]: A smooth structure on a topological manifold can be characterized as a sheaf of local rings, see for example the discussion here. Q: Is there a way to characterize a Riemannian structure on a smooth manifold by a sheaf of functions? A most likely horrible guess to clarify the type of answers I'm thinking about: define a Riemannian manifold to be a locally ringed space that locally looks like the sheaf $(\mathbb R^n, \mathcal H_g)$ where $g$ is some non degenerate symmetric positive definite matrix and $\mathcal H_g$ is the sheaf (is it even a sheaf?) that assigns to open subsets harmonic functions solving the Laplace equation given by $g$. Please forgive my ignorance in the above, this is not my field. Just had to do a little Riemannian geometry today and was thinking whether there's a sheaf-theoretic/functor of points way to think about things. REPLY [15 votes]: Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the sheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$ $$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\}, $$ where $\Delta_{g_i}$ denotes the scalar Laplacian of the metric $g_i$. Long time ago I proved the following result. Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$. If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$. If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class. The strong unique continuation property of harmonic functions shows that this statement is really a statement about the stalks of the sheaves $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.<|endoftext|> TITLE: Symmetric strengthening of the Cauchy-Schwarz inequality QUESTION [26 upvotes]: In this great question by Nathaniel Johnston, and in its answers, we can learn the following remarkable inequality: For all $v,w \in \mathbb{R}^n$ we have \begin{align*} \|v^2\| \, \|w^2\| - \langle v^2, w^2 \rangle \le \|v\|^2 \|w\|^2 - \langle v,w \rangle^2; \quad (*) \end{align*} here, $\langle\cdot,\cdot\rangle$ denotes the standard inner product on $\mathbb{R}^n$, $\|\cdot\|$ denotes the Euclidean norm and $v^2,w^2 \in \mathbb{R}^n$ denote the elementwise squares of $v$ and $w$. Both sides of $(*)$ are nonnegative by the Cauchy-Schwarz inequality, and the LHS gives a non-zero bound for the right RHS, in general. What strikes me is that the RHS and the LHS of $(*)$ have different (linear) symmetry groups: the RHS does not change if we apply any orthogonal matrix $U \in \mathbb{R}^{n \times n}$ to both $v$ and $w$, while this is not true for the LHS. Hence, we can strengthen $(*)$ to \begin{align*} \sup_{U^*U = I}\Big(\|(Uv)^2\| \, \|(Uw)^2\| - \langle (Uv)^2, (Uw)^2 \rangle\Big) \le \|v\|^2 \|w\|^2 - \langle v,w \rangle^2. \quad (**) \end{align*} Unfortunately, I have no idea how to evaluate the LHS of $(**)$. Question. Can we explicitely evaluate the LHS of $(**)$? Or, more generally, is there a version of $(*)$ for which both sides are invariant under multiplying both $v$ and $w$ by (identical) orthogonal matrices? Admittedly, this question is a bit vague since it might depend on one's perspective which expressions one considers to be "explicit" and which inequalities one considers to be a "version" of $(*)$. Nevertheless, I'm wondering whether some people share my intuition that there should be a more symmetric version of $(*)$. Edit. Maybe it is worthwhile to add the following motivating example: If we choose $n = 2$ and $v = (1,1)/\sqrt{2}$, $w = (1,-1)/\sqrt{2}$, then those vectors are orthogonal and the RHS of $(*)$ equals $1$, while the LHS of $(*)$ vanishes since $v^2$ and $w^2$ are linearly dependent. However, the LHS of $(**)$ is also equal to $1$; to see this, choose $U = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} $ and observe that $Uv = (0,1)$, $Uw = (1,0)$. REPLY [7 votes]: $(**)$ is always equality. By homogenuity we may suppose that $\|v\|=\|w\|=1$, then $Uv,Uw$ are any two unit vectors with prescribed inner product $\langle Uv,Uw\rangle =\langle u,v\rangle$. I claim that you may find two two-dimensional vectors with prescribed given inner product such that equality in $(*)$ holds. Then you may make $n$-dimensional examples by adding zero coordinates. Well, just look at GH from MO's proof of (*) and look what we need for equality. If $n=2$, it is sufficient that $F(v,w):=v_1w_1-v_2w_2=0$. Since $F((v_1,v_2),(w_1,w_2))=-F((w_2,w_1),(v_2,v_1))$, we may rotate the vector $(v_1,v_2)$ until it becomes equal $(w_2,w_1)$, at this moment the vector $(w_1,w_2)$ becomes equal to $(v_2,v_1)$ (since the inner product and orientation define the pair of unit vectors upto rotation). By continuity $F$ attains zero value at some intermediate point.<|endoftext|> TITLE: Question about Obstruction Bundle Gluing paper of Hutchings-Taubes QUESTION [6 upvotes]: I'm trying to learn about Embedded Contact Homology. To understand the proof of $d^2=0$, I started by watching Hutchings' lectures on Obstruction Bundle Gluing on YouTube (1, 2, 3) and have now started reading the relevant two-part paper by Hutchings-Taubes to learn more about the details of the gluing argument involved ("Gluing pseudoholomorphic curves along branched covered cylinders" 1, 2). I am stuck at the very beginning of Part I, specifically I'm unable to understand most of the proof of Lemma 1.11 which is about trajectories which are close to breaking (in what follows, I use the notation from the paper). Can someone explain (with more details) why one of (i) or (ii) must occur (it's not completely clear to me that this formally follows from the SFT compactness theorem)? I'm guessing that part of the reason for this is that the curves $u_{\pm}$ are transversally cut out and isolated in $\mathcal M^J(\alpha_+,\beta_+)/\mathbb R$ and $\mathcal M^J(\beta_-,\alpha_-)/\mathbb R$, and so taking $\delta$ sufficiently small would imply that any index $1$ component in the limit is close to one of $u_{\pm}$ and thus, is one of $u_{\pm}$. Also, why must all the other components in all other levels be branched covers of trivial cylinders (I realize they must be curves whose Fredholm indices add up to $0$, but I'm not able to rule out multiply covered curves which are not trivial cylinders from the limit - and these could have index $<0$). Finally, why is it that in case (ii) there are no other levels? REPLY [2 votes]: SFT compactness is, as far as I know, in this situation not directly applicable, since the genera of the curves $u_n$ may be unbounded (maybe I am missing something which ensures genus bounds for them in this situation, or one can use the additional assumptions to make SFT compactness applicable). In any case, there is in dimension 4 a compactness result for $J$-holomorphic currents by Taubes, which is often used for ECH, and could presumably be used here. Assuming for the moment genus bounds on the $u_n$, SFT compactness shows that a subsequence of $u_n$ converges to a holomorphic building $v$, that is "a curve" with potentially several levels $v=(v_1=:v_+, ...,v_k=:v_-)$. If there is no splitting ($k=1,v_+=v_-$), then SFT compactness ensures that a subsequence of curves converges (up to shifts in the $\mathbb{R}$-direction) to a curve with the same asymptotes, that is to some $[v]\in \mathcal{M}^J(\alpha_+,\alpha_-)/\mathbb{R}$. If there is a splitting ($k>1$), note first that the assumptions relating to "parametrization by $C_\pm$" ensure that the parts of $u_n$, whose image is parametrized by $C_\pm$, are both contained in only one level (that is, in this region there is no further splitting possible). If we use then in addition the assumption on $C_0$, we can also conclude that the lowest level $v_-$ has the same asymptotes as $u_-$ and the top level $v_+$ has the same asymptotes as $u_+$ and in the middle we have a piece (consisting potentially of none, one or several levels), whose top asymptotes coincide(!) with the bottom asymptotes (since these coincide with the asymptotes of $u_{\mp}$ (or also of $v_\mp$) at $\pm\infty$ ). Here we used throughout, that $\delta$ can be chosen so small, such that the $\delta$-neighborhood $N_\delta$ of the asymptotic orbits deformation retracts to this collection of the asymptotic orbits. In other words the asymptotes (up to time shifts) are determined by these $C^0$-data. It follows that the $d\alpha$ energy of the whole middle piece is 0 (the $d\alpha$-energy can be computed as difference of the periods of the asymptotic orbits, and depends only on the relative homology class). Any proper nonconstant curve of finite Hofer energy with $d\alpha$ energy zero is necessarily a branched cover over a collection of cylinders. Thus the whole middle piece is a branched cover over a collection of cylinders (potentially in different middle levels). It remains to identify the top and the bottom component as $u_\pm$ (we know already that the images have the same asymptotes). Since the asymptotes are given and the $d\alpha$-energy is computed from the asymptotes, it follows that the sum of the $d\alpha$-energies of the simple curves underlying $v_\pm$ coincides with the sum of $d\alpha$ energies of $u_\pm$. Since moreover the $d\alpha$ energy converges (and the $d\alpha$ energy is additive over the levels), the $d\alpha$ energy of $v$ is equal to the $d\alpha$-energy of the $u_n$'s, which is in turn the same as the sum of the $d\alpha$-energies of $u_\pm$. Therefore all of the components of $v_\pm$, which contribute $d\alpha$-energy, are simple, i.e. they are not nontrivial branched covers of other curves. Thus at most trivial cylinders in the image of $v_\pm$ are (branched) covered. On the other hand, condition 1d ensures now exactly that any (possibly branched) cover over the trivial cylinders with the given asymptotes $(a_i,b_i ...)$ are necessarily unbranched, and thus these components are determined as maps by the asymptotes; hence the components in $v_\pm$ which are (apriori branched) covers of trivial cylinders coincide with the unbranched covers occuring in $u_\pm$. The components of $v_\pm$ contributing nonzero $d\alpha$-energy are therefore parametrized by surfaces of fixed topological type (the same as for $u_\pm$), and we can (as you mentioned) choose $\delta$ small enough, to ensure that also these components coincide (as parametrized curves up to translation) with the corresponding components of $u_\pm$, as the $[u_\pm]$ are isolated in their moduli space. Edit: More on why the ends of $u_+$ coincide with the ends of $v_+$. First I should be more precise above (in paragraph 4) to say, that at that point one only knows, that the ends of $u_+$ coincide (with multiplicity) with the the ends of the simple curve underlying $v_+$. This more precise statement should hold for the following reasons: First, all negative ends of $u_+$ can be distinguished on a $\delta$-level in the target (as $u_+$ is simple and by "unique continuation as almost complex submanifold" or perhaps more directly studying the asymptotes in coordinates). Then there is for each end $e_u$ of $u_+$ negatively asymptotic to some Reeb orbit (almost parametrized by some circle $c$ in $e_u$), a circle $c'$ in the curve $v_+$ (parametrized via $\psi_+$ by $c \subset e_u$), which lies in the same component of the $\delta$-neighborhood $N_\delta$ and is homotopic in $N_\delta$ to $c$. Then the "parametrized component" of $c'$ in $N_\delta \cap v_+$ does not leave $N_\delta$ again by the assumption on $C_0$. Thus $c'$ determines an end $e_v$ of the simple curve underlying $v_+$, which wraps around the Reeb orbit like $c'$ and thus with the same multiplicity as $e_u$. This procedure works for all ends of $u_+$ and all ends of the simple curve underlying $v_+$ are among these. Then one proceeds with the discussion as above, using only what we know about the asymptotics of the simple curves underlying $v_\pm$.<|endoftext|> TITLE: What is the value of this double sum in closed form? QUESTION [8 upvotes]: I encountered the following double sum which requires an evaluation. Is there a closed form for this? $$\sum_{n=0}^{\infty}\frac{\sum_{k=0}^n\binom{n}k^{-1}}{(n+1)(n+2)}.$$ Incidentally, it seems that the following gives (empirically) the same value. Does it? $$\sum_{n=1}^{\infty}\frac1{2^{n-1}(n+1)}\sum_{k=1}^n\frac{\binom{n}{2k-1}}{2k-1}.$$ REPLY [27 votes]: Consider the integral $$I=\int_0^1\int_0^1\frac{zdzdt}{(1-zt)(1-z(1-t))}=\sum_{k,j\geqslant 0} \int_0^1\int_0^1 z^{k+j+1}t^k(1-t)^jdzdt=\\ \sum_{k,j\geqslant 0} \frac1{k+j+2}\cdot \frac{k!j!}{(k+j+1)!},$$ that is your sum (denote $n=k+j$). For evaluating the integral, we first integrate by $t$, get $-2\log(1-z)/(2-z)$. Now denote $1-z=x$ and integrate $2\log x/(1+x)=2\log x(1-x+x^2-x^3+\dots)$ using $\int_0^1 -\log x\cdot x^m dx=\frac1{(m+1)^2}$. We get $2(1-1/4+1/9-1/16+\dots)=\pi^2/6$. As for your second sum, it has indeed the same value which may be obtained similarly. We consider the integral $$ \int_0^1\int_0^1 \frac1x\left(\frac1{1-y\frac{1+x}2}-\frac1{1-y\frac{1-x}2}\right)dxdy=\\ \sum_n \int_0^1y^n dy\cdot 2^{1-n}\int_0^1\frac{(1+x)^n-(1-x)^n}{2x}dx, $$ that is your sum (expand $(1+x)^n-(1-x)^n=2\sum \binom{n}{2k-1}x^{2k-1}$ and integrate). If we integrate first in $x$, we get $-2\log(1-y)/(2-y)$, the same thing as in the first sum.<|endoftext|> TITLE: Cops, Robbers and Cardinals: The Infinite Manhunt QUESTION [13 upvotes]: Cops & Robbers is a certain pursuit-evasion game between two players, Alice and Bob. Alice is in charge of the Justice Bureau, which controls one or more law enforcement officers, the cops. Bob controls a single robber, a guy whose main concern is to evade the cops at any costs. Both deploy their guys on the vertices of a city, a simple graph, and start a manhunt under the following natural rules: On her turn, Alice chooses a (possibly empty) subset of the cops and moves them to some adjacent vertices. On his turn, Bob may either move his robber to an adjacent vertex or stay still. The game ends with a win for Alice whenever any of her cops manages to occupy the same vertex as the robber. Bob wins if this never happens in the game. Cops & Robbers and its variants have been studied on finite graphs so extensively, while little is known in the case of the infinite graphs, particularly those of uncountable cardinality which are the subject of this question. One may take a look at the following references for a fairly comprehensive account of the topic (including some results concerning the infinite version) or watch this and this episodes of PBS Infinite Series show by Kelsey Houston-Edwards: Bonato and Nowakowski, The Game of Cops and Robbers on Graphs, 2011. Florian Lehner, Cops, robbers, and infinite graphs, 2014. Jordan DuBeau and Beth Matys, Cops and Robbers on Infinite Graphs, 2015. The cop number of a graph is the minimum number of the cops that Alice needs to have a winning strategy in the game. A graph with cop number $1$ is called cop-win. Otherwise, it is called robber-win. Before getting into the main question, let's observe that the cop number of a large infinite graph may or may not be large. For instance, $K_{\kappa}$ and $S_{\kappa}$, the complete and star graph of size $\kappa$, are clearly cop-win and therefore have the cop number $1$. In contrary, the totally disconnected graph of size $\kappa$ has the cop number $\kappa$. Some other examples are the infinite path $P_{\infty}$, and Rado graph, $\mathcal{R}$, which have the cop number $\aleph_0$. Definition. We call connected infinite graphs with minimum and maximum possible cop-numbers (i.e. $1$ and $|G|$), law and crime neighborhoods respectively. A Mega City is an infinite connected graph, $M$, with the "dystopian" property that $M$ contains either a law or a crime neighborhood as big as $|M|$ as its sub-graph. An uncountable cardinal $\kappa$ is a Dredd cardinal (named after Judge Dredd) if every connected graph of size $\kappa$ is a Mega City. The first question is about the possible large cardinal strength of the existence of such a Ramsey-like cardinal. Question 1. Is there any Dredd cardinal? What is their large cardinal strength in comparison with other large cardinals of some Ramsey characterization such as weakly compact, Erdős, and Ramsey cardinals? Remark 1. An approach towards question 1 might be through considering $\kappa$-Aronszajn trees and checking the tree property which provides a characterization of weakly compact cardinals. Maybe one can relate the cofinal branches in a $\kappa$-tree to the law or crime neighborhoods in a Mega City of size $\kappa$, given the fact that trees and paths are rather simple cities to analyze for the game of cops and robbers. Also, as Will mentioned in his comments here and here, in the definition of a Mega-City the connectedness assumption is essential for the whole graph and crime neighborhoods in order to avoid easy constructions. The next question is about the possible characterization of infinite connected graphs with arbitrary cop-numbers. Question 2. Let $\kappa$ be an infinite cardinal and $1\leq \lambda\leq \kappa$. Is there a connected graph of size $\kappa$ and cop number $\lambda$? How many graphs of this type are there up to isomorphism (for any given $\kappa$, $\lambda$)? Remark 2. Concerning the question 2 in the special cases such as $\kappa=2^{\aleph_0}$, an approach could be thinking about using cardinal characteristics of the continuum for constructing examples of connected (directed) graphs of size continuum with cop numbers strictly less than $2^{\aleph_0}$. In order to observe this, note that cop number of any graph is less than or equal to its domination number because one may catch the robber in the first move simply by deploying one cop on each vertex in a minimal dominating set of the graph. The idea is that the set-theoretic dominating number, $\mathfrak{d}$, serves as the domination number of a graph, $G$, of size $2^{\aleph_{0}}$ (don't confuse the similar terms in set theory and graph theory with each other). At the request of Monroe, let me clarify the example a little bit more: $G$ is defined by the sequence domination relation between reals (i.e. functions $f:\omega\rightarrow \omega$). In this graph, the vertex labeled by $g$ is adjacent to $f$ (i.e. $g\rightarrow f$) if $g$ dominates $f$ as a sequence (denoted by $f\leq^* g$) that is $f(n)\leq g(n)$ for all but finitely many $n\in \omega$. Now a dominating family, $\mathcal{F}$, of sequences for the entire real numbers is a set of functions from $\omega $ to $\omega$ such that every such function is either in $\mathcal{F}$ or is $\leq^*$-dominated by a member of $\mathcal{F}$ (i.e. $\mathcal{F}$ is a domination set of $G$ in the graph-theoretic sense) and $\mathfrak{d}$ is defined to be the minimum size of such a family (i.e. the domination number of the graph $G$). Consequently we have $c(G)\leq \mathfrak{d}\leq 2^{\aleph_0}$ (which $c(G)$ stands for the cop number of $G$). Finally using forcing one may make $\mathfrak{d}<2^{\aleph_0}$ and so the cop number of the graph $G$ in the generic extension will be strictly smaller than $2^{\aleph_{0}}$. However, one should note that generally, the domination number is a very loose upper bound for the cop number of a graph. (Certainly, the Justice Bureau has much more clever strategies to control the crime rate in the city rather than placing a cop right in the front of every single home!) For instance, the domination number of the dodecahedron graph (of size $20$) is $7$ while its cop-number is only $3$. Here, the following question arises: Question 3. What is the cop-number of $G$, the sequence domination graph of size continuum described in the previous remark? Can it be countable? If not, what is the precise value of this new cardinal characteristic of continuum and how does it relate to the other entities in the Cichoń's diagram? REPLY [4 votes]: Here is an answer to question 2. First note that it suffices to consider the case where $\lambda=\kappa.$ For regular $\kappa,$ you may consider the following graph: The point is that given less than $\kappa$-many points on the graph, there exists a line on which there are no points, except possibly on the center. Similar idea works for singular cardinals, using the following graph. Here $\lambda = cf(\kappa)$ and $(\kappa_\xi: \xi< \lambda)$ is an increasing sequence of regular cardinals cofinal in $\kappa:$<|endoftext|> TITLE: Schur covers of affine 2-transitive groups QUESTION [5 upvotes]: I am interested in Schur covers of minimal 2-transitive groups. A theorem of Burnside gives that every finite 2-transitive group is either almost simple or affine. In the time since, these groups have been classified as a consequence of the classification of finite simple groups (see Section 2 of this paper, for example). Since Schur covers were already determined for the simple groups in order to perform their classification, the Schur covers of almost simple groups are relatively easy to obtain. By contrast, I am having difficulty finding the Schur covers in the affine case. Question: Where can I find descriptions of the Schur covers of minimal affine 2-transitive groups? A good answer looks like: "The Schur cover of PSL(n,q) is SL(n,q), with a few exceptions. See Karpilovsky, The Schur Multiplier, p.246." but with a minimal affine 2-transitive group in place of PSL(n,q). The following appear related, but are not what I'm asking for: "What is the Schur multiplier of the affine linear group AGL(n,q)?" computes the Schur multiplier of AGL(n,q). However, AGL(n,q) is not a minimal 2-transitive group, and I want a Schur cover, not just the multiplier. "Have finite doubly transitive groups been classified?" discusses the classification of affine 2-transitive groups at length. It does not mention Schur multipliers or covering groups. REPLY [2 votes]: Here is an approximate answer. I believe that it is substantially correct, but there might be some small exceptions which I have not thought of. I think that all the examples that are subgroups of ${\rm A \Gamma L}(1,q)$ have trivial multiplier, with a similar argument to that for ${\rm A G L}(1,q)$. The groups of the form $G=N \rtimes H$ with $H = {\rm SL}(n,q)$ and $n \ge 3$ have trivial Schur Multiplier, possibly with a few exceptions. One exception is when $H = {\rm SL}(3,4)$, in which $G$ has nontrivial multiplier $C_4^2$, and the covering group of $G$ is $N \rtimes \hat{H}$, where $\hat{H}$ is the unique covering group of ${\rm SL}(3,4)$. For $G=N \rtimes H$ with $H = {\rm Sp}(n,q)$ with $n \ge 2$ even and and $q$ odd,the multiplier is elementary abelian of order $q$, and the covering group has structure $\hat{N} \rtimes H$, where $\hat{N}$ is a special group of exponent $p$ (where $q$ is a power of the prime $p$) and order $q^{n+1}$. The commutator map in $\hat{N}$ is the symplectic form on $N$ that is preserved by $H$. In a few small cases, probably just with $n=24$, these groups are not minimally $2$-transitive. For example $N \rtimes {\rm SL}(2,3) < N \rtimes {\rm Sp}(2,5)$, $N \rtimes {\rm GL}(2,3) < N \rtimes {\rm Sp}(2,7)$, and $N \rtimes {\rm SL}(2,5) < N \rtimes {\rm Sp}(2,11)$, but in those cases the covering group is still obtained by replacing $N$ by $\hat{N}$. For the examples of form $G=N \rtimes H$ with $H = {\rm Sp}(n,q)$ and $q$ even, and also their subgroups with $H = G_2(q)'$ when $n=6$, what seems to happen is that the covering group again has the form $\hat{N} \rtimes H$ with $|N| = q^{n+1}$, but now $\hat{N}$ is elementary abelian, and is a nonsplit module extension of the trivial module with the natural module for $H$ over ${\mathbb F}_q$. Such module extensions correspond to $H^1(H,N)$, which appears to be $1$-dimensional in these cases. (The first cohomology groups of classical groups on their natural modules are all known - I can check this later.) Again there may be some small exceptions. For example ${\rm Sp}(4,2)$ and ${\rm Sp}(6,2)$ have Schur multipliers of order $2$, so $H$ has to be replaced by its covering group $\hat{H}$ in those cases in the covering group of $G$.<|endoftext|> TITLE: Moduli of stable and semistable $G$-Higgs bundles on curves QUESTION [7 upvotes]: Let $X$ be an irreducible smooth projective curve of genus $g \geq 2$ over $\mathbb{C}$. Let $G$ be a connected reductive affine algebraic group over $\mathbb{C}$. Let $\mathcal{M}_{G,Higgs}^s$ (resp., $\mathcal{M}_{G,Higgs}^{ss}$) be the moduli space of stable (resp., semistable) principal $G$-Higgs bundles on $X$. Is it correct that the codimension of the complement of $\mathcal{M}_{G,Higgs}^s$ inside $\mathcal{M}_{G,Higgs}^{ss}$ has codimension at least $2$? Can anyone give reference for this? Is it true that the same holds for the case of moduli stack of $G$-Higgs bundles also? REPLY [6 votes]: As far as I know this is an open problem outside the case $G=GL_n(\mathbb{C})$ or $G=SL_n(\mathbb{C})$. In those cases, work of Simpson (Moduli of representations of the fundamental group of a smooth projective variety II, Publications Mathématiques de l'IHÉS, Volume 80 (1994), p. 5-79) implies the moduli spaces of Higgs bundles is normal (he proved it for $G=GL_n(\mathbb{C})$, but his proof works for $G=SL_n(\mathbb{C}$)). Hence, the singular locus is in codimension at least 2 (as is clear from the reference, I am considering the case of trivial topological type here; that is, bundles with vanishing rational Chern classes). But in those cases I believe the stable locus is the smooth locus. It is well-known that the stable $GL_n(\mathbb{C})$ or $SL_n(\mathbb{C})$-Higgs bundles are smooth points. The converse should follow since the underlying holomorphic principal bundle is singular except in the $g=2=n$ case (and in that case the strictly polystable Higgs bundles should be seen to be singular directly). For general complex reductive $G$, I believe that Simpson's normality result generalizes (although it is open presently). However, in general, the stable locus is not always the smooth locus (there can be orbifold singularities in the stable locus). So to establish the codimension result along the same lines as above you would have to consider the orbifold singular locus in the stable locus. But there should be a proof of the codimension result from this point-of-view regardless. In fact, I believe the following stronger result holds: the codimension of the stable locus for $G$-Higgs is $\geq 2(g-1)$ for a Riemann surface of genus $g\geq 2$. Here is a rough outline why: Let $N$ be the $G$-principal bundles and $M$ the $G$-Higgs and put a $*$ on each for stable subspace. The smooth locus of $N$ is open and dense and the cotangent bundle of the smooth locus is contained in $M$ as an open dense subset. So $\dim M=2\dim N$ The complement $N-N^*$ consists of polystable and not stable bundles and so correspond to a direct sum of stable subbundles. This locus is stratified by smooth sub-spaces and so the top strata determines the dimension. Call this strata $S$. Then the cotangent bundle of $S$ should determine the dimension of $M-M^*$. Therefore, $\dim(M-M^*)=2\dim(N-N^*)$. We have by a theorem of Biswas, Hoffmann $\mathrm{codim}(N-N^*) \geq g-1$ which implies $\dim(N)-\dim(N-N^*)\geq g-1$ which implies $2\dim(N)-2\dim(N-N^*)\geq 2(g-1)$ which implies $\mathrm{codim}(M-M^*) \geq 2(g-1).$ And so we have the codimension is $\geq 2$ if $g\geq 2$. Step 2 is not obvious since a strictly semistable principal bundle may admit a non-zero Higgs field which make it stable. But I believe the dimension will be right regardless (unless I am missing something Steps 1 and 3 are correct though).<|endoftext|> TITLE: Lipschitz right inverses of Banach space quotients QUESTION [8 upvotes]: Let $X$ be a Banach space and $Y$ a closed subspace of $X$. I am interested in quotients $q:X\to X/Y$ that do not have Lipschitz right inverses (not necessarily linear). Of course, if $Y$ is complemented, then the quotient always has a Lipschitz (bounded linear) right inverse. The only examples I know of that do not have Lipschitz right inverses are: $\ell^\infty / c_0$ by Kalton [1, Theorem 4.2]. A quotient of a certain Càdlàg type space with $C[0,1]$ by Lindenstrauss and Aharoni [2, Remark ii]. I have two questions: Are there any more examples known? Is this perhaps a general feature of non-complemented subspaces of Banach spaces? Or is there an example of a non-complemented subspace $Y$ of some $X$ so that the quotient $q:X\to X/Y$ has a Lipschitz right inverse? [1] Nigel Kalton. Lipschitz and uniform embeddings into ℓ∞. Fund. Math. 212 (2011), no. 1, 53–69. [2] Aharoni, Israel; Lindenstrauss, Joram. Uniform equivalence between Banach spaces. Bull. Amer. Math. Soc. 84 (1978), no. 2, 281–283. REPLY [7 votes]: The answer to (2) is yes and is contained in your reference [2]. That was how Aharoni and Lindenstrauss were able to construct two Lipschitz equivalent Banach spaces that are not isomorphic. Note that their example is non separable. Whether there are two Lipschitz equivalent non isomorphic separable Banach spaces is a famous open problem. Godefroy and Kalton proved that the Aharoni-Lindenstrauss approach cannot work in the separable setting. That is, if a quotient mapping from a separable space has a Lipschitz right inverse, then it has a bounded linear right inverse. This is contained in Godefroy, G.(F-PARIS6-E); Kalton, N. J.(1-MO) Lipschitz-free Banach spaces. (English summary) Dedicated to Professor Aleksander Pełczyński on the occasion of his 70th birthday. Studia Math. 159 (2003), no. 1, 121–141.<|endoftext|> TITLE: graph signal processing QUESTION [8 upvotes]: I have read this article https://arxiv.org/abs/1307.5708 about vertix-frequency analysis on graph. David IShuman in this article claims that,"we generalize one of the most important signal processing tools – windowed Fourier analysis – to the graph setting and When we apply this transform to a signal with frequency components that vary along a path graph, the resulting spectrogram matches our intuition from classical discrete-time signal processing. Yet, our construction is fully generalized and can be applied to analyze signals on any undirected, connected, weighted graph." What's the intuition behind a ''Graph fourier transform'' ? I'm not so much interested in mathematical details or technical applications. I'm trying to grasp what a graph fourier transform actually represents,and what aspects of a graph it makes accessible. To clarify the issue: Graph-structured data appears in many modern applications like social networks, sensor networks, transportation networks and computer graphics. These applications are defined by an underlying graph (e.g. a social graph) with associated nodal attributes (e.g. number of ad-clicks by an individual). A simple model for such data is that of a graph signal—a function mapping every node to a scalar real value The classical Fourier transform is the expansion of a function fin terms of the eigenfunctions of the Laplace operator; i.e., $$ \hat{f} = \langle f, e^{2\pi i \xi t } \rangle$$ Analogously, the graph Fourier transform $\hat{f}$ a function $f \in \mathbb{R}^{N}$ the vertices of graph $G$ the expansion of $f$ in terms of the eigenfunctions of the graph Laplacian. It is defined by $$\hat{f}( \lambda_{l}) = \langle f , \chi_{l}\rangle = \sum_{n=1}^{N} f(n) \chi_{l}^* (n) .$$ I am looking for some simple concrete examples of the ways in which real problems go through graph signal processing and how graph Fourier transforms are obtained. REPLY [8 votes]: "I am looking for some simple concrete examples of the ways in which real problems go through graph signal processing and how graph Fourier transforms are obtained." • A concrete example of a graph Fourier transform, to the Minnesota road network, is presented in Fourier Analysis on Graphs; another example, to genetic profiling for cancer subtype classification, is discussed in Graph SP: Fundamentals and Applications. The graph Fourier transform allows one to introduce the notion of a "band width" to a graph. By analogy with smooth time signals, which have a narrow frequency band width, a graph that exhibits clustering properties (signals vary little within clusters of highly interconnected nodes) will have a narrow band width in the graph Fourier transform. Such a clustered graph would be sparse in the frequency domain, allowing for a more efficient representation of the data. • To obtain the graph Fourier transform you could use the Matlab routine GSP_GFT in the Graph Signal Processing Toolbox.<|endoftext|> TITLE: When is a zero dimensional local ring a chain ring? QUESTION [5 upvotes]: A commutative ring with identity is called a chain ring if all its ideals form a chain under inclusion. I want to know is there any proof for the fact that a zero dimensional local ring is a chain ring whenever its maximal ideal is principal? REPLY [4 votes]: A famous theorem by Kaplansky says that a commutative ring is a principal ideal ring iff all of its prime ideals are principal. By using a zero-dimensional local ring with a principal maximal ideal, you are in that situation. A commutative, local principal ideal ring is well-known to be a chain ring (a.k.a. uniserial ring) as discussed in the wiki.<|endoftext|> TITLE: Isometries and fixed points QUESTION [5 upvotes]: I am new to geometric group theory and I am trying to read a bit to expand my horizons. I have encountered the following theorem: Suppose that $G$ is a group that has a free action by isometries on $\mathbb{R}^n$ with the Euclidean metric. Then $G$ is torsion free. I am wondering about generalisations of it: I think it follows from the proof that if an isometry (that is it the cyclic group generated by it) acts on $\mathbb{R}^n$ and it has a finite orbit, then it has a fixed point. Am I right? I saw somewhere that the theorem holds if one can replace isometry by a continuous function. I think I can prove it for $n=1$ because then a convex set is the same as a connected set, namely, an interval. Does anyone know a reference for the proof? I assume the theorem fails if $\mathbb{R}$ is replaced by $\mathbb{Q}$. Can anyone give a counter example? What happens if $\mathbb{R}$ is replaced by the $p$-adics? Is there a version in characteristic $p$, but maybe instead of torsion free you cannot have orders that are coprime to $p$? REPLY [6 votes]: An isometry must be an affine-linear map. If it has a finite orbit then the average of the orbit is a fixed point. This works equally over $\bf R$ or over $\bf Q$, and should answer questions 1 and 3. Over ${\bf Q}_p$, if we limit to affine-linear maps, the same argument works, as it does in characteristic $p$ for elements order coprime to $p$ (a necessary hypothesis because translations have order $p$). This would also answer questions 4 and 5. But vector spaces over ${\bf Q}_p$ and over finite fields can have isometries that are not affine-linear, and such an isometry might fail to have a fixed point whether or not it has order coprime to $p$. Question 2 is of a rather different flavor. I see that Lee Mosher gave a cohomological proof in the comments. REPLY [3 votes]: Let isometry $\ F:\mathbb R^n\rightarrow \mathbb R^n\ $ and $\ a\in \mathbb R^n\ $ be such that the $t$-fold composition $\ F^t(a)=a\ $ brings $\ a\ $ back to itself for a positive integer $\ t.\ $ Remember that $\ F\ $ is affine. Thus $$ b\ :=\ \frac{\sum_{s=0}^{t-1}F^s(a)}t $$ is a fixed point, $\ F(b)=b.$ REPLY [2 votes]: There is a non-trivial generalization of the result you mention to a whole class of Riemannian manifolds called the Cartan-Hadamard theorem https://en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem. A very nice proof is given in the book of P. Petersen, Riemannian geometry.<|endoftext|> TITLE: Morphism between formal groups and cohomology theories QUESTION [12 upvotes]: Background: $FGL(R)$ will be the category of formal group laws over the ring $R$. If $(E,x_E)$ is an oriented spectrum (i.e. $x_E \in \tilde{E}^2(\mathbb{P}^{\infty})$ is sent to the unit $1 \in \tilde{E}^2(\mathbb{P}^1)$ by the pullback of the inclusion $i: \mathbb{P}^1 \hookrightarrow \mathbb{P}^\infty$), then the pullback of the multiplication $\mu$ as H-space of $\mathbb{P}^\infty$ define an element in $FGL(E^*(*))$ in the following way: $ \mu^*:E^*(*)[[x_E]] \cong E^*(\mathbb{P}^\infty) \to E^*(\mathbb{P}^\infty \times \mathbb{P}^\infty) \cong E^*(*)[[x_1,x_2]] $ where $x_i : = p_i^*(x_E)$ and the isomorphism are given by the collapsing of the Atiyah Hirzebruch spectral sequence. The fgl associated to $(E,x_E)$ is by definition $\mu^*(x_E) \in FGL(E^*(*))$. The Landweber exact theorem 'reverse' this contruction: let $f \in FGL(R)$ be a formal group law over $R$, then by certein hypothesis on $f$ the association $ X \to MU^\bullet(X)\otimes_{\pi_*(MU)} R$ is a cohomology theory on CW-complexes ($MU$ is the complex cobordism spectrum and the $\pi_*(MU)$-module structure on $R$ is given by the classification map $\pi_*(MU) \to R$, since by Quillen's theorem the fgl of the complex cobordism is the universal one). This correspondence given by the Landweber's theorem seems to be only objectwise. Question: Given a morphism of formal group laws $\varphi$ over $R$, which is a power series in $R[[x]]$, is there a way to produce a natural transformation of the related cohomology theories induced by $\varphi$? Is there any reference in the literature where i could find an answer? REPLY [12 votes]: Yes, if $E$ and $E'$ are Landweber exact, then ring maps $E\to E'$ biject with isomorphisms of the associated formal groups, suitably interpreted. One version of this (for the context where $E$ and $E'$ are $2$-periodic) is explained in Proposition 8.43 of this Memoir. There is also a rather terse account in Proposition 11 of this lecture of Lurie.<|endoftext|> TITLE: $R$ a DVR with fraction field $K.$ What are the $R$-submodules of $K^n?$ QUESTION [7 upvotes]: It is known that if $R$ is a DVR with fraction field $K,$ then the $R$-submodules of $K$ are $0,K,x^nR,$ with $n$ any integer and $x$ a generator of the maximal ideal of $R.$ I was wondering if there is a simple structure theorem for the $R$-submodules of $K^n,$ the $n$-dimensional vector space over $K$ (similar to that for finitely generated torsion-free modules over a Dedekind domain). I know that the subroups of $\mathbb{Q}$ are somewhat complicated to describe, but was wondering, in particular, if the situation gets any nicer with, say, $\mathbb{Z}_P$-submodules of $\mathbb{Q}^n,$ where $P$ is a nonzero prime ideal of $\mathbb{Z}.$ REPLY [6 votes]: Passing to the spanned subspace, it is enough to consider those submodule that span $K^n$ as a $K$-module. Then up to composition by some element of $\mathrm{GL}_n(K)$, we can suppose that the submodule $V$ contains $R^n$. So this reduces to classifying submodules of $(K/R)^n$. This is an artinian module, actually a module over $\hat{R}$, the completion of $R$. Write $S=K/R$. Matlis duality $\mathrm{Hom}(-,S)$ yields a natural correspondence between $\hat{R}$-submodules of $S^n$ and the quotient $\hat{R}$-modules of $\hat{R}^n$. In particular, modulo composition by an element of $\mathrm{GL}_n(\hat{R})$, every submodule of $S^n$ can be written as $S^k\times F$, where $F=\prod_{i=k+1}^nF_i\subset S^{n-k}$ has finite length. Actually, we can suppose $F=0$, composing beforehand by another element of $\mathrm{GL}_n(K)$. In particular, Luc's initial claim is correct when $R$ is complete (but fails as soon as it's not complete and $n\ge 2$).<|endoftext|> TITLE: Isometric immersions of $S^2$ to $S^4$ QUESTION [8 upvotes]: There are non-standard isometric immersions of the sphere $S^2(r)$ of radius $01$. The original question about the case $r=1$ is still not answered. REPLY [2 votes]: Here is a simple source of such immersions for $r<1$. Consider geodesic subspheres $\mathbb{S}^2\subset \mathbb{S}^3\subset \mathbb{S}^4$. Take a closed curve $\gamma$ in $\mathbb{S}^2$. Pass to its spherical suspension; it is embedded in $\mathbb{S}^3$. Pass to the spherical suspension again; you get an immersed singular surface $\Sigma^3$ in $\mathbb{S}^4$ which admits an immersion of hemisphere $$\iota\colon\mathbb{S}^3_+\looparrowright\Sigma^3.$$ Note that for $r<1$, there are isometric embeddings $f\colon\mathbb{S}^2(r)\to \mathbb{S}^3_+$ and for the right choice of $f$ and $\gamma$, the composition $\iota\circ f\colon\mathbb{S}^2(r)\to \mathbb{S}^4$ has self-intersections.<|endoftext|> TITLE: Asymptotic rate for $\sum\binom{n}k^{-1}$ QUESTION [5 upvotes]: This MO question prompted me to ask: What is the second order asymptotic growth/decay rate for the sum $$\sum_{k=0}^n\frac1{\binom{n}k}$$ as $n\rightarrow\infty$? REPLY [5 votes]: The following asymptotic expansion is proved in https://arxiv.org/abs/0904.1757 (The Hypercube of Resistors, Asymptotic Expansions, and Preferential Arrangements, by Nicholas Pippenger. Published in Mathematics Magazine 83(N5) (2010), 331-346): $$\sum_{k=0}^n\frac{1}{\binom{n}{k}}=2\left(1+\frac{1}{n}+\frac{2}{n(n-1)}+\ldots \frac{k!}{n(n-1)\cdots(n-k+1)}+O(\frac{1}{n^{k+1}})\right).$$ P.S. This result confirms the answer given in the Gerhard Paseman's comment.<|endoftext|> TITLE: Definitions of $\pi_1 \times \pi_2, \pi_1 \boxplus \pi_2, \pi_1 \boxtimes \pi_2$ QUESTION [15 upvotes]: Let $\pi_i$ be a smooth, admissible (possibly irreducible) representation of $\operatorname{GL}_{n_i}(k)$ for $k$ a $p$-adic field. I have seen the following representations defined in terms of $\pi_1$ and $\pi_2$: $\pi_1 \times \pi_2$ (Rankin–Selberg product?) $\pi_1 \boxplus \pi_2$ (isobaric sum) $\pi_1 \boxtimes \pi_2$ (isobaric product) How are these representations defined exactly? Can they be defined easily in terms of the Local Langlands correspondence? REPLY [15 votes]: The right person to answer this question is probably Dinakar Ramakrishnan (a good working reference is his paper "Modularity of the Rankin–Selberg $L$-series, and multiplicity one for $\mathrm{SL}(2)$"), but since he doesn't seem to use MathOverflow, here is my understanding of this. In each case, I will first discuss the local theory, then the global theory. The isobaric sum $\pi_1 \boxplus \pi_2$ (which Langlands called the "sum operation", and by some is called the Langlands sum) is perhaps the easiest to explain. The key property is that the $L$-function of the isobaric sum is the product of the two $L$-functions: \[L(s,\pi_1 \boxplus \pi_2) = L(s,\pi_1) L(s,\pi_2).\] In terms of representations of $\mathrm{GL}_{n_1}(F)$ and $\mathrm{GL}_{n_2}(F)$, where $F$ is a local field, $\pi_1 \boxplus \pi_2$ is simply the normalised parabolic induction from the Levi subgroup $\mathrm{M} \cong \mathrm{GL}_{n_1}(F) \times \mathrm{GL}_{n_2}(F)$ to $\mathrm{GL}_{n_1 + n_2}(F)$. Via the local Langlands correspondence, in terms of the corresponding $n_1$- and $n_2$-dimensional Weil–Deligne representations $\rho_1$ and $\rho_2$, the isobaric sum simply corresponds to the direct sum $\rho_1 \oplus \rho_2$; note that this is a $n_1 + n_2$-dimensional (reducible) representation of the same group. More precisely, by $\pi_1 \boxplus \cdots \boxplus \pi_r$, I mean the induced representation $\mathrm{Ind}_{\mathrm{P}(F)}^{\mathrm{GL}_n(F)} \bigotimes_{j = 1}^{r} \pi_j$ of $\mathrm{GL}_n(F)$, where $n = n_1 + \cdots + n_r$ and $\bigotimes_{j = 1}^{r} \pi_j$ denotes the outer tensor product of $\pi_1,\ldots,\pi_r$, which is a representation of the Levi subgroup $\mathrm{M} \cong \mathrm{GL}_{n_1}(F) \times \cdots \times \mathrm{GL}_{n_r}(F)$; this is then trivially extended to a representation of the parabolic subgroup $\mathrm{P}$ with Levi subgroup $\mathrm{M}$, then induced to a representation of $\mathrm{GL}_n(F)$. If each $\pi_j$ is essentially square-integrable, then $\pi_1 \boxplus \cdots \boxplus \pi_r$ is called an induced representation of Whittaker type (since these representations have a Whittaker model). These need not be irreducible, but the Langlands quotient theorem states that every irreducible admissible representation of $\mathrm{GL}_n(F)$ is unitarily equivalent to the quotient of an induced representation of Whittaker type. This extends naturally to the global setting: if $\pi_1$ and $\pi_2$ are automorphic representations of $\mathrm{GL}_{n_1}(\mathbb{A}_F)$ and $\mathrm{GL}_{n_2}(\mathbb{A}_F)$, where now $F$ is a global field, then $\pi_1 \boxplus \pi_2$ is an automorphic representation of $\mathrm{GL}_{n_1 + n_2}(\mathbb{A}_F)$; the automorphic forms that for a vector space for this automorphic representation are Eisenstein series induced from the Levi subgroup $M \cong \mathrm{GL}_{n_1} \times \mathrm{GL}_{n_2}$. In particular, a classical Eisenstein series on $\mathrm{GL}_2$ has two characters associated to it (possibly the trivial characters), and this is simply induced from $\mathrm{GL}_1 \times \mathrm{GL}_1$, and the $L$-function of such an Eisenstein series is simply the product of the $L$-functions of the two characters. (As an etymological aside, isobaric means equal pressure, which is somewhat incongruous since $\pi_1$ and $\pi_2$ can have different dimensions. Moreover, the isobaric sum is usually only used to describe the global operation, not the local one, though I personally believe it is appropriate to use it in the local setting.) The point of the isobaric sum is that it allows one to break up (local) representations into parts that cannot be isobarically decomposed any further; these are the essentially square-integrable representations. The global analogue of these isobarically indecomposable representations are cuspidal automorphic representations. For $\pi_1 \boxtimes \pi_2$ and $\pi_1 \times \pi_2$ (and $\pi_1 \otimes \pi_2$), the answer is a little less clear, because authors use these notations interchangeably. To me, $\boxtimes$ ought to denote the outer tensor product, so $\pi_1 \boxtimes \pi_2$ ought to denote a representation of the product of groups $\mathrm{GL}_{n_1}(F) \times \mathrm{GL}_{n_2}(F)$; more precisely, should think of $\boxtimes$ as a map from $\mathcal{R}(\mathrm{GL}_{n_1}(F)) \times \mathcal{R}(\mathrm{GL}_{n_2}(F))$ to $\mathcal{R}(\mathrm{GL}_{n_1 n_2}(F))$, where $\mathcal{R}(\mathrm{GL}_n(F))$ denotes the set of irreducible admissible representations of $\mathrm{GL}_n(F)$. In terms of $\rho_1$ and $\rho_2$, this should simply be the tensor product $\rho_1 \otimes \rho_2$, which is a $n_1 n_2$-dimensional (possibly reducible) representation of the same group. Since the local Langlands correspondence is a theorem, this means there is a representation $\pi_1 \times \pi_2$ (or $\pi_1 \otimes \pi_2$) of $\mathrm{GL}_{n_1 n_2}(F)$ corresponding to the outer tensor product $\pi_1 \boxtimes \pi_2$. Locally, this is reasonably well-understood via the Langlands correspondence: if \[\pi_1 = \boxplus_{j = 1}^{m_1} \pi_{1,j}, \qquad \pi_2 = \boxplus_{k = 1}^{m_2} \pi_{2,k},\] then \[\pi_1 \times \pi_2 = \boxplus_{j = 1}^{m_1} \boxplus_{k = 1}^{m_2} \pi_{1,j} \boxtimes \pi_{2,k}.\] Equivalently, if \[\rho_1 = \bigoplus_{j = 1}^{m_1} \rho_{1,j}, \qquad \rho_2 = \bigoplus_{k = 1}^{m_2} \rho_{2,k},\] then \[\rho_1 \otimes \rho_2 = \bigoplus_{j = 1}^{m_1} \bigoplus_{k = 1}^{m_2} \rho_{1,j} \otimes \rho_{2,k}.\] So everything is reduced to essentially square-integrable representations (in which case understanding this tensor product can be complicated). For the global situation, things are a little murkier, since functoriality is not a theorem in general. In particular, the issue is that if $\pi_1$ and $\pi_2$ are automorphic representations of $\mathrm{GL}_{n_1}(\mathbb{A}_F)$ and $\mathrm{GL}_{n_2}(\mathbb{A}_F)$, where now $F$ is a global field, it is not known if $\pi_1 \times \pi_2$ is an automorphic representation of $\mathrm{GL}_{n_1 n_2}(\mathbb{A}_F)$. That is, we know how to define the global $L$-function $L(s,\pi_1 \times \pi_2)$ of this object (Jacquet, Piatetski-Shapiro, Shalika: "Rankin–Selberg Convolutions") as a product of local $L$-functions, as well as (more or less) how to describe the local components, but we don't know that the global object obtained by gluing together these local components is a genuine automorphic representation. This is the issue of the functorial transfer of the tensor product. Notationally, often $\pi_1 \otimes \pi_2$ is used in place of $\pi_1 \times \pi_2$, which is, I suppose, an artefact of the fact that this corresponds to a genuine tensor product on the Weil–Deligne side. Moreover, when $n_2 = 1$, so that $\pi_2 = \chi$ is a character, then functoriality is a theorem, and we usually write $\pi_1 \otimes \chi$ (but again, $\pi_1 \times \chi$ still appears regularly in the literature). Note that $\pi_1 \times \pi_2$ is called the Rankin–Selberg product or Rankin–Selberg convolution, but that really we only ever mean the $L$-function $L(s,\pi_1 \times \pi_2)$, since in general we don't know the hypothetical automorphic object associated to this $L$-function. In this setting, when this global object is known to be a genuine automorphic representation of $\mathrm{GL}_{n_1 n_2}(\mathbb{A}_F)$, Ramakrishnan denotes this by $\pi_1 \boxtimes \pi_2$ . That is, $\boxtimes$ denotes a map from $\mathcal{A}(\mathrm{GL}_{n_1}(\mathbb{A}_F)) \times \mathcal{A}(\mathrm{GL}_{n_2}(\mathbb{A}_F))$ to $\mathcal{A}(\mathrm{GL}_{n_1 n_2}(\mathbb{A}_F))$, where $\mathcal{A}(\mathrm{GL}_n(\mathbb{A}_F))$ denotes the set of automorphic representations of $\mathrm{GL}_n(\mathbb{A}_F)$, such that $L(s,\pi_1 \boxtimes \pi_2) = L(s,\pi_1 \times \pi_2)$. This is the only usage of $\boxtimes$ in the global setting that I have seen. (I don't believe that $\boxtimes$ is given a name, but if it is, it's definitely not the isobaric product.)<|endoftext|> TITLE: $S^3 \setminus S^1$ doesn't have hyperbolic structure QUESTION [6 upvotes]: I need to prove that $M = S^3 \setminus S^1$ doesn't admit any metric of constantly negative sectional curvature s.t. $M$ is complete respect for this metric. I know that it is consequence of famous Thurston Theorem, but it is quite uncomfortable to use it, so, maybe there is exist direct argument? REPLY [10 votes]: The following contribution comes from conversations with Bill Goldman, any mistakes however are mine alone. Any (geodesically complete) geometric 3-manifold $N=\mathbb{M}/G$ with infinite order elements in its fundamental group (isomorphic to $G$) is covered by an open solid torus admitting the same geometric structure (having model geometry $\mathbb{M}$). Note that $G$ is a subgroup of the isometry group of $\mathbb{M}$ and arises from a discrete faithful homomorphism $\pi_1(N)\to Isom(\mathbb{M})$. To see this, let $T\in G$ be of infinite order. Then the covering space of $N$ given by $\mathbb{M}/\langle T\rangle$ is an open solid torus (coverings of manifolds admitting geometric structures admit geometric structures themselves). Although I have not verified it, I am pretty sure that $\mathbb{M}$ can be any of Thurston's eight geometries excepting only $\mathbb{S}^3$. Here is an example construction (when $\mathbb{M}$ is hyperbolic 3-space $\mathbb{H}^3$). Take a hyperbolic line $L$ (complete geodesic) in $\mathbb{H}^3$ and a hyperbolic translation (loxodromic element) along that line. Call the translation $T$. Consider the cyclic group $\langle T\rangle.$ Then $T$ leaves the line $L$ invariant. Take any unit disk $D$ orthogonal to $L$. Translate $D$ by $T$ along $L$ to obtain the disk $T(D)$. The two disks $D$ and $T(D)$ bound a fundamental domain for $\langle T\rangle$. Fundamental Domain in Ball Model; Image made by Marvin Castellon The quotient of the fundamental domain by $\langle T\rangle$ is homeomorphic to a solid torus (one can do the same thing when $T$ is parabolic). The resulting manifold is geodesically complete. The volume of such a torus must be infinite. For otherwise, by Mostow Rigidity, there would be a unique hyperbolic structure. However, from the construction above we see there are infinitely many distinct structures (pick any non-conjugate loxodromic element). So in your case the manifold is topologically an open solid torus (assuming $S^1$ is unknotted) and so it admits many different complete geometric structures (including many hyperbolic structures).<|endoftext|> TITLE: Domain in which every ideal is generated by at most three elements QUESTION [6 upvotes]: Does anyone know of a domain in which every ideal is generated by at most three elements at least one ideal is generated by no less than three elements? What if three is replaced by n: a positive integer bigger than 3? REPLY [8 votes]: This is too long to be a comment. If we restrict our attention to the case of finitely generated ideals, then there are two classical results by Heitmann and Swan. Theorem 1 (Heitmann 1976) A finitely generated ideal in a Prüfer domain of Krull dimension $n$ needs at most $n + 1$ generators. Theorem 2 (Swan 1984) For every $n \in \mathbb{N}$, there exists a Prüfer domain of Krull dimension $n$ with an ideal $I$ that cannot be generated by less than $n+1$ elements. Unfortunately, we cannot hope to use the above results in order to give an answer to the original question, since a Noetherian Prüfer domain is a Dedekind domain, and it is well-known that in such a domain every ideal is either principal or generated by two elements, see MSE question 597543. References. R. C. Heitmann: Generating ideals in Prüfer domains, Pacific J. Math., Volume 62, Number 1 (1976), 117-126. R. G. Swan: $n$-generator ideals in Prüfer domains, Pacific J. Math., Volume 111, Number 2 (1984), 433-446.<|endoftext|> TITLE: Is the image of the map $A \to \bigwedge^k A$ closed over $\mathbb{R}$? QUESTION [12 upvotes]: Let $V$ a real vector space of dimension $d$. Let $1 TITLE: Singular abelian surfaces that can be defined over $\mathbb Q$ QUESTION [5 upvotes]: An abelian surface $A$ is called singular if it has maximal Picard number $\rho(A) = 4$. By work of Shioda-Mitani, any singular abelian surface $A$ is the product $A = E_1 \times E_2$ of two isogenous elliptic curves with complex multiplication. If both $E_1$ and $E_2$ are defined over $\mathbb Q$, then $A$ is of course defined over $\mathbb Q$. Are there examples of singular abelian surfaces $A$ defined over $\mathbb Q$ which cannot be written as the product of two elliptic curves defined over $\mathbb Q$? REPLY [6 votes]: This is a complement to Joe Silverman's answer and is a bit long for a comment. One seeks to find a $\mathbb{Q}$-curve $E$ so that $\mathbb{Q}(j(E))$ is a quadratic extension of $\mathbb{Q}$. One such $\mathbb{Q}$-curve is $E : y^{2} + \sqrt{2} xy + y = x^{3} + x^{2} + (-2\sqrt{2} - 3) x + \sqrt{2} + 1$, which has CM by $\mathbb{Z}[\sqrt{-6}]$. Moreover, according to the LMFDB database of genus 2 curves, if $C : y^{2} + x^{3} y = x^{3} + 2$, then the Jacobian of $C$ is isogenous to the square of $E$. Is it true that the Jacobian of $C$ is isomorphic to this product? Apparently, Shioda and Mitani proved that if an abelian surface is isogenous to a product of two isogenous elliptic curves with CM, then it is isomorphic to a product of two elliptic curves (which must be isogenous and have CM, necessarily). Note: I don't think one needs to have a $\mathbb{Q}$-curve whose $j$-invariant is defined over a quadratic extension. Another example is $C : y^{2} = x^{5} - x$ whose Jacobian decomposes as the square of a $\mathbb{Q}$-curve defined over $\mathbb{Q}(\sqrt{2})$ which is not a base change from $\mathbb{Q}$, but whose $j$-invariant is $8000$. REPLY [4 votes]: I think that what you want is contained in the theory of $\mathbb Q$-curves, which are elliptic curves defined over $\overline {\mathbb Q}$ that are isogenous to all of their Galois conjugates. Dick Gross studied CM $\mathbb Q$-curves in his thesis, which was published in [1]. So you'd need to use Gross' work to find a CM $\mathbb Q$-curve $E$ such that $\mathbb Q(j(E))$ is a quadratic extension of $\mathbb Q$. [1] Gross, Benedict H. Arithmetic on elliptic curves with complex multiplication. With an appendix by B. Mazur. Lecture Notes in Mathematics, 776. Springer, Berlin, 1980. iii+95 pp. MR0563921<|endoftext|> TITLE: Two questions on "foliation by geodesics" QUESTION [5 upvotes]: I would appreciate if you consider the following two questions on $1$ dimensional foliations whose leaves are geodesic. 1)Assume that $M$ is a Riemannian manifold which is either an open manifold or is a compact manifold with zero Euler characteristic. Does $M$ admit a foliation by geodesics? 2)Assume that $M$ is a Riemannian surface which admit at least one foliation by geodesics. Does there necessarily exist a foliation of $M$ by geodesics which satisfy the "Isocline Locale property"? The Isocline local property is defined as follows: For every $x\in M$ there is locally a geodesic $\alpha $ which is transverse to the foliation and it intersect all leaves with the same angle. REPLY [7 votes]: No to the first question. Let $M$ be a Riemannian $2$-manifold whose universal cover is the hyperbolic plane $H$, and whose fundamental group is not cyclic. For any foliation of $H$ by geodesics (lines), it seems to me that the endpoints of these lines on the boundary circle of $H$ will fill up the whole circle except for two points. These two points will have to be preserved by deck transformations, so any discrete group of isometries of $H$ that preserves the foliation must be cyclic.<|endoftext|> TITLE: Explicit permutation representation of the Schur double cover of the symmetric group QUESTION [11 upvotes]: Main question: How can we describe the double covers $(2\cdot\mathfrak{S}_n)^+$ and $(2\cdot\mathfrak{S}_n)^-$ of the symmetric group $\mathfrak{S}_n$ as permutation groups? (I.e., what sort of set with combinatorial structure do they act faithfully upon?) I am not necessarily asking for the minimal permutation degree (I think this is an open problem; also see PS at bottom), but for some explicit combinatorial construction, reasonably uniform in $n$, that is hopefully reasonably small (perhaps merely exponential in $n$). For a long time, I wrongly believed that one could form such a permutation representation of degree $2^{\lfloor n/2\rfloor}$ (or perhaps twice or four times that, or thereabouts) by taking an appropriate basis in the basic spin representation (and their negatives, and perhaps also times the imaginary unit), but I now realize that this does not work (at least, not as I thought it would). So I wonder if something else can be done. The following question may is closely related to the above (trying to lift some $\mathfrak{S}_n$-set $X$ to a $(2\cdot \mathfrak{S}_n)$-set $\tilde X$ consisting of "signed" elements of $X$): Alternate question: Given a set $X$ on which $\mathfrak{S}_n$ acts faithfully and transitively, with stabilizer $H$, is there some way to detect (merely by looking at $X$) whether $H$ lifts in $2\cdot \mathfrak{S}_n$ to a direct product $\{\pm 1\}\times H$, or equivalently (see below), whether the wreath product $1 \to \{\pm 1\}^X \to \{\pm 1\}\wr_X\mathfrak{S}_n \to \mathfrak{S}_n\to 1$ contains a non-split extension $1 \to \{\pm 1\} \to 2\cdot \mathfrak{S}_n \to \mathfrak{S}_n\to 1$? Indeed, if $2\cdot \mathfrak{S}_n$ acts transitively on a set $\tilde X$, with stabilizer $H$, and the action does not factor through $\mathfrak{S}_n$, then $H$ intersects $\{\pm 1\}$ (center of $2\cdot\mathfrak{S}_n$) trivially, so maps isomorphically to a subgroup of $\mathfrak{S}_n$, which we can also call $H$, and the latter lifts to $2\cdot\mathfrak{S}_n$ as a direct product $\{\pm 1\}\times H$ (and we can see $\tilde X = (2\cdot\mathfrak{S}_n)/H$ as a double covering of $X = \mathfrak{S}_n/H$). But then, it follows from Derek Holt's paper "Embeddings of Group Extensions into Wreath Products" (Quarter. J. Math. Oxford 29 (1978) 463–468), theorem 2, that the wreath product contains our $2\cdot \mathfrak{S}_n$. Conversely, when this is the case, $2\cdot \mathfrak{S}_n$ acts faithfully and transitively (through the wreath product) on $\tilde X := \{\pm 1\}\times X$. For example, if $H$ is generated by an $n$-cycle, and either $n$ is odd or we $n\equiv 2\pmod{4}$ if we are considering $(2\cdot\mathfrak{S}_n)^+$, we get a permutation action of degree $2(n-1)!$ on the set $\mathfrak{S}_n/H$ of cyclic orders with a sign added. But this isn't a great improvement over the regular action ($2n!$) and it's not very explicit, so I'm hoping we can do better. PS: If my computations with Gap are correct, the minimal permutation representation for (Gap's choice) $(2\cdot\mathfrak{S}_n)^-$ is $16,48,80,240,240,480$ for $n=4,5,6,7,8,9$. This is not in the OEIS, I wonder if it's worth adding. PPS: The corresponding values for $(2\cdot\mathfrak{S}_n)^+$ are $8,40,80,240,480,480$ (also not in the OEIS), so apparently they're not the same. (The trick I found to represent $(2\cdot\mathfrak{S}_n)^+$ under Gap is to multiply by a fourth root of unity the generators of $(2\cdot\mathfrak{S}_n)^-$ that lift an odd permutation.) REPLY [5 votes]: According to my calculations, the minimal permutation degrees of $2.A_n$ for $n=5,\ldots, 24$ are as follows. In each case $H$ is a core-free subgroup of $2.A_n$ of largest order. The minimal permutation degree of both versions of $2.S_n$ is at most twice this amount, and I think in each case it is exactly twice it, but I haven't tried to check that. I have only gone up to $n=24$ because that is as far as I have got with exact computations (mainly in Magma). I may be able to do a couple more values of $n$ but I would need to think harder to get much further. G H |H| |G:H| 2A5 5 5 24 2A6 3^2 9 80 2A7 7.3 21 240 2A8 L27 or 2^3.7.3 168 240 2A9 L28.3 1512 240 2A10 L28.3 1512 2400 2A11 M11 7920 5040 2A12 M11 7920 60480 2A13 M11 7920 786240 2A14 M11x3 23760 3669120 2A15 M11x3 23760 55036800 2A16 L27^2.2 or (2^3.7^3)^2.2 56448 370656000 2A17 (L27 or 2^3.7.3) x L28.3 254016 1400256000 2A18 (L28.3)^2 2286144 2800512000 2A19 (L28.3)^2 2286144 53209728000 2A20 M11 x L28.3 11975040 203164416000 2A21 M11 x L28.3 11975040 4266452736000 2A22 M11^2 62726400 17919101491200 2A23 M11^2 62726400 412139334297600 2A24 S12 479001600 1295295050649600<|endoftext|> TITLE: The logic of Buddha: a formal approach QUESTION [63 upvotes]: Buddhist logic is a branch of Indian logic (see also Nyaya), one of the three original traditions of logic, alongside the Greek and the Chinese logic. It seems Buddha himself used some of the features of such a non-standard logic in his philosophical reasoning which makes it important from a philosophical perspective. (see also Trairūpya and Hetucakra) However, in this post, I am going to tackle its mathematical aspects rather than philosophical ones. I came across the term "Buddhist logic" in a personal discussion with an Indian fellow of analytic philosophy background who asked me whether Buddhist/Indian logic could have any applications in the modern (predominantly Greek-logic based) mathematics/physics. Eventually, we came up with the idea that in the absence of a robust formalism and a fully clarified set of axioms for any such "non-standard" logic, there is not much to say about its mathematical properties and (dis)advantages in comparison with other logics, let alone finding any actual applications in daily mathematics, computer science or theoretical physics. My initial search revealed some intriguing pieces of information about Buddhist and Indian logic/mathematics which motivated me to go through a more thorough investigation of the topic: In the following papers, Graham Priest discusses the connection between some features of Buddhist logic such as Catuṣkoṭi and paraconsistent logic. He also provides some formalism for Jania logic a variant of Indian logic corresponding to Jainism. Priest, Graham, None of the above: the Catuṣkoṭi in Indian Buddhist logic. New directions in paraconsistent logic, 517–527, Springer Proc. Math. Stat., 152, Springer, New Delhi, 2015. (MR3476967) Priest, Graham, Jaina logic: a contemporary perspective. Hist. Philos. Logic 29 (2008), no. 3, 263–278. (MR2445859) Douglas Daye has published a series of papers investigating some issues with formalizing Buddhist logic. (See also his follow-up papers: MR0536102 and MR0547735) Daye, Douglas Dunsmore, Metalogical incompatibilities in the formal description of Buddhist logic (Nyāya). Notre Dame J. Formal Logic 18 (1977), no. 2, 221–231. (MR0457129) Prior to Daye, there has been some effort by Staal in the direction of formalizing Buddhist logic: Staal, J. F. Formal structures in Indian logic. Synthese 12 1960 279–286. (MR0131338) Also, some connections between Indian logic and more applied areas of research such as computer science has been investigated: Sarma, V. V. S., A survey of Indian logic from the point of view of computer science. Sādhanā 19 (1994), no. 6, 971–983. (MR1362512) There are several references to logic in the context of Vedic mathematics including in Vyākaraṇa where a scholar named Pāṇini has developed a grammar which "makes early use of Boolean logic, of the null operator, and of context-free grammars, and includes a precursor of the Backus–Naur form (used in the description programming languages)" (cf. Wikipedia) Last but not least, it is worth mentioning that various set-theoretic concepts including the notion of infinity (see the Sanskrit terms Ananta and Purna) has a strong presence in the Buddhist/Indian logic literature where they make a significant distinction between various types (and sizes?) of infinities (i.e. Nitya, Anitya, Anadi, and Anant) in a terminology fairly similar to modern treatment of ordinals. According to Staal, "Rig Veda was familiar with the distinction between cardinal numbers and ordinal numbers". There is also a very detailed treatment of transfinite and infinitesimal numbers in Jain mathematics as well as a variant of cardinal arithmetic and Hilbert's Hotel Paradox in Isha Upanishad of Yajurveda where it starts with the following verses: "That is infinite, this is infinite; From that infinite this infinite comes. If from that infinite, this infinite is removed or added then infinite remains infinite." Along these lines is the classical book Tattvacintāmaṇi which "dealt with all the important aspects of Indian philosophy, logic, set theory, and especially epistemology, ..." (cf. Navya-Nyāya). I am not quite sure if this short list is comprehensive and satisfactory enough to grasp the whole mathematical theory behind Buddhist logic, particularly because I am not of some Buddhist background to be fully aware of terminology (which is a crucial factor in the searching process). Also, most of the mentioned articles deal with the topic only marginally or from a slightly philosophical perspective. So I am looking for the possible help of some native Indian mathematicians or expert logicians who have worked along these lines. Question. What are some other examples of papers on Buddhist/Indian logic in which the axioms and properties of these logics are formally investigated? I am particularly interested in rigorous mathematical papers (rather than historical and philosophical ones) in which some theorems about their logical properties such as expression power, syntax, semantics, number of values, compactness, interpolation, etc., have been discussed (in a possibly comparative way). Any reference to possible implications of these formal systems in the foundation of mathematics especially in connection with the concept of infinity is welcome. REPLY [2 votes]: For the general public there is an interesting essay by Graham Priest: https://aeon.co/essays/the-logic-of-buddhist-philosophy-goes-beyond-simple-truth He starts with functions and relations explaining "catuskoti", meaning "four corners" (with values representing true, false, both true and false, neither true or false). And goes all the way to the inclusion of one more value (ineffable). Apart from the papers already mentioned in the thread, its author mentions his "Plurivalent Logics" where he also refers to N. Vasiliev's "Imaginary Logics".<|endoftext|> TITLE: Real polynomial bounded at inverse-integer points QUESTION [11 upvotes]: Let $p$ be a real polynomial and $N$ be a positive integer. Suppose I tell you that $|p(\frac{1}{k})| \le 1$ for all $k\in\{1,\ldots,N\}$, and also that $p(\frac{1}{N})\le -\frac{1}{2}$ while $p(\frac{2}{N})\ge \frac{1}{2}$. What bounds can you give me on the minimal possible degree $\deg(p)$? An upper bound of $O(\sqrt{N})$ follows by considering the Chebyshev polynomials (which indeed are bounded everywhere in $[0,1]$, not just at the inverse-integer points). On the other hand, the best lower bound I could show was $\deg(p)=\Omega(N^{1/4})$. This follows by restricting our attention to the interval $[\frac{1}{N},\frac{1}{\sqrt{N}}]$, which has no point more than about $1/N$ away from an inverse-integer point, and where (by assumption) $p$ also attains a large first derivative somewhere. We then apply standard results from approximation theory about polynomials bounded at discrete points, due to, e.g., Ehlich, Zeller, Coppersmith, Rivlin, and Cheney. Unfortunately the original papers seem to be paywalled, but the idea here is just to say that either $|p(x)|=O(1)$ in the entire interval $[\frac{1}{N},\frac{1}{\sqrt{N}}]$, in which case we can directly use Markov's inequality to lower-bound its degree, or else $p$ goes on some crazy excursion in between two of the discrete points at which it's bounded (say $\frac{1}{k}$ and $\frac{1}{k-1}$), in which case it attains a proportionately larger derivative there, so Markov's inequality can again be applied. My question is whether there are any fancier tools from approximation theory that yield a better lower bound on the degree, like $\Omega(N^{1/3})$ or conceivably even $\Omega(\sqrt{N})$. In case it helps: I already tried ransacking the approximation theory literature, but while I found many papers about polynomials bounded at evenly-spaced points, I found next to nothing about unevenly-spaced points (maybe I didn't know the right search terms). I also tried using Bernstein's inequality, which often yields better lower bounds on degree than Markov's inequality. But the trouble is that Bernstein's inequality is only useful if our polynomial attains a large first derivative far away from the endpoints of the interval where we're studying it (i.e., towards the center of the interval). And it seems that that can't be guaranteed here, basically because the interval $[0,1]$ has precious few inverse-integer points that are anywhere close to its endpoint of $1$. REPLY [19 votes]: Part I: $CN^{1/3}$ is enough. Start with $P(x)=\prod_{k\le N^{1/3}}(1-k^2x^2)$. Notice that it vanishes at $1/k$ with $k\le N^{1/3}$, is bounded by $1$ on $[0,N^{-1/3}]$, the degree of $P$ is $2N^{1/3}$ and on the interval $[0,2N^{-1}]$ we have $\log P\ge -2\sum_{k\le N^{1/3}}\frac{4k^2}{N^2}=o(1)$. Now just take the Chebyshev polynomial $Q$ of degree $N^{1/3}$ adjusted to the interval $[0,N^{-1/3}]$. It will jump as you requested between $N^{-1}$ and $2N^{-1}$ and stay below $1$ on $[0,N^{-1/3}]$. The product $P(x)Q(x)$ will then satisfy all your conditions. Part 2: $n0$, then the numerator in $L_k(-y)$ is at most $Cy\frac 1{k^2}(n!)^2$ (the same count except now $y=|-y-0|$ is present and $y+k^2=|-y-k^2|$ is missing) while the denominator is $$ \frac{k!(n-k)!(n+k)!}{2k(k-1)!}=\frac{(n-k)!(n+k)!}{2} $$ whence $$ |L_k'(-y)|\le \left(\frac 1y+\sum_{m\ge 1}\frac 1{y+m^2}\right)|L_k(-y)|\le \frac C{k^2}\frac{(n!)^2}{(n-k)!(n+k)!}\le \frac C{k^2}e^{-k^2/n}\,. $$ Thus we can easily derive from here that $$ |P'(-y)|\le C\sum_{k=0}^n |P(k^2)|\frac 1{k^2+1}e^{-k^2/n}\,. $$ Now choose an integer $M$ such that $n^30$ is some small constant. Consider the nodes $x_k=\frac 1M+\frac{k^2}{M}, k=0,1,\dots,n$. If they were available, that would be the end of the story (we would get an estimate $CM$ for the derivative on $[0, M^{-1}]$). Unfortunately only $\frac 1M$ is surely there and the rest have to be approximated. Note that for any $x\in[0,1]$, the nearest to $x$ inverse integer is within $x^2$ from $x$. So, if we choose $z_k$ to be the closest inverse integers to $x_k$, we'll have $|x_k-z_k|\le x_k^2\le 4\frac{k^4}{M^2}$ for $k>0$. Note that the distance from $x_k$ to any neighboring node is about $\frac kM$, so the shift $4\frac{k^4}{M^2}\le 4\frac{n^3}M\frac kM\le 4c\frac kM$ is small compared to it and we, indeed, get $n+1$ different nodes. Let us now look at how different the corresponding elementary Lagrange polynomials $$ \widetilde L_k(x)=\frac{\prod_{m:m\ne k}(x-z_m)}{\prod_{m:m\ne k}(z_k-z_m)} $$ are from the polynomials $L_k(x)$ built on the canonical nodes $x_k$ on the interval $[0,\frac 1M]$. The numerator: We have $\frac {|x-z_m|}{|x-x_m|}\le 1+\frac {4m^2}{M}$, so the numerator can grow at most $\prod_{m=1}^n (1+\frac {4m^2}{M})\le e^{4n^3/M}\le 2$ times. The denominator: If $k=0$, then we have $\frac{z_m}{x_m}\ge 1-\frac {4m^2}{M}$, so, as above, we see that it can drop at most twice. If $k>0$, then $$ \frac {|z_k-z_m|}{|x_k-x_m|}\ge 1-\frac {4(m^4+k^4)}{M|k^2-m^2|}\ge 1-\frac {4(m^3+k^3)}{M|k-m|} $$ We also have $$ \sum_{m:0\le m\le 2k,m\ne k}\frac {4(m^3+k^3)}{M|k-m|}\le C\frac{k^3}M(1+\log k) $$ and $$ \sum_{m:2k\le m\le n}\frac {4(m^3+k^3)}{M|k-m|}\le C\sum_{m:2k\le m\le n}\frac {m^2}{M}\le C\frac {n^3}{M}<1 $$ Thus, the ratio of the denominators we can have against us is at most $Ck^{Ck^3/M}$. That is played against the decay $k^{-2}e^{-k^2/n}$ and it is clear that the decay wins (say, because the power on $k$ in the growth factor is always below $1/2$; we have plenty of leeway here). For the transition from the value to the derivative, we can change the distances $x-x_m$ even twice without changing the corresponding factor too much, so we are done.<|endoftext|> TITLE: Does $(2n)!$ divide $\det[(i^2+j^2)^n]_{0\le i,j\le n-1}$ for each integer $n>2$? QUESTION [8 upvotes]: For $n=1,2,3,\ldots$ let $a_n$ denote the determinant $\det[(i^2+j^2)^n]_{0\le i,j\le n-1}$. Then $$a_1=0,\ a_2=-1,\ a_3=-17280,\ a_4= 1168415539200.$$ QUESTION: Is it true that $(2n)!\mid a_n$ for all $n=3,4,\ldots$? I even conjecture that $$a_n'=\frac{(-1)^{n(n-1)/2}a_n}{2\prod_{k=1}^n(k!(2k-1)!)}$$ is a positive integer for every integer $n>2$. Note that \begin{gather*}a_3'=1,\ a_4'=559,\ a_5'=10767500,\ a_6'=9372614611500. \end{gather*} The question is similar to my previous question http://mathoverflow.net/questions/302130. But it seems that darij grinberg's method there does not work for the present question. Any comments are welcome! REPLY [6 votes]: The conjecture in my posting has been proved finally! Its solution and further generalization now appear in the following preprint: Darij Grinberg, Zhi-Wei Sun and Lilu Zhao, Proof of three conjectures on determinants related to quadratic residues, arXiv:2007.06453, 2020.<|endoftext|> TITLE: Real-rootedness of some polynomials QUESTION [7 upvotes]: Denote the unsigned Stirling numbers of the first kind by $s(n,j)$. Question. Is it true that the polynomials $$P_n(x)=\sum_{j\geq0}s(n,j)\binom{x}j$$ have only real roots? Note. Obviously, the roots of $F_n(x)=\sum_{j\geq0}s(n,j)\,x^j=\prod_{k=0}^{n-1}(x+k)$ are all real (integers). REPLY [8 votes]: Fix $n$ and consider $P_k(x)=(1+\alpha)(1+2\alpha)\cdots(1+(k-1)\alpha)\cdot{x\choose n}$, where $\alpha$ is the operator sending ${x\choose k}$ to ${x\choose k-1}$. That is, $\alpha$ sends $f(x)$ to $f(x+1)-f(x)$. We'll show by induction that each $P_k(x)$ has all real roots and that the distance between consecutive roots is at least $1$. The claim is trivial for $k=1$. Assume the claim holds for $P_k(x)$. Then $P_{k+1}(x)=(1+k\alpha)P_k(x)$ is the linear combination $kP_k(x+1)-(k-1)P_k(x)$. Let the roots of $P_k(x)$ be $a_1<\cdots1$. Therefore $b_i-b_{i-1}\ge a_{i-1}-(a_{i-1}-1)=1$, which completes the induction.<|endoftext|> TITLE: Additivity of characteristic cycle of holonomic D-module QUESTION [6 upvotes]: Let $\mathcal{M}$ be a holonomic D-module on a complex analytic (or alternatively, algebraic) manifold $X$. One can attach to it (using a good filtration) a characteristic cycle $Ch(\mathcal{M})$ which is an analytic cycle on $T^*X$ whose components are conic Lagrangian subvarieties and their multiplicities are positive integers (see e. g. Def. 1.8.5 in the book “Analytic D-modules and applications” by Bjoern). Is it true that for any short exact sequence of holonomic D-modules $$0\to \mathcal{M}_1\to \mathcal{M}_2\to \mathcal{M}_3\to 0$$ one has $$Ch(\mathcal{M}_2)=Ch(\mathcal{M}_1)+Ch(\mathcal{M}_3).$$ An answer for algebraic D-modules would be equally useful. Apologies if the question is too basic. I am not a specialist. REPLY [4 votes]: Isn’t this Björk’s 3.1.4 (p. 130)?<|endoftext|> TITLE: Isn't a Shapiro-Wilk normality test assuming its conclusion? QUESTION [5 upvotes]: I am currently thinking about formalization of some statistics (in Coq). One thing I don't understand is the logic of e.g. the Shapiro-Wilk test for normality. To explain my problem, let's first look at a Kolmogorov test for normality, which doesn't have this problem. A hypothesis test is in general a contradiction argument: one assumes a certain statistical property of an observation (the null hypothesis) and then shows that with this assumption the observation is highly improbable and concludes that the assumption is likely not true. When I do a KS normality test, I assume as null hypothesis that the distribution is not normal and then show that the distance between the observed distribution and the assumed distribution is so small that this is unlikely. The statistics for the distribution distance derived by Kolmogorov is valid for any continuous distribution, so in essence the logic of such a test is: "distribution is not normal" -> "distribution is continuous" -> observation is unlikely from which one can conclude that either premise is likely false. Now let's look at the Shapiro-Wilk test. The difference to the KS test is, that the distribution of the W statistic given in the 1965 paper by Shapiro and Wilk applies only to normal distributions (otherwise one could use the Shapiro-Wilk test for any distribution). So a normality argument based on W statistics has the logic: "distribution is not normal" -> "distribution is normal" -> unlikely where the first premise is the null hypothesis and the second premise is required for applying the W statistics and coming to the "unlikely" conclusion. Again one can conclude from this that either premise is likely false, but in this case this is not that helpful. A non normality test (assuming normality as null hypothesis) would of cause work. Can someone please cut this knot for me? Added (and later edited) How do people work in practice with statistical methods requiring normality tests? In the abstract of reference [1] it is said: "normal distribution ... is an underlying assumption of many statistical procedures". So people do a test with a null hypothesis that the data is normal, and in case the data is not normal this hypothesis is rejected and the method requiring normal distributed data cannot be used. What happens if the test does not reject the null hypothesis of normality? I would think many people then apply the methods requiring normal distributed data without much further thought, since the scientific procedure for checking if the data is not not normal was followed. Is this justified? From a logic point of view not, because after a Shapiro-Wilk test we know nothing at all in case the null hypothesis is accepted. Also as Iosef pointed out (I hope I got him right) statistics claims nothing in this case. What I wanted to say above is this: in case the null hypothesis is accepted - and I would say this is a frequent use case - some tests really say nothing at all, while other tests still give some information. What I still don't understand is The connection between a Shapiro-Wilk test and the applicability of methods which require normal distributed data. Reference [1] claims that there is such a connection, but I see this only in a negative sense - maybe it is meant in this way. If it is possible to know more than nothing at all in case the null hypothesis is accepted and if other tests (like KS) are better in this respect compared to Shapiro-Wilk. How to convince a formal logic system like Coq that some methods requiring (approximately) normal distributed data are applicable - as far as I understood Iosef normality tests are always negative, so they can only show that such methods cannot be applied, but not that they can be applied. References Mohd Razali, Nornadiah & Yap, Bee. (2011). Power Comparisons of Shapiro-Wilk, Kolmogorov-Smirnov, Lilliefors and Anderson-Darling Tests. J. Stat. Model. Analytics. 2. S. S. Shapiro and M. B. Wilk (1965). An Analysis of Variance Test for Normality (Complete Samples). Biometrika Vol. 52, No. 3/4 (Dec., 1965), pp. 591-611 REPLY [3 votes]: In statistics, the test is usually named after the the null hypothesis. For instance, in the tests for independence, the null hypothesis is that certain random variables are independent -- see e.g. https://newonlinecourses.science.psu.edu/stat500/node/56/ . So, in your case, a normality test should assume normality as the null hypothesis $H_0$. Indeed, the Wikipedia article "Normality test", referred to in a comment by the OP, has this: "data are tested against the null hypothesis that it is normally distributed", and then, as you wrote, there is no problem. (In particular, if the Shapiro--Wilk test, based on the normality assumption, says $H_0$ is not rejected, then it means that your data set does not rule out normality. In statistics, one usually cannot positively prove anything specific with certainly; rather, one may only decide to reject or not to reject a specific hypothesis. In particular, in the case of an independence test, the standard goal is not to definitely prove independence. That is quite impossible with usually finite data sets. Indeed, to verify the independence of (say) two real-valued random variables (r.v.'s) $X$ and $Y$ with a joint cumulative distribution function (cdf) $F_{X,Y}$ and marginal cdf's $F_X$ and $F_Y$ means to verify the system $F_{X,Y}(x,y)=F_X(x)F_Y(y)$ of infinitely many equations, for all pairs of real $x,y$, with infinitely many unknowns that are all the values of $F_{X,Y}(x,y),F_X(x),F_Y(y)$. On the other hand, it is incomparably easier to (statistically) disprove independence of such r.v.'s $X$ and $Y$: it suffices to (statistically) show that $F_{X,Y}(x,y)\ne F_X(x)F_Y(y)$ for at just one pair of real $x,y$. Added: The important feature of a test is, not how it is called, but its power, and in this regard, the Shapiro--Wilk test for independence does seem to "pass the test". Indeed, in the abstract of the paper Power Comparisons of Shapiro-Wilk, Kolmogorov-Smirnov, Lilliefors and Anderson-Darling Tests, cited in the OP's comment, we find: "Results show that Shapiro-Wilk test is the most powerful normality test, followed by Anderson-Darling test, Lilliefors test and Kolmogorov-Smirnov test." As for the standard logic of testing, it is not " 'distribution is normal' -> 'distribution is likely normal' ", as suggested in a comment by the OP. Rather, it is like this: Assume for a moment that $H_0$ is true. If the most powerful test we know (of the prescribed significance level) tells us that the data contradict this null assumption, then we'll reject $H_0$; otherwise, we won't reject $H_0$. I can see no problem with this logic. In your case, the null hypothesis $H_0$ is that the distribution is normal. More added: Concerning Table 6 on page 605 [2]: The misleading term "Level" in the head of that table is not the significance level. It actually is the level of the quantiles of the distribution of the test statistic $W$. The use of that table is illustrated right after it, at the top of page 606 in [2]: for $n=7$, the "tabulated 50\% point" (that is, the 50th percentile of the distribution of $W$) is $0.928$. This is less than the value $0.9530$ of $W$ on the sample of size $n=7$ considered in that example. In view of what is said on page 593 in [2], "the mean values of $W$ for non-null distributions tends [sic -- I.P.] to shift to the left of that for the null case", it appears that the test rejects the null hypothesis $H_0$ for smaller values of $W$, less than a chosen critical value. The illustrative piece in [2] concludes thus: "Referring to Table 6, one finds the value of $W$ to be substantially larger than the tabulated 50\% point, which is 0.928. Thus there is no evidence, from the $W$ test, of non-normality of this sample." In other words, the $p$-value of the test in that situation was found to be very large, $>50\%$; therefore, the null hypothesis, of normality, is not rejected. No claim of proving or establishing normality is made (and such a claim would be quite ridiuculous here, given such a small sample size as $n=7$). One may also note that no use of a prescribed significance level is made here; instead, the $p$-value (of the strength of evidence) is used. So, so far I see nothing really objectionable in [2], except for the misleading use of the term "level".<|endoftext|> TITLE: The number of distinct closed subgroups of a compact monothetic group QUESTION [6 upvotes]: Let $G$ be a connected compact separable Hausdorff metric group, which is monothetic, i.e., has a dense subgroup generated by a single element. Such a group is necessarily Abelian. Question: Can the cardinality of the set of all closed subgroups of $G$ be uncountable? Thank you. REPLY [5 votes]: Not only the answer is yes (as indicated in Adam's answer), but there's a full picture: $G$ is a compact abelian group with a homomorphism with dense image $\mathbf{Z}\to G$. "With dense image" means an epimorphism in the category of locally compact abelian groups. This means that its Pontryagin dual $\hat{G}$ is a discrete abelian group comes with a monomorphism into the circle group $\mathbf{R}/\mathbf{Z}$, where monomorphism simply means injective homomorphism. In other words, $G$ is obtained by considering the Pontryagin dual of any subgroup of the circle group endowed with the discrete topology. That $G$ is metrizable means that $\hat{G}$ is countable. Pontryagin duality yields a bijection between the set of closed subgroups of $G$ and the set of subgroups of $\hat{G}$. Now discrete abelian groups with countably many subgroups were characterized by Boyer (1956) and are pretty rare: these are abelian groups $H$ with a finitely generated subgroup $A$ such that $H/A$ is isomorphic to $\bigoplus_{p\in I}C(p^\infty)$, where $I$ is a finite set of primes (no multiplicity allowed). Here $C(p^\infty)$ is the Prüfer (quasi-cyclic) group $\mathbf{Z}[1/p]/\mathbf{Z}$. So it's enough to consider any group not of this form, and embeddable in the circle group. For a subgroup $H$ of the circle group, there are three ways to fail to satisfy the above criterion: to have infinite $\mathbf{Q}$-rank: for instance, one considers free $\mathbf{Z}$-module of infinite countable rank: this yields $G$ to be the infinite-dimensional torus as in Adam's answer; to have $p$ torsion for infinitely many primes $p$. For instance, for some infinite set $J$ of primes $H=\bigoplus_{p\in J}\mathbf{Z}/p\mathbf{Z}$, and then $G=\prod_{p\in J}\mathbf{Z}/p\mathbf{Z}$. to have twice the same prime occurring in the Prüfer part. This holds if $H\simeq\mathbf{Z}[1/p]^2$, so $G=\hat{H}\simeq S^2$ where $S$ (the Pontryagin dual of $\mathbf{Z}[1/p]$) is called "solenoid" (unlike the last two previous examples it's not straightforward here that $G$ has uncountably many closed subgroups, since $S$ itself has only countably many ones, but still it's true).<|endoftext|> TITLE: Generator of $K_0(C_0(\mathbb{C}))$ QUESTION [7 upvotes]: $\newcommand{\C}{\mathbb{C}}\newcommand{\Z}{\mathbb{Z}}$ I know from Bott-periodicity that $K_0(C_0(\mathbb{C}))\simeq \Z$, is there any easy way to compute an explicit generator of $K_0(C_0(\mathbb{C}))$? By this I mean a projection $q$ in $M_k(C_0(\mathbb{C})^+)$ such that $[q]-[p_n]$ generates $K_0(C_0(\mathbb{C}))$. Moreover given an element $x\in K_0(C_0(\mathbb{C}))$ is there an easy way to compute $m\in \Z$ such that $m([q]-[p_n])=x$ (this equates to finding the inverse to the isomorphism $\Z\stackrel{\sim}{\rightarrow}K_0(C_0(\mathbb{C}))$)? I tried digging into the isomorphism $K_0(\C)\stackrel{\sim} {\rightarrow}K_0(C_0(\mathbb{C}))$ that one obtains from Bott-periodicity but things get pretty convoluted. In the end I got a generator that doesn't facilitate any further calculations I'd like to make (like calculating the inverse of the isomorphism). I'm trying to compute this generator to make some calculations regarding a endomorphism in $K_0(C(\C P^1))\simeq K_0(C_0(\C))\oplus K_0(\C)\simeq \Z\oplus \Z$ REPLY [11 votes]: The group $K_0(C_0(\mathbb{C}))$ is generated by by the class $[p_{Bott}] - [1]$ where $p_{Bott} \in M_2(C_0(\mathbb{C})^\sim)$ is the so-called "Bott projection" given by $$ p_{Bott}(z) = \frac{1}{1+|z|^2} \begin{pmatrix} |z|^2 & z \\ \overline{z} & 1 \end{pmatrix}. $$ This class comes from the tautological line bundle on $S^2 \simeq \mathbb{C}P^1$ after identifying the one-point compactification of $\mathbb{C}$ with the 2-sphere $S^2$ and performing some Serre-Swan-ification. It turns out that this class is essentially what makes Bott periodicity run. Edit: I should probably also mention that $[1]$ here is the class of the trivial line bundle on $S^2$. This appears since $p_{Bott}(\infty) = (\begin{smallmatrix} 1 & 0 \\ 0 & 0\end{smallmatrix})$.<|endoftext|> TITLE: Formula for Goldman Lie bracket of surface QUESTION [5 upvotes]: Let $\Sigma_{g}$ be a closed oriented surface of genus $g$, Goldman defined a Lie algebra structure on the free module generated by the free homotopy classes of loops on $\Sigma_{g}$. Roughly speaking, the Lie bracket is defined via intersection and concatenation of loops. In detail, see his paper Invariant functions on Lie groups and Hamiltonian flows of surface group representations . I want to know that are there any formulas to compute this Lie bracket for small nonzero $g$? For example, what is the Goldman Lie algebra of torus? Any help or any references would be very appreciated. REPLY [10 votes]: I would recommend looking at the work of Moira Chas to start. Here are two interesting papers of hers to read: The Goldman bracket and the intersection of curves on surfaces. Combinatorial Lie bialgebras of curves on surfaces She even has an app on her website that computes the bracket for you: Goldman Bracket. How to use the app: in the first box you input the relation defining the surface group (so for a torus abAB or a pair-of-pants abc). The next two boxes are for words in the surface group representing (free) homotopy classes of closed curves. For the syntax, lower case letters like a,b,c are for generators of the surface group and capital letters like A,B,C are for the inverses of a,b,c, respectively. I believe the app expects cyclically reduced words.<|endoftext|> TITLE: Generating function of $SO(N)$ random matrix QUESTION [7 upvotes]: I am interested in the generating function of $SO(N)$ random matrix, that is, I want to compute $$ Z_N[J]=\int dM e^{{\rm Tr} (J^T M)}, $$ where $dM$ is the $SO(N)$ Haar measure, and $J$ is an arbitrary $N\times N$ matrix. From this generating function, I can generate all correlations $\langle M_{ij}M_{kl}\cdots\rangle$ by taking derivatives with respect to the elements of $J$. Due to the invariance of the measure, one sees that $Z[J]=Z[U^TJV]$ with $U,V\in SO(N)$, and thus $Z$ only depends on the singular values of $J$. (Stated otherwise, $Z$ only depends on the $N$ invariants ${\rm Tr}((J^TJ)^n)$, $n=1,...,N$). Finally, at least for $N=2$ and $N=3$, one can show that $Z$ is also invariant under permutations of the singular values of $J$ (maybe it can be generalized for all $N$ ?). It is not too hard to compute explicitly $Z_2[J]$, which is given in terms of a Bessel function of the sum of the two singular values of $J$. Is there a way (or has it been done in the literature) to compute $Z_N$ for any $N$ ? I would already be happy with $Z_3$, which I cannot manage to compute explicitly. EDIT : Here is an attempt, which kind of works for $N=2$, but for which I am stuck for $N=3$. If we define a Laplacian $\Delta=\sum_{ij}\frac{\partial^2}{\partial J_{ij}^2}$ (with $J_{ij}$ the elements of $J$), one shows easily that $$ \Delta Z[J]=N Z[J]. \tag{1} $$ If we call $\lambda_i$ the singular values of $J$ (with $\lambda_1>\lambda_2>\ldots$), using the fact that $Z[J]=Z[\lambda_1,\lambda_2,\ldots]$, one shows (at least for $N=2$ and $N=3$, but it might be generalizable to $N\geq4$) that $$ \Delta Z=\frac{1}{D}\sum_{i}\frac{\partial}{\partial \lambda_i}\left(D\frac{\partial}{\partial \lambda_i}Z\right), $$ where $D=\prod_{i< j}(\lambda_i^2-\lambda_j^2)$ is related to the Jacobian to go from $J_{ij}$ to $\lambda_i$. This equation looks nice enough, so my hope is that a solution exists, I am not quite sure how to find it for $N=3$. In the case $N=2$, we can compute $Z_2$ exactly via its definition, and it reads $Z_2[\lambda_1,\lambda_2]=I_0(\lambda_1+\lambda_2)$, with $I_\nu$ the modified Bessel function of the first kind. One checks that this is indeed a solution of Eq. (1). Unfortunately, even in that case, it is not clear to me how to find this solution starting from Eq. (1) only. Given that $D=\lambda_1^2-\lambda_2^2$, it is tempting to defined $u=(\lambda_1+\lambda_2)/2$ and $v=(\lambda_1-\lambda_2)/2$. Then Eq. (1) is solved by separation of variables and we find a family of solution $Z_{2,\mu}$ (where I have already use the fact that $Z_N[0]=1$) : $$ Z_{2,\mu}[u,v]=I_0(\sqrt{\mu}u)I_0(\sqrt{4-\mu}v). $$ Clearly the solution to my problem corresponds to $\mu=4$, but it is not clear to me what is the rigorous argument to pick this value of $\mu$ (since $Z_N>0$ $\forall J$, we must have $\mu\geq 4$ as $I_0$ can be negative for imaginary variables; but how to select $\mu=4$ as the only viable solution ?). One way to solve this issue of $\mu$ is to use the fact that $Z_2[J=\lambda Id_2]$ can be computed explicitly: $Z_2[J=\lambda Id_2]=I_0(2\lambda)$, which unambiguously selects $\mu=4$. Since we can always compute the expansion of $Z_N[\lambda Id_N]$ explicitly at least for small $\lambda$, this kind of argument might be enough to fix the constant also for $N>2$. For $N=3$, a few special cases can be computed explicitly, which might help too. Any insight for the solution of Eq. (1) would be greatly appreciated. REPLY [6 votes]: For the $n=3$ case, it is best to use quaternionic methods to express $SO(3)$ as a quotient of $S^3$. Explicitly, for $a\in S^3$ one can check that the matrix $$ A = \left[\begin{matrix} a_1^2-a_2^2-a_3^2+a_4^2 & 2a_1a_2-2a_3a_4 & 2a_1a_3+2a_2a_4 \\ 2a_1a_2+2a_3a_4 & -a_1^2+a_2^2-a_3^2+a_4^2 & -2a_1a_4+2a_2a_3 \\ 2a_1a_3-2a_2a_4 & 2a_1a_4+2a_2a_3 & -a_1^2-a_2^2+a_3^2+a_4^2 \end{matrix}\right] $$ lies in $SO(3)$. (This is based on an identification $\mathbb{H}\simeq\mathbb{R^4}$ with $1\in\mathbb{H}$ corresponding to $(0,0,0,1)\in\mathbb{R}^4$.) This construction comes from a group homomorphism, so the Haar measure on $SO(3)$ corresponds to the natural round measure on $S^3$, suitably rescaled. Thus, we just want to integrate the function $\exp(f)$ over $S^3$, where $$ f(a) = (a_1^2-a_2^2-a_3^2+a_4^2)\lambda_1 + (-a_1^2+a_2^2-a_3^2+a_4^2)\lambda_2 + (-a_1^2-a_2^2+a_3^2+a_4^2)\lambda_3. $$ Using $\sum_ia_i^2=1$, we get $f(a)=\sum_i\lambda_i-2g(a)$, where $$ g(a) = (a_2^2+a_3^2)\lambda_1 + (a_1^2+a_3^2)\lambda_2 + (a_1^2+a_2^2)\lambda_3.$$ This is at least visibly invariant under permutations of $\lambda$. Next, because $a_4$ does not appear in $g(a)$, it is natural to use a stereographic parametrisation $\sigma\colon\mathbb{R}^3\to S^3$, as follows: $$ \sigma(x_1,x_2,x_3) = \frac{(2x_1,2x_2,2x_3,\|x\|^2-1)}{\|x\|^2+1} $$ The Jacobian determinant of $\sigma$ turns out to be $8/(\|x\|^2+1)^3$. This leads to an expression for $Z_3$ as an integral over $\mathbb{R}^3$ that retains all the natural symmetries, but it does not seem easy to evaluate either numerically or symbolically.<|endoftext|> TITLE: A conjecture on writing a function as a sum of uncountably many points QUESTION [7 upvotes]: Define the sum of the non-negative numbers $\{r_s \mid s \in S\}$ $S$ uncountable to be $$\sup _{D \subseteq S} \sum _{d \in D} r_d$$ ($D$ being finite), which exists if this supremum is finite. Define a point function to be a function from $[0, 1]$ to $\mathbb R$ that is $0$ everywhere except for a single point, where it takes a positive value. Suppose we have an uncountable family of point functions $f_r: [0, 1] \to \mathbb R$ indexed by $r \in [0, 1]$. Define the pointwise sum function $S[a, b]: [0, 1] \to \mathbb R$ as $$S[a, b] (x) = \sum _{r \in [a, b]} f_r (x) \ .$$ It can be shown that if $S[0, 1]$ is well defined, then so is $S[0, a]$ for any $a$ such that $0 \le a < 1$. Assume that if $S[0, 1]$ is well defined. Does it follow that for Lebesgue almost every $a \in [0, 1]$ the function $S[0, a]$ is discontinuous at at least one point? REPLY [7 votes]: OK, here goes as promised. Fix $\delta>0$ and consider the set $A_\delta$ of all points $a$ such that $S[0,a]$ is continuous and $\max f_a>\delta$ (this maximum is just that exceptional positive point value in your case but we can do arbitrary non-negative not identically $0$ functions). If we could find $a_n,a\in A_\delta$ such that $a_1>a_2>\dots\to a$, then $S[0,a_n]$ would be a decreasing sequence of continuous functions converging to the continuous limit $S[0,a]$ pointwise but not uniformly (because $\max(S[0,a_n]-S[0,a])\ge\max f_{a_n}>\delta$). Thus, for every $a\in A_\delta$, there exists an open interval $(a,b_a)$ free from points in $A_\delta$. Since we can place at most countably many disjoint open intervals on $\mathbb R$, $A_\delta$ is countable for each $\delta>0$. Hence the set of $a$ for which $S[0,a]$ is continuous (which is, say $\cup_{k\ge 1}A_{1/k}$) is also countable and, thereby, of Lebesgue measure $0$. Just made it community wiki not to collect reputation from homeworks :-)<|endoftext|> TITLE: is the conditional expectation faithful? QUESTION [6 upvotes]: Let $G$ be locally compact group and let $H$ be a open subgroup in $G$. Then the full group $C^*$-algebra of $H$, $C^*(H)$, is a subalgebra of $C^*(G)$ and there is a conditional expectation $$E\colon C^*(G)\to C^*(H),$$ which is induced by restriction $f\in L^1(G) \mapsto f_{|H}\in L^1(H)$ of functions which are integrable w.r.t. the left Haar measure on $G$, see Rieffel, induced representations of $C^*$-algebras, Proposition 1.2. Note that $E$ is'n faithful in general, there is an example in this paper in the section above definition 3.: Consider $G$ a nonamenable discrete group and $H$ an open subgroup consisting of the identity element of $G$. Then there are nonzero elements $c$ in the kernel of the left regular representation of $G$ (since $G$ is not amenable) and they satisfy $E(c^*c)=0$. My Question: Now, let $G$ be a locally compact amenable group and $H$ be an open compact (amenable) subgroup. Is then $E$ faithful? I think yes (I have considered some examples), but I am stuck with a proof. If I additionally assume $G$ (and $H$) to be discrete I can prove it considering $E$ as a conditional expectation $C_r^*(G)\to C_r^*(H)$ and then it is $\tau_G=\tau_H\circ E$, where $\tau_G$ and $\tau_H$ are the canonical faithful tracial states on $C_r^*(G)$, $C_r^*(H)$ respectively. It follows that $E$ must be faithful. For the more general case, I thought about trying a similar strategy using the fact that $C_r^*(G_1)$ of a locally compact group $G_1$ which contains a non-trivial amenable open subgroup $H_1$ has a tracial state $\tau^{G_1}$ satisfying $$\tau^{G_1}( \lambda_{G_1}(f))=\int_{H_1}f(s)d\mu(s),$$ see corollary 4.1 in 'embedding theorems in group $C^*$-algebra' by Lee for this fact. If one can check that the tracial state $\tau^{H_1}$ is faithful, then this together with $\tau^{G_1}=\tau^{H_1}\circ E$ implies that $E$ is faithful. But I am stuck with proving faithfulness of the trace. Probably I am on the wrong track..Other strategies regarding my question are welcome. REPLY [2 votes]: As $G$ and $H$ are amenable, we have that $C^*(G) = C^*_r(G) \subseteq VN(G)$ the group von Neumann algebra. So let's prove the stronger result that $E:VN(G)\rightarrow VN(H)$ is faithful. As $H$ is open, it is also closed (write $H$ as the complement of the union of cosets of $H$). It follows that $G/H$ is discrete topologically, and hence also in measure. So $L^2(G) = \ell^2(G/H) \otimes L^2(H)$ (or, $L^2(G)$ is the $\ell^2$ direct sum of $L^2(H)$, one for each coset). Let $U:L^2(H)\rightarrow L^2(G)$ be the inclusion, an isometry. $U^*:L^2(G) \rightarrow L^2(H)$ is just restriction. Then $E(x) = U^* x U$. To see this, compute $(f*\xi|\eta)$ for $f\in L^1(G)$ and $\xi,\eta\in L^2(H)\subseteq L^2(G)$. Let $x\in VN(G)$ with $0 = E(x^*x) = U^*x^*xU$, so $xU=0$. We want to show that $x^*x=0$, that is, $x=0$. As $x\in VN(G)$ it commutes with all right translation operators $\rho(s)$ for $s\in G$. Thus $0 = \rho(s)xU = x \rho(s)U$ and so $x\rho(s)\xi=0$ for all $s\in G, \xi\in L^2(H)$. The result now follows from the observation that the linear span of such vectors $\rho(s)\xi$ is dense in $L^2(G)$ (compute the orthogonal complement).<|endoftext|> TITLE: Is $\mathbb{R}^\omega \cong (\mathbb{R}^\omega \setminus \{x\})$? QUESTION [10 upvotes]: Is $\mathbb{R}^\omega \cong (\mathbb{R}^\omega \setminus \{x\})$, where $\mathbb{R}^\omega$ is given the product topology, and $x\in\mathbb{R}^\omega $? REPLY [9 votes]: The answer is yes. Any two homotopy equivalent Hilbert manifolds are homeomorphic. The countable product of lines is homeomorphic to the separable Hilbert space, see p.174 of [Bessaga C., Pelczynski A., Selected Topics in Infinite-Dimensional topology] where much more information is given. More generally, the complement in $\mathbb R^\omega$ of the union of any countable family of compact subsets is homeomorphic to $\mathbb R^\omega$, see e.g. Theorem V.6.4 in Bessaga-Pelczynski's book.<|endoftext|> TITLE: If the cardinality of $B(X)$, the space of operators on $X$, is continuum, must $X$ be separable? QUESTION [8 upvotes]: Does there exits any non-separable Banach space $X$ such that the size (cardinal number) of $B(X)$, bdd linear operators on $X$, is just of the continuum? REPLY [6 votes]: An example of a non-separable Banach space $X$ with $|B(X)|=\mathfrak c$ is any non-separable Banach space $X$ whose dual $X^*$ is $w^*$-separable and has cardinality $|X^*|=\mathfrak c$. This follows from the observation that the map $B(X)\to B(X^*)$, $T\mapsto T^*$, is injective and hence for a countable $w^*$-dense set $D$ in $X^*$ we have $$|B(X)|\le |B(X^*)|\le |(X^*)^{D}|=|X^*|^\omega=\mathfrak c^\omega=\mathfrak c.$$ A ZFC-example of a non-separable Banach space $X$ whose dual space $X^*$ is $w^*$-separable and has cardinality $|X^*|=\mathfrak c$ is the Banach space $X=C(K)$ of continuous functions on the Alexandroff two-arrow space $K$. The $w^*$-separability of the dual space $X^*$ was proved by Corson, see Theorem 12.43 in this book. The equality $|X^*|=\mathfrak c$ can be seen analyzing the structure of (probability) measures on the compact space $K$.<|endoftext|> TITLE: Regularity of geodesics QUESTION [13 upvotes]: If $M$ is a graph of a $C^1$ function $f:\mathbb{R}^n\to\mathbb{R}$, is it true that the length minimizing geodesics on $M$ are $C^1$? I expect a counterexample. For a related discussion see Metric angles in Riemannian manifolds of low regularity and Existence and uniqueness of geodesics in low regularity. REPLY [3 votes]: A comment: a length minimising geodesic $(u, f(u))$ on $M:={\rm graph}(f)$, which it is convenient to give by its projection $u:[0,1]\to\mathbb{R}^n $ on the domain of $f$, can be identified with a critical point $u\in H^1([0,1],\mathbb{R}^n)$ of $$E(u):={1\over 2}\int^1_0 \|\dot u \|^2+\big(\nabla f(u )\cdot \dot u \big)^2 dt= {1\over 2}\int^1_0 \|\dot u \|^2+\big(\partial_t f(u )\big)^2 dt.$$ There could be a simple argument to show regularity of the minimisers.<|endoftext|> TITLE: Does $\det\left[\left(\frac{i^2-\frac{p-1}2!\,j}p\right)\right]_{1\le i,j\le(p-1)/2}$ vanish for every prime $p\equiv3\pmod 4$? QUESTION [10 upvotes]: For any odd prime $p$, let $D_p$ denote the determinant $$\det\left[\left(\frac{i^2-\frac{p-1}2!\times j}p\right)\right]_{1\le i,j\le (p-1)/2},$$ where $(\frac{\cdot}p)$ is the Legendre symbol. Then \begin{gather*}D_3=0,\ D_5=-1,\ D_7=D_{11}=0,\ D_{13}=-8, \\ D_{17}=-72,\ D_{19}=D_{23}=0,\ D_{29}=-2061248.\end{gather*} QUESTION: Let $p$ be an odd prime. Is it true that $D_p=0$ if and only if $p\equiv 3\pmod4$? In 2013 I formulated this problem and conjectured that the answer is yes. I have verified this for all primes $p<2300$. By a result of L. Mordell [Amer. Math. Monthly 68(1965), 145-146], for any prime $p>3$ with $p\equiv3\pmod4$ we have $$\frac{p-1}2!\equiv(-1)^{(h(-p)+1)/2}\pmod p,$$ where $h(-p)$ is the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$. Any ideas towards the solution? REPLY [7 votes]: Warning: this is mod $p$ answer only. Thus it answers the original question for $p=4k+1$ but $p=4k+3$ remains open. Denote $\alpha=-(\frac{p-1}2)!$ and apply the formula given by Darij Grinberg in the comment here https://mathoverflow.net/a/302143/4312 where $n=(p-1)/2$, $x_j=\alpha\cdot j$, $y_j=j^2$ for $j=1,\dots,n$. Note that all elementary symmetric polynomials $e_k$, $1\leqslant k\leqslant n-1$, for $y$'s are equal to 0 (mod $p$ of course), since they are coefficeints of a polynomial $(t^2-1^2)(t^2-2^2)\dots (t^2-n^2)=t^{p-1}-1$. Thus only two summands remain: $e_0(x)e_n(y)$ and $e_n(x)e_0(y)$. They go with the same (and non-zero) coefficient, thus $p$ divides your determinant if and only if $p$ divides $e_n(x)+e_n(y)=\prod x_i+\prod y_i=-\alpha^{n+1}+\alpha^2=\alpha^2(\alpha^{n-1}-1)$. We have $\alpha^2=(-1)^{n+1}$ (from above polynomial or from Wilson theorem). If $p=4k+3$, $n=2k+1$, this gives $\alpha^2=1$ and indeed $\alpha^{n-1}-1=0$. If $p=4k+1$, $n=2k$, this gives $\alpha^2=-1$ and $\alpha^{n-1}=\alpha^{2k-1}=\pm \alpha\ne 1$.<|endoftext|> TITLE: Does separability of the strong operator topology imply separability of the underlying space? QUESTION [7 upvotes]: Let $X$ be a Banach space and $B(X)$ be the space of bounded operators on $X$. Suppose that the strong operator topology on $B(X)$ is separable and that the cardinal number of $B(X)$ is continuum. Question. Can we conclude the norm topology on $X$ is separable? Remark. It was proved by Tomek Kania that the assumption $|B(X)|=\mathfrak{c}$ is not solely enough to get the separability assertion. REPLY [10 votes]: Let $y$ be a norm-one vector in $X$. Consider the evaluation map $T\mapsto Ty$, which is a (linear) surjection from $B(X)$ onto $X$. This map is SOT-norm continuous. Indeed, suppose that $T_n\to T$ in SOT. In particular, $\|T_ny - Ty\|\to 0$ as $n\to \infty$. A continuous image of a separable space is separable, hence $X$ is separable.<|endoftext|> TITLE: Are lattice points in thin spherical shells uniformly distributed? QUESTION [6 upvotes]: Consider the spherical shell (annulus) $$A(R,r) = \{ x \in \mathbb{R}^3 : R \leq | x|\leq R+r \}.$$ Think of the limit $R \to \infty$. Assume that $r$ depends on $R$ as $r(R) = R^{-\delta}$. We are interested in the distribution of lattice points in $A(R,r)$. From results on the Gauss circle problem in three dimensions (see e.g. Ivic, Krätzel, Kühleitner, and Nowak - Lattice points in large regions and related arithmetic functions: Recent developments in a very classic topic) I know that the number of lattice points in the ball $B(R)$ is given by the volume of the ball, up to an error (the lattice point discrepancy) which is bounded by $\mathcal{O}(R^{\frac{42}{32}+\epsilon})$, for all $\epsilon > 0$. So by taking the difference we can obtain the number of lattice points in the annulus for $\delta >0$ not too large. We find that the number of lattice points is of order $R^{2}r$ (surface of the sphere times width of the shell). In Bourgain, Sarnak, and Rudnick - Local statistics of lattice points on the sphere, I found the reference to Duke - Hyperbolic distribution problems and half-integral weight Maass forms & Golubeva–Fomenko - Asymptotic distribution of lattice points on the three-dimensional sphere, showing that the the lattice points exactly on the sphere are uniformly distributed, if $R^2 \neq 0,4,7 \bmod 8$. (However there are only order $R^{1-\epsilon}$ lattice points exactly on the sphere.) My question is: Is anything known about the distribution of lattice points in the thin spherical shell? For example, consider a spherical cap on $B(R)$ and 'fatten' it up to a radius $R+r$. We now have a segment of the annulus (if you want, the intersection of a cone with the annulus) and ask whether the number of lattice points in this segment is to leading order given by the area of the spherical cap times $r$. If yes, does this stay true if the solid angle defining the cap goes to zero (not too fast) as $R \to \infty$? If it is not true for all such segments, could I at least say that it is true for most segments, or for a specific sequence of radii R? Another possible way of asking could be: consider summing a continuous function over $x/ |x|$, $x \in A(R,r)$ (and normalize by the number of points). Does the sum converge to the integral w.r.t. the uniform measure on the unit sphere? Is anyone aware of results in this direction? REPLY [10 votes]: Yes, they are equidistributed as long as $\delta<11/16$ and $r=R^{-\delta}$ and $R\to\infty$. Without loss of generality, we shall assume that $\delta>-1$ (i.e. $r TITLE: Is this a pseudodifferential operator? QUESTION [5 upvotes]: Let $M$ be a non-compact manifold and $D$ a first-order self-adjoint elliptic differential operator on $M$. Then is the bounded operator $$A:=\sqrt{(D^2+1)^{-1}}:L^2(M)\rightarrow H^1(M)$$ a pseudodifferential operator? More precisely, is there a pseudodifferential operator $P:C_c^\infty(M)\rightarrow C_c^\infty(M)$ of order $-1$ such that $P$ extends to $A$? Here I am defining the inner product on $H^1(M)$ by $$\langle s,t\rangle_{H^1}=\langle s,t\rangle_{L^2}+\langle Ds,Dt\rangle_{L^2}.$$ REPLY [2 votes]: Yes, it is a classical pseudo-differential operator of order $-1$ with principal symbol $\vert p_D(x,\xi)\vert^{-1}$ where $p_D$ is the principal symbol of $D$; it is also possible to prove that you have an asymptotic expansion for the (total) symbol $q$ of $1/\sqrt{1+D^2}$, providing $$ q-\sum_{0\le j TITLE: Regularity of the Jacobian of a $W^{2,n}$ Sobolev mapping QUESTION [5 upvotes]: Given a mapping in the Sobolev space $f\in W^{2,n}_{\rm loc}(\mathbb{R}^n,\mathbb{R}^n)$ I would like to know what is the Sobolev regularity of the Jacobian $J_f=\operatorname{det} Df$. It is well known and easy to prove that if $u,v\in W^{1,p}\cap L^\infty(\mathbb{R}^n)$, then $uv\in W^{1,p}\cap L^\infty$. Indeed, product of a bounded and an $L^p$ function is in $L^p$ and the same argument applies to the derivatives $\partial_i(uv)=(\partial_i u)v+v\partial_i\in L^p$. Now if $u\in W^{1,n}$ than $u$ has very high integrability (Trudinger's inequality) so if $u,v\in W^{1,n}$ (no longer bounded), then $uv\in W^{1,n}$ must belong to some Orlicz-Sobolev space slightly larger than $W^{1,n}$. Thus my question is: Let $u_1,\ldots,u_n \in W^{1,n}(B^n(0,1))$. Find an optimal (or close to optimal) Orlicz-Sobolev space $W^{1,P}$ for some Young function $P$ such that $u_1\cdot\ldots\cdot u_n\in W^{1,P}$. In fact I would like to know if one can find $P$ so that it satisfies the so called divergence condition: $$ \int_0^1 \frac{P(t)}{t^{n+1}}\, dt =\infty. $$ is satisfied. Since the derivatives of $f\in W^{2,n}(\mathbb{R}^n,\mathbb{R}^n)$ belong to $W^{1,n}$ such a result will imply that $J_f=\det Df\in W^{1,P}.$ REPLY [2 votes]: Here is an answer from Andrea Cianchi: A form of Hölder's inequality in Orlicz spaces asserts that, if $f_1\in L^{A_1},\ldots,f_n\in L^{A_n}$, and $B$ is such that $$ A_1^{-1}(t)\cdots A_n^{-1}(t)\leq cB^{-1}(t) \quad \text{for $t\geq 0$},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$ for some constant $c$, then $f_1 f_2\cdots f_n\in L^B$ and $$ \Vert f_1 f_2\cdots f_n\Vert_{L^B}\leq C\Vert f_1\Vert_{L^{A_1}}\cdots\Vert f_n\Vert_{L^{A_n}}, $$ for some constant $C$. If the domain has finite measure, then (1) is only required for sufficiently large $t$. Now it $u_1,\ldots,u_n\in W^{1,n}$, then $u_i\in\exp L^{n'}$ for every $i$ (Trudinger's inequality). In view of the condition (1), with $A_i(t)=t^n$ and $A_j(t)=e^{t^{n'}}$ for $j\neq i$, the product rule yields that $$ \nabla(u_1\cdots u_n)\in L^P $$ if $$ t^{1/n}(\log t)^{1/n'}\cdots(\log t)^{1/n'}\leq cP^{-1}(t) $$ for large $t$ (if the domain has finite measure), where $(\log t)^{1/n'}$ appears ($n-1$)-times. Thus $P$ has to fulfill $$ t^{1/n}(\log t)^{\frac{(n-1)^2}{n}}\leq cP^{-1}(t) $$ so the best possible choice of $P$ is $$ P(t)=t^n(\log t)^{-(n-1)^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) $$ for large $t$. The divergence condition is only satisfied for $n=2$. Therefore we have: If $f\in W^{2,n}_{\rm loc}(\mathbb{R}^n,\mathbb{R}^n)$, then $J_f=\det Df\in W^{1,P}_{\rm loc}$, where $P$ is given by (2).<|endoftext|> TITLE: Translating first order statements about symmetric groups into the language of numbers and back QUESTION [5 upvotes]: A question I was asked recently lead me to the following question. For every closed first order formula $\theta$ in the group signature consider the set $N_\theta$ of natural numbers $n$ such that the symmetric group $S_n$ satisfies $\theta$. Question. Can $N_\theta$ always be defined by a first order formula in the signature of natural numbers? Update. The original question has been answered below by Noah Schweber, but it occurred to me that I am mostly interested in the converse translation. So here is Converse question. Given a first order formula that defines a set $M$ of natural numbers, is there always a first order formula in the group signature defining the set of symmetric groups $\{S_n\mid n\in M\}$? REPLY [8 votes]: If by "signature of natural numbers" you mean the usual signature of arithmetic $\{+,\times\}$ or similar, the answer is yes. This is because using this language we can talk in a first-order way about finite objects: in the same manner as Godel describes strings of symbols, we can explicitly write something like "For all finite groups," or so on. So really, the only issue here is that we be able to talk uniformly about the symmetric groups, but it's not hard to show that there is a primitive recursive (hence a fortiori definable) map sending $n$ to a code for a group isomorphic to $S_n$. Now, one might still be worried by the issue of expressing "$\models$" in the relevant way. However, this is fine: the satisfaction relation for finite structures is in fact uniformly definable in the language of arithmetic. This is because a sentence is satisfied by a structure if and only if appropriate Skolem functions exist, and these functions themselves are finite objects and so can be quantified over. And now for the overkill answer: Clearly for each $\theta$, the set $N_\theta$ is computable; and every computable set is definable. EDIT: The answer to the updated question is no. For any sentence in the language of groups, the set of indices for symmetric groups satisfying it is computable. Now pick some formula of arithmetic defining a non-computable set. Even if we restrict attention to computable sets of natural numbers, the answer is still no. For each $\theta$, we can put a bound on the amount of time it takes to tell whether $S_n\models\theta$; any computable set not computable within this time bound will give a counterexample. (A back-of-the-napkin calculation: each quantifier corresponds to a search through all the elements, so telling whether $S_n$ satisfies an $m$-quantifier sentence in prenex normal form with a length-$l$ matrix probably takes something like $(n!)^k\cdot l$-many steps.)<|endoftext|> TITLE: Is the instanton homology for webs and foams a categorified Chern-Simons? QUESTION [9 upvotes]: In their paper, Kronheimer and Mrowka constructed an instanton homology $J^{\#}$ for webs and foams and conjectured that for planar webs, $\dim J^{\#}=\#\text{ of Tait colorings}$. According to my limited understanding, $J^{\#}$ is a sort of a TQFT (with $\mathbb{F}_2$ coefficient), or a functor from the category of webs and foams to the category of $\mathbb{F}_2$ vector spaces and morphisms. There is a familiar TQFT in one lower dimension which is clearly related to the number of Tait colorings : namely, the (Chern-Simons) quantum invariant for the second symmetric representation of $SU(2)$. In the classical limit $q=1$, it is exactly the number of Tait colorings. I know this is probably just a far-fetched speculation, but I am curious if these two TQFTs can be somehow related. More precisely, is it possible to categorify the latter such that it becomes $J^{\#}$ when reduced to the $\mathbb{F}_2$ coefficient? REPLY [9 votes]: It’s not clear what you mean by categorify here: the J# invariant doesn’t have a grading in their definition. Just taking dimension to get a number then doesn’t seem to yield a polynomial invariant. However, they have a variation of this invariant for SU(3) representations whose dimension yields the number of Tait colorings of cubic graphs. It is bigraded, so has an Euler characteristic (Kronheimer discusssed this at the Georgia Topology conference last summer). The natural conjecture is that this might be related to the Penrose number, which is a signed count of Tait colorings. The exact triangle for J# (which should also exist for SU(3)) involves three graphs which also relate the Penrose number. However, the Penrose number is only defined up to sign (depending on a framing), whereas I think the SU(3) invariant should have an absolute grading. So such a relation might be too much to expect.<|endoftext|> TITLE: Is there a converse to the Brauer–Nesbitt theorem? QUESTION [7 upvotes]: $\DeclareMathOperator\Tr{Tr}$Say that we have an algebra $R$ over $\mathbb{C}$. If, for two finitely generated (edit: and semisimple) $R$-modules $M, N$ we know that $\Tr_M(r)=\Tr_N(r)$ for all $r\in R$, where we consider $r$ as the linear endomorphism of the corresponding module, then we know that $M\cong N$ by the Brauer–Nesbitt theorem. My question is, assuming that we have specific values for traces (some homomorphism $g: R \rightarrow \mathbb{C}$ with $g(rr')=g(r'r)$ for every $r, r'\in R$), do we know that there exists some module $M$ with $Tr_M(r)=g(r)$ for every $r$? If not, can we somehow efficiently describe the functions that can appear as traces? REPLY [8 votes]: Not always — e.g. $g(1)$ should be an integer. The desired description is given in Helling, H., Eine Kennzeichnung von Charakteren auf Gruppen und assoziativen Algebren, Commun. Algebra 1, 491-501 (1974). ZBL0288.16019. Edit as requested: To a linear $g:R\to\mathbf C$ with $g(rs)=g(sr)$ Helling attaches $g_n:R^n\to\mathbf C$ by $$ g_n(r_1,\dots,r_n)= \sum_{\sigma\in\mathfrak S_n}(-1)^\sigma\prod_{(i_1,\dots,i_l):\text{ cycle of }\sigma}g(r_{i_1}\!\cdots r_{i_l}) $$ and proves (Satz p. 496): $g$ is a character iff $g_n=0$ for some integer $n$.<|endoftext|> TITLE: Dirichlet/Chebotarev's Theorem for natural/analytic density QUESTION [9 upvotes]: A little bird told me that Dirichlet/Chebotarev's Theorem is true not only for Dirichlet density, but also for analytic/natural density. I'm having trouble finding a citation for this. Does anyone know of one? REPLY [5 votes]: There is a nice treatment of this and discussions of the error term in Serre's book "Lectures on $N_X(p)$". He also gives various applications but I don't think he proves it. I should note that the natural density version holds over number fields but not over global function fields. Serre discusses this in his book.<|endoftext|> TITLE: Find closest integers in Euclidean rings QUESTION [9 upvotes]: Assume that $K$ is a number field such that $\mathcal{O}_K$ is a norm-Euclidean ring. I am looking for an efficient algorithm that given an element $x\in K$, find an integer $y\in\mathcal{O}_K$ that is closest to $x$ with respect to the norm, i.e. $N(x-y)$ is minimum (at most the Euclidean minimum $m(K)$). For $K=\mathbb{Q}[i]$ and $\mathcal{O}_K=\mathbb{Z}[i]$ is the Gaussian ring, it is easy by rounding, i.e., for $x=a+ib\in K$, then $y=a'+ib'$ with $a'$ and $b'$ are the integers nearest to $a$ and $b$ respectively. In this case the Euclidean minimum is $m(K)=\frac{1}{2}$, i.e. $N(x-y)\leq\frac{1}{2}$. For Eisenstein ring $\mathbb{Z}[w]$ where $w^2+w+1=0$, then for given $x=a+bw\in \mathbb{Q}(w)$, the norm is $N(x)= a^2+b^2-ab$ and it is known that $m(K) =\frac{1}{3}$ and rounding does not work as in the case of Gaussian integers. A naive approach can be done as follows: $$N(x-y)= (a-a')^2+(b-b')^2-(a-a')(b-b')=(a-a'-\frac{b-b'}{2})^2 + \frac{3}{4}(b-b')^2$$ and let $b'$ is the rounding of $b$, then $a'$ is the rounding of $a-\frac{b-b'}{2}$, but what we obtain is that $N(x-y)$ is upper-bounded by $\frac{7}{16}$, not $m(K)=\frac{1}{3}$ as desired. Could anyone please give me some references for such algorithms for Euclidean rings? Many thanks in advance! REPLY [5 votes]: For in depth information I would suggest the well regarded survey The Euclidean Algorithm in Algebraic Number Fields by Franz Lemmermeyer. There are a fair number of known norm Euclidean rings $\mathcal{O}_K$, though not a huge number. Only a few are commonly encountered so I would ask which ones you really care about and suggest looking at those specifically. Let me start with some specifics and then make some general remarks. COMPLEX QUADRADIC NORM-EUCLIDEAN RINGS There are exactly five of these, they come from $\mathbb{Q}[\sqrt{-m}]$ for $m=1,2,3,7,11.$ In the first two cases the ring has basis $\{1,\sqrt{-m}\}$ and embedded in $\mathbb{C}$ is tiled by recatangles. For those two your natural rounding works. For the other three a convenient basis is any two of $\{1,\frac{-1+\sqrt{-m}}2\,\frac{1+\sqrt{-m}}2\}.$ Here is the picture for your case of $m=-3.$ The small dot is $$2\frac6{13}-\frac5{13}\omega=2\frac{11}{13}-\frac5{13}\alpha=2\frac{6}{13}\alpha-2\frac{11}{13}\omega.$$ Naive rounding takes it to to $2,3$ and $2\alpha-3\omega=2-\omega$ respectively. However the actual tiling makes it clear where the closest thing is. The cases $m=-7,-11$ are similar but there is not the rotational symmetry. In these cases $x+y\sqrt{-m}$ has norm $|x^2+my^2|$ so can't be small unless both $x$ and $y$ are, this is why the naive rounding is almost optimal. REAL QUADRADIC NORM-EUCLIDEAN RINGS There are exactly $16$ of these. They come from $\mathbb{Q}[\sqrt{m}]$ for $m=2, 3, 5, 6, 7, 11, 13, 17, 19, 21, 29, 33, 37, 41, 57, 73.$ The norm sends $x+y\sqrt{m}$ to $|x^2-my^2|$ so the smallest results may only come from large $x,y.$ Consider the case $m=2$ and a point $B=\frac{p}{q}+\frac{p'}q\sqrt{2} \in \mathbb{Q}[\sqrt{2}].$ Certainly the minimum norm of $A-B$ for $A=s+t\sqrt{2}\in\mathbb{Z}[\sqrt{2}]$ is a positive fraction $\frac{r}{q^2},$ but it might be tricky to find. If we manage to find $r=1,$ that is surely optimal. For $B=\frac5{11}+\frac{16}{33}\sqrt{2}$ each of $0,1,\sqrt{2}$ and $1+\sqrt{2}$ gives a norm of about $\frac14$. The exact norms are $\frac{287}{1089},\frac{188}{1089},\frac{353}{1089}$ and $\frac{254}{1089}$ respectively. There are only two results that good for $A=x+y\sqrt{2}$ with $-2000 \leq x \leq 2000.$ However they are $A=-37-26\sqrt{2}$ and $A=-13+10\sqrt{2}$ both giving a norm (for the difference) of $\frac{56}{1089}.$ I suspect that is closest but don't actually know. Often one of the four obvious choices is as good as any other choice, and few cases I found are as extreme as this one. COMMENTS I like the question but wonder about the motivation. One motivation might be to compute $\gcd(a,b)$ efficiently with the Euclidean algorithm. For example in $\mathbb{Z}$ the move is $\gcd(a,b)\rightarrow \gcd(b,r)$ where $a=bq+r$ and $0 \lt r \lt b.$ It is known that in a certain sense the worst case is consecutive Fibonacci numbers. For example $$(55,34) \rightarrow (34,21) \rightarrow(21,13) \rightarrow (13,8) \rightarrow(8,5) \rightarrow (5,3) \rightarrow(3,2) \rightarrow(2,1)$$ is $7$ moves. However, $\gcd(b,r)=\gcd(b,b-r)$ and the second seems likely to be more efficient when it is smaller. Hence: $$(55,34) \rightarrow(34,13) \rightarrow(13,5) \rightarrow(5,2) \rightarrow(2,1)$$ is only $4$ moves. Essentially we rounded $\frac{55}{34}$ up to the nearest integer $2$ and wrote $55=34\cdot 2-13$ instead of $55=34 \cdot 1 +21.$ Before going on I'll note that there is sometimes no loss in using the larger of the two remainders, The remainder $13$ is quite a bit smaller that $21$ but an equally short computation is $$(55,34) \rightarrow(34,21) \rightarrow(21,8) \rightarrow(8,3) \rightarrow(3,1).$$ I rather suspect that it is never better to use the larger of $r,b-r$ but I have no proof. If we were trying to find $\gcd(r+s\sqrt{2},t+u\sqrt{2})$ we would go to $\gcd(t+u\sqrt{2},v+w\sqrt{2})$ where $v+w\sqrt{2}=r+s\sqrt{2} -(a+b\sqrt{2})(t+u\sqrt{2})$ and $a+b\sqrt{2} \in \mathbb{Z}[\sqrt{2}]$ somehow chosen to be close to $\alpha+\beta\sqrt{2}=\frac{r+s\sqrt{2}}{t+u\sqrt{2}}\in \mathbb{Q}[\sqrt{2}].$ Could we get less iterations by working hard to make $v+w\sqrt{2}$ have a very small norm relative to $|r^2-2s^2|$, perhaps at the cost of $v,w$ being larger than any of $r,s,t,u?$ It isn't clear. It could be worse. The naive approach does make $v,w$ smaller than $r,s,t,u$ and the norm at each stage is an integer no more than $\frac14$ the previous norm. So the easy to compute naive approximation does give exponential convergence.<|endoftext|> TITLE: Smallness of cut-off functions at critical Sobolev regularity QUESTION [7 upvotes]: Consider the class of functions $$X:=\{f\in \mathcal{C}_0^{\infty}(\mathbb{R})\;s.t.\;f\equiv 1 \mbox{ in a neighbourhood of}\;\;x=0\}$$ Is it true that, for every $\varepsilon > 0$, I can find $f\in X$ such that $\|f\|_{H^{1/2}(\mathbb{R})}<\varepsilon$? For $s\in[0,1/2)$, it is easy to show that the analogous question has affirmative answer. Indeed, given $f\in X$ and $n>0$, define $f_n(x):=f(nx)$. Then $f_n\in X$, and $\|f_n\|_{H^{s}(\mathbb{R})}\to 0$ as $n\to +\infty$. Instead, when $s>1/2$, the analogous question has negative answer. Indeed, by Sobolev embedding one has $$\|f\|_{H^s}\geqslant C\|f\|_{L^{\infty}}\geqslant C|f(0)|= C$$ for any $f\in X$. In the critical case $s=1/2$ I think the answer is affirmative, but I'm not able to prove it. Thank you for any suggestion. REPLY [4 votes]: It is well known and easy to verify that (Exercise 14 p. 309 in [1]) $$ \log\Big|\log\sqrt{x^2+y^2}\Big|\in H^1(B^2(0,e^{-1})) $$ so the trace of this function on the $x$-axis belongs to the trace space $$ f(x)=\log\Big|\log|x|\Big|\in H^{1/2}((-e^{-1},e^{-1})). $$ Let $$ f_t(x)=\begin{cases} 0 & \text{if } f(x)\leq t\\ f(x)-t & \text{if } t\leq f(x)\leq 2t\\ t & \text{if } f(x)\geq 2t \end{cases} $$ be a truncation of the function $f$ between the levels $t$ and $2t$, $t>0$. Then $f_t\in H^{1,2}$ and $\Vert f_t\Vert_{1/2}\leq\Vert f\Vert_{1/2}$. Indeed, the space $H^{1/2}(\mathbb{R})$ is equipped with the norm $$ \Vert u\Vert_{1/2}= \Vert u\Vert_2+ \left(\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{|u(x)-u(y)|^2}{|x-y|^2}\, dx\, dy\right)^{1/2}. $$ Since $|f_t|\leq|f|$ and $|f_t(x)-f_t(y)|\leq |f(x)-f(y)|$, it immediately follows that $\Vert f_t\Vert_{1/2}\leq\Vert f\Vert_{1/2}$. Therefore $$ \left\Vert\frac{1}{t}f_t\right\Vert_{1/2}\leq \frac{1}{t}\Vert f\Vert_{1/2}\to 0 \quad \text{as $t\to\infty$.} $$ The function $t^{-1}f_t$ equals $1$ near $0$ and it has compact support so approximating this function by convolution we can obtain a function $g_t\in C_0^\infty(\mathbb{R})$ such that $g_t=1$ near $0$ and $\Vert g_t\Vert_{1/2}<\varepsilon$, provided $t$ is sufficiently large. [1] L. C. Evans, Partial differential equations. Second edition. Graduate Studies in Mathematics, 19. American Mathematical Society, Providence, RI, 2010.<|endoftext|> TITLE: Invariance of an integral over SO(3) under permutation of parameters QUESTION [5 upvotes]: The integral $$Z_3(\lambda_1,\lambda_2,\lambda_3)=\frac{1}{2}\int_{-1}^1 I_0\left[\tfrac{1}{2}(\lambda_1-\lambda_2) (1-x)\right] I_0\left[\tfrac{1}{2} (\lambda_1+\lambda_2)(1+x)\right]\,e^{\lambda_3 x}\,dx,$$ with $\lambda_1,\lambda_3,\lambda_3\in\mathbb{R}$ and $I_0$ a Bessel function, arises in the evaluation of an integral over SO(3). From that derivation it follows that $Z_3(\lambda_1,\lambda_2,\lambda_3)$ is invariant upon interchange of $\lambda_1$ and $\lambda_3$. I am unable to rewrite this integral in a manifestly permutation-invariant form, can someone help me out? REPLY [7 votes]: Using the variable change $x=2t-1$, the integral $Z_3(\lambda_1,\lambda_2,\lambda_3)$ can be put into the form \begin{equation} Z_3(\lambda_1,\lambda_2,\lambda_3)=\int_{0}^1 I_0\left[(\lambda_1-\lambda_2) (1-t)\right] e^{-\lambda_3(1-t)}I_0\left[(\lambda_1+\lambda_2)t\right]\,e^{\lambda_3 t}\,dt \end{equation} or, alternatively, $$ Z_3(\lambda_1,\lambda_2,\lambda_3)=Z(1)$$ where \begin{equation} Z(T)=\int_{0}^T I_0\left[(\lambda_1-\lambda_2) (T-t)\right] e^{-\lambda_3(T-t)}I_0\left[(\lambda_1+\lambda_2)t\right]\,e^{\lambda_3 t}\,dt \end{equation} $Z(t)$ can be considered as a convolution integral. Its Laplace transform is \begin{equation} \mathcal{L}[Z](s)=\frac{1}{\sqrt{(s+\lambda_3)^2-(\lambda_1-\lambda_2)^2}}\frac{1}{\sqrt{(s-\lambda_3)^2-(\lambda_1+\lambda_2)^2}} \end{equation} The square of the denominator of this transform is a monic polynomial which roots are $\left\lbrace -\lambda_3+\lambda_1-\lambda_2, -\lambda_3-\lambda_1+\lambda_2, \lambda_3-\lambda_1-\lambda_2,\lambda_3+\lambda_1+\lambda_2\right\rbrace$. This set is invariant upon interchange of the parameters. Thus, $Z_3(\lambda_1,\lambda_2,\lambda_3)$ must share the same property. If an integral form needed, $Z_3(\lambda_1,\lambda_2,\lambda_3)$ can be written as the inverse Laplace transform of this function which is permutation invariant: $$Z_3(\lambda_1,\lambda_2,\lambda_3)=\frac 1{2i\pi}\int_{c-i\infty}^{c+i\infty}\frac{e^s\,ds}{\sqrt{P(s)}}$$ where $c>\max(r_1,r_2,r_3,r_4)$, $r_i$ are the roots given above and $$P(s)=s^4-2s^2(\lambda_1^2+\lambda_2^2+\lambda_3^2)-8s\lambda_1\lambda_2\lambda_3+(\lambda_1-\lambda_2-\lambda_3)(\lambda_2-\lambda_3-\lambda_1)(\lambda_3-\lambda_1-\lambda_2)(\lambda_1+\lambda_2+\lambda_3) $$<|endoftext|> TITLE: Characterizing $\mathbf{R}$ as an ordered group QUESTION [14 upvotes]: A standard characterization of $\mathbf{R}$ uses the order and the field structure: any linearly ordered field that is archimedean and complete is isomorphic to $(\mathbf{R}, +, \times, <)$ as an ordered field. Is there a similar characterization of $\mathbf{R}$ as an ordered group? Is any linearly ordered group that is archimedean and complete isomorphic to $(\mathbf{R}, +, <)$ as an ordered group or is some other assumption needed? Any reference is welcome. REPLY [29 votes]: The linearly ordered group $(\mathbb{Z},+,\le)$ is a counterexample, but that is probably not what the OP had in mind. To give a detailed description of the situation, let us use the following notation: By a linearly bi-ordered group we mean a tuple $(G,\cdot,\le)$ where $(G,\cdot)$ is a group and $\le$ is a linear order on $G$ such that $ac \le bc$ and $ca \le cb$ whenever $a,b,c \in G$ such that $a \le b$. We use the notion linearly ordered group as shorthand or linearly bi-ordered group. An isomorphism between two linearly ordered groups $(G,\cdot,\le)$ and $(H,\cdot,\le)$ is a group isomorphism $\varphi: (G,\cdot) \to (H,\cdot)$ such that both $\varphi$ and $\varphi^{-1}$ are increasing. A linearly ordered group $(G,\cdot,\le)$ (whose neutral element we denote by $e$) is called Archimedean if, for all $a,b > e$ there exists an integer $n \in \mathbb{N}$ such that $a^n \ge b$. We call a linear order an a set $S$ complete if every non-empty subset of $S$ that is bounded above has a supremum in $S$ (equivalently, every non-empty subset of $S$ that is bounded below has an infimum in $S$). Note that this property is sometimes called conditionally complete (instead of complete) in the literature. Theorem 1. Let $(G,\cdot,\le)$ be an Archimedean linearly ordered group. Then $(G,\cdot)$ is isomorphic to an ordered subgroup of $(\mathbb{R},+,\le)$ (i.e. a subgroup of $(\mathbb{R},+)$ which carries the order inherited from $\mathbb{R}$). In particular, $(G,\cdot)$ is commutative. This result can, for instance, be found in Theorem 1 in Section IV.1 of Fuchs, L., Partially ordered algebraic systems, Oxford-London-New York-Paris: Pergamon Press. IX, 229 p. (1963). ZBL0137.02001. There, the theorem is attributed to Hölder. As kindly pointed out by user Alec Rhea in the comments, there is a related result by Hahn which gives a description of all commutative linearly ordered groups. Next we note that linearly ordered groups whose order is complete are automatically Archimedean: Theorem 2. Let $(G,\cdot,\le)$ be a linearly ordered group and assume that the order $\le$ on $G$ is complete. Then $(G,\cdot,\le)$ is Archimedean. Proof. Let $e$ denote the neutral element of $(G,\cdot)$, let $a,b > e$ and assume for a contradiction that $a^n < b$ for all $n \in \mathbb{N}$. Then the set $S := \{a^n: \, n \in \mathbb{N}\}$ has a supremum $s$ in $G$. We have $a^{-1}s < s$, so $a^{-1}s < a^n$ for some $n \in \mathbb{N}$. Consequently, $s < a^{n+1} \le s$, which is a contradiction. In an earlier version of this post, a more complicated proof of Theorem 2 was given. The above version of the proof was kindly pointed out by user Emil Jeřábek in the comments. Remark 3. Note that the existence of inverse elements is esssential not only for the proof, but also for the validity of Theorem 2. Indeed, the set $[0,\infty) \times [0,\infty)$, endowed with componentwise addition and the lexicographical order, is an example of a linearly ordered semigroup (whose composition operation is strictly monotone in both components) which is order complete but not Archimedean. By combining Theorems 1 and 2 we arrive at the following corollary which, I think, answers the question of the OP: Corollary 4. Let $(G,\cdot,\le)$ be a linearly ordered group. If the order $\le$ on $G$ is complete, then $(G,\cdot,\le)$ is isomorphic to one of the three linearly ordered groups $(\{0\},+,\le)$, $(\mathbb{Z},+,\le)$ and $(\mathbb{R},+,\le)$. Proof. According to Theorem 2 $(G,\cdot,\le)$ is Archimedean, so it is isomorphic to an ordered subgroup $(H,+,\le)$ of $(\mathbb{R},+,\le)$ due to Theorem 1. If $H$ has only one element, then obviously $H = \{0\}$, so assume that $H$ has at least two elements. Now we distinguish two cases: First case: $h_0 := \inf \{h \in H: \, h > 0\} > 0$. Then it is easy to see that $H = h_0 \mathbb{Z}$, so $(H,+,\le)$ is isomorphic to $(\mathbb{Z},+,\le)$. Second case: $\inf \{h \in H: \, h > 0\} = 0$. Then one readily checks that the set $H$ is dense in $\mathbb{R}$, and the completeness of the order on $H$ implies that $H = \mathbb{R}$. This proves the corollary. -- Note on edits made 2018-06-10. -- I rewrote the answer and consolidated the various edits from the previous versions in order to make this post better readable for future visitors. I also incorporated various suggestions by users Alec Rhea, Emil Jeřábek and YCor, so let me thank them for their comments!<|endoftext|> TITLE: Is this lower bound for a norm of some complex matrices true? QUESTION [15 upvotes]: Let $A = [a_{ij}]_{n\times n}$ be a Hermitian matrix, such that $|a_{ij}| =1$ for $i \neq j$, and $a_{ii} = 0$ for each $i$. I am interested in a tight lower bound of $\|A\|_*:=\sum_{i=1}^n |\lambda_i(A)|$, where $\lambda_i(A)$'s are eigenvalues of $A$. Note that, by minimizing $\sum_{i=1}^n |\lambda_i|$ over two constraints $\sum_{i=1}^n \lambda_i = 0$ and $\sum_{i=1}^n \lambda_i^2= n(n-1)$, one can obtain $\sqrt{2n(n-1)}$ as a lower bound. But it seems that isn't tight. On the other hand, if $A := J - I$ (all ones matrix minus identity), then $\sum_i |\lambda_i(A)| = 2(n-1)$. Is it true that $2(n-1)$ is actually a lower bound (for large enough matrices, say $n \geq 10$) ? Remarks: As Alex's answer below, the minimum of trace norm of such matrices may be less than $2(n-1)$, even for arbitrarily large matrices. But, as a comment of @fedja, the minimum is $(2+o(1))n$ as $n\to\infty$. Added: In the particular case, when $a_{ij}=±1$, the lower bound holds. See this answer below, for an overview of the proof. REPLY [2 votes]: In the special case $a_{ij} = \pm 1$, $2(n-1)$ is a lower bound, for every $n>0$. As a comment by T. Tao above, the problem resembles the sharp Littlewood conjecture on the minimum of the $L^{1}$-norm of polynomials (on the unit circle in the complex plane) whose absolute values of coefficients are equal to $1$. In the special class of polynomials with $\pm 1$ coefficients, Klemes proved the sharp Littlewood conjecture (see here). The proof of Klemes gives us the following equality, for an $n\times m$ matrix $A$ with singular values $\sigma_1,\ldots,\sigma_{r}$ and for $ 0 \leq p \leq 2 $: \begin{equation*} \sum \limits_{i=1}^{r} \vert \sigma_{i} \vert^p= C_p \int_{0}^{\infty} \log \left(1+\sum \limits_{k=1}^{r} S_{k}(A^*A) t^{k} \right)t^{-\frac p2 -1}dt, \end{equation*} where $S_k(A^*A)$ stand for the sum of the determinat of $k\times k$ principle submatrices of $A^*A$ and $C_p$ is a constant depenting on $p$. When $A$ is Hermitian, singular values are equal to eigenvalues and by obtaining a "good" lower bound for $S_{k}(A^2)$, when $A$ is a matrix of the form described in the question, we can establish the lower bound in the special case $a_{ij} = \pm 1$.<|endoftext|> TITLE: Embeddings of linear orders in $\wp(\omega)/Fin$ under Martin's axiom QUESTION [9 upvotes]: We know that, under MA, every linear order $(X,\le)$ with $|X|<\mathfrak c$ embedds in $\wp(\omega)/Fin$. Does this hold for linear orders with cardinality $\mathfrak c$? REPLY [9 votes]: This is actually independent of $MA+\neg CH$. For example, under $MA+OCA$ there are no gaps in $\mathcal{P}(\omega)/fin$ of type $(\mathfrak{c}, \mathfrak{c}^\ast)$, which excludes the possibility that every linear order of size $\mathfrak{c}$ embeds. This is because assuming $MA$, the linear order $L=(2^{<\mathfrak{c}}, <_\text{lex})$ has size $\mathfrak{c}$ and $2^{\mathfrak{c}}>\mathfrak{c}$ gaps of type $(\mathfrak{c}, \mathfrak{c}^\ast)$; hence any embedding of $L$ into $\mathcal{P}(\omega)/fin$ witnesses the existence of a gap of type $(\mathfrak{c}, \mathfrak{c}^\ast)$ in $\mathcal{P}(\omega)/fin$. That said, the consistency of $MA+$"every linear order of size $\mathfrak{c}$ embeds into $\mathcal{P}(\omega)/fin$" is attributed to H. Woodin (see this question "$\mathfrak{c}$-universal linear order") To construct such a model, you use the following two facts, Any $c.c.c.$ forcing which adds a generic gap of type $(\omega_1, \omega_1^{\ast})$ to a ground-model linear order adds an $\omega_1$-branch to some ground-model Souslin-tree. The partial order which specializes a Souslin-tree with finite conditions does not add a branch to any ground-model $\omega_1$-tree. and construct an iteration which interleaves Souslin-free forcings with those that split gaps in the $\mathfrak{c}$-universal linear order you are building.<|endoftext|> TITLE: Prove that these are polynomials QUESTION [8 upvotes]: Define the functions $$F_n(q)=\frac1{(1-q)^{2n}}\sum_{k=0}^n(-q)^k\frac{2k+1}{n+k+1}\binom{2n}{n-k} \prod_{j=0,\,j\neq k}^n\frac{1+q^{2j+1}}{1+q}.$$ The numbers $\frac{2k+1}{n+k+1}\binom{2n}{n-k}$ belong to a family of Catalan triangle of which the special case $k=0$ yields the Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$. I have raised this question earlier which has not yet received an answer. Hence, I would like to propose a (milder) problem. My hope here is to see a better proof than my lengthy argument for $F_n(q)$ are all polynomials. REPLY [14 votes]: First notice that $\frac{2k+1}{n+k+1}\binom{2n}{n-k} = \frac{2k+1}{2n+1}\binom{2n+1}{n-k}$. To prove that $F_n(q)$ is a polynomial it is enough to show that $$\sum_{k=0}^n(-1)^k(2k+1)\binom{2n+1}{n-k}\frac{q^k}{1+q^{2k+1}}$$ has the zero $q=1$ of multiplicity $2n$. Plugging $q=e^t$, one needs to show that \begin{split} &\frac{e^{-t/2}}{2}\sum_{k=0}^n(-1)^k(2k+1)\binom{2n+1}{n-k}\mathrm{sech}(\frac{2k+1}2t)\\ =& \frac{e^{-t/2}}{2}\sum_{k=0}^n(-1)^k(2k+1)\binom{2n+1}{n-k} \sum_{l\geq 0} \frac{E_{2l}(\frac{2k+1}2t)^{2l}}{(2l)!}\\ =& \frac{e^{-t/2}}{2} \sum_{l\geq 0} \frac{E_{2l}t^{2l}}{2^{2l}(2l)!}\sum_{k=0}^n (-1)^k\binom{2n+1}{n-k} (2k+1)^{2l+1} \end{split} has the zero $t=0$ of multiplicity $2n$, where $E_{2l}$ are Euler numbers. So, the problem boils down to proving that $$(\star)\qquad\sum_{k=0}^n (-1)^k\binom{2n+1}{n-k} (2k+1)^{2l+1} = 0\quad\text{for all }l TITLE: When will the real numbers be Borel? QUESTION [22 upvotes]: In set theory Borel sets are important, but we don't actually care about the sets. We can about the Borel codes. Namely, the algorithm to generate a given Borel set starting with the basic open sets (intervals with rational end points, or some other canonical basis) and taking countable unions, complements, and so on. There are many proofs and arguments that look like "A Borel set $X$ is in a certain ideal if and only if it does not contain a real in a certain generic extension". And while that does give us some useful information about ground model sets via generic extensions, it is not $X$ itself that contains that generic real, but rather the "re-computation of $X$ from its code". But what about the set $X$? Clearly you cannot add elements to sets with forcing. When does $X$ remain Borel? Or just generally, when does $\Bbb R$ itself stay Borel in its generic extension? Is there some relatively simple and non-trivial condition on a generic real $c$ for which ground model Borel sets remain Borel? What about when adding uncountably many reals? Clearly, if we collapse $2^{\aleph_0}$ to $\aleph_0$, then all the sets of reals become countable and thus Borel. But I'm looking for something a bit more robust, such as a simple criterion to check against Cohen, Sacks, Random, etc. REPLY [17 votes]: I think that Groszek and Slaman's result (see https://www.jstor.org/stable/421023?seq=1) gives a satisfying answer to your question. Groszek and Slaman's result says that given any inner model $M$ of $ZFC$, $\mathbb{R}\subset M$ if and only if there is a perfect set $A\subset M$. A immediate conclusion of the result is that for any inner model $M$ of $ZFC$ and an $M$-Borel set $A$, $A$ remains to be Borel in any extension $N$ of $M$ with $\mathbb{R}^M\neq \mathbb{R}^N$ if and only if $A$ is countable in $N$.<|endoftext|> TITLE: Largest rank assumed by infinitely many elliptic curves QUESTION [16 upvotes]: One of the most interesting questions in Mathematics concerns the Mordell-Weil rank of the group of rational points on elliptic curves $E/\mathbb{Q}$, namely whether this quantity is bounded as one varies over all elliptic curves defined over the rationals (or some other number field $K$). It is known that there are infinitely many elliptic curves with small rank (say rank at most 2), and that there exist elliptic curves over $\mathbb{Q}$ with rank as large as 28 (due to Elkies). What is the largest positive integer $r$ such that it is known that there are infinitely many elliptic curves over the rationals with rank at least $r$? REPLY [12 votes]: As suggested by @JoeSilverman, I am turning my comment into an answer. According to the table maintained by Dujella at http://web.math.pmf.unizg.hr/~duje/tors/generic.html, the current record is 19, due to Noam Elkies in 2006. This is apparently obtained by a family of elliptic curves over an elliptic curve over $\mathbb Q$ with positive rank such that the generic fiber has Mordell-Weil group of rank 19. By a theorem of Neron (or a stronger version due to Silverman [Thm. C in "Heights and the specialization map for families of abelian varieties", J. Reine Angew. Math. 342 (1983), 197-211]), for all but finitely many of the infinitely many rational points of the base elliptic curve, the specialization map from the Mordell-Weil group of the generic fiber to that of the fiber above that point is injective, so that the rank of the latter is also at least 19.<|endoftext|> TITLE: If $A \in \text{End}(\bigwedge^k \mathbb{R}^d)$ equals $\bigwedge^k B$ for some complex matrix $B$, does it have a real source? QUESTION [5 upvotes]: Let $1 TITLE: local ring all whose non-maximal ideals are finitely generated QUESTION [5 upvotes]: Let $(R, \mathfrak m)$ be a commutative local ring such that every non-maximal ideal is finitely generated. Then, is $R$ Noetherian i.e. is $\mathfrak m$ finitely generated ideal ? It is easy to see that the answer is yes when $R$ is integral domain by considering an ideal $r\mathfrak m$ for $r\in \mathfrak m $ and noting $r\mathfrak m\ne \mathfrak m $ and $r\mathfrak m \cong \mathfrak m$ (as $R$-modules) . If $\mathfrak m$ is not finitely generated in a ring as above, then $\mathfrak m$ is an $R$-module all whose proper submodules are finitely generated, so from 1 (Proposition 1.1, Proposition 1.3 ) and 2 (Proposition 1.2) , one can see the following : $ann_R (\mathfrak m)$ is a prime ideal, $\mathfrak m$ is divisible as $R/ann_R (\mathfrak m)$-module and it is either torsion-free or every element is a torsion. The ring (possibly non-commutative) $End_R (\mathfrak m)$ is local i.e. the set of non-units forms an ideal. Also, $Ass_R (\mathfrak m)=\{P\}$ is singleton and that single associated prime is the set of all zero-divisors of $\mathfrak m$ . If the associated prime ideal is $P=ann_R (\mathfrak m)$, then $\mathfrak m$ is torsion-free over $R/ ann_R (\mathfrak m)$ and in that case $\mathfrak m $ is isomorphic to the fraction field of $R/ann_R (\mathfrak m)$ as $R$-modules and $ann_R (\mathfrak m)$ is not maximal and $\mathfrak m$ is not Artinian as $R$-module. If $ann_R(\mathfrak m) \notin Ass_R(\mathfrak m)$, then the associated prime ideal is maximal and every proper non-maximal ideal of $R$ has finite length, so in particular $\mathfrak m$ is an Artinian $R$-module. In any case, $ann_R (\mathfrak m)$ is not a maximal ideal. Modules Whose Proper Submodules Are Finitely Generated, WILLIAM D. WEAKLEY Rings with an almost Noetherian ring of fractions,Efraim Armendariz REPLY [12 votes]: There exists no such non-noetherian local ring. Below I assume by contradiction that we have such a ring. (a) The first observation is a particular case Proposition 1.2(a) in your reference to Armendariz: for every $r\in R$, we have $r\mathfrak{m}\in\{\mathfrak{m},\{0\}\}$. Indeed, $r\mathfrak{m}$ is a quotient of $\mathfrak{m}$; if it's nonzero, it's quotient by a proper submodule and hence is also not noetherian, which implies $r\mathfrak{m}=\mathfrak{m}$. (b) case when $R$ is a domain (I just expand your argument). Choose $r\in\mathfrak{m}\smallsetminus\{0\}$. Since $R$ is a domain, by (a) we have $r\mathfrak{m}=\mathfrak{m}$, and in particular so $r\in r\mathfrak{m}$, and since $R$ is a domain this implies $1\in\mathfrak{m}$, a contradiction. (c) let us check that $R$ is necessarily of Krull dimension 0. Indeed if $P$ is a nonmaximal prime ideal, then $P\neq\mathfrak{m}$ and hence is finitely generated, so $R/P$ is a counterexample to (b). (d) Now we use Proposition 1.2(b) in Armendariz: $P=\mathrm{Ann}_R(\mathfrak{m})$ is prime (as you've already mentioned). The argument is easy: indeed, for $x,y\notin P$, by (b) we have $x\mathfrak{m}=y\mathfrak{m}=\mathfrak{m}$, which implies $xy\mathfrak{m}\neq 0$, so $xy\notin P$. (e) Combining (c) and (d), the only option for $\mathrm{Ann}_R(\mathfrak{m})$ is that it's equal to $\mathfrak{m}$. Thus $\mathfrak{m}^2=\{0\}$. Then $\mathfrak{m}$ is an infinite-dimensional vector space over the field $R/\mathfrak{m}$, and all its hyperplanes are ideals. In particular they fail to be finitely generated, and this is a contradiction. Edit 1: the argument can be extended to show that there is no commutative ring at all with these conditions (non-noetherian such that all non-maximal ideals are finitely generated). That is, assuming $R$ local is unnecessary. In other words, a commutative ring is noetherian if and only if all its non-maximal ideals are finitely generated ideals. Indeed, (a),(b),(c),(d) work with no change for every given infinitely generated maximal ideal $\mathfrak{m}$. Let us adapt (e): (e') for every infinitely generated maximal ideal $\mathfrak{m}$, by combining (c) and (d), its annihilator is another maximal ideal $\mathfrak{m}'$. Then $\mathfrak{m}$ can be viewed as a $R/\mathfrak{m}'$-vector space. Hence the lattice of ideals contained in $\mathfrak{m}'$ can be identified to the lattice of $R/\mathfrak{m}'$-vector subspaces of $\mathfrak{m}$. In particular, the condition that all its elements except whole $\mathfrak{m}$ are noetherian, implies that $\mathfrak{m}$ has finite dimension (as vector space over $R/\mathfrak{m}'$), hence is finitely generated as an ideal, a contradiction. We deduce every maximal ideal is finitely generated, and hence (since all non-maximal ones are also finitely generated by assumption) that $R$ is noetherian. Edit 2: As mentioned by Keith in the comments, the non-local case can be handled in an even easier way: let $R$ be a ring (commutativity is unnecessary) in which every non-maximal left ideal is finitely generated, and having at least two maximal left-ideals. If $\mathfrak{m}$ is a maximal left-ideal and $\mathfrak{m}'$ is another one, then $\mathfrak{m}\cap \mathfrak{m}'$ is finitely generated, and $\mathfrak{m}/(\mathfrak{m}\cap \mathfrak{m}')\simeq (\mathfrak{m}+\mathfrak{m}')/\mathfrak{m}'=R/\mathfrak{m}'$ is a simple module, so $\mathfrak{m}$ is also finitely generated.<|endoftext|> TITLE: Category theory for a set/model theorist QUESTION [17 upvotes]: I am looking for a book or other reference which develops category theory 'from the ground up' assuming a healthy background in set and model theory, not one in homological algebra or Galois theory etc. Up until now I have been satisfied by the heuristic that most of what takes place in category theory can be translated into model theoretic terms and vice-verse, and because my research is not directly related to either and model theory has always felt like breathing to me I have always just studied model theory. This casual perspective is no longer sufficient; my current research is into field extensions, specifically how to canonically and 'internally' extend the field of fractions of the Grothendieck ring of the ordinals into the Surreal numbers. To this end I have purchased 'Galois theories' by Francis Borceux and George Janelidze, and from the preface it looks like chapter seven is what I need since none of the extensions I care about are Galois (they fail to be algebraic or normal), and they promise a Galois theorem in the general context of descent theory for field extensions without any Galois assumptions on the extensions. I believe I still have loads of descent happening so this is right up my alley, but they give the comment that "The price to pay [for dropping these assumptions] is that the Galois group or the Galois groupoid must now be replaced by the more general notion of a precategory". This, along with other comments like 'pulling back along this morphism in the dual category of rings yields a monadic functor between the corresponding slice categories' and many others which clearly have some precise meaning that is completely hidden to me have convinced me that it's time to start seriously learning some category theory. Are there any good references on 'category theory in the large' for someone with a decent background in set/model theory? I enjoy thinking about 'large scale' mathematics and have always sensed that 'pure' category theory was exactly this, so a reference with an eventual eye towards higher category theory would be appreciated. REPLY [2 votes]: Tim gave a very good answer. I shall add two references. Sebastien Vasey, Accessible categories, set theory, and model theory: an invitation. Suggested Readings.<|endoftext|> TITLE: What is the relationship between $O_G$-parameterized $\infty$-categories and $\infty$-categories enriched in $Top_G$? QUESTION [6 upvotes]: Let $G$ be a finite group. Barwick et al define a $G-\infty$-category to be a fibration over the orbit category $O_G$ of transitive $G$-sets. But in the non-$\infty$-land, the natural guess at where I should work to do $G$-equivariant homotopy theory is a category enriched in the category $Top_G$ of $G$-spaces. In the $\infty$ world, this setting can be generalized directly using Gepner and Haugseng's notion of enriched $\infty$-category. Question: Is there a comparison functor between $\infty$-categories fibered over $O_G$ and $\infty$-categories enriched in $Top_G$? Is this an equivalence, perhaps after passing to certain subcategories? EDIT: Maybe this is what Marc is driving at in the comments, but think of it this way. A category fibered over $O_G$ is a functor $O_G^{op} \to Cat$, which is a functor $O_G^{op} \times \Delta^{op} \to Top$ satisfying some conditions. A category internal to $Top_G = Fun(O_G^{op}, Top)$ is a simplicial object in $Top_G$, i.e. a functor $\Delta^{op}\times O_G^{op} \to Top$, satisfying some conditions. This leads me to post a Revised Question: Are categories fibered over $O_G$ the same thing as categories internal to $Top_G$? Which ones correspond to enriched categories? REPLY [3 votes]: It seems that the answer to the first part of the revised question is yes: categories parameterized over $O_G$ are the same as categories internal to $Top_G$. Let's think through this carefully. A category object should satisfy Segal conditions and a univalence (aka completeness in the sense of Rezk) condition. Admittedly the latter might not be entirely standardized for internal categories, but let's pick a formulation and run with it: A category parameterized over $O_G$ is a functor $X: O_G^{op} \times \Delta^{op} \to Top$ which is, levelwise in $O_G$, a complete Segal space. So we have Segal conditions: $X(G/H,[n]) \to X(G/H,[1]) \times_{X(G/H,[0])} \dots \times_{X(G/H,[0])} X(G/H,[1])$ is an equivalence. Univalence: $X(G/H,[0]) \to Map(E_\bullet,X(G/H,\bullet))$ is an equivalence of spaces. Here $E_\bullet$ is the (nerve of the) walking isomorphism, and $Map$ the $Top$-enriched homset of simplicial spaces. A category internal to $Top_G$ is a functor $X: O_G^{op} \times \Delta^{op} \to Top$ satisfying Segal conditions: $X(-,[n]) \to X(-,[1]) \times_{X(G/H,[0])} \dots \times_{X(-,[0])} X(-,[1])$ is an equivalence. Univalence: $X(-,[0]) \to GMap(E_\bullet, X(-,\bullet))$ is an equivalence of $G$-spaces. Here $E_\bullet$ is included with trivial $G$-action, and $GMap$ denotes the $Top_G$-enriched homset of simplicial $G$-spaces. Comparison: These two notions are indeed equivalent. We see this as follows. Since limits in a presheaf category are computed levelwise, the Segal conditions in the two cases are equivalent. $GMap(E_\bullet, X(-,\bullet))(G/H) = Map(E_\bullet, X(G/H,\bullet))$ because for each $[n]$, $E(G/H,[n])$ is constant as a function of $G/H$ (for me, this implication requires a short calculation). Thus the univalence conditions are also equivalent. Enriched categories? So I guess that a $G$-enriched category is an internal category to $GTop$ with trivial $G$-action on the space of objects? This should translate to a parameterized category where the "core" -- the maximal subfibration which is fibered in groupoids -- is constant?<|endoftext|> TITLE: How much do we need to add to the generating set of the symplectic group to get $SL(2n,2)$? QUESTION [6 upvotes]: Here $Sp(2n,\mathbb{F}_2)$ means the group of matrices preserving the form $\Omega = \left( \begin{array}{cc} 0&I \\ -I&0& \end{array} \right)$, i.e. the symplectic group over an even characteristic ($-1=1$). It is well known that, $Sp(2n,\mathbb{F}_2) \leq SL(2n,\mathbb{F}_2)$. Additionally both groups are generated by transvections, but for $Sp(2n,\mathbb{F}_2)$ the transvections need to preserve the form $\Omega$. Say $S$ is a generating set such that $Sp(2n,\mathbb{F}_2) = \langle S \rangle$, and $X \in SL(2n,\mathbb{F}_2)$ but$X \notin Sp(2n,\mathbb{F}_2)$, then does $\langle S, X \rangle = SL(2n, \mathbb{F}_2)$ hold? It's clear that $SL(2,\mathbb{F}_2) = Sp(2,\mathbb{F}_2)$, yet I'm not sure about the general case. It's tedious, but possible to show it for $n=2$ with elementary methods, but the induction seems unclear. I am thankful for suggestions. REPLY [8 votes]: For general $q$, it's not quite maximal. For $n>1$, the maximal $M$ subgroup of $G={\rm SL}(2n,q)$ containing $H={\rm Sp}(2n,q)$ contains $H$ as a subgroup of index $\gcd(q-1,n)$. So, as Noam D. Elkies points out in a comment, in the case $q=2$ it is indeed maximal for all $n>1$. We have $M = \langle H, X \rangle$, where $X$ contains firstly scalar matrices in ${\rm SL}(2n,q)$, and secondly, when $n$ is even and $q$ is odd, there is an element in the conformal group (i.e. maps the fixed form to a scalar multiple of itself) that induces an outer automorphism of $H$ of order $2$. (When $n$ is odd there is no such conformal element with determinant $1$.) You can find the details in the book "The subgroup structure of the finite classical groups" by P. Kleidman and M. Liebeck. The structure of $M$, which is the normalizer of $H$ in $G$ is described in Proposition 4.8.3, page 166 (although to understand that you need to know a lot of notation), and the maximality of $M$ in $G$ is done in Section 7.8, page 245.<|endoftext|> TITLE: Recovering topological invariants of symplectic manifold from the group of Hamiltonian diffeomorphisms? QUESTION [14 upvotes]: There is a famous result of Banyaga stating that if two closed symplectic manifolds $(M_1, \omega_1)$ and $(M_2, \omega_2)$ have isomorphic groups of Hamiltonian diffeomorphisms $\mathrm{Ham}(M_1, \omega_1)\simeq \mathrm{Ham}(M_2, \omega_2)$ then there exists a diffeomorphism $f:M_1\rightarrow M_2$ such that $f^*\omega_2=\lambda \omega_1$ for some $\lambda\in\mathbb{R}^{*}$. In other words, a symplectic structure on a closed manifold is determined (up to rescaling) by the group of Hamiltonian diffeomorphisms. The result of Banyaga tells us that, in principle, it should be possible to understand the topology of $M$ only from $\mathrm{Ham}(M, \omega)$. However, as far as I understand, it's pretty hard to actually give an algorithm for doing that. So my question is: are there any explicit procedures for recovery of algebro-topological invariants of $M$ from $\mathrm{Ham}(M, \omega)$? I am particularly interested in recovery of the fundamental group $\pi_1(M)$. A weaker question would be: is there any explicit condition on $\mathrm{Ham}(M, \omega)$ which guarantees that $M$ is simply-connected? I am aware of one comparatively weak result in this direction. Banyaga has constructed a map (called flux map) $f:\pi_1(\mathrm{Symp}_0(M, \omega))\rightarrow H^1(M, \mathbb{R})$. He has also shown that there is an isomorphism $$ \mathrm{Symp}_0(M, \omega)/\mathrm{Ham}(M, \omega) \simeq H^1(M, \mathbb{R})/\mathrm{ker}\:f. $$ Therefore, if we know both $\mathrm{Symp}^0(M, \omega)$ and $\mathrm{Ham}(M, \omega)$ we can at least recover a quotient of $H^1(M, \mathbb{R})$ (so we have a lower bound for its rank, for example). P.S.: A clear exposition of some results on the group of Hamiltonian diffeomorphisms can be found in Polterovich's book "The geometry of the group of symplectic diffeomorphisms" (chapter 14 is particularly relevant to the question). REPLY [5 votes]: Let $M$ be compact and connected. For every closed $1$-form $\alpha$ on $M$ consider the Roger $2$-cocyle $\Psi_\alpha$ on the Lie algebra $\operatorname{ham}(M, \omega)$ defined by $$ \Psi_\alpha(X_f, X_g) = \int_M f \, \alpha(X_g) \, \frac{\omega^n}{n!}. $$ It was stated by Roger and proven by Janssens & Vizman that this map yields a bijection between $\operatorname{H}^1_{dR}(M)$ and the Lie algebra cohomology $\operatorname{H}^2(\operatorname{ham}(M, \omega), \mathbb R)$. This shows that you don't need to know the full group of Hamiltonian diffeomorphism but that you can actually extract the first de Rham cohomology of $M$ from the Lie algebra $\operatorname{ham}(M, \omega)$. In applications, it is however often more feasible to calculate the de Rham cohomology of $M$ than the Lie algebra cohomology of Hamiltonian vector fields. A similar result holds for non-compact $M$, where $\operatorname{H}^2(\operatorname{ham}(M, \omega), \mathbb R)$ is isomorphic to the product of $\operatorname{H}^1_{dR, c}(M)$ and a $1$-dimensional subspace spanned by the Kostant-Souriau cocycle.<|endoftext|> TITLE: Constructively, is the unit of the “free abelian group” monad on sets injective? QUESTION [19 upvotes]: Classically, we can explicitly construct the free Abelian group $\newcommand{\Z}{\mathbb{Z}}\Z[X]$ on a set $X$ as the set of finitely-supported functions $X \to \Z$, and so easily see that the unit map $X \to \Z[X]$ (inclusion of generators) is injective. Constructively, this construction of $\Z[X]$ doesn’t work for arbitrary $X$. The unit map $X \to \Z[X]$ should send $x \in X$ to the function $1_x : X \to \Z$, sending $y \in X$ to $1$ if $y = x$ and $0$ otherwise; but this definition-by-cases requires that $X$ has decidable equality (i.e. for every $x,y \in X$, either $y = x$ or $y \neq x$). Similarly, most other classical explicit constructions of $\Z[X]$ rely at some point on decidable equality of $X$. The abstract-nonsense syntactic construction of $\Z[X]$ as a free model of an algebraic theory works fine constructively; but from this construction, it’s not clear that the unit map $X \to \Z[X]$ is injective. Is there some constructive proof that the unit map $X \to \Z[X]$ is injective for arbitrary $X$, or can it fail? [Note this is a self-answered question. It came up since the fact was required here; asking several experts, the answer seems to be not well-known. After a couple of days, I worked out an answer via an explicit construction, and then also got a reply pointing to a slightly less explicit answer in the literature; so I’m writing it up here to make the answer and construction more prominently available.] REPLY [2 votes]: Here's yet another construction of the free abelian group on an arbitrary set, and a proof of the claim. I haven't seen the proof in Mines, Richman, Ruitenburg so I don't know how similar this is to that. We define the free abelian group on a set $A$ to be a quotient $X/\sim$. The set $X$ is the set of all triplets $(n, f, x)$ where $n$ is a natural number, $f : [n] \to A$, where $[n] = \{0, 1, \dots, n-1\}$, and $x \in \mathbb {Z}^n$. The equivalence $(n, f, x) \sim (m, g, y)$ holds if there is some $k \in \mathbb {N}$, $r : [k] \to A$ and maps $i : [n] \to [k]$, $j : [m] \to [k]$ such that $f = r \circ i$, $g = r \circ j$, $i^* (x) = j^* (y)$, where $i^* : \mathbb {Z}^n \to \mathbb {Z}^k$ is the linear map that takes the basis element $e^{(n)}_t \in \mathbb {Z}^n$ to $e^{(k)}_{i (t)} \in \mathbb {Z}^k$, and similarly for $j^*$. This relation is transitive since if also $(m, g, y) \sim (p, h, z)$ via $[m] \xrightarrow {i'} [\ell] \xleftarrow {j'} [p]$, then the pushout of $(j, i')$ is some finite set $[u]$ with maps $[k] \xrightarrow {i''} [u] \xleftarrow {j''} [\ell]$, and then $i'' \circ i : [n] \to [u]$, $j'' \circ j' : [p] \to [u]$ exhibit an equivalence $(n, f, x) \sim (p, h, z)$. Note the crucial use of the existence of pushouts in the category of finite sets. Sums are defined by $(n, f, x) + (m, g, y) = (n+m, f \oplus g, (x,y))$, where $f \oplus g : [n+m] \simeq [n] \sqcup [m] \to A$ is the concatenation of $f$ and $g$, and similarly $(x, y) \in \mathbb {Z}^{n+m}$ is concatenation. I won't bother verifying the abelian group laws. More abstractly, the free abelian group generated by $A$ as a set is the colimit over the category of finite multisets in $A$ of the free abelian groups generated by them, where a finite multiset in $A$ is a map $[n] \to A$. This suggests a generalization to arbitrary Lawvere theories. To prove the claim that $A \to \mathbb {Z} [A]$ is injective, note that the generator corresponding to $a \in A$ is given by $t (a) = (1, (0 \mapsto a), (1)) \in X$. If $t (a) \sim t (b)$ via the maps $[1] \xrightarrow {i} [k] \xleftarrow {j} [1]$ and $h : [k] \to A$, then $i^* (1) = j^* (1) \in \mathbb {Z}^k$ implies that $i (0) = j (0)$, and hence $a = h (i (0)) = h (j (0)) = b$.<|endoftext|> TITLE: Are bipartite Moore graphs Hamiltonian? QUESTION [9 upvotes]: This is motivated by a computer-generated conjecture that bipartite distance-regular graphs are hamiltonian. I decided to check the case of Moore graphs first. The cycles and complete bipartite graphs are hamiltonian (trivial). The girth-6 graphs are the incidence graphs of the projective planes. The classical finite-field planes are hamiltonian by Singer, but there are lots of nonclassical planes. And what about the girth-8 and girth-12 case? EDIT: Corrected the girth-6 case as pointed out by Gordon Royle. REPLY [2 votes]: This recent paper of Sato and Suzuki shows that the graphs corresponding to some classical generalized quadrangles are indeed Hamiltonian: Sato, H. & Suzuki, H. Graphs and Combinatorics (2018). https://link.springer.com/article/10.1007/s00373-018-1940-6<|endoftext|> TITLE: Eilenberg-Watts theorem for the derived category QUESTION [7 upvotes]: Let $A$ and $B$ be $k$-algebras. And for convenience let's say $k$ is a field and both $A$ and $B$ are finite-dimensional. A well known theorem independently discovered by Eilenberg and Watts states that every $k$-linear, right exact, cocontinuous functor $F: A\mathsf{-Mod}\to B\mathsf{-Mod}$ is of the form $M\otimes_A-$ for some bimodule ${}_B M_A$ (in fact $M=F(A)$). A theorem of Rickard (with some refinements by other people like Keller) also states that if there is an equivalence $D^b(A) \to D^b(B)$ there is also an equivalence of the form $X\otimes_A^\mathbb{L}-$ for some $X\in D^b(B\otimes_k A^{op})$. A reasonable question to ask is then: Is every equivalence $D^b(A) \to D^b(B)$ itself of the form $X\otimes_A^{\mathbb{L}}-$ for some bounded complex of bimodules $X$? or more generally: Is every exact functor $D^b(A) \to D^b(B)$ of triangulated categories which commutes with all direct sums (plus maybe some other reasonable condition) of the form $X\otimes_A^\mathbb{L}-$ for some bounded complex of bimodules $X$ ? REPLY [4 votes]: Your first question seems to be a famous open question due to Rickard, see https://academic.oup.com/jlms/article-abstract/s2-43/1/37/888935?redirectedFrom=PDF . It is known in several special cases, such as for hereditary algebras, see http://home.ustc.edu.cn/~xwchen/Personal%20Papers/A%20note%20on%20standard%20equivalence.pdf . A very recent approach can be found here: https://www.sciencedirect.com/science/article/pii/S000187081830207X .<|endoftext|> TITLE: Correspondence between proof-theoretic ordinals and fast growing functions? QUESTION [8 upvotes]: For theories with well known proof-theoretic-ordinals, (what) is there a correspondence between their proof-theoretic-ordinal and (ordinal indexed families of?) fast growing functions provable total in a given theory ? REPLY [11 votes]: Yes: for many theories, $\alpha$ is the proof-theoretic ordinal of T exactly when T proves that $f_\beta$ is total for all $\beta<\alpha$, but does not prove $f_\alpha$ is total. (Where $f_\alpha$ is the $\alpha$-th function in the fast-growing hierarchy.) Avigad has argued that the correct definition of proof-theoretic ordinals is in terms of provably total functions - specifically, that the proof theoretic ordinal of T is $\geq\alpha$ exactly when T proves the totality of all functions which are "$\prec\alpha$-recursive" functions, where $\prec\alpha$-recursive means that the function is given by a program together with a timer which uses ordinal notations $\prec\alpha$. With some care about the encoding, there's a tight connection between proving fast-growing functions total and proving all $\prec\alpha$-recursive functions total, so in some sense one can take proving the totality of fast-growing functions to be the definition of the proof-theoretic ordinal. As usual with proof-theoretic ordinals, all reasonable definitions are going to be equivalent for nice theories (which includes all strong enough theories which have appeared naturally elsewhere in logic), but there are artificial theories that make the various definitions no longer equivalent.<|endoftext|> TITLE: Map Lifting lemma and Etale fundamental group QUESTION [7 upvotes]: In algebraic topology, we have the map lifting lemma which says that given a covering space $p:(\tilde{X},\tilde{x})\rightarrow (X,x)$ and a map $f:(Y,y)\rightarrow (X,x)$ with $Y$ connected and locally path-connected, a lift $\tilde{f}:(Y,y)\rightarrow(\tilde{X},\tilde{x})$ of $f$ exists if and only if $f_*(\pi_1(Y,y))\subset p_*(\pi_1(\tilde{X},\tilde{x}))$. I want to ask whether analogues of the above are true for finite etale maps and etale fundamental groups. REPLY [8 votes]: The answer is yes, at least for finite covers (or even pro-finite covers). The key is to carefully translate everything into the language of finite $\pi_1$-sets using the Galois correspondence. The canonical reference is of course SGA 1, Expose V - in particular $\S5,6$ and 7 once you're familiar with the basics. By the universal property of fiber products, the existence of such a lifting $\tilde{f}$ is equivalent to the existence of a section of the covering map (ie finite etale morphism) $p_Y : Y\times_X\tilde{X}\rightarrow Y$ which contains the geometric point $(y,\tilde{x})$ (such a section if it exists, is unique). By Galois theory, the pointed covering $p : (\tilde{X},\tilde{x})\rightarrow (X,x)$ is equivalent to the data of the pointed set $(p^{-1}(x),\tilde{x})$, together with the monodromy action of $\pi_1(X,x)$ on the underlying set $p^{-1}(x)$. The pullback $p_Y$ is (via the Galois correspondence) given by the pointed set $(p_Y^{-1}(y),(y,\tilde{x}))$ together with the monodromy action of $\pi_1(Y,y)$ acting on the underlying set, which is described by the map $$\pi_1(Y,y)\stackrel{f_*}{\rightarrow}\pi_1(X,x)\rightarrow Aut(p^{-1}(x))\cong Aut(p_Y^{-1}(y))\qquad (*)$$ where the last isomorphism is induced by the natural bijection $p_Y^{-1}(y)\cong p^{-1}(x)$. (In short, the pullback of a cover over $X$ given by a $\pi_1(X)$-set $T$ by any morphism $Y\rightarrow X$ is the cover over $Y$ given by pulling back the $\pi$-set structure on $T$ via the morphism $\pi_1(Y)\rightarrow\pi_1(X)$. (What else could it be?) See $\S6$ in SGA 1 Expose V) Under these identifications, $\pi_1(\tilde{X},\tilde{x})$ is naturally identified with the subgroup of $\pi_1(X,x)$ defined by $Stab_{\pi_1(X,x)}(\tilde{x})$. The map $p_Y$ admits a section containing the geometric point $(y,\tilde{x})$ if and only if the point $(y,\tilde{x})\in p_Y^{-1}(y)$ is fixed by the monodromy action of $\pi_1(Y,y)$ (Note that a section of $p_Y$ is a morphism of finite etale covers over $Y$, and hence corresponds via the Galois correspondence to a $\pi_1(Y,y)$-set morphism from the singleton set with trivial $\pi$-action to $p_Y^{-1}(y)$. Since this morphism is a morphism of $\pi_1(Y,y)$-sets, and hence must commute with the $\pi_1(Y,y)$-action, it implies that the image of this map is fixed by $\pi_1(Y,y)$). However, by the definition of this monodromy action in $(*)$, we see that this is true if and only if $f_*\pi_1(Y,y)\subset Stab_{\pi_1(X,x)}(\tilde{x}) = \pi_1(\tilde{X},\tilde{x})$.<|endoftext|> TITLE: Four circles on the sphere QUESTION [40 upvotes]: Consider configurations consisting of 4 distinct circles on the sphere. Two configurations are equivalent if they can be mapped onto each other by a homeomorphism of the sphere. How to enumerate/classify such configurations? Equivalent problem: classify the arrangements of 4 hyperbolic planes in the hyperbolic space, up to homeomorphisms of the space. Before voting to close this question as trivial, you may look at the classification of generic configurations which we obtained by brute force: Each region bounded by more than 3 sides is labeled by the number of its boundary sides. This is used to show that all configurations are non-equivalent. Questions: Is this new? Is there a scientific method to obtain this? Is there any structure on these 35 configurations? There is a large research area about hyperplane arrangements in a Euclidean space. How about hyperbolic space? There is also a large body of research on hyperbolic tetrahedra. But it is always assumed that the tetrahedron is compact (or has only vertices at infinity). We encountered this question in our studies of the Heun and Painlevé VI equations with real coefficients. (See Appendix II in the linked paper). Projective monodromy groups associated to these equations are generated by 4 reflections in circles. EDIT. It seems that the problem is of purely topological nature: for any collection of Jordan curves on the sphere, such that each pair intersects transversally at at most two points, there exists an equivalent configuration of circles: A. Bobenko, B. Springborn, Variational principles for circle patterns and Koebe's theorem. Trans. Amer. Math. Soc. 356 (2004), no. 2, 659–689. EDIT2. The previous remark is incorrect (thanks to Ivan Izmestiev for his comment). A counterexample with 5 curves is contained in this paper: MR3216670 Kang, Ross J.; Müller, Tobias Arrangements of pseudocircles and circles, Discrete Comput. Geom. 51 (2014), no. 4. EDIT3. From the very last Notices AMS (October 18) I learned that the number of arrangements of $n$ circles on the plane has been studied, and that this number even has a name A250001 in the Online encyclopedia of integer sequences. However, even for n=4 it is not easy to derive the result for the sphere from the result for the plane (the only way to do this that I see is by brute examination of of equivalences of configurations). REPLY [3 votes]: Not an answer; don't have enough rep to comment. You may be able to impose some structure by using the operation of crossing/uncrossing two circles which are otherwise adjacent (That is, they currently do not cross and crossing them such that there is a new region with two boundaries, or conversely removing that region). For example, $X$ is only adjacent to $W$, but $W$ is adjacent to $S$, $V$, $X$, and $X'$. This would break it into families - $T$ cannot be made from $X$. I'm not sure whether this structure is interesting, but it was a thought I had upon seeing your classification. I can't provide much more of the structure, as it's rather difficult to work out by hand.<|endoftext|> TITLE: Projective resolutions of finite-dimensional representations of infinite groups QUESTION [9 upvotes]: Let $G$ be a group and let $V$ be a finite-dimensional complex representation of $G$. Question: Under what circumstances can I find a projective resolution $$ \cdots \longrightarrow P_3 \longrightarrow P_2 \longrightarrow P_1 \longrightarrow V \longrightarrow 0$$ of $\mathbb{C}[G]$-modules such that each $P_i$ is finitely generated? I believe that this condition is usually expressed by saying that $V$ is of type $FP_{\infty}(\mathbb{C})$. One obvious first case is where $V$ is the trivial representation $V = \mathbb{C}$. If a projective resolution as above exists for this $V$, then $G$ is said to be of type $FP_{\infty}(\mathbb{C})$. The groups I am interested in are all of type $FP_{\infty}(\mathbb{C})$. It would be really wonderful if this condition was sufficient for these resolutions to exist for all finite-dimensional representations. Here is a specific example that I don't know how to do and that is typical among the ones I care about: $$G = SL(n,\mathbb{Z}) \quad \text{and} \quad V = \mathbb{C}^n.$$ REPLY [3 votes]: The answer is yes. In the answer below I use $M$ instead of $V$ and $k$ is any field. Thm 2 of https://www.tandfonline.com/doi/abs/10.1080/00927870600796110 shows that if G is $FP_\infty$ over $k$, then $kG$ has a free resolution as a bimodule by finitely generated free bimodules in each dimension. If you tensor this resolution with $M$ over $kG$ you get a free resolution of $M$ with the finiteness properties you want. Tensoring with $M$ gives a resolution because its homology is $Tor^{kG}(M,kG)$. It is easy to check that $(kG\otimes_k kG)\otimes_{kG} M\cong kG^{\dim M}$ as a left $kG$-module so that the free resolution is finitely generated in each degree. The basis as a $kG$-module is the tensors $1\otimes 1\otimes b$ with $b$ running over a basis of $M$.<|endoftext|> TITLE: Citations graphs: what is known? QUESTION [6 upvotes]: There has been much research related to web graphs and social graphs. They can be thought of as a kind of random graphs, but the point is that they are different from the well-known Erdős–Rényi model. Question: Consider citations graphs in some field of research (i.e. the graphs with vertices given by publications and edges by citations). Are there any known mathematical models for them? Or are there empirical observations on the mathematical properties of such graphs (similar to those mentioned below)? Here are some of the main properties of web graphs. I wonder about something similar for citation graphs. The diameter is quite small (~6), i.e. the Law of Six degrees of separation "Sparsity" - adjacency matrix is a sparse matrix, Power law of distribution of degrees of vertices: the number of vertices with $d$ edges scales like $C/d^{2.1}$ And there is a suggestion for the model of how web graphs are growing, namely Barabási–Albert model, with the key idea of Preferential attachment, which means that the more connected a node is, the more likely it is to receive new links. Nodes with higher degree have stronger ability to grab links added to the network. Intuitively, the preferential attachment can be understood if we think in terms of social networks connecting people. Here a link from A to B means that person A "knows" or "is acquainted with" person B. Heavily linked nodes represent well-known people with lots of relations. See e.g. https://www.mccme.ru/free-books/dubna/raigor-4.pdf REPLY [6 votes]: A great deal is known about citation graphs. For example, an analysis of the citation graph for the computer science literature answers the three questions in the OP as: • the connected component has a diameter of 18; • the sparsity is characteristic of a "small world network"; • the power law degree distribution has exponent 1.71. Concerning growth models of citation graphs, see for example A geometric graph model of citation networks with linearly growing node-increment.<|endoftext|> TITLE: Is there a physical/geometric proof for L^2 boundedness of Bourgain's maximal function along the squares? QUESTION [12 upvotes]: One problem that has bugged me for some time (though I only seriously thought about it for a month several years ago) is to give a physical proof of the L^2 boundedness of Bourgain maximal function for averages along the squares. To be precise, define the averages: for a function $f: \mathbb{Z} \to \mathbb{C}$, let $$ A_Nf(x) := N^{-1} \sum_{n=1}^N f(x-n^2) $$ and the maximal function $$ Mf(x) := \sup_{N \in \mathbb{N}} |A_Nf(x)| $$ for $x \in \mathbb{Z}$. Question: Without resorting to the circle method, can one prove that $M: \ell^2(\mathbb{Z}) \to \ell^2(\mathbb{Z})$? In particular, I suspect that there a physical proof analogous to the proofs for $L^2$ boundedness of Stein's spherical maximal function theorem as sketched out by Laba here: https://ilaba.wordpress.com/2009/05/23/bourgains-circular-maximal-theorem-an-exposition/ or as in Schlag's work: https://math.uchicago.edu/~schlag/papers/dukecircles.pdf In particular, I am happy with a physical proof of restricted weak-type $L^2$ boundedness. My suspicion is that Bourgain's proof, which uses the circle method, suggests that we should consider understand what happens on arithmetic progressions, decompose the characteristic function of a set into how arithmetic progressions and combine them in some way. However I could not see how to successfully deploy this strategy... REPLY [3 votes]: Here is a proof based on Doob inequality for martingales. We restrict ourself on a bounded domain $[-L,L]\cap \mathbb{Z}$ and we will extend $L\rightarrow \infty$. For any $x$ let $N_x$ the maximum integer such that $$Mf(x)=\frac{1}{N_x}\sum_{n=0}^{N_x} f(x-n) $$ Proposition : the segments $[x-N_x,x]$, $x\in [-L,L]$ are interlocking ie: if $[y-N_y,y]\cap [x-N_x,x]$ then $[y-N_y,y]\subset [x-N_x,x]$ or $[x-N_x,x]\subset [y-N_y,y]$. Proof : Let us assume that $y-N_y TITLE: Does $n^2$ divide $\det\left[\left(\frac{i^2+2ij+3j^2}n\right)\right]_{1\le i,j\le n-1}$ for each odd integer $n>3$? QUESTION [13 upvotes]: For any odd integer $n>1$ and integers $c$ and $d$, define $$(c,d)_n:=\det\left[\left(\frac{i^2+cij+dj^2}n\right)\right]_{1\le i,j\le n-1},$$ where $(\frac{\cdot}n)$ is the Jacobi symbol. It is easy to show that $(c,d)_n=0$ if $(\frac dn)=-1.$ QUESTION: Does $n^2$ divide $(2,3)_n$ for each odd integer $n>3$? My computation led me to conjecture that $n^2\mid (2,3)_n$ for any odd integer $n>3$ and also $n^2\mid (6,15)_n$ for any odd integer $n>5$. For any positive integer $n\equiv\pm5\pmod{12}$, we have $(\frac 3n)=-1$ and hence $(2,3)_n=0$. Also, I observe that $(2,3)_n=0$ if $n>3$ and $n\equiv 3\pmod 6$. But I don't see why $(2,3)\equiv0\pmod{n^2}$ for each positive integer $n\equiv\pm1\pmod{12}$. Note that $$\frac{(2,3)_{11}}{11^2}=-2^3\ \ \text{and}\ \ \frac{(2,3)_{37}}{37^2}=2^{44}\times467^2.$$ I also have some other questions concerning $(c,d)_n$. For example, I conjecture that $(6,1)_n=0$ for any positive integer $n\equiv 3\pmod4$. For more such questions, one may consult my preprint On some determinants with Legendre symbol entries. Any ideas helpful to the question ? REPLY [3 votes]: This is not an answer, either. Just some attempts to attack the problem. Let $r(n)$ be the square-free part of $(2, 3)_n$, i.e. $r(n)$ is square-free and we have $(2, 3)_n = r(n) B^2$ for some integer $B$. We have the following table of the values of $r(n)$: n r(n) 11 -2 23 2 37 1 47 -6 59 -2 61 1 71 -2 73 -5 83 6 97 -3 107 -6 109 -1 131 6 157 -7 167 2 179 -6 181 7 191 3 193 -1 227 -2 229 -5 239 2 241 7 251 -10 253 -1 263 -1 277 -1 313 -13 337 -1 347 2 349 -7 359 -10 373 -7 383 -1 397 5 Note the composite number $253 = 11 \times 23$, with $r(253) = r(11)r(23)$. This seems to be true in general (identities in $\mathbb{Q}^\times/\mathbb{Q}^{\times 2}$): n r(n) 11*23 -1 = r(11) * r(23) 11*37 -2 = r(11) * r(37) 11*47 3 = r(11) * r(47) 11*59 1 = r(11) * r(59) 23*37 2 = r(23) * r(37) 23*47 -3 = r(23) * r(47) This observation might be proved by using Chinese remainder theorem to rewrite the matrix as some sort of tensor product (not sure about the details, but seems doable). For non-square-free $n$, however, the determinant seems always zero. This suggests that we could concentrate on the case where $n = p$ is a prime number. In this case, the original conjecture (i.e. $p^2$ divides $(2, 3)_p$) is implied by the following two statements: $p$ divides $(2, 3)_p$; $p$ does not divide $r(p)$. Now 1. might be easier (than the original problem) to prove, since we can then do calculations in $\mathbb{F}_p$, and write the Jacobi symbol as $x^{p-1}$. This is again just a very rough idea. As to 2., the thing to notice is that, although $(2, 3)_n$ is extremely large, the numbers $r(n)$ are surprisingly small. I don't see any explanation to this phenomenon. Concerning generalizations of these statements: Statement 2. seems to be quite general. For almost arbitrary choice of parameters $(c, d)$, one always observes very small values of $r(n)$. Statement 1., however, depends strongly on the choices of $(c, d)$. So far I only see this phenomenon occuring for the following choices: "good" (c, d) (2, 3) (4, 12) (6, 27) (8, 48) ...... (6, 15) (12, 60) (18, 135) (24, 240) ...... Of course, a pair $(c, d)$ is not quite different from $(ac, a^2d)$, since the columns are just permuted, and the determinants only possibly differ by sign. So essentially there are only two "primitive" cases: $(c, d) = (2, 3), (6, 15)$. These two are essentially different, as can be seen from the difference between the $r(n)$ values. I have no idea how to find more examples of "good" pairs $(c, d)$. UPDATE: I think it might be more natural to look at the $n \times n$ matrix, by adding the indices with $i = 0$ or $j = 0$. Let us call the determinant of this larger matrix $[2, 3]_n$. It appears that in most cases $[2, 3]_n$ is a multiple of $(2, 3)_n$. In particular, when $n = p$ is a prime number for which $(2, 3)_p$ is non-zero, then we have $[2, 3]_p = \frac{p - 1}{2}(2, 3)_p$. When $n = pq$ is a product of two primes for which $(2, 3)_n$ is non-zero, then we have $[2, 3]_n = \frac{p-1}{2}\frac{q-1}{2}(2,3)_n$. The advantage of considering $[2, 3]_n$ is the following obvious formula: $$ [2, 3]_{nm} = [2, 3]_n^m [2, 3]_m^n, $$ whenever $n, m$ are coprime. This then reduces the statements above to the case of prime numbers.<|endoftext|> TITLE: Are there any advanced level math audiobooks? QUESTION [12 upvotes]: I am aware of a few interesting math podcasts but haven’t come across any interesting advanced math audiobook. Are there some? REPLY [3 votes]: The Great Courses has a selection of video courses that include graduate level mathematics, for example this course on Discrete Mathematics or this one on differential equations, or one on number theory. These are audio+video, but you might still be able to get useful information out if you only listen to the audio.<|endoftext|> TITLE: How does the Bernstein-Zelevinsky construction of irreducibles from supercuspidals parallel the representations of the Weil-Deligne group? QUESTION [12 upvotes]: In the Corvallis article Number Theoretic Background, here is what John Tate has to say on the local Langlands correspondence for a $p$-adic field $F$: So, granting a correspondence between irreducible supercuspidal representations of $\operatorname{GL}_n(F)$ and continuous irreducible $n$-dimensional complex representations of the local Weil group $W_F$, the general correspondence between irreducible representations and $\Phi$-semisimple representations of the Weil-Deligne group $W_F'$ should follow. I understand the (statements of the) Bernstein-Zelevinsky classification of smooth irreducible representations of $\operatorname{GL}_n(F)$. But how are $\Phi$-semisimple representations of $W_F'$ built out of continuous irreducible representations of $W_F$? REPLY [5 votes]: Such representations are sums of tensor products of irreducible representations of $W_F$ with the $n$-dimensional representation in which the monodromy operator acts by the nilpotent matrix with a single Jordan block (in Jordan normal form) and the Frobenius acts diagonally by elements $q^{n/2},q^{n/2-1},\dots, q^{-n/2}$. The tensor product corresponds to the construction of discrete series representations, and the sum to principal series. The Weil-Deligne group is sometimes defined as just $W_F \times SL_2$, in which case this is an irreducible representation of $W_F$ tensor the $n$-dimensional irreducible representation of $SL_2$. In this case, the classification follows immediately from the classification of irreducible representations of $SL_2$. Otherwise, the Weil-Deligne group is defined using a monodromy operator or a one-dimensional unipotent group. In this case, it is not hard to check that the same classification holds by decomposing into irreducible representations of the Weil group and examining how the monodromy operator moves between them (and this is why the two definitions can be used interchangeably) REPLY [5 votes]: Let $\rho$ be a representation of the Weil-Deligne group which is semi-simple on $W_F$ and algebraic on $\mathbb C$. Then $\rho$ is a direct sum of indecomposable such representations (uniquely up to isomorphism and ordering of the factors). If $\rho$ is indecomposable, then it is (isomorphic to) the tensor product of a unique irreducible representation of $W_F$ with the representation $(\mathbb C^{t+1},\rho_{q^{t/2},t},N_t)$ for some $t$ (where as usual $\rho_{q^{t/2},t}$ sends the Frobenius to the diagonal matrix with shifted eigenvalue by increasing powers of $q$ and $N_t$ is the maximal monodromy matrix of the correct size). Hence, taking the representations $(\mathbb C^{t+1},\rho_{q^{t/2},t},N_t)$ and the irreducible representations of $W_F$ as building blocks, you build everything. Compared to the Bernstein-Zelevinsky classification, building Weil-Deligne representations out of irreducible representations is the easy part.<|endoftext|> TITLE: Are sifted (2,1)-colimits of fully faithful functors again fully faithful? (And a de-categorified variant) QUESTION [7 upvotes]: 1) Suppose that I have a sifted diagram of categories $\mathcal{C}_i$, another of the same shape $\mathcal{D}_i$, and that I have a system $F_i:\mathcal{C}_i\to\mathcal{D}_i$ commuting with the morphisms in each diagram. Then I get a functor $$(\operatorname{colim} F_i):\operatorname{colim} \mathcal{C_i} \to \operatorname{colim}\mathcal{D}_i.$$ Suppose I know that each $F_i$ is fully faithful. Under what further assumptions may I conclude that $(\operatorname{colim} F_i)$ is fully faithful? For instance, if I would assume the $F_i$'s have right adjoints, so that fully faithfulness is equivalent to the unit for the adjunction being an isomorphism, does it help? A question which seems like a de-categorified analog of this would be: 2) Fix some co-complete category $X$, and consider some sifted diagram of objects $V_i$ in $X$, another of the same shape $W_i$ in $X$. Suppose I have a system of split monomorphisms $f_i:V_i\to W_i$. Under what conditions do I know that the natural morphism, $$(\operatorname{colim} f_i): \operatorname{colim} V_i \to \operatorname{colim} W_i$$ is also a split monomorphism? I really care about the answer to (1), but I offer (2) in my attempt to find an analogous proof one category number down, where there may be more examples/techniques/counter-examples. Here, in motivating the attempted analogy between (1) and (2), I am trying to replace the 2-category $\operatorname{Cat}$ of categories by a 1-category $X$, and replace a fully faithful functor by a split monomorphism (I am imagining that the $F_i$'s have a right adjoint, which might behave like a splitting to make the analogy work, though it's a stretch). I am aware of https://math.stackexchange.com/questions/1116101/colimit-preserves-monomorphisms-under-certain-conditions, where it is explained that filtered colimits preserve monomorphisms, in the 1-categorical setting. I am hoping, perhaps too optimistically, that if we move from filtered colimits to sifted colimits, we still have a chance to preserve split monomorphisms even though we (I suppose?) no longer preserve arbitrary monomorphisms. I am further hoping that if so, the argument would generalize to the 2-categorical setting. I'd be most grateful for any hints, or references. Disclaimer: I have simplified the above setup a bit to make a cleaner question. In case it could mislead an expert though, let me say that the version of (1) I need involves the $\mathcal{C}_i$ being Cauchy-complete, the $\mathcal{D}_i$ being presentable, and really the two colimits are taken in their respective 2-categories, in Cauchy-complete categories for the diagram of $\mathcal{C}_i$ and in presentable categories for the $\mathcal{D}_i$ (so in particular the functors appearing in each kind of colimit are suitable colimit preserving; the shape of the indexing diagram is the same, though). I assume these are technicalities and that if the question has a nice answer, I can do that kind of adaptation myself. REPLY [5 votes]: Reflexive coequalizers are examples of sifted colimits in a 1-category, and groupoids are examples of reflexive coequalizers. But quotients don't preserve monomorphisms in general. This gives a counterexample to question (2) in the category Set: Let $V = S \rightrightarrows S$ denote the trivial groupoid acting on a set $S$ with two elements, and $W = S\times \mathbb Z/2\mathbb Z \rightrightarrows S$ the action groupoid of $\mathbb Z/2\mathbb Z$ acting on the same set $S$. There is an inclusion map $V\to W$, which is injective on both objects and morphisms. But the induced map on the quotient sets is not injective. In the $(2,1)$-category setting, according to this answer of Denis Nardin, one must truncate the simplicial diagram at the 2-simplices to get a sifted diagram (note that a reflexive coequalizer diagram is the simplicial diagram truncated at the 1-simplices). To get a counterexample to question (1), first take the corresponding 2-truncated Cech simplicial sets of the diagrams above, i.e. $$S \rightrightarrows^{\rightarrow} S \rightrightarrows S$$ and $$S\times \mathbb Z/2 \times \mathbb Z/2 \rightrightarrows^\rightarrow S\times \mathbb Z/2 \rightarrow S$$ Then take the category of sheaves (of vector spaces, say) together with the pushforward functor on these diagrams of sets (thought of as discrete spaces), to get truncated simplicial diagrams of categories (all of which are direct sums of copies of Vect). The diagrams are levelwise fully faithful, but the induced functor on the colimit is identified with the pushforward $Shv(S) \to Shv(pt)$ which is not fully faithful.<|endoftext|> TITLE: Why is the number of Hamiltonian Cycles of n-octahedron equivalent to the number of Perfect Matching in specific family of Graphs? QUESTION [6 upvotes]: In OEIS A003436, it is written that the number of inequivalent labeled Hamilton Cycles of an n-dimesnional Octahedron is the same as the number of Perfect Matchings in a the complement of the Cycle Graph on 2n vertices (i.e. a complete graph $K_{2n}$ without the cycle $C_{2n}$). I can verify this relationship, but it seems very arbitrary to me. Unfortunatly, OEIS does not give further details. Why are these number of these two properties (inequivalent Hamiltonian Cycles and Perfect Matchings, respectively) equivalent in these two different objects (n-dimensional Octahedron and Complement of Cycle Graph $C_{2n}$)? And hint or reference to the literature is very much apprechiated. REPLY [9 votes]: The $n$-dimensional analogue of the octahedron is the complement of a perfect matching of its vertex set. (Every vertex is joined to every other vertex except its antipode.) If you take a Hamilton cycle in the $n$-dimensional octahedron then you can think of that as a labelling of the $2n$-cycle, and the non-edges of the $n$-octahedron give a perfect matching of the complement of this graph. This paragraph was added in answer to Nicodean's comment below. Label the vertices of the $n$-dimensional octahedron $1^+,1^-,...,n^+,n^-$, where the vertices $i^+$ and $i^-$ are antipodes. A labelled Hamilton cycle in the octahedral graph is then exactly a $2n$-cycle made from the $2n$ symbols $i^+,i^-$ with the property that the symbols $i^+$ and $i^-$ are never adjacent. Giving a perfect matching on the complement of a $2n$-cycle is the same as giving a matching of the elements of a $2n$-cycle so that adjacent elements are never matched. To do this, I choose the symbols making up my matching to be the symbols $1^+,1^-,...,n^+,n^-$, where the symbols $i^+$ and $i^-$ are pairs that are matched. The condition that adjacent symbols are never matched is exactly the same condition that I had on the $2n$-cycles that describe the Hamilton cycles in the $n$-octahedral graph. One reason why this seems hard is that the matching sort of stays the same, while the $2n$-cycle whose complement it lives in is the thing that changes when you change the Hamilton cycle.<|endoftext|> TITLE: What is the polynomial functor for the Bag monad QUESTION [7 upvotes]: I may be wrong, but we should be able to write the Bag monad in a polynomial form. The bag monad, is exectly the multiset monad whose category of algebras are the commutative monoids. Another name for the bag monad is a container. Containers are synonymous with polynomail functors. The data that defines a container are precisely the following morphisms in a locally Cartesian closed category $C$: $$ 1 \xleftarrow{\text{f}} X \xrightarrow{\text{g}} Y \xrightarrow{\text{h}} 1 $$ Where $1$ is the terminal object in $C$. This defines an endofunctor for which there is a monad. Specifically, the endofunctor is : $$ C/W \xrightarrow{\text{f^* }} C/X \xrightarrow{\Pi_g} C/Y \xrightarrow{\Sigma_h} C/Z $$ We are interested in endofunctors so $W$ and $Z$ are $1$ in $Set$. What is the polynomial form of the bag monad? REPLY [13 votes]: The bag monad is not polynomial. Any polynomial endofunctor must preserve pullbacks: $f^*$ and $\Pi_g$ preserve all limits since they’re right adjoints, while $\Sigma_h$, being just the forgetful functor from a slice category, is well known (and easily seen) to preserve all connected limits. However, the bag monad doesn’t preserve pullbacks. Write $B$ for the bag monad, and consider the sets $X = \{a,b\}$, $Y = \{y,z\}$, and view their product as a pullback, $X \times Y = X \times_1 Y$. Then the canonical map $B(X \times Y) \to B(X) \times_{B(1)} B(Y)$ fails to be injective, since $\{(a,y),(b,z)\}$ and $\{(a,z),(b,y)\}$ are distinct in $B(X \times Y)$ but have the same image in $B(X)$ and $B(Y)$. This example — and the fact that commutativity forms an obstacle to being polynomial, and similar sorts or representation — has appeared notably before in the literature, in for instance the note 3-computads do not form a presheaf category by Michael Makkai and Marek Zawadowski, and (essentially) in Carlos Simpson’s paper Homotopy types of strict 3-groupoids. REPLY [11 votes]: The free multiset functor is not polynomial in the standard sense; it is though in a categorified sense if you somehow keep track of the different ways two expressions are the same thanks to commutativity: for that you need to pass from the category Set to the 2-category of groupoids. See Data Types with Symmetries and Polynomial Functors over Groupoids.<|endoftext|> TITLE: Five-dimensional manifolds fibering over a fixed hyperbolic surface QUESTION [7 upvotes]: I am aware of the classical work by Smale and Barden computing the diffeomorphism type of smooth simply connected 5-manifolds in D. Barden, Simply connected five-manifolds, Ann. of Math. (2) 82 (1965), 365–385 and S. Smale, On the structure of 5-manifolds, Ann. of Math. (2) 75 (1962), 38–46. Has there been any effort to give invariants for 5-dimensional smooth manifolds fibering over a fixed hyperbolic surface? REPLY [10 votes]: When you say "fibering" do you means as a smooth fiber bundle? Smooth fiber bundles with fiber $F$ and base $B$ are classified by homotopy classes of maps from $B$ to the classifying space $B\mathrm{Diff}(F)$. If you case $F$ is a $3$-manifold so $\mathrm{Diff}(F)$ is somewhat better understood, see Hatcher's survey. The following two examples show what challenges exist. If $F=S^3$, then by Hatcher's proof of the Smale conjecture, or by recent results of Bamler and Kleiner discussed here the group $\mathrm{Diff}(S^3)$ deformation retracts to $O(4)$. Then we just need to give a diffeomorphism classification of the total spaces of linear $S^3$-bundles over the surface $B$. In the orientable case (i.e., when the structure group is $SO(4)$ rather than $O(4)$) such bundles are completely classified in Classification of Oriented Sphere Bundles Over a $4$-Complex by Dold and Whitney, and the bundle is completely determined by its second Stiefel-Whitney class in $H^2(B:\mathbb Z_2)\cong\mathbb Z_2$. One still has to see whether the total spaces are non-diffeomorphic, which is true (Hint: from the long exact sequence of the (pair $5$-disk bundle over $B$, its boundary) note that the inclusion of the boundary is injective on the 2nd cohomology, so the 2nd Stiefel-Whitney class of the disk bundle has nonzero restriction of the boundary iff $w_2\neq 0$. Finally, $w_2$ is a homotopy invariant for closed manifolds, so the two sphere bundles are not even homotopy equivalent). If $F$ is closed orientable hyperbolic $3$-manifold, then by Gabai's proof of Smale conjecture $\mathrm{Diff}(F)$ deformation retracts to a finite group $G:=\mathrm{Iso}(F)$. This can also be done via Ricci flow (see the above mentioned preprint of Bamler-Kleiner). Thus the bundle is classified by the homotopy classes of maps from $B$ to $BG=K(G,1)$, i.e., by the set of conjugacy classes of homomorphisms from $\pi_1(B)$ to $G$. I am not sure if anyone classified their total spaces up to diffeomorphism, but e.g. if $G=1$, then there is only the trivial bundle.<|endoftext|> TITLE: Are there non-projective, but algebraic, hyperkahler varieties? QUESTION [5 upvotes]: Let $k$ be an algebraically closed field of characteristic zero. I am not sure what the right definition of a hyperkahler variety over $k$ is, but I think the following might be close enough. Definition. A smooth proper connected scheme $X$ over $k$ is a hyperkahler variety (over $k$) if $\mathrm{h}^{2,0}(X) = 1$, the etale fundamental group $\pi_1^{et}(X)$ of $X$ is trivial, $\omega_X$ is trivial, and (added later) the generator of $\mathrm{H}^{2,0}(X)$ is everywhere non-degenerate. With this definition, it is not clear whether every hyperkahler variety over $k$ is projective. If $X$ is a two-dimensional hyperkahler variety over $k$, then $X$ is projective. But what about $\dim X > 2$? Is every hyperkahler variety over $k$ projective? PS. Please feel free to correct my definition of a hyperkahler variety over $k$. REPLY [2 votes]: With the correct definition of hyperkähler (which as Jason said requires $H^0(X,\Omega^2_X)$ to be generated by a holomorphic symplectic form), there are examples constructed by Yoshioka in section 4.4 here. The field $k=\mathbb{C}$. These are non-Kähler compact complex simply connected holomorphic symplectic manifolds which are bimeromorphic to a projective hyperkähler manifold via a Mukai flop. They are therefore Moishezon, but certainly not projective. They are also discussed for example in the book by Gross-Huybrechts-Joyce, example 21.7.<|endoftext|> TITLE: Categorification of monotone maps via tilting modules? QUESTION [8 upvotes]: It is well known that for the algebra of $n \times n$-upper triangular matrices over a field the number of tilting modules is equal to the Catalan number $C_n$. This is just the (hereditary) Nakayama algebra with Kupisch series $[n,n-1,...,2,1]$ and can be viewed as the "mother" of all Nakayama algebras with a linear quiver because it has each such algebra as a quotient. The cyclic analogue of this algebra is the Nakayama algebra with Kupisch series $[n,2n-1,2n-2,....,n+2,n+1]$ for $n \geq 2$, which can be viewed as the "mother" of all Nakayama algebras with a cyclic quiver of finite global dimension because it has each such algebra as a quotient. Now let $A$ be this Nakayama algebra with Kupisch series $[n,2n-1,2n-2,....,n+2,n+1]$. This is an algebra with global dimension 2 (while the algebra of upper triangular matrices has global dimension 1). I wondered what the tilting modules over this algebra are. The problem seems to contain the problem of classifying the tilting modules over the algebra of upper triangular matrices as a special case because the indecomposable modules with projective dimension one in $A$ just behave like modules over the algebra of upper triangular matrices and thus the number of 1-tilting modules of $A$ should be also equal to the Catalan numbers. The number of tilting modules of $A$ starts with 1,3,10,35,126 and this suggests that the number of tilting modules equals $\binom{2n-1}{n}$ which are the monotone maps $\{1,...,n \} \rightarrow \{1,...,n \}$, see https://oeis.org/A001700. This leads to the following guess: There is a natural bijection from the set of monotone maps $\{1,...,n \} \rightarrow \{1,...,n \}$ to the set of tilting modules of $A$. Note that the monotone maps $f$ with $f(i) \leq i$ are counted by the Catalan numbers (see for example exercise 78. in the book "Catalan numbers" by Richard Stanley) and thus the above bijection (if it exists) should restrict to a bijection between monotone maps $f$ with $f(i) \leq i$ and the 1-tilting modules of $A$ (at least if it is a nice bijection). I wanted to ask whether there is a quick proof of this guess in case it is true using some advanced tools. I am able to translate the problem into a purely combinatorial problem but it looks very complicated at the moment and maybe there is an easy trick to obtain such a bijection or maybe this is even known. The combinatorial translation gives the problem where $n$ points (corresponding to the indecomposable summands of the basic tilting module) are drawn into two triangles (whose points correspond to the 2-rigid indecomposable modules in the Auslander-Reiten quiver of the algebra). There is one bigger triangle with $\frac{n(n+1)}{2}$ points and one smaller triangle with $\frac{n(n-1)}{2}$ points so that both triangles have together $n^2$ points. Here the tilting modules for $n=3$ (maybe someone can see how they correspond to monotone sequences?): https://www.docdroid.net/YwBhi0k/monotonetilting.pdf Here the configurations where the red market points only occur in the smaller triangle or on the leftmost boundary of the bigger triangle count the 1-tilting modules so that for n=3 we get 5 1-tilting modules. REPLY [3 votes]: I'm missing some details, but I think the answer is lurking in Buan-Krause (2004) and Adachi (2016). Neither of these papers asks exactly the same question as Mare. Let Q be the oriented $n$-cycle, let $k[Q]$ be the path algebra and let $k[[Q]]$ be the completion of the path algebra. The Buan-Krause paper studies tilting modules for $k[[Q]]$; the Adachi paper studies Nakayama algebra $k[Q]/I$ where all the indecomposable projectives have length $\geq n$ (such as the OP's) but studies "basic $\tau$-tilting" and "basic proper support $\tau$-tilting" rather than "tilting". I'm not completely clear on the relation between these concepts, but I hope the OP is. In any case, both these papers biject the modules they study to $n$-tuples $(a_1, \ldots, a_n)$ of nonnegative integers with $\sum a_i = n$. These correspond easily to monotone functions $f: [n] \to [n]$; put $a_i = \# f^{-1}(i)$. These papers give bijections to the variants of tilting modules they study in Theorem D of Baun-Krause, and Theorems 2.16 and 2.19 in Adachi.<|endoftext|> TITLE: The group of Isometries of Shahshahani metric QUESTION [6 upvotes]: Let $$M=\{(x_1,x_2,\ldots,x_n)\in \mathbb{R}^n\mid x_i>0,\;i=1,2,\ldots,n\}$$ For $X=(x_1,x_2,\ldots,x_n)\in M$ put $|X|=\sum_{i=1}^n x_i$. We consider the Shahshahani Riemannian metric $g$ on $M$ with diagonal tensor metric $g_{ii}=\frac{|X|}{x_i}$. What is the dimension and the precise structure of the group of all isometries of $(M,g)$? REPLY [6 votes]: The point is that, if you set $x_i = {u_i}^2$ where $u_i>0$, this becomes a diffeomorphism of $M$ with itself with the property that, in the $u$-coordinates, the Shahshahani metric becomes $$ g = 4({u_1}^2+\cdots+{u_n}^2)\bigl({\mathrm{d}u_1}^2+\cdots + {\mathrm{d}u_n}^2\bigr). $$ Clearly, this metric is just the flat metric in the $u$-coordinates on the positive orthant times the squared '$u$-distance' from the origin. When $n\ge2$, any rotation in the $u$-coordinates would preserve the metric on the entire $\mathbb{R}^n$, but it wouldn't preserve the positive orthant, which is $M$. When $n=2$, this is a flat metric: Set $z = u_1 {+} i\,u_2$. then $g = 4 z\bar z\, \mathrm{d}z\circ\mathrm{d}\bar z =\mathrm{d}w\circ\mathrm{d}\bar w$ where $w = z^2 = (u_1 + i\,u_2)^2 = ({u_1}^2{-}{u_2}^2)+ i\,(2u_1u_2)$. Thus, $w:M\to\mathbb{C}$ isometrically embeds $(M,g)$ as the upper half-plane in $\mathbb{C}$ when $\mathbb{C}$ is given its standard metric. The global isometries are the translations by a real number in the $w$-coordinate together with reflection in the imaginary axis. Meanwhile, the Lie algebra of Killing fields has dimension $3$ instead of $1$. However, when $n>2$, the metric is not flat, and the Killing fields are the infinitesimal generators of the obvious $\mathrm{SO}(n)$-action, as can be seen by direct calculation or conversion to 'polar coordinates'. Meanwhile, globally on $M$, you only have the rotations and reflections (in the $u$-coordinates) that preserve the positive orthant, and this is just the permutations in the $u_i$, which is the permutations in the $x_i$.<|endoftext|> TITLE: projective plane cubics with exactly 9 real points QUESTION [8 upvotes]: It is not hard to construct such curves explicitly, e.g. my favourite example is a curve $U$ singular at $(3:4:5i)$ and also passing through $(1:0:0)$, $(0:1:0)$, $(0:0:1)$, $(1:1:0)$, $(1:0:1)$, $(0:1:1)$. (Amusingly, one more real point on $U$ is also rational, namely $(3^4:4^4:5^4)$; two more are quadratic irrational). Is there an easier argument demonstrating that such curves exist? A reference would be good enough, too. Edit: It is immediate from Bezout theorem that a curve $W$ with more than 9 real points must contain a real oval (i.e. a curve in $\mathbb{RP^2}$), just look at the intersection of $W$ and its complex conjugate $W^*$. A similar argument (see Noam's answer) shows that if it has 8 real points it must also have 9 of them (counting multiplicities, as usual). REPLY [14 votes]: Let $P_1,\ldots,P_8$ be "random" real points (which could even be rational). Then the space of cubic polynomials vanishing on all $P_i$ has dimension $10 - 8 = 2$. Let $(C_1, C_2)$ be a basis of this real vector space. Then the cubic curves $C_1=0$ and $C_2=0$ meet at $P_1,\ldots,P_8$ and at some ninth point $P_9$, which is automatically real (and also rational if $P_1,\ldots,P_8$ were). The cubic $C_1 + i C_2 = 0$ then passes through $P_1,\ldots,P_9$ and has no other real point.<|endoftext|> TITLE: Minimal combinatorial data needed to define a polytope QUESTION [9 upvotes]: Suppose I give a list of vertices $(v_1, v_2, ..., v_n)$, and a list of "adjacencies", i.e. pairs of vertices $(v_i,v_j)$. Does it exists a unique polytope that has this vertices and realises the adjacencies with 1-dimensional faces? I do not know if that is important, but each and every vertex has exactly d adjacent vertices, and I'd expect the polytope to be simple. I should add some information. The data I gave above is represented by a graph of which I know 1) it's connected 2) every vertex has the same valence d. I may come up with further constraints but these are the first I can think of. REPLY [6 votes]: In fact, elaborating on Guillermo Pineda-Villavicencio's answer, Jürgen Richter-Gebert's universality theorem for 4-polytopes shows that even in four dimensions, deciding whether a graph is realized by the vertices and edges of a simple polytope is equivalent to the existential theory of the reals. I am fairly sure that the proof of this theorem can be extended to graphs of degree 4, so for an arbitrary semi-algebraic set defined by inequalities, you can in polynomial time find a graph so that the polytope is realizable if and only if this set is non-empty (although you should check this before you cite it in a paper). See Realization spaces of 4-polytopes are universal by Richter-Gebert and Ziegler, although I suspect this theorem might have a better exposition in Richter-Gebert's monograph Realization Spaces of Polytopes.<|endoftext|> TITLE: Automorphisms of Erdős spaces QUESTION [6 upvotes]: A surjective homeomorphism $h:X\to X$ is minimal if $\overline{\{h^n(x):n\in \mathbb N\}}=X$ for every $x\in X$. In other words, the orbit of each point is dense. Does either of the Erdös spaces $\mathfrak E$ or $\mathfrak E_c$ have a minimal homeomorphism? The Erdös spaces are defined as: $\mathfrak E=\{x\in \ell^2:x_n\in \mathbb Q\text{ for all $n<\omega$}\}$; and $\mathfrak E_c=\{x\in \ell^2:x_n\in \mathbb P\text{ for all $n<\omega$}\},$ $\ell^2$ is the Hilbert space, $\mathbb Q$ is the set of rational numbers, and $\mathbb P=\mathbb R\setminus \mathbb Q$. REPLY [6 votes]: The answer to both questions is affirmative. Theorem 1. The complete Erdos space $\mathfrak E_c$ has a self-homeomorphism whose every orbit is dense in $\mathfrak E_c$. Proof. We use a known result of Kawamura, Oversteegen and Tymchantyn, that the complete Erdos space $\mathfrak E_c$ is homeomorphic to the set $E$ of enpoints of the Lelek fan. So, it suffices to construct a self-homeomorphism $h$ of the Lelek fan $L$ such that the orbit of each point $x\in E$ is dense in $E$. To construct such a homeomorphism, represent the Lelek fan as the inverse limit of a sequence $(L_n)$ of finite geometric graphs such that the preimage $p_n^{-1}(x)$ of any end-point $x$ of the graph $L_n$ under the projection $p_n:L_{n+1}\to L_n$ has some fixed odd number of points, depending only on $n$, but not on $x$ (the odd number will be used in the proof of Theorem 2). In this case we can construct a sequence of homeomorphisms $(h_n:L_n\to L_n)_{n=1}^\infty$ such that for every $n\in\mathbb N$ we have $p_n\circ h_{n+1}=h_n\circ p_n$ and the $h_n$-orbit of any end-point $x$ of $L_n$ coincides with the (finite) set of end-points of the graph $L_n$. Then the limit of the sequence $(h_n)$ is a required homeomorphism $h$ of the Lelek fan such that the $h$-orbit for any end-point $x$ of $L$ is dense in the set of end-points $E$ of $L$. Theorem 2. The rational Erdos space $\mathfrak E$ has a self-homeomorphism whose every orbit is dense in $\mathfrak E$. Proof. We shall use Corollary 5.4 of this paper of Dijkstra and van Mill. This Corollary states that the rational Erdos space $\mathfrak E$ is homeomorphic to the product $\mathfrak E_c\times\mathbb Q^\omega$. Using the argument of the proof of Theorem 1, we can construct a self-homeomorphism $h_1$ of $\mathfrak E_c$ and a self-homeomorphism $h_2$ of the Cantor cube $2^{\omega}$ such that each orbit of the homeomorphism $h:\mathfrak E_c\times 2^\omega\to\mathfrak E_c\times 2^\omega$, $h:(x,y)\mapsto (h_1(x),h_2(y))$, is dense in $\mathfrak E_c\times 2^\omega$. Using Mycielski-Kuratowski Theorem (19.1 in this book of Kechris), we can find a topological copy $C\subset 2^\omega$ of the Cantor set such that the sets $h^n(C)$, $n\in\mathbb Z$, are pairwise disjoint. Then take a topological copy $D\subset C$ of the space $\mathbb Q^\omega$ in $C$ and observe that the space $M=\bigcup_{n\in\mathbb Z}h^n(D)$ is meager and each non-empty closed-and-open set in $M$ is of type $F_{\sigma\delta}$, but not $G_{\delta\sigma}$. By a theorem of van Engelen, up to a homeomorphism, $\mathbb Q^\omega$ is a unique meager zero-dimensional metrizable space whose every closed-and-open set is of type $F_{\sigma\delta}$ but not $G_{\delta\sigma}$. This characterization of van Engelen implies that the space $M$ is homeomorphic to $\mathbb Q^\omega$. Then the product $E:=\mathfrak E_c\times M$ is homeomorphic to the rational Erdos space $\mathfrak E$ and the restriction of the homeomorphism $h$ to $E$ has the required property: the $h$-orbit of any point of $E$ is dense in $E$.<|endoftext|> TITLE: Does Easton forcing preserve measurable cardinals? QUESTION [9 upvotes]: The question is in the title. For Easton's theorem see Wikipedia. Loosely speaking we can use forcing to manipulate the powerset function on regular cardinals as much as we like given we satisfy the following restrictions: (a) $\kappa<\lambda$ implies $2^\kappa\le 2^\lambda$ and (b) $\kappa\kappa^+$ is strictly stronger than a measurable cardinal. One needs $\kappa$ to be $\kappa^{++}$-tall. (See my paper, Hamkins, Joel D., Tall cardinals, Doi:10.1002/malq.200710084, Math. Log. Q. 55, No. 1, 68-86 (2009). ZBL1165.03044, blog post.) So it will be necessary to start with a larger large cardinal situation if this is desired. Meanwhile, one can achieve a large number of positive instances when one begins with a supercompact cardinal. In many of these cases it is often easier to use the Easton-support iteration rather than the Easton product forcing. There is a whole literature on combining Easton's theorem with large cardinals, and perhaps people can post further references. Let me mention a few papers of Brent Cody, who first looked at this topic in his disseration (under my supervision). Cody, Brent, Easton’s theorem in the presence of Woodin cardinals, Arch. Math. Logic 52, No. 5-6, 569-591 (2013). ZBL1305.03039. Cody, Brent; Friedman, Sy-David; Honzik, Radek, Easton functions and supercompactness, Fundam. Math. 226, No. 3, 279-296 (2014). ZBL1341.03069. Cody, Brent; Gitman, Victoria, Easton’s theorem for Ramsey and strongly Ramsey cardinals, Ann. Pure Appl. Logic 166, No. 9, 934-952 (2015). ZBL1371.03064. REPLY [7 votes]: Suppose $GCH$ holds, $U$ is a normal measure on $\kappa$ and $j: V \to M$ is an ultrapower embedding. Let $F$ be an Easton function on regular cardinals and $P_F$ be the corresponding Easton forcing. Also let $G_F$ be $P_F$-generic filter. First note that we can assume that $dom(F)$ contains regular cardinals $\leq \kappa$ (as the rest is $\kappa^+$-closed and does not have any effects on measurability of $\kappa$). If $\kappa \notin dom(j(F)),$ then $\kappa$ remains measurable in $V[G_F].$ This is due to Kunen-Paris: Boolean extensions and measurable cardinals, Annals of Mathematical Logic 2 Issue 4 (1971) pp 359-377. Also if we force with $Add(\lambda, 1)$ for all regular $\lambda < \kappa$, then $\kappa$ fails to be measurable in $V[G_F].$ See pages 375-376 of the above paper of Kunen-Paris (the argument works if we also force at all regular $\lambda \leq \kappa$).<|endoftext|> TITLE: The symmetric group theory of natural numbers QUESTION [38 upvotes]: Sometimes it is not easy to formulate a correct question. Here is a better version of this question (I still do not know if it is optimal, but it is better than the previous one). We say that a set $X$ of natural numbers is symmetric group definable if there exists a first order formula $\theta$ in the group signature such that a symmetric group $S_n$ satisfies $\theta$ if and only if $n\in X$. Of course finite sets and sets with finite complements are symmetric group definable. It is not completely trivial to find any infinite set of natural numbers with infinite complement which is symmetric group definable. But it is not a difficult exercise to show that the set of numbers $n$ such that either $n$ or $n-1$ is prime is such a set. Question. Is any of the following sets symmetric group definable? 1) the set of even numbers 2) the set of prime numbers A more vague question Is there a characterization of the Boolean algebra of symmetric group definable sets? Update Noam D. Elkies' answer below shows that the Boolean algebra contains many sets. Noah Schweber's answer of the previous question suggests looking at the computational complexity of $X$. Clearly, checking whether $n$ belongs to a symmetric group definable set $X$ takes time at most $(n!)^k$ where $k$ is the number of quantifiers in the corresponding formula $\theta$. So we have a couple of even better questions. Question 1 Is the converse true, that is every set of numbers recognizable by deterministic a Turing machine in time $\le (n!)^k$ for some $k$ symmetric group definable? Here the size of a number $n$ is $n$ that is we consider the unary representation of natural numbers. Question 2 Let us replace $S_n$ by, say, $SL_n(\mathbb{F}_2)$ and similarly define $SL_n(\mathbb{F}_2)$-definable sets of numbers. Will the Boolean algebra of all $SL_n(\mathbb{F}_2)$-definable sets coincide with the Boolean algebra of all deterministic polynomial time decidable sets of natural numbers? Question 3 Can the statements that "Primes are in P" be proved this way? REPLY [6 votes]: Almost the same first-order sentence that Noam used to define transpositions defines, in $SL_n(\mathbb F_2)$, a unipotent transvection (i.e. with a single $2\times 2$ Jordan block, and the rest $1 \times 1$). The sentence is "$g$ has order $2$, and its product with any conjugate of itself has order at most $4$". Indeed, if $g$ has order $2$ then $g$ is unipotent, with all Jordan blocks of size at most $2$. If it has one of size at most $2$ then $g$ times any conjugate of $g$ fixes a codimension two subspace, hence has order $1,2,3$, or $4$. On the other hand, if $g$ has $k$ Jordan blocks, we can view it as a unipotent matrix with a single Jordan block over $\mathbb F_{2^k}$, so using $$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 0 \\ x & 1 \end{pmatrix}=\begin{pmatrix} 1+x & 1 \\ x & 1\end{pmatrix} $$ we can produce any conjugacy class in $SL_2(\mathbb F_{2^k})$, and in particular an element of order $2^k+1>4$. I suspect from appropriate finite configurations of transvections one can define the disjoint union of the set of nonzero vectors and the set of nonzero linear functions (it is not possible to separate these because of the inverse-transpose automorphism) but I wasn't able to make it work yet.<|endoftext|> TITLE: Sum of divisors below threshold QUESTION [8 upvotes]: Let $\sigma(n)$ denote the sum of divisors of $n$, that is, $$ \sigma(n) = \sum_{d | n} d. $$ It is known that $\sigma$ can have values as large as order $n \log \log n$. However, obviously the sum is reduced when we restrict it to divisors which are smaller than some threshold $D$. Question 1: How large can $D$ be (as a function of $n$), such that $$ \sum_{\substack{1 \leq d \leq D,\\ ~d | n}} d \leq n, $$ for all (sufficiently large) $n$? One admissible value for $D$ is roughly $n / \exp(c \log n / \log \log n)$, since the number of divisors of $n$ is bounded by $\exp(c \log n / \log \log n)$, and the sum above is obviously dominated by $D$ times the number of divisors of $n$. However, my feeling is that the "right" order for $D$ should rather be around $n/\log n$ or even closer to $n$. Question 2: Can we further increase the bound for $D$ from Question 1 if we only consider those $n$ for which $\varphi(n) > \varepsilon n$, for some fixed constant $\varepsilon$? REPLY [11 votes]: Put $$ B= C \frac{\log (10\sigma(n)/n)}{\log \log (10 \sigma(n)/n)} $$ for a suitably large positive constant $C$. Then I claim that the desired inequality holds with $$ D = \frac{n}{(\log n)^B}, $$ for all large $n$. Since $\sigma(n)/n \ll \log \log n$, it follows that one may always take $$ D = n \exp\Big( -C \frac{\log_2 n \log_3 n}{\log_4 n}\Big), \tag{1} $$ with $\log_j$ denoting the $j$-th iterated logarithm, and in fact this is in general the best possible. If $\sigma(n)/n$ is bounded by $1/\epsilon$ (as in Question 2) obviously one can obtain a stronger result, as $B =C \log(1/\epsilon)/\log \log (1/\epsilon)$ is now permissible. Now for the proof. Put $$ A = \frac 12 \log \log (10\sigma(n)/n), \text{ and } \alpha= \frac{A}{\log \log n}. $$ Note that (writing $d=n/k$) $$ \sum_{\substack{ d|n \\ d\le D}} d =n \sum_{\substack{k|n \\ k > (\log n)^B}} \frac{1}{k} \le n (\log n)^{-B\alpha} \sum_{k|n} \frac{1}{k^{1-\alpha}} = ne^{-AB} \sum_{k|n} \frac{1}{k^{1-\alpha}}. \tag{2} $$ For large $n$ we have $$ \sum_{k|n} \frac{1}{k^{1-\alpha}} \ll \prod_{p|n} \Big(1 + \frac{1}{p^{1-\alpha}}\Big) \ll \exp\Big( \sum_{p|n} \frac{1}{p^{1-\alpha}}\Big). $$ To estimate this, divide the primes $p|n$ into three ranges: $p\le (\log n)^{1/A}$, $(\log n)^{1/A} \le p \le (\log n)$ and $p> \log n$. The contribution of the first range is $$ \sum_{\substack{ p|n \\ p\le (\log n)^{1/A}} } \frac{e}{p} \ll \log (\sigma(n)/n). $$ The second range gives $$ \le \sum_{(\log n)^{1/A} \le p\le (\log n)} \frac{1}{p^{1-\alpha}} \le e^A \sum_{(\log n)^{1/A} \le p\le (\log n)} \frac 1p \ll e^A \log A. $$ The final range gives (since the number of distinct primes dividing $n$ is $\ll \log n/\log \log n$) $$ \ll \sum_{\substack{p|n \\ p>\log n}} \frac{1}{p^{1-\alpha}} \le \frac{1}{(\log n)^{1-\alpha}} \sum_{p|n} 1 \ll \frac{e^A}{\log \log n}. $$ Combining all these bounds, and using it in (2) we find $$ \sum_{\substack{d|n \\ d\le D}} d \ll n e^{-AB} \exp\Big( O(\log (10\sigma(n)/n))\Big) \le n, $$ upon taking $C$ suitably large. This completes the proof that the claimed choice for $D$ works. Let me now quickly say why the general form (1) is optimal. Take $n$ to be the lcm of all the integers up to some point (which is roughly $\log n$ by the prime number theorem). Then, roughly speaking, $$ \sum_{\substack{k |n \\ k >(\log n)^u}} \frac{1}{k} \gg \sum_{\substack{ (\log n)^{u+1} \ge k \ge (\log n)^u \\ p|k \implies p\le (\log n)}} \frac 1k \gg \rho(u+1) \log \log n, $$ where $\rho$ denotes the Dickman function. That is, the divisors of $n$ basically correspond to $\log n$ smooth numbers, and I have invoked the asymptotic for smooth numbers here. Since $\rho(u)$ behaves like $u^{-u}$ we see that $u$ has to be about as large as $\log_3 n/\log_4 n$ in order for $\rho(u+1) \log \log n$ to become negligible. This shows that the range in (1) cannot be improved (apart from the constant $C$).<|endoftext|> TITLE: An interesting Markov chain with uniform marginals QUESTION [7 upvotes]: Consider the Markov chain $(\theta_n, \phi_n)$ on $S^1 \times S^1$ constructed in the following way. For $\xi_n$ a sequence of i.i.d. normal random variables and $\kappa > 0$ a fixed number, we set $$ \theta_{n+1} = \theta_n + \kappa \xi_n\;,\qquad \phi_{n+1} = \arg((3+2\sqrt 2)\cos \phi_n,\sin \phi_n) + \theta_n\;, $$ where $arg(v)$ denotes the angle of the vector $v$ with $(1,0)$. For every $\kappa > 0$, it has a unique invariant measure $\mu_\kappa$ and it is obvious that its first marginal $\pi_\theta^* \mu_\kappa$ is just Lebesgue measure $\lambda$ for every $\kappa$. Of course, one has $\lim_{\kappa \to \infty} \mu_\kappa = \lambda \otimes \lambda$, so that $\lim_{\kappa \to \infty} \pi_\phi^*\mu_\kappa =\lambda$. More surprisingly, it follows from an explicit but lengthy calculation that one also has $\lim_{\kappa \to 0} \pi_\phi^*\mu_\kappa =\lambda$. This begs the question: is it true that $\pi_\phi^*\mu_\kappa =\lambda$ for every $\kappa > 0$? I've checked it numerically to rather high accuracy for several values of $\kappa$ and it seems to be the case. Given the simplicity of the statement, it feels like there should be a rather simple argument if it is true, but it eludes me at the moment. According to computer simulations, the statement seems independent of the specific value $3+2\sqrt 2$ appearing in the definition above although i) it appears to fail when the value is replaced by a negative number and ii) the specific value $3+2\sqrt 2$ simplifies the calculation of the limit $\kappa \to 0$ somewhat. REPLY [11 votes]: Since I was requested to elaborate, here goes. First, let's look at the automorphism of the unit circle induced by this mapping (written in the least revealing way). With $z=e^{it}$, as usual, we have $2\cos t=z+z^{-1}, 2i\sin t=z-z^{-1}$, so for positive $3+\sqrt 2$ (I absolutely loved this red herring) the direction is that of $z+\delta z^{-1}$ with $|\delta|<1$. Normalizing, we get $$ \Psi(z)=\frac{z+\delta z^{-1}}{|z+\delta z^{-1}|} $$ and, most importantly, $$ \Psi(z)^2=\frac{(z+\delta z^{-1})^2}{|z+\delta z^{-1}|^2}= \frac{z+\delta z^{-1}}{z^{-1}+\delta z}=\frac{z^2+\delta}{1+\delta z^{2}}\,. $$ Now the full transformation for random variables $U=e^{i\theta}$, $W=e^{i\varphi}$ is $U,W\mapsto ZU, U\Psi(W)$ where $Z=e^{i\xi_n}$ is something smeared a bit and independent of $U,W$. Since it doesn't matter for the convergence to the limiting distribution with what joint distribution to start, we'll start with independent $U,W$ uniformly distributed over the circle. Claim 1: All four pairs $(\pm U,\pm W)$ have the same distribution at every iteration step. Proof: It is clearly true in the beginning and, since $\Psi$ is odd, this property is preserved by the mapping. This alone immediately kills all moments $\mathcal E [U^kW^\ell]$ with $k$ or $\ell$ odd. Claim 2: If $k,\ell\ge 0$ with $k+\ell>0$, then $\mathcal E [U^{2k}W^{2\ell}]=0$. Proof: It is clearly true in the beginning. Now just recall that $\Psi^2$ is even analytic, so $\Psi(z)^{2\ell}=\sum_{m\ge 0} a_{\ell,m}z^{2m}$ and, thereby, $$ \mathcal E [U_{\text{new}}^{2k}W_{\text{new}}^{2\ell}]= \mathcal E [(ZU)^{2k}(U\Psi(W))^{2\ell}]= \mathcal E [Z^{2k}]\sum_{m\ge 0}a_{\ell,m}\mathcal E [U^{2k+2\ell}W^{2m}]=0\,. $$ In particular $\mathcal E[W^{2\ell}]=0$ for $\ell>0$ (and, by conjugation, for $\ell<0$ too) and we are done. The above recursion can also be used to find the Fourier coefficients of the limiting distribution with $k<0,\ell>0$ row by row.<|endoftext|> TITLE: Effective bound on the expansion of the $j$-invariant QUESTION [11 upvotes]: The $j$-ivariant has the following Fourier expansion $$j(\tau)=\frac 1q +\sum_{n=0}^{\infty}a_nq^n=\frac{1}{q}+744+196884q+21493760q^2+\cdots.$$ Here is $q=e^{2\pi i \tau}$. Is there some simple effective bound on the coefficients $a_n$? Backround. This question comes from On the “gap” in a theorem of Heegner. Let $D$ be a negative discriminant such that $h(D)=1$. We want to show that $J=j(\sqrt{D})$ generates a cubic extension of $\mathbf Q$. Since we have at our disposal a monic cubic polynomial with rational coefficients, the modular equation $\Phi_2(X,j)$, whose root is $J$, and the other two roots are non-real, it is sufficient to show that $J$ is not an integer. In this case $j=j\left(\frac{-1+\sqrt D}{2} \right)$ is also an integer. Set $$t=e^{2\pi i(-1+\sqrt D)/2}.$$ Then $$J=\frac{1}{t^2}+744+196884t^2+O(t^4),$$ and $$j^2-1488j+160512-J=42987520t+O(t^2).$$. On the left there is an integer. However the right side tends to zero as $|D|$ gets large. Stark asserts that $|D|>60$ is enough for the RHS to be less than 1. Why is it enough? REPLY [9 votes]: By a variation of Elkies's answer we can even get $a_n TITLE: Is the measurable space $(\omega_1,\mathcal{P}(\omega_1))$ separable? QUESTION [8 upvotes]: Here $\omega_1$ is the first uncountable ordinal, and $\mathcal{P}(\omega_1)$ denotes the power set of $\omega_1$. Separable means countably generated as a $\sigma$-algebra. REPLY [11 votes]: Whether $\mathcal P(\omega_1)$ is separable is independent of ZFC. If $2^{\aleph_0} \neq 2^{\aleph_1}$ (which is consistent with ZFC -- it is implied by CH for example), then $\mathcal P(\omega_1)$ (which has size $2^{\aleph_1}$) is larger than any countably generated $\sigma$-algebra (which has size at most $2^{\aleph_0}$). On the other hand, $MA + \neg CH$ (which is also consistent with ZFC) implies that every size-$(<\!\mathfrak{c})$ subset $X$ of the real line is a "$Q$-set." This means that every subset of $X$ is a relative $G_\delta$ (a countable intersection of open subsets of $X$). Suppose $X$ is such a set with $|X| = \aleph_1$. Then the induced topology on $X$ generates the $\sigma$-algebra $\mathcal P(X)$. That is, the $\sigma$-algebra $\mathcal P(X)$ is generated by a countable collection of subsets of $X$, namely, the basic open sets of $X$ as a subspace of $\mathbb R$. Re-indexing the points of $X$ with $\omega_1$, we see that $\mathcal P(\omega_1)$ must be countably generated too. The same argument shows that $\mathcal P(\omega_2)$, $\mathcal P(\omega_{42})$, $\mathcal P(\omega_{\omega^2+137})$, and every such set in between can, consistently, be countably generated as well. All you need to do is live in a model of set theory where Martin's Axiom holds and the continuum is at least $\aleph_{\omega^2+138}$.<|endoftext|> TITLE: Compactification of 6d (2, 0) SCFT on 4-manifolds QUESTION [12 upvotes]: This question is about the 6d (2, 0) superconformal field theory (also called 'theory X' by some people). This SCFT, which can be considered as a relative quantum field theory (see here for a definition), exhibits a great deal of mathematical structure which is cogently expressed in the diagram below (taken from slides for D. Ben-Zvi's talk). In particular, its compactification on 2-torus is $N=4$ Yang--Mills theory. Note that the S-duality of the compactification of a topological twist of $N=4$ SYM (the so-called Kapustin--Witten TQFT) on a closed Riemann surface encodes geometric Langlands duality. My question is: has there been any recent progress on understanding the compactification of theory X on general 4-manifolds (marked as '???' in the diagram)? REPLY [7 votes]: You could have a look at https://arxiv.org/abs/1806.02470 and references therein. EDIT (taking into account the comment): compactification of the $\mathcal{N}=(2,0)$ 6d superconformal field theory on a 4-manifold (with an appropriate topological twist along the 4-manifold), produces a 2d (6=4+2) $\mathcal{N}=(2,0)$ superconformal field theory. Taking its chiral part, one should get a chiral algebra of operators, or vertex operator algebra (VOA), which is a well-defined mathematical object. So a weak version of the problem is to produce a VOA from a 4-manifold. This expected VOA is known in some examples and some strategy, used in the linked paper, is to construct more general examples by trying to identify the effect on the VOA of basic topological operations on the 4-manifold. The character of the VOA (or the elliptic genus of the 2d superconformal field theory) is expected to be related to Vafa-Witten invariants of the 4-manifold. In general, Vafa-Witten invariants should be some counts of solutions of some gauge theoretic PDE (similar to Donaldson invariants). The required analysis dealing with the non-compactness of the relevant moduli spaces does not seem to have been done. Recent progress includes a definition of Vafa-Witten invariants via algebro-geometric techniques for projective complex surfaces, by Tanaka and Thomas (see https://arxiv.org/abs/1702.08487 ).<|endoftext|> TITLE: p-adic L-functions for (dual of) fine Selmer Groups QUESTION [6 upvotes]: If $R(E/\Bbb Q_{\infty})$ is the fine Selmer group and $Y(E/\Bbb Q_{\infty})$ is its dual, then we know that $Y(E/\Bbb Q_{\infty})$ is a finitely generated $\Lambda$-module and by a theorem of Kato, it is also torsion. My understanding is that it should thus make sense to want to attach a $p$-adic L-function. Can we say what this $p$-adic L-function is? Or does it just follow (trivially) from the Iwasawa main conjecture? REPLY [2 votes]: You leave out a bit of information, which I will fill in here; I hope correctly. Let $E/\mathbb{Q}$ be an elliptic curve and let $\mathbb{Q}_{\infty}$ be the cyclotomic $\mathbb{Z}_p$-extension for an odd prime $p$. The fine (also called strict) Selmer group is the kernel of the usual $p$-primary Selmer group to the local cohomology of $E[p^{\infty}]$ at all places above $p$. Kato proves indeed in his theorem 13.4.1 in Astérisque 295 that its dual $Y$ over $\mathbb{Q}_\infty$ is a finitely generated, torsion $\Lambda$-module where $\Lambda$ is the usual Iwasawa algebra for $\mathbb{Q}_\infty/\mathbb{Q}$. His formulation of the main conjecture 12.10 states that the characteristic series of $Y$ (denoted by $\mathbf{H}^2(T)$ in Kato) should be equal to that of $\mathbf{H}^1(T)/Z(T)$ where, first, $\mathbf{H}^1(T)$ is the projective limits of global Galois cohomology $H^1(G_S(\mathbb{Q}_n), T)$ unramified outside the usual set $S$ of bad places and places above $p$, and secondly $Z(T)$ is its submodule generated by certain zeta elements; and $T$ could be the Tate module of $E$. Further $\mathbf{H}^1(T)$ is a torsion-free $\Lambda$-module of rank $1$. In fact it is usually free, but in the presence of a $p$-isogeny defined over $\mathbb{Q}$ on $E$ it may be isomorphic to the maximal ideal in $\Lambda$ instead. Let us suppose that it is free. Then there is a particular element $z_{\gamma}$ of Kato's Euler system that generates $Z(T)$. So we can call $z_{\gamma}$ the "$p$-adic $L$-function of $Y$" if we wish. If the reduction is good ordinary, we know that its image under the composition of mapping it into the singular local cohomology $\mathbf{H}^1_s(T)$ and the Coleman map to $\Lambda$ is the usual $p$-adic $L$-function constructed directly from modular symbols. In the supersingular and multiplicative case, we have similar links. Then the main conjecture is equivalent to the usual formulations and known in quite a few cases now. A long time ago I did some computations of the characteristic series of $Y$ and in particular its leading term. It is conjectured and often known that $\mu=0$ for $Y$. In ther large majority of cases one can show that the leading term is a unit, which implies that it is just $T^{r-1}$ times a unit where $T$ is now the variable in $\Lambda=\mathbb{Z}_p[\![T]\!]$ and $r$ is the rank of $E$ over $\mathbb{Q}$. I found a case where the rank jumps from $2$ to $6$ in the first layer of the $\mathbb{Z}_3$-extension, which gives $T$ times a simple cyclotomic factor. For all I know it could be that the Tate-Shafarevich part of the fine Selmer group is always bounded and so there is never anything more interesting. For rank $0$ curves, this would say that the Euler system $z_{\gamma}$ is a generator of $\mathbf{H}^1(T)$ in a very large number of cases even when the $p$-adic $L$-function is genuinely interesting. (This is all very long ago and my recollection of things is not so precise anymore. Probably there are people out there knowing more about the Euler system that could correct me.)<|endoftext|> TITLE: Embedding of a complex submanifold in projective space QUESTION [5 upvotes]: Suppose you have a projective manifold $M$, a very ample bundle $\scr L$ and a transverse holomorphic section $s \in H^0(\scr L)$. Then the zero set of $s$ is a complex submanifold $S_M$. Can we have a embedding of the the projective manifold $M$ in some projective space such that image of $S_M$ will not be contained in a hyperplane? You can assume $M$ has complex dimension $2$. REPLY [7 votes]: For $M=P^2$ the projective plane and $L=S_M$ a line in $P^2$, there are simple examples of projective embeddings of $P^2$ where $L$ does not lie in a hyperplane. To see this consider e.g. a generic 5-dimensional linear system $D$ in the 20-dimensional space of quintic curves on P^2. Then $D$ provides an embedding of P^2 into $P^5$, and $D$ is disjoint from the 14-dimensional space of reducible quintics that contain $L$ (which is isomorphic to the space of quartic curves), because $5+14 < 20$. Therefore no hyperplane section (no divisor in $D$) contains $L$. (Here $Pic(M)=Z$ so this contradicts the comment by abx; the point is the linear system embedding $M$ here is not be complete.) The same reasoning applies to any divisor $S$ of any projective variety $M$ in $P^n$ with $dim(M)>1$. To see this pick a finite set $T$ in $S$ with $|T| > 2 dim(M)+1$. Then for $d$ sufficiently large, the space of divisors of degree $d$ on $M$ passing through $T$ has codimension $|T|$ in the space of all divisors of degree $d$. (This is elementary.) A generic linear system $D$ of degree $d$ and dimension $N= 2 dim(M)+1$ will then contain no such divisors, and hence no divisors containing $S$; and it will embed $M$ in $P^N$ since it gives a generic projection of the Veronese embedding $M$ of degree $d$.<|endoftext|> TITLE: Sign of permutation induced by modular exponentiation QUESTION [9 upvotes]: Given a prime number $p$ and a primitive root $a$ modulo $p$, let $\sigma_{a,p}$ denote the permutation of the set $\{1, \dots, p-1\}$ which maps $b$ to $a^b$ modulo $p$. Question: Let $p$ be fixed. Does the following hold?: If $p$ is congruent to $1$ modulo $4$, then for precisely half of the primitive roots $a$ modulo $p$, the permutation $\sigma_{a,p}$ is odd, and for the other half of the primitive roots it is even. If $p$ is congruent to $3$ modulo $4$, then either all permutations $\sigma_{a,p}$ are odd, or all of them are even -- depending on whether $p$ divides $\frac{p-1}{2}!-1$ or not. REPLY [9 votes]: Your second guess is also correct. At first, we write down the sign of the permutation $\sigma_a$ as the product $\prod_{1\leqslant i TITLE: How were modular forms discovered? QUESTION [54 upvotes]: When modular forms are usually introduced, it is by: "We have the standard action of $SL(2,\mathbb Z)$ on the upper half-plane, so let us study functions which are (almost) invariant under such transformations". But modular forms were discovered in the 19th century, before group theory was available. Therefore, I'd like to ask how were modular forms originally discovered? In his book The 1-2-3 of modular forms D. Zagier writes that this was the reason: Proposition 21 Let $f(z)$ be a modular form of weight $k$ on some group $\Gamma$ and $t(z)$ a modular function with respect to $\Gamma$. Express $f(z)$ locally as $\Phi(t(z))$. Then the function $\Phi(t)$ satisfies a linear differential equation of order $k+1$ with algebraic coefficients After the proof (actually, 3 of them) there is an example: The function $\sqrt[4] E_4$ is formally a mod. form of weight one, which means: $\sqrt[4] E_4(z)=F(\frac 1 {12}, \frac 5 {12}, 1; 1728/j(z))=1+\frac {60} {j(z)}+...$ He states that this is a classical identity found in the works of Fricke and Klein. This, however, sparks a few questions: 1) I know that hypergeometric series were investigated in that time, but how did they find this one and similar ones? 2) Did they find something special about the resulting functions, like the $SL(2,\mathbb Z)$ transformation law? REPLY [58 votes]: You don't need the language of group theory to talk about some aspects of groups. For example, number theorists going back to Fermat were studying the group of units mod $m$ (including things like the order of a unit mod $m$) and geometers were studying groups of motions in space long before anyone defined a "group". The first modular forms (of level $4$, not level $1$) were found by Gauss in his work on the arithmetic-geometric mean around 1800, and it would take until the end of the 19th century for the term "modular form" to be introduced, in 1890. I provided links for this near the end of my answer here, but I don't want to discuss this further. Instead I want to show how one reason for interest in modular forms in the 19th century was the construction of nonconstant meromorphic functions on Riemann surfaces of higher genus. This will explain why someone might care about functions satisfying a transformation rule like $f((a\tau+b)/(c\tau+d)) = (c\tau+d)^kf(\tau)$. Compact Riemann surfaces of genus 1 arise in the form $\mathbf C/L$ where $L$ is a discrete subgroup of $\mathbf C$. Meromorphic functions on $\mathbf C/L$ are the elliptic functions, which in various forms were studied throughout much of the 19th century (by Jacobi, Weierstrass, et al.). We'd like to find a similar story for compact Riemann surfaces of genus greater than $1$. By letting a discrete group $Γ \subset {\rm SL}_2(\mathbf R)$ act on the upper half-plane $\mathfrak h$ by linear fractional transformations, the coset space $\Gamma\setminus\mathfrak h$ will be a compact Riemann surface (with finitely many points missing) and the higher-genus Riemann surfaces arise in this way. A natural collection of discrete subgroups of ${\rm SL}_2(\mathbf R)$ is ${\rm SL}_2(\mathbf Z)$ and its finite-index subgroups, such as the congruence subgroups. When I write $\Gamma$ below, you could consider it to be one of these groups of integral matrices. How can we create nonconstant meromorphic functions on $\Gamma\setminus\mathfrak h$? That is the same thing as creating nonconstant meromorphic functions $F : \mathfrak h \rightarrow \mathbf C$ that are $\Gamma$-invariant: $F(\gamma\tau) = F(\tau)$ for all $\gamma \in \Gamma$ and $\tau \in \mathfrak h$. If we aren't clever enough to be able to write down $\Gamma$-invariant functions directly, we can still make progress by finding some non-invariant functions as long as they are non-invariant in the same way: if $$ f\left(\frac{aτ + b}{cτ + d}\right) = (cτ + d)^kf(τ) $$ and $$ g\left(\frac{aτ + b}{cτ + d}\right) = (cτ + d)^kg(τ) $$ for some "weight" $k \in \mathbf Z$ and all $(\begin{smallmatrix} a& b\\c &d\end{smallmatrix}) ∈ \Gamma$ and $\tau ∈ \mathfrak h$, then the ratio $f(τ)/g(τ)$ is $\Gamma$-invariant: $$ \frac{f((a\tau+b)/(c\tau+d))}{g((a\tau+b)/(c\tau+d))} = \frac{(cτ + d)^kf(τ)}{(cτ + d)^kg(τ)} = \frac{f(\tau)}{g(\tau)}. $$ As long as $f$ and $g$ are not constant multiples of each other (essentially, the space of modular forms $f$ and $g$ live in is more than one-dimensional), the ratio $f/g$ will be a nonconstant $\Gamma$-invariant function. But why should we use fudge factors of the form $(cτ + d)^k$? Suppose for a function $f : \mathfrak h \rightarrow \mathbf C$ that $f(γτ)$ and $f(τ)$ are always related by a nonvanishing factor determined by $γ ∈ Γ$ and $τ ∈ \mathfrak h$ alone, not by $f$: $$ f(γτ) = j(γ, τ)f(τ) $$ for some function $j : \Gamma × \mathfrak h \rightarrow \mathbf C^\times$. If also $g(γτ) = j(γ, τ)g(τ)$ then $f(γτ)/g(γτ) = f(\tau)/g(\tau)$, so $f/g$ is $\Gamma$-invariant. We want to figure out how anyone might imagine using the fudge factor $j(\gamma,\tau) = (c\tau+d)^k$, where $γ = (\begin{smallmatrix} a& b\\ c& d\end{smallmatrix})$. Since $(γ_1γ_2)τ = γ_1(γ_2τ)$, we have $f((γ_1γ_2)τ) = f(γ_1(γ_2τ))$. This turns the above displayed equation into $$ j(γ_1γ_2,τ)f(τ) = j(γ_1,γ_2τ)f(γ_2τ). $$ Since $f(γ_2τ) = j(γ_2,τ)f(τ)$, the above equation holds if and only if the "cocycle condition" $$j(γ_1γ_2, τ) = j(γ_1,γ_2τ)j(γ_2,τ)$$ holds, which looks a lot like the chain rule $(f_1 ◦ f_2)'(x) = f_1'(f_2(x))f_2'(x)$. This suggests a natural example of the function $j(\gamma,\tau)$ using differentiation: when $γ = (\begin{smallmatrix} a& b\\ c& d\end{smallmatrix})$ and $\tau \in \mathfrak h$, set $$ j(γ, τ ) := \left(\frac{aτ + b}{cτ + d}\right)' = \frac{a(cτ + d) − c(aτ + b)}{(cτ + d)^2} =\frac{ad − bc}{(cτ + d)^2}, $$ and for $γ ∈ {\rm SL}_2(\mathbf R)$ this says $j(γ, τ) = 1/(cτ + d)^2$. When $j(γ,τ)$ satisfies the cocycle condition, so does $j(γ,τ)^m$ for each $m \in \mathbf Z$, which motivates the consideration of the transformation rule for modular forms with factors $(cτ + d)^k$, at least for even $k$ (use $k = -2m$). The top answer at https://math.stackexchange.com/questions/312515/what-is-the-intuition-between-1-cocycles-group-cohomology shows how the same cocycle condition arises when trying to write down line bundles on an elliptic curve over $\mathbf C$.<|endoftext|> TITLE: Is every set smaller than a regular cardinal, constructively? QUESTION [8 upvotes]: Constructively, my only interest in regular cardinals is in terms of the "$\Sigma$-universes" they generate. By a $\Sigma$-universe, I mean a collection of triples $(X,Y,f: X \to Y)$ closed under base change, composition, and isomorphism -- here $X,Y$ are sets and $f: X \to Y$ is a function between them. A $\Sigma$-universe can be viewed as a category where the morphisms are pullback squares. Let's say that a $\Sigma$-universe $U$ is essentially small or representable if, when viewed as a category in this way, it has a weakly terminal object. In ZFC, representable $\Sigma$-universes are (almost) in bijection with regular cardinals. The bijection sends a regular cardinal $\kappa$ to the collection of functions with fibers of size $<\kappa$. The regularity of $\kappa$ corresponds to closure of the $\Sigma$-universe under composition. Let's say there are enough representable $\Sigma$-unvierses if every function $f: X \to Y$ is contained in some representable $\Sigma$-universe. In ZFC, there are enough representable $\Sigma$-universes because there are arbitrarily large regular cardinals. Question: Is it true constructively that there are enough representable $\Sigma$-universes? I assume this may depend on what one means by "constructively", but I don't know what the appropriate dividing lines might be. If not, are there natural conditions on a constructive set theory that ensure the existence of enough representable $\Sigma$-universes? Is it true constructively that the poset of representable $\Sigma$-universes is directed? How about if I have a set-indexed family of representble $\Sigma$-universes -- can I find another one containing them all? REPLY [4 votes]: I think there are enough representable $\Sigma$-universes in any regular locally cartesian closed category with disjoint coproducts and $W$-types. One can show the category of sets has $W$-types in $\mathbf{ZF}$ and even $\mathbf{IZF}$. So I don't think any form of choice or existence of regular ordinal is necessary for this particular statement, although there are related statements that do require the existence of regular sets (I believe some of the results in algebraic set theory are like this). It's not provable in $\mathbf{CZF}$ that $\mathbf{Set}$ has $W$-types but it follows from axioms that are often added such as $\bigcup-\mathbf{REA}$ and holds in the type theory interpretation of set theory as long as the type theory has enough inductive types (if I recall correctly this means each universe of small types is closed under $W$-types). Given $f : A \to B$, for each $b \in B$ we write $A_b$ for the fibre over $b$. We then define $U$ to be $W$-type defined as the smallest set closed under the following operations. $U$ contains an element $\ast$ If $b \in B$ and $g : A_b \to U$, then $U$ contains an element $\sup(b, g)$. We then define a function $\operatorname{Br}$ from $U$ to sets by recursion as follows. It's a little tricky to formalise this in a predicatively acceptable way, but I think this can be done using the notion of paths like in Theorem 2.1.5 in Benno van den Berg's thesis (essentially branches are maximal paths). $\operatorname{Br}(\ast) = 1$ If $b \in B$ and $g : A_b \to U$, then $\operatorname{Br}(\sup(b, g)) = \Sigma_{a \in A_b} \operatorname{Br}(g(a))$ We can think of $W$-type as sets of well founded trees, and in this case $\operatorname{Br}(u)$ is the set of branches of the tree $u$. We then take the universe to be the projection $\pi_0 : \Sigma_{u \in U} \operatorname{Br}(u) \to U$. Note that we can define a map $t : B \to U$ as follows. Given $b \in B$, define $t(b)$ to be $\sup(b, \lambda x.\ast)$. Then for each $b$, $\operatorname{Br}(t(b))$ is isomorphic to $A_b$, and so $f$ is a pullback of $\pi_0$ along $t$. Next we show that if $u \in U$ and $h : \operatorname{Br}(u) \to U$, then we can define $s(u, h) \in U$ such that $\operatorname{Br}(s(u, h)) \cong \Sigma_{x \in \operatorname{Br}(u)} \operatorname{Br}(h(x))$. We do this by recursion (and check it works by induction). If $u = \ast$, then $\operatorname{Br}(u)$ has one element, say $0$. Take $s(u, h)$ to be $h(0)$. If $u = \sup(b, g)$, then for each $a \in A_b$, $h : \Sigma_{a \in A_b} \operatorname{Br}(g(a)) \to U$ restricts to a morphism $h_a : \operatorname{Br}(g(a)) \to U$. Take $s(u, h)$ to be $\sup(b, \lambda a.s(g(a), h_a))$. (The way to visualise this is that are given a function from branches of a tree to trees, and we glue each tree to the end of its corresponding branch to get a bigger tree.) We can now use this to show that pullbacks of $\pi_0 : \Sigma_{u \in U} \operatorname{Br}(u) \to U$ are closed under composition. Every pullback is isomorphic to one where the codomain is an object $Y$, the bottom of the pullback is a map $t : Y \to U$ and the map is the projection $\Sigma_{y \in Y} \operatorname{Br}(t(y)) \to Y$. If we are given two composable maps $X \to Y$ and $Y \to Z$ that are both pullbacks of $\pi_0$, say along $t : Z \to U$ and $r : Y \to U$, then the maps are isomorphic to ones where $Y = \Sigma_{z \in Z} \operatorname{Br}(t(z))$ and $X = \Sigma_{z \in Z} \Sigma_{w \in \operatorname{Br}(t(z))} \operatorname{Br}(r(z, w))$. We can then use the map $s$ constructed above to witness the composition as a pullback of $\pi_0$. Namely, for each $z \in Z$, $r$ restricts to a map $r_z : \operatorname{Br}(t(z)) \to U$. We can then take $t' : Z \to U$ to be $t'(z) := s(t(z), r_z)$. The composition is then the pullback along $t'$. I think the other two parts follow from the existence of enough representable $\Sigma$-universes. Because if we have a family of universes, we can take the coproduct of all the universes in the family and then construct a universe for that.<|endoftext|> TITLE: Langlands correspondence for higher local fields? QUESTION [21 upvotes]: Let $F$ be a one-dimensional local field. Then Langlands conjectures for $GL_n(F)$ say (among other things) that there is a unique bijection between the set of equivalence classes of irreducible admissible representations of $GL_n(F)$ and the set of equivalence classes of continuous Frobenius semisimple complex $n$-dimensional Weil-Deligne representations of the Weil group of $F$ that preserves $L$-functions and $\epsilon$-factors. This statement has been proven for all one-dimensional local fields. My question is: is it possible to formulate any meaningful analogue of Langlands conjectures for higher local fields (e.g. formal Laurent series over $\mathbb{Q}_p$)? If this is possible, conjectures for $GL_1$ should probably be equivalent to higher local class field theory (it says that for an $n$-dimensional local field $F$, there is a functorial map $$ K_n(F)\rightarrow \mathrm{Gal}(F^{ab}/F) $$ from $n$-th Milnor K-group to the Galois group of maximal abelian extension, which induces an isomorphism $K_n(F)/N_{L/F}(K_n(L))\rightarrow \mathrm{Gal}(L/F)$ for a finite abelian extension $L/F$). Frankly, I do not even know what should be the right definition of Weil group of a higher local field (nor did my literature search give any results) but maybe other people have figured it out. REPLY [4 votes]: Similar and related questions are: On Geometric Langlands Correspondence Langlands conjectures in higher dimensions Kapranov's analogies probably more... I've already wrote on subj here: https://mathoverflow.net/q/131884 Let me just add a few words. 1) Abelian case of higher dimensional Langlands (=class field theory) developped by A.N. Parshin (1975) and K.Kato (1977) and later on by Fesenko and others (survey 2000). 2) Around 1992-5 Mikhail Kapranov wrote quite a speculative paper "Analogies between the Langlands correspondence and topological quantum field theory" (See: Kapranov's analogies ) One of his ideas is the following: Instead correspondence between Rep(Gal) <-> Rep( G(Adelic) ), one should consider "higher represenations" (representations not into category of vector spaces, but to k-categories). So Kapranov's idea: n-th dimensional k-representations of dimension r of Galois group should correspond to (n-k)-representations of GL_r(n-Local Field) That will include Parshin-Kato abelian case as a subcase as Kapranov explained. However actual work with higher representations is somewhat elusive, so difficult to transform insights into precise theorems/conjectures.<|endoftext|> TITLE: Dense abstract free subgroups in a free profinite group QUESTION [5 upvotes]: Let $\langle a, b \rangle = F_2$ be a two-generator free group and $\hat{F_2}$ be its profinite completion. Is there an element $c\in \hat{F_2}$ such that $\langle a, b, c\rangle \le \hat{F_2}$ is isomorphic to the 3-generator abstract free group $F_3$? I posted the same question on Math StackExchange(link), but my question hasn't been answered yet. I think the construction or the proof of nonexistence would be simple, but I couldn't make it yet.. apologies if I violated some etiquette. Edited: Assuming there is such an element $c$, I tried to make a continuous epimorphism $f:\hat{F_2} \to \hat{H}$, where $H$ is a 3-generator free group with $H=\langle h_1, h_2, h_3\rangle$ and $f(a)=h_1, f(b)=h_2, f(c)=h_3$. Then, since we have a projection from $\hat{F_3}$ to $\hat{F_2}$, we attain a contradiction. However, I cannot convince myself $f$ is well constructed. It is impossible; see comments by William Chen. REPLY [3 votes]: Yes. Here's a recipe to get such a group. Find a profinite group $K$ with three elements $a',b',c'$ such that $\langle a',b'\rangle$ is dense and $(a',b',c')$ is a free family. If so, we can "pull them back" to $\hat{F_2}$. Namely considering the unique homomorphism $\hat{F_2}\to K$ mapping $a\mapsto a'$, $b\mapsto b'$; density implies that it's surjective and hence we can choose $c$ as any preimage of $c$. Let's now find $K$ (there are many alternative ways). Consider the three matrices $$A=\begin{pmatrix}4 & 5\\3 & 4\end{pmatrix},B=\begin{pmatrix}7 & 8\\6 & 7\end{pmatrix},C=\begin{pmatrix}4 & 9\\3 & 7\end{pmatrix}.$$ We'll successively view them as real, and as 3-adic matrices. These are loxodromic elements in $\mathrm{SL}_2(\mathbf{R})$, with pairwise disjoint pairs of fixed points in the projective line. Hence they have the property that for any large enough $n$, say $n\ge N_0$, the powers $A^n,B^n,C^n$ freely generate a free group. This was the use of real numbers. Now view them as 3-adic matrices. Observe that all of them belong to the subgroup $K$ of $\mathrm{SL}_2(\mathbf{Z}_3)$ of those matrices $\begin{pmatrix}x & y\\z & t\end{pmatrix}$ such that $x-1,t-1,z$ are in $3\mathbf{Z}_3$ (i.e. the inverse image of the upper unipotent subgroup in reduction modulo 3). The profinite group $K$ is actually a pro-$3$-group. Its elementary abelianization is given by the homomorphism $f:\begin{pmatrix}1+3a & b\\3c & 1+3d\end{pmatrix}\mapsto (b,c)\in (\mathbf{Z}/3\mathbf{Z})^2=:L$. A basic fact in pro-$p$-groups is that a subgroup $\Gamma$ of $K$ is dense iff $f(\Gamma)=L$. Since $f(A)=(1,2)$, $f(B)=(2,2)$ and $f(C)=(1,0)$ which are pairwise non-collinear, this applies to the subgroup generated by any of the three pairs in $\{A,B,C\}$. Moreover, for any $n$ with $n$ coprime to 3 and any $u\in\{A,B,C\}$, the cyclic subgroups generated by $u$ and $u^n$ have the same closure in $K$. If furthermore $n\ge N_0$, we deduce that $\langle A^n,B^n\rangle$ is still dense and $(A^n,B^n,C^n)$ freely generates a free subgroup. So this is the desired pair $(a',b',c')$.<|endoftext|> TITLE: Show a sequence of sums involving Catalan Numbers converges QUESTION [5 upvotes]: Let $C_n$ be the $n$-th Catalan Number and let $\mathcal{O}_{s,j} = {{2s-j-1}\choose{j}} C_{s-j}^2$. Then we want to consider $\mathcal{E}_s = \sum_{j=0}^{s-1} (-1)^j\mathcal{O}_{s,j}$. We want to show that $$\frac{\mathcal{E}_s}{C_s^2} \to 0 {\text { as }} s\to \infty.$$ Does anyone have any ideas of how to get this result? We check the first 1000 numerically and I believe it to be true. For some quick background, $C_s^2$ counts a set of link diagrams and $\mathcal{E}_s$ comes from counting a subset this larger set. So, $\frac{\mathcal{E}_s}{C_s^2}$ is a probability and thus positive for all $s$. I am a topologist, so this kind of question is pretty out of my area and any help would be great. A few of the things we have tried: Bounding above by some "simple" functions (like $\frac{2}{s}$), but we can't get a handle on comparing them to $\mathcal{E}_s$. Bounding above by more complicated function:$$ f(s) = \sum_{j=0}^{s-1} (-1)^j \frac{s^{j}}{j!8^j}$$ It turns out that $f(s) = e^{-s/8} \frac{\Gamma(s+1,-s/8) }{\Gamma(s+1)} $ which is the regualized Gamma function, and we know this goes goes to zero. Again, we can't seem to get a handle on showing $f(s)\geq \frac{\mathcal{E}_s}{C_s^2}$. Ratio test for sequences: We tried to show $$\frac{\frac{\mathcal{E}_{s+1}}{C_{s+1}^2} }{\frac{\mathcal{E}_s}{C_s^2} } <1$$ which reduces to the problem $$\frac{\mathcal{E}_{s+1}}{\mathcal{E}_s} <16$$ Here, we thought we could get somewhere but the bounds we need to use to get anywhere end up giving exactly 16, not a value strictly less. Again, for the first 1000, it seems to be no larger than 14, numerically. So, if anyone happens to know a better way to bound a sum of lots of factorials, I would love to hear about it. Thanks! REPLY [10 votes]: By 'magic' and a computer (see the book "A=B" by Petkovsek, Wilf and Zeilberger https://www.math.upenn.edu/~wilf/AeqB.html) the numbers $\mathcal{E}_s$ satisfies the recurrence $\sum_{k=0}^3 P_k(s) \mathcal{E}_{s+k} = 0$, with polynomials $P_k$ given by $P_0(s) = 2 s^3+s^2-8 s+5$, $P_1(s) = -26 s^3-93 s^2-82 s-30$, $P_2(s) = -26 s^3-141 s^2-226 s-81$ and $P_3(s) = 2 s^3+17 s^2+40 s+16$. The claim now follows by a result of Poincare: The characteristic equation $2-26 z-26 z^2+2 z^3=0$ has only the simple roots $z_0=-1$, $z_2=7+4\sqrt{3}$ and $z_3=7-4\sqrt{3}$. Poincare now tells us that either $\mathcal{E}_s$ is eventually $0$ (in which case everything is fine!) or $\mathcal{E}_{s+1}/\mathcal{E}_s$ converges to one of the roots $z_i$, $i=1$, $2$ or $3$. Since these are all less than $16 = \lim_{s\rightarrow \infty} C_{s+1}^2/C_s^2$ the result follows.<|endoftext|> TITLE: For a proof of the three-square theorem without using Dirichlet's theorem on primes in arithmetic progressions QUESTION [13 upvotes]: The three-square theorem states that $n\in\mathbb N=\{0,1,2,\ldots\}$ is the sum of three squares if and only if it is not of the form $4^k(8m+7)$ ($k,m\in\mathbb N$). This was first proved by Legendre during 1797-1798. In 1837 Dirichlet proved his famous theorem on primes in arithmetic progressions. Now I can only find proofs of Legendre's three-square theorem using Dirichlet's theorem, see, e.g., M. B. Nathanson's book "Additive Number Theory-The Classical Bases" (GTM 164, Springer, 1996). Legendre's original proof did not involve Dirichlet's theorem which was proved later. A proof without using Dirichlet's theorem might be self-contained and hence suitable for the course of elementary number theory. QUESTION: Where can I find a proof of the three-square theorem without using Dirichlet's theorem? REPLY [13 votes]: I am not sure if Legendre's proof was complete (see this related MO entry), but surely Gauss gave a proof without Dirichlet's theorem (which came decades later). A nice transparent proof (without Dirichlet's theorem) can be found in Serre's book "A course in arithmetic". Very roughly, the proof goes as follows. Assume that $n$ satisfies the necessary local conditions. Step 1: There exists a positive integer $t$ such that $t^2 n$ is the sum of three squares. This relies on the theory of quadratic forms over $\mathbb{Q}$. Step 2: The minimal $t$ that works in Step 1 is $t=1$. This relies on the following simple but key property: for any $(x_1,x_2,x_3)\in\mathbb{Q}^3$ there exists $(y_1,y_2,y_3)\in\mathbb{Z}^3$ such that $\sum (x_i-y_i)^2<1$.<|endoftext|> TITLE: Largest ranks achieved by abelian varieties of fixed dimension QUESTION [5 upvotes]: This is a follow-up to this earlier question on elliptic curves: Largest rank assumed by infinitely many elliptic curves Let $g \geq 1$ be an integer. For each $g$, what is known about the largest positive integer $r(g)$ for which we know there exist infinitely many abelian varieties $A$ of dimension $g$ which are not products of lower dimensional abelian varieties (for example, not a product of elliptic curves) such that the Mordell-Weil rank of $A$ is at least $r(g)$? In the first question, it was essentially answered by Michael Stoll that $r(1) \geq 19$, and in another comment it was mentioned that recent work by Bjorn Poonen and various coauthors, as well as earlier heuristics due to Andrew Granville, indicate that $r(1) = 21$ exactly (that is, there ought to be only finitely many elliptic curves with Mordell-Weil rank exceeding 21). Moreover, if there are any known constructions for such $A$'s for fixed genus $g \geq 2$, do they arise as Jacobians of curves of genus $g$, or is there some other construction? REPLY [9 votes]: Here is a construction (as far as I know, originally due to Mestre) that gives a lower bound $r(g) \ge 4g + 5$. Write $$ \prod_{i=1}^{4g+6} (x - t_i) = h(x)^2 - f(x)\,, $$ where the $t_i$ are independent indeterminates, $h$ is monic of degree $2g+3$ and $f$ has degree $2g+2$. Then there are $4g+6$ points $P_i = (t_i, h(t_i))$ on the hyperelliptic curve $C$ given by $y^2 = f(x)$. Generically, the only relation between them in the Jacobian is that $$ \sum_{i=1}^{4g+6} P_i - (2g+3) D_\infty = \operatorname{div}(y - h(x)) \,,$$ where $D_\infty$ is the sum of the two points at infinity. So most specializations of the $t_i$ will produce a curve whose Jacobian (is simple and) has Mordell-Weil rank at least $4g+5$. It is certainly possible to tweak the construction to get somewhat higher rank. (I had used it in some early papers of mine to get high-rank examples of genus 2 Jacobians over $\mathbb Q$.) But I doubt that one can get superlinear growth in this way. Anyway, the question is whether we should expect $r(g)$ to grow faster than linearly. In view of Joe Silverman's comment below, it should be added that Néron already had a construction that produces infinitely many (again hyperelliptic) Jacobians of curves of genus $g$ with rank $\ge 3g + 6$. This is the Corollaire on page 163 of Néron's paper.<|endoftext|> TITLE: Jordan form on an invariant vector subspace QUESTION [6 upvotes]: Let $\mathbb{F}$ be a field and $V$ an $\mathbb{F}$-vector space. Let $\operatorname{T}\in\mathrm{End}(V)$ be an $\mathbb{F}$-linear operator. It is well known that if $\dim V<\infty$ then $\operatorname{T}$ has a Jordan canonical form, i.e., it is similar to the direct sum of a number of Jordan blocks. If a certain eigenvalue (i.e., root of a polynomial) is not inside $\mathbb{F}$ then in the corresponding Jordan block it is represented by its $\mathbb{F}$-matrix representation (that is, the Jordan block is a block matrix). In infinite dimensions the existence of Jordan canonical form is not guaranteed. I consider only those operators for which it exists. Question: Let $V$ have arbitrary dimensions, and assume that $\operatorname{T}$ does have a Jordan canonical form, i.e., it is similar to the direct sum of (an arbitrary set of) Jordan blocks. Let now $W\subset V$ be an $\mathbb{F}$-vector subspace which is $\operatorname{T}$-invariant, $\operatorname{T}W\subset W$. Is it true that the restriction $\operatorname{T}|_W\in\mathrm{End}(W)$ also has a Jordan canonical form? Thank you. REPLY [8 votes]: The question can be restated as follows: let $F$ be a field $T$ be an endomorphism of a vector space over $F$. Suppose that it decomposes as a direct sum of finite-dimensional $T$-stable subspaces. Does the same property hold for every $T$-stable subspace? In turn, it can be formalized as follows, in a pure language of commutative algebra: let $V$ be a $F[t]$-module. Suppose that $V$ is a direct sum of finite length submodules. Does every submodule share the same property? It's true and follows as a particular case of the following Theorem Let $R$ be a PID. Let $\mathcal{C}_R$ be the class of $R$-modules that decompose as a (possibly infinite) direct sum of finite length submodules (and hence of finite length cyclic submodules). Then $\mathcal{C}_R$ is stable under taking submodules. This is a theorem of Kulikov (1941) when $R=\mathbf{Z}$, in which case the proof can be found in L. Fuchs' book "Infinite abelian groups, Vol 1", §17-18. When $R=k[t]$, polynomial ring in one indeterminate over a field $k$, this exactly yields your claim. I don't know if anybody wrote the result for arbitrary PID (even for such polynomial rings, for which it should not be easier), but the proof in Fuchs' book extends with essentially no change to arbitrary modules over PIDs. The main lemma in Fuchs' book (Theorem 17.1) is the following (in the case of $\mathbf{Z}$). For an irreducible element $r$ in $R$, an $R$-module is called a $p$-module $M$ if for every $x\in M$ there exists $n$ such that $p^nx=0$. Also, the (possibly infinite) number $h(x)=\sup\{n:x\in p^nM\}$ is called the height of $x$. Lemma Let $R$ be a PID and $p$ an irreducible element. A $p$-module $M$ is a direct sum of cyclic modules if and only if one can write $M$ as union of an ascending chain of submodules $M_1\le M_2\le\dots$, $\bigcup_nM_n=M$, such for every $n$, we have $\sup_{x\in M\smallsetminus\{0\}}h(x)<\infty$. It clearly implies, for a given PID $R$, the above theorem. So, for completeness, let me write the proof of the lemma; it essentially consists in a copy of the proof in Fuchs' book. Up to reindexing (adding redundancies if necessary), we can first suppose that $M_n\cap p^n M=\{0\}$ for all $n$. Second, we can suppose that $(M_n)$ is maximal for this property (namely, among sequences $(N_n)_{n\ge 1}$ satisfying $N_n\cap p^n M=\{0\}$ for all $n$ with the ordering $(N_n)\le(M_n)$ if $N_n\le M_n$ for all $n$). Denote by $M[p]$ the kernel of $m\mapsto pm$. For every $n\ge 1$, we choose a maximal independent subset $J_n$ in the subgroup $M_n[p]\cap p^{n-1}M$, and define $J=\bigsqcup_{n\ge 1} J_n$. For every $c\in J$, choose $a_c\in M$ with $p^{h(c)}a_c=c$. We claim that $M=\bigoplus_{c\in J}Ra_c$. The first part is that this is indeed a direct sum. First, since all nonzero elements of the $R$-module $\langle J_n\rangle$ generated by $J_n$ have height $n-1$, we see that the $\langle J_n\rangle$ generate their direct sum, so $J$ is a $R/pR$-free family in $M[p]$. Next suppose that we have a nontrivial combination, namely $\sum_{i\in I} r_ca_c=0$, with $I$ a nonempty finite subset of $J$ and $r_c\notin p^{h(c)+1}R$ for all $i$. If $r_c\notin p^{h(c)}R$ for some $c\in I$, we multiply everything by $p$. So we can suppose that $r_c=p^{h(c)}R$ for all $i$, namely $r_c=p^{h(c)}s_c$ with $s_c\notin pR$. Hence $\sum_{c\in I}s_cc=0$, contradicting that $J$ is a $R/pR$-free family in $M[p]$. Now let us show that $M=\sum_{c\in J}Ra_c$. The first step is to show that $\langle J\rangle=M[p]$. Clearly $\langle J_n\rangle=M_n[p]\cap p^{n-1}M$. Suppose by induction that every element in $M_r[p]$ belongs to $\langle J\rangle$ (which is clear for $r=1$), and consider $b\in M_{r+1}[p]\smallsetminus M_r$. The maximality of $(M_n)$ and $b\in M_r$ implies that $(M_r+Rb)\cap p^rM\neq\{0\}$. So we can find $0\neq c=g+kb\in p^rM$, with $g\in M_r$ and $k\in R$. Then $kb\neq 0$; hence $k\notin pR$. Multiplying by $k'$ with $kk'-1 \mod pR$, we can suppose that $k=1$, so we now assume that $c=g+b$. We have $c\in M_{r+1}$ and $h(c)\ge r$, and thus $h(c)=r$ since $M_{r+1}\cap p^{r+1}M=\{0\}$. Also $h(pc)\ge r+1$, and $pc=pg\in M_r$; if nonzero it would be a contradictio to $M_r\cap p^rM=\{0\}$, so $pc=0$. So $pg=0$ as well. By induction, $g\in\langle J\rangle$, and hence $b=c-g$ as well. Thus $\langle J\rangle=M[p]$. By another induction, we prove that every $m\in M$ with $p^nm=0$ belongs to $M'=\sum_{c\in J}Ra_c$; this holds for $n=1$ and we can now consider the case of $b$ with $p^{n+1}b=0$, with $n\ge 1$; we have to show that $b'\in M'$. By the previous paragraph, we have $p^nb\in\langle J\rangle$. We write $$p^nb=m_1c_1+\dots +m_jc_j+n_1d_1+\dots +n_kd_k,$$ with $c_1,\dots,d_k\in J$, $h(c_i)\ge n$, $h(d_i)\le n-1$. We can write $m_ic_i=p^nm'_ia_{c_i}$, and hence we have $$p^n(b-m'_1a_{c_1}-\dots -m'_ja_{c_j})=n_1d_1+\dots +n_kd_k\in M_{n}.$$ Since $M_{n}\cap p^nM=\{0\}$, we deduce that $p^nb'=0$, with $b'=b-m'_1a_{c_1}-\dots -m'_ja_{c_j}$. By induction, we have $b'\in M'$, and $b\in M'$ follows.<|endoftext|> TITLE: Find a formula for the recurrent sequence $q_{n+1}=q_n(q_n+1)+1$ QUESTION [7 upvotes]: Find an analytic formula for the recurrent sequence $$q_{n+1}=q_n(q_n+1)+1,\;\;q_0\in\mathbb N.$$ (The question was asked on 03.05.2018 by M. Pratsovytyi, see page 109 of Volume 1 of the Lviv Scottish Book). REPLY [9 votes]: If we denote $A_n=q_n+1/2$, then $$A_n=A_{n-1}^2+5/4$$ with $A_0=q_0+1/2\ge 3/2$ by $q_0\in\mathbb{N}$. Further, $$\log A_n=2\log A_{n-1}+\log\left(1+\frac{5}{4A_{n-1}^2}\right),$$ namely $$\frac{1}{2^n}\log A_n-\frac{1}{2^{n-1}}\log A_{n-1}=\frac{1}{2^n}\log\left(1+\frac{5}{4A_{n-1}^2}\right).$$ Thus $$\log A_n=2^n\left(\log A_{0}+\sum_{k=1}^{n}\frac{1}{2^k}\log\left(1+\frac{5}{4A_{k-1}^2}\right)\right).$$ Clearly, $$A_n> A_{n-1}^{2^1}> A_{n-2}^{2^2}>\cdots A_0^{2^n}\ge (3/2)^{2^n}.$$ Thus, $$0<\sum_{k\ge n+1}\frac{2^n}{2^k}\log\left(1+\frac{5}{4A_{k-1}^2}\right)<\sum_{k\ge n+1}\frac{2^n}{2^k}\frac{5}{4A_{k-1}^2}<\frac{5}{4A_{n}^2}.$$ Hence note that $A_0>3/2$ we obtain that $$1-\frac{5}{4A_n^2}A_0=3/2$. For the computing of $\kappa$, it follows from above that $$A_n^{1/2^n}<\kappa TITLE: Question on a generalisation of a theorem by Euler QUESTION [22 upvotes]: We call an integer $k\geq 1$ good if for all $q\in\mathbb{Q}$ there are $a_1,\ldots, a_k\in \mathbb{Q}$ such that $$q = \prod_{i=1}^k a_i \cdot\big(\sum_{i=1}^k a_i\big).$$ Euler showed that $k=3$ is good. Is the set of good positive integers infinite? REPLY [30 votes]: Here I present the joint solution for $k=4$ by Guang-Liang Zhou and me. Actually, we not only show that $k=4$ is good, but also obtain a stroger result: Each rational number $q$ can be written as $abcd$ with $a,b,c,d\in\mathbb Q$ such that $a+b+c+d=1$. (We may also replace the number $1$ by any nonzero rational integer, but that's equivalent to the current version.) Let $x=-81q/32$. If $q\not\in\{4/81,-8/81\}$, then $x\not\in\{1/4,-1/8\}$, hence $$a(x)+b(x)+c(x)+d(x)=1$$ and $$a(x)b(x)c(x)d(x)=-\frac{32}{81}x=q,$$ where \begin{gather*}a(x)=-\frac{(1+8x)^2}{9(1-4x)},\ \ b(x)=\frac49(1-4x), \\ c(x)=\frac{2(1-4x)}{3(1+8x)},\ \ d(x)=\frac{12x}{(1-4x)(1+8x)}.\end{gather*} For $q=4/81$, clearly $$\frac4{81}=\frac{4}3\times\frac13\times\left(-\frac13\right)\times\left(-\frac13\right)\ \ \text{with}\ \frac 43+\frac13+\left(-\frac13\right)+\left(-\frac13\right)=1.$$ Similarly, for $q=-8/81$ we have $$-\frac{8}{81}=\frac{1}3\times\frac23\times\frac23\times\left(-\frac23\right)\ \ \text{with}\ \frac 13+\frac23+\frac23+\left(-\frac23\right)=1.$$<|endoftext|> TITLE: Why does passage to DG categories cure non-locality of derived categories? QUESTION [10 upvotes]: In the famous book 'Residues and duality', the author notes that one of the principal difficulties in constructing the exceptional inverse image functor $f^{!}$ is that the derived category of quasicoherent sheaves is not a local object. In order to resolve this difficulty, the author first defines it for residual complexes, a procedure he finds 'clumsy' and 'roundabout'. In his notes on Geometric Langlands, D. Gaitsgory forcefully explains the necessity of considering DG enhancements to define the derived category for general prestacks The problem is that the gluing procedure alluded to above, is not defined for triangulated categories. (This was the problem that Hartshorne had to confront in his “Residues and duality”; this is why that book is so thick, instead of being just 10 pages long.) Now, the advatnage of the ∞-category language is that gluing can be defined for DG-categories. Is there a high-level explanation as for why the difficulties occuring while trying to glue derived categories (as triangulated categories) are resolved if we pass to DG categories? In other words, how could someone trying to solve this problem come up with the idea to introduce DG enhancements? REPLY [4 votes]: Just a quick answer :-) Triangulated categories lack intrinsicness (is that even a word?), and also the 2-category of triangulated categories is extremely poorly behaved, whereas the 2-category of DG-categories behaves better under 2-categorical constructions (it admits several shapes of 2-co/limits and bi-co/limits). There is a notion of DG-accessible and DG-presentable category, whereas no (interesting) triangulated category whatsoever can be accessible (they can't be concrete categories). The 2-category of DG-categories cures the non-functoriality of the cone construction by defining $Cone(f)$ as a suitable weighted colimit (and the weight is essentially the additive analogue of a topological cone construction). There is no such thing in the triangulated world.<|endoftext|> TITLE: Properly outer automorphisms on type II$_1$ von Neumann algebras QUESTION [9 upvotes]: Let $M$ be a von Neumann algebra with separable predual. Let us assume that $M$ is of type II$_1$, meaning that it is finite but has no type I part. Let $\tau$ be a faithful normal tracial state on $M$. First, I would like to choose some free ultrafilter $\omega$ on $\mathbb N$ and consider the tracial ultrapower $$ M^\omega = \ell^\infty(\mathbb N, M)/\{ (x_n)_n \mid \tau(x_n^*x_n)\to 0 \}. $$ Next, let $\alpha$ be an automorphism on $M$. It is called properly outer, if for every non-zero projection $p\in M^\alpha$, the induced automorphism on $pMp$ is not inner. Note that each automorphism gives rise to $\alpha^\omega\in\operatorname{Aut}(M^\omega)$ via pointwise application, and restricts to an automorphism on the relative commutant $M^\omega\cap M'$. Theorem: (Connes) If $M$ is the hyperfinite II$_1$-factor, then $\alpha$ is (properly) outer if and only if $\alpha^\omega$ is (properly) outer on $M^\omega\cap M'$. My question: Is the above also true if $M$ is a hyperfinite type II$_1$ von Neumann algebra, i.e., not assumed to be a factor? So is an automorphism $\alpha$ properly outer if and only if $\alpha^\omega$ is properly outer on $M^\omega\cap M'$? I am sure that somebody must have worked this out somewhere, but I cannot seem to find it. I would be grateful either for the argument showing/disproving this statement or a reference with the answer. EDIT: I realized that it is not necessary for $\alpha$ to assume that it preserves the trace $\tau$. The question makes perfect without it. REPLY [2 votes]: The answer to your question is yes, although I haven't found a precise reference for this fact. Let $M$ be a hyperfinite, type II$_1$ von Neumann algebra with separable predual. Later on I will use that it is possible to decompose $M$ as $Z(M) \overline{\otimes} \mathcal{R}$, where $Z(M)$ is the centre of $M$ and $\mathcal{R}$ is the hyperfinite II$_1$ factor (this is Theorem 1.5 in Chapter XVI of Takesaki's third book on operator algebras) Definition 1. An automorphism $\alpha \in \text{Aut}(M)$ is centrally (non-)trivial if $\alpha^\omega$ is (non-)trivial on $M^\omega \cap M'$. It is properly centrally non-trivial if none of its restriction to $pM$, for some $\alpha$-invariant, non-zero, central projection $p \in M$, is centrally trivial. The proof is composed of two parts: Every properly outer automorphism on $M$ is properly centrally non-trivial. If $\alpha \in \text{Aut}(M)$ is properly centrally non-trivial, then $\alpha^\omega$ is properly outer on $M^\omega \cap M'$. Part 1 is addressed in the following claim. Claim 1. If $\alpha$ is an outer automorphism of $M$, then $\alpha$ is centrally non-trivial. Proof. If the restriction of $\alpha$ to $Z(M)$ is non-trivial, then the conclusion is obvious, as $Z(M) \subset M^\omega \cap M'$. Suppose that $\alpha$ is trivial on $Z(M) \cong L^\infty(X)$, for some measure space $X$. Then $\alpha$ can be decomposed as \begin{equation} \alpha = \int_X^\oplus \alpha^x, \end{equation} where $\alpha^x \in \text{Aut}(\mathcal{R})$ for every $x \in X$ (a good reference for disintegration of von Neumann algebras and of their automorphisms is Chapter IV of Takesaki's first book on operator algerbas; see in particular Theorem 8.23 for the passage above). The claim follows by contradiction using the following three results: $\alpha$ is centrally trivial if and only if $\alpha^x$ is centrally trivial for almost every $x \in X$ (I'm not sure who originally proved this, but it is proved as Theorem 9.14 here). $\beta \in \text{Aut}(\mathcal{R})$ is centrally trivial if and only if it is inner (Connes). $\alpha$ is inner if and only if $\alpha^x$ is inner for almost every $x \in X$ (this is due to Lance, Theorem 3.4). The second part of the proof, namely that if $\alpha$ is properly centrally non-trivial then $\alpha^\omega$ is properly outer on $M^\omega \cap M'$ was proved by Ocneanu (see Lemma 5.6). One has to backtrack a bit in the paper to make sense of the proof, because of all the notation and definition appearing in the lemma (the setting there is much more general than what we need), but the argument is similar to the one given by Connes to prove the analogous result for $\mathcal{R}$ (Proposition 2.1.2 in Outer conjugacy classes of automorphisms of factors).<|endoftext|> TITLE: An Optimization Problem with Complex Variables, regarding Eigenvalues of Circulant Matrices QUESTION [6 upvotes]: Let $S$ be a finite subset of the complex unit circle and $1 \in S$. For each $n \in \mathbb N $, define $f_n\colon S^{n-1}\to\mathbb R$ by $$f_n(x) := \sum_{w^{n}=1}|x_1w+ x_2w^2\cdots+x_{n-1}w^{n-1}|$$ Denote $z^*_n := \min_{S^{n-1}} f_n$. Is it true that for sufficiently large $n$s (which depend on $S$), we have $z_n^* = 2(n-1)$? Remarks: $f_n(x)$ is sum of the absolute values of eigenvalues of a circulant matrix generated by $(0,x_1,x_2,\ldots,x_{n-1})$. $f_n(x) = 2(n-1)$, when $x$ is all ones vector. Above question stems from this question and may be related to the Littlewood problem, as a comment of Tao to the last linked question. REPLY [3 votes]: The answer is no if there exists $z \in S$ with $z \neq 1$ and $\vert 1 - z\vert \leq \frac{1}{2}$ and $\bar{z} \in S$ : Let $$p(x) = \sum_{k=0}^{n-1} b_k x^k$$ and $x_k = p(w^k)$ , where $w=e^{\frac{2 \pi i}{n}}$ . Then the $n b_k$ are the eigenvalues of the circulant matrix. Now choose $$p(x) = 1 - \frac{1}{n} \sum_{k=0}^{n-1} x^k + (z-1) \frac{1}{n} \sum_{k=0}^{n-1} w^k x^k + (\bar{z}-1) \frac{1}{n} \sum_{k=0}^{n-1} w^{-k} x^k$$ Then $p(1)=0$, $p(w)=\bar{z}$, $p(w^{-1})=z$ and $p(w^k)=1$ otherwise. Furthermore we have $b_0 = 1 - \frac{1}{n}(3 - 2 Re (z))$ and $b_k = \frac{1}{n}(-1 + 2 Re ((z-1)w^k)) \leq 0$ for $k \neq 0$. Therefore $$\sum_{k=0}^{n-1} \vert b_k\vert = 2 b_0 < 2 (1 - \frac{1}{n})$$<|endoftext|> TITLE: Path algebras are formally smooth QUESTION [6 upvotes]: In Ginzburg's notes Lectures on Noncommutative Geometry, he claim that the path algebra of a quiver is formally smooth (See Section 19.2). I have two questions. First, how to show this claim and which criterion of formal smoothness we are going to check? Secondly, does this claim hold for the path algebra of a quiver with relations as well? Any help or hints would be very appreciated. REPLY [8 votes]: Assuming standard results on lifting idempotents, it's not hard to check that a path algebra $kQ$ satisfies the lifting property that Ginzburg uses to define formal smoothness in Definition 19.1.1. If $B$ is an algebra with a nilpotent ideal $I$, we need to check that every homomorphism $\varphi:kQ\to B/I$ lifts to a homomorphism $\tilde{\varphi}:kQ\to B$. Let $\{e_1,\dots,e_n\}$ be the idempotents in $kQ$ corresponding to the vertices of $Q$. Then $\{\varphi(e_1),\dots,\varphi(e_n)\}$ is a set of orthogonal idempotents in $B/I$ whose sum is $1$. These lift to a set $\{f_1,\dots,f_n\}$ of orthogonal idempotents in $B$ whose sum is $1$. For each arrow $a_s$ of $Q$ from vertex $i$ to vertex $j$, pick an arbitrary lift $\widetilde{\varphi(a_s)}$ of $\varphi(a_s)$, and let $b_s=f_i\widetilde{\varphi(a_s)}f_j$, so $b_s$ is also a lift of $\varphi(a_s)$. Then there is a unique map $\tilde{\varphi}:kQ\to B$ with $\tilde{\varphi}(e_i)=f_i$ and $\tilde{\varphi}(a_s)=b_s$ for all $i$ and $s$, and this is a lift of $\varphi$. This doesn't work for quivers with relations. For example, if $Q$ is an acyclic quiver and $A=kQ/I$ with $I$ a nonzero admissible ideal, then the identity map $A\to kQ/I$ does not lift to a homomorphism $A\to kQ$.<|endoftext|> TITLE: Lowest Dimension for Counterexample in Topological Manifold Factorization QUESTION [16 upvotes]: Bing gave a classical example of spaces $X, Y, Z$ such that $X \times Y = Z$, where $X$ and $Z$ are manifolds but $Y$ isn't. The space $Z$ in his example has dimension four. Is it known if this is best possible? In other words, if $X \times Y =Z$ where $X$ is a manifold and $Z$ is a 3-manifold, then is $Y$ a manifold? In Bing's example, $Z$ is not compact. Is there a compact example in dimension $4$? In the example, $X$ is the real line, so one can also ask if it possible to get a four dimensional example where $X$ is a surface, or where $X$ is compact. REPLY [9 votes]: As asked Dusan Repovs (who is an expert in the theory of topological manifolds), and he sent me the following answer: This is indeed best possible result, since whenever a product of two spaces is a topological manifold, both factors must be generalized manifolds - which in dimensions below 3 are topological manifolds. Ref.: A.Cavicchioli, F.Hegenbarth and D.Repovš, Higher-Dimensional Generalized Manifolds: Surgery and Constructions, EMS Series of Lectures in Mathematics, European Mathematical Society, Zurich, 2016. To the second question: there are also compact examples (in dimensions >3): e.g. take the product of the 3-sphere $S^3$ modulo the Fox-Artin wild arc $A$ and $S^1$. This product is homeomorphic to $S^3\times S^1$. Ref.: R.J.Daverman, Decompositions of Manifolds, Academic Press, Orlando, 1986. Added in Edit: Answering a comment of John Samples, Dusan Repovs pointed out that the Chapter 29 of Daverman's book contains the following fact: for any $n,m>2$ there are a generalized $n$-manifold $X$ and a generalized $m$-manifold $Y$ which are not topological manifolds, but their product $X\times Y$ is a topological manifold.<|endoftext|> TITLE: Is every (left) graded-Noetherian graded ring (left) Noetherian? QUESTION [6 upvotes]: I call a $\mathbb{Z}$-graded (non-commutative, associative, unital) ring $A$ (left) graded-Noetherian if every homogeneous (left) ideal is finitely generated, and (left) Noetherian if it is (left) Noetherian as a ring. In the commutative setting, I think I can prove that a graded-Noetherian ring is Noetherian. This basically follows from Hilbert's basis theorem (graded-Noetherianity suffices to show that $A_0$ is Noetherian and that $A$ is finitely generated over $A_0$). Since Hilbert's basis theorem fails noncommutatively the same line of reasoning will not work in the noncommutative setting. Q: Is every (left) graded-Noetherian graded ring (left) Noetherian? I asked this question in MSE a few days ago and got no replies at all, not sure at all where it belongs. REPLY [6 votes]: The answer is yes, by Corollary 2.2 in C. Nastasescu, F. Van Oystaeyen, Graded rings with finiteness conditions II, Comm. Algebra 13 (1985), 605-618. More generally, we have the following. Let $G$ be a commutative group. Every epimorphism $\psi\colon G\twoheadrightarrow H$ of commutative groups gives rise to the $\psi$-coarsening functor $\bullet_{[\psi]}$ from the category of $G$-graded rings to the category of $H$-graded rings. We say that $\psi$-coarsening preserves noetherianness if whenever $R$ is a $G$-graded ring that is noetherian as a $G$-graded ring, then the $H$-graded ring $R_{[\psi]}$ is noetherian as an $H$-graded ring. Now, it follows from the aforementioned result by Nastasescu and Van Oystaeyen (as well as by a result independently proven two years earlier by Goto and Yamagishi): $\psi$-coarsening preserves noetherianness for every epimorphism $\psi$ whose source is $G$ if and only if $G$ is of finite type. For details on the yoga of coarsening cf. this article.<|endoftext|> TITLE: A Muirhead Like Inequality QUESTION [7 upvotes]: I am looking for a proof of the inequality as follow: Let $n$ be an integer number $n \ge 2$ and $x_1, \cdots, x_n$ and $y_1,\cdots, y_n$ are nonegative real numbers such that $(x_1,\cdots, x_n)$ majorizes $(y_1,\cdots, y_n)$; Let $0 \leq a_1, a_2,\cdots,a_n \leq 1$ then $$\sum_{\text{sym}} {x_1}^{a_1}{x_2}^{a_2}\cdots {x_n}^{a_n}\leq \sum_{\text{sym}} {y_1}^{a_1}{y_2}^{a_2}\cdots {y_n}^{a_n}$$ Note: The inequality above is not Muirhead's Inequality. Example: Let $0 \leq a_i \leq 1$ then $4^{a_1}1^{a_2}+ 4^{a_2}1^{a_1} \le 3^{a_1}2^{a_2}+ 3^{a_2}2^{a_1}$ $5^{a_1}5^{a_2}2^{a_3}+5^{a_1}5^{a_3}2^{a_2}+5^{a_2}5^{a_1}2^{a_3}+5^{a_2}5^{a_3}2^{a_1}+5^{a_3}5^{a_1}2^{a_2}+5^{a_3}5^{a_2}2^{a_1} \leq 4.5^{a_1}4^{a_2}3.5^{a_3}+4.5^{a_1}4^{a_3}3.5^{a_2}+4.5^{a_2}4^{a_1}3.5^{a_3}+4.5^{a_2}4^{a_3}3.5^{a_1}+4.5^{a_3}4^{a_1}3.5^{a_2}+4.5^{a_3}4^{a_2}3.5^{a_1}$ See also: Muirhead's Inequality Group multiplication of permutations REPLY [8 votes]: $\newcommand{\al}{\alpha} \newcommand{\be}{\beta} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\Var}{\operatorname{\mathsf Var}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}} \newcommand{\tsi}{\tilde\si}$ For $x=(x_1,\dots,x_n)\in(0,\infty)$, let \begin{equation*} f(x):=2\sum_{\si\in S_n}\prod_{k=1}^n x_k^{a_{\si(k)}}, \end{equation*} where $S_n$ is the set of all permutations of the set $\{1,\dots,n\}$. By the symmetry and the Schur--Ostrowski criterion, it suffices to show that \begin{equation*} x_1\le x_2\overset{\text{(?)}}\implies f_1(x)-f_2(x)\ge0, \tag{1} \end{equation*} where $f_j(x)$ is the partial derivative of $f(x)$ in $x_j$. Consider the bijection $S_n\ni\si\leftrightarrow\tsi\in S_n$ defined by the formula \begin{equation*} \tsi(k):= \left\{ \begin{aligned} \si(2)&\text{ if }k=1,\\ \si(1)&\text{ if }k=2,\\ \si(k)&\text { otherwise}. \end{aligned} \right. \end{equation*} Then \begin{align*} f(x)&:=\sum_{\si\in S_n}x_1^{a_{\si(1)}}x_2^{a_{\si(2)}}\prod_{k=3}^n x_k^{a_{\si(k)}} + \sum_{\tsi\in S_n}x_1^{a_{\tsi(1)}}x_2^{a_{\tsi(2)}}\prod_{k=3}^n x_k^{a_{\tsi(k)}} =\sum_{\si\in S_n}Q P, \tag{2} \end{align*} where \begin{equation*} Q:=Q_\si(x):=x_1^{b_1}x_2^{b_2}+x_1^{b_2}x_2^{b_1},\quad P:=P_\si(x):=\prod_{k=3}^n x_k^{b_k}, \end{equation*} \begin{equation*} b_k:=b_{\si;k}:=a_{\si(k)}. \end{equation*} Denoting by $Q_j$ the partial derivative of $Q$ in $x_j$, we have \begin{equation*} Q_1-Q_2=b_1 (x_1 x_2)^{b_1-1}(x_2^{b_2-b_1+1}-x_1^{b_2-b_1+1}) +b_2 (x_1 x_2)^{b_2-1}(x_2^{b_1-b_2+1}-x_1^{b_1-b_2+1})\ge0. \end{equation*} Now noting that $P$ does not depend on $(x_1,x_2)$, we see that (1) follows from (2), as desired.<|endoftext|> TITLE: Presentation of special linear group over localizations of the integers QUESTION [8 upvotes]: I am looking (for $n,k\in{\mathbb Z}$) for a presentation (in the best of all worlds concretely, as a list of relators) for the group ${\rm SL}_n(R)$ for $R={\mathbb Z}[\frac{1}{k}]=\{\frac{a}{k^l}\mid a\in {\mathbb Z}\}$. A search in MathSciNet found a paper of Behr and Mennicke(A presentation of the groups PSL(2,p), Can. J. Math. 20, 1432-1438 (1968)) that gives a presentation for the special case of $n=k=2$; (add the diagonal matrix $(2,\frac{1}{2})$ as extra generator and describe its conjugation action on suitable generators of ${\rm SL}_n({\mathbb Z})$); but a reference search did not yield a further result. Similarly mathOverflow carried the same question question for other rings, but not $R$. (A follow-up question would be the same question for ${\rm Sp}$) REPLY [4 votes]: The question appears to be rather difficult. I'll give some literature references and discuss how it can be approached, and where the difficulties lie. First, the results mentioned by Luc Guyot in the comments can also be found in Section II.1.4 of Serre's book "Trees". More precisely, there is an amalgam decomposition of $SL_2(\mathbb{Z}[1/p])$ on p.80, and after that a description how to successively write $SL_2(\mathbb{Z}[1/(p_1\cdots p_k)])$ as amalgams of congruence subgroups of $SL_2(\mathbb{Z})$. Then he gives the presentations for $SL_2(\mathbb{Z}[1/2])$ and $SL_2(\mathbb{Z}[1/3])$ and leaves the case $SL_2(\mathbb{Z}[1/6])$ as an exercise. The general procedure to obtain a presentation of a group $SL_n(\mathbb{Z}[1/(p_1\cdots p_k)])$ would be the following: the group acts on a symmetric space which is a product of $SL_n(\mathbb{R})/SO(n)$ and a copy of the Bruhat-Tits building for $SL_n(\mathbb{Q}_{p_i})$ for each prime $p_i$ that is being inverted. If one can determine a fundamental domain for this action, together with presentations for the stabilizer groups, then one can put all these data together to a presentation of the group. (Of course it's more complicated. The case $SL_2$, where the corresponding buildings are trees is discussed in detail in Serre's book. I'd have to dig for good references outlining the general case...) Alternatively, one can add each of the primes to be inverted successively, then only the action of the group on a single building has to be considered, but then the stabilizers are more complicated. So, where are the difficulties. First, determining the fundamental domain plus all the stabilizer groups isn't easy. Next, knowing presentations for the stabilizer subgroups isn't easy. Finally, the group $SL_n(\mathbb{Z}[1/(p_1\cdots p_k)])$ isn't usually the amalgam of the stabilizer subgroups because the quotient of the symmetric space might have nontrivial topology. For instance, in the tree case, the quotient might contain nontrivial loops, in which case we get HNN-extensions instead of simple amalagams. All these questions have to be solved to get a presentation; I guess this is one of the reasons why there appear to be no general results on this question in the literature. Something more can be said, though. The technique of using the action on the symmetric space, computing fundamental domains and stabilizers and so forth is also used to compute group homology. There is this paper of Williams and Wisner which contains partial information on the abelianization of $SL_2(\mathbb{Z}[1/k])$ for general $k$: F. Williams and R.J. Wisner. Cohomology of certain congruence subgroups of the modular group. Proc. Amer. Math. Soc. 126 (1998), 1331--1336. (link to paper on journal website) In particular, Corollary 4.4 of their paper states that the abelianization of $PSL_2(\mathbb{Z}[1/n])$ is a subgroup of an explicit torsion group. Note that nontriviality of the abelianization would already tell us something about necessary generators for the group. Unfortunately, they don't determine the subgroup or its order precisely, because that would require identification of a lot of higher differentials in the spectral sequence they are considering... There is also an explicit GAP-implementation of an algorithm constructing resolutions for $SL_2(\mathbb{Z}[1/m])$ to compute homology: B.A. Tuan and G. Ellis. The homology of $SL_2(\mathbb{Z}[1/m])$ for small m. J. Algebra 408 (2014), 102–108. (link to paper on journal website) The paper contains a table listing the abelianizations for small $m$. Maybe it's possible to extract some preliminary information on the shape of the presentation from the resolution constructed by their algorithm. Note that they also use the technique suggested by Serre: write the group as amalgam and then use that to produce the resolution. The size and representatives for elements in $H_2$ might also (from the point of view of Steinberg groups and K-theory) say something about relations in the presentation.<|endoftext|> TITLE: The étale topos of a scheme is the classifying topos of which groupoid? QUESTION [10 upvotes]: [Sent here from Math.StackExchange by suggestion of an user.] By a theorem of Joyal and Tierney, every Grothendieck topos is the classifying topos of a localic groupoid. It has been proved (e.g. C. Butz. and I. Moerdijk. Representing topoi by topological grupoids. Journal of Pure and Applied Algebra 130, 223-235, 1998) that topoi "with enough points" admit actually a representation as classifying topoi of topological groupoids. Now my question is the following: take a well-known topos, as the étale topos for a scheme. This is the classifying topos of a localic groupoid, acrtually a topological groupoid since it is "coherent" and thus has enough points by Deligne's theorem; but which groupoid is this in concrete? Do you know if someone has ever investigated that? Thank you in advance. P.S.: I have been suggested to look at the proof and try to reconstruct the particular case, which I am going to try. REPLY [4 votes]: I'm not familiar with Joyal and Tierney's construction, but Butz and Moerdijk's construction seems fairly straightforward to me. Also, doing the etale topos is a bit beyond my reach, so here's a description for the case of the classifying topos $\mathcal E_T$ of some first-order theory $T$. Select a set $\mathcal M$ of enough points of the topos. A point of the topos $\mathcal E_T$ is a model $M$ of the theory $T$. Here we can take $\mathcal M$ to be the set of all models of cardinality at most the size of the language of $T$ (cardinality of a model $M$ can be determined by adding up the sizes of all definable sets in $M$). An object of your topological groupoid is a model $M$ from the set $\mathcal M$, equipped with an enumeration by a fixed set $I$ which is "infinitely redundant": every element of the model gets counted infinitely many times. The enumeration is a technicality in order to get the topology to work out correctly. A morphism of your topological groupoid is an isomorphism, which need not respect the enumerations. The topology is such that for each $i \in I$ and each definable set $D$, the set of all enumerated models $f: I \to M$ such that $f(i) \in D$ is an open set. So basically, Butz and Moerdijk express a topos $\mathcal E$ as equivariant sheaves on the topological groupoid of points of $\mathcal E$, cut down in size to make it small, and fattened up a bit (passing to an equivalent groupoid) with enumerations in order to be able to define the topology.<|endoftext|> TITLE: Spectral algebraic geometry vs derived algebraic geometry in positive characteristic? QUESTION [26 upvotes]: Let $R$ be a commutative ring. Then there is a forgetful functor from the $\infty$-category of simplicial commutative $R$-algebras to the $\infty$-category of connective $E_{\infty}$-algebras over $R$. It's well-known that this functor admits left and right adjoint. Moreover, it's an equivalence if $R$ contains the field of rational numbers $\mathbb{Q}$. If I understand correctly, this means that SAG and DAG are (more or less) equivalent over such $R$. On the other hand, if $R$ does not contain $\mathbb{Q}$ (say $R$ is a finite field), then the forgetful functor above is not necessarily an equivalence. This implies that SAG and DAG are very different in this situation. I wonder if there are any examples of problems where using SAG (resp. DAG) is more appropriate? Such examples would help to illustrate the difference between SAG and DAG in positive characteristic. P.S.: related but not identical questions: $E_{\infty}$ $R$-algebras vs commutative DG $R$-algebras vs simplicial commutative $R$-algebras, Why do people say DG-algebras behave badly in positive characteristic? REPLY [27 votes]: I'll try to answer this question from the topological viewpoint. The short summary is that structured objects in the spectral setting have cohomology operations and power operations, which forces spectral algebraic geometry to be different from derived algebraic geometry.$\newcommand{\FF}{\mathbf{F}}\DeclareMathOperator{spec}{Spec}\newcommand{\Eoo}{E_\infty} \newcommand{\Z}{\mathbf{Z}} \newcommand{\GL}{\mathrm{GL}} \newcommand{\N}{\mathbf{N}} $ Let us illustrate this by a few examples. The simplest example is that in positive characteristic, any discrete ring $A$ has a Frobenius endomorphism. Simplicial commmutative rings are freely generated under sifted colimits by finitely generated polynomial $A$-algebras, so in positive characteristic, simplicial commutative rings have Frobenius endomorphisms. However, in general, $\Eoo$-rings have nothing like a Frobenius endomorphism. For a few more examples, consider the discrete ring $\FF_2$, regarded as an $\Eoo$-algebra over the sphere, and as a discrete simplicial commutative ring, as in my comment. Note that the initial object in the category of simplicial commutative rings is the usual ring $\Z$. In the spectral world, the fiber product $M:=\spec\FF_2\times_{\spec \mathbb{S}} \spec \FF_2 = \spec(\FF_2\otimes_\mathbb{S} \FF_2)$ lives over $\spec \FF_2$, and $\pi_\ast \mathcal{O}_M$ can be identified with $\pi_\ast(\FF_2\otimes_\mathbb{S} \FF_2)$, which is the dual Steenrod algebra. In contrast, the fiber product $N:=\spec \FF_2\times_{\spec \Z} \spec \FF_2 = \spec(\FF_2\otimes^\mathbb{L}_\Z \FF_2)$ in the world of derived algebraic geometry is not the dual Steenrod algebra! Instead, $\pi_\ast \mathcal{O}_N$ can be identified with $\pi_\ast(\FF_2\otimes^\mathbb{L}_\Z \FF_2) = \mathrm{Tor}^\ast_\mathbf{Z}(\FF_2, \FF_2) = \FF_2[x]/x^2$ with $|x| = 1$. This is already one example of how the spectral and derived worlds diverge in positive characteristic. Another example comes from considering the affine line. In the classical world, the affine line $\mathbf{A}^1_X$ is flat over the discrete scheme $X$. Taking the functor of points approach, the affine line is defined as $\spec$ of the free commutative algebra on one generator (i.e., the polynomial ring $\Z[x]$). In the spectral world, the free $\Eoo$-algebra on one generator over an $\Eoo$-ring $R$ is given by $\bigoplus_{n\geq 0} (R^{\otimes n})_{h\Sigma_n} =: R\{x\}$ (see the note at the end of this answer). For instance, this means that $\mathbb{S}\{x\}$ is a direct sum of the spectra $\Sigma^\infty B\Sigma_n$. Suppose that $R = \FF_2$, again regarded as a discrete $\Eoo$-ring. Then $$\pi_n\FF_2\{x\} = \bigoplus_{k\geq 0} H_n(\Sigma_k; \FF_2),$$ where each $\Sigma_k$ acts trivially on $\FF_2$. In particular, $\FF_2\{x\}$ is not flat over $\FF_2$. (Recall that if $A$ is an $\Eoo$-ring, then an $A$-module $M$ said to be flat over $A$ if $\pi_0 M$ is flat over $\pi_0 A$ and the natural map $\pi_\ast A\otimes_{\pi_0 A} \pi_0 M\to \pi_\ast M$ is an isomorphism. In particular, any flat module over a discrete ring must be concentrated in degree zero.) This failure corresponds exactly to the existence of Steenrod operations. More generally, the homotopy groups $\pi_\ast R\{x\}$ carry all information about power operations on $E_\infty$-$R$-algebras. On the other hand, the free simplicial commutative ring on one generator over $\FF_2$ is just the polynomial ring $\FF_2[x]$. This is certainly flat over $\FF_2$. Yet another example (technically not "positive characteristic", but is still a good example illustrating the difference between the spectral and derived worlds) along these lines comes from contemplating the definition of the scheme $\mathbf{G}_m$. Classically, this is defined as $\spec$ of the free commutative algebra on one invertible generator. In the spectral world, a natural candidate for this functor already exists: it is the functor known as $\GL_1$. If $R$ is an $\Eoo$-ring, one can define the space $\GL_1(R)$ as the pullback $\Omega^\infty R \times_{\pi_0 R} (\pi_0 R)^\times$, i.e., as the component of $\Omega^\infty R$ lying over the invertible elements in $\pi_0 R$. (One can similarly define $\mathrm{SL}_1(R)$.) We run into the same problem --- the resulting spectral scheme $\GL_1$ is not flat over the sphere spectrum. The space $\GL_1(R)$ is an infinite loop space, and is very mysterious in general. In the derived world, one can define the functor $\mathbf{G}_m$ as $\spec$ of the free simplicial commutative ring on one invertible generator, i.e., as $\spec \Z[x^{\pm 1}]$. In fact, one can then extend the input of $\mathbf{G}_m$ to $\Eoo$-rings, by defining $\mathbf{G}_m(A)$, for $A$ an $\Eoo$-ring, to be $\mathrm{Map}_{\text{infinite loop}}(\Z, \GL_1(A))$. Then, there is a map of schemes $\mathbf{G}_m\to \GL_1$ which is an equivalence over the rationals. This point of view (that cohomology operations and power operations separate the spectral and derived worlds) also helps ground one's intuition for why simplicial commutative rings and $\Eoo$-rings agree rationally: the rational Steenrod algebra is trivial (the rational sphere is just $\mathbf{Q}$)! Of course, this doesn't comprise a proof. Note that in Lurie's books, the affine line is not defined via the free $\Eoo$-ring on one generator; as we saw above, that $\Eoo$-ring is a bit unwieldy. Instead, $\mathbf{A}^1_R$ is defined to be $\spec(R\otimes_\mathbb{S} \Sigma^\infty_+ \mathbf{N})$, where $\mathbf{N}$ is now regarded as a discrete topological space. As $\Sigma^\infty_+ \mathbf{N}$ is flat over $\mathbf{S}$, this sidesteps the non-flatness problem mentioned above. Replacing $\N$ with $\Z$ above, one similarly sidesteps the non-flatness issue for $\mathbf{G}_m$. I want to comment briefly on a surprising example where the spectral and derived worlds agree. Instead of studying spectral algebraic geometry with $\Eoo$-rings, let us work in the context of spectral algebraic geometry with $E_2$-rings. (In the hierarchy of $E_k$-rings, this is the minimum structure one can/should impose before it is reasonable to say that there is some level of commutativity. For instance, if $A$ is an $E_k$-ring, then $\pi_0 A$ is a commutative ring once $k\geq 2$.) The theory of algebraic geometry over $E_k$-rings was studied by Francis in his thesis (see http://www.math.northwestern.edu/~jnkf/writ/thezrev.pdf). In the derived world, the free commutative algebra with $p=0$ is just $\FF_p$. Surprisingly, the free ($p$-local) $E_2$-ring with $p=0$ is also $\FF_p$! This result is due to Hopkins and Mahowald. Other than the actual proof (which is very enlightening, and a nice application of $E_2$-Dyer-Lashof operations; see https://arxiv.org/abs/1411.7988 and https://arxiv.org/abs/1403.2023), I don't have any conceptual explanation for why this result should morally be true. Note, however, that the free ($p$-local) $E_2$-ring with $p^n=0$ is not discrete (this is due to Jeremy Hahn, see https://arxiv.org/abs/1707.00956) for $n>1$.<|endoftext|> TITLE: Given a representation-infinite algebra, when is every AR component infinite? QUESTION [5 upvotes]: Let $A$ be a finite dimensional algebra over an algebraically closed field $K$. The Auslander-Reiten quiver $\Gamma_A$ of $A$ is a means of presenting the category of finitely generated right $A$-modules. The Auslander-Reiten quiver is a locally finite quiver whose vertices are indecomposable modules (up to isomorphism) and whose arrows are irreducible morphisms between indecomposable modules. It is given the structure of a translation quiver via the Auslander-Reiten translate $\tau$. Two distinct vertices/modules of $\Gamma_A$ are said to belong to the same component if there exists a finite path/composition of irreducible morphisms between them. If $A$ is representation-finite (i.e. the category of finitely generated right $A$-modules contains finitely many non-isomorphic indecomposable objects), then $\Gamma_A$ consists of finitely many components, which each have finitely many vertices. However if $A$ is representation-infinite, then necessarily $\Gamma_A$ is infinite. It is easy to construct an example of a representation-infinite algebra $A$ for which there exists a finite component of $\Gamma_A$. (Edit: the example I provide in this question is false, as indicated by the answer given by Jeremy Rickard.) For example, consider the algebra $KQ/I$, where $KQ$ is the path algebra of the quiver $$ Q\colon \; 1 \begin{smallmatrix} \alpha \\ \rightarrow \\ \rightarrow \\ \beta \end{smallmatrix} 2 \begin{smallmatrix} \gamma \\ \rightarrow \\ \color{white} \gamma \end{smallmatrix} 3 \begin{smallmatrix} \delta \\ \rightarrow \\ \color{white} \delta \end{smallmatrix} 4$$ and $I$ is the ideal generated by the set $\{\beta\gamma, \gamma\delta\}$. This is representation-infinite because it contains the Kronecker quiver as a subquiver. But the relation $\gamma\delta$ 'cuts off' the right-hand side of the quiver, which means we have a finite Auslander-Reiten component $$\begin{matrix} S(4) & & & & S(3) \\ & \searrow & & \nearrow & \\ & & P(3)=I(4) & & \end{matrix}$$ (which is essentially $\mathrm{mod}\;K\mathbb{A}_2$) in $\Gamma_A$, where $S(v)$, $P(v)$ and $I(v)$ are the simple, indecomposable projective, and indecomposable injective modules corresponding to the vertex $v$ in $Q$ respectively. (One can also use the fact that this is a string algebra to easily compute $\Gamma_A$.) My question is this: When is it the case that for a representation-infinite algebra $A$, every component of $\Gamma_A$ contains infinitely many vertices? Is this for example true if the algebra is self-injective? (I imagine the proof of the latter question is straightforward if true, but please point me to a reference if one exists.) REPLY [7 votes]: If $A$ is connected and has infinite representation type, then every component of its Auslander-Reiten quiver is infinite. See, for example, Theorem 5.4 in Assem, Simson and Skowronski’s Elements of the Representation Theory of Associative Algebras, Volume 1. The example you give is not a complete component. It continues (infinitely) to the right, starting with an irreducible map from $S(3)$ to $P(2)$.<|endoftext|> TITLE: Closures of torus orbits in flag varieties QUESTION [8 upvotes]: Consider the Lie group $G=SL_n(\mathbb C)$ with Borel subgroup $B$ and maximal torus $T\subset B$. I'm interested in the (Zariski) closures of $T$-orbits in the flag variety $F=G/B$. Now, as far as I can tell, for a generic point in $F$ this closure is the toric variety associated with the permutahedron. Further, let us choose an element $w$ of the Weyl group and let $X_w\subset F$ be the corresponding Schubert variety. Then for a generic point in $X_w$ the closure seems to be the toric variety associated with the convex hull of vertices of the permutahedron corresponding to $w'\ge w$ with respect to the Bruhat order, i.e. the convex hull of the weight diagram of the corresponding Demazure module in an irrep with a regular highest weight. Questions. 1) Is this last description accurate? 2) My main question. Are, in fact, all orbits of this form? More precisely, let $S$ be the set of all points in $F$ that are generic in some Schubert variety in the above sense. Is it then true that any point outside of $S$ can be mapped to a point in $S$ by the action of the Weyl group? References to literature are much appreciated. REPLY [7 votes]: A point in the Grassmannian $ x \in G(k, n) $ defines a matroid $ M = M(x)$. Associated to this matroid is a matroid polytope $P(M)$. The torus orbit closure through $ x $ is the toric variety associated to $ P(M) $. Similarly, if we take a point $ x $ in the flag variety, then the can associate a flag matroid $M(x) $ and a flag matroid polytope $ P(M)$. Again the torus orbit closure through $ x $ is the toric variety associated to $ P(M) $. Thus the classification of torus orbit closures in the flag variety is given by realizable flag matroids. So the answer your second question is no. The theory of flag matroids (in arbitrary type) and their polytopes was developed in the excellent classic paper by GGMS: I. M. Gelfand, R. M. Goresky, R. D. MacPherson and V. V. Serganova, Combinatorial Geometries, Convex Polyhedra, and Schubert cells, Adv. Math., 63 (1987), 301–316. Alex Yong's lecture notes are also quite helpful: https://faculty.math.illinois.edu/~ayong/Math595TheGrassmannian/Grlecture2matroids.pdf<|endoftext|> TITLE: What does an ideal correspond to in the internal language of sheaves? QUESTION [5 upvotes]: Suppose I have a sheaf $\mathcal F$ in some topos $\mathrm{Sh}(\mathcal C)$. Then this becomes the sheaf of rings from algebraic geometry when described as a ring in the internal language of the topos. Whilst in the internal language we were to describe a ring ideal, what would this correspond to externally? REPLY [2 votes]: I assume that you mean that $\mathcal{F}$ is a sheaf of rings. What's internally an ideal of $\mathcal{F}$ is externally simply a sheaf of ideals. In case that the topos in question is the little Zariski topos of a scheme and $\mathcal{F}$ is the structure sheaf $\mathcal{O}_X$, you could ask a follow-up question: How can we characterize quasicoherent sheaves of ideals in the internal language? The answer is that a sheaf $\mathcal{I}$ of ideals is quasicoherent if and only if, from the internal point of view, $$ \forall f : \mathcal{O}_X. \forall s : \mathcal{O}_X. (\text{$f$ invertible} \Rightarrow s \in \mathcal{I}) \Longrightarrow \bigvee_{n \geq 0} (f^n s \in \mathcal{I}). $$ A proof of this characterization is given in Section 8 of these notes of mine, more precisely Corollary 8.5.<|endoftext|> TITLE: Homotopy limit of model categories in the category of categories QUESTION [5 upvotes]: Say $$\mathcal{C'}\to \mathcal{C}\leftarrow \mathcal{D}$$ is a diagram of model categories and (e.g. Left) Quillen functors. I want to write down a (hopefully simple) model category $\mathcal{D}'$, or at least a category with weak equivalences, such that its $\infty$-categorical localization is the homotopy limit of the localizations of this diagram in the $(\infty,1)$ category of $(\infty, 1)$ categories. Is there a nice way to do this? I'm willing to impose any reasonable niceness conditions on the categories in the diagram. REPLY [2 votes]: Philippe Gaucher is right. This problem was solved by Julie Bergner, here. I recently asked a question that summarized some of her work on this problem. The point is that the homotopy limit of your diagram is a category $M$ whose objects are 5-tuples $(x_1,x_2,x_3,u,v)$ with $x_1 \in C'$, $x_2 \in D$, $x_3\in C$, and $F(x_1) \stackrel{u}{\to} x_3 \stackrel{v}{\gets} G(x_2)$ in $C$, where $F$ and $G$ are the two functors in your diagram. The morphisms in this category of 5-tuples are obvious. This category $M$ can be given a model structure where the weak equivalences and cofibrations are levelwise (on each $x_i$), and that model structure can be localized if desired to force $u$ and $v$ to be weak equivalences in the local objects of $M$. Bergner then proves $M$ has the correct homotopy type, meaning that, upon passage to complete Segal spaces (i.e. $(\infty,1)$-categories), it becomes the actual homotopy pullback of the diagram. She has to assume the model categories she starts with are combinatorial, but this seems a standard assumption now from the $\infty$-categorical perspective (i.e. assuming presentability). Bergner uses a right Bousfield localization, so you need to assume right properness, or pass to right semi-model categories like Barwick does in this paper. The difference between a semi-model structure and a full model structure is invisible to the underlying $(\infty,1)$-category. EDIT (in answer to comments): Bergner uses the notation $L_DX$ for the category I called $M$ above. It's the lax homotopy limit. The homotopy limit is the full subcategory where the maps $u$ and $v$ have been forced to be weak equivalences. She does not claim it has a model structure in general, but it does in some special cases, e.g. if $L_DX$ is right proper and combinatorial. This occurs if each of your categories $C,C',D$ is combinatorial and has all objects fibrant, for example. This assumption can be avoided (and Bergner points this out, right after Theorem 3.2 in the linked paper) by using Barwick's method of right Bousfield localization without right properness. The result is a right semi-model structure on $Lim_DX$, and such categories have associated $(\infty,1)$-categories just like model categories do. And Bergner proves that the associated $(\infty,1)$-category is the homotopy limit in the category of $(\infty,1)$-categories, as you'd expect (working in the model of Complete Segal Spaces).<|endoftext|> TITLE: Can two set theories extending Z be different and yet bi-interpretable? QUESTION [11 upvotes]: At Hamkins - Different set theories are never bi-interpretable, it is mentioned that different set theories extending ZF are never bi-interpretable. Where different means "not theoretically equivalent", i.e. there must be a theorem that one has and the other doesn't. Question: Would that same result hold for Z, i.e. is it the case that any different [in the same sense mentioned in that article] theories extending Z are never bi-interpretable? More generally: even if the above fails, the question is about whether this result needs the full strength of ZF, and if not then what would be the least fragment of ZF for which this result holds? REPLY [9 votes]: UPDATE (January 30, 2022): The first question was answered (in the negative) by Hamkins and Freire for $\mathrm{Z}$ (Zermelo set theory) and $\mathrm{ZF}^{-}$ ($\mathrm{ZF}$ without powerset). Their paper "Bi-interpretation in weak set theories" was recently published in the Journal of Symbolic Logic. See here for a preprint of their paper, and here for recent blog of Hamkins about this topic. The second question remains open. What follows is my old answer (June 19, 2018). To my knowledge the two questions you are asking are wide open; indeed the second question is one of the two open questions posed at the end of this 2016 paper of mine, which gives many examples (and non-examples) of theories $T$ that satisfy the principle "different consistent extensions of $T$ are not bi-interpretable". The aforementioned paper was published as: A. Enayat, “Variations on a Visserian theme,” in Liber Amicorum Alberti: a tribute to Albert Visser, Jan van Eijck, Rosalie Iemhoff and Joost J. Joosten (eds.) Pages, 99-110. ISBN, 978-1848902046. College Publications, London, 2016.<|endoftext|> TITLE: Riemannian metric on the sphere with at least one negative sectional curvature at every point QUESTION [8 upvotes]: Does there exist a Riemannian metric on the $n$-sphere ($n > 2$) such that at each point some (but not every) sectional curvature is negative? For $n=2$ it is easily seen that such a metric cannot exist. REPLY [6 votes]: Kazhymurat's answer is definitely correct -- every sphere has such a metric. But, much more explicitly, if your sphere is odd-dimensional, then there are even homogeneous metrics (called Berger metrics) with some planes of negative sectional curvature. Since these metrics are homogeneous, this happens at every point. It is even possible to arrange for these metrics to have arbitrarily negative scalar curvature. Besse's "Einstein manifolds" book may be a good starting point reference.<|endoftext|> TITLE: Cutting a convex body into two congruent pieces QUESTION [5 upvotes]: This question is related to How to make a sandwich from just one piece of bread?, asked on Feb 23 '17 by erz, and it goes as follows: Question. If a convex closed and bounded region $C$ in the plane $\mathbb{R}^2$ can be cut along some straight line into two congruent pieces, must $C$ have either axial or central symmetry? To be specific about what $cutting$ means, a $piece$ consists of the closure of the set of all points of $C$ that lie on the same side of the cutting line. Obviously, if $C$ has either axial or central symmetry, then it can be cut so. The question can be phrased generally in $\mathbb{R}^n$, where we would cut a convex body along a hyperplane and consider $n$ kinds of symmetry, for example central, axial, and mirror in $\mathbb{R}^3$. I believe the answer is $yes$. Is it perhaps known already? In retrospect (see Wlod AA's very nice answer): my intuition was way off. REPLY [7 votes]: Here is a counter-example (in the complex plane $\mathbb C=\mathbb R^2.)\ $ Let $$\ P\ := \ \{ (x\ y)\in\mathbb C : 0\le x\le 1\quad\&\quad 0\le y\le 1-x^2\} $$ Then, $$ C\,\ :=\,\ P\,\cup\, i\!\cdot\! P $$ The imaginary line is the requested cut.<|endoftext|> TITLE: Moduli 'space' of stacks? QUESTION [8 upvotes]: In algebraic geometry, we are frequently interested in parametrizing geometric objects. Formally, parametrization of geometric objects having some property can be viewed as a functor $F:Sch\rightarrow Set$ from the category of schemes to the category of sets (which assigns to a scheme the set of families of geometric objects over this scheme). The problem of finding the moduli space of geometric objects can then be considered (up to distinction between fine and coarse moduli spaces) as the problem of finding an object corepresenting $F$. It sometimes happens that $F$ can not be corepresented within the category of schemes (as in the case of the moduli problem for smooth projective curves). We are then forced to extend our category somehow; for smooth projective curves, it is enough to extend the category of schemes to the category of Deligne--Mumford stacks. My question is: what are examples of geometrically interesting moduli problems for stacks which are not representable within the category of stacks? If there are such examples, what are the resulting extensions of the category of stacks? P.S.: I am being deliberately ambiguous about what kind of stacks we consider (DM, Artin, etc.) because I do not know a priori where interesting examples will come from. REPLY [7 votes]: Fix group $G$ (could be a finite group, could be an algebraic group). The collection of all stacks isomorphic to $BG$ is naturally a $2$-stack. Its $\pi_1$ is $Out(G)$, and its $\pi_2$ is $Z(G)$. The $k$-invariant in $H^3(Out(G),Z(G))$ which classifies the $2$-stack up to isomorphism is the obstruction to the existence of a short exact sequence of groups $$ 0 \to Z(G) \to E \to Out(G) \to 0 $$ It's not very easy to come up with an example of a group $G$ where the obstruction is non-zero (by which I mean it's not very easy to compute the obstruction). I seem to remember that the obstruction is non-zero for the dihedral group with $16$ elements.<|endoftext|> TITLE: Probability of generation of ${\mathbb Z}^2$ QUESTION [28 upvotes]: What is the probability that three pairs $(a,b) $ , $(c,d) $ and $(e,f) $ of integers generate $\mathbb Z^2$? As usual the probability is the limit as $n\to \infty$ of the same probability for the $n\times n$ square. It is well known that for $\mathbb Z $ the probability of two numbers to generate is $6/\pi^2$. REPLY [37 votes]: According to Proposition 1 in the paper G. Maze, Gérard, J. Rosenthal, U. Wagner: Natural density of rectangular unimodular integer matrices, Linear Algebra Appl. 434, No. 5 (2011), 1319-1324, ZBL1211.15044, the probability that $n$ random vectors generate $\mathbb{Z}^{n-1}$ is $$p_n = \prod_{j=2}^n \zeta(j)^{-1}.$$ For $n=2$ this gives $p_2=\zeta(2)^{-1}=6/\pi^2$, whereas for $n=3$ we obtain $$p_3= \zeta(2)^{-1} \zeta(3)^{-1} \simeq 0.505739038$$ REPLY [32 votes]: Let me convert my comments to an answer. Let $u_n$ be the probability that a triple in $([0,n-1]^2)^3$ generates $\mathbb{Z}^2$, and let $v_n$ be the probability that a triple in $((\mathbb{Z}/n)^2)^3$ generates $(\mathbb{Z}/n)^2$. Certainly $v_n\geq u_n$, and I think that $v_n$ should be asymptotic to $u_n$. Using the Chinese Remainder Theorem and the structure of $(\mathbb{Z}/p^k)^2$ we see that $v_n$ is the product of $v_p$ for all primes dividing $n$. A little linear algebra gives $v_p=(1-p^{-2})(1-p^{-3})$. Thus, the expected density is $$ v_\infty = \prod_p (1-p^{-2})^{-1}(1-p^{-3})^{-1} = (\zeta(2)\zeta(3))^{-1} \simeq 0.5057390381 $$ agreeing with the answer that Francesco Polizzi just entered while I was typing this.<|endoftext|> TITLE: Reference Request: Theoretical Mixing Times Research in Machine Learning / Artificial Intelligence (AI) QUESTION [12 upvotes]: I'm doing a PhD in probability theory, focusing mostly on mixing times. It's a pure maths PhD, considering precise models and showing rigorous mixing results. I'm also interested in stuff like machine learning. A brief overview of some google results suggests that MCMC is used quite a lot in these fields, however any references that I could find are of the following (approximate) form: we want to simulate a complicated distribution; write a piece of code to sample this and run it 1,000,000 times. The questions aren't about using rigorous probabilistic methods to find the mixing time. Maybe this is because in general people doing MC/AI aren't so interested in this? I don't know, so I thought I'd ask here.. Specifically, I'm asking for references to work in AI/ML that uses rigorous probabilistic methods to shows bounds on mixing times (runtimes for MCMC algorithms). As mentioned above, it's possible that such a reference doesn't exist. If this is the case, any suggestions of ways that mixing times could be applied to AI/ML would be equally desirable. (Just because there isn't work in an area yet doesn't mean it can't be started!) REPLY [11 votes]: The question as asked is rather broad, because there are several works in ML/AI dedicated to mixing time analysis, as well as to detecting if mixing has happened. I would not draw too sharp a boundary about whether the work is in the ML domain or in a closely related domain. Though, I agree, in ML, often MCMC is used with Bayesian methods, and given the complexity of the distributions involved, often rigorous mixing time bounds are not derived / not possible to derive there. However, there are some recent works in ML/AI, where the distributions involve enough structure, so as to permit a rigorous mixing time analysis. A (biased) list of some references to begin with: N. Anari, S. O. Gharan, and A. Rezaei. Monte Carlo Markov chain algorithms for sampling strongly Rayleigh distributions and determinantal point processes. Conference on Learning Theory, 2016. C. Li, S. Sra., and S. Jegelka. Fast mixing markov chains for strongly Rayleigh measures, DPPs, and constrained sampling. Advances in Neural Information Processing Systems (NIPS) 2016. C. Li, S. Jegelka, and S. Sra. Polynomial time algorithms for dual volume sampling. Advances in Neural Information Processing Systems (NIPS) 2017. Y. T. Lee and S. Vempala. Convergence Rate of Riemannian Hamiltonian Monte Carlo and Faster Polytope Volume Computation. 2017. (this is more a math paper though). J. Gorham and L. Mackey. Measuring sample quality with Stein's method. Advances in Neural Information Processing Systems (NIPS) 2017.<|endoftext|> TITLE: Is algebraic $K$-theory a motivic spectrum? QUESTION [21 upvotes]: I've received conflicting messages on this point -- on the one hand, I've been told that "forming a natural home for algebraic $K$-theory" was one motivation for the development of motivic homotopy theory. On the other hand, I've been warned about the fact that algebraic $K$-theory isn't always $\mathbb A^1$-local. By "algebraic $K$-theory,", I mean the algebraic $K$-theory of perfect complexes of quasicoherent sheaves (I think -- let me know if I should mean something else). I'm pretty sure that Thomason and Trobaugh show that algebraic $K$-theory always (in the quasicompact, quasiseparated case) satisfies Nisnevich descent. Under certain conditions, algebraic $K$-theory is $\mathbb A^1$-local and has some kind of compatibility with $\mathbb G_m$ which should make it $\mathbb P^1$-local. I think Weibel (already Quillen) calls this "the fundamental theorem of algebraic $K$-theory". So putting this together, let $S$ be a scheme, and let $SH(S)$ be the stable motivic ($\infty$-)category over $S$. Questions: Is algebraic $K$-theory of smooth schemes over $S$ representable as an object of $SH(S)$? How about if we put some conditions on $S$ -- say it's regular, noetherian, affine, smooth over an algebraically closed field? Heck, what if we specialize to $S = Spec(\mathbb C)$? Does it make a difference if we redefine $SH(S)$ to be certain sheaves of spectra over the site of smooth schemes affine over $S$ or something like that? REPLY [14 votes]: Let me assume that $S$ is a regular Noetherian scheme (for example a field). Then algebraic K-theory is a motivic spectrum, and in fact it is represented by the $\mathbb{P}^1$-spectrum that is $BGL_\infty\times\mathbb{Z}$ in each level (so it is a $\mathbb{P}^1$-periodic motivic spectrum). This is theorem 4.3.13 in Morel, Fabien; Voevodsky, Vladimir, $\bf A^1$-homotopy theory of schemes, Publ. Math., Inst. Hautes Étud. Sci. 90, 45-143 (1999). ZBL0983.14007. A correct proof of the above result can be found in this survey (thanks to Marc Hoyois for pointing out that the original proof was incorrect). The proof of the fact that algebraic K-theory is a motivic spectrum goes exactly how you described: It satisfies Nisnevich descent on qcqs schemes by Thomason-Trobaugh's main theorem. It is $\mathbb{A}^1$-invariant on regular Noetherian schemes (which every scheme smooth over $S$ is ) by Quillen's fundamental theorem of K-theory It is $\mathbb{P}^1$-periodic by the projective bundle formula (also in Thomason-Trobaugh) When $S$ is not regular noetherian the situation is more complicated. You can still represent $K$ by some version of the infinite Grassmannian (proposition 4.3.14 in Morel-Voevodsky), but this object won't be $\mathbb{A}^1$-invariant anymore. What you can consider is the homotopy K-theory presheaf $KH=Sing_* K$. This object is indeed represented by $BGL_\infty\times \mathbb{Z}$. This result has been announced in Voevodsky's ICM address and proven in Cisinski, Denis-Charles, Descent by blow-ups for homotopy invariant $K$-theory, Ann. Math. (2) 177, No. 2, 425-448 (2013). ZBL1264.19003. (thanks to Marc Hoyois for the reference to this result)<|endoftext|> TITLE: Negation of CH implied by lots of special subtrees? QUESTION [5 upvotes]: In the following, I focus on trees of height $\omega_1$: if there exists a nonspecial tree any of whose $\aleph_1$-subtrees is special, must CH fail? Some neither consistent nor coherent thoughts: Notice that the tree must have cardinality at least $\aleph_2$ and this is a fragment of $MA_{\aleph_1}$. It is known that it is possible to specialize Aronszajn trees with forcings that add no reals, hence obtaining a model of GCH along with all Aronszajn trees are special. But in this case, we are dealing with fat trees and maybe a possible attack is to think about forcings that specialize fat trees without adding reals. REPLY [6 votes]: Your hypothesis is (implied by) the negation of Rado's conjecture, and this is known to be consistent with CH. Rado's conjecture is a combinatorial statement about instances of compactness in chromatic numbers of certain graphs, but Todorčević proved that it is equivalent to the statement that any tree, all of whose subsets of size $\aleph_1$ are special, is itself special. Todorcevic, S., On a conjecture of R. Rado, J. Lond. Math. Soc., II. Ser. 27, 1-8 (1983). ZBL0524.03033. So a counterexample to Rado's conjecture is a nonspecial tree, all of whose subsets of size $\aleph_1$ are special. In particular, all of its $\aleph_1$-subtrees are special. In the same paper Todorčević also proves that Rado's conjecture implies that for every regular $\kappa>\omega_1$, every stationary subset of $\kappa\cap \operatorname{Cof}(\omega)$ reflects. Since the usual forcing to add a nonreflecting stationary set does not add reals, we can start in a model of CH and force the failure of Rado's conjecture, and consequently a tree as in your question, while preserving CH.<|endoftext|> TITLE: Can one construct a regular neighborhood without an ambient space? QUESTION [6 upvotes]: If I understand my PL topology correctly (and please correct me if I don't), if $K$ is a $k-$complex and $n\ge 2k+2$, then any two PL embeddings $a,b\colon K\to \mathbb{R}^n$ are isotopic. Therefore, the regular neighborhoods of $a(K)$ and $b(K)$ are homeomorphic, and we can think about them as "the" regular neighborhood of $K$. Is it possible to describe this regular neighborhood without reference to a particular embedding? For instance, if $K$ is a manifold, then the regular neighborhood is homeomorphic to a product $K\times \mathbb{R}^{n-k}$, and I'd like something similar for complexes that aren't manifolds. REPLY [4 votes]: One might think of the abstract regular neighbourhood as a canonical way of thickening $K$ to a handle decomposition of sufficiently high dimension. If $\dim K = 1$, then $K$ has a canonical $n$-dimensional thickening $M$ as soon as $n\ge 3$, where vertices and edges are replaced by 0-handles and 1-handles. Of course $M$ is a handlebody. Note that $M$ is canonical because we implicitly require $M$ to be orientable, that is $w_1(M)=0$. If $\dim K = 2$, then analogously $K$ has a canonical 5-dimensional thickening $W$ that has $w_1(W) = w_2(W)=0$. This can be proved by hand with standard low-dimensional topology techniques, see for instance Lemma 3.3 in this paper of Hambleton, Kreck, and Teichner. Vertices, edges, and faces are thickened to 0-, 1-, and 2-handles. As they say in this paper, the general case might be covered by this 1966 paper of Wall, but I must admit I haven't read it.<|endoftext|> TITLE: Implications of gauge symmetry breaking on the spectral side of geometric Langlands? QUESTION [7 upvotes]: Let $G$ be a complex reductive algebraic group and $X$ be a smooth compact complex curve. It's easy to see that the space of vacua in B-twisted $N=4$ SUSY Yang--Mills theory is $\mathfrak{h}^*[2]/W$ (where $\mathfrak{h}$ is the Cartan subalgebra of $\mathfrak{g}$ and $W$ is the Weyl group). There is a theorem due to Elliott and Yoo stating that the full category of boundary conditions compatible with the vacuum $0 \in \mathfrak{h}^*[2]/W$ is equivalent to $IndCoh_{\mathcal{N}_G}(LocSys_G)$, the category of ind-coherent sheaves on the moduli space of principal $G$-bundles equipped with flat connection on $X$ with singular support contained in the global nilpotent cone. Note that the former is the spectral side of geometric Langlands correspondence due to Arinkin--Gaitsgory. This theorem gives a physical interpretation to $IndCoh_{\mathcal{N}_G}(LocSys_G)$ (and makes it clear that it's the right category to consider in geometric Langlands correspondence, as opposed to the category of quasi-coherent sheaves $QCoh(LocSys_G)$). Elliott and Yoo formulated following conjecture expressing the effect of gauge symmetry breaking on the categories under consideration: the full subcategory of objects in $IndCoh(LocSys_G)$ compatible with a vacuum $u\in \mathfrak{h}^*/W$ is equivalent to $IndCoh_{\mathcal{N}_L}(LocSys_L)$ where $L$ is the stabilizer of $u$. My question is: what are the mathematical implications of this conjecture in the context of geometric Langlands program? REPLY [6 votes]: We discussed some conjectural implications in Section 4.2 of the paper. I wouldn't say that the category of sheaves with nilpotent singular support was necessarily the "right" category to consider from a gauge theoretic point of view. Instead I'd say that if one considers the equivalence for the whole category of coherent objects (or its completion) the categories on both sides of the equivalence form a flat family over the moduli space of vacua with the equivalence taking place compatibly with this family, which is to say equivariantly for the action of the algebra of local operators on the two sides. Of course there's a missing ingredient here: a concrete understanding of the action of the algebra of local operators on the category of ind-coherent / renormalized D-modules on $\mathrm{Bun}_G$. One possible implication we proposed was that, at least for $G = \mathrm{GL}_n$, these equivalences should "assemble" in the limit $n \to \infty$ to define an equivalence of factorization categories over $\mathbb{C}$. Physically we can think of this as the family of twisted theories on arbitrary collections of parallel D3 branes in type IIB factorizing nicely whenever two coincident branes are moved apart. The existence of this factorization category (and the prediction that duality preserves the factorization structures) is, however, a purely mathematical expectation. It's distinct from the factorization structure on the curve that usually appears in the geometric Langlands literature.<|endoftext|> TITLE: Computer program for counting graph homomorphisms QUESTION [5 upvotes]: I would like to ask is there a computer program for counting graph homomorphisms? REPLY [9 votes]: EDIT: I have since found that the Digraphs package sometimes counts homomorphisms incorrectly. Perhaps this problem has been fixed in more recent versions though. I use Minion for counting homomorphisms instead. The Digraphs package for GAP has several functions for finding homomorphisms of various types between graphs, including functions that will find all such homomorphisms, or all homomorphisms up to symmetries of the target graph. Once you have them all you can just count them to find the number. I really recommend this package, it seems to perform very well. For instance its chromatic number routine seems to be much faster than the chromatic number routine in SAGE.<|endoftext|> TITLE: Is a $G$-invariant character $\theta$ of $H$ extendible to $G$? QUESTION [8 upvotes]: Let $G/H\cong PSL(2,11)$, and $\theta$ be an irreducible $\mathbb{C}$-character of $H$. Suppose $\theta$ is invariant in $G$ and $\theta(1)=9$. Question: Is $\theta$ extendible to $G$? REPLY [12 votes]: The answer is yes (and some of the comments were moving in the right direction): Let $T$ be a transversal to $H$ in $G,$ and let $\sigma$ afford the representation of $H.$ For each $t \in T,$ there is a matrix $M_{t} \in {\rm GL}(9,\mathbb{C})$ such that $\sigma(tht^{-1}) = M_{t}\sigma(h)M_{t}^{-1}$ for all $h \in H,$ and $M_{t}$ is unique up to scalar multiples (by Schur's Lemma) , while any non-zero scalar multiple will have the same property. Extend this to $G$ by letting $M_{th} = M_{t}\sigma(h)$ for each $t \in T, h \in H.$ (It might be convenient here for anyone interested in full detail, to assume that have multiplied $\sigma$ by a suitable power of the linear character $ \lambda = {\rm det} \sigma,$ so that ${\rm det} \sigma$ may be assumed to have multiplicative order a power of $3$). Note also that for $x,y \in G,$ there is a scalar $\alpha(x,y) \neq 0$ such that $M_{xy} = \alpha(x,y)M_{x}M_{y} (\ast).$ Notice then $M_{x}^{|G|}$ is a scalar matrix for each $x \in G.$ Multiplying each $M_{x}$ by a suitable scalar (and we can still keep $M_{h} = \sigma(h)$ for each $h \in H$, and $M_{th} = M_{t}\sigma(h)$ for $t \in T,h \in H$), we may, and do from now on, assume that each $M_{x}$ has determinant a $3$-power root of unity. It follows from $(\ast)$ (on taking determinants), that $\alpha(x,y)$ is a $3$-power root of unity for all $x,y \in G.$ This gives a $2$-cocycle for ${\rm PSL}(2,11)$ of $3$-power order. Now we can finish in either of two ways: the Schur multiplier of the perfect group ${\rm PSL}(2,11)$ is well-known to have order $2.$ But a more general argument is to note that a perfect group with a cyclic Sylow $p$-subgroup always has a Schur multiplier of order prime to $p$ ( and this may be applied here with $p =3$ ).<|endoftext|> TITLE: Rationality of the moduli space of genus g curves QUESTION [6 upvotes]: I'm not an expert in this topic, so please excuse my negligence. I'd also appreciate references to the literature. Throughout, I will work over the complex numbers, although the analogous questions could be asked over other fields, and they may even be more interesting in that situation. Let $M_g$ denote the moduli space of smooth projective curves of genus $g$. Is $M_g$ rational (i.e., birational to a projective space) for a general integer $g\geq 1$? This is true when $g=1$, as $M_1$ is then just the moduli space of elliptic curves. I'm not sure how to show this for genus $2$, but note that as genus $2$ curves are hyperelliptic, the moduli space $M_2$ can be identified with $(\mathbf{P}^1)^6/(\mathrm{SL}_2\cdot\Sigma_6) = \mathbf{P}^6/\mathrm{SL}_2$, so it's at least unirational. (I think that this is even true when we work over a field of characteristic not $2$.) This suggests that the first step in proving rationality would be showing that $M_g$ is in general birationally equivalent to quotient of $\mathbf{P}^N$ by a reductive group, which implies that it is unirational. Rationality seems harder. The article https://arxiv.org/abs/0804.1509 shows that $M_3$ is also rational, and my thoughts about unirationality seem to be confirmed in the introduction. What about genus greater than $3$? REPLY [7 votes]: Your question is a plausible guess; indeed, Severi conjectured that $\mathcal{M}_g$ is unirational for all $g$. He proved this conjecture for $g\leq 10$. Sernesi was first to prove unirationality of $\mathcal{M}_g$ in genus 12, then Chang and Ran proved it in genera 11 and 13, Verra in genus 14. Bruno and Verra have shown that $\mathcal{M}_{15}$ is rationally connected. The picture is quite different for higher genera. For instance, Harris and Mumford proved that $\bar{\mathcal{M}_g}$ is a variety of general type for $g\geq 24$; Farkas did this for $g=22$. Farkas proved that the Kodaira dimension of $\bar{\mathcal{M}_{23}}$ is at least 2 and conjectures that this is, in fact, an equality. A good exposition on this topic can be found in Farkas' "Birational aspects of the geometry of $\bar{\mathcal{M}_g}$".<|endoftext|> TITLE: The dual of a dual in a rigid tensor category QUESTION [8 upvotes]: For a rigid tensor category $\cal{C}$, can it happen that, for some $X \in {\cal C}$, we have that $X$ is not isomorphic to $(X^{*})^*$, for $*$ denoting dual? If so, what is a good example. REPLY [4 votes]: Another way of showing this isn't true in general (and this is a fairly systematic way to handle this kind of questions) is to look at string diagrams, ie at the free rigid (strict, say) monoidal category generated by one object $X$. It is not hard to see that morphisms in that category can be represented by "planar open oriented tangles", i.e. collections of intervals embedded in a rectangle so that the endpoints of each interval are attached to the top or bottom edge of the rectangle, modulo planar isotopy. In that category, $Hom(X,X^{**})$ is simply empty. ANother way to say this is that if this statement was always true, then it would be a formal consequence of the axioms of rigid monoidal categories, and one would be able to write down such an isomorphism purely in terms of the duality data, and it's easy to see there is'nt even a candidate map (let alone an isomorphism). It might be worth mentionning that the statement that does generalize to any rigid category, is the fact that there are canonical isomorphism $${}^*(X^*)\cong X \cong ({}^*X)^*.$$<|endoftext|> TITLE: Creativity and the mechanization of elementary geometry QUESTION [8 upvotes]: In plane geometry, it is customary to say that checking proofs is a mechanical process but that finding new theorems is a creative activity. Citing J. Hadamard, "logic only sanctions the conquests of intuition." We have now computer programs that can check the correctness of geometric statements (at least first order statements) and interactive geometry softwares are used routinely in geometry classes to teach elementary geometry. We can use these programs to discover new theorems, but is there any program that is able to find new geometric statements on its own? More precisely, Is there any new and interesting result in plane geometry that has been produced by a computer program without any human intervention? The answer may state such a result together with the relevant reference. Of course, the software is always made by humans, but the result output by the machine should not be known prior to the start of the program and no intervention should happen during the execution of the program. Also, new and interesting is subjective. By new, I mean some result that is not an easy corollary of some theorem prior to the XXIe century. By interesting, probably the real test would be to produce several results sufficiently different from each others to show creativity and fit the taste of the many geometers on mathoverflow. REPLY [5 votes]: A new theorem discovered by computer prover (1989) Concerning Pappus lines and Leisenring lines there exists a set of interesting theorems. Here we show that these theorems can be easily proved by our computer prover which is implemented on the basis of Wu's method for mechanical theorem proving in geometries. Using this prover we discovered, furthermore, a new theorem: The six intersection points of every Pappus line and its corresponding Leisenring line are collinear.<|endoftext|> TITLE: Enrichments vs Internal homs QUESTION [7 upvotes]: Consider the definition of existence internal homs for a general monoidal category category $\cal{C}$, mainly the existence of an adjoint for the functor $$ X \otimes -: \cal{C} \to \cal{C}, $$ for each object $X$ in $\cal{C}$. Denoting this functor by $$ hom_X(-):\cal{C} \to \cal{C} $$ it is tempting to ask if the functor $$ hom: {\cal C} \times {\cal C} \to {\cal C}, ~~~~~~~ (X,Y) \mapsto hom_X(Y), $$ gives an enrichment of $\cal{C}$ over itself. Is this correct? Moreover, is the existence of an enrichment of $\cal{C}$ over itself equivalent to the existence of internal homs? More generally, when people speak of internal homs for a category, not necessary monoidal, are they just talking about an enrichment of the category over itself? Is this what usually understood by "a category with internal homs"? REPLY [2 votes]: I see that this question has an already accepted answer but I think that may be of interest. There is a notion of category with internal hom with no reference to a monoidal structure, that is the notion of a closed category. This is a category with an internal-hom functor and few other natural and extranatural transformations satisfying coherence conditions. The best reference on the subject is Eilenberg and Kelly's Closed categories. In this paper the authors show how you can enrich in any closed category, without any monoidal structure, and personally I think is one of the best references for enriched categories (even if it is not the most complete). I would strongly suggest to take a look to the article, personally I have learned and understood more enriched-category theory from that paper than other reference. I hope this may help.<|endoftext|> TITLE: Is there a computable homeomorphism between two different Cartesian powers of the computable real numbers? QUESTION [12 upvotes]: It's well know that it is surprisingly difficult to prove that $\mathbb{R}^n$ and $\mathbb{R}^m$ are not homeomorphic for $n\neq m$. Commonly proofs go through Brouwer's fixed point theorem, which is 'computably false' for dimensions greater than one: let $K$ denote the computable real numbers. For $n>1$, there are computable functions from $\left( [0,1]\cap K \right)^n$ to itself with no computable fixed point. That by itself may not seem very bad, since the function may still extend to a continuous function on $[0,1]^n$ and have an incomputable fixed point, but a corollary of this is that there is a computable retraction of $\left( [0,1]\cap K \right)^n$ onto its boundary. Such a function clearly can't be extended to a continuous function on $[0,1]^n$. So we can see that the topological behavior of $K^n$, even when restricted to computable functions, is very different from the topological behavior of $\mathbb{R}^n$. On the other hand, $K$ is homeomorphic to $\mathbb{Q}$ (although not computably so) and $\mathbb{Q}^n$ is homeomorphic to $\mathbb{Q}^m$ for any $n$ and $m$, so $K^n$ is homeomorphic to $K^m$ for any $n$ and $m$. So the question is: Is there a computable homeomorphism between $K^n$ and $K^m$ for some $n\neq m$? If there is one I would assume we need $n,m>1$. REPLY [4 votes]: I believe the full picture to be rather open, but Takayuki Kihara has shown that there is a computable embedding of $K^4$ into $K^3$ in T. Kihara: The Brouwer invariance theorems in reverse mathematics arXiv 2002.10715 The argument is outlined in the middle of p. 13. It uses Ovrekov's "constructive map of the square into itself, which moves every constructive point", from which it follows that $\mathrm{WKL}$ fails for the computable numbers. Kihara's paper then shows that for any model of $\mathbb{R}$ in which $\mathrm{WKL}$ fails, there is a topological embedding from $\mathbb{R}^m$ into $\mathbb{R}^3$ for any $m$. Whether we can even have a computable injection of $K^3$ into $K^2$ is listed as an open question here: http://www.math.mi.i.nagoya-u.ac.jp/~kihara/questions.html Although I don't have much to go on with, I suspect that asking about mutual embeddability might be the cleaner question here compared to computable homeomorphism.<|endoftext|> TITLE: Example of convex n-gon that cannot be decomposed into k congruent convex polygons QUESTION [6 upvotes]: I asked a related question here on MO without any answers yet. The question is in the title - give an example of a convex $n$-gon that cannot be subdivided into $k>1$ congruent convex polygons. Even better, give a family that solves this for all combinations of $(n,k)$. Intuitively, any generic $n$-gon should work, but the crux is in the details - I am very curious about what methods one can use to rigorously prove that a subdivision is impossible. There are of course many variations, e.g., drop the convex restriction, and remove the restriction that the pieces are polygons. REPLY [6 votes]: Here is a cheap argument for a generic $n$-gon with $n>6$. Assume that it is decomposed into $k$ congruent convex $p$-gons. Then the dimension of the subspace of $\mathbb R$ generated by the angles of the $n$-gon over $\mathbb Q$ is generically $n$ while all of them are in the linear span of angles of the $p$-gon whence $p\ge n$. That part does not require convexity or connectedness but the next ones do. Let $s$ be the number of inside points where several vertices meet and $t$ be the number of inside points where several vertices meet on the side of another polygon (including the side of the big one). The angle count yields $$ 2\pi s+\pi t+ \pi(n-2)=\pi k(p-2) $$ i.e., $$ 2s+t+n-2=k(p-2)\,. $$ On the other hand at least $3$ polygons meet by a vertex at those $s$ points and at least $2$ at those $t$ points, so the vertex count gives $$ 3s+2t+n\le kp $$ so $s+t\le 2k-2$ and, since $p\ge n$, we get $4(k-1)\ge (k-1)(n-2)$, so when $n\ge 7$, we have no chance. You can improve this argument a bit but I will also be very interested in an argument that would show that in the generic case $n=3$, $k=m^2$ is the only real option for convex partitions (and $n=3$ is the only real option with the connectedness and convexity restrictions dropped). Note that some decompositions are highly non-trivial. For instance, you can split the equilateral triangle into $5$ congruent polygonal pieces (not connected though)<|endoftext|> TITLE: A question on the sine function QUESTION [13 upvotes]: The Fejer-Jackson-Gronwall inequality involving the sine function is as follows: $$\sum_{k=1}^n\frac{\sin kx}k>0\quad\text{for all}\ n=1,2,3,\ldots\ \text{and}\ 00. $$ We may assume that $m\ge 3$, the cases for $m=1$ and $2$ already being known. Suppose that $K\ge 2$ is a natural number such that $\pi/(K+1) < x \le \pi/K$. Since $\sin(kx) \ge 0$ for all $1\le k\le K$ and $(\sin(kx))^m \ge -1$ for $k \ge K+1$, we find that $F_m(x;n) \ge 0$ if $n \le K$, and if $n \ge K+1$ then \begin{align*} F_m(x;n) &\ge \sum_{k=1}^{K} \Big( \frac{\sin(kx)}{k} \Big)^m - \sum_{k =K+1}^{\infty} \frac{1}{k^m} \\ &\ge \sum_{k=1}^{K} \Big( \frac{\sin(kx)}{k}\Big)^m - \int_{K}^{\infty} \frac{dt}{t^m} \\ &= \sum_{k=1}^{K} \Big( \frac{\sin(kx)}{k}\Big)^m -\frac{1}{(m-1)K^{m-1}}. \end{align*} Note that if $0\le t\le \pi/2$ then $\sin t \ge 2t/\pi$. Using this for $k\le K/2$ above, we conclude that $$ F_m(x;n) \ge \sum_{k\le K/2} \Big( \frac{2k x}{\pi k} \Big)^m - \frac{1}{(m-1)K^{m-1}} \ge \Big\lfloor \frac K2 \Big\rfloor \Big(\frac{2}{(K+1)}\Big)^m - \frac{1}{(m-1) K^{m-1}}. $$ Since $K\ge 2$ we have $2/(K+1) \ge 4/(3K)$ and $\lfloor \frac K2 \rfloor \ge K/3$, so that $$ F_m(x;n) \ge \frac{1}{K^{m-1}} \Big( \frac{1}{3} \Big(\frac 43\Big)^m - \frac{1}{m-1} \Big) > 0 $$ since $m\ge 3$. This completes the proof of one inequality. We now turn to the other inequality, which is more involved. We rewrite this inequality as $$ G_m(x;n) := \sum_{k=1}^{n} (-1)^{k-1} \Big( \frac{\sin (kx)}{k} \Big)^m > 0, $$ which we want to establish for all $m\ge 2$ and all $n$ (the case $m=1$ following from the Jackson inequality). Once again suppose that $K\ge 2$ is such that $\pi/(K+1) < x\le \pi/K$. Since $\frac{\sin y}{y}$ is decreasing and non-negative in $0 \le y\le \pi$, by pairing up two consecutive terms we see that $G_m(x;n)$ is positive if $n\le K$ (if there is an unpaired term at the end it is non-negative), and henceforth we assume that $n \ge K+1$. We now record a useful estimate for tails of our sum, which we shall prove later: for any $0 < x\le \pi/2$ and any integers $A x^m \Big( \frac{1}{2} - \frac{1}{\pi} - \frac{1}{\pi^2} \Big) > 0. $$ This finishes the first case when $K \ge m$. Now we turn to the second case $2 \le K < m$, which only happens for $m\ge 3$. As noted earlier, we may suppose that $n\ge K+1$, and write $$ G_m(x;n) = x^m \sum_{k=1}^{K} (-1)^{k-1} \Big( \frac{\sin (kx)}{k x}\Big)^{m} + \sum_{k=K+1}^{n} (-1)^{k-1} \Big( \frac{\sin (kx)}{k}\Big)^m. $$ As noted earlier $(\sin y)/y$ is non-negative and decreasing on $[0,\pi)$, and therefore by pairing consecutive terms in the first sum (and if there is a last term left unpaired it is non-negative) and discarding all but the first two terms, we see that the first sum above is $$ \ge (\sin x)^m - \Big(\frac{\sin 2x}{2} \Big)^m = (\sin x)^m (1- (\cos x)^m) \ge (\sin x)^{m} (\sin x)^2 \ge \Big( \frac{2}{\pi} x\Big)^m \Big( \frac{2}{K+1} \Big)^2, $$ where we used that $\sin t \ge 2t/\pi$ for $t\in [0, \pi/2]$. As for the second sum, using our inequality for tails this is bounded in size by $$ \frac{m}{2(m-1)} \frac{x}{(K+1)^{m-1} } + \frac{1}{(K+1)^{m-1}} \le x^m \Big( \frac{3}{4 \pi^{m-1}} + \frac{1}{\pi^m} \Big). $$ Since $K x^m \Big( \Big(\frac{2}{\pi}\Big)^m \frac{4}{m^2} - \frac{3}{4 \pi^{m-1}} - \frac{1}{\pi^m} \Big) >0, $$ since $m\ge 3$. This completes the proof of the second case. Lastly it remains to verify the estimate for tails. This is proved by partial summation. The desired sum is $$ x^m \sum_{k=A}^{B} (-1)^{k-1} \Big(\frac{\sin (kx)}{kx} \Big)^m = x^m \int_{A^-}^{B^+} \Big( \frac{\sin (yx)}{yx} \Big)^{m} d\Big( \sum_{A \le k < y} (-1)^{k-1} \Big), $$ and integrating by parts (and note that $|\sum_{A \le k < y} (-1)^{k-1} | \le 1$ always) we can bound this in magnitude by $$ x^m \Big| \frac{\sin (Bx)}{Bx} \Big|^m + x^m \int_A^B \Big| \frac{d}{dy} \Big( \frac{\sin (yx)}{yx} \Big)^m \Big| dy. $$ Now \begin{align*} \Big| \frac{d}{dy} \Big( \frac{\sin (yx)}{yx}\Big)^m \Big| &= m \Big| \frac{\sin (yx)}{yx} \Big|^{m-1} \Big| \frac{\cos (yx)}{y} - \frac{sin (yx)}{ y^2x} \Big| \\ & \le m \Big| \frac{(\sin yx)^{m-1} (\cos yx)}{y^m x^{m-1}} \Big| + \frac{m}{y^{m+1} x^m} \le \frac{m}{2 y^mx^{m-1} } +\frac{m}{y^{m+1} x^m}, \end{align*} where in the last step we used that $|(\sin t)^{m-1} (\cos t)| \le |\sin t \cos t| \le 1/2$. Using this we conclude that our desired sum is $$ \le \frac{1}{B^m} + \int_A^B \Big( \frac{mx}{2y^m} + \frac{m}{y^{m+1}} \Big) dy < \frac{mx}{2(m-1) A^{m-1}} + \frac{1}{A^m}, $$ which completes the proof.<|endoftext|> TITLE: Reference request: tensor induction QUESTION [7 upvotes]: While working on a problem, I constructed something which looked like an induced representation, but with a tensor product instead of a direct sum. Here is a special case. Let $G$ be a group, with $H$ a subgroup of index $2$. Choose $s \in G$ which is not in $H$. For $(\pi,V)$ a representation of $H$, define a representation $\sigma$ of $G$ with underlying space $V \otimes V$ as follows: if $h \in H$, define $\sigma(h) = \pi(h) \otimes \pi(shs^{-1})$. If $g \in G, \not\in H$, define $\sigma(g)$ on generators by $$\sigma(g) v \otimes w = \pi(gs^{-1})w \otimes \pi(sg)v$$ If we had used $V \oplus V$ instead of $V \otimes V$, then following the above construction would have produced the representation $\operatorname{Ind}_H^G(\pi)$. So this is like a tensor product version of induced representation. I found some papers which talk about "tensor induction," and I believe the above construction is a very simple case of that. What is tensor induction, generally speaking? Is it adjoint to some functor, or satisfy a universal property? REPLY [6 votes]: An introduction to tensor induction is given in §13 of Curtis, Charles W.; Reiner, Irving, Methods of representation theory with applications to finite groups and orders. Volume 1., Wiley Classics Library; New York et al.: John Wiley & Sons. (1990). ZBL0698.20001. Some papers also contain brief introductions to tensor induction, for example: Isaacs, I. M., Character correspondences in solvable groups, Adv. Math. 43, 284-306 (1982). ZBL0487.20004. (Both Curtis-Reiner and Isaacs have more references on tensor induction.) A universal property of tensor induction is given in my paper (Thm. 2.1): Ladisch, Frieder, Corestriction for algebras with group action., J. Algebra 439, 438-453 (2015). ZBL1330.16026. arXiv:1409.3166. The universal property is in terms of the natural multilinear map from the induced module into the tensor induced module. As tensor induction is not an additive functor, it can not be a (left or right) adjoint of a functor. In particular, there is no analog of Frobenius reciprocity.<|endoftext|> TITLE: Formality of the 2nd ordered configuration space of a closed Riemann surface QUESTION [12 upvotes]: If $X$ is a smooth manifold, we define its kth ordered configuration space as $$F_kX:=\{(x_1, \ldots,x_k) \; | \; x_i \neq x_j \,\, \mathrm{if} \, \, i \neq j\},$$ in other words, $F_kX = X^k - \Delta,$ where $\Delta$ is the big diagonal. I'm aware of the following two results about the formality (in the sense of rational homotopy theory) of $F_kX$. In the sequel, $g$ will be an integer $\geq 1$. Theorem 1 ([B94, p. 133]) If $\Sigma_g$ is a closed Riemann surface of genus $g$, then $F_k \Sigma_g$ is not formal for $k>2$. Theorem 2 ([LS04, p. 1048]) If $X$ is a closed, connected formal space such that $$H^1(X, \, \mathbb{Q})=H^2(X, \, \mathbb{Q})=0,$$ then $F_2 X$ is formal. Let us now consider $F_2 \Sigma_g =\Sigma_g \times \Sigma_g - \Delta$. Then Theorem 1 does not apply because $k=2$; on the other hand, Theorem 2 does not apply either, even if $\Sigma_g$ is formal (it is a compact Kähler manifold), because the assumptions on the cohomology are not satisfied. So let me ask the following Question. Is the 2nd configuration space $F_2 \Sigma_g$ a formal topological space? I am by no means an expert in rational homotopy theory (actually, at the moment I am using it as a black box allowing me to perform some group theoretical computations), so I apologize in advance if the answer turns out to be trivial for the experts in the field. Every reference to the relevant literature will be highly appreciated. Bibliography. [B94] R. Bezrukavnikov: Koszul DG-algebras arising from configuration spaces, Geometric and functional analysis 4 (1994), 119-135. [LS04] P. Lambrechts, D. Stanley: The rational homotopy type of configuration spaces of two points, Ann. Inst. Fourier 54 (2004), 1029-1052. REPLY [10 votes]: I believe that $F_2 \Sigma_g$ is formal. Here's a sketch of an argument, unfortunately I haven't checked the details. Apply the results of Morgan, "The algebraic topology of smooth algebraic varieties". To be specific, $F_2 \Sigma_g$ is the complement of a smooth divisor in the compact Kähler manifold $\Sigma_g^2$. Then Morgan's results say that the de Rham algebra of $F_2 \Sigma_g$ is equivalent to the algebra of forms on $\Sigma_g^2$ with logarithmic singularities along $\Delta$, and the latter algebra is quasi-isomorphic to the $E_2$ term of the Leray spectral sequence for $F_2\Sigma_g \to \Sigma_g^2$. This can be represented by a cdga given by the mapping cone of $s^{-2}H \to H\otimes H$, where $H = H^\bullet(\Sigma_g,\mathbf Q)$. But now this mapping cone is in turn quasi-isomorphic to the cohomology of $F_2\Sigma_g$, via the natural projection $H^{\otimes 2} \to H^\bullet(F_2\Sigma_g,\mathbf Q)$. A potential snag is that Morgan reasons a lot with minimal models in his paper. But I believe that he doesn't actually need any of that for what I said above (this part, I think, is only involved in putting mixed Hodge structures on various rational invariants). In particular the argument shouldn't require any simple connectedness/nilpotence.<|endoftext|> TITLE: Which matrices can be realized as the Dirichlet-to-Neumann map for a given domain? QUESTION [11 upvotes]: Consider Poisson equation $\nabla \cdot (\sigma(x)\nabla u)=0$ in a domain $D$, where $\sigma(x)$ is the spatially dependent conductivity. On the boundary we have $n$ electrodes (Dirichlet BC $u=\text{const}$ on each electrode). And the rest of the boundary is insulating material $du/d\vec n=0$ (Neumann BC). The electrodes do not have any contact impedance. The Dirichlet-to-Neumann map or Poincare–Steklov operator is the map from voltages to total current on electrodes for a given conductivity distribution (and is a linear map according to Ohm's law and Kirchoff's law (and therefore a matrix)). Given an arbitrary $n\times n$ matrix, is there a way to check if there exist a conductivity distribution which has this matrix as a Dirichlet-to-Neumann map? I am looking for a computationally fast one hopefully, I don't want to compute the conductivity distribution. Maybe there are some domain independent restrictions? (I know there are because I've seen them before but I can't find it again). But I was hoping you could improve them when you know the domain. Further I also know that the conductivity is bounded above by some constant. Is there a way to find what restriction on the DtN map this gives? REPLY [7 votes]: In dimensions 3 and higher, and without any constraints on $\sigma$, one can apparently obtain any symmetric matrix $A = (a_{ij})$ such that $a_{ij} < 0$ when $i \ne j$ and $a_{ii} = -\sum_{j \ne i} a_{ij}$, as suggested in Denis Serre's answer. That answer already explains why these conditions are necessary. To see that they are also sufficient, one can approximate electrodes connected by non-crossing wires of given conductance (in the limiting case $\sigma$ is zero outside the wires, and constant in each wire). In dimension 2, I believe, there are additional geometric constraints (for example, it is not possible to connect the electrodes with non-crossing wires). I bet this has been studied, but I do not know any references. REPLY [6 votes]: This question (for two dimensional domains) was answered by Curtis, Ingerman and Morrow, "Circular Graphs and planar Resistor Networks" (1998). Let $a$ be the $n \times n$ response matrix. As already noted, we must have $a_{ij} = a_{ji}$ and $a(1\ 1\ \cdots\ 1)^T=0$. The additional condition is that, if $i_1$, $i_2$, ..., $i_k$ and $j_1$, $j_2$, ..., $j_k$ appear in circular order around the boundary of the network, and $A'$ is the submatrix with rows $i_r$ and columns $j_s$, then $(-1)^k \det A' \geq 0$.<|endoftext|> TITLE: When is a morphism of smooth schemes over a strictly henselian base etale QUESTION [5 upvotes]: Let $X,Y$ be affine schemes over a strictly henselian base-ring $R$. Assume $X$ and $Y$ are $R$-smooth and consider a $R$-morphism $f$ between $X$ and $Y$. I would like to use that $f$ is étale given that it is étale at every rational point of the special fiber. So the questions is: is this true? Thanks! REPLY [5 votes]: As Piotr pointed out, the answer to your question is no as stated. I just wanted to add a further comment (too long for an actual comment) I guess what you're thinking is that it suffices to check etaleness at closed points, which is true. However, it would be wrong to assume that all closed points of $X$ should lie on the special fiber. This is true if $X\rightarrow\text{Spec }R$ is proper, but false in general. This is the key point in Piotr's counterexample. The ramification locus of his map $f : X = \mathbb{A}^1_R\rightarrow\mathbb{A}^1_R = Y$ given by $f(x) = x+tx^n$ is a union of closed points which do not lie on the special fiber. If you consider his map, and extend it naturally to a map $\overline{X} = \mathbb{P}^1_R\rightarrow\mathbb{P}^1_R = \overline{Y}$, there you would be able to detect the ramification locus on the special fiber of $\overline{X}$. Indeed, the ramification locus is the divisor determined by the $(n-1)$th roots of $\frac{-1}{nt}$ in $\text{Frac}(R)$ (viewed as points on the generic fiber, the corresponding divisor being their closure in $X$). These elements of $\text{Frac}(R)$ are not integral over $R$, so the corresponding divisor intersects the special fiber of $\overline{X}$ at $\infty$. However, when you restrict back to $X = \mathbb{A}^1_R$, the divisor no longer intersects the special fiber. Indeed, since taking the closure of the $(n-1)$th roots of $\frac{-1}{nt}$ inside $\overline{X}$ only adds the $\infty$ of the special fiber, upon restricting to $X$, the ramification divisor becomes a disjoint union of closed points, all lying on the generic fiber. This is also an interesting example of a nice (regular Noetherian) scheme of dimension 2 for which not all closed points have local rings of the same dimension. I believe this failure is stated as saying "the scheme is not equicodimensional" (it seems to be rather difficult to find references for this, but IIRC once upon a time I found a discussion of it in EGA...somewhere). For this, you can even take the scheme $\text{Spec }\mathbb{Z}_{(p)}[x]$.<|endoftext|> TITLE: When does $\pi:M\to M/G$ have homotopy lifting property? QUESTION [7 upvotes]: Let M be an n dimensional topological manifold, may be non-compact. Suppose there is an action of a group G on M, the orbits are closed, but may not be bounded. Consider the projection $\pi: M \to M/G$, do we have path lifting and homotopy lifting property? If not, what other conditions should we add? I know when M is a smooth manifold, and G acts smoothly, freely, and properly, then $\pi$ is a smooth normal covering. But we consider manifold which may not be smooth and we only need lifting properties, the condition is too strict. Consider the example $\pi: \mathbb{R}^3\to \mathbb{R}^3/Z_2$, $\mathbb{R}^3/Z_2$ is the cone over $\mathbb{R}P^2$, not a topological manifold. In this example, do we have path lifting and homotopy lifting property? REPLY [10 votes]: See Armin Rainer: Orbit projections as fibrations, Czechoslovak Math. J. 59 (2009), No. 2, 529--538, arXiv. Abstract. The orbit projection $\pi : M → M/G$ of a proper $G$-manifold $M$ is a fibration if and only if all points in $M$ are regular. Under additional assump- tions we show that $\pi$ is a quasifibration if and only if all points are regular. We get a full answer in the equivariant category: $\pi$ is a $G$-quasifibration if and only if all points are regular.<|endoftext|> TITLE: Torsion subgroup of the group of points of an elliptic curve over local field QUESTION [5 upvotes]: Let $K$ be a local field with residue field $k$ and $E/K$ an elliptic curve. I'm interested for which $K$ and $E$ the group of torsion points on the curve is finite. I can prove that this group is finite for an $n$-dimensional local field $K$ such that its $(n-1)$'th residue field has zero characteristics i.e. a finite extension of $\mathbb Q_p$. The proof goes by induction on $n$. Basis. Let $K$ be a one-dimensional local field i.e. $k = \mathbb F_q$,$q=p^r$. We use the standard notation $E_0(K) = \{P\in E(K)|\tilde{P}\in \tilde{E_{ns}}(k)\}$ and $E_1(K) = \{P\in E(K)|\tilde{P} = \tilde{O}\}$. Then we have an exact sequence of abelian groups: $$ 0\to E_1(K) \to E_0(K) \to \tilde{E_{ns}}(k) \to 0. $$ We know that $E_1(K) \cong \hat{E}(\mathcal M)$ where $\hat{E}$ is the formal group associated with $E$, hence $E_1(K)$ has no non-trivial elements of order $m$. From the exact sequence we see that $E_0(K)[m]$ injects to $\tilde{E_{ns}}(k)$ so all non-trivial torsion elements of $E_0(K)$ go to non trivial torsion elements of $\tilde{E_{ns}}(k)$. But $\#\tilde{E_{ns}}(k)$ is finite so the number of such torsion elements is finite. By Kodaira-Néron theorem the index of $E_0(K)$ in $E(K)$ is finite so we have proved that the torsion of $E(K)$ is finite. Step. Suppose we have proved the theorem for all $(n-1)$-dimensional local fields. Let $K$ be an $n$-dimensional local field i.e. $k$ is an $(n-1)$-dimensional local field. Then if the curve has good reduction the same argument reduces the case to $k$. If not $\tilde{E_{ns}}(k)$ is isomorphic either to $k^*$ or $k^+$ which have finite torsion. Q.E.D. I have two hypotheses. The first is that this is true for all local field of finite dimension. The second is that the only fields that have this property are subfields of higher local fields. In particular, having the property that the torsion of $E(K)$ is finite depends only on the field but not on the curve. Can you prove these statements or provide a counterexample? EDIT There are some mistakes in my proof. First of all, there might exist $p$-torsion in $E_1(K)$ where $p$ is the characteristics but it is finite because $E_1(K)$ is a pro-$p$-finite group such that $p^r$-torsion is finite for every $r$. But I don't know how to generalize this to the case of higher-dimensional fields because now $E_1(K)$ is not at all pro-$p$-group. REPLY [3 votes]: This is more of an answer to the stream of comments than to the question directly. The comments concern the statement that for an elliptic curve $E$ defined over $\mathbb{F}_q((t))$, the torsion subgroup of $E(\mathbb{F}_q((t)))$ is finite. Vesselin Dimitrov claimed that this appears in Clark, Pete L.; Xarles, Xavier Local bounds for torsion points on abelian varieties. Canad. J. Math. 60 (2008), no. 3, 532–555. (doi: 10.4153/CJM-2008-026-x, Internet Archive)) The OP pointed out that it doesn't. I just looked back myself: the OP is right, but Vesselin is also morally right. The statement of that result does not appear in the paper, but Section 3 studies a standard three term filtration on $A(K)$ for $K$ a complete DVF with finite residue field and $A_{/K}$ an abelian variety. To show that $A(K)[\operatorname{tors}]$ is finite, it suffices to show that the torsion subgroups of the successive quotients of the filtration are finite. The point of this section is to do better than that: i.e., give explicit upper bounds on the the torsion subgroups of the successive quotients of the filtration in terms of invariants of $A$ like its dimension and reduction type. In case $K = \mathbb{F}_q((t))$ this is not achieved -- rather, a counterexample is given to the form of the bounds obtained in the characteristic $0$ case -- but nevertheless the successive quotients are shown to have finite torsion subgroups, so $A(K)[\operatorname{tors}]$ is finite. The part that needs separate attention in positive characteristic is the torsion in the formal group: for this, see Proposition 3.2. I also wanted to mention that a more recent (or recently published; the paper was accepted several years ago!) paper gives a treatment of the structure theory of (compact, commutative, second countable) $\mathbb{F}_q((t))$-analytic Lie groups, which in particular gives another proof of the finiteness of the torsion subgroup in the case of an abelian variety. See Theorem 5.2 of Clark, Pete L.; Lacy, Allan; There are genus one curves of every index over every infinite, finitely generated field. J. Reine Angew. Math. 749 (2019), 65–86. (doi: 10.1515/crelle-2016-0037 ;arXiv: 1405.2108)<|endoftext|> TITLE: distributivity of termspace forcing QUESTION [6 upvotes]: Laver introduced the concept of termspace forcing. If $\mathbb P * \dot{\mathbb Q}$ is a two-step iteration, then we can order the $\mathbb P$-names for elements of $\dot{\mathbb Q}$ by putting $\dot q_1 \leq \dot q_0$ when $1 \Vdash \dot q_1 \leq \dot q_0$. This defines the termspace partial order $T(\mathbb P,\dot{\mathbb Q})$. The key fact is that the identity map is a projection from $\mathbb P \times T(\mathbb P,\dot{\mathbb Q})$ to $\mathbb P * \dot{\mathbb Q}$. It is easy to see that if $\Vdash$ "$\dot{\mathbb Q}$ is $\kappa$-closed," then $T(\mathbb P,\dot{\mathbb Q})$ is $\kappa$-closed. Question: If $\Vdash$ "$\dot{\mathbb Q}$ is $(\kappa,\lambda)$-distributive," is $T(\mathbb P,\dot{\mathbb Q})$ $(\kappa,\lambda)$-distributive? Recall that $(\kappa,\lambda)$-distributive means the forcing adds no functions from $\kappa$ to $\lambda$. REPLY [11 votes]: Here is a ZFC counterexample. Let $\mathbb{P}$ be the forcing to add a Cohen real $c$, and let $\omega_1=\bigsqcup_n S_n$ be a partition of $\omega_1$ into disjoint stationary sets $S_n$. Let $\dot{\mathbb{Q}}$ be the forcing to kill the stationarity of all $S_n$, for $n\in c$, the generic Cohen real added by the $\mathbb{P}$ forcing. It is not difficult to see that $\dot{\mathbb{Q}}$ is forced to be $(\omega,\infty)$-distributive, since we are really just shooting a club through the other $S_m$'s. The term forcing for $\dot{\mathbb{Q}}$ over $\mathbb{P}$ will have to add a generic for $\mathbb{Q}$ over any Cohen real in the extension, and so it will have to kill the stationarity of all the $S_n$'s. Thus, it will have to collapse $\omega_1$. So it cannot be $\omega$-distributive. A modification to the argument, using higher cardinals instead of $\omega_1$, will have the property that $\mathbb{P}$ is small relative to the distributivity, as you asked for in the comments. For example, let $\kappa$ be any uncountable regular cardinal and consider an $\omega$-partition of the cofinality $\omega$ ordinals up to $\kappa^+$. The $\mathbb{Q}$ forcing should kill parts of the partition, depending on the digits of $c$, and this will be fine, but the term forcing will have to kill them all, since it must add generics over any Cohen real in the extension. So $|\mathbb{P}|<\kappa$ and $\dot{\mathbb{Q}}$ is $(\kappa,\infty)$-distributive, but the term forcing collapses $\kappa^+$ and hence is not $\kappa$-distributive. Here is a more extreme example, which makes the same point perhaps more clearly. Let $S\subset\omega_1$ be a stationary co-stationary set. Let $\mathbb{P}$ be the lottery sum $\{\text{yes}\}\oplus\{\text{no}\}$ of two trivial forcing notions, so that the generic filter selects either the point yes or the point no, and both of these are generic. Let $\dot{\mathbb{Q}}$ be the forcing that kills the stationarity of $S$, if the generic filter for $\mathbb{P}$ selected yes, and otherwise kills the stationarity of the complement of $S$. This forcing is $\omega$-distributive. But the term forcing has to add a generic for both the generics for $\mathbb{P}$, and so it will have to kill the stationarity both of $S$ and its complement. So it will collapse $\omega_1$.<|endoftext|> TITLE: Can the number of countable models of a complete first-order theory decrease after adding constants? QUESTION [11 upvotes]: If $T$ is a countable complete first-order theory with infinite models, the number of countable models it has, $I(T,\omega)$, must be an element of $N=\{1,3,4,5,6,7,\dots,\omega,\omega_1,2^\omega\}$ (although we don't know if $\omega_1$ can happen). For which pairs $n,m\in N$ does there exist a countable complete theory $T$ with $n$ countable models but $m$ countable models after adding finitely many constants to the theory? Countably many new constants? In particular can we have $m TITLE: Why torsion is only defined for linear connection on TM? QUESTION [5 upvotes]: The concept of curvature is defined for any linear connection on any vector bundle $E \to M$, but the concept of torsion is only defined for connection on the tangent bundle $TM$ of a manifold $M^n$, or for a connection obtained as the pullback of a connection on a vector bundle $E \to M$ isomorphic to $TM$ via an isomorphism $\theta \colon TM \to E$ equivalent to a solder form. Why is that so ? If torsion can be interpreted as the twist of a moving frame along a curve, the same phenomena should occur for a connection on any vector bundle. Is there a way to define a notion of torsion for any vector bundle ? REPLY [4 votes]: The torsion of a connection on the tangent bundle is a zeroth order invariant tensor. There is no zeroth order invariant tensor associated to a connection on a principal bundle on a smooth real manifold, since in any coordinates, the connection is $A=g^{-1} \, dg + \operatorname{Ad}_g^{-1} \Gamma_i(x) \, dx^i$, and we can then change those coordinates to get $x=0$ at our chosen point, $g=1$ there, and $\Gamma_i(0)=0$, by replacing $g$ by $h(x)g$ so that we replace $A$ by $g^{-1} \, dg + \operatorname{Ad}_g^{-1} \operatorname{Ad}_{h(x)}^{-1}(\Gamma_i(x) \, dx^i+dh \, h^{-1})$. We pick $h(x)$ so that $dh \, h^{-1}(0)=-\Gamma_i(0) dx^i$. So all connections look the same at a point, to first order. You can only feel that the connection is not flat at second order. This doesn't work on the frame bundle (the principal bundle associated to the tangent bundle) because you can't change $g$ to $h(x)g$ independent of how you change coordinates $x$.<|endoftext|> TITLE: Algebraic vs analytic normality QUESTION [11 upvotes]: Let $X$ be a complex algebraic variety. We can ask if $X$ is normal as an algebraic variety, but also, if its analytification is normal as a complex analytic space. Is there a relationship between the two? Do we have $$\text{algebraic normality} \implies\text{analytic normality}$$ or $$\text{analytic normality} \implies\text{algebraic normality}$$ or both or neither? REPLY [11 votes]: Francesco Polizzi's answer is perfectly fine, but let me try to explain the technique which helps to relate a lot of "local" properties of locally finite type schemes over $\mathbf C$ to their counterparts in the world of complex analytic spaces. For example, the proof that I will explain below actually shows that a locally finite type scheme $X$ over $\mathbf C$ is reduced/normal/Cohen-Macauley/Gorenstein/regular iff it's analytification $X^{an}$ has the same property. Moreover, exactly the same method works in the non-archimedean situation. The first difficulty in showing equivalence of analytic and algebraic definitions of normality is that we can't directly relate the algebraic local ring $\mathcal O_{X,x}$ to its analytic analogue $\mathcal O_{X^{an},x}$ for a closed point $x\in X$. The second difficulty is that for a non-closed point $x\in X$ we don't even have any analogue of $\mathcal O_{X,x}$ in the analytic world. So, we should be able to prove that we can test normality (regularity/reducedness/so on) of a locally finite type $\mathbf C$-scheme only on closed points. (This is a subtle question, for instance, this is a celebrated theorem of Serre that we can check regularity on closed points for an arbitrary noetherian scheme, though the locus of regularity is not always open). What helps us to overcome both difficulties is the notion of an excellent ring. I don't want to recall the definition here, you can read it here (and you can read a self-contained exposition of the main results about excellent rings in Chapter 13 of Matsumura's book "Commutative algebra"). But let me try to explain the main features of excellent rings: 1) If $A$ is a local excellent ring with a maximal ideal $\mathfrak m$, then the completion map $A \to \hat A_{\mathfrak m}$ is regular (meaning that it is flat and all fibers are geometrically regular). 2) The normalization morphism $\operatorname{Spec} A^{norm} \to \operatorname{Spec} A$ is a finite morphism. 3) Any essentially finite type algebra (a direct limit of finite type algebras) over an excellent ring is excellent (This is a real theorem, see Thm. 77 in Matsumura's book for a proof). 4) For any analytic space $X$ and a point $x$, a local ring $\mathcal O_{X,x}$ is excellent. This is a serious theorem, it is rather hard even to prove that this ring is noetherian (see the proof of Oka's theorem in Grauert-Remmert's book "Analytic Coherent Sheaves"). Unfortunately, I don't know any good reference for the excellence of this ring. Let me just emphasize that the proof that I know uses regularity and noetherianness of $\mathcal O_{\mathbf C^n,0}$ as an input for this fact. With all these said, let me move to the real arguments. First of all, let me address the second issue from above. The definition of normality is local on both sides, so wlog we can assume that $X=\operatorname{Spec} A$. Then using (2) one can easily check that the locus where $X$ is normal is open in $X$ (this locus is exactly the locus where the natural morphism $X^{norm}\to X$ is an isomorphism). Also, you may look at EGA IV$_3$ 7.8.6 for a generalization of this result. Now to finish the argument we need to note that $X$ is a locally finite type scheme over complex numbers, so if an open subset of $X$ contains all closed points it must be equal to the whole scheme $X$. This solves the second problem and allows to restrict our attention only to closed points on the algebraic side. Now, the key idea comes into play. We want to relate normality of $\mathcal O_{X,x}$ to normality of $\mathcal O_{X^{an}, x}$ for a closed point $x\in X$. As I pointed out before, we can't do this directly, but what we can do is to relate normality of $\mathcal O_{X,x}$ (resp. $\mathcal O_{X^{an},x}$) to normality of the completion $\hat{\mathcal O_{X,x}}$ (resp. $\hat{\mathcal O_{X^{an},x}}$) and then prove that $\hat{\mathcal O_{X,x}}$ is canonically isomorphic to $\hat{\mathcal O_{X^{an},x}}$. This step is not formal or trivial at all, it uses rather deep results from commutative algebra (and it crucially relies on the fact that $\mathcal O_{X,x}$ and $\mathcal O_{X^{an},x}$ are excellent). Let me formulate the main ingredients that we are still missing to complete the proof. Theorem 1 (Serre's Criterion of Normality): Let $A$ be a noetherian local ring, then $A$ is normal iff it has properties (R_1) and (S_2). Theorem 2 (Theorem 23.9 from Matsumura's book "Commutative Ring Theory"): Let $A\to B$ be a flat local homomorphisms of local noetherian rings, then (i) If $B$ satisfies (R_i)(resp. (S_i)), so does A. (ii) If both $A$ and the fibre ring $B\otimes_A k(\mathfrak p)$ over any prime ideal $\mathfrak p$ of satisfy (R_i)(resp. (S_i)), so does B. Remark: For the definition of properties (R_i) and (S_i) you can see this section on Stack's Project. Informally, Serre's criterion says that a noetherian scheme, which is regular in codimension one and such that every function from a codimension 2 open subset can be extended to the whole scheme, is normal. Now, let's apply the above theorems to the morphisms $\mathcal O_{X,x} \to \hat{\mathcal O_{X,x}}$ and $\mathcal O_{X^{an},x} \to \hat{\mathcal O_{X^{an},x}}$. We know that both rings $\mathcal O_{X,x}$ and $\mathcal O_{X^{an},x}$ are excellent as we discussed above. Thus all fibers of these two morphisms are geometrically regular by the property (1) from above! So, they automatically satisfy properties (R_i) and (S_i) for any i! Hence, Theorem 1 and Theorem 2 together imply that $\mathcal O_{X,x}$ (resp. $\mathcal O_{X^{an},x}$) is normal if and only if $\hat{\mathcal O_{X,x}}$ (resp. $\hat{\mathcal O_{X^{an},x}}$) is. Aha, so the last part is to prove that $\hat{\mathcal O_{X,x}}$ is canonically isomorphic to $\hat{\mathcal O_{X^{an},x}}$. This will imply that normality of $\mathcal O_{X,x}$ is equivalent to normality of $\mathcal O_{X^{an},x}$. The last step is to actually show that $\hat{\mathcal O_{X,x}}$ is isomorphic to $\hat{\mathcal O_{X^{an},x}}$. This is rather trivial and basically follows from the definition of analytification. Nevertheless, let me explain it (or at least the main idea) in a clear way. From the very definition of the analytification functor, we have a canonical morphism (in the category of locally ringed topological spaces) $$ i:X^{an} \to X. $$ This gives us a morphism $\mathcal O_X \to i_*\mathcal O_{X^{an}}$. Consider the stalk of this morphism at x to get a morphism $$ \mathcal O_{X,x} \to \mathcal O_{X^{an},x}. $$ Both rings are local and the morphism is local. Hence, it is continuous in $\mathfrak m_x$-adic topology. Thus it induces a morphism on the completions (in a sense of commutative algebra) $$ \hat{\mathcal O_{X,x}} \to \hat{\mathcal O_{X^{an},x}}. $$ In order to check that it is an isomorphism it suffices to prove that $\mathcal O_{X,x}/\mathfrak m_x^n \to \mathcal O_{X^{an},x}/(\mathfrak {m}^{an}_x)^n$ is an isomorphism for any integer $n$. The latter fact follows basically from the very definition of the analytification functor (it is also a good exercise).<|endoftext|> TITLE: Topological obstructions to existence of immersion QUESTION [12 upvotes]: Let $M$ be a smooth, non-compact manifold. a) Can one always find a smooth, compact manifold $N$ with $\dim(N) = \dim(M)$ and a smooth embedding $i: M \to N$ ? b) If not, are there some concrete general topological obstructions ? c) What if we require $i$ only to be an immersion ? Variants of these questions have thourougly been studied, for example, one can construct many examples of Riemannian manifolds $(M,g)$ for which there does not exist a conformal embedding/immersion into a compact Riemannian manifold of the same dimension. Even in the non-Riemannian setting, but requiring additionally that $i(M) = N \setminus \partial N$, there are obstructions to the existence of such an embedding $i: M \to N$ (for example, this is never possible if $\pi_1(M)$ is infinitley generated). But what about these slightly more general questions ? REPLY [11 votes]: This is regarding (c) in the 2-dimensional case. First, let's consider the orientable case. Every orientable (connected) noncompact surface is a covering space of a genus 2 surface, and hence has an immersion to a compact surface. In fact, this also holds for closed connected orientable surfaces of negative euler characteristic, which excludes the 2-sphere and 2-torus. To see this, one may appeal to the classification of noncompact surfaces by Ian Richards. He proves that an orientable surface $S$ is determined by its space of ends $B(S)$ (Richards calls this the ``ideal boundary" of the surface), together with a closed subset $B'(S)$ which is the subset of non-planar ends (that admit no planar neighborhood). First let's consider the case of a planar surface with Cantor set ideal boundary. This is realized as the boundary of a tubular neighborhood of a cubic tree graph: In turn, this graph covers a bipartite cubic graph with two vertices, whose thickened boundary is a genus 2 surface: Each vertex of the cubic graph is dual to an pants, and each edge is dual to a curve of the pants decomposition. Moreover, the graph admits a Tait coloring, a 3-coloring of the edges so that each vertex is adjacent to edges with three distinct colors. Hence the cubic tree admits such a coloring by pullback. Moreover, one sees that a bipartite 3-edge colorable cubic graph has boundary of a thickening which is a cover of this surface. Moreover, this holds true if we allow the graph to have rays, corresponding to isolated points in the ideal boundary of the surface. To see, this, truncate each ray to get a graph with vertices of degree 3 and 1 (leaves). Truncate the corresponding boundary of thickening to get a surface with a boundary component for each leaf. This surface embeds in a subsurface of a cover of the genus 2 surface, with each boundary component mapping to one of the 3 curves on the genus 2 surface. Hence, the non-compact surface covers the genus 2 surface by attaching an infinite half-cylinder to each boundary component (coming from half of the annulus covering space corresponding to the curve). For the general planar case, we let $B(S)$ be a subset of the Cantor set, the ideal boundary of the cubic tree, and take the convex subgraph of the cubic tree whose limit set is $B(S)$. This is a cubic bipartite graph with rays, and hence covers the genus 2 surface. To get the positive genus case, insert bigons into the tree to add genus to the ends of the ideal boundary of positive genus $B'(S)$ (or to give the . One may do this by taking the convex subtree of the cubic tree with boundary $B'(S)$, and inserting a bigon into every edge of this subtree inside of the subgraph spanned by $B(S)$. It is easy to extend the Tait coloring to this graph. If $B'(S)=\emptyset$, but $S$ has finite genus $g$, then insert $g$ bigons into the tree to get a graph whose thickening is the appropriate surface. This preserves bipartite and Tait colorability, and hence the thickened boundary still covers the genus 2 surface. For the non-orientable case, I'll describe the result without going into details. There is a unique closed non-orientable surface $\Sigma= P\#P\#P$ of Euler characteristic -1, Dyck's surface, the connect sum of three projective planes. One may also decompose $\Sigma=P\# T$, the connect sum of a projective plane and a torus (in fact, this is unique up to isotopy). In any case, this manifold bounds an orbifold handlebody, with spine $L$ an orbigraph which is a loop with an edge attached (topologically a tadpole graph), and order 2 cone point at one end (looking like a lollipop). Any cubic graph with a perfect matching is a cover of $L$, where the non-matching edges cover the cycle (non-canonically, after some choice of orientation of each component), and each edge of the perfect matching maps as a 2-fold cover of the single orbifold edge in the tadpole graph by folding in half. Now, any tree has a perfect matching, so choose one for the cubic tree. Take the convex subset of the cubic tree corresponding to $B(S)$, giving a planar graph with the same end space. Let $B'(S)$ denote the subset of non-planar ends, and for each edge in the convex hull of $B'(S)$, add a bigon to each edge (and extending the perfect matching) to get a graph whose tubular neighborhood has the same set of ends and non-planar ends. Then let $B''(S)$ denote the non-orientable ends, and take the convex hull of $B''(S)$ to get the subgraph with non-orientable ends. Then insert a tadpole tail to each unmatched edge in this subgraph. One may see then that a tubular neighborhood (in the orbifold sense, i.e. connect summing with a $P^2$ for each tail/leaf) will give a surface with the correct set of ends, and which covers the Dyck surface (an extra bit of argument is needed for the isolated ends as before). One special case to consider is where $B''(S)=\emptyset$, $B'(S)\neq \emptyset$, but $S$ is non-orientable. In this case, there is a parity depending on the number of crosscaps of subsurfaces $A$ such that $S\backslash A$ is orientable. In this case, we add on $0$ or $1$ tails to get the appropriate parity of crosscaps.<|endoftext|> TITLE: On Applications of Forcing in Domain Theory QUESTION [10 upvotes]: An interesting feature of domain theory is to use partial orders in order to provide a mathematical model for the computational approximation in a potentially infinite computational process (e.g. computing all digits of $\pi$) that lies beyond the power of our finite-resource computers. Roughly speaking, from a domain-theoretic perspective, the pieces of information in a computational process form a partial order. Under certain conditions, an infinite set of compatible pieces of partial information may converge to the ultimate answer beyond the computer's computational power. So the computer will never know the final answer (e.g. all digits of $\pi$) but it deals with its approximations. For instance, at a certain point it finds out that $\pi\in I=[3.14, 3.15]$ and after a few more steps of computing it comes up with the fact that $\pi\in J=[3.141, 3.142]$. So the interval $J$ can be viewed as a piece of information that extends its predecessor $I$ via inverse inclusion order, $J\subseteq I$. This approach may sound quite familiar to the set theorists as it reminds the way they deal with conditions in a forcing notion. The ultimate object is the generic filter (and whatever made out of it) which lives in the forcing extension but not in the ground model. However, people living in the ground model have a degree of access to such a "transcendental" object through its approximations that exist in their world. Based on the presented analogy, one may expect forcing and domain theory to share some similar concepts and techniques. But my search didn't reveal that much along these lines, except a master thesis of Håkon Briseid (under the supervision of Dag Normann), titled "Generic Functions in Scott Domains". Its abstract reads as: In this thesis we will apply forcing to domain theory. When a Scott domain represents a function space, each function will be a filter in the basis of the domain. By using partially ordered basis as the forcing relation, each generic filter $G$ yields a model of $ZFC$ in which $G$ is a function, given some other model of $ZFC$ containing this basis. Such generic functions are the main concern of this thesis. ... My question is about the existence of possibly deeper connections between these two branches of mathematical logic. Question. What are other examples of papers and theses relating set theory and domain theory through applying forcing in domain theoretic theorems and constructions? REPLY [3 votes]: Note that forcing techniques are frequently used in computability theory, e.g., to establish the existence of degrees with certain properties. I refer the reader to Odifreddi's Handbook of Computability Theory Volume 2 chapter XII or his paper series "Forcing and Reducibilities" parts 1 and 2 (part 3 is mostly about set theory) . Or simply search for "1-generic Turing degrees." This has also been used in complexity theory to build sets/languages with certain complexity theoretic properties. Some of these uses could be formalized in domain theory but that, as the other answer points out, the reasons for using domain theory rather than just working more informally as one does in computability theory have to do with the intended applications and most of the applications domain theory is used for are less concerned with the kind of properties of the infinite object one could prove using forcing methods. In particular, forcing arguments can be thought of as applications of (a version of) the Baire Category Theorem. The space F of maximal compatible sets of forcing conditions can be thought of as a topological space with the basic open sets given by the conditions which (on reasonable assumptions) is a Baire space letting us infer that there is an element G of F meeting any countable collection of dense sets of conditions. If we need G to be definable/computable in some fashion we restrict our attention to sufficiently definable dense sets of conditions. In either case what we demonstrate is that it's possible to build an infinitary object that meets some countable collection of conditions when we already know we can meet any finite collection of those conditions. As such a forcing argument is necessarily about establishing some infinitary property of G and not the usual concerns about domain theory involving reasoning about the correctness of the program. For instance, I'm sure one could recast the trivial forcing argument building a language outside of $P$ (force with finite partial functions and considering dense sets that are $P$-time decidable) as an argument in domain theory if one viewed a chain through the partial order from domain theory as determining the language but I just don't see why one would want to bother with the domain theory part.<|endoftext|> TITLE: Asymptotic behavior of an integral depending on an integer QUESTION [10 upvotes]: A friend of mine, obtained a lower bound for the trace norm of matrices described in this question (for the special case $a_{ij} = \pm 1$). That lower bound is $ \frac{f(n)}{2\pi}$ where $$ f(n) := \int_0^\infty \log\left( \frac{(1+t)^n +(1-t)^n}{2} +n(n-1) t(1+t)^{n-2}\right)t^{- 3/2} \ \mathrm{d}t $$ Numerical computaions suggest that $$ f(n) = 4 \pi n + o(n) $$ How to justify it? Moreover, is it possible to obtain a good rate of convergence? REPLY [7 votes]: This is an improvement of my previous post. I claim that $$4\pi n-6\pi1-(n-1)t,$$ which can be verified for $t<1/(n-1)$ and $t\geq 1/(n-1)$ separately, we see that $$f(n)-2\pi(n-2)>\int_0^\infty \log\bigl(1+(n-1)^2t\bigr)\,t^{-3/2}\,dt=2\pi(n-1).$$ Hence the lower bound $f(n)>4\pi n-6\pi$ follows also.<|endoftext|> TITLE: Is there infinitely many prime $p$ such that the normalized trace of Frobenius $\frac{a_p(E)}{2\sqrt{p}}$ is arbitrarily small (but not zero)? QUESTION [5 upvotes]: I just encountered the following problem and failed to find proper reference, could anyone kindly help to give me some illustration? For an elliptic curve $E$ without complex multiplication (just like in the Sato-Tate conjecture), let $a_p(E)$ be the trace of Frobenius, given an arbitrary small positive number $\epsilon$, is there infinitely many prime numbers $p$ such that $|\frac{a_p(E)}{2\sqrt{p}}|<\epsilon$? With the proof of Sato-Tate conjecture, the answer should certainly be yes, but what I am wondering is, without the result of Sato-Tate conjecture, do we still have that conclusion? If it is, could you kindly give me some reference? I already knew the supersingular case by Elkies theorem. With many thanks. REPLY [4 votes]: The answer is that the proof of such theorems do indeed require the full force of Sato-Tate. For example, if you take a general elliptic curve E defined over a number field $F$ which is not CM, then your problem is still open.<|endoftext|> TITLE: Almost isometric linear maps QUESTION [17 upvotes]: Say that a linear map $\varphi : B(\mathcal H) \rightarrow B(\mathcal H)$ is a $\epsilon$-almost isometric if $$ 1 - \epsilon \leq \|\varphi(a)\| \leq 1+\epsilon, \quad \forall a\in B(\mathcal H), \text{s.t. }\|a\|=1, $$ where $B(\mathcal H)$ denotes the set of bounded operators on a Hilbert space $\mathcal{H}$. Does there exist a constant $C$ such that for every $\epsilon>0$ and every $\epsilon$-almost isometric linear map $\phi$, there exists an isometric linear map $\psi : B(\mathcal H) \rightarrow B(\mathcal H)$ such that $\|\varphi - \psi\| < C\epsilon$? If that is too hard or false, what about in finite dimension? REPLY [6 votes]: The answer is yes if you assume $\varphi$ is surjective, and you're only looking for a function $f(\epsilon)$ which tends to zero as $\epsilon$ tends to zero. Let's call a function $\varphi$ satisfying your given condition an $\epsilon$-isometric, linear map. Theorem: For every $\epsilon > 0$ there is a $\delta > 0$ such that for every $\mathcal{H}$ and every linear, surjective, $\delta$-isometric map $\varphi : B(\mathcal{H})\to B(\mathcal{H})$, there is a linear isometry $\psi : B(\mathcal{H})\to B(\mathcal{H})$ such that $\|\varphi - \psi\| \le \epsilon$. Proof: Suppose otherwise and fix $1/k$-isometric, linear, surjective maps $\varphi_k : B(\mathcal{H}_k)\to B(\mathcal{H}_k)$ which remain $\epsilon$-far away from any isometry. Then we may define a map $$ \Phi : \prod B(\mathcal{H}_k) / \bigoplus B(\mathcal{H}_k) \to \prod B(\mathcal{H}_k) / \bigoplus B(\mathcal{H}_k) $$ in the obvious way. Note that $\Phi$ is surjective, linear and isometric. By a theorem of Kadison, then, $\Phi$ has the form $\Phi(x) = u \rho(x)$, where $u$ is a unitary and $\rho$ is a Jordan $^*$-isomorphism. (Recall that a Jordan $^*$-isomorphism is a linear, $^*$-preserving map $\rho$ satisfying $\rho(ab + ba) = \rho(a)\rho(b) + \rho(b)\rho(a)$.) We may find unitaries $U_k\in B(\mathcal{H_k})$ such that the sequence $(U_k)$ represents $u$. Then, multiplying by $u^*$ on the left, we may assume that $\Phi = \rho$. Then $\varphi_k$ is an $\epsilon_k$-approximate Jordan $^*$-isomorphism in the sense that $$ \| \varphi_k(ab + ba) - \varphi_k(a)\varphi_k(b) - \varphi_k(b)\varphi_k(a)\| \le \epsilon_k \|a\|\|b\| $$ where $\epsilon_k\to 0$. By a result of Ilišević and Turnšek, there is an actual Jordan $^*$-isomorphism $\psi_k : B(\mathcal{H}_k)\to B(\mathcal{H}_k)$ such that $\|\psi_k - \varphi_k\|\to 0$, and since Jordan $^*$-isomorphisms are isometries, this is a contradiction.<|endoftext|> TITLE: On the structure of a finite group of order $144$ QUESTION [7 upvotes]: Let $G$ be a finite group of order $144$. Suppose there is an irreducible $\mathbb{C}$-character $\theta$ such that $\theta(1)=9$. QUESTION: Prove $G\cong A_4\times A_4$. By using Magma, we know there is only one group of order $144$ with an irreducible $\mathbb{C}$-character $\theta$ of degree $9$. Now I want to prove this result without using Magma. REPLY [11 votes]: Since $G$ has an irreducible character of degree $9 = |G|_{3},$ we have $O_{3}(G) = 1,$ so $G$ has more than one Sylow $3$-subgroup. If $G$ has only $4$ Sylow $3$-subgroups, then $G$ has a normal subgroup of order divisible by $6$ with a normal Sylow $3$-subgroup ( consider the permutation action of $G$ on $N_{G}(S)$ for $S \in {\rm Syl}_{3}(G)$ in the case $[G:N_{G}(S)] = 4.$ The image in $S_{4}$ has order dividing $24,$ so the kernel of the action has order divisible by $6$, and (being a subgroup of $N_{G}(S)$) has a normal Sylow $3$-subgroup, which remains normal in $G$, contrary to $O_{3}(G) = 1).$ Hence $G$ does indeed have $16$ Sylow $3$-subgroups, and a self-normalizing Sylow $3$-subgroup. By Burnside's normal $p$-complement theorem, $G$ has normal $3$-complement of order $16.$ In particular, $G$ is solvable (we could also see this from Burnside's $p^{a}q^{b}$-theorem, but that seems like overkill). Let $Q = O_{2}(G),$ which is a Sylow $2$-subgroup of $G.$ Since $O_{3}(G) =1,$ we have $Q = F(G).$ Now $G/Q$ acts faithfully as a group of linear transformations on $Q/\Phi(Q),$ so that $G/Q$ is isomorphic to a subgroup of ${\rm GL}(m,2)$ for some $m \leq 4.$ Since $[G:Q]$ is divisible by $9,$ we must have $m =4,$ and $Q$ is elementary Abelian. Since $GL(4,2)$ contains no element of order $9,$ we see that $G/Q$ is elementary Abelian of order $9.$ Further, the action of a Sylow $3$-subgroup of $GL(4,2)$ on the natural module gives the desired structure for $G.$<|endoftext|> TITLE: Prescribing the dimension of intersections of sub-vector spaces QUESTION [6 upvotes]: I asked this question on Mathematics Stackexchange, but got no answer. Let $K$ be a field and $n$ a positive integer. To a finite dimensional $K$-vector space $V$, equipped with a family $V_1,\dots,V_n$ of subspaces, we attach the map $d:\mathcal P(\{1,\dots,n\})\to\mathbb N$ defined by $$ d(S)=\dim\left(\bigcap_{s\in S}V_s\right). $$ [Here $\mathcal P(\{1,\dots,n\})$ denotes the set of subsets of $\{1,\dots,n\}$.] Then we have $(1)\quad S\subset T\implies d(S)\ge d(T)$, $(2)\quad d(S\cap T)\ge d(S)+d(T)-d(S\cup T)$ for all $S,T\in\mathcal P(\{1,\dots,n\})$. [Here $S\subset T$ means that each element of $S$ belongs to $T$.] Conversely, given $d:\mathcal P(\{1,\dots,n\})\to\mathbb N$ satisfying $(1)$ and $(2)$, is there a family $(V,V_1,\dots,V_n)$ as above inducing $d$ in the way just described? The answer is No in general, a counterexample being given as follows: $K=\mathbb F_2, n=4$, $$ d(\varnothing)=2=\dim V,\quad d(\{i\})=1=\dim V_i, $$ and $d(S)=0$ if $S$ has at least two elements. [We use the fact that there are only three lines through the origin in $\mathbb F_2^2$.] There is an obvious analog for any finite field. In view of the above observations, it seems natural to ask: Assume that our field $K$ is infinite, and that $d:\mathcal P(\{1,\dots,n\})\to\mathbb N$ satisfies $(1)$ and $(2)$. Is there a family $(V,V_1,\dots,V_n)$ as above such that $$ d(S)=\dim\left(\bigcap_{s\in S}V_s\right) $$ for all $S\ ?$ REPLY [9 votes]: I find this easier to think about if we dualize everything: replace $d(S)$ with $\dim V - d(S)$ and each subspace with its annihilator in $V'$. Recall that the annihilator of a sum is the intersection of the annihilators. Thus your question is the following: Assume $d:\mathcal{P}(\{1,\dots,n\}) \to \mathbb{N}$ is increasing and submodular (this is just the cool name for the condition $d(S \cup T) + d(S\cap T) \leq d(S) + d(T)$) and satisfies $d(\emptyset)=0$ and $d(\{1,\dots,n\}) \leq \dim V$. Then does there exists a collection of subspaces $V_i$ of $V$ such that $\dim \sum_{i \in S} V_i = d(S)$ for each $S \subset \{1,\dots,n\}$? If we further assume that $d(\{x\}) \leq 1$ for each $x$ then this is exactly one of the equivalent definitions of a matroid, and your question is whether every matroid is linear. Apparently the answer is no.<|endoftext|> TITLE: Thomason-Trobaugh Theorem QUESTION [6 upvotes]: Let $X$ be a scheme and $U$ be an open subscheme. The proof of the Thomason-Trobaugh Theorem implies that under some mild assumptions, for any perfect complex $F$ on $U$, we have that $F\oplus F[1]$ can be extended to a perfect complex on $X$. I'm just wondering whether there exists examples where $F$ is a perfect complex on $U$ but $F$ itself cannot be extended to $X$? I've found an example when $X$ is the cone $xy-z^{2}=0$ and $U$ is the complement of the origin. Is there an example for smooth $X$? REPLY [5 votes]: All perfect complexes on an open $U$ of a smooth variety $X$ extend to perfect complexes on $X$. In fact, I believe a much more general statement is true: For an open substack $U$ of a separated Noetherian algebraic stack $X$, all bounded complexes of coherent sheaves on $U$ extend to bounded complexes of coherent sheaves on $X$. Since for smooth varieties, bounded complexes of coherent sheaves are all quasi-isomorphic to a perfect complex, this implies my first claim. For a single coherent sheaf, this is Corollary 15.5 in Champs Algébriques, Laumon-Moret-Bailly (see also 27.22 in the Stacks Project). Here is a sketch of a proof for a general complex (for simplicity, take $f:U\rightarrow X$ an open immersion of varieties). The key claims are: Every quasi-coherent sheaf $\mathcal{F}$ on $X$ is an filtered colimit of coherent subsheaves $\mathcal{F}_i,$ $i$ in some index set. and If $\mathcal{F}$ and $\mathcal{F}_i$ are as in 1), then for any coherent sheaf $\mathcal{G}$, a map $\mathcal{G}\rightarrow\mathcal{F}$ factors through one of the $\mathcal{F}_i.$ Given these two claims, the proof proceeds as follows. Take a bounded complex $C$ of coherent sheaves on $U$. Denote the (non-derived) pushforward of $C$ along $f$ by $f_*C$. We will find a subcomplex $C'$ of $f_*C$ of coherent sheaves, whose pullback along $f$ still gives the original complex $C$. This subcomplex will be constructed one term at a time, starting from the left. So assume that we have around found $C'$ with $C'_i\subseteq f_*C$ coherent for $i<0$ and $f^*C'\cong C.$ Now we need to refine our subcomplex to a subcomplex $C''$ with $C''_0$ coherent. To do this, express $C'_0$ as a filtered colimit of coherent subsheaves $C'_{0,i}$ using Claim 1. By Claim 2, the map $C_0\rightarrow f^*C'_0$ factors through some $f^*C'_{0,i}$. Also by Claim 2, the map $C'_{-1}\rightarrow C'_0$ factors through some $C'_{0,j}.$ Because our colimit is filtered, we can take some $C'_{0,k}$ which contains both $C'_{0,i}$ and $C'_{0,j}.$ Then take $C''_i$ to be $C'_i$ for $i\neq 0$ and to be $C'_{0,k}$ for $i=0.$ Iterating this procedure gives us the desired bounded complex of coherent sheaves on $X$. Claim 1 turns out be more annoying to write out a formal proof for than I thought... so I'll just refer you to 27.22.3 in the Stacks Project. Here is a proof of Claim 2 though: There exist a finite collection of sections $s_i\in\mathcal{G}(U_i),$ $U_i$ an affine open, that together generate $\mathcal{G}$ as a $\mathcal{O}_X$-module. The image of each of these sections is then contained in some $\mathcal{F}_i$, and so by filteredness there will be some $\mathcal{F}_i$ that contains the images of all the $s_i$. The map $\mathcal{G}\rightarrow\mathcal{F}$ will factor through this $\mathcal{F}_i$.<|endoftext|> TITLE: Binomial coefficients in discrete valuation rings QUESTION [6 upvotes]: Let $V$ be a complete discrete valuation ring whose residue field is a finite field $k=\mathbf{F}_q$. Let $\pi\in V$ be a uniformizer. For any integer $d,n\ge 0$, define: $${\pi^d \choose n} := \frac{\pi^d\cdot(\pi^d -1)\cdot\ldots\cdot(\pi^d-n+1)}{n!}.$$ Is ${\pi^d\choose n}$ an element of $V$? For exactly what integers $n\ge 0$ is ${\pi^d\choose n}$ a unit? Example. If $V = \mathbf{Z}_p$, $\pi = p$, then ${p^d\choose n}$ is zero unless $0\le n\le p^d$, and its $p$-adic valuation is $d-v_p(n)\ge 0$ for $1\le n\le p^d$. In all cases, the answer to the first question is yes, and the answer to the second question is: for $n = 0, c\cdot p^d$, with $(c,p) = 1$. REPLY [6 votes]: If $p=2=n$, $d=1$, $V=\mathbf{Z}_2[\sqrt{2}]$ and $\pi=\sqrt{2}$, then $$ \binom{\pi^d}{n} = \frac{\sqrt{2}(\sqrt{2} - 1)}{2} = 2^{-1/2}\cdot \mathrm{unit} \notin V. $$<|endoftext|> TITLE: An ellipse through 12 points related to Golden ratio QUESTION [17 upvotes]: I am looking for a proof of the problem as follows: Let $ABC$ be a triangle, let points $D$, $E$ be chosen on $BC$, points $F$, $G$ be chosen on $CA$, points $H$, $I$ be chosen on $AB$, such that $IF$, $GD$, $EH$ parallel to $BC$, $CA$, $AB$ respectively. Denote $A'=DG \cap EH$, $B'=FI \cap GD$, $C'=HE \cap IF$, then 12 points: $D$, $E$, $F$, $G$, $H$, $I$ and midpoints of $AB'$, $AC'$, $BC'$, $BA'$, $CA'$, $CB'$ lie on an ellipse if only if $$\frac{\overline{BC}}{\overline{BE}}=\frac{\overline{CB}}{\overline{CD}}=\frac{\overline{CA}}{\overline{CG}}=\frac{\overline{AC}}{\overline{AF}}=\frac{\overline{AB}}{\overline{AI}}=\frac{\overline{BA}}{\overline{BH}}=\frac{\sqrt{5}+1}{2}$$ Note: This problem don't appear in AMM, and I don't have a solution for this problem, but there is the same configuration appear in: Problem 12050 in AMM UPDATE 07-Dec-2020: Let $X$ is the centroid of $\triangle ABC$ and $X_1, X_2, X_3$ be three points lie on $XA, XB, XC$ such that: $$\frac{XX_1}{XA}= \frac{XX_2}{XB}= \frac{XX_3}{XC}=\sqrt{2}\Phi^2$$ Where $\Phi=\frac{\sqrt{5}-1}{2}$ Amazingly, three points $X_1$, $X_2$, $X_3$ also lie on the Ellipse. So we should call the ellipse is: 15 Point Golden Ellipse. REPLY [8 votes]: Using an affine tranformation mapping the ellipse into a circle we can and do assume that the configuration is realized using a circle $(\Gamma)$ (more particularly than an ellipse). The proportionalities are conserved. (The triangle remains so far general, constrained to admit the given configuration of $12$ points on the same circle.) The following arguments are written not with the intention of finding the shortest path to the solution, but finding the most "symmetrical path". Let us now fix notations in the posted picture, that allow to name "things" precisely. We denote by $A''_C$, $A''_B$ the mid line in the triangle $\Delta AB'C'$, and use similar notations $B''_A,B''_C$, and $C''_A,C''_B$ for the other points obtained by permuting letters. (Further pictures with additional points will follow, one can use them to visualize the position of these six points introduced so far.) Because of $$ A''_BA''_C \ \|\ B'C' \ \|\ BC \ \|\ B''_CC''_B \ \|\ B''_AC''_A \ , $$ The corresponding trapezes, inscribed in the circle $(\Gamma)$ are isosceles, that have two different bases among the corresponding parallel segments, which are chords on the circle, $$ A''_BA''_C\ ,\ IF\ ,\ B''_CC''_B\ ,\ B''_AC''_A\ . $$ (So they all have a common line segment bisector.) Since the trapez with bases $B''_AC''_A$ and $B''_CC''_B$ is isosceles, we obtain $B''_AB''_C=C''_AC''_B$, so $A'C'=A'B$, so by symmetry we have the congruence of segments $$ A'B'=B'C'=C'A'\ . $$ We proceed similarly with the bases $A''_BA''_C$ and $B''_CC''_B$ to obtain the first equality in $$ A''_BC''_B=A''_CB''_C=B''_AC''_A\ , $$ the second one by symmetry, so taking the doubles $$ AB=AC=BC\ . $$ So the triangles $\Delta ABC$ and $\Delta A'B'C'$ are equilateral, and have parallel sides. Then there exist the intersection $X=AA'\cap BB'\cap CC'$. The points $I,F$ are on the circle $(\Gamma)$, as $A''_C,A''_B;B''_C,C''_B;A''_B,A''_C$, so the line segment bisector of it, the same as for $BC$, $\mu_A$ say, contains $A$ and the center of $(\Gamma)$. So $A'$ is on $\mu_A$. So $AA'=\mu_A$, $BB'=\mu_B$, $CC'=\mu_C$ are the side line bisectors, so $X$ is the center of symmetry of the equilateral triangle $ABC$. And of $\Delta A'B'C'$. With this choice, it is clear that all six points $A''_C,A''_B;B''_C,C''_B;A''_B,A''_C$ are on a circle. We have to introduce the constraint on $I$ (and $F$) being on the same circle to finish. The adjusted picture, with a "truly central" $X$ is: We have inserted also the points $S,T$, the mid points of $A''_BA''_C$, and of $IF$ and $B'C'$. Fix the unit to be equal to the length of the side of $\Delta ABC$. Let $x$ be the length of $AI$. Then: $$ IT=\frac 12 x\ ,\ AT = \frac{\sqrt3}2x\ ,\ ST=\frac 12AT = \frac{\sqrt3}4x\ ,\ AX=\frac 23\cdot\frac {\sqrt 3}2\ ,\ $$ $$ TX=\frac {\sqrt 3}2\left(\frac 23-x\right)\ ,\ TC=\sqrt 3TX=\frac 32\left(\frac 23-x\right)\ ,\ $$ $$ SA''_C=\frac 12 TC=\frac 34\left(\frac 23-x\right)\ ,\ $$ From $XT^2+TI^2 =XI^2=XA''_C{}^2=(XT+TS)^2+A''_CS^2$ we get the equation: $$ \frac 34\left(\frac 23-x\right)^2 + \frac14 x^2 = \frac 34\left(\frac 23-\frac x2\right)^2 +\frac 9{16}\left(\frac 23-x\right)^2\ . $$ The computer solves now, using sage: sage: var('x'); sage: eq = 3/4*(2/3-x)^2 + x^2/4 - 3/4*(2/3-x/2)^2 - 9/16*(2/3-x)^2 sage: eq.factor() 1/4*x^2 + 1/4*x - 1/4 sage: solve( eq==0, x ) [x == -1/2*sqrt(5) - 1/2, x == 1/2*sqrt(5) - 1/2] and we consider only the positive solution $x=\frac 12(\sqrt 5-1)$ which corresponds to the value in the posted question. All the above shows one direction, and moreover the fact that in case $(*)$ of a circle configuration we have a perspectivity of two equilateral triangles $\Delta ABC$ and $\Delta A'B'C'$, centered in $X$, the common symmetry center of the triangles. For the reverse direction, we start with this only possible case $(*)$, use the same calculus, starting from the known value of $x$.<|endoftext|> TITLE: On $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$ QUESTION [7 upvotes]: Euler's totient function $\varphi$ is multiplicative, and it plays important roles in number theory. QUESTION: Is it true that for each integer $m>6$ we have $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$ for some positive integer $n$? For every $m=7,\ldots,10^4$, I have found the smallest positive integer $n$ such that $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$ (cf. http://oeis.org/A248007). For example, for $m=10$ the least positive integer $n$ with $\varphi(m)\varphi(n)\equiv0\pmod{m+n}$ is $14$; in fact, $$\varphi(10)\varphi(14)=4\times 6\equiv0\pmod{10+14}.$$ I formulated the above question in 2014 and conjectured that the answer should be positive. I have ever mentioned this question to some number theorists, but there is no substantial progress on the question. I don't think that the question is very difficult. Any ideas towards its solution? REPLY [2 votes]: If there are counterexamples, they are likely to be square free and very large. Here's why. Suppose $ t^2 $ divides $ m$ and $ t$ is larger than $O( \log p ),$ with $ p$ the largest prime divisor of $m$ (it is enough that $ t$ is larger than $ m/\phi(m) )$. Then $n= t\phi(m) - m$ is a positive multiple of $t^2$, so $t$ divides $ \phi(n)$, and $n,m$ then solves the desired congruence. One can fine tune this to smaller $t$ for certain $m$, but the point now is to look at solving the problem for a given and mostly squarefree $m$. Edit 2018.06.27 GRP A couple more thoughts. Let $t$ be a common factor of $m$ and $\phi(m)$. If $n=\phi(t)\phi(m) - m$ is positive, this is another solution which works. Otherwise, $m$ and $\phi(m)$ are mostly coprime, and we seek an integer $t$ such that $n$ above is positive and so that $n$ is a multiple of $t$. (More generally, for $d$ a large divisor of $\phi(m)$ we could instead look at $n=\phi(t)d -m$ being a multiple of $t$, but I won't do that now.) By running through $t$ (and thus $\phi(t)$) being 3-smooth numbers, we may encounter an $n$ having enough prime factors of the form 6k+1 distinct from factors of $m$ and factors $\phi(m)$ that will yield a solution. As remarked by Stanley above, it is not clear how to guarantee enough of these factors of $n$ to appear. END Edit 2018.06.27 GRP Gerhard "Sure Euler Could Solve It" Paseman, 2018.06.24.<|endoftext|> TITLE: Is the category 2-Vect monoidal closed? QUESTION [9 upvotes]: Kapranov and Voevodsky introduced the following 2-category $2{\rm Vect}$ of "2-vector spaces" over a field $K$: 0-cells are natural numbers $\langle N\rangle,\langle M\rangle...$ 1-cells $\langle N\rangle\to \langle M\rangle $ are strictly-positive-integer valued matrices of size $M\times N$ 2-cells $A\to B$ are matrices whose entries are matrices, whose $(i,j)$ entry is a matrix having size $A_{ij}\times B_{ij}$. So for example, this is a 2-cell $\langle 2\rangle\to \langle 2\rangle$ where all the entries in $\alpha$ belong to the field $K$. Since this is the best of possible worlds, this categorifies correctly (at least part of) the idea of a vector space as a "set of tuples made of scalars $a_i\in K$". In fact, this is a way to make the following intuitive idea work: just as the category of vector spaces is equivalent to the category of natural numbers, where $\hom(n,m)$ is the space of $m\times n$ $K$-valued matrices, so the 2-category of 2-vector spaces is equivalent to the 2-category of natural numbers, where $\hom(n,m)$ is the category of $K$-linear functors ${\rm Vect}^m \to {\rm Vect}^n$. Now. The 2-category of 2-vector spaces has two monoidal structures $\boxtimes$ and $\boxplus$ where $\langle N\rangle\boxtimes \langle M\rangle = \langle N\cdot M\rangle$; the integer-valued matrix $A\boxtimes B$ has in its $(ik,jl)$ entry the integer $A_{ik}B_{jl}$; given 2-cells $\alpha,\beta$ the 2-cell $\alpha\boxtimes\beta$ is the entry-wise tensor product of matrices. and $\boxplus$ (the monoidal sum) is defined similarly, so that $\langle N\rangle ⊞ \langle M\rangle = \langle N+M\rangle$, $A ⊞ B$ is the direct sum of matrices, and $\alpha ⊞ \beta$ is the entry-wise direct sum of matrices (in fact $2{\rm Vect}$ a rig category with respect to $\boxtimes$ and $\boxplus$). Question: Is the $\boxtimes$-structure also closed? Representing the category $2{\rm Vect}(\langle mn\rangle,\langle p\rangle)$ as the category of linear functors $\underbrace{{\rm Vect}^n \times\dots\times{\rm Vect}^n}_{m \text{ times}} \to {\rm Vect}^p$ does not seem to help, and before going deep down the computations I would like to know if there is a conceptual reason for this category to (not) be closed. REPLY [8 votes]: As Tim says, for a conceptual ("coordinate-independent") picture we should start with $\text{Bim}(k)$, the 2-category of $k$-algebras, $k$-bimodules between them, and $k$-bimodule homomorphisms, which we can think of as a reasonably well-behaved 2-category of $\text{Mod}(k)$-modules. By Eilenberg-Watts we can equivalently think of this 2-category as the 2-category whose objects are categories of the form $\text{Mod}(A)$, $A$ a $k$-algebra, and whose morphisms are $k$-linear cocontinuous functors. This 2-category is closed monoidal with monoidal structure given by tensor product, which has a universal property: cocontinuous $k$-linear functors out of $\text{Mod}(A \otimes_k B)$ correspond to $k$-linear functors out of $\text{Mod}(A) \times \text{Mod}(B)$ which are cocontinuous in each variable. Every object is dualizable, with dual given by $A \mapsto A^{op}$, and hence the closed structure is given by $[A, B] = A^{op} \otimes_k B$. This reflects the fact that $\text{Mod}(A^{op} \otimes_k B)$ is precisely $(A, B)$-bimodules over $k$, which are precisely the morphisms $A \to B$ in this 2-category. KV considers the sub-2-category of this consisting of $k$-algebras of the form $k^n$ and finitely generated projective $(k^n, k^m)$-bimodules over $k$, which are described explicitly by $n \times m$ matrices of finite-dimensional $k$-vector spaces. These morphisms are singled out by the property of also being dualizable (that is, they have adjoints; see this blog post), and these $k$-algebras are 2-dualizable in addition to being dualizable, but this does not exhaust all 2-dualizable objects. In general the 2-dualizable objects are the $k$-algebras $A$ which are finite-dimensional and separable over $k$ (and this definition, unlike the KV definition, should satisfy Galois descent); the algebras $k^n$ exhaust these iff $k$ is separably closed. These can be used to define 2-dimensional topological field theories via the cobordism hypothesis. For example, if we take $A$ to be a twisted group algebra for a finite group $G$ with twist defined by a $2$-cocycle $\alpha \in H^2(G, k^{\times})$, then the resulting TFT is 2d Dijkgraaf-Witten theory with gauge group $G$ twisted by $\alpha$.<|endoftext|> TITLE: Divergence of Groups and Metric Spaces QUESTION [9 upvotes]: Several papers, including this and this claim that divergence of finitely generated groups and metric spaces have been introduced by Misha Gromov in his paper "Asymptotic invariants of infinite groups". What is the precise reference and how did Gromov call this invariant? REPLY [4 votes]: Here is (what I believe is) the relevant paragraph of the second "this", of which I am one of the co-authors. This paragraph contains a reference to a particular passage from Gromov's Asymptotic Invariants paper including a citation to the particular paragraph in that paper where the relevant discussion occurs. In symmetric spaces of non-compact type, the order of the divergence of geodesic rays is either exponential (when the rank is one) or linear (when the rank is at least two). This inspired an initial thought that in the presence of non-positive curvature the divergence must be either linear or exponential. See [Gro93] for a discussion—an explicit statement of this conjecture appears in 6.B2, subsection “Geometry of ∂T and Morse landscape at infinity,” Example (h). In particular, Gromov stated an expectation that all pairs of geodesic rays in the universal cover of a closed Riemannian manifold of non-positive curvature diverge either linearly or exponentially [Gro93].<|endoftext|> TITLE: Young tableaux for exceptional Lie algebras QUESTION [9 upvotes]: Irreducible representations for the $A$-series Lie algebras are labelled Young diagrams, with a basis of each given by Young tableaux. Moreover, analogues exist for the $B,C$, and $D$ series. Does such a description exist for the exceptional Lie algebras $$ \frak{g}_2 \subseteq \frak{f_4} \subseteq\frak{e}_6 \subseteq\frak{e}_7 \subseteq\frak{e}_8? $$ REPLY [4 votes]: Young diagrams for the exceptional Lie algebras are considered in the book https://press.princeton.edu/titles/8839.html (Group Theory: Birdtracks, Lie's, and Exceptional Groups, by Predrag Cvitanovic).<|endoftext|> TITLE: An explicit reconstruction of a matrix from its minors QUESTION [17 upvotes]: $\newcommand{\End}{\operatorname{End}}$ $\newcommand{\GL}{\operatorname{GL}}$ $\newcommand{\Cof}{\operatorname{cof}}$ Let $V$ be a $d$-dimensional real vector space. ($d \ge 4$). Fix an odd integer $2 \le k \le d-2$. Define $H_{>k}=\{ A \in \End(V) \mid \operatorname{rank}(A) > k \}$. Consider the map $$ \psi:H_{>k} \to \End(\bigwedge^k V) \, \,, \, \, \psi(A)=\bigwedge^{k}A. $$ $\psi$ is a smooth injective immersion. Question: Is there an explicit formula for $\psi^{-1}$? Comments: Since $\operatorname{rank}(\bigwedge^kA) = \binom {\operatorname{rank}(A)}{k} $, we might start by considering each rank separately. The limitation $k$ is odd is not essential. When $k$ is even, $\psi(A)=\psi(B)$ implies $A=\pm B$ (assuming $A,B \in H_{>k}$), so the inverse is well-defined up to a sign. In fact, the following more general property holds: Let $V$ be a vector space over an arbitrary field $F$. Then, for $A,B \in H_{>k}$, $\psi(A)=\psi(B)$ iff $A=\lambda B$ when $\lambda^k=1$. So, we can ask the question in this more general setting: Is there an explicit formula, which given an element in $\text{Image }(\psi)$, produces a source element? Part of the problem is that we do not have a closed-form description of $\text{Image }(\psi)$, and it is clear that in general we don't expect $\psi^{-1} $ to have a continuous extension to all $\End(\bigwedge^k V)$ (or even to the space of all endomorphisms of rank bigger than $k$, which is where $\text{Image }(\psi)$ lies, but I am not sure about that.) I am OK with a formula which uses an inner product and orientation structures. Even though we don't need them in order to define $\psi$, they somehow appear naturally when trying to compute $\psi^{-1}$. Indeed, in the special case $k=d-1$, $\psi(A) \in \text{GL}(\bigwedge^{d-1}V)$ can be identified with the cofactor matrix of $A$. (The identification is done using the Hodge dual $\star:\bigwedge^{d-1}V \to V$). Then we have $ \Cof(\Cof A)=(\det A)^{d-2}A$. Now, if $\Cof A=B$, then $\Cof B=(\det A)^{d-2}A$, and $\det(B)=(\det(A))^{d-1}$. Since $k=d-1$ is odd, we can take the (unique) $ d-1 $-th root, so $(\det(B))^{\frac{1}{d-1}}=\det(A)$. Thus, $$ A=(\Cof)^{-1} B=(\det B)^{\frac{2-d}{d-1}} \Cof B $$ gives the formula for $\psi^{-1}$. Trying to generalize this derivation using the Hodge dual for general $k$ seems to hit a wall. REPLY [7 votes]: Assume for simplicity that $kl=d+1$. Then I claim that $V \otimes \det V$ appears as a summand of $\left( \bigwedge^k V\right)^{\otimes l}$ with multiplicity $l-1$. The maps $\left( \bigwedge^k V\right)^{\otimes l} \to V \otimes \det V$ are easier to write down - the $i$'th maps sends $$ (v_{1,1} \wedge \dots \wedge v_{1,k}) \otimes \dots \otimes (v_{l,1}\wedge \dots \wedge v_{l,k})$$ to $$\sum_{j=1}^{k} (-1)^{ il+j} v_{i,j} \otimes (v_{1,1} \wedge \dots \wedge v_{1,k} \wedge v_{2,1} \dots \wedge v_{i,j-1} \wedge v_{i,j+1} \wedge \dots \wedge v_{l,k})$$ and we have the relation that the $l$ maps sum to zero. The reverse maps are similar, except that the $i$th map is a sum over all the ways to divide the $kl$ vectors evenly into $l$ boxes, with the first vector in the $i$'th box. Now if we have an endomorphism of $\wedge^k V$, we can let it act on $\left( \bigwedge^k V \right)^{\otimes l}$, compose with one of these maps on each side, and obtain an endomorphism of $V$. If we apply this to an element of $GL(V)$, we get the individual element back, times its determinant, times some fixed scalar depending on the choice of map. We can divide by the scalar and by the $\frac{1}{ {d-1 \choose k-1}}$ power of the determinant of the endomorphism of $\bigwedge^k V$ to obtain the original matrix exactly.<|endoftext|> TITLE: A permutation problem QUESTION [7 upvotes]: Here I ask a question on permutations of $n$ distinct real numbers. QUESTION: Let $a_1,a_2,\ldots,a_n\ (n>1)$ be (pairwise) distinct real numbers. Is there a permutation $b_1,\ldots,b_n$ of $a_1,\ldots,a_n$ with $b_1 = a_1$ such that the $n−1$ numbers $$|b_1 −b_2|,\ |b_2 −b_3|,\ \ldots,\ |b_{n−1} −b_n|$$ are pairwise distinct ? Actually, I formulated this question in 2013 and conjectured that the answer should be positive. If $a_1$ is the smallest (or largest) number among $a_1,\ldots,a_n$, then it is easy to construct a desired permutation $(b_1,\ldots,b_n)$. In fact, when $a_1 1$ distinct real numbers, and let $T$ be any tree with vertices $v_1,\ldots,v_n$, where $v_1$ is a leaf (i.e., $\deg_T(v_1) = 1$). Then there is a bijection $f$ from the vertex set $V(T)$ of the tree $T$ to $A = \{a_1,\ldots,a_n\}$ with $f (v_1) = a_1$ such that those $|f (v_i)−f (v_j)|$ with $v_i$ and $v_j$ adjacent are pairwise distinct. Any ideas towards the final solution of my question? Comments on the above general conjecture are also welcome! REPLY [11 votes]: A simple brute-force of trees yields a counter-example with $n = 7$. Let $\{a_1, \ldots, a_7\} = [7]$, $T$ be a path of four vertices $v_1, v_2, v_3, v_4$, with $v_4$ adjacent to $v_5, v_6, v_7$. If we fix $f(v_1) = 4$, the conjecture is failed. I have not come up with a simple proof yet (other than trying all options of $f$), may update later. UPD: ok, here is the proof. Consider all options of $f(v_2)$: $f(v_2) = 3$ ($5$ is symmetrical). Note that in the set $\{1, 2, 5, 6, 7\}$ each element has another one with absolute difference $1$, hence for all options of $f(v_4)$ there is an index $j \in \{3, 5, 6, 7\}$ such that $|f(v_4) - f(v_j)| = |f(v_1) - f(v_2)| = 1$. $f(v_2) = 2$ ($6$ is symmetrical). The only element of $\{1, 3, 5, 6, 7\}$ not at distance $2$ from another element is $6$. But if $f(v_4) = 6$, there are distinct $i, j \in \{3, 5, 6, 7\}$ with $f(v_i) = 5$ and $f(v_j) = 7$, hence $|f(v_4) - f(v_i)| = |f(v_4) - f(v_j)| = 1$. $f(v_2) = 1$ ($7$ is symmetrical). The only element of $\{2, 3, 5, 6, 7\}$ not at distance $3$ from another element is $7$. Let $f(v_4) = 7$ and $f(v_3) = x$, but then $f(v_i) = 8 - x$ for a certain $i \in \{5, 6, 7\}$, and $|f(v_2) - f(v_3)| = |f(v_4) - f(v_i)| = x - 1$.<|endoftext|> TITLE: $\pi((n+1)^2)-\pi(n^2) \le \pi(n)$ for all $n \ge 370$? QUESTION [6 upvotes]: There are some conjectures of the form: There always exist at least $X$ prime numbers between $A$ and $B$. Examples: Bertrand's postulate: for every $n>1$ there is always at least one prime $p$ such that $n < p < 2n$. Legendre's conjecture: there is a prime number between $n^2$ and $(n + 1)^2$ for every positive integer $n$. Brocard's conjecture: there are at least four prime numbers between $p_{n}^2$ and $p_{n+1}^2$. Oppermann's conjecture: there is at least one prime number between $n(n-1)$ and $n^2$. If we denote by $\pi(x)$ the prime-counting function we can rewrite the above conjectures in the following form: Bertrand's postulate: $\pi(2n)-\pi(n) \ge 1$ for $n>1$ Legendre's conjecture: $\pi(n+1)^2)-\pi(n^2) \ge 1$ Brocard's conjecture: $\pi(p_{n+1}^2)-\pi(p_{n}^2) \ge 4$ Oppermann's conjecture: $\pi(n^2)-\pi(n(n-1)) \ge 1$ I computed and saw that $f(n) = \pi(n^2)+\pi(n)+2-\pi((n+1)^2)$ is increasing when $n$ increasing and $f(n)\ge 0$ for all $n=1, 2, \dots, 18700$ (equivalent to $n^2=1, 4, 25 \cdots , 3.5\times 10^8)$. Graph of $(n,f(n))$ where $f(n) = \pi(n^2)+\pi(n)-\pi((n+1)^2); \; 370 \le n \le 1.1\times10^4$ So I proposed two conjecture as follows: Conjecture 1: For every positive integer $n$, the number of primes between $n^2$ and $(n + 1)^2$ is less than the number of primes between $1$ and $n$ add $2$: $$\pi((n+1)^2)-\pi(n^2) \le \pi(n)+2.$$ Conjecture 2: For every positive integer $n$ greater than $369$, the number of primes between $n^2$ and $(n + 1)^2$ is less than the number of primes between $1$ and $n$: $$\pi((n+1)^2)-\pi(n^2) \le \pi(n).$$ Could you give a remark, comment, reference, or proof? Noting that if the conjecture is true, it is stronger than a special case of the Second Hardy–Littlewood conjecture but this conjecture is not contradictory with the K-Tuple conjecture. PS: In my computation I see that: $$\lim_{n \to +large } \frac{\pi((n+1)^2)-\pi(n^2)}{\pi(n)}=1$$ What do You think with this equality? REPLY [16 votes]: It is a folklore conjecture that for $y\le x$ one has $$ \pi(x+y) -\pi(x) = \int_{x}^{x+y} \frac{dt}{\log t} + O(y^{\frac 12} x^{\epsilon}). $$ This is only relevant for $y \ge x^{\epsilon}$, and is stronger than RH. In the case of primes in progressions, such a conjecture may be attributed to Montgomery. Probabilistic considerations might suggest a stronger error term like $O(y^{\frac 12} \log x)$, but this is known to be false thanks to the work of Maier. But the conjecture as stated above is widely believed. See, The distribution of prime numbers for example for a discussion of this and related work (in particular the discussion around (3.7) there). See also Montgomery and Soundararajan where refined asymptotics for moments of primes in short intervals are made and justified heuristically; these conjectures state that primes in short intervals have an appropriate Gaussian distribution, and imply the conjecture above. Of course we are very far from the conjecture mentioned above, but if true it implies for large $n$ that $$ \pi((n+1)^2) - \pi(n^2) = \int_{n^2}^{(n+1)^2} \frac{dt}{\log t} +O(n^{\frac 12+\epsilon}) = \frac{n}{\log n} + O(n^{\frac 12+\epsilon}), $$ since $\log t = 2\log n + O(1/n)$ throughout the interval $n^2 \le t\le (n+1)^2$. Now we know that $\pi(n)$ is asymptoically $\text{li}(n)$ which has the asymptotic expansion $n/\log n + n/(\log n)^2 +\ldots$. The secondary term of $n/(\log n)^2$ dominates the error term in the conjectured asymptotic, and so one should certainly expect that for large $n$ one has $$ \pi((n+1)^2) - \pi(n^2) \le \pi(n). $$ An analogous question formulated for $\pi((n+1)^3) - \pi(n^3)$ is known, as mentioned by GH from MO in the comments, by Huxley's version of the prime number theorem in short intervals. Now the question is only asking for an upper bound on $\pi((n+1)^2) -\pi(n^2)$, and one does have unconditional upper bounds by sieves. The Brun-Titchmarsh theorem would give a bound like $4\pi (n)$ for this quantity, and one can do somewhat better than this. The best result that I know is due to Iwaniec from whose work (see Theorem 14 there) it follows that $$ \pi((n+1)^2) - \pi(n^2) \le \Big( \frac{36}{11}+ o(1)\Big) \frac{n}{\log n}. $$<|endoftext|> TITLE: The Gauss Circle Problem asymptotic in dimension QUESTION [5 upvotes]: The circle problem in $k$ dimensions: "For $n>0$, how many points $z\in \ \mathbb{Z}^k$ have $\|z\|^2\leq n$?" For large $n$, the answer is $\approx n^{k/2}\cdot \operatorname{Vol}(B^k(0,1))+\Omega(n^{k/2-1})$. This approximation is very bad in limit with dimension. For example, if $n$ grows linearly with $k$ then the approximation term dominates the volume term. Do results exist for this regime, or at least for $n$ small relative to $k$? REPLY [5 votes]: This problem was studied by J.E. Mazo and A.M. Odlyzko in Lattice Points in High-Dimensional Spheres. If $n$ grows with $k$ faster than linearly then the volume asymptotic still applies. If it increases slower than linearly then for most locations of the center of the circle there will be no lattice points inside it. The linear increase mentioned in the OP is a borderline case. For $n=\alpha k$ the asymptotics of the number of points $N$ inside the circle is $$\lim_{k\rightarrow\infty}N=e^{ck},$$ with $c$ a coefficient of order unity that depends on $\alpha$ and on the location of the center of the circle. (For a circle centered at the origin and $\alpha=1$ the value is $c=1.418938538$.)<|endoftext|> TITLE: Calculation of the Schur multiplier of $\mathbb Z^2$ QUESTION [6 upvotes]: Consider a projective representation of $\mathbb Z^2$ with $U(1)$ coefficients. I would like to find the covering group corresponding to this representation. For this, one needs to find the appropriate central extension by $\mathbb Z^2$ of the Schur multiplier $H^2(\mathbb Z^2, C^\times)$. I have some confusion in calculating this object. If I use the definition of the Schur multiplier given in terms of the homology group $H_2(\mathbb Z^2,\mathbb Z)$, I obtain the group $\mathbb Z$. This is in contradiction to the result $H^2(\mathbb Z^2, C^\times) = U(1)$ that I have seen quoted elsewhere, and which I expect to obtain on certain physical grounds (there is a separate physical interpretation of this problem). I would like to know which result is correct. REPLY [7 votes]: I think that we have $$H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z}) \simeq \mathbf{Z}, \quad H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})\simeq \mathbf{C}^{\times}.$$ In general, by the Universal Coefficient Theorem there is an isomorphism $$H^2(G, \, \mathbf{C}^{\times})=H_2(G, \, \mathbf{Z})^*$$ Now, it can happen that an infinite group is not isomorphic to its dual. For instance, in your case we have $$\mathbf{C}^{\times}=H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})=H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z})^*=\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}).$$ The isomorphism $\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}) \simeq \mathbf{C}^{\times}$ is explicitly given by associating to every element $h \in \mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times})$ its value in $1$, namely $h \mapsto h(1)$.<|endoftext|> TITLE: Reference request: Coxeter length and irreducible characters QUESTION [8 upvotes]: Let $S_n$ be the symmetric group on $\{1,2,\ldots, n\}$ and $\ell$ the Coxeter length on $S_n$. There is a well-known formula to compute this length, namely for a $\pi \in S_n$ we have $$\ell(\pi) = |\{1 \leq i < j \leq n \mid \pi(i) > \pi(j)\}|.$$ Furthermore, we know that $$\sum_{\pi \in S_n} sign(\pi) \ell(\pi) = 0$$ for all $n \geq 4$. So I was wondering: What about other characters or class functions? Given a class function $f \in Cl_{\mathbb{Q}}(S_n)$, what can we say about $$\sum_{\pi \in S_n} f(\pi)\ell(\pi)\,\,\, ?$$ What about the case where $f$ is an irreducible character? Given that the length is already known for a long time, I was rather sure that this problem has already been solved. However, I couldn't even find $$\sum_{\pi \in S_n} \ell(\pi) fix(\pi) = \frac{n!n(n-1)}{4} - \frac{(n+1)!}{6}$$ in the OEIS, where $fix(\pi)$ denotes the number of fixed points of $\pi$ (and yes, the two summands there are exactly the sum for the two irreducible characters that make up $fix$). So, after this long introduction, my question is: Did someone already consider this question and am I just not able to find their work? Does anyone here know of a paper mentioning or even solving this problem, is there an application of such a solution in the study of the Coxeter length or other combinatorial functions? Can you suggest any paper, author, book,... where I might find something about this topic, is there maybe a standard textbook on combinatorial representation theory? REPLY [4 votes]: A big thanks to Axel Hultman, who sent me an email to tell me that he basically answered my question in his 2014 paper Permutation statistics of products of random permutations (theorem 6.2). Once again I'm really amazed by the fast and good feedback you get here on MathOverflow. Should anyone be interested in this topic or know of related research, feel free to leave me a message or send me an email, I'm always open for discussion.<|endoftext|> TITLE: How to define transfinite derivatives of a function? QUESTION [5 upvotes]: There are all manners of theories generalizing the notion of derivative. Amongst them is the fractional calculus, a rich theory which gives a sense to the derivation and integration of non-integer (i.e. rational, real, complex) order, that are the real/complex number powers of the differentiation ($D$) and integration ($J$) operators: $Df(x)=\frac{d}{dx}f(x)$ $Jf(x)=\int_{0}^{x}f(x)dx$ Along these lines, one may think about transfinite iteration of $D$ and $J$ as well. While $D^{(n)}$ and $J^{(n)}$ are quite well-defined and well-behaved operators for the natural number $n$, I haven't found any convenient definition of $D^{(\alpha)}$ or $J^{(\alpha)}$ for the ordinal $\alpha\geq \omega$ in the literature if there is any. Due to the similarity between natural and ordinal numbers, it seems the only difficulty is to formulate a definition of the differentiation operator in the limit steps like $D^{\omega}$ or $D^{\omega+\omega}$. One straightforward (but not necessarily well-defined, natural or fruitful) way to do so is to think about $D^{\omega}$ as a functional limit of $D^{n}$s in a certain function space. Though, I am not sure if it is the most clever approach. Anyway, it is somehow "natural" to expect that any $D^{\omega}$ operator demonstrates certain properties like: $D^{\omega}x^{n}=0$ for every $n\in \omega$. Also, the spaces of smooth and analytic functions, $C^{\infty}$ and $C^{\omega}$, don't seem to capture the essence of the very notion of the $\omega$-the derivative of a function, particularly because they don't suggest a clear way of calculating transfinite successor differentiation operators, $D^{\omega+1}$, $D^{\omega+2}$, $D^{\omega+3}$, etc. Question. Is there any paper in which transfinite derivatives of (real/complex) functions are defined/used? If so, what sort of applications do they have? Update. Due to the answer that Andrés mentioned in his comment, it turned out that defining $D^{\omega}$ operator as the limit of $D^{n}$s gives rise to a trivial notion. So maybe a more direct approach is needed here. REPLY [2 votes]: If there is a such a notion of transfinite derivative, I don't think it will be very robust. Here's the problem I see with indexing repeated differentiation with ordinals. We would almost certainly want the following natural property: $$D^{\beta} \circ D^{\alpha} = D^{\alpha+\beta}$$. This would give us, for example, that $$D^{\omega} \circ D = D^{1+\omega} = D^{\omega}$$ and $$D \circ D^{\omega} = D^{\omega+1}$$ But now let's try some of the nicest analytic functions we have: $sin(t)$ and $cos(t)$. We should expect $D^{\omega}sin(t) = sin(t)$ and $D^{\omega}cos(t) = cos(t)$, since the ordinal $\omega$ is "0 (mod 4)", if anything. But then $$sin(t) = D^{\omega}sin(t) = (D^{\omega} \circ D)sin(t) = D^{\omega}cos(t) = cos(t)$$ which is a problem. To make this objection more precise: what is the largest "natural class" of functions $\mathbb{C} \rightarrow \mathbb{C}$ for which we can define transfinite derivatives? Let's assume that by "natural class", we mean it is a collection of functions closed under scaling, addition of functions, and composition of functions. Then there is no natural class $\mathcal{A}$ containing both $e^t$ and all polynomials for which we can make a reasonable definition of transfinite derivative. Why? Because if $e^t, it \in \mathcal{A}$, then $cos(t) = \frac{e^{it}+e^{-it}}{2} \in \mathcal{A}$ and $sin(t) = \frac{e^{it}-e^{-it}}{2i} \in \mathcal{A}$, and we have already argued there is no reasonable way to define the $\omega$-th derivative of $sin(t)$ and $cos(t)$. Here's a rough sketch of an idea using non-standard analysis (as motivated by Dávid Natingga's comment): There probably is a reasonable notion of taking "non-standardly" many derivatives, however. One way of thinking about this is to start with a (standard) analytic function $f$ whose Taylor coefficients are form a nice definable sequence (like $a_n = \frac{1}{n!}$) which decays sufficiently quickly. Then for a non-standard natural number N, we can look at the non-stanard polynomial $p_N(x) = \sum_{n=0}^{N} a_n x^n$. If we plug in a standard real number $r$ into $p_N(x)$ and take the standard part, we will just get $\sum_{n=0}^{\infty} a_n r^n = f(r)$, roughly because the terms $a_M r^M$ for M non-standard are infinitessimal. Now let $M$ be some non-standard natural number (here we imagine N is much bigger than M), we can certainly define $p_{N}^{(M)}(x) = D^{M}\sum_{n=0}^{N} a_n x^n$. Keeping $M$ fixed, as long as $N$ is big enough, the standard part of $p^{(M)}_N(r)$ for $r$ a standard real number does not depend on $N$. So we can define $f{(M)}(r) = p_N^{(M)}(r)$ for any $N$ sufficiently big. This gives us a well-defined notion of of the $M$-th derivative $f^{(M)}$ of a function $f$ with a "nice" Taylor series. (Caveat: I don't really do non-standard analysis, so there may be some conceptual errors here)<|endoftext|> TITLE: Enquiry on a Diophantine problem QUESTION [9 upvotes]: Let $x,y, z$ be relatively prime integers with $xyz \neq 0$. Suppose that $$x^{m/n} + y^{m/n} = z^{m/n}$$ where $m,n$ are relatively prime integers with $mn \neq 0$. Does it necessarily follow that $x,y,z$ are perfect $n$-th powers ? REPLY [13 votes]: Yes, and much more is true. There are no non-trivial dependencies between radicals. See Besicovitch A. S., On the linear independence of fractional powers of integers // J. London Math. Soc. 15 (1940), 3–6.<|endoftext|> TITLE: Does Fermat's last theorem hold in the Grothendieck ring of the ordinals? QUESTION [5 upvotes]: Inspired by this question and its answer, I am curious whether or not Fermat's last theorem holds in the Grothendieck ring of the ordinals under Hessenberg (commutative) operations. The excellent answer to the other question by Emil Jeřábek uses the fact that the sum of the coefficients of an ordinals Cantor normal form is $0$ iff the ordinal itself is $0$, which is no longer true in the Grothendieck ring. The 'Cantor normal forms' now have integer coefficients instead of natural number coefficients, so for example $\omega-1$ is nonzero with coefficients summing to $0$. REPLY [8 votes]: It does hold. Assume for contradiction that $x^n+y^n=z^n$ is a solution with $x,y,z$ nonzero, and put $w=xyz$. It suffices to find a ring homomorphism $f$ to $\mathbb Z$ such that $f(w)$ is nonzero, as then $f(x)^n+f(y)^n=f(z)^n$ is a nontrivial solution in $\mathbb Z$ contradicting the Fermat–Wiles theorem. Now, your ring is isomorphic to the ring of multivariate integer polynomials with the indeterminates being ordinals of the form $\omega^{\omega^\alpha}$. We thus get an evaluation homomorphism to $\mathbb Z$ whenever we fix all the indeterminates to some integers. So, it suffices to show that a nonzero integer multivariate polynomial (the $w$ above) does not compute a constant zero function on $\mathbb Z$. This follows by induction on the number of variables, using the fact that a nonzero univariate polynomial can only have finitely many integer roots.<|endoftext|> TITLE: Restricted Lie algebras with no nonzero proper restricted subalgebras QUESTION [6 upvotes]: Let $L\neq 0$ be a restricted Lie algebra over a field $F$ of characteristic $p>0$. If $F$ is algebraically closed, then it is known that $L$ has no nontrivial restricted subalgebras if and only if $L$ is 1-dimensional. Over arbitrary fields of positive characteristic, is there any description of restricted Lie algebras with the previous property? It is clear that $L$ is generated (as a restricted Lie algebra) by a single element $z$. But what can be said about $z$? REPLY [2 votes]: For an element $a$ of $L$, denote by $\langle a \rangle_p$ the restricted subalgebra generated by $a$. Let $\mathbb{F}[t,\sigma]$ be the ring consisting of all polynomials $f=\sum_{i\geq0}\alpha_i t^{i}$ with respect to the usual sum and multiplication defined by the condition $t\cdot\alpha=\alpha^p t$ for every $\alpha \in \mathbb{F}$. Then $L$ has no non-zero proper restricted subalgebras if and only if $L\cong \mathcal{L}/\langle \bar{f}\rangle_p$, where $\mathcal{L}=\langle x\rangle_p$ is a free cyclic restricted Lie algebra and $\bar{f}=\sum_{i\geq0}\alpha_ix^{[p]^i}$ is an element of $\mathcal{L}$ such that $f=\sum_{i\geq0}\alpha_it^{i}$ is an irreducible element of the ring $\mathbb{F}[t,\sigma]$. The proof can be find here: https://www.tandfonline.com/doi/abs/10.1080/03081087.2019.1708238?journalCode=glma20<|endoftext|> TITLE: Class of maps in localized category may not be a set QUESTION [7 upvotes]: In one of the very first sentences in Hovey's "Model Categories", Ist chapter, we read that One can always invert these "weak equivalences" formally , but there is a foundational problem with doing so, since the class of maps between two objects in the localized category may not be a set. I thought that in a model category fibrations and cofibrations provide an explicit description of weak equivalences and how to invert them - but it seems that there is more stuff going on. So can someone provide me an example with a category with weak equivalences, such that the class of maps between two objects in the category with formally inverted weak equivalences is not a set? REPLY [6 votes]: When Hovey says that we "can always invert these 'weak equivalences' formally", the weak equivalences he's talking about aren't necessarily weak equivalences in a model category. He's introducing and motivating model categories by saying that they solve a problem: sometimes we'd like to formally invert some class of arrows, but there's a risk that we can't for foundational reasons. However, if these weak equivalences are part of a model structure then we can use the model structure to show that the localization exists after all. So if we want an example of a category of weak equivalences whose localization has the foundational problems Hovey's referring to then the weak equivalences can't be part of a model structure. One example along these lines is Example 4.15 of Krause's "Localization theory for triangulated categories". This is an example of a category with weak equivalences whose localization isn't locally small.<|endoftext|> TITLE: How are MTCs permuted by the Galois action on the little disk operad? QUESTION [14 upvotes]: There is a well-studied action of $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$ on (some version of) the $E_2$ operad; see for example this MO question. Modular tensor categories are examples of $E_2$-algebra objects in $\mathrm{Cat}$, the 2-category of linear categories. How does the $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$-action on $E_2$ act on the set of MTCs? (I assume it does preserve modularity.) To be pedantic, I mean the following. Suppose $(\mathcal C,\otimes,\dots)$ is an MTC. Then we have an action of $E_2$ on the underlying category $\mathcal C$, which I could write as $(\otimes,\dots) : E_2 \to \mathrm{End}(\mathcal C)$. Now choose $\gamma \in \operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$, and use it to define an automorphism $\gamma : E_2 \to E_2$. Define a new MTC $\mathcal C^\gamma$ by declaring that the underlying category $\mathcal C$ is unchanged, but the braided tensor structure is modified by $\gamma$ to $(\otimes,\dots) \circ \gamma : E_2 \to E_2 \to \mathrm{End}(\mathcal C)$. I assume that in general $\mathcal{C} \not\cong \mathcal{C}^\gamma$ as braided monoidal categories. So what is the braided monoidal structure on $\mathcal{C}^\gamma$? In case it makes it easier, I care most about the case where $\gamma$ is in the cyclotomic Galois group $\operatorname{Gal}(\mathbb Q^{cyc}/\mathbb Q)$. Actually, most (all?) MTCs over $\mathbb C$ in fact can be defined over $\mathbb Q^{cyc}$. So it wouldn't surprise me if $\mathcal{C}^\gamma$ only depends on the image of $\gamma$ along the map $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q) \to \operatorname{Gal}(\mathbb Q^{cyc}/\mathbb Q)$. REPLY [7 votes]: The Galois action on the profinite or pro-$\ell$ completion of $E_2$ comes, as Dan says, from an action of the corresponding version of the Grothendieck-Teichmüller group. However, to the best of my knowledge none of those acts on MTC's. Rather, they act on braided tensor categories which are themselves pro-finite or pro-unipotent in an appropriate sense. However, even in that case although the action might be higly non-trivial this still produces a braided tensor category equivalent to the one you started with (at least if your category is actually ribbon, but I thin this is true in general). For the pro-algebraic version $GT(k)$, $k$ a field a characteristic 0, those can roughly speaking be identified with those braided tensor categories over $k[[\hbar]]$ which are symmetric monoidal modulo $\hbar$. This includes, e.g., finite dimensional modules over the formal version $U_{\hbar}(\mathfrak g)$ of quantum groups, but typically not their rational or specialized versions. The Galois action you get for $k=\mathbb{Q}_\ell$ is discussed in some extent at the end of Kassel-Rosso-Turaev's book "Quantum groups and knot invariants", and in Furusho's paper Galois action on knots II: Proalgebraic string links and knots.<|endoftext|> TITLE: On the Diophantine equation $x^{4}+y^{4}=z^p$ QUESTION [5 upvotes]: Do there exist integers $x,y,z$ with $xyz\neq 0$, such that $$x^4 + y^4 = z^p$$ where $p\geq 5$ is some prime ? If yes, are there infinitely many of them ? And if there exists infinitely many of them, what are their parametrizations ? REPLY [7 votes]: There are no primitive solutions, even when one of the fourth powers is replaced by a square. This was proved by Bennett, Ellenberg, Ng (see Int. J. Number Theory 6 (2010), 311-338). Non-primitive solutions there are plenty, see Noam Elkies's comment for an example. P.S. I found the paper online here, I don't know how long the link lasts.<|endoftext|> TITLE: A counterexample for the Mean Ergodic Theorem in $L_\infty$ QUESTION [6 upvotes]: The so-called Mean Ergodic Theorem goes back to von Neumann for Hilbert spaces. Later on, versions of this result in reflexive Banach spaces have also appeared (see, e.g., the book by Krengel, Ergodic Theorems, section 2.1. At the beginning of the same book (end of p.4), Krengel mentions that such a result is not true for $L_\infty$. More precisely, he claims that letting $\Omega=[0,1[$, endowed with the Lebesgue measure, and considering the measure-preserving transformation $\tau\omega:=\omega+\alpha$ (mod 1) where $\alpha$ is irrational, one can find a suitable (highly discontinuous) function $f\in L_\infty(\Omega)$ such that $A_nf:=(f+f\circ\tau+f\circ\tau^2+\ldots+f\circ\tau^{n-1})/n$ does not converge to $\bar f:=\int_0^1f$ in the $L_\infty(\Omega)$ norm. He leaves this fact as an exercise. Does anyone know how to show this? REPLY [6 votes]: Let $I_n$ be a collection of rapidly-shrinking intervals with $|I_n|=1/n!$, say. Let $f_0=0$. Inductively define $$ f_n(x)=\begin{cases} (-1)^n&\text{if $x\in\bigcup_{i=0}^{n-1}\tau^i(I_n)$;}\\ f_{n-1}(x)&\text{otherwise.} \end{cases} $$ Let $f(x)=\lim_{n\to\infty}f_n(x)$. This limit exists by the first Borel-Cantelli lemma (as the set of points that switch infinitely often has measure 0). Notice that the intervals $I_n$ decrease so rapidly that there exists a positive measure subset, $J_n$ of $I_n$ so that $\bigcup_{i=0}^{n-1}\tau^i(J_n)$ is disjoint from $\bigcup_{j=n+1}^\infty \bigcup_{i=0}^{k-1}\tau^i(I_j)$. Now for each $N$, $A_Nf$ takes the value $-1$ on each odd $J_n$ with $n\ge N$ and $1$ on each even $J_n$ with $n\ge N$. On the other hand, by the Birkhoff ergodic theorem, $A_Nf$ converges almost everywhere to a constant, $c$, so that $\|A_Nf-c\|_\infty\ge 1$ for each $N$. By the way, there is context for this kind of question in the work of Jones, Bellow, Wierdl, Rosenblatt and collaborators, where they introduce the concept of "strong sweeping out" in subsequence ergodic theorems. In the background of my answer is the Rokhlin lemma, which can be used if one needs more precise control on how the bad parts fill out the space.<|endoftext|> TITLE: Is there an "injective version" of the Bergner model structure? QUESTION [13 upvotes]: The Bergner model structure on $sCat$ (simplicially enriched categories) has a "projective" flavor: fibrations are "levelwise" while cofibrations satisfy stringent relative "freeness" conditions. This seems to be something inherent to modeling $\infty$-categories using strictly-associative composition. Similarly, Barwick and Kan's model structure on relative categories and Horel's model structure on simplicially-internal categories are induced projectively, and not very many objects are cofibrant. Is this really a trade-off we're stuck with? Question 1: Is there a model structure $M$ on $sCat$ other than the Bergner model structure such that the identity functor is a left Quillen equivalence $sCat_M \to sCat_{Bergner}$? More generally, is there a model $M$ of the homotopy theory of $\infty$-categories whose underlying category is $sCat$ which doesn't have fewer cofibrant objects than $sCat_{Bergner}$? Question 2: Is there such an $M$ with a reasonable supply of cofibrant objects? Say, such that every ordinary category is cofibrant? Or perhaps at least such that every ordinary Reedy category is cofibrant? Question 3: As above, but for relative categories or simplicially-internal categories? Two data points: If $C$ is a category and $M$ is a model category, then $M^C$ with levelwise weak equivalences has the correct homotopy theory. But from thinking about the Bergner model structure, you might think you'd have to consider the category of simplicial functors $C' \to M'$ where $C'$ is the standard resolution of $C$ and $M'$ is the simplicial localization of $M$. The fact that you don't have to do this hints at some alternate model category -type structure where $C$ is already cofibrant and $M$ is fibrant -- maybe this would be a model structure on relative categories. A similar phenomenon happens with operads, if I recall correctly (which are also strict monoids for the Kelly tensor product on symmetric sequences) -- the usual model structure (Berger-Moerdijk, I believe) is much more stringent in its cofibrancy requirements than what is needed in practice. As hinted at in the comments, it's a good idea to ask whether the existing "projective" model structures $M$ are left proper. For Left properness of a model structure depends only on the weak equivalences (saying that the co-base change Quillen adjunctions between co-slice categories are Quillen equivalences). A model structure with all objects cofibrant is left proper. Well, it turns out that the Bergner, Lack, Barwick-Kan, and Horel model structures are all left proper! (I'm not sure about the Berger-Moerdijk model structure.) So this leaves open the possibility of model structures with the same weak equivalences and all objects cofibrant. REPLY [4 votes]: EDIT: As Simon Henry points out in the comments, I've been too cavalier with accessibility issues. But since cofibrant generation of the flat maps is the only condition missing in the almost theorem below, one can take any set of flat maps containing generators for the original cofibrations, and its cofibrant closure gives the class of cofibrations for a cofibrantly-generated model structure. In particular, if weak equivalences are stable under finite coproducts, one can ensure that any desired set of objects is cofibrant in the resulting model structure. It turns out there's a pretty formal affirmative answer using the concept of a flat map (dual to sharp maps). Almost Theorem: Let $\mathcal K$ be a left proper, combinatorial model category with weak equvialences stable under filtered colimits. Then there is another "flat" model structure on $\mathcal K$ with the same weak equivalences and cofibrations the flat morphisms of the original model structure. Remark: In a left proper model category, every cofibration is flat. So the flat model structure has more cofibrations than the original one (or else coincides with the original). Moreover, the property of being left proper depends only on being the weak equivalences. So the flat model structure is the one with the given weak equivalences and a maximal number of cofibrations. Remark: A morphism $\emptyset \to X$ is flat if and only if $X \amalg (-)$ preserves weak equivalences. So under the mild condition that the weak equivalences of $\mathcal K$ are stable under finite coproducts, every object of the flat model structure is cofibrant. Almost Proof Sketch: By Jeff Smith's theorem, you have to check that (1) the flat maps are stable under cobase change, transfinite composition, and retracts, (2) similarly for acyclic flat maps, and (3) every morphism with the right lifing property with respect to flat maps is a weak equivalence. The first two are pretty straightforward diagram chases (though the transfinite composition part does seem to require the weak equivalences to be stable under filtered colimits). For the last one, by left properness every cofibiration in the original model structure is flat, so every morphism lifting against flat maps is a trivial fibration in the original model structure and hence a weak equivalence. So this begs the question: what are the flat maps in, say, the Bergner model structure? I think I've identitifed some types, but I'm not sure how to systematically identify them all.<|endoftext|> TITLE: Under what condition we get back path from signatures in rough path theory? QUESTION [5 upvotes]: A link to wikipedia for rough pat theory is: https://en.wikipedia.org/wiki/Rough_path It appears path and signatures has one to one mapping in many cases. I understand that the signature is not dependent on initial condition and if there is any retraces or double back on itself in the path it cancels out in the signature. Therefore we can never recover the starting point of a path or any segment where the path retraces or doubles back on itself from the signature. The question is given a signature when can we get back the path from which it is derived? It is assumed that the initial condition is zero or known and there is retraces or double back on itself in the entire path (never crosses itself). Any comments would be highly appreciated. Any reference or work in this area would be helpful. Is there a formal procedure to get back the path from the signature? REPLY [6 votes]: Loops don't get canceled out in the signature. (You might like to compute the signature of a circle. The second iterated integral is nonzero; in fact, by Green's theorem, it gives you the area inside the circle, maybe up to a factor of $\frac{1}{2}$ or something.) What does get cancelled out is any segment where the path retraces or doubles back on itself. The natural question is which paths have trivial signature. The idea of signature was essentially introduced by K. T. Chen in 1958. Chen has a notion of "irreducible" path, i.e. one which cannot be written as $\alpha \cdot \gamma \cdot \gamma^{-1} \cdot \beta$ where $\cdot$ is concatenation and $\gamma^{-1}$ is the reversal of $\gamma$. Chen showed that for piecewise regular paths (piecewise $C^1$ with non-vanishing derivative), an irreducible path with trivial signature is trivial. So any path with trivial signature is a finite sequence of "retracings". Chen, Kuo-Tsai, Integration of paths. A faithful representation of paths by non- commutative formal power series, Trans. Am. Math. Soc. 89, 395-407 (1959). ZBL0097.25803. More recently, in a very famous paper, Hambly and Lyons considered the case of a bounded variation path. They introduce the notion of a "tree-like path", where in some sense every piece of the path is doubled back; but there might be infinitely many "pieces". They prove that a path of bounded variation has trivial signature iff it is tree-like, and conjectured that this should still hold with less regularity assumptions. Hambly, Ben; Lyons, Terry, Uniqueness for the signature of a path of bounded variation and the reduced path group, Ann. Math. (2) 171, No. 1, 109-167 (2010). ZBL1276.58012. If you look up papers citing Hambly and Lyons, you can find more recent progress. It appears that "trivial signature iff tree-like", appropriately defined, has been proved for weakly geometric paths. Boedihardjo, Horatio; Geng, Xi; Lyons, Terry; Yang, Danyu, The signature of a rough path: uniqueness, Adv. Math. 293, 720-737 (2016). ZBL1347.60094. And for simple paths. Boedihardjo, Horatio; Ni, Hao; Qian, Zhongmin, Uniqueness of signature for simple curves, J. Funct. Anal. 267, No. 6, 1778-1806 (2014). ZBL1294.60063. There has also been work on how to explicitly reconstruct a path, up to tree-like equivalence, from its signature. Geng, Xi, Reconstruction for the signature of a rough path, ZBL06775247.<|endoftext|> TITLE: Prove: If $P_n$ is $n$-$th$ prime number then $P_{n+m} \ge P_n+P_m$ QUESTION [6 upvotes]: Let $n > 1$ and $m > 0$ be two integers and $P_n$ be the $n^{th}$ prime. Prove: $$P_{n+m} \ge P_n + P_m .$$ Can you give a hint, reference, comment, or proof? REPLY [13 votes]: This is an expanded version of my previous answer. It shows, among other things, that the OP's conjecture contradicts the $k$-tuple conjecture for $k=459$. 1. Let $r\geq 0$ be a fixed integer. I claim that the following two statements are equivalent (for integral variables $x,y,n,m$): $$\forall x,y\geq 2:\pi(x+y)\leq\pi(x)+\pi(y)+r\tag{1}$$ $$\forall n,m\geq 2: P_n+P_m-1\leq P_{n+m+r-1}\tag{2}$$ $(1)\Rightarrow(2)$: Let $n,m\geq 2$ be arbitrary, and apply $(1)$ for $x:=P_n-1$ and $y:=P_m-1$, which are at least $2$. We get $\pi(P_n+P_m-2)\leq\pi(P_n-1)+\pi(P_m-1)+r=n+m+r-2$, whence $P_n+P_m-2 TITLE: In what sense bibundles are called as generalized morphisms QUESTION [5 upvotes]: Definition : Let $\mathcal{G}$ and $\mathcal{H}$ be Lie groupoids. A bibundle from $\mathcal{G}$ to $\mathcal{H}$ is a manifold $P$ together with two maps $a_L:P\rightarrow \mathcal{G}_0,a_R:P\rightarrow \mathcal{H}_0$ such that there is a left action of $\mathcal{G}$ on $P$ with respect to an anchor $a_L$ and a right action of $\mathcal{H}$ on $P$ with respect to an anchor $a_R$. $a_L:P\rightarrow \mathcal{G}_0$ is a principal $H$-bundle. $a_R$ is $\mathcal{G}$ invariant. the actions of $\mathcal{G}$ and $\mathcal{H}$ commutes. I am trying to understand in what sense these are called generalized morphisms between Lie groupoids. There is already a notion of generalized morphsim between Lie groupoids from Ieke Moerdijk's article Orbifolds as groupoids. Definition : A generalized morphism from a Lie groupoid $\mathcal{G}$ to a Lie groupoid $\mathcal{H}$ is a morphism of Lie groupoids $\mathcal{G}'\rightarrow \mathcal{H}$ where $\mathcal{G}'$ is a Lie groupoid morita equivalent to $\mathcal{G}$. The name generalized morphisms seems reasonable for this but I do not see in what sense a bibundle is said to be a generalized morphism. Any comments that helps to understand in what sense bibundles are called as generalized morphisms are welcome. Is it that given a bibundle there is a generalised morphism in the sense I have defined above and given a generalised morphism there exists a bibundle associated to that? I could not find any reference. I am also not very comfortable/happy with the point that $a_R$ is just $\mathcal{G}$ invariant. I was guessing that the condition would be $a_R$ is a principal $\mathcal{G}$ bundle just similar to the second condition where it says $a_L:P\rightarrow \mathcal{G}_0$ is a principal $\mathcal{H}$ bundle. Is there any reason why the definition do not ask $a_R$ to be principal $\mathcal{G}$ bundle?? What is the speciality of just $a_L$? I would be very comfortable even if the definition is just the following : Definition : Let $\mathcal{G}$ and $\mathcal{H}$ be Lie groupoids. A bibundle from $\mathcal{G}$ to $\mathcal{H}$ is a manifold $P$ together with two maps $a_L:P\rightarrow \mathcal{G}_0,a_R:P\rightarrow \mathcal{H}_0$ such that there is a left action of $\mathcal{G}$ on $P$ with respect to an anchor $a_L$ and a right action of $\mathcal{H}$ on $P$ with respect to an anchor $a_R$. $a_L$ is $\mathcal{H}$ invariant $a_R$ is $\mathcal{G}$ invariant. the actions of $\mathcal{G}$ and $\mathcal{H}$ commutes. To make sense of question of action of $\mathcal{G}$ and $\mathcal{H}$ being commuttaive, it is just sufficient to have $a_R(g.p)=a_R(p)$ and $a_L(p.h)=a_L(p)$. The definition of bibundle given is none of what I have expected. It is somewhere in between. What is the significance there? Any comments are welcome. https://ncatlab.org/nlab/show/bibundle says, A bibundle is a groupoid principal bundle which is equipped with a compatible second groupoid action “from the other side”. I am somehow ok with this definition. But the question of seeing bibundles as generalized morphisms is still not clear. EDIT : It says in Page no $12$ of Orbifolds as stacks? that, to a $\mathcal{G}-\mathcal{H}$ bibundle $P$ one can associate a functor $\{\mathcal{G}-bundles\}\rightarrow \{\mathcal{H}-bundles\}$. Is this can be the reason for calling bibundles as generalized morphism where morphism of $\mathcal{G}$ to $\mathcal{H}$ is just a morphism of Lie groupoids $\mathcal{G}\rightarrow \mathcal{H}$ where as if you go one step above the ladder you have category of $\mathcal{G}$ bundles and your bibundle map is giving a morphism in that above step. Am I misunderstanding something here? Just for convenience, I am attaching diagram REPLY [2 votes]: As I mentioned in my other answer, Lie groupoids give rise to stacks, and the stack that a Lie groupoid $X$ gives rise to is the stack of principal $X$-bundles. Then in analogy with the Eilenberg-Watts theorem in the theory of rings, a map of such stacks is given by a bibundle. Eilenberg-Watts says that a suitable functor of module categories $Mod_R \to Mod_S$ for rings $R$ and $S$ is, up to isomorphism, nothing other than one whose effect on objects is tensoring an $R$-module with an $R$-$S$-bimodule (which is then an $S$-module). Likewise, a map from the stack of principal $X$-bundles to the stack of principal $Y$-bundles always arises by "tensoring" an $X$-bundle by an $X$-$Y$-bibundle. The definition is not quite so symmetric, since what you get out should be a principal bundle again, so you need the bibundle to be principal on one side, for the $Y$-action. The short answer is that generalised morphisms should satisfy a universal property for bicategories, and taking bibundles as morphisms gives a bicategory of Lie groupoids that has that universal property. But this is somewhat unsatisfactory from the conceptual side.<|endoftext|> TITLE: Positive definite matrices diagonalised by orthogonal matrices that are also involutions QUESTION [5 upvotes]: Let $A$ be a positive definite matrix. Then, $A$ is diagonalized by an orthogonal matrix $P$. I want to know when this matrix is also an involution, i.e., $P^2 = I$. If there is any characterization of such $A$, please kindly share. Thank you. REPLY [5 votes]: Let $n \times n$ matrix $\rm A$ be symmetric and positive definite. Since $\rm A$ is symmetric, it is diagonalizable. Hence, there exists a (non-singular) matrix $\rm P$ such that $\mathrm A = \mathrm P \,\mbox{diag} (\lambda_1, \lambda_2, \dots, \lambda_n) \,\mathrm P^{-1}$, where $\lambda_1, \lambda_2, \dots, \lambda_n > 0$. Suppose $\rm P$ is orthogonal and involutory — and, thus, symmetric. Hence, $$\mathrm A = \mathrm P \,\mbox{diag} (\lambda_1, \lambda_2, \dots, \lambda_n) \,\mathrm P$$ Since $\rm P$ is symmetric and involutory, it has a spectral decomposition and its eigenvalues are $\pm 1$ $$\mathrm P = \mathrm V \,\mbox{diag} (\sigma_1, \sigma_2, \dots, \sigma_n) \,\mathrm V^\top$$ where $\mathrm V$ is an orthogonal matrix and $\sigma_1, \sigma_2, \dots, \sigma_n = \pm 1$. Thus, we have the parameterization $$\mathrm A = \mathrm V \,\mbox{diag} (\sigma_1, \sigma_2, \dots, \sigma_n) \,\mathrm V^\top \,\mbox{diag} (\lambda_1, \lambda_2, \dots, \lambda_n) \,\mathrm V \,\mbox{diag} (\sigma_1, \sigma_2, \dots, \sigma_n) \,\mathrm V^\top$$<|endoftext|> TITLE: "Rotated" version of the Atiyah-Hirzebruch spectral sequence QUESTION [16 upvotes]: Let $G$ be a group, $X$ a topological space with $G$-action. For an Abelian group $A$, let $\mathcal{C}^n(X,A)$ be the group of $n$-cochains on $X$ with $A$ coefficients. We can treat this as a $G$-module (Abelian group with compatible $G$-action) with the $G$-action inherited from the $G$-action on $X$. Now for any $G$-module $B$, we can introduce the Abelian group of "group cochains" $\mathcal{C}^m(G, B)$ that are used to define group cohomology. For example, we can define $\mathcal{C}^m(G, B) = \mathrm{Hom}_G(F_n, B)$, where $$\cdots F_n \to F_{n-1} \to \cdots \to F_0 \to \mathbb{Z} \to 0$$ is a projective $\mathbb{Z}[G]$-resolution of the integers. In particular, $\mathcal{C}^m(G, \mathcal{C}^n(X,A))$ defines a double complex. Now for any double complex there are two spectral sequences we can use to compute the total cohomology. In this case, the total cohomology is equivalent to the equivariant cohomology $H^{\bullet}_G(X,A) = H^{\bullet}(X//G,A)$. One of the two spectral sequences can be identified as the Serre spectral sequence associated with the fibration $X \to X//G \to BG$. The other one does not appear to have a name. My question now is, suppose that we now replace ordinary cohomology with generalized cohomology, that is, we want to compute $\hat{H}^{\bullet}(X//G)$ for some generalized cohomology theory $\hat{H}$. The Serre spectral sequence generalizes to the Atiyah-Hirzebruch spectral sequence. But does the other, unnamed, spectral sequence have any analogous generalization? REPLY [14 votes]: Good question. I think the answer is yes. The unnamed spectral sequence is usually referred to as the isotropy spectral sequence. For a group $G$ acting on $X$ and an abelian group $A$ of coefficients as in your question, it has $E_2$-page given by the sheaf cohomology $H^p(X/G; \mathcal{H}^q)$ of the orbit space, and converges under favourable circumstances to the cohomology $H^*(EG\times_G X;A)$ of the homotopy orbit space. If $X$ is a regular $G$-complex, then the sheaf $\mathcal{H}^q$ of coefficients takes the value $H^q(G_\sigma;A)$ over a simplex $[\sigma]$ of the orbit space $X/G$. Summarising, the spectral sequence goes from cohomology of the orbit space with coefficients in the cohomology of the isotropy subgroups, to the equivariant cohomology. This can be identified with the Leray spectral sequence of the map $EG\times_G X\to X/G$ from the homotopy orbit space to the (genuine) orbit space, given by projecting $EG$ to a point. This is not a fibration in general, but one still gets a spectral sequence, at the expense of replacing cohomology with local coefficients by sheaf cohomology. A decent reference for the Leray spectral sequence of map is the book of Bott and Tu. Now, the Leray spectral sequence of a map works just as well for generalized cohomology theories. The details appear in the thesis of Richard Cain, Cain, R.N., The Leray spectral sequence of a mapping for generalized cohomology, Commun. Pure Appl. Math. 24, 53-70 (1971). ZBL0205.53002. So for a generalized cohomology theory $F^*$ and a regular $G$-complex $X$, you should get a spectral sequence with $E_2$-page $H^p(X/G; \mathcal{F}^q)$ converging to $F^*(EG\times_G X)$, where $\mathcal{F}^q$ takes the value $F^q(G_\sigma)$ over a simplex $[\sigma]$ of $X/G$.<|endoftext|> TITLE: How (non-)computable is set theory? QUESTION [30 upvotes]: Here is a naive outsiders perspective on set theory: A typical set-theoretical result involves constructing new models of set theory from given ones (typically with different theories for the original model and the resulting model). Typically, from a meta-perspective we are allowed (encouraged, or even required) to assume that the models are countable. To the extent that this view is correct, set-theoretic constructions correspond to partial, multivalued operations $T : \subseteq \{0,1\}^\mathbb{N} \rightrightarrows \{0,1\}^\mathbb{N}$ which are defined on sequences coding a model of the original theory, and are outputting a sequence coding a model of the desired theory. These are multivalued, because the constructions may involve something like "pick a generic filter for this forcing notion". Question: Are these operations typically computable, and if not, how non-computable are they? The Weihrauch degrees (https://arxiv.org/abs/1707.03202) provide a framework for classifying non-computability of operations of these types. An answer, however, could also take forms like "For arguments like this, a resulting model is typically computable in the join of the original model and a 1-generic." REPLY [7 votes]: For what it's worth, you might also consider the sets computable by Ordinal Turing Machines (see Koepke's paper, Turing Computations on Ordinals, arXiv:math/0502264v1 [math.LO] 13 Feb 2005). A central theorem regarding Ordinal Turing Machines is the following: A set $x$ of ordinals is ordinal computable from a finite set of ordinal parameters if and only it is an element of the constructible universe $L$. Relative to this type of Turing machine, the non-computable sets are just the non-constructible sets. It would be interesting to study the degree theory of the non-constructible sets relative to Ordinal Turing Machines.<|endoftext|> TITLE: Bochner formula in different forms QUESTION [10 upvotes]: I am looking for a reference (better a book) that contain integral Bochner formulas for domains with boundary (I need it for 1-forms and functions only). For example I will need the following formula: $$\int\limits_\Omega |\Delta f|^2 -|\mathrm{Hess}f|^2 +\langle\mathrm{Ric}(\nabla f),\nabla f\rangle =\int\limits_{\partial\Omega} H\cdot|\nabla f|^2,$$ where $H$ denotes mean curvature of $\partial \Omega$ and $f$ vanish on the boundary, but I will also need its analog for 1-forms and yet general boundary condition. So far I found "Spin geometry" by Lawson and Michelsohn --- based on II/§5, it is easy to derive any formula I need (but I am sure someone did it somewhere). REPLY [4 votes]: Here's a reference to a paper; Theorem 3 gives a general formula for differential forms with no constraints on the boundary values. I'm not sure if this shows up in a book yet. Raulot, S.; Savo, A. A Reilly formula and eigenvalue estimates for differential forms. J. Geom. Anal. 21 (2011), no. 3, 620--640. MR2810846. Here is a direct link to the article.<|endoftext|> TITLE: Closed balls vs closure of open balls QUESTION [26 upvotes]: We work in a separable metric space $(X,d)$. With $\overline{B}(x,r)$ I denote the closed ball around $x$ of radius $r$, and with $cl \ B(x,r)$ I denote the closure of the open ball. Clearly, we always have $cl \ B(x,r) \subseteq \overline{B}(x,r)$, and it is easy to construct examples where the inclusion is strict. Question: Can we have this locally everywhere, meaning does there exists a separable metric space $(X,d)$, a point $x \in X$ and a real $R$ such that for all $r$ with $R > r > 0$ we find that $cl \ B(x,r) \subsetneq \overline{B}(x,r)$ ? My naive attempt to construct this by brute-force violates the separability criterion. I am tempted to believe that we can have the difference only for countably many radii, but so far I was not able to prove this. Restrictions on $(X,d)$ I'd be willing to accept (and that I would particularly enjoy in a positive example) are completeness and local compactness. This question here is related, it is asking about spaces where $cl \ B(x,r) = \overline{B}(x,r)$ holds always. REPLY [28 votes]: The following theorem (or its corollary) implies negative answer to the original question. Theorem. For any point $x$ of a metric space $(X,d)$ the set $R_x:=\{r>0:cl(B(x,r))\ne \bar B(x,r)\}$ has cardinality $|R_x|\le w(X)$, where $w(X)$ stands for the weight of the topology of the metric space $(X,d)$. Proof. Fix a base $\mathcal W$ of the topology of the space $X$ of cardinality $|\mathcal W|=w(X)$. For every $r\in R_x$ choose a point $x_r\in\bar B(x,r)\setminus cl(B(x,r))$ and find a basic neighborhood $W_r\in\mathcal W$ of $x_r$, which is disjoint with the open ball $B(x,r)$. Assuming that $|R_x|>w(X)=|\mathcal W|$, we can find two distinct real numbers $r<\rho$ in $R_x$ such that $W_r=W_\rho$. Then $x_r\in W_r=W_\rho$ and $x_r\in \bar B(x,r)\subset B(x,\rho)$. So, $x_r\in W_\rho\cap B(x,\rho)=\emptyset$, which is a desired contradiction showing that $|R_x|\le w(X)$. Corollary. For any point $x$ of a locally compact (more generally, locally separable) metric space $(X,d)$ there exists $\varepsilon>0$ such that the set $\{r\in(0,\varepsilon):cl(B(x,r))\ne\bar B(x,r)\}$ is at most countable.<|endoftext|> TITLE: Is every odd positive integer of the form $P_{n+m}-P_n-P_m$? QUESTION [16 upvotes]: I am looking for a comment, reference, remark, or proof of three conjectures as follows: Conjecture 1: Let $x$ be an odd positive integer. Then there exist two integers $n, m \ge 2$ so that $$x=P_{n+m}-P_n-P_m,$$ where $P_n$ is the $n^{th}$ prime. My calculations the conjecture 1 true for $x=1, 3,\dotsc, 10^7-1$ and $x = 5.4371349x10^7$, $\cdots, 5.4375349x10^7$ A general of the conjecture above as follows (but weaker): Let every integer $r_0$ exist positive integer $x_0$ such that every odd number $x \ge x_0$ has the form $x=P_{n+m+r_0}-P_n-P_m$, where $m, n \ge 2$ (PS: inspired from the comment of Lev Borisov be low) Or simpler: Conjecture 2: Is every odd positive integer of the form $P_{c}-P_a-P_b$ ? See also: Goldbach's weak conjecture A stronger comjecture 2 Conjecture 3 (Maillet's conjecture) Is every even positive integer of the form $P_{a}-P_b$ I have just computed the conjecture 3 is true with $x=2, 4, \cdots, 10^6$ and $3873$ numbers $x = $$9.82197492x10^8$$,$ $\cdots$$,$ $9.82226054x10^8$ Example: $2=5-3$; $4=7-3$; $6=13-7$; $8=13-5$; $10=17-7$ See also: Goldbach's strong conjecture REPLY [2 votes]: This isn't an answer to conjecture 1, just an elaboration of things others mentioned. There is every reason to think that the answer to conjecture 1 is yes and even that for fixed $m \gt 1$ we have that for each odd integer $x \geq 3$ there are infinitely many $n$ with $p_{n+m}-p_n-p_m=x.$ We don't know that there is even one, however we can say with good precision how the number of solutions should be expected to grow as $n \rightarrow \infty.$ Computations always (seem) to conform to this with good fidelity as far as checked. I'll explain that a bit and then point out that that fixing $m=2$ would likely not be the most fruitful way to find a solution. So with $m=3$ and $p_3=5$ solving $p_{n+3}-p_n-5=15$ amounts to finding two primes $p_N$ and $p_n$ with $p_N-p_m=20$ $N-m=3.$ We do not know that $p_N-p_m=20$ infinitely often. For each even integer g define $\pi_g(X)$ to be the number of pairs $(p_N,p_n)$ with $p_N-p_n=g$ and $p_N \leq X.$ Then $\pi_2(X)$ is the number of twin primes up to $X$ We don't know that this grows without bound but can expect that it is assymptotic (in some sense which could be made precise) to $C_2\frac{X}{\ln^2X}$ and that the constant $C_2=\prod_p(1-\frac1{(p-1)^2})$ with the product over the odd primes. There is a similar constant $C_g$ for each even $g.$ Namely $C_g=C_2\prod_p\frac{p-1}{p-2}$ where the product is over odd primes which divide $g.$ So $p_n+20$ is expected to be prime more often that $p_2$ but about as often as $p+10.$ i.e. $\pi_{20}(X) \sim \pi_{10}(X)\sim \frac{4}{3}\pi_2(X).$ That is a prediction which holds up quite well as far as checked. For an amazing exposition of this read Heuristic Reasoning In The Theory of Numbers by Polya. But our goal had an extra condition. Want something like to have $p_n,p_n+6,p_n+14,p_n+20$ to all be prime but $p_n+h$ be composite for $h=2,4,8,10,12,16,18.$ It is possible to similarly predict how often that happens up to $X.$ Summing over finite number of cases would give a prediction for the number of solutions of the given problem. $m=2$ is a little easier but I wanted to use a different value. But where is it most fruitful to look for solutions to $p_{n+m}-p_n-p_m=x$? Here is a graph for $p_{n+m}-p_n-p_m=101$ with $n+m\lt 200.$ I find $168$ solutions. here is a graph. Of the $168$ solutions, $61$ of them have $m \in\{34,35,36,37,38\}$ and the ratio $\frac{n}{m}$ ranges from $3$ to $5.2$ Using $p_k \sim {k}{\ln{k}}$ one might be able to argue that for fixed $r=\frac{n}{m}$ ( really $r$ in some small range like $[3,5]$) there is a narrow range of $m$ values that would be worth searching first. Perhaps the best $r$ (given $x$) could be estimated. I would have guessed $r \sim 1$ is best, but that seems not to be the case based on this one computation. Perhaps the optimal range is past $m+n=200.$<|endoftext|> TITLE: Elementary equivalence of monoidal categories =? QUESTION [9 upvotes]: Recall that, in model theory, two models $M_1$ and $M_2$ of the same signature are elementary equivalent if $ M_1 \models \phi \Leftrightarrow M_2 \models \phi $ for every first order formula $\phi$ in the language. For example, we can talk about two rings being elementary equivalent. I'd like to know what it means for two monoidal categories $\mathcal C_1$ and $\mathcal C_2$ to be elementary equivalent. Now, before someone jumps at me and tells me it's trivial, let me insist that I'm only going to accept an answer if it satifies the principle of equivalence (i.e., it shouldn't be evil). So, what I'm asking is: What, exactly, are the formulas that I'm allowed to check? For example, the following formula is certainly not ok: $\exists X \in \mathcal C, (X\cong 1) \wedge (X\not = 1)$ The following formula is also not ok: $\exists X\in \mathcal C,$ the associator $a_{X,X,X}:(X\otimes X) \otimes X \to X\otimes (X \otimes X)$ is an identity morphism. REPLY [5 votes]: The question "what are the formulas that I'm allowed to check" is answered by specifying a first-order language in which to talk about monoidal categories. The language determines the formulas that are available to you. The most natural language is a two-sorted language with sorts $O$ (for objects) and $A$ (for arrows), with function symbols $s,t: A\to O$ (source and target), $\circ\colon O\times O\to O$ (composition), $i\colon O\to A$ (identity maps), $\otimes_O\colon O\times O\to O$ (tensor product on objects), $\otimes_A\colon A\times A\to A$ (tensor product product on arrows), and a constant symbol $I\in O$. Of course, you have to decide what to do with non-composable arrows - one possibility is to replace the binary function symbol $\circ$ with a ternary relation symbol $\circ$ defining the graph of composition ($\circ(f,g,h)$ holds iff $f$ and $g$ are composable with composition $h$), while another possibility is to include "dummy elements" $*_O$ and $*_A$ in $O$ and $A$ which are the outputs of any terms which don't make sense. As you noted, this language allows you to write down formulas which don't satisfy the principle of equivalence. In fact, any language which is expressive enough to define the relations $X\cong Y$ and $X\neq Y$ on objects will fail the principle of equivalence. There is a standard variant of first-order logic which does not have a primitive symbol $=$ for equality (this goes by the obvious name first-order logic without equality); you could try to fix this issue by removing the ability to talk about equality, at least on the object sort. But in any reasonable first-order language for categories, the relation "$f$ and $g$ are composable" will be definable on arrows, and then equality of objects $X = Y$ can be expressed by "the identity arrows $i_X$ and $i_Y$ are composable". This seems like a real obstruction to me. It's conceivable that there's a logic $L$ which has expressivity similar to first-order logic but is only capable of expressing formulas which satisfy the principle of equivalence. I don't know such a logic, but I'm imagining one which removes equality on objects and builds the relation on arrows "$f$ and $g$ are composable" directly into the syntax. i.e. a dependent type theory where an arrow $f$ has a type $S_f\to T_f$, and the term $f\circ g$ is only well-formed when $S_f = T_g$. In any case, based on the obstruction above, I'm convinced the logic $L$ would have to have some features not present in first-order logic. Given such a logic $L$ (and maybe the intent of your question was really "what is the logic $L$"?), you could ask about the $L$-equivalence of monoidal categories. But it would be incorrect to call this notion of equivalence elementary equivalence, since the term elementary refers specifically to first-order logic.<|endoftext|> TITLE: Relation between monads, operads and algebraic theories (Again) QUESTION [18 upvotes]: This question (as the title obviously suggests) is similar to, or a continuation of, this question that was asked years ago on MO by a different user. The present question, though, is different from the old one in some ways: The old question focussed on reference request. My question instead, while clear and readable references are welcome, is more focussed on getting some quick and dirty intuitive understanding. Most probably what I'm looking for is already buried in Tom Leinster's comprehensive monograph Higher operads, higher categories, but currently I'm not planning on reading it (or going in a detailed way through other technical material on the topic). I already have an idea of how algebraic theories (also called Lawvere theories) and monads differ from each other. I would also like to understand how operads fit into this picture. In the old question the semantic aspect was not considered (yes, there's the expression "model of theories" in that question, but just in the sense of "way of understanding the notion of mathematical algebraic theory", not in the more technical sense of semantics). In what follows I may be missing some hypothesis: let me know or just add them if necessary. Here is what I remember about the algebraic theories vs monads relationship. Algebraic theories correspond one-to-one to finitary monads on $\mathbf{Set}$, and the correspondence is an equivalence of categories. If one wants to recover an equivalence on the level of all monads on the category of sets, then one has to relax the finitary condition on the other side, so getting a generalization of the notion of Lawvere theory to some notion of "infinitary Lawvere theory". Furthermore, given a (just plain old, or possibly non-finitary) Lawvere theory $\mathcal L$ and the corresponding monad $\mathscr T$, a model of (also called algebra for) $\mathcal L$ in $\mathbf{Set}$ corresponds to a $\mathscr{T}$-algebra (necessarily in $\mathbf{Set}$), and this correspondence extends to an equivalence of the categories of algebras $$\mathrm{Alg}^{\mathcal{L}}(\mathbf{Set}):=\mathrm{Hom}_{\times}(\mathcal{L},\mathbf{Set})\simeq \mathrm{Alg}^{\mathscr T}(\mathbf{Set})=:\mathbf{Set}^{\mathscr T}\,.$$ There is also the variant with arities. While usual Lawvere theories / finitary monads have finite ordinals as arities, and the corresponding non-finitary versions have sets as arities, one can introduce the notions of Lawvere theory with arities from a category $\mathfrak{A}$ and monads with arities from $\mathfrak{A}$, and again one has an equivalence between these notions, and an equivalence between the algebras in $\mathbf{Set}$. a) How do operads fit into all this? Are operads somehow more general or less general objects than monads/theories (with all bells and whistles)? There is an asymmetry between the notion of semantics for Lawvere theories and for monads (In what follows I will drop the finitary assumption). Namely, every Lawvere theory $\mathcal{L}$, as remarked above, is essentially a monad $\mathscr{T}_{\mathcal{L}}$ on $\mathbf{Set}$; though it can have models in every (suitable) category $\mathcal C$: just define the category of models of $\mathcal L$ in $\mathcal C$ to be $\mathrm{Hom}_{\times}(\mathcal L,\mathcal C)$. On the other hand, a monad $\mathscr T$ on $\mathbf{Set}$, by definition, only has models (aka algebras) in $\mathbf{Set}$. To remedy this, one introduces $\mathcal{V}$-enriched monads $\mathscr T$, where $\mathcal{V}$ is a given monoidal category, and $\mathscr T$-algebras are now objects of $\mathcal{V}$ (endowed with some further structure). If I get it correct, there is now an equivalence of categories $\mathrm{Hom}_{\otimes}(\mathcal{L},\mathcal V)\simeq \mathcal{V}^{\mathscr{T}}$. But models of $\mathcal L$ are related to each other: you can take, again if I get it correct, (nonstrict?) $\otimes$-functors $\mathcal{V}\to\mathcal{V}'$ intertwining the (strict?) tensor functors $\mathcal{L}\to\mathcal{V}$ and $\mathcal{L}\to\mathcal{V}'$. b) Can we read these "inter-model" relationships in the language of (enriched) monads? c) How does the above semantic aspect go for operads and how, roughly, does the translation go from there to monads/theories (and viceversa)? d Is there a "natural" symultaneous generalization of all the three things (theories, monads, and operads)? Does also the notions of semantics have a "natural" common generalization? Edit. I had written the above (comprising questions a),...,d) ) some time ago, and just posted it now. But I forgot that I also wrote essentially the same questions in a perhaps more systematic way! I'm now going to post it below, without deleting the above paragraphs. Q.1 What is the most general monads/theories equivalence to date? (possibly taking into account at the same time: arities, enrichment, and maybe sorts) Q.2 What is the most general monads/algebras adjunction? (again, possibly throwing arities, enrichment, and maybe sorts, simultaneously into the mix) $$\mathrm{Free}^T:\mathcal{E}\rightleftarrows \mathrm{Alg}^T(\mathcal{E})=:\mathcal{E}^T:U^T$$ where $T$ is a monad on $\mathcal{E}$, $U^T$ is a forgetful functor and $\mathrm{Free}^T$ a "free $T$-algebra" functor. Q.3 Which is the relation between the different notions of semantics (for monads and theories)? Can enrichment "cure" this asymmetry? This question was further explained a bit in question b) above. Q.4 How far are operads from being algebraic theories? There's an article by Leinster in which it is shown that the natural functor $$G:\mathbf{Opd}\to \mathbf{Mnd}(\mathbf{Set}),\quad P\mapsto T_P$$ where $$T_P(X):=\amalg_{n\in \mathbb{N}}P(n)\times X^{\times n}$$ is not an equivalence, and it is shown that the essential image of $G$ is given by "strongly regular finitary monads" on $\mathbf{Set}$, or equivalently by those Cartesian monads $T$ admitting a Cartesian monad morphism to the "free monoid monad". In the case of symmetric operads, $G$ becomes an equivalence. So, question Q.4 is about such a functor $G$ in the most general setting (arities, enrichment,...), and can be seen as asking: which properties does such a functor $G$ have? What is a characterization of its essential image? What is the (essential) fiber of $G$ over a $T\in G(\mathbf{Opd})$? Q.5 How does the notion of semantics for operads (well, it's algebras for an operad) relate to the notion of semantics for monads? (this is similar to question Q.3, but featuring operads instead) Q.6 In the light of the above, what is an operad, intuitively? An operad $P$ has, in general, "more structure" than its associated monad $T_P$ ($G$ not injective). Also, $P$ is "more rigid" an object than $T_P$ ($G$ not full). So what do these extra things amount to? This must be some sort of detail, because clearly both monads and operads "want" to be a formalization of the intuitive notion of "algebraic theory". Q.7 Do all the above considerations naturally extend to the case of $T$-categories instead of $T$-algebras? Maybe one has to consider colored operads? Or sorts? Q.8 Which further structure on a monad/theory describes an equational theory? Same question in arities, enriched, or sorts, flavor. This is about the distinction between e.g. the (logical formal) equational theory of groups and the Lawvere theory of groups (or, equivalently, the corresponding monad). So, what further structure on a theory/monad can be seen as a "presentation" of it. Also, which properties does the functor $$\{ \textrm{equational theories}\}\to \mathbf{Mnd}$$ have? Ess. surjective? Full? (I think not). Faithful? (again, no...). REPLY [2 votes]: I'm just throwing a few references on the table, as this is the best I can do now. I guess if you mention monads with arities you are familiar with Weber's Familial 2-functors and parametric right adjoints, but one never knows. Colored operads also arise as suitable monoids, but instead of taking finite sets and bijections you employ in a suitable way the construction of the free strict symmetric monoidal category on a small category $A$ (exercise: this construction gives what you expect for operads if $A=1$ is the terminal category https://www.cl.cam.ac.uk/~mpf23/papers/PreSheaves/Esp.pdf and the beautiful Memoir http://www.ams.org/books/memo/1184/ I also guess one should mention that, surprisingly, many unpleasant asymmetries one encounters when working with operads and their associated analytic monads are solved passing to the $\infty$-world: https://arxiv.org/abs/1712.06469 this paper by Kock, Gepner and Haugseng develops quite a bit of $(\infty,2)$-category theory in order to prove the claim that "$\infty$-operads are analytic monads". Now I'm not the right person to give a more precise account, but the introduction of the paper contains a few slogans (in centered, italic typeface) for which there isn't an equally clean 1-categorical analogue.<|endoftext|> TITLE: Curious anti-commutative ring QUESTION [9 upvotes]: Has anyone seen the ring $\Lambda[x_0, x_1, x_2, \ldots]/(x_i x_j - (i+1) x_0 x_{i+j})$ in some natural context? Here $\Lambda[x_0, x_1, x_2, \ldots]$ is the (graded-)commutative algebra (either over the integers or the integers localized at 2) freely generated by elements $x_0,x_1,x_2,\ldots$ of odd homological degrees, so that $x_i x_j = - x_j x_i$. In particular, we only get $2 x_i^2 = 0$, not $x_i^2 = 0$. I probably shouldn't write $\Lambda$ here: in characteristic $2$ or integrally, $\Lambda$ usually adds the relation $x_i^2 = 0$. ADDED NOTE: In the meantime, I have found a better way of solving the problem in which this arose, so it is merely a curiosity now. I am happy to delete it if people wish. Forgive me for posting this to 'Commutative algebra', but as a topologist, commutative means $x y = (-1)^{(\deg x)(\deg y)} yx$, and my $x_i$ are in odd degrees. SECOND NOTE: This algebra has now shown up in another context, so Vladimir's answer below has been quite useful. Thanks to Vladimir and MO. REPLY [6 votes]: I noticed this now, and I want to remark that the underlying abelian group can in fact be described very precisely. To do that, note that: (1) the defining relations easily imply that the abelian group of elements of degree $d\ge 2$ in this algebra is certainly generated by $x_0^{d-1}x_k$, $k\ge 0$, and (2) as discussed in the comments, there are the relations $(i+j+2)x_0x_{i+j}=0$ that follow from the defining relations and anticommutativity; effectively, these give just one relation for each $n$, namely $(n+2)x_0x_n=0$. Now let me (inspired by typical Gröbner bases calculations) consider the following two chains of equalities: $$ x_ix_jx_k=(i+1)x_0x_{i+j}x_k=(i+1)(i+j+1)x_0^2x_{i+j+k} $$ and $$ x_ix_jx_k=(j+1)x_ix_0x_{j+k}=-(j+1)x_0x_ix_{j+k}=-(j+1)(i+1)x_0^2x_{i+j+k}. $$ They imply that $$ (i+1)(i+2j+2)x_0^2x_{i+j+k}=0 $$ for each choice of $i$ and $j$ with $i+j\le n$. In particular, if $n\ge 1$, we may take $i=n-1$, $j=1$, obtaining $$ n(n+3)x_0^2x_n=(n-1+1)(n-1+2+2)x_0^2x_n=0. $$ But $(n+2)x_0x_n=0$ implies $(n+1)(n+2)x_0^2x_n=0$, so by subtraction we see that $2x_0^2x_n=0$. Moreover, no further relations can be obtained in a similar way, because once we have the 2-torsion property, we have $$ (i+1)(i+2j+2)x_0^2x_n=(i+1)ix_0^2x_n=0, $$ since $(i+1)i$ is always even. In fact, using a version of Gröbner bases (or rewriting systems) for ideals in free anticommutative algebras, one can see that the system of all the defining relations thus obtained, namely $$ \begin{cases} x_ix_j=(i+1)x_0x_{i+j},\\ (n+2)x_0x_n=0,\\ 2x_0^2x_n=0 \end{cases} $$ is complete, and so your ring as an abelian group : is freely generated by $1$ in degree $0$, is freely generated by $x_0,x_1,\ldots$ in degree $1$, is the product of cyclic groups of orders $2,3,\ldots$ generated by $x_0^2, x_0x_1, x_0x_2, \ldots$ respectively in degree $2$, is a product of countably many cyclic groups of order $2$ generated by $x_0^d, x_0^{d-1}x_1, x_0^{d-1}x_2, \ldots$ in each degree $d\ge 3$ .<|endoftext|> TITLE: What is the Katz-Sarnak philosophy? QUESTION [30 upvotes]: It has been recently mentioned by a speaker (his talk is completely not relevant to random matrix theory/RMT though) that modern statistics, especially random matrices theory, will help solving some number theoretic problems. I was quite intrigued and ask for more explanation but the speaker himself said he did not understand the philosophy well enough but point me to a few keywords to search for, one among which is "Katz-Sarnak philosophy". After a few searches, I figured out that all sources seem to point to [KS]. The major results in [KS] seems like saying that the class of general linear (compact) groups have the same n-level correlations. While some researchers [Kowalski][Miller] do mention that they applied KS philosophy, but what they have done seems very different...So for number theorists and experts on elliptic curves: (1) When you refer to "Katz-Sarnak philosophy", what kind of thinking/technique do you actually mean? (2) Is there a formalism/explanation of this philosophy in language of RMT? I asked this because it might be helpful to understand it in this perspective (at least to a probabilist). Any inputs are highly appreciated. Reference [KS]Katz, Nicholas M., and Peter Sarnak, eds. Random matrices, Frobenius eigenvalues, and monodromy. Vol. 45. American Mathematical Soc., 1999.Google books [Kowalski]http://blogs.ethz.ch/kowalski/2008/07/30/finding-life-beyond-the-central-limit-theorem/ [Miller]Miller, Steven J. "One-and two-level densities for rational families of elliptic curves: evidence for the underlying group symmetries." Compositio Mathematica 140.4 (2004): 952-992. REPLY [18 votes]: I'm going to give an answer that discusses some things that the other answers don't go into as much detail on. In particular let me try to explain why the results you mention on classical groups having the same $n$-level correlations are in fact relevant to the number theory situation. It will build off KConrad's answer in some ways. The Katz-Sarnak book proves theorems about zeroes of $L$-functions in function field setting. These are defined by exact analogues of the definitions of $L$-functions, over another field. Of course, as KConrad points out, these aren't the same as $L$-functions over a number field, and many of the families considered in the Katz-Sarnak book don't match the families of interest over number fields. But some match, and this has been rectified in later work. One key advantage of the function field setting is that the situation is conceptually clearer. The $L$-function is a polynomial, and by the Grothendieck-Lefschetz trace formula it is the characteristic polynomial of some naturally defined matrix (the action of a Frobenius element on a Tate module or etale cohomology group). So we have no difficulty interpreting the zeroes as eigenvalues of a matrix. Moreover, this matrix lies in some matrix group, the monodromy group of the family. These matrices are over an $\ell$-adic field and so can't naturally be viewed as unitary matrices. But Deligne (with Weil first in some cases) showed that if we base change to the complex numbers, they are conjugate to unitary matrices. So in fact the zeroes of the $L$-function are eigenvalues of elements of the maximal compact subgroup of some algebraic matrix group. So in this setting, the question of why we should expect the zeroes to behave like eigenvalues of random unitary matrices is not mysterious at all - it's the simplest possible behavior they could have, given that they are eigenvalues of unitary matrices. In fact Deligne did more and showed that these Frobenius conjugacy classes are equidistributed in the maximal compact subgroup of the monodromy group. In other words, the statistics of any continuous function of the set of zeroes at all matches the statistics for random matrices in some compact group. Such a powerful result doesn't come without a cost, and it has two. First, we may not know what the group is. Second, this statistical equivalence only shows up in the limit as we take the finite field size to infinity. In this limit, the number of zeroes is fixed, so we can take the same compact group each time. However, in the number field settings, people always consider limits where some parameter controlling the number of zeroes grows. The first cost was dealt with by Katz-Sarnak by computing the group in some important cases. They found that, while a priori it could be any group at all, in fact it was usually a classical group (which is something Katz had already noticed in some of his earlier investigations). So they focused attention on the maximal compact subgroups of classical groups. The second was and is more difficult to deal with. However, since the transfer from the function field case to the number field case is conjectural anyways, one merely has to guess how to transfer statistics from finitely many zeroes to infinitely many zeroes, and then perform the calculations to do the transfer. They guessed that taking the limit of distribution of the smallest $k$ eigenvalues in a family of matrices in larger and larger groups would match the distribution of the smallest $k$ zeroes in $L$-functions over number fields. The first few chapters of the Katz-Sarnak book are then devoted to calculating these limits, which is the part that looks like honest random matrix theory. So Katz and Sarnak combined algebraic geometry - the results of Deligne and the monodromy group calculations - with random matrix theory - these calculations in the limit as the size of the classical group grows - to obtain precise predictions for the statistics of zeroes of $L$-functions over number fields. These predictions have then been expanded and refined in later work. I guess this is all a bit removed from your motivating question, which is how that random matrix theory can help in number theory. Unfortunately I don't know how much cutting edge research in random matrix theory is really needed, because we know of no rigorous relationship between $L$-functions over number fields and any kind of random matrix, so thus far random matrix theory has mostly been used to give predictions, and the predictions we already have are hard enough to try to prove that any property of random matrices found by more powerful techniques might be even further out of reach over number fields. However, on a positive note, I would suggest you check out Terry Tao's latest blog post, which is on a question involving random matrices, still unsolved, motivated by number theory (in fact, possibly motivated by a joke I made in a talk?).<|endoftext|> TITLE: Can we make 101 almost perfect banknotes from 100? QUESTION [14 upvotes]: Disclaimer. The practical execution of the algorithm in question might be illegal in certain jurisdictions, and is thus strongly discouraged by the poser of the problem. This recent post on the Muffin problem made me think of the following question. Can we cut 100 banknotes into pieces of size at least $10\%$ each, and reassemble them into 101 banknotes of size $100\pm2\%$ each? So each original banknote is cut into at most $10$ pieces of substantial size, and each new banknote also consists of at most $10$ pieces. The patterns on these newly formed banknotes should match, so we also demand, say, that no part of a banknote appears twice on a new banknote. Of course, these numbers are quite ad hoc, I'm happy to see any similar result. Note that if we don't require each piece to be at least $10\%$, then it is easy to make the trick by cutting each banknote into only two (sometimes very unequal) parts. I also wonder if non-vertical cuts might help, but I would like to keep the pieces simply connected regions bounded by Jordan curves. Also, is there some implication between this question and the Muffin problem? REPLY [2 votes]: This video http://thekidshouldseethis.com/post/62804856022 shows a possible solution (well, not for banknotes but for chocolate and one piece is only about 4% in size). This is a variant of Paul Curry's paradox described in Martin Gardner's book "Mathematics, Magic and Mystery": http://store.doverpublications.com/0486203352.html In fact some real money was done from this paradox: a plastic version of Gardner's square was made in China and sold in stores. See Gardner's interview https://www.jstor.org/stable/25653710 However, a better way to make money is provided by Banach-Tarski paradox: A gold sphere can be chopped into a finite number of chunks, and these chunks can then be put together again to yield two gold spheres, each of which has the same size as the one that just went into pieces. See https://arxiv.org/abs/math/0202309 (The Banach-Tarski paradox or what mathematics and religion have in common, by Volker Runde). The history of this theorem can be colourfully traced through the paper https://arxiv.org/abs/1710.05659 (From Poland to "Petersburg": The Banach-Tarski Paradox in Bely's Modernist Novel, by Noah Giansiracusa and Anastasia Vasilyeva). P.S. Even if we forget about preserving the pattern on the banknote, the Banach-Tarski paradox doesn't work for planar banknotes. As mentioned in https://www.sciencedirect.com/science/article/pii/S016800720300126X (On the Warsaw interactions of logic and mathematics in the years 1919–1939, by Roman Duda): This is the famous Banach–Tarski paradox. By the way, one can ask about money: can a banknote produce two of its kind? It is a problem in applied mathematics, but the answer is, unfortunately negative: no bounded set in the plane can have such a paradoxical decomposition [41, footnote 1 on p. 218]. And [41] is Adolphe Lindenbaums article "Contributions à l'étude de l'espace métrique": http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-fmv8i1p16bwm<|endoftext|> TITLE: If $\mathcal C$ has amalgamation, does $Ind(\mathcal C)$ have amalgamation? QUESTION [7 upvotes]: Recall that a category $\mathcal C$ has amalgamation if every span admits a cocone. If $\mathcal C$ has amalgamation, then does $Ind(\mathcal C)$ have amalgamation? The "obvious way to show this" would be to induct on the presentability ranks of objects being amalgamated, presenting them as colimits of chains of objects of lower presentability rank. But this doesn't work -- it seems one can only prove amalgamation over finitely-presentable objects this way (i.e. for a span $B \leftarrow A \to C$ where $A$ is finitely-presentable, there is a cocone). EDIT and maybe not even that! See the comments So I suspect the answer is no. But I don't have a counterexample. Note that if $\mathcal C$ has functorial amalgamation, then something like the above proof attempt should work to show that $Ind(\mathcal C)$ does too. But I don't want to assume that amalgamation is functorial. REPLY [5 votes]: A counterexample is given in the paper Disjoint Amalgamation in Locally Finite AEC by Baldwin, Koerwien, and Laskowski (link). They're actually interested in finding AECs in which amalgamation holds for models up to size $\aleph_k$ but not for models of size $\aleph_{k+1}$ for $k\in \omega$. Since you're interested in (essentially) a special case of this, I'll give a slightly simpler version of their construction, which is itself based on an earlier construction by Laskowski and Shelah (link). Of course, there might be a much simpler purely categorical example. Consider the first-order language $L = \{f_n\mid n\in \mathbb{N}\}$, where the $f_n$ are binary function symbols. Let $K$ be the class of all finite $L$-structures which contain no independent subset of size $3$, i.e. such that there is no triple $(a,b,c)$ such that $a\notin \langle b,c\rangle$, $b\notin \langle a,c\rangle$, and $c\notin \langle a,b\rangle$, where $\langle X\rangle$ is the substructure generated by $X$. Now consider the category $\mathcal{C}$ whose objects are $K$ and whose arrows are embeddings between structures in $K$. Claim: $\mathcal{C}$ has the amalgamation property. Suppose $f\colon A\to B$ and $g\colon A\to C$ are embeddings between structures in $K$, and identify $A$ with $f(A)\subseteq B$ and $g(A)\subseteq C$. Let $D$ be the disjoint union of $B$ and $C$ over $A$. To make $D$ into an $L$-structure we just need to define $f_n(b,c)$ when $b\in B\setminus A$ and $c\in C\setminus A$. Enumerate $D$ as $d_0,\dots,d_m$, set $f_n(b,c) = d_n$ for $n\leq m$ and $f_n(b,c) = b$ for $n>m$. Then $D\in K$, since for any triple $(d_1,d_2,d_3)$ either all $d_i$ are contained in $B$ or all $d_i$ are contained in $C$, in which case the triple is not independent by hypothesis, or we have WLOG $d_1\in B\setminus A$ and $d_2\in C\setminus A$, in which case $d_3\in \langle d_1,d_2\rangle$ by construction. Claim: $\mathsf{Ind}(\mathcal{C})$ does not have the amalgamation property. We can identify objects in $\mathsf{Ind}(\mathcal{C})$ with $L$-structures such that every finitely generated substructure is in $K$ and arrows in $\mathsf{Ind}(\mathcal{C})$ with embeddings between such structures. Now it's possible to find an infinite structure $A$ and extensions $B$ and $C$, all in $\mathsf{Ind}(\mathrm{C})$, with elements $b\in B\setminus A$ and $c\in C\setminus A$, such that $\langle Ab\rangle\subseteq B$ is not isomorphic over $A$ to a substructure of $C$, and similarly $\langle Ac\rangle\subseteq C$ is not isomorphic over $A$ to a substructure of $B$. That is, the quantifier-free type of $b$ over $A$ is not realized in $C$, and vice versa. Now suppose there is an amalgamation $D$ of $B$ and $C$ over $A$, and pretend all embeddings are inclusions. Then we must have $b\in D\setminus C$ and $c\in D\setminus B$. Since $\langle b,c\rangle$ is finite, there is some $a\in A\setminus \langle b,c\rangle$. Then $(a,b,c)$ is independent. Indeed, $\langle a,b\rangle\subseteq B$, so $c\notin \langle a,b\rangle$, and $\langle a,c\rangle\subseteq C$, so $b\notin \langle a,c\rangle$. It follows that $\langle a,b,c\rangle\notin K$, so $D$ is not in $\mathsf{Ind}(\mathcal{C})$, which is a contradiction.<|endoftext|> TITLE: Knots of fixed genus with arbitrarily large volume QUESTION [9 upvotes]: Consider all knots with fixed genus $g\ge 2$ (I am considering the classical 3-genus). Do there exist infinite families of genus $g$ knots with arbitrarily large volume? The answer seems like it should definitely be yes, but I can’t seem to find any references. REPLY [10 votes]: The result of Brittenham for genus 1 knots pointed out by Sam Nead was generalized to all genus in Theorem 8.2 of this paper: Stoimenow, A., Realizing Alexander polynomials by hyperbolic links, Expo. Math. 28, No. 2, 133-178 (2010). ZBL1196.57009.<|endoftext|> TITLE: Why did Euler consider the zeta function? QUESTION [26 upvotes]: Many zeta functions and L-functions which are generalizations of the Riemann zeta function play very important roles in modern mathematics (Kummer criterion, class number formula, Weil conjecture, BSD conjecture, Langlands program, Riemann hypothesis,...). Euler was perhaps the first person to consider the zeta function $\zeta(s)$ ($1\leq s$). Why did Euler study such a function? What was his aim? Further, though we know their importance well, should we consider that the Riemann zeta function and its generalizations happen to play key roles in modern number theory? REPLY [42 votes]: This history is described in Euler and the Zeta Function by Raymond Ayoub (1974). In his early twenties, around 1730, Euler considered the celebrated problem to calculate the sum $$\zeta(2)=\sum_{n=1}^\infty \frac{1}{n^2}.$$ This problem goes back to 1650, it was posed by Pietro Mengoli and John Wallis computed the sum to three decimal places. Ayoub conjectures that it was Daniel Bernoulli who drew the attention of Euler to this challenging problem. (Both lived in St. Petersburg around 1730.) Euler first publishes several methods to compute the sum to high accuracy, arriving at $$\zeta(2)=1.64493406684822643,$$ and finally obtained $\pi^2/6$ in 1734. (We know this date from correspondence with Bernoulli.) It was published in 1735 in De summis serierum reciprocarum. $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\text{etc.}=\frac{p^2}{6}$, thus the sum of this series multiplied by 6 equals the square of the circumference of a circle that has diameter 1. [Notice that the symbol $\pi$ was not yet in use.] The generalization to $\zeta(s)$ with integers $s$ larger than two followed in "De seribus quibusdam considerationes". In 1748, finally, Euler derives a functional equation relating the values at $s$ and $1-s$ and conjectures that it holds for any real $s$. (Euler's functional equation is equivalent to the one proven a century later by Riemann.).<|endoftext|> TITLE: Boolean ultrapower of V[G] by G QUESTION [6 upvotes]: In Joel David Hamkins's "Well-founded Boolean Ultrapowers as Large Cardinal Embeddings", it is mentioned that if $U \in \mathbf{V}$ is an ultrafilter of a complete Boolean algebra $\mathbb{B}$ and $U$ is non-generic over $\mathbf{V}$, then either $U$ misses a countable maximal antichain of $\mathbb{B}$ in $\mathbf{V}$, or it misses a maximal antichain of size a measurable cardinal. What I am confused about is, if we force in $\mathbf{V}$ with an atomless forcing notion that does not add any countable sequence of elements of $\mathbf{V}$ (eg. any atomless countably-closed forcing), we get to a generic extension $\mathbf{V}[G]$ with $G$ non-generic over $\mathbf{V}[G]$, and all countable maximal antichains of $\mathbb{B}$ in $\mathbf{V}[G]$ are already in $\mathbf{V}$. So $G$ meets all countable maximal antichain in $\mathbf{V}[G]$, and since $\mathbb{B}$ remains atomless, $G$ is not $\mathbf{V}[G]$-generic. Does this imply the existence of a measurable cardinal in $\mathbf{V}[G]$? REPLY [11 votes]: I share your view that this is a subtle point. To illustrate it, my co-author Dan Seabold and I had pointed to the case of adding a Cohen subset to $\omega_1$ (see example 44 in Boolean ultrapowers paper, the paper you mention). If you use the natural tree order $2^{<\omega_1}$, then this partial order has very few countable maximal antichains, and every ultrafilter, which is to say maximal filter, in the partial order will meet all of them. But the Boolean completion of this order, on the other hand, has many more countable maximal antichains, and no ultrafilter in $V$ will meet them all; furthermore, ultrafilters in the partial order do not generally generate ultrafilters in the Boolean algebra. It was this example that convinced us that the Boolean ultrapower was at bottom about the complete Boolean algebra and not about the underlying partial order. (Meanwhile, section 8 of the paper tries to recover what one can about the Boolean ultrapower from the partial order. See also the tutorial lecture series I gave for the Young Set Theory workshop in Bonn 2011.) So, a central issue for your example is the non-completeness of the Boolean algebra $\newcommand\B{\mathbb{B}}\B$ in the extension. The general phenomenon, I claim, is that a Boolean algebra $\B$ is never complete after forcing with it. In truth I was confused about this issue for years, and I am pleased now to have a reasonably clear account of it, in the theorem below. Only a special case of this result is proved in the BU paper. An important secondary issue is that in many cases, the generic filter $G$ will not even generate an ultrafilter on the completion of $\B$ in the extension. Theorem. No nontrivial Boolean algebra $\B$ is complete after forcing with $\B$. Proof. Suppose $\B$ is a non-atomic Boolean algebra and that $G\subset V$ is $V$-generic filter. We consider $\B$ in the forcing extension $V[G]$, and I claim that $\B$ is not complete in $V[G]$. Some special cases of this are a bit easier to see, and so let us warm up with these cases before giving the general argument. For example, consider the case where the forcing adds a new real $u\subset\omega$. In this case, let $A\subset\B$ be any countably infinite antichain in $\B$. (Every nontrivial Boolean algebra has a countably infinite antichain, by a result of Tarski.) Enumerate $A$ as $\{a_n\mid n\in\omega\}$ and by completeness, let $a=\bigvee_{n\in u} a_n$. Note that $a$ has nonzero overlap with $a_n$ if and only if $n\in u$, and so we can reconstruct $u$ in $V$ from $a$ and $A$, contradicting the fact that $u$ is new. An essentially similar idea works if the forcing adds a new subset $u\subset\kappa$ and $\B$ has an antichain of size $\kappa$ in $V$. Another common case occurs when the generic filter $G$ is generated by a linearly ordered set. This happens, for example, whenever one forces with a tree, for then the generic filter is a branch, which is linearly ordered. Lemma 43 in the Boolean ultrapower paper shows that a linearly ordered set in a complete Boolean algebra can never generate an ultrafilter. The argument is this: $1=b_0>b_1>\cdots>b_\alpha>\cdots$ be a continuous descent co-initial in the ultrafilter. Let $d_\alpha=b_\alpha-b_{\alpha+1}$ be the corresponding difference antichain, which is maximal. Let $x=\bigvee\{d_\alpha\mid\alpha\text{ odd}\}$, so $\neg x=\bigvee\{d_\alpha\mid\alpha\text{ even}\}$. Neither $x$ nor $\neg x$ is above any particular $b_\alpha$, contradicting the assumption that they generate an ultrafilter. So if the generic filter $G$ is generated by a linearly ordered set, as it often is, then the Boolean algebra cannot be complete in the extension. And worse: the generic filter $G$ does not even generate an ultrafilter on the Boolean completion of $\B$ in the forcing extension $V[G]$! Finally, I claim that a similar idea allows one to handle the general case. Assume $G\subset\B$ is $V$-generic for a nontrivial forcing notion $\B$, and assume $\B$ is complete in the extension $V[G]$. In the ground model $V$, let us fix an arbitrary nontrivial splitting assignment of all conditions, $$b=(b)_0\vee (b)_1,$$ with $(b)_0$ and $(b)_1$ incompatible and nonzero. The filter $G$ determines a certain descent through these conditions: we start with $p_0=1$, and then given $p_\alpha\in G$, it must be that $G$ selects either $(p_\alpha)_0$ or $(p_\alpha)_1$, which we define as $p_{\alpha+1}$. At limits, we take the infimum $p_\lambda=\bigwedge_{\alpha<\lambda}p_\alpha$ and continue, as long as this remains nonzero. This process yields a continuous descent $\vec p=\langle p_\alpha\mid\alpha<\gamma\rangle$ in $G$, with $\bigwedge_\alpha p_\alpha=0$. This descent $\vec p$ cannot be in the ground model, since the difference antichain $d_\alpha=p_\alpha-p_{\alpha+1}$ would be a maximal antichain missed by $G$. By the presumed completeness of $\B$ in $V[G]$, let $$a=\bigvee\left\{(p_\alpha)_0\strut\mid (p_\alpha)_1\in G\right\}.$$ That is, the condition $a$ is the join of the conditions going to the left, when the generic filter opted to go right. Notice that this is simply the join of part of the difference antichain, since $(p_\alpha)_0=p_\alpha-(p_\alpha)_1=d_\alpha$, when $(p_\alpha)_1\in G$. Because of this, given $p_\alpha\in G$, it follows that $(p_\alpha)_0\leq a\iff (p_\alpha)_1\in G$. Thus, from $a$ and the splitting function, we can reconstruct the descent $\vec p$ in the ground model, contrary to our earlier observation that it was not in the ground model. So $\B$ is not complete in $V[G]$. $\Box$.<|endoftext|> TITLE: Forall exists formula QUESTION [6 upvotes]: This is a reference request. Let $A$ be a formula in the language of rings which is of the form $\forall_{x_1}\dots\forall_{x_n}\exists_{y_1}\dots\exists_{y_m} F$, where $F$ is quantifier-free. I once read that if $A$ is valid for all finite fields, then it is valid for $\mathbb C$. Where would I find a proof of this statement? REPLY [11 votes]: I don't know a reference but it's not that hard to prove. Let's call your formula $\varphi$. Then by classical arguments, the models of $\varphi$ are closed under directed union, in particular since $\varphi$ holds in any finite field, it holds in any $\overline{\mathbb{F}_p} = \displaystyle\bigcup_{n<\omega}\mathbb{F}_{p^n}$, the algebraic closure of the field with $p$ elements, which is the directed union of its finite subfields. But then by Los's theorem, $\varphi$ holds in $\displaystyle\prod_{p\in \mathbb{P}}\overline{\mathbb{F}_p}/\mathcal{U}$ for any $\mathcal{U}$ ultrafilter on $\mathbb{P}$. Picking a non-principal ultrafilter yields a field isomorphic to $\mathbb{C}$, so the formula holds in $\mathbb{C}$ too.<|endoftext|> TITLE: Strengthened supercongruences for Ramanujan-type formulas for $1/\pi^k$ QUESTION [5 upvotes]: The question below is again a follow-up of an old question. Motivation: Zhi-Wei Sun listed a number of supercongruences attached to Ramanujan-type $1/\pi$ formulas in the arXiv paper which can be found here. A typical example of these congruences can be found on p. 18 of this paper: $$\begin{split}\sum_{k=0}^{p-1}\frac{(1/2)_k(1/3)_k(2/3)_k}{(1)_k^3}\left(-\frac{1}{500^2}\right)^k(14151k+827)\equiv\\ 827\left(\frac{-3}{p}\right)p+\frac{13}{150}p^3B_{p-2}\left(\frac{1}{3}\right)\pmod{p^4},p>5\end{split}$$ which is stronger than the formula (24) in W. Zudilin's paper. $B_n(x)$ is a Bernoulli polynomial and $p$ is a prime number. Bernoulli polynomials and Euler polynomials occur quite frequently in other supercongruences conjectured by Sun. It is well known that Bernoulli numbers and Euler number also occur in the values of $\zeta$ and $L_{-4}$ at negative integers, which implies that one can replace Bernoulli polynomials and Euler polynomials by the non-zero values of degree 1 $L$-functions at negative integers. Experiment: When one compared the supercongruences raised by Sun with the factorization of $L$-functions of hypergeometric motives attached to Ramanujan $1/\pi$ formulas(see M. Watkins' table on p.29), one will see Euler numbers(or Bernoulli numbers) in the supercongruences where the L-functions attached to hypergeometric motives have $L_{-4}$(or $\zeta$) as a factor. Example: We denote the truncated sum $$\sum_{k=0}^{p-1}\frac{(1/2)_k(1/4)_k(3/4)_k}{(1)_k^3}\left(\frac{1}{99}\right)^{4k}(26390k+1103)$$ by $S_p$. Since the $L$-function attached the hypergeometric motive has $L_{-8}$ as a factor, we denote $$T_p=\left(S_p-1103\left(\frac{-2}{p}\right)p\right)/(p^3L_{-8}(3-p)).$$ Sun's strengthened supercongruences implies $$T_p\equiv\frac{x}{y}\pmod{p}$$ for every prime $p>5$ and integers $x,y$. The integers $x,y$ in the system of linear congruences can be determined by lattice reduction. We then determine $x=-5,y=1089$. It seems that the strengthened supercongruences can be found in every Ramanujan-type formula $1/\pi^k$ with the same method, e.g.,$$\begin{split}\sum_{k=0}^{p-1}\frac{(1/2)_k(1/3)_k(2/3)_k(1/6)_k(5/6)_k}{(1)_k^5}\left(-\frac{1}{80^3}\right)^k(5418k^2+693k+29)\equiv\\ 29\left(\frac{5}{p}\right)p^2-\frac{35}{216}p^5L_5(4-p)\pmod{p^6},p>5,\end{split}$$ where $L_5$ in the congruence is the Dirichlet $L$-function attached to the primitive real character with period $5$. We also note that $L_5$ is a factor of the $L$-function attached to the hypergeometric motive of the formula (29) in Zudilin's paper. Questions: Why do the values of degree 1 $L$-factor at negative integers occur in the strengthened supercongruences? REPLY [2 votes]: For the two examples in your question I conjecture $p$-adic expansions that begin as: $$ S_p = 1103 \left(\frac{-2}{p}\right)p - \frac{5}{1089} L_{-8,p}(2) p^3 + \cdots, $$ and $$ S_p = 29 \left(\frac{5}{p}\right) p^2 - \frac{35}{216} L_{5,p}(3) p^5 + \cdots, $$ where $L_{-8, p}(k)$ is the $p$-adic analogue of $L_{-8}(k)$, and $L_{5, p}(k)$ the $p$-adic analogue of $L_{5}(k)$. As $$L_{-8, p}(2) \equiv L_{-8}(3-p) \pmod{p},$$ and $$L_{5,p}(3) \equiv L_5(4-p) \pmod{p},$$ the above $p$-adic expansions imply the supercongruences in the question. For details and support of the conjectured $p$-adic expansions, see http://arxiv.org/abs/1910.01961)<|endoftext|> TITLE: To compare the total, base and fiber spaces of two fiber bundles QUESTION [8 upvotes]: Consider the following commutative diagram of the fiber bundles $% F\rightarrow E\rightarrow B$ and $F^{\prime }\rightarrow E^{\prime }\rightarrow B^{\prime }$ where $B^{\prime }$ is simply connected space (but $B$ no need simply connected) and $F$, $F^{\prime }$, $B$, and $B^{\prime }$ are path-connected spaces. $\require{AMScd}$ \begin{CD} F @>{}>> E @>{}>> B \\ @VVV @VVV @VVV\\ F' @>{}>> E' @>{}>> B' \end{CD} If \begin{equation*} H^{\ast }\left( B^{\prime };% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( B;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \end{equation*} and \begin{equation*} H^{\ast }\left( E^{\prime };% %TCIMACRO{\U{211a} } %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( E;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \end{equation*} are isomorphisms, then is \begin{equation*} H^{\ast }\left( F^{\prime };% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( F;% %TCIMACRO{\U{211a} } %BeginExpansion \mathbb{Q} %EndExpansion \right) \end{equation*} an isomorphism? I want to use Zeeman comparison theorem but the local coefficient system of the fiber bundle $F\rightarrow E\rightarrow B$ has not simple. REPLY [2 votes]: Let $f$ be the map between the two Serre fibrations. The question then asks: if $f$ is a rational homotopy equivalence on the base and total spaces, then is it a rational homotopy equivalence on the fibers? In other words, is the induced map $f_\mathbf{Q}:F_\mathbf{Q}\to F'_\mathbf{Q}$ a homotopy equivalence? As pointed out above, the answer to this question is no in general. Conceptually, the reason for this negative answer can be boiled down to the fact that rationalization preserves (homotopy) colimits, but fiber sequences are (homotopy) limits. Concretely, this means that the fiber of the rationalized map $E_\mathbf{Q}\to B_\mathbf{Q}$ (which maps, via the homotopy equivalence $F_\mathbf{Q}$, to $E'_\mathbf{Q}\to B'_\mathbf{Q}$) is not necessarily $F_\mathbf{Q}$. (However, if $E$ and $B$ are simply connected, the map from $F_\mathbf{Q}$ to the fiber of $E_\mathbf{Q}\to B_\mathbf{Q}$ is indeed a homotopy equivalence.) Similarly, the fiber of the map $E'_\mathbf{Q}\to B'_\mathbf{Q}$ is not necessarily $F'_\mathbf{Q}$. This means that one cannot use the long exact sequence in homotopy to deduce that $f_\mathbf{Q}:F_\mathbf{Q}\to F'_\mathbf{Q}$ induces isomorphisms on rational homotopy, i.e., is a rational homotopy equivalence (hence giving a negative answer to your question in general, but a positive answer for simply connected base and total spaces).<|endoftext|> TITLE: How many facets does the convex hull of all the roots of a root system have? QUESTION [10 upvotes]: Let $V$ be an $n$-dimensional Euclidean vector space with inner product $\langle\cdot,\cdot\rangle$ and $\Phi$ an irreducible crystallographic root system in $(V,\langle\cdot,\cdot\rangle)$. Question 1: Is there a root-theoretic formula for the number of facets of $\mathrm{ConvHull}(\Phi)$? Surely this must be known but I have not been able to find a reference. Some thoughts about this question: Consider instead the dual polytope $\mathcal{P}:=\{v\in V\colon \langle v,\alpha \rangle \leq 1 \textrm{ for all $\alpha \in \Phi$}\}$. Counting facets of $\mathrm{ConvHull}(\Phi)$ is the same as counting vertices of $\mathcal{P}$. Since $\mathcal{P}$ is $W$-invariant, it is "enough" to understand the intersection of $\mathcal{P}$ with the dominant cone. This intersection is the polytope with facets: $\langle v, \alpha_i\rangle \geq 0$ for all simple roots $\alpha_i$, $1 \leq i \leq n$; $\langle v, \theta\rangle \leq 1$ for the highest weight $\theta$ of $\Phi$. But the polytope given by those inequalities is well-known: it is just the fundamental alcove $A_0$ (I think technically it is the fundamental alcove for the dual root system $\Phi^\vee$? I always get tripped up by the distinction between $\Phi$ and $\Phi^\vee$). Note that $A_0$ is a simplex. Explicitly, the vertices of $A_0$ are $0$ together with $\frac{1}{a_i}\omega_i$ for $1\leq i \leq n$, where $\omega_1,\omega_2,\ldots,\omega_n$ are the fundamental coweights (i.e., the dual basis to the basis of simple roots), and $a_1,a_2,\ldots,a_n$ are the integer coefficients determined by writing $\theta = a_1\alpha_1+a_2\alpha_2+\cdots+a_n\alpha_n$. For example, the minuscule coweights (i.e., those $\omega_i$ with $a_i=1$) are a subset of the vertices of $A_0$. So we understand the vertices of $A_0$ and $\mathcal{P}=W(A_0)$. We should be almost done. How many $W$-orbits does $\frac{1}{a_i}\omega_i$ have? That's easy: its stabilizer is $W_i$, the Weyl group of the maximal parabolic root system obtained by removing node $i$. So $\#W(\frac{1}{a_i}\omega_i) = \#W/\#W_i$. So the number of vertices of $\mathcal{P}$ is given by the nice root-theoretic formula $\sum_{i=1}^{n} \#W/\#W_i$, right? Not quite. You can check that this already doesn't work for $\Phi=B_2$: $\mathrm{ConvHull}(\Phi)$ has $4$ facets, but the formula would give $8/2+8/2=8$ as an answer. The problem: not every $\frac{1}{a_i}\omega_i$ is actually a vertex of $\mathcal{P}$. Question 2: Which of the $\frac{1}{a_i}\omega_i$ are actually vertices of $\mathcal{P}$? Of course an appropriate answer to Question 2 would yield an answer to Question 1 by the above discussion. REPLY [4 votes]: These are studied under the name root polytopes by various authors in particular in the context of abelian ideal in root posets. See for example Cellini Triangulations of root polytopes and Cellini-Marietti Root polytopes and Borel subalgebras where facet ideals of abelian ideal are studied. As Sam pointed out in the comments, the latter shows that the orbits of the facets correspond to the simple roots that do not disconnect the extended Dynkin graph give a facet description of the root polytope.<|endoftext|> TITLE: Rings with all non-prime ideals finitely generated QUESTION [9 upvotes]: Motivated by this question, I would like to ask: If all non-prime ideals in a ring are finitely generated, then is the ring Noetherian? Can we at least say anything in the local case? Note that for zero-dimensional rings, the answer is yes by the linked question. So we only need to think about rings of positive dimension. NOTE: All our rings are commutative with unity. REPLY [8 votes]: No. If $\mathbb Z_{p^{\infty}}$ is a Prufer group for prime $p$, then its endomorphism ring is isomorphic to the ring $\mathbb Z_p$ of $p$-adic integers. Hence $\mathbb Z_{p^{\infty}}$ is a $\mathbb Z_p$-module, and we can form the idealization $R:=\mathbb Z_p\oplus \mathbb Z_{p^{\infty}}$. The ideals of this ring are known: they are of the form $0\oplus H$ for a subgroup $H\leq \mathbb Z_{p^{\infty}}$ or of the form $(p^k)\oplus \mathbb Z_{p^{\infty}}$ for some nonnegative integer $k$. (See Example 1 of Paul A. Froeschl, Chained rings. Pacific J. Math. Volume 65, Number 1 (1976), 47-53.) In particular, $R$ is a local ring with exactly one non-finitely generated ideal $I:=0\oplus \mathbb Z_{p^{\infty}}$. This ideal is prime because $R/I$ is isomorphic to $\mathbb Z_p$, a domain.<|endoftext|> TITLE: Confusion about Montgomery's pair correlation conjecture QUESTION [7 upvotes]: This question will be based roughly on the Bourgade–Keating review on Zeta function and eigenvalue asymptotics (BK): https://link.springer.com/chapter/10.1007/978-3-0348-0697-8_4 To set up the question, we consider the Riemann zeta function $\zeta(s)$ with zeros on the critical line $\frac{1}{2} + i t_n$. The unfolded zeros are defined as: $w_n = \frac{t_n}{2\pi} \log \frac{t_n}{2\pi}$. The name "unfolded" is justified by the fact that distribution of $w_n$ is asymptotically uniform along the critical axis. In terms of these unfolded zeros, we can define the following integral: \begin{equation} R_{2,\zeta}(f,W) = \int_{-\infty}^{\infty} f(x) \frac{1}{W} \sum_{\substack {j\neq k \\ w_j,w_k \leq W}} \delta(x-w_j+w_k) dx \end{equation} Suppose the $w_n$'s had no pair correlation, the integral above would simply equal to $\int f(x) dx$. However, $w_n$'s clearly have pair correlations! So deviation of the integral from $\int f(x) dx$ roughly measures the pair correlations between the unfolded zeros $w_n$, up to some cutoff $W$. For $\zeta$ functions Montgomery proved an important theorem about this integral stated in KS as follows: Theorem 1 (Montgomery): Assume the Riemann Hypothesis. Then for test functions f(x) such that: $$ \hat f(\tau) = \int_{-\infty}^{\infty} e^{2 \pi i x \tau} f(x) dx$$ has support in $(-1,1)$, the following limit exists: $$ \lim_{W \rightarrow \infty} R_{2,\zeta}(f,W) = \int_{-\infty}^{\infty} f(x) R_2(x) dx $$ with $$R_2(x) = 1 - (\frac{\sin(\pi x)}{\pi x})^2 $$ The confusion I have is: this $R_2(x)$ seems to carry unnecessary information. To show that, suppose I Fourier transform the integral: \begin{equation} \int_{-\infty}^{\infty} f(x) R_2(x) dx = \int_{-\infty}^{\infty} \hat f(\tau) K(-\tau) d \tau \end{equation} Where $K(\tau)$ is the Fourier transform of the pair correlation function, sometimes referred to as the spectral form factor. Then since $\hat f(\tau)$ only has support on $(-1,1)$, I can restrict the $\tau$ integral to $(-1,1)$. But that would mean the integral is only sensitive to $K(\tau)$ for $\tau \in (-1,1)$. Within that domain, there are other functions that produce the same $K(\tau)$. For example, following equation (26) in BK, we can define: $$ \tilde R_2(x) = 1 - \frac{1}{2 (\pi x)^2} \quad R_2(x) = 1 - (\frac{\sin(\pi x)}{\pi x})^2 $$ These functions have the same Fourier transform within $(-1,1)$. Montgomery's Theorem is usually stated as a connection between random matrix theory and zeta function because $R_2(x)$ is precisely the pair correlation of Wigner random matrix ensembles. However, the calculation above would suggest that $\tilde R_2(x)$ would do just as well within the domain of interest. Hence, I feel that $R_2(x)$ carries unnecessary information about the pair correlations, and it now seems rather artificial that it matches the random matrix results. This seems like a simple enough question. But I haven't found any explanation of it in various review articles on RMT-zeta function connections. So I would like some help from the experts: Why did Montgomery put the RMT correlation function in his theorem if it carries unnecessary information? Was it just an inspired guess or is there something deep I am missing? Note: The referenced review article seem to be on the boundary between math and physics. But since my question is more about the mathematical side, I thought it would be best to pose it here. If the moderators feel this is more fitting for other Physics SE or Math SE, please help me move the question to the right place. Thanks REPLY [5 votes]: Assuming the Riemann Hypothesis Montgomery consider the function $$F(\alpha)=F(\alpha,T)=\Bigl(\frac{T}{2\pi}\log T\Bigr)^{-1} \sum_{0<\gamma, \gamma'\le T} T^{i\alpha(\gamma-\gamma')}w(\gamma-\gamma')$$ where $w(x)=4/(4+x^2)$. He proved results only in the range $-1+\varepsilon\le\alpha\le 1-\varepsilon$. Therefore he restricted $\widehat r(\alpha)$ to this range and asserts $$\sum_{0\le \gamma,\gamma'\le T} r\Bigl((\gamma-\gamma')\frac{\log T}{2\pi}\Bigr) w(\gamma-\gamma')=\Bigl(\frac{T}{2\pi}\log T\Bigr)\int_{-\infty}^{+\infty}F(\alpha)\widehat{r}(\alpha).$$ From this result he obtains some consequences. For example, at least $2/3$ of the zeros are simple, that are proved conditionally on the Riemann hypothesis. Then he give reasons to assume that $F(\alpha)\sim1$ for $|\alpha|\ge1$ and make his conjecture that $1-((\sin\pi u)/\pi u)^2$ is the pair correlation function for the zeros of the zeta function. This leads then to F. J. Dyson famous remarks. So, you have to distinguish between what Montgomery proved under Riemann hypothesis and the conjecture of the correlation that goes beyond the Riemann hypothesis. Montgomery's conjecture was reinforced by the statistics of Odlyzko. There is also another alternative hypothesis which postulates the normalized zeros tends to be separated by integers or half integers. Odlyzko's figure was restricted to the range $0\le \gamma-\gamma'\le 6$. In 2013 I duplicated this statistics but using the range $0\le \gamma-\gamma'\le 40$. My figures revealed a surprising structure for this range, more in the mood of the alternative hypothesis. I used Pratt zeros for this statistics, nevertheless it will be useful that somebody with more computer expertise than I cofirm my results. The figure is obtained from the zeros in the range 20.000.000 to 30.000.000 more or less.<|endoftext|> TITLE: Is restriction a closed map? QUESTION [6 upvotes]: Originally asked on MSE. Let $X$ be a normal (or even metrizable) topological space and let $Y$ be a closed subset of $X$. Let $C(X)$ be the linear space of all continuous scalar functions on $X$ endowed with the compact-open topology. Consider a map $R:C(X)\to C(Y)$ defined by $Rf=f|_Y$. This map is obviously linear, and it is easy to see that it is continuous. By Tietze theorem it is also surjective. It seems that this is in fact a quotient map. However, what interests me is the following question: Let $B$ be a closed convex balanced subset of $C(X)$. Is $RB$ closed in $C(Y)$? Under additional assumption that $B$ is weakly compact, this is true, since then $B$ is compact in the pointwise topology, and so $RB$ is pointwise compact, and so pointwise closed, and so closed in $C(Y)$. I can also show that this is wrong if we don't require $Y$ to be closed in $X$. REPLY [7 votes]: The answer to the main question is negative: Consider the compact subset $X=[0,1]\cup \{2\}$ of the real line and let $Y=\{2\}$ be a singleton in $X$. In the function space $C(X)$ consider the closed convex balanced subset $$B:=\{f\in C(X):\sup_{x\in [0,1]}|f(x)|\le 1,\;f(0)=0,\;f(2)=\int_0^1 f(t)dt\}.$$ It is easy to see that for the restriction operator $R:C(X)\to C(Y)$, the image $R(B)=\{f\in C(Y):|f(2)|<1\}$ is not closed in $C(Y)$.<|endoftext|> TITLE: Is $C^{\infty}(E)$ a projective Frechet $C^{\infty}(M)$-module for a $C^{\infty}$-fiber bundle $E\to M$ with compact fiber? QUESTION [7 upvotes]: The question is a special case of a previous question. Let $M$ be a compact smooth manifold, then it is clear that $C^{\infty}(M)$ is a Frechet algebra with pointwise multiplication and a collection of semi-norm defined by $p_{\alpha}(f):=\sup_{\beta\leq\alpha}||\partial^{\beta}(f)||$. Now let $E\to M$ be a $C^{\infty}$-fiber bundle with compact base and compact fiber. Then it is clear that $C^{\infty}(E)$, the space of smooth functions on the total space of the fiber bundle, is a Frechet $C^{\infty}(M)$-module. My question is: is $C^{\infty}(E)$ always a projective Frechet $C^{\infty}(M)$-module? I think this question is trivial for experts. Please let me know if there is any references or it the question is not suitable for mathoverflow. REPLY [9 votes]: The answer is Yes. Basically, one can realize $C^\infty(E)$ as a direct summand in a larger projective $C^\infty(M)$-module. Perhaps this really is trivial to experts, but to me the argument occurred only after staring sufficiently long at related arguments in this article, which came up in your other question: Ogneva, O. S., Coincidence of the homological dimensions of the Fréchet algebra of smooth functions on a manifold with the dimension of the manifold, Funct. Anal. Appl. 20, 248-250 (1986); translation from Funkts. Anal. Prilozh. 20, No. 3, 92-93 (1986). ZBL0626.46057. Namely, start by recalling the isomorphism $C^\infty(U \times F) = C^\infty(U) \hat{\otimes} C^\infty(F)$, where $\hat{\otimes}$ is the projective tensor product of Fréchet spaces. This means that $C^\infty(U\times F)$ is a free $C^\infty(U)$-module (freely generated by $C^\infty(F)$, in the category of Fréchet $C^\infty(U)$-modules), and hence projective. By the result of Ogneva, for any open $U\subset M$, $C^\infty(U)$ is projective over $C^\infty(M)$. Hence, $C^\infty(U\times F) \cong C^\infty(U) \otimes_{C^\infty(M)} C^\infty(M\times F)$ is also projective over $C^\infty(M)$, being the tensor product of projective modules. Next, consider a countable, locally finite open cover of $(U_i)$ of $M$, trivializing the fiber bundle $E\to M$ as $E|_{U_i} \cong U_i\times F \to U_i$, where $F$ is the typical fiber, and let $(\chi_i)$ be a partition of unity subordinate to this cover. Such a cover and corresponding partition of unity certainly exist if the base $M$ is compact, but also more generally even if $M$ is not compact, but satisfies a suitable countability condition (second countable, paracompact). Now, the countable direct product $\prod_i C^\infty(U_i \times F)$ is still Fréchet and projective over $C^\infty(M)$, and it has $C^\infty(E)$ as a direct summand, as evinced by the inclusion/projection maps \begin{align*} C^\infty(E) \to \prod_i C^\infty(U_i \times F) &\colon f \mapsto (f|_{E_{U_i}}) , \\ \prod_i C^\infty(U_i \times F) \to C^\infty(E) &\colon (f_i) \mapsto \sum_i \chi_i f_i . \end{align*}<|endoftext|> TITLE: Functoriality for wrong way maps QUESTION [9 upvotes]: In the K-theory formulation of the index theorem one defines the topological index in terms of the so called wrong way maps. Those maps are defined for embeddings of compact manifolds $i:X \to Y$: see section 2.2 (page 16) in Landweber - K-theory and elliptic operators article for more details. I would like to understand why this construction is functorial, i.e. why $(j \circ i)_!=j_! \circ i_!$ where $i:X \to Y$ and $j:Y \to Z$ are embeddings of compact manifolds. Forgive me if this question is too elementary for this site. REPLY [4 votes]: You will find a careful proof of this in Eldon Dyer's monograph "Cohomology theories". It is true for any multiplicative cohomology theory $h^*$ and smooth maps $i:X\to Y$ and $j:Y\to Z$ which are $h^*$-oriented, meaning with a choice of Thom class for their stable normal bundles. The definition of the Umkehr is on page 53, and the functoriality result is stated as Theorem 8 on page 57.<|endoftext|> TITLE: Must a manifold covered by $ S^n $ admit a metric of constant positive sectional curvature? QUESTION [10 upvotes]: Suppose that the smooth manifold $ M $ has the n-sphere for its universal cover (in the topological sense). Does there exist a Riemannian metric on $ M $ (not necessarily compatible with the covering map) for which all sectional curvatures of $ M $ are constant, equal to 1? REPLY [7 votes]: Updated: If $n = 3$, then the answer to your question is yes. This is essentially the Thurston Elliptization Conjecture which was settled by Perelman's proof of the Geometrization Conjecture. In my initial answer, I had hoped to use the obstruction $\alpha$ to construct a manifold which was finitely covered by $S^n$ which did not even admit positive scalar curvature metrics. As is discussed in the comments, this construction cannot work for lens spaces. In fact, it cannot happen for $n \geq 5$ at all, as was shown in Fake spherical spaceforms of constant positive scalar curvature by Kwasik & Schultz. In Paul Seigel's answer, he mentions that the answer to your question is no. Concretely, in Free Metacyclic Group Actions on Homotopy Spheres, Petrie shows that there is a smooth free action of $\mathbb{Z}_p\rtimes\mathbb{Z}_q$ on $S^{2q-1}$ where $p$ is odd and $q$ an odd prime. Such a quotient cannot admit a constant positive sectional curvature metric. However, by the result of Kwasik & Schultz, it does admit a constant positive scalar curvature metric. Allow me to expand a little bit on the first part on Paul Siegel's answer before heading in a slightly different direction in the search for examples. There is a homomorphism $$\alpha : \Omega_n^{\text{spin}} \to KO(S^n) = \begin{cases}\mathbb{Z} & n \equiv 0\bmod 4\\ \mathbb{Z}_2 & n \equiv 1, 2 \bmod 8\\ 0 & \text{otherwise}\end{cases}$$ defined by Milnor. It was shown by Milnor and Adams that when $n = 1, 2 \bmod 8$ and $n \neq 1, 2$, there is an exotic $n$-sphere $\Sigma$ such that $\alpha(\Sigma) \neq 0$. In all other dimensions, exotic spheres have value zero. It was later shown by Hitchin that if a closed spin manifold $X$ admits a metric of positive scalar curvature, then $\alpha(X) = 0$. So, if $X$ is an $n$-dimensional closed spin manifold which admits a metric of positive scalar curvature and $n \equiv 1, 2 \bmod 8$, $n \neq 1, 2$, then $X\#\Sigma$ does not admit a metric of positive scalar curvature (connected sum is the addition operation in $\Omega^{\text{spin}}_n$ and $\alpha$ is a homomorphism). If $S^n \to X$ is the $k$-fold universal cover (here $S^n$ denotes the standard smooth sphere), then the universal cover of $X\#\Sigma$ is $S^n\# k\Sigma$. The group of exotic spheres is finite (except possibly in dimension four), so let $d$ be the order of $\Sigma$. If $d \mid k$, then $S^n\# k\Sigma = S^n$ and you'd have an example of the phenomena you're looking for. Note, this doesn't help when $n \equiv 2 \bmod 8$ as $S^n$ only covers $\mathbb{RP}^n$ which is not orientable, and hence not spin. When $n \equiv 1 \bmod 8$, then $S^n$ is the universal cover of infinitely many manifolds, e.g. lens spaces. In particular, for any $k$, there is an closed orientable manifold $X$ which has $S^n$ as its $k$-sheeted cover, all that's left to know is when such an $X$ can be chosen to be spin. I don't know how to do this off the top of my head, but the answers to this question seem promising. I believe that examples can be constructed this way, but I have never taken the time to construct one, sorry.<|endoftext|> TITLE: How continuous can a bijection between line and plane be? QUESTION [13 upvotes]: Is there a bijection $f$ from $[0, 1]$ to $[0, 1]^2$ such that the set of points of discontinuity of $f$ has measure zero? If not, could it be dense/comeager? REPLY [17 votes]: Yes! Let $X$ be the Cantor middle third set. $X \subset [0,1]$ is closed, measure zero, and $|X| = 2^{\aleph_0} = |X^c|$. Notice there is a continuous bijection $g: X^c \rightarrow (X^c \times \{0\})$ given by $g(x) = (x,0)$. It's easy to see the caridnality of $(X^c \times \{0\})^c$ is $2^{\aleph_0}$, and $X$ has cardinality $2^{\aleph_0}$, so there is also a (not-continuous) bijection $h: X \rightarrow (X^c \times \{0\})^c$. Now define $f(x) = g(x)$ if $x \not\in X$ and $f(x) = h(x)$ if $x \in X$. We can check that $f$ is continuous on $X^c$ which is an open measure 1 set. It might be more interesting to know about the set of points of discontinuity of $f^{-1}$. Perhaps a property similar to Lebesgue covering dimension can give some constraints.<|endoftext|> TITLE: Atiyah's May 2018 paper on the 6-sphere QUESTION [50 upvotes]: A couple years ago Atiyah published a claimed proof that $S^6$ has no complex structure. I've heard murmurs and rumors that there are problems with the argument, but just a couple months ago he apparently published a follow-up which fleshes out the details. He writes: In [1] I gave a proof of the long-standing conjecture that the 6-dimensional sphere has no complex structure. In this paper I will present the proof in a more transparent manner. I use the example of the 6-sphere to shed new light on many problems of physics. In the future I expect these ideas will provide a different perspective, with substantial benefits in all areas. Unfortunately I can't find a version of the new paper in front of a pay wall, and even if I could I would wonder if he had addressed any of the possibly existent problems with the original. Does anyone have more information (or at least references)? REPLY [16 votes]: Since the question has been "hanging on" for a while, I think it makes sense to give an outline of Atiyah's argument in the paper. Note that the paper is short (page 1 introduction, page 5-6 reference and further work). The main arguments in the paper are in pages 2-4. (page2) First he introduced the "conformal sphere", which I guess Atiyah meant $\mathbb{S}^{6}$ as a conformal manifold with a conformal metric. The `standard' conformal sphere $S(\infty)$ in the Minkowski 8-space $M=(x,y,t)$ can then be viewed as the limit of smaller spheres $S(c)$, where the equation $$|x|^2+|y|^2=c^2 t^2, x\in \mathbb{R}^{4}, y\in \mathbb{R}^{3}, c\in \mathbb{R}^{+}$$ The conformal sphere does not have a Riemannian metric but it has a conformal structure induced from $S(c)$ with hyperbolic metrics (inherited from $M$?). Atiyah defined a notion natural, which he defined as " Natural always means compatible with the appropriate symmetry group, which here is the Lorentz group of automorphisms of Minkowski space". In particular here $S(\infty)$ inherit the action of $SO(1,7)$ instead of $SO(8)$. Then Atiyah suggest this is the main source of "confusion" he wanted to dispel. (page 3) Atiyah suggested to use the isomorphism involving quotient of Lie groups for the conformal sphere: $$ \mathbb{S}^{6}\cong \textrm{Spin}^{+}(7,1)/(\mathbb{Z}/2\times \textrm{Spin}(6)) $$ The $\mathbb{Z}/2$ factor acts by interchanging $(x, t)\rightarrow (-x,-t)$. In particular, any two points on $\mathbb{S}^{6}$ can be transformed to "standard antipodal pair" $(e,0), (-e,0), e\in \mathbb{R}^{3}$. (Not sure how to prove this) Atiyah quotes a version of index theorem which Bott proved for homogeneous vector bundles. If I recall correctly this is an independent piece work of Bott produced when he was trying to verify the index theorem. So it can be viewed as special case of index theorem. However, it is not entirely clear what he meant exactly here. Atiyah then claim the index in the representation ring for the round $6$-sphere lands in $R(\textrm{Spin}(7))$. For the conformal sphere it lands in $R(\mathbb{Z}/2\times \textrm{Spin}(6))$. (I guess) by a deformation argument Atiyah then reduce the case to $\Gamma=\mathbb{Q}_{8}$, the quarterion group of order 8. It is pointed out that index theorem only needs an almost complex structure (correct). Atiyah then make a bold font claim that "$\Gamma$ acts on the conformal sphere $\mathbb{S}^{6}$ without invoking any additional symmetry". This I do not really follow. Does he mean that no other group structure in the identification $$ \mathbb{S}^{6}\cong \textrm{Spin}^{+}(7,1)/(\mathbb{Z}/2\times \textrm{Spin}(6)) $$ is actually used??? (page 4) Atiyah suggests this "fact" is related to CPT theorem. Further the presence of $\Gamma=\mathbb{Q}_{8}$ can be understood from the point of view of complex conjugations: We have $$ 0\rightarrow \mathbb{Z}/2\rightarrow \Gamma\rightarrow \mathbb{Z}/2\times \mathbb{Z}/2\rightarrow 0 $$ And the first $\mathbb{Z}/2$ action can be interpreted as conjugation on $\mathbb{C}^{3}$ preserving conjugation. It is pointed out that the index would be equal to $(-1)^{F}$, which is a topological invariant. For Minkowski space it would be odd. (Not sure how relevant is this...). (The claimed proof) Assume there is a hypothetical complex structure on the conformal sphere $\mathbb{S}^{6}=S(\infty)$, which he viewed as limit of $S(c)$ with hyperbolic metric. Note that this sphere is homeomorphic to the standard round sphere. Apply the $\Gamma$-equivariant version of the theorem to $\mathbb{S}^{6}$ with the complex structure. Since the central extension $\mathbb{Z}/2$ acts by complex conjugation , it is an abelian representation (meants a representation of $V$). However, I am not sure why the fact $\mathbb{S}^{6}$ having the hypothetical complex structure means the first $\mathbb{Z}/2$ action is trivial for the representation. Since the theorem does not depend on the metric, with the choice of round metric on $\mathbb{S}^{6}$ the index becomes an element in the representation ring of $\Gamma$. But then $\Gamma$ acts freely on the round sphere, so its index is the regular representation of $\Gamma$, which is not abelian. In particular it cannot be identified with a representation of $V$. The crucial part of this paper left unaddressed (to me) are: How the existence of the complex structure on $\mathbb{S}^{6}$ "forced" it to have an abelian index when using (the equivariant version) of index theorem? Regardless of (1), does Atiyah implicitly assume the complex structure on $\mathbb{S}^{6}$ has to be "compatible" with its structure as a homogeneous space? In order words, how is the reduction from $\mathbb{Z}/2\times Spin(6)$ to $\Gamma$ related to the complex structure given by (purportedly some wierd metric)? What would happen if it does not? Assume (1) and (2) are addressed. Now why an almost complex structure on $\mathbb{S}^{6}$ would be incompatible with the claims in (1) and (2)? In other words, what differentiates $i$ from $J$ for the purpose of $\mathbb{Z}/2$ actions, other than the fact that the central extension inside of $\Gamma$ can be identified with the complex conjugation? It does not seem Atiyah used theorems like Nirenberg-Newlander in any significant way. It seems to me what Atiyah is really claiming is that a "special class of metric" on the conformal sphere $\mathbb{S}^{6}$ cannot give it a nice enough complex structure with respect to the existing homogeneous space structure. It would be interesting to check if the almost complex structure on $\mathbb{S}^{6}$ is compatible with the action of $\Gamma$, for example. Otherwise I suspect there may be an arithmetic mistake somewhere in the proof.<|endoftext|> TITLE: iterated limit sets of a countable subset of real numbers QUESTION [6 upvotes]: Let $A\subset \mathbb{R}$ be a closed subset, and $A'$ be the sets of limit points. We know that if $A$ is a countable set, $A'$ is a proper subset of $A$. Is it possible to find a subset( closed and countable) so that $A^{(n)}$ is non-empty for any $n\in \mathbb N$? We denote by $A^{(n)}=(A^{(n-1)})'$. REPLY [13 votes]: Yes, just take a copy of the ordinal $\omega^\omega$ in the reals. This has Cantor Bendixson rank exactly $\omega$. One way to see this is first to understand how to make a closed set last for exactly $n$ steps. A convergent sequence lasts one step; a convergent sequence of convergent sequences lasts two steps, and so on. At each step, take a convergent sequence of sets of the previous type. Put them together by considering a copy of the $n^{th}$ set in the interval $[n,n+1]$. In this way, applying $n$ derivatives the desired set does this separately in each interval, and the $n^{th}$ interval last for $n$ steps before disappearing. So the whole set will last for $\omega$ steps. If you want a compact example, you can squeeze the example into an open interval and add a limit point (which will add one final isolated point, disappearing in one additional step). If you think about it, the ordinal $\omega^\omega$ is just like this, since $\omega^\omega$ is the limit of the ordinals $\omega^n$, which lasts for $n$ iterations of the derivative. Every countable ordinal has an order isomorphism into the rationals, by Cantor's universality theorem for the countable endless dense linear order. By taking the closure of such an order, you get countable closed sets in the reals of any desired Cantor Bendixson rank. It may be interesting to note that this kind of of problem was the origin of Cantor's discovery of the ordinals, since before he knew the ordinals, he was playing with sets like this, and they lead naturally to the concept of ordinal.<|endoftext|> TITLE: Uniqueness of quasi-inverses in infinity categories QUESTION [7 upvotes]: I've been trying to learn some of the basic language of infinity-category theory (in the sense of Lurie), and in particular, to understand which basic statements in (1-)categories have analogues in the infinity-categorical setting. Specifically, I'm trying to understand how equivalences behave in infinity categories, motivated by the following question. Vague Question: in an infinity-category $\mathcal C$, to what extent is it true that a quasi-inverse to a morphism $f$, if it exists, is unique up to a contractible space of choices? Here is how I believe this question should be made precise. If we let $I$ denote the nerve of the undirected interval category (the category with two objects and exactly one morphism between each pair of objects), then there is an inclusion $\Delta^1\hookrightarrow I$ coming from the inclusion of the directed interval category in the undirected interval category. Thus we have a restriction map $$\mathcal C^I\rightarrow\mathcal C^{\Delta^1}$$ of simplicial sets (in fact a functor of infinity-categories which is a fibration for the Joyal model structure), where informally the left-hand side is the infinity-category of quasi-inverse pairs $(g,h)$ of morphisms in $\mathcal C$ (with a family of homotopies realising their quasi-inverseness), the right-hand side is the infinity-category of arrows in $\mathcal C$, and the functor takes a pair $(g,h)$ to its first component $g$. Precise Question: if $f$ is a morphism in an infinity-category $\mathcal C$, is it true that the fibre of the map $\mathcal C^I\rightarrow\mathcal C^{\Delta^1}$ over the point $f$ of $\mathcal C^{\Delta^1}$ is either empty or a contractible Kan complex? I am interested both in answers to the precise question and the vague questions above, especially if there are alternative precise formulations which make the answer clearer. Remark: The condition that $f$ admit a quasi-inverse should be be equivalent to the assertion that it be an isomorphism in the homotopy category of $\mathcal C$. This is asserted in this nlab page, and something similar in the language of topological categories is proved in the setting of topological categories in Proposition 1.2.4.1 of Higher Topos Theory. REPLY [4 votes]: A possibly simpler way of proving what you are after is using marked simplicial set. Recall that marked simplicial sets are pairs $(X,S)$ where $X$ is a simplicial set and $S\subseteq X_1$ is a set of 1-simplices of $X$ containing all degenerate 1-simplices. If $X$ is a simplicial set we will denote the minimal and maximal marking by $X^\flat$ and $X^\sharp$ respectively If $X,Y$ are two marked simplicial sets, we can form additional simplicial sets $\mathrm{Map}^\flat(X,Y)$ and $\mathrm{Map}^\sharp(X,Y)$ by $$ \mathrm{Hom}_{\mathrm{sSet}}(K,\mathrm{Map}^\flat(X,Y))=\mathrm{Hom}_{\mathrm{sSet}^+}(K^\flat\times X,Y)\,,$$ $$ \mathrm{Hom}_{\mathrm{sSet}}(K,\mathrm{Map}^\sharp(X,Y))=\mathrm{Hom}_{\mathrm{sSet}^+}(K^\sharp\times X,Y)\,.$$ In HTT.3.1.3.7 a simplicial model structure is constructed on the category of marked simplicial sets such that the fibrant objects are precisely the $\infty$-categories with the equivalences marked. The important part here will be that For any anodyne morphism of simplicial sets $A\to B$ the map $A^\sharp\to B^\sharp$ is a marked trivial cofibration. This is because of the definition of marked trivial cofibration (HTT.3.1.3.3) and the fact that for any simplicial set $A$ and every $\infty$-category $C$ $$ \mathrm{Map}^\sharp(A^\sharp,C)=\mathrm{Map}(A,\mathrm{Core}(C))$$ If $f:X\to Y$ is a marked trivial cofibration and $C$ is a fibrant object (i.e. an $\infty$-category with the equivalences marked), then the map $$f^*:\mathrm{Map}^\flat(B,C)\to \mathrm{Map}^\flat(A,C)$$ is a trivial fibration (HTT.3.1.3.3) Then we can factorize the arrow you want to study as $$C^J=\mathrm{Map}^\flat(J^\sharp,C)\to \mathrm{Map}^\flat((\Delta^1)^\sharp, C)\to \mathrm{Map}^\flat((\Delta^1)^\flat,C)=C^{\Delta^1}$$ The first arrow is a trivial fibration and the second is precisely the inclusion of the subcategory of $C^{\Delta^1}$ spanned by the equivalences.<|endoftext|> TITLE: Bi-Lipschitz extension QUESTION [8 upvotes]: Given a bi-Lipschitz homeomorphism $\Phi:\mathbb{B}^n(0,1)\to\mathbb{R}^n$, (that is a bi-Lipschitz map onto the image), can one find a bi-Lipschitz homeomorphism $\Psi:\mathbb{R}^n\to\mathbb{R}^n$ such that $$ \Psi|_{\mathbb{B}^n(0,\frac{1}{2})}=\Phi|_{\mathbb{B}^n(0,\frac{1}{2})}\ \ ? $$ REPLY [2 votes]: Let me describe a baby case of Sulivan's construction which is beautiful. Assume I have a homeomorphism $h$ from pentagon to itself that sends each side to itself. Let us assume that the pentagon has right angles in the conformal disc model of Lobachevsky plane. Extend the map $h$ to the disc applying reflections in the sides of pentagon, so the obtained map $\tilde h$ commutes with and any reflection in a side of pentagon. Note that $\tilde h$ is bi-Lipschitz and it is identity on the boundary of the disc. So we can extend $\tilde h$ by identity map outside of disc. It is problematic to do the same for general $h$ and higher dimensions, but all this seem to be solved by Sullivan.<|endoftext|> TITLE: Cayley graph of $A_5$ with generators $(1,2,3,4,5),(1,4,3,2,5)$ QUESTION [20 upvotes]: The Cayley graph of $A_5$ with two generators of order 5 seems rather complicated. What is its graph genus (orientable or non-orientable)? The best I could get by trial and error is an embedding without crossings on a sphere with 10 crosscaps. Running the following in Sage works in theory, but takes too long: A5 = groups.permutation.Alternating(5) S = [(1,2,3,4,5),(1,4,3,2,5)] d = A5.cayley_graph(generators=S) bt = d.to_undirected() bt.genus() In Sage we can also use JSMol to get a not-so-helpful view of the graph. REPLY [2 votes]: According to Sagemath documentation, the time complexity of their algorithm is $$ \mathcal{O}\left(|V| \prod_{v \in V} (d(v) - 1)!\right). $$ (Note that in this instance this evaluates to $6^{60}$.) However, quick google search reveals that approach via integer linear programming or SAT solvers might be viable for this particular case. I suggest you contact the authors of the following articles: Stronger ILPs for the Graph Genus Problem, 27th Annual European Symposium on Algorithms (ESA 2019) A Practical Method for the Minimum Genus of a Graph: Models and Experiments, International Symposium on Experimental Algorithms, SEA 2016: Experimental Algorithms, pp 75-88<|endoftext|> TITLE: Characterizing pseudo-differential operators as a subalgebra of continuous endomorphisms of tempered distributions QUESTION [7 upvotes]: I'm aware that the following question is at best a refined version of at least 2 questions which are already on this site. I think it is justified however in that it is more precise and has some new content in it. If, however, anyone decides after reading this question that it is a duplicate I apologize in advance. Fix a positive integer $n$. For every real number $m$ define the symbol class $S^m \subset C^{\infty}(\mathbb{R}^n_x \times \mathbb{R}^n_{\xi})$ consisting of functions whose derivatives are all bounded in the $x$-direction and which grows at most like $|\xi|^m$ in the $\xi$ direction (this is not a precise definition but hopefully for those who are familiar its clear what class of symbols i'm using). Any such symbol (element of $S^m$ for $m \in \mathbb{R}$), defines a pseudo-differential operator which acts continuously on the space of Schwartz functions $\mathcal{S} := \mathcal{S}(\mathbb{R}^n$) and extends to a continuous endomorphism of tempered distributions $\mathcal{S}'$. Here are some facts about this construction: The map from the symbols to endomorphisms of $\mathcal{S}$ is one to one. And so it endows the symbols with a non-commutative multiplication coming from composition (this multiplication can also be phrased without reference to $\mathcal{S}$ and is given by a certain combination of Fourier transform and convolution). Call this algebra $\Psi DO$. The composition defined above respects the increasing filtration (by order) defined by setting $\Psi DO^{\le m}$ to be all operators that come from symbols in $S^{l}$ for $l \le m$. For every real number $1 \lt p \lt \infty$ there's a decreasing filtration (the Sobolev filtration) indexed by the real numbers (say $s \in \mathbb{R}$) on $\mathcal{S}'$ where the $s$-filtered piece is the Sobolev space $W^{s,p} \subset \mathcal{S}'$ (it also has the nice property that $\bigcap_{s\in \mathbb{R}} W^{s,p} = \mathcal{S}$ and $\bigcup_{s\in \mathbb{R}} W^{s,p} = \mathcal{S}'$). The order filtration on $\Psi DO$ respects the Sobolev filtrations on $\mathcal{S}'$ (for every $1 \lt p \lt \infty$). This is just the (rather non-trivial statement) that every $P \in \Psi DO^{\le m}$ gives a bounded linear operator $P : W^{s,p} \to W^{s-m,p}$ for all $p \in (0,\infty), s \in \mathbb{R}$. All operators in $\Psi DO$ are pseudo-local, that is they do not increase the microsupport (or wave front sets) of distributions. The subalgebra $\Psi DO^{- \infty} := \bigcap_m \Psi DO^{m}$ is a (filtered) two sided ideal and the quotient $\Psi DO / \Psi DO^{-\infty}$ is complete for the induced filtration. My question is whether these properties characterize $\Psi DO$'s, more precisely: Question: Let $\mathcal{A} \subset End(\mathcal{S}')$ be a subalgebra of continuous endomorphisms of the space of tempered distributions. For every $p \in (0,\infty)$ the Sobolev filtration on $\mathcal{S}'$ induces an increasing filtration on $\mathcal{A}$ by setting $\mathcal{A}^{\le m, p} := \{ P \in \mathcal{A} | P: W^{s,p} \to W^{s-m,p} , \forall s \in \mathbb{R}\}$. Suppose $\mathcal{A}$ satisfies the following 4 properties inspired from the above discussion: (Pseudo-locality) All operators in $\mathcal{A}$ are pseudolocal (microsupport non-increasing). (Exhaustion & Strictness) For every $p \in (0,\infty)$ the induced filtration is strictly increasing, i.e. $\mathcal{A}^{\le m,p} \subsetneq \mathcal{A}^{\le l,p}$ whenever $m \lt l$, and exaustive, i.e. $\mathcal{A}^{\infty}:= \bigcup_m \mathcal{A}^{\le m ,p} = \mathcal{A}$ (Constancy in $p$) For all $1 \lt p \lt q \lt \infty$ the induced filtrations agree. In other words $\mathcal{A}^{\le m,p} = \mathcal{A}^{\le m, q}$ for all $m \in \mathbb{R}$. (Completeness) The quotient $\mathcal{A}/\mathcal{A}^{- \infty}$ is complete for the induced filtration. Is it true that $\mathcal{A} = \Psi DO$ ? If not perhaps its true if we require that $\mathcal{A}$ be the smallest subalgebra satisfying the above conditions? REPLY [2 votes]: Let me try: Consider $S^m_{1,0}$ and $SG^{m,0}$, where the second class of SG-symbols $SG^{m_\psi,m_e}$ is defined by the estimates $$|\partial_x^\alpha \partial_\xi^\beta a(x,\xi)| \lesssim_{\alpha,\beta} \langle x\rangle^{m_e-|\alpha|} \langle \xi\rangle^{m_\psi-|\beta|}.$$ Clearly, $SG^{m,0}$ is a subset of $S^m_{1,0}$, therefore $L^p$-boundedness follows from the $L^p$-boundedness of Kohn-Nirenberg pseudos. I haven't thought about the completeness, but I don't see a big difference between Kohn-Nirenberg and SG there.<|endoftext|> TITLE: Solutions to the Diophantine equation $x^2+3y^2+3z^2=n$ QUESTION [5 upvotes]: For a fixed positive integer $n$, the Diophantine equation $$x^2 + y^2 + z^2 = n$$ was studied by Gauss in Disquisitiones Arithmeticae. As is known, this equation is intimately connected to the quaternion algebra $B_{-1,-1}$, characterized by the relations $$ i^2 = -1, j^2 = -1, k^2 = (ij)^2 = -1. $$ More precisely, it is equivalent to the norm equation $$ \operatorname{nrd}(xi + yj + zk) = n, $$ where $\operatorname{nrd}$ denotes the reduced norm of the (integral) pure quaternion $xi + yj + zk$. For me, the most interesting fact here is that for $n \not \equiv 0 \pmod 4$ the number of solutions to this equation is some multiple the Hurwitz class number $H(n)$. In the special case $n \equiv 7 \pmod 8$, it is a multiple of zero. Now, I wonder if the same phenomenon was observed for other Diophantine equations of this kind. For example, in the quaternion algebra $B_{-1, -3}$ defined by the relations $$ i^2 = -1, j^2 = -3, k^2 = (ij)^2 = -3, $$ we can consider the norm equation $$ \operatorname{nrd}\left(ix + \frac{1+j}{2}y + \frac{i + k}{2}z\right) = n, $$ which is equivalent to the Diophantine equation $$ x^2 + y^2 + z^2 + xz = n. $$ Is there a formula for the number of its solutions? (see the OEIS sequence A014453) Is it true that, for some $n$, it is equal to the multiple of the (Hurwitz?) class number of some number field, say $\mathbb Q(\sqrt{-n})$? Perhaps, a Diophantine equation $$ x^2 + 3y^2 + 3z^2 = n $$ was studied in detail? I'd be thankful for any references. REPLY [7 votes]: One of the beautiful (and sometimes flummoxing) aspects of the theory of ternary quadratic forms is that you can find answers and questions coming from many different points of view! Using modular forms will give you an answer along the lines that Henri Cohen suggests. I'd like to put in a pitch for a quaternionic approach, which has other features. A reference is my book http://quatalg.org. The theorem of Gauss is Theorem 30.1.3. You can mimic the proof of this theorem for your ternary quadratic forms, to express the number of primitive representations in terms of (Hurwitz) class numbers, because the associated quaternion orders have type number one. Here are a few more details--if you'd like a more complete write up, just ask. The main input is Theorem 30.4.7, which sums the number of optimal embeddings = number of primitive representations over the right class set of an order, expressing this in terms of the class number and a local correction factor. The two quadratic forms you list arise as the reduced norm on the trace zero submodule of orders $\mathcal{O}$ in the quaternion algebra $B=(-1,-3\,|\mathbb{Q})$ of discriminant $3$. The first form $x^2+y^2+z^2+xz$ arises from $\mathcal{O}$ a maximal order one: this order is in fact Euclidean under the reduced norm, and so has class number $1$, which means the average is just over the one term. The second form $x^2+3y^2+3z^2$ arises from an order of reduced discriminant $36$ and index $12$; I compute using Magma that this order has class number $2$ but type number $1$, which means that the sum is over two terms that are equal, so we just need to divide the right-hand side by $2$. So to finish, we need to compute the local embedding numbers, which provides a correction term depending on $n$ modulo a power of $2$ and $3$. This gets a bit technical, so I would just stare at the numerical answer provided by Will Jagy. For the first form, the order is maximal so the local embedding numbers are in 30.5 (just a correction at $3$); unfortunately for the second form requires a separate calculation (not covered by 30.5 or 30.6). Happy to say more if you'd like!<|endoftext|> TITLE: How many simple closed geodesics in a given primitive homology class? QUESTION [5 upvotes]: It is well-known that an essential closed curve on a hyperbolic surface (possibly with boundary) is homotopic to a unique closed geodesic. Moreover, if the curve under consideration is simple, then so is the geodesic homotopic to it. A reference is "A primer on mapping class groups" of Farb and Margalit (propositions 1.3 and 1.6). It is proved here that for a torus with 1 puncture $\Sigma_{1, 1}$ (endowed with a complete hyperbolic metric) every primitive homology class $h \in H_1(\Sigma_{1, 1}, \mathbb{Z})\approx \mathbb{Z}^2$ contains a unique simple closed geodesic. This can be surprising for a beginner like me since the preimage of $h$ under abelianization map $\mathrm{ab}:\pi_1(\Sigma_{1, 1})\approx F_2\rightarrow H_1(\Sigma_{1, 1}, \mathbb{Z})$ is infinite. Every homotopy class in this preimage contains a closed geodesic yet only one contains a simple closed geodesic. My question is: are there examples of hyperbolic surfaces of different topology such that every primitive homology class contains exactly one simple closed geodesic? What are the results/references in this general direction? REPLY [6 votes]: The thrice punctured sphere has no simple closed geodesics. The four-times punctured sphere has a unique simple geodesic in each homology class. In general, it is a result of I. Rivin that the number of simple closed geodesics of length bounded above by $L$ grows like $L^{6g - 6 + 2 c},$ where $c$ is the number of punctures. We can restrict to closed surfaces for simplicity, so $c=0.$ Then, the number of homology classes where there is a simple closed geodesic of length $\leq L$ grows no faster than $L^{2g},$ which means that for uniqueness you have to have $g<2,$ which rules out every hyperbolic surface.<|endoftext|> TITLE: Are all totally ramified $\mathbb{Z}_p$-extensions of local fields come from (relative) Lubin-Tate formal groups? QUESTION [8 upvotes]: The setup is as follows: $k/\mathbb{Q}_p$ is a finite extension, $\mathfrak{p}$ is the maximal ideal of $\mathcal{O}_k$, $q=\#(\mathcal{O}_k/\mathfrak{p})$ $k'/k$ is a finite unramified extension of degree $d$ It's known that for a relative Lubin-Tate formal group $\mathcal{F}$ relative to $k'/k$ with parameter $\xi$ ($\xi\in\mathcal{O}_k$ with $\mathrm{ord}_{\mathfrak{p}}(\xi)=d$), it gives an abelian extension tower $k'(\mathcal{F}[\mathfrak{p}^n])$ of $k$ with degree $[k'(\mathcal{F}[\mathfrak{p}^n]):k']=(q-1)q^{n-1}$ and each $k'(\mathcal{F}[\mathfrak{p}^n])/k'$ is totally ramified. My question is, if a tower $\{k_n'\}$ is given such that for any $n$, $k_n'/k$ is abelian, $k_n'/k'$ is totally ramified, and $[k_n':k']=(q-1)q^{n-1}$, then can we find a relative Lubin-Tate formal group $\mathcal{F}$ such that $k_n'=k'(\mathcal{F}[\mathfrak{p}^n])$? If not, are there any criteria for it? In fact I'm considering the following special case: let $K$ be an imaginary quadratic field, $p$ be a prime split in $K$, $\mathfrak{p}$ be a prime of $K$ above $p$, $H$ be the Hilbert class field of $K$, $w$ be a prime of $H$ above $\mathfrak{p}$. Let $H_n$ be the ring class field of $K$ of conductor $p^n$, then $w$ is totally ramified over $H_n/H$ and we have $[H_n:H]=2(p-1)p^{n-1}/\#\mathcal{O}_K^\times$. I'd like to know the answer of above question for $k=K_{\mathfrak{p}}\cong\mathbb{Q}_p$, $k'=H_w$, $k_n'=H_{n,w}$, i.e. whether the anti-cyclotomic tower $\{H_{n,w}\}$ comes from (relative) Lubin-Tate formal group. (Of course it's true for cyclotomic tower and the $\mathbb{Z}_p$ extension tower unramified outside $\mathfrak{p}$. And it looks like that we are in trouble when $d_K=-3,-4$.) EDIT: If we assume $k=\mathbb{Q}_p$ (as in the special case I'm considering) and $p\geq 3$, then by the fact $\mathbb{Q}_p^{\mathrm{ab}}=\mathbb{Q}_p^{\mathrm{unr}}(\mu_{p^\infty})$ and that the open subgroup of $\mathrm{Gal}(\mathbb{Q}_p^{\mathrm{unr}}(\mu_{p^\infty})/\mathbb{Q}_p^{\mathrm{unr}})\cong\mathbb{Z}_p^\times$ of index $(p-1)p^{n-1}$ is unique, we can conclude that $\mathbb{Q}_p^{\mathrm{unr}}k_n'=\mathbb{Q}_p^{\mathrm{unr}}(\mu_{p^n})$ for any $n$. Can we obtain more information from this? REPLY [4 votes]: After discussing with local people, I come out with a proof of my question under the assumption $p\nmid\#(\mathcal{O}_k^\times)_{\mathrm{tors}}$. For example, if $k=\mathbb{Q}_p$ and $p\geq 3$. This assumption ensures that for any $n\geq 1$, the open subgroup of $\mathcal{O}_k^\times$ of index $(q-1)q^{n-1}$ is unique. First we choose any relative Lubin-Tate formal group $\mathcal{F}_\xi$ relative to $k'/k$ of parameter $\xi$, where $\xi\in\mathcal{O}_k$ with $\mathrm{ord}_{\mathfrak{p}}(\xi)=d$. Then $k^{\mathrm{ab}}=k^{\mathrm{unr}}(\mathcal{F}_\xi[\mathfrak{p}^\infty])$ with Lubin-Tate character $\chi:k^{\mathrm{ab}}/k^{\mathrm{unr}}\to\mathcal{O}_k^\times$ which is an isomorphism. For any $n\geq 1$, the fields $k^{\mathrm{unr}}/k'$ and $k_n'/k'$ are linear disjoint over $k'$, since one is unramified and another one is totally ramified. Note that $k^{\mathrm{unr}}k_n'$ is a subextension of $k^{\mathrm{ab}}/k^{\mathrm{unr}}$ of degree $(q-1)q^{n-1}$, this forces that $k^{\mathrm{unr}}k_n'=k^{\mathrm{unr}}(\mathcal{F}_\xi[\mathfrak{p}^n])$ and the natural restriction map $\mathrm{Gal}(k^{\mathrm{unr}}(\mathcal{F}_\xi[\mathfrak{p}^n])/k^{\mathrm{unr}})\to\mathrm{Gal}(k_n'/k')$ is an isomorphism. Therefore $k^{\mathrm{unr}}k_\infty'=k^{\mathrm{ab}}$ and the natural map $$ \mathrm{Gal}(k^{\mathrm{ab}}/k^{\mathrm{unr}}) \hookrightarrow\mathrm{Gal}(k^{\mathrm{ab}}/k') \twoheadrightarrow\mathrm{Gal}(k_\infty'/k')\qquad(*) $$ is an isomorphism. The local class field theory assert the following diagram commutes ([1], Chapter I, Proposition 1.8) $$ \begin{matrix} \mathcal{O}_k^\times & \hookrightarrow & k^\times \\ \chi^{-1}\uparrow\cong~~ & & ~~~\downarrow\mathrm{Art} \\ \mathrm{Gal}(k^{\mathrm{ab}}/k^{\mathrm{unr}}) & \hookrightarrow & \mathrm{Gal}(k^{\mathrm{ab}}/k) \end{matrix} $$ Note that $\mathrm{ord}_\mathfrak{p}(\xi)=d$, so $\mathrm{Art}(\xi)$ is in fact contained in $\mathrm{Gal}(k^{\mathrm{ab}}/k')$. Hence we can define $\alpha$ to be the image of $\mathrm{Art}(\xi)$ under the composition map $$ \mathrm{Gal}(k^{\mathrm{ab}}/k')\twoheadrightarrow \mathrm{Gal}(k_\infty'/k')\xrightarrow{(*)^{-1}} \mathrm{Gal}(k^{\mathrm{ab}}/k^{\mathrm{unr}})\xrightarrow[\cong]{\chi^{-1}} \mathcal{O}_k^\times\hookrightarrow k^\times, $$ and if we define $\varpi:=\xi\alpha^{-1}$ then we have $\varpi\in\ker(k^\times\xrightarrow{\mathrm{Art}}\mathrm{Gal}(k^{\mathrm{ab}}/k)\twoheadrightarrow \mathrm{Gal}(k_\infty'/k))$. We claim that a relative Lubin-Tate formal group $\mathcal{F}_\varpi$ relative to $k'/k$ of parameter $\varpi$ is what we want to find. In fact, by [1], page 11, paragraph 3, for any $n\geq 1$, the kernel of $k^\times\to\mathrm{Gal}(k'(\mathcal{F}_\varpi[\mathfrak{p}^n])/k)$ equals $\langle\varpi\rangle\cdot(1+\mathfrak{p}^n)$, hence the kernel of $k^\times\to\mathrm{Gal}(k'(\mathcal{F}_\varpi[\mathfrak{p}^\infty])/k)$ equals $\langle\varpi\rangle$. This forces that $k_\infty'=k'(\mathcal{F}_\varpi[\mathfrak{p}^\infty])$, and by the uniqueness of the open subgroup of $\mathcal{O}_k^\times$ of given index, we must have $k_n'=k'(\mathcal{F}_\varpi[\mathfrak{p}^n])$. References [1] E. de Shalit. Iwasawa Theory of Elliptic Curves with Complex Multiplication. Academic Press, 1987. I still want to know a proof which doesn't need the assumption.<|endoftext|> TITLE: What is the complexity of determining if a knot group is $\mathbb{Z}$? QUESTION [7 upvotes]: It is known from the work of Waldhausen that the isomorphism problem for knot groups is decidable. What is then: The complexity of determining if a knot group is $\mathbb{Z}$? .i.e. same as the unknot. The complexity of the isomorphism algorithm in general. Regards, Prathamesh REPLY [8 votes]: The knot group is $\mathbb{Z}$ if and only if the knot is unknotted: see https://math.stackexchange.com/questions/1478171/knot-group-and-the-unknot So, your first question is on the complexity of deciding whether a knot is the unknot. This is known to be in the intersection of NP and co-NP, see the Wikipedia article on this any many other interesting facts.<|endoftext|> TITLE: In a publication, should an `\ldots` always be followed by a period? QUESTION [14 upvotes]: The Latex command \ldots is often used to denote "and so forth". For instance, $$ \pi \approx 3.1415\ldots $$ When a sentence ends with an \ldots, one is faced with the conundrum of how to properly terminate it. The "cleanest" way would of course be to add a terminating "period", thus signaling the reader that this grammatical unit has now reached completion. For instance, $$ \rm{\ we\ have \ } \pi \approx 3.1415\ldots. \rm{\ However,} $$ We now have four dots at the end of the sentence, which seems like an excessive amount. The alternative however leaves a lingering feeling of dissatisfaction as the sentence does not feel finished: $$ \rm{\ we\ have \ } \pi \approx 3.1415\ldots \rm{\ However,} $$ What is the best practice in such situations? Is there a "norm" to handle such a case when preparing a manuscript for publication? REPLY [5 votes]: There are two possible correct answers. The first one the the mathematician's: the ellipsis for some implied content thing that isn't written, while the full stop marks the end of a sentence. Since each of the two typographical signs has its own functions, both should be present. This is logically correct, but typographically wrong. :) The second one is the typographer's, and it is better illustrated by the following three situations: Whenever a sentence contains an ellipsis right before the full stop, the full stop is no longer written. Whenever a sentence ends with an abbreviation (containing dots, such as "U.S.A." or "etc."), the full stop is no longer written (see also the links therein). Whenever a sentence ends with a quotation that itself ends with a full stop, the "outer" full stop is no longer written (it is a matter of debate, though, whether the remaining full stop should be inside or outside the quotation marks). Follow the links in the answers given here and see also this answer citing the Chicago Manual of Style. To conclude, whenever a full stop would logically follow another typographical sign that itself ends with a dot, the full stop is no longer written (it "fuses" with the one preceding it).<|endoftext|> TITLE: Does $0^{\#}$ imply the failure of upward directedness in the set generic universe over $L$? QUESTION [11 upvotes]: Question 1. Suppose that $0^{\#}$ exists. Are there set generic filters $g,h$ over $L$ ($g,h \in V$) such that for any $f$ ($\in V$) which is generic over $L$ we have that $L[g] \cup L[h] \not \subseteq L[f]$? In other words: Does $0^{\#}$ imply the failure of the upward directedness of the set generic universe over $L$? I have the following partial result: Corollary 2. Suppose $0^{\#}$ exists. Let $\kappa$ be the least innaccessible cardinal of $L$. Then there are $g,h$ which are $\mathbb{C}$-generic over $L$ ($\mathbb{C}$ is Cohen forcing) such that for any $f$ which is $\mathbb{P}$-generic over $L$ for some $\mathbb{P} \in J_{\kappa}$ we have $L[g] \cup L[h] \not \subseteq L[f]$. That observation was inspired by: Proposition 3. (Mostowski?) Let $M$ be a countable transitive model of $\mathrm{ZFC}$ and let $\mathbb{C}$ be Cohen forcing. Then there are $\mathbb{C}$-generic filters $g,h$ over $M$ such that for any transitive $\mathrm{ZFC}$ model $N$ with the same ordinal height of $M$ we have $M[g] \cup M[h] \not \subseteq N$. In particular there is no set generic filter $f$ over $M$ such that $M[g] \cup M[h] \subseteq M[f]$. Proof. Let $\alpha = M \cap \mathrm{Ord} < \omega_{1}$ and fix a bijection $f \colon \omega \to \alpha$. Define $E \subseteq \omega \times \omega$ via $$ (m,n) \in E \iff f(m) \in f(n). $$ Let $z \in ^{\omega}2$ code $E$ in an absolute fashion. E.g. we may let $$ z(k) = 1 \iff k = 2^{m}\cdot3^{n} \wedge (m,n) \in E. $$ Clearly any transitive model having $z$ as an element has $E$ as an element as well and, since $E$ is a well-order of order type $\alpha$, must also have $\alpha$ as an element. It now suffices to construct Cohen reals $c,d$ over $M$ such that they combined code the real $z$. We construct $c,d \in ^{\omega}2$ as follows: Since $M$ is countable, we may fix an enumeration $(D_{k} \mid k < \omega)$ of all dense subsets of $\mathbb{C}$ that are elements of $M$. Let $c_{0} \in D_{0}$ and let $d_{0} = (0^{\mathrm{length}(c_{0})}) ^{\frown} (1) ^{\frown} z(0) ^{\frown} y_{0}$ for some $y_{0}$ such that $d_{0} \in D_{0}$. Given $c_{k}, d_{k}$ for some $k < \omega$ we let $$ c_{k+1} = c_{k} ^{\frown} (0^{\mathrm{length}(d_{k})- \mathrm{length}(c_{k})}) ^{\frown} (1) ^{\frown} x_{k+1} $$ for some $x_{k+1}$ such that $c_{k+1} \in D_{k+1}$ and $$ d_{k+1} = d_{k} ^{\frown} (0^{\mathrm{length}(c_{k+1}) - \mathrm{length}(d_{k})}) ^{\frown} (1) ^{\frown} y_{k+1} $$ for some $y_{k+1}$ such that $d_{k+1} \in D_{k+1}$. We then let $c = \bigcup_{k < \omega} c_{k}$ and $d = \bigcup_{k < \omega} d_{k}$. The initial segments of $c$ and $d$ look as follows $$ \begin{array}{cc|c|c|c|c} c = & c_{0} & 0 \ldots 0 & 1 \ x_{1} & 0 \ldots 0 & \ldots \\ d = & 0 \ldots 0 & 1 \ z(0) \ y_{0} & 0 \ldots 0 & 1 \ z(1) \ y_{1} & \ldots \end{array} $$ and the blocks of $0$'s in $c$ and $d$ now allow us to reconstruct $z$ from $c,d$ via a recursive function. (Q.E.D.) Proof of Corollary 2. Since $0^{\#}$ exists, $\kappa$ exists and is countable in $V$. Since all dense subsets of $\mathbb{C}$ that $L$ can see are in $J_{\omega_{1}^{L}} \subseteq J_{\kappa}$, there are only countably many such dense sets. Just like before we fix a real $z$ that codes the countable ordinal $\kappa$ and construct $g,h$ which are $\mathbb{C}$-generic over $L$ such that they combined code $z$. Suppose there were some $\mathbb{P} \in J_{\kappa}$ and some $f$ which is $\mathbb{P}$-generic over $L$ with $L[g] \cup L[h] \subseteq L[f]$. $g,h,f$ are, of course, $\mathbb{C}$(respectively $\mathbb{P}$)-generic over $J_{\kappa}$ and we would have $g,h \in J_{\kappa}[f]$. But now $J_{\kappa}[f]$ would have to contain $\kappa$ as an element. Contradiction! (Q.E.D.) Side note: The proof above only uses that there is some $\kappa$ which is worldly in $L$ but countable in $V$ -- the assumption that $0^{\#}$ exists is overkill. It's just the first setting that came to mind when I thought about an inner model that has a lot of generic filters in $V$. REPLY [4 votes]: One can omit $0^\sharp$, as well as the need for any large cardinals, if one simply limits the class of forcing extensions. For example, perhaps one wants to consider only amalgamation and non-amalgamation in the generic multiverse of Cohen-real extensions $L[c]$, or in the generic multiverse of proper-forcing extensions $L[G]$ or $\omega_1$-preserving forcing extensions. Theorem. If $\omega_1^L$ is countable in $V$, then there are $L$-generic Cohen reals $c$ and $d$, whose corresponding extensions $L[c]$ and $L[d]$ are not amalgamated by any $L$-generic Cohen real $e$. Indeed, there are such $c$ and $d$ for which there is no $\omega_1$-preserving extension $L[G]$ extending both $L[c]$ and $L[d]$. Proof. Assume $\omega_1^L$ is countable in $V$. In particular, in $V$ we can enumerate all the dense subsets of Cohen forcing $\text{Add}(\omega,1)$ in $L$ in an $\omega$-sequence in $V$. From this enumeration, we can easily build $L$-generic Cohen reals. Using the proof of proposition 3 in the question, we can code any real coding $\omega_1^L$, and this real cannot be added in any $\omega_1$-preserving forcing over $L$. $\Box$. Let me also add a mention of our recent paper: M. E. Habič, J. D. Hamkins, L. D. Klausner, J. Verner, and K. J. Williams, Set-theoretic blockchains, ArXiv e-prints, pp. 1-23, 2018. (under review) Abstract. Given a countable model of set theory, we study the structure of its generic multiverse, the collection of its forcing extensions and ground models, ordered by inclusion. Mostowski showed that any finite poset embeds into the generic multiverse while preserving the nonexistence of upper bounds. We obtain several improvements of his result, using what we call the blockchain construction to build generic objects with varying degrees of mutual genericity. The method accommodates certain infinite posets, and we can realize these embeddings via a wide variety of forcing notions, while providing control over lower bounds as well. We also give a generalization to class forcing in the context of second-order set theory, and exhibit some further structure in the generic multiverse, such as the existence of exact pairs.<|endoftext|> TITLE: Generalized Smith Theorem for the torsion of cokernels QUESTION [8 upvotes]: Let $R$ be a (commutative) domain and let $Q$ be its fraction field. Consider a morphism $f\colon R^n \to R^m$, i.e. a matrix $A \in M(m,n;R)$, and let $K= \operatorname{coker} f$. Let $I_k=(\det \operatorname{min}_k(A))$ be the ideal of $R$ generated by all determinants of minors of $A$ of size $k$. I wonder if \begin{equation} \operatorname{Tor}_1^R(K,Q/R) \simeq \prod_{k=1}^{\operatorname{rk} A} \frac{I_{k-1}}{I_{k}}. \end{equation} If $R$ is a PID, the above isomorphism holds by Smith Normal Form. Does it hold for other classes of rings? I'm particularily interested in the case of $R$ is an order in a quadratic imaginary extension of $\mathbb{Q}$. REPLY [3 votes]: As Mohan already observed, the Smith Norm Formal Theorem can be extended in your sense to the class of Dedekind domains. This is somehow the best we can get: Claim. Let $R$ be a Noetherian domain which is not a Dedekind domain. Then there is a maximal ideal $\mathfrak{m}$ of $R$ such that the dimension of $\mathfrak{m}/\mathfrak{m}^2$ over $R/\mathfrak{m}$ is at least $2$. In particular, $R/\mathfrak{m} \times R/\mathfrak{m}$ and $R/\mathfrak{m} \times \mathfrak{m}/\mathfrak{m}^2$ are not isomorphic as $R$-modules although we have $I_0 = R, I_1 = \mathfrak{m}$ and $I_2 = \mathfrak{m}^2$ for an inclusion map $\mathfrak{m} \times \mathfrak{m} \subset R \times R$. By an inclusion map, I mean any $R$-module homomorphism $f: R^n \rightarrow R^2$ with image $\mathfrak{m} \times \mathfrak{m}$. Proof. By hypothesis, we can find a maximal ideal $\mathfrak{m}$ such that $\mathfrak{m}R_{\mathfrak{m}}$ cannot be generated by a single element as an ideal of the localization $R_{\mathfrak{m}}$ at $\mathfrak{m}$ [3, Theorem 11.2]. Thus the dimension of $\mathfrak{m}R_{\mathfrak{m}}/\mathfrak{m}^2R_{\mathfrak{m}}$ over $R/\mathfrak{m}$ is at least two. As we have $\mathfrak{m}/\mathfrak{m}^2 \simeq (\mathfrak{m}/\mathfrak{m}^2)_{\mathfrak{m}} \simeq \mathfrak{m}R_{\mathfrak{m}}/\mathfrak{m}^2R_{\mathfrak{m}}$, the result follows. As a result, the Smith Normal Form Theorem can be extended in your sense to an order of a number field if and only if this order is maximal. Let us conclude with a remark on two obvious generalizations. If $R$ is any commutative ring with identity, OP's isomorphism trivially holds whenever $m = 1$. The existence of a Smith Normal Form also trivially guarantees the existence of OP's isomorphism. By definition, every Elementary Divisor Ring (EDR) ensure this existence, independently of $m$ and $n$, and there are EDRs which aren't principal ideal rings. The ring of algebraic integers is one instance because of the principal ideal theorem, see also this MO post for further examples. Edit. It turns out that the isomorphism holds more generally for Prüfer domains of finite character, i.e., Prüfer domains in which every nonzero element is contained in only finitely many maximal ideals. This is [2, Proposition 1]. For an example of one of these rings that is neither Bézout nor Noetherian, see [1]. [1] W. Heinzer, "Quotient overrings of integral domains", Mathemafika 17, 1970 [2] L. Levy, "Invariant Factor Theorem for Prüfer Domains of Finite Character ", 1985. [3] H. Matsumura, "Commutative ring theory", 1989.<|endoftext|> TITLE: What is $e^{- \zeta_{\Delta} '(0)}$ for a $\Delta$ the Laplacian of a manifold? QUESTION [5 upvotes]: For a connected, finite graph $G$, let $\lambda_1, \ldots, \lambda_n$ denote the nonzero eigenvalues of the graph Laplacian. We define $\zeta_G = \Sigma_{i = 1}^n \lambda_i^s$. Then Kirkoffs Matrix-Tree theorem can be reformulated as saying that $e^{ - \zeta_G'(0)} / |G| = \tau(G)$, where $\tau$ is the number of trees of the graph $G$. (See here for more: http://www.math.ucsd.edu/~fan/lattice.pdf ) If $M$ is a Riemannian manifold, and $\Delta$ is it's Laplacian, we can define a similar zeta function: $\zeta_{\Delta}$. ( https://en.wikipedia.org/wiki/Minakshisundaram%E2%80%93Pleijel_zeta_function ) Question: Is there a reasonable meaning that can be assigned to $e^{- \zeta_{\Delta}'(0)} / Vol(M)$? I think (but am not sure) that this quantity doesn't make much sense as stated. The reason is because: $\zeta_{\Delta}(s) \sim (4 \pi s)^{-n/2} \Sigma_{m = 0}^{\infty} a_m s^m$ (see application one on the wikipedia page) , which has no derivative at zero. Thank you! REPLY [3 votes]: If I am not mistaken, this cannot be true in general. For example, one may scale the metric for $\mathbb{S}^{1}$ proportionally and the determinant of the Laplacian remains the same independent of the length of the circle. For dimension 2, the determinant of the Laplacian obtains its maximum when $M$ has constant sectional curvature. If we rescale the metric proportionally at all places, similar phenomenon happens. One may work with a sphere or a torus as $\mathbb{R}^{2}/L$ with $L$ being a lattice to check explicitly how the $\zeta$-function behaves under dilation. There is (allegedly) a related conjecture by Nicolas Bergeron, Mehmet Haluk Sengun, Akshay Venkatesh, which claims analytic torsion grows exponentially with respect to the covering of 3-manifolds. I am not entirely sure whether the statement is true (I heard people found counter-examples and the conjecture has to be modified). I think the motivation of $\tau(S)$ may be from elsewhere (like Weyl's paper).<|endoftext|> TITLE: A cube is placed inside another cube QUESTION [15 upvotes]: I known following problem with two square $$Area(1)+Area(3)=Area(2)+Area(4).$$ My question. Is this problem true for two cubes? We place a cube $XYZT.X'Y'Z'T$ into another cube $ABCD.A'B'C'D',$ is it true that \begin{align*} & Vol(ADD'A'.XTT'X')+Vol(BCC'B'.YZY'Z')\\ =&Vol(ABCD.XYZT)+Vol(A'B'C'D'.X'Y'Z'T')\\ =&Vol(CDD'C'.ZTT'Z')+Vol(ABB'A'XYY'X'). \end{align*} My question. If this is true for two cubes then is this true for two similar rectangular prism? My question. And are these problems true for two hypercubes and two similar hyperrectangles in $n$-dimesion Euclidean space? REPLY [10 votes]: The answer to your question is: no Its also easy to give a concrete counter-example: Take the cube with vertices $$ \left(\frac{273}{340},\,\frac{79}{68},\,\frac{13}{20}\right) , \left(\frac{407}{340},\,\frac{57}{68},\,\frac{27}{20}\right) , \left(\frac{239}{340},\,\frac{789}{884},\,\frac{337}{260}\right) , \left(\frac{249}{340},\,\frac{621}{884},\,\frac{217}{260}\right) , \left(\frac{263}{340},\,\frac{1195}{884},\,\frac{289}{260}\right) , \left(\frac{417}{340},\,\frac{573}{884},\,\frac{231}{260}\right) , \left(\frac{431}{340},\,\frac{1147}{884},\,\frac{303}{260}\right) , \left(\frac{441}{340},\,\frac{979}{884},\,\frac{183}{260}\right) $$ inside a $0-2$-cube. Then the three volumes you ask for are $\frac{24421}{8840}\neq \frac{24563}{8840}\neq \frac{24491}{8840}$. Here is a picture of the configuration: The outer cube has volume 8 and the inner cube has volume $1/8$. As a sanity check, lets see if everything adds up: $$\frac{24421}{8840} + \frac{24563}{8840} + \frac{24491}{8840} = \frac{14695}{1768}$$ and $$\frac{14695}{1768} + \frac{1}{8} -8 =\frac{193}{442}.$$ Since the edges of the outer cube are not coplanar with the edges, we have now double counted all the 12 tetrahedra, that arise from the fact that these pairs of edges are non-coplanar. Their volumes are $$\frac{471}{8840} , \frac{11}{170} , \frac{59}{2210} , \frac{29}{680} , \frac{49}{1768} , \frac{29}{8840} , \frac{29}{680} , \frac{29}{8840} , \frac{59}{2210} , \frac{49}{1768} , \frac{11}{170} , \frac{471}{8840} $$ and sure enough the sum of these numbers is $\frac{193}{442}$. Here is a picture of all those tetrahedra: ...and a picture of a single one: By the way: those two cube are concentric, so your conjecture does not work even in the concentric situation.<|endoftext|> TITLE: For which theories does ZFC without global choice prove the existence of a proper class monster model? QUESTION [11 upvotes]: Proper class sized monster models are typically formulated in a class theory like $NBG$ and they can reasonably be formalized in $ZFC$ with some kind of global choice, but for some theories you don't need either classes as a first-order object or global choice to exhibit proper class sized monster models. For a complete first-order theory $T$, a class model, $\mathfrak{C}$, is a class $C$ (as in the class of sets satisfying some formula maybe with parameters) with relations and functions also given by formulas maybe with parameters, such that $\mathfrak{C}$ models $T$ in the obvious way (although there is a subtlety here with $ZFC$ proving that $\mathfrak{C}$ models $T$ uniformly vs. individually proving that it models each finite set of sentences in $T$). A proper class monster model is a class model $\mathfrak{C}$ such that $C$ is a proper class and which is 'proper class saturated,' i.e. for every subset $A\subset C$, $\mathfrak{C}$ realizes every type in $S_1(A)$ (I'm not going to worry about homogeneity right now). For uncountably categorical theories and other extremely nice theories there clearly are proper class monster models. For an uncountably categorical theory we can take an Ehrenfeucht-Mostowski model with $Ord$ for its spine (which, it should be noted, at least gives us the existence of a proper class model for any theory, which isn't a priori obvious). You should be able to do something similar with $\omega$-stable $\omega$-categorical theories in light of the coordinatization theorem (there's a finite collection of totally categorical strongly minimal sets that models of this theory are prime over). I'm pretty sure that if you take a class sized model of a unidimensional theory and take an appropriate ultrapower of it (which is well-defined for set sized index sets, using the index set and ultrafilter as parameters) that the ultrapower will be proper class saturated. It is plausible that the same can be done for stable theories in general (or at least nice enough stable theories), although I haven't tried to think about the details. On the other hand there are a couple of amusing examples of unstable theories with simple to describe proper class monster models. The surreal numbers are a proper class monster model of $RCF$ (and therefore $DLO$ in the reduct) and if we define an edge relation on the class of all sets by $xEy$ iff $x\in y$ or $y\in x$, then we get a proper class monster model of the theory of the random graph. Is there a general method for exhibiting proper class monster models in $ZFC$ without global choice? If not are there nice families of theories for which we can exhibit proper class monster models? Is there a model $V$ of $ZFC$ and a theory $T$ that does not have a proper class monster model in $V$? EDIT: There were requests for details on the EM model with $Ord$ for a spine construction. There might be some subtlety that I wasn't aware of regarding the model satisfying the theory but I'm fairly confident that you can literally construct the class. The construction is not completely uniform in that it needs the EM functor or at least a well-ordering of the language as a parameter in the end. First run the typical EM functor construction: Given a consistent theory $T$ with infinite models, find a complete Skolemization $T^{sk}$ and then find an EM type of a non-constant indiscernible sequence in $T^{sk}$. To construct an EM model, assuming we Skolemized the theory the 'dumb' way, it's sufficient to consider terms of the form $f(\overline{a})$ where $f$ is a Skolem function and $\overline{a}$ is a strictly decreasing tuple of indiscernibles, because any more complicated Skolem terms and terms with permutations of variables are equivalent to some base Skolem function corresponding to a formula of the form $y=t(\overline{x})$. So to construct the base class of the model, first let $$C_0 = \{\left:f\text{ an }n\text{-ary Skolem fn, }\alpha\in Ord\text{ with CNF }\omega^{\beta_0}+\dots+\omega^{\beta_{n-1}}\},$$ with the intended interpretation that $\left$ is equal to $f(\beta_0,\dots,\beta_{n-1})$. You can define Skolem function application by setting $f(\left,\dots,\left)$ equal to $\left$ where you choose $g$ and $\gamma$ algorithmically from the list $f,f_0,\dots,f_{n-1}$ and the set of the exponents in the CNFs of $\alpha_0,\dots,\alpha_{n-1}$ (basically by the same kind of reasoning as why we don't need Skolem terms that are more complicated than just single Skolem functions applied to indiscernibles). Then you define an equivalence relation on $C_0$ by $\left \sim \left$ for $\alpha = \omega^{\beta_0}+\dots+\omega^{\beta_{n-1}}$ and $\gamma = \omega^{\delta_0}+\dots+\omega^{\delta_{m-1}}$ by looking at a set of indiscernibles in the original indiscernible sequence with the same order type as $\{\beta_0,\dots,\beta_{n-1},\delta_0,\dots,\delta_{m-1}\}$ and then checking equality of the original corresponding terms. Let $\left[ \left\right]$ be the equivalence class of $\left$ (which may be a proper class). Then define the actual base class of the model by the typical kind of trick to avoid picking representatives: $$ C = \{ \left[\left\right] \cap V_\beta : \left \in C_0 , \, \beta \text{ minimal s.t. } \left[\left\right] \cap V_\beta \neq \varnothing \}.$$ (This only matters if you want to avoid picking a well-ordering of the set of Skolem functions, otherwise you can just pick representatives.) Since the Skolemized theory is complete the definition of Skolem function application on $C_0$ is consistent with $\sim$, so it defines functions on $C$. Then you can define predicates using a dumb encoding trick: Let $f_0$ be the 0-ary Skolem function for the formula $y=y$ and let $f_1$ be the 0-ary Skolem function for the formula $y\neq f_0$. Then for any $n$-ary predicate symbol $P$, let $f^\ast_P$ be the Skolem function corresponding to the formula $\left(P(\overline{x})\rightarrow y = f_0 \right) \wedge \left( \neg P(\overline{x}) \rightarrow y = f_1 \right)$. Then the predicate $P$ is defined by $P(\overline{x})$ if and only if $f^\ast_P(\overline{x})=\left[\left\right]$. (EDIT2: Although note that this also works for formulas in general, so by Skolemizing a theory we've actually also constructed a truth predicate for models of the Skolemized theory.) Clearly we already have constants and functions defined in terms of the Skolem functions. And then pretty much by construction this is a class model of the $\forall \exists$ part of $T^{sk}$ (uniformly? since it has bounded quantifier complexity and you can define truth in ZFC for sentences with bounded quantifier complexity), which entails all of $T^{sk}$. Also it seems like this should be uniform in the parameters $T^{sk}$ and the EM type. And really I think the only parameter you need is a well-ordering of the language, because given that you can canonically pick $T^{sk}$ and the EM type by expanding $\mathcal{L}$ with Skolem functions in the typical way to $\mathcal{L}^{sk}$ and extending the well-ordering to $\mathcal{L}^{sk}$; adding constants $I=\{c_i: i < \omega\}$ for a sequence of indiscernibles; adding the Skolem function axioms, $\forall \overline{x} (\exists y \varphi(y;\overline{x}) \rightarrow \varphi(f_\varphi(\overline{x});\overline{x}))$, and the indiscernible sequence axioms, $\varphi(\overline{c})\leftrightarrow \varphi(\overline{c}^\prime)$ and $c_0 \neq c_1$, for all $\varphi$ and strictly increasing tuples $\overline{c},\overline{c}^\prime$; and then picking a completion by going through the $\mathcal{L}^{sk}_I$ sentences $\{\varphi_i : i < \lambda \}$ according to the well-order and adding $\varphi_i$ if it is consistent (according to some fixed proof system) and $\neg \varphi_i$ otherwise. EDIT2: I didn't see this question before. Joel's argument works the same here: Once you pick a well-ordering of the language $\mathcal{L}$ you can encode it in ordinals and consider the inner model $L[\mathcal{L}]$ where global choice holds and you can construct class sized models easily. I think the EM functor construction accomplishes something slightly different in that having an EM functor for a theory (i.e. a Skolemization and then an EM type in the Skolemized theory) seems weaker than having a well-ordering of the language, which might be useful. EDIT3: I realized that the random graph, $DLO$, and maybe some others have a property in common that allows this to work: There is a canonical procedure for, given a set of parameters $A$, constructing a complete $|S_1(A)|$-type whose component $1$-types hit every type in $S_1(A)$. For the random graph we can take an element for each subset of $A$ and add in no connections between the new elements and for $DLO$ we can add in each cut. Any theory with this property has a proper class monster model because we can just iterate it along $Ord$. I think it's really close to something you can do with an arbitrary stable theory, specifically if every type in $S_1(A)$ is stationary you can take a product type, $\bigotimes_{p\in S_1(A)} p$ (I think?). The problem I'm having with this is that types over the new set of parameters won't necessarily be stationary, so I'd like to use strong types but it's not clear to me that there's a canonical way of passing from a set $A$ to its algebraic closure in $T^{eq}$. It seems like it might need some amount of global choice to do. But maybe that's how to get a counterexample? Something involving a model of $ZFC$ where global finite choice or global choice for pairs fails. REPLY [6 votes]: By adapting Kanovei and Shelah's construction of a definable hyperreal field, (EDIT: After actually looking at their paper I feel that I should mention that they basically pointed out this application of their construction at the end of their paper.) I believe I can show that every structure $\mathfrak{A}$ has a definable proper class monster model elementary extension $\mathfrak{C}$ in ZFC without any assumption of global choice (and furthermore the definition is uniform in $\mathfrak{A}$ and does not require a choice of a well-ordering of $\mathfrak{A}$), but please tell me if there's some mistake in my reasoning. It will be proper class saturated in the sense that given any set sized structures $\mathfrak{B}_0 \prec \mathfrak{B}_1\equiv \mathfrak{A}$ and an elementary embedding $f:\mathfrak{B}_0\prec\mathfrak{C}$ there is an elemenetary embedding $g:\mathfrak{B}_1 \prec \mathfrak{C}$ extending $f$. I think you can define a truth predicate on $\mathfrak{C}$ given the elementary diagram of $\mathfrak{A}$, but you don't need it to construct $\mathfrak{C}$. (I'm basing this off of Keisler's exposition (section 1G) of Kanovei and Shelah's result, rather than the original paper.) Lemma: There is a uniformly definable family of linear orders $(A_\kappa,\sqsubset_\kappa)$ for each infinite cardinal $\kappa$ and uniformly definable functions $f_\kappa : A_\kappa \rightarrow 2^{2^\kappa}$ such that $f_\kappa(a)$ is a non-principal ultrafilter on $\kappa$ for each $a\in A_\kappa$ and every non-principal ultrafilter on $\kappa$ is $f_\kappa(a)$ for some $a\in A_\kappa$. Proof: For each infinite cardinal $\kappa$ let $<_\kappa$ be the lexicographical ordering on $2^\kappa$ (i.e. $a <_\kappa b$ if $a(\alpha) < b(\alpha)$ for the first $\alpha$ at which they disagree where $<$ is the standard order on $2=\{0,1\}$). Then let $\sqsubset_\kappa$ be the lexicographic ordering on ${\left(2^\kappa\right)}^\left|2^\kappa\right|$ relative to $<_\kappa$, i.e. the set of functions from $\left|2^\kappa\right|$, the initial ordinal with the same cardinality as $2^\kappa$, to $2^\kappa$ (this is uniform because we don't need to choose a bijection between $2^\kappa$ and $\left|2^\kappa\right|$). Finally let $A_\kappa$ be the set of all functions $g\in {\left(2^\kappa\right)}^\left|2^\kappa\right|$ whose range is a non-principal ultrafilter on index set $\kappa$. Then $f_\kappa(g)=\text{range}(g)$. $\square$ So now we're going to define a proper class length elementary chain of elementary extensions of $\mathfrak{A}$. Let $\mathfrak{C}_0 = \mathfrak{A}$. For limit ordinals $\lambda$, let $\mathfrak{C}_\lambda = \bigcup_{\alpha<\lambda} \mathfrak{C}_\alpha$. Otherwise let $\mathfrak{C}^\prime_\alpha$ be the finite support iterated ultrapower of $\mathfrak{C}_\alpha$ using $A_{\aleph_{\alpha}}$. If $h:\mathfrak{C}_\alpha \rightarrow \mathfrak{C}^\prime_\alpha$ is the natural embedding, let $\mathfrak{C}_{\alpha+1}$ be $\mathfrak{C}_\alpha \cup \left( \mathfrak{C}_\alpha^\prime \backslash h(\mathfrak{C}_\alpha) \right)$ and define all the atomic predicates accordingly. So then let $\mathfrak{C}= \bigcup_{\alpha\in\mathbf{Ord}}\mathfrak{C}_\alpha$. If $\mathfrak{B}_0 \prec \mathfrak{B}_1$ are set sized and $\mathfrak{B}_0 \prec \mathfrak{C}_\alpha$ for some $\alpha$, then by the typical kind of argument there exists an ultrafilter $U$ on some cardinal $\aleph_\beta$ such that $\mathfrak{B}_1$ embeds into $\prod_{U}\mathfrak{C}_\alpha$ in a way that fixes $\mathfrak{B}_0$. By padding the index set we may assume that $\beta \geq \alpha$, so we have that at some point in the chain, $\mathfrak{C}_\beta$ contains an elementary substructure isomorphic to $\prod_{U}\mathfrak{C}_\alpha$ in a way that fixes $\mathfrak{C}_\alpha$, so we get that $\mathfrak{B}_1$ can be embedded into $\mathfrak{C}_\beta$ in a way that fixes $\mathfrak{B}_0$. More or less $\mathfrak{C}$ is the finite support iterated ultrapower of $\mathfrak{A}$ along the linear order $\sum_{\kappa\in\textbf{InfCard}}(A_\kappa,\sqsubset_\kappa)$, similar to what Joel suggested. We also get a certain amount of homogeneity in that any automorphism of some $\mathfrak{C}_\alpha$ extends to an automorphism of all of $\mathfrak{C}$ (in a way that's uniformly definable in terms of the automorphism, because really any expansion of the theory at some $\mathfrak{C}_\alpha$ uniformly extends to all of $\mathfrak{C}$). I think that at certain strong limit cardinals (i.e. $\aleph_\alpha = \beth_\lambda$ for limit ordinal $\lambda$), $\mathfrak{C}_{\alpha}$ will be a special model of the theory, and therefore resplendent, so given any partial elementary map $h:A\rightarrow B$ for sets $A,B\subset \mathfrak{C}$, I think it can be extended to an automorphism of some $\mathfrak{C}_{\alpha}$ for sufficiently large $\aleph_\alpha = \beth_\lambda$ (in a non-uniform way that required choosing an ordering of $\mathfrak{C}_{\alpha}$), which can then be extended to a definable (with parameters) automorphism of $\mathfrak{C}$.<|endoftext|> TITLE: The partial preorder on $\mathbb N$ generated by the finite axioms of choice QUESTION [10 upvotes]: Let $\mathsf C_n$ denotes the statement: for any family $\mathcal F$ of $n$-element sets there exists a choice function (i.e., a function $f:\mathcal F\to\bigcup\mathcal F$ such that $f(F)\in F$ for all $F\in\mathcal F$). It is known that $\mathsf C_2\Rightarrow \mathsf C_4$ in ZF. This fact suggests introducing a partial preorder $\preceq$ on the set $\mathbb N$ of positive integers defined by $n\preceq m$ if $\mathsf C_m\Rightarrow \mathsf C_n$ in ZF. Also we can write that $n\cong m$ if $\mathsf C_n\Leftrightarrow \mathsf C_m$. It is easy to show that $n\preceq m$ if $n$ divides $m$. So, $1\preceq n$ for any $n\in\mathbb N$ and $2\preceq n$ for any even number $n$. On the other hand, $\mathsf C_2\Rightarrow \mathsf C_4$ implies that $2\cong 4$. What else is known about the partial preorder $\preceq$? Maybe there exists a precise (arithmetic) description of this preorder. A more specific question: is $2^n\cong 2$ for any $n\in\mathbb N$? I know that similar questions were studied by Mostowski, Tarski, Truss, Jech so maybe the answer is already known? REPLY [12 votes]: In Jech's "The Axiom of Choice", at the end of Chapter 7, he formulates the following condition on two natural numbers $n>m$: (S) There is no decomposition of $n$ into $p_1+\ldots+p_s=n$ such that $p_i>m$ is a prime number for all $i$. And he goes on to prove that if $\mathsf{C}_k$ holds for all $k\leq m$, then (S) implies $\mathsf{C}_n$, and moreover if (S) fails, then there is a model of $\mathsf{C}_k$ for all $k\leq m$, but $\lnot\mathsf{C}_n$. So, for example, if the Goldbach conjecture is true, then $\mathsf{C}_2$ implies nothing more than $\mathsf{C}_4$, since in that case every even number other than $2$ is the sum of two primes, and other than $4$ these primes have to be odd. So this is now a question about number theory, rather than set theory, and I will let the experts in that field make their remarks. Emil Jeřábek notes in the comments that every $n\geq 8$ is the sum of some amount of $3$s and $5$, both prime and both odd, so $\mathsf{C}_2$ does not extend its reach beyond $\mathsf{C}_4$.<|endoftext|> TITLE: Which motivic spectra are dualizable? QUESTION [11 upvotes]: Let $S$ be a scheme, and $SH(S)$ the stable motivic category over $S$. Which objects of $SH(S)$ are dualizable with respect to the smash product? All I can find on this question is an old abstract of Röndigs stating that for a $k$ a perfect field, every smooth projective scheme over $k$ is dualizable in $SH(Spec(k))$. Is every compact object of $SH(S)$ dualizable? Every finite cellular object? How about every smooth and proper $S$-scheme? Do extra assumptions about $S$ make a difference (such as assuming $S$ is smooth and/or affine or $S = Spec(k)$ for a perfect field $k$)? REPLY [15 votes]: If $X$ is noetherian of dimension $>0$, there is always a compact object of $SH(X)$ which is not dualizable: e.g. $j_\sharp$ of the sphere spectrum where $j:U\to X$ is any dense open immersion with non-empty complement (in each connected component). However, smooth and proper $X$-schemes always provide a source of dualizable objects. In fact, $f_*$ preserves dualizable objects for any smooth and proper map $f:Y\to X$ (this is a consequence of relative purity, of the projection formula, and of proper base change). To see how few dualizable objects there are, it is enlightening to look at classical sheaves. Let $X$ be a topological space, and $D(X)$ be the derived category of sheaves (of complexes of abelian groups, or of spectra...) over $X$. Then the dualizing objects of $D(X)$ are precisely the locally constant sheaves with values in perfect complexes. For schemes over $\mathbf{C}$, the Betti realization functor $$SH(X)\to D(X(\mathbf C))$$ commutes with the six operation, hence, in particular, with tensor product, and therefore preserves dualizable objects. This means that any sheaf on $X(\mathbf C)$ which is not locally constant and which is of geometric origin must provide an idea of how to construct a non-dualizable compact object of $SH(X)$. If you want to work over a more general base, you may replace $D(X(\mathbf C))$ with other target of realizations ($\ell$-adic sheaves, or arithmetic $D$-modules).<|endoftext|> TITLE: Counterexample for the Skolem-Noether Theorem QUESTION [5 upvotes]: If a division ring is finite-dimensional over its center then we can apply Skolem-Noether theorem (which asserts that every endomorphism is a conjugation). Can someone give a counterexample of the Skolem-Noether theorem when the division ring is infinite-dimensional over its center? REPLY [6 votes]: The following statement is taken from Cohn's 1995 book Skew Fields: Theory of General Division Rings (Cambridge EOM 57), proposition 2.3.4 on page 69, which is attributed to Köthe: Let $k$ be a commutative field with an automorphism $\alpha$ and put $E = k(\!(t;\alpha)\!)$ [the division ring of $\alpha$-twisted Laurent series over $k$]. Given any automorphism $\beta$ of $k$ such that $\alpha\beta = \beta\alpha$, extend $\beta$ to $E$ by the rule $t^\beta = t$. Then $\beta$ is an inner automorphism of $E$ if and only if $\beta = \alpha^r$ for some $r\in\mathbb{Z}$. Cohn adds: For example, if $k = F(s)$, where $F$ is any field of characteristic $0$, and $\alpha\colon s\mapsto s+1$ the $\beta\colon s\mapsto s+\frac{1}{2}$ is an outer automorphism of $E = k(\!(t;\alpha)\!)$. For completeness of MathOverflow, let me recall the definition of the division ring $E = k(\!(t;\alpha)\!)$ of $\alpha$-twisted (or "skew") Laurent series: its elements are of the form $\sum_{i=i_0}^{+\infty} t^i c_i$ with $c_i \in k$ [here I'm following Cohn's convention of using right-multiplication by $k$], addition being defined componentwise and multiplication by distributing and using the rule $c t^n = t^n c^{\alpha^n}$ [where, of course, $c^{\alpha^n}$ denotes the image of $c$ under the $n$-th power of $\alpha$].<|endoftext|> TITLE: Is every abelian variety a subvariety of a Jacobian? QUESTION [24 upvotes]: Let $k$ be an infinite field and $A$ be an abelian variety over $k$. Can $A$ be embedded into a Jacobian variety $J$ over $k$? In these notes by William Stein this is stated without proof in remark 1.5.8; it is attributed to personal conversation with Brian Conrad. Unfortunately I was unable to locate or come up with any proof. REPLY [7 votes]: You can find a detailed proof here (theorem 1.2) in the case of principally polarized abelian varieties. One reduces to this case using the Zarhin's trick. The assumption of $k$ being infinite should not be necessary (see remark 1.3 in the paper)<|endoftext|> TITLE: Convergence of a sequence by iteration QUESTION [5 upvotes]: Let $F:\mathbb R^d\to\mathbb R$ be a convex function. Assume that $F$ has a uniformly bounded gradient, $|\sup_{x\in\mathbb R^d}\nabla F(x)|<+\infty$. Define the sequence as follows: Take an arbitrary $x_0\in\mathbb R^d$ and set $x_{n+1}:=\nabla F(x_n)$ for $n\ge 0$. I am very interested in the convergence of $(x_n)_{n\ge 1}$. First this sequence is bounded by definition. Second, it follows from definition that $$(x_{n+1}-x_n)\cdot (x_n-x_{n-1}) \quad =\quad \big(\nabla F(x_n)-\nabla F(x_{n-1})\big)\cdot (x_n-x_{n-1}) \ge 0.$$ If $d=1$, then $(x_n)_{n\ge 1}$ is monotone and thus convergent. But this is not clear for $d>1$ (even for $d=2$)! Any comments or references will be very much appreciated! PS: This problem does not seem deep, but I still have not found a complete answer after discussing with people. REPLY [4 votes]: (Skip the initial two – or even three – paragraphs for the actual solution). The equation $$x_{n+1} - x_n = \nabla F(x_n) - x_n$$ seems to be a discretised version of $$x'(t) = \nabla F(x(t)) - x(t).$$ The vector field $\nabla F(x) - x$ is conservative, with scalar potential $|x|^2/2 - F(x)$ which is confining (that is, it goes to $\infty$ at infinity). Thus, the gradient flow will converge to the set of critical points of the potential. In a typical situation, this set is discrete, in which case the answer for the continuous problem is affirmative (although the limit need not be unique: it may depend on the initial value). If, however, you choose a sufficiently nasty potential, the gradient flow may fail to converge. Now let us try a similar approach for the discrete equation: write $G(x) = |x|^2/2 - \nabla F(x)$, so that $x_{n+1} - x_n = -\nabla G(x_n)$. This is the gradient descent sequence for the function $G$, with a constant parameter. It is well-known that gradient decent may fail to converge if the parameter is too large, compared to the second derivatives of $G$. However, in our case the second derivative of $G$ in any direction is bounded by $1$, which is precisely what we need. Here the actual solution begins: Write $G(x) = |x|^2/2 - F(x)$, so that $x_{n+1} = x_n - \nabla G(x_n)$. By convexity of $F$, $$ G(x) \le G(x_n) + \nabla G(x_n) \cdot (x - x_n) + |x - x_n|^2/2 . $$ Setting $x = x_{n+1}$ leads to $$ \begin{aligned} G(x_{n+1}) & \le G(x_n) + \nabla G(x_n) \cdot (-\nabla G(x_n)) + |\nabla G(x_n)|^2/2 \\ & = G(x_n) - |\nabla G(x_n)|^2/2 . \end{aligned} $$ It follows that $(G(x_n))$ is a (stricly) decreasing sequence, and therefore it converges to a certain limit. In other words, $\sum_n |\nabla G(x_n)|^2 < \infty$, that is, $\sum_{n = 1}^\infty |x_{n+1} - x_n|^2 < \infty$. In particular, $$\lim_{n \to \infty} |\nabla G(x_n)| = \lim_{n \to \infty} |x_{n+1} - x_n| = 0.$$ This implies that the set of accumulation points of $(x_n)$ is connected (see below). Additionally, assuming continuity of $\nabla F$, for every accumulation point $x^* = \lim_{n \to \infty} x_{k_n}$ we have $\nabla G(x^*) = \lim_{n \to \infty} \nabla G(x_{k_n}) = 0$. To summarise: If $\nabla F$ is continuous and the set of critical points of $|x|^2/2 - F(x)$ is discrete, then the answer is affirmative: $(x_n)$ converges to a critical point of $F$. Let me remark why the set of accumulation points of a bounded sequence $(x_n)$, satisfying the condition $\lim_{n \to \infty} |x_{n+1} - x_n| = 0$, is connected. Suppose that it is not, that is, it can be divided into two closed parts $F_1$ and $F_2$, whose distance is positive. Let $3 \delta$ denote the distance between $F_1$ and $F_2$. Choose and arbitrary $N > 0$ large enough, so that $|x_{n + 1} - x_n| < \delta$ when $n > N$. There exist elements $x_i, x_k$ such that $N < i < k$, $\operatorname{dist}(x_i, F_1) < \delta$ and $\operatorname{dist}(x_k, F_2) < \delta$. For $j = i, i+1, \ldots, k$ we have $\operatorname{dist}(x_j, F_1) + \operatorname{dist}(x_j, F_2) \ge 3 \delta$, each summand changes by at most $\delta$ when $k$ is replaced by $k + 1$. It is therefore easy to see that for some $k \in \{i + 1, i + 2, \ldots, k - 1\}$ we must have $\operatorname{dist}(x_k, F_1) \ge \delta$ and $\operatorname{dist}(x_k, F_2) \ge \delta$. In other words, for any $N$ there is $k > N$ such that $\operatorname{dist}(x_k, F_1) \ge \delta$ and $\operatorname{dist}(x_k, F_2) \ge \delta$. By choosing a convergent subsequence of these elements $x_k$, we find an accumulation point of $(x_n)$ away from $F_1 \cup F_2$, a contradiction. OK, I cheated: the actual solution starts here. What happens if the set of critical points of $G(x) = |x|^2/2 - F(x)$ is not discrete? It turns out that $(x_n)$ may fail to converge! It is difficult to describe a counterexample, and I did not check every detail; roughly speaking, the graph of the function $G(x)$ described below is an infinite spiral ramp around the unit disk. [Edit: I re-wrote the following construction.] We suppose that $d = 2$. We use polar coordinates $(r, t)$ rather than Cartesian ones $(x, y)$. We also write $P_n$ (rather than $x_n$) for the sequence of points defined in the question: $P_{n+1} = P_n - \nabla G(P_n)$. Step 1. We divide $\mathbb{R}^2$ into five regions: 'interior': the unit disk, $D_0 = \{r \le 1\}$; 'ramp': the area between two spirals, $D_1 = \{1 + e^{-t} \le r \le 1 + 2 e^{-t} , \, t \ge 0\}$; 'walls': everything between the consecutive turns of the ramp, $D_2 = \{1 + 2 e^{-t} < r < 1 + e^{-t + 2 \pi} , \, t \ge 2 \pi\}$; 'circus': $D_3 = \{r < 10\} \setminus (D_0 \cup D_1 \cup D_2 \cup D_3) = \{1 + 2 e^{-t} < r < 10 , \, 0 \le t < 2 \pi\}$; 'exterior': $D_4 = \{r \ge 10\}$. Our goal is to define $G$ in such a way that: (A) the second order derivatives of $G$ are small; (B) a particle placed on the ramp and following the recurrence equation $P_{n+1} = P_n - \nabla G(P_n)$ stays on the ramp forever. Step 2. First of all, we set $G = 0$ in the interior $D_0$ and the exterior $D_4$. Step 3. We now define $G$ on the ramp $D_1$ in such a way that condition (B) is satisfied: for a fixed $t \ge 0$, if $1 + e^{-t} \le r \le 1 + 2 e^{-t}$, we let $G$ to be a linear function of $r$: $$ G = c (e^{-2 t} + a(t) (r - 1 - e^{-t})) , $$ where $a(t) > 0$ is extremely small. More precisely, we require that if the point $P$ lies on the inner edge of the ramp: $r = 1 + e^{-t}$, then the point $$ P' = P - \nabla G(P) $$ also lies on the same curve, a little bit down the ramp. Denote the polar coordinates of $P$ by $t$ and $r = 1 + e^{-t}$, and the polar coordinates of $P'$ by $t'$ and $r' = 1 + e^{-t'}$. At $r = 1 + e^{-t}$ we have $\partial_r G = c a(t)$ and $\partial_t G = -2 c e^{-2 t} + c a(t) e^{-t}$. Our condition reads $$ r' \cos(t' - t) = r - \partial_r G(P) , \quad r' \sin(t' - t) = -r \partial_t G(P) , $$ that is, $$ (1 + e^{-t'}) \cos(t' - t) = (1 + e^{-t}) - c a(t) , \quad (1 + e^{-t'}) \sin(t' - t) = c (1 + e^{-t}) (2 e^{-2 t} - a(t) e^{-t}) . $$ Given a small $c > 0$, by the inverse mapping theorem, it looks like we can choose $t'$ and $a(t)$ in such a way that the above condition holds, and $$ \begin{gathered} t' - t \sim 2 c e^{-2 t} , \\ a(t) \sim 2 e^{-3t} , \\ a'(t) \sim -6 e^{-3 t} , \\ a''(t) \sim 18 e^{-3 t} . \end{gathered} $$ (I did not work out the details here). The above properties of $a(t)$ imply that the second order derivatives of $G$ are bounded by a constant times $c$ on $D_1$. Step 4. We now build the walls for our ramp: we define $G$ in $D_2$. For $t \ge 2 \pi$ and $1 + 2 e^{-t} < r < 1 + e^{-t + 2 \pi}$ we let $G$ to be the cubic polynomial in $r$ which matches the values and the derivatives of $G$ with respect to $r$ at the boundary values $r_0 = 1 + 2 e^{-t}$ and $r_1 = 1 + e^{-t + 2 \pi}$, defined already in Step 3. The formula for $G$ can be given explicitly in terms of $t$, $a(t)$ and $a(t - 2 \pi)$. The important thing is that the second derivative of $G$ with respect to $r$ is bounded by the ratio of the 'height of the wall' (which is at most $c e^{-2 t} (e^{2 \pi} - 1)$) to the square of the 'width of the wall' (which is $e^{-t} (e^{2 \pi} - 2)$) plus the ratio of $a(t)$ to the 'width of the wall'. Thus, it is bounded by a constant times $c$. Similarly, one can bound second order derivatives of $G$ in any direction at all points of $D_2$ (and again I did not work out the details). Step 5. We need to extend $G$ to the circus $D_3$. Since this is a nice domain, and we have already defined $G$ elsewhere so that it has second order derivtives bounded by a constant times $c$, we can define $G$ in $D_3$ in such a way that the bound on the second order derivatives remains the same (possibly with a larger constant). Step 6. Now we are ready to eventually choose $c$: we make it small enough, so that the second derivatives of $G$ are in fact bounded by $1$. And we eventually define $F = r^2/2 - G$. Now $F$ is convex (because the second derivatives of $F$ in any direction belong to the interval $(0, 2)$) and $F(r e^{i t}) = r^2 / 2$ for $r \ge 10$. There is one more step needed to assert that $\nabla F$ is bounded: we modify $F(r e^{i t})$ for $r \ge 10$ and we set $F(r e^{i t}) = 10 (r - 10)$ for $r \ge 10$. This of course does not affect convexity of $F$. End of the construction. By property $(B)$, if $P_0$ corresponds to $t = 0$ and $r = 1 + e$ (which lies on the inner edge of the ramp $r = 1 + e^{-t}$), then $P_n$ corresponds to $t_n$ and $r_n = 1 + e^{-t_n}$ for some $t_n > 0$. Additionally, $t_{n+1} - t_n = t_n' - t_n \sim 2 c e^{-2 t_n}$, and hence $t_{n+1} - t_n \to 0$ and $t_n \to \infty$. Therefore, $(P_n)$ is not convergent. I am still not satisfied with the above description (and the quality of the image). If you are able to simplify it, feel free to edit this answer.<|endoftext|> TITLE: Big ideas and big ways of thinking in statistics? QUESTION [8 upvotes]: I'm moving to a new university for the fall semester, and I'll be teaching a statistics class for the first time. I'm familiar enough with doing statistics (my dissertation in math ed was a mixed-methods study with some quantitative components). However, I've never taught it before, so I don't yet have that level of knowledge about key statistics concepts and/or the "right" ways to think about them. To give you a for-example: When I teach calculus, I know that I want people to leave understanding (among other things) that the subject is really mostly about creating and refining approximations (the limit process; linear approximation; the Riemann sum; etc. etc.). So, in an effort to bootstrap my understanding a little bit, I thought I'd ask here: What are the big ways of thinking that are really useful for understanding and properly using statistics? If it's helpful to know, our book is OpenIntro's Intro Stats with Randomization and Simulation. Thanks in advance! (FYI, I've crossposted this to /r/math.) REPLY [2 votes]: The American Statistical Association has resources for statistics education, available here: http://www.amstat.org/asa/education/home.aspx ...including, in particular, resources for undergraduate education. Perhaps the "Guidelines for Assessment and Instruction in Statistics Education (GAISE) College Report" (available under the "Guidelines and Reports" heading) might be helpful. Oh, here is a direct link, why not: http://www.amstat.org/asa/files/pdfs/GAISE/GaiseCollege_Full.pdf This report seems relevant for your question because it lists 9 "big" goals for students. From there you can follow up to get more detail about those goals; recommendations for how to help students work toward the goals; pointers to further resources; etc. Good luck!<|endoftext|> TITLE: Example of a ring with non-finitely generated unit group? QUESTION [13 upvotes]: The well known Dirichlet's unit theorem states that the unit group of a maximal order in a quadratic number field is finitely generated of rank blah blah blah. I think it's pretty naive to expect a most radical generalization to hold, namely: Generalized Dirichlet unit theorem. The unit group of an associative unital ring with finitely generated additive group is finitely generated. Is there any concrete counterexample? REPLY [7 votes]: At Lee Mosher's request, here is how Borel-Harish-Chandra is applied, which is the bulk of the argument (with some additional work in the case with torsion as described in Aurel's post). Let $A$ be a finitely generated (associative unital) ring whose underlying additive group is free abelian of finite rank. From left multiplication we get a representation of $A$ on $B=A\otimes_\mathbf{Z}\mathbf{R}$, and thus an injective ring homomorphism $i:A\to M=\mathrm{End}(B)\simeq M_d(\mathbf{R})$, where $\mathrm{End}$ means as real vector space; it maps into $E=E(\mathbf{R})$, the set of endomorphisms as algebra. Inside $E\times E$, consider the set $G$ of pairs $(x,y)$ such that $xy=1$. This is a Zariski closed submonoid of $E\times E$, and can be identified to the automorphism group of $B$; it lies inside $M_{2d}(\mathbf{R})$. For $x\in A^\times$, consider $j(x)=(x,x^{1})$. Then the group homomorphism $j$ maps injectively $A^\times$ into $G(\mathbf{Z})$. We claim that this is surjective. Indeed, consider $(x,y)\in E(\mathbf{Z})\times E(\mathbf{Z})$ with $xy=1$. Then $x$ preserves the lattice of integral points, i.e., $x$ induces an automorphism of $A=B(\mathbf{Z})$. [Note: we had to introduce $E\times E$ instead of $E$ to describe $G$ as a closed subgroup, which is necessary to apply Borel-Harish-Chandra.] Thus we have $A\simeq G(\mathbf{Z})$. Borel and Harish-Chandra precisely proved (Annals of Math, 1962) that $G(\mathbf{Z})$ is finitely generated (and even finitely presented, as mentioned in Borel's ICM proceeding of 1962) for every $\mathbf{Q}$-defined subgroup of $\mathrm{GL}_k$. [Remark: $G(\mathbf{Z})$ is possibly not a lattice in $G(\mathbf{R})$; however $G(\mathbf{Z})=H(\mathbf{Z})$ for some $\mathbf{Q}$-defined normal subgroup $H$ of $G$, such that $G^0/H^0$ is a $\mathbf{Q}$-split torus, and $H(\mathbf{Z})$ is a lattice in $H(\mathbf{R})$.]<|endoftext|> TITLE: fractional Brownian Motion driven stochastic integrals QUESTION [5 upvotes]: We consider a stochastic process $\left(X_{t}\right)_{t\geq 0}$, defined as an integral process, s.t. $$X_{t}=\int_{0}^{t}u_{s}\,dB_{s}^{H}.$$ With a fractional Brownian motion $B^H_{t}$. If $H\neq\frac{1}{2}$, the stochastic integral can not be defined in the classical Itô sense, due to Bichteler-Dellacherie theorem. Using the classical Young theory, $X_{t}$ is well defined, if the trajectories of $u_{t}$ has finite $q$ variation, if $q<\frac{1}{1-H}$. Question 1: Is it possible to define $X_{t}$ in such a way, that the trajectories of $u_{t}$ don't have to be restricted w.r.t. there regularity? As far as I know, it is possible to use Rough Path Theory, to extend Young's classical result. Up to which extent, is it possible to extend Young's theory by rough path theory? Edit: More precisely, given an integral $\int_{}{}fdg$, with $f$ having finite $q$-variation and $g$ having finite $p$ variation. According Young (Link to Young's classical paper), the following holds: The integral $\int_{}{}fdg$ is well defined if (Y1) there are no common discontinuities and (Y2) if $\frac{1}{p}+\frac{1}{q}>1$. So, how does the transition from Young to RPT affect condition (Y2)? Question 2: Which classical stochastic analysis tools are available using the rough path approach? More precisely are there substitutes of the following classical tools? Itô formula Burkholder inequality (Upper bounds for moments of $X^{*}_{t}=\underset{s\leq t}{\text{sup}}\,X_{s}$ ) Question 3: Is it possible to extend Young's approach using other tools? Regularity structures Malliavin Calculus (Skorohod integral) White Noise Analysis ... REPLY [5 votes]: Yes, RPT allows you to define a notion of stochastic integral against fBm for a class of integrands that is larger than what Young's theory allows. Assuming that you're really interested in solving SDEs, so that $u$ locally looks again like $B^H$, you can go down to $H > {1\over 4}$. Below that things break down, and there's actually a good reason for that. The resulting notion of integration does satisfy the usual chain rule, but moment bounds are much trickier and there's nothing as sharp as Burkholder's inequality. The best kind of moment bounds that I'm aware of are those from the article by Cass, Litterer and Lyons. Regarding other tools, regularity structures are a generalisation of RPT, so they will give you essentially the same notion of integration, although in some special cases you can get a bit more mileage out of it because of its greater flexibility. WNA and Skorokhod integration also give you some notion of stochastic integral that generalises Itô's integral, but this is completely different and the fact that it is a "reasonable" notion of "integral" (approximable by some kind of Riemann sums) is very specific to the case $H = {1\over 2}$.<|endoftext|> TITLE: Can $S_n$ be partitioned into subsets containing an involution and satisfying $∀σ≠τ, ∃j$ s.t. $σ(j)≠τ(j),σ^{−1}(j)=τ^{−1}(j)$? QUESTION [6 upvotes]: Background Let $\sigma, \tau \in S_n$. We will say that $\sigma$ and $\tau$ are locally orthogonal and write $\sigma \perp \tau$ if there exists $j \in \{1, 2, \ldots, n\}$ such that $\sigma(j) \neq \tau(j)$ but $\sigma^{-1}(j) = \tau^{-1}(j)$. (I will explain this terminology in the Motivation section below.) We will call a subset $S \subset S_n$ exclusive if $\sigma \perp \tau$ for any $\sigma, \tau \in S$, $\sigma \neq \tau$. Observe that any exclusive subset of $S_n$ can contain at most one involution. We say $\sigma, \tau \in S_n$ are Knuth equivalent if they have the same insertion tableau under the Robinson-Schensted correspondence. This is the same as $\sigma^{-1}$ and $\tau^{-1}$ being dual Knuth equivalent, i.e., having the same recording tableau. Involutions are precisely those permutations that have the same recording and insertion tableaux, so each Knuth equivalence class contains exactly one involution, as does each dual Knuth equivalence class. Questions Does there exist a partition of $S_n$ into exclusive subsets such that each subset contains an involution? Does either Knuth equivalence or dual Knuth equivalence imply local orthogonality? Either would imply a positive answer to Q1. EDIT: A simple counterexample to Q2 is the pair of permutations $\sigma = (132)(4)$ and $\tau = (1)(234)$, which are Knuth equivalent but do not satsify $\sigma(j) = \tau(j)$ or $\sigma^{-1}(j) = \tau^{-1}(j)$ for any $j$. Q1 remains open. Has this notion of local orthogonality for elements of $S_n$ been defined anywhere before (presumably under some other name)? I have verified Q1 by brute force for $n \leq 11$. Motivation The notion of local orthogonality appears in a more general context as part of a necessary condition for multipartite quantum correlations. A positive answer to Q1 would imply that in the variant of the 100 prisoners problem (sometimes called the locker puzzle) in which $n$ prisoners may only open 2 drawers, the classical best solution cannot be improved upon using shared quantum entanglement between the prisoners. (Note that this is different from the quantum version of the game considered by Avis and Broadbent in which the prisoners are allowed to open a superposition of drawers.) REPLY [3 votes]: Yes. For each permutation $\sigma\in S_n$, choose (arbitrarily) from each nontrivial cycle of $\sigma$ a point $x$ such that $x < \sigma(x)$, and let $f(\sigma)$ be the set of pairs $(x,\sigma(x))$ so obtained. Then $f$ is a map from $S_n$ to the set of sets of disjoint ordered pairs $(i, j)$ with $i,j\in\{1,\dots,n\}$, $i TITLE: representing an uncountable free group as a union of an increasing sequence of countable subgroups QUESTION [6 upvotes]: Let $(G_\alpha)$ and $(K_\alpha)$ $(\alpha<\aleph_1)$ be strictly increasing chains of countable sets such that if $\alpha$ is a limit, then $G_\alpha=\bigcup_{\beta<\alpha}G_\beta$ and $K_\alpha=\bigcup_{\beta<\alpha}K_\beta$. Assume $\bigcup_{\alpha<\aleph_1}G_\alpha=\bigcup_{\alpha<\aleph_1}K_\alpha$. Does there exist a club $C$ such that $K_\gamma=G_\gamma$ for all $\gamma\in C$? In the paper ``The Abelianization of Almost Free Groups" (the end of the proof of Lemma 2.5 on page 1801), this assertion is made where the $G_\alpha$ and $K_\alpha$ are subgroups of a free group with some additional properties, but the author makes the assertion I want without any explanation, so I assume the reason must be simple and may not depend on the group theory. REPLY [15 votes]: The set $C$ of $\gamma$ such that $K_\gamma = G_\gamma$ is closed, because if $\alpha$ is a limit point of $C$, then $$G_\alpha = \bigcup_{\beta < \alpha } G_\beta = \bigcup_{\beta <\alpha, \beta \in C} G_{\beta} = \bigcup_{\beta <\alpha, \beta \in C} K_{\beta} = \bigcup_{\beta<\alpha} K_\beta= K_\alpha.$$ So it suffices to show that $C$ is unbounded. Given any $\alpha_1$, because $K_{\alpha_1}$ is countable and is contained in $\bigcup_{\alpha<\aleph_1} G_\alpha$, we have $K_{\alpha_1} \subseteq G_{\alpha_2}$ for some $\alpha_2$. Similarly we have $G_{\alpha_2}\subseteq K_{\alpha_3} \subseteq G_{\alpha_4} \subseteq \dots$ so $$G_{\lim_{n \to \infty} }\alpha_n = K_{\lim_{n \to \infty} } \alpha_n$$ and thus $C$ contains an element greater than $\alpha_1$. So $C$ is unbounded.<|endoftext|> TITLE: Cyclic quadrilateral in metric space QUESTION [5 upvotes]: Consider a metric space $(\Bbb M,d).$ If $X,Y,Z\in \Bbb M.$ We define cosin of angle by $$\cos(\angle YXZ)=\frac{d(X,Y)^2+d(X,Z)^2-d(Y,Z)^2}{2d(X,Y)\cdot d(X,Z)}.$$ If we have four points $A,$ $B,$ $C$ and $D$ in $\Bbb M$ satify the indentity $$d(A,B)\cdot d(C,D)+d(A,D)\cdot d(B,C)=d(A,C)\cdot d(B,D).$$ Then we say that $A,$ $B,$ $C,$ $D$ are concyclic. My Question 1. If $A,$ $B,$ $C$ and $D$ are concyclic then can we show that $$\cos(\angle BAC)=\cos(\angle BDC).$$ My Question 2. If $A,$ $B,$ $C$ and $D$ are concyclic then can we show there is a point $O$ such that $$d(O,A)=d(O,B)=d(O,C)=d(O,D).$$ My Question 3. If two questions 1 and 2 are not true then what is condition of $\Bbb M$ such that these are true? REPLY [4 votes]: For a counterexample to 2, we can take the points $A,B,C,D$ at $0,30,60,90$ degrees on the unit circle, and then add a bump to perturb the $x$-axis. In more detail: Consider the surface $z=f(x,y)^2$, where $$f(x,y)=\max\left(0,\ 0.1 - \left(x-0.5\right)^2 - y^2 \right),$$ and distances from shortest paths on the surface. Take points $$A=\big(1,0,0\big),\ B=\big(\frac{\sqrt{3}}{2},\frac{1}{2},0\big),\ C=\big(\frac{1}{2},\frac{\sqrt{3}}{2},0\big),\ D=\big(0,1,0\big).$$ The distances between $A,B,C$ and $D$ are all the same as in the $xy$-plane, since the surface is flat away from $(0.5,0,0)$. So $A,B,C,D$ are concyclic in this surface. Taking $O=(0,0,0)$, the distances $OB,OC,OD$ are also as in the $xy$-plane, and $O$ is the unique point on the surface with $OB=OC=OD$. But the shortest path from $O$ to $A$ has to pass by the bump, so $OA>OB$, and there is no point with $OA=OB=OC=OD$.<|endoftext|> TITLE: Localizing $\mathrm{CombModCat}$ at the Quillen equivalences QUESTION [15 upvotes]: Let $\mathrm{CombModCat}$ be the category of combinatorial model categories with left Quillen functors between them. By Dugger's theorem and the appendix of Lurie's "Higher Topos Theory" it ought to be true that its localization at the class of Quillen equivalences is equivalent to the homotopy category of presentable $\infty$-categories with left adjoint $\infty$-functors between them: $$ \mathrm{CombModCat}\big[\text{QuillenEquivs}^{-1}\big] \;\simeq\; \mathrm{Ho}\big( \mathrm{Presentable}\infty\mathrm{Cat} \big) $$ Has this been made explicit anywhere, in citable form? Something close is made explicit in Olivier Renaudin, "Theories homotopiques de Quillen combinatoires et derivateurs de Grothendieck" (arXiv:math/0603339) (thanks to Mike Shulman for the pointer!), where it is shown that the 2-categorical localization of the 2-category version of $\mathrm{CombModCat}$ is equivalent to the 2-category of presentable derivators with left adjoints between them. [edit:] By corollary 2.3.8, this implies that the 1-categorical localization of $\mathrm{CombModCat}$ is equivalent to the 1-categorical homotopy category of presentable derivators with left adjoints between them. The latter clearly ought to be equivalent to the homotopy category of $\mathrm{Presentable}\infty\mathrm{Cat}$, but is that made explicit anywhere? REPLY [6 votes]: The answer is affirmative and is provided by the paper Combinatorial model categories are equivalent to presentable quasicategories. Among other things, it proves that the relative categories of combinatorial model categories, left Quillen functors, and left Quillen equivalences; presentable quasicategories, cocontinuous functors, and equivalences; and other models for homotopy locally presentable categories and homotopy cocontinuous functors are all weakly equivalent to each other as relative categories. This also implies the equivalence of underlying quasicategories. Additionally, combinatorial model categories can be assumed to be left proper and/or simplicial. There is also an analogous statement for derivators, which must be formulated as an equivalence of (2,1)-categories due to the truncated homotopical nature of derivators.<|endoftext|> TITLE: Math journal publishing work related to combinatorics, probability, counting problems etc.? QUESTION [15 upvotes]: I'm a high school student. My peer and I have done some work on the Ballot Theorem counting problem and Catalan Numbers. We have come up with a new proof to the Ballot Theorem and we demonstrate the duality between the Ballot Theorem and Catalan Numbers. Or at least we hope the proof is new, we've scourged the net and not found work similar enough to ours. We'd like to send it a journal for a review process, and also try our shot at publishing. What are our chances at getting published, and which journal should we try in? REPLY [14 votes]: You may want to try Combinatorics, Probability & Computing or The Electronic Journal of Combinatorics. Your chances at getting published depend on the quality of your paper.<|endoftext|> TITLE: Points of elliptic curves over cyclotomic extensions QUESTION [19 upvotes]: Let $E$ be an elliptic curve over $\mathbb Q$. Let's look at the group of points of this elliptic curve over $\mathbb Q(1^{1/\infty})$ which we get after adding all roots of unity to $\mathbb Q$. It is easy to prove that it is not finitely generated and the theorem of K.Ribet asserts that its torsion is finite. What else can we say about it? Does it look more like $\bigoplus \mathbb Z$ or $\mathbb C/\Gamma$? Is it divisible? (I don't think so). I apologize that the questions are not precise, I just need to know which results do we have in this direction. REPLY [3 votes]: As for the torsion subgroup, Michael Chou has classified the possible subgroups that may occur as $E(\mathbb{Q}^{\text{ab}})_{\text{tors}}$ for an elliptic curve $E/\mathbb{Q}$. You can find a preprint here.<|endoftext|> TITLE: Which primes $\mathfrak{p} \in \mathbb{Z}[i]$ can be represented as $\mathfrak{p} = x^2 + 2y^2$? QUESTION [8 upvotes]: Consider the quadratic form $x^2 + 2y^2$ over the ring $\mathbb{Z}[i]$. It is irreducible, so I wanted to know which primes could be represented by this quadratic form. $$ \mathfrak{p} = x^2 + 2y^2 $$ There's two slightly different questions here. One about the integer $\mathfrak{p} \in \mathbb{Z}[i]$ and one about the ideals $(\mathfrak{p}) \subseteq \mathbb{Z}[i]$. This is still a PID so I can write just one element. This hopefully in analogy to Fermat's theorem on primes as the sum of two squares, where it is solved by a congruence condition. The analogous question for $x^2 + y^2 = (x+iy)(x-iy)$ is degenerate. REPLY [3 votes]: Just elaborating on the very nice comments and answer by Cherng-tiao Perng. As Noam Elkies said, $$ \mathbb{Q}[i,\sqrt{-2}]=\mathbb{Q}[\exp(2\pi i/8)]. $$ As is well known, the Galois group of the $8$-th cyclotomic field over $\mathbb{Q}$ is isomorphic to $(\mathbb{Z}/8\mathbb{Z})^{*}=\{1,3,5,7\}$. Let $\sigma_j$ be the element in $Gal(\mathbb{Q}[\exp(2\pi i/8)]/\mathbb{Q})$ that sends $\exp(2\pi i/8)$ to $\exp(2\pi ij/8)$, $j\in \{1,3,5,7\}$. As is well known, $\sigma_5$ fixes $i$, sends $\sqrt{2}$ to $-\sqrt{2}$, $\sqrt{-2}$ to $-\sqrt{-2}$. $\sigma_3$ fixes $\sqrt{-2}$, sends $i$ to $-i$, $\sqrt{2}$ to $-\sqrt{2}$. $\sigma_7$ fixes $\sqrt{2}$, sends $i$ to $-i$, $\sqrt{-2}$ to $-\sqrt{-2}$. Let $x=a+ib$, $y=c-id$. If $\mathfrak{p}\in\mathbb{Z}[i]$ may be written as $$ \mathfrak{p}=x^2+2y^2=(x+y\sqrt{-2})(x-y\sqrt{-2})=(a+ib+\sqrt{-2}c+\sqrt{2}d)(a+ib-\sqrt{-2}c-\sqrt{2}d), $$ then $$\mathfrak{p}=(a+ib+\sqrt{-2}c+\sqrt{2}d)\sigma_5((a+ib+\sqrt{-2}c+\sqrt{2}d)).$$ Since the primes $\mathfrak{p}$ in $\mathbb{Z}[i]$ are rational primes $p=3,7\bmod 8$ or primes with norms that are rational primes $p=1,5\bmod 8$, we consider these cases. Suppose $p=1,5\bmod 8$ so that $p=\mathfrak{p}\mathfrak{\bar{p}}$. Since $\mathfrak{\bar{p}}=\sigma_3(\mathfrak{p})$, then if $$ \mathfrak{p}=(a+ib+\sqrt{-2}c+\sqrt{2}d)\sigma_5((a+ib+\sqrt{-2}c+\sqrt{2}d)), $$ then $$ \mathfrak{\bar{p}}=\sigma_3((a+ib+\sqrt{-2}c+\sqrt{2}d))\sigma_7((a+ib+\sqrt{-2}c+\sqrt{2}d)) $$ and $$ p=\prod_{j\in\{1,3,5,7\}}\sigma_j((a+ib+\sqrt{-2}c+\sqrt{2}d)). $$ As is well known, for odd primes $p$, this only happens if $p=1\bmod 8$. As $\{a_0+a_1i+a_2\sqrt{2}+a_3\sqrt{-2}| a_0,a_1,a_2,a_3\in\mathbb{Z}\}$ is a subring of $\mathbb{Z}[\exp(2\pi i/8)]$, i.e., $$ a_0+a_1i+a_2\sqrt{2}+a_3\sqrt{-2}=a_0+a_1i+(a_2+a_3)\exp(2\pi i/8)+(a_3-a_2)\exp(2\pi i3/8), $$ as is pointed out in the other answer, we have to also check, when $p=1\bmod 8$ with $p=\mathfrak{p}\mathfrak{\bar{p}}$, where $\mathfrak{p}$ is an element and not an ideal, which of $\mathfrak{p}^{\prime}\in\mathfrak{p}\{1,-1,i,-i\}$ may be written as $$ \mathfrak{p}^{\prime}=\prod_{j\in\{1,5\}}\sigma_j(a_0+a_1i+a_2\sqrt{2}+a_3\sqrt{-2}), $$ and as the other answer indicates, and we may verify, for example, using PARI/GP rnfisnorm(rnfisnorminit(y^2+1,x^2-2,1+4*y)) [Mod(Mod(-1, y^2 + 1)*x + Mod(y + 2, y^2 + 1), x^2 - 2), 1] rnfisnorm(rnfisnorminit(y^2+1,x^2-2),y*(1+4*y)) [Mod(Mod(11/2*y - 3/2, y^2 + 1)*x + Mod(8*y - 2, y^2 + 1), x^2 - 2), 1] and also prove using congruences, that for such situations it only happens when $\mathfrak{p}^{\prime}$ is of the form $a+ib$, $a,b\in\mathbb{Z}$, $a$ odd, $b=0\bmod 4$. Suppose that $p=3\bmod 8$. Then there are $x,y\in\mathbb{Z}$ so that $$ p=x^2+2y^2. $$ Suppose that $p=7\bmod 8$. Then there is $x\in\mathbb{Z}$, $y\in i\mathbb{Z}$ so that $$ p=x^2+2y^2. $$<|endoftext|> TITLE: $x f'$ bounded by $x^2f $ and $f''$? QUESTION [11 upvotes]: Consider the Hilbert space of functions $f \in L^2(\mathbb R)$ such that $x^2f \in L^2(\mathbb R) $ and $ f'' \in L^2(\mathbb R).$ I am wondering whether it is true that $xf'\in L^2(\mathbb R)$ as well? It seems natural and perhaps some sort of interpolation should yield the claim but I fail to see how to show this. REPLY [15 votes]: By integration by parts, $$\int (xf')^2 = \int (x^2f') f' = -\int 2xf'f - \int x^2 f'' f.$$ $$-\int 2xf'f = \int f^2=\|f\|^2.$$ By the Cauchy-Schwarz inequality $$\bigg|\int x^2 f'' f\bigg|\le \|x^2f\|\|f''\|.$$ The conclusion is proved.<|endoftext|> TITLE: Spin-H structures QUESTION [10 upvotes]: Let us define a Spin-H structure as a reduction of a SO(n)-bundle by the group: $$Spin^H (n)=Spin(n) \times SU(2)/\{ 1,-1\}$$ The Spin-H structures are analogous to the well-known Spin-C structures but for the Hamilton numbers instead of the complex numbers. Can we prove that an oriented riemannian manifold admits a Spin-H structure if and only if it is Spin? REPLY [13 votes]: As mentioned in Arun Debray's answer, a closed orientable smooth manifold $M$ is spin${}^h$ if and only if there is a principal $SO(3)$-bundle (or equivalently, an orientable real rank three bundle) $P$ such that $w_2(M) = w_2(P)$. If $M$ is spin, then $w_2(M) = 0$. Taking $P$ to be the trivial bundle, we see that $w_2(M) = 0 = w_2(P)$, so $M$ is spin${}^h$. If $M$ is spin${}^c$, then $w_2(M)$ has an integral lift $c \in H^2(M; \mathbb{Z})$. There is a complex line bundle $L$ with $c_1(L) = c$ and hence $w_2(L) = w_2(M)$; one choice of such an $L$ is the complex line bundle associated to a spin${}^c$ structure. Now taking $P = L \oplus \varepsilon^1$, we see that $w_2(M) = w_2(L) = w_2(L\oplus\varepsilon^1) = w_2(P)$, so $M$ is spin${}^h$. For $M = \mathbb{CP}^2$ with its standard spin${}^c$ structure, this is the example given in Arun Debray's answer. If $M$ is a closed oriented Riemannian four-manifold, then $w_2(M) = w_2(\Lambda^+) = w_2(\Lambda^-)$ so taking $P = \Lambda^+$ or $\Lambda^-$ shows that $M$ is spin${}^h$. Note, this can also be deduced from the previous point as all closed orientable smooth four-manifolds are spin${}^c$. If $M$ is the Wu manifold, $SU(3)/SO(3)$, let $P$ be the principal $SO(3)$-bundle $SO(3) \to SU(3) \to SU(3)/SO(3)$. Then $w_2(M) = w_2(P)$, so $M$ is spin${}^h$. This is a notable example as $M$ is not spin${}^c$. I learnt this from Xuan Chen, a former student of Blaine Lawson. In conclusion, every spin manifold is spin${}^c$, but not conversely (as $\mathbb{CP}^2$ demonstrates), and every spin${}^c$ manifold is spin${}^h$, but not conversely (as $SU(3)/SO(3)$ demonstrates). Added Later: In this paper, Aleksandar Milivojevic and I prove the following: The primary obstruction to the existence of a spin$^h$ structure on an orientable manifold $M$ is $W_5(M)$, the fifth integral Stiefel-Whitney class of $M$. It is worth noting that in the spin$^c$ case the primary (and only) obstruction is $W_3(M)$. Every orientable manifold of dimension $\leq 7$ is spin$^h$. In every dimension $\geq 8$, there exist infinitely many homotopy types of closed, simply connected manifolds which are not spin$^h$. An explicit example of an orientable manifold which is not spin$^h$ is the ten-dimensional manifold $(SU(3)/SO(3))\times(SU(3)/SO(3))$.<|endoftext|> TITLE: Why should I care about the Jones polynomial? QUESTION [34 upvotes]: The invention of the Jones polynomial led to hundreds of papers and a Fields medal. However, as far as I can tell it had few consequences in topology. After all, after Thurston’s work we already had algorithms to completely classify knots, so by itself a new invariant seems to be of limited value. Given all the excitement, however, I suspect that I might be missing something. Why was this so important? REPLY [21 votes]: There have been some topological applications of the Jones polynomial and its various generalizations. I believe that these applications increased the interest in these invariants by topologists. One application was to the Tait conjectures. Jones used his polynomial to give lower bounds on the bridge number of links; see Proposition 15.6 of Jones, V. F. R., Hecke algebra representations of braid groups and link polynomials, Ann. Math. (2) 126, 335-388 (1987). ZBL0631.57005. See also the Morton-Franks inequality Theorem 15.1 in the paper estimating the braid index in terms of the HOMFLY polynomial. Another source of interest is the volume conjecture which has received much attention due to the possible connection with geometric invariants of knots. There seem to be many other connections of the colored Jones polynomials with other more geometric invariants (cf. the slope conjecture and noncommutative A-polynomial). As mentioned in the other answers, the biggest development was the definition of Khovanov homology, a categorification of the Jones polynomial, which led to a new proof of the Milnor conjecture. Khovanov homology is known to distinguish the unknot, and can be used to give invariants of transverse knots.<|endoftext|> TITLE: Algebraic power series over $\mathbb{F}_2$ as roots of polynomials of special form QUESTION [9 upvotes]: Let $F = \mathbb{F}_2$ be the field with two elements. I will denote the rings of polynomials and formal power series over $F$ as $F[t]$ and $F[[t]]$ respectively. Suppose that $x \in F[[t]]$ is algebraic over $F[t]$ (there exists a non-zero polynomial $P$ with coefficients in $F[t]$ such that $P(x) = 0$). Is it true that there exists a non-zero polynomial $Q(y) = \sum\limits_{k = 0}^m q_k y^{2^k}$ with coefficients $q_k$ from $F[t]$ such that $Q(x) = 0$ (the difference with is that all non-zero coefficients in $Q$ correspond to the powers of two)? (I encountered this statement being used without proof in literature, so it most probably is true.) REPLY [11 votes]: Getting rid of the power series, your question boils down to showing that any polynomial $P(x)$ divides some polynomial $Q(x)$ which has the form $Q(x) = \sum_{i} c_i x^{2^i}$. Polynomials $Q$ which have this property are known as additive polynomials. They have some nice alternative descriptions: for instance, if $K$ is a field of characteristic $2$ then a polynomial $Q(x) \in K(x)$ with distinct roots is additive if and only if the set of roots of $Q(x)$ in $\overline{K}$ form a $\mathbb{F_2}$-vector space. (This is a nice exercise, or see Basic Structures of Function Field Arithmetic by Goss for a proof.) Using this description, it's easy to construct $Q(x)$. Let $K = \mathbb{F_2}(t)$ and let $V$ be the $\mathbb{F_2}$-subspace of $\overline{K}$ generated by the roots of $P$. We set $Q(x) = (\prod_{\alpha \in V} (x-\alpha))^{2^N}$ for $N$ chosen to be large enough so that $P(x)$ divides $Q(x)$. Then $Q(x) \in K(x)$ because the finite vector space $V$ is stable under the action of $\mathrm{Gal}(\overline{K}/K)$, and $Q(x)$ is additive by the fact of the previous paragraph + Frobenius (Note that everything here generalizes in the obvious way if you replace $\mathbb{F}_2$ by $\mathbb{F}_q$ for any prime power $q$.) REPLY [7 votes]: In fact for any integral domain $R$, if $x$ is any element that satisfies a polynomial equation of degree $n$ over $R$, and $S$ is any set of numbers of size greater than $n$, then $x$ satisfies a polynomial equation of the form $\sum_{i \in S} c_i x^i$ for $c_i \in R$ not all zero. This is just because $\{x^i | i\in S\}$ cannot be linearly independent over the field of fractions of $R$, so some linear relation holds over the field of fractions, and we can clear denominators.<|endoftext|> TITLE: Irreducibility of the unramified principal series QUESTION [6 upvotes]: Let $G = \operatorname{GL}_n(F)$ with the usual Borel subgroup $P = TU$. Let $\chi = \chi_1 \otimes \cdots \otimes \chi_n$ be an unramified character of $T$. Suppose that $\chi$ is regular, which is to say that $\chi_i \neq \chi_j$ for $i \neq j$. Equivalently, $\chi \neq w.\chi$ for all $w \in W(T,G)$. Then $I(\chi) = \operatorname{Ind}_{TU}^G \chi \delta^{1/2}$ is irreducible if and only if $\chi_i \neq \chi_j | \cdot |_F$ for all $i \neq j$. Is this true without the regularity assumption on $\chi$? This is claimed in Example 4.2 of Prasad and Raghuram's notes on representation theory for $\operatorname{GL}_n$, but I wasn't sure if they were making an underlying assumption of regularity. REPLY [5 votes]: Suppose $F$ is a non-archimedean local field. This regularity assumption is not needed, nor is the assumption that $\chi$ is unramified. The sufficient part for irreducibility is part of Theorem 4.2 of Bernstein and Zelevinsky - Induced representations of reductive $\mathfrak p$-adic groups. I and the other part is part of Theorem 6.1 in Zelevinsky - Induced representations of reductive $\mathfrak p$-adic groups. II. On irreducible representations of $\mathrm{GL}(n)$. Edit: To respond to Vincent's comment, let me say what I know (very limited): Note that when $F$ is a finite field, $\delta$ and $|\cdot|_F$ are trivial and the assertion follows from Mackey theory that $$\mathrm{Hom}_G(I(\chi),I(\chi))=\bigoplus_{w\in N_G(T)/T}\mathrm{Hom}_T(\chi,{}^w\chi)$$ When one tries to adapt this to (any) local field, the normalization factor $\delta$ and $|\cdot|_F$ should naturally arise. In general $I(\chi)$ is no longer semisimple and one needs some tricks to know a bit more than the $\mathrm{Hom}$. Beyond this I know nothing about how the case $F$ archimedean has in common with the $p$-adic case.<|endoftext|> TITLE: When does doubling the size of a set multiply the number of subsets by an integer? QUESTION [29 upvotes]: For natural numbers $m, r$, consider the ratio of the number of subsets of size $m$ taken from a set of size $2(m+r)$ to the number of subsets of the same size taken from a set of size $m+r$: $$R(m,r)=\frac{\binom{2(m+r)}{m}}{\binom{m+r}{m}}$$ For $r=0$ we have the central binomial coefficients, which of course are all integers: $$R(m,0)=\binom{2m}{m}$$ For $r=1$ we have the Catalan numbers, which again are integers: $$R(m,1)=\frac{\binom{2(m+1)}{m}}{m+1}=\frac{(2(m+1))!}{m!(m+2)!(m+1)}=\frac{(2(m+1))!}{(m+2)!(m+1)!}=C_{m+1}$$ However, for any fixed $r\ge 2$, while $R(m,r)$ seems to be mostly integral, it is not exclusively so. For example, with $m$ ranging from 0 to 20000, the number of times $R(m,r)$ is an integer for $r=2,3,4,5$ are 19583, 19485, 18566, and 18312 respectively. I am seeking general criteria for $R(m,r)$ to be an integer. Edited to add: We can write: $$R(m,r) = \prod_{k=1}^m{\frac{m+2r+k}{r+k}}$$ So the denominator is the product of $m$ consecutive numbers $r+1, \ldots, m+r$, while the numerator is the product of $m$ consecutive numbers $m+2r+1,\ldots,2m+2r$. So there is a gap of $r$ between the last of the numbers in the denominator and the first of the numbers in the numerator. REPLY [33 votes]: Put $n=m+r$, and then we can write $R(m,r)$ more conveniently as $$ R(m,r) = \frac{(2n)!}{m! (n+r)!} \frac{m! r!}{n!} = \frac{\binom{2n}{n} }{\binom{n+r}{r}}. $$ So the question essentially becomes one about which numbers $n+k$ for $k=1$, $\ldots$, $r$ divide the middle binomial coefficient $\binom{2n}{n}$. Obviously when $k=1$, $n+1$ always divides the middle binomial coefficient, but what about other values of $k$? This is treated in a lovely Monthly article of Pomerance. Pomerance shows that for any $k \ge 2$ there are infinitely many integers with $n+k$ not dividing $\binom{2n}{n}$, but the set of integers $n$ for which $n+k$ does divide $\binom{2n}{n}$ has density $1$. So for any fixed $r$, for a density $1$ set of values of $n$ one has that $(n+1)$, $\ldots$, $(n+r)$ all divide $\binom{2n}{n}$, which means that their lcm must divide $\binom{2n}{n}$. But one can check without too much difficulty that the lcm of $n+1$, $\ldots$, $n+r$ is a multiple of $\binom{n+r}{r}$, and so for fixed $r$ one deduces that $R(m,r)$ is an integer for a set of values $m$ with density $1$. (Actually, Pomerance mentions explicitly in (5) of his paper that $(n+1)(n+2)\cdots (n+r)$ divides $\binom{2n}{n}$ for a set of full density.) Finally let me show that $R(m,r)$ is not an integer infinitely often when $r \ge 2$ is fixed. Let $p$ be a prime with $r

TITLE: Additive basis of order 2 QUESTION [5 upvotes]: Can we find $\alpha>1$ such that $u=(\lfloor n^\alpha\rfloor)_{n\geqslant0}$ is an additive basis of order $2$ (i.e. $\forall x\in\mathbb{N}, \exists(n,m)\in\mathbb{N}^2, x=u_n+u_m$) ? Remark : This question has been asked previously on math.SE, but no response has been provided yet. REPLY [4 votes]: For $1<\alpha<\frac32$, $(\lfloor n^{\alpha}\rfloor)_{n\geqslant0}$ is an asymptotic basis of order 2. I finally found these two articles: J-M. Deshouillers, Un problème binaire en théorie additive, Acta Arith. 25 (1974), 393-403 S.V. Konyagin, An additive problem with fractional powers, Mathematical Notes, 2003, 73:4, 594–597<|endoftext|> TITLE: What is closed homology? QUESTION [9 upvotes]: Bott & Tu in Differential forms in Algebraic Topology write in Remark 5.17, pg.52 The two Poincare duals of a compact orientated submanifold correspond to two homology theories - closed and compact homology. Closed homology has now fallen into disuse, while compact homology is known these days as the homology of singular chains...in general Poincare duality sets up an ismorphism between closed homology and de Rham cohomology, and between compact homology and compact de Rham cohomology. What is closed homology, and why did it fall into disuse? Did a better alternative turn up? REPLY [10 votes]: I think that closed homology is another name for Borel-Moore homology, which certainly has the isomorphism you suggest: https://en.m.wikipedia.org/wiki/Borel-Moore_homology It still is used quite frequently.<|endoftext|> TITLE: primary decomposition for nonabelian cohomology of finite groups QUESTION [18 upvotes]: Let $G$ be a finite group, and let $M$ be a group on which $G$ acts (via a homomorphism $G\to \operatorname{Aut}(M)$). If $M$ is abelian, hence a $\mathbb{Z}G$-module, there is a primary decomposition $$ H^k(G;M)=\bigoplus_p H^k(G;M)_{(p)} $$ for each $k>0$, where $p$ ranges over the primes dividing $|G|$ and $H^k(G;M)_{(p)}$ denotes the $p$-primary component of $H^k(G;M)$. It follows that if $H^1(G;M)$ is nonzero, then $H^1(P;M)$ is nonzero for some $p$-Sylow subgroup $P$ of $G$. See Section III.10 of K.S. Brown's "Cohomology of Groups". If $M$ is nonabelian, then $H^1(G;M)$ is still defined, but is not a group in general, so that asking for a decomposition such as above would not make any sense. But one could still ask the following weaker question: With $G$ and $M$ as above, suppose that $H^1(G;M)$ is non-trivial. Then must $H^1(P;M)$ be non-trivial for some $p$-Sylow subgroup $P$ of $G$? Edit: Tyler Lawson's answer below is very informative, but unfortunately is not complete because the cohomology of a free product is not necessarily isomorphic to the direct product of the cohomologies. So the question still seeks an answer. REPLY [7 votes]: It turns out the answer is no. Here I'll sketch a counter-example in which $H^1(G;M)$ is non-trivial (in fact infinite), while $H^1(H;M)$ is trivial for all proper subgroups $H TITLE: How to prove positivity of determinant for these matrices? QUESTION [19 upvotes]: Let $g(x) = e^x + e^{-x}$. For $x_1 < x_2 < \dots < x_n$ and $b_1 < b_2 < \dots < b_n$, I'd like to show that the determinant of the following matrix is positive, regardless of $n$: $\det \left (\begin{bmatrix} \frac{1}{g(x_1-b_1)} & \frac{1}{g(x_1-b_2)} & \cdots & \frac{1}{g(x_1-b_n)}\\ \frac{1}{g(x_2-b_1)} & \frac{1}{g(x_2-b_2)} & \cdots & \frac{1}{g(x_2-b_n)}\\ \vdots & \vdots & \ddots & \vdots \\ \frac{1}{g(x_n-b_1)} & \frac{1}{g(x_n-b_2)} & \cdots & \frac{1}{g(x_n-b_n)} \end{bmatrix} \right ) > 0$. Case $n = 2$ was proven by observing that $g(x)g(y) = g(x+y)+g(x-y)$, and $g(x_2 - b_1)g(x_1-b_2) = g(x_1+x_2 - b_1-b_2)+g(x_2-x_1+b_2-b_1) > g(x_1+x_2 - b_1-b_2)+g(x_2-x_1-b_2+b_1) = g(x_1-b_1)g(x_2-b_2)$ However, things get difficult for $n \geq 3$. Any ideas or tips? Thanks! REPLY [25 votes]: To complement Fedor's answer, here is more explicit proof. Let the original matrix be $G$. Let $D_x :=\text{Diag}(e^{x_1},\ldots,e^{x_n})$. Then, we can write \begin{equation*} G = D_x C D_b,\quad\text{where}\ C = \left[ \frac{1}{e^{2x_i}+e^{2b_j}}\right]_{i,j=1}^n. \end{equation*} To prove that $\det(G)>0$ it thus suffices to prove that $\det(C)>0$. Notice now that $C$ is nothing but a Cauchy matrix, and by explicitly writing its determinant out (under the hypotheses on $x$ and $b$) we can easily conclude that $\det(C)>0$. Remark. The above argument actually proves that $\text{sech}(x-y)$ is a Totally positive kernel (because the $k(x,y) := 1/(x+y)$ is known to be a TP kernel).<|endoftext|> TITLE: Minimum dimension of faithful representation of mapping class groups? QUESTION [6 upvotes]: Let $\Sigma_{g}$ be a closed orientable surface of genus $g$. Let $d_g$ denote the minimum dimension of a faithful representation of the mapping class group of $\Sigma_g$. For $g=1$, the mapping class group is $SL(2, \mathbb{Z})$ so $d_g=2$. For $g=2$, this paper proves that $d_g \leq 64$. For what $g \geq 3$ (if any) is it known that $d_g$ is finite? This question has been asked here 3 years ago; maybe something has changed since then. Is it known that $d_2=64$? REPLY [7 votes]: There has been no progress on this question for many years. In particular, the precise value of $d_2$ is not known and it is not known if $d_g$ is finite for any $g \geq 3$. The only related paper is this one by Korkmaz, which proves that any representation of the genus $g$ mapping class group to $GL(n,\mathbb{C})$ has abelian image for $n \leq 2g-1$ (this slightly improves results of Funar and Franks-Handel). Of course, for $n=2g$ there is the classical symplectic representation.<|endoftext|> TITLE: Question about taking the Zariski closure in $\mathbb{A}_{\mathbb{R}}^n$ QUESTION [8 upvotes]: Let $\mathbb{A}_{\mathbb{R}}^n$ be $\mathbb{R}^n$ endowed with the Zariski topology, where closed sets are algebraic sets (in $\mathbb{R}^n$) defined by real polynomials. Suppose $V \subseteq \mathbb{A}_{\mathbb{R}}^n$ is an irreducible affine variety. Let $U$ be an open (with respect to the usual topology) ball $U$ around a non-singular point of $V$ and of small enough radius. Does it then follow that the Zariski closure of $(V \cap U)$ is $V$? I thought it should be true (maybe not?), but I was wondering how I can show this. Any comments are appreciated. Thank you! REPLY [4 votes]: As noted in this answer to a previous question of yours, for any subvariety $W \subseteq \mathbb A_\mathbb R^n$, we have $$\dim_{\mathbb R} W(\mathbb R) \leq \dim W$$ (where $\dim W$ denotes the dimension in the sense of scheme theory, written there as $\dim_\mathbb C W(\mathbb C)$), with equality if $W$ has a smooth $\mathbb R$-point. In particular, if $W \subseteq V$ is the Zariski closure of $V \cap U$, then applying the above to both $V$ and $W$ gives $$\dim_\mathbb R W(\mathbb R) \leq \dim W \leq \dim V = \dim_\mathbb R V(\mathbb R).$$ But $W(\mathbb R)$ contains the full-dimensional subset $V(\mathbb R) \cap U$ of $V(\mathbb R)$, hence all dimensions must be equal. Since $V$ is irreducible and $\dim W = \dim V$, this forces $W = V$. $\square$<|endoftext|> TITLE: If $A,B$ are upper triangular matrices such that $AX=XA\implies BX=XB$ for upper triangular $X$, is $B$ a polynomial in $A$? QUESTION [34 upvotes]: A professor of mine told me that this is true, but he doesn't remember what the proof was or where to find it, and I haven't been able to find a source for it yet. As such I am looking for one here. In the theorem as stated, $\mathbb{F}$ is any field and $T_n(\mathbb{F})$ denotes the algebra of upper triangular $n\times n$ matrices over $\mathbb{F}$. Theorem: Let $A,B\in T_n(\mathbb{F})$ be such that for all $X\in T_n(\mathbb{F})$, $$AX=XA\implies BX=XB$$ Then $B=p(A)$ for some $p\in \mathbb{F}[t]$. Does anyone know of a source for this result? I have searched Google, MSE, MO, and the like to no avail. If we replace $T_n(\mathbb{F})$ by $M_n(\mathbb{F})$, the question is answered in this paper. Unfortunately, the argument doesn't seem to translate directly, as I can't find a way to force the $M_i$ maps to be upper-triangular. Also, I have already asked this question here on MSE. As the question is for an undergraduate research project, it felt appropriate to ask it here as well. Thanks for any help! Edit on 9 July, 2018: It's probably worth mentioning that the following theorem is false, so an appeal to Jordan form won't work (at least, not as easily as we'd hope it would). Fake Theorem: If $A\in T_n(\mathbb{F})$, then there exists an invertible $T\in T_n(\mathbb{F})$ and a permutation matrix $P$ such that $P^{-1}T^{-1} ATP$ is in Jordan form. An explicit counterexample is $$A=\left[\begin{array}{cccc} 0&1&0&0\\ &0&0&1\\ & &0&1\\ & & &0\end{array}\right]$$ and a more detailed demolishing of this theorem is given here, where the authors prove that if $n\geq 12$ and $\mathbb{F}$ is infinite, then there are infinite sets of nilpotent matrices in $T_n(\mathbb{F})$, none of which are conjugate (in $T_n(\mathbb{F})$) to any of the others. I mention this because I thought it was true for longer than I'd like to admit, and a few other people I've talked to thought it was true as well until told otherwise. REPLY [30 votes]: This is false! Let $$A = \begin{bmatrix} 0&0&0&1 \\ &0&1&0 \\ &&0&0 \\ &&&0 \\ \end{bmatrix}.$$ Imposing that $XA=AX$ for upper triangular $X$ gives linear equations on the $10$ entries of $X$. Solving them, I get that this occurs precisely for $X$ of the form $$X=\begin{bmatrix} a&0&\ast&\ast \\ &b&\ast&\ast \\ &&b&0 \\ &&& a \\ \end{bmatrix}.$$ In turn, an upper triangular matrix commutes with all such $X$ if and only if it is of the form $$B=\begin{bmatrix} c&0&0&d \\ &c&e&0 \\ &&c&0 \\ &&&c \\ \end{bmatrix}.$$ But such a $B$ is only a polynomial in $A$ if $d=e$.<|endoftext|> TITLE: Hausdorff dimension of the graph of an increasing function QUESTION [7 upvotes]: Let $f$ be a continuous, strictly increasing function from $[0,1]$ to itself with $f(0)=0, f(1)=1$. Let $\Gamma_f$ denote its graph. What can be said about the Hausdorff dimension of $\Gamma_f$? In particular, is it true that it is always 1? If not, is there a link between $\dim_H(\Gamma_f)$ and $\dim_H(\mu)$, where $\mu$ is the measure whose distribution function is $f$? (That is, $f(x)=\mu[0,x]$.) I would appreciate some examples if there's no general answer. Specifically, I think something has to be known when $f$ is the Minkowski question mark function, but Google wasn't much help here, unfortunately. REPLY [11 votes]: Theorem 1. The Hausdorff dimension of the graph $\Gamma_f$ of $f$ equals $1$. Proof. Take a partition of $[0,1]$ by intervals of length $1/n$. Since the function is increasing you can cover the graph by boxes of size $1/n\times (f((k+1)/n)-f(k/n))$, $k=0,1,\ldots,n-1$ located above the intervals of partition. The diameters of each of the boxes is bounded by $1/n+(f((k+1)/n)-f(k/n))$ and the sum of diameters is bounded by $$ \sum_{n=0}^{n-1} \frac{1}{n}+\left(f\left(\frac{k+1}{n}\right)-f\left(\frac{k}{n}\right)\right)=2. $$ Since the diameters of the boxes covering the graph converge to $0$ as $n\to\infty$, it follows that the Hausdorff $\mathcal{H}^1$ measure of the graph is bounded by $2$. $\Box$ Note that the the argument used in the proof shows also that the length of the graph is bounded by $L(\Gamma_f)\leq 2$. The graph of $f$ can be parametrized by a curve $\gamma:[0,1]\to\mathbb{R}^2$, $\gamma(t)=(t,f(t))$. In fact a stronger result is true: Theorem 2. $\mathcal{H}^1(\Gamma_f)=L(\Gamma_f)\leq 2$. This is a consequence of a more general results about curves in metric spaces that we will discuss now. If $\gamma:[a,b]\to (X,d)$ is a curve in a metric space, then its length is defined by $$ L(\gamma)={\rm Var}(\gamma)= \sup\left\{\sum_{i=1}^{n-1}d(\gamma(t_i),\gamma(t_{i+1}))\right\} $$ where the supremum is over all $n$ and all partitions $a=t_1<\ldots TITLE: Cofibrations in a category of fibrant objects QUESTION [8 upvotes]: There is an obvious (?) notion of cofibration in a category of fibrant objects, namely a morphism which satisfies the left lifting property with respect to all trivial fibrations. I don't seem to be able to find a place in the literature where this concept is discussed. For a concrete question: suppose $A$ is a cofibrant object in a category of fibrant objects $\mathcal{C}$, and $X_\bullet\to X$ is a simplicial resolution of the object $X$ of $\mathcal{C}$. Does the simplicial set $X_\bullet(A)$ represent the homotopy type of the mapping space $\mathop{Hom}(A,X)$ in the simplicial localization of $\mathcal{C}$? REPLY [2 votes]: Let $\mathcal{C}$ be a category. A cylinder, $\mathbf{I}$, on $\mathcal{C}$ is a functor (cylinder functor) $$I:\mathcal{C} \longrightarrow \mathcal{C}$$ together with three natural transformations $$e^{0}: 1_{\mathcal{C}} \Longrightarrow I , e^{1}: 1_{\mathcal{C}} \Longrightarrow I, \sigma: I \Longrightarrow 1_{\mathcal{C,}}$$ such that $\sigma e^{0}= \sigma e^{1}= 1,$ with $1: 1_{\mathcal{C}} \Longrightarrow 1_{\mathcal{C}}.$ A morphism $i:A\rightarrow X$ of Category with cylinder $\equiv$ $(\mathcal{C}, I,e^{0},e^{1},\sigma)$ is a cofibration if and only if the diagram $\require{AMScd}$ \begin{CD} A @>>e^{0}_{A}> I(A)\\ @V i V V @VV I(i) V\\ X @>>e^{0}_{X}> I(X) \end{CD} is weak pushout in $\mathcal{C}.$ K. H. Kamps and T. Porter, Abstract Homotopy and Simple Homotopy Theory.<|endoftext|> TITLE: Intersection of free/affine submodules, comparison with vector spaces QUESTION [7 upvotes]: If $W_1,W_2 \subset V$ are finite-dimensional $k$-vector spaces of dimensions $d_1, d_2 \leq d$, respectively, then $d_1 + d_2 > d$ suffices to guarantee $W_1 \cap W_2 \neq \{0\}$. There are similar results for affine subspaces of a spaces of a $k$-vector space, $E_1,E_2 \subset V$. I'm looking for analogous results for submodules of a free $R$-module $N_1,N_2 \subset M$ of finite ranks $r_1,r_2 \leq r$, and for "affine submodules" (i.e. torsors/cosets/translates of submodules), $A_1,A_2 \subset R$. The main question I'm interested in is what can be said about the intersections $N_1\cap N_2, A_1\cap A_2$ (i.e. are they nonzero or nonempty?) given certain conditions on the subspaces (i.e. they are (translates of) submodules of certain rank). I currently have in mind $\mathbb{Z}$-modules, but I am also interested in modules over other rings (probably all Noetherian). In addition to intersections, I'd be interested to read more general information. The notion of greatest common divisor is evidently at play here. I would be interested in good references/resources or in direct statements/proofs of useful results. E.g. Consider the translates of rank-2 $\mathbb{Z}$-modules $$A_1 = (u_1,v_2,w_1)+\langle (a_1,b_1,c_1),(a_2,b_2,c_2) \rangle$$ $$A_2=(u_2,v_2,w_2)+\langle (a_3,b_3,c_3),(a_4,b_4,c_4) \rangle \subset \mathbb{Z}^3$$ What is $A_1 \cap A_2$? Lastly I have a terminology question, relating to an important distinction between the module case and the vector space case. Is there a name for the property of a submodule being maximal with respect to its rank? ("primitive"?) For example $\langle (2,4) \rangle \subset \mathbb{Z}^2$ is a free rank-1 submodule, but it is properly contained in the free rank-1 submodule $\langle (1,2) \rangle$, which is not properly contained in another rank-1 submodule. Thank you. REPLY [8 votes]: There is a natural generalization of the aforementioned dimension-based reasoning to modules $M$ over commutative domains or commutative Noetherian reduced rings. Let $M$ be a module over a commutative domain $R$ and let $K$ be the fraction field of $R$. We set $\text{rk}(M) \Doteq \dim_K(M \otimes_R K)$. Claim. Let $R$ be a commutative domain. Let $M$ be an $R$-module of finite rank over $R$ and let $N_1, N_2 \subseteq V$ be submodules of $M$. Then we have $\text{rk}(N_1) + \text{rk}(N_2) = \text{rk}(N_1 \cap N_2) + \text{rk}(N_1 + N_2)$. Proof. Let $f: N_1 \oplus N_2 \rightarrow M$ be the $R$-linear map defined by $f(n_1, n_2) = n_1 + n_2$. It suffices to apply the exact functor $(-)\otimes_R K$ to the exact sequence $0 \rightarrow N_1 \cap N_2 \rightarrow N_1 \oplus N_2 \xrightarrow[]{f} N_1 + N_2 \rightarrow 0$ and then use the Rank-Nullity Theorem. If $R$ is any commutative reduced Noetherian ring, then its total ring of fractions $K$ is a product $\prod_i K_i$ of finitely many fields $K_i$. We set $\text{rk}(M) \Doteq (\dim_{K_i}(M \otimes_R K_i))_i$, a finite-dimensional vector with integer coordinates. The same claim will hold true if we replace "domain" by "reduced Noetherian ring" and if vectors are ordered component-wise. About the terminology. Let $R$ be a commutative domain, $M$ a torsion-free module over $R$ and $N$ an $R$-submodule of $M$. Let us call $N$ a primitive submodule of $M$ if there is no non-invertible element $d \in R$ such that $N = dN'$ for some submodule $N'$ of $M$. If $R$ is moreover Noetherian, then every submodule $N$ is contained a primitive submodule of $M$. If $M$ is free of finite rank and if $R$ is an elementary divisor ring (e.g., a principal ideal domain), then a finitely generated submodule $N$ is a primitive submodule of $M$ if and only if the minimal number of generators of $M/N$ is strictly less that $\text{rk}(M)$, if and only if $N$ has a generating set containing a basis vector of $M$, or primitive vector of $M$. In the theory of orders in quadratic number fields, the terminology primitive ideal is standard and has a meaning which is close to our definition. About affine submodules. Let $R$ be any commutative ring with identity and let $M$ be an $R$-module. Let $N_1, N_2$ be two submodules of $M$. Let $a_1, a_2 \in M$ and set $A_i = a_i + N_i$ for $i = 1,2$. Then $A_1 \cap A_2$ is non-empty if and only if $a_1 - a_2 \in N_1 + N_2$. If $A_1 \cap A_2$ is not empty, then it is an affine $R$-module of the form $s + N_1 \cap N_2$ where $s$ is any element of $A_1 \cap A_2$. From now on, we assume that $R$ is a domain or a reduced Noetherian ring with total ring fractions $K$. In this case, the rank of $N_1 \cap N_2$ is subject to the relation of the above claim. If $M$ is moreover torsion-free of finite rank and $N_1$ and $N_2$ are both finitely generated, then deciding whether $A_1 \cap A_2$ is $\{0\}$ boils down to deciding whether $AX = B$ has a solution $X \in R^n$ for some $m$-by-$n$ matrix $A$ over a field and some $B \in R^n$. This can be handled by linear algebra if $B = 0$. Let us assume now that $R$ is a domain and both $N_1$ and $N_2$ are finitely generated over $R$. Compute a minimal generating set of $V_1 \cap V_2$ is manageable if $R$ is an elementary divisor ring (see also this MO post), e.g., a principal ideal ring. In this case, we can use the Smith Normal Form of $A$ to solve $AX = 0$. If $M$ is free of finite rank over $R$, we can do the same for the equation $AX = B$ under the same assumptions. As the localization at any prime $\mathfrak{p}$ of a Dedekind ring $R$ is principal ideal domain, we obtain a Smith Normal Form of $A$ over $R_{\mathfrak{p}}$ for any such $\mathfrak{p}$. This yields an algorithm to decide on the existence of solutions for the system $AX = B$: we only need to check existence for the minimal prime ideals containing the Fitting ideal of $A$, see [5, Proposition 20.7 and subsequent comment]. Addendum. Here is an example of what can be achieved by sage. K = CyclotomicField(4) R = K.maximal_order() # R = Z[i], the ring of Gaussian integers. The ring R is Euclidean. i = K.gen(); # Definition of the intersection problem of two affine planes in R^3 b_1 = vector([4 * i + 6, -7 * i , -19 + 17 * i]) b_2 = vector([-1, 35, 0]) A_1 = Matrix(R, [[ 0, 2 - i], [ 3 - 2 * i, -i ], [ 0, 1 + 7 * i]]) A_2 = Matrix(R, [[ 2 - i, 1 + 2 * i], [ 0, 1 - i], [ 1, 0]]) # Concatenating A_1 with -A_2 A_2 = - A_2 C = [ c for c in A_1.columns() ] C += [ c for c in A_2.columns() ] A = Matrix(R, C).transpose(); # Finding one solution # Note: solve_right() may find a solution with coordinates in the fraction field of R but not in R # I don't know if in the latter case this indicates that there is no solution. A.solve_right(b_2 - b_1); # In our example, it finds (7*zeta4 + 8, -3*zeta4 - 2, 0, 0) where zeta4 stands for i. # Using is_submodule() to check existence of solutions M = R^3 N = M.submodule(A.transpose()) # This is the submodule generated by the columns of A. L = M.span(Matrix(R, b_1 - b_2)) ans = L.is_submodule(M); ans; # The answer is yes, our linear system has at least one solution in R^3. # Using intersection() to describe the whole solution set N_1 = M.span(A_1.transpose()) N_2 = M.span(A_2.transpose()) N_1.intersection(N_2); # This is the rank 1 R-module generated by [4*zeta4 + 7 -7*zeta4 - 35 17*zeta4 - 19] in our example # Using kernel() to describe the whole solution set kerA = kernel(A.transpose()) solution_dim = len(kerA.basis()) # This is 1 in our example. basis_vector = kerA.basis()[0]; x_1 = vector([ basis_vector[index] for index in range(0, 2)]) n_1 = Matrix(R, A_1) * x_1; # This vector generates N_1 \cap N_2 over R in our example. # Using smith_form() to describe the whole solution set # With the invariant factor matrix and the two change-of-basis matrices # you can decide whether the linear system has a solution and describe the full set of solutions A.smith_form(); Edit. As explained above, the resolution of linear systems of the form $AX = B$ over an arbitrary commutative case is particularly relevant to the question when each module $M_i \, (i = 1, 2)$ is given with an explicit embedding in some free module $R^{k_i}$. I just realized that there exist general theorems by means of which one can decide on the existence and uniqueness of solutions. MacCoy's Theorem [2, Th. 51] addresses the problem as to whether the system $AX = 0$ has a non-trivial solution $X \in R^n$ for $A$ an $m \times n$ matrix over an arbitrary commutative ring $R$: a non-trivial solution exists if and only if the determinantal rank of $A$ is less than $n$. [3, Proposition 1] shows that $AX = B$ has a solution in $R^n$ if and only if the localized system at $\mathfrak{m}$ has a solution in $R_{\mathfrak{m}}^n$ for every maximal ideal $\mathfrak{m}$ of $R$, an arbitrary commutative ring. [3, Theorem 6] and [4, Theorem 10] give (the same) necessary and sufficient conditions for the existence of solutions of the system $AX = B$ when $R$ is a Prüfer domain, generalizing a result of E. Steinitz [1] established for Dedekind rings. [1] E. Steinitz, "Rechteckige systeme and moduln in algebraischen zahlkorpern", 1912. [2] N. MacCoy, "Rings and Ideals", 1948. [3] P. Camion, L. Levy and H. Mann, "Linear equations over a commutative ring", 1971. [4] J. Hermida and T. Sanchez-Giralda, "Linear equations and commutative rings", 1986. [5] D. Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", 2004.<|endoftext|> TITLE: Invariant subspace in infinite dimensions QUESTION [5 upvotes]: Let $A(t)$ be a family of skew self-adjoint operator defined on some Hilbert space $H$ with common domain $D(A).$ The dependence on $t$ is in the strongly continuous sense, i.e. for all $x \in D(A)$ the map $t \mapsto A(t)x$ is continuous. Consider the initial value problem $$\varphi'(t)=A(t)\varphi(t)$$ with $\varphi(0)=\varphi_0.$ Assume that there is a dense subspace $X$ in $H$ that is contained in the domain of $A(t)$ for all $t$ and $A(t)X \subset X.$ I ask: Let $\varphi_0 \in X$. Does this imply that $\varphi(t) \in X$ for all $t>0?$ This sounds very natural but I do not have any tools/ideas to show this at the moment. REPLY [5 votes]: In general, the answer is "no" even in the autonomous case, i.e. in the case where $A := A(t)$ does not depend on $t$. First note that if $A$ generates a $C_0$-semigroup on $H$, if $X$ is a dense subspace of $H$ and if $X$ is invariant under this $C_0$-semigroup, then $X$ is even dense in $D(A)$ (which we endow with the graph norm); see for instance [Engel and Nagel: One-Parameter Semigroups for Linear Evolution Equations, Springer (2000), Proposition II.1.7]. Now let $H = L^2((0,1))$ and let $B$ be the Laplace operator on $H$ with periodic boundary conditions, i.e. \begin{align*} D(B) & = \{u \in H^2((0,1)): \, u(0) = u(1) \text{ and } u'(0) = u'(1)\}, \\ Bu & = u''. \end{align*} Note that $B$ is self-adjoint. Set $A := iB$ and let $X$ be the space of test functions on $(0,1)$. Then $AX \subseteq X$, but $X$ is not dense in $D(A)$ with respect to the graph norm; thus, $X$ is not invariant under the semigroup $(e^{tA})_{t \ge 0}$ generated by $A$.<|endoftext|> TITLE: Centraliser of an absolute Galois group QUESTION [17 upvotes]: Let $K$ be a finite extension of $\mathbb{Q}_p$. Is the centraliser of $\operatorname{Gal}(\overline{K}/K)$ in $\operatorname{Gal}(\overline{\mathbb{Q}_p} / \mathbb{Q}_p)$ trivial ? If yes, how can I show it ? REPLY [15 votes]: For an extension $L/\mathbf{Q}_p$, let $G_L$ denote the absolute Galois group $\mathrm{Gal}(\overline{L}/L)$. If $\sigma \in G_{\mathbf{Q}_p}$ acts centrally on $G_K$, then it also acts centrally on the subgroup $G_L \subset G_K$ for any finite $L/K$. But then it also acts trivially on the abelianization of $G_L$. By class field theory, the abelianization of $G_L$ may be identified with the pro-finite completion of $L^{\times}$, and moreover this isomorphism is compatible with the action of $G_{\mathbf{Q}_p}$. Hence $\sigma$ must also act trivially on $L^{\times}$ for any finite $L/\mathbf{Q}_p$, and thus $\sigma$ is trivial.<|endoftext|> TITLE: What about the action on this space? QUESTION [7 upvotes]: I think it is true that there is no free-action of $\mathbb{Z}_p(p\neq 2$) on product of $\mathbb{CP}^n(n$ odd) and $\mathbb{S}^{2m}$. But I don't know how to prove it. Any solution will be helpful. Thanks in advance. REPLY [12 votes]: I think the following will work for $p > 3$. I am not sure if it can be made to work for $p =3$, but maybe it can. I will the appeal to the representation theory of $\mathbb{Z}/p\mathbb{Z}$, but probably this is overkill. Given a continuous action of $\mathbb{Z}/p\mathbb{Z}$ on your space $X = \mathbb{CP}^n\times S^{2m}$, the rational cohomology groups are all representations of $\mathbb{Z}/p\mathbb{Z}$. There are two irreducible representations of $\mathbb{Z}/p\mathbb{Z}$ over $\mathbb{Q}$, the trivial representation and a non-trivial representation of dimension $p-1$. Each cohomology group can be expressed as a direct sum of copies of these representations. On the other hand, the Künneth theorem shows that the only non-zero cohomology groups of $X$ have dimension 1 or 2, and that these occur in even degree. Since $p-1 > 2$, the cohomology groups must be direct sums of copies of the trivial representation. In other words, each element $g$ of $\mathbb{Z}/p\mathbb{Z}$ acts trivially on the cohomology of $X$. It follows easily now that the Lefschetz number of any such element is positive, and therefore that $g$ has a fixed point.<|endoftext|> TITLE: Kuznetsov trace formula, orthogonality of Bessel functions QUESTION [9 upvotes]: Sorry if this is a vague question. I remember from my younger days that before proving his trace formula, Kuznetsov had a pretty result on orthogonality of Bessel functions. The formulas that I am going to write are WRONG, but the whole point is that I would like to remember the correct one (or a pointer to a reference). If $J_{\nu}$ denotes the $\nu$th $J$-Bessel function, then if $n$ and $m$ are positive integers, then $=0$ when $n\ne m$ for the scalar product $\int_0^\infty f(x)g(x)dx/x$ (I know this is wrong, let me continue). Thus, as for Fourier series, one can think of expanding reasonable functions into linear combinations of $J_n$ by computing the scalar product to get the coefficients. This does not work because the $J_n$ are not complete, contrary to the $e^{inx}$. Thus, in addition, Kuznetsov adds the "continuous spectrum" functions which are $J_{i\nu}-J_{-i\nu}$ for $\nu\in\Bbb R$, and shows that these are orthonormal, orthogonal to the $J_n$, and now form a complete set. I know that this is very weak form of the Selberg trace formula, but could someone give me the correct formula ? REPLY [13 votes]: The Bessel functions $J_\ell$ for $\ell\geq 1$ odd are pairwise orthogonal on the positive axis with respect to the measure $dx/x$. They correspond to the holomorphic spectrum (of various even weights $\ell+1$) of $L^2(\Gamma\backslash H)$. The orthogonal complement of the span of these $J_\ell$'s is continuously (and orthogonally) spanned by the functions $J_{2it}-J_{-2it}$ with $t>0$. This corresponds to the (weight zero and tempered) Maass and Eisenstein spectrum of $L^2(\Gamma\backslash H)$ (of various Laplace eigenvalues $1/4+t^2$). For more details, I recommend Sections 9.3-9.4 in Iwaniec: Introduction to the spectral theory of automorphic forms. The Bruggeman-Kuznetsov formula is not a weak form of the Selberg trace formula, in fact in many situations it is a more refined (or more suitable) tool than the Selberg trace formula. It can be interpreted as a relative trace formula, see e.g. the paper of Knightly and Li in Acta Arithmetica. REPLY [7 votes]: I was acquainted with Nikolay Vasil'evich Kuznetsov while worked in Vladivostok, 1990s. And he was very kind to mee, too. He tought me that many asymptotics for Bessel functions are not valid, many from Erdelyi's book on asymptotics, and some formulas for integral transforms. And as you I do not remember all his lessons, unfortunately. But I think the answer to your question is in this paper or this one. Fortunately they are digitized officially now by the Steklov Institute project on mathnet.<|endoftext|> TITLE: Are there invariants of cell complexes similar to the Euler characteristic? QUESTION [9 upvotes]: The Euler characteristic is an invariant (under homeomorphism) of manifolds that can be computed from a cellulation by (weighted) counting of different kinds of objects, namely \begin{equation} \chi=\textrm{#vertices}-\textrm{#edges}+\textrm{#faces}-\ldots \end{equation} Are there any other (independent) invariants that are computed in a similar way? I would allow for arbitrary weights (not only $\pm 1$) and arbitrary kinds of "objects" (not only $d$-cells, but also things like "corners", "cycles consisting of $3$ edges", "pairs of a $3$-cell and a $1$-cell that is part of its boundary"). In particular, in odd dimensions where the Euler characteristic is trivial by Poincare duality, is there any sort of replacement for it? REPLY [2 votes]: As pointed out in the comments, every characteristic class in $H^d(BO(d), G)$ provides a $G$-valued locally computable invariant of $d$-manifolds, by pulling back via the classifying map of the tangent bundle. As argued by Levitt and Rourke, there are local combinatorial formulas on trianglulations for dual simplicial $d-i$-cycles representing any degree-$i$ characteristic classes, whose values on a $d-i$-simplex only depend on the star of that simplex. To be precise, for this to work we need to add some decorations to the triangulation, namely a local ordering. So every degree-$d$ characteristic class yields a local combinatorial formula for a simplicial $0$-cycle, whose summation over vertices is an invariant of the type in the question. The "objects" are the different stars of vertices, to which we associate elements of $G$ according to the local formula. Summing up all the $G$-values on the vertices corresponds to evaluating the invariant. Concrete local formulas are in fact known for many cases. For $G=\mathbb{Z}_2$, the relevant characteristic classes are generated by degree-$d$ polynomials (with $\mathbb{Z}_2$ sum and cup product) of Stiefel-Whitney classes. For the latter, combinatorial formulas are given by Goldstein and Turner, and a combinatorial cup product for cycles can be easily obtained via its geometric interpretation as intersection. For $G=\mathbb{Z}$ on oriented manifolds, we additionally need the Bockstein homomorphism for the short exact sequence $$\mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}_2$$ and the Pontryagin classes. The former only involves simplex-wise operations and the co-boundary operator. The Pontryagin classes are the only case for which no satisfactory local formulas exist to date. $\mathbb{Q}$-valued formulas have been described by Gaifullin, but not very explicitly. I'm not aware of any converse argument that every local formula for a $0$-cycle invariant corresponds to a characteristic class, however I don't think there are any known examples for invariants which don't.<|endoftext|> TITLE: Is there a discrete lattice analogue of conformal transformations? QUESTION [12 upvotes]: There is a simple discrete combinatorial analogue of manifolds and homeomorphisms: Replace manifolds by simplicial complexes and homeomorphisms by Pachner moves. Equivalence classes of manifolds under homeomorphism are in one-to-one correspondence with equivalence classes of simplicial complexes under Pachner moves (not mentioning technical details, e.g. that we should add "piece-wise linear" to both manifolds and complexes). A very similar construction can be made for spin manifolds and spin homeomorphisms: In e.g. https://arxiv.org/abs/1505.05856 they give a discrete representant of the second Stiefel-Whitney class for $3$ dimensional simplicial complexes as a $1$-cycle in the complex. Then a discrete representation of the spin structure is given by a $2$-cycle that has this $1$-cycle as boundary. My question: Is there any construction similar in spirit for manifolds and conformal maps (in $2$ dimensions)? I.e. is there any kind of $2$ dimensional combinatorial structure and a set of local moves such that equivalence classes of the combinatorial structure under local moves correspond to equivalence classes of manifolds under conformal maps? One could for example still use simplicial complexes and restrict the set of moves such that a combinatorial analogue of "angles are preserved" holds, or add a new kind of structure to the combinatorial description as in the case of spin structures above. REPLY [3 votes]: Thurston produced a constructive discretisation of the Riemann mapping theorem which proceeds as follows: given a region $D$, approximate it by densely packing circles into its interior (as a fragment of hexagonal lattice), then transform the resulting arrangement to fill a disc (by using the circle packing theorem). Wikipedia illustrates this better than I can:<|endoftext|> TITLE: Generalized projective spaces, spheres, and exotic spheres QUESTION [6 upvotes]: I like to explore and ask for proper references for the relations between generalized projective spaces, spheres, and exotic spheres: The real projective space $\mathbb{RP}^1 \simeq S^1,$ is homeomorphic and diffeomorphic to 1-sphere $S^1.$ The complex projective space $\mathbb{CP}^1 \simeq S^2,$ is homeomorphic and diffeomorphic to 2-sphere $S^2.$ The quaternion projective space $\mathbb{HP}^1 \simeq S^4,$ is homeomorphic and diffeomorphic to 4-sphere $S^4.$ I am looking for general constructions of generalized projective spaces, and their relations to spheres, and exotic spheres. For example, we can consider generalized projective spaces homeomorphic to $S^8,S^{16},\dots, S^{2^n},$ etc., for $n\in \mathbb{N}$(yes ?). Are they necessarily diffeomorphic to the standard spheres? Are there generalized projective spaces of exotic spheres? Can one consider the quotient manifold of generalized projective spaces to construct spheres, and/or exotic spheres? (e.g. The 8-dimensional $\mathbb{HP}^{2}$ has a circle action, by the group of complex scalars of absolute value 1 acting on the other side (so on the right, instead of on the left action). Therefore we can obtain the quotient manifold $\mathbb{HP}^{2}/\mathrm{U}(1)$ with U(1) for the circle group. It has been shown that this quotient is the 7-sphere, a result of Arnold (1996), also later by Witten and Atiyah.) Thank you in advance! REPLY [5 votes]: While I think it is difficult to give a solid answer to this question, you may perhaps be interested in Projective planes, Severi varieties and spheres from 2002, by Atiyah and Berndt. For example they show that $\mathbb{OP}^2 / Sp(1) \simeq S^{13}$ naturally follows the $\mathbb{HP}^2/ U(1) \simeq S^7$ diffeomorphism that you mention, and they give a fairly-extensive discussion of the context.<|endoftext|> TITLE: Relations between coefficients of expansions of a rational function at 0 and infinity QUESTION [12 upvotes]: This question goes in the bucket of "this must be well known, but I don't see it and am not sure where to look it up." Given two Laurent power series $A(t)=\sum_{k>N}a_kt^k$ and $B(t)=\sum_{k>M}b_kt^k$ for $a_k,b_k\in F$ a field, we say that these are expansions at 0 and $\infty$ of a rational function $f$ if in the field $F((t))$, we have that $f(t)=A(t)$ and $f(t^{-1})=B(t)$. Now, think about $\,a_k$ and $\,b_k$ as formal variables: Are there any relations between $a_k$ and $b_k$, or any other way of expressing algebraically that $A(t) = B(t^{-1})$? It seems to me that there can be no relations in the strong sense that I can fix any finite number of $a_*$ and $b_*$ and complete to get such compatible expansions, but maybe there's some more subtle relation I've missed. EDIT: After feeding the answer below, I see what I should have thought of before: this is actually a topological property. We can endow the ring generated by $a_k$ and $b_k$ with a topology such that the ideal generated by the $m\times m$ minors of the Hankel matrix for each $m$ give a basis of neighborhoods of $0$. A map of this ring to the base field $F$ (with the discrete topology) is continuous if and only if one of these neighborhoods is killed, and we have a rational function that we have the two expansions of. REPLY [15 votes]: One common criterion that is used to algebraically encode whether a power series $\sum_{n\geq 0} a_nx^n$ represents a rational function is that the infinite Hankel matrix $(a_{i+j-1})$ be of finite rank. If this is the kind of criterion you are looking for then for your pair of series $A(t)$ and $B(t)$ satisfy $A(t)=B(t^{-1})$ if and only if the (bi infinite) Hankel matrix associated to $$C(t)=A(t)-B(t^{-1})$$ is of finite rank. That is, express $A(t)-B(t^{-1})$ as a power series $\sum_{n\in \mathbb Z}c_n x^n$ and then check that the bi infinite matrix $(c_{i+j-1})$ has finite rank. This is the same as saying that certain determinants with entries of the form $a_k-b_{-k}$ vanish. More explicitly this is encoding the fact that if, for example, $A(t)$ looks like $$\sum_{n\geq 0}\left(P_1(n)\alpha_{1}^n+\cdots P_k(n)\alpha_k^n\right)t^n$$ then $B(t)$ looks like $$-\sum_{n\geq 0}\left(P_1(-n)\alpha_{1}^{-n}+\cdots P_k(-n)\alpha_k^{-n}\right)t^n.$$ In other words as formal power series we have a "vanishing relation" $$\sum_{n\in \mathbb Z} P(n)\alpha^nz^n=0$$ This last relation is a nonrigorous way of expressing what I said above, but I like thinking about it as a relation in the same sense as in Brion's formula from polyhedral generating functions/equivariant localization (and it can be made rigorous using the same methods).<|endoftext|> TITLE: An inequality related to area and sidelengths of a polygon $Area(A_1A_2....A_n) \le \frac{1}{n}cotg{\frac{\pi}{n}} \sum_{i=1}^nA_iA_{i+1}^2$ QUESTION [6 upvotes]: $\DeclareMathOperator\Area{Area}\DeclareMathOperator\cotg{cotg}$I am looking for a proof (or a reference) of an inequality related to area and the sidelengths of a polygon as follows: Let $A_1A_2\cdots A_n$ be an arbitrary polygon, then: $$\Area(A_1A_2\cdots A_n) \le \frac{1}{4}\cotg{\frac{\pi}{n}} \sum_{i=1}^nA_iA_{i+1}^2$$ This is a generalization of Weitzenböck's inequality. You can see a stronger version at Strengthened version of Isoperimetric inequality with n-polygon. Geometric meanings: $$\Area(A_1A_2\cdots A_n) \le \frac{\Area(1)+\Area(2)+\dotsb +\Area(n)}{n}.$$ PS: I found this inequality long time ago, that time I thought this is old inequality. But today, I think this is new because I can not see any reference for the inequality. REPLY [8 votes]: Cauchy–Schwarz tells you that $$\sum_{i=1}^n \lvert A_iA_{i+1}\rvert^2\geq \frac{P^2}{n}$$ where $P$ is the perimeter of the polygon. Then we need the inequality $$P^2\geq 4n\tan(\pi/n)A$$ which is the classical isoperimetric inequality for polygons with many proofs in the literature, analytic, geometric and algebraic. See Fan, Taussky, and Todd - An algebraic proof of the isoperimetric inequality for polygons article and its references, for example. REPLY [4 votes]: Presumably the indices $i$ in $A_i$ are taken mod $n$, so "$A_{n+1}$" is to be identified with $A_1$. This must be a known isoperimetric inequality, but it's easier to prove than to find in the literature. Fix the area $\cal A$ of the $n$-gon. By a standard compactness argument there exists an $n$-gon $A_1 A_2 \ldots A_n$ that minimizes $\sum_{i=1}^n (A_i A_{i+1})^2$. We first show that this polygon is convex (but possibly with $A_{i-1} A_i A_{i+1}$ collinear for some $i$). Indeed if it is not we can replace it by the convex hull, with each side $A_j A_k$ of the convex hull divided into $k-j$ equal subsegments; this both increases the area and decreases $\sum_{i=1}^n (A_i A_{i+1})^2$, so we can shrink the polygon back to area $\cal A$ and make the sum of its sides' squares even smaller. Given convexity, fix all but one of the vertices, say $A_2$. Then $A_2$ is limited to a line parallel to $A_1 A_3$, and we readily see (as by choosing coordinates that make $A_1,A_3 = (\pm 1, 0)$ ) that $(A_1 A_2)^2 + (A_2 A_3)^2$ is minimized when $(A_1 A_2) = (A_2 A_3)$. Thus the minimizing $n$-gon has all sides equal, say with each $(A_i A_{i+1}) = s$; and then $ns^2$ is minimized when $ns$ is $-$ but $ns$ is the perimeter, and the usual isoperimetric inequality for $n$-gons then finishes the proof that the area-$\cal A$ polygon with the smallest $\sum_{i=1}^n (A_i A_{i+1})^2$ is regular.<|endoftext|> TITLE: The number of co-circular four tuples QUESTION [5 upvotes]: Let $A,B ⊂ \mathbb{R}$ such that $|A| = |B| = n$. What is the best-known upper bound on the number of four-tuples in $A \times B$ where the four points are co-circular, they lie on the same circle? REPLY [2 votes]: Incidence geometry can be used to prove that any finite set $P \subset \mathbb R^2$ with $O(|P|^{1/2})$ points on a single circle determines $O(|P|^3)$ cocircular quadruples. Since your grid point set satisfies this condition, it follows that there are $\ll n^6$ cocircular quadruples. Is this tight? The argument is similar to that which uses the Szemeredi-Trotter Theorem to show that a set of points in the plane with no more than $|P|^{1/2}$ on a line determine $O(|P|^2 \log |P|)$ collinear triples. Here is a rough proof. For a fixed point set $P$, let $C_k$ denote the set of all circles containing at least $k$ points from $P$. You can use Pach-Sharir's generalisation of Szemeredi-Trotter for pseudolines to prove that $$|C_k| \ll \frac{|P|^3}{k^5} + \frac{|P|}{k} \ll \frac{|P|^3}{k^5}, $$ where the last inequality follows from the assumption that $k \ll |P|^{1/2}$. Actually there is a hidden assumption here that $k$ is larger than some absolute constant, but then we can count cocircular quadruples on such poor circles by some trivial arguments. Now, dyadically decompose the number of cocircular quadruples and apply the bound on $|C_k|$ to complete the proof. $$|\{ \text{all cocircular quadruples}| \leq \sum_{j \geq 3} \sum_{C : 2^{j-1} \leq | C \cap P| < 2^j} |C \cap P|^4 \ll \sum_j (2^j)^4 \frac{|P|^3}{(2^j)^5} \ll |P|^3.$$<|endoftext|> TITLE: Holomorphic functions with equal inverse images of unit circle QUESTION [5 upvotes]: Let $f,g:\mathbb{C} \to \mathbb{C}$ be holomorphic and have the property $f^{-1}(S)=g^{-1}(S)$ where S is the unit circle centered at 0. What can be said about $f$ and $g$. REPLY [5 votes]: An exhaustive study of this question is contained in the paper MR0825840 Stephenson, Kenneth Analytic functions sharing level curves and tracts. Ann. of Math. (2) 123 (1986), no. 1, 107–144 (freely available online).<|endoftext|> TITLE: Chromatic numbers of infinite abelian Cayley graphs QUESTION [15 upvotes]: The recent striking progress on the chromatic number of the plane by de Grey arises from the interesting fact that certain Cayley graphs have large chromatic number; namely, the graph whose vertices are the ring of integers of a certain number field K endowed with a complex absolute value ||, and in which x and y are adjacent if and only if |x-y| = 1. This made me realize I know very little about the chromatic numbers of infinite Cayley graphs, and too my surprise, I wasn't able to find much in the literature! So here's a sample question. Let A be a finitely generated free abelian group, let S be a finite subset of A closed under negation, and let G be the Cayley graph whose vertices are the elements of A and where a,b are adjacent if and only if |a-b| lies in S. For every integer N, G has a quotient G/N, a finite Cayley graph whose vertices are A/NA and whose edges are given by the images of S in A/NA. (Maybe better to take N large enough so that no element of S lies in NA, so G/N has no loops.) Evidently a coloring of G/N pulls back to G, so we get an inequality of chromatic numbers $\chi(G) \leq \chi(G/N)$. My question is: is it always the case that $\chi(G) = \inf_N \chi(G/N)$? In other words: if G has a k-coloring, does it necessarily have a periodic k-coloring? (You might think of $\inf_N \chi(G/N)$ as the chromatic number of a profinite Cayley graph.) I don't even know how to do this for A=Z! REPLY [7 votes]: Here's an answer for $A = \mathbb Z$ (assuming that $S$ is finite): Let $c$ be an optimal (i.e. using the minimal number of colours) colouring of $\mathrm{Cay}(\mathbb Z,S)$ and let $k > \max S$. The degrees in $\mathrm{Cay}(\mathbb Z,S)$ are bounded, hence $c$ uses a finite number of colours. This means there are only finitely many possibilities for the sequence $(c(n+i))_{0 \leq i \leq k}$ and hence there are $n_1,n_2 \in \mathbb Z$ such that $n_2 > n_1 + k$ and $c(n_1+i) = c(n_2 + i)$ for $0 \leq i \leq k$. Without loss of generality assume that $n_1 =0$, $n_2=l$ Define a colouring $d$ by $d(n) = c(n \mod l)$. This clearly is periodic. Since $k > \max S$ and $c(i) = c(l + i)$ for $0 \leq i \leq k$, the neighbours of any vertex $n$ have the same colours in $d$ as the neighbours of $(n \mod l)$ have in $c$, whence the colouring is proper.<|endoftext|> TITLE: Examples of probability measures with `fake' decay QUESTION [7 upvotes]: To be concise, I am wondering whether there are natural examples of probability measures $\mu$ compactly supported on the real line which satisfy $\mu(I) \lesssim l_n^\alpha$ for all intervals $I$ with $|I| = l_n$, for some $0 \leq \alpha \leq 1$ and a certain sequence $l_1 \geq l_2 \geq \dots \to 0$, but for which the general inequality $\mu(I) \lesssim |I|^\alpha$ fails. Often, in order to lower bound the Hausdorff dimension of a certain set $X$ by a constant $\alpha$, one applies Frostman's lemma, finding a natural probability measure $\mu$ supported on $X$ with $\mu(I) \lesssim |I|^\alpha$. Often, one constructs $X$ as a fractal, as a limit of sets $X_n$, each a union of intervals of length $l_n$ (i.e. $X_1 \supset X_2 \supset \dots \to X$). Often, one obtains $\mu(I) \lesssim l_n^\alpha$ for all intervals $I$ of length $l_n$, and then tries to 'cheat out' the general bound from these estimates. My question is whether there are natural counterexamples of 'fake' $\alpha$ dimensional probability measures for which one can obtain the bound $\mu(I) \lesssim l_n^\alpha$ for all intervals $I$ of length $l_n$, where $l_1 \geq l_2 \geq \dots$ is a sequence converging to zero, but for which the bound $\mu(I) \lesssim |I|^\alpha$ does not hold for all intervals. REPLY [2 votes]: Later on in my research I answered this question in the negative: Consider a sequence of dyadic scales $\{ r_k \}$, such that $r_k/r_{k+1} \geq 4$. Consider a sequence of dyadic lengths $\{ r_k \}$, and construct a set $X$ by a Cantor-type decomposition, defined as $\lim_{k \to \infty} X_k$ where $X_k$ is a union of side-length $r_k$ intervals. We set $X_0 = [0,1]$, and $r_0 = 1$. Given $X_k$, which is a union of sidelength $r_k$ intervals, we define $X_{k+1}$ arbitrarily, such that for any sidelength $r_k$ interval in $X_k$, $X_{k+1}$ contains a single sidelength $r_{k+1}^{1/2}$ interval. If we let $N_k$ to be the covering number of $X_k$ by $r_k$ intervals, then $N_0 = 1$, and $N_{k+1} = N_k(r_k/r_{k+1}^{1/2})$. Thus $$ N_k = \frac{\left( r_1 \dots r_{k-1} \right)^{1/2}}{r_k^{1/2}} $$ If we set $\mu(I) = 1/N_k$ for each side-length $r_k$ interval $I$ selected in $X_k$, and $\mu(I) = 0$ otherwise, then $\mu$ satisfies the mass distribution principle and thus extends to a Borel probability measure. If $l_k = r_k^{1/2}$, then for any side-length $l_{k+1}^{1/2}$ interval, either $\mu(I) = 0$ or $$ \mu(I) = \frac{l_{k+1}}{(r_1 \dots r_k)^{1/2}} \leq \frac{l_{k+1}}{r_k^{k/2}}. $$ If $r_{k+1} \leq r_k^{4k}$, then $l_{k+1}^{1/4} = r_{k+1}^{1/8} \leq r_k^{k/2}$, so $\mu(I) \leq l_k^{3/4}$ for each index $k$ and side-length $l_k$ interval $I$. Thus at the scales $l_k$, the set $X$ looks like it has dimension at least $3/4$. But as $k \to \infty$, $$ H^{1/2}_{r_k}(X) \leq N_k \cdot r_k^{1/2} \leq (r_1 \dots r_{k-1})^{1/2} \to 0 $$ Thus $\dim(X) \leq 1/2$, so we cannot possibly obtain a general bound $\mu(I) \leq_\varepsilon l^{3/4-\varepsilon}$ for all scales $l$.<|endoftext|> TITLE: Example of group cohomology not annihilated by exponent of $G$? QUESTION [12 upvotes]: Is there an example of a finite group $G$ and an action on $M=\mathbb{Z}^n$ such that $H^2(G,M)$ has exponent greater than the exponent of $G$? (Especially, can we have $G=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ and some free $\mathbb{Z}$ module $M$ with $G$ action, such that $H^2(G,M)$ has elements of order $4$?) REPLY [21 votes]: For each finite group $G$ there is a $G$-module $M$ that is a free abelian group of finite rank such that $H^2(G,M)=\mathbb{Z}/|G|$. Proof: Let $I$ be the augmentation ideal of $\mathbb{Z}G$. Then $H^1(G,I)=\mathbb{Z}/|G|$. $\,\,I$ is a finitely generated $\mathbb{Z}G$-module (it's f.g. even as free abelian group). Hence there is a short exact sequence of $\mathbb{Z}G$-modules $$0 \to M \to F \to I \to 0$$ where $F$ is a free $\mathbb{Z}G$-module of finite rank. In particular, $F$ is a free abelian group of finite rank and so is the subgroup $M$. The long exact cohomology sequence now yields $H^2(G,M) \cong H^1(G,I)$. q.e.d. Added: If $\Omega^n(\mathbb{Z})$ denotes the kernel of $P_{n-1} \to P_{n-2}$ of a projective resolution $P \to \mathbb{Z}$ then the same argument shows $H^n(G,\Omega^n(\mathbb{Z}))=\mathbb{Z}/|G|$.<|endoftext|> TITLE: Distributing $N$ points on the sphere so that the sum of their mutual distances is maximized? QUESTION [7 upvotes]: Generaliation the result in our paper for sum and similarly my previous question for product. I have a question: My question: Distributing $N$ points on the sphere so that the sum of their mutual distances is maximized? REPLY [6 votes]: This is actually an open problem, the only proven optimal configurations I have found are the universal configurations with N=1,2,3,4,6,12, and the dipyramid with N=5. Some work for N=7 is done here, which is compatible with this. Some bounds are given in the paper Sums of distances between points of a sphere.<|endoftext|> TITLE: Homotopy pullback of a homotopy pushout is a homotopy pushout QUESTION [15 upvotes]: Let's assume that we have a cube of spaces such that everything commutes up to homotopy. The following holds: - The right square is a homotopy pushout and - all the squares in the middle are homotopy pullbacks. I was told that (in Top) the left square is also a homotopy pushout then. Do you know any source where this is stated? (Or: is this actually true?) REPLY [17 votes]: This is Mather's second cube theorem, see Theorem 25 in Mather, Michael, Pull-backs in homotopy theory, Can. J. Math. 28, 225-263 (1976). ZBL0351.55005.<|endoftext|> TITLE: Why should we study derivations of algebras? QUESTION [14 upvotes]: Some authors have written that derivation of an algebra is an important tools for studying its structure. Could you give me a specific example of how a derivation gives insight into an algebra's structure? More generally: Why should we study derivations of algebras? What is the advantage of knowing that an algebra has a non-trivial derivation? REPLY [7 votes]: Here are some algebraic aspects why one could be interested in derivations. First, derivations actually appear pretty prominently in algebraic geometry. For a ring homomorphism $R\to S$, one can consider $R$-linear derivations on $S$, with image in some $S$-module $M$. There is a universal such module, the module of Kähler differentials $\Omega_{S/R}$, with a universal derivation $d:S\to\Omega_{S/R}$. This is a fairly important object to study in algebraic geometry, e.g., in the definition and study of smooth varieties, or in the algebraic versions of de Rham cohomology. (A chapter on derivations can be found in any algebraic geometry textbook.) On the other hand, and a bit more specialized, there are also reasons to study locally nilpotent derivations on algebras. Since locally nilpotent derivations are related to actions of the additive group $\mathbb{G}_a$ on varieties, the locally nilpotent derivations also play a significant role in the study of quotients of $\mathbb{G}_a$-actions. In particular, (as found on the Wikipedia page) some counterexamples or Hilbert's 14th problem arise as kernels of locally nilpotent derivations. (survey paper here) Locally nilpotent derivations are also related to various open questions on the structure of affine space $\mathbb{A}^n$ (like the Zariski cancellation problem, see e.g. the Makar-Limanov invariant which can distinguish affine space from some affine varieties diffeomorphic to it). There is a whole book about the algebraic theory of locally nilpotent derivations: G. Freudenburg. Algebraic theory of locally nilpotent derivations. Encyclopaedia of Mathematical Sciences, 136. Invariant Theory and Algebraic Transformation Groups, VII. Springer-Verlag, Berlin, 2006. REPLY [2 votes]: The two examples which I can give are the following. One is the algebra of octonions. The derivations of this algebra form a 14-dimensional Lie algebra. From this fact you can conclude that the automorphism group of the octonions is 14-dimensional. An automorphism is obtained from a given derivation by taking the $exp$ function. The exponential function doesn't work over finite fields. There are formulas for derivations of octonions e.g. $[L_x,L_y]+ [L_x,R_y]+[R_x,R_y]$ is a derivation for every pair of octonions $x,y$. In this formula $L_x$ and $R_x$ denote left and right multiplication by the octonion $x$. You can find more in Schafer's book or Barton-Sudbery https://arxiv.org/abs/math/0203010. We can define the Jordan algebra $h_3(\mathbb O)$ of symmetric matrices of octonions. In this algebra $[L_x,L_y]$ is a derivation. The algebra of derivations is 52-dimensional in this case. It is the $f_4$ Lie algebra. I wonder whether the $e_6$, $e_7$ and $e_8$ Lie algebras can be obtained as derivations of some algebra.<|endoftext|> TITLE: Properly elliptic surface with no multiple fibers and without a section QUESTION [5 upvotes]: I am aware that if an elliptic surface contains multiple fibers, then it has no section. Is the converse false? In particular, I am looking for an example of a projective, properly elliptic surface (Kodaira dimension 1), fibered over $\mathbb{P}^1$, with no multiple fibers and no section. REPLY [5 votes]: Please confer Corollary 2.2 of the following with $d$ equal to $3$ and with $n$ equal to $2$. Jason Starr A pencil of Enriques surfaces of index 1 with no section https://arxiv.org/pdf/math/0602639.pdf This proves that for every integer $e\geq 2$, for a very general hypersurface $X$ in $\mathbb{P}^2\times \mathbb{P}^1$ in the complete linear system of $\mathcal{O}_{\mathbb{P}^2\times \mathbb{P}^1}(3,e)$, there is no rational section of the projection, $$\text{pr}_2|_X:X\to \mathbb{P}^1.$$ Since the locus of multiple curves in the complete linear system $\mathcal{O}_{\mathbb{P}^2}(3)$ has codimension $3$, for a general $X$ in the complete linear system, there are no multiple fibers. By adjunction, the dualizing sheaf of $X$ equals $\mathcal{O}_{\mathbb{P}^2\times \mathbb{P}^1}(0,e-2)|_X$. Thus, for $e\geq 3$, the dualizing sheaf is the pullback of an ample sheaf by $\text{pr}_2|_X$. In that case, the Kodaira dimension equals $1$.<|endoftext|> TITLE: Expected time of distinguishability of a series of Poisson processes bounded by each other QUESTION [5 upvotes]: Consider a system of $n$ "bounded" Poisson processes over the integers, $X_1, \ldots X_n$, all incrementing at rate $\lambda$. Initially all the processes begin at $0$. The process $X_i$ is inactive until $X_{i+1} - X_i > 1$, at which point it begins incrementing itself at rate $\lambda$. Whenever $X_{i+1} - X_i \leq 1$, the process $X_i$ stops, waiting for $X_{i+1}$ to increment itself, before becoming active again. ($X_n$ is always active.) We can see that eventually every $X_i$ will have a unique value. My question is regarding the expected time $E[\mathcal{T}]$ of the first time this event occurs: what is the first time all processes have a unique value? A simple bound seems to be $E[\mathcal{T}] = O(n^2)$. However, my simulations indicate that this takes linear time in $n$. I'd love to see an analysis that shows something along the lines of $E[\mathcal{T}] = O(n)$, or any other asymptotic analysis for this system. REPLY [3 votes]: The process you are looking at is called TASEP (totally asymmetric simple exclusion process); though your initial conditions are unusual. A more conventional version of your question would be to consider the TASEP with step initial conditions (one particle at every negative integer) and ask for a typical time that the first $n$ particles are at distances at least, say, 2 from each other. TASEP is a determinantal process and it is exactly solvable in a very strong sense, and a huge literature on asymptotic results for TASEP or its generalizations is available. The results in https://arxiv.org/pdf/0807.1713.pdf in particular, suggest that the time of interest should indeed scale linearly with $n$; e. g. Theorem 1 say that the typical time before the $n$-th particle moves scales linearly, and the intuition (corroborated by Theorem 2) is that once they all have started moving, they will spread out quickly. Note that Tracey and Widom are looking at ASEP which is a lot harder (non-determinantal), so their methods would be an overkill for TASEP. Maybe you should look into Johansson's papers they cite or look for a pre-2008 survey on TASEP.<|endoftext|> TITLE: Faltings theorem and number of singularities QUESTION [6 upvotes]: The Faltings theorem states that the number of rationals over an algebraic curve is finite if the genus is greater than 1. The genus decreases by increasing the number of singularities. My question is this. Should one count only the singularities that are rational points or all the singularities over the complex field? REPLY [12 votes]: The definition of the geometric genus in terms of (d-1)(d-2)/2 minus the contributions of the singularities is not a great one. It's better to give a more intrinsic definition, as the dimension of the space of global section of the canonical line bundle of the normalization (or the first sheaf cohomology of the normalization). In particular, this definition makes it possible to compute what the contributions of different types of singularities are. In particular, this definition is straightforwardly invariant under change of base field. So it's possible to define and compute it just over an algebraically closed field. This is what most references do (as most introductions are focused more on the pure algebraic geometry than in the arithmetic applications) which is probably the source of your contribution. So because singular points not defined over the base field correspond to multiple points over an algebraically closed field, they actually have a larger contribution than singular points defined over the base field.<|endoftext|> TITLE: Insights from disproofs after counterexamples have been given QUESTION [6 upvotes]: Some conjectures are disproved by a single counter-example and garner little or no further interest or study, such as (to my knowledge) Euler's conjecture in number theory that at least $n$ $n^{th}$ powers are required to sum to an $n^{th}$ power, for $n>2$ (disproved by counter-example by L. J. Lander and T. R. Parkin in 1966). But other conjectures retain interest, even after being disproved by counter-example, and lead to analytic disproofs and insights. Question What are some conjectures, first known to be false through counter-example, but whose subsequent analytic disproofs shed novel insights into the problem? More generally, what are the properties of known false conjectures that nevertheless garner significant interest and efforts of mathematicians to produce analytic disproofs? REPLY [2 votes]: The first two things that come to my mind are actually the dual of what you asked for, namely conjectures where the counterexample did not come first, and the search for an explicit counterexample led to interesting developments. Brosnan and Belkale disproved a conjecture of Kontsevich about polynomially countable graphs, but their argument did not lead to a specific counterexample. An explicit counterexample was provided by Dzmitry Doryn. That $\pi(x) > \mathrm{li}(x)$ for all $x$ was disproved by Littlewood, but without giving an explicit counterexample. I believe that there is still no explicit counterexample, but the search for one has inspired some interesting work.<|endoftext|> TITLE: CW Product via Whitehead map QUESTION [5 upvotes]: Product CW-complexes are defined via characteristic maps rather than from attaching maps, so via maps from $\mathbb D^n$ rather than from $\mathbb S^{n-1}$, because we have the propriety that $\mathbb D^n\times\mathbb D^m=\mathbb D^{m+n}$. I want to define products in a synthetic way, only manipulating objetcs up to homotopy (so no discs, anywhere, even during the construction as a provisional step, because there are nothing more than trivial things up to homotopy) so I need to use only the spheres. So basically if $X$ and $Y$ are CW-complexes and $Z$ is the product, to attach one new cell to $Z_{p-1}$ means attaching a product of cells $\mathbb S^{n-1}\to X_{n-1}$ and $\mathbb S^{m-1}\to Y_{m-1}$, where $n+m=p$, and this should give a function $\mathbb S^{n+m-1}\to Z_{p-1}$. We actually already have such a map when the CW-complexes $X$ and $Y$ are the spheres $\mathbb S^{n}$ and $\mathbb S^{m}$, which is the Whitehead map. But note that there I mean the Whitehead product not defined as the attaching map of the product of two cells, because it would be a construction using discs in a implicit way. I take it defined from the homotopy pushout definition of the join of two spheres. Indeed we have then (for a CW product of spheres) $Z_{n+m-1}=\mathbb S^{n}\vee\mathbb S^{m}$ and this implies that the wanted attaching map is of type $\mathbb S^{n+m-1}\to \mathbb S^{n}\vee\mathbb S^{m}$, where by definition $\mathbb S^{n+m-1} = \mathbb S^{n-1} *\mathbb S^{m-1}$. Now I want this bit to be generalizable to every product of finite CW-complex (it can have just a finite number of cells if nedded). So instead of a simple map $\mathbb S^{n+m-1}\to \mathbb S^{n}\vee\mathbb S^{m}$ want some function $\mathbb S^{n+m-1}\to Z_{p-1}$ where obviously $Z_{p-1}$ can be more complex than just a wedge. The definition of this map would look like that of the map $\mathbb S^{n+m-1}\to \mathbb S^{n}\vee\mathbb S^{m}$, so we should look a the pushout span $\mathbb S^{n-1}\leftarrow \mathbb S^{n-1}\times\mathbb S^{m-1}\to \mathbb S^{m-1}$ but then the attaching maps of the cells are not trivial like they were for products of sphere. This looks very closely like the Whitehead product but I guess there is some non trivial things to consider at this point and I feel like I can't find the good way to tackle the problem. I would like to know if some work about this had already be done (in a synthetic "up-to-homotopy" way), I'd be very glad that you have some reference about this problem. Thank you very much REPLY [7 votes]: The attaching map of the product of cells is sometimes described as an exterior join construction. Let $F:D^n\to Cf\subseteq X$ be the characteristic map of an $n$-cell of $X$ with attaching map $f:S^{n-1}\to X_{n-1}$, and let $G:D^m\to Cg\subseteq Y$ be the characteristic map of an $m$-cell of $Y$ with attaching map $g:S^{m-1}\to Y_{m-1}$. To describe the attaching map of the product of these cells, note that there is a homeomorphism $$ S^{n+m-1} \approx D^n\times S^{m-1} \cup S^{n-1}\times D^m \subseteq D^n\times D^m. $$ Under this homeomorphism, the attaching map of the product cell is $$ F\times g \cup f\times G: D^n\times S^{m-1} \cup S^{n-1}\times D^m\to Cf\times Y_{m-1} \cup X_{n-1}\times Cg, $$ the cofibre of which is $Cf\times Cg$. The exterior join construction is perhaps not very well documented in the literature, but here are some references: Baues, Hans Joachim, Iterierte Join-Konstruktionen, Math. Z. 131, 77-84 (1973). ZBL0244.55016. Marcum, Howard J., Fibrations over double mapping cylinders, Ill. J. Math. 24, 344-358 (1980). ZBL0459.55009. Stanley, Donald, On the Lusternik-Schnirelmann category of maps, Can. J. Math. 54, No. 3, 608-633 (2002). ZBL1002.55002.<|endoftext|> TITLE: Is there a function from a Suslin tree to itself which send compatible elements to incompatible elements? QUESTION [8 upvotes]: We say $S$ is a Suslin forest if adding a minimum to $S$ we have a Suslin tree. So a Suslin Forest is essentially a Suslin tree $S$ in which we drop the requirement for $S$ to have a single root. Notice that every uncountable subset of a Suslin tree is a Suslin forest. Given a Suslin forest $S$, I would like to know wether it is possible or impossible to define a function $f:S\to S$ such that for every $x,y\in S$ with $x TITLE: $(2x^2+1)(2y^2+1)=4z^2+1$ has no positive integer solutions? QUESTION [7 upvotes]: Equation $$(2x^2+1)(2y^2+1)=4z^2+1$$ has no solutions in the positive integers. Its true? REPLY [13 votes]: By popular demand, I turn my comment to an answer: There are no solutions according to Theorem 6 in Kashihara: Explicit complete solution in integers of a class of equations $(ax^2−b)(ay^2−b)=z^2−c$, Manuscripta Math. 80 (1993), 373-392.<|endoftext|> TITLE: Integral structures via lattices QUESTION [5 upvotes]: I am looking at the paper "p-adic Groups" by Bruhat (in the Boulder Proceedings, 1965). I have a question about one of the statements. Let $k$ be the quotient field of a complete discrete valuation ring $\mathcal{O}$. Let $V$ be a vector space over $k$. Let $L$ be a lattice in $V$. Choose a basis for $L$. Bruhat states the following in pp. 63-64: (1) The algebra $\mathcal{O}[GL]:=\mathcal{O}[g_{ij},(det(g_{ij}))^{-1}]$ is an $\mathcal{O}$-structure for $GL(V)$. (2) More generally, if $G$ is a linear algebraic group over $k$, then given a faithful rational representation $\rho:G \rightarrow GL(V)$, the image of $\mathcal{O}[GL(V)]$ in $k[G]$ is an $\mathcal{O}$-structure for $G$. (3) Any $\mathcal{O}$-structure for $G$ may be obtained in this way. How is (3) proven? REPLY [3 votes]: Here is a proof of (3). Hopefully there are no gaps. I will write $K$ instead of $k$, which I usually reserve for the residue field. Just to set the terminology: $G$ is an algebraic group over $K$. An $\mathcal{O}$-structure for $G$ is a finitely generated Hopf $\mathcal{O}$-subalgebra $A$ of $K[G]$ such that $K\cdot A=K[G]$. (This is the same as saying that $\mathcal{G}:=\mathrm{Spec}\, A$ is a flat group $\mathcal{O}$-scheme of finite type with generic fiber $G$.) If $\rho:G\to\mathrm{GL}(V)$ is a faithful representation and $L\subseteq V$ is an $\mathcal{O}$-lattice, then the $\mathcal{O}$-structure of $G$ induced by $\rho$ is the $\rho^*(\mathcal{O}[\mathrm{GL}(L)])$, where $\rho^*:K[\mathrm{GL}(V)]\to K[G]$ is the $K$-algebra homomorphism adjoint to $\rho$. Also, recall that giving a representation $\rho:G\to \mathrm{GL}(V)$ is the same as endowing $V$ with a (left) $K[G]$-comodule structure $\Delta_\rho:V\to K[G]\otimes V$. And now to the proof. We are given an $\mathcal{O}$-structure $A\subseteq K[G]$ for $G$ and we wish to show that it is induced by some faithful representation $\rho:G\to \mathrm{GL}(V)$ and some $\mathcal{O}$-lattice $L\subseteq V$. Claim. There exists a finitely generated $\mathcal{O}$-submodule $L$ of $A$ such that $L$ generates $A$ as an algebra and $\Delta(L)\subseteq A\otimes L$ ($\Delta$ is the comultiplication). Sketch of proof. The analogous statement for fields is well-known, e.g. see Section 3.3 in Waterhouse's "Introduction to Affine Group Schemes". If $A$ is projective over $\mathcal{O}$, then $A$ has an $\mathcal{O}$-basis (because $\mathcal{O}$ is local) and the argument in Waterhouse applies without change. I think that $A$ must always be projective over $\mathcal{O}$, but in case it is not, one can use Corollary 1.5 in this paper by Thomason. $\square$ Now, since $L$ is finitely generated torsion-free and since $\mathcal{O}$ is a valutation ring, $L$ is free. Write $V=L\cdot K$. Then $\Delta (V)\subseteq K[G]\otimes V$ and thus $V$ is naturally a $K[G]$-comdule, corresponding to a representation $\rho :G\to \mathrm{GL}(V)$. This representation is faithful because $V$ generates $K[G]$. I claim that $\rho$ and $L$ induce the $\mathcal{O}$-structure $A$. Note first that since $\Delta(L)\subseteq A\otimes L$, the morphism $\rho$ extends to a representation of the group $\mathcal{O}$-scheme $\mathrm{Spec}\, A$ into $\mathrm{GL}(L)$, hence $\rho^*(\mathcal{O}[\mathrm{GL}(L)])\subseteq A$. To see the converse, fix an $\mathcal{O}$-basis $\{v_1,\dots,v_t\}$ to $L$ and write $$ \Delta v_i=\sum_j a_{ij} \otimes v_j $$ with $a_{ij}\in A$; the $a_{ij}$ are uniquely determined by the $v_i$. The Hopf algebra axioms imply that $v_i=\sum_j a_{ij}\varepsilon(v_j)$ (where $\varepsilon:A\to\mathcal{O}$ is the counit), and so $A$ is generated by the $a_{ij}$. We use the basis $\{v_i\}$ to identify $\mathcal{O}[\mathrm{GL}(L)]$ with $\mathcal{O}[x_{11},x_{12},\dots,x_{tt},y]/(\det(x_{ij})y-1)$ and $K[\mathrm{GL}(V)]$ with $K[x_{11},x_{12},\dots,x_{tt},y]/(\det(x_{ij})y-1)$. The tautological representation of $\mathrm{GL}(V)$, denoted $\tau$, corresponds to the $K[\mathrm{GL}(V)]$-comodule structure on $V$ determined by $$\Delta_\tau(v_i)=\sum_j x_{ij}\otimes v_j.$$ Now, the fact that $\rho=\tau\circ\rho$ implies that $\Delta_\rho=(\rho^*\otimes \mathrm{id}_V)\circ \Delta_\tau$. This, together with the previous equations, imply that $\rho^*(x_{ij})=a_{ij}$. Since the $a_{ij}$ generate $A$, we see that $ A\subseteq \rho^*(\mathcal{O}[\mathrm{GL}(L)]) $. We conclude that $$ A=\rho^*(\mathcal{O}[\mathrm{GL}(L)]). $$<|endoftext|> TITLE: Is $\Bbb S^2 \times \Bbb S^4$ symplectic? QUESTION [10 upvotes]: I'm playing around with products $M = \Bbb S^{n_1} \times \Bbb S^{n_2}$, and a quick computation using the Künneth formula tells us that if $(n_1,n_2)$ is not $(1,1)$ or $(2,4)$, $M$ is not symplectic (WLOG $1 \leq n_1 \leq n_2$, of course). The $(1,1)$ case is obviously symplectic, but I couldn't decide about the $(2,4)$ case. So: Is $\Bbb S^2 \times \Bbb S^4$ symplectic? Edit: after some time I came back to those calculations. I had missed the obvious case $(n_1,n_2) = (2,2)$. The proof given in the answers can be adapted to show that products of the form $\Bbb S^2 \times \Bbb S^{n_2}$ for even $n_2>2$ are not symplectic. The conclusion of what happened here is the Theorem: Let $1 \leq n_1 \leq n_2$ be natural numbers. Then $\Bbb S^{n_1}\times \Bbb S^{n_2}$ is symplectic if and only if $n_1=n_2=1$ or $n_1=n_2=2$. REPLY [27 votes]: There is no symplectic form on $\mathbb{S}^2\times\mathbb{S}^4$. More generally, there is no symplectic form on $M\times \mathbb{S}^{2n}$, if $n>1$ and $M$ is compact, see Symplectic structures on $M\times \mathbb{S}^{2n}$ . REPLY [27 votes]: No. Note that $H^2(S^2\times S^4,\mathbb R)$ is one dimensional, spanned by $\pi^*\alpha$, where $\pi:S^2\times S^4\to S^2$ is the projection, and $\alpha$ is a volume form on $S^2$. Suppose $\omega$ is a symplectic form on $S^2\times S^4$. Then $[\omega]=c[\pi^*\alpha]$ for some $c\in\mathbb R^\times$. Then $[\omega^3]=c^3[\pi^*\alpha^3]=0$, contradicting the requirement that $\omega^3$ is everywhere nondegenerate.<|endoftext|> TITLE: Weyl's Branching Rule for $SU(N)$-Setting QUESTION [5 upvotes]: On the Wikipedia page for restricted representations https://en.wikipedia.org/wiki/Restricted_representation there is presented a number of explicit "branching rules". In particular, there is the Weyl's branching rule from U(N) to U(N-1) given in terms of signatures $f_1 \geq \cdots \geq f_N$, for $f_i \in \mathbb{N}$, labelling irreps of U(N). I would guess that this generalises directly to the case of branching from $SU(N)$ to $SU(N-1)$ but cannot find a reference. Can someone suggest a reference? REPLY [3 votes]: Every irrep of $SU(n)$ extends to irreps of $U(n)$, and conversely, the restriction of any irrep of $U(n)$ to $SU(n)$ remains irreducible. If your dominant weight of $SU(n)$ is $(a_1,\ldots,a_{n-1})$ then extend it to $(\sum_{i=1}^{n-1} a_i, \sum_{i=2}^{n-1} a_i, \ldots, a_{n-1}, 0)$, apply the $U(n)$ restriction, take differences $f_i-f_{i+1}$ of the resulting signatures.<|endoftext|> TITLE: Examples of "miraculous" proofs QUESTION [39 upvotes]: Concerning the proof that $\zeta(3)$ is irrational, Van der Poorten famously noted that "Apéry's incredible proof appears to be a mixture of miracles and mysteries". Indeed, many ideas introduced in Apéry's proof such as $\zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {\binom {2k}{k}k^{3}}}$ and the recurrence $n^3u_n + (n-1)^3 u_{n-2} = (34n^3-51n^2+27n-5)u_{n-1}, n\geq2$ amazed contemporary mathematicians (although the fast-converging series was already derived several years earlier by Hjortnaes). What are other examples of "miraculous" proofs, whose ingredients amazed mathematicians of their time? In particular, proofs such that: a large portion of the Theorems involved in the proof represent entirely new ideas, these Theorems are applicable to a wide area of mathematics, the general atmosphere among contemporary mathematicians was a mixture of surprise and awe ("where did this come from?"). REPLY [4 votes]: I am surprised that nobody has mentioned the Riemann hypothesis and its consequences. I can never hide my fascination that somebody could manage to prove that the distribution of prime numbers and many other problems in number theory, all boil down to proving that the roots of a seemingly nice function in the complex plane have equal real parts. I wish I could some day travel back in time and ask him really, what kind of sorcery is this?<|endoftext|> TITLE: Finite group representation as $\mathrm{Aut}(\Gamma)$ action $H^1(\Gamma,\mathbb{Z})$ of graph? QUESTION [9 upvotes]: Let $\Gamma$ be a finite graph, then $H^1(\Gamma,\mathbb{Z})\cong \mathbb{Z}^{g(\Gamma)}$ can be viewed as a $\mathrm{Aut}(\Gamma)$ module. Conversely, given a finite group $G$, and a $G$-module $\mathbb{Z}^n$, does there always exist a finite graph $\Gamma$ such that the $G$ module $\mathbb{Z}^n$ arises as $G\overset{f}{\to}\mathrm{Aut}(\Gamma)\curvearrowright H^1(\Gamma,\mathbb{Z})$ for some $f\colon G\to \mathrm{Aut}(\Gamma)$? REPLY [10 votes]: No. An action of a group $G$ on a graph $\Gamma$ induces a homomorphism $G\to \mathrm{Out}(\pi_1(\Gamma))=\mathrm{Out}(F_n)$. So a representation $G\to \mathrm{GL}_n({\mathbb Z})$ can come from an action on a graph only if it lifts to a homomorphism $G\to \mathrm{Out}(F_n)$ over the quotient homomorphism $\mathrm{Out}(F_n)\to \mathrm{GL}_n({\mathbb Z})$. This paper of Zimmermann gives an example of a finite cyclic subgroup of $\mathrm{GL}_n({\mathbb Z})$ that does not lift.<|endoftext|> TITLE: A geometric proof of Krull's Principal ideal theorem QUESTION [8 upvotes]: Krull's height theorem states that in a Noetherian, local ring $(A,\mathfrak m)$, for any $f \in \mathfrak m$, the minimal prime ideal containing $(f)$ is at most height $1$. This is a very geometric statement and is essentially saying that hypersurfaces can cut down the dimension by at most one. However, I have never seen a geometrically motivated proof. All the proofs I have seen essentially muck around with symbolic powers of prime ideals and this seems very ad-hoc/unmotivated to me. Surely a geometrical statement should have a geometric proof! Does someone have a geometric way of seeing why this theorem should be true or what is going on? Or what is going on geometrically with the standard proof and symbolic powers? REPLY [7 votes]: Anon's answer gives a beautiful geometric proof when $A$ is a variety. Below I am trying to give some geometric interpretation of the usual algebraic proof. First a disclaimer: I'm not an algebraist, so the explanation below will be a learner's perspective, probably from an analytic perspective, and thus may seem idiosyncratic to experts. I am going to start by interpreting two key ingredients used in the proof. 1. Symbolic power Let $\frak p$ be a prime ideal of $A$. I think of the localization $A_{\frak p}$ as capturing the behavior of functions on a neighborhood of the generic point of $V({\frak p})$. To see what I mean, take for example $A=k[x,y]/(xy,y^2)$ and ${\frak p}=(y)$. Geometrically $Spec A$ is the $x$-axis plus some fuzz of order 2 at the origin, and $V({\frak p})$ is just the $x$-axis. Now look at $y\in A$. We have $y$ is nonzero in $A$ but becomes zero in $A_{\frak p}$ (where $y=xy/x=0$.) The geometric explanation is that $y$ is indeed zero on a neighborhood of $(x_0,0)$ for any $x_0\neq0$, because there is no fuzz around that point. The only reason for $y\neq0$ in $A$ is that it vanishes only to order 1 near the origin, which is captured by the fuzz (of order 2) there. But this happens at a single point in $V({\frak p})$ (a so called embedded prime), so that behavior is not generic (in the colloquial sense of the word), so we can still say that $y$ vanishes on a neighborhood of the generic point of $V({\frak p})$, which explains why it vanishes in $A_{\frak p}$. Generalizing this example a bit, if we take $A=k[x,y]/(xy^n,y^{n+1})$, we see that $Spec A$ is the $x$-axis with multiplicity $n$ (in other words, with fuzz of order $n$ in the $y$ direction), plus some fuzz of order $n+1$ in the $y$ direction at the origin. Again let ${\frak p}=(y)$. Then ${\frak p}^{(m)}$ (the symbolic power) consists of functions that vanish to order $m$ at the generic point of $V({\frak p})$. Thus for $m TITLE: On a certain integral representation for Hurwitz zeta functions QUESTION [5 upvotes]: A recent question On a certain integral representation for Dirichlet L-functions referenced an integral representation of $\zeta(s)$ due to Jensen that was new to me: $$ (s-1)\zeta(s)=\frac{\pi}{2(s-1)}\int_{-\infty}^\infty\frac{(1/2+it)^{1-s}}{\cosh^2(\pi t)}\, dt. $$ Looking for analogs for the Hurwitz zeta function, I found some interesting integral representations due to Hermite in Whittaker and Watson, based on Plana's Theorem. (Also found in Wikipedia under the Abel-Plana formula). Exercises there include still more representations of $\zeta(s)$ due to Jensen, published in L'Intermediaire des Mathematicians, 1895. In that publication I found the result above; towards the proof Jensen only says "proven with the aid of Cauchy's Theorem." Internet searching eventually led me to (23) on p. 92 of Series Associated with Zeta and Related Functions, by Srivastava and Choi, Kluwer, 2001: for $a>1/2$, $$ \zeta(s,a)=\frac{\pi}{2(s-1)}\int_{-\infty}^\infty\frac{(a-1/2+it)^{1-s}}{\cosh^2(\pi t)}\, dt. $$ No proof is given. For $a=1$ this reduces to Jensen. Can anyone supply details of the proof? A side note, the journal L'Intermediaire des Mathematicians was new to me. It seems to be a 19th century print analog of MathOverflow, consisting of questions posed by researchers, and answers to questions from previous issues. Jensen was writing in reference to a question of Cesaro. REPLY [7 votes]: All these are special cases of the Abel--Plana formula: under simple regularity and growth condition which are easily given, for any $a$ with $\Re(a)>-1/2$ we have $$\sum_{n\ge 0}f'(n+a)=-\dfrac{\pi}{2}\int_{-\infty}^\infty \dfrac{f(a-1/2+it)}{\cosh^2(\pi t)}\,dt$$ plus some correction terms due to the possible poles of $f'$ in the right half-plane, if any.<|endoftext|> TITLE: Can a big set always look small? QUESTION [10 upvotes]: For a set $C\subset \mathbb R^2$, define its visibility from a point $x$ as $vis_C(x)=\{\varphi\in \mathbb S^1\mid \exists t>0~~x+t*\varphi\in C\}$, where $\mathbb S^1$ denotes the unit circle. Say that $C$ looks $m$-big from $x$ if $\mu(vis_C(x))/2\pi=m$, where $\mu$ is the (outer) Lebesgue measure on $\mathbb S^1$. If $C$ is at most $0.99$-big from every $x$, then is $\mu(C)=0$? Here again $\mu$ can be the Lebesgue measure, but I'm interested in any similar results as well. My motivation is to solve this problem. The question also seems somewhat related to this one. Note that the converse is false, as if $C$ is a line, then it looks $1/2$-big from every $x\in \mathbb R^2\setminus C$, and if we take the union of countably many horizontal lines, we can construct a set that looks $1$-big. REPLY [4 votes]: Set $C$ cannot have positive measure. Indeed, if $C$ had positive measure, then by the Lebesgue density theorem there would be a ball $B(x,r)$ such that more than 99% of $B(x,r)$ is in $C$. However, $\mu\bigl(B(x,r)\cap C\bigr)\leq vis_C(x) \cdot \mu\bigl(B(x,r)\bigr)$ by Fubini's theorem. So, $vis_C(x)> 0.99$.<|endoftext|> TITLE: Moments of area of random triangle inscribed in a circle QUESTION [27 upvotes]: The $2m$th moment of the (random) area of the triangle whose vertices are three independent, uniformly distributed random points on the unit circle appears to be $((3m)!/(m!)^3)/16^m$. Can anyone prove this? Better yet, can anyone give a conceptual explanation for why this moment should be rational? (If this observation is not new, references would be appreciated.) This question was inspired by John Baez's posts https://johncarlosbaez.wordpress.com/2018/07/10/random-points-on-a-sphere-part-1/ and https://johncarlosbaez.wordpress.com/2018/07/12/random-points-on-a-sphere-part-2/. REPLY [9 votes]: Let $A_d$ be the area of a triangle whose vertices are chosen uniformly at random from a unit sphere in $\mathbb{R}^d$. I claim that for $d\ge 2$ and $m\ge 1$: $$ E(A_d^{2m})=\frac{3}{4^m} \prod _{q=1}^{m-1} \frac{3 d+6 m-2 q-6}{d+2 m-2 q-2}\prod _{q=1}^m \frac{d+2 m-2 q-1}{d+2 m-2 q}\\ = \frac{3\ \Gamma \left(\frac{d}{2}\right)^2 \Gamma \left(\frac{d-1}{2}+m\right) \Gamma \left(\frac{3 d}{2}+3 m-3\right)}{4^m\ \Gamma \left(\frac{d-1}{2}\right) \Gamma \left(\frac{d}{2}+m-1\right) \Gamma \left(\frac{d}{2}+m\right) \Gamma \left(\frac{3 d}{2}+2 m-2\right)} $$ For $d=2$ (and any $m\ge 1$) this simplifies to the established formula: $$E(A_2^{2m}) = \frac{(3m)!}{16^m\ (m!)^3}$$ For $m=1$ (and any $d\ge 2$) it simplifies to: $$ E(A_d^2)=\frac{3(d-1)}{4d} $$ To prove the general formula, first note that the squared area of a triangle can be described in terms of a Grammian determinant: $$ A_d^2 = \frac{1}{4} \det{\left(s_i \cdot s_j\right)} $$ where the triangle has vertices $v_0, v_1, v_2$ and: $$ s_i = v_i - v_0, \: i=1,2 $$ For $d\ge3$, we can always rotate a triangle with vertices on the sphere into the configuration: $$\begin{array}{rcl} v_0 & = & e_0 \\ v_1 & = & \cos(\theta_1)\, e_0 + \sin(\theta_1)\, e_1 \\ v_2 & = & \cos(\theta_2)\, e_0 + \sin(\theta_2)\cos(\phi_2)\, e_1 + \sin(\theta_2)\sin(\phi_2)\, e_2 \end{array}$$ The expectation values of the even moments can then be expressed as an integral over three coordinates of a suitably weighted version of the squared area raised to a power: $$ E(A_d^{2m}) = \frac{(d-2)\ \Gamma \left(\frac{d}{2}\right)}{2 \pi ^{3/2}\ \Gamma \left(\frac{d-1}{2}\right)} \int_{0}^\pi \int_{0}^\pi \int_{0}^\pi (A_d^2)^m \sin(\theta_1)^{d-2} \sin(\theta_2)^{d-2} \sin(\phi_2)^{d-3} \,d\theta_1\,d\theta_2\,d\phi_2 $$ where the vertex $v_1$ is a representative of a $(d-2)$-sphere of radius $\sin(\theta_1)$ over which it can be rotated while keeping $v_0$ fixed, and $v_2$ is a representative of a $(d-3)$-sphere of radius $\sin(\theta_2)\sin(\phi_2)$ over which it can be rotated while keeping $v_0, v_1$ fixed, and the weights incorporate the measures of these spheres with respect to the whole $(d-1)$-sphere. For $m=1$, the integrand expands as a sum of products of non-negative integer powers of sines, and the integral can be carried out explicitly to obtain: $$ E(A_d^2)=\frac{3(d-1)}{4d} $$ For $m\ge 2$, we can integrate by parts to obtain the recursion relation: $$ E(A_d^{2(m+1)}) = \frac{(d-1) (3 d+4 m)}{4 d^2} E(A_{d+2}^{2m}) $$ The general formula then follows by induction. Although we derived this formula for even moments, it also gives correct values for odd moments using half-integer values for $m$, including the average area if we set $m=1/2$: $$ E(A_d) = \frac{3\ \Gamma \left(\frac{3 (d-1)}{2}\right) \Gamma \left(\frac{d}{2}\right)^3}{2\ \Gamma \left(\frac{d-1}{2}\right)^2 \Gamma \left(\frac{d+1}{2}\right) \Gamma \left(\frac{3 d}{2}-1\right)} $$ For example: $$\begin{array}{rcl} E(A_2) & = & \frac{3}{2\pi} \\ E(A_3) & = & \frac{\pi}{5} \end{array}$$<|endoftext|> TITLE: Forcing and Family Contentions: Who wins the disputes? QUESTION [24 upvotes]: The famous game-theoretic couple, Alice & Bob, live in the set-theoretic universe, $V$, a model of $ZFC$. Just like many other couples they sometimes argue over a statement, $\sigma$, expressible in the language of set theory. (One may think of $\sigma$ as a family condition/decision in the real life, say having kids or living in a certain city, etc.) Alice wants $\sigma$ to be true in the world that they live but Bob doesn't. In such cases, each of them tries to manipulate the sequence of the events in such a way that makes their desired condition true in the ultimate situation. Consequently, a game of forcing iteration emerges between them as follows: Alice starts by forcing over $V$, leading the family to the possible world $V[G]$. Then Bob forces over $V[G]$ leading both to another possible world in which Alice responds by forcing over it and so on. Formally, during their turn, Alice and Bob are choosing the even and odd-indexed names for forcing notions, $\dot{\mathbb{Q}}_{0}$, $\dot{\mathbb{Q}}_{1}$, $\dot{\mathbb{Q}}_{2}$, $\cdots$, in a forcing iteration of length $\omega$, $\mathbb{P}=\langle\langle\mathbb{P}_{\alpha}: \alpha\leq\omega\rangle, \langle\dot{\mathbb{Q}}_{\alpha}: \alpha<\omega\rangle\rangle$, where the ultimate $\mathbb{P}$ is made of the direct/inverse limit of its predecessors (depending on the version of the game). Alice wins if $\sigma$ holds in $V^{\mathbb{P}}$, the ultimate future. Otherwise, Bob is the winner. Question 1. Is there any characterization of the statements $\sigma$ for which Alice has a winning strategy in (the direct/inverse limit version of) the described game? How much does it depend on the starting model $V$? Clearly, Alice has a winning strategy if $\sigma$ is a consequence of $ZFC$, a rule of nature which Bob can't change no matter how tirelessly he tries and what the initial world, $V$, is! However, if we think in terms of buttons and switches in Hamkins' forcing multiverse, the category of the statements for which Alice has a winning strategy seems much larger than merely the consequences of $ZFC$. I am also curious to know how big the difference between the direct and inverse limit versions of the described game is: Question 2. What are examples of the statements like $\sigma$, for which Alice and Bob have winning strategies in the direct and inverse limit versions of the described game respectively? Update. Following Mohammad's comment, it seems a variant of this game in which Alice and Bob are restricted to choose certain types of forcing notions (e.g. c.c.c. or proper) might be of interest as well. So the following version of the question 1 arises: Question 3. Is there any characterization of the statements $\sigma$ for which Alice has a winning strategy in the described game restricted to forcing notions from the class $\Gamma$ where $\Gamma$ is the class of all c.c.c./proper/... forcings? REPLY [20 votes]: I like this question a lot. It provides an interesting way of talking about some of the ideas connected with the maximality principle and the modal logic of forcing. Let me make several observations. First, Alice can clearly win, in one move, with any forceably necessary statement $\sigma$, which is a statement for which $\newcommand\possible{\Diamond}\newcommand\necessary{\Box}\possible\necessary\sigma$ holds in the modal logic of forcing, meaning that one can force so as to make $\sigma$ remain true in all further forcing extensions. She should simply force to make $\necessary\sigma$ true, and then Bob cannot prevent $\sigma$ in any further extension, including the limit model. Under the maximality principle, all such statements are already true. Let me point out that there are some subtle issues about formalization in the question. For example, the strategies here would be proper class sized objects, and so one must stipulate whether one is working in ZFC with only definable classes or whether one has GBC or KM or whatever and whether global choice holds. Another difficulty concerns the determinacy of the game, since even open determinacy for class games is not a theorem of GBC. Meanwhile, here is something positive to say. I shall consider only the direct-limit version of the game. Theorem. Suppose there is a $\omega$-closed unbounded class of cardinals $\kappa$ such that the statement $\sigma$ is forced by the collapse forcing of $\kappa$ to $\omega$. Then Alice has a winning strategy in the $\sigma$ game. Proof. Let $C$ be the class $\omega$-club of such $\kappa$, and let Alice simply play always to collapse the next element of $C$ above the size of the previous forcing played by Bob. It follows that the limit forcing $\mathbb{P}$ will collapse all the cardinals up to an element $\kappa\in C$, and since $\kappa$ will have cofinality $\omega$, it will also collapse $\kappa$ itself. Since the forcing will also have size $\kappa$, in the direct limit case, it follows that the forcing is isomorphic to $\text{Coll}(\omega,\kappa)$, and so $\sigma$ holds in the model $V[G]$, so Alice has won. $\Box$ For example, if the GCH holds, then CH will be such a statement $\sigma$, even though this is a switch, because the collapse forcing will collapse $\kappa$ and the CH will hold in $V[G]$, as a residue of the GCH in $V$. Thus, the GCH implies that Alice can win the CH game. A dual analysis is: Theorem. If the class of cardinals $\kappa$ of countable cofinality for which the collapse forcing $\text{Coll}(\omega,\kappa)$ forces $\sigma$ is stationary, then Alice can defeat any strategy of Bob in the $\sigma$ game. Proof. For any strategy for Bob, there is a club of cardinals $\theta$ such that $V_\theta$ is closed under the strategy, in the sense that if Alice plays a poset in $V_\theta$ then Bob's strategy will reply with a strategy in $V_\theta$. So by the stationary assumption of the theorem, there is a $\kappa$ of countable cofinality that is closed under the strategy. Alice can now play so as to collapse more and more of $\kappa$, and Bob will always reply with a poset below $\kappa$. So the limit forcing will again be the collapse of $\kappa$, which forces $\sigma$. So Alice can defeat this strategy. $\Box$ For example, if the GCH fails on a $\omega$-closed unbounded class of cardinals (this contradicts SCH), then Alice can win with $\neg$CH. And if it fails on a stationary class of such cardinals with countable cofinality, then Bob cannot win the $\neg$CH game. Theorem. From suitable consistency assumptions, it is consistent with GBC that the CH game is not determined with respect to class strategies. Proof. Using the Foreman-Woodin theorem that it is relatively consistent that GCH fails everywhere, we can perform additional forcing by first adding a generic class of cardinals, and then forcing certain instances of GCH by collapsing cardinals. The result will be a model of GBC where the class of cardinals $\kappa$ of cofinality $\omega$ at which the GCH holds is both stationary and co-stationary. By the theorem above, considered from either Alice's or Bob's perspective, either player can defeat any strategy of the other player. So the game is not determined. $\Box$ I guess the argument isn't just about CH, but rather any statement $\sigma$ such that there is a stationary/co-stationary class of $\kappa$ of countable cofinality such that $\sigma$ holds after the collapse of $\kappa$. In such a case, neither player can have a winning strategy. The ideas appear to culminate in answer to question 1. Theorem. The following are equivalent for any statement $\sigma$. Alice has a winning strategy in the $\sigma$ game. The class of cardinals $\kappa$ of countable cofinality, such that the collapse forcing of $\kappa$ to $\omega$ forces $\sigma$, contains a class $\omega$-club. Proof. Statement 2 implies statement 1 by the first theorem above. Conversely, if statement 2 fails, then there is a stationary class of such $\kappa$ where $\sigma$ fails in the collapse extension. In this case, Bob can defeat any strategy for Alice, by Bob's analogue of the second theorem. $\Box$<|endoftext|> TITLE: Does Vizing's conjecture hold for the infinite graphs? QUESTION [8 upvotes]: In finite graph theory, there are many (in)equalities which relate the integer value of a certain graph invariant (e.g. domination or chromatic number) for the product of two finite graphs (e.g. tensor, Cartesian, strong product, etc.) to the value of the same invariant for each component. Such (in)equalities are often formulated as $i(G*H)\sim f(i(G), i(H))$ where $i$ is the operator assigning a certain integer-valued invariant to the graph, $*$ is a particular type of graph product, $\sim$ is the (in)equality sign, and $f$ is a function from $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$. For instance, see Vizing's conjecture. However, such (in)equalities seem to be quite vulnerable to the finiteness condition and may fail so badly if one brings infinite graphs and transfinite-valued graph invariants into the play as a direct generalization of the original statement. A typical example is Hajnal's proof (cf. MR0815579) of the failure of Hedetniemi's conjecture for infinite graphs. (See also Assaf Rinot's result along these lines.) Here in this post, I look for more results of this type, namely the instances of the failure of the (open/proved) (in)equalities of the described form in the infinite case. Of course, some of these conjectures might be wide open in their finite case, yet have some relatively easy counterexamples in the infinite form. Question 1. What are examples of finite graph (in)equalities of the form $i(G*H)\sim f(i(G), i(H))$ which (may) fail for the infinite graphs (up to consistency)? I am particularly interested in the case of Vizing's conjecture ($VC$): Question 2. Can $VC$ fail for the infinite graphs? Another line of thought might be considering a case where a statement fails in the finite case but holds for infinite graphs. Particularly, this might be the case for "Ramsey-type" graph invariants which require the underlying structures to be very large to behave nicely. Question 3. What are examples of graph (in)equalities of the form $i(G*H)\sim f(i(G), i(H))$ which fail for the finite graphs but hold for certain class of infinite graphs? REPLY [3 votes]: Sorry for the confusion, I am being slow today. Letting $\gamma(G)$ be the domination number of $G$, we have the trivial inequality $$\gamma(G\times H) \geq {\rm max}(\gamma(G),\gamma(H)).$$ This is because a dominating set for $G\times H$ will project onto a dominating set in either factor. So if either domination number is infinite then ${\rm max}(\gamma(G), \gamma(H)) = \gamma(G)\gamma(H)$ and Vizing's conjecture holds. If both domination numbers are finite, but $G$ or $H$ is infinite, there could be a real question here. Domination number can strange for infinite graphs. E.g., I can give an infinite graph with domination number $k$ which is the union of an increasing sequence of finite graphs each of which has domination number $1$.<|endoftext|> TITLE: What is the minimal dimension of a complex realising a group representation? QUESTION [26 upvotes]: This question is inspired by this one, which was about representations that can be realised homologically by an action on a graph (i.e., a 1-dimensional complex). Many interesting integral representations of groups arise from a group acting on a simplicial complex that is homotopy equivalent to a wedge of spheres, by applying homology. A classical example is the action of groups of Lie type on spherical buildings. On homology this gives an integral form of the Steinberg representation. One may ask if there exists a complex of lower dimension than the Tits building that realises the (integral) Steinberg representation in this way. I expect that the answer is No, but how to prove it? More generally, given an integral $G$-representation that can be realised as the homology of a spherical complex with an action of $G$, is there an effective lower bound on the dimension of such a complex? One obvious lower bound is given by the minimal length of a resolution by permutation representations. Is this something that has been studied? REPLY [3 votes]: This does not answer Greg's question, but it is related. You can realize any $\mathbb{Z}G$-module you that like as $H_1$ of a based 2-complex, or as $H_2$ of a 3-complex if you insist that the complex should be simply-connected. Furthermore, you can require $G$ to act freely on the complex except for fixing the base point. Given a $\mathbb{Z}G$-module $M$, take a presentation for $M$, i.e. an exact sequence $F_1\rightarrow F_0\rightarrow M\rightarrow 0$ in which each $F_i$ is a free $\mathbb{Z}G$-module. Now you can realize $F_0$ as the 2nd homology of a wedge of 2-spheres with $G$ acting freely except on the basepoint. You can attach a disjoint union of 3-balls permuted freely by $G$ (with $H_0$ isomorphic to $F_1$) in such a way that the cellular chain complex is just $F_1\rightarrow F_0\rightarrow 0\rightarrow \mathbb{Z}$, with the given map from degree 3 to degree 2. $H_3$ of this space is of course the kernel of the map $F_1\rightarrow F_0$, while $H_2$ is isomorphic to $M$. With a wedge of 1-spheres and attached 2-cells everything works the same way (the Hurewicz theorem tells you that $\pi_1$ surjects onto $H_1$), except that the 2-dimensional complex will probably have fundamental group a lot bigger than the abelian group $M$.<|endoftext|> TITLE: Holonomy groups of compact Riemannian symmetric spaces QUESTION [6 upvotes]: Let $M$ be a compact Riemannian symmetric space. By the classification of Cartan, it belongs to the table of homogeneous spaces given in the Wikipedia page: https://en.wikipedia.org/wiki/Symmetric_space In the Berger classifiction of holonomy groups https://en.wikipedia.org/wiki/Holonomy#The_Berger_classification the symmetric case is omitted because the holonomy group can easily be read off the Cartan classification in the symmetric space case. How does this "reading off" work exactly. Can someone point to me a list of the holonomy groups of the compact symmetric spaces? REPLY [6 votes]: At the request of the OP I put my comment as an answer: in general, the holonomy group and the isotropy group have the same identity component (this is a theorem of E. Cartan). So if you assume that $M$ is simply-connected, they are equal. You can see a proof (for instance) in section 10.79 of Arthur Besse's Einstein manifolds.<|endoftext|> TITLE: Existence of non-null-homotopic map from $M^n$ to $S^{n-1}$ QUESTION [14 upvotes]: Let $M^n$ be compact, connected, oriented $n$-dimensional smooth manifold without boundary, the Hopf degree theorem states that the homotopy class of continuous maps from $M^n$ to $S^n$ is classified by its degree. What about continuous map from $M^n$ to $S^{n-1}$ (assume $n>0$) ? I am mostly interested in the case $n>2$. More specially, when does there exist a non null-homotopic map in the case $n=3$? If $H^2(M,\Bbb Z) \not=0$ then it's true as $[M,\Bbb{C}P^{\infty}]=H^2(M,\Bbb Z)$ and $\Bbb CP^1 \cong S^2$. This is not necessary as $M=S^3$ shows. REPLY [12 votes]: For $n=3$, there are good general results about $[X,S^2]$ and $[X,S^3]$ (for arbitrary spaces $X$) in the paper The principal fibration sequence and the second cohomotopy set by Laurence Taylor. For $n\geq 4$ I claim that there is an exact sequence $$ H^{n-2}(M;\mathbb{Z}) \xrightarrow{\text{Sq}^2\circ\rho} H^n(M;\mathbb{Z}/2) \simeq\mathbb{Z}/2 \to [M,S^{n-1}] \to H^{n-1}(M;\mathbb{Z}) \to 0 $$ The first map here is the composite of the reduction map $\rho\colon H^{n-2}(M;\mathbb{Z})\to H^{n-2}(M;\mathbb{Z}/2)$ with the Steenrod operation $\text{Sq}^2\colon H^{n-2}(M;\mathbb{Z}/2)\to H^n(M;\mathbb{Z}/2)$. It could be analysed further using the universal coefficient theorem and the theory of Wu classes, but I will not go into that here. To prove the claim, standard calculations show that for $n\geq 4$ we have $\pi_{n-1}(S^{n-1})=\mathbb{Z}\iota$ and $\pi_n(S^{n-1})=\mathbb{Z/2}\eta$ and $\pi_{n+1}(S^{n-1})=\mathbb{Z}/2\eta^2$. Also, the homotopy groups of the connective real $K$-theory spectrum $kO$ are $(\mathbb{Z}\iota,\mathbb{Z}/2\eta,\mathbb{Z}/2\eta^2,0,\dotsc)$. Now let $B$ be the $(n-1)$'th space in the $\Omega$-spectrum for $kO$, so $[X,B]=kO^{n-1}(X)$ for all spaces $X$. Let $F$ be the fibre of the unit map $S^{n-1}\to B$. From the long exact sequence of the fibration we find that $\pi_i(F)=0$ for $i\leq n+1$. By induction over the cells of $X$, we deduce that the map $[X,S^{n-1}]\to [X,B]$ is bijective whenever $X$ is a CW complex of dimension at most $n$. In particular, we can take $X=M$ to see that $[M,S^{n-1}]=kO^{n-1}(M)$, so we now have a calculation in stable homotopy theory. I will formulate the rest of the argument in terms of the Atiyah-Hirzebruch spectral sequence, but it could be reformulated in terms of obstruction theory if desired. The spectral sequence has the form $$ E_2^{pq} = H^p(M;kO^q) \Longrightarrow kO^{p+q}(M), $$ with differentials $d_r\colon E_r^{pq}\to E^{p+r,q-r+1}$. Here $kO^q$ is the same as $\pi_{-q}kO$, so it is trivial unless $q\leq 0$, so the spectral sequence is concentrated in the fourth quadrant. Also, the cohomology of $M$ is concentrated in degrees zero to $n$, so the spectral sequence is concentrated in columns zero to $n$. The target group $kO^{n-1}(M)$ has a filtration with quotients $E_\infty^{p,n-1-p}$. The only possible nonzero terms are $E_\infty^{n-1,0}$ (which is a subquotient of $E_2^{n-1,0}=H^{n-1}(M;\mathbb{Z})$) and $E_\infty^{n,-1}$ (which is a subquotient of $E_2^{n,-1}=H^n(M;\mathbb{Z}/2\eta)$). We therefore have a short exact sequence $$ E_\infty^{n,-1} \to kO^{n-1}(M) \to E_\infty^{n-1,0} $$ Now recall that $d_r$ has bidegree $(r,1-r)$. It follows that all differentials starting at $(n-1,0)$ or $(n,-1)$ have endpoint in the region $p>n$ where the spectral sequence is zero. Similarly, all differentials ending at $(n-1,0)$ or $(n,-1)$ have starting point in the region $q>0$ where the spectral sequence is zero, with the exception of the differential $$ d_2\colon H^{n-2}(M;\mathbb{Z}) = E_2^{n-2,0} \to E_2^{n,-1} = H^n(M;\mathbb{Z}/2\eta). $$ Thus, $E_\infty^{n,-1}$ is the cokernel of the above differential, whereas $E_\infty^{n-1,0}$ is just the same as $E_2^{n-1,0}=H^{n-1}(M;\mathbb{Z})$. By considering the Posnikov tower of $kO$, one can also check that the nontrivial differential is the composite of the reduction map $\rho\colon H^{n-2}(M;\mathbb{Z})\to H^{n-2}(M;\mathbb{Z}/2)$ with the Steenrod operation $\text{Sq}^2\colon H^{n-2}(M;\mathbb{Z}/2)\to H^n(M;\mathbb{Z}/2)$.<|endoftext|> TITLE: What is the “free symmetric monoidal category” 2-monad? QUESTION [5 upvotes]: I have come across an n-category cafe post where someone describes a monad that generates symmetric monoidal categories. Can someone give details, like what is the base category, what exactly is the endofunctor and natural isomorphisms for “free symmetric monoidal category” 2-monad? Could this generate the category of finite dimensional hilbert spaces and unitary maps? I have been reading this post by Jeffery Morton. He describes the following “free symmetric monoidal category” 2-monad: The bosonic Fock space is then ⊕nC⊗sn, the direct sum of all symmetric tensor products of some number of copies of this space. One way to say this is that the symmetric tensor product of a space V with itself is the equalizer of two maps V⊗V→V⊗V, namely the identity and the swap map. Likewise, V⊗sn is the equalizer of all the permutation automorphisms that appear because V⊗n is automatically a representation of Sn. So the symmetric product is the trivial representation. Since C⊗sn≅C, this is just a sum of a bunch of 1-dimensional spaces, each of which describes an n-particle system, which again has only one state. The only thing to say about this state is that it has n particles in it. Jamie’s original paper explains this by means of a monad on Hilb, which is essentially the “free commutative monoid” monad: the Fock space is the free commutative monoid on C. This fact gives a bunch of special maps, including a bialgebra structure on the Fock space, and the raising and lowering operators can be constructed out of this. The commutation relations are a consequence of that. Now, groupoidifying this is a categorification, so this description has to be weakened. To start with, we take a groupoid describing a system with only one configuration (the “it’s there” state for our particle). This will be the trivial groupoid 1, with one object and only the identity morphism. Then we want to take the “groupoidified Fock space”. > Since groupoids live in a 2-category, the equivalent of the “free commutative monoid” monad turns out to be a bit weaker, namely the “free symmetric monoidal category” 2-monad. We get a “direct sum” (i.e. in Span(Gpd), the disjoint union) of a bunch of objects which show up as certain 2-limits. In particular, we freely generate a bunch of objects like (⊗...⊗), and we must get not EQUATIONS, but ISOMORPHISMS corresponding to all the switch maps. This is essentially where the groupoid of finite sets and bijections come from: think of as the groupoid which contains exactly the 1-element set - the free symmetric monoidal category this generates is the groupoid which contains all finite sets and their bijections. REPLY [6 votes]: There is a 2-monad $P$ on $\mathrm{Cat}$ whose strict algebras are symmetric strict monoidal categories, and whose pseudo-algebras are "unbiased" symmetric monoidal categories. On objects, $PA$ is the category whose objects are finite lists of objects of $A$, and in which a morphism $(a_1,\dots,a_n)\to (b_1,\dots,b_m)$ consists of a bijection $\sigma : \{1,\dots,n\} \to \{1,\dots,m\}$ (so that in particular $n=m$) and morphisms $f_i : a_i \to b_{\sigma i}$ in $A$. The unit $A \to P A$ sends $a$ to the 1-element list $(a)$, and the multiplication $P P A \to P A$ removes parentheses. Is this what you're looking for?<|endoftext|> TITLE: How to visualize a Witt vector? QUESTION [16 upvotes]: As the question title asks for, how do others "visualize" Witt vectors? I just think of them as algebraic creatures. Bonus points for pictures. REPLY [10 votes]: You can view the spectrum of the ring Witt vectors, in the sense of scheme theory, pretty reasonably. If $R$ is $p$-torsion free, then $\mathrm{Spec}(W_n(R))$ is $n+1$ copies (or $n$ if you use the traditional indexing) of $\mathrm{Spec}(R)$ glued together in a suitable way along their fibers over $p$. But there are two qualifications. First, there is some "Frobenius twisting" involved, which is impossible to visualize because the Frobenius morphism is impossible to visualize, as far as I know. Second, each component is not simply glued transversally to the previous ones, but there is some higher order gluing. A simple example is $\mathrm{Spec}(W_n(\mathbf{Z}))$. It consists of $n+1$ copies of $\mathrm{Spec}(\mathbf{Z})$, numbered $0$ to $n$, where the $k$-th copy is glued to the $(k-1)$-st copy modulo $p^k$. So copy $1$ is glued to copy $0$ transversally. Copy $2$ is glued to copy $1$ tangentially, but only to order $1$, and so on. There is a little picture of $W_1$ on page 5 of my paper The basic geometry of Witt vectors, I. The affine case.<|endoftext|> TITLE: Surjectivity of norm map on subspaces of finite fields QUESTION [5 upvotes]: It is basic that the norm map $N:\mathbf{F}_{q^n}^* \to \mathbf{F}_q^*$ is surjective for finite fields. In fact $N(x) = x^{(q^n-1)/(q-1)}$. How well does this simple fact extend to subspaces? A basic example is an intermediate extension $\mathbf{F}_{q^d}$. On $\mathbf{F}_{q^d}^*$ we have $$N(x) = \left(x^{(q^d-1)/(q-1)}\right)^{(q^n-1)/(q^d-1)} = \left(x^{(q^d-1)/(q-1)}\right)^{n/d}$$ since the term in the brackets is in $\mathbf{F}_q^*$ and $(q^n-1)/(q^d-1) \equiv n/d \pmod {q-1}$. So $N$ is surjective on $\mathbf{F}_{q^d}^*$ if and only if $(n/d, q-1) = 1$. In particular $N$ fails to be surjective on a subspace of dimension $n/2$ whenever $n$ is even and $(n/2, q-1) > 1$. As a sort of converse note that if $(n,q-1)=1$ then $N$ is surjective on every one-dimensional subspace. Is it true that if $V \leq \mathbf{F}_{q^n}$ is a $\mathbf{F}_q$-rational subspace of dimension $>n/2$ then $N$ is surjective on $V$? Equivalently, if $\dim_{\mathbf{F}_q} V > n/2$, can we always find $x^{q-1} \in V$? REPLY [5 votes]: The number of elements with norm $a$ is $$\frac{1}{q-1} \sum_{\chi: \mathbb F_q^\times \to \mathbb C^\times} \sum_{x \in V} \chi(Nx) \overline{\chi(a)}$$ The summand vanishes unless $\chi$ has order dividing $n$ so there are at most $gcd(n,q-1)$ terms. One of the terms has size $q^{\dim V}$ so it is sufficient that the other terms have size $ \gcd(n,q-1)-1$. This is automatically satisfied for $\dim V \geq n/2+1$ and is satisfied for $\dim V= (n+1)/2$ as long as $q$ is sufficiently large with respect to $n$. This method is the same as Felipe Voloch's answer, because the eigenvalues in the curve he writes down are Gauss sums, except that we get a savings by dropping the unnecessary multiplicative characters. It is likely possible to improve the dependence on $q$ somewhat by using etale cohomology theorem (the key thing is not to view this as a $1$-dimensional variety over $\mathbb F_{q^n}$ but as an $n$-dimensional variety over $\mathbb F_q$). But this will probably lead to some loss in $n$ so I don't know whether tit will improve the $\dim (n+1)/2$ case.<|endoftext|> TITLE: about morphisms of affine formal schemes $\mathrm{Spf}(B)\to \mathrm{Spf}(A)$ QUESTION [6 upvotes]: It is well known that there is a correspondence between homomorphism of rings $A\to B$ and morphism of affine schemes $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$. Question: (1) In analogy, is there anything similar between homomorphism of adic rings and morphisms of affine formal schemes $\mathrm{Spf}(B)\to \mathrm{Spf}(A)$? (2)What could be good references for this? REPLY [6 votes]: See EGA I, sec. 10. For the specific question you mention, see paragraph 10.2. There is further issues on the cohomology of formal schemes in EGA III, sect 3.4. Also, the existence theorem is treated in the volume "Fundamental Algebraic Geometry: Grothendieck’s FGA explained", Mathematical Surveys and Monographs 123 (2006), A.M.S.<|endoftext|> TITLE: Homology of bar complex vs homology of indecomposables QUESTION [5 upvotes]: $\require{AMScd}$ Background: This question is about the bar and cobar constructions, and their relationship with the indecomposables of a dg-algebra. A brief summary of the bar and cobar constructions on ncatlab can be found at [1]. I will also mention the more general case of $A_\infty$ algebras; this case is discussed in this survey [2] or in Proute's thesis [3]. I posted a related question a few months ago in the mathoverflow post [4]. There I called the dg-module of indecomposables the ``linearized chain complex.'' It seems that the former is more standard terminology so I've switched for the time being. Setup: Let $(A,\partial,\epsilon)$ be an augmented dg-algebra over a field $k$, i.e. a dg-algebra $(A,\partial)$ along with dg-algebra map $\epsilon:A \to k$. Here $k$ is a dga with trivial grading and differential. The module of indecomposables $(IA,\partial)$ of $(A,\partial,\epsilon)$ is a dg-module whose underlying graded $k$ vector-space is $IA = \ker(\epsilon)/\ker(\epsilon)^2$. The differential $\partial$ on $A$ descends to a differential $\partial$ on $IA$. Furthermore, the correspondence $(A,\partial,\epsilon) \mapsto (IA,\partial)$ is functorial. That is, given a map: $$f:(A,\partial,\delta) \to (B,\partial,\epsilon)$$ of augmented dg-algebras, i.e. a dg-algebra map with $\epsilon \circ f = \delta$, we get a map of dg-modules: $$If:(IA,\partial) \to (IB,\partial)$$ The bar complex $BA$ of $(A,\partial,\epsilon)$ to a dg-coalgebra whose underlying graded vector-space is: $$ BA := k \oplus BA^+ \qquad BA^+ := \bigoplus_{i=1}^\infty \ker(\epsilon)[1]^{\otimes i} $$ The differential combines multiplication and differentiation from $(A,\partial)$ (see [1] for an actual definition) and the coaugmentation $\iota:k \to BA$ is just the inclusion. Similarly, the cobar complex $\Omega C$ of an augmented dg-coalgebra $C$ is an augmented dg-algebra admitting an analogous description to that of $BA$, as a certain tensor algebra. Both constructions are functorial, and there are generalizations of to $A_\infty$ algebras and coalgebras (see [2] or [3]). The functors $B$ and $\Omega$ form an adjoint pair ($B$ is right adjoint to $\Omega$) and we have natural adjoint maps $\Omega B A \to A$ and $C \to B \Omega C$, which are quasi-isomorphisms of augmented dg-algebras and coaugmented dg-coalgebras, respectively. There is a reference for this in the post [5]. In particular, we get a natural map of dg-modules between the indecomposables: $$ I\Omega B A \to IA $$ I believe that it is true that $I\Omega C \simeq \text{coker}(\kappa)$ for any coaugmented dg-coalgebra $(C,\partial,\kappa)$ (an actual isomorphism of chain groups, not just a quasi-isomorphism) so that the above map yields a canonical map of dg-modules and a corresponding map on homology: $$ BA^+ \to IA \qquad H(BA^+) \to H(IA) $$ Question: Under what circumstances is the natural map $I\Omega B A \to IA$ an isomorphism on homology? When is it surjective on homology? I'm particularly interested in the case where the underlying graded algebra of $(A,\partial,\epsilon)$ is the free graded commutative algebra over some generators. Any partial answers are much appreciated. Examples: In some cases this map is simple to understand. For instance, when $A$ is an algebra with the trivial differential and grading, then $H(IA) = IA$ and I believe that the map on homology $H(BA^+) \to H(IA)$ is just projection onto $H^1(BA^+) \simeq IA$. Another context where something similar happens is when $A = \Omega C$ where $(C,\partial,\kappa)$ is a coaugmented dg-coalgebra (or even an $A_\infty$ coalgebra). Then the map: $$\text{coker}(\iota) \simeq BA^+ \to IA \simeq \text{coker}(\kappa)$$ is just the $A_\infty$ quasi-isomorphism on the cokernels of the coaugmentations induced by the quasi-isomorphism of $A_\infty$ coalgebras of $B\Omega C \to C$, itself induced by the adjunction map $\Omega B \Omega C \to \Omega C$. Since $A_\infty$ quasi-isomorphisms descend to isomorphisms on homology, we know in particular $H(BA^+) \simeq H(IA)$ under the map of interest. REPLY [4 votes]: The functor of indecomposables is the left adjoint of a Quillen adjunction between dg-algebras and dg-modules. (For a general reference, see Section 12.1.3 of the book Algebraic Operads by Loday and Vallette, though this was certainly known before the book – I just happen to have it on my desk.) As such it preserves quasi-isomorphisms between cofibrant algebras. The bar-cobar construction on $A$ is always cofibrant, so a sufficient condition for your claim is that $A$ should be cofibrant. Since you are interested in the case of free symmetric algebra, if $A$ is free symmetric on some space of generators $V$ equipped with a filtration $$0 = F_0 V \subset F_1 V \subset \dots \subset V = \bigcup_{i \ge 0} F_i V$$ such that $\partial(F_i V) \subset S(F_{i-1}(V))$ (i.e. the differential of an element in $F_i V$ is a product of elements in lower filtration), then $A$ is cofibrant. In fact you can assume that the filtration is indexed by an ordinal, not just integers, if you need. The homology of $I \Omega B A$ is the Hochschild homology of $A$ with constant coefficients $HH_*(A;\Bbbk)$. (Again, see the book of Loday and Vallette.) I wouldn't expect your map to be a quasi-isomorphism if $A$ is not cofibrant – it certainly can happen, but it's basically a coincidence. As for surjective on cohomology... I have no idea.<|endoftext|> TITLE: Does a spectral gap lift to covering spaces? QUESTION [17 upvotes]: Let $M$ be a complete Riemannian manifold. Denote $\Delta_M\ge0$ the unique self-adjoint extension of the Laplace-Beltrami operator in $L^2(M)$ and $\sigma(\Delta_M)\subset [0,\infty)$ its spectrum. Further define: $$\lambda(M):=\inf\{\mu\in\sigma(\Delta_M)\vert~\mu\neq 0\}$$ Question: Let $N$ be a complete Riemannian manifold with $\lambda(N)>0$. If $p\colon \hat N\rightarrow N$ is a finite sheeted Riemannian covering, do we also have $\lambda(\hat N)>0$? I have asked the same question on math.stackexchange without receiving an answer. Please see there for some examples and my attempts to answer the question. REPLY [2 votes]: This is not an answer, but just somewhere useful you could start looking. Its good to think about when the quotient of fundamental groups has more than one generator. The Rayleigh principle should imply that $\lambda_1(M,g)\geq \lambda_1(\tilde{M},\tilde{g})$, since your $\lambda_1(M,g)=\inf_{f\in H^1(M), ||f||_{2,g}=1}\int_M|df|^2$, and this integral is multiplicative under finite covers. (So this infimum upstairs can only be smaller, as there may be $H^1(M)$ functions which weren't lifts by $p$). In this paper, https://projecteuclid.org/euclid.tmj/1178224610, they prove that when the covering $p:\tilde{M}\rightarrow M$ satisfies $\pi_1(M)/p_\ast\pi_1(\tilde{M})\simeq \mathbb{Z}_k$, there exists a metric on $M$ such that $\lambda_1(M,g)=\lambda_1(\tilde{M},\tilde{g})$. But this may not hold general finite covers.<|endoftext|> TITLE: The binary product of two presentable objects QUESTION [6 upvotes]: The binary product of two $\lambda$-presentable objects (in a locally presentable category) is $\mu$-presentable for some regular cardinal $\mu \geq \lambda$ (because all objects are $\mu$-presentable for some regular cardinal $\mu$). I don't see any reason for $\mu$ to be equal to $\lambda$. Even if $\lambda^2=\lambda$ and even if the binary product of two finite sets is finite for example. Is there one ? Or is there a regular cardinal big enough $\mu$ such that the class of $\mu$-presentable objects is closed under binary product ? REPLY [3 votes]: Unfortunately, no example for the first question is coming to mind at the moment. For the second question, let $\lambda$ be such that the product functor $\times: \mathcal C \times \mathcal C \to \mathcal C$ is $\lambda$-accessible [1]. Let $\mu \rhd \lambda$ be such that the binary product of $\lambda$-presentable objects is $\mu$-presentable. I claim that $\mu$-presentable objects are closed under binary products. To see this, let $A,B$ be $\mu$-presentable. Then $(A,B)$ is $\mu$-presentable in $\mathcal C \times \mathcal C$ [2]. Write $(A,B) = \varinjlim_{i \in I} (A_i, B_i)$ where $I$ is $\lambda$-filtered and $\mu$-small [3]. Because $\times$ is $\lambda$-accessible, we have $$A \times B = \varinjlim_I A_i \times B_i \qquad (\ast)$$ If $C = \varinjlim_{j \in J} C_j$ is a $\mu$-filtered colimit, then we have $$\mathcal C(A \times B, C) = \mathcal C(\varinjlim_{i \in I} A_i \times B_i, \varinjlim_{j \in J} C_j) = \varprojlim_{i \in I} \mathcal C(A_i \times B_i, \varinjlim_{j \in J} C_j) \\ = \varprojlim_{i \in I} \varinjlim_{j \in J} \mathcal C(A_i \times B_i, C_j) \\ = \varinjlim_{j \in J} \varprojlim_{i \in I} \mathcal C(A_i \times B_i, C_j) \\ = \varinjlim_{j \in J} \mathcal C(\varinjlim_{i \in I} A_i \times B_i, C_j) = \varinjlim_{j \in J} \mathcal C(A \times B, C_j) $$ as desired. In the first line, we have used $(\ast)$. In the second line, we have used that $A_i \times B_i$ is $\mu$-presentable and $J$ is $\mu$-filtered. In the third line we have used that $I$ is $\mu$-small and $J$ is $\mu$-filtered. In the fourth line, we have used $(\ast)$ again. [1] In fact any $\lambda$ will do here: recall that a right adjoint functor is $\lambda$-accessible iff its left adjoint preserves $\lambda$-presentable objects. The left adjoint of $\times$ is the diagonal functor $\Delta: \mathcal C \to \mathcal C \times \mathcal C$, which preserves $\lambda$-presentable objects for every $\lambda$. [2] To see this, suppose that $A = \varinjlim_{k \in K} A_k$ and $B = \varinjlim_{l \in L} B_l$ where $K,L$ are $\mu$-small and $\lambda$-filtered. Then we claim that $(A,B) = \varinjlim_{(k,l) \in K \times L} (A_k, B_l)$; then we can take $I = K \times L$, which is also $\mu$-small and $\lambda$-filtered. It suffices to show that the projection $\pi: K \times L \to K$ is cofinal (and dually). Now, for $k \in K$, $k \downarrow \pi = (k \downarrow K ) \times L$, which is a product of filtered categories and so filtered itself, and in particular nonempty and connected. So indeed $\pi$ is cofinal. [3] Thanks to Mike Shulman, who points out below that this is Rmk 2.15 in Adámek and Rosický -- and rather, $A \times B$ is a retract of such a colimit -- but this is sufficient for our purposes.<|endoftext|> TITLE: Basel problem and inversive geometry QUESTION [7 upvotes]: An interesting solution of the Basel problem is given in https://www.tandfonline.com/doi/abs/10.1080/00029890.2018.1452473 (Basel Problem: A Solution Motivated by the Power of a Point, by Kapil R. Shenvi Pause). In the integral $$\zeta(2)=\int_0^1\int_0^1\frac{dx\,dy}{1-xy}$$ let's make a change of variables $$x=\cos{\phi}-\tan{\theta}\sin{\phi},\;\;y=\cos{\phi}+\tan{\theta}\sin{\phi}. \tag{1}$$ The Jacobian of this transformation is twice the inverse of $1/(1-xy)$ and hence $$\zeta(2)=2\int\int d\theta\,d\phi.$$ The integration domain in the last integral is determined by the conditions $0\le x,y\le 1$, which gives $$-\frac{\phi}{2}\le\theta\le\frac{\phi}{2},\;\;\;\phi-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}-\phi. \tag{2}$$ Geometrically (2) is a quadrilateral in the $(\theta,\phi)$ plane with mutually orthogonal diagonals of lengths $\pi/3$ and $\pi/2$. Therefore $$\zeta(2)=2\;\frac{1}{2}\frac{\pi}{2}\frac{\pi}{3}=\frac{\pi^2}{6}.$$ Very nice, isn't it? In the paper the transformation of variables (1) is motivated by an interpretation of $1/(1-xy)$ in terms of inversive geometry. Can this approach be extended to $$\zeta(n)=\int_0^1\cdots \int_0^1\frac{dx_1\cdots dx_n}{1-x_1\cdots x_n}?$$ In particular can we calculate in this manner $\zeta(2n)$? What is the relation (if any) of (1) with the Beukers-Kolk-Calabi change of variables? For the latter see https://pdfs.semanticscholar.org/35be/01e63c0bfd32b82c97d58ccc9c35471c3617.pdf P.S. The Beukers-Kolk-Calabi change of variables is $$x=\frac{\sin{\phi}}{\cos{\theta}},\;\;\;y=\frac{\sin{\theta}}{\cos{\phi}}.$$ Its Jacobian is $1-x^2y^2$ and it is applied to the integral $$\zeta(2)=\frac{4}{3}\int_0^1\int_0^1\frac{dx\,dy}{1-x^2y^2}.$$ In this case the integration domain becomes isosceles right triangle $0\le \phi,\;0\le \theta,\;\phi+\theta\le \pi/2$. REPLY [3 votes]: It doesn't quite answer your questions, but there is another change of variables that Zagier and Kontsevich use to evaluate $$\int_{0}^{1} \int_{0}^{1} \frac{1}{4\sqrt{xy} \ (1-xy)} \ dx \ dy.$$ (See page 8 of https://www.maths.ed.ac.uk/~v1ranick/papers/kontzagi.pdf). The above integral one hand is equal to the series $$\sum_{n=0}^{\infty} \frac{1}{4(n+1/2)^2}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$$ and on the other hand equal to $\pi^2/8$ using the change of variables $$x=\frac{\xi^2(1+\eta^2)}{1+\xi^2}, \quad y=\frac{\eta^2(1+\xi^2)}{1+\eta^2}.$$ It turns out the Jacobian Determinant for this transformation is $$\left|\frac{\partial(x,y)}{\partial(\xi,\eta)} \right|=\frac{4\sqrt{xy}(1-xy)}{(1+\xi^2)(1+\eta^2)}.$$ Like the Shenvi Pause change of variables and the Eugenio Calabi change of variables that you mention, the Jacobian Determinant contains the denominator of the integrand. Perhaps making some kind of trigonometric substitution for $\eta$ and $\xi$ could recover the Shenvi Pause change of variables. There is, however, a $k$-dimensional generalization for the above change of variables. To evaluate $$\int_{(0,1)^k} \frac{1}{2^k\sqrt{x_1 \dots x_k}(1-(-1)^k x_1 \dots x_k)} \ dx_1 \dots \ dx_k,$$ which is equal to $$\sum_{n=0}^{\infty} \frac{(-1)^{nk}}{2^k(n+1/2)^k}=\sum_{n=0}^{\infty} \frac{(-1)^{nk}}{(2n+1)^k},$$ put $$x_i= \frac{\xi^2_i(1+\xi^2_{i+1})}{1+\xi^2_i}, \quad i \in \lbrace 1, \dots, k \rbrace,$$ where $\xi^2_{k+1}:=\xi^2_{1}.$ It turns out $$\left|\frac{\partial(x_1, \dots, x_k)}{\partial(\xi_1, \dots, \xi_k)} \right|=\frac{2^k \sqrt{x_1 \dots x_k}(1-(-1)^k x_1 \dots x_k)}{(1+\xi_1^2) \dots (1+\xi_k^2)}.$$ It is quite similar to Calabi's change of variables because of the structure (cyclical indexing) and the Jacobian Determinant once again containing the denominator of the integrand. In my published article (http://www.ams.org/journals/qam/0000-000-00/S0033-569X-2018-01499-3/), I discuss how the generalized Calabi's change of variables and the generalized Zagier/Kontsevich's change of variables lead to integrals whose evaluations are remarkably similar and a non-traditional representation of $\zeta(k)$ for even $k.$ Some trigonometric substitution for $\xi_i$ might provide a generalized version of the Shenvi Pause's change of variables.<|endoftext|> TITLE: Fulton's deformation to the normal cone vs Verdier's QUESTION [6 upvotes]: Let $X$ be a smooth variety over a field $k$, and let $Y$ be a smooth subvariety. In the literature, I've seen two versions of the deformation to the normal cone: Verdier's version: $\tilde{X}_Y^\mathrm{Ver} := \operatorname{Bl}_{Y\times \{0\}}(X\times \Bbb A^1_k) - \operatorname{Bl}_Y(X)$ Fulton's version: $\tilde{X}_Y^\mathrm{Ful} := \operatorname{Bl}_{Y\times \{0\}}(X\times \Bbb P^1_k) - \operatorname{Bl}_Y(X)$ Evidently, $\tilde{X}_Y^\mathrm{Ver}$ is just $\tilde{X}_Y^\mathrm{Ful}$ with the subvariety $X\times \{\infty\}\subseteq (X\times \mathbb P^1_k)-(Y\times \{0\})\subseteq \tilde{X}_Y^\mathrm{Ful}$ removed. My Question: In what sort of situation is it necessary to use one version and not the other? REPLY [5 votes]: The following answer was emailed to me by Claude Sabbah. I have received his permission to post it here. Whenever you need to get some object on $Y$ from an object existing on the normal cone (in fact normal bundle in your case), you need to proceed by pushforward. As it is better to use a proper pushforward, the version $\tilde{X}_Y^{\mathrm{Ful}}$ is best suited to the question. This is the way Fulton defines Chern classes for example. He needs to use a projection formula that only holds when the fibre of the projection is a projective space, not an affine space. On the other hand, Verdier is not interested in pushing forward to $Y$. The specialization of a perverse sheaf is defined on the normal bundle, and one recovers the nearby cycles by restricting to a section $Y \times \{1\}$, if it exists. Together with Fourier-Sato transform, this also gives the vanishing cycle functor.<|endoftext|> TITLE: Who first chose the names Alice and Bob for players A and B? QUESTION [13 upvotes]: Who first chose the names Alice and Bob for the players (or observers) A and B? REPLY [16 votes]: Allow me to mention that since the players in effect adopt the roles of the quantifiers $\forall$ and $\exists$, as Bob has a winning strategy just in case for every move for Alice, there is a reply by Bob, and so on, some logicians have preferred to use alternative names that would better highlight this connection. When I was a graduate student attending lectures of Adrian Mathias in Berkeley, he used the names Abelard and Éloïse, after the famous couple and their love letters at court (made more interesting by the fact that Abelard was a logician). The advantage, you see, is that the names begin with A and E, aligning with $\forall$ and $\exists$. Another example, of course, would be Adam and Eve, although Mathias preferred Abelard and Éloïse. In truth, however, I recall that Mathias was not fully satisfied with either of these examples, and sought additional famous couples, whose names begin with A and E. Does anyone know any? Let me collect here the examples of A/E names of famous pairs contributed in the comments and elsewhere. Abelard and Éloïse, the famous lovers. Abelard was a logician Adam and Eve, from the creation myth Anna and Elsa, from the Disney film, Frozen Arwen and Elessar (otherwise known as Aragorn), from the Tolkien saga. Connected with infinity and eternal life. Albert and Elizabeth, British royals Ares and Enyo, Greek gods of war Please add more in the comments!<|endoftext|> TITLE: Mordel's conjecture for function fields in positive characteristic QUESTION [7 upvotes]: Manin proves Mordel's conjecture for function fields in characteristic zero.his proof has a gap but Coleman fill this gap and restate Manin proof in a more modern language.both of them work over characteristic zero.has anyone used Manin ideas to prove theorem for positive characteristic ? REPLY [11 votes]: Implicitly, my proof (Inv. Math. 104 (1991) 643-646) is an extension of Manin's approach to char p. I first extended Manin's ideas to char p for elliptic curves (Comp. Math. 74 (1990) 247-258) and realized the connection with p-descent so, for the Mordell conjecture, I just used the p-descent directly. See also my papers with Buium (Crelle 460 (1995) 117-126 and Comp. Math. 103 (1996) 1-6).<|endoftext|> TITLE: Random 3-manifolds in $R^4$ QUESTION [8 upvotes]: Consider following program: Generate random 3-manifold embedded in $R^4$. Perform its triangulation. Put it to Regina and calculate what manifold it is. Assuming that we have good algorithm for random submanifolds in point 1. then we can conclude which 3-manifolds of of complexity 5,6,7,8 etc are embeddable in $R^4$. For example from this paper I can see that there are 175 3-manifolds of complexity 7. Those which were not obtained in this process we can assume are not embeddable in $R^4$ with some probability. Possible choices for algorithm in point 1 are: a) zero of four variables polynomial; b) random embedded 1-surgery; c) gluing cubes; d) drilling small hole cubes in big cube; e) boundary of regular neighborhood of 2-complex in $R^4$ (added 2018-08-23) The questions are: A. What are achievements in finding good polynomial of four variables hoping to obtain interesting 3-manifold as its zero ? B. What could be the algorithm for finding random loop in $M$ embedded in $R^4$ to perform embedded surgery ? C. Is Regina accepting command line execution with some input in TXT file containing triangulation and producing result (or LOG) in other TXT file ? Related questions are: D. What could be other ideas for producing random 3-submanifolds of $R^4$ ? E. How could we generate random slice knots and what manifolds we obtain by repeating 1-surgery on slice knots ? F. Is it known which 2-dimensional CW-complexes are embeddable in $R^4$ ? Such CW-complex can be seen as few words in set of generators which are forming bouquet of circles. I am hoping all 3-manifolds embeddable in 4-space are boundaries of regular neighborhoods of some 2-complex. EDIT 2018-07-30 Regarding last question. I have been able to find embedding of 2-complex with one word in 4-space. So I thought to use this as starting point. Assuming that this 2-complex is defined as 2-skeleton in $\mathbb R^4$. Related question is F2. Is it known algorithm for finding regular neighborhood of 2-skeleton in $R^4$ ? If I have it then I find its triangulated boundary as 3-manifold I want. The 3-simplex belongs to boundary when it belongs to only one 4-simplex. EDIT 2018-08-23 In this question I found reference to the paper: Dranišnikov, A. N.; Repovš, Dušan, Embedding up to homotopy type in Euclidean space, Bull. Aust. Math. Soc. 47, No. 1, 145-148 (1993). ZBL0796.57011. In this paper there is construction of embedding of any 2-complex in $R^4$ up to homotopy type. It is described as simpler proof of Stallings theorem from 1965. Therefore I am planning to convert that construction to simplicial complex in $R^4$. Next construct its regular neighborhood, its boundary will be 3-manifold which I would like to recognize using Regina or other software. What is not clear for me is why embedding of 2-complex in $R^4$ listed as open issue number 5.3 on Kirby's open problem list. Regards, REPLY [5 votes]: Regarding A: As far as I know, there's only some special cases and no big familiy of interesting examples known. Regarding C: Yes, Regina has a fairly good Python interface. We don't have every feature of the C++ library implemented in Python, but quite a bit is. Regarding D: Perhaps the most sensible way to define a "random submanifold of $R^4$" would be to start with the standard triangulation of $S^4$, and sequentially do barycentric subdivision. A "random" submanifold could be a vertex-normal solution to the normal 3-dimensional submanifold equations. This is something I do with Regina systematically and it's one of the more fruitful ways of generating 3-dimensional submanifolds of $S^4$. Unfortunately, you generate the simplest 3-manifolds the most often, much like with random knot algorithms. It's unclear to me how to make this more effective, other than throwing enormous computational power at the problem. Regarding E: I don't think people have any great algorithms for producing random slice knots. You either have to restrict to fairly specific knot families or live with cripplingly slow algorithms.<|endoftext|> TITLE: Reference Request: Specialization map in Huber's Context QUESTION [5 upvotes]: The specialization map $sp:\mathfrak{X}_\eta\to \mathfrak{X}_{red}$ has an important role in rigid analytic geometry. I tried looking in Huber's papers ("Continuous Valuations", "A generalization of formal schemes and rigid analytic varieties", and "Etale cohomology of Rigid Analytic Varieties and Adic Spaces") browsing for a treatment of this map and I didn't find much. Ideally, I would like to find a statement like the following: Proposition: Given a Tate Huber-pair $(A,A^+)$ consider the topological space $Spa(A,A^+)$. Then there is a map of topological spaces $sp: Spa(A,A^+)\to Spf(A^+)$, this map is a spectral map of spectral spaces and it is a closed map. I know that in a traditional Noetherian formal scheme context (or probably even topologically finite type over a valuation ring context) this proposition is a consequence of interpreting the $Spa(A,A^+)$ as the limit of admissible blow-ups, but I was wondering if this was known and written in a non-Noetherian context like perfectoid spaces or even in a non-sheafy context. Any reference to the specialization map in a Huber-like approach will be much appreciated. REPLY [4 votes]: In case you're still interested: Bhatt recently proved that for any Tate-Huber pair $(A,A^+)$, the topological space $\mathrm{Spa}(A,A^+)$ is homeomorphic to an inverse limit of admissible blowups in the expected manner (Theorem 8.1.2 here). For fun I'll summarize the construction (which probably won't seem too surprising). Fix a topologically nilpotent unit $\varpi \in A^+$, and let $I$ denote the category of proper birational maps of schemes $f_i:X_i \to \mathrm{Spec}(A^+)$ which restrict to isomorphisms over the open subset $\mathrm{Spec}(A) \subset \mathrm{Spec}(A^+)$. This is cofiltered, because any $X_i \times_{\mathrm{Spec}(A^+)}X_{i'}\to \mathrm{Spec}(A^+)$ is still in $I$. Let $\overline{X_i} \subset X_i$ denote the vanishing loci of $\varpi$, so these are also cofiltered, and they all map compatibly to $\mathrm{Spec}(A^+/\varpi)$. Now, there is a canonical map $\Phi: \mathrm{Spa}(A,A^+) \to \lim_{\leftarrow} \overline{X_i}$ given by the following recipe: for any point $x \in \mathrm{Spa}(A,A^+)$, let $K_x$ and $K_x^+$ be the associated residue field and valuation subring, respectively, so there are canonical maps $A \to K_x$ and $A^+ \to K_x^+$. We then get canonical compatible maps $\mathrm{Spec}(K_x) \to X_i$ for all $i$, simply because the loci in the $X_i$'s where $\varpi \neq 0$ all identify with $\mathrm{Spec}(A)$ by definition, so we can lift the given map $\mathrm{Spec}(K_x)\to \mathrm{Spec}(A)$ uniquely. But we also have a map $\mathrm{Spec}(K_x^+) \to \mathrm{Spec}(A^+)$, so by the valuative criterion of properness this lifts uniquely along each $f_i$, giving a compatible system of maps $r_i : \mathrm{Spec}(K_x^+) \to X_i$. Applying the $r_i$'s to the unique closed point of $\mathrm{Spec}(K_x^+)$ then gives the desired point in $\lim \overline{X_i}$. Anyway, Bhatt proves that $\Phi$ is always a homeomorphism. The specialization map $\mathrm{sp}$ you want is just the composition of $\Phi$ with the natural map $\lim \overline{X_i} \to \mathrm{Spec}(A^+/\varpi)$. Of course, you don't need to go through all of this to define the specialization map, but it does seem to be the easiest way to prove the following: Fact: The map $\mathrm{sp}$ is continuous, spectral, and closed. Proof sketch: Continuity and spectrality follow from Bhatt's result together with some basic nonsense about limits of spectral spaces. Closedness can be deduced from the following useful criterion: any quasicompact, spectral, specializing map of locally spectral spaces is closed.<|endoftext|> TITLE: Homotopy pushout independent of factorization and symmetric in cofibration category QUESTION [8 upvotes]: $\require{AMScd}$In Algebraic homotopy, Baues defines the notion of homotopy pushout in a cofibration category in the following way: a commutative diagram \begin{CD} A @>k>> C \\ @AfAA @AAhA\\ B @>g>> D \end{CD} is a homotopy pushout if for one factorization $B\hookrightarrow W\stackrel{\sim}\to A$ the induced map $W\cup_B D\to C$ is a weak equivalence. He then says "This easily implies that for any factorization $B\hookrightarrow V\stackrel{\sim}\to A$ of $f$, the map $V\cup_B D\to C$ is a weak equivalence. Thus in the definition we could have replaced 'some' by 'any' or used $g$ in place of $f$." This is page 9 of the book, and no theory has been developed yet, except the construction of cylinders. Q1. "If one factorization works, all of them do." I managed to reduce this to the following: suppose in addition that there is trivial cofibration $W\to V$ that is a 'morphism of factorizations' in the obvious sense. One can then carry out the pushouts for both factorizations and get an induced map from $W\cup_D B\to V\cup_D B$. Unless I am missing something this is a pushout of $W\to V$, and then a weak equivalence. If this is true, I have a proof, since by taking pushouts and factoring I can connect any two factorizations by a diamond of factorizations $W\to Q\leftarrow V$ where both maps are trivial cofibrations. Is there a simpler way to proceed? Q2. "One can resolve the other variable". It also says "...or used $g$ in place of $f$." Does the phrasing suggest this is a consequence of the fact any factorization for $f$ works if one does? This is not clear to me. I think an argument similar to the one for Q1 should work in proving one can connect everything using a pushout of two factorizations (one of $f$ and one of $g$), much like one does when balancing Tor or Ext. Edit. Answer to Q2 is "yes": This follows easily by using Q1 and taking a few pushouts and using C1 and C2. For completeness, here are the axioms of a cofibration category, which is endowed with a class of cofibrations and of weak equivalences. C1. Cofibrations and weak equivalences contain the isomorphisms, weak equivalences satisfy 2 out of 3 and cofibrations are closed under compositions. C2. Pushouts exist along cofibrations, and cofibrations are stable under them. Moreover, weak equivalences are stable under pushouts along cofibrations in both directions. C3. Maps can be factored into a cofibration followed by a weak equivalence. C4. Every object admits a trivial cofibration into an object $R$ which is fibrant, in the sense every trivial cofibration $R\to Q$ splits. REPLY [2 votes]: (The following was intended as a comment to Karol's answer, that after using the "Brown Type Factorization" trick, we can prove the result by applying a result in the book, which does not assume cofibrantness, but due to the word count, I think it might be more appropriate to present it as an answer.) First, using the Brown Type Factorization trick, we can reduce to showing the following case : suppose the factorization involving $W,V$ are weakly equivalent, that is, if we have a diagram : $$\begin{align}B\rightarrow V\rightarrow A\\||\quad\;\downarrow i \;\quad||\\B\rightarrow W\rightarrow A\end{align}$$ (with the middle horizontal arrow being a weak equivalence) then $W\cup_BD\to C$ is a weak equivalence iff $V\cup_BD\to C$ is a weak equivalence. By (C1), it then suffices to show $i\cup_B 1_D:V\cup_BD\to W\cup_B D$ is a weak equivalence. Now the result follows from Lemma 1.2(b) in chapter II in the book applied to the diagram $$\begin{align}V\leftarrow B\rightarrow D\\i\downarrow\quad\;||\quad\quad||\\W\leftarrow B\rightarrow D\end{align}$$<|endoftext|> TITLE: Homotopy theories of operads QUESTION [16 upvotes]: I know of three homotopy theories of colored operads. The (derived) localization category of Berger-Moerdijk's model structure on the category of strict simplicial (or topological) operads, with weak equivalences given by strict maps $O\to O'$ which induce weak equivalences on spaces of operations. The "dendroidal" notion of infinity-operad, introduced by Moerdijk and Weiss and further studied in this paper by Cisinski and Moerdijk. Lurie's infinity-operads, which are infinity categories fibered over the nerve of the category $\mathrm{Set}^*$ of pointed sets, satisfying certain Segal-style properties. If all is well in the world, these three homotopy theories (viewed as e.g. $\infty$-categories) should be equivalent (or there should be a good reason for them not to be). But I can't find references for any equivalences between them. This paper seems to compare (1) and (2), via a Quillent adjunction, which it does not show is an equivalence. The obvious functor (1) $\implies$ (3) is written down in Lurie's Higher Algebra, but it is not (as far as I can tell) shown to be an equivalence. Is more known about comparisons between these homotopy theories? I'm specifically interested in the comparison between (1) and (3). Are the corresponding $\infty$-categories equivalent, perhaps under additional restrictions? Is the functor (1) $\implies$ (3) fully faithful? Are there known examples where this functor is not an equivalence? REPLY [15 votes]: The answer is yes: see the paper of Chu-Haugseng-Heuts, "Two models for the homotopy theory of ∞-operads", arXiv:1606.03826. In brief, already Cisinski and Moerdijk ("Dendroidal sets and simplicial operads", arXiv:1109.1004) proved a Quillen equivalence between simplicial operads and dendroidal sets. In the paper of Cisinski and Moerdijk that you link to, they prove an equivalence between dendroidal sets and complete dendroidal Segal spaces. What Chu-Haugseng-Heuts prove is an equivalence between complete dendroidal Segal spaces and Barwick's complete Segal operads, and already before this Barwick had proved that complete Segal operads were equivalent to Lurie's ∞-operads.<|endoftext|> TITLE: Is a torsion free sheave of rank one on a reducible curve the pushforward of a line bundle on a normalization? QUESTION [7 upvotes]: Let $X$ be a nodal curve, possibly reducible. Then can any torsion free sheaf of rank one on $X$ be expressed as $\pi_*(L)$, where $L$ is a line bundle on a partial normalization of $X$? This looks standard, but I just can not find a reference that precisely state this conclusion. REPLY [3 votes]: Short proof: A given rank $1$, torsion-free sheaf $F$ is the direct image of a sheaf $L$ on the partial normalization $Y := \operatorname{Spec} \operatorname{Hom}(F,F)$. When $X$ is nodal, $Y$ is Gorenstein, and a theorem of Vasconcelos implies that $L$ is a line bundle. In more detail, I first claim the sheaf $\operatorname{Hom}(F,F)$ of endomorphisms is a sheaf of finite commutative $\mathcal{O}_{X}$-algebras. Since $F$ is torsion-free, $\operatorname{Hom}(F,F)$ injects into its generic fiber which is just the ring of rational functions since $F$ is generically trivial of rank $1$. We conclude that $\operatorname{Hom}(F, F)$ is commutative. The $\mathcal{O}_{X}$-algebra is also finitely generated (fix generators and track where they go) and integral (use Hamilton-Cayley to produce a polynomial that kills a given local section), so we conclude that it is finite. Essentially by construction, $Y := \operatorname{Spec} \operatorname{Hom}(F,F)$ is a partial normalization of $X$ that carries a sheaf $L$ whose direct image is $F$. Furthermore, $L$ satisfies $\operatorname{Hom}(L,L)=\mathcal{O}_{y}$. Since $X$ was nodal, $Y$ is nodal as well, and we conclude by Theorem (3.1) of Vasconcelos's paper "Reflexive modules over Gorenstein rings". Note: This argument applies to a curve with arbitrary singularities, although we can only conclude that $L$ is a line bundle when $Y$ is Gorenstein (otherwise we can only assert that $\operatorname{Hom}(L,L) = \mathcal{O}_{Y}$). The partial normalization $Y$ is always Gorenstein when $X$ has at worst $A_{n}$ singularities, but otherwise $X$ admits sheaves that are not direct images of line bundles. For example, if $X$ is the Zariski closure of the curve $\{x^3-y^4=0\}$, then the maximal ideal $(x,y)$ is not the direct image of a line bundle.<|endoftext|> TITLE: Is there a PL, or topological, bordism hypothesis? QUESTION [14 upvotes]: The bordism hypothesis says that the $(\infty, n)$-category of smooth, framed $n$-bordisms, $(n-1)$-dimensional boundaries, and corners down to points, is freely generated symmetric monoidal with duals upon a single object. The proof sketch by Lurie uses Morse theory extensively, and has since been formalised and extended by several authors. Morse theory is a technique from real differential topology, and it is related to handle decompositions. Piecewise linear (PL) manifolds are not equivalent to smooth manifolds from dimension 5 upwards. Topological manifolds famously already depart from PL and smooth manifolds in dimension 4. I don't know whether PL manifolds or topological have a well-developed analogue of Morse theory, but they have handle decompositions (except for 4d topological manifolds), so I would expect a lot of the technical procedures from the bordism hypothesis proof to work. I do not understand whether it is possible to define a $(\infty,n)$-category of bordisms in these cases, as opposed to a mere 1-category. Then the awkward question is: How do the $(\infty,n)$-categories of PL and topological bordisms look like? They cannot possibly be equivalent to the smooth one, or otherwise we would have trivially proven that extended TQFTs do not detect PL or smooth structures? But then, which part of the construction goes awry? What's so special about the smooth categories that the bordism hypothesis works there? REPLY [10 votes]: This is addressed in Remark 2.4.30 of Jacob's paper. The PL case has a very nice description but the topological case does not. In particular, there's no difference between framed bordisms in the PL and smooth case. So the framed part of the cobordism hypothesis goes through with no changes. This means that not only does $\mathrm{O}(n)$ act on the core of the bordism category, the larger group $\mathrm{PL}(n)$ does. So say unoriented PL TFTs valued in $\mathscr{S}$ are classified by $\mathrm{PL}(n)$-homotopy fixed points in the core $\mathscr{S}^\times$. In the topological case, Jacob argues that one should not expect a good similar description. Specifically, he says that the cobordism hypothesis is about handle decompositions of bordisms, but there are topological manifolds (such as Freedman's $E_8$ manifold) which do not have any handle decomposition at all. It's not obvious to me why there couldn't be some description of topological bordisms by starting with the smooth bordism category and then adding some new generators (for example, a new generator corresponding to the fake 4-ball). But at any rate there's no good result known in this direction and it certainly seems like it would be very difficult.<|endoftext|> TITLE: Paul Mahlo's Original Large Cardinal Paper QUESTION [6 upvotes]: It is well known that Paul Mahlo (1883-1971) developed a systematic hierarchy of inaccessible cardinals of the type $\pi_{a,b}$ where $\pi_{1,b}$ enumerates the strongly innacessible cardinals, $\pi_{2,b}$ enumerate the fixed points of $\pi_{1,b}$ and so on. My question is, where can I find the original paper of Mahlo in English? If this doesn't exist, are there any good expositional articles on this accessible online? REPLY [5 votes]: Just to show what to expect from a machine translation, I spent a little time on two paragraphs of the 1924 Mahlo paper, which I OCR'd and passed to Google Translate. With all the formulas it requires considerable post-processing, but if one is motivated this should be doable in an afternoon. On a property of a sub-type of the continuum by Paul Mahlo in Recklinghausen If one considers a set of subsets of the continuum $C$ of this type, which is so simply ordered that each subset contains all of its preceding real subsets, it can be easily proved that its order type is always a sub-type of the continuum; it therefore contains partial types of regular starting numbers or their inverses of at most $\omega$ and $^\ast\omega$. In the following we show that such arrangements of subsets of $C$ of the same type can also contain the partial types $\omega_1$ and $^\ast\omega_1$. In doing so we first connect to a final order of the $\omega$-sequences, belonging to continued fractions of positive irrational numbers $x> 1$. [...] After these auxiliary considerations, we now deal with a sub-type of the continuum $C$. We denote by $M$ a nowhere dense perfect part of $C$, but always presuppose this $M$ as bounded; all $M$'s therefore have the same type, and every $M$ has $c$ two-sided and $a$ one-sided limits. There exist nowhere dense perfect subsets $M'$ of $C$, that contain $M$ as nowhere dense part, where each one-sided limit of $M$ is a two-sided limit of $M'$. One can easily construct $a$ perfect sets $M_0$, $M_1$, $\ldots$, $M_n$, $\ldots$ ($n <\omega$), so that each $M_n$ contains only two-sided limits of $M_{n + 1}$. The type of elementary sets of all $M_n$ is called $\pi$; it is independent of the particular choice of $M_n$.<|endoftext|> TITLE: Ellipsoid minimizing Banach-Mazur distance to convex body QUESTION [9 upvotes]: Given a (symmetric) convex body $K \subset \mathbb{R}^n$ (equivalently, given a norm on $\mathbb{R}^n$), there is a unique ellipsoid of maximal volume in $K$, called the John ellipsoid. The John ellipsoid can be described as a ``canonical ellipsoid'' associated to a convex body, and reading around there seem to be a few other notions of canonical ellipsoid. On a recent research project of mine, it turned out to be important to associate a different type of ellipsoid to $K$, namely the ellipsoid which minimizes the Banach-Mazur distance to $K$, suitably normalized. More directly, let $E$ be the ellipsoid contained in $K$ which minimizes the value $\lambda \geq 1$ for which $K \subset \lambda E$. In my paper I proved a volume ratio inequality for this ellipsoid $E$ for dimension two and applied it to a problem on quasiconformal mappings (the preprint is at https://arxiv.org/pdf/1703.05891.pdf ). At the time, I asked a few convex geometry people what was known about this ellipsoid, or if it had been studied before, as the John ellipsoid has been. They didn't really have anything to say on the matter, nor did I find anything in standard references, so I didn't dwell on it. Now, however, there are some junior mathematicians working on thesis projects, etc. who have been talking to me and want to use my work. So this is making me want to revisit the question of attribution. I'm curious about the following: Has the ellipsoid minimizing Banach-Mazur distance to a convex body in $\mathbb{R}^n$ ever appeared or been studied in an important/useful/systematic way? Is there a best name to give to the "ellipsoid which minimizes Banach-Mazur distance to $K$" without having to say this every time? One candidate could be simply ``Banach-Mazur ellipsoid'', but the answer to 1. might suggest a different name. REPLY [3 votes]: Yes, it has appeared in the works of several experts in Functional Analysis. In the book Banach-Mazur distances and finite dimensional operator ideals, by N. Tomczak-Jaegermann, it is proved that in finite dimensional Banach spaces that have enough symmetries, the distance ellipsoid coincides with John's ellipsoid. In particular, it is unique. The author comments, on p.134, that in general "extremal ellipsoids may be very far from distance ellipsoids". In fact, maximal or minimal volume ellipsoids share many contact points with the (boundary of the ) corresponding convex body, whereas distance ellipsoids may share as few as only two contact points. A preprint called Remarks and examples concerning distance ellipsoids, by Dirk Praetorius, provides a result of this nature. In this paper, the authors mention a result by B. Maurey, asserting that if a space $X$ does not have a unique (up to homothety) distance ellipsoid, then there is a subspace which has the same distance to a Hilbert space as the whole space and which has a unique distance ellipsoid. Unfortunately, Maurey never published this result. A nice corollary is that for two-dimensional Banach spaces, the distance ellipse is unique. "Banach-Mazur ellipsoid" is possible, but a terminology which is widely accepted by experts is "distance ellipsoid", e.g. - all references given above.<|endoftext|> TITLE: What is the weakest negative curvature condition ensuring a manifold is a $K(G,1)$? QUESTION [16 upvotes]: The only statement I'm sure of is that any hyperbolic or Euclidean manifold is a $K(G,1)$ (i.e. its higher homotopy groups vanish), since its universal cover must be $\mathbb H^n$ or $\mathbb E^n$. But for example, if a complete Riemannian manifold $M$ satisfies one of the following, can I conclude that $M$ is a $K(G,1)$? $M$ has sectional curvature bounded above by some negative number. $M$ has negative sectional curvature. $M$ has nonpositive sectional curvature. $M$ has sectional curvature bounded above by $f(\operatorname{vol}(M))$ (where $f: \mathbb R \to \mathbb R$ is some function depending only on the dimension of $M$ that I don't know). $M$ has scalar curvature bounded above by some negative number. $M$ has negative scalar curvature. $M$ has nonpositive scalar curvature. $M$ has scalar curvature bounded above by $f(\operatorname{vol}(M))$. Do the answers change if I assume that $M$ is compact? Have I left out a relevant condition of some kind? REPLY [6 votes]: Let me summarize the information in the comments in a CW post. Feel free to edit. For "weaker" notions of curvature, negative curvature seems to not imply that a manifold is a $K(G,1)$. As Deane Yang pointed out, Lohkamp showed that for each $d \geq 3$, there are numbers $a(d) > b(d) > 0$ such that every manifold $M$ of dimension $d$ admits a complete metric $g$ with $-a(d) < \operatorname{Ric}(M,g) < -b(d)$. I believe this implies an analogous result for scalar curvature. The only loophole I can see is that there might be a smaller interval $a(d) \geq a' \geq b' \geq b(d) > 0$ such that if $M$ admits a metric $g$ with $-a'\leq \mathrm{Ric}(M,g) \leq -b'$, then $M$ is a $K(G,1)$. (Possibly $a',b'$ might depend on further parameters such as $\operatorname{diam}(M,g)$ or $\operatorname{vol}(M,g)$). For instance, if $M$ admits a metric of constant negative Ricci curvature, does this imply that $M$ is a $K(G,1)$? Igor Belegradek points out below that the answer is no in this case as shown by Yau. For sectional curvature, the story is different. As several people pointed out, the Cartan-Hadamard theorem says that any manifold admitting a complete metric of nonpositive sectional curvature is a $K(G,1)$. We may ask if this can be improved to allow a small amount of positive curvature. As Igor Belegradek pointed out, "small amount" can't be specified in terms of volume, since $R S^2 \times g(R) S^1$ has constant volume $a$ for appropriate $g(R)$, but by choosing $R$ sufficiently large, it has arbitrarily small positive curvature. But as Igor Belegradek also pointed out, Fukaya and Yamaguchi showed that there is a positive number $\epsilon(d,D)$ dependent only on the dimension $d$ and diameter $D$, such that any compact Riemannian manifold $M$ with $-1 \leq \operatorname{sec}(M) < \epsilon(\operatorname{dim}(M), \operatorname{diam}(M))$ is a $K(G,1)$. The lower bound on the curvature is necessary; Fukaya says that Gromov constructed metrics on $S^3$ with fixed diameter and arbitrarily small sectional curvature. I don't know if Yamaguchi - Fukaya's result holds for complete Riemannian manifolds (as Igor Belegradek points out the question doesn't even make sense in this case). Another direction which might be interesting would be to control "small amounts of positive curvature" in some other way. For instance, rather than controlling the $L^\infty$ norm of the sectional curvature, one might ask for control over some averaged version of it -- this might allow the curvature to become very positive at a point so long as it's not very positive in a large region. Somehow the necessity of the lower curvature bound in Fukaya's result suggests to me that something like this might be a good idea.<|endoftext|> TITLE: The Tall Tale of Terminating Transfinite Towers QUESTION [10 upvotes]: The transfinite tower of iterative automorphisms of a group $G$ is simply definied to be the following chain of the groups where $G_{\alpha+1}=Aut(G_{\alpha})$ for each ordinal $\alpha$ and the direct limit is taken at the limit stages: $G\rightarrow Aut(G)\rightarrow Aut(Aut(G))\rightarrow\cdots\rightarrow G_{\alpha}\rightarrow G_{\alpha+1}\rightarrow\cdots$ The tower terminates when a fixed point is reached, namely one of the groups in the chain is isomorphic to its automorphism group by the natural map. Simon Thomas has proved that the automorphism tower of every centerless group eventually terminates. Later, Hamkins completed Thomas' result by showing that the automorphism tower terminates for every group: Thomas, Simon, The automorphism tower problem, Proc. Am. Math. Soc. 95, 166-168 (1985). ZBL0575.20030. Hamkins, Joel David, Every group has a terminating transfinite automorphism tower, Proc. Am. Math. Soc. 126, No. 11, 3223-3226 (1998). ZBL0904.20027. Hamkins' theorem gives a sense to the natural definition of the notion of terminating number of a group, $\tau(G)$, that is the least ordinal where the automorphism tower of $G$ terminates. My first question is about the minimum power of $ZFC$ that is needed to carry out Thomas-Hamkins' proof: Question 1. How much $ZFC$ is needed to prove that the automorphism tower terminates for every group, $G$, and so $\tau(G)$ is well-defined? Particularly, is $AC$ used anywhere in Hamkins or Thomas' results (which Hamkins' proof is partially based on)? If so, is this use of $AC$ essential? If yes, are the following two statements equivalent? The automorphism tower terminates for every group. The Axiom of Choice. My next question is about the relation between the terminating number of the direct product of two groups and the terminating number of each component: Question 2. What is the relation between $\tau (G\times H)$ and $\tau (G)$, $\tau(H)$? Is there an upper bound for $\tau (G\times H)$ expressible in terms of $\tau (G)$, $\tau(H)$? For instance, is it true to say $\tau (G\times H)\leq Max (\tau (G), \tau(H))$ or $\tau (G)+\tau(H)$ or $\tau (G).\tau(H)$ ...? The "Max" bound in the above question is inspired by the fact that for finite groups, $G, H$, whose orders are relatively prime, we have $Aut(G\times H)\cong Aut(G)\times Aut(H)$. If one somehow manages to keep this pattern through the entire chain then the automorphism tower of $G\times H$ terminates after $Max (\tau (G), \tau(H))$ steps. In particular, computing $\tau(G^n)$ (and comparing it with $\tau(G)$) could be of interest as well. For instance, in the special case that $G$ is a cyclic group of order $p$, one has $Aut(G^n)\cong GL_{n}(\mathbb{F}_p)$ and so $\tau (G^{n})=\tau (GL_{n}(\mathbb{F}_p))+1$. REPLY [12 votes]: Here is a relevant paper, Kaplan, Itay; Shelah, Saharon, The automorphism tower of a centerless group without choice, Arch. Math. Logic 48, No. 8, 799-815 (2009). ZBL1192.03026. At least for a centerless group, it seems that every automorphism tower stabilizes below the Lindenbaum number of the power set of finite sequences of the group. While I don't quite know how to move from a group to a centerless group without choice, let me point out that $G\mapsto\operatorname{Aut}(G)$ is a function on the universe, and by replacement there is some $\alpha$ such that $V_\alpha$ is a closure point of that function. I suspect that an argument would work in a choiceless setting from this point.<|endoftext|> TITLE: Is the Lisca-Matic bound (aka slice-Bennequin bound) strictly stronger than the Bennequin bound? QUESTION [10 upvotes]: The Bennequin bound [1] says that, for a transverse knot (or later link) $K$ in $S^3$, $$\mathrm{sl}(K) \le - \chi(\Sigma)$$ for any Seifert surface $\Sigma$ for $K$, where $\mathrm{sl}$ is the self-linking number. Lisca and Matic proved [2, Theorem 3.4] that for a Legendrian knot $K$ $$|r(K)| + \mathrm{tb}(K) \le 2g(\Sigma) - 1$$ for any smooth surface $\Sigma \subset B^4$ with boundary $K$, where $\mathrm{tb}$ and $r$ are the Thurston-Bennequin and rotation numbers. EDIT: This was actually proved by Rudolph [3]. I have several questions here. (a) Why is the Lisca-Matic bound always stated in terms of Legendrian knots? The combination $|r| + \mathrm{tb}$ is exactly the self-linking number of the transverse push-off, with one of the two orientations. Wouldn't it be simpler to just state it in terms of transverse knots? (b) Isn't the Lisca-Matic bound strictly stronger than the Bennequin bound? Why isn't it more popular to quote it? Maybe it's about the extension to links (which, in the case of the Bennequin bound, is due to Eliashberg)? It seems that Lisca and Matic only state the result for knots in their paper, but it seems to me that the proof should carry over. For knots/links in more general 3-manifolds, the Bennequin bound applies as long as you have a tight contact structure, while to state Lisca-Matic you need some sort of filling. [1] Bennequin, "Entrelacements et équations de Pfaff", In Third Schnepfenried Geometry Conference, Vol. 1, Schnepfenried, 1982, 87–161. Astérisque 107. Paris: Société Mathématique de France, 1983. [2] Lisca, P., and G. Matić. “Stein 4-manifolds with Boundary and Contact Structures.” 55–66. Topology and Its Applications 88, nos. 1–2, 1998. [3] Rudolph, Lee, "Quasipositivity as an obstruction to sliceness." Bull. Amer. Math. Soc. (N.S.) 29 (1993), no. 1, 51–59. REPLY [5 votes]: Yes, the slice-Bennequin bound is stronger than Bennequin bound: (a) For example, the knot $K \# - K$ is ribbon, so the slice-Bennequin implies that any transverse representative must have negative self-linking number. But the Bennequin bound only implies it is less than $4 g(K) - 1$, which is positive if $K$ is nontrivial. (b) Given any transverse representatives $T, -T$ of a knot and its mirror, we can take a transverse connected sum to obtain a knot satisfying $sl(T \# - T) = sl(T) + sl(T) + 1 \leq -1$. Let $T$ be the positive torus knot $T(p,q)$. This maximizes the Bennequin bound so we can assume $sl(T) = pq - p - q$. But this implies that the self-linking number of any transverse representative of the negative torus knot $T(p,q)$ is as most $p + q - pq - 2$. As Marco points out, one reason to state the result in terms of Legendrian knots is because we have a good local model for attaching a 2-handle along a Legendrian knot. There is a theory for attaching handles along transverse knots, described in David Gay's thesis, but it is trickier. The approach of Lisca-Matic is to use the genus bound coming from the global adjunction inequality in a Kahler surface of general-type to deduce a slice-genus bound. Finally, complex points and Lai indices give another, very satisfying conceptual relation between the Bennequin-Eliashberg bound and the Adjunction Inequality that is more "transverse" than "Legendrian". A generic real, immersed surface $\Sigma$ in a complex surface $X$ will have finitely many complex points --- i.e. points where the tangent plane is a complex line. These can be either positive or negative (according to whether the orientation of $\Sigma$ agrees with the complex orientation) and either elliptic or hyperbolic. The Lai indices of $\Sigma$ count these points with signs: $I_+(\Sigma) = e_+ - h_+ \qquad I_-(\Sigma) = e_- - h_-$ If $\Sigma$ is embedded, these satisfy the relations $I_+(\Sigma) + I_-(\Sigma) = \chi(\Sigma) + [\Sigma]^2$ $I_+(\Sigma) - I_-(\Sigma) = \langle c_1(X),[\Sigma] \rangle$ The adjunction inequality is then equivalent to the inequality $I_-(\Sigma) \leq 0$. There is an analogous story for a convex surface $\Sigma$ in a contact manifold $(M,\xi)$. The tangencies of $\xi$ along $\Sigma$ are the singularities of the characteristic foliation. These can be either positive or negative and either elliptic or hyperbolic. We can then define Lai indices exactly as above. If $\Sigma$ is a Seifert surface bounded by a transverse knot $K$, then $sl(K) = I_-(\Sigma) - I_+(\Sigma)$ Moreover, the Bennequin-Eliashberg bound is equivalent to the inequality $I_-(\Sigma) \leq 0$. Eliashberg proves this directly by showing how to perturb the surface to eliminate all negative elliptic points. See the books Stein manifolds and holomorphic mappings by Forstneric and Surgery on contact 3-manifolds and Stein surfaces by Ozbagci-Stipsicz.<|endoftext|> TITLE: $\mathbb{C}^{*}$-actions on Fano $3$-folds QUESTION [8 upvotes]: I am looking for an example of a smooth Fano $3$-fold $X$ over $\mathbb{C}$, with a non-trival $\mathbb{C}^{*}$-action, which satisfies the following properties: There is a $\mathbb{C}^{*}$-action such that the fixed point set is finite. $Aut(X)$ contains no copy of $(\mathbb{C^{*}})^{2}$. The rank of the Picard group is at least $2$. (Note that if we relax any of these three conditions, then there are examples. If we relax 1. then we can take $X = \mathbb{P}^{1} \times S$, where $S$ is a del Pezzo surface with discrete automorphism group. If we relax 2. then there are toric examples. If we relax $3.$, then the Mukai-Umemura Fano $3$-fold works. Hence, I believe that there is a reasonable chance that such an example should exist.) REPLY [3 votes]: Choose $X$ to be the blow-up of a smooth quadric $Q\subset \mathbb{P}^4$ at a curve $\Gamma\subset Q$ being a smooth normal rational quartic curve. In coordinates, you can for instance choose $\Gamma$ to be the image of $$\mathbb{P}^1\to \mathbb{P}^4, [u:v]\mapsto [u^4:u^3v:u^2v^2:uv^3:v^3]$$ and $Q$ to be given by $$x_2^2+x_1x_3-2x_0x_4=0$$ so the action of $\mathbb{C}^*$ given by $[u:v]\mapsto [u:\xi v]$ extends to a unique action of $\mathbb{C}^*$ on $\mathbb{P}^4$, which has only $5$ fixed points, four of them being on $Q$ and $2$ of them on $\Gamma$, namely $[1:0:0:0:0]$ and $[0:0:0:0:1]$. You can check in coordinates that the action has finitely many fixed points on the blow-up. Moreover, the group $\mathrm{Aut}^{\circ}X$ cannot contain $(\mathbb{C}^*)^2$ as this group would come from an action on $Q$ preserving $\Gamma$. (In fact, we can check that $\mathrm{Aut}^{\circ}X$ is $\mathrm{PGL}_2$ has index $2$ in $\mathrm{Aut} X$).<|endoftext|> TITLE: Reference request: A collection of topologies on $\mathbb{N}$ formed via series QUESTION [6 upvotes]: First, some quick notation: for any series $\sum_{n=1}^\infty a_n$ whose terms are positive real numbers, and for any subset $M = \{m_1, m_2,...\} \subseteq \mathbb{N}$, we write $\sum_M a_n$ to mean $a_{m_1} + a_{m_2} + ....$. Then, for each series, one can associate a topology on $\mathbb{N}$ by declaring a proper subset $M\subsetneq \mathbb{N}$ to be closed iff $\sum_M a_n < \infty$. So, e.g., in the $\sum \frac{1}{n}$-topology, the set of even numbers is neither open nor closed (both $\sum \frac{1}{2n+1}$ and $\sum \frac{1}{2n}$ diverge), but the set of squares is closed ($\sum \frac{1}{n^2}$ converges). There is, of course, an interplay between the topological properties $\mathbb{N}$ inherits and the analytic properties of $\sum a_n$. For instance, $\mathbb{N}$ gets the cofinite topology iff $\liminf a_n > 0$. Much less trivially, one can show that this association generates precisely $|\mathbb{R}|$ many topologies on $\mathbb{N}$ (distinct up to homeomorphism). In fact, for $0\leq p < q \leq 1$, the spaces obtained from $\sum \frac{1}{x^p}$ and $\sum \frac{1}{x^q}$ are not homeomorphic. I am not a general topologist by training so my question is probably very simple. Has anything like this appeared in the literature before? MathSciNet doesn't seem to turn up anything, but maybe there is some terminology I need to know. Maybe these form a subclass in some well studied class of topological spaces? In case it helps, here are some curious data points about these topological spaces. They are homogeneous. More generally, the homeomorphism group acts transitively on the set of $k$-element subsets for any fixed $k$. Every subset is closed or dense. (This is equivalent to the topology being downward-closed: every subset of a proper closed set is closed.) If $\lim_{n\rightarrow \infty} a_n = 0$ but $\sum a_n$ diverges, then any non-empty open set is homeomorphic to $\mathbb{N}$. REPLY [4 votes]: What you've described as a topology, is actually a family of analytic P-ideals on $P(\mathbb{N})$ called the "summable ideals." They are studied for different reasons, and the general search terms to use are "summable ideal" or more generally "analytic P-ideal".<|endoftext|> TITLE: The skew monoidal structure induced by a functor QUESTION [6 upvotes]: $\require{AMScd}$Let $\cal K$ be a 2-category, and $j : A\to B$ one of its 1-cells. Assume that the induced map $$ j^* : {\cal K}(B,B)\to {\cal K}(A,B) $$ precomposing with $j$ has a left adjoint $j_!$, the left extension along $j$. Given 1-cells $x,y\in {\cal K}(A,B)$, it is possible to obtain canonical maps $\gamma_{xyz} : j_! (j_!x\circ y) \circ z \to j_!x\circ j_!y\circ z$, a map that in turn is determined by a map $\bar\gamma : j_!(j_!x\circ y) \to j_!x\circ j_! y$ via the universal property of $j_!$, so that $\gamma = \bar\gamma * z$; in ${\cal K}=\bf Cat$, a sufficient condition for this to be invertible is that left extensions along $j$ are absolute. $\sigma : j_!(j)\to 1_B$ the counit of the density comonad of $j$; in ${\cal K}=\bf Cat$, this is invertible if and only if $j$ is dense. $\eta : x \to j_!(x)\circ j$, the $x$-component of the unit of $j_!\dashv j^*$; in ${\cal K}=\bf Cat$, this is invertible if and only if $j$ is fully faithful. I want to prove that the following two diagrams of 2-cells are commutative: $$ \begin{CD} j_!(j_!x\circ y)\circ j @>\gamma_{xyj}>> j_!x\circ j_!y\circ j\\ @A\eta_{j_!x\circ y}AA @| \\ j_!(x)\circ y @>>j_!x * \eta_y> j_!x\circ j_!y\circ j \end{CD} \begin{CD} j_!(j_!j\circ x)\circ y @>j_!\sigma * y>> j_!x\circ y \\ @| @AA\sigma * j_!x\circ y A \\ j_!(j_!j\circ x)\circ y @>>\gamma_{jxy}> j_!j\circ j_!x \circ y \end{CD} $$ But, as I expected when I began, this is going to be quite painful. Any advice to make the proof simpler and clearer? A conceptual proof in ${\cal K}=\bf Cat$, that can be adapted to a generic $\cal K$, is welcome! Edit (Jul 23, 2018). What I am trying to prove is that the claim that appears as Theorem 3.1 in "Monads need not be endofunctors" holds in a generic 2-category. REPLY [2 votes]: There is a fast and down to earth proof. We use the letters $R$ and $L$ to indicate the right and left adjoints of morphisms. For the first one, just call $a=j_!x\circ y$ and $b=j_!x\circ j_!y$, we have one morphism $\phi:a\to j^*b$ with left adjoint $L\phi:j_!a\to b$. Then your diagram becomes $$\begin{CD} j^*j_!a @>j^*L\phi>> j^*b\\ @A\eta AA@|\\ a@>\phi>>j^*b \end{CD}$$ which is an easy $1$ categorical fact about adjunction. It's just the right adjoint diagram of $L\phi\circ\text{id}=\text{id}\circ L\phi$, with $\eta=R\text{id}$ and $\phi=RL\phi$. For the second one, observe that you can prove commutativity before composing with $y$: every arrow has the form $\phi\ast y$ for some $\phi$. Now just use adjunction of $j^*$ and $j_!$: your diagram becomes $$\begin{CD} j_!j\circ x @>Rj_!(\sigma\ast\text{id})>> j^*j_!x\\ @|@A\sigma\ast\text{id} AA\\ j_!j\circ x@>\text{id}\ast\eta>>j_!j\circ j^*j_!x \end{CD}$$ We can now unpack $$Rj_!(\sigma\ast\text{id}):j_!j\circ x\to j^*j_!x$$ as $$j_!j\circ x\xrightarrow{\sigma\ast\text{id}} x\xrightarrow{\eta} j^*j_!x$$ To check that this unpacking is right, just observe that $R\eta=\text{id}$. Hence the diagram above becomes $$\begin{CD} x @>\eta>> j^*j_!x\\ @A\sigma\ast\text{id} AA@A\sigma\ast\text{id} AA\\ j_!j\circ x@>\sigma\ast\eta>>j_!j\circ j^*j_!x \end{CD}$$ which is obvious.<|endoftext|> TITLE: How can I functorially dualise in a symmetric monoidal $(\infty,1)$-category with duals? QUESTION [16 upvotes]: If $\mathcal{C}$ is a symmetric monoidal $(\infty,1)$-category with duals, then there should be a functor $$ d: \mathcal{C} \longrightarrow \mathcal{C}^{op} $$ such that $d(x)$ is dual to $x$ for all objects $x \in \mathcal{C}$. How can one construct such a functor? Of course, the above is only a very weak formulation of what one would actually expect. In fact, it should be possible to choose $d$ such that there is a coherent homotopy $d^{op} \circ d \simeq \operatorname{Id}_{\mathcal{C}}$. By 'coherent' I mean that $\mathcal{C}$ should have homotopy fixed-point data for the involution $$ (\ )^{op}: \mathit{Cat}_\infty^{\otimes} \longrightarrow \mathit{Cat}_\infty^{\otimes}. $$ And while I'm on it, let me give the most general version of the quesion: Is there a canonical (or maybe even essentially unique) trivialisation of the $(\ )^{op}$ involution when restricted to the full subcategory $\mathit{Cat}_\infty^{\otimes,d}\subset \mathit{Cat}_\infty^{\otimes}$ of those symmetric monoidal $(\infty,1)$-categories in which every object is dualisable? I am aware of a solution to this question in the $1$-categorical context; it goes as follows: For a symmetric monoidal $1$-category $\mathcal{C}$ one constructs a category $D(\mathcal{C})$ where objects are tuples $(x,y,e,c)$ with $x,y$ two objects of $\mathcal{C}$ and $(e,c)$ is an evaluation-coevaluation pair that exhibits them as dual. The morphisms $\varphi:(x,y,e,c) \to (x',y',e',c')$ in this category are pairs $(f:x \to x',g:y' \to y)$ such that $f$ is dual to $g$ with respect to the duality data specified. This category admits canonical projections $$ \mathcal{C} \longleftarrow D(\mathcal{C}) \longrightarrow \mathcal{C}^{op}. $$ If $\mathcal{C}$ has duals, one can use essential uniqueness of duals to show that both projections are categorical equivalences. Using this construction it should be rather straight-forward to show all claims made above for the special case of $1$-categories. (Except for uniqueness of the trivialisation.) Now the problem is that I have no idea how to generalise this proof to the $\infty$-categorical situation. For instance, for the two morphisms $f$ and $g$ to be dual would then be additional structure instead of a property. I would be happy to use the cobordism hypothesis in dimension $1$, if that is of any help. (It can be used to construct $d$ on the maximal subgroupid $\mathcal{C}^\sim \subset \mathcal{C}$, but it seems hard to use it to say anything about duals of non-invertible morphisms. As a side-note: Where can I find a proof or even a proof sketch of the cobordism hypothesis in dimension $1$?) I'm interested in answers at any stage of generality. REPLY [10 votes]: One way to construct the duality functor ${\cal C} \to {\cal C^{\rm op}}$ is through the notion of a pairing of $\infty$-categories (see HA, Definition 5.2.1.5). In particular, in this case we're talking about a self-pairing on ${\cal C}$, which by definition is a right fibration $\mu:{\cal M} \to {\cal C} \times {\cal C}$, classified by a functor $b: {\cal C}^{\rm op} \times {\cal C}^{\rm op} \to {\cal S}$ to spaces. We say that $\mu$ is left representable if for every $x \in {\cal C}$ the functor $y \mapsto b(x,y)$ is representable in ${\cal C}$, and right representable if for every $y \in {\cal C}$ the functor $x \mapsto b(x,y)$ is represetable in ${\cal C}$. If $\mu$ is a pairing which is both left and right representable then it determines an adjunction between ${\cal C}$ and ${\cal C}^{\rm op}$. We say that $\mu$ is a perfect pairing if it is both left and right representable and the associated adjunction is an equivalence. One can then show that the data of a perfect pairing $\mu: {\cal M} \to {\cal C} \times {\cal C}$ is in fact equivalent to the data of an equivalence ${\cal C} \to {\cal C}^{\rm op}$. The advantage of this point of view is that the $\mathbb{Z}/2$-action on the space of equivalences ${\cal C} \to {\cal C}^{\rm op}$ becomes explicit: it is given by sending $\mu: {\cal M} \to {\cal C} \times {\cal C}$ to ${\rm swap} \circ \mu: {\cal M} \to {\cal C} \times {\cal C}$, where ${\rm swap}:{\cal C} \times{\cal C} \to {\cal C} \times {\cal C}$ is the equivalence that swaps the two components. In particular, to construct an equivalence ${\cal C} \stackrel{\simeq}{\to} {\cal C}^{\rm op}$ which is self-dual in a homotopy coherent manner is equivalent to constructing a perfect pairing ${\cal M} \to {\cal C} \times {\cal C}$ together with a $\mathbb{Z}/2$-action on ${\cal M}$ which lifts the swap action on ${\cal C} \times {\cal C}$. Note that this swap action can itself be encoded as a right fibration $p:{\rm Sym}({\cal C}) \to {\rm B}(\mathbb{Z}/2)$, where ${\rm B}(\mathbb{Z}/2)$ is the groupoid with one object whose endomorphism group is $\mathbb{Z}/2$, and the fiber of $p$ over this one object is ${\cal C} \times {\cal C}$. One can then encode a lift of the swap action to ${\cal M}$ by a right fibration $\widetilde{\cal M} \to {\rm Sym}({\cal C})$ whose restriction to ${\cal C} \times {\cal C}$ is ${\cal M}$. In general, we may refer to right fibrations $\widetilde{\cal M} \to {\rm Sym}({\cal C})$ as symmetric pairings, and say that a symmetric pairing is perfect when its base change to ${\cal C} \times {\cal C} \subseteq {\rm Sym}({\cal C})$ is perfect. The notion of a perfect symmetric pairing then encodes a duality on ${\cal C}$, i.e., a self-dual equivalence $d:{\cal C} \to {\cal C}^{\rm op}$ (with all the higher coherences taken into account). Now in the case of a symmetric monoidal $\infty$-category with duals, the pairing encoding the duality is the right fibration ${\cal M} \to {\cal C} \times {\cal C}$ classified by the functor $b_{\cal C}: {\cal C} \times {\cal C} \to {\cal S}$ sending $(x,y)$ to ${\rm Map}_{\cal C}(x \otimes y,1_{\cal C})$. There is a direct way to construct this as a symmetric perfect pairing. Indeed, suppose that $\pi:{\cal C}^{\otimes} \to {\rm Fin}_*$ is the coCartesian fibration encoding the symmetric monoidal structure on ${\cal C}$, and let ${\cal C}^{\otimes}_{\rm act} \to {\rm Fin}$ be its active part, i.e., the base change to the category ${\rm Fin}$ of finite sets via the functor $I \mapsto I_+$. Then we can identify ${\rm B}(\mathbb{Z}/2)$ as the subcategory of ${\rm Fin}$ consisting of the object $\left<2\right>^{\circ} = \{1,2\} \in {\rm Fin}$ and all its automorphisms, and the base change of ${\cal C}^{\otimes}_{\rm act}$ to ${\rm B}(\mathbb{Z}/2) \subseteq {\rm Fin}$ is naturally equivalent to ${\rm Sym}({\cal C})$. If $1_{\cal C} \in ({\cal C}^{\otimes}_{\rm act})_{\left<1\right>^{\circ}}$ now denotes the unit then the right fibration $$ ({\cal C}^{\otimes}_{\rm act})_{/1_{\cal C}}\times_{{\rm Fin}} {\rm B}(\mathbb{Z}/2) \to {\cal C}^{\otimes}_{\rm act} \times_{{\rm Fin}}{\rm B}(\mathbb{Z}/2) \simeq {\rm Sym}({\cal C}) $$ is a symmetric pairing whose underlying pairing is $b_{\cal C}$ above. When ${\cal C}$ has duals this symmetric pairing is perfect, yielding the associated self-dual equivalence $d:{\cal C} \to {\cal C}^{\rm op}$.<|endoftext|> TITLE: Expectation inequality for sampling without replacement QUESTION [8 upvotes]: Is the following proposition correct? $X_1, X_2, X_3$ are uniformly at random sampled from a finite set $\mathcal X$ without replacement. $f : \mathcal X^2 \rightarrow \mathbb R_{\ge0}$ is symmetric: $ f(x, y) = f(y, x) $, then: $$ \mathbb E_{X_1, X_2, X_3} f(X_1, X_2) f(X_1, X_3) f(X_2, X_3) \le ( \mathbb E_{X_1, X_2} f^2(X_1, X_2) )^{3/2} $$ I tried to use Hoeffding's result $$ \mathbb E f\left( \sum_{i = 1}^n X_i \right) \le \mathbb E f\left( \sum_{i = 1}^n Y_i \right) $$ ($X_i$ are uniformly at random sampled without replacement, $Y_i$ are uniformly at random sampled with replacement, $f$ is convex and continuous) by combining two elements from set $\mathcal X$ to form a new set: $\{ ( X_i, X_j ) : i \ne j, X_i, X_j \in \mathcal X \}$. However, the sampling process for new set is no longer uniformly at random so I cannot use Hoeffding's result. Since items are sampled uniformly, this is equivalent to: $$ \left( \dfrac{ \sum_{1 \le i < j < k \le n} f_{ij} f_{ik} f_{jk} }{\binom{n}{3}} \right)^2 \le \left( \dfrac{ \sum_{1 \le i < j \le n} f^2_{ij} } {\binom{n}{2}} \right)^3 $$ For $n = 3$, this is: $$ \left(f_{12} f_{13} f_{23}\right)^2 \le \left( \dfrac{ f_{12}^2 + f_{13}^2 + f_{23}^2}{3} \right)^3 $$ which follows from the inequality between the geometric mean and the root-mean-square: $$ \left(abc\right)^{1/3} \le \sqrt{\dfrac{a^2 + b^2 + c^2}{3}} $$ For $n=4$, this is: $$ \left(\frac{f_{12}f_{13}f_{23}+f_{12}f_{14}f_{24}+f_{13}f_{14}f_{34}+f_{23}f_{24}f_{34}}{4}\right)^2 \leq \left(\frac{f_{12}^2+f_{13}^2+f_{14}^2+f_{23}^2+f_{24}^2+f_{34}^2}{6}\right)^3 $$ which follows from https://artofproblemsolving.com/community/user/12908: \begin{align} \left(abd+ace+bcf+def\right)^2 &= \Big(a(bd+ce)+(bc+de)f\Big)^2 \\ &\le \left(a\sqrt{(b^2+c^2)(d^2+e^2)}+f\sqrt{(b^2+d^2)(c^2+e^2)}\right)^2 \\ &\le \left(\sqrt{(a^2+f^2)\big((b^2+c^2)(d^2+e^2)+(b^2+d^2)(c^2+e^2)\big)}\right)^2 \\ &= (a^2+f^2)(b^2+c^2)(d^2+e^2)+(a^2+f^2)(b^2+d^2)(c^2+e^2)\\ &\le \left(\frac{a^2+f^2+b^2+c^2+d^2+e^2}{3}\right)^3+\left(\frac{a^2+f^2+b^2+d^2+c^2+e^2}{3}\right)^3 \\ &= 16\left(\frac{a^2+b^2+c^2+d^2+e^2+f^2}{6}\right)^3 \end{align} REPLY [2 votes]: Let $\lambda_1,\dots,\lambda_n$ be eigenvalues of the symmetric matrix $(f_{ij})$, where $f_{ii}=0$ by definition. They are real, $\sum \lambda_i=0$ and the inequality rewrites as $$ \left(\frac{\sum \lambda_i^3}{n(n-1)(n-2)}\right)^2\leqslant \left(\frac{\sum \lambda_i^2}{n(n-1)}\right)^3, $$ or $(\sum \lambda_i^3)^2\leqslant \frac{(n-2)^2}{n(n-1)}(\sum \lambda_i^2)^3$. Now fix $\sum \lambda_i^2=S$ and $\sum \lambda_i=0$ and maximize $\sum \lambda_i^3$. When $\sum \lambda_i^3$ is maximal, the gradient vectors $(1,1,\dots,1),2(\lambda_1,\lambda_2,\dots,\lambda_n),3(\lambda_1^2,\lambda_2^2,\dots,\lambda_n^2)$ must be linearly dependent by Lagrange multipliers theorem. In other words, there should exist number $A,B,C$ not all equal to 0 such that $A+B\lambda_i+C\lambda_i^2=0$ for all $i$. Therefore $\lambda$'stake at most 2 different values. Without loss of generality $a+b=n$, $a$ $\lambda$'s are equal to $b$, $b$ $\lambda$'s are equal to $-a$ (for some $a\in \{1,2,\dots,n-1\}$), and the inequality rewrites as $(ab^3-ba^3)^2\leqslant \frac{(n-2)^2}{n(n-1)}(ab^2+ba^2)^3$, or $(b-a)^2\leqslant \frac{(n-2)^2}{n-1}ab$. This is clear from $|b-a|\leqslant n-2$, $ab\geqslant n-1$.<|endoftext|> TITLE: General principles which lead to good questions in many concrete situations QUESTION [12 upvotes]: I believe that in various fields of mathematics there are general principles which might lead to good questions and good results in many concrete situations. I would like to have a list of such principles. Below is my own list. 1) Given a Banach space, one may ask what is its dual. In many concrete situations this question has interesting and important answers. (For example the dual space of the space of continuous functions on a compact space is the space of measures on it.) 2) Given a functor between two categories, one may ask whether is has right/ left adjoint. One may also ask whether it has right/ left derived functors. (For example the simple operation of push-forward of sheaves of abelian groups has left adjoint - pull back- which is somewhat less obvious.) 3) Given a metric space, one may ask what is its completion. (Concrete example which I like: consider the set of isometry classes of $n$-dimensional closed Riemannian manifolds of diameter at most $D$ and sectional curvature at least $\kappa$ equipped with the Gromov-Hausdorff metric. Points of its completion are compact metric spaces which are so called Alexandrov spaces with curvature bounded below.) REPLY [3 votes]: Look for situations where two effects are competing, and study which effect "wins," or whether they balance each other in some sense. For example, people talk about certain dispersive PDEs being subcritical (dispersion wins over the nonlinear effects) or supercritical (nonlinear effects win). The critical/borderline case is often the most interesting. Other examples arise in combinatorics, where some structure is guaranteed to appear if the size of the problem is sufficiently large, and one may ask where this guarantee begins, i.e. at what size the complexity/unlikeliness of that structure is exactly balanced with the pigeonhole principle.<|endoftext|> TITLE: to compare cohomologies of fibers of two fiber bundles QUESTION [6 upvotes]: Consider the following commutative diagram of the fiber bundles $% F\rightarrow E\rightarrow B$ and $F^{\prime }\rightarrow E^{\prime }\rightarrow B^{\prime }$ where $B^{\prime }$ is simply connected space (but $B$ is not simply connected space) and all spaces are path-connected spaces. $\require{AMScd}$ \begin{CD} F @>{}>> E @>{}>> B \\ @VVV @VVV @VVV\\ F' @>{}>> E' @>{}>> B' \end{CD} Suppose that \begin{equation*} H^{\ast }\left( B^{\prime };% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( B;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \end{equation*} and \begin{equation*} H^{\ast }\left( E^{\prime };% %TCIMACRO{\U{211a} } %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( E;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \end{equation*} are isomorphisms. If \begin{equation*} H^{i }\left( F;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right)=0 \end{equation*} for all $i \geq n$ ($n$ fixed), then \begin{equation*} H^{i }\left( F^{\prime };% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right)=0 \end{equation*} for all $i \geq n$? REPLY [11 votes]: No. Let $B'$ be any space, and take $E'=PB'$ and $F'=\Omega B$. The Kan-Thurston theorem gives a map $f\colon B\to B'$ such that $H^*(f;\mathbb{Q})$ is an isomorphism but $\Omega B$ is discrete, so $H^i(\Omega B;\mathbb{Q})=0$ for $i>0$. The diagram $\require{AMScd}$ \begin{CD} \Omega B @>{}>> PB @>{}>> B \\ @VVV @VVV @VVV\\ \Omega B' @>{}>> PB' @>{}>> B' \end{CD} satisfies most of your hypotheses, but $H^*(\Omega B')$ need not be bounded above. One problem with the above example is that we have fibrations, but these need not be fibre bundles. If necessary this can be fixed by a detour into simplicial sets and simplicial groups. Another problem with the above example is that the space $F=\Omega B$ is disconnected. If you want an example where absolutely everything is connected, we can proceed as follows. We can assume that we have actual fibre bundles, and then let $\Sigma_BE$ denote the fibrewise unreduced suspension of $E$, which is a fibre bundle over $B$ with fibre $\Sigma F$, which is always connected. We can describe $\Sigma_BE$ as the homotopy pushout of $B\xleftarrow{}E\xrightarrow{}B$, and from this we see that the map $\Sigma_BE\to\Sigma_{B'}E'$ is a homology equivalence provided that $B\to B'$ and $E\to E'$ are homology equivalences. Thus, we can apply this procedure to the previous counterexample to obtain a new counterexample in which everything is connected. In general, if you know that a result is true for simply connected spaces, and you want to check whether that assumption can be relaxed, you should ask yourself whether the Kan-Thurston theorem gives counterexamples.<|endoftext|> TITLE: How long iterations of $x \to (p \!\!\! \mod \!\! x)$ can be? QUESTION [9 upvotes]: Suppose that $p$ is a prime number and $x_1$ is an integer in range $[1, p - 1]$. If $x_k \not = 1$, define $x_{k + 1} := p \!\!\! \mod \!\! x_k$. Clearly, $x_{k + 1} < x_k$ and $x_{k + 1} > 0$ (because $p$ is prime), so there exists some $n$ (depending on $x_1$) such that $x_n = 1$. Is it true that $n = O(\log p)$? For reference, it is quite easy to prove that $n = O(\sqrt p)$. Let $q_i := [\frac{p}{x_i}]$, i. e. $p = q_i x_i + x_{i + 1}$ for every $i \in [1, n - 1]$. Because $q_i x_i + x_{i + 1} = p = q_{i + 1} x_{i + 1} + x_{i + 2}$ for $i \in [1, n - 2]$ and $x_i > x_{i + 1}, x_{i + 1} > x_{i + 2}$, $q_i$ is increasing. On the other hand, $x_i$ is decreasing. Also $q_i x_i < q_i x_i + x_{i + 1} < p$ for every $i \in [1, n - 1]$. Therefore $q_i < \sqrt{p}$ or $x_i < \sqrt{p}$ for each $i \in [1, n - 1]$. Because $q_i$ increases, there can be at most $[\sqrt{p}]$ $i$-s with $q_i < \sqrt{p}$. The same way there can be at most $[\sqrt{p}]$ $i$-s with $x_i < \sqrt{p}$. Therefore $n \leqslant 2[\sqrt{p}]$. Some ideas: Maybe it makes sense to think backwards: i. e. how long can a sequence $y$ be if $y_1 = 1$ and $y_{i + 1}$ is some divisor of $p - y_i$ that is bigger than $y_i$ (in this notation $x_i = y_{n + 1 - i}$)? One more way to look at this is to notice that the question asks to estimate how deep can a rooted tree on $p - 1$ vertices be, with root in $1$ and edges between $x$ and $p \!\!\! \mod \!\! x$ for $x = 2, 3, \ldots, p - 1$. Maybe it is possible to prove that this tree is somewhat balanced? Two heuristics suggest that the number of steps is indeed $O(\log p)$: first one assumes that $p \!\!\! \mod \!\! x$ is a random integer number in range $[1, x - 1]$ and second one uses "backwards" point of view and assumes that $p - y_i$ is divisible by $t$ with probability $t^{-1}$. Of course, both heuristics don't make too much sense, but they show that $n = O(\log p)$ is at least a reasonable enough statement to consider. In fact, any ideas on proving any bound better than $O(\sqrt{p})$ are appreciated. REPLY [2 votes]: Here is an idea to pursue. Fix $p$. I will call $p \bmod x$ (your dynamic) $d(x)$, and I look at $H$, the set of $x$ less than $p$ with $2*d(x) \gt x$. The dynamic can stay out of $H$ only a logarithmic number of times, so now we ask how attractive $H$ can be as a dynamic. $H$ looks like a collection of intervals. Dividing everything by $p$, the set is like $(1/2,2/3)$ union $(1/3,2/5)$ union intervals of the form $(1/k, 2/(2k-1))$, for enough $k$ until you get bored. This is some constant fraction of $p$, so pretty large in logarithmic terms. However, we don't have to be afraid of $H$. If $d$ visits $H$ and not $H$ alternately, $d$ still reaches $1$ in a logarithmic number of steps. (Indeed, we can set a number in $(1,\log p)$ as a goal to finish in logarithmic time, so I am not going to worry about the very end of the dynamic.) So let us consider when $x$ is in $H$ and $d(x)$ is also in $H$. Suppose $x$ and $x+1$ are in $H$. Then $d(x+1)-d(x) \gt 1$ when $x \lt p/2$. So the set of $x$ less than $p/2$ which visit $H$ at least twice in a row is less than half the number of $x$ less than $p/2$ which visit $H$ at least once under $d$. If we start from a point less than $p/2$, then the chance it lands in $H$ is small, and the chance it lands in $H$ twice in a row by iterating $d$ gets very small. The idea is to analyze this region below $p/2$, and show that landing in $H$ at least $k$ times in a row is less than $(1/2)^k$ as likely as landing once. I leave the detail to others. Now we turn our attention to $(2x \gt p)$. The question now is how long can one iterate $d$ and stay above $p/2$. But this happens for no $x$ less than $p$. So the ticket is to show the subsets of $H$ which are visited by $d$ at least $k$ times in a row grows exponentially small. Gerhard "Seems Better Than Square Root" Paseman, 2018.07.18.<|endoftext|> TITLE: What is the modal logic of outer multiverse? QUESTION [16 upvotes]: The mathematical multiverse could be viewed as a gigantic Kripke model with models of $ZFC$ as possible worlds connected to each other via a certain accessibility relation. The modal logic associated with this multiverse has been studied in the case that the accessibility relation is set forcing. According to this interpretation, a statement $\varphi$ in the language of set theory is considered possible if it is true in some set forcing extension and necessary if holds in all of them. For more details see: Hamkins, Joel David; Löwe, Benedikt, The modal logic of forcing, Trans. Am. Math. Soc. 360, No. 4, 1793-1817 (2008). ZBL1139.03039. Hamkins, Joel David, The set-theoretic multiverse, Rev. Symb. Log. 5, No. 3, 416-449 (2012). ZBL1260.03103. However, set forcing extension is just a special case of interesting and well-behaved extensions that a model of $ZFC$ might have. If $M$ is an inner model of $N$ then we call $N$ an outer model of $M$. Recall that any model of $ZFC$ could be considered an inner model of its generic extension and so a forcing extension is a special case of an outer model. Of course, there might be outer models of a given model of $ZFC$ which don't arise as a set forcing extension of the ground. A classical example is $L\subseteq L[0^{\sharp}]$. Now let's ease the conditions in the above interpretations of the possibility and necessity across the multiverse so that $\varphi$ is possible if it is true in some outer model and is necessary if it holds in all of them. Philosophically speaking, in such an interpretation of modalities, a statement is possible if it could be realized by expanding our world-view and is necessary if it holds no matter how we expand our universe. (Here, Plato's Allegory of the Cave might be relevant where the cave or tight universe is our ground model). In the view of the fact that the modal logic of set forcing multiverse is fully determined by Hamkins and Löwe to be S4.2 (under certain conditions), I wonder what could be said about the modal logic of outer model multiverse. Question. What is the modal logic of the outer multiverse; the set-theoretic multiverse equipped with the outer model relation (rather than set forcing) as the accessibility bridge between the universes? REPLY [21 votes]: I've noticed that recently you have asked a few questions about my work, and so let me thank you; you are kind to take an interest. This particular question can be seen as part of the subject of set-theoretic potentialism, which my co-author Øystein Linnebo and I recently investigated in our paper: J. D. Hamkins and Ø. Linnebo, The modal logic of set-theoretic potentialism and the potentialist maximality principles, to appear in Review of Symbolic Logic, 2018. arxiv:1708.01644 In that paper, we consider several different notions of set-theoretic accessibility, as below, from forcing accessibility to Grothendieck-Zermelo potentialism to rank extensions or transitive extensions or submodel potentialism and others. In each case, for each concept of accessibility we determine the corresponding modal logic of that concept of potentialism. In other related work, W. Hugh Woodin and I recently looked at the case of top-extensional set-theoretic potentialism, proving that the modal logic is S4. J. D. Hamkins and W. H. Woodin, The universal finite set, under review. (arxiv:1711.07952) (You may also be interested in my more philosophical remarks on this topic at the end of my paper, The modal logic of arithmetic potentialism and the universal algorithm.) Nevertheless, despite all that, I am very sorry to say that none of these cases are exactly the case you asked about, which is outer-model accessibility. Outer-model potentialism is certainly a natural case of potentialism, and so let me try to tell you what I know about it. Like forcing potentialism, this would be a case of width potentialism and height actualism, since outer models increase only the width of the universe and not the height. First, let us fix a countable model of ZFC plus V=L, say, and consider it in the context of all its outer models. Since we can force so as to destroy any stationary set in this model, while preserving others, we see that there is an infinite family of independent buttons, "the $n^{th}$ stationary set in the $L$-least partition of $\omega_1^L$ is no longer stationary." This can be made true in an outer model (by forcing) and once true, remains true in all further outer models; and the statements can be controlled independently. And since we also have a family of independent switches arising from the GCH patterns, it follows by the main modal logic analysis (as in Structural connections...), it follows that the modal logic of outer-models is contained within S4.2. And it certainly contains S4, because it is reflexive and transitive. So the answer to your question this a modal logic between S4 and S4.2. I don't actually know which side it will end up on, or if it will end up strictly between, and I think this is a good question. If you hope to prove that the model logic is S4.2, then usually one does this by proving an directedness or amalgamation theorem. The problem here, however, is that we already know that outer model possibility is not directed, since you can move from a model to outer models in incompatible ways that cannot be amalgamated (this is due to Mostowski). So that avenue of showing (.2) is valid is closed off. I don't know if (.2) is valid for this notion of possibility or not, but I am inclined against it. I think there will be fundamentally incompatible possibily necessary statements, which is to say, railway switches, and this will be incompatible with S4.2. If you hope to prove that only S4 is valid, then you could follow some of the recent work on the universal algorithm and universal finite sets. The general consequence of the existence of these finite sequences with the universal extension property is that they cause the existence of railyard labelings, which then cause the modal logic to be contained in S4. I don't know if there is any universal finite sequence phenomenon for outer models, and this also is an interesting question. So I believe that the exact modal logic of outer-model potentialism is an open question.<|endoftext|> TITLE: A question on symmetric matrices QUESTION [8 upvotes]: $\newcommand{\R}{\mathbb{R}}$ The question is Is there a constructive (say, parametric) description of the set (say $M_n$) of all symmetric matrices $A\in\R^{n\times n}$ such that all the diagonal entries of $A$ are $0$ and the matrix $(A-tI_n)^2$ is diagonal for some real $t$? Here, of course, $I_n$ is the $n\times n$ identity matrix. E.g., if $A=I_n-\frac1n\,1_n1_n^T$, then all the diagonal entries of $A$ are $0$ and $(A-\frac12\,I_n)^2=\frac14\,I_n$, so that $A\in M_n$. Here, as usual, $1_n1_n^T$ is the matrix in $\R^{n\times n}$ all of whose entries are $1$. Even a necessary condition for a matrix $A$ to be in the set $M_n$ could be useful, if it is close enough to sufficiency. REPLY [6 votes]: Partial answer: As Christian Remling noted in remarks, it suffices to deal with the case that $(A-tI_{n})^{2} = \lambda I_{n}.$ Since $A$ is symmetric, we can only have $\lambda = 0$ when $A = tI_{n}$ for some $t$. But since the diagonal entries of $A$ are all $0,$ this only happens when $A = 0.$ Hence we may suppose that $\lambda \neq 0.$ Then $A = tI_{n} + \sqrt{\lambda}T$ for some matrix $T$ with $T^{2} = I_{n}.$ If $T = I_{n}$ we again see that $A$ is a scalar matrix, so the zero matrix. Hence the only non-zero possibilities for $A$ arise when $T^{2} = I \neq T.$ Now trace $T$ = $r-s$ where $T$ has $r$ eigenvalues $1$ and $s$ eigenvalues $-1$ Also, since $tI_{n} + \sqrt{\lambda}T$ has all diagonal entries zero, it follows that all diagonal entries of $T$ must be equal. Hence each diagonal entry of $T$ is $\frac{r-s}{n}$ and $t + \frac{(r-s)\sqrt{\lambda}}{n} = 0.$ Now $r = s$ gives $t = 0,$ so that $A = \sqrt{\lambda}T.$ If $r \neq s,$ then $\lambda = \frac{n^{2}t^{2}}{r-s}.$ Hence the problem is reduced in essence to determining which symmetric matrices $T$ of order two have all diagonal entries equal. But if $T$ is such a matrix, then $\frac{I+T}{2}$ and $\frac{I-T}{2}$ are mutually orthogonal idempotent symmetric matrices, each with all diagonal entries equal. Now the problem is reduced to finding symmetric idempotent matrices $E$ with all diagonal entries equal (for we may take $T = 2E-I$ if we find such an $E$). If $E$ has rank $m,$ then note that each diagonal entry of $E$ is $\frac{m}{n}.$ We may obtain such an $E$ or each positive divisor $d$ of $n$: take $E$ to be the direct sum (in the obvious sense) of $n/d$ copies of $\frac{J_{d}}{d},$ where $J_{d}$ is the $d \times d$ matrix with all entries $1.$ At the moment, I don't see how to determine all possibilities for such an $E$. Later edit: (...but someone else might). One observation which might turn out to be relevant is that if $E$ is such an idempotent matrix, and $X$ is a "signed permutation matrix", that is, a matrix with exactly one non-zero entry in each row and each column, that entry being $\pm 1$), then $XEX^{t}$ is another such matrix. I don't know if there are other orthogonal matrices leaving this set of idempotents invariant. Even later edit: Note that if $E,F$ are mutually orthogonal symmetric idempotent matrices with all diagonal matrices with all entries equal, then $E+F$ is symmetric idempotent with all diagonal entries equal. This allows us to produce symmetric idempotent matrices with all diagonal entries equal of every rank for some values of $n$. For example, when $n = 2^{r},$ we may consider $Y = \frac{1}{\sqrt{n}} X,$ where $X$ is the character table of an elementary Abelian $2$-group of order $2^{r}.$ Let $u_{i}$ be the $i$-th column of $Y$. Then $E_{i} = u_{i}u_{i}^{t}$ is a symmetric idempotent matrix of rank $1$ with all diagonal entries $\frac{1}{n}.$ For $j \neq i,$ it is easy to see that $E_{i}E_{j} = E_{j}E_{i} = 0.$ Hence for $1 \leq k \leq n,$ we may take the sum of $k$ distinct $E_{i}$'s to get an idempotent symmetric matrix with all diagonal entries $\frac{k}{n}.$<|endoftext|> TITLE: A weak form of the Erdős-Turán conjecture QUESTION [9 upvotes]: This question is motivated by the answer of Gowers to the question Erdos Conjecture on arithmetic progressions. Question. (1)-Suppose $A \subset \mathbb{N}$ is such that Lim$_n$ $log(n) \cdot |A \cap \{ 1, \dots, n\}|\over{n}$=1. Does $A$ contain arithmetic progressions of arbitrary large finite length? (2) What if we replace Lim with Limsup? REPLY [7 votes]: The answer to both questions is almost certainly yes, but this has not been proven. It remains an open question even for progressions of length 3 (although the current bounds are pretty close). For progressions of length greater than 3, the best bounds are due to Gowers, which stated in your terms give: if $$ \limsup \frac{(\log\log N)^{c_k}\lvert A\cap \{1,\ldots,N\}\rvert}{N}>0,$$ where $c_k>0$ is some constant depending only on $k$, then $A$ contains infinitely many arithmetic progressions of length $k$. For progressions of length 3 the $(\log\log N)^{c_k}$ can be replaced by $\log N(\log\log N)^{-4}$. For progressions of length 4 it becomes $(\log N)^{c}$, which is a recent result of Green and Tao. Obtaining even this kind of bound, let alone answering your question, for progressions of length 5 or longer seems quite far out of each at the moment. Based on the constructions we know, something like the following is probably true: if $$ \limsup \frac{\exp((\log N)^{c_k})\lvert A\cap \{1,\ldots,N\}\rvert}{N}>0$$ then $A$ contains infinitely many arithmetic progressions of length $k$.<|endoftext|> TITLE: $G\cong C_4\times A_5$ or $C_2\times C_2\times A_5$? QUESTION [5 upvotes]: Let $G$ be a finite group of order $240$. If $G\cong C_4\times A_5$ or $C_2\times C_2\times A_5$, then the all degrees of irreducible $\mathbb{C}$-characters of $G$ are $ [1,1,1,1,~3,3,3,3,3,3,3,3, ~4,4,4,4,~5,5,5,5 ]. $ Conversely, Suppose that $G$ is non-solvable, and the all degrees of irreducible $\mathbb{C}$-characters are $ [1,1,1,1,~3,3,3,3,3,3,3,3, ~4,4,4,4,~5,5,5,5 ]. $ Question: $G\cong C_4\times A_5$ or $C_2\times C_2\times A_5$? REPLY [8 votes]: $G$ is non-solvable, so must have $A_{5}$ as a composition factor (as no other non-Abelian simple group has less than $168$). Hence $F(G)$ can have order at most $4$. If $G$ has no component, then $F(G) = F^{\ast}(G)$ has order $4$, and is centralized by all elements of order $5$, contrary to $C_{G}(F^{\ast}(G)) \leq F^{\ast}(G).$ Hence $G$ has a component (quasi-simple subnormal subgroup). This is isomorphic to either $A_{5}$ or ${\rm SL}(2,5).$ If it is (isomorphic to) ${\rm SL}(2,5),$ it is then normal (it has index two), and ${\rm SL}(2,5)$ has an irreducible character of degree $6$. By Clifford's theorem, $G$ has an irreducible character of degree at least $6,$ which is precluded here. Hence $G$ has a normal subgroup $L$ isomorphic to $A_{5}.$ Each irreducible character of degree $3$ of $A_{5}$ must extend to an irreducible character of degree $3$ of $G$ (since $G$ has no irreducible character of degree $6$ or more). Hence no element of order $5$ in $L$ can be conjugate to all its non-identity powers. Let $S$ be a Sylow $5$-subgroup of $L$. Then by a Frattini argument, $G = LN_{G}(S) = LC_{G}(S)$ (since $[N_{G}(S):C_{G}(S)] = [N_{L}(S):C_{L}(S)] = 2).$ Now $G$ induces inner automorphisms on $F^{\ast}(G) = L \times O_{2}(G)$ so $G = F^{\ast}(G)$ has one of the listed structures.<|endoftext|> TITLE: Prove that the matrix $[\Gamma(\lambda_{i}+\mu_{j})]$ is nonsingular QUESTION [10 upvotes]: Let A is a $n\times n$ matrix given by \begin{align*} a_{ij} = [\Gamma(\lambda_{i}+\mu_{j})] \end{align*} where $0 < \lambda_{1} < \ldots < \lambda_{n}$ and $0 < \mu_{1} < \ldots < \mu_{n}$ are real positive numbers and $\Gamma$ denotes the Gamma function given by $\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt$ \, for Re(z)>0. We need to show that matrix A is non-singular. I have no idea how to start. Any hint or solution will be appreciated. Thanks in advance. REPLY [5 votes]: This is merely a reformulation of Abdelmalek Abdesselam's answer, in a somewhat different language and with different references. It should be a comment to that answer, but it is unfortunately too long. Long story short: see Karlin, Total positivity, formula (2.10) in Section 1.2, with $u(t) = t$ and $\sigma(dt) = \mathbb{1}_{(0, \infty)}(t) t^{-1} e^{-t} dt$. The kernel $K(x,y)$ is said to be totally positive on $X \times Y$, where $X, Y \subseteq \mathbb{R}$, if $$ K\pmatrix{x_1&x_2&\cdots&x_n\\y_1&y_2&\cdots&y_n} := \det \left|\matrix{K(x_1,y_1)&K(x_1,y_2)&\cdots&K(x_1,y_n)\\K(x_2,y_1)&K(x_2,y_2)&\cdots&K(x_2,y_n)\\\vdots&\vdots&&\vdots\\K(x_n,y_1)&K(x_n,y_2)&\cdots&K(x_n,y_n)\\}\right| \geqslant 0 $$ whenever $x_1 < x_2 < \ldots < x_n$ and $y_1 < y_2 < \ldots < y_n$ (and, of course, $x_1, x_2, \ldots, x_n \in X$, $y_1, y_2, \ldots, y_n \in Y$). It is strictly totally positive if strict inequality holds. A standard reference for totally positive kernels is Karlin's book Total positivity (Stanford, 1968). Our goal is thus to prove that the kernel $G(\mu,\nu) = \Gamma(\mu + \nu)$ is strictly totally positive. It is known that the kernel $e^{x y}$ is strictly totally positive on $\mathbb{R} \times \mathbb{R}$; see, for example, Example (i) in Section 2.1 of Karlin's book. Substituting $t = e^y$, we see that $K(x, t) = t^x$ and $\check{K}(t, x) = t^x$ are strictly totally positive on $\mathbb{R} \times (0, \infty)$ and $(0, \infty) \times \mathbb{R}$, respectively. Define $\sigma(dt) = t^{-1} e^{-t} dt$ on $(0, \infty)$. Observe that $$ G(\mu,\nu) = \Gamma(\mu + \nu) = \int_0^\infty t^{\mu + \nu - 1} e^{-t} dt = \int_0^\infty K(\mu, t) \check{K}(t, \nu) \sigma(dt) . $$ The basic composition formula (as it is called by Karlin, see (2.5) in Section 1.2 in his book; Karlin's reference for this formula is problem 68 in Pólya and Szegő, Aufgaben und Lehrsdtze aus der Analysis, vol. 1) tells us that $$ \begin{aligned} & G\pmatrix{\mu_1&\mu_2&\cdots&\mu_n\\\nu_1&\nu_2&\cdots&\nu_n} = \idotsint\limits_{0 TITLE: Understanding two proofs in Dwyer and Kan article "Simplicial Localizations" QUESTION [6 upvotes]: I can't understand the proofs of propositions 2.6 and 4.2 in https://www3.nd.edu/~wgd/Dvi/SimplicialLocalizations.pdf We have a category $C$ and a family of maps $W$, and we define the standard resolution $F_*C$ of $C$ as in definition 2.5 and the simplicial localization $LC$ of $C$ with respect to $W$ as in definition 4.1. There is then a natural map $\varphi_*:F_*C \rightarrow C$ defined in 2.5. Proposition 2.6 states "the map $\varphi_*:F_*C \rightarrow C$ is a weak homotopy equivalence". Proposition 4.2 states "$\pi_0LC=C[W^{-1}] $". In 2.6 the proof they suggest involves a "contracting homotopy" which I can't define. How can I write this explicitly? And how is it related to the weak equivalence of categories? In 4.2 they give no indication and I tried to extend the equivalence in 2.6 to an equivalence between the simplicial localization and the ordinary localization, in order to obtain the result simply applying $\pi_0$, but I don't know how to write it down formally. REPLY [4 votes]: It's not so bad to prove 4.2 directly in terms of generators and relations. Dwyer and Kan define $LC = F_\ast C [F_\ast W^{-1}]$ where $F_n$ is the $(n+1)$st iteration of the free category comonad. So $\pi_0 LC (X,Y) = F_0 C [ F_0 W^{-1}] / F_1 C [ F_1 W^{-1}]$. Now, $F_0 C[F_0 W^{-1}]$ is the set of paths in $C$ which can go backward along maps in $W$, modulo the equivalence relation identifying $w \bar w = 1$ and $\bar w w = 1$ where $w \in W$ and $\bar w$ denotes $w$ going backwards. There is an obvious map from here to $C[W^{-1}](X,Y)$, so we just have to check that the equivalence relation imposed by the maps from $F_1 C[F_1 W^{-1}]$ is the same as the equivalence relation defining $C[W^{-1}](X,Y)$. The elements of $F_1 C[F_1 W^{-1}](X,Y)$ are like those of $F_0 C[F_0 W^{-1}](X,Y)$, but with an extra layer of parentheses thrown in -- the two maps to $F_0 C[F_0 W^{-1}](X,Y)$ are the map which removes the parentheses, and the map which takes composites within parentheses. These simply impose the relations saying that a path in $C$ is equivalent to its composite, and a backwards path in $W$ is equivalent to its composite, along with concatenations of these rules with each other. These are precisely the relations that hold in $C[W^{-1}](X,Y)$ (in addition to the identifications already made in $F_0 C[F_0 W^{-1}](X,Y)$), so the map $\pi_0 LC(X,Y) \to C[W^{-1}](X,Y)$ is indeed a bijection.<|endoftext|> TITLE: Cohen generics over the ground model still Cohen over other generic extensions? QUESTION [6 upvotes]: Let $M$ be a countable transitive model of (enough of) ZFC. I'm looking for notions of forcing $\mathbb{P}$ such that if $G$ is $M$-generic for $\mathbb{P}$, then $c$ is a Cohen real over $M$ if and only if $c$ is Cohen real over $M[G]$. Ideally, $\mathbb{P}$ should be a "typical" poset for adding reals in an obvious way, e.g., $\mathbb{P}$ is equivalent to the quotient of the Borel sets in $\mathbb{R}$ by some non-trivial $\sigma$-ideal (in $M$). One such example is Sacks forcing, see Todd Eisworth's answer here: Is a model of set theory determined by the Cohen reals over it? Are there other examples? REPLY [7 votes]: Another example of a forcing like this is the forcing to add an infinitely equal real. I'll sketch the proof, although it is quite similar to the one for Sacks forcing. Recall that conditions in the poset $\mathbb{P}$ are partial functions $p:\omega\to\omega$ with coinfinite domains and satisfying $p(n)\leq 2^n$, ordered by inclusion. We want to show that any dense open set in the extension by $\mathbb{P}$ contains a dense open set from the ground model (the conclusion about preserving Cohenness then follows). So fix a condition $p$ and a name $\dot{D}$ for a dense open subset of $2^{<\omega}$. We will find a function $f:2^{<\omega}\to 2^{<\omega}$ in $V$, satisfying $t\subseteq f(t)$, and a condition $q\leq p$ such that $q\Vdash \mathrm{ran}(f)\subseteq\dot{D}$. The condition $q$ will be the fusion of a sequence of conditions $p=p_0\geq_0 p_1\geq_1 p_2\geq_2\dots$, where $r\geq_n t$ means that $r\geq t$ and they have the same first $n$ many points missing from their domains. We construct the function $f$ alongside and we will ensure that $p_{n+1}$ forces that $f$ maps the $n$th level of the tree into $\dot{D}$. So suppose we are given $p_n$. There are only finitely many conditions $\bar{p}_n^j\leq p_n$ where we have only filled in the first $n$ many blanks in $p_n$. Running through all of these and then finally putting those blanks back, we find a condition $p_{n+1}\leq_n p_n$ and function values $f(s)$ for $s$ of length $n$ such that any extension of $p_{n+1}$ obtained by filling in the first $n$ blanks forces that $f$ maps the $n$th level into $\dot{D}$. It follows that $p_{n+1}$ forces the same thing. Both this and the Sacks example work by showing that every new dense open set of Cohen conditions contains an old dense set (even dense open), and this suffices as mentioned by Joel in the comments. One can also look at the sets of reals directly and restate this as: every new open dense (or comeager) set contains an old coded comeager set. In other words, the meager ideal of the extension is generated by the old meager ideal. The following is Lemma 6.3.21 from the Bartoszyński--Judah book. Theorem: Let $\mathbb{P}$ be a proper poset. The following are equivalent: The meager ideal of the extension by $\mathbb{P}$ is generated by the meager ideal of the ground model. $\mathbb{P}$ is $\omega^\omega$-bounding and for every suitable countable elementary substructure $X$ and every Cohen real $c$ over $X$, any condition $p\in X\cap\mathbb{P}$ can be extended to an $X$-master condition which forces that $c$ remains Cohen over $X[\dot{G}]$. So, if $\mathbb{P}$ is a proper $\omega^\omega$-bounding forcing which doesn't satisfy (1), we can find a model $X$ (or its collapsed version $M$), a Cohen real $c$ over $X$ and a $P$-generic $G$ over $X$ such that $c$ is not Cohen over $X[G]$. In this sense, (1) is a necessary condition for this class of forcings to preserve Cohenness over arbitrary countable models.<|endoftext|> TITLE: Geometric/combinatorial depiction of algebraic identity? QUESTION [17 upvotes]: I'm looking for a geometric or combinatorial depiction of the algebraic identity $$ xyz = \frac{1}{24} \Big\{(x+y+z)^3 - (x-y+z)^3 - (x+y-z)^3 + (x-y-z)^3 \Big\}. \label{*}\tag{$*$} $$ Here is the kind of thing I'd like. For the simpler identity $xy = \frac{1}{4} \big\{(x+y)^2 - (x-y)^2 \big\}$ we can rearrange to $(x+y)^2 = (x-y)^2 + 4xy$. Now, if $x>y>0$, we can take a square with side length $x-y$, and $4$ rectangles of size $x \times y$, and put them together to make a square of side length $x+y$. Just put the little square in the middle and the rectangles around its sides. My idea was to rearrange $\eqref{*}$ into $$ (x+y+z)^3 = (-x+y+z)^3 + (x-y+z)^3 + (x+y-z)^3 + 24xyz . $$ Then, suppose $x,y,z>0$ and they satisfy triangle inequalities. Now three cubes of edge lengths $-x+y+z$, $x-y+z$, and $x+y-z$, plus $24$ "bricks" of size $x \times y \times z$, have the same volume as a cube of edge length $x+y+z$. Unfortunately it's not generally possible to stack the 3 little cubes plus $24$ bricks into a big cube. (Try $(x,y,z)=(11,13,17)$. The only way to get the right areas of faces of the big cube is for each face of the big cube to have exactly one face of a little cube, plus $4$ faces of bricks. And the little cubes have to be centered on the big cube faces; they can't be in the corners or the middles of the edges. But there are $6$ big cube faces and only $3$ little cubes.) This is a bit open-ended, but can anyone suggest a different way to illustrate the identity, especially if it can be depicted in a graphic? Maybe a different algebraic rearrangement of $\eqref{*}$, or another shape besides cubes? REPLY [6 votes]: This shows the identity $$(a+b+c)^3 = a^3 + b^3 + c^3 + 3a(a+b)(b+c) + 3c(b+c)(a+b)$$ which builds on Jairo’s answer. Each summand represents a block in the cube, e.g. $a(a+b)(b+c)$ represents the block $(0,a) \times (0,a+b) \times (a,a+b+c)$, and multiplication by 3 represents cyclic permutation through the axes. The cube is a sum of 9 smaller blocks, though two are not visible in the drawing.<|endoftext|> TITLE: Can one compute the fundamental group of a complex variety? Other topological invariants? QUESTION [19 upvotes]: Given a system of polynomial equations with rational coefficients, is there an algorithm to compute the geometric fundamental group of the variety defined by these equations? I'm interested in both the affine case and the projective case. By "compute" I mean "express in terms of generators and relations." Can one at least compute the unipotent or profinite completion in some manner? Some possible approaches: If the equations define a smooth, projective variety, then one can compute the Hodge cohomology, which tells you the rational de Rham cohomology. By formality, one can use the rational de Rham cohomology to compute the unipotent completion of the fundamental group. Another approach to computing the cohomology is by looking at numbers of points over various finite fields and using the Weil conjectures. Is there, for example, a way to get a CW complex homotopy equivalent to the complex points of the variety? REPLY [11 votes]: There are algorithms to compute triangulations of real algebraic varieties, and thus of complex algebraic varieties. Sadly, the algorithms tend to be doubly exponential in the size of the input data. For more, see: Basu, Saugata, Algorithms in real algebraic geometry: a survey, ZBL06843372. In fact, I don't think this is particularly new, so you can also look at the very nice book: Basu, Saugata; Pollack, Richard; Roy, Marie-Françoise, Algorithms in real algebraic geometry, Algorithms and Computation in Mathematics 10. Berlin: Springer (ISBN 3-540-33098-4/hbk). x, 662 p. (2006). ZBL1102.14041.<|endoftext|> TITLE: A reference to infinite version of the Sunflower Lemma QUESTION [9 upvotes]: Please help me to find a proper reference to the following infinite version of the Sunflower Lemma. Lemma. Let $n\in\mathbb N$. Every infinite family of $n$-element sets contains an infinite subfamily $\mathcal F$ such that $A\cap B=\bigcap\mathcal F$ for any distinct sets $A,B\in\mathcal F$. Browsing through Internet I could find only finite and uncountable versions of the Sunflower Lemma (but not the countable one). REPLY [13 votes]: I found it in the book Komjáth, Péter; Totik, Vilmos, Problems and theorems in classical set theory, Problem Books in Mathematics. New York, NY: Springer (ISBN 0-387-30293-X/hbk). xii, 514 p. (2006). ZBL1103.03041. It's stated on p. 107 as the first item in the chapter on $\Delta$-systems: An infinite family of $n$-element sets ($n\lt\omega$) includes an infinite $\Delta$-subfamily. A proof is given on p. 421. No reference is given.<|endoftext|> TITLE: Model category of diagrams with the colimit detecting the weak equivalences QUESTION [5 upvotes]: Let $I$ be a small category and $\mathcal{K}$ be a combinatorial model category. Is it known a model category structure on the functor category $\mathcal{K}^I$ such that a map of diagrams $D\to E$ is a weak equivalence if and only if $\mathrm{colim} D \to \mathrm{colim} E$ is a weak equivalence of $\mathcal{K}$ ? (so not the objectwise weak equivalences, and by $\mathrm{colim}$, I mean the colimit) I have no trace of a thing like that in the nLab or in the MathReview. I don't know what keyword to use in fact. Since there is no reason for a colimit of and objectwise weak equivalence to be a weak equivalence, it is not possible to see it as a localization of the projective or the injective model category structure. REPLY [3 votes]: Following the suggestion of Mike Shulman to check the condition for existence of the left-induced model structure: Under the additional assumptions that $I$ be connected and that $I$-colimits commute with pull-backs, the so-called left-acyclicity condition is indeed fulfilled. I just did the exercise figuring out some assumptions we have to impose so that it becomes really easy. Presumably, they can be weakened quite considerably, but I will leave this to others. (E.g., a very particular kind of pull-back is enough, which is more likely to commute with colimits, I guess.) Let us call a morphism $\varphi$ in $K^I$ a cofibration (resp. a weak equivalence) if $\mathrm{colim}(\varphi)$ is a cofibration (resp. a weak equivalence) in $K$. From Garner, Kedziorek, Riehl, Lifting accessible model structures, arXiv:1802.09889, Corollary 2.7, and the assumption that the model category $K$ is combinatorial (or more generally, accessible) it follows that we can left-induce the model structure along $\mathrm{colim}\colon K^I\to K$ if and only if the left-acyclicity condition holds: If a morphism $\varphi$ in $K^I$ satisfies the right-lifting property against all cofibrations in $K^I$, then it is a weak equivalence. Let $\varphi\colon A\to B$ be a morphism in $K^I$ satisfying the right-lifting property against all cofibrations in $K^I$. We have to show that it is a weak equivalence. That is, we have to show that $\mathrm{colim}(\varphi)\colon \mathrm{colim}(A)\to\mathrm{colim}(B)$ is a weak equivalence in $K$. But for this, it is enough that $\mathrm{colim}(\varphi)$ satisfies the right-lifting property agains all cofibrations in $K$. So, let $\gamma\colon c\to d$ be a cofibration in $K$ and suppose we are given any pair of morphisms $c\to \mathrm{colim}(A)$ and $d\to \mathrm{colim}(B)$ such that the following diagram commutes. $$\require{AMScd}\begin{CD} c @>>> \mathrm{colim}(A)\\ @V\gamma VV @VV\mathrm{colim}(\varphi)V\\ d @>>> \mathrm{colim}(B) \end{CD}\quad\quad(\ast)$$ I will denote the constant diagram-functor $K\to K^I$ by underlining. Consider $C := \underline c\times_{\underline{\mathrm{colim}(A)}}A$ and $D := \underline d\times_{\underline{\mathrm{colim}(B)}}B$, where $A\to \underline{\mathrm{colim}(A)}$ and $B\to \underline{\mathrm{colim}(B)}$ are the universal morphisms. Then we get a commutative square $$\begin{CD} C @>>> A\\ @V\underline{\gamma}\times\varphi VV @VV\varphi V\\ D @>>> B \end{CD}\quad\quad(\ast\ast)$$ in $K^I$, where the horizontal morphisms are the natural projections. Now, assuming that $\mathrm{colim}(\underline c\times_{\underline{\mathrm{colim}(A)}}A) = c$, naturally in $c$ and $A$, (e.g., if $I$ is connected and $I$-colimits commute with pull-backs,) this commutative square gets mapped to $(\ast)$ under $\mathrm{colim}$. In particular, $\underline{\gamma}\times\varphi$ is a cofibration. Thus, $\mathrm{colim}(\varphi)$ satisfies the right-lifting property agains all cofibrations as soon as $\varphi$ does, but then $\varphi$ is a weak equivalence, as claimed. I think through a refinement of the argument, one can get rid of the connectedness-assumption. But I don't know whether this really depends on some sort of continuity.<|endoftext|> TITLE: Understanding what it means to be ''of general type'' QUESTION [9 upvotes]: I'm attempting to understand the Bombieri-Lang Conjecture: If $X$ is a smooth projective variety of general type defined over a number field, then the set of rational points of $X$ is not dense. I don't understand what it means for a variety to be ''of general type''. I know it's when the variety's Kodaira dimension is maximal, but this doesn't mean much to me. Is there an equivalent condition, or more intuitive way to visualise Kodaira dimension? REPLY [9 votes]: $X$ is of general type iff it is birational to its canonical model $X^c={\rm Proj}(\oplus _{m\geq 0}H^0(mK_X))$. Here $X^c$ has canonical singularities and $mK_{X^c}$ is a very ample Cartier divisor for some $m>0$. Thus there is an embedding $f:X^c\to \mathbb P ^N=|mK_{X^c}|$ and $\omega _{X^c}^{\otimes m}=\mathcal O _{\mathbb P ^N}(1)|_{X^c}$ under this embedding. This is a very useful characterization. One of the main consequences/results about varieties of general type is that (up to birational isomorphism) they have good moduli spaces. Let $v = K_{X^c}^{\dim X}$ be the canonical volume which is given by the top self intersection of the canonical divisor on the canonical model. In any fixed dimension, these volumes belong to a discrete set and for any fixed dimension and fixed volume, canonical models are parametrized by a quasi-projective variety (see https://arxiv.org/abs/1503.02952 for some state of the art results, details and references). Eg if $\dim X =1$, then $v=2g-2$ and the corresponding moduli space has dimension $3g-3$. An important feature is that by the easy addition theorem, $X$ can not be covered by by varieties not of general type, in particular by rational curves or abelian varieties (which tend to have many rational points).<|endoftext|> TITLE: On the Large Cardinal Strength of Normal Moore Space Conjecture QUESTION [9 upvotes]: In his seminal 1937 paper, Jones [1] proved the following result about Moore spaces: Theorem. (Jones) If $2^{\aleph_0}<2^{\aleph_1}$ then all separable normal Moore spaces are metrizable. Then he came up with the idea that maybe the separability condition in the above theorem could be removed. This led him to formulate the following famous conjecture: Normal Moore Space Conjecture (NMSC). All normal Moore spaces are metrizable. Long story short, NMSC turned out to be of high large cardinal strength and so far beyond the power of $ZFC$ to decide. In 1980, Nyikos [2] proved the consistency of NMSC using Product Measure Extension Axiom (PMEA) which is known to be consistent assuming the consistency of strongly compact cardinals. Later a more direct proof of the consistency of NMSC has been presented by Dow et al [3] using the stronger assumption of the consistency of supercompact cardinals. Interestingly, Fleissner [4] used a core model argument to prove the fact that the consistency of NMSC implies the consistency of measurable cardinals as well. Thus NMSC lies between strongly compacts and measurables in the consistency strength order of the large cardinals hierarchy. I wonder whether the direct implication power of NMSC is as strong as its large cardinal strength or not. Question 1. Does NMSC directly imply the existence of any large cardinals, say those of some topological nature such as weakly compacts? To be more precise, let me add that I look for the theorems of the form NMSC $\Rightarrow \exists \kappa$ Large (i.e. at least strongly inaccessible) if there is any. Update. According to Will's answer, it turned out that NMSC doesn't contain any direct large cardinal strength. In other direction, one may strengthen NMSC as follows: Complete Normal Moore Space Conjecture (CNMSC). All normal Moore spaces are completely metrizable. Here is my second question: Question 2. Is CNMSC consistent? What are upper and lower bounds for its large cardinal strength? Strongly compacts and measurables or more? References. Jones, F. Burton, Concerning normal and completely normal spaces, Bull. Am. Math. Soc. 43, 671-677 (1937). ZBL0017.42902. Nyikos, Peter J., A provisional solution to the normal Moore space problem, Proc. Am. Math. Soc. 78, 429-435 (1980). ZBL0446.54030. Dow, Alan; Tall, Franklin D.; Weiss, William A. R., New proofs of the consistency of the normal Moore space conjecture. I & II, Topology Appl. 37, No. 1, 33-51 (1990). ZBL0719.54038. Fleissner, William G., If all normal Moore spaces are metrizable, then there is an inner model with a measurable cardinal, Trans. Am. Math. Soc. 273, 365-373 (1982). ZBL0498.54025. REPLY [14 votes]: If $\text{NMSC}$ is consistent, then so is $\text{NMSC}+\text{"there are no strongly inaccessible cardinals"}$. This is because if $V \models \text{NMSC}$, then $V_\kappa \models \text{NMSC}$ for any inaccessible cardinal $\kappa$. (A proof sketch is given below.) In particular, if $\kappa$ is the first strongly inaccessible cardinal, then $V_\kappa \models \text{NMSC}+\text{"there are no strongly inaccessible cardinals"}$. [If you think through the definition of a Moore space, it's not too hard to see that if $(X,\tau)$ is some topological space in $V_\alpha$, then the statement "$(X,\tau)$ is a normal Moore space" can be decided by looking at $V_{\alpha+\omega}$. Likewise, the fact that $(X,\tau)$ is metrizable can be decided by looking at $V_{\alpha+\omega}$. Thus the truth/falsity of the NMSC for $(X,\tau)$ is the same in $V$ and in $V_\kappa$. Because this is so for every topological space, the truth/falsity of the NMSC is the same in $V$ and in $V_\kappa$.]<|endoftext|> TITLE: If $G$ is a paracompact topological group, then is $G \times G$ paracompact? QUESTION [11 upvotes]: If $G$ is a paracompact topological group, then is $G \times G$ paracompact? This question is raised by Gepner and Henriques (first paragraph of 2.2). Of course, this is not true for arbitrary paracompact spaces, as shown by the Sorgenfrey plane. Actually -- what's an example of a non-paracompact group? REPLY [12 votes]: I am at a topology conference today, and among the many good topologists here is Jan van Mill, a leading expert on topological groups. I ran your question by him, thinking he might know the answer off the top of his head. He did -- the answer is that if $G$ is a paracompact topological group, then $G \times G$ need not be paracompact. The construction of such a group can be found in Yinhe Peng and Liuzhen Wu, "A Lindelöf group with non-normal square" (link) Their main theorem is that there is a Hausdorff group $G$ such that $G$ is Lindelöf but $G \times G$ is not. Every Hausdorff group is regular, every regular Lindelöf space is paracompact, and every Hausdorff paracompact space is normal. Thus their group $G$ is paracompact, while $G \times G$ is not. More information on Peng and Wu's work can be found in these slides of Yinhe Peng's.<|endoftext|> TITLE: What is the coskeleton tower of a quasi-category? QUESTION [14 upvotes]: I was giving a talk in a seminar, and I mistakenly said that the coskeleton tower of a quasi-category was its Postnikov tower. Someone corrected me, but a discussion then ensued about what, precisely, this tower is. It appears to be homotopy-invariant, and each $k$-coskeleton looks like it is somehow related to something along the lines of a weak $(k,1)$-homotopy-category for $k\geq 2$, but the other members of the seminar said that the $(k,1)$-homotopy-category is constructed differently and doesn't seem to be equivalent. What's interesting is this tower seems to converge to $X$ when $X$ is a quasicategory. Has this tower been studied before? Does it have a name? REPLY [13 votes]: It turns out the answer is yes: $k$-coskeletalization of a quasicategory models truncation of an $(\infty,1)$-category to a $(k-1,1)$-category. Let's collect some easy observations. We have an adjunction $sk_k \dashv cosk_k : sSet \to sSet$. $cosk_k$ preserves the property of being a quasicategory, i.e. descends to a functor $cosk_k : qCat \to qCat$. To see this, note that $sk_k(\Lambda^i[n] \to \Delta[n])$ is an isomorphism for $k \leq n-2$, a horn inclusion for $i \geq n$, and $\Lambda^i[n] \to \partial \Delta[n]$ for $k=n-1$, which can be extended to a horn inclusion by postcomposing $\partial \Delta[n] \to \Delta[n]$. So if $X$ is a quasicategory, we can transpose any horn lifting problem for $cosk_k X$ to one for $X$, and use this analysis to solve the lifting problem there. Thinking about this a bit more reveals that $cosk_k$ also preserves inner fibrations between quasicategories -- although not between arbitrary simplicial sets. $cosk_k$ preserves equivalences between quasicategories. Equivalently, $cosk_k$ preserves homotopies between functors between quasicategories. This follows from the fact that the walking isomorphism is 0-coskeletal and $cosk_k$ preserves binary products (in fact, all limits). If $X$ is an $k$-coskeletal quasicategory, then its hom-spaces are $(k-1)$-coskeletal Kan complexes, and in particular they are $(k-1)$-truncated spaces. To see this, use the model $Hom^R_X(x,y)$. For a simplicial set $A$, maps $A \to Hom^R_X(x,y)$ are represented by certain maps $A \ast \Delta[0] \to X$. In particular, a map $\partial \Delta[n-1] \to Hom^R_X(x,y)$ is a map $\partial \Delta[n-1] \ast \Delta[0] \to X$. But there is a degeneracy condition which allows us to fill the last remaining face to obtain a map $\partial \Delta[n] \to X$. In turn this can be uniquely filled (for $n \geq k$ if $X$ is $k$-coskeletal) to obtain a map $\Delta[n] = \Delta[n-1] \ast \Delta [0] \to X$, i.e. a map $\Delta[n-1] \to Hom^R_X(x,y)$. Because $Hom^R_X(x,y)$ is a Kan complex, the fact that it is $(k-1)$-coskeletal implies that it is $(k-1)$-truncated. The map $Hom^R_X(x,y) \to Hom_{cosk_k X}^R(x,y)$ is an isomorphism on $(k-1)$-skeleta. So it is the $(k-1)$-coskeletalization map, i.e. the $(k-1)$-truncation map. An $n$-simplex of $Hom^R_X(x,y)$ is an $(n+1)$-simplex of $X$ satisfying some condition involving certain faces, so it depends only on the $(n+1)$-skeleton of $X$. Since $k$-coskeletalization doesn't change the $(n+1)$-skeleton for $n \leq k-1$, this is clear.<|endoftext|> TITLE: Existence of an antiderivative function on an arbitrary subset of $\mathbb{R}$ QUESTION [18 upvotes]: Let $f:\mathbb{R}\to \mathbb{R}$ be continuous at $x$ for every $x\in I$ where $I\subset \mathbb R$ could be arbitrary. Does there always exist a function $F:\mathbb{R}\to \mathbb{R}$ differentiable on $I$ and $F'(x) = f(x)$ for every $x \in I$? The definition of a primitive is naturally defined on an interval. what sort of weaker result can we obtain under weaker hypotheses?. If I is an interval or an open set, the answer to the question is positive. If f is locally Lebesgue integrable,the answer to the question is also positive. I have already asked the question here https://math.stackexchange.com/questions/2855483/existence-of-an-antiderivative-function-on-an-arbitrary-subset-of-mathbbr REPLY [4 votes]: As requested, I turn my comments into an answer. First of all, we replace $f$ by its lower semi-continuous envelope: $$ g(x) = \liminf_{y \to x} f(x) . $$ Observe that $g(x) = f(x)$ for $x \in I$ and $g$ is continuous at every $x \in I$. Furthermore, $g$ is lower semi-continuous, and hence Borel measurable. If $g$ is continuous at $x$, then there is a neighbourhood $U_x$ of $x$ such that $g$ is bounded in $U_x$. Let $U$ be the union of $U_x$ over all $x \in I$. Consider a connected component $(a, b)$ of $U$. Then $g$ is locally bounded on $(a, b)$ (for any compact subinterval of $(a, b)$ can be covered by finitely many sets $U_x$ with $x \in I$), and thus we can define $$ F(x) = \int_{(a+b)/2}^x g(y) dy $$ for $x \in (a, b)$. Clearly, $F$ is differentiable at every point of continuity $x \in (a, b)$ of $g$, and $F'(x) = g(x)$. In particular, $F'(x) = g(x) = f(x)$ for all $x \in I \cap (a, b)$. We define $F$ as above on every connected component $(a, b)$ of $U$, and we set $F(x) = 0$ for $x \notin U$. By construction, $F'(x) = f(x)$ for every $x \in I \cap U = I$, as desired.<|endoftext|> TITLE: What is the time complexity for solving Diophantine equations of degree 2? QUESTION [5 upvotes]: Manders and Adleman mention that the computational complexity for binary quadratic Diophantine equations is NP-complete. Has a more specific complexity been claimed for polynomials of the form $Axy + Bx + Cy = D$ where the coefficients are nonnegative integers? The only algorithm I have encountered so far is Alpern's method for solving $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, so I use that one and set $A=0, C=0$. I am looking for the most efficient algorithm (or the lowest complexity claim for such an algorithm) for solving this "simpler" form. REPLY [2 votes]: $Axy+Bx+Cy=D$ is $(Ax+C)(Ay+B)=AD+BC$ so it essentially equivalent to factoring $AD+BC$.<|endoftext|> TITLE: Locales as spaces of ideal/imaginary points QUESTION [15 upvotes]: I posted this question on MSE a few days ago, but got no response (despite a bounty). I hope it will get more answers here, but I'm afraid it might not be appropriate as I'm not sure it's actually research-level. Please do tell me if it's not appropriate and if possible tell me how to modify the question so that it may become appropriate (if such modifications exist) I recently saw a video of a presentation of Andrej Bauer here about constructive mathematics; and there are two examples of locales he mentions that strike me : he explains quickly what the space of random reals might be, by saying that it's the locale of reals that are in every measure $1$ subset of $[0,1]$ (for instance): as he says, of course there are no such reals, but that should not stop us from considering the space of these reals, which may have interesting topological properties even if it has no points. Similarly in a constructive setting (or classical setting without AC) some rings may have no prime/maximal ideals, and so their spectrum as usually conceived is uninteresting. But that should not stop us from studying the space of prime/maximal ideals with the Zariski topology, even if it has no points. My questions are related to these examples specifically and to generalizations: Is the first example of random reals in any way connected to the random reals one mentions in forcing ? e.g. is forcing to add some random reals in any way connected to considering the topos of sheaves on the locale of random reals ? Has the second example been extensively studied ? What sort of properties can we get from the study of this "Zariski locale" ? Is there some form of general theory of locales as spaces of imaginary points ? For instance is this how one usually sees locales intuitively; or better is there some actual theory (more than a heuristic) of constructing pointless (or with few points) spaces of objects that we'd like to exist but don't actually exist ? This is very vague so I'll give a further example of what one might envision: if two first-order structures $A$ and $B$ aren't isomorphic but $A\cong_p B$, we might want to study the space of isomorphisms of $A$ and $B$, which would ideally be a pointless locale. One could say something similar about generic filters of a poset when one is trying to do some forcing : from the point of view of the small model, these generic filters don't exist: we could envision a space of generic filters. In these four cases we have some objects that don't exist (random reals, maximal ideals, isomorphisms) but that we can define and that in some very vague sense ought to exist, and so we construct the space of these objects; but it turns out that this space can have no points at all: is there a general theory of this sort of thing ? These questions are very vague so I hope they're appropriate. I'll appreciate answers with references, but I'd also very much like answers that themselves provide some intuition (though a bit more technical than what I've expressed in the question), and some thoughts. REPLY [4 votes]: Here are some comments on intuition. You can think of locales as being analogous to affine schemes and frames as being analogous to commutative rings; from this point of view the existence of locales with few or no points is no more surprising than the existence of affine schemes with few or no global points (morphisms from $\text{Spec } \mathbb{Z}$). In fact we have the following table of analogies: Frame : commutative ring Locale : affine scheme Coproducts / joins : addition Products / meets : multiplication Open of a locale : function on an affine scheme Point of a locale : global point of an affine scheme Sierpinski space : affine line etc. Now, consider the following construction: there is an affine scheme which deserves to be called the "classifying scheme of square roots of $-1$," in that morphisms from $\text{Spec } R$ into this affine scheme correspond to square roots of $-1$ in $R$. Of course this scheme is just $\text{Spec } \mathbb{Z}[x]/(x^2 + 1)$. Note that it has no points, in the sense of no morphisms from $\text{Spec } \mathbb{Z}$, because $-1 \in \mathbb{Z}$ has no square roots. However, the "theory of a square root of $-1$" is nevertheless "consistent," in the sense that $\mathbb{Z}[x]/(x^2 + 1)$ is not the zero ring, and this "theory" does have "models," just in more complicated rings than $\mathbb{Z}$. The quotation marks are meant to emphasize the following analogies between presenting a commutative ring by generators and relations and presenting a locale as the classifying locale of a propositional geometric theory: Propositional geometric theory : collection of variables and polynomial identities between them Model of a theory : collection of elements of a commutative ring satisfying some polynomial identities Classifying locale of a theory : Spec of a commutative ring presented by generators and relations Somewhat more explicitly, when we present a commutative ring by generators and relations as $\mathbb{Z}[x_1, \dots x_n]/(f_1, \dots f_m)$, the resulting affine scheme has the universal property that maps from $\text{Spec } R$ into it correspond exactly to solutions of the system of equations $f_1 = \dots = f_m$ in $R$. This is exactly analogous to the universal property of the classifying locale of a theory.<|endoftext|> TITLE: Any 3-manifold can be realized as the boundary of a 4-manifold QUESTION [7 upvotes]: We know "Any closed, oriented $3$-manifold $M$ is the boundary of some oriented $4$-manifold $B$." See this post: Elegant proof that any closed, oriented 3-manifold is the boundary of some oriented 4-manifold? I heard this statement is true: (1) Any closed 3-manifold is a boundary of some compact 4-manifold. See also this paper p.2's 3rd paragraph uses the fact: (2) Any 3-manifold $M$ can be realized as the boundary of a 4-manifold $B$. In particular, we know that all 3-manifolds can be triangulable. However for 4-manifolds, there are simply connected non-triangulable manifolds (such as the E$_8$ manifold). (Note: a closed 4-manifold is triangulable if and only if it's smoothable.) See this MO post: Not all manifolds can be triangulated (3) For any 3-manifold $M_3$ that can be realized as the boundary of a 4-manifold $B_4$, the $M_3$ must be triangulable. So must the $M_3$ be the boundary of a triangulable 4-manifold $B_4$? (4) Are there any non-triangulable 4-manifold $B_4'$ with a 3-dimensional boundary (i.e. $B_4'$ is not closed)? Then would the 3-manifold boundary $M_3'$ be triangulable (if $M_3'$ is non-triangulable, isn't that leads to a contradiction)? Can one show these (1), (2) and explain them as intuitively as possible? REPLY [3 votes]: In this question it is shown how to see that two smooth manifolds are topologically cobordant if and only if they are smoothly cobordant, which answers some subset of the questions (the second manifold can be $S^3,$ though I am usually abusively think of it as being $\emptyset$), and in arbitrary dimension [the fact that any three-manifold bounds is more elementary and does not require the algebraic machinery].<|endoftext|> TITLE: When does a map of spaces deloop a closed subgroup inclusion? QUESTION [7 upvotes]: I believe Kan showed that any connected CW complex is the delooping of a topological group. I'm interested in the relative question: Question: Let $Y \to X$ be a map of connected CW complexes. Under what conditions does there exists a topological group $G$ and a closed subgroup $H \subseteq G$ such that $BH \to BG$ is homotopy equivalent to $Y \to X$? I'd also be interested in removing the connectedness assumption and asking about realization as the delooping of a closed subgroupoid of a groupoid. I have a sneaking suspicion that any map can be so realized. For instance, if $X = BG$ for $G$ finite, you might think that $H$ has to be a subgroup of $G$, but I think this is not so -- for instance, you could realize various larger groups as subgroups of $G \times \mathbb R^\infty$. If it simplifies things to assume, say, that $X$ deloops a compact group, or even a compact Lie group or a finite group, I'd love to hear about it. Motivation: In equivariant homotopy theory, most concepts are native to homotopy theory, except for closed subgroups and homogeneous spaces. It would be nice to understand their origins purely homotopy-theoretically. REPLY [5 votes]: Actually, this turns out to be easier than I thought, using the nice description of Kan's loop group functor $G$ by Danny Stevenson here. The answer is indeed that any map of connected spaces is the delooping of a closed subgroup inclusion. It suffices to find a simplicial model of $Y \to X$ such that $GY \to GX$ is an injection of simplicial sets. If $X$ is a simplicial set with one vertex, then Stevenson gives the formula $$G(X)_n = \pi_1(Dec_n X / X_{n +1})$$ Here $Dec$ is the decalage construction, with $Dec_n X = X_{\bullet + n +1}$. Now, $Dec_n X$ deformation retracts via an extra degeneracy onto $X_n$, i.e. it is homotopically discrete. The map $X_{n+1} \to Dec_n X$ is a cofibration, so its cofiber $Dec_n X / X_{n+1}$ is a homotopy cofiber. This is the homotopy cofiber of a map of discrete spaces, so we easily compute that $Dec_n X / X_{n+1} = \vee^{X_{n+1} / X_n} S^1$ (where $X_{n+1} / X_n$ just denotes a quotient of sets). Thus $$\pi_1(Dec_n X / X_{n+1}) = \langle X_{n+1} / X_n \rangle$$ is the free group on the set of nondegenerate 1-simplicies of $Dec_n X$. Crucially, this formula is functorial in $X$. Therefore, it suffices to find a simplicial model of $Y \to X$ such that $Y$ and $X$ each have only one vertex. $Y \to X$ is injective. $Y \to X$ sends nondegenerate simplices to nondegenerate simplices. But (1) and (2) are easy to arrange (first obtain (1) by choosing fibrant models for $X$ and $Y$ and then throwing away all but one vertex; then the injective model of $Y \to X$ produced by the small object argument will still have property (1)). Moreover, (2) implies (3).<|endoftext|> TITLE: Statistical independence of eigenvectors of real symmetric Gaussian random matrices QUESTION [5 upvotes]: What is known about the statistical independence of the eigenvectors of a real symmetric matrix with independent Gaussian entries with zero mean, and finite variance? The matrix elements are not assumed to have same variance. I see some results for Wigner matrices in literature, where the entries are i.i.d. standard Gaussian (except diagonal) - though even in this case, whether the eigenvectors are in fact statistically independent is not entirely clear to me (though I suspect that to be the case for Wigner matrices). So, does there exist results regarding statistical independence of eigenvectors of random real symmetric matrices with non-identical, but statistically independent Gaussian entries? Any references for this in literature would be helpful. REPLY [2 votes]: A precise answer exists for the Gaussian Orthogonal Ensemble (all variances the same): then the eigenvectors are the columns of an orthogonal matrix which is uniformly distributed with the Haar measure; they are therefore not independent --- they cannot be because they must be orthogonal to one another. In the limit $n\rightarrow\infty$ of large $n\times n$ matrices $M$, any finite subset of elements of the eigenvectors does become statistically independent [1] with a Gaussian distribution (mean zero, variance $1/n$). This "central limit" result does not require that the elements of $M$ have identical distributions. [1] How Many Entries of a Typical Orthogonal Matrix Can Be Approximated by Independent Normals? (2006).<|endoftext|> TITLE: How to visualize the Frobenius endomorphism? QUESTION [14 upvotes]: As the question title asks for, how do others "visualize" the Frobenius endomorphism? I asked some people in real life and they said they didn't know and that I could go and ask on MO and possibly get miseducated. So that's what I am doing. Bonus points for pictures. REPLY [16 votes]: Probably the best way to think about the Frobenius morphism, and more generally purely inseparable finite morphisms of schemes, is as some sort of foliation. For example, Ekedahl proved the following result: Proposition [Ek, Prop. 2.4]. Let $Y$ be a smooth variety over a perfect field $k$ of characteristic $p > 0$. Then there is a $1$-$1$-correspondence between finite flat height $1$ purely inseparable morphisms $Y \to X$ to a smooth variety $X$ and subbundles $\mathscr E \subseteq \mathcal T_{Y/k}$ stable under Lie brackets and $p$-th powers, associating to $Y \to X$ the subbundle $\mathcal T_{Y/X} \subseteq \mathcal T_{Y/k}$. Here, a purely inseparable morphism $Y \to X$ has height n¹ if and only if there exists a morphism $X \to Y^{(p^n)}$ such that the composition $Y \to X \to Y^{(p^n)}$ is the $p^n$-th power relative Frobenius. Among height $1$ morphisms, the Frobenius $Y \to Y^{(p)}$ is clearly terminal, hence corresponds to the subbundle $\mathcal T_{Y/k}$ (this is also clear from the description of the correspondence). If $Y$ is smooth of dimension $n$ over a perfect field $k$, then the fibres of relative Frobenius $F_{Y/k} \colon Y \to Y^{(p)}$ are isomorphic to the 'thick point' $\operatorname{Spec} k[x_1,\ldots,x_n]/(x_1^p,\ldots,x_n^p)$. What does this look like? So we are to think of purely inseparable morphisms as 'infinitesimal foliations': they define a subbundle of the tangent bundle, and 'integrating' this subbundle should give the 'leaves' of the foliation. Here is a very crude picture: We see that the tangent space of the leaves is the entire space, and they form some sort of thick covering of $Y$, whose fibres are Artin local schemes of length $p^n$ (as above). ¹This seems to be standard terminology. If $Y \to X$ has height $n$, it also has height $m$ for any $m \geq n$. Probably a better definition would be that $Y \to X$ has height $\leq n$. References. [Ek] Ekedahl, Torsten, Foliations and inseparable morphisms. In: Algebraic geometry, Proc. Summer Res. Inst., Brunswick/Maine 1985, Part 2. Proc. Symp. Pure Math. 46, No. 2, 139-149 (1987). ZBL0659.14018.<|endoftext|> TITLE: A balanced tree-like presentation of $S_3$ QUESTION [8 upvotes]: Does the 6-element group $S_3$ have a finite (balanced) semigroup presentation of the form $$\langle a_1,...,a_n\mid a_1=u_1, a_2=u_2,...,a_n=u_n\rangle$$ where $u_1,u_2,...,u_n$ are semigroup words? Let us call such a semigroup presentation {\it tree-like}. Edit: By a semigroup word, I mean a word without inverses of letters. By a semigroup presentation, I mean a presentation in the class of semigroups (not groups). For example, $\langle a,b\mid a^3=a, b=ab^2a\rangle$ defines $S_3$ in the class of groups but a "bigger" semigroup in the class of semigroups. In that semigroup, $b$ does not divide $a$, so it is not a group. Motivation 1. The cyclic group of order $n$ has a tree-like semigroup presentation $\langle a\mid a=a^{n+1}\rangle$. Matt Brin noted that the 8-element quaternion group has the tree-like presentation $\langle a,b,c \mid a=bc, b=ca, c=ab\rangle$. The groups $S_m, m>3$ do not have tree-like presentations because they do not even have balanced (same number of relations and generators) presentations at all since their Schur multipliers are non-trivial. The group $S_3$ has a balanced presentation. Motivation 2. Every tree-like semigroup presentation corresponds to a (closed) subgroup of the R. Thompson group $F$. For example as shown by Guba the Brin tree-like presentation of the quaternion group corresponds to a copy of the Thompson group $F_9$ (the group of piecewise linear homeomorphisms of $[0,1]$ with slopes of the form $9^k$ and break points of the derivative from $\mathbb{Z}[1/9]$). REPLY [7 votes]: Campbell, C. M.; Mitchell, J. D.; Ruškuc, N. On defining groups efficiently without using inverses. Math. Proc. Cambridge Philos. Soc. 133 (2002), no. 1, 31–36 shows among other things that a group with a balanced group presentation has a balanced semigroup presentation. The proof gives a tree like presentation. Look at the proof of Prop 2.4 and 2.5 and note R' and R'' are empty when the group presentation was balanced. See https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/on-defining-groups-efficiently-without-using-inverses/ED8A580F336B47B2162DB5E3A5FA8459<|endoftext|> TITLE: Trivializing unitary cocycles in abelian von Neumann algebras that are uniformly close to the trivial one QUESTION [6 upvotes]: Suppose $M$ is an abelian von Neumann algebra, carrying a (point-ultraweakly) continuous action $G\curvearrowright M$ of a locally compact, second-countable group. Let $y: G\to\cal U(M)$ be an ultraweakly continuous unitary cocycle, i.e., $y_{gh}=y_g\cdot (g.y_h)$ for all $g,h\in G$. Suppose that $y$ is uniformly close to the trivial cocycle, i.e., $$ \sup_{g\in G} \|y_g-1\|=:\varepsilon<1. $$ My Question: Is $y$ always a coboundary? This means $y_g=v\cdot(g.v)^*$ for some $v\in\cal U(M)$. Can one say more about the possible choices for $v$, for example $\|v-1\|\leq C(\varepsilon)$ for some universal constant, where $\lim_{\varepsilon\to 0} C(\varepsilon)=0$? Ideas: I suspect that this should hold for $C(\varepsilon)=\varepsilon(1+\frac{1+\varepsilon}{1-\varepsilon})$. (The block of text below also gives a justification for amenable groups) This would be implied by the following statement: (I guess an addendum to the above question is whether what follows is always true.) (Conjectured) Lemma: Let $S$ be the ultraweak closure of the convex hull of $\{y_g\}_{g\in G}$. Then there exists an element $s\in S$ with $s=y_g\cdot (g.s)$ for all $g\in G$. First let us convince ourselves why this would solve the problem. Indeed, if such an element exists, then $s^*s=g.(s^*s)$ follows immediately for all $g\in G$, i.e., the element $s^*s$ is fixed by the action. Moreover, since $s$ is a limit of convex combinations of the $y_g$, it follows that $\|s-1\|\leq\varepsilon$. Therefore it is invertible and $v=s|s|^{-1}$ is a unitary that trivializes $y$ as desired. Furthermore $$ \|v-1\| = \|s|s|^{-1}-s+s-1\| \leq \|s\| \underbrace{ \||s|^{-1}-1\| }_{\leq \varepsilon/(1-\varepsilon)} + \|s-1\| \leq \frac{(1+\varepsilon)\varepsilon}{1-\varepsilon}+\varepsilon. $$ Proof of the above lemma for amenable groups: (Note: The following argument appears to work in any von Neumann algebra) Suppose $G$ is amenable. Let $\mu$ be the left-invariant Haar measure. Then there is a sequence of (non-zero) functions $f_n\in L^1(G)$ with $0\leq f_n\leq 1$ such that $\displaystyle 0=\lim_{n\to\infty} \max_{g\in K} \frac{\|f_n-g.f_n\|_1}{\|f_n\|_1}$ for all compact sets $K\subseteq G$. Set $$ s_n = \|f_n\|^{-1} \int_G f(g)y_g~d\mu(g) \in M. $$ Evidently $s_n\in S$, and a brief calculation (using the fact that $y$ is a cocycle) shows for every compact set $K\subseteq G$ that $$ \max_{g\in K} \| s_n-y_g\cdot(g\cdot s_n) \| \leq \max_{g\in K} \frac{\|f_n-g.f_n\|_1}{\|f_n\|_1} \stackrel{n\to\infty}{\longrightarrow} 0. $$ Thus we may obtain the desired element $s\in S$ as some cluster point of $(s_n)_n$ in the ultraweak topology. Problem: Although I believe I can generalize the above argument further to the case where the $G$-action is amenable, I am not so happy with this approach. I actually believe that amenability of $G$ might be a red herring for the specific setup of this question, especially due to $M$ being abelian. For example, for some arbitrarily badly-behaved $G$, consider the case where the $G$-action on $M$ is trivial (so highly non-amenable). Then $y$ has a disintegration into a family of group homomorphisms $M\to\mathbb T$, but by assumption these take value in $\{z\mid |z-1|<1\}$, so in fact $y=1$ and the claim trivially holds. This toy example leads me to suspect that I am going at it the wrong way, and that an element $s\in S$ as required by the above lemma may always exist by some more clever Hahn-Banach trickery on compact convex sets. REPLY [7 votes]: With your uniformly closeness assumption, $b(g) := \sqrt{-1}\log y_g$ is a usual additive cocycle (i.e., $b(gh) = b(g) + g\cdot b(h)$) which is moreover real and bounded. I think any bounded cocycle into $L^\infty(X)$ is a coboundary. By exponentiating it, one gets the unitary element $v$. Here's a standard proof for the case $G$ being countable. The general locally compact second countable case will follow by considering a countable dense subgroup. We assume $b$ is real and put $\xi(x) := \sup_{h \in G} b(h)(x)$. It is not difficult to see $\xi$ is invariant under the the affine order-preserving action $\rho_g\colon \zeta \mapsto g\cdot\zeta + b(g)$. Indeed, $$\begin{array}{lcl}(\rho_g\xi)(x) & = & \xi(g^{-1}x) + b(g)(x) \\ & = & \sup_h b(h)(g^{-1}x)+b(g)(x) \\ & = & \sup_h b(gh)(x) \\ & = & \xi(x). \end{array}$$ This means $b(g) = \xi - g\cdot\xi$. In case $G \curvearrowright X$ is a topological minimal action and $y_g \in {\mathcal U}(C(X))$, one can take $v$ to be continuous. For this, see the proof of Theorem 2.6 in [N. S. Ormes, Pacific J. Math. 195 (2000), no. 2, 453–476], which I replicate here. For every $\epsilon>0$, the subset $K := \{ x \in X : \xi(x) \le \epsilon\}$ is closed and $x\in gK \Leftrightarrow \xi(x) \le b(g)(x) + \epsilon$. By definition, $\bigcup_g gK = X$, and so $K$ has non-empty interior $K^\circ$ by the Baire category theorem. One has $X = \bigcup_g gK^\circ$ by minimality and $b(g) \le \xi \le b(g)+\epsilon$ on $gK^\circ$. This proves continuity of $\xi$.<|endoftext|> TITLE: Is there a classification of polynomial Poisson brackets? QUESTION [11 upvotes]: As an example, consider the following Poisson bracket on ${\mathbb R}^n$: $$\{x_i, x_{i+1}\} = x_ix_{i+1}(x_i+x_{i+1}),\\ \{x_i, x_{i+2}\} = x_ix_{i+1}x_{i+2}.$$ The indices are taken modulo $n$, and the "distant" variables commute. The Jacobi identity holds but does not look obvious. This bracket appears in the study of Volterra lattice or discrete KdV. Is there a classification of Poisson brackets whose values on coordinate functions are given by polynomials in the coordinates? Or are some large families of such brackets known? What if we require that the cyclic permutation of coordinates is a Poisson map, as in the above example? REPLY [2 votes]: This is a late partial answer, but I agree with the previous answer in that I don't think there are many general classifications of polynomial Poisson structures available. The only classification for higher-order Poisson structures that I have been able to find is the case of (homogeneous) quadratic Poisson structures in dimension 3 due to Dufour and Haraki (MR1086519, available here), which are classified into 14 families of Poisson structures. This suggests that a classification in higher degrees or in higher dimensions will be quite involved.<|endoftext|> TITLE: Current vs Varifold QUESTION [10 upvotes]: I know the basic definitions concerning current and varifold, and they are generalization of submanifolds. What are their respective pros and cons? What are their crucial similarities and differences? REPLY [8 votes]: A comparison of current versus varifold representations of images that discusses their pros and cons: Current- and Varifold-Based Registration of Lung Vessel and Airway Trees Varifolds are weaker objects than currents due to the lack of orientation of the tangent vector of the momenta used to represent a shape. For some objects where the orientation matters, such as velocity fields, only current representations are acceptable. For other objects, such as the airway trees studied in this paper, the orientation is not essential and varifolds increase the robustness when matching line segments with uncertain tangent orientation. When matching a target to a template an error in the orientation of the tangent fails to produce a good match for the current representation (left panel), while the undetermined orientation of the varifold representation does produce a good match (right panel). Image matching results using current and varifold representations. The circles show the branches that had opposite tangent orientations between the template and the target. The current-based registration fails (left panel) while the varifold-based registration succeeds (right panel).<|endoftext|> TITLE: Non-triangulable 4-manifold as a boundary of some 5 manifold QUESTION [10 upvotes]: We know that there are non-triangulable 4-manifolds, such as the E$_8$ manifold. Can E$_8$ manifold be a boundary of some 5-manifold $M_5$? Can such a $M_5$ be triangulable or non-triangulable? What are the possible $M_5$ (s)? I wonder whether the non-triangulable 4-manifold can always be a boundary of some 5-manifold $N_5$? Then, can such a $N_5$ be triangulable or non-triangulable? See also some background Proofs of Rohlin's theorem (an oriented 4-manifold with zero signature bounds a 5-manifold), Any 3-manifold can be realized as the boundary of a 4-manifold, Not all manifolds can be triangulated: In which dimensions? REPLY [15 votes]: Your questions are answered by Hsu in his paper 4-Dimensional Topological Bordism. In particular, associated to any closed oriented topological 4-manifold $X$ is a signature $\sigma(X) \in \Bbb Z$ and the Kirby-Siebenman class $\text{ks}(X) \in H^4(X;\Bbb Z/2) = \Bbb Z/2$. Hsu proves that these two invariants precisely classify 4-dimensional oriented topological manifolds up to bordism (and in particular $\text{ks}$ is an invariant of topological bordism). A 4-manifold is triangulable if and only if it is smoothable. (This is true but not obvious; it invokes the 3D Poincare conjecture) If $X$ is smoothable then $\text{ks}(X) = 0$ (but the converse is not necessarily true). The signature is an expected obstruction to finding a null-bordism, but Kirby-Siebenmann is less obvious. There is a manifold $F\Bbb{CP}^2$ ($F$ standing for Fake) which is homotopy equivalent to $\Bbb{CP}^2$ but not homeomorphic; it has $\text{ks} = 1$ and signature 1, and so is not smoothable. In particular, $F\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$ has signature 0 but nonzero Kirby-Siebenmann invariant, and thus by Hsu's results it is not null-bordant. There is a formula for spin 4-manifolds: $\text{ks}(X) = \sigma(X)/8 \pmod{2}$. In particular, $\text{ks}(X_{E8}) = \sigma(E8)/8 = 1$. Thus the $E8$-manifold is not null-bordant. However, there are many that are: for instance, take $X_{E8} \# X_{E8}$. This is still not smoothable by Donaldson's theorem (the intersection form of a smooth 4-manifold, if positive/negative definite, is diagonalizable; $E8 \oplus E8$ is not). $\text{ks}$ is additive, and so $\text{ks}(X_{E8} \# X_{E8}) = 0$, and this manifold is null-bordant. As mentioned by Jim Conant in the comments above, because $X_{E8} \# X_{E8}$ is not triangulable, neither is the null-bordism.<|endoftext|> TITLE: p-torsion in Picard group of simply-connected variety of char p QUESTION [5 upvotes]: Let $k$ be an algebraically closed field of char $p>0$ and $X$ be a proper smooth variety over $k$ that is simply connected. Then we know $Pic(X)$ does not contain any $\ell$-torsion for $\ell \not= p$, so Picard variety $Pic^0$ is trivial and $Pic(X)$ is a finitely generated abelian group. Do we have $Pic(X)[p]=0$ as well? What if $X$ is only assumed to be tamely simple connected? REPLY [7 votes]: What follows is more relevant to TKe's question in the comments than it is to the original question. Regarding the original question, I believe that Enriques surfaces in characteristic 2 provide the simplest and best-studied example of this phenomenon. However, this phenomenon does happen in every characteristic. For every prime integer $p\geq 5$, let $k$ be an algebraically closed field of characteristic $p$, and let $\mu_{p,k}$ denote the $k$-group scheme $\text{Spec}(A)$ with multiplication map $m$ where $$A:= k[t]/\langle t^p - 1 \rangle, \ \ m^*(t) = t\otimes t\in A\otimes_k A.$$ Denote by $W$ the projective $k$-scheme $\text{Proj}(B)$ with an $\mathcal{O}(1)$-linearized action $s$ of $\mu_{p,k}$ where $$B=k[x_0,x_1,\dots,x_{p-1}], \ \ s^*(x_\ell) = t^\ell x_\ell.$$ As a $k$-scheme, $W$ is projective space of dimension $p-1$. The action $s$ is free on the complement of the $p$ coordinate points. Denote by $q:W\to V$ the geometric quotient of the action $s$. This morphism is flat away from the images of the coordinate points. Denote by $Z\subset V$ the image of the set of the coordinate points, and denote by $U$ the open complement of $Z$ in $V$. As the target of a flat $k$-morphism from a smooth $k$-scheme, also $U$ is a smooth $k$-scheme. By SGA 2, for a sufficiently ample, sufficiently general hypersurface $T$ in $V$ that is disjoint from the finite scheme $Z$, the restriction map on fundamental groups, resp. Picard groups, from $V$ to $T$ is an isomorphism, and the same holds for the restriction map from $W$ to $q^{-1}(T)$. Since $W$ is simply connected, also $q^{-1}(T)$ is simply connected. Thus, every finite, étale $T$-scheme, say $\nu:T'\to T$, admits a $T$-morphism from $q^{-1}(T)$. If $T'$ is integral, then the induced field extension $k(T)\to k(T')$ is a subextension of the field extension $k(T)\to k(q^{-1}(T))$. Since this field extension has degree $p$, it follows that $\nu$ is an isomorphism. Therefore $T$ is algebraically simply connected. On the other hand, the standard character of $\mu_{p,k}$, i.e., inclusion of $\mu_{p,k}$ in $\mathbb{G}_{m,k}$, induces a $\mathbb{G}_{m,k}$-torsor on $T$. Geometrically, this is the quotient of $q^{-1}(T)\times_{\text{Spec}(k)}\mathbb{G}_{m,k}$ by the diagonal action of $\mu_{p,k}$. This $\mathbb{G}_{m,k}$-torsor is equivalent to an invertible sheaf. That invertible sheaf is $p$-torsion, since the standard character of $\mu_{p,k}$ is $p$-torsion.<|endoftext|> TITLE: Is every complete Boolean algebra isomorphic to the quotient of a powerset algebra? QUESTION [11 upvotes]: Is every complete Boolean algebra isomorphic to a quotient, as a Boolean algebra, of some powerset algebra $\wp(X)$? It is not true for arbitrary Boolean algebras, see the comments, or see my MathSE question. I am also aware of the Loomis-Sikorski Theorem: Every $\sigma$-complete Boolean algebra is isomorphic to the a quotient $\mathbf{F}/\mathcal{I}$, where $\mathbf{F}$ is a $σ$-field of sets and $\mathcal{I}\subseteq\mathbf{F}$ is a $\sigma$-ideal. So this can be viewed as asking whether there is an equivalent theorem for complete Boolean algebras. REPLY [20 votes]: This variation of the question comes from the comments on the original question. The question is whether all (complete) BAs are isomorphic in the category BA to a quotient of a powerset algebra. The Sikorski extension theorem guarantees that every complete BA is a quotient of a power set algebra. The Theorem. (Sikorski) Let $A$ be a subalgebra of a Boolean algebra $B$, and let $f:A\to C$ be a homomorphism from $A$ to a complete Boolean algebra $C$. Then $f$ can be extended to a homomorphism $\widehat{f}:B\to C$. Application. Let $A=C$ be a complete Boolean algebra, and represent it as a subalgebra of a power set BA $B={\mathcal P}(X)$. (Use Stone duality or Birkhoff's subdirect representation theorem for this.) The Sikorski theorem guarantees that that the identity function ${\sf id}\colon A=C\to C$ can be extended to a homomorphism $\widehat{\sf{id}}: B={\mathcal P}(X)\to C$. Since $\widehat{\sf id}$ extends the identity function, it is surjective. \\\ This argument shows that any complete BA is in fact a retract of a power set BA. The converse is easily seen to be true, so the class of complete BA's is exactly the class of retracts of power set BA's. Also, from the comments on the original question: There are incomplete BA's that are quotients of power set BA's. E.g. ${\mathcal P}(\omega)/\textrm{fin}$. Infinite free BA's are not quotients of power set algebras.<|endoftext|> TITLE: Equivariant Mobius inversion QUESTION [15 upvotes]: I'll first explain what Mobius inversion says, and then state what I am fairly sure the equivariant version is. I can write out a proof, but I also can't believe this hasn't been done already; this is a request for references to where it has already been done. Ordinary Mobius Inversion Let $P$ be a finite poset with minimal element $0$. Let $u$ be a function $P \to \mathbb{Z}$ and define $v: P \to \mathbb{Z}$ by $v(p) = \sum_{q \geq p} u(q)$. Mobius inversion aims to recover $u(0)$ from the values of $v$. It says that $u(0) = \sum_{q \in P} \mu(q) v(q)$. The function $\mu : P \to \mathbb{Z}$ can be described topologically: Let $(0,q)$ be the poset $\{ r \in P : 0 < r < q \}$ and let $\Delta((0,q))$ be the order complex, which is the simiplicial complex whose faces are totally ordered subsets of $(0,q)$. Then $\mu(q)$ is the reduced ordered characteristic of $\Delta((0,q))$. The equivariant situation Let $P$ be a finite poset with minimal element $0$ and let $G$ be a group acting on $P$. For each $p \in P$, let $U(p)$ be a finite dimensional $\mathbb{C}$-vector space. Define $V(p) : = \bigoplus_{q \geq p} U(p)$, so $V(0) = \bigoplus_p U(p)$. Let $G$ act on $V(0)$, with $g U(p)= U(gp)$. My goal is to recover the class of $U(0)$, in the representation ring $Rep(G)$, from the $V(p)$'s. For $p \in P$, let $G_p$ be the stabilizer of $p$, so $U(p)$ and $V(p)$ are $G_p$-reps. Let $\mu_{eq}(q)$ be the equivariant reduced Euler characteristic of $q$, meaning the sum $\sum (-1)^j [\tilde{H}^j(\Delta((0,q)))]$ computed in the representation ring $Rep(G_q)$. Let $G\backslash P$ be a set of orbit representatives for $G$ acting on $P$. Then I claim that $$[U(0)] = \sum_{q \in G \backslash P} \mathrm{Ind}_{G_q}^G \left[ \mu_{eq}(q) \otimes V(p) \right]$$ in $Rep(G)$. Has anyone seen this before? REPLY [4 votes]: Sami Assaf and I prove this in section 5 of our paper Specht modules decompose as alternating sums of restrictions of Schur modules. It is surprising that we couldn't find a reference!<|endoftext|> TITLE: Relationship between irreducible representations of the Schur covering group and elements of $H^2(G,U(1))$ QUESTION [6 upvotes]: Let $G$ be a finite group and let $D(g)$ be a projective representation of $G$ i.e. \begin{equation} D(g) D(h) = e^{i \omega(g,h)} D(gh) \end{equation} These can be classified by the equivalence relation $\omega(g,h) \sim \omega(g,h)+\theta(g)+ \theta(h) - \theta(gh)$ subject to the condition $\omega(g,h)+\omega(gh,l)-\omega(h,l)-\omega(g,hl) = 0$. The distinct equivalent classes of projective representations are labeled by elements of $H^2(G,U(1))$. I know that for every such finite group, there is atleast one finite covering group C with the property that every projective representation of G can be lifted to an ordinary representation of C [1]. My question is about the relationship between the irreducible representations (irreps) of C and the group $H^2(G,U(1))$. Specifically, is the following statement true?: Every irrep $\Gamma_i$ of C can be associated an element of $\nu \in H^2(G,U(1))$ like $\Gamma^\nu_i$. The group property of $H^2(G,U(1))$ is reflected in the Clebsch-Gordan decomposition of tensor product of irreps of C: \begin{equation} \Gamma^\nu_i \otimes \Gamma^\mu_j \cong \bigoplus_k \Gamma^{\nu+\mu}_k \end{equation} I have noticed that this is true for all cases I have seen when $ H^2(G,U(1)) \cong \mathbb{Z}_2$ like $G = \mathbb{Z}_2 \times \mathbb{Z}_2$, $C = D_8$ but I am unsure if this is true in general. [1] https://en.wikipedia.org/wiki/Schur_multiplier#Relation_to_projective_representations PS: I am a physicist and my interest in the above question comes from its use in condensed matter physics. REPLY [4 votes]: The answer to your question is Yes. Consider your covering group $C$ as a central extension: $$1 \to N \to C \to G \to 1$$ and suppose it is given by a 2-cocycle $\alpha \in H^2(G, N)$. Then for any homomorphism $f: N \to \mathbb{C}^\times$, we get a 2-cocycle $f \circ \alpha \in H^2(G,\mathbb{C}^\times)$ by composition of maps, and this yields a homomorphism $$\operatorname{Hom}(N,\mathbb{C}^\times) \to H^2(G,\mathbb{C}^\times).$$ You are assuming that $C$ is a "representation group", which amounts to the property that this homomorphism is an isomorphism. Thus, any projective representation of $G$ lands in a commutative diagram $$\require{AMScd} \begin{CD} 1 @>>> N @>>> C @>>> G @>>> 1 \\ @. @VVV @VVV @VVV \\ 1 @>>> \mathbb{C}^\times @>>> GL(V) @>>> PGL(V) @>>> 1 \end{CD}$$ The 2-cocycle for your projective representation corresponds to the 1-dimensional representation of $N$ given by the leftmost vertical arrow. Any tensor product of representations of $C$ induces a tensor product of 1-dimensional representations of $N$. Tensor product of 1-dimensional representations of $N$ yields the group law on $\operatorname{Hom}(N,\mathbb{C}^\times)$, and by our assumption, this is the group law on $H^2(G, \mathbb{C}^\times)$. I found all the statements I needed in this exposition by Mendonca.<|endoftext|> TITLE: Obstruction of spin-c structure and the generalized Wu manifods QUESTION [13 upvotes]: Bockstein homomorphim and obstruction of spin-c structure: Let $w_2$ be the Stiefel Whintney class of manifold $M$. Let the Bockstein homomorphim $\beta$ be the $$ H^2(\mathbb{Z}_2,M) \to H^3(\mathbb{Z},M), $$ such that $\beta(w_2)$ is the integral cohomology class. (1) Is this true that for certain dimensions of $M=M^d$, say some dimension $d$, the existence of such a nontrivial $\beta(w_2)$ indicates the obstruction of the spin$^c$ structure of $M$? How do we show this? (2) Wu manifold seems to be a 5-dim manifold with an obstruction of spin-c structure, which is $$ SU(3)/SO(3). $$ Does a more general manifold (for an integer $n$) $$ SU(n)/SO(n) $$ admits spin, or spin-c structures, or do $SU(n)/SO(n)$ have obstructions for them? REPLY [16 votes]: Define the Wu manifold $W(n) = SU(n)/SO(n)$, the inclusion $SO \to SU$ given by thinking of $\Bbb C^n = \Bbb R^n \otimes \Bbb C$ (that is, including real matrices into complex matrices). Note that $W(1) = *$, $W(2) = S^2$, and $W(3)$ is what is usually called the Wu manifold. There is a natural map $W(n) \to W(n+1)$, including $SU(n) \to SU(n+1)$ and then 'quotienting by more'; in fact, because $SO(n+1) \cap SU(n) = SO(n)$, it is injective. A long exact sequence calculation shows that $\pi_1 W(n) = 0$ for all $n$, while $\pi_2 W(n) = \Bbb Z/2$ for $n \geq 3$ and $\Bbb Z$ for $n = 2$; the natural map $\pi_2 W(n) \to \pi_2 W(n+1)$ is a surjection for all $n \geq 2$ and an isomorphism for $n \geq 3$. In particular, we have that the map $H^2(W(n+1);\Bbb Z/2) \to H^2(W(n);\Bbb Z/2)$ is an isomorphism for all $n \geq 2$. To determine if $TW(n+1)$ is spinnable or not (that is, if $w_2$ is zero or not), it suffices to check that of its restriction to $W(n)$. Because $TW(n+1)\big|_{W(n)} = TW(n) \oplus NW(n)$, the cruciual inductive step is calculating $w_2(NW(n))$. Now notice that the map $W(n) \to W(n+1)$ factors through $SU(n+1)/SO(n) =: E(n+1).$ In fact, this fits into a fiber sequence $S^n \to E(n+1) \to W(n+1)$, and so the map $W(n) \to W(n+1)$ comes equipped with a section of this sphere bundle. Allow me to phrase the necessary result in unnecessary generality: given a manifold $M$, a fiber bundle $F \to E \to B$, and an embedding $M \xrightarrow{i_E} E$ that projects to an embedding $i_B: M \to B$, then if we write $T^F E$ for the fiberwise tangent bundle, $N(i_E) = N(i_B) \oplus i_E^* T^FE$. When the fiber bundle is furthermore the unit sphere bundle of a vector bundle $V$ over $B$, then $T^F E \oplus \Bbb R = \pi^*V$. Before continuing, then, we identify the rank $n+1$ vector bundle $V$ giving rise to the sphere bundle $S^n \to SU(n+1)/SO(n) \to SU(n+1)/SO(n+1)$. If $G$ is a topological group with closed subgroup $SO(n)$, then we have a fibration sequence $G/SO(n) \to BSO(n) \to BG$; the first map classifies the $SO(n)$-bundle $G \to G/SO(n)$, and the associated sphere bundle is $G \times_{SO(n)} (SO(n)/SO(n-1)) = G/SO(n-1)$, as desired. In particular, the vector bundle $V$ corresponding to $W(n+1)$ is classified by the fiber inclusion $W(n+1) \to BSO(n+1)$. Running the long exact sequence of homotopy groups of a fibration, we see that this is an isomorphism on $\pi_2$ for all $n$, and in particular, $w_2(V) \neq 0$. Lastly, the normal bundle to $W(n)$ in $E(n+1)$ is trivial: there is an $SU(n)$-equivariant trivialization of the normal bundle to $SU(n)$ in $SU(n+1)$ (using the usual push-around-using-group-structure-argument), which descends to a trivialization of the normal bundle of $SU(n)/SO(n)$ in $SU(n+1)/SO(n)$, as desired. Plugging all this together, we get the formula $\text{triv} = NW(n) \oplus i_E^*\pi^*V = NW(n) \oplus i_B^* V$; and because the map $i_B^*: H^2(W(n+1);\Bbb Z/2) \to H^2(W(n);\Bbb Z/2)$ is an isomorphism for $n+1 \geq 3$, we see that $w_2(NW(n)) = w_2(i_B^*V) \neq 0$ for all $n \geq 2$. Because this is precisely the difference between $w_2 TW(n+1)$ and $w_2 TW(n)$, and $H^2(W;\Bbb Z/2) = \Bbb Z/2$, we see that whether or not $W(n)$ is spin alternates in $n$, for $n \geq 2$. As $W(2) = S^2$, we see that $W(n)$ is spin for even $n$ and fails to be spin for odd $n$. Lastly, because one may verify that $\beta: H^2(W(3);\Bbb Z/2) \to H^3(W(3);\Bbb Z/2)$ is nontrivial by hand using a calculation of $H^*(W(3))$ and the Bockstein long exact sequence, we see via naturality that $\beta$ is nontrivial on the degree 2 class of $W(n)$ for all $n \geq 3$; in particular, $W(2n+1)$ is not even spin^c for $2n+1 \geq 3$.<|endoftext|> TITLE: Poincare duality spaces vs. manifolds via lifting maps, the obstruction theory and the role of simply connectedness QUESTION [11 upvotes]: Suppose that we are given a topological space $X$: assume for simplicity that $X$ is compact we want to adress the following question: Is it true that one can find a manifold $M$ which is homotopy equivalent to $X$? Necessary condition $X$ must satisfy Poincare duality, namely there must be a class in the top homology of $X$ such that the cap product with this class induces isomorphism between homology and cohomology. As one could expect this is not enough. So let us assume that $X$ indeed satisfy Poincare duality. Then one can associate to $X$ the so called Spivak normal fibration. This fibration is classified by the map $f:X \to BG$ where $G$ is a space of self-homotopy equivalences of the sphere. Second assumption: Let us assume that $X$ is simply connected. Then one can find a manifold homotopy equivalent to $X$ which is: -smooth iff the map $f$ lifts to a map $X \to BO$ -piecewise linear iff the map $f$ lifts to a map $X \to BPL$ -topological iff the map $f$ lifts to a map $X \to BTop$. (Here as far as my knowledge goes, $Top$ is the space of homoeomorphisms of the sphere, $PL$ the space of piecewise linear homeomorphisms of the sphere and so on-but please correct me if I'm wrong). Problem of lifting maps leads to the obstruction theory which gives classes in cohomology $H^{n+1}(X,\pi_n(F))$ where $F$ is a fiber of our fibration which we would like to lift (in our case $F$ is equal to $G/O,G/PL,G/Top$ respectively). Question 1 It is general knowledge of the obstruction theory that it works better in the simply connected case: in the non simply connected case one has to deal with local coefficients. In our problem of finding a manifold in the given homotopy type in the non simply connected case there is a further obstruction: is it right to think that the presence of this obstruction is due to the fact that one has to work with local coefficients? Question 2 What is known about the homotopy groups of the fibers $G/O$ (resp. $G/PL$, $G/Top$)? Is it somehow possible to interpret the various theorems of the form ,,up to some dimension/in some dimension two notions of manifolds (e.g. smooth, PL, Top) coincide'' in this language, i.e. as vanishing of the homotopy groups of the fiber up to some dimension? Question 3 Due to the theorem of Sullivan every manifold of dimension different than $4$ has unique Lipschitz atlas: how does it fit into this picture? My question is very broad, it is rather some kind of a big picture: nevertheless I hope that it would be interesting to someone who is less familiar with this theory. REPLY [14 votes]: Let's talk first about the smooth and simply connected case. As you say, Poincare duality for $X$ yields a spherical fibration, or map $X\to BG$. A lifting $X\to BO$ is necessary but not sufficient for a manifold structure. Such a lifting, or "normal structure", determines an element of the surgery obstruction group ($L$-group). A necessary and sufficent condition (assuming the dimension is not too small) for existence of a manifold structure compatible with the given normal structure is the vanishing of this surgery obstruction. If we say PL or topological instead of smooth, then the relevant normal structure is a lift to $BPL$ or $BTop$ instead of $BO$. In the non-simply-connected case there are two important differences. First, the surgery obstruction groups are different. (But they depend only on $\pi_1(X)$ and the orientation character $\pi_1(X)\to \lbrace\pm 1\rbrace$ and the mod $4$ class of the dimension.) Second, in order for any of this to work you need to assume that $X$ satisfies Poincare duality in a strong sense (cap product isomorphisms between cohomology with twisted coefficients and homology with twisted coefficients). In the following sense it is correct to say that the extra complications in the non-simply-connected case are because of local coefficents: In the simply-connected case Poincare duality (say, in the case when the dimension is a multiple of $4$) produces a quadratic form over $\mathbb Z$, and the $L$-groups are what they are because of some stable classification of such forms, and this is a powerful enough invariant because for simply connected spaces homology has a powerful influence on homotopy theory (sorry for the vagueness). But in the general case one needs twisted homology (with coefficients in $\mathbb Z[\pi_1(X)]$-modules) to get that power, so one needs to think about twisted duality and about quadratic forms over $\mathbb Z[\pi_1(X)]$.<|endoftext|> TITLE: What is the best reference for motives? QUESTION [12 upvotes]: I want to learn about homotopy theory on number fields, and I heard that the theory of motives made it possible, so I want to know what is a good textbook for motive theory. To be honest, I don’t know algebraic geometry (I only read Hartshorne). So please tell me what is needed to get some knowledge about Algebraic Geometry (SGA? Or other book?). REPLY [5 votes]: A wealth of information is contained in: Uwe Jannsen, Steven L. Kleiman, Jean Pierre Serre, “Proceedings of Symposia in Pure Mathematics, Vol 55, Parts 1 and 2”. Amer Mathematical Society (February 1, 1994) This is from before the Voevodsky era. But I think you should know about the classical story before plunging into the derived category of motives. See especially the paper on “Classical motives” by Scholl in the above reference. However, since you mention homotopy theory, you certainly should work your way towards $\mathbb{A}^1$-homotopy theory à la Voevodsky and others.<|endoftext|> TITLE: Differential equation changing sign almost everywhere QUESTION [22 upvotes]: Conjecture: Let $f:\mathbb{R}→\mathbb{R}$ be an everywhere differentiable function and assume that $f(x)+f′(x)∈ \{-1,1\}$ almost everywhere and $f'(0)=0$. Then is $f$ necessarily a constant function? Can you give me a counter-example? I have already asked the question here on MathSE. REPLY [38 votes]: Your conjecture is true and there is no counterexample. Suppose, contrary to the above claim, that your conjecture is false. Define $$g(x) = f(x) + \int_0^x f(y) dy,$$ so that $g'(x) = f'(x) + f(x)$. Thus, $g$ is everywhere differentiable, $g'(x) \in \{-1,1\}$ almost everywhere, and $g'$ is not a constant. (For if $g'$ was constant $\pm 1$, we would have $f(x) = \pm 1 + c e^x$, which, together with $f'(0) = 0$, would imply that $f$ is constant). This beautiful result of J.A. Clarkson from 1947 asserts that if $\alpha < \beta$, then $$E(\alpha, \beta) := \{x : g'(x) \in (\alpha, \beta)\}$$ is either empty or it has positive Lebesgue measure. Since $g'$ is not constant, it takes at least two values, and by Darboux's theorem, in fact it takes all values in some interval $(\alpha, \beta)$. Changing this interval to a smaller one if necessary, we may assume that $(\alpha, \beta) \cap \{-1, 1\} = \varnothing$. Since $E(\alpha, \beta)$ is non-empty, it has positive Lebesgue measure, and therefore $g'(x)$ cannot belong to $\{-1, 1\}$ almost everywhere — a contradiction.<|endoftext|> TITLE: A non integrable distribution which is totally geodesic QUESTION [7 upvotes]: Is there a non integrable $2$ dimensional distribution $D$ of a $3$ dimensional Riemannian manifold such that the distribution is totally geodesic in the following sense: Every geodesic whose tangent vector of its intitial point is tangent to the distribution then the tangent vector at all its points is tangent to $D$, too. REPLY [2 votes]: Take $\mathbb{R}^3$ with the distribution which is the kernel of the one-form $dz - y dx$. This is the standard example of a contact structure on $\mathbb{R}^3$. See https://en.wikipedia.org/wiki/Contact_geometry .<|endoftext|> TITLE: Minimize spectral norm under diagonal similarity QUESTION [5 upvotes]: Let $A$ be a real square matrix of size $n \times n$. Is there an upper bound on the minimum spectral norm under diagonal similarity, i.e., $$ s(A) = \min_{D} \lVert D^{-1} A D\rVert_2, $$ where $D$ is a non-singular, diagonal real matrix. Also, is there are a relation between $s(A)$ and the spectral radius $\rho(A)$? For the numerical radius $r(A) = \max_{\lVert x \rVert_2 = 1} \lVert \langle Ax, x\rangle \rVert$, it is $$\rho(A) \leq r(A) \leq \lVert A\rVert_2 \leq 2 r(A).$$ I was hoping that it may be possible to minimize $r(D^{-1} A D)$. REPLY [5 votes]: You cannot get an upper bound in general in terms of the spectral radius $\rho(A)$. Counterexample: if $$ A = \begin{bmatrix} x & 1 \\ -x^2 & -x \end{bmatrix} $$ then $\rho(A) = 0$ and $s(A) = 2|x|$. (This $A$ is essentially the most general $2\times 2$ matrix whose eigenvalues are both $0$.)<|endoftext|> TITLE: Is there a relationship between norms/transfers in equivariant homotopy theory and norms in the Tate construction / ambidexterity? QUESTION [6 upvotes]: In homotopy theory, the word "norm" is commonly used in two different ways (well, surely there are other ways, but these two have a particular familial resemblence). Let $G$ be a finite group. A $G$-spectrum $E$ can be restricted to an $H$-spectrum for any subgroup $H \subseteq G$, and there is an "inclusion of fixed points" map $E^G \to E^H$. If $E$ is "genuine" then there is also a "wrong-way" map in the other direction, called the transfer. There is a similar multiplicative story: if $E$ is a $G$-ring spectrum, then for any map of finite $G$-sets $T \to S$, there is an "inclusion of fixed points" map $(\wedge^S E)^G \to (\wedge^T E)^G$. If the multiplicative structure on $E$ is "genuine" then there is also a "wrong-way" map in the other direction, called the norm. Let $G$ be a finite group acting on a spectrum $E$. Then there is a natural composite map $E^{hG} \to E \to E_{hG}$, but there is also a "wrong-way" map in the other direction, called the norm. If the norm is invertible, the action is ambidextrous, and the obstruction to ambidexterity (i.e. the cofiber $E^{tG}$ of the norm) is the Tate construction. Question: Is there a relationship between these two uses of the word "norm" (or perhaps between "norm" in the second sense and "transfer")? I think this is more than a coincidence of terminology partly because Nikolaus and Scholze use norm data in the second sense to describe cyclotomic spectra, a sort of equivariant spectrum, which data is normally encoded in terms of transfers. But I haven't studied Nikolaus-Scholze closely enough to extract what's going on. REPLY [4 votes]: Here's a bit more of an organized answer: First you have to decide how you'd like to model the notion of a genuine $G$-spectrum. There are at least ways that it's currently in vogue to do this, each of which is convenient for different purposes. (My groups are finite below, and I won't write down functors that aren't homotopically meaningful- so everything is 'derived'.) One could begin with the homotopy theory of $G$-spaces, where weak equivalences are detected by checking on all fixed points and then 'invert' $\Omega^{\rho}$ or $\Sigma^{\rho}$, where $\rho$ is the regular representation. There are lots of ways of doing this, but no matter how you do it you will be able to present any $G$-spectrum as a homotopy colimit $X = \mathrm{hocolim}\, S^{-n\rho}\wedge\Sigma^{\infty}X_{n\rho}$ where the $X_{n\rho}$ are pointed $G$-spaces. This way of thinking about things makes it easy to define symmetric monoidal, accessible functors $\mathrm{Sp}^G \to \mathcal{C}$ since they'll be determined by what they do to suspension spectra. For example, from this point of view you can define geometric fixed points by doing what you expect: commuting with the formation of suspension spectra, commuting with homotopy colimits, and being symmetric monoidal. The other type of fixed points, genuine fixed points, are essentially uniquely determined by the property that the functor $(-)^G: \mathrm{Sp}^G \to \mathrm{Sp}$ is exact and satisfies $\Omega^{\infty}(X^G) = (\Omega^{\infty}X)^G$. (This functor is not symmetric monoidal- for example, it doesn't even send the unit to the right place, by the tom Dieck splitting). Alternatively, one can use that every stable homotopy theory is canonically enriched in spectra, and take the spectrum of maps from $S^0$ to $X$. Now, in this setting I sketched an answer to your question about transfers (the classical answer) which came from the map $\mathrm{E}G_+ \to S^0$. This induces a map $(\mathrm{EG}_+ \wedge X)^G \to X^G$. Now, why is it the case that $(\mathrm{EG}_+ \wedge X)^G$ is the same as the homotopy orbits? (A case of the 'Adams isomorphism'). You can prove this inductively using the standard bar filtration of $\mathrm{EG}_+$, i.e. inductively define a map $(\mathrm{sk}_j\mathrm{E}G_+ \wedge X)_{hG} \to (\mathrm{sk}_j\mathrm{E}G_+ \wedge X)^G$ and show it's an equivalence. The base case and inductive case reduce to the statement that there's a natural equivalence $(T_+ \wedge X)_{hG} \to (T_+ \wedge X)^G$ where $T$ is a finite free $G$-set. In other words, $(T/G)_+ \wedge X = \mathrm{map}_{\mathrm{Sp}}((T/G)_+, X)$ (using the self-duality of finite sets in $\mathrm{Sp}$) is naturally equivalent to $\mathrm{map}_{\mathrm{Sp}^G}(S^0, T_+ \wedge X)$. This, finally, follows because every finite $G$-set is self dual in $\mathrm{Sp}^G$: embed into some $G$-representation and produce a collapse map as in the proof of Atiyah duality to define the duality datum. Another choice is to define the homotopy theory of genuine G-spectra as 'spectral Mackey functors', following Guillou-May and Barwick-Dotto-Glasman-Nardin-Shah, just take the homotopy theory of product preserving functors $\mathrm{Span}(G) \to \mathrm{Sp}$ where the former denotes the $(2,1)$-category of finite $G$-sets with mapping groupoids given by the groupoid of spans. And here I'm working $\infty$-categorically, so these are homotopy coherent functors. The value of the functor on the orbit [G/H] is the H-fixed points. In particular, there are natural inclusions of both the orbit category and its opposite into the category of spans. The map $[G] \to [*]$ and its 'opposite' automatically, by functoriality, gives the maps $X_{hG} \to X^G$ and $X^G \to X^{hG}$. The identification of the composite as the usual norm is then a kind of `double-coset formula', i.e. it follows from the formula for composition of spans coming from pullbacks. Finally, there is the paper of Glasman which codifies and makes precise ideas of Greenlees and Greenlees-May: G-spectra can be described by their collection of geometric fixed points, and gluing data having something to do with Tate spectra. From this point of view it's easiest to say what the transfer maps here are in the case of something like $G=C_p$ (so I don't have to talk about generalized Tate constructions and so on). From this point of view, a $C_p$-spectrum is a triple $(X, X^{\Phi C_p}, X^{\Phi C_p} \to X^{tC_p})$ where $X$ is a Borel $C_p$-spectrum. The genuine fixed points are then defined as the pullback of $X^{hC_p} \to X^{tC_p} \leftarrow X^{\Phi C_p}$. This automatically produces a map $X_{hC_p} \to X^{C_p}$ factoring the 'norm/trace' since the Tate spectrum is the cofiber of the norm/trace. All of the above is well-documented, with references enough to suit any taste. There's Lewis-May-Steinberger, the 'Alaska notes', an online book of Schwede, the first few sections of Hill-Hopkins-Ravenel and their appendices, that paper of Glasman's, etc. etc.<|endoftext|> TITLE: Equidistribution of CM points in the principal genus QUESTION [14 upvotes]: It is well known that as the negative discriminant $-D$ goes to infinity, the number of quadratic forms of discriminant $-D$ belonging to the principal genus also goes to infinity. Can we say anything about the asymptotic equidistribution of the corresponding CM points on the classical modular curve? I know that Duke '88 proved equidistribution for all CM points and that this has since been refined in various ways, but I couldn't find any literature addressing this question. REPLY [10 votes]: Vesselin Dimitrov gave a nice answer, but let me point out that for the OP's question one does not need the result of Harcos-Michel (2006). Instead, the original work of Duke (1988) or alternatively Duke-Friedlander-Iwaniec (1993) suffices. Indeed, one only needs to detect cancellation in Weyl sums twisted by genus class characters, i.e. the real characters of the class group of $Q(\sqrt{-D})$. If $\chi$ is such a character, and $g$ is a Maass cusp form of level 1 (or an Eisenstein series of level 1 participating in the spectrum), then one needs a subconvex estimate $L(\tfrac{1}{2},g\otimes\chi)\ll|D|^c$ for some fixed $c<1/2$, with an implied constant depending polynomially on the Laplace eigenvalue of $g$. As $\chi$ is a genus class character, the $L$-value factors as $L(\tfrac{1}{2},g\otimes(\tfrac{\cdot}{D_1}))L(\tfrac{1}{2},g\otimes(\tfrac{\cdot}{D_2}))$, where $D_1$ and $D_2$ are certain fundamental quadratic discriminants satisfying $D_1D_2=-D$. As a result, the problem reduces to a subconvex bound $L(\tfrac{1}{2},g\otimes(\tfrac{\cdot}{D_i}))\ll|D_i|^c$, and this is covered by the earlier results mentioned above when $g$ is a Maass cusp form, and by Burgess's classical estimate when $g$ is an Eisenstein series. For more details, see these concise notes (along with errata which only concern the equidistribution of closed geodesics on the modular surface).<|endoftext|> TITLE: If $R$ is an etale extension of $\mathbb Z$, then $R = \mathbb Z^n$? QUESTION [10 upvotes]: Related question. Let $A$ be a ring, and let $B$ be an $A$-algebra which is projective and finite as an $A$-module. Then the trace $B \rightarrow A$ can be defined. Let's say that $B$ is separable over $A$ if the trace induces an isomorphism of $A$-modules $B \rightarrow \operatorname{Hom}_A(B,A)$. For the purposes of this question, let's say that $B$ is an etale $A$-algebra if it is finite, projective, and separable. Is it true that every etale $\mathbb Z$-algebra is of the form $\mathbb Z^n$? This is stated in these notes, example 1.1.13(ii). It does not seem so easy to prove. If this is true, then combined with the much easier result, "if $L/K$ is an unramified extension of number fields, then $\mathcal O_L$ is an etale $\mathcal O_K$-algebra," we would obtain an algebraic proof that there are no nontrivial unramified extensions of $\mathbb Q$. I had thought that the only known proofs of this result relied on Minkowski's bound for the discriminant. Perhaps any proof that etale $\mathbb Z$-algebras are of the form $\mathbb Z^n$ must rely on this fact. REPLY [4 votes]: Let $X$ be a normal integral scheme. Then $\pi_1^\mathrm{et}(X)$ is isomorphic to $\mathrm{Gal}(K(X)^\mathrm{un}/K(X))$, where $K(X)^\mathrm{un}$ is the compositum of all finite extensions $F$ of $K(X)$ (in a fixed algebraic closure of $K(X)$) for which $X$ is unramified in $F$. Applying this to $X = \mathrm{Spec}(\mathbf{Z})$, it follows the result you are asking about is equivalent to Minkowski's theorem (that there are no nontrivial unramified extensions of $\mathbf{Q}$).<|endoftext|> TITLE: Numerical minimization spectral norm under diagonal similarity QUESTION [5 upvotes]: This question is a follow up. Let $A$ be a real square matrix of size $n \times n$. How to determine the minimum spectral norm under diagonal similarity, i.e., $$ s(A) = \inf_{D} \lVert D^{-1} A D\rVert_2, $$ where $D$ is a non-singular, diagonal real matrix. As it is unlikely to find an analytical upper bound, I would like to ask how $s(A)$ could be determined numericallly. REPLY [4 votes]: Here is a better, more direct solution. This problem can be cast as a Generalized Eigenvalue Problem as is shown by Boyd, El Ghaoui, Feron, and Balakrishnan on page 39 (§3.3) of Linear Matrix Inequalities in System and Control Theory: $$ s(A) = \inf \left\{\gamma \mid A^*PA < \gamma^2 P \textrm{ for diagonal } P > 0 \right\} $$ Previous answer Unfortunately behind a paywall, but following my own comment about chasing literature on matrix balancing, I found the following old paper that solves your problem (EDIT: As noted by Sebastian, this paper actually only provides a solution for a restricted case), not only for the operator norm, but for a variety of other norms. T. Ström. Minimization of norms and logarithmic norms by diagonal similarities. Computing, March 1972, Volume 10, Issue 1–2, pp 1–7.<|endoftext|> TITLE: Why is every deformation of the universal enveloping algebra of a complex semisimple Lie algebra trivial? QUESTION [13 upvotes]: I have read in these lecture notes that every deformation $U_h(\mathfrak{g})$ of $U(\mathfrak{g})$ is trivial, i.e. isomorphic to $U(\mathfrak{g})[[h]]$ as associative $\mathbb{C}[[h]]$-algebras. Why is this true? The reason they cite is that $HH^2(U(\mathfrak{g}),U(\mathfrak{g}))=0$. Now, I know how to compute Hochschild cohomology of simpler algebras, but several problems arise for me in this setup: I am unsure how to proceed in the computation of $HH^2(U(\mathfrak{g}),U(\mathfrak{g}))$ (or more generally, replacing $U(\mathfrak{g})$ with $T(V)$, the tensor algebra of a vector space). How can I conclude that this is $0$? Why does the "trivial" deformation look like a power series ring? They say that the multiplication in $U_h(\mathfrak{g})$ looks like "multiplication modulo $h$" in $U(\mathfrak{g})$; what does this mean precisely? The authors instead deform $U(\mathfrak{g})$ as a Hopf algebra, but don't elaborate on what such a deformation should look like, what cohomology theory classifies such deformations, etc. Where can I find a resource that concretely answers these questions? An English translation of Drinfel'd's paper "Quasi-Hopf algebras" would be a good starting point, but I haven't been able to find one. REPLY [3 votes]: Means that as a $\mathbb C[[\hbar]]$ associative algebra $U_\hbar(\mathfrak g)$ is isomorphic to $U(\mathfrak g)[[\hbar]]$. Each element can be understood as a formal power series $\sum_{n=0}^\infty \hbar^n X_n$ where each $X_n\in U(\mathfrak g)$ and the product is given by the usual product formula of formal power series $$\sum_{n=0}^\infty \hbar^n X_n \cdot \sum_{n=0}^\infty \hbar^n Y_n=\sum_{n=0}^\infty \hbar^n \sum_{j=0}^n X_j\cdot Y_{n-j}$$ Means that $U_\hbar(\mathfrak g)$ is a $\mathbb C[[\hbar]]$ Hopf algebra such that $U_\hbar(\mathfrak g)/\hbar U_\hbar(\mathfrak g)$ is again a Hopf algebra isomorphic (as Hopf algebra) to $U(\mathfrak g)$ . To see this as a deformation problem in a cohomological manner you should look for what is called Gerstenhaber-Shack cohomology. You can have a look at the original papers on the subject: 1 Gerstenhaber and Shack, Bialgebra cohomology, deformations and quantum groups, Proc. Nat. Acad. Sci. USA 87 (1990). [2] Gerstenhaber and Shack, Algebras, bialgebras, quantum groups, and algebraic deformations, Contemp. Math. 134 (1992). I have the impression (I ask for specialists to correct me on this point) that in general computing GS cohomology is very hard and very few general results are known and that is why these works for quite some time were somewhat not really considered in the qg community. As an aside, difficulties in finding the English version of the paper by Drinfel'd were already addressed here: English version of “Quasi-Hopf Algebras”. It is often difficult to find the very short lived Leningrad Journal of Math but with help from you library you should succeed.<|endoftext|> TITLE: How do mathematicians and physicists think of SL(2,R) acting on Gaussian functions? QUESTION [20 upvotes]: Let $\mathcal{N}(\mu,\sigma^2)$ denote the Gaussian distribution on $\mathbb{R}$: $$ \mathcal{N}(\mu,\sigma^2)(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}.$$ A Gaussian distribution is defined by its mean $\mu\in\mathbb{R}$ and its standard deviation $\sigma>0$. Thus the upper-half plane is the parameter space for the normal family of distributions. Let $z = \mu+i\sigma$ denote an element of the upper-half plane, to be interpreted as the parameters for the Gaussian $\mathcal{N}(\mu,\sigma^2)$. Let $G\in SL(2,\mathbb{R})$, the group of two by two real matrices with unit determinant. Let $g = \begin{pmatrix}a & b\\ c&d\end{pmatrix}$ be an element of $G$. This group acts on the upper-half plane by fractional linear transformations: $$g\cdot z = \frac{az+b}{cz+d}.$$ Thus $G$ acts on Gaussians via $$ g\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}(Re(g\cdot z),\, (Im(g\cdot z))^2).$$ We now describe this action in more detail. If $g = \begin{pmatrix}a & b\\ 0&d\end{pmatrix}$, then $g$ acts on a Gaussian as follows: $$ g\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(\frac a d \mu + \frac b d,\, \frac {a^2} {d^2} \sigma^2\right).$$ For an arbitrary $g$, it follows that $$\begin{align*} Re(g\cdot z) &= \frac{ac(\mu^2+\sigma^2) + bd + (ad+bc)\mu}{(c\mu+d)^2+c^2\sigma^2}\\ Im(g\cdot z) &= \frac{\sigma}{(c\mu+d)^2+c^2\sigma^2}, \end{align*} $$ and the action is $$ g\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(\frac{ac(\mu^2+\sigma^2) + bd + (ad+bc)\mu}{(c\mu+d)^2+c^2\sigma^2},\left(\frac{\sigma}{(c\mu+d)^2+c^2\sigma^2}\right)^2\right).$$ $G$ has the Iwasawa decomposition $KAN$. Thus $g = k(g)a(g)n(g)$, where $$\begin{align*} k(g) &= \begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}\end{align*}\\ a(g) = \begin{pmatrix}e^t & 0\\ 0&e^{-t}\end{pmatrix}\\ n(g) = \begin{pmatrix}1 & u\\ 0&1\end{pmatrix}, $$ where $\theta$, $t$, and $u$ are analytic functions of the coefficients of $g$ (see Keith Conrad's notes). It follows that $N$ acts on a Gaussian by moving its mean: $$ n(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left( \mu + u\,, \sigma^2\right),$$ and $A$ acts on a Gaussian by dilating the mean and standard deviation: $$ a(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left( e^{2t}\mu\,, e^{4t}\sigma^2\right).$$ $K$'s action is much more interesting: $$ k(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(\frac{\cos\theta\sin\theta(\mu^2+\sigma^2) - \cos\theta\sin\theta + (\cos 2\theta)\mu}{((\sin\theta)\mu+\cos\theta)^2+(\sin^2\theta)\sigma^2}, \left(\frac{\sigma}{((\sin\theta)\mu+\cos\theta)^2+(\sin^2\theta)\sigma^2}\right)^2\right).$$ The standard normal $\mathcal{N}(0,1)$ is fixed by $K$. For any other Gaussian $\mathcal{N}(\mu,\sigma^2)$, moving through values of $\theta$ will cause oscillations of the mean and standard deviation, creating a "wobbling" of the Gaussian. In particular, if $\mu=0$ and $\theta = \pi/2$, then $$ k(g)\cdot \mathcal{N}(0,\sigma^2) = \mathcal{N}\left(0,\frac 1 {\sigma^2}\right). $$ The right-hand side is the Fourier transform of $\mathcal{N}(0,\sigma^2)$. For an arbitrary $\mathcal{N}(\mu,\sigma^2)$, with $\theta = \pi/2$ we have: $$ k(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(-\frac \mu {\mu^2+\sigma^2},\frac 1 {\sigma^2}\right), $$ which means that for any $z=\mu+i\sigma$ on the upper semi-circle of radius $\rho$: $$ k(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(- \frac \mu {\rho^2} ,\frac 1 {\sigma^2}\right), $$ and again we have the Fourier transform, combined with reflection about the imaginary axis and dilation of the mean by the square of the curvature of the semi-circle. Now define $$ g_\mu = \begin{pmatrix}1 & \mu\\ 0&1\end{pmatrix} \begin{pmatrix}0 & -1\\ 1&0\end{pmatrix}\begin{pmatrix}1 & -\mu\\ 0&1\end{pmatrix}. $$ Then $$g_\mu\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(\mu,\frac 1 {\sigma^2}\right).$$ Questions: What are some physical interpretations of this? What are some good resources for exploring similar links among probability and geometry? REPLY [13 votes]: The physics application I am aware of is not quite the one in the OP, but similar in spirit: in ray optics the SL(2,R) matrix $$g=\begin{pmatrix} A & B \\ C & D \end{pmatrix}$$ describes the effect of a lens on a Gaussian beam in the paraxial approximation. (The determinant of $g$ is unity if the refractive index remains unchanged.) The fractional linear transformation is called the "ABCD law" in that context. The SL(2,R) matrix $g$ is called the "ABCD matrix" in the ray optics community, which here at MO sounds a bit silly. The complex parameter $q$ that is transformed has real and imaginary parts given by $$\frac{1}{q}=\frac{1}{R}-\frac{i\lambda}{\pi w^2},$$ where $\lambda$ is the wave length, $w$ the spot size, and $R$ the radius of curvature of the beam. The corresponding wave profile is $$u(x,y,z)=\frac{1}{w(z)}\exp\left(-\frac{x^2+y^2}{w(z)^2}\right)\exp\left(\frac{i\pi(x^2+y^2)}{\lambda R(z)}\right).$$ As the beam propagates along the $z$-axis through a lens, $\lambda$ remains the same (if the refractive index does not vary), but $w$ and $R$ change according to the fractional linear transformation $$g\cdot q=\frac{Aq+B}{Cq+D}.$$ This Wiki lists examples of transfer matrices $g$ for various optical elements.<|endoftext|> TITLE: Error in Hoffman-Kunze (normal operators on finite-dimensional inner product space with a cyclic vector) QUESTION [7 upvotes]: I'm teaching a second course in advanced linear algebra, following the second half of Hoffman-Kunze. I have come across what I believe to be an error, but I want confirmation (or refutation) by experts. Here is the setup: Let $V$ be an $n$-dimensional inner product space over either $\mathbb C$ or $\mathbb R$. [Here the inner product is always positive definite.] Let $T$ be a normal linear operator on $V$. This means $T$ commutes with its adjoint $T^*$. In Hoffman-Kunze Section 9.6, page 355, they (correctly) prove that if $W$ is any $T$-invariant subspace of $V$, then it is also $T^*$-invariant, and consequently the orthogonal complement $W^{\perp}$ is also both $T$-invariant and $T^*$-invariant. Moreover, the restriction of $T$ to both $W$ and $W^{\perp}$ is again normal. [They don't say it this way, but that is implicit in their proof of Theorem 19.] Next, they say that "one can now easily prove" a *strengthened cyclic decomposition theorem", which is Theorem 20 at the bottom of page 355. I didn't find this so easy (perhaps I am missing something), but I did indeed find a proof. The relevant fact here is that there is an orthogonal direct sum decomposition $V = Z(v_1, T) \oplus \cdots \oplus Z(v_r, T)$ where $Z(v, T) = \mathrm{span} \{ v, Tv, T^2v, \cdots \}$ is the $T$-cyclic subspace of $V$ generated by $v$. It's the "Corollary" at the top of page 356 that I do not believe. It says: Corollary. If $A$ is a normal matrix over $\mathbb R$ (respectively, $\mathbb C$) then there exists a real orthogonal (respectively, unitary) matrix $P$ such that $P^{-1} A P = P^* A P$ is in rational canonical form. This means that one can find an orthonormal basis of $V$ in which the matrix for $T$ is in rational canonical form. Since the $Z(v_i, T)$'s are orthogonal, this means that we can find an orthonormal basis of each $Z(v_i, T)$ with respect to which the matrix of the restriction $T_i$ of $T$ to $Z(v_i, T)$ is the companion matrix for the minimal polynomial $p_i$ of $T_i$. But this seems to be equivalent to saying that, if $S$ is a normal operator on an $m$-dimensional inner product space $W$ that admits a cyclic vector $w$, then we can choose that cyclic vector so that $\{ w, Tw, \ldots, T^{m - 1} w \}$ is orthonormal. It is easy to see that this definitely cannot hold if the field is $\mathbb R$, $m > 2$, and $S$ is self-adjoint, for the following reason. Since $S$ is self-adjoint, it is orthogonally diagonalizable by the spectral theorem. Let $\{ w_1, \ldots, w_m \}$ be an orthonormal basis of eigenvectors of $S$, with $S w_i = \lambda_i w_i$. Then the $w$ we seek is of the form $w = c_1 w_1 + \cdots + c_m w_m$ for some scalars $c_1, \ldots, c_m \in \mathbb R$. But then we have $$\langle w, T^2 w \rangle = \sum_{i=1}^m \lambda_i^2 c_i^2 = \langle Tw, Tw \rangle.$$ But we want the left hand side to be zero and the right hand side to be one, which gives a contradiction. Does my argument look correct? I can't find a mistake. Does anyone else agree that the Corollary on page 356 of Hoffman-Kunze is false? Thanks! REPLY [7 votes]: I haven't thought carefully about your argument, but I agree that the corollary must be false. Suppose that $A$ is a normal $2 \times 2$ real matrix. The corollary claims that $A$ is orthogonally similar to a matrix $B$ in rational canonical form. Then $B$ is also normal, and is either diagonal (hence $A$ is symmetric), or else $$ B = \begin{bmatrix} 0 & a \\ 1 & b \end{bmatrix} $$ for some $a, b$. Normality of $B$ implies that $a = \pm 1$, so $\det A = \det B = \pm 1$. So the Corollary entails that every nonsymmetric real normal $2 \times 2$ matrix has determinant $\pm 1$. But $\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$ is a counterexample.<|endoftext|> TITLE: How does one complexify a real $n$-dimensional Riemannian manifold $(M,g)$? QUESTION [7 upvotes]: If $V$ is a real vector space, then the complexification of $V$ is formally defined as $V^{\mathbb{C}}=V\otimes_{\mathbb{R}}\mathbb{C}$. Is there an analogous complexification operation for a real $n$-dimensional Riemannian manifold $(M,g)$? Idea: The notion of complexification exists for Lie groups, so perhaps one can "complexify" a real Riemannian manifold by realizing it as a Lie group (or the quotient of one). It seems that under complexification of a real manifold some additional information must be added to determine a complex structure. The reason I ask this is because I am looking through the Riemannian holonomy section of this article and it states that "the complexified holonomies $SO(n,\mathbb{C})$, $G_2(\mathbb{C})$, and $Spin(7,\mathbb{C})$ may be realized from complexifying real analytic Riemannian manifolds." What precisely does complexifying a real analytic Riemannian manifold mean in this context? Any help would be much appreciated! REPLY [9 votes]: I believe the following is meant: Every smooth (real) manifold $M$ has a (unique) real-analytic structure compatible with the smooth structure. So, cover $M$ with real-analytic charts, i.e. whose transition functions are real-analytic diffeomorphisms $$ \phi_{ij}:=\phi_j^{-1}\circ\phi_i: U_{ij}:=\phi_i^{-1}(\phi_i(U_i)\cap\phi_j(U_j))\to U_{ji} $$ One can find open subsets $U_i^{\mathbb{C}}\subseteq\mathbb{C}^n$ with $U_i^{\mathbb{C}}\cap\mathbb{R}^n=U_i$ and $U_{ij}^{\mathbb{C}}\cap\mathbb{R}^n=U_{ij}$ such that the (real-analytic) $\phi_{ij}$ extend to biholomorphisms $\phi_{ij}^{\mathbb{C}}:U_{ij}^{\mathbb{C}}\to U_{ji}^{\mathbb{C}}$ satisfying the usual cocycle conditions. Then the complexification $M^{\mathbb{C}}$ is defined as a quotient space of the disjoint union, $\left(\coprod_i U_i^{\mathbb{C}}\right)/\sim$, where $z_i\sim z_j$ iff $z_i\in U_{ij}^{\mathbb{C}}$ and $z_j = \phi_{ij}^{\mathbb{C}}(z_i)$ (this works because of the cocycle conditions). The maps $U_i^{\mathbb{C}}\hookrightarrow\coprod U_i^{\mathbb{C}}$ induce coordinate charts $U_i^{\mathbb{C}}\to M^{\mathbb{C}}$ with biholomorphic transition functions. This and the details around it are part (of the proof of) Bruhat-Whitney's theorem* on the existence of $M^{\mathbb{C}}$. Moreover, complexification is functorial in the obvious way. By Grauert, $M^{\mathbb{C}}$ is in fact a Stein manifold. *F. Bruhat and H. Whitney, Quelques propriétés fondamentales des ensembles analytiques-réels, Comment. Math. Helv. 33, 132-160 (1959).<|endoftext|> TITLE: What are examples of non-equivalent virtualizations of a large cardinal? QUESTION [5 upvotes]: This is a follow up to my previous question concerning virtual large cardinals, that are generally weaker axioms of infinity obtained from ordinary large cardinals through the so-called virtualization process. If $A$ is a large cardinal property characterized by the existence of suitable embeddings. A cardinal is virtually $A$ if the embeddings characterizing $A$ exist in some set-forcing extensions. Such a reformulation of a large cardinal axiom is called a virtualization of $A$. Remarkable cardinal is an instance of a virtual large cardinal axiom. By definition, it is the virtualization of a certain characterization of supercompact cardinals presented by Magidor (See ZBL0263.02034). Later Gitman and Schindler proved that the virtualization of a particular characterization of strong cardinals gives rise to an equivalent version of remarkable cardinals as well: Theorem. The followings are equivalent: (1) $\kappa$ is remarkable. (2) (Virtualization of Magidor's characterization of supercompacts) For every $\lambda>\kappa$, there is $\overline{\lambda}<\kappa$ such that in a set-forcing extension there is an elementary embedding $j:V_{\overline{\lambda}}^{V}\rightarrow V_{\lambda}^{V}$ with $j(crit(j))=\kappa$. (3) (Virtualization of a characterization of strong cardinals) For every $\lambda>\kappa$ there is $\alpha>\lambda$ and a transitive $M$ with $V_{\lambda}\subseteq M$ such that in a set forcing extension there is $j:V_{\alpha}^V\rightarrow M$ with $crit(j)=\kappa$ and $j(\kappa)>\lambda$. Thus, virtualization doesn't preserve the consistency strength order of large cardinal axioms strictly because there are characterizations of totally different large cardinals whose virtualizations are equivalent. Conversely, one may ask whether different characterizations of the same large cardinal axiom are always equivalent. According to what Joel mentioned in his comment here, the answer is negative! So: Question 1. What are examples of large cardinal axioms which virtualization of their different characterizations gives rise to essentially different virtual large cardinals? Can Vopěnka's principle (which has many characterizations) be such a large cardinal axiom with non-equivalent virtual forms? As the consistency strength of the virtualization of a certain large cardinal axiom highly depends on one's choice of its characterization, it is reasonable to ask about the possible criteria which guides us to choose the right characterization to virtualize. Question 2. What are the criteria for choosing a certain characterization of a large cardinal to virtualize among many others? REPLY [15 votes]: An important feature which separates the notion of virtual large cardinals from the related notion of generic large cardinals is that we only consider embeddings on set-sized structures. Since most large cardinals are characterized by embeddings of the entire universe, there will always be some arbitrary choices made in the characterization we pick because we need to reduce to a set-sized embedding. It is true that Schindler originally thought of remarkable cardinals as virtual supercompacts using the Magidor characterization. But in fact, the most natural characterization of supercompact cardinals in terms of set-sized embedding: $\kappa$ is supercompact if for every $\lambda>\kappa$, there is $\alpha>\lambda$ and a transitive model $N$ closed under $\lambda$-sequences with an elementary embedding $j:V_\alpha\to N$ with critical point $\kappa$, when virtualized gives a remarkable cardinal! (Note that $N$ is assumed to be closed under $\lambda$-sequences in $V$.) So remarkable cardinals is not an example you are looking for. Vopenka's Principle comes closer. Bagaria showed that Vopenka's Principle is equivalent to the existence of a $C^{(n)}$-extendible cardinal for every $1\leq n<\omega$. $C^{(n)}$ is the collection of all cardinal $\alpha$ such that $V_\alpha\prec_{\Sigma_n} V$. A cardinal $\kappa$ is $C^{(n)}$-extendible if for every $\alpha\in C^{(n)}$ above $\kappa$, there is an elementary $j:V_\alpha\to V_\beta$ with critical point $\kappa$ and $\beta\in C^{(n)}$. It is not difficult to see that we can always obtain a $C^{(n)}$-extendibility embedding with the additional property that $j(\kappa)>\alpha$. The proof makes use of the Kunen inconsistency and it turns out that this is fundamental. Kunen's inconsistency does not hold for virtual embeddings in the sense that in a forcing extension we can have embeddings $j:V_\lambda\to V_\lambda$ with $\lambda$ much greater than the supreumum of the critical sequence of $j$. This has the effect of messing up some properties of virtual $C^{(n)}$-extendible cardinals and virtual Vopenka's Principle. We showed with Joel Hamkins that virtual $C^{(n)}$-extendible cardinals with the assumption that $j(\kappa)>\alpha$ are not equivalent to virtual $C^{(n)}$-extendible cardinals, call them virtual weakly $C^{(n)}$-extendible cardinals, without this assumption (see here). However the two notions are equiconsistent, so the consistency strength is not affected by the different characterizations. The consequence for virtual Vopenka's Principle is that it is no longer equivalent to the existence of a virtual $C^{(n)}$-extendible cardinal for every $1\leq n<\omega$, but to the existence of the virtual weak $C^{(n)}$-extendibles. It seems that the only instance of equivalent characterizations producing different virtual notions all revolve around the consequences of the failure of the Kunen Inconsistency. In practice the most robust embedding characterizations for virtualizing have the form $j:V_\alpha\to V_\beta$, namely the target model has the form $V_\beta$. In my opinion, the reason that it appears that strong cardinals and supercompact cardinals have equivalent virtual versions is that strong cardinals simply don't have a virtual characterization because they don't have a "robust" embedding characterization for virtualizing. A discussion of this can be found in our joint paper with Ralf Schindler (see here).<|endoftext|> TITLE: Discrete Hadwiger–Nelson problem variant QUESTION [5 upvotes]: The Hadwiger–Nelson problem asks for the minimum number of colors required to color the plane such that no two points at distance 1 from each other have the same color. We could build the following discrete variant on the infinite two dimensional integer grid (the nodes are points with integer coordinates). [Discrete Hadwiger–Nelson problem variant] Given an integer $U$ (which is the equivalent of the unit distance in the Hadwiger–Nelson problem), and an integer tolerance distance $t < U$; we consider the infinite induced subgraph $G_{U,t}$ of the integer grid in which two integer points $(x_1,y_1), (x_2, y_2)$ are linked if their distance $d$ is in the interval $[U-t, U+t]$ ( $$|U - ((x_1 - x_2)^2 + (y_1 - y_2)^2)^{1/2}| <= t$$ What is the chromatic number of $G_{U,t}$ ? Is this problem known? If it is known, can you give some references? REPLY [3 votes]: This problem is equivalent to coloring all points of the plane such that no two points at distance $1\pm\varepsilon$ have the same color (if $t$ is small enough compared to $U$). This has been posed here by Benoît Kloeckner and shown by me that we need at least $6$ colors. For completeness, the argument for your variant. Color every point of the plane to the color of the integer point nearest to it. This new coloring avoids distance $U$ if $t\ge \sqrt 2/2$, and every color class is a nice region, for which it is known that we need at least 6 colors, see Townsend, Woodall. Accidentally, just when searching for the links to the papers, I've discovered that Benoît's problem has been studied before a lot, see this paper for a nice historical summary.<|endoftext|> TITLE: Counting Hamiltonian cycles in $n \times n$ square grid QUESTION [10 upvotes]: I wonder if anyone has counted these curves, either exactly or asymptotically? Let $S_n$ be an $n \times n$ subset of $\mathbb{Z}^2$ consisting of $n^2$ lattice points: a lattice square. Define a rectilinear filler curve for $S_n$ to be a simple closed curve that passes through each of the $n^2$ lattice points, and is composed entirely of vertical and horizontal edges. So the curve is what is called a "rectilinear" or "orthogonal" polygon in the literature. Every turn of such a curve is $\pm 90^\circ$. I'd like to know the number $f(n)$ of distinct filler curves for $S_n$, distinct up to rotations and reflections. So if $C_1$ can be rotated and/or reflected to lay on top of $C_2$, then $C_1$ and $C_2$ are not distinct. $f(2) = 1$, and $f(4) = 2$:           $f(n)=0$ when $n$ is odd, as can be seen as follows. View a filler curve $C$ as composed of unit-length segments connecting lattice points; call these the edges of $C$ (so two incident edges can be collinear). Each horizontal line $y = m + \frac{1}{2}$ for $m$ an integer crosses an even number of edges of $C$; similarly for vertical lines. So the total number of edges $E$ of $C$ is even. In Euler's relation $V-E+F=2$, $F=2$ (interior & exterior of $C$). So $V=E$. So $V$ must be even. But $V=n^2$ for $n$ odd is odd. Already I don't know what is $f(6)$. It is easy to see the growth of $f$ is exponential in $n$, but I don't know more. In particular, I do not see how to recursively connect $f(n)$ to $f(n-2)$. REPLY [3 votes]: This is a summary of the sprawling comments above. This is OEIS sequence http://oeis.org/A209077 (number of isomorphism classes of Hamiltonian circuits on $2n \times 2n$ grid). Ed Wynn discussed a method of computing these in https://arxiv.org/abs/1402.0545 and extended the known values from $1 \le n \le 4$ to $1 \le n \le 10$. In particular, for the 6 by 6 grid (where $n=3$) which was asked about in the question, there are 149 of these circuits up to reflections and rotations.<|endoftext|> TITLE: Upperbounding a sum of Legendre-Symbols QUESTION [6 upvotes]: Let $p$ be a prime with $p\equiv 3 \mod 4$, for any $\mathcal{I} \subset \lbrace 0,...,p-1 \rbrace $ and any $\mathcal{J} \subset \lbrace 0,...,p-1 \rbrace $ with $\vert\mathcal{I}\vert \leq \sqrt{p} $ and $\vert\mathcal{J}\vert \leq \sqrt{p} $, I am looking for an upperbound on the following expression \begin{align*} \left\vert \sum _{i\in\mathcal{I}} \sum _{j\in\mathcal{J}} \left( \frac{i-j}{p} \right) \right\vert \leq c(p) \end{align*} Where $\left( \frac{a}{p} \right) $ is the Legendre Symbol defined by \begin{align*} \left( \frac{a}{p} \right) = \begin{cases} 1 &\quad \text{ if } a \text{ is a quadratic residue modulo } p \text{ and } a \not\equiv 0 \mod p \\ -1 &\quad \text{ if } a \text{ is a quadratic non-residue modulo } p \\ 0 &\quad \text{ if } a \equiv 0 \mod p \end{cases} \end{align*} Essentially I am looking for some expression for $c(p)$. Some brute force attempts I have calculated on MATLAB in small dimensions suggest that for $p>7$ holds $c(p)/p > 1/2$, explicitly the values are for $p=7 \;,\; c(p) = 3 $, for $p=11 \; ,\; c(p) = 6$, for $p=19 \; ,\; c(p) = 12 $ and for $p=23 \; , \; c(p) = 13 $. I could prove that when $\vert\mathcal{I}\vert >\sqrt{p}$ and $\vert\mathcal{J}\vert > \sqrt{p} $ then $\left\vert \sum _{i\in\mathcal{I}} \sum _{j\in\mathcal{J}} \left( \frac{i-j}{p} \right) \right\vert < \vert\mathcal{I}\vert\vert\mathcal{J}\vert $. But this doesn't really help. REPLY [10 votes]: If $A$ and $B$ are two subsets of ${\Bbb Z}/p{\Bbb Z}$ with $|A|$ and $|B|$ being bigger than $p^{\alpha}$ (for some $\alpha>0$) then one expects that $$ \Big| \sum_{a\in A} \sum_{b\in B} \chi(a+b) \Big| \le p^{-\delta} |A||B|, $$ for some $\delta >0$ (depending only on $\alpha$). Here $\chi$ denotes a non-principal character $\pmod p$. We are very far from knowing anything like this. Writing the multiplicative character in terms of additive characters, and then using Cauchy--Schwarz, one can show that $$ \Big| \sum_{a\in A} \sum_{b\in B} \chi(a+b) \Big| \le \sqrt{p |A||B|}, $$ which is useful when $|A||B| >p$. In general, this is the best bound known. Improving it even by a constant factor would have a nice consequence: a conjecture of Sarkozy asserts that the set of quadratic residues $\mod p$ cannot be expressed as the sumset $A+B$ where $A$ and $B$ both have size at least $2$. Amazingly, even this very weak conjecture is unknown. For recent work on this problem (generalized to sumsets involving three or four sets), and references to related work, see this paper of Brandon Hanson (which appeared in Acta Arith). There are a couple of situations in which non-trivial bounds are known: when $A$ is an interval (this is due to Friedlander and Iwaniec), and when $A$ has small doubling (due to Chang).<|endoftext|> TITLE: Obstructions for the lifting problem after a pull-back QUESTION [8 upvotes]: This is a cross-post from a MSE question which received no answers. Beware that the notation here is a little different. Consider the following lifting problem(s): $\require{AMScd}$ \begin{CD} & & & & E\\ & & & @VV{p}V\\ Y @>{g}>> X @>{f}>> B \end{CD} Let's work with the following assumptions and notations (edited the indexing error pointed out by Aleksandar Milivojevic): The map $p: E \rightarrow B$ is a Hurewicz fibration with path-connected base $B$ and $(d-2)$-connected fiber $F$ for some $d \geq 2$. In case $d=2$ we require $\pi_1(F)$ to be abelian. For every $k \geq 0$, the action of the fundamental group $\pi_1(F)$ on the homotopy group $\pi_k(F)$ is trivial. As a consequence, there is a well-defined action [Davis-Kirk, Proposition 6.62] of $\pi_1(B)$ on $\pi_k(F)$. $X,Y$ are finite CW complexes, $g$ is a cellular map, and $\dim(X) = d < \dim(Y)$ with the same $d$ in (1). For each $k \geq 0$, write $\rho(f,k): \pi_1(X) \rightarrow Aut(\pi_k(F))$ for the induced action of $\pi_1(X)$ from (2). Consider this as a local coefficient system over $X$, so that there are well-defined cohomology groups $H^*(X;\pi_k(F)_{\rho(f,k)})$ with local coefficients. There is a similar situation with $\rho(fg,k) : \pi_1(Y) \rightarrow Aut(\pi_k(F))$. Under these conditions, there is a well-understood obstruction theory. Using [Davis-Kirk, Theorem 7.37] and the remarks later: First of all, $f$ can be lifted over the $(d-1)$-skeleton $X_{d-1}$. Second, no matter which lift over the $(d-1)$-skeleton we choose, the obstruction class for lifting it further over the $d$-skeleton is unique. This primary obstruction $z_f$ is an element of $H^d(X;\pi_{d-1}(F)_{\rho(f,d-1)})$. For dimension reasons $z_f$ is the only obstruction for lifting $f$ along $p$. The problem of lifting $fg$ along $p$ has a similar obstruction theory to above. There is again a primary obstruction $z_{fg} \in H^d(Y;\pi_{d-1}(F)_{\rho(fg,d-1)})$, but now there might be higher obstructions lying in $H^{k+1}(Y;\pi_k(F)_{\rho(fg,k)})$ for $k \geq d$. Now suppose $z_f \neq 0$ but $z_{fg} = 0$. In other words (by the naturality of the obstruction classes), $z_f$ lies in the kernel of the induced map $$g^*: H^d(X;\pi_{d-1}(F)_{\rho(f,d-1)}) \rightarrow H^d(Y;\pi_{d-1}(F)_{\rho(fg,d-1)}) \, .$$ Does this vanishing imply that $fg$ can be lifted over the whole $Y$ (not just its $d$-skeleton)? If not, can we at least say that there is a unique nonzero higher obstruction class for lifting $fg$? My intuition is that the lifting problem for $fg$ should not be harder than the lifting problem for $f$. Davis, James F.; Kirk, Paul, Lecture notes in algebraic topology, Graduate Studies in Mathematics. 35. Providence, RI: AMS, American Mathematical Society. xvi, 367 p. (2001). ZBL1018.55001. REPLY [10 votes]: First, note that since you are assuming $F$ is $d-1$-connected, the primary obstruction lies in $H^{d+1}$, not $H^d$. Now, consider the diagram $\require{AMScd}$ \begin{CD} & & & & S^5\\ & & & @VV{p}V\\ S^4 @>{g}>> S^3 @>{id}>> S^3 \end{CD} where $p$ represents the nontrivial element in $\pi_5(S^3) \cong \mathbb{Z}_2$ converted into a fibration, and $g$ represents the nontrivial element in $\pi_4(S^3) \cong \mathbb{Z_2}$. From the long exact sequence in homotopy groups for a fibration, we see that the fiber $F$ of $p$ is simply connected and $\pi_2(F) \cong \mathbb{Z}$. The primary (and only) obstruction to lifting $id$ lies in $H^3(S^3; \mathbb{Z})$ and does not vanish, since otherwise the identity on $S^3$ would factor through $S^5$ and hence be trivial. Since $H^3(S^4;\mathbb{Z}) = 0$, the "primary obstruction" $z_{id \circ g}$ to lifting $id \circ g$ vanishes. However, there are higher obstructions to lifting this map, that do not vanish. Indeed, if there were a lift, then we would have that the non-trivial map $S^4 \to S^3$ factors through $S^5$, meaning it would be trivial.<|endoftext|> TITLE: Exponential map of a Formal Group Scheme QUESTION [8 upvotes]: Let $k$ be a field of characteristic $0$ and let $\mathfrak{g}$ be a finite dimensional Lie algebra over $k$. $\mathfrak{g}$ corresponds to a formal group scheme $\mathcal{G} = \text{Spf} (U(\mathfrak{g})^{*})$. In fact, there is an equivalence of categories between finite-dimensional Lie algebras and infinitesimal formal group schemes over $k$ with finite-dimensional tangent space. This is elaborated on here by Akhil Mathew. There should be an exponential map $\text{exp} : \mathfrak{g} \rightarrow \mathcal{G}$. This is mentioned at the end of the blog post above, but I can't figure out how to construct it explicitly. Any help on this would be much appreciated. REPLY [8 votes]: $\newcommand{\g}{\mathfrak{g}}$ In a way this is tautological, in the sense that $\mathcal G$ can be seen as a formal exponentation of $\g$, although I don't think there is an actual exponential map from one to the other in general. Rather, there is a map (in fact an isomorphism of formal schemes) from the formal completion of $\g$ at the origin to $\mathcal G$. Namely, the dual $U(\g)^*$ can be identified as a (topological) algebra with the algebra $\widehat S(\g^*)$ of formal power series on $\g^*$, ie with the formal completion of $\mathcal O(\g)$ w.r.t to the ideal of function vanishing at the origin. This identification, should indeed be thought as pulling back functions through the exponential map: indeed, the induced coproduct on $\widehat S(\g^*)$ is expressed through the Baker-Campbell-Hausdorff formula $$\forall x,y \in \g, \Delta(f)(x\otimes y)=f(BCH(x,y))$$ where BCH is the Lie formal power series defined on formal variables $a,b$ by $$BCH(a,b):=\log(e^ae^b).$$<|endoftext|> TITLE: Moyal $\star$-product inverse? QUESTION [8 upvotes]: On a 2n-dimensional phase-space with coordinates $x$ and $p$, the Moyal product can be written explicitly as $$g(x,p) \star h(x,p) = g(x,p) e^{\frac{i}{2}\left( \overleftarrow{\partial_x} \cdot \overrightarrow{\partial_p} - \overrightarrow{\partial_x} \cdot \overleftarrow{\partial_p}\right)} h(x,p) \, .$$ Assume now that we have something like $f(x,p)=g(x,p)\star h(x,p),$ where we have to solve for $g(x,p)$. I suppose it's not always possible, but how would one go about solving that when it is? And what would the invertibility conditions even be? I can't seem to find the answer anywhere. Maybe even a simpler problem, $f(x,p)=g(x,p)\star g(x,p) $ with $f(x,p)$ known. How would one take the $\star$-square root and when would it be possible? REPLY [4 votes]: The inversion is conveniently described in terms of the Fourier transform $$g(x,p)=\int dy\,e^{-iyp}G(x+y/2,x-y/2).$$ Then the composition $f(x,p)=g(x,p)\star h(x,p)$ is a matrix multiplication [1], $$F(x,y)=\int dz\, G(x,z)H(z,y).\qquad(\ast)$$ So to find $h$ if $f$ and $g$ are given one would first calculate the Fourier transforms $F$ and $G$, $$G(x+y/2,x-y/2)=\frac{1}{2\pi}\int dp\,e^{iyp}g(x,p),$$ then solve the integral equation $(\ast)$ for $G$, and finally transform back to $g$. Whether this is doable will of course entirely depend on the details of the particular problem. But this is the general recipe. [1] Map of Witten's $\star$ to Moyal's $\star$, Itzhak Bars, 2001.<|endoftext|> TITLE: Fell's trick for Lie groups QUESTION [5 upvotes]: Let $\Gamma$ be a countable discrete group and $\lambda$ the left regular representation of $\Gamma$ on $l^2(\Gamma)$. Let $\rho:\Gamma\rightarrow U(H)$ be a unitary representation of $\Gamma$ on some separable infinite-dimensional Hilbert space $H$, and consider the representation $\lambda\otimes\rho$ of $\Gamma$ on $l^2(\Gamma)\otimes H$. Now let $H_0$ be the same Hilbert space as $H$ but equipped with the trivial representation of $\Gamma$; then one can show that, as $\Gamma$-representations, $$l^2(\Gamma)\otimes H\cong l^2(\Gamma)\otimes H_0,$$ where $\Gamma$ still acts by the left regular representation on the first factor on the right. For example one isomorphism is the map $\delta_g\otimes u\mapsto\delta_g\otimes(\rho(g)\cdot u)$, where $\delta_g:\Gamma\rightarrow\mathbb{C}$ is the function taking value $1$ at $g\in\Gamma$ and $0$ elsewhere. I would like to know whether this can be generalised to the case of $G$ a Lie group instead of $\Gamma$. More specifically: Question 1: Given a unitary representation $\rho:G\rightarrow U(H)$, is there an isomorphism $$L^2(G)\otimes H\cong L^2(G)\otimes H_0,$$ where $G$ acts on $L^2(G)$ by the left regular representation and $H_0$ is the same space as $H$ but with the trivial representation? I'm not sure if the left-regular representation of $G$ on $L^2(G)$ is the right representation to use here, since the proof in the discrete case relies on the fact that the $\delta_g$ form a basis for $l^2(\Gamma)$, which doesn't seem to have an analogue in the continuous case. If this is hard to establish, I would also like to ask: Question 2: Can one replace $L^2(G)$ by another representation in question 1 so that the conclusion is true? REPLY [7 votes]: Fell absorption works for all locally compact groups and all strongly continuous unitary representations; the intertwining map that you give in the discrete case can be generalized to $$ W: L^2(G)\otimes_2 H \to L^2(G)\otimes_2 H $$ given by $$ \langle W(\xi\otimes f), (\eta\otimes g) \rangle := \int_G f(s)\overline{g(s)} \langle \pi(s)\xi,\eta\rangle_H $$ (my inner products being linear in the first variable.) If you are worried about this being well-defined on the whole of $L^2(G)\otimes_2 H$: first show that it is well-defined for $\xi,\eta \in C_c(G)$; then show by direct calculation that $$ \Vert W(\xi\otimes f) \Vert^2 = \langle W(\xi\otimes f), W(\xi\otimes f) \rangle = \Vert \xi\otimes f\Vert^2$$ for all $\xi\in C_c(G)$ and all $f\in H$; so we have a linear map defined on a dense subspace which is isometric.<|endoftext|> TITLE: Equivariant non symmetric operads QUESTION [5 upvotes]: The definition of a symmetric $G$-Operad is basically a $G$ object in the category of symmetric operads. As far as I understand there is not a good notion of the non symmetric case. I would like to know what are the main drawbacks of defining it as a $G$ object in the category of non symmetric operads. REPLY [3 votes]: I see no issues at all defining a $G$-equivariant non-symmetric operad $O$ as a $G$-object in the category of non-symmetric operads. Depending on your setting -- pointed or unpointed, reduced or non-reduced -- you might need to specify that the action fixes the unit in $O(1)$. For example, see Section 2.4 of my paper with Javier Gutierrez on $G$-equivariant symmetric operads. With this definition, $G$-equivariant non-symmetric operads would sit inside $G$-equivariant symmetric operads, just like the non-equivariant setting. There's lots of operadic questions you could investigate. I do agree that the homotopy theory would be less interesting, for the reason Denis pointed out (in the symmetric setting, you need to look at subgroups of $G \times \Sigma_n$, so without the $\Sigma_n$ actions you are left with only subgroups of $G$). Basically, it should work like it does for $G$-spaces. Anyway, don't let any of this stop you from investigating. I'm sure there's lots you can prove about $G$-equivariant non-symmetric operads!<|endoftext|> TITLE: Intersections in $\mathbb{P}^1\times\mathbb{P}^1$ QUESTION [5 upvotes]: Let $F$ be an algebraically closed field and $\mathbb{P}^1$ the projective line over $F$. Suppose $V_1, V_2$ are two 1-dimensional subvarieties of the 2-dimensional variety $\mathbb{P}^1\times\mathbb{P}^1$. Now $V_1$ and $V_2$ do not necessarily intersect, as the simple example $V_i=\{x_i\}\times\mathbb{P}^1$ with $x_1\neq x_2$ shows. But are there any conditions on $V_1$ and $V_2$ known, under which they do intersect? REPLY [11 votes]: The curve $V_i$ is given by the vanishing of a polynomial $F_i(x_1,x_2,y_1,y_2)$ that is homogeneous in $x_1,x_2$ of degree $d_{i,1}$ and homogeneous in $y_1,y_2$ of degree $d_{i,2}$. Then counting intersection points with multiplicities, $$ V_1 \cdot V_2 = d_{1,1}d_{2,1} + d_{1,2}d_{2,2}. $$ So $V_1$ and $V_2$ always intersect except in the case that they are both horizontal slices or they are both vertical slices, as in the example that you give.<|endoftext|> TITLE: Homology spectral sequence for function space QUESTION [6 upvotes]: The question is in the title. Suppose that $X$ and $Y$ are two pointed connected CW-complexes. I was wondering if there exists a spectral sequence computing the homology of the function space $$H_{\ast}(map_{\ast}(X,Y);k) $$ where $k$ is a fixed field. Could we say something interesting in the case when $H_{\ast}(X;k)$ is trivial. For simplicity we can assume that $X$ and $Y$ are simply connected. REPLY [8 votes]: An alternative spectral sequence for $H_∗(map_∗(X,Y);k)$ can be constructed using the approach in this paper of mine (apols for self-promotion). Actually, this spectral sequence is discussed already in an earlier paper of Bendersky and Gitler. This spectral sequence converges in the same cases as the Anderson spectral sequence that is alluded to in John Klein's answer. An explicit comparison of the two spectral sequences was given by Podkorytov. See also the paper of Ahearn and Kuhn that gives a nice introduction to the spectral sequence, and studies its structure in detail. I think that the approach in this paper can answer the question on what happens when $H_*(X;k)=0$. Indeed, the spectral sequence is based on the following model for the chains on $map_*(X, Y)$. Let $C_*(Y)$ denote the (reduced) complex of singular chains on $Y$ and $C^*(X)$ denote the reduced singular cochains on $X$. Let $\mathcal E$ be the category of non-empty finite sets and surjective functions, and let $\mathcal E^t$ be the "twisted arrow category" of $\mathcal E^t$. An object of $\mathcal E^t$ is a surjection of sets $i\twoheadrightarrow j$ and a morphism from $i\twoheadrightarrow j$ to $s\twoheadrightarrow t$ consists of sujections $i\twoheadrightarrow s$ and $t\twoheadrightarrow j$ that make the evident square commute (note the twist in the directions of the arrows). There is an evident contravariant functor from $\mathcal E^t$ to chain complexes that sends $i\twoheadrightarrow j$ to $C^*(X^{\wedge j}) \otimes C_*(Y^{\wedge i}) \otimes k$. The point is that the homotopy limit of this functor is quasi-isomorphic to $C_*(map_*(X,Y);k)$ when $X$ is a finite q-dimensional complex and $Y$ is $q-1$-connected. Now, suppose $H_*(X;k)=0$. Then $C^*(X^{\wedge j})\otimes k$ is an acyclic complex for all $j$, and therefore the homotopy limit is acyclic as well. So we may conclude the following Claim: Suppose $X$ is a finite $q$-dimensional complex and $Y$ is a $q-1$-connected space. Suppose furthermore that $H_*(X;k)$ is trivial. Then $H_*(map_*(X,Y);k)$ is trivial. Remark 1: it is not necessary to assume that $X$ is simply connected. Remark 2: The conclusion should hold whenever either one of the spectral sequences converges. The condition that $X$ is q-dimensional and $Y$ is $q-1$-connected is the standard condition that guarantees convergence. But when $k={\mathbb Z}/p$ there also are results about "exotic convergence" of these types of spectral sequences that may be relevant.<|endoftext|> TITLE: "Long-standing conjectures in analysis ... often turn out to be false" QUESTION [41 upvotes]: The title is a quote from a Jim Holt article entitled, "The Riemann zeta conjecture and the laughter of the primes" (p. 47).1 His example of a "long-standing conjecture" is the Riemann hypothesis, and he is cautioning "those who blithely assume the truth of the Riemann conjecture." Q. What are examples of long-standing conjectures in analysis that turned out to be false? Is Holt's adverb "often" justified? 1 Jim Holt. When Einstein Walked with Gödel: Excursions to the Edge of Thought. Farrar, Straus and Giroux, 2018. pp.36-50. (NYTimes Review.) REPLY [3 votes]: Fuglede's conjecture was open for 30 years (1974-2004) only to be proven false by T. Tao for dimensions $d\geq 5$ with a counter-example arising from a set with an exponential orthonormal basis (a spectrum) in a finite abelian group which does not tile by translation. Interestingly, the largest dimension for which the conjecture is open now is $d=2$.<|endoftext|> TITLE: $\mathbb{P}^1$-bundle over compact base QUESTION [5 upvotes]: 1) Is it possible to construct a $\mathbb{P}^1$-bundle $P\to B$ where $B$ is a proper variety and $P$ is not $\mathbb{P}(V)$ for a rank 2 vector bundle $V\to B$? If we drop the properness assumption on $B$, I only know of a couple examples. 2) Does a Zariski locally trivial projective bundle come from the projectivization of a vector bundle? (This feels like it should be standard, but I'm ignorant.) REPLY [4 votes]: 1) There are many examples. As abx mentioned, they are called Severi-Brauer varieties, and they are related to 2-torsion classes in the Brauer group of $B$. If you want a geometric construction, to get a relatively simple example consider a general cubic 4-fold $X \subset \mathbb{P}^5$ containing a plane $\Pi \subset X$. Let $F(X)$ be the Hilbert scheme of lines on $X$ and $$ P \subset F(X) $$ be the closure of the locus of lines in $X$ that intersect $\Pi$ but are not contained in $\Pi$. Then it has a structure of a $\mathbb{P}^1$-bundle over $B$, which is the double covering of $\mathbb{P}^2$ branched over a sextic curve (thus $B$ is a K3-surface). 2) This is true, only if $B$ is smooth. For singular surfaces there are counterexamples. The simplest I know is for $$ B = (\mathbb{P}^1 \times \mathbb{P}^1)/(\mathbb{Z}/2), $$ where the group acts diagonally, and nontrivially on each factor.<|endoftext|> TITLE: Can the Induction axiom in the Peano arithmetic be replaced by the irrationality of $\sqrt{2}$? QUESTION [13 upvotes]: This is inspired by the Alexander Shen's post here: https://www.facebook.com/groups/mathpuz/permalink/1058782384297603/ (the post is in Russian, but it is easy Russian, and google translate should work fine). All proofs of irrationality of $\sqrt{n}$ where $n$ is not a perfect square that I know are using (explicitly of implicitly) the induction axiom of the Peano arithmetic. Question: Can the induction axiom be replaced by the axiom that $\sqrt{2}$ is irrational? REPLY [23 votes]: One can prove that $\sqrt{2}$ is irrational and indeed that any particular $\sqrt{n}$ is irrational (if $n$ is not a perfect squares) assuming only the principle of induction for $\Delta_0$ formulas, which is much weaker than the full strength of PA. In this sense, the answer is negative. The classical proof of the irrationality of $\sqrt{2}$ relies on the fact that every rational number has a representation in lowest terms, and this can be proved with a very weak induction principle, using no unbounded quantifiers. Namely, every $p/q$ has a lowest-term representation, since if this is true for all pairs of numbers whose numerator is smaller than $p$, then it is true for $p/q$, since otherwise we would immediately reduce to a case with a smaller numerator. Thus, we never need to embark on unbounded searches to find the lowest-terms representation of a rational number. And once one has the lowest-term representations, one can easily prove the classical result that $\sqrt{2}$ is irrational, since otherwise one gets $2q^2=p^2$ and so on, as we all know, ultimately showing that both $p$ and $q$ are even, a contradiction. For the case of $\sqrt{n}$, for fixed $n$, then one can also prove this using just the existence of lowest-terms representations. Basically, $\sqrt{n}$ is irrational, unless $n$ is a perfect square, since otherwise $nq^2=p^2$ and the prime exponent parities don't work out, as you know. To prove the universal statement about all such $n$, however, that is, the statement "for all numbers $n$, the square root $\sqrt{n}$ is irrational unless $n$ is a perfect square," it seems that one uses the fundamental theorem of arithmetic, asserting that every number has unique prime factorization, which can be proved using $\Sigma_1$ induction. Namely, if every number smaller than $n$ has a unique factorization into primes, then $n$ also has this, since otherwise one can factor $n$ and reduce to the smaller case. (Here, we have to search a little bit, but not much, to find the code of the factorization.) But once you have the factorization into primes, then again the classical argument shows as above that $\sqrt{n}$ is irrational unless $n$ is a perfect square. I'm not sure if one can prove the fundamental theorem using only $\Delta_0$-induction. And because we moved up to $\Sigma_1$ induction to get the fundamental theorem, it is conceivable to me that one might be able to have a bizarre model of $\Delta_0$-induction in which the universal claim about all $\sqrt{n}$ was not true, even if it was true in every standard-finite instance, and this would be an interesting answer to your question. I would want the proof theorists to weigh in on this. Since PA proves the consistency of $\Sigma_0$ induction, and indeed of $\Sigma_n$-induction for any fixed $n$, it follows by the incompleteness theorem (as mentioned in the comment of Dan Piponi) that we cannot axiomatize PA by adding the assertion "$\sqrt{2}$ is irrational" or "$\sqrt{n}$ is irrational for all non-perfect squares $n$" to any finite list of axioms. The essence of PA is the induction axiom for increasingly difficult arithmetic properties. No bounded amount of complexity for induction, limited to $\Sigma_n$ induction for fixed $n$, is sufficient to capture the full power of PA.<|endoftext|> TITLE: Why are the coefficients of the modular equation so large? QUESTION [10 upvotes]: The modular equation $\Phi_n(X,Y)$ is a polynomial in $\mathbf Z[X,Y]$ relating the modular invariant $j$ and the functions $j\left(\frac{a\tau+b}{c\tau+d}\right)$, where $ad-bc=n$. For example, we have identically $$\Phi_n(j(n\tau),j(\tau))=0.$$ It is often asserted that the coefficients of $\Phi_n(X,Y)$ are astronomically large even for small values of $n$. Why is this so? Is there some lower bound on the coefficients? REPLY [15 votes]: Paula Cohen, On the coefficients of the transformation polynomials for the elliptic modular function, Math. Proc. of the Cambridge Phil. Soc. 95, 389–402 (1984). The largest coefficient $c_{m}$ in absolute value of $\Phi_{m}$ is of order $$c_{m}\simeq m^{km},$$ with $k$ a number of order unity ($k=9$ for $m=2^n$). So for $m=2^{10}=1024$ one has $c_m\simeq 10^{27743}$ --- astronomically large, indeed.<|endoftext|> TITLE: Embeddings of Boolean algebras in $\wp(\omega)/Fin$ QUESTION [9 upvotes]: If we assume MA+¬CH, then every boolean algebra with cardinality smaller than the continuum embeds in ℘(ω)/Fin. A proof of this result can be found in Theorem 1.1, Chapter 8 of the book "Hausdorff gaps and limits". In the paper "R. Frankiewicz, Some remarks on embeddings of boolean algebras and topological spaces II, Fund. Math, 126 (1), 1985, 63-68.", the author showed that it is consistent with MA+¬CH the existence of a boolean algebra with cardinality equal to the continuum which does not embed in ℘(ω)/Fin. Is it consistent with $MA+\neg CH$ that every boolean algebra with cardinaly equal to the continuum embeds in ℘(ω)/Fin? REPLY [4 votes]: In the paper Embedding of Boolean algebras in $Ρ(ω)/$fin the following partial result is proved: Theorem There is a model of $ZFC$ with arbitrarily large continuum in which each Boolean algebra $B$ of cardinality $\leq 2^{\aleph_0}$ can be embedded into $P(ω)/$fin. In addition, Martin's axiom for $σ$-linked orderings holds in the model. You may also look at On automorphisms of Boolean algebras embedded in P (ω)/fin, where a model as above is constructed in which the following additional property holds: every automorphism of $B$ extends to an automorphism of $P(ω)/$fin.<|endoftext|> TITLE: A recurrence relation for $\zeta(2n)$ - reference request QUESTION [8 upvotes]: In the textbook https://www.springer.com/gp/book/9783034851688 (Klassische elementare Analysis, by M. Koecher) the following elegant recurrence relation is proved for $\zeta(2n)$ (on p. 157): $$\left(n+\frac{1}{2}\right)\zeta(2n)=\sum\limits_{m=1}^{n-1}\zeta(2m)\,\zeta(2n-2m). \tag{1}$$ In fact (1) is equivalent to Euler's recurrence relation for Bernoulli numbers (independently found by Ramanujan) $$(2n+1)B_{2n}=-\sum\limits_{m=1}^{n-1}\binom{2n}{2m}B_{2m}\,B_{2n-2m}. \tag{2}$$ Why, In contrast to (2), (1) can seldom be found in the literature (I was able to find only https://link.springer.com/article/10.1007/s00591-007-0022-2 that mentions (1))? Are there any other references that discuss (1)? P.S. In addition to juan's answer. G.T. Williams was not the first to state the result in this form. It can be found at least in N. Nielsen, Handbuch der theorie der gammafunktion, Leipzig: Druck und Verlag von B.G. Teubner, 1906, p. 49. I found this reference thanks to the paper "Some identities involving the Riemann zeta function. II." by R. Sitaramachandrarao and B. Davis, Indian J. Pure Appl. Math. 17(10):1175–1186, 1986. https://www.insa.nic.in/writereaddata/UpLoadedFiles/IJPAM/20005a50_1175.pdf This reference also has (1) and proves (among others) an interesting generalization of (1): $$4\sum\limits_{i+j+k=n}\zeta(2i)\zeta(2j)\zeta(2k)=(n+1)(2n+1)\zeta(2n)-6\zeta(2)\zeta(2n-2),$$ where $n\ge 3$ and the sum extends over all ordered triples $(i,j,k)$ of positive integers satisfying $i+j+k=n$. REPLY [8 votes]: I have this recurrence in my collection of problems for my lessons in Analytic Number Theory. I have there the reference: P. Ribenboim, Classical Theory of Algebraic Numbers, Springer, New York, 2001, p. 503. I add a note that related formulas can be found in S. Sekatskii, Novel integral representations of the Riemann zeta-function and Dirichlet eta-function, close expressions.... https://arxiv.org/abs/1606.02150 (I have not checked this paper, but contains some interesting similar relations). After several hours I find the formula in Theorem 1 of G. T. Williams, "A new method of evaluating $\zeta(2n)$", Amer. Math. Soc, 60, (1953) 19--25. The author of this paper think he is the first to state the result in this form.<|endoftext|> TITLE: Does the Serre spectral sequence of the Fadell-Neuwirth fibration collapse if there is a cross-section? QUESTION [9 upvotes]: I had asked a vague question in MSE where a useful pointer to the Leray-Hirsch theorem was mentioned by Mike Miller in the comments, but received no answers. Here I will specialize to an interesting case to get a well-posed question. Let $M$ be a closed manifold of dimension $\geq 2$ and consider the Fadell-Neuwirth fibration $f_k\colon \text{PConf}_k(M) \rightarrow M$ where $\text{PConf}_k(M)$ is the ordered configuration space ($f_k$ forgets all but one point) for some $k \geq 2$. The associated cohomology Serre spectral sequence has the form $$E_2^{p,q} = H^p(M ; H^q(\text{PConf}_{k-1}(M - \{pt\}) ; \mathbb{Z})) \Rightarrow H^{p+q}(\text{PConf}_k(M))$$ If $f_k$ has a cross-section $s \colon M \rightarrow \text{PConf}_k(M)$, does the spectral sequence collapse on the $E_2$ page? Remark 1: There are Serre fibrations with a cross-section that do not exhibit a collapse on $E_2$. Remark 2: There is always a section of $f_k$, provided that $M$ has a positive first Betti number (seems to be a somewhat forgotten fact from the original Fadell-Neuwirth paper, Corollary 5.1). This suggests many cases to look at but I haven't seen an explicit computation addressing the issue. REPLY [3 votes]: Nice question! Here is a partial answer which was too long for a comment. A necessary condition for collapse at $E_2$ is that the homomorphism $i^*: H^*(\operatorname{PConf}_k(M))\to H^*(\operatorname{PConf}_{k-1}(M-\{{\rm pt}\}))$ induced by fibre inclusion is an epimorphism. This is seen by identifying this homorphism with the edge homomorphism of the spectral sequence. If we now assume that $H^*(M)$ is torsion-free, and that the system of local coefficients on $M$ is trivial, so that $$E_2^{p,q}\cong H^p(M)\otimes H^q( \operatorname{PConf}_{k-1}(M-\{{\rm pt}\})),$$ then this necessary condition is also sufficient. This follows from the multiplicative nature of the spectral sequence. A reference is McCleary's book on spectral sequences, Theorems 5.9 and 5.10. Now I claim that when $k=2$ and $M$ is an oriented closed manifold with torsion-free cohomology, the spectral sequence always collapses at $E_2$ (regardless of whether the first Fadell-Neuwirth fibration admits a section). Let $\Delta\in H^{\dim M}(M\times M)$ denote the diagonal class (i.e., the Poincaré dual of the diagonal submanifold $d: M\to M\times M$). Then a theorem of Cohen and Taylor gives an algebra isomorphism $$ H^*(\operatorname{PConf}_2(M))\cong \frac{H^*(M\times M)}{(\Delta)}. $$ (This can be seen using the long exact sequence of the pair $(M\times M,\operatorname{PConf}_2(M))$, together with excision and the Thom isomorphism for the normal bundle of the diagonal.) On the other hand, letting $\mu\in H^{\dim M}(M)$ denote the fundamental class (dual to $1\in H^0(M)$), we get an algebra isomorphism $$ H^*(M-\{{\rm pt}\})\cong \frac{H^*(M)}{(\mu)}. $$ The fibre inclusion homomorphism $i^*:H^*(\operatorname{PConf}_2(M))\to H^*(M-\{{\rm pt}\})$ is induced by the homomorphism $H^*(M\times M)\to H^*(M)$ which sends $1\times x$ to $x$, and is therefore surjective. I suspect the answer to your question to be no in general, but I don't have a good enough understanding of the differentials to produce an example. Perhaps the following paper may be helpful in this regard: Shih, Weishu, Homologie des espaces fibrés, Publ. Math., Inst. Hautes Étud. Sci. 13, 93-176 (1962). ZBL0105.16903.<|endoftext|> TITLE: 2-morphisms for Bord(n) QUESTION [5 upvotes]: I am currently reading in Boundary Conditions for Topological Quantum Field Theories, Anomalies and Projective Modular Functors, and have a (I guess) pretty basic question for my understanding of the (∞, n) category of Cobordisms... Their (informal) definition of Bord(n) is A genuine example of an (∞, n)-category with n > 0 is given by Bord(n), the ∞-category of cobordisms, which can be informally described as consisting of having points as objects, 1-dimensional bordisms as 1-morphisms, 2-dimensional bordisms between bordisms as 2-morphisms, and so on until we arrive at n-dimensional bordisms as n-morphisms, from where higher morphisms are given by diffeomorphisms and isotopies: more precisely, the (n + 1)-morphisms are diffeomorphisms which fix the boundaries, (n + 2)-morphisms are isotopies of diffeomorphisms, (n + 3)-morphisms are isotopies of isotopies, and so on. I am wondering about the 2-morphism (and therefore any higher morphism): Lets take the interval I=[0,1] as the bordism between two points {0} and {1}, and similarly I'=[3,4]. What would be the 2-morphism between I and I' (or equivalently, what would be the endomorphisms of I)? (informally). The reason why I am asking simply is that I cannot think of any 2 dimensional compact manifold M with boundary where the boundary is the disjoint union of two intervals, I and I'... Thanks in advance! Alex :) REPLY [3 votes]: The endomorphisms of the interval are, at this informal level of discussion, surfaces with $S^1$ boundary. More generally, if you want to talk about $\hom(M,N)$, where $M$ and $N$ are $k$-dimensional bordisms, then certainly $M$ and $N$ had better have the same domain and codomain $B = \partial M = \partial N$ (since they had better be $k$-morphisms between the same $(k-1)$-morphisms). A bordism from $M$ to $N$ is then a manifold with boundary $M \cup_B N$.<|endoftext|> TITLE: Decay estimate for the heat equation: $\sup_{t>0}\int_{\mathbb{R}} t^\alpha |u_x|^2\ dx$ QUESTION [5 upvotes]: Let $u$ be a solution of the heat equation $$u_t - u_{xx} = 0, \quad t>0, x \in \mathbb{R}$$ with initial data $u(0,\cdot) = u_0$. Fix $\alpha >0$. How can I estimate (without using explicitly the heat kernel) $$\sup_{t>0}\int_{\mathbb{R}} t^\alpha |u_x|^2 \ dx,$$ in terms of the initial data? Could you point out a reference where such an estimate is obtained? Is it fair to call what we obtain a decay estimate? REPLY [4 votes]: I'm sorry for the late answer, but joined MathOverflow just this week. The Fourier Splitting method, developed by María Elena Schonbek in the 80's asserts that "decay is determined by the low frequencies of the solutions" for many dissipative linear and nonlinear equations (heat, fractional heat, Navier-Stokes, dissipative quase-geostrophic, amongst many others). For $v_0 \in L^2 (\mathbb{R} ^n)$, assume there is a unique $-\frac{n}{2} < r^{\ast} (v_0) < \infty$, called the decay character, such that \begin{equation} r^{\ast} (v_0) = \lim _{\rho \to 0} \rho ^{-2r-n} \int _{B(\rho)} \large|\widehat{v}_0 (\xi) \large|^2 \, d \xi \end{equation} i.e., the decay character measures the "order" of $v_0$ at the origin in frequency space (see Bjorland and Schonbek, Adv. Diff. Eq. 2009; Niche and M.E. Schonbek, J. London Math. Soc. 2015). Then, for a large family of dissipative operators, the decay character provides sharp estimates for decay of linear (and some nonlinear) equations. In the specific case of the heat equation in $\mathbb{R}^n$, we have that for the solution $v$ with initial data $v_0$ \begin{equation} C_1 (1 + t)^{- \left( \frac{n}{2} + r^{\ast} \right)} \leq \Vert v(t) \Vert _{L^2} ^2 \leq C_2 (1 + t)^{- \left(\frac{n}{2} + r^{\ast} \right)}, \end{equation} for some absolute constants $C_1, C_2 > 0$, see Theorem 6.5 in Bjorland and M.E. Schonbek (op.cit.) and Theorem 2.10 in Niche and M.E. Schonbek (op. cit.). Now take $u_0 \in \dot{H} ^1 (\mathbb{R})$, i.e. $\partial_x u_0 \in L^2 (\mathbb{R})$. For $r^{\ast} = r^{\ast} (\partial_x u_0)$ the estimate above implies \begin{equation} C_1 (1 + t)^{- \left( \frac{1}{2} + r^{\ast} \right)} \leq \Vert u(t) \Vert _{\dot{H}^1} ^2 \leq C_2 (1 + t)^{- \left( \frac{1}{2} + r^{\ast} \right)}. \end{equation} If $u_0 \in H^s (\mathbb{R}), s > 0$, then $u_0 \in L^2 (R)$ as well and from Theorem 2.11 in Niche and M.E. Schonbek we have $r^{\ast} (\partial_x u_0) = s + r^{\ast} (u_0)$ so \begin{equation} C_1 (1 + t)^{- \left( \frac{1}{2} + s + r^{\ast} (u_0) \right)} \leq \Vert u(t) \Vert _{\dot{H}^1} ^2 \leq C_2 (1 + t)^{- \left( \frac{1}{2} + s + r^{\ast} (u_0) \right)}. \end{equation} Note that the decay character does not always exists, there are $v_0 \in L^2 (\mathbb{R}^n)$ which oscillate a lot near the origin (in frequency space) for which $r^{\ast} (v_0)$ is not defined, see the article by Brandolese in SIAM J. Math. Anal. 2016 to find a precise characterization in terms of Besov spaces for when the decay character exists and for other results.<|endoftext|> TITLE: Is the action of $SO(n)$ on the sphere $S^{n-1}$ ballanced? QUESTION [5 upvotes]: A subset $B$ of a group $G$ is called balanced if $gBg^{-1}=B$ for all $g\in G$. An action of a group $G$ on a metric space $X$ is called ballanced if for each non-empty balanced subset $B\subset G$ and each $x\in X$ the set $Bx$ coincides with an open or closed ball centered at $x$. This means that the topology (and a large piece of the metric structure) of $X$ can be recovered from the ballanced action of the group $G$ on the set $X$. The term "ballanced" was coined as a mix of two words: "ball" and "balanced". Fact. The standard action of the group $SO(3)$ on the sphere $S^2$ is ballanced. Proof. Take any non-empty balanced subset $B\subset SO(3)$ and observe that any element $g\in B$ is a rotation of the sphere by some angle $\alpha$. Being balanced, the set $B$ contains all possible rotations of the sphere by the angle $\alpha$. Looking at spherical equilateral triangles with base points $x,y$ on the sphere and the angle $\alpha$ at the vertex, we can see that the points $y$ fill a closed spherical disk $D_g$ centered at $x$. The union $\bigcup_{g\in B}D_g$ of such spherical disks is an open or closed spherical disk in $S^2$, which coincides with $Bx$. Problem 1. For which $n$ the action of the group $SO(n)$ on the sphere $S^{n-1}$ is ballanced? Is it ballanced for $n=5$? Is it ballanced for all odd $n$? Remark. The action of $SO(n)$ on $S^{n-1}$ is ballanced for $n\in\{1,3\}$ but not ballanced for any even $n$ (since the singleton $B=\{-1\}$ consisting of the map $-1:x\mapsto -x$ is balanced in $SO(n)$ but $Bx=\{-x\}$ is not a ball centered at $x$). So, the question actually concerns odd $n$. But for even $n$ we can ask a local version of the ballanced property. An action of a topological group $G$ on a metric space $X$ will be called locally ballanced if there exists a neighborhood $U\subset G$ of the unit such that for any non-empty balanced subset $B\subset U$ and any $x\in X$ the set $Bx$ is an open or closed ball centered at $x$ in the metric space $X$. Problem 2. For which $n$ the action of the group $SO(n)$ on the sphere $S^{n-1}$ is locally ballanced? Is it locally ballanced for $n=4$? Is it locally ballanced for all $n\ge 3$? REPLY [8 votes]: If $n$ is even, then the action is not locally ballanced. You can choose $B$ to be the conjugacy class of an element of $SO(n)$, as close as you like to the identity, that does not have $1$ as an eigenvalue. Now whatever $x$ is, $x$ will not belong to $Bx$. On the other hand, if $n$ is odd then every element of $SO(n)$ does have $1$ as an eigenvalue, so if $B$ is any conjugacy class then $x\in Bx$. Furthermore, $Bx$ is preserved by all elements of $SO(n)$ that fix $x$, and is connected, and (if $n>1$) does not consist of $x$ alone; and this makes it a closed ball centered at $x$. Therefore for general balanced nonempty $B$ the set $Bx$ is a union of such balls.<|endoftext|> TITLE: Information from the derived categories of coherent sheaves QUESTION [6 upvotes]: Recently I read some results about derived categories of coherent sheaves, and see one use Fourier-Mukai transforms to prove that the derived categories of coherent sheaves of a scheme, under some conditions, determine the scheme. More precisely, I have seen (Bondal and Orlov) Let $X$ and $Y$ be smooth projective varieties. If there is an exact equivalence $\text{D}^{\text{b}}(X)\simeq\text{D}^{\text{b}}(Y)$ and the canonical sheaf of $X$ is ample or anti-ample, then $X\simeq Y.$ (corollary of above plus some discussions on elliptic curves) Let $C$ be a smooth projective curve and $Y$ be a smooth projective variety. Then $\text{D}^{\text{b}}(X)\simeq\text{D}^{\text{b}}(Y)$ if and only if $X\simeq Y.$ (Gabriel) Let $X$ and $Y$ be smooth projective varieties. Then $\text{Coh}(X)\simeq\text{Coh}(Y)$ if and only if $X\simeq Y.$ My question is, do these results really help determine a variety? I mean, passing the question to the level of derived version will make it easier? Is there any concrete applications? Thanks!! REPLY [4 votes]: If I understand the question correctly, it is about what information about a smooth projective variety you can extract from its derived category of coherent sheaves. Some result in this direction (besides those mentioned in the question statement) the dimension of a variety is uniquely determined by its derived category the canonical ring (and thus Kodaira dimension) is uniquely determined by the derived category. a smooth projective surface which is not elliptic, K3 or an abelian variety, is determined up to isomorphism by the derived category. For K3 surfaces, the Hodge structure on $H^*(X, \mathbb{Z})$ is determined up to isometry by the derived category (note that $H^2(X, \mathbb{Z})$ is NOT determined up to Hodge isometry by the derived category; if it were, then the derived category would be a perfect invariant for K3 surfaces by Global Torelli). However, the derived category is not a perfect invariant. For example: abelian variety and its dual have equivalent derived categories. A nice place to read about this stuff is Hyubrecht's "Fourier--Mukai transforms in algebraic geometry". As for the applications, I do not know much about them but here is one example: Bridgeland had a conjecture describing the group of autoequivalences of the derived category of a K3 surface. This conjecture has been proved in the Picard rank 1 case by Bayer--Bridgeland; Sheridan--Smith inferred from this that the symplectic Torelli group of some K3 surfaces is infinitely generated. In general it is very hard to say anything about symplectic Torelli group. Though one should note that in this example, there were a lot of extra non-trivial ideas (like homological mirror symmetry) so maybe it is not fair to say that this result is an application of theorems about derived categories.<|endoftext|> TITLE: Models of arithmetic in a signature with exponentiation but not addition and multiplication QUESTION [8 upvotes]: Let $\mathcal{L}_{\mathrm{exp}}$ be the language with signature $(0, ^\prime, <, \mathrm{exp})$ (with $0$ interpreted as zero, $^\prime$ as successor, and $\mathrm{exp}(x)$ as $2^x$) and let $\mathsf{TA}_{\mathrm{exp}}$ be true arithmetic in this signature; i.e. $\{ \phi \in \mathsf{Sent}(\mathcal{L}_{\mathrm{exp}}): \mathbb{N} \vDash \phi \}$. What is known about the models of $\mathsf{TA}_{\mathrm{exp}}$? Does the analogue of Tennenbaum's theorem hold? I've loooked around a bit, but haven't been able to find anything; pointers to the literature would be greatly appreciated. REPLY [7 votes]: $\mathsf{TA}_{\exp}$ does have recursive nonstandard models. In fact, even the considerably stronger theory $\mathrm{Th}(\mathbb N,+,2^x)$ (Presburger arithmetic with exponentiation) has recursive nonstandard models. This follows from results of Semënov, elaborated in this paper by Point: the theory is decidable (with an explicitly given axiom set), and it has elimination of quantifiers in a language expanded with $<$, congruence predicates, and the logarithm function $\ell(x)=\lfloor\log_2(x)\rfloor$. It is a general fact that any decidable theory has a recursive model, and even a model with decidable satisfaction predicate. Now here we have the minor problem that this construction may happen to produce the standard recursive model; thus, we apply it to the theory expanded with a new constant $c$, along with axioms making $c$ nonstandard. We need to make sure the theory is still decidable, and one way to do that is to use the quantifier elimination result to exhibit a recursively axiomatized complete type. I didn’t check this in detail, but I believe it is possible. Alternatively, one may do something similar directly for the theory of $\mathsf{TA}_{\exp}=\mathrm{Th}(\mathbb N,<,2^x)$. I did verify the details, but unfortunately, the full proof is too long to present here, so I will only state the results with a few hints. Theorem: $\mathrm{Th}(\mathbb N,0,{}',<,2^x)$ is decidable, and it can be explicitly axiomatized by a. the axioms of a discrete linear order with a least element and no largest element; b. the axioms $x<\exp(x)$, $x0\,\exists y\,(\exp(y)\le x<\exp(y'))$; c. the axioms $\exp(\overline k)=\overline{2^k}$ and $\overline k TITLE: Kunneth formula for semidirect product QUESTION [5 upvotes]: I wonder if the following Kunneth formula for semidirect product is valid $$ H^n(N\rtimes_\phi G;\mathbb{Z}) = \sum_{i+j=n} H^i(G; H^j(N;\mathbb{Z})),$$ where $H^*$ is the group cohomology and $G$ has a proper action on $H^j(N;\mathbb{Z})$ as induced by $\phi$. (For direct product, $G$ has no action on $H^j(N;\mathbb{Z})$ and the above reduces to the standard Kunneth formula.) https://arxiv.org/abs/math/0406130 only showed above when $N$ has a form $\mathbb{Z}^k$. REPLY [7 votes]: Just to collect the references I wrote in comments, and more. They each give (in)finite (non)abelian counterexamples, as well as general explanations for failure of collapse of the LHS spectral sequence with semi-direct products: Charlap, L. S.; Vasquez, A. T., The cohomology of group extensions, Bull. Am. Math. Soc. 69, 815-817 (1963). ZBL0122.02803. Benson, D. J.; Feshbach, M., On the cohomology of split extensions, Proc. Am. Math. Soc. 121, No. 3, 687-690 (1994). ZBL0819.20058. Totaro, Burt, Cohomology of semidirect product groups, J. Algebra 182, No. 2, 469-475 (1996). ZBL0862.20038. Siegel, Stephen F., On the cohomology of split extensions of finite groups, Trans. Am. Math. Soc. 349, No. 4, 1587-1609 (1997). ZBL0951.20039. See also references within.<|endoftext|> TITLE: What English translations are there of work done by the Italian school of algebraic geometry? QUESTION [5 upvotes]: What English translations are there of work done by the Italian school of algebraic geometry? Perhaps I'm being too spoiled here, given that mathematical French, German, Italian are much easier to pick up on the fly than say, mathematical Russian or Japanese, for a native English speaker. REPLY [5 votes]: Here is a complete bibliography. It includes translations, but not in English (many in French, German, and Spanish). From this bibliography I notice that only Guido Fubini and Beniamino Segre occasionally published in English. The collected papers of Giacomo Albanese, which I have not accessed, may well contain English translations. In answer to the OP, assuming this bibliography is indeed complete, and with the possible exception of Albanese,$^\ast$ I would conclude that, no, there are no published English translations of original papers from the Italian school of algebraic geometry. $^\ast$ update: I located the table of contents of Albanese's collected papers, and it has only the Italian originals, no translations.<|endoftext|> TITLE: Reference for: $p$-primary component of $\pi^S_k$ is $\Bbb Z_p$ when $k=2l(p-1)-1$ QUESTION [8 upvotes]: I remember coming across this result some time ago but I am having trouble finding a reference for it. It goes something like this: Let $p$ be a(n odd?) prime, then the $p$-primary component of $\pi^S_k$ is $\Bbb Z_p$ when $k=2l(p-1)-1$ for $l=1,\dots,p-1$ and is trivial for all other $k<2p(p-1)-2$. This is what I have written down on the back of an envelope. I checked this with Wikipedia's table and it seems to be true. What is the reference in which it is proved? And if its simple could you overview it as an answer here? My guess is that it is proven by Toda but his papers are difficult to search through. REPLY [6 votes]: This follows easily from Theorem 4.4.20 of Ravenel's book, Complex cobordism and stable homotopy groups of spheres". The elements in $Ext^i$ with $i>1$ have total degrees greater than or equal to $2p(p−1)−2$, so we only have to look at $Ext^1$, and we have $$\pi _{ql-1}(S^0)\cong Ext ^{1,ql}\cong Z/p \mbox{ where $q=2(p-1)$}$$ other $Ext^1$ groups vanishing in the degree range you give.<|endoftext|> TITLE: Is a smooth intersection of hypersurfaces equidimensional? QUESTION [5 upvotes]: Let $X$ be a smooth projective complex algebraic variety. Let $V_i$, for $i=1,\dots, n$, be a collection of (smooth) connected hypersurfaces such that, for all $I\subseteq [n]$, the intersection $\cap_{i \in I} V_i$ is smooth. Is the intersection $\cap_{i=1,\dots, n} V_i$ equidimensional? Added later: If we take $V_i$ to be smooth effective (possible non ample) divisors, is the intersection $\cap_{i=1,\dots, n} V_i$ equidimensional? REPLY [5 votes]: If the intersection $\cap_{i=1}^nV_i$ is irreducible, then it is equidimensional. Otherwise let $r TITLE: Cayley graph properties QUESTION [5 upvotes]: Consider an infinite graph that satisfies the following property: if any finite set of vertices is removed (and all the adjacent edges), then the resulting graph has only one infinite connected component. So, obviously, the Cayley graph for the group $\mathbb Z \times \mathbb Z$ w.r.t. the standard generating set is an example. Obviously, the Cayley graph for a free group is not an example. I have a question: what is the name of such a property? Has it been studied? And the next question: which are the Cayley graphs with this property? REPLY [6 votes]: Quoting the answer given by YCor in the comments. For a graph with finite valency, it is called "one-ended graph" and it's been extensively studied (in the case of Cayley graphs, see for instance Stallings theorem). If one allows infinite valency vertices, it is sometimes called "one-ended graph" (but there's then an inequivalent alternative definition of one-ended, replacing "finite" with "bounded").<|endoftext|> TITLE: Model theory of Banach algebras QUESTION [5 upvotes]: Let us consider the (metric) theory of Banach algebras. I have a sentence encoding the (possible) openness of multiplication in a given Banach algebra: $$(\forall x) (\forall y) (\forall \varepsilon > 0)( \exists \delta > 0)(\forall z)\; \big(\|z-xy\| < \delta \Rightarrow (\exists u)(\exists v)[\|u-x\|<\varepsilon \& \|v-y\|<\varepsilon \;\&\; z = uv]\big)$$ This sentence looks quite first-order to me. Suppose that a Banach algebra $A$ satisfies the above formula. Can we directly conclude that some ultrapower of $A$ would also satisfy it? By an ultrapower, I mean the metric (Banach-space) ultrapower. I know that people in C*-algebras have mastered such methods but I am not sure to what extent a similar machinery is available in the more general setting of Banach algebras. (I know that this holds for all ultrapowers if we shift $(\forall x)(\forall y)$ after $(\exists \delta > 0)$ but the reason is not model-theoretic and the condition is too strong for me.) Should it hold, I would be most grateful for pointing out the relevant literature touching this topic. REPLY [2 votes]: What makes this tricky is that metric or continuous logic (in the sense of this document) doesn't directly allow for quantification over $\varepsilon$'s and $\delta$'s like that. You can still express it with a schema, but what's difficult in this case is that you have quantification over elements of the structure before quantifying over $\varepsilon$ and $\delta$. There's also some subtlety with how you formalize implication in this context. Generally speaking a statement is only going to be preserved by ultrapowers/metric elementary equivalence if it's uniformly true in the structure. This fact is baked into the formalism in some places, such as the requirement to give a modulus of uniform continuity for a given function. If a function in a metric structure fails to be uniformly continuous, then in some elementary extension it won't even be a function. More generally purely 'topological' properties are often too fragile to be preserved under ultrapowers unless they are actually true in some uniform way. EDIT2: This is not as straightforward as I thought. I can tell you roughly when this is not going to work. If there is an $\varepsilon > 0$ such that for every $\gamma > 0 $ there exists vectors $x,y$ with $\| x\| = \| y\|=1$ such that the $\delta$ needed to satisfy your condition is $<\gamma$, then in an ultrapower (over a countable index set) the condition will fail. So I would guess that in order for the condition to be satisfied in ultrapowers of the structure you actually need the stronger form you mentioned where the $(\forall x)(\forall y)$ quantifiers are after the $(\exists \delta > 0)$ quantifier. And I mean 'need' in a strong sense, i.e. if you have a Banach algebra such that your condition is true in all ultrapowers then the stronger form of the condition (although restricted to norm 1 vectors) actually holds in the first place. EDIT: I think that maybe you can construct a counterexample from the proof of theorem 4.12 here, specifically some kind of product algebra of the algebras $\ell_1(\mathbb{Z}_{n!})$ with the convolution algebras. EDIT3: This reference is still relevant but I was being too optimistic about openness of multiplication being preserved under products.<|endoftext|> TITLE: Unique factorisation of prime geodesics? QUESTION [10 upvotes]: In T. Sunada's 1985 paper ``Riemannian coverings and isospectral manifolds'', it is noted that for a compact Riemannian manifold $M$, prime (closed and non self-intersecting) geodesics behave like primes in a number field. That is, given a finite normal (i.e. ``Galois'') cover $M\to N$ of Riemannian manifolds, the preimage of prime geodesic $\mathfrak{p}$ in $N$ is a disjoint union $\mathfrak{P}_1,...,\mathfrak{P}_k$ of prime geodesics in $M$. The degree $f_i$ of the covering map $\mathfrak{P}_i\to \mathfrak{p}$ of circles should be thought of as the inertia degree of $\mathfrak{P}_i/\mathfrak{p}$, and the Frobenius $\sigma_{\mathfrak{P}_i}\in \text{Deck}(M/N)$ acts on $\mathfrak{P}_i$ by a $1/f_i$ rotation. Note $[M:N]=\sum f_i$. Other similar properties also hold, for instance there is also a Chebatorev density theorem. My question is: is there an analogue of unique factorisation of ideals in this context? I don't know what the right analogue of a non-prime ideal is here: perhaps $\mathfrak{p}^2$ should be $\mathfrak{p}$, but the geodesic loops round twice, but I can't think of the right definition for $I$ in general. REPLY [2 votes]: These questions are answered in Darin Brown's very nice paper: Brown, Darin, Lifting properties of prime geodesics, Rocky Mt. J. Math. 39, No. 2, 437-454 (2009). ZBL1170.53022. (available free).<|endoftext|> TITLE: What is a tensor category? QUESTION [16 upvotes]: A monoidal category is a well-defined categorical object abstracting products to the categorical setting. The term tensor category is also used, and seems to mean a monoidal category with more structure, usually the structure of an abelian cateogry, but I can't find a precise definition. So I ask question: What is a tensor category? REPLY [13 votes]: There is no single accepted definition of “tensor category” that matches all uses. Almost always it means abelian (or a similar cocomplete condition) and k-linear. Usually it also means rigid. Often it also means the unit object is simple. Occasionally it also means symmetric. You just have to look at the definition used in each particular paper. REPLY [13 votes]: There seem to be many different definitions in the literature, based on individual papers. But, I think that might change, now that the textbook Tensor Categories, by Etingof, Gelaki, Nikshych, and Ostrik, has appeared. They define a tensor category as follows: Let $k$ be an algebraically closed field, and $C$ a locally finite $k$-linear abelian rigid monoidal category. If the bifunctor $\otimes: C\times C\to C$ is bilinear on morphisms, then $C$ is called a multitensor category. Assume that $C$ is indecomposable (i.e. not equivalent to a direct sum of nonzero multitensor categories). If $End_C(1) \cong k$ then $C$ is called a tensor category. Of course, I've also seen tensor category used to mean monoidal category, often in papers to do with braidings. But, generally, tensor means more than monoidal. This is also true in homotopy theory: a tensor model category has to satisfy more than a monoidal model category (it needs the functors $X\otimes -$ and $-\otimes X$ to preserve weak equivalences, for cofibrant X; see this paper of mine with Yau). Anyway, I agree with Noah that you should try to figure it out from context, and asking questions like this is a good way to make sure people are being careful with the terminology, so that we don't end up with even more definitions! For myself, I'll only use "tensor category" for what Etingof, Gelaki, Nikshych, and Ostrik mean.<|endoftext|> TITLE: Lattices of PU(n,1) with large abelianization QUESTION [7 upvotes]: I am interested in properly discontinuous cocompact subgroups of the group $PU(n,1)$ of automorphisms of the complex hyperbolic space $H^n_{\mathbb{C}}$, says for $n=2,3$. Is there such a lattice $G$ whose abelianisation $G' = G/[G,G]$ has positive rank? How large can the rank of $G'$ be? In dimension $n=1$, the situation is that any such a lattice is the fundamental group of a hyperbolic surface, so the rank of its abelianisation, i.e the rank of the first homology group of the surface, is twice the genus. So, it can be arbitrary large. REPLY [3 votes]: The other answers give fine examples, but I found a reference to the 1981 thesis of Livné which constructs complex hyperbolic lattice which admits a surjective holomorphic map to a Riemann surface. This is detailed in chapter 16 of Deligne, Pierre; Mostow, George Daniel, Commensurabilities among lattices in $\text{PU}(1,n)$, Annals of Mathematics Studies. 132. Princeton, NJ: Princeton University Press. 183 p. $ 49.95/ hbk; $ 19.95/pbk; £ 33.50/hbk; £ 15.00/pbk (1993). ZBL0826.22011. Since Riemann surfaces have covers with arbitrarily large betti numbers, so do the corresponding ball quotients. The point is here that if one has a holomorphic map $\phi: X \to Y$, $X$ and $Y$ compact, $Y$ a Riemann surface, then $\pi_1(X)$ must surject a finite-index subgroup of $Y$. Otherwise $\phi_\#(\pi_1(X))$ would induce an infinite cover $\tilde{Y}\to Y$ and lift $\tilde{\phi}: X\to \tilde{Y}$ so that $\phi$ factors through $\tilde{\phi}$. But the map $\tilde{\phi}$ must be constant, since it maps a compact complex manifold to a non-compact Riemann surface (by the open mapping theorem, restricted to 1-dimensional complex subspaces of $X$, $\tilde{\phi}$ is open if non-constant, and hence $\tilde{\phi}(X)$ is both open and compact in $\tilde{Y}$, a contradiction). Thus, $\pi_1(X)$ must surject $\pi_1(\tilde{Y})$ for $\tilde{Y}\to Y$ a finite-sheeted cover. Then covers of $\tilde{Y}$ with arbitrarily large betti number induce such covers of $X$. Some generalizations to other examples are given by Deraux using a forgetful map. If the lattice is arithmetic, then covers induced from a map to a Riemann surface will usually not be congruence covers. In the arithmetic case, once one has a (congruence) cover with positive betti number, one can find further (congruence) covers with arbitrarily large betti numbers, as hinted at in Venkataramana's answer. Another perspective on this is given here.<|endoftext|> TITLE: Splitting of primes in cyclotomic $\mathbb{Z}_p$-extension QUESTION [6 upvotes]: Let $p$ be a prime, $K$ be a finite extension of $\mathbb{Q}$ and $K_{\infty}$ be a cyclotomic $\mathbb{Z}_p$-extension of $K$ i.e. Gal$(K_{\infty}/K) \cong \mathbb{Z}_p$, the group of $p$-adic integers under addition. Then What can we say about the prime decomposition in $K_{\infty}$ for any prime of $\mathbb{Q}$ for e.g. ramification index, order of the decomposition subgroup, finitely decomposed or not ? The only result I know in this direction is that such extensions are unramified outside of $p$. Can we say something more when $K=\mathbb{Q}$ ? REPLY [5 votes]: For any $\mathbb{Z}_p$-extension the ramification is concentrated among the primes above $p$. Those that are ramified are totally ramified. For the cyclotomic $\mathbb{Z}_p$-extension all places above $p$ are totally ramified. That is because you obtain it as a subextension of $\bigcup_n K(\mu_{p^n})$. For all unramified places the decomposition group is of finite index.<|endoftext|> TITLE: Is KL divergence $D(P||Q)$ strongly convex over $P$ in infinite dimension QUESTION [12 upvotes]: By KL divergence I mean $D(P||Q) = \int dP \log(\frac{dP}{dQ})$. I am looking for the conditions under which this strong convexity is true and possible references. I could not find an answer for infinite dimensions. I am specifically interested in the case where the dimension is uncountable. Note that this strong convexity is equivalent to the strong convexity of negative entropy function $H(P) = \int dP \log(dP)$. REPLY [6 votes]: $\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon}$ Take any probability measures $P_0,P_1$ absolutely continuous with respect (w.r.) to $Q$. We shall prove the following: Theorem 1. For any $t\in(0,1)$, \begin{align*} \De:=(1-t)H(P_0)+tH(P_1)-H(P_t) \ge\frac{(1-t)t}2\,\|P_1-P_0\|^2, \end{align*} where $\|P_1-P_0\|:=\int|dP_1-dP_0|$ is the total variation norm of $P_1-P_0$, \begin{equation*} H(P):=D(P||Q)=\int \ln\frac{dP}{dQ}\,dP, \end{equation*} and, for any elements $C_0,C_1$ of a linear space, $C_t:=(1-t)C_0+tC_1$. Thus, by "A third definition[8] for a strongly convex function", indeed $D(P||Q)$ is strongly convex in $P$ w.r. to the total variation norm. We see that the lower bound on $\De$ does not depend on $Q$. Proof of Theorem 1. Take indeed any $t\in(0,1)$. Let $f_j:=\frac{dP_j}{dQ}$ for $j=0,1$, so that $f_t=\frac{dP_t}{dQ}$. By Taylor's theorem with the integral form of the remainder, for $h(x):=x\ln x$ and $j=0,1$ we have \begin{equation*} h(f_j)=h(f_t)+h'(f_t)(f_j-f_t)+(f_j-f_t)^2\int_0^1 h''((1-s)f_t+sf_j)(1-s)\,ds, \end{equation*} whence \begin{align*} \de&:=(1-t)h(f_0)+th(f_1)-h(f_t) \\ &=(1-t)t\,(f_1-f_0)^2\, \int_0^1\Big(\frac t{(1-s)f_t+sf_0}+\frac{1-t}{(1-s)f_t+sf_1}\Big)(1-s)\,ds \\ &=(1-t)t\,(f_1-f_0)^2\, \int_0^1\Big(\frac t{f_{u_0(t,s)}}+\frac{1-t}{f_{u_1(t,s)}}\Big)(1-s)\,ds, \end{align*} where $$u_j(t,s):=(1-s)t+js.$$ So, \begin{equation} \De=\int\de\,dQ=(1-t)t\,\int_0^1(1-s)\,ds\,[tI(u_0(t,s))+(1-t)I(u_1(t,s))], \tag{1} \end{equation} where \begin{equation*} I(u):=\int\frac{(f_1-f_0)^2}{f_u}\,dQ. \end{equation*} Next, take any $u\in(0,1)$. Then $P_1$ is absolutely continuous w.r. to $P_u$. Introducing $g_u:=\dfrac{dP_1}{dP_u}=\dfrac{f_1}{f_u}$, we have \begin{multline*} I(u)=\frac1{(1-u)^2}\,\int\frac{(f_1-f_u)^2}{f_u}\,dQ =\frac1{(1-u)^2}\,\int(g_u-1)^2\,dP_u \\ \ge\frac1{(1-u)^2}\,\Big(\int|g_u-1|\,dP_u\Big)^2 =\frac1{(1-u)^2}\,\|P_1-P_u\|^2=\|P_1-P_0\|^2. \tag{2} \end{multline*} Note also that for any $t\in(0,1)$ and $s\in(0,1)$ we have $u_0(t,s)\in(0,1)$ and $u_1(t,s)\in(0,1)$ and hence, by (2), $I(u_j(t,s))\ge\|P_1-P_0\|^2$ for $j=0,1$. Now Theorem 1 follows by (1). Remark. The constant factor $\frac12$ in the lower bound in Theorem 1 is the best possible one. Indeed, assuming that $P_1$ is absolutely continuous w.r. to $P_0$ and introducing $f:=\frac{dP_1}{dP_0}$, after some rather straightforward manipulations we get \begin{equation} \De=\int k(t,f)\,dP_0, \tag{*} \end{equation} where $\De$ is as before and \begin{equation} k(t,f):=t f \ln f-(1-t+t f)\ln(1-t+t f). \end{equation} Take now any $h\in(0,1)$ and let $f$ take values $1-h,1+h$ each on a set of $P_0$-measure $1/2$, so that $\|P_1-P_0\|=h$. Then, in view of (*), for each $t\in(0,1)$, \begin{equation} \De=\frac12\,k(t,1-h)+\frac12\,k(t,1+h)\sim \frac{(1-t)t}2\,h^2=\frac{(1-t)t}2\,\|P_1-P_0\|^2 \end{equation} as $h\downarrow0$, which confirms the optimality claim.<|endoftext|> TITLE: What are the reflective subcategories of the category of presentable categories? QUESTION [11 upvotes]: I am actually interested in the $\infty$-categorical case, but the same question is meaningful in the $1$-categorical situation as well. A nice property of presentable $\infty$-categories is that if you have one such $\mathcal{C}$ and you have a full subcategory $\mathcal{D}\subseteq\mathcal{C}$ of it, then: Theorem: The inclusion $\mathcal{D}\subseteq\mathcal{C}$ is reflective (i.e. has a left adjoint) if and only if $\mathcal{D}$ is closed under (small) limits and sufficiently filtered colimits. This almost looks like the adjoint functor theorem, but note that we don't know that $\mathcal{D}$ is presentable in advance. In $1$-categories this is proved here, and I bet it is true $\infty$-categorically as well. Let's assume this for a moment. Now, let $\mathbf{Pr}^L$ be the $\infty$-category of presentable $\infty$-categories and left adjoint (equivalently, colimit preserving) functors. I have a full subcategory $\mathcal{C}\subseteq \mathbf{Pr}^L$ and I want to know that there is a left adjoint to the inclusion, under some general closure conditions on $\mathcal{C}$. This would have been the case if only $\mathbf{Pr}^L$ was itself presentable... Of course, there are some serious size issues here! The $\infty$-category $\mathbf{Pr}^L$ is huge (i.e. more than large, which is itself more than small) and I don't think it even has all large limits. What would be the poset $0 \to 1$ to the power of "a lot"? (at least its limit in the $\infty$-category of large categories is not presentable). Also, presentable $\infty$-categories are locally small, this doesn't seem to allow large limits. To conclude, I don't think $\mathbf{Pr}^L$ is "presentable" in the sense that we replace "small" with "large" and "large" with "huge". Keeping "small" and replacing only "large" with "huge" also doesn't seem to work for trivial reasons, since $\mathbf{Pr}^L$ can't be generated from a small set by small colimits, otherwise, it would have been at most large. Nevertheless, I still hope there might be a positive answer to the question in the title: Question: Is there a characterization of reflective subcategories of $\mathbf{Pr}^L$ in terms of closure properties? Perhaps this can be achieved by showing that $\mathbf{Pr}^L$ is "presentable" in some new sense, which takes into account the three sizes small-large-huge. REPLY [9 votes]: Some ideas, building off of Simon Henry and Ivan di Liberti's remarks: $Pr^L$ is in fact essentially a large (not huge) category (in either the ordinary or $\infty$ context). That is, let $\lambda$ be the size of (a skeleton of) the universe of small sets (i.e. "small" means $\lambda$-small, and $\lambda$ is inaccessible). Choose one representative from each equivalence class of objects of $Pr^L$ [1]. I claim that there are $\lambda$-many of these, that the number of isomorphism classes of morphisms between any two of them is $\lambda$, and between any two of these the space of natural isomorphisms (even natural transformations) has at most $\lambda$-many cells. All told, $Pr^L$ localized at the equivalences of categories is equivalent to a large quasicategory with $\lambda$-many simplices. I've detailed a size estimate at the end. $Pr^L$ has small limits and colimits. They are easy to compute: in both $Pr^L$ and $Pr^R$, small limits are computed at the level of underlying categories. $Pr^L$ is built from the presentable categories $Pr^L_\kappa \simeq \kappa-Cocts$ in the following manner.[3] We have a large chain of presentable categories $Pr^L_\kappa$, and $Pr^L$ is the colimit. The linking functors $Pr^L_\kappa \to Pr^L_\mu$ are left adjoints, so preserve colimits, since the corresponding functor $\kappa-Cocts \to \mu-Cocts$ is given by $Ind_\kappa^\mu$, the free completion under $\mu$-small, $\kappa$-filtered colimits, which is left adjoint to the forgetful functor. Here $\kappa-Cocts$ is the category of small categories with $\kappa$-small colimits and functors that preserve them. Generation properties. One kind-of surprising consequence of this is that $Pr^L_\kappa$ is closed under small colimits in $Pr^L$ (not being a full subcategory, it is certianly not coreflective). In particular, $Pr^L$ is not generated under small colimits by any small subcategory. For example, even though every small category is a colimit of copies of the walking arrow category $[2]$, it's not the case that every object of $Pr^L$ is a colimit of copies of $Set^{[2]}$. On the other hand, each object $C$ of $Pr^L$ is a localization of a presheaf category $Psh(C_0) \in Pr^L_0$, and is in fact the geometric realization of a simplicial object $C_\bullet$ where each $C_n\in Pr^L_0$ is a presheaf category (but the simplicial maps are only in $Pr^L_\kappa$ where $\kappa$ is the presentability rank of $C$). For that matter, $Set^{[2]}$ is a strong generator in $Pr^L$ (i.e. the functor $Pr^L(Set^{[2]},-)$ reflects equivalences). These are weaker senses in which $Pr^L$ is "small-generated". I'm not sure about the dual properties. Classically, $Pr^L_\kappa$ is closed in $Pr^L$ under $\kappa$-small PIE limits. And an arbitrary product of objects of $Pr^L_\kappa$ is again in $Pr^L_\kappa$, but if $(F_\alpha : C \to D_\alpha)$ is a family of functors in $Pr^L_\kappa$, the induced functor $C \to \Pi_\alpha D_\alpha$ typically is not in $Pr^L_\kappa$. So $Pr^L_\kappa$ is not closed in $Pr^L$ under small limits. It's possible that $Pr^L$ is generated by, say, $Pr^L_\omega$ under small limits -- I'm not sure. I think at least that $Pr^L_0$ is closed under small limits, though. $Pr^L$ has localizations with respect to sets of morphisms. Let $S$ be a set of morphisms, and let $\kappa$ be large enough so that all $S$ are between objects of $Pr^L_\kappa$. Then for each $\mu \geq \kappa$, we may localize $Pr^L_\mu$ at $S$ because $Pr^L_\mu$ is presentable. These localizations cohere: to see this, it suffices to check that if $X$ is local in $Pr^L_\mu$ with respect to $A \to B$, then $X$ is still local in $Pr^L_\nu$ with respect to $A \to B$ for $\nu \geq \mu$. This is not hard to see: it follows from the fact that every functor in $Pr^L_\nu$ from $A$ to $X$ or $B$ to $X$ is a colimit of functors in $Pr^L_\mu$. It's not clear to me if there's a slick characterization of those localizations which arise from localizing at a set of maps. If $Pr^L$ were presentable, these would just be the accessible localizations. But it's not. These localizations have the property that the localization functor preserves $Pr^L_\kappa$ for sufficiently large $\kappa$, and restricts to an accessible localization on $Pr^L_\kappa$. But other localizations might have this same property -- for instance, one might localize at a saturated class of morphisms whose intersection with each $Pr^L_\kappa$ is generated by a set -- such a localization wouldn't automatically exist, but it might in some cases. Here is a size estimate for $Pr^L$: Objects: The equivalence classes of objects are surjected onto by the following data: A (small) regular cardinal $\kappa$. A small, skeletal, $\kappa$-cocomplete category $C$. The bijection sends $(C,\kappa)$ to $Ind_\kappa(C)$. There are $\lambda$-many regular cardinals, and $\lambda$-many $\kappa$-complete small categories, for a total of $\lambda \times \lambda = \lambda$ many presentable categories up to equivalence. Morphisms: If $C$ is $\kappa$-cocomplete and $\mathcal D$ is presentable, then isomorphism classes of left adjoint functors $Ind_\kappa(C) \to \mathcal D$ are in bijection with the following data: An isomorphism class of functors $F: C \to \mathcal D$ preserving $\kappa$-small colimits. The bijection sends a functor $F$ to its left Kan extension. Note that $\mathcal D$ has at most $\lambda$ many isomorphism classes of objects, each of which has a small number of morphisms between them. Since $C$ is small, this bounds the number of isomorphism classes of functors by $\lambda^{|C|} = \lambda$ since $\lambda$ is inaccessible. 2-morphisms: Now if $F,G: \mathcal C \to \mathcal D$ are left adjoint functors, the number of natural transformations (and in particular the number of natural isomorphisms) between them is bounded by the number of objects of $\mathcal C$ times the number of morphisms between the corresponding images in $\mathcal D$, for a $\lambda$-sized sum of small sets, which has size at most $\lambda$ [2]. [1] This is actually a subtle point -- if you keep an object for each isomorphism class of presentable categories, then for each $\mathcal C \in Pr^L$, you have additional data saying how many copies of each isomorphism class of $\mathcal C$ there are. Then you have to ask what the maximum allowed number of copies is. This will depend on foundational choices like precisely how you defined the notion of presentable categories. I take the perspective that counting size the way I do above is "what you really want to do", although I suppose if you want to think about $Pr^L$ as an $\infty$-category without localizing at the equivalences, you might have some technical issues. [2] Actually, we can do better: we can choose $\kappa$ such that $F,G$ are both Kan extended from the $\kappa$-small objects. Then any natural transformation $F \Rightarrow G$ is determined by its values on the $\kappa$-small objects. So there are actually a small number of natural transformations between $F$ and $G$. Thus when viewed as a 2-category or $(\infty,2)$-category, $Pr^L$ is large, locally large, but locally locally small. Even when viewed as a (2,1)-category or $(\infty,1)$-category, the largeness of $Pr^L$ is "not so bad": $\pi_0$ of a homspace can be large, but its connected components are small. [3] Alexander Campbell points out that in the ordinary context, $\kappa-Cocts$ and thus the equivalent category $Pr^L_\kappa$ is not actually presentable due to strictness issues. So it's probably safest to just interpret everything I'm saying here in the $\infty$ context.<|endoftext|> TITLE: Irreducible representations and invariant subspaces QUESTION [5 upvotes]: Consider two invertible n-by-n matrices, $n>2$, $X$ and $Y$ over a finite field $k$ (say for simplicity $k=\mathbb Z/ \mathbb Z_2$). Is there any reasonable way to check that there is no proper subspace $V\subset k^n$ such that $V$ is invariant under the action of both $X$ and $Y$, i.e. $XV\subset V$ and $YV \subset V$? Any known classes of examples of such $X$ and $Y$ that do not have a proper subspace $V\subset k^n$ such that $V$ is invariant under the action of both $X$ and $Y$? REPLY [2 votes]: Firstly, about "known classes" of examples. Most obviously, if $X$ itself has irreducible characteristic polynomial, in which case it does not admit invariant subspaces. A slightly more interesting example is where $X$ and $Y$ are both diagonalizable, but the eigenbasis of $Y$ consists entirely of vectors will all coordinates non-zero with respect to the eigenbasis of $X$. Another interesting example is the matrix $X$ whose $(i,i+1)$th entry is $1$ and all other entries are $0$, and $Y=X^T$ (the transpose of $X$). If you are interested in small values of $n$ (up to $4$), it may still be possible to classify all such pairs. This would be based on the fact that the invariant subspaces for $X$ depend only on its similarity class type, which has a classification (for each $n$) independent of the finite field.<|endoftext|> TITLE: Can a primitive root-permutation of $A=\{1, 2, \ldots, p-1 \}$ be a cycle of length $p-1$ only for finitely many $p$? QUESTION [6 upvotes]: Let $p$ be an odd prime, $g$ a primitive root of $p$ and $A=\{1, 2, \ldots, p-1 \}$. Obviously, $\sigma_g(p)=\begin{pmatrix} 1 & 2 & \ldots & {p-1} \\ g^1\pmod{p} & g^2\pmod{p} & \ldots & g^{p-1}\pmod{p} \end{pmatrix}$ is a permutation of $A$. I observed that "almost always" $\sigma$ is a product of cycles whose length is strictly less than $p-1$. If $p=3$ and $g=2$, then $\sigma_2(3)=\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}=(1\quad2)$ is a cycle of length $3-1=2$. If $p=5$ and $g=3$, then $\sigma_3(5)=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix}=(1\quad 3\quad 2\quad 4)$ is a cycle of length $5-1=4$. But these two cases seem to be just exceptions. For other (small) values of $p$ and every primitive root $g$ of $p$, there are no cycles of length $p-1$. I can prove that $g=2$ and $g=\frac{p+1}{2}$ cannot produce such a big cycle, but I do not have any idea how to attack the problem for other values of $g$. I conjecture that for every $g$ there is not such a big cycle (for sufficiently large $p$) My question is: Do only finitely many primes $p$ exist, such that for some primitive root $g$ the permutation $\sigma_g(p)=\begin{pmatrix} 1 & 2 & \ldots & {p-1} \\ g^1\pmod{p} & g^2\pmod{p} & \ldots & g^{p-1}\pmod{p} \end{pmatrix}$ is a cycle? REPLY [3 votes]: As pointed out in the comments, primitive roots with a single cycle appear to be rather common. The standard heuristic argument suggests that there are infinitely many such $p$, and a bit more. The share of $N$-cycles among all permutations on $N$ symbols is $1/N$. By the Borel-Cantelli lemma, given an infinite sequence $\{\sigma_i\}$ of independent uniformly distributed random permutations on $N_i$ symbols, if $\sum 1/N_i$ diverges then with probability one infinitely many $\sigma_i$ constitute a single cycle. Since the series of reciprocal primes diverges, we may expect that given a random sequence $\{g_i\}$ of primitive roots mod $p_i$ over all primes, with probability one there are infinitely many $\sigma_i=\sigma_{g_i}(p_i)$ that constitute a single cycle. Of course, since uniform distribution and independence are only heuristic, this is not a proof!<|endoftext|> TITLE: Definition of algebraic de Rham cohomology of non-smooth affine variety QUESTION [8 upvotes]: $\newcommand{\Hdr}{H_{\mathrm{dRh}}}$ $\newcommand{\spec}[1]{\mathrm{spec}(#1)}$ $\require{amsmath}$ Let $A = k[x_1,\ldots,x_n]$ the polynomial ring over a field $k$ of characteristic zero and $I \subseteq A$ an ideal of $A$. Let $Y= V(I) = \spec{A/I}$ and $X = \spec{A}$. According to R. Hartshorne, On the de Rham cohomology of algebraic varieties, Publications mathématiques de l’I.H.É.S., tome 45 (1975), p. 5-99 the algebraic de Rham cohomology of $Y$ is defined as follows: Let $\hat{X}$ be the ringed space with $Y$ as a topological space and the sheaf derived from $\varprojlim_r A/I^r = \hat{A}$ as structure sheaf. Additionally let $\widehat{\Omega_X^\bullet}$ be the completion of $\Omega_{X|k}^\bullet$ (the exterior powers of the ordinary Kähler-differentials) at $I$, so that $\widehat{\Omega_X^\bullet} = \Omega_{X|k}^\bullet \otimes_A \hat{A}$. The exterior derivative extends to $d^p:\widehat{\Omega_X^p} \to \widehat{\Omega_X^{p+1}}$. Now Hartshorne defines $$\Hdr^p(Y) = \mathbb{H}^p(\hat{X}, \widehat{\Omega_X^\bullet})$$ with $\mathbb{H}$ standing for the hypercohomology. If $Y$ is a smooth $k$-scheme, then by Hartshorne's Proposition 1.1, one can replace $X$ by $Y$ and get $$(*) \quad \quad \Hdr^p(Y) = \mathbb{H}^p(Y, \Omega_{Y|k}^\bullet) = H^p(Y, \Omega_{Y|k}^\bullet)$$ where $\Omega_{Y|k}^\bullet$ is the complex of exterior powers of the ordinary Kähler-differentials of $Y/k$ and at the right stands ordinary cohomology, because of affine-ness of the situtation. QUESTION: Does this equality (*) still hold, when $Y$ is no longer a smooth scheme over $k$? The answers to algebraic de Rham cohomology of singular varieties seem to suppose so, but I could not find a proof by searching with google or on my own. REPLY [13 votes]: (Synthesis of answers from comments, posted as community-wiki answer for convenience.) If $k = \mathbb{C}$ then algebraic de Rham cohomology, defined a la Hartshorne using the completion of $X$ along $Y$, is always isomorphic to singular cohomology. This is a theorem of Hartshorne, stated as Theorem 1.6 of his 1972 Manuscripta Math survey article Algebraic de Rham cohomology, and the proof is in Chapter IV of his 1975 PMIHES paper On the de Rham cohomology of algebraic varieties. There are examples, as in this question, of (non-smooth) affine $\mathbb{C}$-varieties $Y$ for which the "naive de Rham cohomology" $\mathbb{H}^\bullet(Y, \Omega^\bullet_Y)$ does not agree with singular cohomology. The Arapura--Kang curve $x^5 + y^5 + x^2 y^2 = 0$ is one such example. Hence the Arapura--Kang curve, or any other example as in (2), must also be an example where naive de Rham cohomology disagrees with Hartshorne's definition. In particular, when this question uses the notation $H^\bullet_{\mathrm{dR}}(Y)$ to mean $\mathbb{H}^\bullet(Y, \Omega^\bullet_Y)$, this is misleading and non-standard, since this definition is not equivalent to the standard definition due to Hartshorne.<|endoftext|> TITLE: Non-separable metric probability space QUESTION [7 upvotes]: Let us say a metric probability space $(X,\rho,\mu)$ has property (*) if: the support of $\mu$ is contained in a separable subspace of $X$. Questions: 1. Is there a standard name for this property? Is it true that continuum+choice implies property (*) -- is there a reference? Is it true that if (either? both?) continuum+choice don't hold, (*) can fail? Again, reference please! REPLY [3 votes]: Iosif Pinelis has given an answer to question 1 and partial answers to 2 and 3. Since he advised me to turn my comments into an answer, here it is. I will deal with the case where the axiom of choice holds, and mention the case where it does not only at the end. I will start with some definitions. A measure $\mu$ on $(X,\mathcal{P}(X))$ is called $\kappa$-additive if for all pairwise disjoint families of sets $(S_i)_{i \in I}$ such that $|I| < \kappa$ (or, speaking loosely, such that the cardinality of the family is $< \kappa$), we have $\mu\left(\bigcup_{i \in I} S_i\right) = \sum_{i \in I}\mu(S_i)$. So "countably additive" is equivalent to $\aleph_1$-additive (Not $\aleph_0$-additive, notice the strict inequality. Finite additivity is $\aleph_0$-additivity.). A real-valued measurable cardinal is a cardinal $\kappa$ such that there exists a $\kappa$-additive probability measure $\mu$ on $(\kappa, \mathcal{P}(\kappa))$ that vanishes on singletons, i.e. $\mu(\{x\}) = 0$ for all $x \in \kappa$. There are two kinds of real-valued measurable cardinals. Measurable cardinals are real-valued measurable cardinals such that there exists a probability measure taking only the values $0$ and $1$ (this can be defined equivalently in terms of the existence of an ultrafilter closed under intersections of cardinality $< \kappa$). Atomlessly measurable cardinals are those such that the measure $\mu$ is atomless, i.e. if $\mu(S) > 0$, there exists a $T \subseteq S$ such that $0 < \mu(T) < \mu(S)$. The following facts were essentially proved by Ulam in his paper Zur Masstheorie in der allgemeinen Mengenlehre, although with different terminology. They are proved in a modern way in Jech's Set Theory: Third Millennium Edition, Chapter 10. A real-valued measurable cardinal is never a successor cardinal, and never the union of a strictly smaller family of strictly smaller sets, i.e. a real-valued measurable cardinal is weakly inaccessible. A measurable cardinal is not embeddable in the powerset of a strictly smaller cardinal, i.e. measurable cardinals are strongly inaccessible. Atomlessly measurable cardinals embed in the continuum, so have cardinality less than or equal to it. A key fact is that the smallest real-valued measurable cardinal is the smallest cardinal having a countably-additive measure $\mu$ vanishing on singletons. This is how Ulam originally phrased things, so if $\kappa$ is the smallest real-valued measurable, he called cardinals $< \kappa$ unmeasurable and cardinals $\geq \kappa$ measurable. However, in the modern terminology, $\kappa^+$ is not real-valued measurable. Because the notion of "cardinals not admitting a countably-additive measure vanishing on singletons" is such a useful notion, and we can't use the term "unmeasurable" for it, Fremlin introduced the name measure-free for cardinals smaller than the first real-valued measurable cardinal. Now we can approach the problem of whether non-separable measures exist. Any set $X$ can be given the discrete metric, and the Borel sets of the topology determined by this metric are exactly $\mathcal{P}(X)$. If $X$ has a countably additive measure $\mu$ vanishing on singletons, then $\mu$ does not have separable support, because separable subsets of a discrete space are countable sets, and $\mu$ vanishes on every countable set by countable additivity. So if there is a real-valued measurable cardinal, then there is a metric space without the property (*) in the question. As pointed out in Iosif's answer, Billingsley proves the converse as Theorem 2 of Appendix 3 in the first edition of Convergence of Probability Measures and its Russian translation, making nontrivial use of the paracompactness of metric spaces. All together, a metric space has no inseparable measures iff all its discrete subspaces have measure-free cardinality. The continuum hypothesis enters as follows. If there exists an atomlessly measurable cardinal $\kappa$, we have $\kappa \leq 2^{\aleph_0}$, and as real-valued measurable cardinals are weakly inaccessible, we not only have cardinals in between $\aleph_1$ and $2^{\aleph_0}$, but a weakly-inaccessible-cardinalful of them (in fact later results showed that there are even more than this). So if the continuum hypothesis holds, or even if the continuum is any smallish sort of cardinal, like $\aleph_3$ or $\aleph_{\epsilon_0 + 1}$ or $\aleph_{\omega_9}$, there are no atomlessly measurable cardinals. But there may still be measurable cardinals, which will necessarily be strongly inaccessible. So the continuum hypothesis implies that metric spaces that have cardinals smaller than the first inaccessible (in fact, again, we can actually go larger than the first inaccessible too, but it is hard to describe exactly how high we can go before reaching the first measurable cardinal), such as $2^{\aleph_0}$, $2^{2^{2^{\aleph_5}}}$ and even $\beth_{\omega_1}$, do not have inseparable measures. However, Silver, Solovay and Kunen showed that if the existence of measurable cardinals is consistent with ZFC, then they are compatible with the generalized continuum hypothesis. Therefore the answer to question 2 is no. And for 3, of course, in the absence of the continuum hypothesis, the reals, equipped with the discrete metric, may be a counterexample (if atomlessly measurable cardinals exist). Billingsley, like many other authors of his era (such as Schaefer in Topological Vector Spaces and Gilman and Jerison in Rings of Continuous Functions), describes the question of whether real-valued measurable cardinals exist as an "unsolved problem". However, the existence of real-valued measurable cardinals is better understood as an axiom that exceeds the consistency strength of ZF(C), thanks to set-theoretic results that have been proven since then. That is to say, one can show that the nonexistence of real-valued measurable cardinals is consistent with ZFC (by taking V=L), but the consistency of real-valued measurable cardinals cannot be proven, assuming only the consistency of ZFC. The reason for this is that if $\kappa$ is measurable, $V_\kappa$ is set-sized model of ZFC, so measurable cardinals prove the consistency of ZFC. Solovay showed in his article Real-Valued Measurable Cardinals that the existence of atomlessly measurable cardinals and the existence of measurable cardinals are equiconsistent, so the existence of real-valued measurable cardinals also implies the consistency of ZFC. I will now briefly discuss what can happen if choice does not hold. In Solovay's model (defined in an much earlier article) where the axiom of choice does not hold, but dependent choice does, and all subsets of $\mathbb{R}$ are Lebesgue-measurable, Lebesgue measure itself defines a probability measure on $([0,1], \mathcal{P}([0,1]))$ vanishing on singletons, which is therefore a non-separable measure on $[0,1]$, equipped with the discrete metric. So in some sense, the absence of choice makes things worse (we only need an inaccessible cardinal, not a measurable one to build this model).<|endoftext|> TITLE: Mathematical Evidence Backing $|\mathbb{R}|=\aleph_2$ QUESTION [30 upvotes]: The "true" size of the real line, $\mathbb{R}$, has been the subject of Hilbert's first problem. Due to the Goedel and Cohen's work on the inner and outer models of $\text{ZFC}$, it turned out to be undecidable based on the current set of the axioms of mathematics. However, as Woodin stated in a panel discussion, this independence result may indicate that there are some missing mathematical principles that need to be discovered and added to $\text{ZFC}$ in order to give us enough mathematical tool to settle the question of the true size of the continuum in the "correct" way. So in some sense, Hilbert's first problem is still open and the search for an answer to the question of determining the true size of the set of real numbers is still ongoing. It is worth mentioning that amongst various transfinite values that $|\mathbb{R}|$ can take up to consistency, set theorists have isolated one particularly special one, namely $\aleph_2$, as the most likely value of the continuum. It is often said that both Goedel and Cohen favored $\aleph_2$ as the "true" size of the continuum as well but I have seen no direct reference to any original quote of them concerning this. In the case of Cohen, who had a strong background in analysis (mostly related to the Littlewood Conjcture), it is particularly important to know whether his choice of $\aleph_2$ (if real) was based on some possibly deep understanding of the mathematical machinery in analysis or not. Question 1. What are references to some original works/interviews of Goedel and Cohen in which they clearly expressed their opinion about the true size of the continuum? What were their reasons to believe in such a specific value for $|\mathbb{R}|$? Though, the story of considering $|\mathbb{R}|=\aleph_2$ goes far beyond mere speculations of Goedel and Cohen. There are actually some mathematical theorems which could be interpreted as a justification for such an assumption. For instance, Woodin's $\Omega$-logic argument in favor of $2^{\aleph_0}=\aleph_2$ initially convinced him to believe that $\aleph_2$ is the true value of continuum, however, later he changed his mind in favor of the Ultimate $L$ principle which implies $2^{\aleph_0}=\aleph_1$. Also, some forcing axioms such as PFA and Martin's Maximum (which successfully settle many independent statements of ordinary mathematics) directly imply the value $\aleph_2$ for the continuum. Here the following natural question arises: Question 2. What are some other examples of mathematical evidence backing $|\mathbb{R}|=\aleph_2$? Maybe some machinery in certain parts of mathematics which work more smoothly if one assumes $2^{\aleph_0}=\aleph_2$ rather than $\aleph_1$ or any other cardinal $\geq\aleph_3$? Maybe some mathematical objects (including the real line itself) start to behave "nicely" or demonstrate some "regularity properties" if the continuum is exactly $\aleph_2$? Please provide references to the results (and quotes from mathematicians) if there is any. Evidence from mathematical disciplines other than set theory are especially welcome. REPLY [6 votes]: You may also look at Judah's paper Was Godel right. Judah intensively discusses why the actual evidences accumulated by 30 years of forcing considerations suggest that the most reasonable size for the continuum is $\aleph_2$. See also Martin's Axiom and the Continuum.<|endoftext|> TITLE: Set theory bootstrapping QUESTION [11 upvotes]: Let $\mathcal{L}$ be the first order language of ZFC set theory, and let $\mathcal{L}_{\infty,\infty}$ be the usual infinitary extension of the language allowing arbitrary long disjunctions/conjunctions and quantifications. What happens if one replaces the usual axiom schema of replacement for ZFC by a new schema over (the class of) formulas from $\mathcal{L}_{\infty,\infty}$ instead of from the first-order theory? Does one get a "standard" model of set theory? Do set theorists believe that this improved version of replacement is true? If so, why still work over the less expressive language $\mathcal{L}$? If not, why not? Does the connection between replacement and transfinite induction completely disappear under this extension? Motivation: I was thinking about Skolem's paradox, and thought that working in $\mathcal{L}_{\infty,\infty}$ might resolve part of the problem---namely that first order language just wasn't expressive enough to talk about important aspects of big sets. This seemed to be backed up by other things I read. But then, it struck me that if we are working with a more expressive language, why limit our axioms to the simpler language? REPLY [6 votes]: The idea of studying ZF and its subsystems formulated in infinitary languages, to my knowledge, seems to have begun and ended with the work of Klaus Gloede, in the 1970s. The following paper of Gloede provides a useful synopsis of his work, which began with his doctoral dissertation (it is available here behind a paywall, the first two pages are freely visible, they consist of the table of contents and the first page of the paper). K. Gloede, Set theory in infinitary languages. ⊨ISILC Logic Conference (Proc. Internat. Summer Inst. and Logic Colloq., Kiel, 1974), pp. 311–362. Lecture Notes in Math., Vol. 499, Springer, Berlin, 1975.<|endoftext|> TITLE: "König's theorem" for $T_2$-spaces? QUESTION [5 upvotes]: For any topological space $(X,\tau)$ we define a matching to be a collection of non-empty and pairwise disjoint open sets. We define the matching number $\nu(X,\tau)$ to be the smallest cardinal $\kappa$ such that for every matching ${\cal M}\subseteq \tau$ we have $|{\cal M}|\leq \kappa$. Recall that $D\subseteq S$ is dense if $D$ intersects every non-empty open set. In the language of hypergraphs, dense sets are vertex covers. So we define the vertex covering number $t(X,\tau)$ to be the smallest cardinality that a dense set can have. König's theorem that for all finite bipartite graphs $B$ we have $\nu(B)=\tau(B)$, when using the equivalents of the definitions above for graphs. Let $(X,\tau)$ be a non-empty Hausdorff space. Questions. 1) Do we always have $\nu(X,\tau)\leq t(X,\tau)$? 2) Do we always have $\nu(X,\tau)\geq t(X,\tau)$? REPLY [11 votes]: What you are calling the "matching number" of $X$ is usually called its Souslin number -- the smallest cardinal bounding the size of any collection of pairwise disjoint open subsets of $X$. What you are calling the "vertex covering number" of $X$ is usually called its density. Thus your question is how the Souslin number and the density of a Hausdorff space are related. The answer is that the density is always $\geq$ the Souslin number, but the inequality can be strict. Thus the answer to your first question is yes, and the answer to the second question is no. (Denote by $S(X)$ and $d(X)$ the Souslin number and density of $X$, respectively. To see that $S(X) \leq d(X)$, just observe that any dense subset $D$ of $X$ is at least as large as a collection $\mathcal U$ of pairwise disjoint open subsets of $X$, because $D$ must contain a point from each member of $\mathcal U$. To see that this inequality can be strict, try to show that the space $[0,1]^\kappa$ has countable Souslin number, but fails to be separable for any $\kappa > \mathfrak{c}$.)<|endoftext|> TITLE: Can the real line be embedded in a space $X$ such that all the nonempty open subsets of $X$ are homeomorphic? QUESTION [12 upvotes]: The question is in the title: Q1: Is there a topological space $X$ containing a copy of the real line and having the property that all the nonempty open subsets of $X$ are homeomorphic? Let us say that $X$ is a homeomorphic open set space, or a hoss for short, if all the nonempty open subsets of $X$ are homeomorphic. Such spaces were asked about here, and N. de Rancourt's answer shows that if $D$ is infinite and discrete then $D^\omega$ is a hoss. It follows that every ``ultrametrizable'' space embeds in a hoss. (A space is called ultrametrizable if it is homeomorphic to an ultrametric space. Spaces of the form $D^\omega$ are themselves ultrametrizable, and every other ultrametrizable space embeds in one of this form.) This is just about all I know about hosses and spaces that embed in them -- any other information or insight is welcome. Familiar examples of hosses include the space $\mathbb Q$ of rational numbers and the space $\mathbb R \setminus \mathbb Q$ of irrational numbers. EDIT: User bof has answered my question by finding a $T_1$ space $X$ containing the real line, and with the property that all nonempty open subsets of $X$ are homeomorphic. However, I still wonder if there is a Hausdorff space with this property: Q2: Is there a Hausdorff space $X$ containing a copy of the real line and having the property that all the nonempty open subsets of $X$ are homeomorphic? REPLY [8 votes]: A metrizable example can be constructed as follows. In the plane consider the subset $$\Xi:=\big\{(x,\tfrac{2k+1}{2^n}):k,n\in\mathbb Z,\;x\in\mathbb R\setminus \tfrac1{2^n}\mathbb Z\big\}.$$ It is clear that $\Xi$ contain (countably many) topological copies of the real line. There are at least two ways of proving that any non-empty open subset of $\Xi$ is homeomorphic to $\Xi$. One is more geometric and is due to Volodymyr Mykhaylyuk (from Chernivtsi). He observed that for any open set $U\subset \Xi$ and any connected component $C$ of $U$ the interval $C$ has a base of clopen neighborhoods homeomorphic to the strip $\Xi\cap(\mathbb R\times(-\sqrt{2},\sqrt{2}))$ in $\Xi$. Then $U$ can be decomposed into countably many pairwise homeomorphic clopen sets and the same can be done with the space $\Xi$. Another way is more global. Just to prove a characterization theorem for the space $\Xi$: Theorem. A topological space $X$ is homeomorphic to the space $\Xi$ if and only if 1) $X$ is a metrizable space; 2) for any point $x\in X$ the connected component $C_x$ containing $x$ is homeomorphic to the real line; 3) the family $\mathcal C=\{C_x\}_{x\in X}$ of connected components of $X$ is countable; 4) for any point $x\in X$ and a neighborhood $O_x\subset X$ of $x$ there exists a non-empty set $V=\bigcup\{C\in \mathcal C:C\cap V\ne\emptyset\}$ in $O_x\setminus C_x$ such that $V$ is clopen in $X\setminus C_x$, $C_x\cup V$ is a neighborhood of $x$. The proof of this characterization uses the back-and-forth argument: We enumerate the connected components and at the $n$-th step construct clopen neighborhoods of $n$-th intervals and establish the combinatorial correspondence between these neighborhoods. I will write down the details in a preprint (with many authors with whom I discussed this problem being on a Summer School in Carpathian mountains). The characterization theorem implies the following corollary: Corollary. Let $(C_n)_{n\in\omega}$ be a family of parwise disjoint curves in $\mathbb R^d$ such that each $C_n$ is homeomorphic to $\mathbb R$, is closed and nowhere dense in the union $X:=\bigcup_{n\in\omega}C_n$ and $\sum_{n=1}^\infty lenth(C_n)<\infty$. Then the space $X$ is homeomorphic to $\Xi$. REPLY [4 votes]: Another example of a $T_1$-space containing the real line and having all non-empty open sets homeomorphic can be constructed using some standard facts of Infinite-Dimensional Topology. Namely it is known that each closed locally compact subset $Z$ of the Hilbert space $\ell_2$ is a $Z$-set in $\ell_2$, which implies that the complement $\ell_2\setminus Z$ is homeomorphic to $\ell_2$, being a contractible $\ell_2$-manifold. Then the Hilbert space $\ell_2$ endowed with the topology $\tau$ consisting of all complements to closed locally compact sets has the required property: it contains a topological copy of the real line and has all non-empty open sets homeomorphic.<|endoftext|> TITLE: How rich is the class of vertex- and edge-transitive polytopes? QUESTION [6 upvotes]: There are only a few regular polytopes (five in 3D, six in 4D, three in any dimension above). In contrast, the class of uniform polytopes becomes very rich with higher dimensions. The class of vertex- and edge-transitive polytopes is in the middle between these two classes (in 3D, they are called quasi-regular). Recently I had a result only applying to polytopes from this class, and now I wonder how relevant this result is. Does it only apply to a small finite number per dimension, or only a few infinite families? Or are there many "new" such polytopes in every higher dimension? Note: I am only interested in convex polytopes in Euclidean spaces. REPLY [2 votes]: I can't provide you with a full classification of those, but at least with a subset of convex polytopes which all are being contained. And which itself might be "rich enough"? Just consider any Coxeter group (reflection group) diagram with just a single node (at any position) being ringed. All their corresponding polytopes (according to Wythoff's kaleidoscopical construction rule) would clearly pass: Any such polytope clearly would be uniform and thence is vertex transitive (via construction by a single seed point, i.e. all vertices are each other's images). Moreover each ringed node symbol represents a class of symmetry equivalent edges (transversal to the according mirror class). Thence, when just a single such is being ringed, the resulting polytope is being bound to be edge transitive as well. - Wrt. linear diagrams those are just all the (multi-)rectified versions of the (convex) regular polytopes. But then there are bifurcated diagrams as well. You might increase that set further, by observing that for those (undecorated) Coxeter group diagrams, which have (as a diagram) some further outer symmetry, an according symmetrical decoration by ringed nodes could result within still edge transitive polytopes again. This is because then the being used edge classes (i.e. those node ringings) due to that outer symmetry become unified after all. - This not only allows for such figures as the decachoron, but for instance all multiprisms of polytopes of the above mentioned subclass. Therefore already within 4D you would encompass the infinite set of all regular n-gonal duoprisms. --- rk<|endoftext|> TITLE: Vector valued disc "algebra" QUESTION [5 upvotes]: I am interested in a vector-valued form of the disc "algebra" (which in this setting is not in general an algebra, hence the scare quotes). Let $E$ be a Banach space, and let $A(\mathbb D,E)$ be the space of bounded continuous functions $f:\overline{\mathbb D}\rightarrow E$ which are holomorphic on $\mathbb D$. Has this space been systematically studied? I am particularity interested in different continuity assumptions. If $f$ is only weakly-continouous, i.e. $\mu\circ f\in A(\mathbb D)$ for each $\mu\in E^*$, is there a counter-example showing that $f$ need not be norm continuous? If $E = F^*$ is a dual space, we can ask the same question with $\mu\in F$, i.e. consider weak$^*$-continuity. There are of course different notions of Banach-space valued Holomorphic functions (weakly or weak$^*$-holomorphic) but it is well-known that these all agree and imply norm analytic. Thus it is continuity at the boundary which is of interest. Probably it is just my lack of knowledge about the classical disc algebra which is a problem, and so I'm hoping a few references might help me? REPLY [2 votes]: I believe the following provides a counter-example for "weak implies norm continuity" in the case $E=c_0$. Given any $F:\overline{\mathbb D} \rightarrow c_0$ let $F_n:\overline{\mathbb D} \rightarrow \mathbb C$ be the $n$th coordinate. If $F$ is weakly continuous on $\overline{\mathbb D}$ and analytic on $\mathbb D$, then each $F_n$ is a member of $A(\mathbb D)$ and $\|F_n\|_\infty\leq K$ for some $K$ independent of $n$. Conversely, if these two conditions hold on the $F_n$ then for any $a=(a_n)\in\ell^1 = c_0^*$ we have that $$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle}\ip{a}{F(z)} = \sum_n a_n F_n(z) $$ and so $z\mapsto \ip{a}{F(z)}$ is in $A(\mathbb D)$ as it is the absolutely convergent sum of members of $A(\mathbb D)$. By properties of functions in Hardy spaces (see e.g. page 50 of these notes) we know that $$ F_n(\xi) = \int_{\mathbb T} \frac{z F_n(z)}{z-\xi} \ dz \qquad (\xi\in\mathbb D). $$ We now seek to construct $F$ by constructing a suitable sequence $(F_n)$. That $F(z)\in c_0$ for all $z$ implies that $F_n(z)\rightarrow 0$ pointwise. Under the assumption that $\|F_n\|_\infty\leq K$ for all $n$, for fixed $\xi\in\mathbb D$ we can apply the dominated convergence theorem to the integral above to see that $F_n(z)\rightarrow 0$ pointwise for $z\in\mathbb T$ implies that $F_n(\xi)\rightarrow 0$. We shall use Outer functions (page 35 in the notes linked above). Choose a positive function $h$ on $\mathbb T$ with $\log h$ being integrable; then we can form an outer function $F$ with $|F|$ having nontangential limit $h$ at the boundary, almost everywhere. It is not clear to me that just because $h$ is continuous it will follow that $F$ is in $A(\mathbb D)$ instead of just a Hardy space; so we work a bit harder. Firstly, choose $t\mapsto h_n(e^{it})$ to be the piecewise linear function with $h_n(1) = h_n(e^{i2\pi/n})=1/n$ and $h_n(e^{i\pi/n})=1$. Let $F_n$ be the Outer function with $|F_n|$ having nontangential limit $h_n$ on the boundary, almost everywhere. We can find $r_n<1$ such that if we define $G_n(z) = F_n(r_nz)$ for $z\in\overline{\mathbb D}$ then there will be $a_n, b_n\in\mathbb T$ with $|a_n-1|<\frac1n, |b_n-e^{i\pi/n}|<\frac1n, ||G_n(a_n)|-\frac1n|<\frac1n$ and $||G_n(b_n)|-1|<\frac1n$. Certainly $G_n\in A(\mathbb D)$ with $\|G_n\|\leq 1$. Finally, form $G\in A(\mathbb D, c_0)$ using the sequence $(G_n)$, and observe that $$ \| G(b_n) - G(a_n) \|_{\infty} \geq | G_n(b_n) - G_n(a_n) | \geq 1 - 3/n $$ so as $a_n \rightarrow 1$ and $b_n\rightarrow 1$, we conclude that $G$ cannot be norm continuous. Alternative construction (from a hint from Yemon Choi): Simply directly define $$ F_n(z) = \exp\big(k_n(e^{-i\pi/n}z-1)\big). $$ where $(k_n)$ is a rapidly increasing sequence. Then obviously $F_n$ is in the disc algebra. We have that $$ |F_n(e^{it})| = \exp\big( k_n(\cos(t-\pi/n)-1) \big). $$ For $t\not=0$ this obviously converges to $0$; we chose $(k_n)$ increasing fast enough so that $\exp( k_n(\cos(-\pi/n)-1) )\rightarrow 0$. Thus $F_n(z)\rightarrow 0$ pointwise on $\mathbb T$, and so on $\mathbb D$, as before. The remainder of the argument also runs the same way, as $F_n(1)\rightarrow 0$ while $F_n(e^{i\pi/n}) = 1$.<|endoftext|> TITLE: Discrete approximations of Riemannian manifolds QUESTION [6 upvotes]: MSE crosspost It's known (due to Perelman) that in class of Alexandrov spaces of fixed dimension and bounded from below curvature Gromov-Hausdorff distance separates homeomorphism types — every $\epsilon$-close to $X$ space will be homeomorphic to $X$ for some $\epsilon$. Well, if we have some finite metric space $X_{\delta}$ which is $\epsilon/2$-close to $X$, then $(X_{\delta}, n, C)$ define homeomorphism type of, say, compact Riemannian manifold, where $n$ is dimension and $C$ is lower curvature bound. Now let's fix $C$ once for all (take $-1$, for example) and call finite metric space $X_{\delta}$ a model of a manifold $X$ if for some $\epsilon$ the only manifold $\epsilon$-close to $X_{\delta}$ is $X$ with some metric with curvature bounded below by $-1$. We can define two functions on homeomorphism (diffeo, if dim > 4, thanks to Grove-Peterson-Wu) classes of $n$-dimensional manifolds: $min \, |X_{\delta}|$ and $min \, k: X_{\delta} \to \Bbb R^k$ for isometric embedding into real space with some norm, where minimum is taken over all models. It seems appropriate to me to call first one metric complexity $mCom(X)$ and second one — essential dimension $edim(X)$. Can $edim(X)$ be strictly less than dimension of $X$ [Edit after @Sergio answer: if $X$ is not contractible or sphere]? Are there some bounds on $mCom$ in terms of something like LS category or topological complexity (i. e. minimal cardinality of open cover over which $eval: X^I \to X \times X$ has local sections? What is, for example, $mCom(S^1 \times S^1)$ — or something else $\geq 2$-dimensional — and what is the model? (I guess that for all surfaces answer should be derivable from known results about triangulations et cetera). (I'm totally not an expert in this area, so maybe those questions are either very easy or hopelessly hard; if it's so, I'll gladly accept as an answer putting them into one of these two categories.) REPLY [5 votes]: The cigar gives both a positive answer for 1 and a negative answer for an upper bound in 2. If it is thin enough, a sequence of aligned points along it is a model and it can be embedded in $\mathbb{R}$. Giving a positive answer for 1. If it is thick enough, but very long, you will need a lot of points in your model (there is no upper bound) to differentiate it from the version with a hole in it. Giving a negative answer to an upper bound in 2. For lower bounds of $mCom$ in terms of the topology I would expect a positive answer, since keeping track of the geometry requires more information than keeping track of the topology.<|endoftext|> TITLE: Minimal resolution of local cohomology module QUESTION [8 upvotes]: Let $(R,\mathfrak m)$ be a Noetherian local ring of dimension $d\geq 1.$ Suppose $\lambda(H_{\mathfrak m}^i(R))<\infty$ for some $0\leq i\leq d-1.$ Question Can we say anything about Betti numbers or minimal resolution of $H_{\mathfrak m}^i(R)?$ In particular, can we say anything (whether the Betti numbers are nondecreasing or strictly increasing, direct summand of the syzygies) about $H_{\mathfrak m}^0(R)?$ P.S. Any reference will be extremely helpful. REPLY [5 votes]: It is hard to give a useful answer. I suspect whatever you want/need would be more specific. In particular, details on how such $R$ arises in your research would make it easier to say something more concrete. But anyhow, there are a couple of remarks and references one can point to. First, your ring is not Cohen-Macaulay. So any nonzero finite length module would have infinite minimal resolution. Typically, the Betti numbers are expected to have eventual exponential growth (see Avramov's surveys from the reference below). One of the outstanding open problems in this area is whether the Betti numbers are always eventually non-decreasing for any module. There is a related question of whether the syzygies will eventually have full dimension (equals $\dim R$). I don't expect the fact that your modules are local cohomology to make a big difference. Here is a nice recent paper on both questions mentioned above. The bibliography also contains some useful references, particularly the surveys of Avramov.<|endoftext|> TITLE: Set-theoretical multiverses and their representation as functors? Why *the* multiverse? QUESTION [11 upvotes]: In some related MO questions like The set-theoretic multiverse as a (bi)category it is discussed how one might represent the multiverse (see The set-theoretic multiverse) in a category theoretic way, e.g. by a category having universes as objects (models of ZFC). However it seems not so obvious which morphisms to take for such an approach (e.g. forcing relation or large cardinal embeddings or any other techniques producing new models from old ones). Question 1. What implications do the different choices for a method to produce new set theoretic worlds from old ones have w.r.t. the multiverse / is there a right choice (what are the criteria)? If not, then the unfinished discussion about the right morphisms in a category theoretic representaation of a multiverse seems to just reflect this? Question 2. However, is it essential in this context to talk about the multiverse rather than a multiverse? If it is not obvious which morphisms could be best, could't one regard different multiverses or multiverse representations and treat them equally until one finds justification to single out some i.e. the real multiverse? Question 3. Couldn't it be the better way to think of a multiverse as a functor $$V: T\rightarrow M$$ mapping extensions of ZFC in $T$ to ZFC-models in $M$, after all this seems to be natural, since a multiverse is about variable worlds (and then think about multiverses with certain properties like having a left adjoint $V_*:M\rightarrow T$ or to study the functor category of multiverses $M^T$ or the like)? REPLY [8 votes]: Of course we have been investigating a wide variety of multiverse concepts, and in this sense, yes, we have not just one, but many, multiverses. But to be sure, much of this multiverse analysis has been inspired by the philosophical idea that there is or might be a unique grand multiverse for mathematics: the actual multiverse, but aware of the possibility that this is a mirage. Even when one has the idea, as many of us do, of looking for mathematical truth in various set-theoretic universes or categories, there might remain the question: which such universe or categories are there? What kind of universes can there be? To assert that there are facts-of-the-matter about what kind of universes exist is to take a step in the direction of the multiverse, and this is philosophical issue at stake in the singularity or plurality of the multiverse. Of course it is not satisfactorily answered simply by adopting the multiverse position itself. I was led in my multiverse paper, to which you have linked, by this kind of thinking to formulate axioms expressing what kind of existence principles we would want or expect in such a multiverse. Hamkins, Joel David, The set-theoretic multiverse, Rev. Symb. Log. 5, No. 3, 416-449 (2012). DOI:10.1017/S1755020311000359, ZBL1260.03103, blog post How nice it was to observe that the philosophical issues transform into purely mathematical questions, when one proposes various specific mathematical multiverse conceptions and begins to analyze their mathematical nature. Thus, one gains philosophical insight by means of a purely mathematical investigation. And this investigation has been undertaken in earnest. So the basic approach has been to study various specific multiverse conceptions as toy multiverses of a kind, standing in for the actual larger multiverse that we seek to understand. Let me mention several instances. Victoria Gitman and I looked at the multiverse consisting of the countable computably saturated models of set theory, in our paper: Gitman, Victoria; Hamkins, Joel David, A natural model of the multiverse axioms, Notre Dame J. Formal Logic 51, No. 4, 475-484 (2010). DOI:10.1215/00294527-2010-030), ZBL1214.03035, blog post. This collection of models, it turns out, fulfill all of the multiverse axioms I had identified in my multiverse paper, and more. Woodin had defined the generic multiverse of a model of set theory, which is the smallest collection containing that model and closed under forcing extensions and ground models. This is, of course, a robust multiverse conception, but nevertheless, there are many set-theoretic principles, such as the GCH, that are achievable over a model of set theory, but not necessarily by set forcing. Benedikt Löwe and I had investigated the generic multiverse as a Kripke model with two natural modal operators, an upward-oriented forcing possibility and a downward-oriented ground-model possibility, and we explored the modal validities of this system in our papers on the modal logic of forcing and also in: Hamkins, Joel David; Löwe, Benedikt, Moving up and down in the generic multiverse, Lodaya, Kamal (ed.), Logic and its applications. 5th Indian conference, ICLA 2013, Chennai, India, January 10--12, 2013, Proceedings. Berlin: Springer (ISBN 978-3-642-36038-1/pbk). Lecture Notes in Computer Science 7750, 139-147 (2013). DOI:10.1007/978-3-642-36039-8_13, ZBL1303.03078, blog post. The Vienna approach, advanced by Sy Friedman and his collaborators, including Carolin Antos and others, has focused on the collection of countable transitive models of ZFC, the set-theoretic hyperverse, which is a multiverse conception, in which the concept of well-foundedness is emphasized as absolute. Friedman has identified a program of using the hyperverse as a means of identifying natural set-theoretic axioms, such as the Inner Model Hypothesis. The New York approach, in contrast, in work of myself, Victoria Gitman and Kameryn Williams, has emphasized the non-absolute character of well-foundedness, and has accordingly accommodated ill-founded models of set theory. There have been many recent developments in this area, including the universal finite set: J. D. Hamkins and W. Woodin, The universal finite set, ArXiv:1711.07952, pp. 1-16, 2017. blog post Recently, in my work with Øystein Linnebo and others, we have explored diverse concepts of set-theoretic potentialism, which I view as multiverse perspectives. J. D. Hamkins and Ø. Linnebo, The modal logic of set-theoretic potentialism and the potentialist maximality principles, to appear in Review of Symbolic Logic, 2018. Arxiv:1801.04599, blog post In that paper, we look at various multiverse conceptions of potentialism, exploring collections of models of set theory under the relation of rank-extension and end-extension and many others, investigating in each case the modal validities of that potentialist conception. In summary, there is a huge effort currently underway to investigate diverse multiverse conceptions. Lastly, I notice that you mention the category-theoretic approach to the multiverse, so let me mention that, unfortunately, the category-theoretic perspective has not seemed to play a large role in recent work in this area. Perhaps ideas arising from this perspective will find a fruitful application in the future. I am hopeful about that.<|endoftext|> TITLE: VC dimension, fat-shattering dimension, and other complexity measures, of a class BV functions QUESTION [7 upvotes]: I wish to show that a function which is "essentially constant" (defined shortly) can't be a good classifier (machine learning). For this i need to estimate the "complexity" of such a class of functions. So let $\mathcal X$ be an abstract set (we may assume has metric structure, e.g $\mathbb R^d$). Given $0 < \alpha \ll 1$, define $$ \mathcal H_\alpha := \{h: \mathcal X \rightarrow [0, 1]\text{ s.t } | \exists \bar{h} \in \mathbb [0, 1] \text{ veryfing } |h(x) - \bar{h}| \le \alpha\;\forall x \in \mathcal X\}. $$ Questions (A) What is the VC dimension or fat-shattering dimension of $\mathcal H_\alpha$ ? Good lower and upper bounds thereof would be just as important. (B) I'd also be interested in metric complexity measures for $\mathcal H_\alpha$ (covering number, metric entropy, etc.). Good lower and upper bounds thereof would be just as important. Observations Theorem 2 of this 1997 paper shows that there exists absolute constants $c_1,c_2$ such that for any $\mathcal H \subseteq [0,1]^{\mathcal X}$, the following bounds metric entropy and fat-shattering dimension holds $$ \operatorname{fat}_{\mathcal {H}}(4\epsilon)/32 \le \max_{P}\log_2(\mathcal N(\epsilon,\mathcal H,\mathcal L_1(dP))) \le c_1 \operatorname{fat}_{\mathcal H}(c_2 \epsilon)(\log_2(1/\epsilon))^2. $$ So if we can estimate $\operatorname{fat}_{\mathcal H_\alpha}(\gamma)$, then we're done! Answer to one-dimensional case. A user has given an explicit computation of $\operatorname{fat}_{\mathcal H_\alpha}(\gamma)$ in the simple one-dimensional case $\mathcal X = [0, 1]$. REPLY [3 votes]: I think the family $\mathcal{H}_\alpha$ is ill-suited for your purposes, because it is too rich. For example in the case $\mathcal{X} = [0,1]$ (or anything atomless probability measure space) one can find arbitrarily many $h_\tau\in\mathcal{H}_\alpha$ such that any two of them are at distance $\alpha/2$. This means that $\mathcal{N}(\varepsilon,\mathcal{H}_\alpha,L^1([0,1])) = \infty$ as soon as $\varepsilon\le \alpha/2$. To prove this explicitely, one can use Hadamard matrices of size $N$ (this is certainly a conceptual overkill and could be replaced by a random choice, or a classical non-compactness argument in functional analysis, but it seems the simplest way to proceed). Divide $[0,1]$ into $N$ equal intervals, and for each sequence $\tau=(\tau_i)_{1\le i\le N} \in \{-1,1\}^N$ define $h_\tau$ to be $\frac{1+\tau_i\alpha}2$ on the $i$-th interval. If $\tau,\tau'$ are two lines of a Hadamard matrix, they differ on exactly half their entries, so that $\lVert h_\tau-h_{\tau'}\rVert_{L^1([0,1])}=\frac\alpha2$. There exist arbitrarily large Hadamard matrices and we are done. Note that if you stay at the level of precision $\varepsilon\simeq \alpha$, then your family is essentially reduced to a point (any two $h\in\mathcal{H}_\alpha$ are indistinguishable at this scale). In order not to stay on a negative claim, let me suggest that you use instead another definition of ``essentially constant'', through a more refined measures of variations of a function. The problem with your condition is that it does not see any of the geometry of $\mathcal{X}$ (the class $\mathcal{H}_\alpha$ essentially only depend on the cardinal of $\mathcal{X}$), and you cannot expect anything (unless possibly you capture some geometry with the chosen metric on the considered space of functions - $L^1$ would not do, as the $[0,1]^{n}$ are all isomorphic when endowed with their Lebesgue measures). There are many natural choices: H"older functions, when $\mathcal{X}$ is a metric space (the particular case of Lipschitz function is the most common), As mentioned by Aryeh Kontorovich, BV functions (this notion is quite simple in dimension $1$, but significantly more intricate in higher dimension), and for a rougher notion $p$-BV (in dimension $1$, one replaces the $\lvert f(x_{i+1})-f(x_i)\rvert$ by $\lvert f(x_{i+1})-f(x_i)\rvert^p$, where $1/p<1$ plays the same role as the H"older exponent, smooth ($\mathcal{C}^k$) functions when $\mathcal{X}$ is a domain in $\mathbb{R}^n$ or a manifold, and the many available variations: Sobolev, Besov, etc.<|endoftext|> TITLE: How to find Erdős' treasure trove? QUESTION [36 upvotes]: The renowned mathematician, Paul Erdős, has published more than 1500 papers in various branches of mathematics including discrete mathematics, graph theory, number theory, mathematical analysis, approximation theory, set theory, and probability theory. A complete list of his published works is available in an archive organized by Jerrold Grossman. Besides the published papers, it is quite likely for a very active mathematician like Erdős to have a long list of unpublished/unfinished papers as well. Just like a treasure trove, these notes may contain incredibly valuable mathematical stuff such as conjectures, lemmas, proof ideas, etc., which can easily be expanded to full papers or at least, give rise to some interesting piece of research as a source of inspiration. They could be of some historical significance too. Question. Is there an archive of (even a portion of) Erdős' unpublished/unfinished works? I am particularly interested in those unpublished notes of him which are related to set theory and infinitary combinatorics. Update. Thanks to Ron Graham the question is now fully answered with the identification of the exact location of Erdős' mathematical diaries. Unfortunately, it seems these diaries aren't currently available for public (and even for close friends and colleagues of Erdős). See my below answer for further details. REPLY [17 votes]: Following my private communication with Ron Graham, he kindly shared all what he knew about Erdős' mathematical diaries with me. Here is part of our discussion which I am allowed to share on MathOverflow for the benefit of the public mathematical community. At Ron's request, I removed the name of one of Erdős' co-authors in the below passage. Emphases are mine. I would like to sincerely thank Ron for his attention to my question as well as his permission for publishing part of our private communication here on MathOverflow. Besides the literally thousands of letters that Erdős wrote during his lifetime, the best source of his unpublished mathematical thoughts are contained in his mathematical diaries, which he meticulously kept for most of his life. In them, he would write what he was thinking about, who he was visiting, etc. There are 15 or so of them, kept in what looks like laboratory notebooks. Of course, they are all in Hungarian. When Erdős died, the diaries were given to his close colleague [name removed] (who) still has them but will not let anyone see them. Many of us have tried to change [name removed]'s mind but to no avail.<|endoftext|> TITLE: If $E\oplus_\phi E \cong E\oplus_\psi E,$ does it imply that $\phi= \psi$? QUESTION [8 upvotes]: Let $E\neq \{0\}$ be a Banach space. For each $p\in[1,\infty), $ we define $$E\oplus_p E = \{(x,y): x\in E, y\in E, \|(x,y)\| = \sqrt[p]{\|x\|^p + \|y\|^p}\}.$$ Let $F$ be another Banach space. By $E\cong F,$ I mean that $E$ and $F$ are isometrically isomorphic. Question: Suppose that $p,q\in [1,\infty).$ If $$E\oplus_p E \cong E\oplus_q E\,$$ then is it true that $p=q$? If $E$ is of finite-dimensional, then the question is affirmative. However, I do not know what will happen if $E$ is of infinite-dimensional. I would be glad to see a proof if it is true or a counterexample if it is false. We say that a norm $\phi:\mathbb{R}^2\to\mathbb{R}$ is normalized if $$\phi(0,1) = \phi(1,0) = 1.$$ Also, $\phi$ is monotone if for $|a_1|\leq |b_1|$ and $|a_2|\leq |b_2|,$ then $$\phi(a_1,a_2) \leq \phi(b_1,b_2).$$ We define $$E\oplus_\phi E = \{(x,y): x\in E, y\in E, \|(x,y)\| = \phi(\|x\|, \|y\|) \}.$$ A more general question: Suppose that $\phi,\psi:\mathbb{R}^2\to \mathbb{R}$ are norms that satisfy normalization and monotonicity. Assume that $\phi$ and $\psi$ are not $\ell^\infty$ norm. If $$E\oplus_\phi E \cong E\oplus_\psi E,$$ then is it true that $\phi = \psi?$? REPLY [5 votes]: It was proved by E. Behrends in Studia Math. 55, 71-85 (1976) that apart from $E=\mathbb{R}^2$ with the sup norm, which is isometric to $E$ with the $\ell_1$-norm, a Banach space $E$ admits a decomposition $E_1\oplus_p E_2$ for at most one value of $p$. This theorem is what Misha has outlined above.<|endoftext|> TITLE: Is a weak functor which strictly preserves horizontal composition and which runs between strict bicategories automatically strict? QUESTION [6 upvotes]: Let $\mathbf{B}$ and $\mathbf{B'}$ be strict bicategories and $F: \mathbf{B} \to \mathbf{B'}$ a weak functor which preserves horizontal composition strictly (i.e. $Ff * Fg = F(f * g)$ natural in f and g.) Does this imply $F$ preserves identities strictly, i.e. $1_{F_a} = F(1_a)$? With the unit axiom for weak functors, the strict preservation of $*$ and the strictness of $\mathbf{B}$ and $\mathbf{B'}$ it follows that $$ Ff * 1_{Fa} = Ff = F(f * 1_a) = Ff * F1_a, $$ for arbitrary f. But I don't see how this implies $1_{F_a} = F(1_a).$ But if this isn't true, the claim that a cubical functor (see for instance definition 2 in https://arxiv.org/pdf/1409.2148.pdf ) automatically strictly preserves units would be false. (I've seen this claim as an aside now on at least 3 different occasions, so I might just miss something really simple here.) REPLY [3 votes]: It seems to me that the simplest possible counterexample works. Restrict first to the case when $\mathbf{B}$ and $\mathbf{B}'$ have one object, so we are talking about a strong monoidal functor $F:C\to D$ between strict monoidal categories. Now let $C$ be the terminal strict monoidal category, with one object $I$, only its identity morphism, and $I\otimes I = I$. And let $D$ be the strict monoidal codiscrete category corresponding to the monoid $\{J,E\}$ where $J$ is the unit object and $E$ an idempotent, $E\otimes E = E$; codiscrete means that we have a unique isomorphism $J\cong E$, so that $D$ is equivalent to $C$ as a category. Let $F:C\to D$ be defined by $F(I) = E$. Then $F$ preserves binary tensors strictly, since $F(I\otimes I) = F(I) = E = E\otimes E = F(I)\otimes F(I)$, but it doesn't preserve the identity strictly, $F(I) \cong J$ coherently but $F(I)\neq J$.<|endoftext|> TITLE: Can the bramble number and the strict bramble number of a graph be equal? QUESTION [5 upvotes]: Let $G$ be a connected graph with vertices $V(G)$. A bramble of $G$ is a set of connected subgraphs $H_1,\ldots,H_n$ such that for each $i$ and $j$, $H_i$ touches $H_j$; that is, either $H_i$ intersects $H_j$ in a vertex, or there is an edge in $G$ that connects a vertex of $H_i$ to a vertex of $H_j$. The order of a bramble is the minimum size of a set $S\subset V(G)$ such that $S\cap H_i$ is nonempty for all $i$. Then, the bramble number of $G$, denoted $Br(G)$, is the largest order of any bramble of $G$. A strict bramble of $G$ is a set of connected subgraphs $H_1,\ldots,H_n$ such that for each $i$ and $j$, $H_i$ intersects $H_j$ in a vertex. The order of a strict bramble is similarly the minimum size of a set $S\subset V(G)$ such that $S\cap H_i$ is nonempty for all $i$, and the strict bramble number of $G$, denoted $sBr(G)$, is the largest order of any strict bramble of $G$. (Strict brambles are sometimes called intersecting families, and the strict bramble number is sometimes called the “PI number" for “pairwise intersecting”.) Since any strict bramble is also a bramble, we have $sBr(G)\leq Br(G)$ for all graphs $G$. My question is this: do we ever have $sBr(G)=Br(G)$? Or is it always the case that if there exists a strict bramble of order $k$, there exists a bramble of order $k+1$? Edit: they are equal if G consists of one vertex and no edges. Are there any graphs with two or more vertices for which they are equal? (Thanks to Arun for pointing this out!) (See this post for more discussion on bramble number and strict bramble number.) REPLY [3 votes]: Indeed, for every connected graph $G$ with at least two vertices, we have $sBr(G) TITLE: A recursive formula QUESTION [12 upvotes]: $$a_0 = 1, \ \ a_{n+1} = \ 1+\frac{n *a_n}{n+a_n} , \ \ n=0,1,2,3,4,...$$ I have built the above recursive formula. Some terms of sequence are: 1, 1, 3/2, 13/7, 73/34, 501/209, 4051/1546, 37633/13327, 394353/130922, 4596553/1441729, 58941091/17572114, 824073141/234662231,... Τhe numerators of the fractions are identical to the sequence A000262 in OEIS encyclopedia and the denominators to the A002720. If we take n = {inf,.......,5,4,3,2,1}, ie to start from quite high and finish at 1. Let a (about 1.6768...) the last term. Then 1/(a*exp(1)) converges to: 0.21938393439552027367716377546012164903 ... that is the decimal expansion of -Ei(-1), A099285 in OEIS. My question is: How it is explained, and how does the formula relate to the above sequences? REPLY [15 votes]: For $n\geq 1$, let $p_n$ be the $(n+1)$-th term of A000262, and let $q_n$ be $n$-th term of A002720. Then, according to the description of these two sequences (more precisely by the contributions of Dennis P. Walsh and Paul Berry) $$p_n=\sum_{k=0}^{n-1}\binom{n}{k}\binom{n-1}{k}k!\qquad\text{and}\qquad q_n=\sum_{k=0}^{n-1}\binom{n-1}{k}^2k!\ .$$ We derive some identities. First, \begin{align}p_{n+1}-q_{n+1}&=\sum_{k=0}^{n}\binom{n+1}{k}\binom{n}{k}k!-\sum_{k=0}^{n}\binom{n}{k}^2k!\\[6pt] &=\sum_{k=0}^n\left(\binom{n+1}{k}-\binom{n}{k}\right)\binom{n}{k}k!\\[6pt] &=\sum_{k=1}^n\binom{n}{k-1}\binom{n}{k}k!\\[6pt] &=n\sum_{k=1}^n\binom{n}{k-1}\binom{n-1}{k-1}(k-1)!\\[6pt] &=n\sum_{k=0}^{n-1}\binom{n}{k}\binom{n-1}{k}k!\\[6pt] &=np_n. \end{align} That is, $$p_{n+1}=np_n+q_{n+1}.\tag{1}$$ Second, \begin{align}q_{n+1}-p_n&=\sum_{k=0}^{n}\binom{n}{k}^2k!-\sum_{k=0}^{n-1}\binom{n}{k}\binom{n-1}{k}k!\\[6pt] &=n!+\sum_{k=0}^{n-1}\binom{n}{k}\left(\binom{n}{k}-\binom{n-1}{k}\right)k!\\[6pt] &=n!+\sum_{k=1}^{n-1}\binom{n}{k}\binom{n-1}{k-1}k!\\[6pt] &=n!+n\sum_{k=1}^{n-1}\binom{n-1}{k-1}^2(k-1)!\\[6pt] &=n!+n\sum_{k=0}^{n-2}\binom{n-1}{k}^2k!\\[6pt] &=n\sum_{k=0}^{n-1}\binom{n-1}{k}^2k!\\[6pt] &=nq_n.\end{align} That is, $$q_{n+1}=p_n+nq_n.\tag{2}$$ From $(1)$ and $(2)$, it follows by induction that $$a_n=\frac{p_n}{q_n}.\tag{3}$$ Indeed, $(3)$ holds for $n=1$. Assuming $(3)$ holds for a given $n$, it also holds for $n+1$ in place of $n$: $$a_{n+1}=1+\frac{na_n}{n+a_n}=1+\frac{np_n}{p_n+nq_n}=1+\frac{p_{n+1}-q_{n+1}}{q_{n+1}}=\frac{p_{n+1}}{q_{n+1}}.$$ This answers the second part of the OP's question. More precisely, it would also be desirable to prove that $\gcd(p_n,q_n)=1$, but I have not verified this. Added 1. I can now prove that $\gcd(p_n,q_n)=1$, i.e., the fraction $(3)$ is in lowest terms. Let us proceed by induction. For $n=1$, the statement is clear. We assume therefore that $\gcd(p_n,q_n)=1$ for a given $n$, and we prove that $\gcd(p_{n+1},q_{n+1})=1$. If this is not the case, then there is a prime $\ell$ that divides both $p_{n+1}$ and $q_{n+1}$. From $(1)$ we get $\ell\mid np_n$, hence from $(2)$ we get $\ell\mid n^2 q_n$. As $p_n$ and $q_n$ are relatively prime by the induction hypothesis, this forces $\ell\mid n$, and then from $(2)$ we get $\ell\mid p_n$. However, $$p_n=1+\sum_{k=1}^{n-1}\binom{n}{k}\binom{n-1}{k}k!=1+n\sum_{k=1}^{n-1}\binom{n-1}{k-1}\binom{n-1}{k}(k-1)!\ ,$$ whence $p_n\equiv 1\pmod{n}$, and therefore $p_n\equiv 1\pmod{\ell}$. This contradiction shows that $\gcd(p_{n+1},q_{n+1})=1$, and we are done. Added 2. I can now answer the remaining first part of the question. Briefly, as pointed out by მამუკა ჯიბლაძე in a comment, the observed "property of $a$" is equivalent to the identity $$e E_1(1)=[1/1,1,1/2,1,1/3,1,1/4,1,1/5,1,\dots],$$ where (cf. Wikipedia) $$E_1(x):=\int_x^\infty\frac{e^{-t}}{t}\,dt,\qquad x>0.$$ More generally, it is known that (cf. Wikipedia) $$e^x E_1(x)=[x/1,1,x/2,1,x/3,1,x/4,1,x/5,1,\dots].$$<|endoftext|> TITLE: Why “modding out the homeomorphism” in the category Top makes no rigorous sense? QUESTION [14 upvotes]: We can rigorously talk about Top, the category of all topological space, and also FTop, the category of all finite topological space. So I thought, we can define a category FTop', where we “mod out by homeomorphism” between objects in FTop. i.e. define the equivalence relation ~ in FTop as X ~ Y if they are homeomorphic and define FTop’ as FTop/~. But recently I am told that this “makes no rigorous sense”. Why is that? REPLY [32 votes]: There are a couple of things to watch out for. First, the collection of all spaces which are homeomorphic to a given space $X$ forms a proper class, and so we cannot just naively use set theory to form it. Nevertheless this is not a true obstacle, since we can employ Scott's trick. Second, two spaces can be homeomorphic in many ways, which presents a problem when we try to define morphisms between equivalence classes. Writing $[X]$ for the homeomorphism class of $X$, what could the morphisms $[X] \to [Y]$ be? One might naively try to answer with "equivalence classes of continuous maps $f : X \to Y$ of some sort". However, this will not work so easily. Here's an obvious attemp that does not work. Given $X \cong X'$ and $Y \cong Y'$ and $f : X \to Y$ and $f' : X' \to Y'$, say that $f \cong f'$ if there are homeomorphisms $p : X \to X'$ and $q : Y \to Y'$ such that $f' \circ p = q \circ f$ (as suggested in the comments). Then composition of functions is not a congruence and so composition of morphisms is unclear (as noted in the comments). As far as I can tell, the problem persists with finite spaces as well. If fact, the problem is not specific to topological spaces at all. As soon as there can be multiple isomorphisms between two objects, we'll have a problem. One possible solution (which I initially called "standard" but people in the comments questioned it) is not to quotient the category but to pass to its skeleton, i.e., we form the full subcategory on chosen representatives of homeomorphism classes. In general this may require some form of the axiom of choice, but let's not worry about that today. Supplemental: The discussion in the comments veered in the direction of homotopy type theory, so it might be worth looking at what happens there. If we take topological spaces to live at the level of 0-types (also known as h-sets), then by the structure identity principle homeomorphic spaces are going to be equal. This may seem mistifying at first, for does such a statement not suffer from the same defect regarding morphisms, as above? No, because HoTT is proof-relevant. Concretely, suppose $h : X' \to X$ is a homeomorphism and $f : X \to Y$ is a continuous map. The following does not work in HoTT: Faulty HoTT reasoning: Because $X'$ and $X$ are homeomorphic, by the structure identity principle $X' = X$, and since $f : X \to Y$ then also $f : X' \to Y$. We must pay attention to why $X' = X$ (that's what proof relevance is about, proofs are mathematical objects, they're not just published stories and logicians' fetishes): Because $h : X' \to X$ is a homeomorphism, by the structure identity principle $\mathsf{sip}(h)$ proves $X = X'$. We may therefore transport any construction involving $X$ to one on $X'$ along $\mathsf{sip}(h)$. In particular, given $f : X \to Y$, there is $\mathsf{transport}_{\mathsf{sip}(h)}(f) : X' \to Y$. A little calculation reveals that $\mathsf{transport}_{\mathsf{sip}(h)}(f) = f \circ h$. Note how instead of saying "isomorphic things are equal" we make things proof-relevant by saying "every isomorphism begets a proof of equality".<|endoftext|> TITLE: Do choice principles in all generic extensions imply AC in $V$? QUESTION [17 upvotes]: It's well-known that not all choice principles are preserved under forcing, e.g. in this answer https://mathoverflow.net/a/77002/109573 Asaf shows the ordering principle can hold in $V$ and fail in a generic extension. Indeed, the standard proof for preservation of AC is based on the fact that well-orderability is preserved under surjection, a fact that doesn't seem to have any nice generalization for weaker choice principles at all. So I wonder if we can get any results in the opposite direction. Are there any known results of the form "If all generic extensions satisfy [some weak choice principle] then [some stronger choice principle] holds in $V$"? I take choice principles to include e.g. AC, DC, AC$_{\omega}$, the selection principle, "all infinite sets are Dedekind-infinite," and "(strongly) amorphous sets don't exist." Two conjectures I want to focus on are: Plausible conjecture: AC$_{\omega}$ in all generic extensions implies AC in $V$ (the idea here is that if there's a set in $V$ without a choice function, maybe there's a way to collapse its cardinality to $\omega$ without adding a choice function), and Ridiculous conjecture: If every generic extension has no strongly amorphous sets, then AC holds in $V$ (I can't believe this is true, but I also have no idea what property $V$ can have to prevent forcing amorphous sets). REPLY [4 votes]: Working on a related topic for a new paper with Jonathan Schilhan I've stumbled into a disproof of your ridiculous conjecture. Definition. The class $W$ is the closure of $\{\{a\}\mid a\in V\}$ under well-ordered unions. Sets in $W$ are called "almost well-orderable sets". The axiom $V=W$ is consistent with $\lnot\sf AC$, for example in Gitik's model, and in a large cardinals-free model, Blass' model where all ultrafilters are principal. Theorem. If $V=W$, then there are no amorphous sets. Proof. We define a rank function based on how many steps it requires to generate a set using well-ordered unions of singletons, call this the $W$-rank of a set. If $A$ is amorphous of minimal rank, then it is a well-ordered union of proper subsets. Since $A$ is minimally ranked, those must be finite sets, as an infinite subset of an amorphous set is amorphous. So the union must be infinite, and therefore $A$ can be mapped onto $\omega$, at the very least, which is impossible. $\square$ Theorem. If $V=W$ and $G$ is $V$-generic, then $V[G]=W^{V[G]}$. Proof. Note that there is a surjection from $V$ onto $V[G]$, definable in $V[G]$. Or, formally speaking, for every set in $V[G]$ there is a set in $V$ which maps onto it. Moreover, the closure of $\{\{a\}\mid a\in V[G]\}$ under well-ordered unions, in $V[G]$, contains $V$. So if we show that this closure is itself closed under surjective images, we are done. This too is done by induction on the $W$-rank. If $x=\bigcup_{\alpha<\delta}x_\alpha$ and $f\colon x\to y$ is a surjection, then by the induction hypothesis $f``x_\alpha=y_\alpha$ is in the closure, and of course $y=\bigcup_{\alpha<\delta}y_\alpha$. $\square$ Finally, the wanted corollary. Corollary. If $V=W$, then no generic extension of $V$ contains amorphous sets. On the other hand, over the Cohen model, where $W$ is the class of well-orderable sets, there is a forcing adding an amorphous set (as was shown by Monro). So while the above shows that the ridiculous conjecture is false, it is not entirely clear how ridiculous it was to begin with. Namely, can we refine it to something "more or less" in the spirit of $V\neq W$?<|endoftext|> TITLE: Expressive power of FO with $\mu$ QUESTION [8 upvotes]: Let us consider the first-order logic extended with the least fixed point operator (FO+LFP). That is, together with the usual first-order formulas, we also have formulas of the form: $$\mu X[\overline{y}] . \phi(X, \overline{y})$$ where $X$ (must occur positively in $\phi$) is a "predicate" variable of arity equal to the length of sequence of "parameters" $\overline{y}$. The semantic of this formula (in a given algebraic structure) is the least set $X^*$ such that: $$X^*(\overline{y}) \Leftrightarrow \phi(X^*, \overline{y})$$ For example, if $R$ is a binary relation symbol, then: $$\mu X[y_1, y_2] . R(y_1, y_2) \vee (\exists_z R(y_1, z) \wedge X(z, y_2))$$ defines the transitive closure of $R$. If $A$ is an algebraic structure, let us write $\mathit{Th_{lfp}}(A)$ for the first-order theory of $A$ extended with the least fixed point operator (i.e. the set of all FO+LFP sentences that are true in $A$). Does there exist an algebraic structure $A$ such that both of the following hold: $\mathit{Th_{lfp}}(A)$ is decidable FO+LFP is strictly more expressive than FO over $A$ (i.e. there is a FO+LFP formula that is not equivalent to FO formula over $A$)? An example of a structure that satisfies the first property (but does not satisfy the second) is the structure of rational numbers with the natural ordering $\langle\mathcal{Q}, \leq\rangle$. An example of a structure that satisfies the second property (but does not satisfy the first) is the structure of natural numbers with the natural ordering $\langle\mathcal{N}, \leq\rangle$. One intuition is that there should be no such structure $A$ --- if $A$ defines arbitrary long well-founded orders, then the theory of $A$ should be undecidable; and if it does not define, then LFP seems to be pretty useless. Another intuition is that there might be such a structure, because the above intuition is difficult to formalize, thus may contain essential holes. REPLY [2 votes]: You also might be interested in the paper The Role of Decidability in First Order Separations over Classes of Finite Structures by Steven Lindell and Scott Weinstein. While they don't address the above question directly, they investigate connections between the decidability of the first-order theory of a (family of) structures and the power of first order vs. LFP definability. For example, they prove that "$\text{Th}(\mathcal{C}$) is decidable" implies $\text{FO}(\mathcal{C}) \neq \text{FO+LFP}(\mathcal{C})$ when $\mathcal{C}$ is a proficient family of structures. (Here $\text{Th}(\mathcal{C})$ is the set of first-order sentences satisfied for all structures in $\mathcal{C}$. Proficient simply means that there is no finite bound on the closure ordinals of all LFP formulas.)<|endoftext|> TITLE: Criterion for existence of a homogeneous space associated to a Lie pair QUESTION [8 upvotes]: Recall that every finite-dimensional real Lie algebra $\mathfrak{g}$ is the Lie algebra of a simply-connected Lie group, which is unique up to isomorphism. This statement generalises somewhat to homogeneous spaces. Given a Lie pair $(\mathfrak{g},\mathfrak{h})$ consisting of a finite-dimensional real Lie algebra $\mathfrak{g}$ and a Lie subalgebra $\mathfrak{h}$, is there a simply-connected homogeneous space $G/H$, where $G$ is a simply-connected Lie group with Lie algebra (isomorphic to) $\mathfrak{g}$ and $H$ is a closed connected subgroup of $G$ with Lie algebra $\mathfrak{h}$? We know that $\mathfrak{g}$ is the Lie algebra of a simply-connected group $G$ and that $\mathfrak{h}$ is the Lie algebra of a unique connected subgroup $H$ of $G$. So existence of the homogeneous space requires that $H$ be closed and uniqueness requires that $(\mathfrak{g},\mathfrak{h})$ be effective; that is, that $\mathfrak{h}$ does not contain any proper ideal of $\mathfrak{g}$. My question is whether given $(\mathfrak{g},\mathfrak{h})$ there is an easily checked criterion which would allow me to determine whether or not this is the pair associated to a homogeneous space. In other words, is there a way to check (from the Lie algebraic data) that the unique connected subgroup $H$ generated by $\mathfrak{h}$ of a simply connected group $G$ with Lie algebra $\mathfrak{g}$ is closed. I know of a few criteria applicable to linear groups, but I have some cases in mind to which these criteria do not apply. I am wondering how best to tackle this problem. REPLY [3 votes]: There are some special cases: For $codim_G (H) \le 4$ the Lie subgroup H always is closed. See Mostow G. "The Extensibility of Local Lie Groups of Transformations and Groups on Surfaces" November 1950 Annals of Mathematics 52(3) G is simply connected solvable General answer - see Theorem 15 and around it (pages 183-189) in this article (in English !): A. Malcev, “On the theory of the Lie groups in the large”, Матем. сб., 16(58):2 (1945), 163–190 http://www.mathnet.ru/links/75d4e7ae407f7a41b7dacd28c10a2fba/sm6329.pdf<|endoftext|> TITLE: Another notion of exactness: how to refine it, and where does it fit? QUESTION [25 upvotes]: There are many notions of "exactness" in category theory, algebraic geometry, etc. Here I offer another that generalizes the category of frames, the notion of valuation (from probability theory), and touches on aspects of abelian categories and the Seifert-van Kampen theorem. My intention is to hear comments from the community on how this notion should be refined and improved, other examples and fields where this notion arises, whether this notion fits into a larger theory or extends an existing theory. The rough idea is that an exactness structure on a category is a set of commutative squares, such as pushout-pullbacks, and that a functor is exact when it preserves the chosen squares. In this sense, it is something like a "limit sketch". One last note: I am aware that the term "exact square" already exists—and I'll give it as an example of what I call exact squares—so although I think the name "exact square" is fitting, I would also be happy to hear alternatives. Let $2=\fbox{$\bullet\to\bullet$}$ denote the free arrow category, so $2\times 2$ is the free commutative square. Definition: Let $C$ be a category with an initial object $\bot$. An exactness structure on $C$ is a set $E$ of squares, $e\colon 2\times 2\to C$, called exact squares $$ \begin{array}{ccc} A&\xrightarrow{f}&B\\ \scriptstyle g\textstyle\downarrow\;&e&\;\downarrow \scriptstyle h\\ C& \underset{i}{\to}&D \end{array} $$ satisfying the following conditions: The composite of either projection $2\times 2\to 2$ and any morphism $2\to C$ ("any degenerate square") is exact; The composite of the swap map $\sigma\colon 2\times 2\to 2\times 2$ and any exact square $e\colon 2\times 2\to C$ is exact; The pasting of any two exact squares in $C$ $$ \begin{array}{ccccc} \bullet&\to&\bullet&\to&\bullet\\ \downarrow&&\downarrow&&\downarrow\\ \bullet&\to&\bullet&\to&\bullet \end{array} $$ is exact; and if $e\cong e'$ are isomorphic squares then $e$ is exact iff $e'$ is. We refer to a category with an exactness structure as an exacting category. We say that a functor is exacting if it preserves initial objects and exact squares. We say that an exacting category $(C, \bot, E)$ is normalized if it has a final object and continuous if it has filtered colimits, and similarly morphisms are normalized and/or continuous if they preserve these structures. Let $\mathsf{ExCat}$, $\mathsf{CtsExCat}$, $\mathsf{NrmExCat}$, and $\mathsf{NrmCtsExCat}$ denote the various combinations of these adjectives. Example: If $C$ is an abelian category, then it can be given the structure of a normalized exacting category. The top element is 0, and a square $$ \begin{array}{ccc} A&\xrightarrow{f}&B\\ \scriptstyle g\textstyle\downarrow\;&&\;\downarrow \scriptstyle h\\ C& \xrightarrow{i}&D \end{array} $$ is exact in the present sense iff the sequence $$0\to A\xrightarrow{(f,g)}B\oplus C\xrightarrow{h-i}D\to 0$$ is exact in the sense of chain complexes. Example: The classical Seifert-van Kampen theorem is the statement that the fundamental group functor $\pi_1\colon\mathsf{Top}\to\mathsf{Grp}$ from topological spaces to groups is exact if we choose the exact squares in $\mathsf{Top}$ to be pushout-pullback squares with simply connected pullback, and those in $\mathsf{Grp}$ to be the pushout squares. Example: The category $\mathsf{Cat}$ of categories can be given the structure of a (normalized continuous) exacting category, where a square $$ \begin{array}{ccc} A&\xrightarrow{f}&B\\ \scriptstyle g\textstyle\downarrow\;&&\;\downarrow \scriptstyle h\\ C& \xrightarrow{i}&D \end{array} $$ is called exact iff it is exact in the sense of the nlab, i.e. if $g_!f^*=i^*h_!$ as functors $\mathsf{Psh}(B)\to\mathsf{Psh}(C)$. Example: A frame (a.k.a. a locale), e.g. the poset of open sets in any topological space, has a continuous, normalized exactness structure. It can be regarded as a category in the usual way, its top / bottom elements serve as initial / final objects, and we say that a square is exact if it is both a pullback and a pushout: $$ \begin{array}{ccc} A\cap B&\to&A\\ \downarrow&&\downarrow\\ B& \to&A\cup B \end{array} $$ This is a full and faithful embedding $\mathsf{Frm}\to\mathsf{NrmCtsExCat}$. Indeed any monotone map between the underlying posets of frames $F$ and $F'$ that preserves top and bottom elements and filtered colimits (directed sups), is a map of frames iff it preserves binary meets and binary joins. But this is the case iff it preserves exact squares. [In fact, the functor $\mathsf{Frm}\to\mathsf{CtsExCat}$ is also fully faithful.] Example: The poset $\mathbb{R}^+:=\{r\in\mathbb{R}\mid 0\leq r\}\cup\{\infty\}$ of nonnegative real numbers plus infinity under the usual $\leq$ ordering can be given a normalized continuous exactness structure where a square $$ \begin{array}{ccc} m&\to&n\\ \downarrow&&\downarrow\\ m'& \to&n' \end{array} $$ is exact iff $m+n'=m'+n$. Remark: If $(C,\bot,E)$ is a (continuous) exacting category and $c\in C$ is an object, then the slice category $C_{/c}$ inherits a (continuous) exacting structure. Let $U\colon C_{/c}\to C$ be the forgetful functor. Then $C_{/c}$ inherits an initial object and filtered colimits from $C$, and we take a square $e$ to be exacting in $C_{/c}$ iff $U(e)$ is exacting in $C$. Valuations are a constructive approach to probability theory, which agrees with the usual Kolmogorov definition in nice cases. It does not use $\sigma$-algebras but instead is defined on frames. Here we give the usual definition, except with the present terminology. Note that $\mathbb{R}^+_{/1}$ has as objects the closed interval $[0,1]$. Definition: Let $F$ be a frame. A valuation on $F$ is an exacting functor $\mu\colon F\to\mathbb{R}^+_{/1}$. It is called normalized and/or continuous if it is normalized and/or continuous as an exacting functor. In other words, our terminology "normalized" and "continuous" was chosen to match that of valuations. The above definition situates valuations in a much broader context. Proposition: Any left-exact functor preserves exacting category objects and exacting functors, normalized or not. Moreover, the direct image part of a geometric morphism preserves continuous exacting posets, such as frames and the nonnegative lower reals as described above. Again, my question is "how will the community respond"? In other words, I'm looking for insights into this notion, how it fits with other notions I haven't discussed above, other examples of it, whether it already exists, whether there are additional requirements that should be made, etc. Thanks! REPLY [4 votes]: If you were willing to allow a generalization where "being exact" is structure on a square rather than a property of it, then there would be an example in the homotopy category of a stable $\infty$-category (or stable derivator), where an "exact structure" on a (homotopy) commutative square is a choice of homotopy filling it and making it into a pushout+pullback square. Similarly, one could put a "proof-relevant" exactness structure on Ho(Cat) where an exact structure on a commutative-up-to-isomorphism square is a choice of isomorphism filling it and making it into an exact square in the sense of your example above. If you further relaxed the requirement that exact squares have to actually commute, you could allow non-invertible transformations with the same exactness property.<|endoftext|> TITLE: Visual proof of convergence for Steiner's symmetrization QUESTION [7 upvotes]: I want to find a visual proof of the following fact: For any convex figure in the plane there is a sequence of Steiner's symmetrizations that makes it arbitrary close to a circular disc. All proofs I know require some integral estimates. I would prefer a more visual proof (even if it is more involved). REPLY [3 votes]: Just for illustration, here is an animation of one step of a Steiner symmetrization applied to a rasterized version of a polygon.                                         First: a histogram of vertical lengths. Second: lengths lifted to be symmetric around horizontal.<|endoftext|> TITLE: Higher dimensional scutoids? QUESTION [8 upvotes]: The recent discovery of scutoids in biological structures is fascinating. Two scutoids are depicted below (from Scientists Have Discovered an Entirely New Shape, And It Was Hiding in Your Cells), each with a pentagon on one side (top or bottom) and a hexagon on the opposing side. A scutoid could be viewed as a linking of the vertices of the 2-dimensional associahedron in one plane to those of the 2-D permutohedron / permutahedron in a parallel plane through edges extending in the dimension orthogonal to the planes. (The pentagon is also a 2-D stellohedron / stellahedron.) Is there a natural (hopefully interesting by retaining some optimization properties) generalization of scutoids to higher dimensions? REPLY [2 votes]: It's not clear from the linked article that the hexagon is a permutahedron, so we don't restrict to that case for starters. For our construction of higher dimensional scutoids we need a couple of standard polytope operations: Prism: For a $d$-polytope $P$ let $\mathrm{prism}(P)$ be the $d+1$-dimensional prism over $P$ given by $P \times [0,1]$. Vertex Truncation: For any vertex $v$ of $P$ let $E(v)$ be the edges of $P$ incident to $v$ ordered by length. Let $H$ be any affine hyperplane such that $v$ is the only vertex of $P$ lying on one side of $H$. Then the truncation $\mathrm{trunc}_{v,H}(P)$ is the subpolytope of $P$ that lies on the side of $H$ not containing $v$. A $d$-scutoid is any polytope $Q$ that can be written as $Q = \mathrm{trunc}_{v,H}(\mathrm{prism}(P))$ for some $d-1$-dimensional polytope $P$ (called the base polytope) and some vertex $v$ of $P$. So a $3$-scutoid is combinatorially equivalent to a scutoid (in the sense of the linked article in the OP) when the base polytope is a pentagon. If the above definition of a $d$-scutoid is not refined enough for some purpose one can always restrict to various families of allowable base polytopes. One chain of families that gets us from arbitrary polytopes to associahedra is $$\text{arbitrary polytopes} \supset \text{fiber polytopes} \supset \text{associahedra}.$$ Edit: As @MarkS points out in a comment below, scutoids are not polytopes: for scutoids to pack correctly some of their faces must be curved. (For visualizations see page 9 of the original article in Nature or this video.) Since the $3$-scutoids with pentagonal base polytopes described above are polytopes, they are technically not scutoids, but rather combinatorially-equivalent polytopal approximations. Edit: An earlier version of this answer incorrectly identified the base polytope of a (combinatorial) scutoid as a hexagon instead of a pentagon.<|endoftext|> TITLE: On a surprising property of free theories QUESTION [12 upvotes]: Yesterday I observed (and proved) the following odd fact, which I found very surprising. I'm very curious to know if this was known by some people, or if it follows from some other more general fact, or if anyone had any kind of comments about it. (for example "this is false" would be a very helpful comment ! ). I hope this is not to vague for MO. My observation is the following: $1^{st}$ Version: Let $T$ be a free Lawvere theory (i.e. freely generated by some operation in different arities) then when you see $T$ has a category, and freely add to it an initial object, the resulting category has all finite limits. One can rephrase this in terms of model: $2^{nd}$ version : Let $T$ be a free lawvere theory, or (equivalently) a free symmetric set-operad, then let $C$ be the category of free finitely generated model, then $C$ with a freely added terminal object has all finite colimits. (note: "freely" added terminal object means that it is terminal and it has no morphism out of him except the identity) In fact, one can obtain the same results far more generally: $3^{rd}$ version: Let $T$ be a "free" Cartmell generalized algebraic theory. Here free means that it has no equality axioms, only terms and type introduction rules. Then the contextual category of $T$, together with a freely added initial object, has all finite limits. I would also be interested to know if some example of non-free theory also have this properties (I believe the answer is no, but I don't really know) Maybe I should give an example of those limits to give an idea of how it works, and how 'weird' these limits can be. In fact the following example gives a pretty good idea of how the proof works (though the version for Cartmell theories, or at least the only proof I know, involve a very messy and complicated induction to really makes it into a proof) Let's look at the the Lawvere theory of "magmas", i.e. the theory with just one operation $m$ of arity $2$. I will look at colimits in the category of free finitely generated magma. Coproducts clearly exists, so I will focus on coequalizer. I denote by $M_k$ the free magma on $k$-generators. $M_1 \rightrightarrows M_2$ be the pair of map wich send the generators of $M_1$ to $m(x,y)$ and $m(x,x)$ where $x$ and $y$ are the generators of $M_2$. Then the colimits is $M_1$ where $x$ and $y$ are both sent to the generator of $M_1$. Indeed in a free magma the only way to have $m(f,g)=m(f,f)$ is if $f=g$. $M_1 \rightrightarrows M_2$ sending the generator to $m(x,y)$ and $m(x,m(x,x))$. Then the colimit is $M_1$ again, with $x$ being sent to the generator $t$ and $y$ to $m(t,t)$. Indeed in a free magma $m(f,g)=m(f,m(f,f))$ can hapen only if $g=m(f,f)$. $M_1 \rightrightarrows M_2$ sending the generator to $m(x,y)$ and $m(x,m(x,y))$ then the colimits is the freely added terminal object. Indeed in a free magma the relation $m(f,m(f,g)) = m(f,g)$ implies $m(f,g)=g$ which can never be satisfied in a free magma (for example by counting the number of $m$ that appears in the unique expression of $g$ in terms of the generators). Any colimits involving the freely added terminal object is the freely added terminal object itself (it is the only object that admit a map from it) REPLY [4 votes]: This is a consequence of first-order unification. The main issue is the existence of (co)equalizers, since (co)products are trivially seen to exist. Here is a brief explanation of how to translate calculating (co)equalizers into the language of unification. Your three setups have arrows pointing in different directions, I'll pick the colimit direction for simplicity but the reasoning below applies to all three setups equally well. Work as in your magma examples but in the general case. Think of the generators of the free algebra $F_m$ with $m$-generators as variable symbols $x_1,\ldots,x_m$. Then we can think of morphisms $F_m \to F_n$ as variable substitutions $\{x_1 \mapsto t_1,\ldots,x_m \mapsto t_m\}$, where $t_1,\ldots,t_m$ are terms involving the $n$-variable symbols $y_1,\ldots,y_n$ from $F_n$. Given a parallel pair $$\sigma =\{x_1 \mapsto s_1,\ldots,x_m \mapsto s_m\}, \tau =\{x_1 \mapsto t_1,\ldots,x_m \mapsto t_m\}:F_m \rightrightarrows F_n,$$ and a $$\upsilon = \{y_1 \mapsto u_1,\ldots,y_n \to u_n\}:F_n \to F_k,$$ we have $\upsilon\circ\sigma = \upsilon\circ\tau$ precisely if $s_1^\upsilon = t_1^\upsilon,\ldots,s_k^\upsilon = t_k^\upsilon$ (where I use superscripts to denote the application of a substitution to a term). In other words, when the substitution $\upsilon$ is a unifier for the unification problem $\{s_1 \doteq t_1,\ldots,s_k \doteq t_k\}$. It was shown by J. A. Robinson that if the unification problem $\{s_1 \doteq t_1,\ldots,s_k \doteq t_k\}$ has a unifier, then it has a most general unifier $\nu$ in the sense that any other unifier is obtained from $\nu$ by applying a further substitution on top of it. Translating back into the language of category, this means that this $\nu$ is a coequalizer for the parallel pair $\sigma,\tau:F_m \rightrightarrows F_n$. Robinson privided an algorithm to decide whether a unifier exists and to find a most general unifier if there is one. A much more efficient algorithm was later found by Martelli and Montanari. In response to your final query, yes, there are some non-free theories that allow unification; see Baader and Snyder for some common examples. Robinson, J. A., A machine-oriented logic based on the resolution principle, J. Assoc. Comput. Mach. 12, 23-41 (1965). ZBL0139.12303. Martelli, Alberto; Montanari, Ugo, An efficient unification algorithm, ACM Trans. Program. Lang. Syst. 4, 258-282 (1982). ZBL0478.68093. Baader, Franz; Snyder, Wayne, Unification theory, Robinson, Alan (ed.) et al., Handbook of automated reasoning. In 2 vols. Amsterdam: North-Holland/ Elsevier; 0-444-50812-0 (vol. 2); 0-444-50813-9 (set)). 445-533 (2001). ZBL1011.68126.<|endoftext|> TITLE: Prove that if a group $G$ is generated by all cyclic subnormal subgroups, then every cyclic subgroup is subnormal QUESTION [5 upvotes]: I have already found two definitions for a Baer group. $G$ is a Baer group if it is generated by all cyclic subnormal subgroups. $G$ is a Baer group if every cyclic subgroup is subnormal. I want to prove the equivalence of the two definitions. Obviously, (2) implies (1). Please help me with the converse. REPLY [3 votes]: Claim. If group is generated by cyclic subnormal groups, then every finitely generated subgroup is subnormal. Proof. Let $A, B < G$ be f. g., nilpotent, and subnormal. We want to prove that $C := \langle A, B \rangle$ has this properties. (Then every cyclic will be subnormal, because every subgroup of nilpotent group is subnormal — it's easy). First, note that $C$ lies in Hirsch radical — more or less by definition, therefore (locally, but this doesn't matter because it is f. g.) nilpotent. Now you want to prove that $C$ is subnormal. I'll sketch the proof, it's pretty straightforward. Let's look at some subnormal chain from $A$ to $G$, $A = U_0 \lhd \dots \lhd U_k = G$. First, replace $A$ by $A' = A^C$, so that $A' \lhd C$. Look at partial subnormal chain $A^{U_i}$. Take $B$-invariant subgroups of them; it's also subnormal chain, say, $V_i$, going from $A$ to $A^G$. Finally observe that $V_i \lhd V_{i+1} B$, $B$ subnormal in $V_{i+1} K$. It immediately gives that $V_i B$ is subnormal in $V_{i+1} B$, and because $A^G B$ is subnormal in $G$ we're done. I guess it's written in Baer's articles near late 40s somewhere, but I do not remember particular reference.<|endoftext|> TITLE: The Riemann hypothesis as a problem in analysis QUESTION [15 upvotes]: The recent post("Long-standing conjectures in analysis ... often turn out to be false") prompted me to think about a question which I have not given much though before: to what extent the Riemann hypothesis (RH) may be regarded as a problem in analysis. It may actually be not as silly as it sounds. The particular side of it I am curious about is the following. The Theorem : $\zeta(s)\neq 0$ for $\Re s>1$. may be considered a very weak consequence of RH. However, this statement is only trivial in the context of number theory. One may ask, is it possible to give it a "purely analytic" proof, without using Euler product and other stuff related to the primes? Apparently, it is not possible to formulate this as a precise mathematical question because all the theories used to formalize analysis contain a good deal of arithmetic, but it should be precise enough for practical purposes. (You know number theory when you see it.) Most likely such a proof does not exist yet, but some readers may know more about the relevant things then I do. I am honestly curious because while it definitely looks tough, it may be not entirely implausible. If such a proof is found, it might give us a fresh look on the old problem. REPLY [10 votes]: Andrew Booker is of the opinion that it is the nonvanishing of zeta (or an L-function) to the right of the critical strip which is more fundamental than the Euler product. See Slide 10 of his recent talk https://heilbronn.ac.uk/wp-content/uploads/2018/07/Booker-talk.pdf<|endoftext|> TITLE: Can an injective $f: \Bbb{R}^m \to \Bbb{R}^n$ have a closed graph for $m>n$? QUESTION [19 upvotes]: Question. Suppose $m>n$ are positive integers. Is there a one-to-one $f: \Bbb{R}^m \to \Bbb{R}^n$ such that the graph $\Gamma_f$ of $f$ is closed in $\Bbb{R}^{m+n}$? Remark 1. The answer to the above question is well-known to be negative by basic results in topological dimension theory if the condition that $\Gamma_f$ is closed is strengthened to the continuity of $f$. Remark 2. By elementary topology, a function $f$ from a compact Hausdorff space $X$ to a compact Hausdorff space $Y$ is continuous iff $\Gamma_f$ is closed in $X \times Y$. This fact can be used to show that the answer to the above question is also negative if in the statement of the question, $\Bbb{R}$ is replaced by $[0,1]$, i.e., $f$ is stipulated to be a function from $[0, 1]^m$ to $[0,1]^n$. REPLY [20 votes]: There is no such function. Suppose $f: \mathbb R^m \rightarrow \mathbb R^n$ is an injective function with $\Gamma_f$ closed in $\mathbb R^{m+n}$. For each $i \in \mathbb N$, let $K_i = f^{-1}([-i,i]^n)$. I claim that each $K_i$ is closed and nowhere dense in $\mathbb R^m$. $K_i$ is closed because it is the projection onto $\mathbb R^m$ of the set $\Gamma_f \cap (\mathbb R^m \times [-i,i]^n)$, which is closed in $\mathbb R^m \times [-i,i]^n$, and the projection of a closed subspace of $X \times Y$ onto $X$ is always closed when $Y$ is compact. To see that $K_i$ is nowhere dense, suppose $C$ is a closed subset of $K_i$. Notice that $\Gamma_f \cap (C \times [-i,i]^n)$ is closed in $C \times [-i,i]^n$, and it is the graph of the function $f \!\restriction\! C$. In particular, $f \!\restriction\! C$ is a function into a compact Hausdorff space, and $\Gamma_{f \restriction C}$ is closed. By the closed graph theorem (the one alluded to in the question), $f \!\restriction\! C$ is continuous. This implies that $K_i$ is nowhere dense: otherwise (because we already know $K_i$ is closed) there is a closed ball $C \subseteq K_i$, in which case $f \!\restriction\! C$ is a continuous injection from a topological copy of $[0,1]^m$ into $[0,1]^n$. As the OP already mentioned, this is impossible. (This is because a continuous injection on $C$ would be an embedding (because $C$ is compact), and one cannot embed $[0,1]^m$ into $[0,1]^n$ when $m > n$.) The Baire Category Theorem now provides us with a contradiction. Because each $K_i$ is nowhere dense, it is impossible to have $\mathbb R^m = \bigcup_{i \in \mathbb N}K_i$. On the other hand, $\mathbb R^m = \bigcup_{i \in \mathbb N}K_i$ is implied by the definition of the $K_i$.<|endoftext|> TITLE: How to visualize Dirichlet’s unit theorem? QUESTION [33 upvotes]: As the question title asks for, how do others "visualize" Dirichlet’s unit theorem? I just think of it as a result in algebraic number theory and not one in algebraic geometry. Bonus points for pictures. REPLY [34 votes]: One viewpoint that is useful is to think of this as a special case of the theory of arithmetic groups via extension of scalars. First I need to describe a proof of the Dirichlet unit theorem via Mahler's compactness theorem. Given a number field $K=\mathbb{Q}(\alpha)$, where $p(\alpha)=0, p\in \mathbb{Z}[x]$, $p(x)$ a monic irreducible polynomial, we may think $K$ as an algebra over $\mathbb{Q}$ which is a $d=deg(p(x))$-dimensional vector space. In the obvious $\mathbb{Q}$-basis $\{1,\alpha, \ldots, \alpha^{d-1}\}$, multiplication by $\alpha$ becomes the companion matrix $C(p)$ for the polynomial $p(x)$. Instead, let us choose a $\mathbb{Q}$-basis $\{\alpha_1,\ldots,\alpha_d\}\subset K$ which is also a $\mathbb{Z}$-basis for the ring of integers $\mathcal{O}_K$. Multiplication by elements of $K$ gives a map $\rho: K \to M_d(\mathbb{Q})$. What is the image $\rho(K)$ in $M_d(\mathbb{Q})$? Well, the image is commutative, so it should be contained in $\{ A\in M_d(\mathbb{Q}), A\cdot \rho(\alpha)=\rho(\alpha)\cdot A\}$. Of course, we also know that the image is $\mathbb{Q}[\rho(\alpha)]\subset M_d(\mathbb{Q})$, and one may check that these two spaces agree by linear algebra. Then one may consider the subring $M_d(\mathbb{Z}) \cap \rho(K)$. This is a lattice $\rho(\mathcal{O}_K)\subset \rho(K)$. Now the image of the ring of units is $\rho(\mathcal{O}_K^\times) = \rho(\mathcal{O}_K) \cap GL_d(\mathbb{Q})$. This is equal to $\rho(K)\cap Aut(\mathcal{O}_K)$, where $Aut(\mathcal{O}_K) \cong GL_d(\mathbb{Z})$, making an identification using our preferred basis. Consider the space $X=GL_d(\mathbb{R})/GL_d(\mathbb{Z})$, which may be regarded as the space of $d-$dimensional lattices in $\mathbb{R}^d$ via $A\cdot \mathbb{Z}^n$, $A\in GL_d(\mathbb{R})$. Inside $GL_d(\mathbb{R})$ we have the subspace $Y'=\mathbb{R}[\rho(\alpha)]\cap det^{-1}(\pm 1)=\{A \in det^{-1}(\pm 1) | [A,\rho(\alpha)]=0\}$. Let $Y$ be the image of $Y'$ in $X$. Note that $\mathbb{R}[\rho(\alpha)]\cong K\otimes_{\mathbb{Q}}\mathbb{R}\cong \mathbb{R}^{r_1}\times \mathbb{C}^{r_2}$ as a tensor product of fields. Let's check the conditions of Mahler's compactness theorem. The determinant is $=\pm 1$ on $Y'$ and hence on $Y$. We also need to check that there is a neighborhood $N$ of $0\in \mathbb{R}^d$ so that the only lattice point of $\Lambda\in Y$ inside of $N$ is $0$. Suppose that we have a sequence $A_i\in Y', v_i\in \mathbb{Z}^n$ with $A_i v_i \to 0$. Then $\rho(\alpha)^jA_iv_i = A_i \rho(\alpha)^j v_i \in A_i \mathbb{Z}^n$, and hence $\rho(\alpha_j)A_iv_i=A_i\rho(\alpha_j)v_i$. Hence there is a basis for a sublattice of $\Lambda_i=A_i\mathbb{Z}^n$ given by $\{\rho(\alpha_1)v_i, \rho(\alpha_2)v_i,\ldots, \rho(\alpha_d)v_i\}$ which approaches $0$. But this contradicts that the lattice $\Lambda_i=A_i\mathbb{Z}^n$ has covolume $1$ since $det(A_i)=\pm 1$. Hence $Y$ has compact closure in $X$ by Mahler's compactness theorem. But $Y$ itself is closed, since it is characterized by being the lattices $\Lambda$ such that $\rho(\alpha)\Lambda\subset \Lambda$, which is a closed condition. So $Y$ is compact. This implies that $Y$ is a compact torus in $X$, which is $(\mathbb{R}[\rho(\alpha)]\cap det^{-1}(\pm 1) ) / \rho(\mathcal{O}_K^\times)$. Thus, I think of $\mathcal{O}_K^\times$ as a (multiplicative) lattice (arithmetic subgroup ) in a maximal torus (commutative algebraic subgroup) of $det^{-1}(\pm 1)$ (this Lie group contains $SL_d(\mathbb{R})$ with index $2$). This maximal torus is the $\pm 1$-determinant elements of $\mathbb{R}[\rho(\alpha)]^{\times} \cong (\mathbb{R}^\times)^{r_1} \times (\mathbb{C}^\times)^{r_2}$, where $r_1$ is the number of real places of $K$, and $r_2$ is the number of complex places of $K$. Then Dirichlet's unit theorem follows immediately from this property. To visualize this for a real quadratic number field, note that $X$ is the space of unit lattices in $\mathbb{R}^2$. Modding out by rotation, $SO(2)\backslash X$ is the modular curve $\mathbb{H}^2/PSL_2(\mathbb{Z})$. Then $Y$ in this case is a closed geodesic. Here is a picture from this paper by recent Fields medalist Akshay Venkatesh (and Einsiedler and Lindenstrauss and Michel):<|endoftext|> TITLE: Smoothen a nodal curve QUESTION [6 upvotes]: Let $k$ be a field, let $X/k$ be a nodal curve (which means $X_{\overline{k}}$ is a connected reduced proper curve with at worst nodal singularities.) Does there always exists a proper flat family $\mathcal{X}/\mathrm{Spec}(k[[t]])$ with special fiber $X$ and smooth generic fiber? (Is there a reference for this?) REPLY [12 votes]: This is correct that you can always "deform" a nodal curve into a smooth one. A good reference for this fact is Corollary $7.11$ in these notes by Talpo and Vistoli. UPD: As Qixiao points out Talpo and Vistoli prove the result only under the assumption that all singularities are rational. I'll explain in the UPD section below how to deduce the general result from their one. The proof crucially uses the fact that any nodal singularity is defined over a finite separable extension of a ground field. This allows us to use Galois descent to reduce the situation to the case of rational singularities. The proof boils down to two (rather independent parts): classification of Ordinary Double Points and Deformation Theory of (isolated) singularities. The first part crucially relies on Artin Approximation Theorem which is beautifully explained in Chapter 3.6 of the book "Neron Models" by Bosch, Lutkebohmert and Raynaud. Classification of ordinary double points (and much more) is done in Chapter 3.2 Freitag and Kiehl's book "Etale Cohomology and Weil conjectures". Let me summarize the theorem below: Definition 1: Let $k$ be an arbitrary field (not necessary algebraically closed or even perfect), then we say that a pair $(U,x)$ of a finite type $k$-scheme $U$ and a $k$-rational point $x\in U(k)$ defines an rational ordinary double point, if there is an isomorphism of $\bar{k}$-algebras $\mathcal O_{U\otimes_k \bar{k}, \bar x}\cong \bar{k}[[u,v]]/(uv)$. Definition 2: Let $k$ be an arbitrary field, then we say that a pair $(U,x)$ of a finite type $k$-scheme and a closed point $x\in U$ defines an ordinary double point, if there is a point $\bar x\in U(\bar k)$ which lies over $x$ and such that $(U\otimes_k \bar k, \bar x)$ defines a rational ordinary double point. Theorem 1: (Classification of rational ordinary double points). Let $(U, x)$ be a rational ordinary double point over a field $k$, then there is a pair $(W,y)$ and two etale residually trivial maps $f:(W,y) \to (U,x)$ and $g:(W,y) \to (\operatorname{Spec} k[[u,v]]/(uv), 0)$. Theorem 2: (Classification of ordinary double points). Let $(U, x)$ be an ordinary double point over a field $k$, then the residue field $k(x)$ is finite separable over $k$ and there is a pair $(W,y)$ and two etale maps $f:(W,y) \to (U,x)$ and $g:(W,y) \to (\operatorname{Spec} k[[u,v]]/(uv), 0)$. As for deformation theory, the main thing that we need to understand is tangent-obstruction theory for the deformation problem of local complete intersection, generically smooth finite type $k$-schemes. This is done in Chapters $3-5$ of Vistoli's notes. The main result of these Chapters is the following theorem: Theorem 3: Let $X_0$ be a local complete intersection, generically smooth scheme of finite type over a field $k$, then the tangent space to the deformation problem of $X_0$ is given by $\operatorname{Ext^1}(\Omega^1_{X_0/k}, \mathcal O_{X_0})$ and the obstruction theory to this problem is given by $\operatorname{Ext^2}(\Omega^1_{X_0/k}, \mathcal O_{X_0})$. Now, we are ready to explain the main idea behind the proof given in Talpo-Vistoli's notes. First of all, we deal with a 'standard model' of an affine nodal curve with $1$ (isolated) singularity. Namely, we mean a pair $(\operatorname{Spec} k[x,y]/(xy),0)$. Talpo and Vistoli compute explicitly its miniversal deformation and show that the generic fibre is 'smooth' (doesn't quite make sense, better to speak in terms of completed local rings). Then they relate a miniversal deformation of a general affine nodal curve with $1$ isolated singularity to this very specific case of a standard model (here they crucially use the classification result above and general theory of miniversal deformations developed in their notes before). This is Proposition $7.5$ in their notes. Now we need to 'glue' these local deformations to a formal deformation of $X_0$. In order to do that we use a low degree terms exact sequence associated to a local-to-global spectral sequence for Ext-groups: $$ 0 \to H^1(X_0, \underline{Hom}(\Omega^1_{X_0/k}, \mathcal O_{X_0})) \to \operatorname{Ext}^1(\Omega^1_{X_0/k}, \mathcal O_{X_0}) \to H^0(X_0, \underline{Ext^1}(\Omega^1_{X_0/k}, \mathcal O_{X_0})) \to H^2(X_0, \underline{Hom}(\Omega^1_{X_0/k}, \mathcal O_{X_0})). $$ We identify $\operatorname{Ext}^1(\Omega^1_{X_0/k}, \mathcal O_{X_0})$ with the tangent space to deformations of $X_0$, $H^0(X_0, \underline{Ext^1}(\Omega^1_{X_0/k}, \mathcal O_{X_0}))$ with local deformations (a covering of a curve by affines and deformation of each affine part s.t. these deformations are isomorphic on overlaps) and also we need to identify a map between these groups with a natural map that sends a deformarion to an associated local deformation. But since $X_0$ is a curve we know that the $H^2$ term is automatically zero. So, given a covering of our nodal projective curve $X_0$ by affines and a bunch of deformations of these affines charts that are isomorphic on intersections we can always glue them to a deformation of the whole curve $X_0$. Apply this to the situation where each affine chart $U_i$ has exactly one distinct singular point (then all intersections are smooth and affine, so any deformation of such schemes are trivial). So, we can indeed glue these local "smoothifications" to a global one. I should say explicitly that this means that we can glue to a formal deformation $\mathcal X \to \operatorname{Spf} k[[t]]$ (Note that $\mathcal X$ is only a formal scheme, not a usual scheme. So we must prove that it is algebraizable in order to speak about its generic fiber as a scheme). The last part is to algebraize $\mathcal X$. In general it may be a difficult problem. But here we are in a one-dimensional situation, so it is easily seen that there are no obstruction to deform an ample line bundle on $X_0$ to an ample line bundle $\mathcal X$ and then by Grothendieck's Existence we can algebraize it. So, finally we have a flat projective curve $X\to \operatorname{Spec} k[[t]]$ s.t. the special fibre is isomorphic to our initial curve $X_0$ and it is easy to see (following the construction) that the generic fibre should be smooth (the easiest way is to control all completed local rings. But one should be careful since we want to check that the generic fibre is smooth and not just regular. For details see Corollary 7.11 in the notes). UPD: Let me show how to get of rid of the assumption that all singularities are rational. Let $X_0$ be a reduced projective curve over $k$ with only ordinary double points. We want to find a flat projective relative curve $X\to \operatorname{Spec} k[[t]]$ such that the generic fibre is smooth and the special fibre is isomorphic to $X_0$. According to theorem 2, we can choose a finite separable extension $k'/k$ such that all singularities of $X_0\otimes_k k'$ are rational. If necessarily, extend $k'$ even further to assume that all irreducible components of $X_0\otimes_{k} k'$ of genus zero are actually isomorphic to $\mathbf P^1$ (any form of $\mathbf P^1$ splits after a finite separable extension). This allows us to choose a $\operatorname{Gal}(k'/k)$-invariant divisor $D'$ on $X_0':=X_0\otimes_k k'$ s.t. a pair $(X_0',D')$ is a semistable pointed curve (in a sense of Knudsen). Consider a deformation problem $Def_{k''[[t]]}(X_0',D')$. Step 1: The functor $Def_{k''[[t]]}(X_0',D')$ is pro-representable by $k'[[t]][[T_0,\dots,T_n]]=:A'$ for some n. In order to check that this functor is pro-representable it is sufficient to show that this deformation problem admits a tangent-obstruction theory and that the Automorphism scheme $\underline{\operatorname{Aut}}_{X_0',D'}$ is unramified (This is due to Schlessinger's criterion). The existence of tangent-obstruction theory is standard and, for example, is explained in Chapter 5 here. As for the Automorphism-scheme, I don't know a good reference for this fact. But a proof is very close to the proof of the same fact in the case of not-pointed semi-stable curves (Theorem $1.11$ in a celebrated Deligne-Mumford paper). The last thing to verify is that a ring that pro-represents $Def_{k''[[t]]}(X_0',D')$ is actually isomorphic to power series ring $k'[[t]][[T_0, \dots,T_n]]$. Again, we know that it is enough to show that this deformation problem is unobstructed. By Theorem $5.4$ the obstruction space is given by $$ \operatorname{Ext^2}(\Omega^1_{X_0'/k'}, \mathcal I_D)=\operatorname{Ext^2}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0}). $$ We want to show that this space is zero. Using local-to-global spectral sequence we reduce the question to showing that $$ H^0(X_0,\underline{Ext^2}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})))=H^1(X_0,\underline{Ext^1}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})))=H^2(X_0,\underline{Hom}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})))=0. $$ OK, a local calculation (exercise) shows that $\underline{Ext^2}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})=0$. Since $X_0$ is a curve we also conclude that $H^2(X_0,\underline{Hom}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})))$ vanishes. Finally, note that $\underline{Ext^1}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0}))$ is supported at a closed subscheme of singular points. So, this sheaf is actually a pushforward from some closed $0$-dimensional subscheme, this implies that $H^1(X_0,\underline{Ext^1}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})))=0$ as well. So, our problem is indeed unobstructed. Step 2: Algebraization and descent. Ok, so we've proven that $A'$ pro-represents our deformation functor. Thus there is a universal formal family $\mathcal X' \to \operatorname{Spf} A'$ together with $m$-sections $\mathcal X'(A')$ extending the divisor $D'$ (where $m=\deg D'$). Firstly, note that we can algebraize $\mathcal X'$ to a flat projective scheme $X'$ over $\operatorname{Spec} A'$ by Grothendieck's existence (+deformation theory of vector bundles) since $X_0'$ is a curve (so no obstructions to deform line bundles). And likewise we can lift all sections of $\mathcal X'$ to section $\sigma_i$ of $X'$. Moreover, note that a locus in $\operatorname{Spec} A'$ over which $X'\to \operatorname{Spec} A'$ has only ordinary double points singularities is open, so $(X' \to \operatorname{Spec} A', \sigma_i)$ is a semi-stable pointed curve. We claim that there is a descent of this scheme to a projective semistable curve $X\to \operatorname{Spec} k[[t]][[T_0,...,T_n]]$. Let us denote by $A$ the ring $k[[t]][[T_0,...,T_n]]$. In order to descent $X'$ we need to find an $A'$-ample line bundle $L'$ on $X'$ such that the pair $(X',L')$ has a descent datum w.r.t a morphism $A\to A'$. The desired sheaf is the dualizing sheaf $\omega_{X'/A'}(\sum \sigma_i)$ twisted by sections $\sigma_i$. Two questions naturally arise: Why is it $A'$-ample and why does a pair $(X', \omega_{X'/A'}(\sum \sigma_i))$ has a descent datum? As always, the first question is answered in Knudsen's paper, see Corollary 1.10. As for the second question, let us understand explicitly what it means to have a descent datum. Observe that $(A')^{\operatorname{Gal}(k'/k)}=A$ and $A\otimes_k k'\cong A'$. So $A'\otimes_A A'\cong A\times \operatorname{Gal}(k'/k)$. And a descent datum is "just" an action of the Galois group $\operatorname{Gal}(k'/k)$ on $X'$ s.t. for any $s\in \operatorname{Gal}(k'/k)$ we have an isomorphism $s^*((X', \omega_{X'/A'}(\sum \sigma_i)))\cong (X', \omega_{X'/A'}(\sum \sigma_i))$ + cocycle condition. Note that $X'$ has a natural action of $\operatorname{Gal}(k'/k)$ because the Galois group $\operatorname{Gal}(k'/k)$ acts on a functor $\operatorname{Def}_{k'[[t]]}(X_0', D')$ (because $X_0'$ and $D'$ are both defined over $k$). So it acts on an object that pro-represents this functor as well as on its algebraization. To see that $\omega_{X'/A'}(\sum \sigma_i))$ is preserved by this action just note that the divisor $D'$ was chosen to be $\operatorname{Gal}(k'/k)$-invariant and the formation of dualizing sheaf commutes with etale base change. Step 3: Use the known case of rational singularities to finish the proof. Great, so now we have a flat projective curve $X \to \operatorname{Spec} A$ s.t. the special fibre is isomorphic to $X_0$. Moreover, from the known case of rational singularities we know that $X' \to \operatorname{Spec} A'$ is smooth over a generic fibre. How do we know this? Just choose an arbitrary deformation $Y' \to \operatorname{Spec} k'[[t]]$ s.t. the generic fibre is smooth (it exists because $X_0 '$ has only rational ordinary double points). This defines a formal deformation $\mathcal Y' \to \operatorname{Spf} k'[[t]]$. By the universal property it comes as a pullback of $\mathcal X' \to \operatorname{Spf}A'$ through some morphism $\operatorname{Spf} k'[[t]] \to \operatorname{Spf} A'$. This corresponds to a morphism $\operatorname{Spec} k'[[t]] \to \operatorname{Spec} A'$ and since algebraization commutes with base change we see that some fiber of $X' \to \operatorname{Spec} A'$ must be smooth (as $Y' \to \operatorname{Spec} k'[[t]]$ is smooth over its generic fibre). Smoothness is an open condition, therefore we see that the generic fibre of $X' \to \operatorname{Spec} A'$ must be smooth as well. Since smoothness descends along etale morphisms we conclude that $X \to \operatorname{Spec} A$ is smooth over the generic point. Recall again that locus in $\operatorname{Spec} A$ over which $X$ is smooth is open. So the complement is given by some non-trivial ideal $I$ in $k[[t]][[T_0,\dots,T_n]]=k[[t, T_0,\dots,T_n]]$ and we want to check that some pullback of $X$ is smooth over a generic fibre for some morphism $k[[t]] \to k[[t,T_0,\dots,T_n]]$. By what being said we reduced the question to purely algebraic statement: Given a non-trivial ideal $I$ in $k[[t,T_0,\dots,T_n]]$ there always exists a local $k$-morphism $\phi:k[[t, T_0,\dots,T_n]] \to k[[t]]$ such that $\phi(I)\neq 0$ (This means that the associated morphism $\operatorname{Spec} k[[t]] \to \operatorname{Spec} A$ sends a closed to a closed point and a generic point into the smooth locus $\operatorname{Spec} A -V(I)$. That's exactly what we are looking for). Let us prove this fact by induction. First of all, choose any non-zero element $f\in I$ and forget about $I$. We try to construct a $k$-homomorphism $\phi:k[[t, T_0,\dots,T_n]] \to k[[t]]$ s.t. $\phi(f)\neq 0$. For simplicity of notations reformulate the question for an abstract power-series ring $k[[x_1,\dots, x_n]]$ and we construct a $k$-morphism to $k[[t]]$. Observe that in order to construct $\phi$ it suffices to construct $\psi:k[[x_1, \dots, x_n]] \to k[[y_1, \dots, y_{n-1}]]$ s.t. $\psi(f)\neq 0$ (assuming that $n\geq 2$). So, we will actually construct that morphism instead of $\phi$ (and then we will get the desired $\phi$ as a composition of $\psi$'s). OK, being said this let us do some real work. Note that any element $y\in k[[x_1, \dots, x_n]]$ with a non-trivial linear form and zero constant term can be extended to a regular system of parameters $(y=y_0, y_1, \dots, y_{n-1})$. This implies that $k[[x_1,\dots, x_n]]=k[[y_0, y_1,\dots, y_{n-1}]]$. Hence, if we find an element $y$ with a non-trivial linear form, zero constant term and such that $f\notin (y)$, we can consider a natural projection $$ \psi:k[[x_1,\dots, x_n]]=k[[y_0, y_1,\dots, y_{n-1}]] \to k[[y_1,\dots, y_{n-1}]] $$ and we are done since $\psi(f)\neq 0$. So, the only thing we are left to show is that there always exists $y$ such that $y$ has a non-trivial linear form and $f$ doesn't lie in an ideal generated by $y$. Suppose the contrary, this means that a power series $f$ is divisible by any power series with non-trivial linear form. This is impossible because there infinitely many distinct irreducible power series with non-trivial linear form over any field (for example, $x_1+x_2^m$ for all $m$. Here is the place where we use that $n\geq 2$!). But $k[[x_1,\dots, x_n]]$ is UFD, so it is impossible that an element is divisible by infinitely many distinct (up to units) irreducible polynomials. Hence, the desired $y$ always exists.<|endoftext|> TITLE: Explicit descriptions of a flop QUESTION [8 upvotes]: I want to know how to describe explicitly the flop of the following flop contraction. Because the construction is so natural and simple, I was wondering such descriptions should already exist in the literature. Unfortunately, I could not find them. Here is the construction. Let $\mathbb P^4$ over $\mathbb C$ with homogenous coordinates $[x_1:\cdots:x_5]$, let $Z=\{x_1=x_2=0\} \subset \mathbb P^4$. Set $$\pi: P={\rm Bl}_Z \mathbb P^4 \to \mathbb P^4$$ be the blowup of $Z$. The linear system $|-K_{P}|$ is base point free and let $X \in |-K_P|$ be a general element. By adjunction $K_X=0$. Let $Y = \pi(X) \subset \mathbb P^4$. Then the natural contraction $\pi^-: X \to Y$ is a flop contraction ($X \cap {\rm Exc}(\pi)$ maps generally finite to $Z$, hence $\pi^-$ is a small contraction). Now I want to get an explicit description of its flop $\pi^+: X^+ \to Y$. (For example, can $X^+$ be construct as some hypersurface? ) The above construction can be describe in equations: As $Z \subset Y$ and $Y \in |-K_{\mathbb P^4}|$, we have $$ Y = \{f=\sum_{1 \leq a+b \leq 5} x_1^a x_2^b f_{ab}=0\} \subset \mathbb P^4.$$ Here $a, b \in \mathbb{Z}_{\geq 0}$ and $f_{ab}$ are general polynomials of degree $(5-a-b)$. For $z \in Z$ such that $X_z$ is a curve, we have $$z \in \{x_1=x_2=f_{10}=f_{01}\} \subset \mathbb P^4.$$ In fact, at such point, the relative tangent space $T_{Y/Z}$ has dimension $1$. Thus, as $\deg f_{10}=\deg f_{01}=4$, we have $16$ flopped curves. As a result, the flop $\pi^+: X^+ \to Z$ should also contract $16$ curves. REPLY [3 votes]: You have a threefold hypersurface $Y\subset \mathbb{P}^4$ of the form $V(ax_1 - bx_2)$, for some quartic polynomials $a,b\in\mathbb{C}[\mathbb{P}^4]$. You blow up the plane $Z=V(x_1,x_2)$ in the ambient space, to obtain $X\subset \mathbb{P}^1_{\lambda:\mu}\times\mathbb{P}^4$. As you say, this makes a small resolution of 16 nodes and it amounts to introducing blowup parameters $\tfrac{\lambda}{\mu}=\tfrac{x_1}{x_2}=\tfrac{b}{a}$. To obtain the flop $X^+$ you can blow up the quartic surface $Z^+=V(x_1,b)$ in the ambient space instead. To describe $X^+$ explicitly, consider the standard affine patches $U_i=Y\cap\{x_i=1\}$. Then, over $U_i$, the blowup is given by $Bl_{Z^+}U_i\subset\mathbb{P}^1_{\lambda':\mu'}\times\mathbb{A}^4 \to U_i$ defined by the equations $$ ax_1 - bx_2 = \lambda'x_1 - \mu'b = \lambda'x_2 - \mu'a = 0 $$ (after setting $x_i=1$).<|endoftext|> TITLE: invertible endomorphisms on a space of linear maps between finite-dimensional vector spaces QUESTION [5 upvotes]: This is a linear algebra question that came up in my research, and I feel like there ought to be either a simple proof or a simple counterexample, but I have been unable to find either. Assume $V$ and $W$ are real finite-dimensional inner product spaces, so there is an induced inner product on $\text{Hom}(V,W)$ defined via its canonical identification with $V^* \otimes W$. Suppose $\Phi : \text{Hom}(V,W) \to \text{Hom}(V,W)$ is a linear map with the property that $\langle A , \Phi(A) \rangle > 0$ for every element $A \in \text{Hom}(V,W)$ with rank 1. Equivalently, using the identification $\text{Hom}(V,W) = V^* \otimes W$, this condition means $\langle w , \Phi(v_\flat \otimes w) v \rangle > 0$ for all $v \in V$ and $w \in W$, where $v_\flat := \langle v,\cdot \rangle$. Does it follow that $\Phi$ is invertible? My instinct says no: for instance, if we pick bases of $V$ and $W$ and express $\Phi$ in the resulting basis of $\text{Hom}(V,W)$, then the condition says that all diagonal entries of the matrix for $\Phi$ are positive, bus that's certainly not enough to conclude that $\Phi$ is invertible (at least not without some information about the magnitude of the diagonal entries relative to the non-diagonal entries, which I don't have, as far as I know). But the hypothesis seems to say more than this since it is not tied to any specific bases of $V$ and $W$, e.g. the diagonal entries will be positive for any basis of $\text{Hom}(V,W)$ consisting of rank 1 elements. My intuition and algebraic knowledge are insufficient to say what that means, but it sounds like it means something. Or is there a simple counterexample, e.g. where $V$ and $W$ are both 2-dimensional? REPLY [9 votes]: It does not follow that $\Phi$ is invertible. Consider $V=W=\mathbb{R}^2$ together with the standard inner product. With respect to the standard basis we identify $\operatorname{Hom}(V,W)$ with the space of $2\times 2$ real matrices. We define the map $\Phi$ by \begin{align*} \operatorname{Mat}_{2,2}(\mathbb{R})&\to\operatorname{Mat}_{2,2}(\mathbb{R})\\ \left(\begin{matrix}a&b\\c&d\end{matrix}\right)&\mapsto\left(\begin{matrix}a-b&b-c\\c+d&d+a\end{matrix}\right). \end{align*} Notice that $\Phi$ is not bijective since $\Phi(N)=0$ for \begin{align*} N=\left(\begin{matrix}1&1\\1&-1\end{matrix}\right). \end{align*} On the other hand, for every $A\in\operatorname{Mat}_{2,2}(\mathbb{R})$ with entries $a,b,c,d$ we have \begin{align*} \langle A,\Phi(A)\rangle&=\operatorname{tr}\left(A\Phi(A)^T\right)=\operatorname{tr}\left(\left(\begin{matrix}a&b\\c&d\end{matrix}\right)\left(\begin{matrix}a-b&c+d\\b-c&d+a\end{matrix}\right)\right)\\ &=a^2-ab+b^2-bc+c^2+cd+d^2+da\\ &=\frac12\left((a-b)^2+(b-c)^2+(c+d)^2+(d+a)^2\right). \end{align*} This expression is positive unless $A$ is a scalar multiple of $N$. Note that $\operatorname{rk}(N)=2$. We can conclude that $\langle A,\Phi(A)\rangle$ positive whenever the matrix $A$ has rank $1$.<|endoftext|> TITLE: Determining if some permutation of a vector satisfies a system of linear equations QUESTION [16 upvotes]: Let $A$ be a matrix and $x$ a fixed vector. How can we determine whether or not there exists a permutation matrix $P$ such that $APx=0$? Does this problem reduce to anything well-understood? REPLY [3 votes]: Let $\mathbb P_n$ be the set of $n \times n$ permutation matrices. Given matrix $\mathrm A \in \mathbb R^{m \times n}$ and vector $\mathrm v \in \mathbb R^n$, we would like to find a permutation matrix $\mathrm P \in \mathbb P_n$ such that $$\mathrm A \mathrm P \mathrm v = 0_m$$ The convex hull of $\mathbb P_n$ is the Birkhoff polytope $\mathbb B_n$ (the set of all $n \times n$ doubly stochastic matrices) $$\mathbb B_n := \left\{ \mathrm X \in \mathbb R^{n \times n} \mid \mathrm X \mathrm 1_n = \mathrm 1_n, \mathrm 1_n^\top \mathrm X = 1_n^\top, \mathrm X \geq \mathrm O_n \right\}$$ Thus, a convex relaxation of the original discrete feasibility problem in $\mathrm P \in \mathbb P_n$ is the following continuous feasibility problem in $\mathrm X \in \mathbb B_n$ $$\begin{array}{ll} \text{minimize} & 0\\ \text{subject to} & \mathrm A \mathrm X \mathrm v = 0_m\\ & \mathrm X \mathrm 1_n = \mathrm 1_n\\ & \mathrm 1_n^\top \mathrm X = 1_n^\top\\ & \mathrm X \geq \mathrm O_n\end{array}$$ Let us look for a solution on the boundary of the feasible region. Hence, we generate a (nonzero) random matrix $\mathrm C \in \mathbb R^{n \times n}$ and minimize $\langle \mathrm C, \mathrm X \rangle$ instead. We have the following linear program (LP). $$\begin{array}{ll} \text{minimize} & \langle \mathrm C, \mathrm X \rangle\\ \text{subject to} & \mathrm A \mathrm X \mathrm v = 0_m\\ & \mathrm X \mathrm 1_n = \mathrm 1_n\\ & \mathrm 1_n^\top \mathrm X = 1_n^\top\\ & \mathrm X \geq \mathrm O_n\end{array}$$ Although the vertices of the Birkhoff polytope are doubly stochastic matrices, the introduction of the equality constraints $\mathrm A \mathrm X \mathrm v = 0_m$ likely produces other vertices. We may have to generate several matrices $\rm C$ until we obtain an LP whose minimum is attained at a permutation matrix. Numerical experiment Suppose we are given $$\rm A = \begin{bmatrix} 1 & 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 1 & 1\end{bmatrix}$$ $$\rm v = \begin{bmatrix} 1 & 1 & -1 & 0 & 0\end{bmatrix}^\top$$ Using NumPy to randomly generate matrix $\rm C$ and CVXPY to solve the LP: from cvxpy import * import numpy as np A = np.array([[1,1,1,0,0], [0,1,1,1,0], [0,0,1,1,1]]) v = np.array([1,1,-1,0,0]) (m,n) = A.shape C = np.random.rand(n,n) ones_n = np.ones((n,1)) X = Variable(n,n) # define optimization problem prob = Problem( Minimize(trace(C.T * X)), [ A * X * v == np.zeros((m,1)), X * ones_n == ones_n, ones_n.T * X == ones_n.T, X >= 0 ]) # solve optimization problem print prob.solve() print prob.status # print results print "X = \n", np.round(X.value,2) which outputs the following permutation matrix that exchanges the 2nd and 4th entries: 0.669610896837 optimal X = [[ 1. 0. 0. 0. 0.] [ 0. 0. -0. 1. 0.] [-0. -0. 1. 0. -0.] [ 0. 1. 0. 0. 0.] [ 0. 0. 0. 0. 1.]] I ran the Python script a few (maybe $5$) times until I obtained a matrix $\rm X$ that is (close enough to) a permutation matrix. Unsurprisingly, the script does not produce such nice results for all choices of $\rm C$. Running the script a few more times, I obtained another permutation matrix: 1.46656456314 optimal X = [[ 0. 0. 0. 1. 0.] [ 1. 0. 0. 0. 0.] [-0. -0. 1. 0. -0.] [ 0. 0. -0. 0. 1.] [ 0. 1. 0. 0. 0.]]<|endoftext|> TITLE: Degeneration of smooth curves and Picard-Lefschetz formula QUESTION [5 upvotes]: Let $\pi:\mathcal{C} \to \Delta$ be a family of projective curves of genus $g \ge 2$ over the unit disc $\Delta$, smooth over the punctured disc $\Delta\backslash \{0\}$ and central fiber $\pi^{-1}(0)$ is an irreducible nodal curve with exactly one node. For $t$ close to $0$, denote by $\delta_t$ the vanishing cycle on $H_1(\mathcal{C}_t,\mathbb{Z})$, where $\mathcal{C}_t:=\pi^{-1}(t)$. I am looking for conditions on the central fiber $\pi^{-1}(0)$ such that there exists a $1$-cycle $\gamma \in H_1(\mathcal{C}_t,\mathbb{Z})$ satisfying: $\gamma.\delta_t=1$. Any idea/reference will be most welcome. If I understand correctly, this is true if $\pi$ is a degeneration of elliptic curves. REPLY [8 votes]: Because the symplectic form on $H_1(C_t, \mathbb Z)$ is a perfect pairing, it suffices to check that there is a group homomorphism $H_1(C_t,\mathbb Z) \to \mathbb Z$ that sends $ \gamma$ to $1$, which follows if $\gamma$ is not divisible by any $n>1$ in $H_1(C_t,\mathbb Z)$. Because $\gamma$ is defined as the generator of the kernel of $H_1(C_t,\mathbb Z) \to H_1(C_0,\mathbb Z)$, $\gamma$ is indivisible as long as $H_1(C_0,\mathbb Z)$ is nontorsion, which follows from it being an irreducible nodal curve.<|endoftext|> TITLE: Goldbach's conjecture for the Liouville function QUESTION [8 upvotes]: Is it true that for every even integer $N > 2$, there exist positive integers $a,b$ such that $a + b = N$ and $\lambda(a) = \lambda(b) = -1$ ? Here $\lambda$ is the Liouville function. REPLY [3 votes]: In "The equation $\omega(n)=\omega(n+1)$" (Mathematika 50, 99-101, 2003) it was shown that the equation $\omega(n)=\omega(n+1)$ has infinitely many solutions. Pintz has a series of results on consecutive integers, one of which is that for every $k>k_0$ we have that the equation $\omega(n)=\omega(n+1)=k$ has infinitely many solutions, where $k_0$ is some small integer (may be 5). Unfortunately the closest reference I could find is "Small gaps between products of two primes" by Goldston, Graham, Pintz, and Yildirim. However, I guess a similar application of the GPY-sieve should give that for all sufficiently large $N$ and $k>k_0$ we have that $n+m=N$ with $\omega(n)=\omega(m)=k$ has solutions. Going from $\omega$ to $\Omega$ should not pose any problems, so even a stronger statement should be true and provable with today's methods.<|endoftext|> TITLE: ICM 2018 lecture videos QUESTION [34 upvotes]: Is there a place to watch ICM 2018 plenary lectures (and other lectures if possible)? Here is the official Youtube channel of the ICM but they don't seem to be posting the lectures. https://www.youtube.com/channel/UCnMLdlOoLICBNcEzjMLOc7w Update: The public lectures of Etienne Ghys, Cedric Villani, and Ingrid Daubechies are up! REPLY [5 votes]: There are now several more videos of ICM talks available. I imagine that even more will be posted in the near future. https://www.youtube.com/channel/UCnMLdlOoLICBNcEzjMLOc7w/videos<|endoftext|> TITLE: Algebraic models of non-simply connected spaces in string topology QUESTION [7 upvotes]: I want to find some algebraic models relating string topology to Hochschild and cyclic homologies. If the space $X$ is simply-connected and we are working over rational numbers, we can use Sullivan model or Quillen model of $X$ to do the job. See nLab page or this paper. My question is how about non-simply conncted spaces? We still have Sullivan model. But how to relate them to free loop spaces? Can we still use Hochschild and cyclic homologies to compute loop homology and string homology? For example, the torus $T^{2}$. Its Sullivan model is just $\wedge (x,\,y)$ with zero differential where $|x|=|y|=1$, and it is isomorphic to the Chevalley-Eilenberg cochain algebra of 2 dimensional abelian Lie algebra. But I don't know how to use this to compute loop and string homologies. Any help would be very appreciated. REPLY [10 votes]: In my paper with Zeinalian (https://arxiv.org/abs/1612.04801) we prove that the coHochschild complex of the dg coalgebra of singular chains on a path connected (possibly non-simply connected) space calculates the homology of the free loop space of $X$. In my paper with Saneblidze (https://arxiv.org/abs/1712.02644) we describe a smaller model which might be suitable for your calculations. However, you must be careful: if you use dg algebra models - such as the Sullivan model - under quasi-isomorphisms you loose the information the fundamental group. If you want to calculate something for a non-simply connected space $X$ you may use any connected dg coalgebra $C$ which is weakly equivalent to the dg coalgebra of singular chains on $X$ with Alexander-Whitney coproduct in the following sense: we say a map $f: C \to C'$ between two dg connected coalgebras is a weak equivalence (we also called $\Omega$-quasi-isomorphism) if the map of dg algebras $\Omega f: \Omega C \to \Omega C'$ obtained after applying the cobar functor is a quasi-isomorphism. Note that this notion is stronger than quasi-isomorphism of dg coalgebras.<|endoftext|> TITLE: Exponential law w.r.t. compact-open topology QUESTION [5 upvotes]: It is well-known that if a topological space $Y$ is locally compact (not necessarily Hausdorff), then the map $$ \operatorname{Hom}(X \times Y, Z) \to \operatorname{Hom}(X, Z^Y) $$ (here we use the compact-open topology for $Z^Y$) is bijective for arbitrary topological spaces $X, Z$. Does the converse of this hold? If we impose no restriction on the topolgy for $Z^Y$, $Y$ satisfies this condition (exponentiable in $\mathsf{Top}$) if and only if $Y$ is core-compact. Hence $Y$ satisfies the condition if and only if $Y$ is core-compact and the topology on $Z^Y$ given here coincides with the compact-open topology for every topological space $Z$. Edit I use the definitions given here for the local compactness and the compact-open topology. Under this definition, the map is bijective for every $X, Z$. For Hausdorff (actually sober is enough) spaces, core-compact implies locally compact. REPLY [2 votes]: The answer is yes. Let $C(Y,Z)$ denote the compact-open topology on $Z^Y$ and let $E(Y,Z)$ denote the exponential topology on $Z^Y$. Let $S$ denote the Sierpinski space. In the following, we will identify $S^Y$ with the set of open subsets of $Y$ in the natural way. Claim: If $Y$ is core-compact and $C(Y,S)$ refines $E(Y,S)$, then $Y$ is locally compact. Proof: Suppose not. Then after passing to an open subspace if necessary, there is a point $y \in Y$ such that no open neighborhood of $y$ is contained in a quasicompact set. Because $Y$ is core-compact, there is an open neighborhood $U \ni y$ with $U << Y$. The set $\{V \in S^Y \mid U << V\}$ is an open neighborhood of $Y$ in $E(Y,S)$. Because $C(Y,S)$ refines $E(Y,S)$, there is a quasicompact set $A \subseteq Y$ such that $A \subseteq V \Rightarrow U << V$ for all $V \in S^Y$. We may assume without loss of generality that $A$ is an intersection of open sets (for if $A$ is quasicompact, so is the intersection of all open sets containing $A$). In particular, $A \subseteq V \Rightarrow U \subseteq V$, so $U \subseteq A$. But then $A$ is a quasicompact set containing an open neighborhood of $y$, contradicting the choice of $y$.<|endoftext|> TITLE: Real numbers with given complexity QUESTION [13 upvotes]: This may be an easy question or it may be related to a well known open problem in Computer Science. Let $\alpha>0$. We say that $\alpha$ is computed in time $T(n)$ if there is a Turing machine which for every $n>0$ written in binary produces a finite binary approximation of $\alpha$ with error bounded by $\frac 1{2^n}$. Question. Is there a real number $\alpha$ which can be computed in time $2^{2^{cn}}$ for some $c>1$ but cannot be computed in time $2^{d2^n}$ for any $d>1$? REPLY [11 votes]: Yes. First use the Time Hierarchy Theorem to find a problem in $DTIME(2^{2\cdot 3^{2^n}}) = DTIME({(2^{3^{2^n}})}^2)$ but not in $DTIME(2^{3^{2^n}})$. Let $f(k)$ be the answer (1 for yes, 0 for no) to the k-th instance of this problem. Here the k-th instance is going to correspond to an instance of this problem of size $n=log_2(k)$. So $f(k)$ is computable in time $$2^{2\cdot 3^{2^{log_{_2}(k)}}} = 2^{2 \cdot 3^k}$$ but not in time $$2^{3^{2^{log_{_2}(k)}}} = 2^{3^k}$$ and thus also not in time $2^{d2^k}$ for any d > 1. Finally let $\alpha$ be the real number whose binary expansion is $\sum_{i=1}^{\infty} f(i)\frac{1}{2^i}$. In order to compute the $\alpha$ within $< \frac{1}{2^n}$, you need to know the first n digits, which is the same as computing $f(1), f(2), ... , f(n)$.<|endoftext|> TITLE: Question on Hall's theorem QUESTION [5 upvotes]: Theorem 9.3.1 in Hall's group theory says: Let $G$ be a solvable group and $|G|=m\cdot n$, where $% m=p_{1}^{\alpha _{1}}\cdot \cdot \cdot p_{r}^{\alpha _{r}}$, $(m,n)=1$. Let $% \pi =\{p_{1},...,p_{r}\}$ and $h_{m}$ be the number of $\pi -$Hall subgroups of $G$. Then $h_{m}=q_{1}^{\beta _{1}}\cdot \cdot \cdot q_{s}^{\beta _{s}}$ satisfies the following condition for all $i\in \{1,2,...,s\}$. $% q_{i}^{\beta _{i}}\equiv 1$ (mod $p_{j}$), for some $p_{j}$. This question arises now that if we replace assumption solvable group with $p$-solvable group whether again Theorem is true. In the other words: Is it true the following claim? Or is there any counterexample for the claim? Let $G$ be a $p$-solvable group and $|G|=p^{\alpha }\cdot n$ such that $(p^{\alpha },n)=1$($p\neq 2$). Let $% h_{m}$ be the number of Sylow $p-$subgroups of $G$. Then $h_{m}=q_{1}^{\beta _{1}}\cdot \cdot \cdot q_{s}^{\beta _{s}}$ satisfies the following condition for all $i\in \{1,2,...,s\}$. $q_{i}^{\beta _{i}}\equiv 1$ (mod $p$). REPLY [4 votes]: I think the answer is yes. The number of Sylow $p$-subgroups of $G$ is $[G:N_{G}(P)],$ where $P$ is a Sylow $p$-subgroup of $G$. This number is unchanged if we pass to $G/O_{p}(G),$ so we might as well suppose that $O_{p}(G) = 1.$ Then since $G$ is $p$-solvable, we have $O_{p^{\prime}}(G) = N \neq 1.$ The Schur-Zassenhaus theorem is indeed relevant here, because by the Schur-Zassenhaus Theorem, we have $N_{G^{\ast}}(P^{\ast}) = N_{G}(P)N/N \cong N_{G}(P)/N_{N}(P),$ where $G^{\ast} = G/N$ (though in fact I think the Frattini argument would suffice here). Hence $[G^{\ast} : N_{G^{\ast}}(P^{\ast})] [N:N_{N}(P)] = [G:N_{G}(P)],$ so by induction, it suffices to prove the result for $NP.$ In other words, it suffices to prove the result for $H = PN,$ which is a group with a normal $p$-complement. But for each prime divisor $q$ of $|N|,$ there is a $P$-invariant Sylow $q$-subgroup $Q$ of $N$ which contains a Sylow $q$-subgroup of $C_{H}(P).$ Now we have $N_{PQ}(P) = PC_{Q}(P),$ so by Sylow's Theorem, applied in $PQ$, we have $[Q:C_{Q}(P)| \equiv 1$ (mod $p$). But $[H:N_{H}(P)] $ is the product of the various $[Q:C_{Q}(P)]$ ( as $q$ runs through prime divisors of $|N|),$ so the result follows. Note that $[H:N_{H}(P)] = [N:C_{N}(P)]$ since $N$ is a normal subgroup of order prime to $p$ and $P$ is a $p$-group.<|endoftext|> TITLE: Running most of the time in a connected set QUESTION [29 upvotes]: Let $P$ be a compact connected set in the plane and $x,y\in P$. Is it always possible to connect $x$ to $y$ by a path $\gamma$ such that the length of $\gamma\backslash P$ is arbitrary small? Comments: Be aware of pseudoarc --- it is a compact connected set which contains no nontrivial paths. If $P$ contains an everywhere dense curve, then the answer is "yes"; the same holds if $P$ contains a dense countable collection of curves. REPLY [9 votes]: The answer to this question is positive. A required path $\gamma$ can be constructed inductively using the following Lemma. For any continuum $P\subset\mathbb R^2$, distinct points $x,y\in P$, and $\varepsilon>0$ there are continua $P_1,\dots,P_{k}\subset P$ of diameter $<\varepsilon$, and horizontal or vertical arcs $A_0,\dots,A_{k+1}$ in the plane such that $\{x\}=A_0$, $\{y\}=A_{k+1}$; $\sum_{i=1}^k\lambda(A_i\setminus P)<\varepsilon$; for every $i\in\{1,\dots,k\}$ the continuum $P_{i}$ intersects the arcs $A_{i-1}$ and $A_i$. Proof. Replacing $P$ by a smaller subcontinuum, we can assume that $P$ is irreducible between points $x$ and $y$, which means that any subcontinuum of $P$ that contains $x$ and $y$ coincides with $P$. The irreducibility of $P$ implies that for any rational numbers $a TITLE: Semigroup of differentiable functions on real line QUESTION [13 upvotes]: Let $D(\mathbb R) $ be the set of all differentiable functions $f: \mathbb R \to \mathbb R$. Then obviously $D(\mathbb R)$ forms a semigroup under usual function composition. Can we characterize (up to semigroup isomorphism) all finite subsemigroups of $D(\mathbb R)$ which do not contain any constant function ? REPLY [8 votes]: Yes, we can! ;-) In fact there are only two finite subsemigroups of $D(\mathbb{R})$ which do not contain constant functions. Every finite subsemigroup of $D(\mathbb{R})$ necessarily contains an idempotent $f$, i.e. a function $f$ such that $f \circ f = f$. Let us examine the properties of $f$. Let $A$ be the range of $f$. By continuity of $f$, $A$ is an interval. Since $f(f(x)) = f(x)$ for every $x \in \mathbb{R}$, we have $f(x) = x$ for $x \in A$. If $f$ is non-constant, then $A$ has non-empty interior. We claim that in this case $A = \mathbb{R}$. Suppose, contrary to this claim, that $A$ is bounded from above, and denote the right endpoint of $A$ by $b$. Then $f(b) = \lim_{x \to b^-} f(x) = \lim_{x \to b^-} x = b$, and so $b \in A$. Thus, $f$ attains a local maximum at $b$. Since $f$ is differentiable at $b$, we have $f'(b) = 0$. On the other hand, $f'_-(b) = \lim_{x \to b^-} f'(x) = 1$, a contradiction. We conclude that $A$ is not bounded from above. Similarly, $A$ is unbounded from below. Thus, either $f$ is constant or $f$ is the identity function. It follows that any finite subsemigroup of $D(\mathbb{R})$ contains either a constant function or the identity function. Suppose that $X$ is a finite subsemigroup of $D(\mathbb{R})$ with no constant function and $g \in X$. Then the subsemigroup of $X$ generated by $g$ contains an idempotent, and hence — the identity function. In other words, $g^{\circ n}$ is the identity function for some $n$. Todd Trimble already pointed out in his answer that necessarily $n = 1$ or $n = 2$, and if $n = 2$, then $g$ is decreasing. Here is a shorter variant of his argument that does not require differentiability: $g$ is invertible and continuous, and hence strictly monotone; $g \circ g$ is thus strictly increasing; if $g(g(x)) > x$ for some $x$, then $g^{\circ 2n}(x) > x$, a contradiction; similarly, if $g(g(x)) < x$ for some $x$, then $g^{\circ 2n}(x) < x$; therefore, $g(g(x)) = x$ for all $x$; if $g$ is increasing, then in a similar way $g(x) = x$ for all $x$. Finally, if $g, h \in X$ and none of them is the identity function, then both are decreasing, and so $g \circ h$ is an increasing function in $X$. Therefore, $g \circ h$ is the identity function, and consequently $g = h^-1 = h$. We have thus proved that that any discrete subsemigroup of $D(\mathbb{R})$ with no constant function contains the identity function and at most one strictly decreasing function $g$ such that $g = g^{-1}$. REPLY [5 votes]: This is of course nowhere near a full answer, but in response to Mark Sapir's question, I think the only invertible elements in the monoid of finite order are of order $1$ or $2$ (involutions). A diffeomorphism, by which I mean an invertible element in the monoid of (not necessarily continuously) differentiable functions on $\mathbb{R}$, clearly cannot have zero derivative anywhere, by the chain rule. Nor can the derivative ever change sign, since a derivative function satisfies the intermediate value property: if it is positive at one point and negative at another, then it is $0$ somewhere in between (even in the non-continuously differentiable case). So, by the mean value theorem, a diffeomorphism is either strictly monotone increasing or strictly monotone decreasing. In the increasing case, the only element $f$ of finite order is the identity. For if $x < f(x)$ for any $x$, then by strict monotonicity of $f$ we have $x < f(x) < f f(x) < f f f (x) < \ldots$, and a similar argument applies if $f(x) < x$. In the case where $f$ is decreasing, suppose $f$ has order $n$. Then $f^2 = f \circ f$ is increasing and has order $n/2$ if $n$ is even, and order $n$ if $n$ is odd. By the preceding paragraph, it follows that $n/2 = 1$ or $n = 1$ respectively (and of course the latter case doesn't happen since the identity function is not monotone decreasing).<|endoftext|> TITLE: Is there any sort of higher-order SVD (quadratic and above) for dimensionality reduction? QUESTION [7 upvotes]: (Posted this on math.stackexchange and cross.correlated over more than a week ago, but didn't get an answer, and this is a question in my research so this seems like it might have been the better place to ask...) I'm wondering if there exists any higher-order SVD for dimensionality reduction. Note that I do not mean multilinear PCA, which operates on data tensors, but some form of SVD which can produce, say, a quadratic approximation of a dataset. Intuitively, SVD takes in a set of vectors as a datset, and computes the ranked eigenvalues and eigenvectors which correspond to a reduced subspace and a "weight" representing their importance. Given a decomposition: $$A = U\Sigma V^T$$ Where $A$ is a matrix where the columns are data vectors, the truncated matrix $U$ (keeping only the leftmost $n$ columns; call this truncated version $\mathcal{U}$) provides a reduced approximate basis for the original dataset $A$. However, this approximation is purely linear. It can thus be thought of as a linear approximation of the data on some $n$ reduced variable set. For a particular column of $a_i$ of $A$, $a_i \approx \mathcal{U} x_i$ for some dimensionality-reduced vector $x_i$. I am wondering if it is possible to do something akin to a Taylor expansion here, and recover a higher-order approximation of the data. For example, for a quadratic system of order $n$, I'd like to generate the $\mathcal{U}$ matrix, but also a third-order tensor $\mathcal{W}$ such that a "quadratic" approximation of the data could be written, as, say: $a_i \approx \mathcal{U} x_i + \frac{1}{2} x_i^T (\mathcal{W} ~ \vdots ~ x_i)$ where $\vdots$ is a tensor-vector contraction. (This is a form I made up now for discussion's sake; it is possible that the true form of some quadratic approximation is slightly different). Here, the best $x_i$ the quadratic approximation will be better than the linear one. Ideally, I'd be searching for some general theory which would allow for arbitrary order approximation. Is there any general established method for this (or something similar)? If so, is it tractable? If not, is there a mathematically grounded reason why not? REPLY [2 votes]: A Non-linear Generalization of Singular Value Decomposition proposes a non-linear extension of the singular value decomposition by appending additional columns to the data matrix which are non-linearly derived from the existing columns. Section V gives an application to the logistic map, $X_{n+1}=\lambda X_n(1-X_n)$, where $\lambda$ is an unknown parameter that we wish to retrieve from the data. When SVD is applied to a matrix that contains columns with the time series $X_n$ as well as the square $X_n^2$, the appearance of zero singular values indicates a linear relationship between columns, from which an estimate for $\lambda$ is obtained.<|endoftext|> TITLE: Existence of Solution, System of Equations QUESTION [5 upvotes]: Suppose $P(\lambda, i)$ is the probability that a Poisson random variable with average $\lambda$ is equal to $i$, i.e. $\frac{\lambda^i}{e^{\lambda}i!}$ I think the following system of equations always has solution in $x$ and $y$, non-negative real numbers, for any $\alpha>0$ and $k\in \mathbb{N}_+$ \begin{cases} \alpha=\sum_{i=0}^{\infty}P(x, i)\cdot P(y, k+i) \\ \alpha=\sum_{i=0}^{\infty}P(x, i)\cdot P(y, k+i+1) \end{cases} where the necessary condition $\alpha\leq P(k+1, k+1)$ holds. It is easy to prove that this is indeed a necessary condition, equivalent to the condition that $\alpha=P(\lambda,k+1)$ has a solution. It is also easy to see that solution $y$ of the system is smaller or equal to $\lambda$, the largest solution to the equation $\alpha=P(\lambda,k+1)$. Experiments show that for each fixed $\alpha$ and $k$, there is a solution, but I did not manage to prove it analytically. Is there any analogue for mean value theorem for multidimensional functions? Any suggestion for the proof directions will be appreciated. REPLY [6 votes]: Let $a:=\alpha$ and \begin{equation*} F_k(x,y):=\sum_{j=0}^\infty \frac{x^j}{j!}\frac{y^{k+j}}{(k+j)!}\,e^{-x-y}, \end{equation*} assuming the standard convention $0^0:=1$. We have to consider the existence of a solution in $x$ and $y$ of the system \begin{equation*} a=F_k(x,y)=F_{k+1}(x,y). \tag{1} \end{equation*} We shall prove the following. Theorem 1. Take any natural $k$ and any \begin{equation*} a\in(0,a_k],\quad\text{where}\quad a_k:=\sup_{x,y\ge0}F_{k+1}(x,y). \tag{1.5} \end{equation*} Then the system (1) has a solution $x,y\ge0$. Remark 1. Since $F_k>0$, the condition $a\in(0,a_k]$ is obviously necessary in Theorem 1. Proof of Theorem 1. Note that $F_k(x,y)\ge0$ for any real $x,y\ge0$ and $F_k(x,y)$ is continuous in real $x,y\ge0$. The crucial observation is the identity \begin{equation*} \partial_y F_{k+1}(x,y)=F_k(x,y)-F_{k+1}(x,y) \tag{2} \end{equation*} for real $x,y$. Next, fix for a moment any real $x\ge0$. Then $F_{k+1}(x,0)=0$ and, by dominated convergence, $F_{k+1}(x,\infty-)=0$. So, $F_{k+1}(x,y)$ attains its maximum in $y$ at some real point $y=y_x\ge0$. At this point, we have $\partial_y F_{k+1}(x,y)=0$. So, by (2), \begin{equation*} F_k(x,y_x)=F_{k+1}(x,y_x)=\max_{y\ge0}F_{k+1}(x,y)=:M_{k+1}(x), \tag{3} \end{equation*} for all real $x\ge0$. Next, \begin{align*} M_k(x)&\le\sum_{j=0}^\infty \frac{x^j}{j!}\max_{y\ge0}\frac{y^{k+j}}{(k+j)!}\,e^{-x-y} \\ &=\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\frac{(k+j)^{k+j}}{(k+j)!}\,e^{-k-j} \tag{4} \\ &\ll\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\frac1{\sqrt{k+j}}=E\frac1{\sqrt{k+\Pi_x}} \underset{x\to\infty}\longrightarrow0 \end{align*} by dominated convergence and because $\Pi_x\underset{x\to\infty}\longrightarrow\infty$ in probability, where $\Pi_x$ is a Poisson random variable with parameter $x$. So, $M_k(\infty-)=0$. It is also not hard to see that $F_k(x,y)$ is continuous in real $x\ge0$ uniformly in real $y\ge0$ (see the Appendix), so that $M_k(x)$ is continuous in $x\ge0$. So, $M_{k+1}(x)$ attains its maximum in $x\ge0$ (equal $a_k$, by (1.5)) and takes all values in the interval $(0,a_k]$. Now Theorem 1 follows by (3). $\qquad\Box$ Appendix. Similarly to (2), \begin{equation*} \partial_x F_k(x,y)=F_{k+1}(x,y)-F_k(x,y). \end{equation*} for real $x,y$. Therefore and because $0\le F_k\le1$, we have $|\partial_x F_k(x,y)|\le1$ for real $x,y$, so that $F_k(x,y)$ is indeed continuous in real $x\ge0$ uniformly in real $y\ge0$. Added: Let us now show that \begin{equation*} a_k=c_{k+1},\quad\text{where}\quad c_k:=\frac{k^k}{k!}\,e^{-k} \sim\frac1{\sqrt{2\pi k}} \end{equation*} as $k\to\infty$. To this end, note first that \begin{equation*} c_{k+1}/c_k=(1+1/k)^k/e<1, \end{equation*} and so, $c_k$ is decreasing in $k$. So, recalling (4), we have \begin{equation*} M_k(x)\le\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\,c_{k+j} \le\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\,c_k=c_k=M_k(0). \end{equation*} Thus, in view of (1.5) and (2), \begin{equation*} a_k=\max_{x\ge0}M_{k+1}(x)=M_{k+1}(0)=c_{k+1}, \end{equation*} as desired. In particular, for $k=0,1,2,3$ the values of $a_k$ are $\approx0.367879, 0.270671, 0.224042$.<|endoftext|> TITLE: Atiyah-Patodi-Singer for manifolds with cusps QUESTION [7 upvotes]: Dear Colleagues and Friends, Please let me know if you are aware of any references to the following question. The classical result of Atiyah, Patodi and Singer tells us that if $W$ is a compact oriented Riemannian 4-manifold with boundary $M$ and, moreover, if we assume that near M the metric is isometric to a product, then $$ sign(W)= \frac{1}{3} \int_W p_1 - \eta(M),$$ where $p_1$ is the differential form representing the first Pontryagin class of $W$, and $\eta$ is the eta-invariant of $M$. What about the case when both $W$ and $M$ are hyperbolic manifolds and are allowed to have cusps? Or, say, $W$ and $M$ are Riemannian as above, with infinite ends of finite volume, on which the metric is isometric to a product? Any information will be appreciated. Please excuse my ignorance as differential geometer. Correction: for hyperbolic manifolds with boundary and/or cusps the metric near the boundary is not a product (otherwise we won't have finite volume, I'd suppose). There is a correction term (already in the paper by Long and Reid), which is, however, vanishing due to various reasons (e.g. for totally geodesic boundary its second fundamental form vanishes, and it annihilates the correction term, and for the cusp case we can deduce it from the fact that the volume of cusp section by a small horoball is 0 in the limit, and we integrate over that horoball). REPLY [2 votes]: This type of questions has been investigated systematically by Melrose in the framework of 'c-calculus', where $c$ stands for the cusp. The basic idea, if I recall correctly is to blow up the boundary with the cusp three times (two blow ups for the boundary, an extra blow-up for the cusp on the boundary) and work with the heat operator associated with the Dirac operator on the blow-up space. Normally this does not help you "gain" anything since the metric is arbitrary. In your case, however the metric is product type. So your can explicitly recover the $\eta$-invariant and there will be an extra contribution from the third blow up. The exact meaning of this contribution may be difficult to interpret from the view of spectral flow. I am not entirely sure what is a good reference for this, though. There is no good reference for $c$-calculus (Melrose has a book on $b$-calculus, which is dense). This paper may be the closest thing I can find at the moment (there should be expository articles on this written by Rafe, here is an article by Paul). I am no longer working in the area, so there may be mistakes from my poor memory.<|endoftext|> TITLE: Primality test for specific class of Proth numbers QUESTION [11 upvotes]: Can you provide a proof or a counterexample for the following claim : Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$ Let $N=k\cdot 2^n+1$ such that $n>2$ , $0< k <2^n$ and $\begin{cases} k \equiv 1,7 \pmod{30} \text{ with } n \equiv 0 \pmod{4} \\ k \equiv 11,23 \pmod{30} \text{ with } n \equiv 1 \pmod{4} \\ k \equiv 13,19 \pmod{30} \text{ with } n \equiv 2 \pmod{4} \\ k \equiv 17,29 \pmod{30} \text{ with } n \equiv 3 \pmod{4} \end{cases}$ Let $S_i=S_{i-1}^2-2$ with $S_0=P_k(8)$ , then $N$ is prime iff $S_{n-2} \equiv 0 \pmod N$ . You can run this test here . A list of Proth primes sorted by coefficient $k$ can be found here . I have tested this claim for many random values of $k$ and $n$ and there were no counterexamples . Note that for $k=1$ we have Inkeri's primality test for Fermat numbers . Reference : Tests for primality, Ann. Acad. Sci. Fenn. Ser. A I 279 (1960), 1-19. REPLY [7 votes]: Your criterion is equivalent to: $$(4+\sqrt{15})^{k2^{n-1}}\equiv (4+\sqrt{15})^{\frac{N-1}{2}}\equiv -1 (\bmod N \mathbb{Z}[\sqrt{15}]).$$ The point is that $P_m(8)=(4+\sqrt{15})^m+(4-\sqrt{15})^m$. Moreover, $S_i=(4+\sqrt{15})^{k2^i}+(4-\sqrt{15})^{k2^i}$, which one may prove by induction, using the fact that $x\mapsto x^2-2$ is semi-conjugate to $z\mapsto z^2$ under the relation $x=z+1/z$. Then the condition $S_{n-2} = (4+\sqrt{15})^{k2^{n-2}}+(4-\sqrt{15})^{k2^{n-2}} \equiv 0 (\bmod N)$ is equivalent to $(4+\sqrt{15})^{k2^{n-1}} \equiv -1 (\bmod N)$. By Fermat's little theorem (see Theorem 1 of this paper) in the quadratic number field $\mathbb{Q}[\sqrt{15}]$, for any odd prime $p$, one has $(4+\sqrt{15})^{p-\left(\frac{15}{p}\right)}\equiv 1 (\bmod p)$, where $\left(\frac{15}{p}\right)$ is the Legendre symbol (note that $(4+\sqrt{15})(4-\sqrt{15})=1$, so $4+\sqrt{15}$ is a unit in the ring of integers $\mathbb{Z}[\sqrt{15}]$). One also has by Theorem 2 of this paper that if $l$ is the smallest integer with $(4+\sqrt{15})^l \equiv -1 (\bmod p)$, and $(4+\sqrt{15})^K\equiv -1(\bmod p)$, then $K=lu$, where $u$ is odd. Also, the smallest integer with $(4+\sqrt{15})^e \equiv 1 (\bmod p)$ is $e=2l$. Now, let's show that your criterion implies that $N$ is prime. For contradiciton, let $p < N$ be a prime factor of $N$, then by your criterion, $(4+\sqrt{15})^{k2^{n-1}}\equiv -1 (\bmod p)$. Then $k2^{n-1} = l u$, where $l$ is the smallest exponent $>0$ with $(4+\sqrt{15})^k\equiv -1(\bmod p)$, and $u$ is an odd integer by Theorem 2. So $l=2^{n-1}\delta$, with $\delta | k$, and hence $e \geq 2^n$. Theorem 1 implies that $(4+\sqrt{15})^{p\pm 1}\equiv 1(\bmod p)$. Then $p\pm 1\geq 2^n$, for every prime $p | N$. $N$ is not a square from your congruence conditions on $k$ and $n$ (check $(\bmod 15)$), so it has a factorization $pq$ with $p\geq 2^n-1$, $q\geq p+2$. But then $N = p\cdot q \geq p(p+2)\geq (2^n-1)(2^n+1)=2^n\cdot 2^n-1 > k 2^n +1 =N$ since $k < 2^n$, a contradiction. Going in the reverse direction (following the proof of Theorem 1 of this paper), your congruence conditions imply that $N\equiv 2,8 (\bmod 15)$ and $N\equiv 1(\bmod 8)$, hence $\left(\frac{15}{N}\right)=1, \left(\frac{2}{N}\right)=1,\left(\frac{5}{N}\right)=-1$. We have $(4+\sqrt{15})=\frac{5+\sqrt{15}}{5-\sqrt{15}}$. Hence $$(4+\sqrt{15})^{\frac{N-1}{2}}\equiv -1 (\bmod N)$$ if and only if $$(5+\sqrt{15})^{\frac{N-1}{2}}\equiv - (5-\sqrt{15})^{\frac{N-1}{2}} (\bmod N)$$ if and only if $$(5+\sqrt{15})^{N-1} \equiv - 10^{\frac{N-1}{2}} \equiv - 2^{\frac{N-1}{2}} 5^{\frac{N-1}{2}} \equiv - \left(\frac{2}{N}\right) \left(\frac{5}{N}\right ) \equiv 1 (\bmod N),$$ which holds by Theorem 1 and the properties of Legendre symbols.<|endoftext|> TITLE: Do abelian varieties have Neron models over arbitrary valuation rings? QUESTION [10 upvotes]: Let $\mathcal{O}_K$ be a valuation ring with fraction field $K$. Let $A$ be an abelian variety over $K$. Does $A$ have a Neron model? If $\mathcal{O}_K$ is a discrete valuation ring, then this is proven in the book of Bosch-Lutkebohmert-Raynaud on Neron models. REPLY [5 votes]: As a partial answer to your question, let me sketch a proof of the following result found in 2015 in collaboration with David Holmes. Proposition. For a valuation ring $V$, every abelian $V$-scheme $A$ is the Néron model of its generic fiber in the sense that for every smooth $V$-scheme $S$, we have $A(S) = A(S_K)$ via pullback, where $K := \mathrm{Frac}(V)$. Proof. Firstly, the result should hold: by weak uniformization (which is still conjectural in positive and mixed characteristic), every valuation ring is a filtered direct limit of regular local rings, and over a regular ring every abelian scheme is its own Néron model: the claim basically amounts to extending line bundles on the dual abelian scheme, and line bundles over regular bases do extend. As far as the actual proof goes, the idea is to deduce the claim via a limit argument from the extension result of Weil about rational maps into group schemes. Now for the details. Since $A$ is separated over $V$ and $S_K$ is schematically dense in $S$, the injectivity of $A(S) \rightarrow A(S_K)$ is clear (it follows from EGA I, 9.5.6, as usual). Thus, we need to show that any $V$-map $S_K \rightarrow A$ extends to a $V$-map $S \rightarrow A$. For this, since the extensions are unique by the previous step, we may work Zariski or even étale locally on $S$, so we fix a point $s \in S$ around which we want to extend and may assume that $S$ is affine, connected, and admits an étale $V$-map $S \rightarrow \mathbb{A}^n_V$ for some $n \ge 0$ (we use EGA IV, 17.11.4). Let $v \in \mathrm{Spec}(V)$ be the image of $s$. By replacing $V$ by its localization $\mathcal{O}_{V, v}$, which is also a valuation ring (https://stacks.math.columbia.edu/tag/052K), and spreading out, we may assume that $v$ is the closed point of $\mathrm{Spec}(V)$. Then, by replacing $V$ by its strict Henselization, which is still a valuation ring https://stacks.math.columbia.edu/tag/0ASK, we assume that $V$ is strictly Henselian. In this case, by Hensel's lemma (EGA IV, 18.5.17), every connected component of $S_v$ meets some $V$-point $\mathrm{Spec}(V) \rightarrow S$. By EGA IV, 15.6.5, the fibral connected components of $S$ that meet this $V$-point comprise an open subset of $S$, so, by replacing $S$ by this open, we are reduced to the case when the fibers of $S$ are geometrically connected (we are using the well-known EGA IV, 4.5.13). Let $s_0 \in S$ be the generic point of $S_v$. We claim that $\mathcal{O}_{S, s_0}$ is a valuation ring. For this, since $s_0$ maps to the generic point $a_0$ of the fiber $\mathbb{A}^n_v$ (generizations lift along flat maps!), the stability under étale maps of being a valuation ring (https://stacks.math.columbia.edu/tag/0ASJ) reduces us to considering $\mathcal{O}_{\mathbb{A}^n_V, a_0}$. The latter is a domain whose fraction field is the function field $K(\mathbb{A}^n_K)$ and, since the ideals of $V$ are totally ordered, one sees directly that every rational function in $K(\mathbb{A}^n_K)$, i.e., a quotient of polynomials with $K$-coefficients, can be expressed as a quotient of polynomials with $V$-integral coefficients and at least one coefficient in $V^\times$. Then either this quotient or its inverse lies in $\mathcal{O}_{\mathbb{A}^n_V, a_0}$, so that the latter is indeed a valuation ring (https://stacks.math.columbia.edu/tag/052K again), and hence so is $\mathcal{O}_{S, s_0}$. Now we apply the valuative criterion of properness to obtain a $V$-map $\mathrm{Spec}(\mathcal{O}_{S, s_0}) \rightarrow A$ compatible with the map $S_K \rightarrow A$ that we want to extend. The former spreads out to an affine neighborhood of $\mathrm{Spec}(\mathcal{O}_{S, s_0})$, so we may find a quasicompact open $S_0 \subset S$ containing both $s_0$ and $S_K$ to which the initial map $S_K \rightarrow A$ extends. Since $V$ is a filtered direct limit of normal domains $D$ that are of finite type over $\mathbb{Z}$, we may assume that $A$, $S$, $S_0$, $S_K \rightarrow A$, $S_0 \rightarrow A$ arise by base change from corresponding objects $A'$, $S'$, $S'_0$, $S'_{K'} \rightarrow A'$, $S'_0 \rightarrow A'$ (that are subject to analogous assumptions) defined over some such $D$, where $K' := \mathrm{Frac}(D)$. Since $s$ and $s_0$ have the same image in $\mathrm{Spec}(V)$ and we work locally at $s$, by replacing $S'$ by the preimage of the (open) image of $S_0'$ in $\mathrm{Spec}(D)$ and using the fact that $S'$ inherits geometric connectedness of fibers from $S$, we may assume that $S_0'$ is fiberwise dense in $S'$. Then $S_0'$ covers all the height $\le 1$ points of $S'$: indeed, it covers $S'_K$ and every other height $\le 1$ prime is the generic point of a fiber of $S'$ at a height $\le 1$ prime of $D$. Therefore, by Weil's result Thm. 4.4/1 in Bosch, Lutkebohmert, Raynaud "Neron models," the map $S_0' \rightarrow A$ extends to a map $S' \rightarrow A$. The base change of this extension back to $V$ is the desired extension $S \rightarrow A$ of $S_K \rightarrow A$. QED. To end with some speculations, I am guessing that in the case of semiabelian reduction, Néron lft models should exist over arbitrary valuation rings--it would be interesting to see this worked out (or a counterexample given).<|endoftext|> TITLE: Berkovich space including both archimedean and non-archimedean worlds QUESTION [13 upvotes]: From this Temkin's paper (at the end of section 1.1.3), I know that one may define Berkovich spaces that include both archimedean and non-archimedean worlds. This looks very interesting. Temkin mentions an example: the affine line over $\mathbb Z$. But I was wondering if there are more examples to include both worlds(especially in higher dimensions). REPLY [20 votes]: The definition of analytic space over $\mathbf{Z}$ was given by Berkovich in his foundational book "Spectral theory and analytic geometry over non-Archimedean fields" (see the beginning of section 1.4 and section 1.5). It is quite general, and anyway enough so that analytifications of schemes locally of finite type over $\mathbf{Z}$ make sense. (Let me mention that similar constructions exists over $\mathbb{Z}[1/N]$ or ring of integers of arbitrary number fields.) The construction is actually quite simple. As a set, the affine analytic space of dimension $n$ over $\mathbf{Z}$ is just the set of multiplicative seminorms on $\mathbf{Z}[T_1,\dotsc,T_n]$. Of course, it also comes with a topology (which makes it a Hausdorff locally compact space) and a sheaf of functions. Once you have the affine space, you can take open subsets, closed analytic subsets, etc. and glue to get arbitrary spaces. To understand better what is going on, let me first describe the space of dimension 0, usually denoted by $\mathcal{M}(\mathbf{Z})$. As I said before, it is the set of multiplicative seminorms on $\mathbf{Z}$, so Ostrowski's theorem allows us to describe it explicitly: it contains the trivial absolute value (sending 0 to 0 and everything else to 1), the usual absolute value possibly raised to some power between 0 and 1 (the power 0 giving back the trivial absolute value) and for each prime number $p$, the $p$-adic absolute value raised to some power between 0 and $+\infty$ (the power 0 giving back the trivial absolute value, the power $+\infty$ giving the seminorm induced by the trivial absolue value on $\mathbf{Z}/p\mathbf{Z}$). I will not try to draw a picture here, but it looks like a spider with the trivial absolute value in the middle and a leg for each place. Now, if you want to describe what the affine analytic space of dimension $n$ over $\mathbf{Z}$, you can look at its projection onto $\mathcal{M}(\mathbf{Z})$ and describe the fibers. As you could guess, the fiber over a $p$-adic absolute value (power not 0 nor $+\infty$) will be a honest Berkovich affine analytic space of dimension $n$ over $\mathbf{Q}_p$. (If the power is 0, i.e. the absolute valued is trivial, you get a space over $\mathbf{Q}$ trivially valued and if the power is $+\infty$, you get a space over $\mathbf{F}_p$ trivially valued.) Interestingly enough, over the usual absolute value (or a non-zero power of it), you get something quite close to $\mathbf{C}^n$, namely its quotient by the complex conjugation (with diagonal action). Moreover, the topology and the structure sheaf on this fiber are the usual ones. These spaces have not been studied so much but some basic properties are known. In the references mentioned in Xarles' answer, I proved that they have nice local properties (first reference for the affine line, second for arbitrary spaces): the stalks of the structure sheaf are noetherian local rings (even excellent), the structure sheaf is coherent, etc. In his thesis, Thibaud Lemanissier proved that they are locally path-connected. In on-going joint work, we prove some cohomological properties (vanishing of coherent cohomology on disks, annuli, etc., GAGA-like theorems). Since you asked about examples, let me add one to finish. It illustrates another nice feature of Berkovich spaces over $\mathbf{Z}$: they provide nice parameter spaces. Let me first consider the relative open punctured unit disk $\mathbf{D}^*$ over $\mathbf{Z}$, i.e. the subset of the affine line (with coordinate say $q$) defined by $\{0 < |q| < 1\}$. You can now take a relative $\mathbf{G}_m$ (i.e. the affine line minus 0) over $\mathbf{D}^*$. So over each point $x$ of $\mathbf{D}^*$, you have an analytic $\mathbf{G}_m$ that is defined over the residue field $\mathcal{H}(x)$ of the point (as for schemes, each point has an associated residue field that is a complete valued field). Edit about $\mathbf{G}_m$: Here I really mean the Berkovich analytic line minus 0. Still it behaves as you would expect from scheme theory: the $\mathcal{H}(x)$-rational points of $\mathbf{G}_m$ over $\mathcal{H}(x)$ are exactly $\mathcal{H}(x)^\times$ (and similarly $\mathbf{G}_m(K) = K^\times$ for any extension of $\mathcal{H}(x)$). But you also have an element $q(x)$ in $\mathcal{H}(x)$ with absolute value between 0 and 1. So you can construct the quotient of the multiplication by $q(x)$ and end up with a Tate curve over $\mathcal{H}(x)$. Edit: This is nothing but the Berkovich analogue of $\mathcal{H}(x)^\times/q(x)^\mathbf{Z}$. (Actually $\mathcal{H}(x)^\times/q(x)^\mathbf{Z}$ is exactly the set of $\mathcal{H}(x)$-rational points, but you there are points over arbitrary extensions too.) Note that this works for all fields $\mathcal{H}(x)$, which could be $\mathbf{Q}_p$, $\mathbf{F}_p((t))$, $\mathbf{C}$, etc. The nice point is that this construction can be carried out globally (as opposed to fiber by fiber) in exactly the same way, giving rise to some sort of universal Tate curve (defined as an analytic space over $\mathbf{Z}$) with an analytic global uniformization map. (In joint work with Daniele Turchetti, we do similar constructions for Mumford curves of higher genus.) This answer is already long enough I believe, so I will stop here, but if you have any questions, please ask. And sorry for the shameless self-promotion.<|endoftext|> TITLE: Properties of a "research announcement" QUESTION [21 upvotes]: Some mathematics journals publish "research announcements", a class of publication that before today I had not heard of. An example is Electronic Research Announcements in Mathematical Sciences. I presume that after publishing a research announcement in such a journal presenting a particular result, one can subsequently then publish a full research paper on the same result at a later date, generally in a different journal. Obviously, it is not ordinarily the case that journals will knowingly allow the same result to be published twice. Therefore, research announcements must have some defining properties which make this practice acceptable. My question is the following: what are the properties of the research announcement which allow the subsequent publication of the full research paper describing the same result? Such a question may be of importance; for example, to a journal editor who is handling a submission that describes a result which has been previously published as a research announcement. Perhaps the answer is simple: that the research announcement must contain no proofs. But perhaps the convention is more subtle than this, I'm not sure. I am aware that the practice of publishing research announcements is not widespread, and I am not interested for the purposes of this question in discussing whether anyone ought to publish a research announcement in any particular situation. I believe that this question is best suited to mathoverflow.net, rather than (for example) to academia.stackexchange.com, as I am asking specifically about publication practice in mathematics. REPLY [18 votes]: This is an extended comment on history of research announcements. The principal mathematical journal which did this was Comptes Rendus published by the French Academy. It published (and still publishes) very short notes (1-2 pages), usually without complete proofs or with very short sketches of proofs. The papers were either by the members of the Academy, or approved by the members, and the time of publication was very short. Usually (but not always) several publications on the same topic were followed by a long paper, with complete proofs, covering the same material. Soviet Doklady was analogous. Later, Bull Amer. Math. Soc. followed (it published surveys and research announcements). The Soviet analog, Uspekhi Mat. Nauk (translated as the Russian Math Surveys) also published research announcements, but they were not translated into English. Another similar publication was Abstracts of the talks presented to the AMS. It published very short (few lines) announcements based on the talks actually made in the AMS meetings. The purpose of these publications was manifold. First, to secure priority. Second, to inform the experts, what results you obtained. (For a real expert in the area, a short announcement may be enough to understand what is going on). After you secure priority and inform people on what you have done, you may relax and spend several years of time on writing a complete definitive version. Such publications, (at least in theory) were very quick: the papers of the members of the Academy were not refereed, while other papers were only endorsed by a member. (I remember how once I explained my result to Paul Malliavin, on a conference, and gave him a note. Next morning he told me that it is accepted:-) If you look at the publication lists of such people as Picard, Poincare, Krein or Kolmogorov, you see that about 70% of their publications are CR/Doklady notes. Of the very famous results published only as "research announcements" by their authors, let me mention KAM theory by Kolmogorov, and Krein's "theory of string". As far as I know these authors never published complete papers leaving this to others. Of more modern examples, the works of Sullivan and Douady -Hubbard which made the revolution in holomorphic dynamics were published as research announcements in Comptes Rendus in 1982. Sullivan published a complete version in 1985, when several alternative expositions based on his announcement were already available, while the work of Douady Hubbard is still available only as a preprint. On my opinion it was very important that interested people could learn about this breakthrough in 1982 three years before Sullivan published a complete paper. It was possible for an expert to understand his proof from the short announcement. The papers of Yoccoz and Smirnov (for which Fields medals were awarded) were also published in Comptes rendus, without complete details. EDIT. Nowadays, when we have the arXiv where we can publish anything, with no delay, the role of research announcements declines. On my opinion the arXiv performs both main functions of research announcements: quickly spreads the information and secures priority.<|endoftext|> TITLE: Is this conjecture strictly weaker than P=NP? QUESTION [33 upvotes]: My three computability questions are related to the following group theory question (first asked by Bridson in 1996): For which real $\alpha\ge 2$ the function $n^\alpha$ is equivalent to the Dehn function of a finitely presented group (i.e., what numbers belong to the isoperimetric spectrum)? Clearly $\alpha\ge 1$ for every $\alpha$ in the isiperimetric spectrum, and by Gromov's theorem, the isoperimetric spectrum does not contains numbers from $(1,2)$. In what follows, all functions are bounded by polynomials, so two functions $f(n), g(n)$ are equivalent if $af(n)0$. Recently Olshanskii proved the same statement for all $\alpha\ge 2$ (the paper will appear in the Journal of Combinatorial Algebra). On the other hand if $\alpha$ is in the isoperimetric spectrum, then $\alpha$ can be computed in time at most $2^{2^{c2^{m}}}$ for some $c>0$. If P=NP, then one can reduce the number of 2's to two and bring the upper bound to be equal to the lower bound, completing the description of the isoperimetric spectrum. But the proof in our paper (Corollary 1.4) would give two 2's also if the following seemingly weaker conjecture holds. Conjecture. Let $T(n)$ be the time function of a non-deterministic Turing machine which is between $n^2$ and $n^k$ for some $k$. Then there is a deterministic Turing machine $M$ computing a function $T'(n)$ which is equivalent to $T(n)$ and having time function at most $T(n)^c$ for some constant $c$ (depending on $T$). (For the definition of the time function see this question). Question Is the conjecture strictly weaker than P=NP? Update My note with a reference to Emil's answer is here. REPLY [24 votes]: The conjecture is indeed strictly weaker than $\mathrm{P = NP}$, in the sense that it follows from $\mathrm E=\Sigma^E_2$, which is not known to imply $\mathrm{P = NP}$. Of course, we cannot prove this unconditionally with current technology, as it would establish $\mathrm{P\ne NP}$. Here, $\mathrm E$ denotes $\mathrm{DTIME}(2^{O(n)})$, $\mathrm{NE}$${}=\mathrm{NTIME}(2^{O(n)})$, and $\Sigma^E_2=\mathrm{NE^{NP}}$ is the second level of the exponential hierarchy (with linear exponent), $\mathrm{EH}$. (See e.g. [2]; warning: their notation for $\mathrm{(N)E}$ and $\mathrm{EH}$ is $\mathrm{(N)EXPTIME}$ and $\mathrm{EXPH}$, which are nowadays used for somewhat different classes.) Equivalently, we may define $\Sigma_2^E$ using alternating Turing machines (see [1,§5.3]) as $\Sigma^E_2=\Sigma_2\text-\mathrm{TIME}(2^{O(n)})$. Note that conversely, $\mathrm{P = NP}$ implies $\mathrm{P=PH}$, which implies $\mathrm E = \Sigma^E_2=\mathrm{EH}$ by a padding argument similar to [1,§2.6.2]. To see that $\mathrm E = \Sigma^E_2$ implies the conjecture, let $t$ be the function defined exactly like $T$, but with the input and output integers written in binary, and let $g_t=\{(x,y):y\le t(x)\}$ be its subgraph. Then $g_t\in\Sigma^E_2$: in order to show $(x,y)\in g_t$, we only need to (nondeterministically) guess an input $w$ of the original NTM of length $x$ along with an accepting computation, and (co-nondeterministically) check that there is no accepting computation of length $ TITLE: Fixed points of injective self-maps QUESTION [11 upvotes]: Is it consistent in $\mathsf{ZF}$ that there is a set $X$ with more than $1$ point such that every injective map $f:X\to X$ has a fixed point? REPLY [12 votes]: As Yair suggests, a strongly amorphous set has this property. Recall that an amorphous set is a set which cannot be split into two infinite sets. A strongly amorphous set is a set such that in addition to being amorphous, every partition has only finitely many non-singletons. While we didn't require that it is an infinite set, this is the common assumption, so the existence of an amorphous set contradicts the axiom of choice. One can easily show that an amorphous set cannot be linearly ordered, and that a strongly amorphous set cannot carry a group structure. Now. If $X$ is strongly amorphous and and $f\colon X\to X$ is a permutation (which it has to be if $f$ is injective, since amorphous sets are Dedekind-finite), then the orbits of $f$ are all finite and form a partition of $X$. Therefore all but finitely many points are moved at all. One could argue that perhaps this is too extreme of an example. Well, in Cohen's first model with a Dedekind-finite set of reals, the canonical Dedekind-finite set (that of the generic Cohen reals) is linearly ordered, but has the property that any partition into finite parts is almost entirely singletons (i.e. all but finitely many parts are singletons). The same argument now works on that set as well. If you have a permutation, its orbits must be finite, so almost all of them are singletons. One can also have analogues of strongly amorphous sets for larger cardinals. For example an $\aleph_1$-amorphous set is a set where every subset is countable or co-countable. Being strongly $\aleph_1$-amorphous means that all but countably many parts of any partition are singletons. And it is worth noting that $\sf DC_\kappa$ is consistent with the existence of a $\kappa^+$-amorphous set. So you can't use $\sf DC$ principles to avoid this issue.<|endoftext|> TITLE: Space filling curve whose all level sets are finite (countable) QUESTION [8 upvotes]: Is there a continuous surjective function $f:[0,1] \to [0,1]^2$ such that every level set $f^{-1}(y)$ is a finite set? If the answer is no, what about if we replace the finiteness of level sets by "countable level sets"? REPLY [14 votes]: Recall the definition of the Peano square-filling curve $f:[0,1]\to[0,1]^2$, which is given in terms infinite ternary strings. If $a\in [0,1]$ has a base $3$ representation of the form $0,a_1a_2a_3\dots$, the point $f(a):=(b,c)$ has base $3$ digits resp. $$b_n:={\bf k}^{a_2+a_4+\dots a_{2n-2}}a_{2n-1}$$ $$c_n:={\bf k}^{a_1+a_3+\dots a_{2n-1}}a_{2n}$$ where ${\bf k}$ is the involutory bijection of the set $\{0,1,2\}$ into itself given by $i\mapsto2-i$ (and exponents denote iterated composition, which in this case only depends on the parity of the exponent, since ${\bf k}^2={\bf id}$). As Peano observes, this gives a bijection (actually a homeomorphism wrto product topologies of discrete spaces) $\{0,1,2\}^{\mathbb{N}} \to \{0,1,2\}^{\mathbb{N}}\times \{0,1,2\}^{\mathbb{N}}$. In fact $$a_{2n-1}:={\bf k}^{c_1+c_2+\dots +c_{n-1}}b_{n}$$ $$a_{2n}:={\bf k}^{b_1+b_2+\dots +b_{n}}c_{n}.$$ The point of the whole construction is that, thanks to the effect of the map ${\bf k}$, the above bijection on ternary strings is compatible with the quotient map $\operatorname{val}:\{0,1,2\}^{\mathbb{N}} \to[0,1]$, that takes a ternary string $(a_1,a_2,\dots)$ to its value as a ternary expansion of a real number, $\sum_{n=1}^\infty 3^{-n}a_n$. The latter map $\operatorname{val}$ is surjective but of course not injective, due to points in $[0,1]$ with double representations of "ternary rationals", that is points in the set $T:=\{ m/3^r: r\in\mathbb{N},\ 0 TITLE: An asymptotic formula for this sum QUESTION [6 upvotes]: Let $X$ be a positive real number. Can someone help me by providing an asymptotic formula for this sum. $$\sum_{n \leq X, \; n\, \equiv\, a \mod{b}} \log{n},$$ where $a$ and $b$ are two coprime integers. Thanks in advance. REPLY [10 votes]: The sum $$F(X)=\sum_{n \leq X, \; n\, \equiv\, a \mod{b}} \log{n}=\sum_{p={\rm Int}\,[-a/b]}^{{\rm Int}\,[(x-a)/b]}\log(a+pb)$$ can be approximated in the large-$X$ limit by $$F_\infty(X)=\sum_{p=1}^{(X-a)/b}\log(pb)=\frac{X-a}{b}\log b+\log\Gamma\left(\frac{X-a}{b}+1\right)$$ Here is a plot of $F(X)$ (gold) and $F_\infty(X)$ (blue) for $a=5$, $b=11$.<|endoftext|> TITLE: A set-family game QUESTION [5 upvotes]: Two players, Green and Red, play a zero-sum game. It is parametrized by two integers $n\geq 0, k\geq 0$, and a finite family $F$ of sets of size $n$ (each set may appear multiple times in $F$). Each turn, Green colors an uncolored element green, and then Red colors an uncolored element red. A set in $F$ is considered "good" if it contains at least $k$ green elements. The score of Green is the fraction of good sets (number of good sets divided by $|F|$). Define $g(n,k,F)$ as the maximum score that Green can get in this game, and define: $$ G(n,k) := \min_{F} g(n,k,F) $$ i.e, the maximum score Green can get in a worst-case set-family. What is G(n,k)? Here are some examples. $G(n,0)=1$ for every $n\geq 0$. $G(0,k)=0$ for every $k\geq 1$. Moreover, $G(n,k)=0$ for every $k>n$. Moreover, $G(n,k)=0$ for every $k$ such that $2 k - 2 \geq n$. Proof. Let $F$ contain a single set, i.e, $F=\{\{1,\ldots,n\}\}$. Obviously at most $k-1$ elements will be green, so the single set will not be good and Green's score will be 0. In particular, $G(2,2)=0$. $G(1,1)=1/2$. Proof. Green can use the following strategy: each turn, pick the element that appears in the largest number of uncolored sets. Thus, each couple of turns, the number of sets that become good is at least as large as the number of sets that become bad, so Green's score is at least $1/2$. For tightness, let $F = \{\{1\},\{2\}\}$. Regardless of Green's initial move, Red can make at least one set bad, making Green's score at most $1/2$. I could also prove that, for every $n\geq 1$, $G(n,1)= 1 - 1 / 2^n$. The smallest case that I cannot prove is $G(3,2)$. I can prove it is at least $3/8$ and at most $1/2$ but do not know the exact value. Was this game studied before? Is anything else known about it? EDIT. I could get many upper bounds with the following strategy. The state of each set, during the game, can be described as $(n',k')$, where $n'$ is the number of uncolored elements in the set and $k'$ the number of elements it is missing to be good. So initially the state of all sets is $(n,k)$; when an element of such set is colored red or green, its state changes to $(n-1,k)$ or $(n-1,k-1)$ respectively; etc. To each set we assign a potential $P(n',k')$ based on its state. The potential is determined such that: The potential of a good set is 1: $P(n',0)=1$ for all $n'$. The potential of a bad set is 0: $P(0,k')=1$ for all $k'\geq 1$. The potential increases when an element becomes green and decreases when an element beccomes red, but the increase is at least as high as the decrease: $P(n'-1,k'-1)-P(n',k') \geq P(n',k') - P(n'-1,k')$ for all $k'\geq 1, n'\geq 1$. The potential does not decrease when two elements become red and green simultaneously: $P(n',k')\leq P(n'-2,k'-1)$. Using a simple Google spreadsheet I could numerically calculate some appropriate values of $P$ (I am still working on finding a closed-form formula). Using this $P$, the strategy of Green is: For every element, calculate its potential loss - how much potential will be lost if this element will be colored red. Pick the element with the largest potential loss. By construction of $P$, the potential increase when this element becomes green is weakly larger than the potential decrease when another element becomes red. Therefore, the potential weakly increases. Initially, the total potential is $P(n,k)$ times the number of sets. Finally, the potential exactly equals the number of good sets. Therefore the fraction of good sets is at least $P(n,k)$. In particular, $P(n,1) = 1 - 2^n$ and $P(3,2)=3/8$, which gives the claimed lower bounds. REPLY [3 votes]: $G(3,2)=\frac{1}{2}$. For any set $F$ composed of 3-element sets, let's assume there's a sequence of choices $a_1,b_1,\dots a_k,b_k$ where B gains the upper hand and has two elements chosen out of more than half the sets in $F$, despite A playing optimally. If B can force such a sequence, then A can force the sequence $b_1,?,b_2,?\dots,b_k$ for itself, leaving it with more than half the sets in $F$, despite B playing optimally. Therefore, B can never gain the upper hand at any move in $G(3,2)$, so it can have at most half the sets with two of its chosen elements in them.<|endoftext|> TITLE: Do arithmetic schemes have non-singular alterations? QUESTION [7 upvotes]: Let $X$ be an integral normal flat finite type scheme over $\mathbb{Z}$. Does there exist a proper surjective generically finite morphism of schemes $Y\to X$ with $Y$ an integral regular finite type scheme over $\mathbb{Z}$? I could not find such a result in the literature. REPLY [4 votes]: This is Theorem 8.2 in de Jong's original paper [dJ]. [dJ] de Jong, A. J., Smoothness, semi-stability and alterations. Publ. Math., Inst. Hautes Étud. Sci. 83, 51-93 (1996). ZBL0916.14005.<|endoftext|> TITLE: Bicategory of bimodules over internal monoids QUESTION [7 upvotes]: In a monoidal category $(C,\otimes)$ one can consider internal monoids and bimodules over these monoids (provided some additional requirements like existence of coequalizers hold). These monoids together with bimodules as 1-morphisms and bimodule morphisms as 2-morphisms should constitute a bicategory $\mathsf{Bimod}$. Does anyone know a reference for this result? The only thing I could find so far is this page on nLab, but there is no proof given. REPLY [6 votes]: This example of a bicategory (in the case of monoids in Ab, i.e. rings) appears already in Benabou's original Introduction to bicategories (example 2.5), although he does not provide the detailed proof you seem to be looking for. In searching the 2-categorical literature, you may want to look for constructions of the larger bicategory of profunctors (a.k.a. distributors or bimodules) whose objects are enriched categories, the bicategory of monoids and bimodules being its full sub-bicategory on the one-object categories. I believe the earliest references for this include Benabou's Les Distributeurs and Lawvere's Metric spaces, generalized logic, and closed categories, though on a quick glance I do not see a detailed proof in either of them either. For actually checking axioms of this sort, it is often convenient to exhibit the operations as having a universal property. In the case of bicategories, or more generally pseudo double categories (the bicategory Bimod enlarges to a pseudo double category whose additional morphisms are monoid homomorphisms), such a universal property can be expressed in terms of virtual double categories (a.k.a. fc-multicategories). A construction of the virtual double category of monoids and bimodules can be found in section 5.3 of Leinster's Higher operads, higher categories, though he doesn't give a detailed proof of the conditions under which it is a pseudo double category. A sketch of the latter --- but, again, with details left to the reader --- can be found in Appendix A of Cruttwell-Shulman, A unified framework for generalized multicategories.<|endoftext|> TITLE: Can I bring the Kirillov 2-form on coadjoint orbits to adjoint orbits? QUESTION [5 upvotes]: I tried asking this question on stackexchange and received no response. Given a semisimple Lie group, there is a symplectic structure on the coadjoint orbits arising from the Kirillov 2-form. Can I use this to define a volume form on the adjoint orbits, perhaps pulling it back via a homeomorphism between corresponding adjoint and coadjoint orbits? I know there is another way to get a volume on adjoint orbits, via Haar measure on the Lie group and taking the quotient measure identifying the orbit with the quotient of $G$ by the stabilizer, but I would like to do it ignoring this identification entirely. My end goal is this; given $f:\mathfrak{g} \to \mathbb{R}$, and $X \in \mathfrak{g}$ I would like to define some type of integration $\int_{O(X)} f \cdot d\omega$. As mentioned above, there is a way this is usually done by taking a Haar measure on $G$ and getting some normalized quotient measure $\dot{dg}$ over $G/C_G(X)$, giving the so-called orbital integrals $\int_{G/C_G(X)} f(gXg^{-1})\, \dot{dg}$. But I would like to instead have some general volume form on $O(X)$. The Kirillov-Kostant-Souriau symplectic structure on the coadjoint orbits seems to be the way to do this, but I am unsure about the technical details of such an approach. REPLY [7 votes]: For $G$ semisimple as you assume, the Killing form gives a $G$-equivariant map $\mathfrak g\to\mathfrak g^*$, which identifies adjoint to coadjoint orbits. That is all you need.<|endoftext|> TITLE: Understand the difference between two stacks QUESTION [7 upvotes]: Let us work over $\mathbb{C}$. Let $G$ be a finite group, acting on $\mathbb{A}^1$ via a character, and let $H$ be the kernel of the action. Assume that $\mathbb{A}^1$ is the coarse moduli space of the $[\mathbb{A}^1/G]$, and that $[\mathbb{G}_m/G$ maps onto $\mathbb{G}_m$. The way I picture the stack quotient $[\mathbb{A}^1/G$ is $\mathbb{A}^1$ with a $BG$ at the origin and $BH$ at every point different from the origin. In these terms, how should I think of the difference between $[\mathbb{G}_m/G]$ and $\mathbb{G}_m\times BH$? They ought to be different, but I keep making the mistake of confusing one with the other. Is it possible to write $[\mathbb{G}_m/G]$ as a quotient stack involving some twist of $H$? REPLY [8 votes]: One can see the difference by writing down the functor of points explicitly. For a test scheme $T$, \begin{align} (\mathbb G_m\times BH)(T) & =\{(f,p)\mid f:T\to \mathbb G_m, p:T'\to T\text{ $H$-torsor}\},\\ [\mathbb G_m/G](T) & =\{(q,g)\mid q:\tilde T\to T\text{ $G$-torsor, }g:\tilde T\to \mathbb G_m\text{ $G$-equivariant}\}\\ & =\{(f,p')\mid f:T\to \mathbb G_m,p':\tilde T\to T\times_{f,\mathbb G_m,\chi}\mathbb G_m\text{ $H$-torsor}\}, \end{align} where $\chi :\mathbb G_m\to \mathbb G_m$ is the quotient by $G/H$. Note that $H$-torsors $p$ and $p'$ have different targets! We can define a map $\mathbb G_m\times BH\to [\mathbb G_m/G]$ by pulling back $p$ from $T$ to $T\times_{f,\mathbb G_m,\chi}\mathbb G_m$, but it won't be invertible in general. One can see it by considering a single $G$-orbit on both sides. On the left, we get $pt/H=(G/G)/H$, and on the right $(G/H)/G$. But the two actions don't have to commute, so in general we get two non-isomorphic groupoids. One can also see the difference by computing $K$-theory of both spaces after decomposing $\mathbb A^1$ as $\mathbb G_m\cup pt$.<|endoftext|> TITLE: When can I "draw" a topology in Baire space? QUESTION [13 upvotes]: The motivation for this question is a bit convoluted, so in the interests of conciseness I'm just asking it as a curiosity (and I do find it interesting on its own); if anyone is interested, feel free to email me. I'm playing around with various notions of describing topological spaces, and I've found the following fun: Let Baire space $\mathcal{N}$ be $\mathbb{N}^\mathbb{N}$ with the usual topology. Say that a space $\mathcal{X}=(X,\tau)$ is pictorial if there are $A,B\subseteq\mathcal{N}$ arbitrary and $R\subseteq\mathcal{N}^2$ open such that $$\{\{a\in A: aRb\}: b\in B\}$$ generates a topology $\sigma$ on $A$ with $\mathcal{X}\cong(A,\sigma)$. Of course, we could replace $\mathcal{N}$ with an arbitrary space, or even (distinguishing between "point part" and "set part") an arbitrary pair of spaces, but at present I don't see what that added generality gives us. My question is simply: What spaces are pictorial? Or even better: Are some good sources on this topic/type of problem? I haven't been able to find any, but I'm not familiar with the relevant literature. There are a couple easy observations: All pictorial spaces are separable. (And truly trivially, all pictorial spaces have at most continuum many points and a base of size at most continuum.) Subspaces, countable products, and $T_0$-ifications of pictorial spaces are pictorial. There are Hausdorff non-first-countable pictorial spaces, and there are $T_1$ pictorial spaces such that no point is characterized uniquely by countably many open sets; on the other hand, if $(X,\tau)$ is Hausdorff and pictorial, then for every point $p\in X$ there is a countable family $(U_i)_{i\in\mathbb{N}}$ of open sets characterizing $p$ completely. (A point $p$ is characterized uniquely by countably many open sets if there is a sequence $(U_i)_{i\in\mathbb{N}}$ of open sets such that for all points $q$ we have $\{i: q\in U_i\}=\{i: p\in U_i\}\iff p=q$). However, I have no idea how to prove any nontrivial results about pictorial spaces, the main issue being the difficulty of proving any negative results. For example, I don't know whether the continuous image of a pictorial space need be pictorial: removing points from the $B$-part is tempting at first but the pullback of a subbase of the target may not look anything like the subbase witnessing the pictoriality of the starting space. REPLY [5 votes]: I have two things to offer, the first of which could help with getting better characterizations, the latter should give ample of examples. Since I am not aware of standard terminology, call $(X,\tau')$ a topological weakening of $(X,\tau)$ if $\tau' \subseteq \tau$. Theorem: A space is pictorial iff it is homeomorphic to a topological weakening of a subspace of $\mathcal{N}$. For the forward implication, note that for every $b \in B$ the set $\{a \mid aRb\}$ is open in the subspace topology that $A$ inherits from $\mathcal{N}$, hence the entire topology we induce on $A$ is a subset of its subspace topology. For the converse implication, take the standard coding of open subsets of $\mathcal{N}$ over $\mathcal{N}$, and note that p is contained in the open set coded by q is an open relation on $\mathcal{N}$. So to see that any topological weakening $(A,\tau)$ of a subspace $\mathbf{A}$ is pictorial, just take the underlying set $A$, and for $B$ the set of all codes for extensions of opens in $\tau$. We get plenty of preservation results as corollaries, namely everything that commutes with weakening of the topology and preserves subspaces of $\mathcal{N}$. Theorem: Every sequential $\mathrm{QCB}_0$ space[1] is pictorial. The sequential $\mathrm{QCB}_0$-spaces are precisely those topological spaces that arise as represented spaces in computable analysis. A represented space is just a pair $(X,\rho)$, where $\rho : \subseteq \mathcal{N} \to X$ is a partial surjection onto the set $X$. Whenver $\mathbf{X}$ is a represented space, then so is the space $\mathcal{O}(\mathbf{X})$ of its open subsets[2]. The relationship $x \in U$ is open on $\mathbf{X} \times \mathcal{O}(\mathbf{X})$, which can be pulled back to the level of $\mathcal{N}$. Now just choose a single representative from each $\rho_\mathbf{X}^{-1}(\{x\})$ for $x \in \mathbf{X}$ to construct $A$. For $B$, we can just take the domain of the representation of $\mathcal{O}(\mathbf{X})$. The second observation accounts for all sequential separable spaces "I have come across naturally" as being pictorial. [1] $\mathrm{QCB}_0$ spaces are $T_0$ quotients of countably-based spaces. See https://www.sciencedirect.com/science/article/pii/S0304397501001098 for more. [2] A subset $U \subseteq \mathbf{X}$ of a represented space $(X,\delta)$ is open, if $\delta^{-1}(U)$ is open in $dom(\delta)$.<|endoftext|> TITLE: Is the boundary of a manifold topologically unique? QUESTION [10 upvotes]: Let $X$ be a manifold without boundary and let $Y$ and $Z$ be two manifolds with boundary such that $X$ is homeomorphic to their interiors: $X \cong Y^\circ \cong Z^\circ$. Does it follow that $Y \cong Z$ as well? That is, does there exist a homeomorphism that extends to the boundaries as well? If not, can we at least say that $\partial Y \cong \partial Z$ even if perhaps they are "attached" to their respective interiors differently? I rarely work with manifolds with boundary so I apologize if this is a stupid question. It seems like it should be true but I haven't been able to find a reference for this. It is certainly obvious that we have a homotopy equivalence $Y \simeq Z$ since a manifold with boundary is homotopy equivalent to its interior. REPLY [9 votes]: I don't think this is true, by a classical `swindle' that I think is due to Stallings. You can find variations on this argument eg in Milnor's paper, Two complexes which are homeomorphic but combinatorially distinct (Annals 1961). He refers to some notes of Stallings. Here is how I recall the argument, as applied to your question. I might be off by a sign, depending on the dimension. Consider a manifold $M$ for which there is an h-cobordism $Y$ from $M$ to itself with non-zero Whitehead torsion; then your $X$ will be two copies of $M$, or (taking orientations into account) $X = -M \cup M$. Then obviously $\partial Y = X$. On the other hand, you could take $Z = M \times I$. Claim: $Y^0 \cong Z^0 = M \times R$. Proof: Consider an h-cobordism $Y'$ from $M$ to itself whose torsion is the negative of that of $(Y,M,M)$. Then $$Y \cup_M Y' \cong M \times I \cong Y' \cup_M Y.\quad (*)$$ Now consider $$ \cdots (Y \cup Y') \cup (Y \cup Y') \cup Y \cup (Y'\cup Y) \cup (Y'\cup Y) \cdots $$ By (*), this is $(-\infty,0] \times M \cup Y \cup [0,\infty)\times M = Y^0$. On the other hand, you can put the $Y$ in the middle onto one side, and redistribute the parentheses to see that this is an infinite union of copies of $Y \cup Y' = M \times I$, and hence is $Z^0$. REPLY [9 votes]: Since you say in a comment that you might be satisfied with a homotopy equivalence, let me sketch a proof that the homotopy type of the boundary depends only on the interior. Let $Y$ be the interior of $X$ and let $DY$ be the space of all proper maps $[0,1)\to Y$, suitably topologized. Then $DY$ must be homotopy equivalent to $\partial X$. To see this, first observe that $DY$ is homotopy equivalent to the space $D'Y$ of germs of such maps, where two maps $[0,1)\to Y$ have the same germ if they agree on $[a,1)$ for some $0\le a<1$. Then observe that $D'Y$ depends only a neighborhood of $\partial X$, and that by using a collar neighborhood $C\cong I\times\partial X$ we find $D'Y$ to be homeomorphic to the space of all germs of proper maps $[0,1)\to [0,1)\times\partial X $. The latter space of germs is homotopy equivalent to the space of all proper maps $[0,1)\to [0,1)\times\partial X$. Finally, this last space is the product of (1) the space of all proper maps $[0,1)\to [0,1)$ and (2) the space of all continuous maps $[0,1)\to \partial X$. (1) is contractible. (2) is homotopy equivalent to $\partial X$.<|endoftext|> TITLE: What did the Intuitionists want to do with applied mathematics? QUESTION [9 upvotes]: Oversimplification: Newton & Leibnitz &c build the calculus and other methods that solve a vast number of practical problems. Weierstrass, Dedekind, Cantor &c build a foundation under it dependent on transfinite quantities. Kronecker, Brouwer, etc, were appalled by this. Later, Bishop &c actually demonstrates approaches to founding these techniques on constructive methods. In the long interval before Bishop, what did Intuitionists and/or Constructivists think about practical applications? Did they expect bridges to fall down? Or did they simply believe that mathematics had not yet built a meaningful foundation for the practical methods? REPLY [2 votes]: There is a more general question lurking in the background, which is what do critics of logical foundations generally think about applications? Historically, intuitionism is not the only foundational controversy. Earlier, there were critics of the logical foundations of calculus (recall Berkeley's "ghosts of departed quantities"), and today there are unresolved mathematical difficulties in quantum field theory. The pattern is usually the same. The practitioners, based on intuition and experience, know what they have to do to make sure that "bridges don't fall down," while the critics point out that the practitioners have failed to articulate clearly what the ground rules are. In particular, the critics can sometimes construct calculations that appear to avoid all explicitly forbidden operations, yet yield the wrong answer. To determine what a specific person (e.g., Brouwer) thought, one obviously needs to examine what that particular person said on the subject. But in general, there will be a range of opinions. Some, as you say, will believe that the scientific/engineering theories must be fundamentally correct even though we haven't hammered out all the logical details yet, while others of a more alarmist bent may worry that people are trusting the theories too much, to the point where a "bridge will fall down." Obviously, in the real world, bridges do sometimes fall down, but there are many reasons for this, most of which have nothing to do with inadequate logical hygiene. I would be curious to know if there are any examples of an actual engineering disaster with the following features: The disaster can be traced to a calculation that the engineers mistakenly trusted (and not because there was an inadvertent error or bug). If one were to apply "more rigorous reasoning" then the calculation would have come out differently and the disaster would have been averted. I can imagine that there might be modern examples where the results of some kind of numerical simulation are trusted, but where theoreticians can show that the numerical results do not accurately reflect the behavior of the equations. (If the equations themselves are an inadequate model of physical reality, then that is a different matter, which I'm not concerned with here.) But in general, I think such situations are rare, because theoretical predictions are usually tested experimentally before an actual engineering project that might endanger people's lives is carried out. If I am right about this then most fears of "bridges falling down" because of lack of logical rigor are overblown.<|endoftext|> TITLE: Proof of Green's formula for rectifiable Jordan curves QUESTION [8 upvotes]: $\newcommand{\Ga}{\Gamma}$ I am trying to find a proof of Green's formula for rectifiable Jordan curves $\Ga$ (and the corresponding interior regions $R$). There is a proof by Ridder, followed by very similar proofs by Verblunsky and Potts. The idea in those three papers, which looks quite natural, is to subdivide $R\cup \Ga$ into small regions $R_i\cup \Ga_i$ by vertical and horizontal lines. The problem is how to deal with the "partial" regions $R_i\cup \Ga_i$ that are not rectangles; more precisely, with the line integrals over the corresponding $\Ga_i$'s. To deal with such line integrals rigorously, one first of all needs to parametrize the boundaries $\Ga_i$. However, none of the mentioned three authors (or any other ones known to me) even mentions such a parameterization. In Mathematical Analysis: a Modern Approach to Advanced Calculus, 1957, by Apostol, an apparent attempt is made to make Ridder's approach rigorous. The Green formula in question is stated there as Theorem 10--43. The crucial part in the proof of Apostol's Theorem 10--43 is Theorem 10--42, which tries to deal with the mentioned "partial" regions $R_i\cup \Ga_i$. However, there are a few places in the proof of Apostol's Theorem 10--42 that I don't understand. Let $\Ga$ be a rectifiable Jordan curve bounding the corresponding interior region $R$. First, Apostol takes a lowest and highest points, $p_1$ and $p_2$ on $\Ga$. These two points are then separated by a horizontal line, say $\ell$. Then $p_1$ and $p_2$ are joined by "arcs" $C_1$ and $C_2$ to points $q_1$ and $q_2$ in $R$ close to $p_1$ and $p_2$, respectively, so that $q_1$ is below $\ell$ and $q_2$ is above $\ell$. Question 1. Apostol says such "arcs" (possibly non-rectifiable) exist, without proof. I do not immediately see how to prove this, even though it's probably simple. (The $C_j$'s are never mentioned again in the proof, after they are introduced.) Then the points $q_1$ and $q_2$ are connected by a polygonal line $C\subset R$, which must have an intersection point with $\ell$. Without loss of generality, there are finitely many such points. For each such point $p$, let $[u,v]$ be the maximal segment on $\ell$ such that the open interval $(u,v)$ is contained in $R$. Then Apostol says there must be at least one such maximal segment $[u,v]$ that has an odd number of intersections with the polygonal line $C$; I guess this follows because the total number of the intersections of $C$ with $\ell$ can be proved to be odd, by induction. But then Apostol just says: "Using such a segment, it is clear that $L[u,v]$[$=[u,v]$ -- I.P.] forms, with $\Ga$, two rectifiable Jordan curves $\Ga_1$ and $\Ga_2$, one of which contains $p_1$ (call this one $\Ga_1$) and the other contains $p_2$. Moreover, these curves form the boundaries of two regions $R_1$ and $R_2$, whose union is $R$. The positively oriented boundaries $\Ga(R_1)$ and $\Ga(R_2)$ of $R_1$ and $R_2$ such that $\Ga(R)=\Ga(R_1)+\Ga(R_2)$." There is a picture there, making the above claims plausible. Yet, here is my main question: Question 2. Nothing in that passage is clear to me. How does this "forming" of curves $\Ga_1$ and $\Ga_2$ occur? Are they Jordan curves? If so, why? How are $\Ga_1$ and $\Ga_2$ parameterized? Why is the union of $R_1$ and $R_2$ equal to $R$? What precisely are the roles of the points $q_j$ and the "arcs" $C_j$ in this proof? Can one help me decipher this proof? Or, perhaps even better, is there a completely rigorous, fully detailed proof of this obviously important and natural result? (Of course, there is a general Stokes theorem for manifolds, but it does not seem at all obvious how to build a bridge from there to Jordan curves.) REPLY [7 votes]: One can circumvent the technical difficulties as follows. Consider a large ball $K$ containing $\Gamma$ and any $p>2$. Given a function $f\in L^p(K)$, we can define its Cauchy transform $$ \left(\mathcal{C}f \right)(z)=\frac{1}{\pi}\int_K\frac{f(w)}{z-w}. $$ By Hölder inequality, this is a continuous operator from $L^p(K)$ to $C(K)$. This operator inverts the $\bar{\partial}$ operator, namely, $\mathcal{C}\bar{\partial}\varphi\equiv\varphi$ for any $\varphi \in C^1_0(K)$; see the computation after the equation (8) in Chapter 5 of "Lectures on quasiconformal mappings" by Ahlfors (1966). Now, the composition of $\mathcal C$ and integration over $\Gamma$ is a continuous linear functional on $L^p(K)$: $$ L(f):= \frac{1}{2i}\int_\Gamma \left(\mathcal{C}f\right)(z)dz=\int_K f(z)\psi(z), $$ for some $\psi \in L^{p'}(K)$. It follows that for any $\varphi \in C^1_0(K)$, we have $$ \frac{1}{2i}\int_\Gamma \varphi(z) dz = \int_K\psi(z)\bar{\partial}\varphi(z). $$Hence, it remains to check that $\psi\equiv 1$ inside $\Gamma$ and $\psi \equiv 0$ outside; since $\Gamma$ itself is rectifiable, its area is zero. To this end, it suffices to check that $L(f)=\int_{\text{inside }\Gamma} f$ for any $f\in C_0(K\setminus\Gamma)$. For such an $f$, we can exchange the integrals: $$ L(f)=\frac{1}{2\pi i}\int_\Gamma dz\int_K \frac{f(w)}{z-w}=\int_K f(w)\frac{1}{2\pi i} \int_\Gamma \frac{dz}{z-w}. $$ The inner integral can be interpreted as the increment of $\log (z-w)$ along $\Gamma$, and thus it is equal to the winding number of $\Gamma$ w. r. t. $w$. Therefore, the result follows from the topological statement that any Jordan curve has winding number $\pm 1$ w. r. t. a point inside it, and $0$ w. r. t. a point outside. Some references for this statement are discussed here.<|endoftext|> TITLE: Smallest set of nonzero vectors in $\mathbb F_2^n$ which intersects every 2-dimensional subspace QUESTION [6 upvotes]: What is the smallest set of nonzero vectors in $\mathbb F_2^n$ which intersects every 2-dimensional subspace? For example, for n = 3, the set {001, 010, 011} does the job, and is minimal. For n = 4, {0001,0010,0011,1000,1001,1010,1011} does the job. This is a special case of the Minimum Hitting Set problem (and equivalently, Minimum Set Cover), which is known to be hard in general. However, it is such a special case that I wonder if more can be said. I have not had much luck finding existing results on the question, and trying to find the numbers by computer power does not seem easy, even for modest values of n. My best results for $n$=5 and $n$=6 are sets of size 16 and 44, respectively, but I am not certain those are minimal. That might suggest A129045, but I know the answer for $n$=7 is much less than 134. There are obvious generalizations – if you can answer the question for $k$-dimensional subspaces of $\mathbb F_q^n$, so much the better, but the particular case I have mentioned seems like a good starting point to me. Most of all, I would appreciate links to existing literature! Thanks. REPLY [10 votes]: $2^{n-1}-1$. Let $H$ be a codimension one subspace of the $n$-dimensional vector space $V$. Then the intersection of $S$ with any $2$-dimensional subspace has dimension at least one, so that $H-0$ is a set of the kind you are asking about. Conversely suppose that $S$ is a set of nonzero vectors intersecting every $2$-dimensional subspace. If $S$ does not have every nonzero vector in it, then say that it does not have $v$ and let $L$ be the one-dimensional subspace spanned by $v$. The image of $S$ in $V/L$ intersects every one-dimensional subspace nontrivially, so it has at least $2^{n-1}-1$ elements. EDIT: More generally if $S$ is a set of nonzero vectors intersecting each $d$-dimensional subspace then $S$ has at least $2^{n-d+1}-1$ elements. Proof: Wlog $d$ is minimal. Choose a $(d-1)$-dimensional subspace $W$ of $V$ disjoint from $S$. The image of $S$ in $V/W$ has all the nonzero vectors. And this is best possible by taking $S$ to be the complement of $0$ in a vector subspace of codimension $d-1$.<|endoftext|> TITLE: A linear algebra problem in positive characteristic QUESTION [18 upvotes]: Let $A$ be a symmetric square matrix with entries in $\mathbb{Z}/p\mathbb{Z}$ for a prime $p$ such that all of its diagonal entries are nonzero. Does there exists always a vector $x$ with all coordinates nonzero in the image of $A$? (this is true for $p=2$, but I don't know the answer for other values of $p$) REPLY [29 votes]: This is false. Let $c$ be a quadratic nonresidue modulo $p$. Our matrix will be $(p^2-1) \times (p^2-1)$, with rows and colums indexed by pairs $(x,y) \in \mathbb{F}_p^2 \setminus \{ (0,0) \}$. Our matrix is defined by $$A_{(x_1,y_1) \ (x_2, y_2)} = x_1 x_2 - c y_1 y_2.$$ This is obviously symmetric. Since $c$ is a nonresidue, we have $x^2-cy^2 \neq 0$ for $(x,y) \in \mathbb{F}_p^2 \setminus \{ (0,0) \}$, so the diagonal entries are nonzero. Each column of this matrix is a linear function of $(x,y)$. So every vector in the image of this matrix is a linear function $\mathbb{F}_p^2 \setminus \{ (0,0) \} \longrightarrow \mathbb{F}_p$ and, hence, takes the value $0$ somewhere. We could make a smaller $(p+1) \times (p+1)$ example by just taking one point $(x,y)$ on each line through $0$ in $\mathbb{F}_p^2$. Moreover, I claim that $(p+1) \times (p+1)$ is optimal. In other words, if $A$ is an $n \times n$ matrix with $n \leq p$ and nonzero entries on the diagonal, then some vector in the image of $A$ has all coordinates nonzero. Interestingly, I don't need the symmetry hypothesis. Let $W$ be the image of $A$. Note that $W$ is not contained in any of the coordinate hyperplanes. Let $\vec{u}= (u_1,\ u_2, \ \ldots,\ u_n)$, among all elements of $W$, have the fewest $0$ entries. Suppose for the sake of contradiction that some $u_i$ is $0$. Then there is some other $\vec{v}$ in $W$ with $v_i \neq 0$. Consider the points $(u_j : v_j)$ in $\mathbb{P}^1(\mathbb{F}_p)$ as $j$ ranges over all indices where $(u_j, v_j) \neq (0,0)$. There are fewer then $p+1$ such $j$, so some point of $\mathbb{P}^1(\mathbb{F}_p)$ is not hit, call it $(a:b)$. Then $-b \vec{u} + a \vec{v}$ is in $W$ and has fewer nonzero entries than $\vec{u}$, a contradiction.<|endoftext|> TITLE: Special Cases of Duistermaat-Heckman Formula QUESTION [9 upvotes]: The Duistermaat Heckman localization formula states how integrals over symplectic spaces with Hamiltonian $U(1)$ group actions. $$ \int_M \frac{\omega^n}{n!} e^{-\mu} = \sum_{x_i \text{ fixed}} \frac{e^{-\mu(x_i)}}{e(x_i)} $$ Here $M$ is a symplectic manifold and here are a few invariants: $\omega$ is the symplectic form $\mu$ is the moment map of the $U(1)$ rotation action $x_i$ are the fixed points of the group action $e(x_i)$ is the product of the weights of the rotation at fixed point $x_i$ In a few cases, we can produce formulas an engineer might recognize: $M = \mathbb{R}^2 \simeq \mathbb{C}$ and $\omega = dx \wedge dy$ $U(1)$ the the rotation $z = x_1 + i x_2 \mapsto e^{i\theta}z$, the moment map is $\mu(z) = |z|^2 = x_1^2 + x_2^2$ Certainly $\mathbb{R}^{2n}$ is analogous. The fixed point of the rotation is $x_i = 0$. This leads to the formula for the Gaussian integral: $$ \int (dx \wedge dy) \,e^{-(x^2 + y^2)} = \frac{e^{-\mu(0)}}{e(0)} = \frac{1}{2\pi}$$ I'm not actually sure that $e(0) = 2\pi$ but I'm guessing. There's not too many cases were all the ingredients are known. It is possible to write a manifold (or orbifold or variety) with moment map $\mu$. My understanding is that we can rarely write $\omega$ down explicitly. For the sphere $S^2 = \{ x^2 + y^2 + z^2 = 1\}$ there's a symplectic form $\omega = \frac{dx \wedge dy}{z}$ which is invariant under rotating aroun the $z$ axis. Are there other cases where the DH formula specializes in a particularly nice way? I'm guessing that typically we bypass explicitly computing $\omega$. REPLY [6 votes]: Nice examples are worked out in Audin (2004, §VI.3.d), Arvanitoyeorgos (1999)(pdf), McDuff-Salamon (1998, §5.6). It’s not true that $\omega$ is rarely explicit: e.g. on all coadjoint orbits (including $\smash{S^2}$) it is, and DH gives a formula of Harish-Chandra for their Fourier transforms; see Berline-Vergne (1983), Vergne (1983), or Guillemin-Sternberg (1984, ends of §§33 and 34).<|endoftext|> TITLE: Books with exercises to learn Langlands program, Galois representations, modular forms QUESTION [10 upvotes]: I want to develop a basic background in number theory with a goal towards contributing to some part of the Langlands program (which I know is vast, and has many different aspects; I want to do something involving Shimura varieties, Galois representation, modular forms). I have been reading (but not doing the exercises to save time) various books, papers, going to conferences, seminars, etc but that has not seemed to be helping me retain any information or develop the skills to prove anything. I’ve decided I just need to start writing up solutions to a bunch of exercises. (I confess I didn’t do enough exercises in graduate school and that could be why I have been a failure as a researcher, now I have the luxury of time to go back and fill in gaps in my knowledge). Here are some possibly relevant books that I’ve found that have exercises, and my question is what are some others?: Childress Class field theory
 Neukirch Algebraic Number Theory Neukirch Schmidt Winberg Cohomology of Number Fields
 Silverman Elliptic Curves (both volumes) Helgason Differential Geometry, Lie Groups, and Symmetric Spaces
 Washington Cyclotomic Fields
 Liu Algebraic Geometry and Arithmetic Curves
 Ramakrishnan Valenza Fourier Analysis on Number Fields
 Diamond Shurman First Course in Modular Forms
 Cox Primes of the form x^2 + ny^2
 Gille Szamuely Central Simple Algebras and Galois Cohomology Bump Automorphic Forms and Representations Bourbaki Lie Groups and Lie Algebras Vakil Foundations of Algebraic Geometry Hartshorne Algebraic Geometry What are some others? REPLY [25 votes]: This is perhaps not the type of answer you were looking for (in which case I apologize), but it isn't clear to me that doing all of the exercises in an enormous number of books is the most efficient way to get to your goal of contributing to some aspect of the Langlands program, Shimura varieties, Galois representations, modular forms, etc. The issue, I think, is that (1) the approach you outline is bound to take an enormous amount of time, and (2) while you are reading all of this material it will never be clear which topics are going to be relevant to your future research and which aspects won't be. In general I find that learning from a book is easiest when I have a clear goal (e.g., I want to generalize a known result from quaternion algebras to arbitrary central simple algebras, and this book concerns the structure of central simple algebras). Absent such a goal I find it very hard to figure out what aspects of the relevant theory are most important and wind up retaining very little of what I read. In light of the above, I would recommend that you begin by choosing a single topic from your list, say modular forms. Obviously you can't work on problems about modular forms if you don't know anything about them, so I would read through Diamond and Shurman's First Course in Modular Forms. (Some of the other topics on your list require more in the way of background knowledge.) My goal would not be to memorize the book or even complete all of the exercises. I would just focus on trying to get to the point that you are familiar with the basic objects of study, definitions and fundamental results (i.e., congruence subgroups, Hecke operators, newforms, etc). Once I've gotten to this level I would start looking for a paper about modular forms that looked interesting to me (i.e., the statement of the main theorem makes sense to me, the proof doesn't look too technical, it isn't too long, etc) and start reading the paper. You'll of course need to do a lot of supplementary reading (from textbooks) while working your way through the paper, but I believe that you will retain much more of what you read because you have a clear goal (understand what the author of the paper wrote). Keep in mind that it might take a few iterations of this process before you land on a paper that you like / are able to work through. Once you've read this paper you could try to find a different paper to read, or even try to decide if it would be interesting to generalize the paper (e.g., This paper proved something about integral weight modular forms. I remember vaguely reading something about half-integral weight modular forms; perhaps it would be interesting to learn more about them and see if I can generalize the paper to the setting of half-integral weight modular forms). Also, these topics all interact in a variety of ways, so (to continue the example with modular forms) your work on modular forms could very well lead you to learn a bit about elliptic curves. None of this will get you from 0 to writing important papers about Shimura varieties of course, but hopefully it will help you start learning about whatever area interests you in an efficient manner and build up some confidence in your ability to do research in this area.<|endoftext|> TITLE: The Ungraded Milnor-Moore Theorem QUESTION [5 upvotes]: Let $k$ be a field of characteristic $0$. There is a functor $U$ from Lie-algebras over $k$ to Hopf algebras over $k$ sending a $k$-Lie algebra $\mathfrak{g}$ to its universal enveloping algebra $U(\mathfrak{g})$, and a functor $P$ from the category of Hopf algebras over $k$ to the category of $k$-Lie algebras sending a Hopf algebra $H$ over $k$ to the set $P(H)$ of its primitive elements, which acquires the structure of a Lie algebra. There is an adjoint relationship where $U \dashv P$. We may ask, "for which $H$ is the counit $\epsilon_H : U(P(H)) \rightarrow H$ of this adjunction an isomorphism". The Milnor-Moore theorem says that $\epsilon_H$ is an isomorphism when $H$ is graded, co-commutative, and generated by its primitive elements. But, for my purposes, $H$ is not a graded Hopf algebra, merely co-commutative and generated by its primitive elements. I have only been able to find expositions of the Milnor-Moore theorem which use graded Hopf algebras and graded Lie-algebras, but I am interested in the general case. For instance, I have seen the paper "On the structure of Hopf algebras", by Milnor and Moore, but this seems to work only in the graded case from what I can see. Could someone direct me to the most accessible exposition (that they know of) of an ungraded version of the Milnor-Moore theorem? REPLY [5 votes]: The "ungraded" version of the theorem -which is actually the version for the Hopf algebras- can be found in most of the classical references on the subject, although its statement and proof appears scattered among paragraphs or several different sections. For example see: Sweedler's book: Hopf algebras, Theorem 8.1.5, p 176 (with no particular assumptions on the field) and section 13.1, p. 279, where the version for fields of characteristic $0$ is stated as an exercise. Montgomery's book: Hopf algebras and their actions on rings, Corollary 5.6.4, (3), p.78 and Theorem 5.6.5, p. 79 (for zero characteristic), Abe's book: Hopf algebras, in ch. 2, sect. 5.2, p. 110, theorem 2.5.3, deals with the special case of the irreducible, cocommutative hopf algebras over zero characteristic (i am not sure if the pointed case is contained in that book). Finally, it is worth taking a look at: A primer of Hopf algebras, sect. 3.8, theorem 3.8.2, p.45 for a more modern presentation of the topic.<|endoftext|> TITLE: Finite group with a character having one nonzero absolute value QUESTION [7 upvotes]: Let $G$ be a finite group. Assume that $\chi$ is a complex irreducible character of $G$ of degree $n\geq 2$, with the property that for each element $g\in G$ either $\chi(g)=0$ or $|\chi(g)|=n$. Does it necessarily follow that $\chi$ is imprimitive, i.e., induced from a character of a subgroup? The only such examples I could think of are given by extraspecial groups. Is there a classification of all the finite groups with this property? The questions are partly motivated by this post: Finite groups with a character having very few nonzero values? REPLY [10 votes]: As Geoff said, a group having such an irreducible character is called a group of central type, and Howlett and Isaacs have shown, using the classification of finite simple groups, that such groups are solvable: see Howlett, Robert B.; Isaacs, I. Martin, On groups of central type, Math. Z. 179, 555-569 (1982). ZBL0511.20002. Using this result, we can show that a character $\chi$ of central type is necessarily imprimitive. For suppose that $Z$ is the center of $\chi$, so that $\chi(1)^2 = \lvert G:Z \rvert $, and let $N$ be any normal subgroup of $G$ containing $Z$. Let $\tau$ be a constituent of $\chi_N$. By Clifford theory, $\chi$ is induced from the inertia subgroup of $\tau$ in $G$. So if $\tau$ is not invariant in $G$, we are done. Otherwise, we have $\chi_N = e\tau$ and $\tau$ vanishes outside $Z$. It follows that $e^2 = \lvert G:N \rvert$ and $\lvert G:N \rvert $ is a square. But by solvability of $G$, maximal normal subgroups have prime index. Of course it would be nice to have an argument not depending on CFSG. To the best of my knowledge, groups of central type are not classified. The paper by Howlett and Isaacs contains a non-nilpotent example. There are also nilpotent examples of arbitrarily large nilpotency class: Take the semidirect product of a cyclic group of order $p^{n+1}$ with the Sylow $p$-subgroup of its automorphism group.<|endoftext|> TITLE: homeomorphisms induced by composant rotations in the solenoid QUESTION [10 upvotes]: Let $S$ be the dyadic solenoid. Let $x\in S$, and let $X$ be the union of all arcs (homeomorphic copies of $[0,1]$) in $S$ containing $x$. $X$ is called a composant of $S$. It is well-known that $X$ is a dense first category one-to-one continuous image of the reals, and that $S$ is a homogeneous continuum. Now let $y$ and $z$ be any two points in $S\setminus X$. Is there a self-homeomorphism $h:S\to S$ such that $h[X]=X$ and $h(y)=z$? REPLY [5 votes]: In this paper of J.Kwapisz I have found the following Theorem 1. Any homeomorphism $h$ of the dyadic solenoid $S$ is isotopic to the "affine" homeomorphism of the form $g:x\mapsto \pm(2^n x+b)$ for some $n\in\mathbb Z$ and some $b\in S$. If $h$ preserves the path-connected component $X$ of the neutral element, then so does the affine homeomorphism $g$, which implies that $b\in X$. It follows that for any $y\in S\setminus X$ the image $h(y)$ belongs to $g(y)+X\subset \pm 2^{\mathbb Z}y+X$, which is contained in a countable union of path-connected components and hence cannot be an arbitrary element $z\in S\setminus X$. So, the answer to the original question is negative. The same negative answer holds for homeomorphisms of any (not necessarily dyadic) solenoid.<|endoftext|> TITLE: Can we stay invertible while approximating linear maps in Sobolev spaces? QUESTION [5 upvotes]: Let $\Omega \subseteq \mathbb{R}^n$ be an open bounded domain with a smooth boundary. Fix $1 0$ a.e. Do there exist $u_n \in C^{\infty}\big(\Omega,\text{GL}(\mathbb{R}^n)\big)$ such that $u_n \to A$ in $W^{1,p}_{loc}$? I am also interested in a weaker result: Are there $u_n \in C^{\infty}\big(\Omega,\text{End}( \mathbb{R}^n)\big)$, such that $u_n(x) \in \text{GL}(\mathbb{R}^n)$ a.e. and $u_n \to A$ in $W^{1,p}_{loc}$? I don't really need the $u_n$ to be defined on all $\Omega$. It suffices that for every arbitrarily small ball in $\Omega$, there would be a neighbourhood where such a sequence $u_n$ would be defined. The problem is that it is not always true that $A_x \in \text{GL}( \mathbb{R}^n)$ for every $x \in \Omega$. The rank can fall on a subset of measure zero. If we knew $A(x) \in \text{GL}(\mathbb{R}^n)$ everywhere then the answer would be positive. This follows from the facts that "being invertible" is an open condition, and that continuous Sobolev maps can be approximated uniformly by smooth maps over compact subsets. In more detail, let $K \subseteq \Omega$ be compact. Since we assumed $A \in C\big(\Omega, \text{GL}(\mathbb{R}^n) \big)$, the map $\psi:x \to A_x$, considered as a map $K \to \text{GL}( \mathbb{R}^n)$, is continuous. Thus $\psi(K) $ is compact and $\text{dist}\big(\psi(K),\partial \text{GL}(\mathbb{R}^n)\big)>0$. Now consider each component of $\psi(x)=A_x \in \text{End}(\mathbb{R}^n) $. We can approximate each component of $\psi$ using mollification on an open subset of $\Omega$ containing $K$. Since each component is a continuous function, the mollifications converge uniformly on $K$. This implies that from a certain point in the mollified sequence, $\text{dist}(u_n,A)<\text{dist}\big(\psi(K),\partial \text{GL}(\mathbb{R}^n)\big)$, so the $u_n$ are invertible. REPLY [4 votes]: I will recycle the answer I've been writing into some words of explanation on Alex Gavrilov's example, which is more simple and elegant. We can focus on the first column $p_n$ of an approximating sequence $u_n:=[p_n,q_n]$ of $A(x,y):=\begin{bmatrix}x&-y\\y&x\end{bmatrix}$. If $p_n\in C^0\cap W^{1,1}_{loc}(\mathbb{R}^2,\mathbb{R}^2)$ converges in $W^{1,1}_{loc}(\mathbb{R}^2,\mathbb{R}^2)$, then for at least one $r_0>0$ (actually, for a.e. $r>0$) the sequence restricted to the boundary, $p_{n|\partial B(0,r_0)}$ is in $W^{1,1}(\partial B(0,r_0))$ and converges there, therefore also uniformly. Then for some $n_0$, $$\|p_{n_0|\partial B(0,r_0)}-{\operatorname{id}_{\partial B(0,r_0)}}\|_{\infty,\partial B(0,r_0) }0$ holds at some point $x_0)$ )<|endoftext|> TITLE: One question about the $\eta$ invariant QUESTION [13 upvotes]: This question is from the paper, The Analysis of Elliptic Families II. Dirac Operators, Eta Invariants, and the Holonomy Theorem, Commun. Math. Phys. 107, 103-163 (1986) --- Proposition 2.8. Suppose that $Tr[D^uexp(-tD^2)]=\frac{C_{-n/2}}{t^{n/2}}+\cdots+ \frac{C_{-1/2}}{t^{1/2}}+O(t^{1/2})$ as $t\to0$. Here $D$ denotes the Dirac operator and $D^u$ denotes the $u$-derivative of $D$. (I think one could ignore the exact definition here.) And, we have that $$\Gamma(\frac{s+1}2)\eta(s)=-s\int^\infty_0t^{\frac{s-1}2}Tr[D^uexp(-tD^2)] dt.$$ Q By the asymptotic formula of the "heat kernel", how could we deduce that $$\eta(0)=-2C_{-1/2}/\sqrt\pi.$$ PS: What I do not understand: Why the other terms vanish under the integral? Since the integral is from zero to infinity, is it safe to use the asymptotic formula near $0$? REPLY [7 votes]: Summary: The integral $\int_0^\infty t^{\frac{s-1}{2}} \operatorname{Tr}[ D^u (\exp(-t D^2) ] \, dt$ has a simple pole at $s=0$ with residue $2 C_{-1/2}$, from which your expression for $\eta(0)$ follows. To compute the residue, we compute the integral explicitly (up to a holomorphic remainder) near $t=0$ for $\operatorname{Re} s$ large, then analytically continue. Each term in the $t \to 0^+$ asymptotic expansion of the heat trace corresponds to a pole in $s$ of the integral. To explain, I'm going to phrase this as a more general property of zeta functions defined by a heat trace via the Mellin transform; then at the end we'll return to the eta function. I think these ideas are first due to Seeley in the paper Complex powers of an elliptic operator. A good reference is Gilkey's book Invariance theory, the heat equation, and the Atiyah-Singer index theorem. But I will try to write out all the details here. Let $h(t)$ ($t > 0$) be a smooth function that satisfies these two conditions: (A): $h(t)$ is $O(\exp(-\epsilon t ))$ as $t \to \infty$ (for some $\epsilon > 0$). (B): $h(t) = \displaystyle \sum_{k=0}^K C_{p_k} t^{p_k} + R_K(t)$, where $R_K(t)$ is $O(t^{p_{K+1}})$ as $t \to 0^+$ (for some increasing sequence of powers $p_0 < p_1 < p_2 < \dots \to \infty$). Example/remark: The original question is interested in the case when $h(t) = \operatorname{Tr}[ D^u \exp(-t D^2) ]$, which satisfies those conditions with $p_k = -n/2 + k$, as you said. Note that $\operatorname{Tr}[ P \exp(-t L) ]$ also satisfies the conditions for more general operators $P$ and $L$, where $L$ is a strictly positive elliptic differential operator. (Or if $L$ is nonnegative but has a nontrivial kernel, we can project away from the kernel to get the exponential decay as $t\to \infty$.) The zeta function associated to $h(t)$ is defined using the Mellin transform. The integral in the following definition converges for $s \in \mathbb{C}$ with $\operatorname{Re} s > -p_0$ (see below): $$ \zeta(s) := \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} h(t) \, dt.$$ Note that $\frac{1}{\Gamma(s)}$ is entire (i.e., holomorphic everywhere) since the gamma function $\Gamma(s)$ is nonvanishing. Thus to study poles of $\zeta(s)$, we really just need to study the integral $$ I(s) := \int_0^\infty t^{s-1} h(t) \, dt = I_A(s) + I_B(s),$$ where the two pieces are defined by $$ I_A(s) := \int_1^\infty t^{s-1} h(t) \, dt, ~~~~ I_B(s) := \int_0^1 t^{s-1} h(t) \, dt.$$ We'll deal with the two pieces using the respective conditions above: (A): Since $h(t)$ is $O(\exp(-\epsilon t ))$ as $t \to \infty$, the integral defining $I_A(s)$ converges for all $s \in \mathbb{C}$, and $I_A(s)$ is entire (so we can ignore it for the sake of studying poles). (B): Using the asymptotic expansion, for each $K$ (which defines the point at which we truncate the asymptotic expansion), we have \begin{align*} I_B(s) &= \sum_{k=0}^K C_{p_k} \int_0^1 t^{s-1 + p_k} \, dt + \int_0^1 t^{s-1} R_K(t) \, dt \\ &= \sum_{k=0}^K C_{p_k} \frac{1}{s+p_k} + \text{(holomorphic for $\operatorname{Re} s > -p_{K+1}$ by the estimate on $R_K(t)$)} \end{align*} This shows that $I_B(s)$ is holomorphic for $\operatorname{Re} s > -p_0$ (i.e., $\operatorname{Re} s > \frac{n}{2}$ in your case), and furthermore that $I_B(s)$ possesses an analytic continuation to the half-plane where $\operatorname{Re} s > -p_{K+1}$ with (for $k=0, 1, 2, \dots, K$) a simple pole at $s = -p_k$, where the residue is $C_{p_k}$. Putting all that together, we have that $\zeta(s) = \frac{1}{\Gamma(s)}(I_A(s) + I_B(s))$ has (for $k=0, 1, 2, \dots$) a simple pole at $s= -p_k$, where the residue is $\frac{C_{p_k}}{\Gamma(-p_k)}$. In particular, we will need that the residue of $\zeta(s)$ at $s = \frac{1}{2}$ is $\frac{C_{-1/2}}{\Gamma(1/2)} = \frac{C_{-1/2}}{\sqrt{\pi}}$. (I looked up $\Gamma(1/2)$ on Wikipedia.) Now finally we can return to your question about the eta function. If I understand correctly, your definition of $\eta(z)$ is (this appears to be a derivative of some other eta function) $$ \eta(z) = -z \zeta \left( \frac{z+1}{2} \right).$$ (I hope the new variable $z$ helps avoid confusion.) Since $\zeta \left( \frac{z+1}{2} \right)$ has only simple poles, the factor of $-z$ gives that $\eta(0)$ is $-1$ times the residue of $\zeta \left( \frac{z+1}{2} \right)$ at $z=0$, which is $2$ times the residue of $\zeta(s)$ at $s = \frac{1}{2}$, which we computed above to be $\frac{C_{-1/2}}{\sqrt{\pi}}$. Thus $$\eta(0) = \frac{-2 C_{-1/2}}{\sqrt{\pi}}.$$<|endoftext|> TITLE: A commutative variant of the exterior algebra QUESTION [7 upvotes]: Consider $A = \mathbb{R}[t_1,\ldots,t_{k}]$, the ring of real $k$-variate polynomials in the indeterminates $t_1,\ldots,t_k$. For $y\in\mathbb{R}$, define the element $p(y) \in A$ through $$ p(y) = t_1 y + t_2 y^2 + \cdots + t_k y^k \in A\,. $$ Let $I$ be the ideal in $A$ generated by $\{p(y)^2 \mid y \in \mathbb{R}\}$, and let $S = A/I$ be the quotient algebra of $A$ by $I$. Has $S$ been studied? Does it have a name? REPLY [3 votes]: For $k=\infty$, this algebra appears when studying integrable representations of level 1 of the Lie algebra $\widehat{\mathfrak{sl}}_2$, see, for example, discussion in Section 2 of A. V. Stoyanovsky, B. L. Feigin. Functional models for representations of current algebras and semi-infinite Schubert cells. Functional analysis and its applications, 28 (1994), 55–72. The case of finite $k$ is used when applying when applying this sort of representation theoretic construction to obtain new combinatorial identities. In particular, in the notation of the paper B. L. Feigin, S. A. Loktev. On the finitization of the Gordon identities. Functional analysis and its applications, 35 (2001), 44–51 your algebra has the name $D_2(1,1,1)$.<|endoftext|> TITLE: Determine $2^{\frac{p-1}{4}}\equiv 1\pmod p$ or $2^{\frac{p-1}{4}}\equiv -1\pmod p$ when $p\equiv 1 \pmod 8$ QUESTION [13 upvotes]: Let $p=8k+1\equiv 1\pmod 8$ be a prime, thus $2$ is a quadratic residue module $p$. Euler's criterion show that $$2^{\frac{p-1}{2}}\equiv 1 \pmod p.$$ So we must have $$2^{\frac{p-1}{4}}\equiv \delta(p) \pmod p$$ where $\delta(p)=\pm1$. Now my question is how to determine $\delta(p)$. I calculated many examples. Statistics show that the value of $\delta(p)$ should be related with the parity of $k$ and $h(-p)$, the latter being the class number of $\mathbb{Q}(\sqrt{-p})$. I know little algebraic number theory and hope some experts could help me solve this problem. I am waiting for help, thank you very much! EDIT Now I have formulated my conjecture about $\delta(p)$, which states as follows. Conjecture: $$\delta (p)=(-1)^{\frac{h(-p)}{4}+k}$$ Any ideas to prove or counterexample to disprove this? REPLY [9 votes]: Barrucand and Cohn (MR0249396, Note on primes of type $x^2+32y^2$, class number, and residuacity. J. Reine Angew. Math. 238, 1969, 67--70) have proved that for primes $p \equiv 1 \textrm{ mod } 8$, the condition $h(−4p) \equiv 0 \textrm{ mod } 8$ is equivalent to $−4$ being a $8$-th power mod $p$. This implies your observation because $(-4)^{\frac{p-1}{8}} = (-1)^k 2^{\frac{p-1}{4}}$. I learned this in Merel's article L'accouplement de Weil entre le sous-groupe cuspidal et le sous-groupe de Shimura de $J_0(p)$, J. Reine Angew. Math. 477 (1996), 71--115, where you can find other characterizations e.g. $p$ is of the form $x^2+32y^2$, cf. Théorème 3. In the case $p \equiv 5 \textrm{ mod } 8$, the formula 1.5 in Williams and Currie's article mentioned by Henri Cohen, together with $(\frac{p-1}{2})!^2 \equiv -1 \textrm{ mod } p$ for $p \equiv 1 \textrm{ mod } 4$ (Wilson's theorem) implies your second observation.<|endoftext|> TITLE: $S$-dual of filtered spectra QUESTION [7 upvotes]: I hope this is research level. Suppose $E$ is the direct limit of finite spectra, say $E=\mathrm{colim }\ E_i$, which itself is not finite. I wonder how much and under which conditions the inverse limit $\mathrm{lim}\ D(E_i)$ is a good candidate for playing role of $D(E)$? Here, I write $D$ for the $S$-duality functor. I feel there is problem with the possibility of existence of phantom maps here. I will be very grateful for any advise on this, or possibly some references on the topic. As to what one might expect. For instance, if $f:X\to Y$ is a map between finite spectra then $f_*=0$ if and only if $D(f)_*=0$. Or if we know about dimension of bottom cell of $E$ then that would tell about the dimension of the top cell of the dual spectrum. Here, $f_*$ is the map induced in ordinary homology and I consider CW spectra. And if $D(E)$ can always be identified with the function spectrum $F(E,S^0)$?! I also would like to know what features of duality for finite spectra fail to hold in general. For the moment, homological behavior is very interesting for me. I also wonder if my worry about phantom maps has any point?!? REPLY [4 votes]: Yet more fun examples ... If $A$ is an abelian torsion group and $n \ge 2$, the suspension spectrum of $K(A,n)$ has trivial dual: this is a result of Chun-Nip Lee [Amer.J.Math 114 (1992)]. (This implies the result of T.Y. Lin that Greg mentioned.) An example that I observed in [Progress in Math 215 (2003) (arXiv)] goes as follows: if $K$ and $X$ are finite CW complexes, and the connectivity of $X$ is greater than the dimension of $K$, then the suspension spectrum of the space of continuous maps from $K$ to $X$ is an $S$-dual. (For example, $\Sigma^\infty \Omega^{2n-1}\mathbb CP^{m}/\mathbb CP^n$ is an $S$-dual, for all $m>n$.) In fact, it is an $S$-dual of a filtered spectrum as a functor of $X$, and the dual of this filtration yields the Goodwillie tower. Regarding intuition behind the Segal conjecture, it can be rephrased as saying that a map from a fixed point spectrum to an associated homotopy fixed point spectrum is an appropriate sort of completion. So, for example, there is an easily defined sensible map $\mathbb RP^{\infty} \vee S^0 \rightarrow D(\mathbb RP^{\infty})$ that turns out to be an equivalence after completing the sphere at 2.<|endoftext|> TITLE: What kind of computer tools topologists/geometrists use to visualize the objects they deal with? QUESTION [15 upvotes]: I have recently started to read a bit about geometry and topology. Hopf fibration, Lense spaces, CW complexes, stuff that are discussed in Hatcher's Algebraic Topology and other things that require good visualization. What is apparent to me is that the further I go, the less I understand what is going on. I have searched on YouTube and found some really nice animations for some of these topics but good animations are rare like gems. Advanced stuff in mathematics are less discussed and available on the internet. I have realized that if I want to understand math one day, at some point I should be able to create my own animations. Now, my question is rather directed at people with experience in teaching advanced mathematics or currently doing research in mathematics in areas where geometric intuition is absolutely necessary. What kind of tools do you use? Do you develop them on your own in your research team/group? Can an independent person have access to them? Is it possible for an independent person to develop this kind of tools on their own? Can you think of a situation where you couldn't understand a geometric concept visually but you created an animation that demystified it for you? REPLY [5 votes]: SnapPy is a widely used program within the geometric topology community for studying hyperbolic structures on $3$-manifolds. Here is a YouTube video of Nathan Dunfield showing off some of SnapPy's functionality (creating Dirichlet domains, showing how to visualize the cusps, etc).<|endoftext|> TITLE: Maximize product of sums QUESTION [12 upvotes]: Let $n,k\geq 2$ be positive integers. For each $1\leq i\leq n$, let $I_i$ be a nonempty subset of $\{1,2,\dots,k\}$. Let $P_i=\sum_{j\in I_i}x_j$, and let $P=P_1\cdot P_2\cdot\dots\cdot P_n$. (For example, $P=x_1(x_1+x_2)(x_1+x_3)$.) We want to maximize this expression subject to the constraints $x_i\geq 0$ for all $i$, and $\sum_{i=1}^k x_i=1$. Let $A$ be the value of $P_1$ at the maximum. Let $B$ be the value of $P_1$ at the maximum if we instead maximize the expression $P'=P_2\cdot P_3\cdot\dots\cdot P_n$, subject to the same constraints. Is it true that $A\geq \frac{n-1}{n}B+\frac{1}{n}$? Equality can be obtained, e.g., for any value of $n$, when $P_1=x_1$ and $P_i=x_2$ for $i=2,\dots,n$ (so $A=1/n$ and $B=0$). Another example: $n=2$, $P_1=x_1$ and $P_2=x_1+x_2$. Then $A=1$ and $B$ can be anything in $[0,1]$, so the inequality always holds. REPLY [4 votes]: Yes, that is true. Here is a sketch of the argument. It needs some polishing in places but, I hope, it makes clear what is going on here. We will show that if $P_1,P_2,\dots, P_m$ are of the above form and maximize $F=\sum_k a_k\log P_k$ (so they are viewed as functions of $a=(a_1,\dots,a_m)\in (0,+\infty)^m$), then $\frac{\partial P_1}{\partial a_1}\ge (1-P_1)(\sum_k a_k)^{-1}$ (to be precise, we are talking about the right derivative here). The desired result can be obtained from here by an appropriate integration. Since everything is homogeneous in $a$, we can assume without loss of generality that $\sum_k a_k=1$. Let $da_1$ be a small (infinitesimal) positive increment of $a_1$. Let $dP_i$ be some admissible increments of $P_i$ ($P_i$ are running over some convex polyhedron, so locally the set of admissible increments is an infinitesimal cone in general). Then we have the following increment of $F$ (up to third order terms): $$ dF-da_1\log P_1\approx \sum_k a_k\frac {dP_k}{P_k}+ da_1\frac {dP_1}{P_1}-\frac 12\sum_k a_k\left(\frac {dP_k}{P_k}\right)^2\,. $$ The true increment $dP$ should essentially maximize the RHS. Notice that if you consider the true maximizer $dP=(dP_1,\dots,dP_m)$, then you can compare it with $tdP$ for $t$ close to $1$ and obtain that the linear in $dP$ part of the RHS should be twice as large as the maximum itself (all I'm saying here is that $ut-\frac 12vt^2$ attains its maximal value $\frac {u^2}{2v}$ at $t=u/v$ where the linear part $ut$ equals $\frac{u^2}{v}$). Since we must have $$ \sum_k a_k\frac {dP_k}{P_k}\le 0 $$ because $P$ is the point of maximum and the admissible domain is convex, we conclude that at the true $dP$, we must have $da_1dP_1$ at least as large as twice the maximum of the RHS. Now it remains to estimate that maximum using some particular perturbation. The simplest one will be $x_i\mapsto (1+\tau)x_i$ for $x_i$ participating in $P_1$ and $x_j\mapsto \left(1-\tau\frac{P_1}{1-P_1}\right)x_j$ for $x_j$ not participating in $P_1$. Note that this perturbation is admissible for both small positive and small negative $\tau$ unless one of the variables is $1$, which means that all $P_k=1$, in which case there is nothing to prove. Thus for $dP$ corresponding to this perturbation, we have $$ \sum_k a_k\frac {dP_k}{P_k}=0\,. $$ We also have $\frac{dP_1}{P_1}=\tau$ and $-\frac{P_1}{1-P_1}\tau\le\frac{dP_k}{P_k}\le\tau$ for all $k$. Recalling that for a mean zero real function $f$ on $[0,1]$ squeezed between $-u$ and $v$, we have $\int_0^1 f^2\le uv$, we conclude that for this perturbation $$ \frac 12\sum_k a_k\left(\frac{dP_k}{P_k}\right)^2\le \frac 12\frac{P_1}{1-P_1}\tau^2 $$ whence we can take $\tau=da_1\frac{1-P_1}{P_1}$ and see that the maximum of the RHS is at least $\frac 12 (da_1)^2\frac {1-P_1}{P_1}$. Since $da_1\frac{dP_1}{P_1}$ for the true maximizing increment should be at least twice that large, we get $dP_1\ge (1-P_1)da_1$, as desired.<|endoftext|> TITLE: Why is the number of Perfect Matchings in a triangular grid equivalent to the number of Royal Paths? QUESTION [9 upvotes]: The sequence A006318 at OEIS stands for the Schröder numbers. They describes the number of lattice paths from the southwest corner $(0,0)$ of an $n\times n$ grid to the northeast corner $(n,n)$, using only single steps north, $(0,1)$, northeast $(1,1)$ or east $(1,0)$, that do not rise above the SW–NE diagonal (sometimes called Royal Paths). Also, according to OEIS, they correspond to the number of perfect matchings in number of perfect matchings in a triangular grid of n squares (n = 1, 4, 9, 16, 25, ...). Tomislav Doslic wrote a very nice paper on a similar relation between perfect matchings in a hexagonal grid and the corresponding lattice path. He also mentions the triangular case, and cites the book "Enumerative Combinatorics" vol2, by Stanley, Ex. 6.39 - where I cannot find an answere. Therefore I very much hope for some hints or further literature references on the question given in the title. REPLY [10 votes]: I was able to find the result at Ex. 6.39 s in Stanley's EC Vol.2. The reference given there is to Ciucu, M. "Perfect Matchings of Cellular Graphs" Journal of Algebraic Combinatorics 5 (1996), 87-103 In theorem 4.1, Ciucu computes the generating function of the perfect matchings in a triangular grid of squares and concludes that it is the same as the generating function of Schröder numbers. In the remark that follows the proof of theorem 4.1, a bijection between matchings and paths is also given.<|endoftext|> TITLE: A curious valuation of this sequence QUESTION [12 upvotes]: The sequence $a_n$ given by $$a_n=\sum_{k=0}^n\frac{n!}{k!}$$ is found at A000522 on OEIS with a description: total number of arrangements of a set with $n$ elements. Let $\nu_2(x)$ denote the $2$-adic valuation of the integer $x$. My first question: Is it true that $\nu_2(a_n)=\nu_2(n+13)$? Well, that is what experiments suggest to me. My second question (curiosity) is: What is special about the presence of the number $13$ above? Thanks to darij grinberg the above claim has failed, which prompts me to ask: What is the $2$-adic valuation of $a_n$ then? REPLY [24 votes]: To your first question: No, it is false. For $n = 256 - 13$, the number $a_n$ has $\nu_2\left(a_n\right) = 7 < 8 = \nu_2 \left(n+13\right)$. HOWEVER, it is almost correct: namely, it is correct whenever $128 \nmid n+13$ (so the smallest counterexample is $n = 128 - 13 = 115$). This is the following theorem: Theorem 1. We have $\nu_2\left(a_n\right) = \nu_2\left(n+13\right)$ for all $n \in \mathbb{N}$ that don't satisfy $128 \mid n+13$. Proof of Theorem 1 (sketched). We observe the following facts: Each nonnegative integer $n$ and each positive integer $k$ satisfy $a_{n+k} \equiv a_n \mod k$. (This is proven by induction on $n$. The base case boils down to $a_k \equiv 1 \mod k$, which follows from $a_k = k a_{k-1} + 1$. The induction step uses the recursion $a_n = na_{n-1} + 1$.) Each $n \in \mathbb{N}$ satisfies $a_{n+128} \equiv a_n \mod 128$. (This follows by applying fact 1 to $k = 128$.) Each $n \in \mathbb{N}$ satisfies $128 \nmid a_n$ unless $128 \mid n+13$. (This needs only to be checked for the first $128$ values of $n$, due to fact 2 above.) Combining facts 2 and 3, obtain $\nu_2\left(a_{n+128}\right) = \nu_2\left(a_n\right)$ for each $n \in \mathbb{N}$ unless $128 \mid n+13$. Using fact 4, the Theorem 1 can be proven by induction (with $128$ base cases, unfortunately). $\blacksquare$ So what about the cases when $128 \mid n+13$ ? Their behavior is weird. The smallest such case is $n = 115$, in which $\nu_2 \left(a_n\right) = 9 > 7 = \nu_2 \left( n + 13 \right)$. However, for all larger $n$'s of the form $2^k - 13$, the sign flips: Theorem 2. We have $\nu_2\left(a_n\right) = 7 < \nu_2\left(n+13\right)$ for all $n$ of the form $n = 2^k - 13$ with $k \geq 8$. Proof of Theorem 2 (sketched). We claim that $a_{2^k - 13} \equiv 2^7 \mod 2^8$ for each $k \geq 8$. This is proven by induction on $k$. The induction base ($k = 8$) is verified by computer; the induction step relies on fact 1 from the proof of Theorem 1. $\blacksquare$ Theorem 2 should not suggest that $\nu_2\left(a_n\right)$ is always at most $7$; it is not. Higher values of $\nu_2\left(a_n\right)$ appear when $n+13$ is a multiple of $128$ but not itself a power of $2$. For example, $\nu_2\left(a_{2675}\right) = 15$. Note that the same inductive argument that we used to prove fact 1 can be used to show the following stronger result: Proposition 3. For any nonnegative integer $n$ and any positive integer $k$, we have \begin{align*} a_{n+k}-a_{n} & \equiv% \begin{cases} k, & \text{if }n+k\text{ or }n\text{ is odd};\\ 0, & \text{if neither }n+k\text{ nor }n\text{ is odd}% \end{cases} \mod 2k. \end{align*} This might come useful. Code: It is not immediately obvious how to compute $a_n$ for large values of $n$ efficiently. After all, $a_n \geq n!$, so we'd at least need a large integer type. Fortunately, all we care about are the $2$-adic valuations $\nu_2\left(a_n\right)$ and the remainders of $a_n$ modulo powers of $2$. The former valuations can easily be obtained from the latter remainders, so we only need to care about the remainders. And we can easily compute the remainders of $a_n$ modulo any given integer $k$ by recursion, because the recursive relation $a_n = na_{n-1} + 1$ descends to $\mathbb{Z}/k\mathbb{Z}$. Here, for example, is some simple Sage (or Python2) code that recursively $a_n$ modulo $2^k$ for given $n$ and $k$: def a(n, k): # This gives `a_n` modulo `2^k`. if n == 0: return 1 return (n * a(n-1) + 1) % (2 ** k) This is already quite useful (e.g., you can let it compute "a(256-13, 8)" to check that $\nu_2\left(a_n\right) = 7$ for $n = 256-13$), but not really optimal, because it's recursive and eventually (around $n = 1000$) exceeds Python's maximum recursion depth. So let us replace the recursion by dynamic programming. Here is some better Sage (or Python2) code, which tabulates $\nu_2(a_n)$ for $n = 1, 2, \ldots, m$ given a positive integer $m$: def val2(k): # Return the `2`-adic valuation `\nu_2(k)` of the integer `k`. if k == 0: return # returns ``None``, standing for `-\infty`. res = 0 kk = k while kk % 2 == 0: res += 1 kk = kk // 2 return res def aas(n, k): # Return the list `[a_0, a_1, \ldots, a_{n-1}]` modulo `2^k`. pwk = 2 ** k res = [1] * n for i in range(1, n): res[i] = (i * res[i-1] + 1) % pwk return res def output(m): # Return a LaTeX table of `\nu_2(a_n)` for `n = 1, 2, \ldots, m`. k = 3 # Starting with modulo `2^3`, will later replace by better `2`-adic precision. nums = aas(m+1, k) while (0 in nums): k += 1 nums = aas(m+1, k) vs = [val2(i) for i in nums] res = r"\begin{array}{|c|c|} \hline" + "\n" + r"n & \nu_2(a_n) \\ \hline " for i in range(1, m+1): res += "\n " res += str(i) + r"&" + str(vs[i]) + r" \\" res += " \n" + r"\hline \end{array}" return res If I let it execute "print output(25)", it returns me the LaTeX code for a table of values of $\nu_2(a_n)$... which I'm not quoting here, because Theorem 1 renders it obsolete. A better idea is to tabulate only those values that Theorem 1 does not cover: def output2(m): # Return a LaTeX table of `\nu_2(a_{128n - 13})` for `n = 1, 2, \ldots, m`. k = 3 # Starting with modulo `2^3`, will later replace by better `2`-adic precision. M = 128*m - 13 nums = aas(M+1, k) while (0 in nums): k += 1 nums = aas(M+1, k) vs = [val2(i) for i in nums] res = r"\begin{array}{|c|c|c|} \hline" + "\n" + r"n & \nu_2(a_{128n-13}) & \nu_2(128n) \\ \hline " for j in range(1, m+1): i = 128 * j - 13 res += "\n " res += str(j) + r"&" + str(vs[i]) + r"&" + str(val2(128 * j)) + r" \\" res += " \n" + r"\hline \end{array}" return res Now, executing "print output2(25)" yields \begin{equation} \begin{array}{|c|c|c|} \hline n & \nu_2(a_{128n-13}) & \nu_2(128n) \\ \hline 1&9&7 \\ 2&7&8 \\ 3&8&7 \\ 4&7&9 \\ 5&11&7 \\ 6&7&8 \\ 7&8&7 \\ 8&7&10 \\ 9&9&7 \\ 10&7&8 \\ 11&8&7 \\ 12&7&9 \\ 13&10&7 \\ 14&7&8 \\ 15&8&7 \\ 16&7&11 \\ 17&9&7 \\ 18&7&8 \\ 19&8&7 \\ 20&7&9 \\ 21&15&7 \\ 22&7&8 \\ 23&8&7 \\ 24&7&10 \\ 25&9&7 \\ \hline \end{array} . \end{equation}<|endoftext|> TITLE: A paradox on the deformation of singularities QUESTION [6 upvotes]: Setup: $\pi: \mathcal X \to C$ is a flat morphism from a germ of a smooth curve $(C, o)$. Suppose the special fiber $\pi^{-1}(o) := \mathcal X_o$ has a certain class of singularities, one wants to study if $\mathcal X$ preserves such singularities. It has been shown (see "Deformations of canonical singularities") that the above property holds for canonical singularities; but it fails for klt singularities (see "On the extension problem of pluricanonical forms," Example 4.3). However, I found an one-line argument for both cases, could anyone point out where I was wrong? Here is the argument: First, it is known that $\mathcal X$ is $\mathbb Q$-Gorenstein. Because $\{o\}$ is a divisor on $C$, $\mathcal X_o$ can also be viewed as the pull-back Cartier divisor $\pi^*{o}$, hence it is Cartier, and by adjunction (assuming $\mathcal X$ is CM) $$(K_{\mathcal X}+\mathcal X_o)|{_{\mathcal X_o}} = K_{\mathcal X_o}.$$ Then by the precise inversion of adjunction, $${\rm total~discrepancy}\{\mathcal X_o\} = {\rm total~discrepancy}\{(\mathcal X, \mathcal X_o){\rm~with~center~intersects~} \mathcal X_o\}.$$ Hence the minimal discrepancy of $(\mathcal X, \mathcal X_o)$ near $\mathcal X_o$ is $\geq 0$ in the canonical case and $>-1$ in the klt case. In particular, $\mathcal X$ is canonical and klt respectively. REPLY [4 votes]: Just to add to Hacon's answer. This kind of thing has also been studied quite a bit in the characteristic $p > 0$ side. In fact, it was observed in F-purity and rational singularity by R. Fedder (1983) that $F$-pure singularities don't have this property. $F$-pure was later seen to be the analog of log canonical singularities. The paper F-Regularity Does Not Deform by A. K. Singh (1999) [arxiv] shows that F-regularity doesn't satisfy this property (F-regularity is the analog of KLT singularities). The singularities there are given by rather explicit equations.<|endoftext|> TITLE: Approximation of Wasserstein distance between $p_\theta$ and $p_{\theta + d\theta}$ QUESTION [8 upvotes]: Given a parametric family of distributions $\{p_\theta\mid\theta \in \Theta\}$, one can show that under some regularity conditions, the following approximation is valid $$\operatorname{KL}(p_\theta\parallel p_{\theta + d \theta}) = d \theta^TF(\theta) \, d\theta + \mathcal O(\|d\theta\|^3), $$ where $$F(\theta)_{ij} := \mathbb E_{x \sim p_\theta}\left[\frac{\partial^2}{\partial \theta_i \, \partial \theta_j} \log(p_\theta(x))\right] $$ is the Fisher information matrix of $p_\theta$. A very rough sketch of the proof can be found on wikipedia. Question 1 Is there such an approximation formula for the Wassertein distance or other measures of discrepancy between probability distributions ? Question 2 Same question, specialized to $f$-divergences (of which KL is a particular case). REPLY [2 votes]: A natural field here is Wasserstein information geometry. See Wuchen Li, Guido Montufar: Natural gradient via optimal transport https://arxiv.org/abs/1803.07033 For related applications see my talk https://speakerdeck.com/lwc2017/learning-via-wasserstein-information-geometry<|endoftext|> TITLE: Fourth obstruction, Pontryagin and Euler class QUESTION [11 upvotes]: Assume the first three obstruction classes of a rank 4 vector bundle vanish and look at the fourth obstruction class. This fourth obstruction class can be decomposed as the Euler class and the first Pontryagin class (since $\pi_3(SO_4) \simeq \mathbb{Z} \oplus \mathbb{Z}$). Is there a geometric description of a system of generators in $\pi_3(SO_4)$ which is associated to these classes? Recall that $SO_4$ is double covered by $SU_2 \times SU_2$ and since $SU_2 \cong S_3$, $π_3(SO_4)=\pi_3(S_3) \oplus \pi_3(S_3)= \mathbb{Z} \oplus \mathbb{Z}$. The question is: how do the Euler and Pontryagin classes relate to this double cover? In other words, what is the system of generator $\langle \alpha, \beta \rangle$ of $\mathbb{Z} \oplus \mathbb{Z}$ so that given an element, if one writes it down as $a\alpha+b\beta$ then $a$ would be associated to the Euler class and $b$ to the Pontryagin class REPLY [14 votes]: Geometric generators for $\pi_3(SO(4))$ have been identified in §22 of Steenrod's "Topology of fibre bundles", using the identification of $S^3$ as unit quaternions. Conjugation of quaternions induces an element of $\pi_3(SO(4))$ denoted by $\alpha_3$ and left multiplication induces an element denoted by $\beta_3$. These generate $\pi_3(SO(4))\cong\mathbb{Z}\oplus\mathbb{Z}$. The relation between obstruction classes and characteristic classes is discussed in A. Dold and H. Whitney. Classification of oriented sphere bundles over a 4-complex. Ann. Math. 69 (1959), 667--677. I think their Theorem 2 states that the part of the obstruction class corresponding to the generator $\beta_3$ is exactly to the Euler class. On the other hand, the Pontryagin class of the bundle is $-4d_1-2d_2$ where $d_1$ is the part of the obstruction class corresponding to $\alpha_3$ and $d_2$ is the part of the obstruction class corresponding to $\beta_3$. Of course, then one can identify an actual element of the homotopy group corresponding to the Pontryagin class, but this will not be a generator of $\pi_3(SO(4))$. Note that, contrary to what is implicitly claimed in the question, the obstruction class doesn't actually decompose as sum of Euler class and Pontryagin class (but this is consistent with the index of the Hurewicz map being 2).<|endoftext|> TITLE: Deforming a section to a section without zeros QUESTION [6 upvotes]: Let $M$ be an oriented manifold of dimension $n$. Suppose furthermore that $E$ is an oriented vector bundle of rank $n-1$ over $M$. Let $s$ be a section of $E$ transversal to the zero section in $E$. Thus the zero set $Z$ of $s$ is a union of embedded circles in $M$ (we assume that the zero set lies in the interior of $M$ in case $M$ has non-empty boundary) Under what conditions (if any) is it possible to deform $s$ such that the deformed section $s'$ has no zeros anymore? Note that, in case the rank of $E$ is equal to the rank of $M$, then this is always possible provided the Euler class of $E$ is zero. REPLY [5 votes]: This is a classical problem in obstruction theory. There is indeed an extra obstruction living in $H^n(M,\pi_{n-1}S^{n-2})=H^n(M,\mathbb{F}_2)$. But this obstruction is defined only modulo an indeterminacy comming from $H^{n-2}(M,\pi_{n-2}S^{n-2})$. There are two dual ways to look at it, so let me try to explain both. We can consider the filtration on $M$ by skeleta, and then the picture is as follows: The condition that the Euler class vanishes exactly allows us to deform the section to be non-zero on the $n-2$-skeleton. Indeed, the Euler class can be defined by the cocycle that encode the mapping degrees of attaching maps of $n-1$-cells in some local trivialization of the bundle at each $n-1$-cell. By modifying it along the $n-2$-cells we can replace $e(E)$ by any cohomologous cocycle and so we can re-choose the cocycle to be $0$ onthe nose, allowing to deform the value in the interior of the $n-1$-cells into a non-vanishing one. Then, we can lift the resulting section on the $n-1$-skeleton to a section on $M$, and we get a section with zeros only inside the $n$-cells. We can ask if we can eliminate those zeros. choose a trivialization of the bundle on each cell, then the section on the $n-1$-skeleton give us a collection of maps $S^{n-1}\to \mathbb{R}^{n-1}-0\cong S^{n-2}$ and the collection of homotopy classes is now a function from $n$-cells to $\pi_{n-1} S^{n-2}$ which for $n>4$ is $\mathbb{F}_2$, represented by the stabilization of the Hopf fibration. So we get a cocycle that at each cell tells us if the map from the boundary is null homotopic or not. Note that the invariant in $\mathbb{F}_2$ corresponding to each cell is related to the cohomology operation $sq^2$ which detects the Hopf map. Here is another, more algebraic way to understand this obstruction. We have a map $BSO_{n-1}\to K(\mathbb{Z},n-1)$ classifying the Euler class. DEnote the fiber by $F_1$, then one can compute the cohomology of $F_1$ with coefficients in $\mathbb{F}_2$ by the Leray spectral sequence and one find a cohomology class in degree $n$, as follows. We have a fibration $K(\mathbb{Z},n-2)\to F_1\to BO_{n-1}$ and by definition the fundamental class of $K(\mathbb{Z},n-2)$ has its transgression the Euler class $e$. On the level of chains, we choose a class $y$ in $C^*(F_1)$ such that $d(y)=\pi^*y$ for $\pi:F_1\to BO_{n-1}$. Then, since the cohomology of $K(\mathbb{Z},n-1)$ has the pattern $\mathbb{Z},0,\mathbb{F}_2,0,...$ starting from $n-2$, we see that there is a $\mathbb{F}_2$ component in $H^n(K(\mathbb{Z},n-2))$ and this is exactly the class that detects the Hopf map in $\pi_{n-1}S^{n-2}$. Taken mod 2, this class in cohomology can be idetified with $sq^2(y)$ where $y$ is the foundamental class. Now, we know that $sq^2(w_{n-1})=w_{n-1}w_2$ in $H^*(BO_{n-1},\mathbb{F}_2)$ so in particular the class $sq^2(y)-w_2y$ servives the spectral sequence and gives a cohomology class in $H^n(F_1,\mathbb{F}_2)$, well defined modulo some ambiguities that ill ignore for now. If $E$ is a vector bundle with vanishing euler class, then choosing a way to write the euler class as a coboundary gives us a lift of the map $M\to BO_{n-1}$ along the map $\pi: F_1 \to BO_{n-1}$. Geometrically this is a choice of a surface with boundary the circles of zeros of a section precicely! Then, the next obstruction is the pull-back along the lift of the class $sq^2(y)+w_2(y)$ in the cohomology of $F_1$. Now, at least informally (and there should be some secondary cohomology operations formalism that make this precise) the next obstruction is given by choosing a cochain $y$ with $d(y)=e(E)$ and then the obstruction is something like $sq^2(y)+w_2(E)\cup y \mod 2$. The problem here is that we need cohomology operation on the chain level so again there should be some secondary cohomology operations formalism that makes it precise. Now the ambiguity: Here its pretty straight foward what it is: We can change $y$ by a cocycle, so we need to consider the secondary obstruction modulo the classes of the form $sq^2(a)-w_2(E)a$ for $a\in H^{n-2}(M,\mathbb{Z})=H^{n-2}(M,\pi_{n-2}S^{n-2})$. Using Wu formula, in the case of oriented manifold we can re-write the term $sq^2(a)$ as $v_2 \cup a$ so the indetermenacy of the obstruction is the ideal generated by $v_2(M) + w_2(E)$ where $v_2$ is the second Wu class of the manifold $M$. It is interesting if there's a simple description in terms of the Zeros of the deformed section, but I don't know. Edit: In fact, one can recover the homotopy class of the section on the attaching map of a top cell using the zeros, in a way that was mensioned in the comments above. The zeros are now stably framed cicrcles inside the cell, and the homotopy class corresponding to the cell is the class of this stably framed circle. To see this, just take a point in the unit disc in $\mathbb{R}^{n-1}$, considered as the vector bundle trivialized at the cell, and drag it generically to zero. The fiber at the point in the unit disc is the class of the map from the boundary of the cell, and the preimage at the origin is the zero locus so they are equivalent.<|endoftext|> TITLE: Has incorrect notation ever led to a mistaken proof? QUESTION [84 upvotes]: In mathematics we introduce many different kinds of notation, and sometimes even a single object or construction can be represented by many different notations. To take two very different examples, the derivative of a function $y = f(x)$ can be written $f'(x)$, $D_x f$, or $\frac{dy}{dx}$; while composition of morphisms in a monoidal category can be represented in traditional linear style, linearly but in diagrammatic order, using pasting diagrams, using string diagrams, or using linear logic / type theory. Each notation has advantages and disadvantages, including clarity, conciseness, ease of use for calculation, and so on; but even more basic than these, a notation ought to be correct, in that every valid instance of it actually denotes something, and that the syntactic manipulations permitted on the notation similarly correspond to equalities or operations on the objects denoted. Mathematicians who introduce and use a notation do not usually study the notation formally or prove that it is correct. But although this task is trivial to the point of vacuity for simple notations, for more complicated notations it becomes a substantial undertaking, and in many cases has never actually been completed. For instance, in Joyal-Street The geometry of tensor calculus it took some substantial work to prove the correctness of string diagrams for monoidal categories, while the analogous string diagrams used for many other variants of monoidal categories have, in many cases, never been proven correct in the same way. Similarly, the correctness of the "Calculus of Constructions" dependent type theory as a notation for a kind of "contextual category" took a lot of work for Streicher to prove in his book Semantics of type theory, and most other dependent type theories have not been analogously shown to be correct as notations for category theory. My question is, among all these notations which have never been formally proven correct, has any of them actually turned out to be wrong and led to mathematical mistakes? This may be an ambiguous question, so let me try to clarify a bit what I'm looking for and what I'm not looking for (and of course I reserve the right to clarify further in response to comments). Firstly, I'm only interested in cases where the underlying mathematics was precisely defined and correct, from a modern perspective, with the mistake only lying in an incorrect notation or an incorrect use of that notation. So, for instance, mistakes made by early pioneers in calculus due to an imprecise notion of "infinitesimal" obeying (what we would now regard as) ill-defined rules don't count; there the issue was with the mathematics, not (just) the notation. Secondly, I'm only interested in cases where the mistake was made and at least temporarily believed publically by professional (or serious amateur) mathematician(s). Blog posts and arxiv preprints count, but not private conversations on a blackboard, and not mistakes made by students. An example of the sort of thing I'm looking for, but which (probably) doesn't satisfy this last criterion, is the following derivation of an incorrect "chain rule for the second derivative" using differentials. First here is a correct derivation of the correct chain rule for the first derivative, based on the derivative notation $\frac{dy}{dx} = f'(x)$: $$\begin{align} z &= g(y)\\ y &= f(x)\\ dy &= f'(x) dx\\ dz &= g'(y) dy\\ &= g'(f(x)) f'(x) dx \end{align}$$ And here is the incorrect one, based on the second derivative notation $\frac{d^2y}{dx^2} = f''(x)$: $$\begin{align} d^2y &= f''(x) dx^2\\ dy^2 &= (f'(x) dx)^2 = (f'(x))^2 dx^2\\ d^2z &= g''(y) dy^2\\ &= g''(f(x)) (f'(x))^2 dx^2 \end{align}$$ (The correct second derivative of $g\circ f$ is $g''(f(x)) (f'(x))^2 + g'(f(x)) f''(x)$.) The problem is that the second derivative notation $\frac{d^2y}{dx^2}$ cannot be taken seriously as a "fraction" in the same way that $\frac{dy}{dx}$ can, so the manipulations that it justifies are incorrect. However, I'm not aware of this mistake ever being made and believed in public by a serious mathematician who understood the precise meaning of derivatives, in a modern sense, but was only led astray by the notation. Edit 10 Aug 2018: This question has attracted some interesting answers, but none of them is quite what I'm looking for (though Joel's comes the closest), so let me clarify further. By "a notation" I mean a systematic collection of valid syntax and rules for manipulating that syntax. It doesn't have to be completely formalized, but it should apply to many different examples in the same way, and be understood by multiple mathematicians -- e.g. one person writing $e$ to mean two different numbers in the same paper doesn't count. String diagrams and categorical type theory are the real sort of examples I have in mind; my non-example of differentials is borderline, but could in theory be elaborated into a system of syntaxes for "differential objects" that can be quotiented, differentiated, multiplied, etc. And by saying that a notation is incorrect, I mean that the "understood" way to interpret the syntax as mathematical objects is not actually well-defined in general, or that the rules for manipulating the syntax don't correspond to the way those objects actually behave. For instance, if it turned out that string diagrams for some kind of monoidal category were not actually invariant under deformations, that would be an example of an incorrect notation. It might help if I explain a bit more about why I'm asking. I'm looking for arguments for or against the claim that it's important to formalize notations like this and prove that they are correct. If notations sometimes turn out to be wrong, then that's a good argument that we should make sure they're right! But oppositely, if in practice mathematicians have good enough intuitions when choosing notations that they never turn out to be wrong, then that's some kind of argument that it's not as important to formalize them. REPLY [5 votes]: Edward Nelson’s proof of the inconsistency of Peano arithmetic (and weaker systems) had an error which he saw only after Terence Tao refined some notation. Specifically, Tao reformulated Chaitin’s theorem from Given a theory $T$, there exists an $\ell$ with the property that, if $T$ is consistent, then there does not exist an $x$ such that $T$ can prove $K(x)>\ell$ to Given a theory $T$, there exists an $\ell(T)$ with the property that, if $T$ is consistent, then there does not exist an $x$ such that $T$ can prove $K(x)>\ell(T)$ In particular, this allowed Tao to note that $\ell(T) < \ell(T’)$ is possible even when $T’$ is a restricted version of $T$, contrary to an implicit assumption in Nelson’s proof.<|endoftext|> TITLE: Reference request: Recovering a Riemannian metric from the distance function QUESTION [13 upvotes]: Let $M = (M, g)$ be a Riemannian manifold, and let $p \in M$. Writing $d$ for the geodesic distance in $M$, there is a function $$ d(-, p)^2 : M \to \mathbb{R}. $$ This function is smooth near $p$. Hence for each point $x \in M$ sufficiently close to $p$, we have the Hessian $$ \text{Hess}_x(d(-, p)^2) $$ (defined using the Levi-Civita connection), which is a bilinear form on $T_x M$. In particular, we can take $x$ to be equal to $p$ itself, giving a bilinear form $$ \text{Hess}_p(d(-, p)^2) $$ on $T_p M$. But of course, we already have another bilinear form on $T_p M$, namely, the Riemannian metric $g_p$ itself. And the fact is that up to a constant factor, these two forms are equal: $$ g_p = \frac{1}{2} \text{Hess}_p(d(-, p)^2). $$ I'm looking for a reference for this fact. For the purposes of what I'm writing, it would ideally be a reference that states this fact in the same simple direct terms as above, without involving any other differential-geometric concepts (e.g. normal coordinates). I understand that this is a basic fact of Riemannian geometry, so I've already looked for it in various introductions to the subject, including those by do Carmo, Jost, Lee, and Petersen. But I haven't found it stated in any of those sources (which isn't to say it's not there). I have found more sophisticated stuff about $\text{Hess}_x(d(-, p)^2)$ for points $x$ different from $p$, but not the simple fact I'm looking for. Requests for references often result in people giving their favourite proofs rather than a reference. While that doesn't do any harm (and can be quite interesting), I emphasize that it's a reference I'm looking for, not a proof. REPLY [2 votes]: I believe that the reason why you cannot find the result that you are asking about printed anywhere is that it is, after all, a mere exercise in Riemannian computation. First, it is easy to show that if $f$ is smooth around $p$, then $(Hf)_{ij} = \partial^2_{ij}f - \Gamma_{ij} ^k \partial_k f$ in any system of coordinates around $p$. Now, since your $f = d_p^2$ has radial symmetry, it is natural to continue the work in spherical normal coordinates, i.e. you go in $T_pM$ through $\exp_p ^{-1}$ and there you introduce spherical coordinates $r, \sigma_1, \dots, \sigma_n$, with $n = \dim M$. Since $\Gamma_{ij}^k (p) = 0$ as a consequence of your coordinates being normal, you will have $(Hf)_{ij} (p) = (\partial^2_{ij}f) (p) = (\partial ^2 _{rr} r^2) (p) = 2$ (all the other second-order partial derivatives vanish at $p$ because $f=r^2$ does not contain the variables $\sigma_1, \dots, \sigma_n$). On the other hand, it is known that in normal spherical coordinates the expression of the metric tensor is $g_{ij} = \delta_{ij} + o(r)$, so that $g_{ij} (p) = \delta_{ij} (p)$ (the Kronecker symbol), whence it follows that $(Hf)(p) = 2g(p)$ (the metric evaluated at $p$). See p.114 of I. Chavel, "Riemannian Geometry - A Modern Introduction", 2006, or the more general theorem 2.53 of Cartan on p.83 of S. Rosenberg, "The Laplacian on a Riemannian Manifold", 1997, or Petersen's book cited here.<|endoftext|> TITLE: Induced maps on homotopy groups by self maps of $\mathbb{CP}^n$ QUESTION [11 upvotes]: Let $f:\mathbb{S}^2\to \mathbb{S}^2$ with degree $d$. It is well known that the induced map $$f_\ast:\pi_3(\mathbb{S}^2)=\mathbb{Z}\to \pi_3(\mathbb{S}^2)=\mathbb{Z}$$ is given by multiplication by $d^2$. But, $\mathbb{S}^2=\mathbb{C}P^1$ and $[\mathbb{C}P^n,\mathbb{C}P^n]=\mathbb{Z}$ for $n\geq 1$ and for any $f:\mathbb{C}P^n\to \mathbb{C}P^n$ determined by $d\in \mathbb{Z}$ the induced map $$f_\sharp: H_{2k}(\mathbb{C}P^n)=\mathbb{Z}\to H_{2k}(\mathbb{C}P^n)=\mathbb{Z}$$ is given by multiplication by $d^k$ for $1\leq k\leq n$. Question: is it true that the map $$f_\ast : \pi_{2n+1}(\mathbb{C}P^n)=\mathbb{Z}\to \pi_{2n+1}(\mathbb{C}P^n)=\mathbb{Z}$$ is determined by multiplication by $d^{n+1}$? REPLY [5 votes]: This is just to provide a bit of background to Dmitri's elegant answer. You can do this by induction by recognising that $\mathbb{C}P^n$ is the $n^{th}$ projective plane of the $H$-space $S^1$, and in particular may be constructed using the Hopf construction. This iteratively produces quasi-fibrations $\gamma_n:\ast^nS^1\rightarrow \mathbb{C}P^{n-1}$, starting with $n=1$, where $\ast^nS^1$ is the $n$-fold join of $S^1$, and defines $\mathbb{C}P^n$ as the cofiber $$\ast^nS^1\xrightarrow{\gamma_n} \mathbb{C}P^{n-1}\rightarrow\mathbb{C}P^n.$$ Note that the degree $d$ self map $d:S^1\rightarrow S^1$ is homotopic to the $d$-fold power map (defined with the Lie product), and since $S^1$ is abelian this map is an $A_{\infty}$-map. It follows that there is an induced map $\underline d_n:\mathbb{C}P^n\rightarrow \mathbb{C}P^n$ between the Hopf constructions at each stage. In particular, at the $(n+1)^{th}$ stage we have $\ast^{(n+1)}S^1\cong \Sigma^n\bigwedge^{n+1}S^1\cong S^{2n+1}$ and $\ast^{n+1}d\cong \Sigma^n\bigwedge^{n+1}d\simeq d^{n+1}$ sitting in a diagram of cofibrations $\require{AMScd}$ \begin{CD} S^{2n+1}@>\gamma_{n+1}>> \mathbb{C}P^n@>>>\mathbb{C}P^{n+1}\\ @Vd^{n+1} V V @VV \underline d_n V@VV \underline d_{n+1} V\\ S^{2n+1} @>\gamma_{n+1}>> \mathbb{C}P^n@>>>\mathbb{C}P^{n+1}. \end{CD} Technically, it now remains to confirm that $\underline d_n$ is indeed the map you describe, but its Friday and the pub's open, so I'm going to stop typing here.<|endoftext|> TITLE: What are some open problems in moduli spaces and moduli stacks? QUESTION [7 upvotes]: I would like to know what are the open big and interesting problems related to moduli spaces and moduli stacks ? Thanks in advance for your help. REPLY [5 votes]: The Debarre-de Jong conjecture: if $\mathrm{X}\subset\mathbf{P}^n$ is a smooth hypersurface of degree $d\leqslant n$, then the dimension of the moduli space of lines on $\mathrm{X}$ is the expected one, namely $2n-d-3$. It is known to be true for at least $d\leqslant 6$.<|endoftext|> TITLE: Formalizations of the idea that something is a function of something else? QUESTION [22 upvotes]: I'll state my questions upfront and attempt to motivate/explain them afterwards. Q1: Is there a direct way of expressing the relation "$y$ is a function of $x$" inside set theory? More precisely: Can you provide a formula of first order logic + $\in$, containing only two free variables $y$ and $x$, which directly captures the idea that "$y$ is a function of $x$"? In case the answer to Q1 is negative, here's Q2: Do any other foundations of mathematics (like univalent) allow one to directly formalize the relation "$y$ is a function of $x$"? Or have there been any attempts to formalize (parts of) mathematics with a language where the relation is taken as primitive/undefined? From discussing Q1 with colleagues I've learned that it's hard to convey what my problem is, causing frustration on both sides. I suspect this is to a certain extent because we all only learned the modern definition of function (which is not the answer to Q1) and because neither the people I talk to nor myself are experts in logic/type theory/category theory. So please bear with me in (or forgive me for) this lengthy attempt at an Explanation/Motivation: The relation "$y$ is a function of $x$" between two things $y$ and $x$, was the original (and apparently only) way of using the word function in mathematics until roughly 1925. The things $y$ and $x$ were traditionally called variable quantities, and the same relation was sometimes worded differently as "$y$ depends on $x$", "$y$ is determined by $x$" or "$y$ changes with $x$". This was used as a genuine mathematical proposition: something that could be proved, refuted or assumed. People would say "let $y$ be a function of $x$" just like today we might say "let $U$ be subgroup of $G$". I could cite more than 40 well known mathematicians from Bernoulli to Courant who gave definitions of this relation, but I'll limit myself to quote eight at the end of my question. As far as I can tell, these definitions cannot be directly translated into first order logic + $\in$. Although the word function assumed a different meaning with the rise of set theory an formal logic, the original relation is still used a lot among physicist, engineers or even mathematicians. Think of statements like "The pressure is a function of the volume", "The area of the circle is a function of its radius", "The number of computations is a function of the size of the matrix", "The fiber depends on the base point" etc. This even crops up in scientific communities where I wouldn't expect it. One finds it for instance in Pierce's Types and Programming Languages or Harper's Practical foundations of programming languages. So it seems that something being a function of something else (or something depending on something else) is a very natural notion for many people. In fact, I have the impression that for physicists, engineers and most scientist who apply mathematics, this relation is closer to the ideas they want to express, than the modern notion of function. Yet, I don’t see a direct way of formalizing the idea inside set theory. (The modern notion of a map $f\colon X \to Y$ is not what I'm looking for, since by itself it's not a predicate on two variables.) I know how to capture the idea at the meta-level, by saying that a formula of first order logic is a function of $x$, iff its set of free variables contains at most $x$. But this is not a definition inside FOL. When a physicist says “The kinetic energy is a function of the velocity” he’s certainly making a physical claim and not a claim about the linguistic objects he uses to talk about physics. So this syntactic interpretation of “$y$ is a function of $x$” is not what I’m looking for. I also know a way to encode the idea inside set theory. But I’m not completely happy with it. First, here’s a naive and obviously wrong approach: Let $x\in X$ and $y\in Y$, call $y$ a function of $x$, iff there exists a map $f:X\to Y$, such that $y=f(x)$. Since every such $y\in Y$ is a function of every $x$ (use a constant map $f=(u\mapsto y))$, this is not the right definition. Here’s a better approach: Let $x$ and $y$ be maps with equal domain, say $x:A\to X$ and $y: A \to Y$. Before giving the definition let me switch terminology: Instead of calling $x$ a "map from $A$ to $X$" I'll call it a "variable quantity of type $X$ in context $A$". (This change of terminology is borrowed from categorical logic/type theory. In categorical logic people say “$x$ is a generalized element of $X$ with stage of definition $A$”. But don’t assume from this, that I have a thorough understanding of categorical logic or type theory.) Definition: Let $x$ and $y$ be variable quantities of type $X$ and $Y$ in the same context $A$. We call $y$ a function of $x$, iff there exists a map $f: X\to Y$ such that $y=f\circ x$. (It would be suggestive to switch notation from $f{\circ} x$ to $f(x)$, so we could write $y=f(x)$ when $y$ is a function of $x$. I'll refrain from doing this, since $f(x)$ has an established meaning in set theory.) Since my post is getting long, I won’t explain why this definition captures the original idea quite well. Let me only say why I don't consider it a direct way of capturing it: It seems backwards from a historical perspective. Mathematicians first had the notion of something being a function of something else, and only from there did they arrive at maps and sets. With this approach we first need to make sense of maps and sets, in order to arrive at the original idea. This might not be a strong counter argument, but if I wanted to use the original idea when teaching calculus, I would need a lot of preparation and overhead with this approach. What I’d like to have instead, is a formalization of mathematics where the relation can be used "out of the box". The other thing I don’t like about this (maybe due to my lack of knowledge of categorical logic) has to do with the context $A$ and what Anders Kock calls an “important abuse of notation”. To illustrate: suppose I have two variables quantities $x,y$ of type $\mathbb{R}$ in some context $A$. If I now assume something additional about these variables, like the equation $y=x^2$, this assumption should change the context from $A$ to a new context $B$. This $B$ is the domain of the equalizer $e:B\to A$ of the two maps $x^2,y\colon A\to \mathbb{R}$. The abuse of notation consist in denoting the "new" variable quantities $x\circ e, y\circ e$ in context $B$, with the same letters $x,y$. I suspect this abuse is considered important, since in everyday mathematics it's natural to keep the names of mathematical objects, even when additional assumptions are added to the context. In fact, if I’m not mistaken, in a type theory with identity types there is no abuse of notation involved when changing the context from $A\vdash (x,y) \colon \mathbb{R}^2$ to $A, e\colon y=x^2 \vdash (x,y) \colon \mathbb{R}^2$. So maybe type theorist also already know a language where one can talk of "functions of something" in a way that's closer to how way mathematicians did until the 1920's? Some historical definitions of "$y$ is a function of $x$": Johann Bernoulli 1718, in Remarques sur ce qu’on a donné jusqu’ici de solutions des Problêmes sur les isoprimetres, p. 241: Definition. We call a function of a variable quantity, a quantity composed in any way whatsoever of the variable quantity and constants. (I'd call this the first definition. Leibniz, who initiated the use of the word function in mathematics around 1673, gave a definition even earlier. But his is not directly compatible with Bernoulli's, even though he later approved of Bernoulli's definition.) Euler, 1755 in Institutiones calculi differentialis, Preface p.VI: Thus when some quantities so depend on other quantities, that if the latter are changed the former undergo change, then the former quantities are called functions of the latter ; this definition applies rather widely, and all ways, in which one quantity could be determined by others, are contained in it. If therefore $x$ denotes a variable quantity, then all quantities, which depend upon $x$ in any way, or are determined by it, are called functions of it. Cauchy, 1821 in Cours d'analyse, p. 19: When variable quantities are related to each other such that the values of some of them being given one can find all of the others, we consider these various quantities to be expressed by means of several among them, which therefore take the name independent variables. The other quantities expressed by means of the independent variables are called functions of those same variables. Bolzano, ca. 1830 in Erste Begriffe der allgemeinen Grössenlehre, The variable quantity $W$ is a function of one or more variable quantities $X,Y,Z$, if there exist certain propositions of the form: "the quantity $W$ has the properties $w,w_{1},w_{2}$,", which can be deduced from certain propositions of the form "the quantity $X$ has the properties $\xi,\xi',\xi''$, -- the quantity $Y$ has the properties $\eta,\eta',\eta''$; the quantity $Z$ has the properties $\zeta,\zeta',\zeta''$, etc. Dirichlet, 1837 in Über die Darstellung ganz willkürlicher Functionen durch Sinus- und Cosinusreihen: Imagine $a$ and $b$ two fixed values and $x$ a variable quantity, which continuously assumes all values between $a$ and $b$. If now a unique finite $y$ corresponds to each $x$, in such a way that when $x$ ranges continuously over the interval from $a$ to $b$, ${y=f(x)}$ also varies continuously, then $y$ is called a continuous function of $x$ for this interval. (Many historians call this the first modern definition of function. I disagree, since Dirichlet calls $y$ the function, not $f$.) Riemann, 1851 in Grundlagen für eine allgemeine Theorie der Functionen einer veränderlichen complexen Grösse If one thinks of $z$ as a variable quantity, which may gradually assume all possible real values, then, if to any of its values corresponds a unique value of the indeterminate quantity $w$, we call $w$ a function of $z$. Peano, 1884 in Calcolo differenziale e principii di calcolo integrale p.3: Among the variables there are those to which we can assign arbitrarily and successively different values, called independent variables, and others whose values depend on the values given to the first ones. These are called dependent variables or functions of the first ones. Courant, 1934 in Differential and Integral Calculus Vol. 1, p.14: In order to give a general definition of the mathematical concept of function, we fix upon a definite interval of our number scale, say the interval between the numbers $a$ and $b$, and consider the totality of numbers $x$ which belong to this interval, that is, which, satisfy the relation $$ a\leq x \leq b. $$ If we consider the symbol $x$ as denoting at will any of the numbers in this interval, we call it a (continuous) variable in the interval. If now to each value of $x$ in this interval there corresponds a single definite value $y$, where $x$ and $y$ are connected by any law whatsoever, we say that $y$ is a function of $x$ (It's funny how at after Cauchy many mathematicians talk of values of variables, which is not something we're allowed to do in set theory. (What's the value of a set or of the element of a set?). Yet, if one looks at modern type theory literature, it's full of talk of "values" again.) REPLY [4 votes]: There is some philosophical literature on this topic. My book 'Reasoning with Arbitrary Objects' (1985, Blackwell: Oxford) advocates the traditional view of variables. A new edition will soon come out in OUP with a new introduction in which I attempt to axiomatize the traditional view of variables. Leon Horsten has a recent book 'The Metaphysics and Mathematics of Arbitrary Objects' (2019, Cambridge University Press) in which he develops the theory using the idea of there being an underlying set (relative to which the variables vary).<|endoftext|> TITLE: Whence "Durchschnitt" and "Vereinigung"? QUESTION [10 upvotes]: Today the set-theoretic operations of intersection $\cap$ [German: Durchschnitt] and union $\cup$ [German: Vereinigung] are standard. The modern notations are present in the first edition of van der Waerden's Moderne Algebra (1930). However, the notation is mostly missing from Steinitz' Algebraische Theorie der Körper (1910), which was a precursor of van der Waerden. Instead of Menge (for set), Steinitz uses System. The term Durchschnitt is present in Steinitz, but the word Vereinigung and the symbols $\cap,\cup$ are missing. Has anything been written on the history of set-theoretic language in algebra? REPLY [15 votes]: An extensive discussion of the origin of "Menge" is given in Earliest Known Uses of Some of the Words of Mathematics (scroll down to "Set and Set Theory"). Cantor's (1880) Über unendliche linear Punktmannigfaltigkeiten is one of the earliest uses. It contains the notation $\{\cdots\}$ for a set and introduces the term "Vereinigung" for the union (symbol ${\cal M}$) and Durchschnitt for the intersection (symbol ${\cal D}$) Schröder (1877) had previously used the notation $+$ and $\times$ for union and intersection. This was changed into the presently used symbols $\cup$ and $\cap$ by Peano (1888) to avoid confusion with addition and multiplication in algebra. Calcolo geometrico secondo l'Ausdehnungslehre di H. Grassmann (1888). In this brief work by Schröder (37 pages) the mathematical logic is developed that forms the introduction of the present book. I found it useful to replace the logical symbols $\times,+,A_1,0,1$ used by Schröder by the symbols $\cap,\cup,-A$, ⚪, ⚫ in order to avoid a possible confusion between symbols from logic and from mathematics (a possible confusion noted by Schröder himself). I also introduced the logical symbols $\lt$ and $\gt$, although not strictly necessary...<|endoftext|> TITLE: Lebesgue measure theory applications QUESTION [20 upvotes]: I'm looking for reasonably simple examples of applications of Lebesgue measure theory outside the measure theory setting. I give an example. Theorem: Let $X$ be a differentiable submanifold of $\mathbb{R}^n$ with codimension $\geq 3$. Then $\mathbb{R}^n\setminus X$ is simply conected. Proof. Let $\alpha:S^1\to \mathbb{R}^n\setminus X$ be a closed $C^1$ curve. We want to show that there exists a point $p$ outside $X$ s.t. a linear homotopy between $\alpha$ and $p$ can be contructed. Well, define $F:\mathbb{R}\times S^1\times X\to \mathbb{R}^n $ by $$ F(t,s,x)= (1-t)\alpha(s)+tx. $$ Note that, 1) $F$ collects all the bad lines, i.e., the lines connecting $\alpha(s)$ and $x\in X$. 2) $F$ is $C^1$. 3) $\dim(\mathbb{R}\times S^1\times X)\leq n-1$. So, by Sard's theorem, the set $F(\mathbb{R} \times S^1\times X)$ has zero Lebesgue measure, and therefore its complement is non-empty. This easily implies the result. Also as an example we have the the weak form of the Whitney immersion theorem, for which one can use the same kind of argument in the proof. I want to know more "simple" applications of the type the above mentioned, in areas other than measure theory. But not too complicated ones! Sorry if this question is too basic. REPLY [7 votes]: I do not know what the protocol is when you give another answer; this has nothing to do with my previous example. The most spectacular applications of measure theory that I know come from Margulis' work. For example, suppose $\Gamma \subset SL_3(\mathbb R)$ is a discrete subgroup with compact quotient. Then Margulis shows that every non-trivial normal subgroup of $\Gamma $ has finite index in $\Gamma$. The proof uses measure theory ( and a lot else besides) in a serious way. His proof that such a $\Gamma$ is arithmetic also uses ergodic theory (and measure theory). These purely "algebraic" statements were proved by use of measure theory.<|endoftext|> TITLE: Is there an 11-term arithmetic progression of primes beginning with 11? QUESTION [29 upvotes]: i.e. does there exist an integer $C > 0$ such that $11, 11 + C, ..., 11 + 10C$ are all prime? REPLY [2 votes]: There seems to be an interest in $p$-app ($p$-long prime arithmetic progressions) which start with prime $p$. Thus, I've decided to put some $\LaTeX$ sweat into more information, as elementary as it is. A $p$-app is an arithmetic progression $\ (p+t\!\cdot\! d\ :\ t=0\ldots n\!-\!1)\ $ such that all its terms are primes, and integer $\ d>1.\ $ Then a simple theorem assures us that $$ \prod \mathbf P(p-1)\ |\ d $$ Where $\ \mathbf P(x)\ $ is the set of all primes $\ \le\ x.$ Given any prime $p$ we would like to find all we can about the $p$-app's (do they exist, etc.). Let $\ D:=\prod\mathbf P(n_1)\ $ as above. In particular, we would like to know everything about $\ r\ $ such $\ d=r\cdot D,\ $ where $\ d\ $ is the difference of arbitrary $p$-app. Thus, let $\ q\ $ be any prime not in the said $p$-app. Then $$ d\not\equiv 0\mod q\quad \Longrightarrow\quad \forall_{0\le k< p}\quad p+k\cdot d\ \not\equiv 0 \mod q $$ Notations:   Let $\ \ /_n\ \ $ be the $\mod n\ $ division by non-$0$ integers which are not factors of $\ n,\ $ with the division value in $\ \{0\ldots\ n\!-\!1.$ Thus, $$ d\not\equiv 0\mod q\quad \Longrightarrow\quad \forall_{0\le k< p}\quad k\ \ne (-p)\,\ /_q\,\ d $$ In other words, looking at the division remaining options (and under the established notation), THEOREM 1 $$ d\not\equiv 0\mod q\quad \Longrightarrow\quad p\ \ \le\ \ (-p)\,\ /_q\,\ d\ \ <\ \ q $$ This significantly reduces the number of options for $\ d\ $ when it is (easily!) applied to all primes $\ q>p\ $ at the same time (in the same computer program). APPLICATIONS Let prime $\ p>3\ $ be a younger sibling of prime $ q:=p+2.\ $ Then $$ (-p)\,\ /_q\,\ d\,\ =\,\ -\!1\,\ \mbox{or}\ -\!2 $$ $\qquad$ (Thus, even one prime $\ q\ $ contributes to the computational savings). Let $\ p:=11\ $. The (see above) $\ D=2\cdot 3\cdot 5\cdot 7=210,\ $ and let $\ d := r\cdot D\ $ be the respective difference of an arbitrary $11$-app. Then, by the above THEOREM 1, when $\ d\not\equiv 0 \mod 13\ $ (i.e. $\ d\not\equiv 0 \mod 13)\ $ then $$ 2\,\ /_{13}\ d\ \equiv\ -\!1\,\ \mbox{or}\ -\!2\quad \mod 13 $$ or $$ d\ \equiv\ -\!1\,\ \mbox{or}\ -\!2\quad \mod 13 $$ Since $\,\ d = r\cdot D = r\cdot 210 \equiv 2\cdot r\,\ \mod 13,\ $ and allowing for the divisibility $13\,|\,r,\ $ we finally obtain, $$ r\,\ \equiv\,\ 0\ \mbox{or}\ 6\ \mbox{or}\ 12 \mod 13 $$ This reduces the amount of computation $\frac 3{13}\ $ time. Then, taking into account another prime, $\ q:=17,\ $ we reduce the computation time again $\ \frac 7{17}\ $ times, or for a total saving $$ \frac 3{13}\cdot\frac 7{17}\ =\ \frac {21}{221}$$ times (more then ten times faster), etc.<|endoftext|> TITLE: The soccer splitting problem in arbitrary commutative ring QUESTION [7 upvotes]: There's a folklore problem: Let $x_1, \cdots, x_{23} \in \mathbb{Z}$ be the weights of $23$ soccer players. Now Master Yoda want's to form two soccer teams with $11$ players each. Turns out for any $1 \leq i \leq 23$, one can partition $\{1, \cdots, n \} - \{ i \}$ into two disjont sets $A, B$ with $|A| = |B| = 11$ such that $\displaystyle \sum_{k \in A} x_k = \sum_{k \in B} x_k$. Prove that all numbers must be equal. The solution is well known and is not very hard for $\mathbb{Z}$. I'm wondering replacing $\mathbb{Z}$ by which commutative ring with unit $R$ makes the problem false. If $R = \mathbb{Q}$, then it's also same as $\mathbb{Z}$ (and the answer is affirmative), just multiply everything by the LCM of the numerators to reduce it to the case $R = \mathbb{Z}$. If $R = \mathbb{R}$, then also the problem is true, but you need a lemma by Dirichlet (which is proven by PHP) to reduce it to the case $R = \mathbb{Z}$. If $R = \mathbb{C}$, then also the problem is true. Because if $\displaystyle \sum_{k \in A} z_k = \sum_{k \in B} z_k \Rightarrow \sum_{k \in A} \text{Re}(z_k) = \sum_{k \in B} \text{Re}(z_k)$, and by the previous one $R = \mathbb{R}$ applied to the real components, you get $\text{Re}(z_i) = \text{Re}(z_j)$ for all $i, j$. Similarly you prove the imaginary components are same, so all numbers are same. If $R = \mathbb{Q}[x], \mathbb{C}[x], \mathbb{R}[x], M_{m,n}(\mathbb{Q}), M_{m,n}(\mathbb{Z}), M_{m,n}(\mathbb{C}), M_{m,n}(\mathbb{R}) $, even then the problem is true since you can look at the problem "component wise" and reduce it to the above cases. However I have no idea whether the problem is true when $R = \mathbb{Z}_p$ for some prime $p$ or in some other rings (ring of rational functions etc). Is it true for all rings, or are there some rings for which this problem doesn't hold ? If it's false for some rings, are there any characterizations for such rings ? REPLY [2 votes]: As already indicated, this is a result depending only on the underlying group (rather than ring) and you can find in my earlier question, MO 105400, a comment from me that points to: Martin, G. A. (1988). A class of Abelian groups arising from an analysis of a proof. The American Mathematical Monthly, 95(5), 433-436. JSTOR; Sci-Hub. Here are the relevant portions:<|endoftext|> TITLE: The Stacks project QUESTION [8 upvotes]: I have a question concerning the admirable Stacks Project. Which comparable projects are there: approach-wise: "an open source textbook on algebraic stacks and the algebraic geometry that is needed to define them", i.e. projects as "an open source textbook on X and the Y that is needed to define (and understand) X" technical-wise [Side question: which technical framework(s) is the Stacks Project based on?] tag-system-wise (using the same tag system as the Stacks Project) tag-wise (with a significant number of tags shared with the Stacks Project)? The question concerns also the possible interoperability of several such projects. REPLY [8 votes]: tag-system-wise Anyone can use Gerby, with a similar tag system. The actual tag assignment isn't dealt with by Gerby, you can set up your own conventions if you like. There might be some minor things you'll have to change if you decide to do so. So far I'm not aware of anyone actually doing this, except for the work-in-progress Kerodon, which will be Gerby applied to the works of Jacob Lurie. There's no fixed date to go live yet, maybe in September we'll have a more definite idea about this. Observe that the text will not be open-source (there's no need for this in the Gerby system). tag-wise No projects share tags with the Stacks project (as there are no other projects using a similar tag system), although it is often cited. It's also not quite desirable to "share" tags, as nothing ensures uniqueness of tags then. technical-wise This has been answered before, but feel free to ask follow-up questions (which probably would be better done via e-mail, or chat), as I'm the person responsible for the majority of the implementation (with help from @RaymondCheng).<|endoftext|> TITLE: The differential of the Gauss normal map from a Lie algebraic view point QUESTION [8 upvotes]: Let $S\subset \mathbb{R}^3$ be a smooth surface with the Gauss normal map $N:S\to S^2$. Then for every $x\in S$, the differential $(dN)_x:T_xS\to T_{N(x)}S^2$ can be considered as an endomorphism of the tangent space $T_xS$ since $T_xS$ is parallel to $T_{N(x)}S^2$. So from now on, without any ambiguity and in a unique way, we count $dN$ as an endomorphism of the tangent bundle $TS$ of the surface $S$. So $dN$ defines a linear operator $$dN:\chi^{\infty}(S)\to \chi^{\infty}(S)$$ where $\chi^{\infty}(S)$ is the space of all smooth vector fields on $S$. Under which geometric conditions on $S$ this operator preserve the Lie bracket of $\chi^{\infty}(S)$? Under which conditions on $S$, the range $dN(\chi^{\infty}(S))$ of this operator is a Lie algebra? Of course we can ask the same question for every codimension $1$ submanifold $S$ of $\mathbb{R}^n$. REPLY [8 votes]: ADDED: I checked my calculation of the first displayed equation, and it appears to be correct. If so, it looks to me that it already implies that $A$ is either the identity or zero and therefore the second paragraph isn't even needed. The shape operator is an example of a bundle map $A: S\rightarrow S$, where $S$ is a $d$-dimensional smooth manifold. If it preserves the Lie algebra structure of vector fields, then $[AX,AY] = A[X,Y]$ for any smooth vector fields $X$ and $Y$. For any $p \in S$ choose local coordinates $x = (x^1, \dots, x^d)$ such that $p$ is at the origin. For any $1 \le i,j,k \le d$, let $X = \partial_i$ and $Y= x^j\partial_k$. A straightforward calculation shows that at the origin, $$ [AX,AY] =A^i_jA_k^p\partial_p\text{ and }A[X,Y] = \delta_i^jA_k^p\partial_p. $$ Therefore, at each point in $S$, $A^2 = A$ and therefore $A$ is diagonalizable with eigenvalues $0$ and $1$. Now suppose $A$ is neither the identity nor $0$ at a point $p \in S$. Let $X$ and $Y$ be smooth vector fields that are nonzero at $p$ and satisfy $AX = 0$ and $AY = Y$ at $p$. Then $A[X,Y] = [AX,AY] = 0$ at $p$. Let $f$ be a smooth function such that $Xf(p) \ne 0$. Then on one hand, $$ [AX,A(fY)] = 0. $$ On the other hand, at $p$, $$ [AX,A(fY)] = A[X,fY] = A((Xf)Y + f[X,Y]) = (Xf)AY + fA[X,Y] = (Xf)Y \ne 0. $$ This is a contradiction and therefore $A$ is either the identity or $0$.<|endoftext|> TITLE: Symmetric tensor of Lie algebra of $su(N)$ QUESTION [6 upvotes]: I am interested in knowing the exact form of the anti-commutation of two generators of $su(N)$ lie algebra. Let us denote $T^a$ to be the generator of $su(N)$ lie algebra in the defining representation. Since the number of generators is $n^2-1$, the index takes value in $a=1, ..., n-1$. The normalization of $T^a$ is $$Tr(T^aT^b)=\frac{1}{2} \delta^{ab}$$ The anti commutation of two such generators is $$\{T^a, T^b\}=T^a T^b+T^bT^a=\frac{1}{N}\delta^{ab}I+d^{abc}T^c$$ where $d^{abc}$ is a totally symmetric tensor in all the three indices. In https://pdfs.semanticscholar.org/1101/914fc76a36d4fb0ab0022f8c4ec6295d8d1f.pdf, it was shown that $$d^{abc}d^{abh}=\frac{N^2-4}{N}\delta^{ch}$$ where the repeated indices are to be summed over. In the above, we contract over two indices from each $d$-tensor. My questions are: 1) Is there a simple expression for $$d^{abc}d^{agh}$$ where we only contract one index for each $d$-tensor? (in terms of $N$) 2) Is there a simple expression for $d^{abc}$ itself? (in terms of $N$) REPLY [2 votes]: It is only a partial answer so far: to the best of my understanding, the fully symmetrized version of $d^{abc}d^{agh}$ can be expressed via Kronecker deltas, see Example 5.1 here.<|endoftext|> TITLE: Clebsch–Gordan decomposition for $\mathrm{SU}(2)$, in indices QUESTION [5 upvotes]: Let $\pi_m$, $m \geq 0$, be the unitary irreps of $\mathrm{SU}(2)$. The Clebsch–Gordan decomposition then gives that $$ \pi_m \otimes \pi_n = \bigoplus_{k=0}^{\min(m,n)}\pi_{m+n-2k}.$$ But suppose I want to think of this decomposition as matrices. Evaluating at a point $x \in \mathrm{SU}(2)$, on the left I have $$ (\pi_m(x))_{ij} (\pi_n(x))_{pq}.$$ How do the indices $i$, $j$ and $p$, $q$ correspond to the indices on the big matrix on the right? REPLY [3 votes]: You can find such a formula with indices here: https://en.wikipedia.org/wiki/Wigner_D-matrix#Kronecker_product_of_Wigner_D-matrices,_Clebsch-Gordan_series<|endoftext|> TITLE: Structures of the space of neural networks QUESTION [19 upvotes]: A neural network can be considered as a function $$\mathbf{R}^m\to\mathbf{R}^n\quad \text{by}\quad x\mapsto w_N\sigma(h_{N-1}+w_{N-1}\sigma(\dotso h_2+w_2\sigma(h_1+w_1 x)\dotso)),$$ where the $w_i$ are linear functions (matrices) $\mathbf{R}^{d_{i-1}}\to\mathbf{R}^{d_i}$, the $h_i\in \mathbf{R}^{d_i}$, with $d_0=m$ and $d_N=n$, and $\sigma\colon\mathbf{R}\to\mathbf{R}$ is a non-linear function (e.g. sigmoid), which in the formula above must actually be understood as $\sigma\oplus\dotsb\oplus\sigma$ an appropriate number of times. There are several studies in the literature (e.g. https://doi.org/10.1016%2F0893-6080%2891%2990009-T) proving that the set of such functions (for fixed $N$ and $\{d_i\}$) is dense in other function spaces, such as measurable-function or continuous-function spaces. Other studies focus on how well functions in this set approximate functions in other sets, according to various measures. I would like to know, instead, what kind of mathematical structures the set of such functions enjoys, either for fixed or for variable $N$ and $\{d_i\}$. For example, is it a vector space? (answer seems to be yes if the $\{d_i\}$ aren't fixed) Is it a ring under function composition or under some other operation? Is it a convex set? – And similar questions. I'm thankful to anyone who can provide some literature about such questions. Update – example: this brilliant study by Petersen, Raslan, Voigtlaender https://arxiv.org/abs/1806.08459 answers the question about convexity. REPLY [8 votes]: In information geometry, people study structures of Riemannian manifolds with dual affine connections on sets of neural networks. (The metric measures how close neural networks are in their input-output behaviour.) Riemannian geometry can then be used to study learning algorithms (like gradient descent methods). Here are a couple of references, more can be found by searching for the tags "information geometry" and "neural networks". S. Amari. Information geometry of the EM and em algorithms for neural networks. Neural Networks 8 (1995), 1379-1408. S. Amari, K. Kurata and H. Nagaoka. Information geometry of Boltzmann machines. IEEE Transactions on Neural Networks, vol. 3, no. 2, (1992), 260-271. S. Amari. Information Geometry of Neural Networks — An Overview —. In: Ellacott S.W., Mason J.C., Anderson I.J. (eds) Mathematics of Neural Networks. Operations Research/Computer Science Interfaces Series, vol 8. Springer, Boston, MA, 1997. (link to Springer website, paywall) I don't know about recent developments, all the references above are well before the deep learning revolution.<|endoftext|> TITLE: Coverage of balls on random points in Euclidean space QUESTION [6 upvotes]: We have n points randomly distributed in a d-dimensional unit hypercube. We randomly sample k of those points and center a ball with radius r on each of those k points. Does there exist an estimate of the radius r such that r is the smallest radius expected to cover all the points with those k possibly overlapping balls? REPLY [3 votes]: In an email, Doug Jungreis gave me the following solution to this question. His solution illustrates the case where $n=10^7$, $k=500$, and $\epsilon = 1/100$. It seems to work well in the datasets I'm looking at (even where $d$ is small). I'm quoting his email directly instead of trying to reformat the notation and risk a mistake. Doug makes two independence assumptions described by Greg Kuperberg: The first independence assumption is that if $p$ is a typical point in the cube, then whether one ball covers it has little do with whether another ball covers it. Obviously this assumption is bad in when $d$ is small since points at the end are harder to reach than points in the middle. But easily by the time you get to 50 dimensions, almost everything is comparably in the middle. The second independence assumption is that whether one point is covered has little to do with whether another point is covered. Again, this assumption is bad when $d$ is small, but it's a good assumption by the time you get to $d=50$ with the given parameters. Here is Doug's solution: Here is an estimate if $d$ is large. I'll make the same independence assumptions as Greg. If you choose 2 values $x,y$ in the unit interval, then their expected distance squared is $\int_0^1 \int_0^1 (x-y)^2 dx\, dy = 1/6$, and their expected distance^4 is $\int_0^1 \int_0^1 (x-y)^4 dx\, dy = 1/15$ (if I didn't make a mistake). So the distance squared has mean 1/6 and variance $1/15 - (1/6)^2 = 7/180$. Then the distance squared between two points in the $d$-hypercube has mean $d/6$ and variance $7d/180$, and assuming $d$ is reasonably large, it is very close to normally distributed. You want a 1/100 chance that any of the 10 million points misses all 500 spheres, so a $1/10^9$ chance that a particular point misses all 500 spheres, so a $(1/10^9)^{1/500} = .9594$ chance that a particular point misses a particular sphere. So you want a .9594 chance that two particular points have distance^2 greater than $r^2$. The value .9594 corresponds to 1.744 standard deviations on a normal distribution, so $r^2$ should be roughly 1.744 standard deviations below the mean, i.e., $r^2 = (d/6) - 1.744 \sqrt{7d/180}$.<|endoftext|> TITLE: KL divergence and mixture of Gaussians QUESTION [11 upvotes]: Do we have an exact formula to compute the KL divergence between 2 mixtures of Gaussians (i.e convex combinations of a finite number of Gaussian distributions)? If not exactly known, are there good upperbounds that are known for this quantity? REPLY [8 votes]: There is no closed form expression, for approximations see: Lower and upper bounds for approximation of the Kullback-Leibler divergence between Gaussian mixture models (2012) A lower and an upper bound for the Kullback-Leibler divergence between two Gaussian mixtures are proposed. The mean of these bounds provides an approximation to the KL divergence which is shown to be equivalent to a previously proposed approximation in: Approximating the Kullback Leibler Divergence Between Gaussian Mixture Models (2007)<|endoftext|> TITLE: Reference request: Can iterated torus links be mutated? QUESTION [6 upvotes]: I believe that most iterated torus links cannot be changed non-trivially by a Conway mutation, as follows. If you look at the JSJ decomposition of the double-branched cover, then each satellite torus that appears in the iterated torus construction appears. Thus, you cannot have an essential torus that intersects these satellite tori non-trivially. If you look at the double-branched cover of a Conway mutation sphere, you get a torus; if the mutation is non-trivial, then the torus must be essential, and so it lives in one of the pieces of this JSJ decomposition. Then you analyze the purported essential torus a little more (essentially in the torus knot case) and see that it can't exist. There's a bunch of details in fleshing out this argument. The JSJ decompositions of link complements are actually kind of intricate; I think this is the canonical reference: JSJ-decompositions of knot and link complements in the 3-sphere https://arxiv.org/abs/math/0506523 But that paper doesn't mention mutation. I think this argument should be standard; is it written anywhere? In case it matters, I'm interested in the case of links of algebraic singularities. The paper is pretty algebraic as a whole, and I'd really rather not take an extended detour into JSJ decompositions. REPLY [3 votes]: I don't think that it's that difficult to deduce this for prime iterated cable links. The point being that the preimage of a Conway sphere is an essential torus, so must be isotopic into a Seifert piece of the JSJ decomposition. But this torus must be saturated (foliated) by the Seifert fibers, hence the involution, which restricts to the elliptic involution on the torus, must send the fiber to its reverse. However, for a prime iterated cable link, one may see from the JSJ decomposition (described in Budney) that the boundary tori of the link complement are foliated by non-meridinal Seifert fibers, and that the fibration pulls back to the 2-fold branched cover. Hence the involution preserves the orientation of the fibers on each Seifert piece that meets the fixed point set, a contradiction. One issue with this argument is that not every Seifert-fibered space admits a unique Seifert-fibering, and so one might have one fibering for which the involution reverses orientation, and another for which it preserves it. Examples of closed manifolds with non-unique Seifert fiberings are $S^3$, lens spaces, $T^3$, and a few others. But for irreducible manifolds with non-trivial JSJ decomposition, I think the Seifert-fibering is unique. It seems that these cases have been dealt with mostly in Appendix A.4 of Bonahon-Siebenmann. The only case where this argument fails is for non-prime iterated torus links, such as the key chain link. The issue is here that the Seifert fibering has meridinal slopes on one boundary component, so doesn't pull back to a Seifert fibering of the double branched cover. I still think that you can make the argument work here: the 2-fold cover of non-prime graph manifold links admits a graph manifold structure in the geometers sense: there exists a decomposition into submanifolds with $S^1$-actions, so that the actions commute on the torus interfaces. But now the action might have fixed points, so is not a Seifert fibering. These structures occur on manifolds which are collapsed in the sense of Cheeger-Gromov. Anyway, I think the Conway sphere/torus argument still gives a contradiction in this case, since the 2-fold cover still reverses Seifert fiberings. One just has to check that the Conway torus is still saturated, which I think can be done if it is incompressible.<|endoftext|> TITLE: Are there known examples of sets whose power set is equal in size to power set of larger sets only in absence of choice? QUESTION [8 upvotes]: The question of existence of sets $x,y$ such that $$|x|<|y| \wedge |P(x)|=|P(y)|$$ is known to be independent of $\text{ZFC}$! But are there known examples of sets fulfilling the above condition that necessitates violation of choice? REPLY [11 votes]: Let me add some comments to Asaf's answer. We are looking for situations with $|X|<|Y|$ and $|\mathcal P(X)|=|\mathcal P(Y)|$ (and choice fails). Asaf rightly identifies that a very natural way of approaching this question is via the (soft) $\mathsf{ZF}$ result that whenever $|S|\le^*|T|$, that is, if $S$ is empty or there is a surjection from $T$ onto $S$, then $|\mathcal P(S)|\le|\mathcal P(T)|$. This indicates that it suffices to look for $X,Y$ satisfying $|X|<|Y|$ and $|Y|\le^*|X|$. It may be this is the only way of creating "combinatorial" examples that explicitly exploit a failure of choice. (I understand this opinion is currently rather vague.) Now, that $|Y|\le^*|X|$ means that $Y$ is a quotient of $X$, we can think of $Y$ as the set $X/{\sim}$ of equivalence classes of elements of $X$ under some equivalence relation. So, we are looking for examples of sets $X$ and equivalence relations $\sim$ on $X$ such that $|X|<|X|/{\sim}$. (Note this is only possible if choice fails.) Now, there are many relatively concrete examples of this situation, since we have studied quotients of $\mathbb R$ for a long time. For instance, $\sim$ could be Vitali's equivalence relation $E_0$, in a model where all sets of reals have the Baire property. Actually, it is quite natural to concentrate on quotients of $\mathbb R$: It is still open whether $\mathsf{CH}(X)$, that is, the statement that there are no sets of intermediate cardinality between $X$ and $\mathcal P(X)$, implies that $X$ is well-orderable. It is known that if $\mathsf{CH}(X)$ but $\mathcal P(X)$ is not well-orderable, then $\mathsf{CH}(\mathcal P(X))$ fails rather badly. In the concrete case that $\mathsf{CH}=\mathsf{CH}(\omega)$ holds, but $\mathbb R$ is not well-orderable, this tells us there are many sizes between the cardinalities of $\mathbb R$ and its power set. In concrete situations, we actually find many quotients of $\mathbb R$ of strictly larger size. We have a big advantage here since, in natural scenarios, the cardinality of $\mathbb R/E_0$ is a successor of $\mathbb R$, and many natural equivalence relations $E$ on $\mathbb R$ carry enough information that one can explicitly see that $|\mathbb R/E_0|\le|\mathbb R/E|$. In fact, we know we can embed complicated partial orderings in the family of quotients of $\mathbb R$ by Borel equivalence relations, ordered by Borel reducibility. In natural situations, in particular in models of determinacy, we can replace "Borel reducibility", that is, "Borel cardinality" via Borel injections by actual cardinality. And there is more. A rather concrete way of having $\mathbb R$ fail to be well-orderable is that in fact $\aleph_1\not\le|\mathbb R|$. This gives us that $|\mathbb R|<|\mathbb R\cup\omega_1|$. But (provably in $\mathsf{ZF}$) $\aleph_1\le^*|\mathbb R|$, so $|\mathcal P(\mathbb R\cup\omega_1)|=|\mathcal P(\mathbb R)|$. (By the way, if $\aleph_1\not\le|\mathbb R|$, then $|\mathbb R|<|[\mathbb R]^{\aleph_0}|$, another example suggested by Asaf in comments.) Some quick references: For embeddings of complicated partial orderings in the Borel reducibility poset, see for instance MR3549382. Kechris, Alexander S.; Macdonald, Henry L.. Borel equivalence relations and cardinal algebras. Fund. Math. 235 (2016), no. 2, 183–198. (Also available here.) For some of the remarks about determinacy and Vitali's equivalence relations, see here and MR2777751 (2012i:03146). Caicedo, Andrés Eduardo; Ketchersid, Richard. A trichotomy theorem in natural models of $\mathsf{}^+$. In Set theory and its applications, 227–258, Contemp. Math., 533, Amer. Math. Soc., Providence, RI, 2011. (Also available here.) For results and references on $\mathsf{CH}(X)$ vs. well-orderability of $X$, see MR1954736 (2003m:03076). Kanamori, A.; Pincus, D. Does GCH imply AC locally?. In Paul Erdős and his mathematics, II (Budapest, 1999), 413–426, Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, 2002. (Available here.)<|endoftext|> TITLE: Good papers on stochastic differential equations with applications in finance QUESTION [7 upvotes]: I recently completed reading the book "Stochastic Differential Equations" by Bernt Oksendal which is the first time ever I was exposed to the topic. Now I am interested in pursuing research ( Ph.D.) SDEs and its applications in finance and I would like some help finding some recent papers related to or useful when doing research. I have already looked at some papers on MathSciNet by the same author but I would much appreciate if anyone can suggest some journals or papers/ articles that are relevant and useful in the current times. Thank you in advance! REPLY [5 votes]: As indicated in the comments, the field is very wide, but I understand from the comment of the OP to zab's answer that there is a specific interest in the more narrow subtopic of applications of fractional Brownian motion to quantitative finance. Here are some overviews: Fractional Brownian Motion in Finance (2003) Fractional Brownian Motion and applications to financial modelling (2011) A note on the use of fractional Brownian motion for financial modeling (2013) To get a feel for recent research on this topic, here are some arXiv contributions from the last year or so: Modeling the price of Bitcoin with geometric fractional Brownian motion Pricing European option with the short rate under Subdiffusive fractional Brownian motion regime Option Pricing Models Driven by the Space-Time Fractional Diffusion: Series Representation and Applications The evaluation of geometric Asian power options under time changed mixed fractional Brownian motion Series representation of the pricing formula for the European option driven by space-time fractional diffusion Hedging in fractional Black-Scholes model with transaction costs The 2013 paper referred to above notes that the application of fractional Brownian motion to financial modeling still has several unsolved problems of a foundational nature, so this might a fruitful area of research for someone entering the field (it seems a less mature topic than others).<|endoftext|> TITLE: Counting promenades on graphs QUESTION [6 upvotes]: Let a "promenade" on a tree be a walk going through every edge of the tree at least once, and such that the starting point and endpoint of the walk are distinct. What we mean by isomorphic promenades should be clear (in particular, the underlying trees have to be isomorphic). What is the total number $N(l,r)$ of promenades (up to isomorphism) of length $l$ with $r$ distinct edges (letting the underlying tree vary)? REPLY [3 votes]: In case you find it useful here is a simple computer enumeration for $\ell,r\le 20$ $$ \begin{array}{ c|r|r*{19}{r}} \ell\backslash r& \sum & 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20\\ \hline 1& 1& 1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 2& 1& 0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 3& 3& 1&1&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 4& 4& 0&1&2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 5& 12& 1&3&5&2&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 6& 22& 0&3&7&8&3&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 7& 61& 1&7&20&18&11&3&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 8& 122& 0&4&24&41&33&15&4&1&0&0&0&0&0&0&0&0&0&0&0&0\\ 9& 355& 1&15&69&106&93&47&19&4&1&0&0&0&0&0&0&0&0&0&0&0\\ 10& 765& 0&10&74&192&227&161&71&24&5&1&0&0&0&0&0&0&0&0&0&0\\ 11& 2243& 1&31&221&516&632&464&249&94&29&5&1&0&0&0&0&0&0&0&0&0\\ 12& 5020& 0&16&222&800&1334&1288&815&374&129&35&6&1&0&0&0&0&0&0&0&0\\ 13& 14951& 1&63&677&2260&3732&3665&2522&1290&530&163&41&6&1&0&0&0&0&0&0&0\\ 14& 34599& 0&36&655&3242&7080&8902&7325&4364&1992&736&211&48&7&1&0&0&0&0&0&0\\ 15& 103641& 1&127&2019&9282&20087&25322&21704&13836&7053&2903&986&258&55&7&1&0&0&0&0&0\\ 16& 246070& 0&64&1902&12578&35447&55860&57304&42202&23895&10967&4157&1301&321&63&8&1&0&0&0&0\\ 17& 741510& 1&255&5923&36592&101567&160201&167476&128416&77918&38912&16377&5734&1675&383&71&8&1&0&0&0\\ 18& 1800739& 0&136&5513&48097&170563&330313&409966&362207&244779&133712&61152&23814&7803&2131&463&80&9&1&0&0\\ 19& 5451731& 1&511&17206&140476&492009&953828&1197472&1083274&764487&443648&218859&92624&33690&10342&2663&542&89&9&1&0\\ 20& 13499887& 0&256&15879&180845&799195&1873028&2765328&2869554&2260169&1430391&758236&345860&136814&46726&13559&3296&641&99&10&1\\ \end{array} $$ I used the following basic recursive program in Pari/GP and let it run for 4 minutes : lessOrEqualReverse(P,r) = { \\ Is the promenade P lexicographically less than or equal to its reverse? local(B, x, y, z); B = Vec(0, r); z = 0; forstep(i = #P, 1, -1, y = B[x = P[i]]; if(y==0, B[x] = y = z = z+1); if(y < P[#P+1-i], return(-1)); \\ Return No as an answer if(y > P[#P+1-i], return(1)) \\ return Yes (less than) as an answer ); return(0) \\ return Yes (equal) as an answer } enumeratePromenades(P, A) = { \\ P is a promenade to extend, A holds a list of neighbours for every node in P local(z, Az); z = P[#P]; \\ z is the current last node in P if(z>1 && lessOrEqualReverse(P,#A)>=0, C[#P-1,#A-1]++); \\ Count the current promenade if it doesn't end with the starting node 1 and is preferred to its reverse if(#P>N, return); \\ Don't go any deeper if max length has been reached Az = A[z]; \\ Current neighbour list of last node for(i = 1, #Az, enumeratePromenades(concat(P, Az[i]), A) \\ Extend P with every known neighbour of its last node ); A[z] = concat(Az, #A+1); \\ Temporarily add a newborn node to z's neighbour enumeratePromenades(concat(P, #A+1), concat(A, [[z]])) \\ Extend P with the newborn node } N = 20; C = matrix(N,N); enumeratePromenades([1],[[]]) C The idea is to label the start point with 1, the next point on the promenade with 2, then the next with 1 or 3 according to whether the promenade goes back or reach a new third point, and so on. Then it is only a matter of counting corresponding finite sequences of integers. This seems to be highly related to https://oeis.org/A186952 because when -- unlike what you want -- you distinguish the start point from the endpoint and further allow both to possibly be the same, you get the sequence $\sum_r = 1, 1, 2, 4, 9, 20, 48, 113, 282, 689, 1767, 4435, 11616, 29775, 79352, 206960, 559906, 1482188, 4064235, 10901289, 30265366$ for $0 \le \ell \le 20$, which is the mentioned oeis sequence.<|endoftext|> TITLE: Hölder continuity for operators QUESTION [6 upvotes]: Let $x,y$ be positive real numbers then $$|\sqrt{x}-\sqrt{y}|=\dfrac{|x-y|}{\sqrt{x}+\sqrt{y}}=\sqrt{|x-y|}\cdot \dfrac{\sqrt{|x-y|}}{\sqrt{x}+\sqrt{y}}\leq 1\cdot |x-y|^{\frac{1}{2}}$$ we obtain $1/2$-Hölder continuity for the square-root. I would like to know if $x,y$ are positive Hilbert-Schmidt operators. Does it follow then that for some $C>0$ $$\left\lVert \sqrt{x}-\sqrt{y} \right\rVert_{HS} \le C \left\lVert x-y\right\rVert_{HS}^{\frac{1}{2}}.$$ Sounds natural, but on the other hand, it is less obvious to me how this should follow. One remark however is that if it would hold for finite-rank operators, then a density argument yields the claim. REPLY [13 votes]: Your proposed inequality doesn't even work on diagonal matrices. Let $x\in\mathbb M_n$ be diagonal with entries $x_i\geq 0$. Then $x^{1/2}$ is diagonal with entries $\sqrt{x_i}$. Thus $$ \|x^{1/2}\|_{HS}^2 = \sum_i x_i, \qquad \|x\|_{HS} = \Big( \sum_i x_i^2 \Big)^{1/2}. $$ Taking e.g. $x_i=1/\sqrt n$ we obtain $n/\sqrt n \leq C^2$ so $C\geq n^{1/4}$. The Powers-Stormer inequality (see section 4 of https://projecteuclid.org/euclid.cmp/1103842028) says that $$ \|x^{1/2} - y^{1/2}\|_{HS}^2 \leq \|x - y\|_1 $$ for all $x,y$ positive and Hilbert Schmidt. Here $\|\cdot\|_1$ is the trace-class norm (which takes the value $\infty$ if $x-y$ is not trace-class). In finite-dimensions, we have that $\|x\|_1 \leq n^{1/2} \|x\|_{HS}$ (reduce to the diagonal case and apply Cauchy-Schwarz) and so we see that $C=n^{1/4}$ works in general. REPLY [11 votes]: For the Hilbert-Schmidt norm, the inequality $\|X^{1/2}-Y^{1/2}\|_2 \le C\|X-Y\|_2^{1/2}$ is false in general. Consider for that the case of $n\times n$ positive definite matrices. Put e.g., $Y=0$, then asking for a $C>0$ amounts to requiring \begin{equation*} \text{tr}\,X \le C'\sqrt{\text{tr}\,X^2} \end{equation*} which in general cannot hold for a constant independent of the dimension of $X$. However, for all positive operators $X$ and $Y$, and for any unitarily invariant norm $|\!|\!| \cdot |\!|\!|$ \begin{equation*} |\!|\!|X^p - Y^p|\!|\!| \le |\!|\!| |X-Y|^p |\!|\!|,\quad 0 \le p \le 1. \end{equation*} Thus, in particular for the operator norm, this inequality implies $p$-Hölder continuity (with $C=1$).<|endoftext|> TITLE: Confusion in known result about moduli space of vector bundle of rank 2 degree 0 vector bundles over smooth curve of genus 2 QUESTION [5 upvotes]: Theorem: Let $X$ be a complete, non-singular algebraic curve of genus $2$. Let $U(2, \Theta)$ be the space of $S$-equivalence classes of semi-stable vector bundles of rank $2$ and degree $\Theta$. The group $\Gamma$ of elements of order $2$ in $J$ acts on $PH^0(J^1, L_\Theta^2)$ in a natural way; let $A$ be the associated projective bundle on $J/\Gamma \cong J$. Then $U(2, 0)$ is canonically isomorphic to $A$. In other words, $U(2, 0)$ is canonically isomorphic to the space of positive divisors on $J^1$ algebraically equivalent to $2\Theta$. Moduli of Vector Bundles on a Compact Riemann Surface Author(s): M. S. Narasimhan and S. Ramanan Source: Annals of Mathematics, Second Series, Vol. 89, No. 1 (Jan., 1969), pp. 14-51 Question: Is the projective bundle in theorem comes from some vector bundle on $J$. Means is it $P(E)$ of some bundle $E$ over $J$. REPLY [3 votes]: EDIT: After I posted my answer, I realized that you were very close to answering your own question---the vector bundle you want is the one whose fiber over $\alpha \in {\rm Jac}(C)$ is $H^{0}(\mathcal{O}(2\Theta) \otimes \alpha)^{\vee}.$ However, this is equivalent via Strange Duality to what I have written below. In general, if $C$ is a smooth projective curve of genus $g \geq 2$ $U_{C}(r,0)$ is the moduli space of S-equivalence classes of vector bundles of rank $r$ and degree $0$ on $C,$ and we fix a line bundle $L$ on $C$ of degree $g-1,$ the locus $$\Theta_{L} = \{E \in U_{C}(r,0) : h^{0}(E \otimes L) > 0 \}$$ is an ample divisor on $U_{C}(r,0).$ Recall that the fibers of the determinant map ${\rm det} : U_{C}(r,0) \to {\rm Jac}(C)$ are each isomorphic to ${\rm SU}_{C}(r)$, which parametrizes rank-$r$ vector bundles with trivial determinant. For any two line bundles $L,L'$ of degree $g-1$ on $C,$ the restrictions of $\Theta_L$ and $\Theta_{L'}$ to any fiber of ${\rm det}$ are linearly equivalent; the resulting ample line bundle on ${\rm SU}_{C}(r)$, which is denoted by $\mathcal{L},$ is called the $\textit{determinant bundle},$ and it generates ${\rm Pic}(SU_{C}(r)).$ For $k \geq 1$, the $\textit{Verlinde bundle}$ on the Jacobian ${\rm Jac}(C)$ corresponding to the rank-level pair $(r,k)$ (and our choice of $L$, typically a theta-characteristic) is the vector bundle $$\mathbf{E}_{r,k} := {\rm det}_{\ast}\mathcal{O}(k\Theta_{L})$$ For each $\alpha \in {\rm Jac}(C)$ the fiber of $\mathbf{E}_{r,k}$ over $\alpha$ is isomorphic to $$H^{0}({\rm SU}_{C}(r,\alpha),\Theta_{L}|_{{\rm SU}_{C}(r,\alpha)}) \cong H^{0}({\rm SU}_{C}(r),\mathcal{L}).$$ In the case where $g=r=2,$ the paper of Narasimhan-Ramanan that you are reading proves ${\rm SU}_{C}(2) \cong \mathbb{P}^{3}.$ In particular, the determinant bundle $\mathcal{L}$ on ${\rm SU}_{C}(2)$ is exactly the hyperplane bundle $\mathcal{O}(1)$ on $\mathbb{P}^{3}.$ The map ${\rm det}$ then realizes $U_{C}(2,0)$ as a $\mathbb{P}^3$-bundle over ${\rm Jac}(C)$; this is the projectivization of the rank-4 Verlinde bundle $\mathbf{E}_{2,1}.$<|endoftext|> TITLE: If a group $G$ has decidable word problem, must it have a decidable square problem? QUESTION [9 upvotes]: My question is a refinement of this one about 'efficient' construction of square elements: If the word problem for a (finitely generated, finitely presented) group is decidable, must the 'square problem' (given an element $g$ of $G$, is there an element $h$ with $g=h^2$?) also be decidable? If not, how 'nice' can $G$ be while still having an undecidable square problem? For instance, can $G$ be automatic? (It feels like there should be an argument based on the Dehn function that precludes this, but I'm not immediately seeing it.) Could it even be hyperbolic? REPLY [2 votes]: (This is not really an answer, rather a suggestion that the answer is probably negative.) It is known that for linear groups over integers $GL(n,\mathbb{Z})$ starting from $n=4$ the membership problem is undecidable. (It is decidable for $n=2$. The case $n=3$ is an open problem for what I know.) Consider a finitely generated subgroup $G\subset GL(n,\mathbb{Z})$ with undecidable membership. It is, of course, very much decidable whether $g\in G$ is a square in $GL(n,\mathbb{Z})$ (using a Jordan normal form). There may be several square roots $h_1,h_2,\dots,h_k$, and the problem is to find out if any of them belong to $G$. I suspect that this restricted version of the membership problem is still undecidable in general, although I do not have a proof of this.<|endoftext|> TITLE: Paradoxical Mathematical Objects Pending for Construction QUESTION [33 upvotes]: The possible properties and applications of some mathematical objects have been described far before their rigorous mathematical definition. Some of them even had a seemingly paradoxical description which prevented them from existing at the time. An iconic example is the imaginary number, $i$, which was originally described as $\sqrt{-1}$. However, further advancements in the field and exploring new perspectives towards the subject eventually led to a rigorous non-contradictory mathematical definition of $i$. A definition that serves as a formal introduction of the (already existing) concept to the mathematical literature while satisfying all of its required properties. Similar attempts have been conducted along the lines of providing a solid formal basis for seemingly paradoxical concepts such as negative probability or sets with a negative number of elements (see also multisets): Loeb, D., Sets with a negative number of elements, Adv. Math. 91, No. 1, 64-74 (1992). ZBL0767.05005. Another contemporary example of such paradoxical mathematical objects which are pending for a rigorous definition is the so-called "field with one element". It naturally appears in various occasions and relates different concepts together. For instance, a group can be seen as a Hopf algebra over the "field with one element". See also this related MO question for further information. Question. I am looking for more examples of such deep and useful mathematical concepts with contradictory descriptions which are not rigorously defined yet (or at least have no widely accepted formal definition). References to their possible applications are also welcome. Remark. I would like to emphasize the initial paradoxical description of the possible answers to this question. The important point is that the natural appearance of paradoxical objects often indicates the urgent need for a deep expansion of our mathematical worldview or an essential improvement of the foundations. REPLY [4 votes]: The Dirac delta function is a "function" which is zero almost everywhere but integrates to one. No function can possibly behave like that, but it nevertheless seemed like a useful notion. It was eventually made rigorous by the theory of distributions (or generalized functions). It shows up paradoxically if you try to figure out what the identity element for the convolution operator. You can quickly determine that no such element exist, but if it were to it would have such and such properties. Said properties are exactly the defining properties of the Dirac Delta.<|endoftext|> TITLE: Can we have an infinite sequence of decreasing cardinality all terms of which have equal sized power sets? QUESTION [9 upvotes]: Is the following consistent with $\text{ZF}$? There exists a set $S=\{x_1,x_2,x_3,...\}$ such that: $|x_{i+1}| < |x_i|$ $\forall m,n \in S (|P(m)|=|P(n)|)$ Where cardinality $``||"$ is defined after Scott's. REPLY [14 votes]: Yes. For silly reasons. Suppose that $X$ is a Dedekind-finite set, then $S(X)$, the set of all injective finite sequences from $X$ is also Dedekind-finite. Let $S_n(X)$ denote the subset of $S(X)$ of sequences whose domain is at least $n$. It is easy to see why $S_n(X)$ surjects onto $S(X)$. Simply erase the first $n$ coordinates. This means, by arguments from my answer to your last question, that $S(X)\leq^* S_n(X)$ for all $n$. Since those are Dedekind-finite sets, and the inclusion is strict, they form the wanted sequence. You can even be more clever than this, and for some chain in $\mathcal P(\omega)$ which has order type $\Bbb R$, define a sequence of order type $\Bbb R$ of Dedekind-finite sets, all of which have equipotent power sets. REPLY [7 votes]: This is consistent, at least under a rather tame large cardinal assumption. (One can also produce examples by manipulating Dedekind finite sets, but Asaf's answer addresses this. The answer here works even in the context of $\mathsf{DC}$.) For instance, see MR3612001. Conley, Clinton T.; Miller, Benjamin D. Measure reducibility of countable Borel equivalence relations. Ann. of Math. (2) 185 (2017), no. 2, 347–402. There, Clinton and Ben show that every basis for the nonmeasure-hyperfinite countable Borel equivalence relations under measure reducibility is uncountable. A basis here is a set $B$ such that given any such equivalence relation, there is one below it (in the ordering of measure reducibility) and in $B$. They explain how their arguments give stronger results, for instance, continuum-many pairwise incomparable such relations, or infinite strictly decreasing sequences. They also explain how in $L(\mathbb R)$, under the assumption of determinacy, their results actually give that the corresponding quotients $|\mathbb R/E|$ are decreasing in cardinality. I'm fairly certain that this result (the existence of such decreasing sequence of equivalence relations or, under determinacy, of such a decreasing sequence of cardinals) predates the Conley-Miller paper by more than 10 years, but had a bit of trouble tracking a specific reference. The point is that incomparability results in the theory of countable Borel equivalence relations are typically established via Baire category arguments, so they hold for true cardinality in, say, $L(\mathbb R)$ if determinacy holds, since then all sets of reals have the Baire property. Anyway, the other point is that all the relations under consideration here are such that $|\mathbb R|<|\mathbb R/E|$ (in fact, $|\mathbb R/E_0|<|\mathbb R/E|$, where $E_0$ is the Vitali equivalence relation), so we also have that the power sets of all these cardinals have the same size $2^{\mathfrak c}$, by the argument indicated in the answers to this previous question.<|endoftext|> TITLE: A conformal map whose Jacobian vanishes at a point is constant? QUESTION [13 upvotes]: Let $f:M \to N$ be a smooth weakly conformal map between connected $d$-dimensional Riemannian manifolds, i.e. $f$ satisfies $df^Tdf =(\det df)^{\frac{2}{d}} \, \text{Id}_{TM}$. Assume $d \ge 3$ and that $df=0$ at some point. Is it true that $f$ is constant? A proof for the Euclidean case, can be found in "Geometric Function Theory and Non-linear Analysis", by Iwaniec and Martin. Their proof uses the fact both $M,N$ are Euclidean. If I am not mistaken, a conformal map (in $d\ge3$) is determined by its 2-jet at a point, so it suffices to prove all the second derivatives at the point where $df$ vanishes are zero. (But maybe this wrong; it's possible that the "2-jet determination" only holds for conformal maps whose differentials are everywhere non-zero. I am not sure.) Edit: David E Speyer proved below that if $f$ is conformal and $df_{x_0}=0$ then all the derivatives of $f$ vanish at $x_0$ (for all orders). Thus, if the above statement about the two-jet-determination is true, the only solutions are indeed constants. Does anybody know whether or not this "two-jet-determination" hold in the case where we allow the Jacobian to vanish? REPLY [2 votes]: In a different direction, there are solutions where $f$ is $C^1$ and the metric $g_M$ is $C^0$. Let $\phi : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ be a $C^1$ function such that $\lim_{s \to 0^+} \phi(s) = \lim_{s \to 0^+} s \tfrac{\phi'(s)}{\phi(s)} = 0$. An example is $\phi(s) = \tfrac{1}{-\log s}$. (Or rather, that function near $0$ smoothed out somehow to not blow up at $s=1$.) Write $s$ for the function $\sum x_i^2$ on $\mathbb{R}^n$. Let $M=N=\mathbb{R}^n$ and define $f:M \to N$ by $f(\vec{x}) = \phi(s)\ \vec{x}$. Let $g_N$ be the standard metric and define $g_M = \tfrac{1}{\phi(s)^2} f^{\ast} g_N$ on $M \setminus \{ 0 \}$. I claim that $g_M$ extends continuously to $0$. We compute: The Jacobian of $f$ at the point $\vec{x}$ is $$J = \phi(s) \mathrm{Id} + \phi'(s)\ \vec{x} \ \vec{x}^T.$$ Using the identity $\vec{x}^T \vec{x}=s$, we compute: $$g_M = \tfrac{1}{\phi(s)^2} J J^T = \mathrm{Id} + \left(2 \tfrac{\phi'(s)}{\phi(s)} + \tfrac{s \phi'(s)^2}{\phi(s)^2} \right) \vec{x} \ \vec{x}^T .$$ We will show that the second term zpproaches $0$ as $\vec{x} \to 0$. Note that each entry of $\vec{x}^T \vec{x}$ is $O(|\vec{x}|^2) = O(s)$. So each entry in the second term is bounded by $$2 \tfrac{s \phi'(s)}{\phi(s)} + \left( \tfrac{s \phi'(s)}{\phi(s)} \right)^2$$ and we assumed $\tfrac{s \phi'(s)}{\phi(s)} \to 0$. Unfortunately, the hypothesis that $\tfrac{s \phi'(s)}{\phi(s)} \to 0$ gives $\tfrac{\phi'}{\phi} = o(s^{-1})$ so $\log |\phi| = o(\log s)$ and $\phi$ must approach $0$ more slowly than any $s^{\epsilon}$. So $f(x,0,0,\ldots,0)$ has derivative $0$ at $0$ but is larger than any $x^{1+\epsilon}$. So this method can't be pushed to $C^2$ functions.<|endoftext|> TITLE: What is the interpretation of the Gerstenhaber bracket? QUESTION [8 upvotes]: The homology of an $E_2$-algebra is a Gerstenhaber algebra. How precisely is the Gerstenhaber structure related to the $E_2$-structure? Obviously, the Gerstenhaber product is the commutative product that the $E_2$-product induces in homology. But what precisely is the interpretation of the Gerstenhaber bracket? It cannot be the commutator of the $E_2$-product because that would be zero in homology, right? Thanks for any hints. REPLY [13 votes]: I think the most transparent interpretation is by identifying $E_2$ algebras with brace algebras (which was proved by McClure and Smith). Namely, a brace algebra satisfies the relation $$ab - (-1)^{|a||b|}ba = (-1)^{|a|} d(a\{b\}) - (-1)^{|a|}(da)\{b\} + a\{db\},$$ where $a\{b\}$ is one of the brace operations (so, it witnesses the first-order commutativity of the multiplication). The Gerstenhaber bracket is the antisymmetrization of the first brace: $$[a, b] = a\{b\} - (-1)^{(|a|+1)(|b|+1)} b\{a\}.$$ Note that from the first equation you can see that $[-, -]$ is a $d$-closed operation, so it makes sense on homology. So, the Gerstenhaber bracket is a secondary operation: the product is commutative on homology, but the homotopy might not be (anti)symmetric.<|endoftext|> TITLE: Has the $E_8$-based generating function for squares numbers been proven? QUESTION [24 upvotes]: In his 2004 paper Conformal Field Theory and Torsion Elements of the Bloch Group, Nahm explains a physical argument due to Kadem, Klassen, McCoy, and Melzer for the following remarkable identity. Let $C \in \operatorname{Mat}_8(\mathbb{Z})$ denote the inverse of the Cartan matrix of $E_8$ and $(q)_n = \prod_{i=1}^n (1 - q^i)$. Then $$ \sum_{n=\mathbb{N}} \frac{q^{2n^2}}{(q)_{2n}} = \sum_{v \in \mathbb{N}^8} \frac{q^{vCv}}{\prod_{j=1}^8 (q)_{v_j}} $$ Nahm reports that, as of his writing in 2004, the result identity still not mathematically proven, because the arguments rely on not-yet-proven facts about conformal field theories, but that it has been checked to high orders. It's been 14 years. Has the above identity been proven in that time, either by firming up the foundations of conformal field theory or through other means? REPLY [24 votes]: This identity was actually proven 24 years ago in S.O. Warnaar and P.A. Pearce, "Exceptional structure of the dilute A 3 model: E8 and E7 Rogers-Ramanujan identities" J.Phys. A27 (1994) L891-L898 The proof essentially establishes a finite polynomial identity whose limit under one of the parameters becomes the desired series. I want to remark that Nahm calls this identity a conjecture in the arXiv version of the paper, but in the published version it correctly refers to the article above for a proof.<|endoftext|> TITLE: Solving system of bilinear equations QUESTION [6 upvotes]: Consider a collection of $m$ matrices $A_i$ of size $n\times n$, and a vector $b$ of size $m$. I want to solve the bilinear system $$\left\{ x^T A_i y = b_i : i = 1,\dots,m \right\}$$ in variables $x,y$. Is there an efficient way of doing this? This is both a theoretical and a practical question: the matrices $A_i$ I have in mind are sparse and have size in the thousands. I understand that if $m=1$ then one can consider a vector $z=[x y]$ and run a semidefinite solver on $\|z^T [0\; A_1;A_1^T\; 0]z-2b_1\|$; however the $m$ I have in mind is also in the thousands. REPLY [4 votes]: We have a system of $m$ bilinear equations in $\mathrm x, \mathrm y \in \mathbb R^n$ $$\begin{aligned} \mathrm x^\top \mathrm A_1 \,\mathrm y &= b_1\\ \mathrm x^\top \mathrm A_2 \,\mathrm y &= b_2\\ &\vdots\\ \mathrm x^\top \mathrm A_m \,\mathrm y &= b_m\end{aligned}$$ If matrices $\mathrm A_1, \mathrm A_2, \dots, \mathrm A_m$ are very sparse, perhaps it would not be utterly hopeless to use symbolic methods (e.g., Gröbner bases). However, we can use numerical methods. Note that $$b_i = \mathrm x^\top \mathrm A_i \,\mathrm y = \mbox{tr} \left( \mathrm x^\top \mathrm A_i \,\mathrm y \right) = \mbox{tr} \left( \mathrm y^\top \mathrm A_i^\top \,\mathrm x \right) = \mbox{tr} \left( \mathrm A_i^\top \mathrm x \mathrm y^\top \right) =: \langle \mathrm A_i, \mathrm x \mathrm y^\top\rangle$$ Let $\mathrm Z := \mathrm x \mathrm y^\top$. Hence, we have $m$ linear equality constraints in $\rm Z$ and a constraint on its rank $$\begin{aligned} \langle \mathrm A_1, \mathrm Z \rangle &= b_1\\ \langle \mathrm A_2, \mathrm Z\rangle &= b_2\\ &\vdots\\ \langle \mathrm A_m, \mathrm Z\rangle &= b_m\\ \mbox{rank} (\mathrm Z) &= 1\end{aligned}$$ Since the nuclear norm is a convex proxy for the rank, we solve the following convex program in $\rm Z$ $$\begin{array}{ll} \text{minimize} & \| \mathrm Z \|_*\\ \text{subject to} & \langle \mathrm A_1, \mathrm Z \rangle = b_1\\ & \langle \mathrm A_2, \mathrm Z \rangle = b_2\\ & \qquad\vdots\\ & \langle \mathrm A_m, \mathrm Z \rangle = b_m\end{array}$$ If the optimal solution is rank-$1$, then we have solved the original system of $m$ bilinear equations.<|endoftext|> TITLE: Cycle generating function of permutations with only odd cycles QUESTION [13 upvotes]: Let $\mathrm{ODD}(n)$ be the set of permutations in $\mathfrak{S}_n$ whose cycle lengths are all odd. It is known that $$ \#\mathrm{ODD}(n) = \begin{cases} ((n-1)!!)^2 &\textrm{ if $n$ is even}; \\ n\cdot((n-2)!!)^2 &\textrm{ if $n$ is odd}. \end{cases}$$ Set $\mathcal{C}_{\mathrm{ODD}}(n,t) = \sum_{\pi \in \mathrm{ODD}(n)} t^{\kappa(\pi)}$, where $\kappa(\pi)$ is the number of cycles of $\pi \in \mathfrak{S}_n$. Then it is easy to show that $$ \sum_{n\geq0}\mathcal{C}_{\mathrm{ODD}}(n,t) \cdot \frac{z^n}{n!} = \left(\frac{\sqrt{1-z^2}}{1-z}\right)^t$$ From this one can show that $$ \mathcal{C}_{\mathrm{ODD}}(n,t)= \sum_{i=0}^{\lfloor n/2\rfloor}\frac{1}{2^i\cdot i!}\prod_{j=0}^{n-1-2i}(t+j)\prod_{k=0}^{i-1}(t-2k)(n-2k)(n-2k-1)$$ In particular if we set $t:=2k$ to be an even integer, then $$ \mathcal{C}_{\mathrm{ODD}}(n,2k)=\frac{(n-1)!}{(2k-1)!} \cdot \sum_{j=0}^{k}(-1)^j\binom{k}{j}\prod_{i=0}^{k-j-1}(n+2i)(n+2i+1)\prod_{i=0}^{j-1}(n-2i)(n-2i-1)$$ Thus, for example $\mathcal{C}_{\mathrm{ODD}}(n,2)=2\cdot n!$ and $\mathcal{C}_{\mathrm{ODD}}(n,4)=4n\cdot n!$. Define the polynomial $P_k(x)$ by setting $$ \mathcal{C}_{\mathrm{ODD}}(n,2k) = \frac{(n-1)!\cdot k! \cdot 2^k}{(2k-1)!}\cdot P_k(n)$$ For example, $P_1(x)=x$; $P_2(x)=3x^2$; $P_3(x)=10x^3+5x$; $P_4(x)=35x^4+70x^2$; $P_5(x)=7\cdot 3^2(2x^5+10x^3+3x)$; $P_6(x)=3\cdot 7\cdot 11(2x^6+20x^4+23x^2)$; et cetera Conjecture 1: $P_k(x)$ is a polynomial of degree $k$ with nonnegative integer coefficients, with zero constant term, and which is odd if $k$ is odd and even if $k$ is even. Conjecture 2: We have, $$ \sum_{k \geq 0} \frac{P_k(x)}{(2k-1)!!} \cdot z^k = \frac{1}{2}\cdot\left( \frac{1+z}{1-z}\right)^x$$ (The constant term on the RHS is $1/2$ so take whatever convention for $P_0(x)$ or $(-1)!!$ for that to work.) With regard to Conjecture 2, note that from the above we have $$ \sum_{n\geq0}\mathcal{C}_{\mathrm{ODD}}(n,2k) \cdot \frac{z^n}{n!} = \left(\frac{1+z}{1-z}\right)^k$$ Question: Are these conjectures correct? Are these cycle generating functions studied somewhere? REPLY [9 votes]: Conjecture 2 follows from setting $t=2k$ in the formula $$ \sum_{n\geq 0}\mathcal{C}_{\mathrm{ODD}}(n,t)\cdot\frac{z^n}{n!} =\left(\frac{\sqrt{1-z^2}}{1-z}\right)^t. $$ It then follows easily that $$ \sum_{j\geq 0}P_k(j)x^j = \frac{(2k-1)!!\,x(1+x)^{k-1}}{(1-x)^{k+1}}.\ \qquad (1) $$ By Theorem 3.2 of http://math.mit.edu/~rstan/papers/cycles.pdf we see that all the zeros of $P_k(x)$ are purely imaginary, which implies that $P_k(x)$ has nonnegative coefficients and is either even or odd, depending on the parity of $k$. Clearly also from (1) $P_k(0)=0$. I don't see immediately from (1) why $P_k(x)$ has integer coefficients. Let me also remark that the polynomial $(P_k(x)+P_k(x+1))/(2k-1)!!$ is the Ehrhart polynomial of the standard $k$-dimensional cross-polytope. See Exercise 4.61 of Enumerative Combinatorics, vol. 1, second edition. REPLY [5 votes]: I'll start by addressing conjecture 2. By summing your generating fun over all values of $k$ we obtain $$F(z,w)=\sum_{n\geq 0}\sum_{k\geq 0}C_{\text{ODD}}(n,2k)\frac{z^n}{n!}w^k=\sum_{k\geq 0}w^k\left(\frac{1+z}{1-z}\right)^k=\frac{1-z}{1-w-z-wz}$$ From here we see that $$\sum_{n\geq 1}\sum_{k\geq 1}\frac{P_k(n)}{(2k-1)!!}w^k \frac{z^{n-1}}{(n-1)!}=\frac{1}{2}\frac{d}{dz}\int\left(F(z,w)-1\right)dw$$ $$=\int \frac{w}{(1-w)^2}\frac{1}{\left(1-z(\frac{1+w}{1-w})\right)^2}dw$$ By extracting the coefficients of $z^{n-1}/(n-1)!$ on both sides we have $$\sum_{k\geq 0}\frac{P_k(n)}{(2k-1)!!}w^k=\int \frac{nw}{(1-w)^2}\left(\frac{1+w}{1-w}\right)^{n-1} dw=\frac{1}{2}\left(\frac{1+w}{1-w}\right)^n+\text{constant}$$ which is what we wanted. I just realized that we can also answer conjecture 1 by making use of this identity. Start with the expansion $$\left(\frac{1+w}{1-w}\right)^n=\left(1+\frac{2w}{1-w}\right)^n=1+\sum_{r\geq 1} \binom{n}{r}\left(\frac{2w}{1-w}\right)^r$$ $$=1+\sum_{k\geq 1}w^k\sum_{r\geq 1}2^r\binom{n}{r}\binom{k-1}{r-1}$$ which gives us an explicit formula for $P_k$ $$P_k(n)=\sum_{r\geq 1}2^{r-1}(2k-1)!!\binom{n}{r}\binom{k-1}{r-1}$$ Which immediately tells us that $P_k(n)$, as a linear combination of $\binom{n}{r}$ for $1\le r\le k$, is a polynomial of degree $k$ with no constant term. Combined with the fact that $\left(\frac{1+w}{1-w}\right)^n=\left(\frac{1-w}{1+w}\right)^{-n}$ we have $P_k(n)=(-1)^kP_k(-n)$ which tells us that $P_k$ has the same parity as $k$. Now it remains to establish integrality of the coefficients. The explicit formula for $P_k$ can be rearranged as $$P_k(n)=\sum_{r\geq 1}\binom{k+r-1}{r,r-1,k-r}\frac{(2k-1)!}{2^{k-r}(k+r-1)!}(n)_r$$ and from here it is clear that the coefficients of $P_k$ have nonnegative $p$-adic valuation for any odd prime $p$. It remains to show the following lemma Lemma: For any $k\geq r\geq 1$ we have $$\nu_2\left(\frac{(2k-1)!}{r!(r-1)!(k-r)!}\right)\geq k-r.$$ Proof We make use of the fact that $\nu_2\left(\frac{s!}{\lfloor\frac{s}{2}\rfloor!}\right)=\lfloor\frac{s}{2}\rfloor$. Our expression can be written as $$\nu_2\left(\frac{(2k-1)!}{r!(r-1)!(k-r)!}\right)=\nu_2 \left(\frac{(2k-1)!}{(k-1)!}\frac{\lfloor\frac{r}{2}\rfloor!\lfloor\frac{r-1}{2}\rfloor!}{r!(r-1)!}\right)+\nu_2\left(\binom{k-1}{\lfloor\frac{r}{2}\rfloor,\lfloor\frac{r-1}{2}\rfloor,k-r}\right)$$ the first term is equal to $k-1-\lfloor\frac{r}{2}\rfloor-\lfloor\frac{r-1}{2}\rfloor=k-r$ and the second term is clearly nonnegative. This completes the proof of integrality.<|endoftext|> TITLE: Balls in Lawvere metric spaces QUESTION [11 upvotes]: Let $V$ be the monoidal category $[0,\infty)$ (as a poset) with $+$ and $0$. Lawvere shows that $V$-enriched categories are a more natural generalisation of the notion of a metric space (note no symmetry). Where it turns out many classical theorems about metric spaces (and similar structures like ultrametric spaces) are simply special cases of certain theorems in enriched category theory. Now one of the objects one uses when working with metric spaces are balls. Let $C$ be a $V$-enriched category. Choosing some $v\in V$ we can define the ball centered at some point $x\in C$. As the 'set' of points in $C$ such that there is a morphism $\mathrm{Hom}(x,y)\to v$. This however is not really an ideal definition. It feels very artificial and not very categorical as I would like. What would be a nice categorical way to define an 'open ball' of a Lawvere metric space? Edit: I have confused the direction of the arrows in $V$, this means my second the last previous paragraph should be rephrased: Given a $v\in V$, the ball centered at $x \in C$ is the set of points in $C$ such that $v \to \mathrm{Hom}(x,y)$. This does feel a bit more categorical but it is not quite there yet. REPLY [7 votes]: Luckily for you, Lawvere has already considered this in Taking categories seriously. On page 18, he defines the family: $$\mathcal V^{\mathrm op}\times A\xrightarrow{B} \mathcal{V}^{A^{\mathrm{op}}}$$ defined by $$B(r,c)(a)=\mathcal V(r,A(a,c))$$ Where $B(r,c)$ reads the closed ball of given radius and center, since $$0 \ge B(r,c)(a)\iff r\ge A(a,c)$$ This is quite a cool construction because you can consider closed balls in any $\mathcal V$-enriched category.<|endoftext|> TITLE: Mathematical Writing: Proof Outlines/Overview in a Paper QUESTION [22 upvotes]: While my question topic is that of mathematical writing of papers, which is a broad subject, the particular question is specific. I am writing a paper, in which we have a section called "Outline of Proof". (It's Section 2.) The outline is fairly informal, and we omit some technical details, making approximations. Among these approximations, should we state (and label) important definitions and results (lemmas, equations, etc), with the intent of, later in the paper, referencing thes? This raises the point of redundencies: some people don't like things being stated twice precisely (including in the outline), so wouldn't want anything explained/stated (even in the outline) re-explained/stated. This seems ill-advised to me. When I read a paper, I rarely carefully read the outline: I just read it, and try to get an overview (or 'outline') of the proof; if there are parts that I don't really understand, I don't get hung up on them, trusting that with the more rigorous explanation later I'll be able to make sense of what the authors are saying. So my question is this: (a) is it standard to read an outline of a proof carefully? (b) is it standard (or at least not discouraged) to state in the outline precisely important, even key, results/definitions that will be referred back to in the main body of the paper when giving proofs? REPLY [2 votes]: Having that outline is very useful for editors, to see quickly whom to send the paper for reviewing. Then for referees, to see if they want to review or not the paper, and also to have a quick idea on whether they intend to make a positive/negative report. And finally, for referees again, in case they make a positive report, it's so much easier to copy a bit from that outline part when making your report. Passed this, and getting to regular users of the paper, no rule of course. Just write your paper for yourself, I mean make it pleasant to read for yourself, guess that's the best rule when writing something.<|endoftext|> TITLE: Contractibility of balls in Alexandrov spaces QUESTION [6 upvotes]: Let $X$ be a compact finite dimensional Alexandrov space with curvature bounded below. Does there exist $\varepsilon_0>0$ (depending on $X$) such that for any $\varepsilon \in (0,\varepsilon_0)$ and any point $x\in X$ the open ball $B(x,\varepsilon)$ is contractible? Is similar statement true for closed balls? $B(x,\varepsilon)$ denotes the open ball of radius $\varepsilon$ with the center at $x$. REPLY [7 votes]: Formally speaking the answer is "no". Take a 2-dimensional cone with small total angle. Then for any $\varepsilon>0$ there is a point $x$ close enuf to the tip of the cone such that $B(x,\varepsilon)$ is an annulus.<|endoftext|> TITLE: Are there infinitely many primes of this form? QUESTION [9 upvotes]: The semiprime $87 = 3*29$ has a curious property: it's the fact that both $87^2 + 29^2 + 3^2 = 8419$ and $87^2 - 29^2 - 3^2 = 6719$ are prime numbers. This intrigued me and led me to wonder if there are other semiprimes with the same property, and I found that $21 = 3*7$ is another example, since both $21^2 + 7^2 + 3^2= 499$ and $21^2 - 7^2 - 3^2 = 383$ are prime numbers So the following question arises: Are there infinitely many prime numbers $p$ and $q$, with $p \neq q$, such that both $(pq)^2 + p^2 + q^2$ $(pq)^2 - p^2 - q^2$ are also primes? Does this follows from some known theorem or conjecture? REPLY [18 votes]: If both $p$ and $q$ are $\pm 1\mod 6$, then $(pq)^2 + p^2 + q^2$ is divisible by 3, so that there is only a possibility if one of the primes is 2 or 3. For $q = 2$ (and $p \ne 3$), $(pq)^2 + p^2 + q^2 = 5p^2 + 4$ also is divisible by 3. The question then becomes, are there infinitely many primes $p$ so that both $10p^2 + 9$ and $8p^2 - 9$ are prime? Since it is not known if there is any polynomial of degree greater than 1 that produces infinitely many primes, let alone a pair of polynomials that in an infinite number of primes produce prime numbers, I guess it will be extremely hard to prove that there are infinitely many such pairs $(p, q = 3)$. It may be possible to prove that they don't, though I expect that would be hard as well.<|endoftext|> TITLE: Can a positive polynomial on sphere be represented as the sum of squares of spherical harmonics QUESTION [7 upvotes]: Let $p\in {\mathbb{R}}[x_1,\ldots, x_d]$ be a homogenous polynomial degree $2n$. We know that if $p$ is positive on $[-\pi,\pi]^d$, $p$ is sum of squares polynomial, i.e. $p$ can be witten as sum of squares of $d$ dimensional Fourier harmonics up to degree $n$. My question is if $p$ is positive on the unit sphere $S\subset {\mathbb{R}}^n$ and can be represented as combination of spherical harmonics dimension $d$, then does there exist some spherical harmonic polynomials $g_1,\ldots, g_k$ of degree $n$ such that $p=g_1^2+\cdots g_k^2$ is a sum of squares? i edit the question after Zach Teitler's comment. The interval $[-\pi,\pi]^d$ means we concern the trigonometric polynomials positive on frequency domains. The optimization problems about the polynomials positive on frequency domain $[-\pi,\pi]^d$ can be implemented via SDP approach(Gram matrix Rpresentation). Given a positive polynomial represented as combination of spherical harmonics dimension $d$, Obviously, it is sum of squares of $d$ dimensional Fourier harmonics. Furthermore, it implies the symmetry relationship between $[-\pi,\pi] \times [0,\pi]$ and $[-\pi,\pi] \times [-\pi,0]$ on 2-sphere as an example. May be there is less information on sphere than cube? So, is it the sum of squares of spherical harmonics? edit after Zach Teitler's answer My appoligize, the theroy of positive polynomial on $[-\pi,\pi]^d$ is from Dumitrescu, “Trigonometric Polynomials Positive on Frequency Domains and Applications to 2-D FIR Filter Design,” IEEE Transactions on Signal Processing, 2006 Theorem 1. That is Given $z=[z_1,\ldots, z_d]$ and $z^k=z_1^{k_1}z_2^{k_2}\ldots z_d^{k_d}$, a Hermitian trigonommetric polynomial of degree $n$ that $R(z)=\sum_{k=-n}^n r_k z^{-k}, r_{-k}=r_k^*$ is positive on unit $d$-circle, i.e. $z_i=e^{j\theta_i}, \theta_i \in[-\pi,\pi]^d$ , then it is sum of squares. My question is on the relationship between sum of suqare polynomial and the polynomial that sum of squares spherical harmonics. Thank you very much again! Please feel free to provide any advices. Any comments and references (in English) will also be very welcome ! Thank you very much in advance! REPLY [6 votes]: This Wikipedia page has many references: https://en.wikipedia.org/wiki/Positive_polynomial, including for example Marshall, Murray Positive polynomials and sums of squares. Mathematical Surveys and Monographs, 146. American Mathematical Society, Providence, RI, 2008. B. Reznick, Uniform denominators in Hilbert's seventeenth problem. Math. Z. 220 (1995), no. 1, 75-97. It is not true that positivity on the unit sphere implies representation as a sum of squares of polynomials. For example $M(x,y,z) = x^4 y^2 + x^2 y^4 + z^6 - 3 x^2 y^2 z^2$ is a nonnegative homogeneous form which cannot be written as a sum of squares. (This is called the "Motzkin form". See for example http://www.msri.org/attachments/workshops/327/553_Lecture-notes_week1_Blekherman.pdf.) Note that for a homogeneous form $p$ the following are equivalent: $p \geq 0$ on the unit sphere $p \geq 0$ on a neighborhood of the origin $p \geq 0$ everywhere (globally) so there is no difference between assuming that $p \geq 0$ on the unit sphere, or on a cube such as $[-\pi,\pi]^d$. It is the same hypothesis. In either case, it does not imply that $p$ is a sum of squares of polynomials. If you want to ask about $p$ being a sum of squares of rational functions, or if $p$ is not homogeneous, then that is a different situation.<|endoftext|> TITLE: Is every metric continuum almost path-connected? QUESTION [10 upvotes]: The question was motivated by this question of Anton Petrunin. By a metric continuum we understand a connected compact metric space. Let $p$ be a positive real number. A metric continuum $X$ is called $\ell_p$-almost path-connected if for any points $x,y\in X$ and any $\varepsilon>0$ here exists a family $\big((a_n,b_n)\big)_{n\in\omega}$ of pairwise disjoint open intervals in the unit segment $[0,1]$ and a continuous map $\gamma:[0,1]\setminus\bigcup_{n\in\omega}(a_n,b_n)\to X$ such that $\gamma(0)=x$, $\gamma(1)=y$ and $\sum_{n=0}^\infty d_X(\gamma(a_n),\gamma(b_n))^p<\varepsilon$. It is easy to see that each almost $\ell_p$-connected metric continuum is $\ell_q$-almost connected for any $q\ge p$. By my answer to the question of Anton Petrunin, each plane continuum is almost $\ell_1$-connected. By analogy it can be shown that each continuum in $\mathbb R^3$ is $\ell_2$-connected. Problem. Is there a metric continuum which is not almost $\ell_1$-path connected? not almost $\ell_p$-connected for every $p<\infty$? REPLY [3 votes]: Yes there are such examples. Assume the sequence $\varepsilon_n$ is very fast converging to $0$. Consider a sequence of short $\varepsilon_n$-crooked maps between intervals $\mathbb{J}_n\to \mathbb{J}_{n-1}$. Its inverse limit is a pseudoarc $\mathbb{J}_\infty$; denote by $\phi_n\colon\mathbb{J}_\infty\to \mathbb{J}_n$ the projections. Equip $\mathbb{J}_\infty$ with the maximal metric such that $$|x-y|_{\mathbb{J}_\infty}\le \tfrac1{2^{n/p}}+|\phi_n(x)-\phi_n(y)|_{\mathbb{J}_n}.$$ Assume that $\mathbb{J}_\infty$ is $\ell_p$-almost path-connected. Let $\gamma$ be the path that connects the ends of $\mathbb{J}_\infty$. It can not have more then one jump of size $1$; the jump brakes $\alpha$ in two arcs one of which has diameter at least $1-\varepsilon_1$; this arc can not have more than 2 jumps of length $\tfrac12$, so one of the subarcs has diameter at least $1-\varepsilon_1-2\cdot \varepsilon_2$ and so on. At the end of the day you see that $\mathbb{J}_\infty$ contains a subarc of $\alpha$ of positive diameter. But $\mathbb{J}_\infty$ has no arcs, a contradiction. The given construction is nearly identical to Example 4.2 in my paper on intrinsic isometries.<|endoftext|> TITLE: Partial product of Euler factors QUESTION [11 upvotes]: Let $\mathbb P$ denote the set of prime numbers and for a subset $T\subset \mathbb P$ let $$ \zeta_T(s)=\prod_{p\in T}\frac1{1-p^{-s}}, $$ where $\mathrm{Re}(s)>1$. Is there any $T$ such that $T$ and $T^c={\mathbb P}\smallsetminus T$ are both infinite and $\zeta_T$ has a meromorphic continuation to $\mathbb C$? REPLY [3 votes]: (Not an answer but a long comment.) I think it is a safe bet that there are no such functions, but this possibility will not be easy to rule out because for a Dirichlet series being meromorphic is not a very handy condition. I would like to point out some obvious but probably interesting facts about this hypothetical decomposition. We have $$\zeta(s)=\zeta_{T}(s)\zeta_{T^c}(s),$$ and $\zeta(s)$ has a pole at $s=1$. For both functions $$\zeta_{T}(s)>1,\,\zeta_{T^c}(s)>1,\,s>1$$ hence one of them has a pole at $s=1$ (the "big" one) and another does not (the "small" one). We may assume that $\zeta_{T}(s)$ is the latter. The Dirichlet series $$\zeta_{T}(s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$$ has nonnegative coefficients ($a_n\ge 0$), hence if it does not have any poles on the line $s>0$, then both the series and the product converge absolutely on the halfplane $\Re s>0$. It is not difficult to see that in such a case this function cannot have an analytic continuation beyond $\Re s=0$; the reason for this are the poles of the factors $\frac{1}{1-p^{-s}}$ on this line. For a proof of this fact it is convenient to use the logarithmic derivative $$-\frac{\zeta'_{T}(s)}{\zeta_{T}(s)}=\sum_{p\in T}\frac{\log p}{p^s-1},\,\Re s>0.$$ If $\zeta_{T}(s)$ were meromorphic at $s=0$ then we would have $$-\frac{\zeta'_{T}(s)}{\zeta_{T}(s)}\sim \frac{n}{s},\,s\to+0$$ but RHS grows faster then this (for infinite $T$). It follows that the set $T$ can't be too small, there is an asymptotic like $$\#\{p\in T: p\le x\}\sim n\int_2^x\frac{dt^\sigma}{\log t}$$ ($0<\sigma<1$), which can be proved more or less the same way as the prime number theorem.<|endoftext|> TITLE: Is $K[[x_1,x_2,\dots]]$ an $\mathfrak m$-adically complete ring? QUESTION [16 upvotes]: I asked this question on Mathematics Stackexchange (link), but got no answer. Let $K$ be a field, let $x_1,x_2,\dots$ be indeterminates, and form the $K$-algebra $A:=K[[x_1,x_2,\dots]]$. Recall that $A$ can be defined as the set of expressions of the form $\sum_ua_uu$, where $u$ runs over the set monomials in $x_1,x_2,\dots$, and each $a_u$ is in $K$, the addition and multiplication being the obvious ones. Then $A$ is a local domain, its maximal ideal $\mathfrak m$ is defined by the condition $a_1=0$, and it seems natural to ask Is $K[[x_1,x_2,\dots]]$ an $\mathfrak m$-adically complete ring? I suspect that the answer is No, and that the series $\sum_{n\ge1}x_n^n$, which is clearly Cauchy, does not converge $\mathfrak m$-adically. REPLY [8 votes]: [Edit: The lemma was revised and proved, changed the point of view from series to sequences.] [2nd Edit: The proof of the lemma was improved, and now the argument can show that Cauchy series such as $x_1+(x_{1^3+1}^2+\dots x_{2^3}^2)+(x_{2^3+1}^{3}+\dots+x_{3^3}^3)+\dots$, diverge in the $\mathfrak{m}$-adic topology.] Here is a construction of a Cauchy sequence which does not converge. It is based on the following lemma, a proof of which will is given at the end. Lemma. Let $\mathfrak{m}_c$ denote the maximal ideal of $K[[x_1,\dots,x_c]]$. Then there exist sequences of natural numbers $(r_n)_{n\in\mathbb{N}}$, $(c_n)_{n\in\mathbb{N}}$ and a sequence of elements $(p_n\in(\mathfrak{m}_{c_n})^n)_{n\in\mathbb{N}}$ such that: $\limsup r_n=\infty$ and $p_n$ cannot be written as a sum of $r_n$ terms $\sum_{i=1}^{r_n} a_{i} b_i$ with $a_{i},b_i\in\mathfrak{m}_{c_n}$. In fact, one can take $c_n=n^2$, $p_n=x_1^{n}+\dots+x_{n^2}^{n}$ and let $r_n=\lceil \frac{n^2}{2(n-1)}\rceil-1$ if $\mathrm{char}\,K\nmid n$ and $r_n=1$ when $\mathrm{char}\,K\mid n$. It would be more convenient to replace $K[[x_1,x_2,\dots]]$ with the isomorphic ring $$S:=K[[y_{ij}\,|\,i,j\in \mathbb{N}]].$$ For every $n$, there is a ring homomorphism $\phi_n:S\to K[[x_1,\dots,x_{c_n}]]$ specializing $y_{n1},\dots,y_{n{c_n}}$ to $x_1,\dots,x_{c_n}$ and the rest of the variables to $0$. Let $f_n=p_n(y_{n1},\dots,y_{nc_n})$ and define in $S$ the (formal) partial sums $$ g_t:=\sum_{n=t}^\infty f_n. $$ Then, by construction, $(g_t)_{t\in\mathbb{N}}$ is a Cauchy sequence relative to the $\mathfrak{m}$-adic topology, but it does not converge. Indeed, if $(g_t)_t$ converges in the $\mathfrak{m}$-adic topology, then it also converges relative to the (coarser) grading topology of $S$, and so must converge to $0$. However, if this is so, there exists $t\in\mathbb{N}$ such that $g_t=\sum_{n=t}^\infty f_n\in \mathfrak{m}^2$. In particular, we can write $\sum_{n=t}^\infty f_n=\sum_{i=1}^u a_{i}b_i$ with $a_{i},b_i\in \mathfrak{m}$. Choose $n\geq t$ sufficiently large to have $r_n\geq u$. Applying $\phi_n$ to both sides of the last equality gives $p_n=\sum_{i=1}^u (\phi a_{i})(\phi b_{i})$ with $\phi a_{i},\phi b_i\in\mathfrak{m}_{c_n}$, which is impossible by the way we chose $p_n$. Back to the lemma: Given $f\in K[x_1,\dots,x_c]$, let $D_if$ denote its (formal) derivative relative to $x_i$. The lemma follows from the following general proposition. Proposition. Let $f\in K[[x_1,\dots,x_c]]$ be a homogeneous polynomial of degree $n$, and let $r\in\mathbb{N}$ denote the minimal integer such that $f$ can written as $\sum_{i=1}^ra_ib_i$ with $a_i,b_i\in \mathfrak{m}_c$. Suppose that $D_1f,\dots,D_cf$ have no common zero beside the zero vector over the algebraic closure of $K$. Then $r\geq \frac{c}{2(n-1)}$. Proof. Suppose otherwise, namely, that $f=\sum_{i=1}^r a_ib_i$ with $2r(n-1)0$. Furthermore, $V$ is nonempty because it contains the zero vector (because $a'_1,b'_1,a'_2,b'_2,\dots\in \mathfrak{m}_c$). Thus, there exists a nonzero $v\in \overline{K}^c$ annihilating $a'_1,b'_1,a'_2,b'_2,\dots$. Now, by Leibniz's rule, we have $$D_jf=\sum_i D_j(a'_ib'_i)=\sum_i(D_ja'_i\cdot b'_i + a'_i\cdot D_j b'_i).$$ It follows that $v$ above annihilates all the derivatives $D_1f,\dots,D_cf$, a contradiction! $\square$ If it weren't for the passage from power series to polynomials, the proof would work for non-homogenous polynomials and also give the better bound $r\geq \frac{c}{2}$. [Edit: This is in fact possible, see the comments.] Such a bound would suffice to prove that the Cauchy series $x_1+x_2^2+x_3^3+\dots$ suggested in the question diverges in the $\mathfrak{m}$-adic topology.<|endoftext|> TITLE: Condition on a differential form arising from the theory of elasticity QUESTION [8 upvotes]: Let $D$ be the unit $n$-ball (for concreteness). Let $\beta\in\Omega^1(D;R^n)$ be an $R^n$-valued one-form, having full rank (viewed as a section of $T^*D\otimes R^n$). Under what conditions on $\beta$, does there exist a section $Q$ of $SO(n,R)$ (over $D$), such that $Q\circ\beta$ is closed (hence exact)? The question is non-trivial for the following reason: if there exist such $Q$ and an $f:D\to R^n$, such that $df = Q\circ\beta$, then $\beta^T\circ\beta = df^T\circ df$, and the latter is (up to a musical isomorphism) a flat metric on $D$, whose Riemann curvature tensor vanishes. So in a sense, I have an answer to my question. What I am looking for is a more explicit condition; in particular, I wonder whether there exists a condition that is linear in $\beta$. For the curious, this question came up twice in two different contexts in the theory of elasticity. REPLY [8 votes]: This is really a question of computing the curvature of the Levi-Civita connection of the Riemannian metric $g = \beta^T\circ\beta$. Thus, what one needs to do is first solve the equations $$ \mathrm{d}\beta = -\theta\wedge\beta\qquad\text{and}\qquad \theta^T+\theta=0 $$ for a $1$-form $\theta$ taking values in skew-symmetric $n$-by-$n$ matrices. The Fundamental Lemma of Riemannian geometry guarantees that there is always a unique solution to this system of linear algebraic equations for $\theta$. Then one needs to compute the curvature $2$-form $$ \Theta = \mathrm{d}\theta + \theta\wedge\theta. $$ Then a necessary and sufficient condition for the stated problem to have a solution $Q$ is that $\Theta$ vanish identically. Necessity follows since, if there exists a $Q$ mapping the ball to $\mathrm{SO}(n)$ such that $Q\beta$ is closed, say, equal to $\mathrm{d}x$ for some $\mathbb{R}^n$-valued function on the ball, then one sees that one must have $\theta = Q^{-1}\mathrm{d}Q$, which implies $\Theta \equiv 0$. Sufficiency follows since, if $\Theta\equiv0$, then the overdetermined equation $\theta = Q^{-1}\mathrm{d}Q$ can be solved for $Q$, uniquely up to left translation by a constant element of $\mathrm{SO}(n)$, and then $Q\beta$ will be closed. Note however, that, while $\theta$ is found by solving a system of linear algebraic equations (whose coefficients depend on $\beta$ and $\mathrm{d}\beta$), the expression for $\Theta$ is quadratic in the expression for $\theta$, at least when $n>2$. Thus, asking for a `linear' condition on $\beta$ that detects $\Theta\equiv0$ is asking for too much. (By the way, the condition $\Theta\equiv0$ is, of course, exactly the condition that the Riemann curvature tensor of the metric $g$ be identically zero.)<|endoftext|> TITLE: Cohomology of $\mathbb Z_4$ via the Lyndon-Hochschild-Serre spectral sequence QUESTION [8 upvotes]: I'm trying to understand how to construct the Lyndon-Hochschild-Serre spectral sequence for the cohomology (with integer coefficients) of the central extension $G$ of a group $Q$ by a group $N$, given a representative cocycle of $H^2(Q,N)$ corresponding to such an extension. I will use the example of $\mathbb Z_4$, which is a nontrivial central extension of $\mathbb Z_2$ by $\mathbb Z_2$. I have tried to explain my reasoning below. So I start with the exact sequence $0 \rightarrow \mathbb Z_2 \overset{i}{\rightarrow} \mathbb Z_4 \overset{p}{\rightarrow} \mathbb Z_2 \rightarrow 0$ and compute the $E_2$-page using $E_2^{p,q} = H^p(\mathbb Z_2,H^q(\mathbb Z_2,\mathbb Z))$. In the case of the trivial central extension, where the action of every group on its respective coefficient module is trivial, we get a page which vanishes for $q$ odd and is just $\mathbb Z_2$, except for $E_2^{0,0} = \mathbb Z$, when $q$ is even. This sequence stabilizes on the $E_2$-page giving the results expected by using, say, the Kunneth formula. However, this cannot be the right $E_2$-page for $\mathbb Z_4$, because the higher cohomology groups would then be too large. Therefore we must have a nontrivial action of $\mathbb Z_2$ on the coefficient modules $H^q(\mathbb Z_2,\mathbb Z)$. This is where I am stuck. I have two questions: 1) What is the above action, and is there a systematic way to see it for large $q$? 2)Is there a systematic way to compute differentials in the $E_2$-page given the maps $i,p$ and a representative cocycle for this extension? REPLY [5 votes]: The action of the quotient on the cohomology groups of the normal subgroup is the trivial action, because the normal subgroup is central. (Think of group cohomology as a functor of the group: the conjugation action of $Q$ on $N$ induces the action of $Q$ on $H^*(N;\mathbb{Z})$.) What happens for the spectral sequence that you are thinking of is that there are non-trivial differentials. In this particular example, all the non-trivial differentials are $d_3$. In the $E_2$ page you have that $E_2^{i,j}$ is either zero or order two (except for $E_2^{0,0}$ which is infinite cyclic), and the pairs $(i,j)$ for which the group has order two are those for which $j=0$ and $i>0$ is even, or $j>0$ is even and $i$ is arbitrary. The differential $d_3$ cancels lots of these order two groups in pairs, leaving groups of order two in $E_4^{i,j}$ only in the cases when either $j=0$ and $i>0$ is even or when $j=2$ and $i$ is even. There are lots of tricks for computing differentials. In this case the easiest thing it probably to use the ring structure on each $E_i^{*,*}$, together with the known values of $H^k(\mathbb{Z}_4;\mathbb{Z})$ for $k=0,1,2$.<|endoftext|> TITLE: Union of random intervals with total length equal to infinity QUESTION [14 upvotes]: Let $a_1,a_2,\dots$ be a sequence of positive numbers less than $1$, such that $$\sum_{n=1}^\infty a_i= \infty,$$ and $S^1 = \mathbb{R}/\mathbb{Z}$. Suppose $I_1,I_2,\dots$ be random intervals with respective lengths $a_1,a_2, \dots$in $S^1$ such that the distribution of the centers of $I_n$ (for every $n$) are uniform and independent. It can be shown that with probability $1$, $I = \cup_{n=1}^\infty I_n$ is a full measure subset of $S^1$. Is it true that "With probability $1$, $I = S^1$"? If this is not always true, does there exist a good characterization of the sequences $\{ a_n\}_{n=1}^{\infty}$ with this property? Edit. A more precise question: "What happens in the special case $a_n = \frac1n$?" REPLY [8 votes]: This is a refinement of Iosif Pinelis's answer, so we shall be somewhat brief. For a punchline, jump to the "Added" section below. We claim that if $a_n>c/n$ holds some $c>1$ and for all $n\geq n_0$, then $P(I=S^1)=1$. To see this, fix a large integer $N$ and any interval $J$ of length $1/N$. Then, $$P(J\not\subseteq I)\leq\prod_{n_0\leq n TITLE: Are $\varepsilon$-connected components dense? QUESTION [6 upvotes]: Let $X$ be a connected compact metric space. Given a positive $\varepsilon$ and two points $x,y\in X$ we write $x\sim_\varepsilon y$ if there exists a sequence $C_1,\dots,C_n$ of connected subsets of diameter $<\varepsilon$ in $X$ such that $x\in C_1$, $y\in C_n$ and $C_i\cap C_{i+1}\ne\emptyset$ for all $i 0$. For any closed set $F$, let $F^\ast = \bigcup\{G:G\text{ closed, connected, }G\cap F\neq \varnothing\text{, diam}(G)\leq\varepsilon\}$. Now notice that $\{G:G\text{ closed, connected, }G\cap F\neq \varnothing\text{, diam}(G)\leq\varepsilon\}$ is a closed subset of hyperspace (connectedness, having non-empty intersection with $F$, and having diameter $\leq \varepsilon$ are all closed conditions in hyperspace). This implies that $F^\ast$ is also a closed set, since the union of a closed family of sets is closed (since $X$ is compact). Then some definition chasing tells us that $[x]_{\leq\varepsilon}$ is $\{x\}\cup\{x\}^\ast\cup\{x\}^{\ast\ast}\cup\dots$, so that in particular $[x]_{\leq \varepsilon}$ is a $\sigma$-compact set. Therefore $[x]_\varepsilon$ is $\sigma$-compact as well. As for the density question, I suspect that you can use the boundary bumping theorem to show that $[x]_\varepsilon$ always contains the composant of $x$, which would imply that it's always dense since composants are always dense. The topologist's sine curve gives an example of a non-dense $\varepsilon$-connected component. Specifically the $\varepsilon$-connected component of any point in the line segment is not dense for sufficiently small $\varepsilon$.<|endoftext|> TITLE: On a quantum Riemann Hypothesis QUESTION [33 upvotes]: Here is a revised version: On a revised quantum Riemann hypothesis. Robin's theorem (1984) states that $$ \sigma(n) < e^\gamma n \log \log n$$ for all $n > 5040$ if and only if the Riemann hypothesis is true. Recall that $γ$ is the Euler–Mascheroni constant and $σ(n)$ is the divisor function, given by $$\sigma(n) = \sum_{d\mid n} d.$$ To formulate a quantum Riemann hypothesis, we will use Robin's theorem and the following facts: a natural number $n$ can be encoded into the cyclic group $C_n$ a finite group $G$ can be encoded into the finite index irreducible depth $2$ subfactor $R \subseteq R \rtimes G$ a finite index irreducible subfactor $N \subseteq M$ can be encoded into a planar algebra $\mathcal{P}$. For a justification of the qualifier "quantum", see the following article: Jones, Vaughan. On the origin and development of subfactors and quantum topology. Bull. Amer. Math. Soc. (N.S.) 46 (2009), no. 2, 309--326. Galois correspondences: the divisors $d\mid n$ are $1$-$1$ with the subgroups $H \subseteq C_n$, the subgroups $H \subseteq G$ are $1$-$1$ with the intermediate subfactors $R \subseteq K \subseteq R \rtimes G$, the intermediate subfactors $N \subseteq K \subseteq M$ are $1$-$1$ with the biprojections $b \in [e_1,id]$. The notations match as follows: $n = |G| = [M:N] = |id : e_1|$, $d = |H| = [K:N] = |b : e_1|$. The equality $|G| = |G:H| \cdot |H|$ extends to $|id : e_1| = |id:b| \cdot |b:e_1|$. In general, $|id : e_1|$ is not necessarily an integer, but (by Jones' theorem) can be any element in $$\{4\cos^2(\pi /n)|n=3,4,5,...\}\cup [4,+\infty).$$ Let $\mathcal{P}$ be an irreducible subfactor planar algebra. We define the analog of the set of divisors by $$D(\mathcal{P}) := \{|b : e_1| \text{ with } b \in [e_1,id] \},$$ (which is finite by Watatani's theorem) and the analog of the divisor function by $$\sigma(\mathcal{P}) := \sum_{\beta \in D(\mathcal{P})} \beta.$$ Quantum Riemann Hypothesis (of depth $n$) There is $\alpha_n>0$ such that for every irreducible depth $n$ subfactor planar algebra $\mathcal{P}$ with $\alpha:=|id : e_1|> \alpha_n$, we have $$\sigma(\mathcal{P}) < e^\gamma \alpha \log \log \alpha.$$ Of course, a proof of this quantum Riemann Hypothesis (QRH) is not expected as an answer of this post, because it implies the usual Riemann Hypothesis (RH). For the group case, QRH follows from RH, because for $\sigma(G):=\sigma(\mathcal{P}(R \subseteq R \rtimes G))$, we have $\sigma(G) \le \sigma(|G|)$ by Lagrange's theorem. Idem if $|id:b|$ and $|b:e_1|$ are integers $\forall b \in [e_1,id]$, like the irreducible depth $2$ case, because then $\sigma(\mathcal{P}) \le \sigma(|id : e_1|)$. Let's denote QRH of depth $n$ by QRH$_n$. Then, QRH$_2 \Leftrightarrow$ RH, and we can take $\alpha_2 = 5040$. Question: Does RH imply QRH$_n \ \forall n \ge 2$? Or, do you see a counterexample for some $n$? Bonus question: Assuming QRH$_n$ true $ \forall n \ge 2$, can the sequence $(\alpha_n)$ be bounded? To learn more, you can watch the series Quantum Symmetries and Quantum Arithmetic. The last video finishes on (revised) QRH. REPLY [5 votes]: Can you clarify whether there exists a notion of direct product in this setting with the desired properties? If so, the asymptotics you are predicting only seem consistent with the hypothesis that there are no such objects besides groups. (I guess they do actually exist or you wouldn't ask this question.) The first version of this post suggested taking the direct sum of many factors of size less than $4$, and the OP said that this was invalid because most of them didn't have "bounded depth." Fair enough. But if you take direct products of many subfactors of small indices which have the property that they are "coprime" then it seems you can easily exceed your desired bound unless the only new objects also have rational orders. Here's a very weak explicit version of this idea using a single new object. It's know that $\displaystyle{\sup \frac{\sigma(n)}{e^{\gamma} n \log \log n} = 1}$. Take an $\alpha$ for which the ratio is very close to $1$. How close will be clear below. Let $X$ be the object coming from the cyclic group $C_{\alpha}$. Now let $Y$ be any other object whose index is some number $\eta$ which is not a rational number (assuming it exists), and let $Z = X \times Y$ (assuming it exists), and let $\beta = \alpha \eta$ be the index of $Z$. Then it would seem (by considering the subgroups of $C_{\alpha}$ and those subgroups direct sum with $Y$), and using that $\eta$ is not rational, that $$\sigma(Z) \ge \sigma(X) + \eta \sigma(X) \sim (1 + \eta) e^{\gamma} \alpha \log \log(\alpha) \sim \left(1 + \frac{1}{\eta}\right) e^{\gamma} \beta \log \log(\beta),$$ and the QRH is false.<|endoftext|> TITLE: Torus action implying infinite fundamental group QUESTION [8 upvotes]: Suppose that a $d$-dimensional torus $T$ acts smoothly and effectively on an $n$-dimensional closed manifold $M$. What conditions on $d$ and $n$ imply that $\pi_1(M)$ must be infinite? Consider the case where $d=n-1$. Within Section 4 of this paper by Grove and Ziller, it is shown that if $n\geq 4$ and $T^{n-1}$ acts smoothly and effectively on $M^n$, then $\pi_1(M)$ is infinite. In contrast for $n=2$, there is of course the $S^1$-action on $S^2$, and for $n=3$, there is the $T^2$-actions on $S^3$ and lens spaces. Are there other results or references related to this question? REPLY [6 votes]: Actions of $T^n$ on simply-connected $n+2$-manifolds were constructed in Theorem 4.7 of this paper. However, I haven't checked that the action is smoothable. The authors are only considering locally smooth actions in the paper (see p. 170 of Bredon for the definition of locally smooth); however, the construction yielding Theorem 4.7 is quite explicit, and seems to be at least piecewise-smooth. All such actions were subsequently classified by McGavran. So presumably one could use this classification to determine if there are smooth actions. Addendum: Oops, it looks like there's some gaps in McGavran's arguments that were fixed by Hae Soo Oh at least in dimensions 5 and 6. Moreover, Oh's classification is in the smooth category. Moreover, in Remark (4.7) of this paper, Oh constructs examples of $T^n$ actions on simply-connected $n+2$-manifolds in all dimensions. Since he's working in the smooth category, this seems to show that there are smooth examples (although I don't see immediately in his argument where smoothness is proven).<|endoftext|> TITLE: Simplicial nerve functor commutes with opposites QUESTION [10 upvotes]: There are two "opposite" functors: $$ op_\Delta\colon sSet\to sSet$$ and $$op_s\colon sCat\to sCat.$$ The first takes a simplicial set to its opposite simplicial set by precomposing with the opposite of a functor $\Delta\to \Delta$ which is the identity on objects and takes a morphism $\langle k_0,\ldots,k_n\rangle\colon [n]\to [m]$ (where $k_i$ is the integer that $i$ gets mapped to by this morphism) to the morphism $\langle m-k_n,\ldots,m-k_0\rangle$. For example, the morphism $[1]\to [2]$ that takes $0$ to $0$ and $1$ to $1$ gets mapped to the morphism that takes $0$ to $1$ and $1$ to $2$. The second functor takes a simplicial category to the opposite simplicial category, which is easier to define. It has the same objects but given $x,y\in C^{op}$, the mapping complex $C^{op}(x,y)=C(y,x)$. There is also the simplicial nerve functor $N\colon sCat\to sSet$. I am interested in a proof of the fact that for a given fibrant simplicial category $C$, there is a weak equivalence of quasicategories $op_\Delta\circ N(C)\simeq N\circ op_s(C)$. I'm relatively certain that this is an elementary proof, but I don't feel skilled enough with the simplicial nerve to figure out the details. Does anyone have a proof of this fact? REPLY [6 votes]: (This is a supplement to Harry's answer which doesn't quite fit into a comment.) As Harry mentions, it suffices to show that the cosimplicial objects $\mathfrak{C}((\Delta^\bullet)^\mathrm{op})$ and $(\mathfrak{C}(\Delta^\bullet))^\mathrm{op}$ are isomorphic. This follows from the following observation: Although the simplicial category $\mathfrak{C}(S)$ is rather mysterious for a simplicial set $S$, in the case $S$ is the nerve of a poset, it admits a completely explicit description analogous to the definition of $\mathfrak{C}(\Delta^n).$ Namely, if $P$ is a poset, then $\mathfrak{C}(N(P))$ is the simplicial category whose objects are $P$, and whose hom-simplicial sets are given by $$\mathfrak{C}(N(P))(x,y)=N(\{I\subset P \mid I\text { a finite totally ordered set, }\min I=x,\,\max I=y\}).$$ The composition is induced by the operation of union. One can verify easily that $\mathfrak{C}(N(P))$ is as stated above by directly verifying the universal property of $\operatorname{colim}_{\Delta^n\downarrow N(P)}\mathfrak{C}(\Delta^n)$. With this description, we have a natural (in $P$) isomorphism $\mathfrak{C}(N(P^\mathrm{op}))\cong\mathfrak{C}(N(P)^\mathrm{op})=\mathfrak{C}(N(P))^\mathrm{op}$. This makes the claim at the beginning transparent.<|endoftext|> TITLE: Does this consequence of measurability in terms of games of length $\omega+1$ imply measurability? QUESTION [13 upvotes]: For any two structures $\mathcal{M}$ and $\mathcal{N}$ in the same first-order language $\mathcal{L}$ and any ordinal $\theta$, let $G_\theta(\mathcal{M},\mathcal{N})$ be the two-player game of perfect information of length $\theta$ such that in round $\alpha$ player I plays an element $x_\alpha\in \mathcal{M}$ and then player II plays an element $y_\alpha \in \mathcal{N}$, and player II wins if and only if the augmented structures $(\mathcal{M}, x_\alpha)_{\alpha < \theta}$ and $(\mathcal{N}, y_\alpha)_{\alpha < \theta}$ have the same theory in the language obtained from $\mathcal{L}$ by adding $\theta$ constant symbols. Note that if there is an elementary embedding $j$ of $\mathcal{M}$ into $\mathcal{N}$ then player II has winning strategies in $G_\theta(\mathcal{M},\mathcal{N})$ for all $\theta$ obtained by letting $y_\alpha = j(x_\alpha)$. Conversely, if player II has a winning strategy in $G_\theta(\mathcal{M},\mathcal{N})$ where $\theta$ is the cardinality of $\mathcal{M}$ then we may obtain an elementary embedding of $\mathcal{M}$ into $\mathcal{N}$ by letting $j(x_\alpha) = y_\alpha$ where $(x_\alpha:\alpha < \theta)$ is an enumeration of $\mathcal{M}$. The existence of a winning strategy for player II in the case $\theta = \omega$ is weaker: it is equivalent to the existence of an elementary embedding of $\mathcal{M}$ into $\mathcal{N}$ in every generic extension of $V$ by the poset $\text{Col}(\omega,\mathcal{M})$. Let us define a cardinal $\kappa$ to be $\theta$-strategically measurable if there is an ordinal $\kappa'$ and a transitive set $N$ such that $\kappa < \kappa' \in N$ and letting $\mathcal{M} = (H_{\kappa^+}; \mathord{\in},\kappa, \xi)_{\xi < \kappa}$ and $\mathcal{N} = (N; \mathord{\in}, \kappa', \xi)_{\xi < \kappa}$, player II has a winning strategy in the game $G_\theta(\mathcal{M},\mathcal{N})$. Remarks: $2^\kappa$-strategic measurability is equivalent to measurability. If $0^\sharp$ exists then every Silver indiscernible is $\omega$-strategically measurable in $L$. (Use $j:L \to L$ and the absoluteness of existence of winning strategies for closed games of length $\omega$). Every $(\omega+1)$-strategically measurable cardinal is a Ramsey cardinal and a limit of Ramsey cardinals. (Proof given below.) Questions: Is ($\omega+1$)-strategic measurability equivalent to measurability? If not, what is its consistency strength? Have these games been studied before for $\theta > \omega$? Do they have a name? Proof of remark 3: We use a winning strategy for player II in the game $G_{\omega+1}(\mathcal{M},\mathcal{N})$ (or just the nonexistence of a winning strategy for player I) to build increasing sequences $(M_n, n<\omega)$ and $(\mu_n, n<\omega)$ such that $M_n \prec H_{\kappa^+}$, $\left| M_n\right| = \kappa$, $\mu_n$ is an $M_n$-normal ultrafilter on $\mathcal{P}(\kappa)\cap M_n$, and $M_n, \mu_n \in M_{n+1}$. In round $n$ player I plays an enumeration of $\mathcal{P}(\kappa)\cap M_n$ in order type $\kappa$ and uses player II's response to define $\mu_n$. Then letting $M_\omega = \bigcup_{n<\omega}M_n$ and $\mu_\omega = \bigcup_{n<\omega} \mu_n$, we see that $\mu_\omega$ is a weakly amenable $M_\omega$-normal ultrafilter on $\mathcal{P}(\kappa)\cap M_\omega$. Moreover it is countably complete (in the sense of nonempty intersection) because otherwise player I can win by playing a counterexample to countable completeness in round $\omega$. Since we can take $M_0$ to contain any given subset of $\kappa$, it follows that $\kappa$ is a Ramsey cardinal. Because $M_\omega \prec H_{\kappa^+}$ we can reflect Ramseyness below $\kappa$. (See Gitman, Ramsey-like cardinals for the relationship between Ramsey cardinals and weakly amenable countably complete ultrafilters.) Further remarks: The proof of Ramseyness outlined above resembles the "filter games" introduced by Holy and Schlicht and further studied by Nielsen and Welch. Defining ($\omega+1$)-strategic strongness in an analogous way, I think I can prove that it is equiconsistent with strongness. (And again the argument only requires the nonexistence of a winning strategy for player I. The previous sentence is wrong; I will add more details later.) However, the proof goes through ${\bf\Sigma}^1_4$ generic absoluteness and does not seem to generalize to other strategic large cardinals. REPLY [10 votes]: This answer addresses only the consistency strength of $\omega+1$-strategically measurable. Claim: If there is a $\omega+1$-strategically measurable cardinal then there is an inner model with a measurable cardinal. Proof: A winning strategy for $G_{\omega+1}(H(\kappa^{+}), \mathcal{N})$ for some transitive $\mathcal{N}$ as in the definition of $\omega+1$-measurability implies that there is a winning strategy for the Mycielski game (the "cut and choose" game) on $\kappa$. In the Mycielski game, a sequence of subsets of $\kappa$ $A_0, A_1, \dots, A_n, \dots$ in constructed in the following way. Let us denote $A_{-1} = \kappa$. At step $n < \omega$, player I picks a partition of $A_{n-1}$ into two sets $B_n, C_n$ (so $B_n \cup C_n = A_{n-1}$) and player II chooses one of them to be $A_n$. Player II wins if $|\bigcap_{n < \omega} A_n| > 1$. Given a winning strategy for the game $G_{\omega+1}(\mathcal{M},\mathcal{N})$, where $\mathcal{M}, \mathcal{N}$ are as in the question, we can construct a winning strategy for the Mycielski game as follows. Player II will maintain an auxiliary play in the game $G_{\omega+1}(\mathcal{M},\mathcal{N})$. At step $n < \omega$, let $(B_n, C_n)$ be the partition that player I played. Player II define $x_n = (B_n, C_n)$ and use the winning strategy of $G_{\omega+1}(\mathcal{M},\mathcal{N})$ in order to produce some $y_n \in \mathcal{N}$. Clearly, $y_n = (B_n^\star, C_n^\star)$. Now, player II will pick $A_n$ to be $B_n$ iff $\kappa \in B_n^\star$ (and then $A_n^\star = B_n^\star$. Otherwise, $A_n^\star = C_n^\star$). We want to make sure that player II wins. Indeed, in the constructed play in the game $G_{\omega+1}(\mathcal{M},\mathcal{N})$, we obtained the sequence of steps $\langle x_n, y_n \mid n < \omega\rangle$. Player II can ask what the winning strategy will answer to $x_{\omega} = \langle A_n \mid n < \omega\rangle$. Let $y_\omega$ be the answer. Then, by elementarity $y_n = \langle A_n^\star \mid n < \omega\rangle$. Now, since $\kappa \in \bigcap A_n^\star$, $\bigcap A_n$ cannot be bounded in $\kappa$. It is known that the existence of a winning strategy for player II in the Mycielski game is equiconsistent with a measurable cardinal (see "The evolution of large cardinals axiom in set theory" by Kanamori and Magidor, section 27. The argument is attributed there to Silver and Solovay). QED Let me remark that it is consistent to have a winning strategy for the Mycielski game at accessible cardinals. For example, it is consistent there is a winning strategy for the Mycielski game on $\omega_2$.<|endoftext|> TITLE: Examples of set theory problems which are solved using methods outside of logic QUESTION [10 upvotes]: The question is essentially the one in the title. Question. What are some examples of (major) problems in set theory which are solved using techniques outside of mathematical logic? REPLY [8 votes]: Inspired by Joel David Hamkin's comment---Simon Thomas has provided applications of various super-ridigity theorems (from the ergodic theory of group actions) to the theory of the Borel complexity of countable equivalence relations, for example he shows that the universal countable equivalence relation is not essentially free Thomas, Simon, Popa superrigidity and countable Borel equivalence relations, Ann. Pure Appl. Logic 158, No. 3, 175-189 (2009). ZBL1162.03029.<|endoftext|> TITLE: When is the semidirect product of an elementary abelian group and a cyclic group generated by two elements? QUESTION [9 upvotes]: I am trying to characterize when a semi-direct product of the form $(Z/pZ)^n \rtimes (Z/qZ)$ is isomorphic to a group generated by two elements. Here $p$ and $q$ are distinct odd primes. I would be happy for a reference or even some examples of this happening for $n > 1.$ (I found a similar question here in the case that $p=2$, but the odd case may be considerably different, and I am looking for different structural information any way.) Thanks! REPLY [6 votes]: This answer corroborates YCor's claim according to which the conditions $n_i \le 1$ for $i > 1$ and $n_1 \le 2$ on the irreducible modular representations'multiplicities $n_i$, are necessary and sufficient for $G$ to be two-generated. We actually show a slightly more general result expressed in terms of the geometric multiplicity of the eigenvalue $1$. Let $K = Z/pZ, V = K^n$ and $C = Z/qZ$ with $p$ a prime number and with $q \ge 2$ an integer. We fix a group homomorphism $\varphi: C \rightarrow GL(V)$ and a favored generator $a$ of $C$. Let $G = V \rtimes_{\varphi} C$ the corresponding semi-direct product. The $K$-vector space $V$ is endowed with the structure of $K[X]$-module induced by $X \cdot v = \varphi(a)(v)$. If $V$ is a cyclic $K[X]$-module generated by $w$, then $G$ is generated by $(w, a)$, hence two-generated. The converse holds if $G$ is centerless. Claim 1. Assume that the center of $G$ intersects $V$ trivially, i.e., $\varphi(a)$ has no non-zero fixed vector. If $G$ is two-generated then $V$ is a cyclic $K[X]$-module. Proof. Since $\varphi(a)$ has no non-zero eigenvector associated to $1$, the minimal polynomial $\mu(X)$ of $\varphi(a)$ over $K$ is coprime with $X - 1$. As $\mu(X)$ divides $X^q - 1$, this minimal polynomial divides $\nu(X) = 1 + X + \cdots + X^{q -1}$. Let $(v_1a^{n_1}, v_2a^{n_2})$ be a generating pair of $G$ with $v_i \in V, n_i \in \mathbb{Z}$ for $i = 1,2$. We can reduce this pair to a generating pair of the form $(v, wa)$ with $v,w \in V$ by means of Nielsen transformations (consider the induced transformations on $C^2$). Given $u \in V$, we shall show that $u \in K[X] \cdot v$. Since $(v, wa)$ generates $G$, there is a word $x$ on the alphabet $\{x_1^{\pm 1}, x_2^{\pm 1}\}$ such that $u = x(v, wa)$. Write $x = yx_2^{s}$ where $y$ belongs to the normal closure of $x_1$ in the free group $F(x_1, x_2)$ and $s$ lies in $\mathbb{Z}$. Note that we have $y(v, wa) \in K[X]\cdot v$ because $K[X]\cdot v$ is normalized by $wa$. It follows that $s = kq$ for some $k \in \mathbb{Z}$ and hence $u = y(v, wa) + k \nu(X) \cdot w = y(v, wa)$, which completes the proof. Checking whether $V$ is a cyclic $K[X]$-module can be done algorithmically by computing the Smith Normal Form of $\varphi(a)$ [Theorem 20 of Section 12, 1]. If the center $Z(G)$ of $G$ has a non-trivial intersection with $V$, then $G/(Z(G) \cap V)$ yields a semi-direct product decomposition $V' \rtimes C$ where $V'$ is a finite-dimensional vector space over $K$ of lower dimension. The space $V'$ is obtained as the direct sum of the non-trivial irreducible components of the representation $\varphi$ when $p$ does not divide $q$ as modular representations are completely reducible in this case. If $G$ is moreover two-generated, then $V'$ must be a cyclic $K[X]$-module. In addition, $Z(G)$ must be two-generated. I am indebted to Derek Holt and Geoff Robinson for the latter remark which is expanded below. If $G$ is the semi-direct product $V \rtimes C$ of an arbitrary Abelian group $V$ with an arbitrary cyclic group $C$, then it is easily checked that $Z(G) = (Z(G) \cap V) \times (Z(G) \cap C)$. In the context of the question, $C$ is simple so that $Z(G) \cap C$ is either equal to $C$, in which case $G$ is Abelian, or trivial. If moreover $p$ does not divide $q$, it follows that $Z(G)$ is a direct factor of $G$. Thus, if $G$ is two-generared, so is $Z(G)$. The above claim and the subsequent remark allows us to recover YCor's claim. Assume that $p$ does not divide $q$. Then the following are equivalent: $G$ is two-generated; the multiplicity of any non-trivial irreducible representation in $\varphi$ is at most $1$, the multiplicity of the trivial representation is at most $2$. Proof. In order to show that the condition is sufficient, we can reduce to the case of a cyclic $K[X]$-module by moving out of $V$ one $Z/pZ$-factor of $V$ with trivial $Z/qZ$-action if necessary, noting that $Z/pZ \times Z/qZ \simeq Z/pqZ$. The fact that the condition is necessary follows from Claim 1. Addendum. The following generalization relies on the fact that the restriction of $\varphi(a)$ to the generalized eigenspace associated to $1$ has a Jordan normal form. Claim 2. Let $p$ be a prime number and let $q \ge 2$ be an integer. Let $A = \varphi(a)$ and let $\mu_A$ be the minimal polynomial of $A$ over $K$. Write $\mu_A(X) = (X - 1)^k P(X)$ with $P(X) \in K[X]$ and $P(1) \neq 0$. Let $I \in GL_n(K)$ be the identity matrix and let $E = \ker((A - I)^k)$. Let $\gamma_1(A) = \dim_K(\ker(A - I))$. We set $$ d(A) = \left\{ \begin{array}{cc} \gamma_1(A) + 1& \text{ if } p \text{ divides } q, \\ \gamma_1(A) & \text{ otherwise. } \end{array} \right. $$ Then the group $G = V \rtimes_{\varphi} C$ is two-generated if and only if $V/E$ is a cyclic $K[X]$-module and $d(A) \le 2$. Proof. The abelianization of $G$ surjects onto an elementary Abelian $p$-group of rank $d(A)$. Thus the condition on $d(A)$ is necessary. By Claim 1, the quotient $G/E$ is two-generated if and only if $V/E$ is a cyclic $K[X]$-module. Now, we only need to verify that $G$ is two-generated when $d(A) = 2$ and $p$ does not divide $q$. Indeed, for the other cases, the restriction of $\varphi(a)$ to $E$ has at most one Jordan block so that $V$ is a cyclic $K[X]$-module if $V/E$ is. If $d(A) = 2$ and $p$ doesn't divide $q$, then the restriction of $\varphi(a)$ to $E$ must be the identity, since otherwise its multiplicative order, which is a divisor of $q$, would be a multiple of $p$ (consider the top left $2$-by-$2$ sub-matrix of a Jordan block). We conclude as before, by moving out of $V$ one of the two point-wise $C$-invariant $Z/pZ$-factors. [1] D. Dummit and R. Foote, "Abstract Algebra", 1999.<|endoftext|> TITLE: Dehn-Nielsen-Baer Theorem for surfaces with boundary and punctures QUESTION [5 upvotes]: Let $S=S_{g,b}$ be a compact orientable surface with genus $g$ and $b$ boundary components, such that $\chi(S)=2-2g-b<0$. Let $Q=\{x_1,\ldots , x_n\}$ be a set of $n$ distinguished points in the interior of $S$. Define ${\rm Mod}(S,\{Q\})$ to be the mapping class group of orientation-preserving self-homeomorphisms of $S$ which fix the boundary $\partial S$ pointwise and preserve $Q$ setwise, modulo isotopies of the same type. I believe this is the same as the mapping class group ${\rm Mod}(S-Q)$ of the punctured surface with boundary $S-Q$. Now fixing a basepoint $d\in \partial S$, there are homomorphisms $$ {\rm Mod}(S,\{Q\})\to {\rm Aut}(\pi_1(S-Q,d))\to {\rm Out}(\pi_1(S-Q,d)). $$ Is there a version of the Dehn-Nielsen-Baer theorem which states that the above composition is isomorphic onto the subgroup ${\rm Out}^*(\pi_1(S-Q,d))$ of ${\rm Out}(\pi_1(S-Q,d))$ consisting of outer automorphisms which preserve the individual conjugacy classes represented by the components of $\partial S$, and permute the conjugacy classes represented by simple closed curves around the punctures? We have looked in the following sources, which are all very helpful but seem to only treat the cases of boundary and punctures separately: Farb, Benson; Margalit, Dan, A primer on mapping class groups, Princeton Mathematical Series. Princeton, NJ: Princeton University Press (ISBN 978-0-691-14794-9/hbk; 978-1-400-83904-9/ebook). xiv, 492 p. (2011). ZBL1245.57002. Ivanov, Nikolai V., Mapping class groups, Daverman, R. J. (ed.) et al., Handbook of geometric topology. Amsterdam: Elsevier. 523-633 (2002). ZBL1002.57001. Zieschang, Heiner; Vogt, Elmar; Coldewey, Hans-Dieter, Surfaces and planar discontinuous groups. (Poverkhnosti i razryvnye gruppy)., Moskva: Nauka. 688 p. (1988). ZBL0701.57001. Boldsen, Soren, Different versions of mapping class groups of surfaces, https://arxiv.org/abs/0908.2221 REPLY [2 votes]: The issue here is Dehn twists along curves parallel to the circles of $\partial S$. These usually generate infinite cyclic subgroups of ${\rm Mod}(S,Q)$, the only exceptions being when $S$ is a disk and $Q$ is empty or a single point. If one chooses a basepoint in $\partial S$ then these Dehn twists induce the identity on $\pi_1(S-Q)$ except for a twist along a curve parallel to the component of $\partial S$ containing the basepoint, which induces the inner automorphism which ccnjugates $\pi_1(S-Q)$ by the loop given by this component of $\partial S$. Thus the composition ${\rm Mod}\to {\rm Aut}\to {\rm Out}$ will usually have a nontrivial kernel.<|endoftext|> TITLE: Naming in math: from red herrings to very long names QUESTION [30 upvotes]: The are some parts of math in which you encounter easily new structures, obtained by modifying or generalizing existing ones. Recent examples can be tropical geometry, or the theory around the field with one element. If one works in those areas, one cannot avoid the problem of naming new objects. When working with such a "new" notion, more general than an existing one, you have different options to name it. Either the red herring option, like group without inverses, a brand new name, like monoid, a derived name, like semigroup (which is actually a group without inverses and without an identity element), or no name at all, so a very long name, i.e. the category $M$ of sets with an associative binary operation. Or even you can also decide to use the old name with a new meaning. (The examples I wrote don't pretend to have a historical justification). Although the red herring construction is used everywhere in math, I feel that it is not a good practice. To use the old name with a different meaning can be the origin of a lot of errors. And the option of not giving any name at all is like you elude your responsibility, so if someone needs to use it they will have to put a name to it (maybe your name?). So my preferred options are to choose a derived name or a new name. Derived names are quite common: e.g. quasicoherent, semiring, pseudoprime, prescheme (which is an old term), and they contain some information which is useful, but sometimes they are ugly, and it could seem you don't really want to take a decision: you just write quasi/semi/pseudo/pre in front of the name. But new names can be difficult to invent, to sell and to justify: if you decide to give the name jungle to a proposed prototype of tropical variety, because it sounds to you that in the tropics are plenty of jungles, it is a loose justification and probably will have no future (unless you are Grothendieck). My question is: Which do you think is the best option? In fact, the situation can be worse in some cases: what happens if some name has already been used but you don't agree with the choice? Is it adequate to modify it, or can it be seen as some sort of offense? I could put some very concrete examples, even papers where they introduce red herrings, new meanings for old names, new names and no-names for some objects, all in the same paper. But my point is not to criticize what others did but to decide what to do. REPLY [2 votes]: For me, I think naming is not as important as understanding what it is. We all understand Fourier transform, Banach space, Peter-Weyl theorem, Pauli matrix. If we rename just one of them, what would we call? As someone who used to work as a translator and is very inquiring about how an idea forms and evolves, I have some chances to observe how a same concept is translated to different terms, and how a same object is renamed several times in different contexts. Perhaps this practice is exotic in academical mathematics, and I admit that I have zero experience in graduate research, but cognitively I don't think our brains are that different. When does a perfect name come? When you are in a rush. When you are in a rush, you don't have time to think about correctness, you just want to make it quick to solve another important, urgent problem. Your brain will cut all unnecessary information about the object, leaving just enough bit so you can jump to conclusion, or in this case, a name. Those unnecessary bits may be essential to for the concept to form at the first place, but unfortunately, don't really relate to the surrounding concepts in the sentence. The surviving bits that constitute the new name are the ones closer to the surroundings. So even when the original name is short and accurate, it will be replaced by a new name that fits the context.¹ In research, you are expected to be accurate, and you have plenty of time to learn it, but the principle is the same. My advice is to try using the concept to study many more other concepts, and see how your mind reacts with it when you are in a rush. You can also discuss those other concepts with your colleagues, and see how they complete this sentence when you are stuck: "you mean the ______?" Of course in the realm of cognitive science and linguistic, sometimes you have to accept a sticking bad name. But lucky for us, this is also the realm of math, and unlike jellyfish or pineapple, non-abelian fermion or group without inverses aren't that imaginative, so they will always invoke an unpleasant feeling when reading it. Hopefully one day we can go around them, and let them rest in peace. Here is a challenge, inspired from the small-world network: try explaining a topic by introducing only 6 intermediate terms. This will force you to twist what you already know about it, so that you can view it in a different perspective. You have to be bold to cut off the details that rock your soul, but by then the big picture will emerge. Only after seeing the big picture that naming can become a piece of cake. I have an article for this, you can check it out: Making concrete analogies and big pictures. ¹ This is my own theory, but is inspired from Gentner's Structure Mapping Theory Related: Hyphens after the prefixes “non-” and “anti-” in mathematics<|endoftext|> TITLE: A matrix identity related to Catalan numbers QUESTION [12 upvotes]: Let $$C_n=\frac{1}{2n+1}\binom{2n+1}{n}$$ be a Catalan number. It is well-known that $$(\sum_{n\ge{0}}C_n x^n)^k=\sum_{n\ge{0}}C(n,k)x^n$$ with $$C(n,k)=\frac{k}{2n+k}\binom{2n+k}{n}.$$ It is also known that the Hankel matrix $\left( {{C(i+j,2)}} \right)_{i,j = 0}^{n - 1}$ can be factored in the form $$\left( {{C(i+j,2)}} \right)_{i,j = 0}^{n - 1}=A_{n} A_{n}^T,$$ where $$A_{n}=\left(\binom{2i+1}{i-j}\frac{2(j+1)}{i+j+2}\right) _{i,j = 0}^{n - 1}=\left(\binom{2i+1}{i-j}-\binom{2i+1}{i-j-1}\right) _{i,j = 0}^{n - 1}.$$ Computer experiments suggest that for $k\ge{1}$ $$\left( {{C(i+j,k+2)}} \right)_{i,j = 0}^{n - 1}=A_{n}G_{n,k} A_{n}^T$$ with $$G_{n,k}=\left({g(i,j,k)}\right) _{i,j = 0}^{n - 1},$$ where $$g(i,j,k)= \sum_{m={|i-j|-1}}^{i+j}\binom{k-1}{m}. $$ Is there a simple proof of this identity? REPLY [12 votes]: After unpacking the equation $$\left( {{C(i+j,k+2)}} \right)_{i,j = 0}^{n - 1}=A_{n}G_{n,k} A_{n}^T$$ we see that we want to prove the identity $$C(i+j,k+2)=\frac{k+2}{(2i+2j+k+2)}\binom{2i+2j+k+2}{i+j}$$ $$=\sum_{0\le r\le i,0\le s\le j} \left[\binom{2i+1}{i-r}-\binom{2i+1}{i-r-1}\right]\cdot\left[\sum_{m=|r-s|-1}^{r+s}\binom{k-1}{m}\right]\cdot \left[\binom{2j+1}{j-s}-\binom{2j+1}{j-s-1}\right] \tag{*}$$ The left hand side is the coefficient of $x^{-1-k}$ in $F(x)=(1-x^2)\left(x+\frac{1}{x}\right)^{2i+2j+k+1}$. We will be done if we can show that the right hand side is the coefficient of $x^{-1-k}$ in $$\left[(1-x^2)\left(x+\frac{1}{x}\right)^{2i+1}\right]\cdot\left[\frac{1}{1-x^2}\left(x+\frac{1}{x}\right)^{k-1}\right]\cdot\left[(1-x^2)\left(x+\frac{1}{x}\right)^{2j+1}\right]$$ which is easily seen to also equal $F(x)$. If we fix $r\le i,s\le j$, the contribution that comes from monomials of the form $x^{-1-2r}x^{2r+2s-k+1}x^{-1-2s}$ is $$\left[\binom{2i+1}{i-r}-\binom{2i+1}{i-r-1}\right]\cdot\left[\sum_{m\le r+s}\binom{k-1}{m}\right]\cdot \left[\binom{2j+1}{j-s}-\binom{2j+1}{j-s-1}\right]$$ which is nonzero in the case $r\geq 0,s\geq 0$, or $r\geq -s\geq 1$, or $s\geq -r\geq 1$. The second case can be rewritten with $s'=-s-2$ as $$-\left[\binom{2i+1}{i-r}-\binom{2i+1}{i-r-1}\right]\cdot\left[\sum_{m\le r-s'-2}\binom{k-1}{m}\right]\cdot \left[\binom{2j+1}{j-s'}-\binom{2j+1}{j-s'-1}\right]$$ and the third case can be written with $r'=-r-2$ as $$-\left[\binom{2i+1}{i-r'}-\binom{2i+1}{i-r'-1}\right]\cdot\left[\sum_{m\le s-r'-2}\binom{k-1}{m}\right]\cdot \left[\binom{2j+1}{j-s}-\binom{2j+1}{j-s-1}\right]$$ Combining everything together we see that for $r\geq s$ the total contribution simplifies to $$\left[\binom{2i+1}{i-r}-\binom{2i+1}{i-r-1}\right]\cdot\left[\sum_{m\le r+s}\binom{k-1}{m}-\sum_{m\le r-s-2}\binom{k-1}{m}\right]\cdot \left[\binom{2j+1}{j-s}-\binom{2j+1}{j-s-1}\right]$$ $$=\left[\binom{2i+1}{i-r}-\binom{2i+1}{i-r-1}\right]\cdot\left[\sum_{m= r-s-1}^{r+s}\binom{k-1}{m}\right]\cdot \left[\binom{2j+1}{j-s}-\binom{2j+1}{j-s-1}\right]$$ and similarly for $s\geq r$ we get $$\left[\binom{2i+1}{i-r}-\binom{2i+1}{i-r-1}\right]\cdot\left[\sum_{m\le r+s}\binom{k-1}{m}-\sum_{m\le s-r-2}\binom{k-1}{m}\right]\cdot \left[\binom{2j+1}{j-s}-\binom{2j+1}{j-s-1}\right]$$ $$=\left[\binom{2i+1}{i-r}-\binom{2i+1}{i-r-1}\right]\cdot\left[\sum_{m= s-r-1}^{r+s}\binom{k-1}{m}\right]\cdot \left[\binom{2j+1}{j-s}-\binom{2j+1}{j-s-1}\right]$$ which is exactly the right hand side in $(*)$.<|endoftext|> TITLE: The ¨irreducible¨ representation variety of surface group QUESTION [5 upvotes]: Let $S$ be a closed surface of genus larger than 1, $G$ be a compact, simply connected simple Lie group with finite center. Consider the representation variety $M(S,G)=Rep(\pi_1(S), G)$. Witten´s formula gives a way to calculate the volume of this space. Now Let $X$ be the subset of this variety consisting of irreducible representations of the surface group $\pi_1(S)$ to $G$, and let´s call it the ¨irreducible¨ representation variety of surface group. My questions are: Can we say anything about the distribution of $X$? For example, is it an open submanifold of Zariski dense part of $M(S,G)$? And is it connected? Do we have some formulas, similar to Witten´s, to calculate the volume of $X$ w.r.t symplectic form that restricted from $M(S,G)$? Answers and References are very welcome. Thanks a lot. REPLY [2 votes]: Let $\Sigma$ be a closed orientable surface of genus $g\geq 2$, and $\pi$ its fundamental group. Let $X(\pi, G)=Hom(\pi,G)/G$ be the conjugation quotient for $G$ a compact Lie group whose derived subgroup $DG$ is simply connected. From: Ho, Nan-Kuo; Liu, Chiu-Chu Melissa. Connected components of spaces of surface group representations.II. Int. Math. Res. Not. (2005), no. 16, 959–979, we know $X(\pi,G)$ is connected (irreducible). Moreover, the reducible locus is a proper Zariski closed subset. The fact that a free group of rank $g\geq 2$ surjects onto $\pi$, implies the irreducible locus is non-empty. Thus, the locus of irreducible representations is connected, open, and dense. Since the reducible locus is proper and Zariski closed, any volume calculation for $X(\pi, G)$ will be the same for the irreducible locus. However, even if $G$ is simply-connected and simple (as you postulate), the irreducible locus will generally have orbifold singularities (unless $G=SU(n)$). So it is not a submanifold of the locus of Zariski-dense representations (those whose image is Zariski dense).<|endoftext|> TITLE: Hopf dual of the Hopf dual QUESTION [16 upvotes]: Given any Hopf algebra $A$ over a field $k$, one can also define the Hopf dual $A^*$ of as follows: Let $A^∗$ be the subspace of the full linear dual of $A$ consisting of elements that vanish on some two-sided ideal of $A$ of finite codimension. Then $A^∗$ has a natural Hopf algebra structure. Question: Is the Hopf dual of the Hopf dual of $A$ isomorphic to $A$. It is not obvious for me that it is. If not, then do we know in which cases it is true? REPLY [2 votes]: Regarding your first question: the answer is generally no, the restricted dual of the restricted dual of $A$ is generally not isomorphic to $A$: $$ (A^{\circ})^\circ\ncong A $$ as has already been indicated by the counterexamples of darij grinberg's answer. More counterexamples can be constructed when the restricted dual (or: Sweedler's dual) $A^\circ$ is trivial while $A$ is not: Consider the case when the algebra $A$ has no finite dimensional representations, except the zero vector space $V=\{0\}$. Now, recall that $f\in A^\circ$ $\Longleftrightarrow$ $\dim(A\rightharpoonup f)<\infty$ where the action of the $A\rightharpoonup f$ module is defined by $(a\cdot f)(b)=f(ba)$ for all $a,b\in A$ and $f\in A^\circ$ (this is a standard result, see for example Montgomery's book, Lemma 9.1.1, p. 149). Thus, If $A$ has no finite dimensional representations apart from the zero vector space $V=\{0\}$, then $A^\circ=\{0\}$. Regarding your second question, unfortunately i have no examples or conditions available (and i think it is a difficult task to find something similar in the literature).<|endoftext|> TITLE: Algorithm to decide if the union of a set system covers the power set QUESTION [6 upvotes]: Assume that we have a set system $\mathfrak T = \{\mathcal T_1, \mathcal T_2, \dots, \mathcal T_N \}$ where each $\mathcal T_k$ is a collection of subsets of $[n] := \{1,\dots,n\}$ of the form $$ \mathcal T_k = [m_k, M_k] := \{T \subseteq [n]:\; m_k \subseteq T \subseteq M_k \}. $$ Moreover, we know that $\mathcal T_k$s are mutually disjoint, i.e., $\mathcal T_k \cap \mathcal T_{k'} = \emptyset$ when $k \neq k'$, which implies that $N \le 2^n$. Note also that $\mathcal T_k \subseteq 2^{[n]}$ for each $k$. Question: Is there an algorithm with polynomial time in $(N,n)$ to either find a subset in $2^{[n]}$ which is not in any of the $\mathcal T_k$s when possible, otherwise conclude that $\bigcup \mathfrak T := \bigcup_{k=1}^N \mathcal T_k = 2^{[n]}$, i.e. the set system covers the power set? EDIT: Changed the number of subsets to $N$ instead of $m$. Corrected the question from polynomial time in $n$ to polynomial time in $(N,n)$. Also, fixed the statement regarding the union of elements of $\mathfrak T$. There were many confused statements in the original post. Hopefully, this version is clear. The problem is to see if we can cover the power set, not the ground set $[n]$. That is, all the action is at the level of subsets of $[n]$. REPLY [3 votes]: Note that because $T_k$ are disjoint, it is trivial to check whether they cover $[n]$: just add up their sizes ($|T_k| = 2^{|M_k| - |m_k|}$) and compare the sum with $2^n$: if it is smaller than $2^n$, then there exists a uncovered subset of $[n]$, otherwise there is no such subset. Quite often, if there exists a polynomial time algorithm for decision problem, there also is a polynomial time algorithm for finding a certificate. In the case of this problem it is true as well. Indeed, consider the following procedures: do_cover($n, T$): $~~~~$Returns whether $T$ covers $[n]$. Checks this by simply summing up $|T_k|$. reduce($n, T, b$): $~~~~$Reduces each $T_k$ by leaving only sets that contain $n$ or not $~~~~$if $b = 1$ or $b = 0$ respectively. In case of $b = 1$ element $n$ $~~~~$is also deleted from all sets that belong to $T_k$. $~~~~$Returns reduced $T$ (and does not modify $T$) $~~~~$This reduction preserves disjointness of $T_k$ and the $~~~~$fact that $T_k = [m_k, M_k]$ for some $m_k, M_k \subset [n]$. $~~~~$Reduced $T_k$ may become empty, but that does not matter much. find_uncovered($n, T$): $~~~~$if do_cover($n, T$): $~~~~~~~~$no such element then $~~~~$if n == 0: $~~~~~~~~$just try both possibilities $~~~~$Now, we know for sure that there exists a subset of $[n]$ $~~~~$that is uncovered and want to find this subset. $~~~~$There are 2 possibilities: either it contains element $n$, $~~~~$or it does not. We will just iterate over them. $~~~~$if do_cover($n - 1$, reduce($n, T, 0$)): $~~~~~~~~$return find_uncovered($n - 1$, reduce($n, T, 0$)) $~~~~$else if do_cover($n - 1$, reduce($n, T, 1$)): $~~~~~~~~$return find_uncovered($n - 1$, reduce($n, T, 1$)) $\cup \{ n \}$ $~~~~$else: $~~~~~~~~$can't happen, because there is an uncovered set Clearly, find_uncovered runs in polynomial time of $n$ and $N$ (because do_cover does). Now, note that the same problem is NP-hard if there is no restriction of $T_k$ being disjoint. Indeed, consider some 3-SAT instance with $n$ variables $x_1, x_2, \ldots, x_n$ and $N$ clauses. $T_k$ for $k = 1, 2, \ldots, N$ will consist exactly from such subsets $S$ of $[n]$ such that setting $x_i := (i \in S)$ for all $i = 1, 2, \ldots, n$ will make $k$-th clause false. Clearly, a solution of original problems for such $T_k$ would yield a solution to our 3-SAT instance. P. S. The solution above for the case of disjoint $T_k$ can be implemented quite efficiently if you forego the immutability of $T$ that I used for clarity. Indeed, you only need to keep $m_k, M_k$ and sum of $|T_k|$. All queries and changes that you'll do all end up very small: you just ask whether some element belongs to some set $m_k$ or $M_k$, do similar minor modifications to $m_k$'s and $M_k$'s (don't forget to update the sum of $|T_k|$, when doing so) and check whether the sum of $T_k$ is equal to $2^x$ for some $x$.<|endoftext|> TITLE: "Mächtigkeit" versus "Kardinalität"? QUESTION [7 upvotes]: In Cantor's set theory, is there any difference between the terms Mächtigkeit and Kardinalität ? REPLY [9 votes]: Here is Cantor's Beiträge zur Begründung der transfiniten Mengenlehre (Erster Artikel). Read the bottom four lines on the first page: ",Mächtigkeit' oder ,Cardinalzahl' von $M$ nennen wir $\ldots$". That looks a lot like Cantor intended the two to be names for the same concept. The two lines above that reinforce that idea.<|endoftext|> TITLE: Can we realize a graph as the skeleton of a polytope that has the same symmetries? QUESTION [16 upvotes]: Given a graph $G$, a realization of $G$ as a polytope is a convex polytope $P\subseteq \Bbb R^n$ with $G$ as its 1-skeleton. A realization $P\subseteq \Bbb R^n$ is said to realize the symmetries of $G$, if for each graph-automorphism $\phi\in\mathrm{Aut}(G)$ there is an isometry of $\Bbb R^n$ inducing the same automorphism on the edges and vertices of $P$. Question: Let $G$ be the 1-skeleton of a polytope. Is there a realization of $G$ that realizes all its symmetries? Note that if we know $G$ to be the skeleton of an $n$-polytope, then the symmetric realization does not have to be of the same dimension. In fact, this is not always possible. The complete graph $K_n,n\ge 6$ is realized as a neighborly 4-polytope, but its only symmetric realization is the simplex in $n-1\ge5$ dimensions. However, this is the only example I know of. One way to construct a counter-example would be to find an $n$-polytope with a skeleton $G$ that is not more than $n$-connected (e.g. $n$-regular), but where $\mathrm{Aut}(G)$ is no subgroup of the point group $O(n)$. Its symmetric realization must be in dimension $\ge n+1$, but it cannot be because of Balinski's theorem. REPLY [4 votes]: A counterexample is giving by the edge graph of the dual to the polytope of Bokowski, Ewald and Kleinschmidt, described in Gil Kalai's answer. Let $P$ be the BEK polytope, let $P^{\vee}$ be its dual and let $G$ be the edge graph of $P^{\vee}$. The polytope $P$ is simplicial so $P^{\vee}$ is simple and its edge graph $G$ is $4$-regular. The symmetry $\phi$ in BEK is a symmetry of the simplicial complex $\partial P$, so it induces a symmetry of $G$. By a theorem of Gil Kalai, any $4$-polytope $Q$ with edge graph $G$ is combinatorially isomorphic to $P^{\vee}$. So the dual polytope $Q^{\vee}$ of such a $Q$ would be combinatorially isomorphic to $P$. If $\phi$ were realized by a linear symmetry of $Q$, then this would induce a linear symmetry of $Q^{\vee}$, contradicting the result of Bokowski, Ewald and Kleinschmidt. It remains to check that $G$ is not the edge graph of a polytope in some dimension $d \neq 4$. Since $G$ is $d$-regular, we would have to have $d<4$ and $d \leq 2$ is clearly impossible. According to Mathematica's PlanarGraphQ function, $G$ is not a planar graph, so it is not the edge graph of a $3$-polytope. For those who want to experiment on their own, here are the facets (vertices numbered 0 through 9) of $P$: {{1,2,3,7}, {1,2,4,8}, {9,2,3,8}, {9,2,0,7}, {1,3,4,7}, {2,3,4,8}, {9,3,0,8}, {2,3,0,7}, {1,4,6,7}, {1,5,6,8}, {9,0,6,8}, {9,5,6,7}, {4,5,6,7}, {1,4,5,8}, {0,5,6,8}, {9,0,5,7}, {1,2,6,7}, {1,2,6,8}, {9,2,6,8}, {9,2,6,7}, {3,4,5,7}, {3,4,5,8}, {3,0,5,8}, {3,0,5,7}, {1,4,5,6}, {1,2,3,4}, {9,0,5,6}, {9,2,3,0}} and the edges of $G$ (vertices numbered 1 through 28). {{5, 1}, {6, 2}, {6, 3}, {7, 3}, {8, 1}, {8, 4}, {9, 5}, {11, 7}, {13, 9}, {13, 12}, {14, 2}, {14, 10}, {15, 10}, {15, 11}, {16, 4}, {16, 12}, {17, 1}, {17, 9}, {18, 2}, {18, 10}, {18, 17}, {19, 3}, {19, 11}, {19, 18}, {20, 4}, {20, 12}, {20, 17}, {20, 19}, {21, 5}, {21, 13}, {22, 6}, {22, 14}, {22, 21}, {23, 7}, {23, 15}, {23, 22}, {24, 8}, {24, 16}, {24, 21}, {24, 23}, {25, 9}, {25, 10}, {25, 13}, {25, 14}, {26, 1}, {26, 2}, {26, 5}, {26, 6}, {27, 11}, {27, 12}, {27, 15}, {27, 16}, {28, 3}, {28, 4}, {28, 7}, {28, 8}} The symmetry $\phi$ switches $1 \leftrightarrow 9$, $7 \leftrightarrow 8$, $0 \leftrightarrow 4$ and fixes $2$, $3$, $5$, $6$. Here is a drawing of $G$:<|endoftext|> TITLE: Homotopy orbits, spectra and infinite loop spaces QUESTION [11 upvotes]: Let $X$ be an (naive) $O(n)$-spectrum (I'm choosing to work with orthogonal spectra). I've recently come across the following results, $$(S^{n-1} \wedge X)_{hO(n)} \simeq X_{hO(n-1)}$$ and $$\Omega^\infty (X_{hG)}) \simeq (\Omega^\infty X)_{hG} $$ where $G$ is a compact Lie group. I've spent the last few days trying to find sources for these results - and trying to prove them myself to little avail. Any help/suggestions or references would be greatly appreciated. Added The answer by Tyler has shown that the second is false. This raises the question about how close can we get, in the following sense: If $X$ is $n$-connected, then how connected is the map $$ (\Omega^\infty X)_{hG} \to \Omega^\infty(X_{hG})\,? $$ REPLY [13 votes]: Both of these are false. The first is close to true: if $S(n-1)$ is the unit sphere in $\Bbb R^{n}$ with its standard $O(n)$-action, then we can identify $S(n-1)$ with $O(n) / O(n-1)$ and so get the identification $$ (S(n-1)_+ \wedge X)_{hO(n)} = EO(n)_+ \wedge_{O(n)} O(n)/O(n-1)_+ \wedge X = EO(n)_+ \wedge_{O(n-1)} X. $$ To show that the first one is false, let's take $X = \Sigma^\infty O(n)_+$. Then the first equation would give an equivalence between $\Sigma^\infty S^{n-1}$ and $\Sigma^\infty O(n)/O(n-1)_+ \cong \Sigma^\infty S(n-1)_+$. These have nonisomorphic homology for all $n$. For the second one, we can take $X$ to be the desuspension of the Eilenberg-MacLane spectrum $H\Bbb Z$ with trivial $O(n)$-action. Then $\Omega^\infty X$ is contractible and so its homotopy orbit space is $BG$ (or contractible, if you use the based orbit space), while $$ \begin{align*} \pi_n \Omega^\infty (\Sigma^{-1} H\Bbb Z)_{hG} &= \pi_n (\Sigma^{-1} H\Bbb Z_{hG})\\ &= \pi_{n+1} (H\Bbb Z_{hG})\\ &= \pi_{n+1} (H\Bbb Z \wedge \Sigma^\infty BG_+)\\ &= H_{n+1}(BG, \Bbb Z) \end{align*} $$ which is typically very different from either $\pi_n BG$ or $\pi_n (*)$.<|endoftext|> TITLE: What does it mean to say the first Goodwillie derivative of $TC$ is $THH$? QUESTION [7 upvotes]: A paradox: Goodwillie calculus considers only finitary functors. $TC$ isn't finitary. Yet in some sense $\partial(TC) = \partial(K) = THH$ is the crux of the Dundas-Goodwillie-McCarthy theorem. (Here, a finitary functor is one preserving filtered colimits[1]. $\partial$ denotes the first Goodwillie derivative. $TC,K,THH$ are respectively topological cyclic homology, algebraic $K$-theory, and topological Hochschild homology, regarded as functors from $E_1$-ring spectra to spectra.) Obviously I don't fully understand that last point, which is just a rough idea I think I've read somewhere, and that's what I want to ask about. Questions: What does it mean to say that the first Goodwillie derivative of $TC$ is $THH$? Is there a general formalism for Goodwillie calculus of non-finitary (but, say, accessible) functors? If so, how much of the usual theory goes through? If the answer to (2) is "yes", does it specialize in the case of $TC$ to recover the answer to (1)? [1] At any rate, in Goodwillie calculus one always requires one's functor to commute with sequential colimits. Any functor that commutes with sequential colimits and $\aleph_1$-filtered colimits commutes with all filtered colimits. $TC$ is defined from $THH$ (which commutes with filtered colimits) via a countable limit and therefore $TC$ commutes with $\aleph_1$-filtered colimits. I conclude that if $TC$ doesn't commute with filtered colimits, then it already doesn't commute with sequential colimits, and so standard Goodwillie calculus doesn't apply to it. REPLY [12 votes]: This answer addresses Question 1. Let "ring" mean associative unital ring spectrum, say in the $A^\infty$ sense. For a functor $F$ from rings to spectra (such as $TC$), differentiating $F$ at the ring $R$ means finding the best excisive approximation to the functor from rings-having-$R$-as-a-retract to spectra, $$ (R\to S\to R)\mapsto fiber (F(S)\to F(R)). $$ The universal excisive functor takes values in $R$-bimodules. (That is, a spectrum object for the category of rings over $R$ can be encoded in an $R$-bimodule.) Call it $\Omega_R$. If $S$ is $n$-connected relative to $R$ then $\Omega_R(S)$ is related to $fiber(S\to R)$ by a roughly $2n$-connected map of bimodules. The derivative of $F$ at $R$ must be $L\circ \Omega$ for some linear functor $L$ from $R$-bimodules to spectra. Thus to name the derivative of $F$ at $R$ you have to name a certain such linear functor. If $F$ is what I call analytic (or even if it is what I call stably excisive) then this means that $fiber (F(S)\to F(R))$ is related by a roughly $2n$-connected map to $L(S\to R)$. If $F$ is finitary then its derivative at $R$ is a finitary linear functor, and therefore the corresponding map $L$ is given by tensoring over $R\wedge R^{op}$ with a fixed bimodule. It happens that the derivative of $TC$ is finitary (even though $TC$ is not) and the fixed bimodule in question is $R$ itself. Tensoring an $R$-bimodule $M$ with $R$ gives $THH(R;M)$, so this means that in a stable range $fiber(TC(S)\to TC(R)$ looks like $THH(R;fiber (S\to R))$. There is an unfortunate clash of terminology. When differentiating a functor of spaces at $X$ you get an excisive functor from spaces-containing-$X$-as-a-retract to spectra. In Calculus 3 I called this the "differential" of $F$ at $X$, denoting it by $D_XF$. If $F$ is finitary then $D_XF$ is as well, and then it must be given by fiberwise smash product with a fixed parametrized spectrum over $X$. I denoted the fiber of that parametrized spectrum at $x\in X$ by $\partial_xF(X)$ and called it the (partial) derivative. So the differential is given by smashing with the derivative. In the setting of functors of rings, the "spectra" made out of objects over $R$ correspond to $R$-bimodules, whereas in the setting of functors of spaces the "spectra" made out of objects over $X$ correspond to parametrized spectra over $X$. In the latter setting "derivative" means the parametrized spectrum that you smash with to give the differential. By analogy in the former setting it ought to mean the bimodule you tensor with to give the differential. In those terms the derivative of $TC$ at $R$ is the bimodule $R$.<|endoftext|> TITLE: Are there examples of conjectures supported by heuristic arguments that have been finally disproved? QUESTION [48 upvotes]: The idea for this comes from the twin prime conjecture, where the heuristic evidence seems just so overwhelming, especially in the light of Zhang's famous result from 2014 about Bounded gaps between primes and its subsequent improvements. Are there examples of conjectures with some more or less "good" heuristic arguments, but where those arguments have finally not been "strong" enough? I don't mean heuristic in the sense that some conjecture holds for small numbers, e.g. the fact that $li\ x-\pi (x) $ was thought to be always positive, before Littlewood showed that not only it eventually changes sign, but does so infinitely often later on. So my question is different from the question about eventual counterexamples. Likewise, the fact that the first $10^{15}$ or so zeros of the zeta function "obey" RH does tell us something, but not a whole lot compared with infinity. So again this is not what I mean by heuristic. REPLY [6 votes]: I think Hauptvermutung (the "main conjecture" in German) is a good example. It certainly is supported by very plausible heuristic arguments, and nobody had any doubt for half a century.<|endoftext|> TITLE: What are the possible Stiefel-Whitney numbers of a five-manifold? QUESTION [19 upvotes]: On a compact five-manifold, the Stiefel-Whitney number $w_2w_3$ can be nonzero. An example is the manifold $SU(3)/SO(3)$, and also another example is a $\mathbb{CP}^2$ bundle over a circle where the holonomy is given by the complex conjugation automorphism of $\mathbb{CP}^2$. My question is whether the Stiefel-Whitney number $w_1^2w_3$ can be nonzero on a compact five-manifold. I haven't found either a simple proof that it is zero, or a simple example where it is nonzero. REPLY [30 votes]: Recall that on a closed $n$-manifold $M$, there is a unique class $\nu_k$ such that $\operatorname{Sq}^k(x) = \nu_kx$ for all $x \in H^{n-k}(M; \mathbb{Z}_2)$; this is called the $k^{\text{th}}$ Wu class. The Wu classes can always be written in terms of Stiefel-Whitney classes due to Wu's theorem which states that $w = \operatorname{Sq}(\nu)$. The first Wu class $\nu_1$ is $w_1$, so $\operatorname{Sq}^1(x) = w_1x$ for all $x \in H^4(M; \mathbb{Z}_2)$ where $M$ is a closed five-manifold. Now, by the Wu formula, we have $\operatorname{Sq}^1(w_3) = w_1w_3$. So $$w_1^2w_3 = \operatorname{Sq}^1(w_1w_3) = \operatorname{Sq}^1(\operatorname{Sq}^1(w_3)) = 0$$ as $\operatorname{Sq}^1\circ\operatorname{Sq}^1 = 0$ by the Ádem relations. In fact, $w_2w_3$ is the only Stiefel-Whitney number which can be non-zero and therefore defines an isomorphism $w_2w_3 : \Omega_5^O \to \mathbb{Z}_2$. To see this, note that the Stiefel-Whitney numbers of a closed five-manifold are $$w_1^5, w_1^3w_2, w_1^2w_3, w_1w_4, w_1w_2^2, w_2w_3, w_5.$$ It has already been established that $w_1^2w_3 = 0$ and $w_2w_3$ can be non-zero. Note that $w_5 = 0$ as it is the mod $2$ reduction of the Euler characteristic. One of the properties of Steenrod squares is that $\operatorname{Sq}^k(x) = 0$ if $k > \deg x$. In particular, if we consider $\operatorname{Sq}^k : H^{n-k}(M;\mathbb{Z}_2) \to H^n(M; \mathbb{Z}_2)$, we see that if $k > n - k$ (i.e. $k > \frac{n}{2}$), it must be the zero map and hence by Poincaré duality, we must have $\nu_k = 0$. So on a five-manifold, the classes $\nu_3$, $\nu_4$, and $\nu_5$ must be zero. As can be found here (see also this note), we have \begin{align*} \nu_3 &= w_1w_2\\ \nu_4 &= w_4 + w_1w_3 + w_2^2 + w_1^4\\ \nu_5 &= w_1w_4 + w_1^2w_3 + w_1w_2^2 + w_1^3w_2. \end{align*} As $\nu_3 = w_1w_2 = 0$, we have $w_1^3w_2 = 0$ and $w_1w_2^2 = 0$. Now note that $\nu_5 = w_1w_4 + w_1^2w_3 + w_1w_2^2 + w_1^3w_2 = w_1w_4$ so $w_1w_4 = 0$. Finally, as $\nu_4$ is zero, so is $w_1\nu_4 = w_1w_4 + w_1^2w_3 + w_1w_2^2 + w_1^5 = w_1^5$.<|endoftext|> TITLE: Number guessing game with lying oracle QUESTION [14 upvotes]: You are probably already familiar with the usual number guessing game. But for concreteness I restate it. The usual game The Oracle chooses a positive integer $n$ between 1 and 1024 (or any power of 2). At each turn, you make a guess $g$, and the Oracle tells you whether $n \leq g$ or $n > g$. The game ends when you can determine what $n$ is. It is pretty easy to see that the optimal strategy requires exactly 10 turns. The modified game In this game, the game runs almost exactly the same as the usual game. Except that there is a probability $p \in [0,1/2)$ (which is known to you) that the Oracle lies. More precisely, at each turn, the Oracle flips a (biased) coin and determines, independently of previous actions and other things in the game, whether to lie to you or not. When she lies she gives the exact opposite of the correct answer for the comparison $n \leq g$ or $n > g$. The game ends when you can determine, to a previously-agreed-upon confidence level, what is the answer $n$. (For argument sake, say that you can say with 95% probability what the value of $n$ is.) To model this, imagine you starting with 0 knowledge, so that each number between 1 and 1024 is equally likely. At each step you can update the probability distribution using the usual Bayesian updating procedure. The game ends when one of the numbers has probability 95% or higher. Question Can we estimate the expected number of guesses before the conclusion of the game, as a function of the lying probability $p$? (... and of the desired confidence level, and the number of bins?) Obviously, if the confidence level is anything above 50%, and if $p = 0$, the game reduces the usual game. As $p \to 1/2$, the expected number of guesses tend to infinity (at $p = 1/2$ the Oracle just responds randomly). So in particular, what are the asymptotics of the number of guesses as $p \to 0$ and as $p \to 1/2$? Numeric Data The simulations below using the following naive strategy for guesses: At each step, based on the current "prior" probability, guess the number $g$ such that the difference $|P(n \leq g) - P(n > g)|$ is minimized. In the case $p = 0$ this reduces (one can check) to the binary search method. This choice is made to "maximize information gained" from that step. I don't have a proof that this is the optimal strategy. From numerical simulations, with 1024 numbers, 95% confidence interval, and 3000 trials, p = 2^-2 avg = 56.806666666666665 p = 2^-3 avg = 23.315 p = 2^-4 avg = 16.112666666666666 p = 2^-5 avg = 13.047666666666666 p = 2^-6 avg = 11.633 p = 2^-7 avg = 10.604333333333333 p = 2^-8 avg = 10.285333333333334 p = 2^-9 avg = 10.156666666666666 p = 2^-10 avg = 10.069 p = 2^-11 avg = 10.029666666666667 p = 2^-12 avg = 10.011333333333333 p = 2^-13 avg = 10.005666666666666 p = 2^-14 avg = 10.0 p = 2^-15 avg = 10.0 For the other end, as $p \to 1/2$, as expected simulations take much longer, and so far I have the following data, based on 10000 simulation runs. p = 0.5 - 2^-3 avg = 234.0019 min=48 max=763 p = 0.5 - 2^-4 avg = 937.3372 min=236 max=3437 p = 0.5 - 2^-5 avg = 3750.5765 min=896 max=13490 Frequentist versus Bayesian James Martin brought up a very good point about frequentist versus Bayesian. Let me illustrate that with data from two runs using the strategy described above, with $p = 2^{-8}$. Starting game with 1024 bins. Running game now with secret answer: 857 Guessing 512 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0 My best guess is that the secret answer is 513 with probability 0.00194549560546875 Guessing 767 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000002 My best guess is that the secret answer is 768 with probability 0.003875850705270716 Guessing 895 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999999 My best guess is that the secret answer is 768 with probability 0.007721187532297775 Guessing 831 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000004 My best guess is that the secret answer is 832 with probability 0.015441436261097105 Guessing 863 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999986 My best guess is that the secret answer is 832 with probability 0.030880965812145108 Guessing 847 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000013 My best guess is that the secret answer is 848 with probability 0.0617618727298829 Guessing 855 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 0.9999999999999987 My best guess is that the secret answer is 856 with probability 0.12256258895055303 Guessing 859 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999982 My best guess is that the secret answer is 856 with probability 0.24704724796624977 Guessing 857 Is secret number (857) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999997 My best guess is that the secret answer is 856 with probability 0.4903092482458648 Guessing 856 Is secret number (857) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 0.9999999999999998 My best guess is that the secret answer is 857 with probability 0.9881889766208873 I did it! And it only took me 10 tries; the Oracle lied 0 times. Note that in binary, the number 857 is 1101011001 with more or less even distributions of zeros and ones. For 513 whose distribution is more extreme: Starting game with 1024 bins. Running game now with secret answer: 513 Guessing 512 Is secret number (513) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0 My best guess is that the secret answer is 513 with probability 0.00194549560546875 Guessing 767 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000033 My best guess is that the secret answer is 513 with probability 0.003875733349545551 Guessing 639 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999948 My best guess is that the secret answer is 513 with probability 0.007721187532297778 Guessing 575 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999996 My best guess is that the secret answer is 513 with probability 0.015323132493622977 Guessing 543 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000004 My best guess is that the secret answer is 513 with probability 0.030180182919776057 Guessing 527 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000004 My best guess is that the secret answer is 513 with probability 0.05857858737666496 Guessing 519 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.999999999999996 My best guess is that the secret answer is 513 with probability 0.1106260320552282 Guessing 515 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 1.0000000000000002 My best guess is that the secret answer is 513 with probability 0.19905831824030176 Guessing 513 Is secret number (513) bigger than guess? Oracle truthfully responds false Update probabilities Sanity check, total probabilities sum to 0.9999999999999991 My best guess is that the secret answer is 513 with probability 0.3315926624554764 Guessing 385 Is secret number (513) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 1.0000000000000002 My best guess is that the secret answer is 513 with probability 0.6614346507956576 Guessing 512 Is secret number (513) bigger than guess? Oracle truthfully responds true Update probabilities Sanity check, total probabilities sum to 0.9999999999999979 My best guess is that the secret answer is 513 with probability 0.9902152355556972 I did it! And it only took me 11 tries; the Oracle lied 0 times. Incidentally, this also explains (partly) the drop between $2^{-13}$ and $2^{-14}$ seen in the numerical results above. When $p = 2^{-13}$, when $n = 513$, after 10 truthful responses the confidence that 513 is the correct answer is "only" 94%. But the 95% threshold is breached when $p = 2^{-14}$. REPLY [5 votes]: If $p$ is small enough that you can upper bound the number of lies by some $k_{max}$, the problem is called Ulam's game. Some coding theoretic approaches work in this case, Berlekamp has worked on this. See David DesJardin's PhD thesis, e.g. See also, Rivest, Ronald L.; Meyer, Albert R.; Kleitman, Daniel J.; Winklmann, K. Coping with errors in binary search procedures 10th ACM Symposium on Theory of Computing. doi:10.1145/800133.804351. and Pelc, Andrzej (1989). "Searching with known error probability Theoretical Computer Science. 63 (2): 185–202. doi:10.1016/0304-3975(89)90077-7. See also this recent paper on arxiv for a good review of related results including lies as well as inversions and other disorder here<|endoftext|> TITLE: Prove $\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1$ QUESTION [6 upvotes]: The question is to prove: $$ \int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1. $$ Numerically it seems to hold true. So I have made some attempts to prove this analytically but have all failed. I also wonder if there is a systematic approach to solve this kind of problem. Thanks for your help. REPLY [7 votes]: Here is a proof of the inequality for $x\geq 2$. For the remaining range, see the Added section below. Let $\lambda:=1-1/\sqrt{2}$, then by convexity we have $$\frac{1}{\sqrt{1+s^2}}\leq 1-\lambda s^2,\qquad 0\leq s\leq 1.$$ Using this bound we can estimate \begin{align*}\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}\,ds&<\int_0^1\frac{1-\lambda s^2}{e^{sx}}\,ds+\int_1^\infty\frac{1-\lambda}{e^{sx}}\,ds\\[6pt] &=\frac{1}{x}-\lambda\left(\int_0^1\frac{s^2}{e^{sx}}\,ds+\int_1^\infty\frac{1}{e^{sx}}\,ds\right)\\[6pt] &=\frac{1}{x}-2\lambda\frac{1-e^{-x}x-e^{-x}}{x^3} .\end{align*} On the other hand, for $x>0$ we also have $$\arctan\left(\frac{1}{x}\right)=\int_0^{1/x}\frac{1}{1+s^2}\,ds>\int_0^{1/x}(1-s^2)\,ds=\frac{1}{x}-\frac{1}{3x^3},$$ hence it suffices to verify that $$\lambda(1-e^{-x}x-e^{-x})>\frac{1}{6},\qquad x\geq 2.$$ This is straightforward, so we proved the original inequality for $x\geq 2$. Added. One can cover the remaining range $1\leq x<2$ as follows. Let $f(x)$ denote the LHS and $g(x)$ denote the RHS in the original inequality. These two functions are decreasing, hence it suffices to verify the following $20$ numeric inequalities: $$f(1+(n-1)/20) TITLE: What is a true invariant of $G$-crossed braided fusion categories? QUESTION [9 upvotes]: Definition. An invariant of a (spherical) fusion category with extra structure is a number or a set or tuple of numbers preserved under (appropriate) equivalences. (Spherical) fusion categories have well-known invariants like their global dimension, their rank, the categorical and Frobenius dimensions of their simple objects, or the Frobenius-Schur coefficients. Ribbon fusion categories have some more invariants like the $S$-matrix, the twist eigenvalues, and all the invariants of its symmetric centre. Graded fusion categories have e.g. the size of the group and the dimension as an invariant. An example for something that is not an invariant is an $F$-matrix. It depends on the choice of basis vectors for trivalent morphism spaces, and that is not preserved by an equivalence. Summing over the appropriate elements of the $F$-matrices yields the Frobenius-Schur coefficients, though, and they are invariants. A (spherical) $G$-crossed braided fusion category (short: $G\times$-BFC) is a $G$-graded fusion category with a compatible $G$-action and a crossed braiding. This implies e.g. that its trivial degree is a ribbon fusion category. All this gives us access to the invariants I've already mentioned, but I want to know whether a $G\times$-BFC has any new invariants. Question. Are there any invariants of $G\times$-BFCs that are not invariants of the underlying $G$-graded fusion category, or of the trivial degree? For example, does the crossed braiding contain information beyond the trivial degree, and what information does the $G$-action possess? Bonus question. Can those new invariants be expressed diagrammatically, i.e. in the graphical calculus of $G\times$-BFCs? REPLY [6 votes]: Modular tensor categories give (via Reshetikhin-Turaev) a 321 oriented TFT. This gives a huge source of invariants, in particular any closed oriented 3-manifold gives a numerical invariant of MTCs. For example, the rank is the value of the 3-torus, while the global dimension is inverse of the square of the value of the 3-sphere. More elaborate invariants like the S-matrix and T-matrix also come up this way: the vector space assigned to the torus has a natural basis and the S-matrix and T-matrix are the value of the TFT on the mapping cylinders of the generators of the mapping class group of the torus. To any spherical G$\times$-BFC you get a 321 G-HTFT, that is a TFT defined on oriented bordisms endowed with a map to BG. So completely analogously to the above you get an invariant of G$\times$-BFC assigned to any closed oriented 3-manifold endowed with a map to BG. Similarly you can get more elaborate invariants by looking at bordisms with boundary. I'm not sure which ones are the most useful, but that's a big source of invariants. REPLY [4 votes]: This is not a direct answer to the question, but rather a longer comment. My claim is most of the information is contained in the $G$-extension of the trivial degree BFC. Let $\mathcal F=\bigoplus_{g\in G} \mathcal F_g$ be a $G\times$-BFC and assume that the braiding on $\mathcal F_e$ is non-degenerate and that the $G$-grading is faithful. Then the Drinfel'd center of $\mathcal F$ is braided equivalent to $\mathcal F^G\boxtimes \mathcal F_e^{\mathrm{rev}}$ by DMNO and the fact that $(\mathcal F^G)_G\cong \mathcal F$. Additionally, there are central functors $\tilde\iota\colon\mathcal F_e^\mathrm{rev}\to \mathcal F$ and $\mathcal F^G\to \mathcal F$. The $G\times$-BFC can be recovered from the central functor $\tilde\iota$. As a tensor functor $\tilde\iota$ is just the embedding $\iota\colon\mathcal F_e\hookrightarrow \mathcal F$, which is determined by the $G$-graded structure of $\mathcal F$ as a fusion category. So the only new additional information which $\mathcal F$ as a $G\times$-BFC has, is the braiding on $\mathcal F_e$ and the above central structure on the embedding $\mathcal F_e\hookrightarrow \mathcal F$. In many cases, this structure is unique. But, I am not sure which new invariants the central structure gives. The only thing which comes to my mind is the "crossed" S-matrix. And I am not even certain if this is a true invariant of the $G\times$-BFC, considering my above discussion. BTW, I showed that such a central structure always yields a $G$-crossed braided extension in the unitary setting [Prop. 2.4, https://arxiv.org/abs/1803.04949], namely: Let $\mathcal D$ be a unitary modular tensor category and $\mathcal F=\bigoplus_g \mathcal F_g$ a (faithfully) $G$-graded unitary extension of $\mathcal F_e=\mathcal D$ together with a central structure $\iota\colon \mathcal D^\mathrm{rev}\to \mathcal F$ on the canonical inclusion functor. Then $\mathcal F$ is naturally a $G$-crossed braided extension of $\mathcal D$. In the degenerate case, information seems to be lost. For example $\mathrm{Vec}$ is a $G\times$-BFC with $\mathrm{Vec}^G=\mathrm{Rep}(G)$, but $Z(\mathrm {Vec})\cong \mathrm {Vec}$ so the information about the equivariantization is not contained in the fusion category anymore, if you want.<|endoftext|> TITLE: Is this sum of cycles invertible in $\mathbb QS_n$? QUESTION [13 upvotes]: I am interested the following element of the group algebra $\mathbb{Q}S_n$: \begin{align} \phi_n=2e+(1\ 2)+(1\ 2\ 3)+\dotsb+(1\ldots n) \end{align} where $e$ is the identity permutation. My question is whether $\phi_n$ is a unit. For small $n$ I can see numerically that $\phi_n$ is a unit, but I have no idea how to prove it for general $n$. As a linear map from $\mathbb{Q}S_n$ to itself, for $n<10$, $\phi_n$ seems to be diagonalisable with positive integer eigenvalues, and I have no idea why this should be. REPLY [16 votes]: $\newcommand{\cyc}{\operatorname{cyc}} \newcommand{\id}{\operatorname{id}} \newcommand{\BB}{\mathbf{B}} \newcommand{\AA}{\mathbf{A}} \newcommand{\kk}{\mathbf{k}} \newcommand{\ww}{\mathbf{w}} $ PART 1 OF 3 [This is part of a long answer, which I had to split into 3 posts. Go to part 1. Go to part 2. Go to part 3.] The answer that follows is an elaboration on parts of the following preprint: Adriano Garsia, On the Powers of Top to Random Shuffling. In particular, the proofs I will give are taken from Section 3 of the preprint (but I include more details). Unlike Garsia, I mean the exposition below to be elementary, relying only on the content of a typical "abstract algebra 1" class plus the definition of a group ring. (I even plan on reusing some of this post in my next semester's combinatorics class.) 1. Notations Let $\kk$ be any commutative ring. For each integer $n$, we let $\left[ n\right]$ denote the set $\left\{ 1,2,\ldots,n\right\} $ (which is empty if $n\leq0$). Fix a nonnegative integer $n$. Let $S_n$ be the $n$-th symmetric group; this is the group of all $n!$ permutations of $\left[ n\right] $. We define multiplication on $S_n$ in the continental way (i.e., if $\alpha\in S_n$ and $\beta\in S_n$ are two permutations, then $\alpha\circ\beta$ is the permutation that sends each $i\in\left[ n\right] $ to $\alpha\left( \beta\left( i\right) \right) $). The identity permutation in $S_n$ will be called $\id $. (You are calling it $e$ instead.) The one-line notation of a permutation $\sigma\in S_n$ is defined to be the $n$-tuple $\left( \sigma\left( 1\right) ,\sigma\left( 2\right) ,\ldots,\sigma\left( n\right) \right) $. For any $k$ distinct elements $i_1, i_2, \ldots, i_k$, we let $\cyc_{i_1, i_2, \ldots, i_k}\in S_n$ be the permutation that sends $i_1, i_2, \ldots, i_{k-1}, i_k$ to $i_2, i_3, \ldots, i_k, i_1$, while leaving all other elements of $\left[ n\right] $ unchanged. (You are calling this permutation $\left( i_1, i_2, \ldots, i_k \right)$. But that notation would clash with my one-line notation.) Consider the group ring $\kk S_n$. We define an element $\AA \in\kk S_n$ by \begin{equation} \AA = \sum_{i=1}^{n} \cyc_{1,2,\ldots,i}. \label{eq.defA} \tag{1.1} \end{equation} For each $k\in\left\{ 0,1,\ldots,n\right\} $, we let $B_k$ be the set of all permutations $\sigma\in S_n$ satisfying $\sigma^{-1}\left( k+1\right) <\sigma^{-1}\left( k+2\right) <\cdots<\sigma^{-1}\left( n\right) $. That is, $B_k$ is the set of all permutations $\sigma$ such that the numbers $k+1,k+2,\ldots,n$ occur in this order in the one-line notation of $\sigma$. For example, if $n=3$, then \begin{align*} B_0 & =\left\{ \left( 1,2,3\right) \right\} ,\\ B_1 & =\left\{ \left( 1,2,3\right) ,\left( 2,1,3\right) ,\left( 2,3,1\right) \right\} ,\\ B_2 & =\left\{ \left( 1,2,3\right) ,\left( 1,3,2\right) ,\left( 2,1,3\right) ,\left( 2,3,1\right) ,\left( 3,1,2\right) ,\left( 3,2,1\right) \right\} =S_3, \end{align*} where we represent each permutation $\sigma\in S_3$ by its one-line notation. Note that if $k\geq n-1$, then the chain of inequalities $\sigma^{-1}\left( k+1\right) <\sigma^{-1}\left( k+2\right) <\cdots <\sigma^{-1}\left( n\right) $ is vacuously true for every $\sigma\in S_n$. Thus, \begin{equation} \text{if }k\geq n-1\text{, then } B_k = S_n. \label{eq.Bk=Sn} \tag{1.2} \end{equation} For each $k\in\left\{ 0,1,\ldots,n\right\} $, we define an element $\BB_k$ of $\kk S_n$ by \begin{equation} \BB_k = \sum_{\sigma\in B_k}\sigma. \label{eq.defBk} \tag{1.3} \end{equation} Theorem 1. Let $n\geq1$. Then, $\BB_1 = \AA$. Theorem 2. For each $a\in\left\{ 0,1,\ldots,n-1\right\} $, we have \begin{equation} \BB_a \AA = a \BB_a + \BB_{a+1}. \end{equation} Theorem 3. We have \begin{equation} \prod_{i\in\left\{ 0,1,\ldots,n-2,n\right\} }\left( \AA-i\right) = 0 \label{eq.t3.claim} \tag{1.4} \end{equation} in the ring $\kk S_n$. (The factors in this product commute, since they are all polynomials in $\AA$.) Corollary 4. The element $\id + \AA$ of $\kk S_n$ is invertible if the numbers $1,2,\ldots,n-1,n+1$ are invertible in $\kk$. Corollary 5. All the $n+1$ elements $\BB_0,\BB_1,\ldots,\BB_n$ are polynomials (with integer coefficients) in $\AA$. Corollary 6. For each $a\in\left\{ 0,1,\ldots,n-1\right\} $, we have \begin{equation} \AA \BB_a = a \BB_a + \BB_{a+1} . \end{equation} Corollary 4 is your first claim, because $\id + \AA$ is exactly your $\phi_{n}$ (for $\kk =\mathbb{Q}$). Moreover, Theorem 3 shows that (again for $\kk = \mathbb{Q}$) the operator "left multiplication by $\AA$" on $\kk S_n$ is diagonalizable with eigenvalues being contained in the set $\left\{ 0,1,\ldots,n-2,n \right\}$ (indeed, it can also be shown that the spectrum is precisely this set; see Garsia's preprint for that and for the dimensions of the eigenspaces); therefore, the operator "left multiplication by $\id + \AA$" has positive integer eigenvalues. So this proves your second claim too. Corollary 6 and Theorem 3 are the equations (3.1) and (3.5) from Garsia's preprint, except that he uses $\BB_1$ instead of $\AA$ (but this is equivalent due to Theorem 1). The element $\BB_1$ of $\kk S_n$ is best known by its action on representations of $S_n$; namely, it acts as a so-called "top-to-random shuffle" or "random-to-top shuffle" (I don't remember precisely which one it is). It is also known as the "Tsetlin library" to probability theorists. Further references on the elements $\BB_0,\BB_1,\ldots,\BB_n$ of $\kk S_n$ include: the appendix of Nolan R. Wallach, Lie Algebra Cohomology and Holomorphic Continuation of Generalized Jacquet Integrals, Advanced Studies in Pure Math. 14 (1988) Representations of Lie Groups, Hiroshima, 1986, pp. 123--151. (At a quick glance, it proves the above Theorem 3 too.) Persi Diaconis, James Allen Fill and Jim Pitman, Analysis of Top to Random Shuffles, Combinatorics, Probability and Computing 1 (1992), pp. 135--155. Manfred Schocker, Idempotents for derangement numbers, Discrete Mathematics, vol. 269 (2003), pp. 239--248. (His $\Xi_{n,k}$ is our $\BB_k$.) F. Hivert, J.-G. Luque, J.-C. Novelli, J.-Y. Thibon, The (1-E)-transform in combinatorial Hopf algebras, arXiv:0912.0184v2, Journal of Algebraic Combinatorics, March 2011, Volume 33, Issue 2, pp. 277--312. (The formula (125) in this paper corresponds to our Theorem 3, but the notations aren't for the faint of heart.) A. B. Dieker, Franco Saliola, Spectral analysis of random-to-random Markov chains, arXiv:1509.08580v3, Advances in Mathematics 323 (2018), pp. 427--485. (This analyzes the much more complicated "random-to-random shuffle", which corresponds to the element $\omega\left( \AA \right) \cdot \AA$ of $\kk S_n$, where $\omega$ is the $\kk$-algebra anti-automorphism of $\kk S_n$ that sends each $g \in S_n$ to $g^{-1}$.) Darij Grinberg, The signed random-to-top operator on tensor space. (Couldn't help mentioning this -- it studies the kernel of the action of $\AA$ on the $n$-th tensor power of a vector space.) There is probably plenty of probabilistic literature that I just don't know about. 2. Proof of Theorem 1 We warm ourselves up with proving the easy Theorem 1. Combinatorially, it simply says that the set $B_1$ contains the $n$ permutations $\cyc_{1,2,\ldots,i}$ for $i \in \left[ n \right]$ and no others (and that these $n$ permutations are distinct). If you find this obvious, move on to the next section; I'll do nothing to convince you otherwise. Proof of Theorem 1. Let $i\in\left[ n\right] $. Then, the permutation $\cyc_{1,2,\ldots,i}$ has one-line notation $\left( 2,3,\ldots,i,1,i+1,i+2,\ldots,n\right) $. Thus, the numbers $2,3,\ldots,n$ occur in this order in the one-line notation of $\cyc_{1,2,\ldots,i}$; in other words, we have \begin{equation} \left( \cyc_{1,2,\ldots,i}\right) ^{-1}\left( 2\right) <\left( \cyc_{1,2,\ldots,i}\right) ^{-1}\left( 3\right) <\cdots <\left( \cyc_{1,2,\ldots,i}\right) ^{-1}\left( n\right) . \end{equation} In other words, $\cyc_{1,2,\ldots,i}\in B_1$ (because we have defined $B_1$ to be the set of all permutations $\sigma\in S_n$ satisfying $\sigma^{-1}\left( 2\right) <\sigma^{-1}\left( 3\right) <\cdots<\sigma^{-1}\left( n\right) $). Now, forget that we fixed $i\in\left[ n\right] $. We thus have shown that $\cyc_{1,2,\ldots,i}\in B_1$ for each $i\in\left[ n\right] $. Hence, we can define a map \begin{align*} \mathbf{f}:\left[ n\right] & \to B_1,\\ i & \mapsto\cyc_{1,2,\ldots,i}. \end{align*} Consider this map $\mathbf{f}$. On the other hand, define a map \begin{align*} \mathbf{g}:B_1 & \to\left[ n\right] ,\\ \sigma & \mapsto\sigma^{-1}\left( 1\right) . \end{align*} If $i\in\left[ n\right] $, then \begin{align*} \left( \mathbf{g}\circ\mathbf{f}\right) \left( i\right) & =\mathbf{g} \left( \underbrace{\mathbf{f}\left( i\right) }_{=\cyc_{1,2,\ldots,i}} \right) =\mathbf{g}\left( \cyc_{1,2,\ldots,i}\right) \\ & =\left( \cyc_{1,2,\ldots,i}\right) ^{-1}\left( 1\right) \qquad\left( \text{by the definition of }\mathbf{g}\right) \\ & =i=\id \left( i\right) . \end{align*} Thus, $\mathbf{g}\circ\mathbf{f}=\id $. On the other hand, let $\sigma\in B_1$ be arbitrary. We shall show that $\left( \mathbf{f}\circ\mathbf{g}\right) \left( \sigma\right) =\id \left( \sigma\right) $. In fact, $\sigma\in B_1$ and therefore $\sigma^{-1}\left( 2\right) <\sigma^{-1}\left( 3\right) <\cdots<\sigma^{-1}\left( n\right) $ (by the definition of $B_1$). Let $i=\mathbf{g}\left( \sigma\right) $. Thus, $i=\mathbf{g}\left( \sigma\right) =\sigma^{-1}\left( 1\right) $ (by the definition of $\mathbf{g}$). But $\sigma^{-1}$ is a permutation of $\left[ n\right] $; thus, the numbers $\sigma^{-1}\left( 2\right) ,\sigma ^{-1}\left( 3\right) ,\ldots,\sigma^{-1}\left( n\right) $ are precisely the elements of $\left[ n\right] $ distinct from $\sigma^{-1}\left( 1\right) $. In view of $\sigma^{-1}\left( 1\right) =i$, this rewrites as follows: The numbers $\sigma^{-1}\left( 2\right) ,\sigma^{-1}\left( 3\right) ,\ldots,\sigma^{-1}\left( n\right) $ are precisely the elements of $\left[ n\right] $ distinct from $i$. In other words, these numbers $\sigma^{-1}\left( 2\right) ,\sigma^{-1}\left( 3\right) ,\ldots ,\sigma^{-1}\left( n\right) $ are precisely the numbers $1,2,\ldots ,i-1,i+1,i+2,\ldots,n$ in some order. Moreover, we know what this order is: it is the increasing order (since $\sigma^{-1}\left( 2\right) <\sigma ^{-1}\left( 3\right) <\cdots<\sigma^{-1}\left( n\right) $). Thus, the numbers $\sigma^{-1}\left( 2\right) ,\sigma^{-1}\left( 3\right) ,\ldots,\sigma^{-1}\left( n\right) $ are precisely the numbers $1,2,\ldots,i-1,i+1,i+2,\ldots,n$ in this order. Combining this with $\sigma^{-1}\left( 1\right) =i$, we conclude that \begin{equation} \left( \sigma^{-1}\left( 1\right) ,\sigma^{-1}\left( 2\right) ,\ldots,\sigma^{-1}\left( n\right) \right) =\left( i,1,2,\ldots ,i-1,i+1,i+2,\ldots,n\right) . \end{equation} Hence, $\sigma^{-1}=\cyc_{i,i-1,\ldots,1}$. Therefore, $\sigma=\left( \cyc_{i,i-1,\ldots,1}\right) ^{-1}=\cyc_{1,2,\ldots,i}$. Comparing this with \begin{equation} \left( \mathbf{f}\circ\mathbf{g}\right) \left( \sigma\right) =\mathbf{f}\left( \underbrace{\mathbf{g}\left( \sigma\right) }_{=i}\right) =\mathbf{f}\left( i\right) =\cyc_{1,2,\ldots ,i}\qquad\left( \text{by the definition of }\mathbf{f}\right) , \end{equation} we obtain $\left( \mathbf{f}\circ\mathbf{g}\right) \left( \sigma\right) =\sigma=\id \left( \sigma\right) $. Now, forget that we fixed $\sigma$. We thus have shown that $\left( \mathbf{f}\circ\mathbf{g}\right) \left( \sigma\right) =\id \left( \sigma\right) $ for each $\sigma\in B_1$. In other words, $\mathbf{f}\circ\mathbf{g}=\id $. Combining this with $\mathbf{g}\circ\mathbf{f}=\id $, we conclude that the maps $\mathbf{f}$ and $\mathbf{g}$ are mutually inverse. Hence, the map $\mathbf{f}:\left[ n\right] \to B_1$ is a bijection. Now, \begin{equation} \AA=\underbrace{\sum_{i=1}^{n}}_{=\sum_{i\in\left[ n\right] } }\underbrace{\cyc_{1,2,\ldots,i}} _{\substack{=\mathbf{f}\left( i\right) \\\text{(by the definition of }\mathbf{f}\text{)}}}=\sum_{i\in\left[ n\right] }\mathbf{f}\left( i\right) =\sum_{\sigma\in B_1}\sigma \end{equation} (here, we have substituted $\sigma$ for $\mathbf{f}\left( i\right) $ in the sum, since the map $\mathbf{f}:\left[ n\right] \to B_1$ is a bijection). Comparing this with \begin{equation} \BB_1 = \sum_{\sigma\in B_1}\sigma\qquad\left( \text{by the definition of }\BB_1 \right) , \end{equation} we obtain $\BB_1 = \AA$. This proves Theorem 1. $\blacksquare$ REPLY [11 votes]: $\newcommand{\cyc}{\operatorname{cyc}} \newcommand{\id}{\operatorname{id}} \newcommand{\BB}{\mathbf{B}} \newcommand{\AA}{\mathbf{A}} $ PART 2 OF 3 [This is part of a long answer, which I had to split into 3 posts. Go to part 1. Go to part 2. Go to part 3.] 3. Proof of Theorem 2 The proof of Theorem 2 is bijective, and, as most bijective proofs, it involves a lot of combinatorial verification that is quick if you know your way around the symmetric groups but turns into awkward drudgery when you try to write it up. Garsia conveniently sweeps it under the rug in his preprint with the magic broom of "easily seen", and I cannot blame him for doing that in a preprint that he did not even publish himself (also, he gives two proofs for Theorem 2 in that preprint). I have less of a good excuse to skip it; if I'm already copying all the ideas from Garsia, I guess I should at least improve on the exposition. So I'm going to give an undergrad-friendly version of the proof, with all the details included. (It's an undergrad-friendly theorem after all -- you just need to know what a group ring is.) For this whole section, let us fix $a \in \left\{ 0,1,\ldots,n-1 \right\}$. Thus, $0\leq a\leq n-1$, so that $n-1\geq0$ and thus $n\geq1$. Also, from $a \in \left\{ 0,1,\ldots,n-1 \right\}$, we obtain $a+1 \in \left[ n \right]$. For every $i\in\left[ n\right] $, we define a permutation $z_i\in S_n$ by \begin{equation} z_i=\cyc_{1,2,\ldots,i}. \label{eq.defzi} \tag{3.1} \end{equation} For every $b\in\left[ n\right] $, $i\in\left[ n\right] $ and $j\in\left[ n\right] $, we define a subset $B_{b,j\to i}$ of $B_{b}$ by \begin{equation} B_{b,j\to i}=\left\{ \sigma\in B_{b}\ \mid\ \sigma\left( j\right) =i\right\} . \label{eq.defBbji} \tag{3.2} \end{equation} The next two lemmas will build up bijections between certain pieces of $B_a$ and $B_{a+1}$. Lemma 7. Let $i\in\left[ a\right] $. Let $j\in\left[ n\right] $. Then, the map \begin{align*} B_{a,1\to i} & \to B_{a,j\to i},\\ \alpha & \mapsto\alpha\circ z_j \end{align*} is well-defined and a bijection. Proof of Lemma 7. The definition of $z_j$ yields $z_j =\cyc_{1,2,\ldots,j}$. Hence, $z_j\left( j\right) =1$ and therefore $z_j^{-1}\left( 1\right) =j$. Also, from $z_j =\cyc_{1,2,\ldots,j}$, we obtain $z_j^{-1}=\left( \cyc_{1,2,\ldots,j}\right) ^{-1} =\cyc_{j,j-1,\ldots,1}$. From $i\in\left[ a\right] $, we obtain $i\leq a\leq n-1\leq n$ and thus $i\in\left[ n\right] $. The definition of $B_{a,1\to i}$ yields \begin{equation} B_{a,1\to i}=\left\{ \sigma\in B_{a}\ \mid\ \sigma\left( 1\right) =i\right\} . \end{equation} The definition of $B_{a,j\to i}$ yields \begin{equation} B_{a,j\to i}=\left\{ \sigma\in B_{a}\ \mid\ \sigma\left( j\right) =i\right\} . \end{equation} Let $\alpha\in B_{a,1\to i}$. We shall show that $\alpha\circ z_j\in B_{a,j\to i}$. We have $\alpha\in B_{a,1\to i}=\left\{ \sigma\in B_{a}\ \mid \ \sigma\left( 1\right) =i\right\} $. In other words, $\alpha\in B_a$ and $\alpha\left( 1\right) =i$. From $\alpha\in B_a$, we obtain \begin{equation} \alpha^{-1}\left( a+1\right) <\alpha^{-1}\left( a+2\right) <\cdots <\alpha^{-1}\left( n\right) \label{pf.l7.0} \tag{3.3} \end{equation} (by the definition of $B_a$). Moreover, for each $k\in\left\{ a+1,a+2,\ldots,n\right\} $, we have \begin{equation} \alpha^{-1}\left( k\right) \in\left\{ 2,3,\ldots,n\right\} . \label{pf.l7.1} \tag{3.4} \end{equation} [Proof: Let $k\in\left\{ a+1,a+2,\ldots,n\right\} $. We must show that $\alpha^{-1}\left( k\right) \in\left\{ 2,3,\ldots,n\right\} $. Indeed, assume the contrary. Thus, $\alpha^{-1}\left( k\right) \notin\left\{ 2,3,\ldots,n\right\} $, so that $\alpha^{-1}\left( k\right) \in\left[ n\right] \setminus\left\{ 2,3,\ldots,n\right\} =\left\{ 1\right\} $. Thus, $\alpha^{-1}\left( k\right) =1$. Hence, $k=\alpha\left( 1\right) =i\leq a$. This contradicts $k>a$ (which follows from $k\in\left\{ a+1,a+2,\ldots,n\right\} $). This contradiction completes our proof of \eqref{pf.l7.1}.] Recall that $z_j^{-1}=\cyc_{j,j-1,\ldots,1}$. Thus, the one-line notation of $z_j^{-1}$ is $\left( j,1,2,\ldots ,j-1,j+1,j+2,\ldots,n\right) $. This reveals that the map $z_j^{-1}$ is strictly increasing on the interval $\left\{ 2,3,\ldots,n\right\} $. Now, the $n-a$ integers $\alpha^{-1}\left( a+1\right) ,\alpha^{-1}\left( a+2\right) ,\ldots,\alpha^{-1}\left( n\right) $ belong to the interval $\left\{ 2,3,\ldots,n\right\} $ (by \eqref{pf.l7.1}) and are listed in strictly increasing order (by \eqref{pf.l7.0}). Hence, applying the map $z_j^{-1}$ to them preserves this strictly increasing order (since the map $z_j^{-1}$ is strictly increasing on the interval $\left\{ 2,3,\ldots ,n\right\} $). In other words, \begin{equation} z_j^{-1}\left( \alpha^{-1}\left( a+1\right) \right) a$ (which follows from $k\in\left\{ a+1,a+2,\ldots ,n\right\} $). This contradiction completes our proof of \eqref{pf.l7.6}.] Recall that $z_j=\cyc_{1,2,\ldots,j}$. Hence, the one-line notation of $z_j$ is $\left( 2,3,\ldots,j,1,j+1,j+2,\ldots ,n\right) $. This reveals that the map $z_j$ is strictly increasing on the subset $\left[ n\right] \setminus\left\{ j\right\}$ of its domain. Now, the $n-a$ integers $\beta^{-1}\left( a+1\right) ,\beta^{-1}\left( a+2\right) ,\ldots,\beta^{-1}\left( n\right) $ belong to this subset $\left[ n\right] \setminus\left\{ j\right\} $ (by \eqref{pf.l7.6}) and are listed in strictly increasing order (by \eqref{pf.l7.5}). Hence, applying the map $z_j$ to them preserves this strictly increasing order (since the map $z_j$ is strictly increasing on the subset $\left[ n\right] \setminus\left\{ j\right\}$). In other words, \begin{equation} z_j\left( \beta^{-1}\left( a+1\right) \right) a+1$ (which follows from $k\in\left\{ a+2,a+3,\ldots,n\right\} $). This contradiction completes our proof of \eqref{pf.l8.1}.] Recall that $z_j^{-1}=\cyc_{j,j-1,\ldots,1}$. Thus, the one-line notation of $z_j^{-1}$ is $\left( j,1,2,\ldots ,j-1,j+1,j+2,\ldots,n\right) $. This reveals that the map $z_j^{-1}$ is strictly increasing on the interval $\left\{ 2,3,\ldots,n\right\} $. Now, the $n-a-1$ integers $\alpha^{-1}\left( a+2\right) ,\alpha^{-1}\left( a+3\right) ,\ldots,\alpha^{-1}\left( n\right) $ belong to the interval $\left\{ 2,3,\ldots,n\right\} $ (by \eqref{pf.l8.1}) and are listed in strictly increasing order (by \eqref{pf.l8.0}). Hence, applying the map $z_j^{-1}$ to them preserves this strictly increasing order (since the map $z_j^{-1}$ is strictly increasing on the interval $\left\{ 2,3,\ldots ,n\right\} $). In other words, \begin{equation} z_j^{-1}\left( \alpha^{-1}\left( a+2\right) \right) a+1$ (which follows from $k\in\left\{ a+2,a+3,\ldots ,n\right\} $). This contradiction completes our proof of \eqref{pf.l8.6}.] Recall that $z_j=\cyc_{1,2,\ldots,j}$. Hence, the one-line notation of $z_j$ is $\left( 2,3,\ldots,j,1,j+1,j+2,\ldots ,n\right) $. This reveals that the map $z_j$ is strictly increasing on the subset $\left[ n\right] \setminus\left\{ j\right\}$ of its domain. Now, the $n-a-1$ integers $\beta^{-1}\left( a+2\right) ,\beta^{-1}\left( a+3\right) ,\ldots,\beta^{-1}\left( n\right) $ belong to this subset $\left[ n\right] \setminus\left\{ j\right\}$ (by \eqref{pf.l8.6}) and are listed in strictly increasing order (by \eqref{pf.l8.5}). Hence, applying the map $z_j$ to them preserves this strictly increasing order (since the map $z_j$ is strictly increasing on the subset $\left[ n\right] \setminus\left\{ j\right\}$). In other words, \begin{equation} z_j\left( \beta^{-1}\left( a+2\right) \right) a+1$. Hence, $a+1<\alpha\left( 1\right) \leq n$. Thus, both $a+1$ and $\alpha\left( 1\right) $ are elements of the interval $\left\{ a+1,a+2,\ldots,n\right\} $. From $\alpha\in B_a$, we obtain \begin{equation} \alpha^{-1}\left( a+1\right) <\alpha^{-1}\left( a+2\right) <\cdots <\alpha^{-1}\left( n\right) \end{equation} (by the definition of $B_a$). In other words, if $u$ and $v$ are two elements of the interval $\left\{ a+1,a+2,\ldots,n\right\} $ satisfying $un$. Then, for each $a,b\in\mathbb{N}$, we have \begin{equation} \BB_{a}\BB_{b}=\sum_{r=\max\left\{ a,b\right\} }^{a+b} \dfrac{a!b!}{\left( r-a\right) !\left( r-b\right) !\left( a+b-r\right) !}\BB_{r}. \end{equation} (The fraction $\dfrac{a!b!}{\left( r-a\right) !\left( r-b\right) !\left( a+b-r\right) !}$ on the right hand side of this is an integer, because it equals $\dbinom{a}{r-b}\dbinom{b}{r-a}\left( a+b-r\right) !$.) Theorem 13 is actually easily derived from Propositions 11 and 12 using the following fact: Proposition 14. For each $k\in\mathbb{N}$, let $X^{\underline{k}}$ denote the polynomial $X\left( X-1\right) \cdots\left( X-k+1\right) $ in the polynomial ring $\kk \left[ X\right] $. Then, for each $a,b\in \mathbb{N}$, we have \begin{equation} X^{\underline{a}}X^{\underline{b}}=\sum_{r=\max\left\{ a,b\right\} } ^{a+b}\dfrac{a!b!}{\left( r-a\right) !\left( r-b\right) !\left( a+b-r\right) !}X^{\underline{r}}. \label{eq.p13.claim} \tag{6.1} \end{equation} Note that Garsia uses the notation $\left( X\right) \downarrow_{k}$ instead of $X^{\underline{k}}$. Proposition 14 is a standard exercise on induction. Its inductive proof is found in the solution of Exercise 7 in: Darij Grinberg, UMN Math 5705: Enumerative Combinatorics, Fall 2018: Homework 1 (where I also show (in Claim 3 (a)) that the coefficients $\dfrac{a!b!}{\left( r-a\right) !\left( r-b\right) !\left( a+b-r\right) !}$ are integers). Note that I state it for a real number $x$ instead of the indeterminate $X$, but the computation does not depend on what $X$ is. Proposition 14 can also be derived from a standard combinatorial identity: Lemma 15. If $a,b\in\mathbb{N}$ and $x\in\mathbb{N}$, then \begin{equation} \dbinom{x}{a}\dbinom{x}{b}=\sum_{r=\max\left\{ a,b\right\} }^{a+b}\dbinom {r}{a}\dbinom{a}{r-b}\dbinom{x}{r}. \end{equation} Proof of Lemma 15 (sketched). Fix an $x$-element set $X$. Let us count all pairs $\left( A,B\right) $ consisting of an $a$-element subset $A$ of $X$ and a $b$-element subset $B$ of $X$. There are two ways to count these pairs: The first way is, we choose $A$ first and then choose $B$. This clearly gives us $\dbinom{x}{a}\dbinom{x}{b}$ many options. The second way is, we first choose the size $r$ of the union $U:=A\cup B$; then, we choose this $U$ itself (there are $\dbinom{x}{r}$ many choices here); then we choose $A$ as a subset of $U$ (there are $\dbinom{r}{a}$ many choices here), and finally we choose $B$ as a $b$-element subset of $U$ that must contain $U\setminus A$ as a subset (because otherwise, $A\cup B$ would not be $U$) (there are $\dbinom{a}{r-b}$ many choices here, since choosing such a $B$ is tantamount to choosing its complement $U\setminus B$, which is merely required to be an $\left( r-b\right) $-element subset of $A$). This gives us $\sum_{r=\max\left\{ a,b\right\} }^{a+b}\dbinom{r}{a}\dbinom{a}{r-b} \dbinom{x}{r}$ many options. Comparing the two results, we obtain Lemma 15. $\blacksquare$ Proof of Proposition 14 (sketched). Fix $a,b\in\mathbb{N}$. Both sides of \eqref{eq.p13.claim} are polynomials with integer coefficients. Hence, we WLOG assume that $\kk =\mathbb{Z}$. Now, let $x\in\mathbb{N}$. For each $k\in\mathbb{N}$, set $x^{\underline{k} }=x\left( x-1\right) \cdots\left( x-k+1\right) $; this is of course the result of substituting $x$ for $X$ in the polynomial $X^{\underline{k}}$. Note that \begin{equation} k!\cdot\dbinom{x}{k}=x^{\underline{k}}\qquad\text{for each }k\in \mathbb{N}. \label{p13.pf.xk} \tag{6.2} \end{equation} Lemma 15 yields \begin{equation} \dbinom{x}{a}\dbinom{x}{b}=\sum_{r=\max\left\{ a,b\right\} }^{a+b}\dbinom {r}{a}\dbinom{a}{r-b}\dbinom{x}{r}. \end{equation} Multiplying both sides of this equality by $a!b!$, we obtain \begin{align*} a!b!\dbinom{x}{a}\dbinom{x}{b} & =a!b!\sum_{r=\max\left\{ a,b\right\} }^{a+b}\dbinom{r}{a}\dbinom{a}{r-b}\dbinom{x}{r}\\ & =\sum_{r=\max\left\{ a,b\right\} }^{a+b}\underbrace{a!b!\dbinom{r} {a}\dbinom{a}{r-b}}_{=\dfrac{1}{\left( r-a\right) !\left( r-b\right) !\left( a+b-r\right) !}\cdot r!}\dbinom{x}{r}\\ & =\sum_{r=\max\left\{ a,b\right\} }^{a+b}\dfrac{1}{\left( r-a\right) !\left( r-b\right) !\left( a+b-r\right) !}\cdot\underbrace{r!\dbinom{x} {r}}_{\substack{=x^{\underline{r}}\\\text{(by \eqref{p13.pf.xk})}}}\\ & =\sum_{r=\max\left\{ a,b\right\} }^{a+b}\dfrac{1}{\left( r-a\right) !\left( r-b\right) !\left( a+b-r\right) !}x^{\underline{r}}. \end{align*} Comparing this with \begin{equation} a!b!\dbinom{x}{a}\dbinom{x}{b}=\underbrace{a!\dbinom{x}{a}} _{\substack{=x^{\underline{a}}\\\text{(by \eqref{p13.pf.xk})}} }\underbrace{b!\dbinom{x}{b}}_{\substack{=x^{\underline{b}}\\\text{(by \eqref{p13.pf.xk})}}}=x^{\underline{a}}x^{\underline{b}}, \end{equation} we obtain \begin{equation} x^{\underline{a}}x^{\underline{b}}=\sum_{r=\max\left\{ a,b\right\} } ^{a+b}\dfrac{a!b!}{\left( r-a\right) !\left( r-b\right) !\left( a+b-r\right) !}x^{\underline{r}}. \label{pf.p13.1} \tag{6.3} \end{equation} Forget that we fixed $x$. Thus, we have proven \eqref{pf.p13.1} for each $x\in\mathbb{N}$. But it is known that if two polynomials $P,Q\in \mathbb{Z}\left[ X\right] $ satisfy $P\left( x\right) =Q\left( x\right) $ for each $x\in\mathbb{N}$, then they satisfy $P=Q$. Applying this to $P=X^{\underline{a}}X^{\underline{b}}$ and $Q=\sum_{r=\max\left\{ a,b\right\} }^{a+b}\dfrac{a!b!}{\left( r-a\right) !\left( r-b\right) !\left( a+b-r\right) !}X^{\underline{r}}$, we conclude that \begin{equation} X^{\underline{a}}X^{\underline{b}}=\sum_{r=\max\left\{ a,b\right\} } ^{a+b}\dfrac{a!b!}{\left( r-a\right) !\left( r-b\right) !\left( a+b-r\right) !}X^{\underline{r}} \end{equation} (because \eqref{pf.p13.1} shows that these $P$ and $Q$ satisfy $P\left( x\right) =Q\left( x\right) $ for each $x\in\mathbb{N}$). This proves Proposition 14. $\blacksquare$ Now for the second remark. The first proof Garsia gives for his Theorem 1.2 (our Theorem 13) uses the so-called Solomon's Mackey formula (his Theorem 1.1). While he cites Solomon's original paper for its proof, we have by now some better sources (better because Solomon's paper works in the general setting of Coxeter groups, which requires some work before it can be translated back into the language of permutations): Corollary 2.4 in Franco V. Saliola, Hyperplane arrangements and descent algebras (errata). Theorem 1 in Mark Wildon, On Bidigare's proof of Solomon's theorem (errata).<|endoftext|> TITLE: Roots of lacunary polynomials over a finite field QUESTION [13 upvotes]: If $P$ is a polynomial over the field $\mathbb F_q$ of degree at most $q-2$ with $k$ nonzero coefficients, then $P$ has at most $(1-1/k)(q-1)$ distinct nonzero roots. Does this fact have any standard name / reference / proof / refinements / extensions? REPLY [5 votes]: This is just an uncertainty principle for the cyclic group $\mathbb F_q^*$; see, e.g. this paper by Tao.<|endoftext|> TITLE: Immersions of the hyperbolic plane QUESTION [24 upvotes]: Is it possible to isometrically immerse the hyperbolic plane into a compact Riemannian manifold as a totally geodesic submanifold? Any nice examples? Edit: Although I did not originally say so, I was looking for injective immersions or at least for immersions that do not factor through a covering onto a compact surface. Thank you for your answers and comments, they've been very helpful. REPLY [3 votes]: Definitely the simplest examples are the covering maps from the upperf half space to a compact hyperbolic surface that Bryant mentioned, although these are not injective. A slight variation is to consider the main diagonal embedding $ \imath: \mathbb{H}^2 \to \mathbb{H}^2 \times \mathbb{H}^2$ and then use different projections on each factor. For example, the first can be the orbit projection $\pi : \mathbb{H} \to \mathbb{H}^2/G$ where $G \subset PSL(2, \mathbb{R})$ contains no translations in the real axis and the second could be $ \pi \circ T$ where $T$ is any of such translations. Then $(\pi, \pi \circ T) \circ \imath $ is one to one. On the other hand, every compact hyperbolic 3-manifold has plenty of totally geodesic immersed $\mathbb{H}^2$ by just projecting totally geodesics $\mathbb{H}^2 \subset \mathbb{H}^3$, as Bryant pointed it is not clear whether the projection could be made injective. This should be a comment but I am not able to comment yet :)<|endoftext|> TITLE: Dominated convergence 2.0? QUESTION [7 upvotes]: During my research, I came across the following question. Let $(f_n)_n$ be a sequence in $C^2([0,1])$ converging pointwise to $g \in L^1([0,1])$. Assume that: $\forall n\in\mathbb N, f_n'' TITLE: On a special type of normed linear spaces QUESTION [17 upvotes]: Let $(V,\|.\|)$ be a normed linear space such that for every group $(G,*)$, every function $f:G \to V$ satisfying $$ \|f(x*y)\|\ge \|f(x)+f(y)\|,\qquad\forall x,y\in G,\tag{Z} $$ is a group homomorphism i.e. $f(x*y)=f(x)+f(y),\forall x,y\in G$. Then is it true that the norm on $V$ comes from an inner product? REPLY [3 votes]: Not a full answer: the space should be strictly convex. In fact, if you make your condition with the restricted demand that the group is $\mathbb{Z}$, then it is actually equivalent to the norm being strictly convex. Recall that the norm is not strictly convex if there exist distinct $x,y\in V$ such that $$ (*) \quad \forall~t\in [0,1],\quad \|tx+(1-t)y\|=1. $$ Assuming this is the case we will be done by letting $f:\mathbb{Z}\to V$ be defined by $$ f(n)= \begin{cases} nx & n\neq\pm 1 \\ y & n= \pm 1 \end{cases} $$ Assuming the norm is strictly convex, using the observation that $f(-n)=-f(n)$ and scaling $x=f(1)$ to be of norm 1 it is enough to show by an induction on $n\in \mathbb{N}$ that for $y=f(n+1)-f(n)$ we have $(*)$. This follows from the following two lines: $$ \|f(n+1)\|\geq \|f(n)+f(1) \|=n+1 $$ $$ 1 =\|f(1)\| =\|f((n+1)+(-n))\| \geq \|f(n+1)-f(n)\| $$ from which you deduce first that $\|f(n+1)\|=n+1$ and then that both $x$ and $y$ are on the intersection of the unit sphere and the sphere of radius $n$ around $f(n+1)$. Let us go back now to a general group $G$ and assume, in view of the above, that $V$ is strictly convex. I claim that $g\mapsto |g|:=\|f(g)\|$ is a conjugation invariant seminorm on $G$ (recall that a seminorm on $G$ is a function $|\cdot|:G\to [0,\infty)$ satisfying for every $g,h\in G$, $|gh|\leq|g|+|h|$) which is homogeneous (that is it satisfies for every $n\in \mathbb{Z}$, $|g^n|=|n|\cdot |g|$). The fact that $|\cdot |$ is a seminorm is easy: $$ \|f(xy)\|\leq \|f(xy)-f(y)\|+\|f(y)\| \leq \|f(x)\|+\|f(y)\| $$ and the fact that it is homogeneous follows from the case $G=\mathbb{Z}$ discussed above. The fact that $|\cdot |$ is conjugation invariant actually follows formally from the previous two facts: for $g,h\in G$, the inequality $$ |ghg^{-1}|=|gh^ng^{-1}|/n \leq (|g|+|h^n|+|g^{-1}|)/n=|h|+2|g|/n$$ shows, by taking the limit on $n\to \infty$, that $|ghg^{-1}|\leq |h|$, but substituting in this inequality $ghg^{-1}$ for $h$ and $g^{-1}$ for $g$ we get the reverse inequity. Let my now make the remark that for a conjugation invariant seminorm $|\cdot |$ on a group $G$ it makes sense to define its kernel $K TITLE: Parsing the definition of center of an algebra in a higher-categorical setting QUESTION [6 upvotes]: I'm having trouble parsing a definition in Lurie's "Rotation Invariance in Algebraic $K$-Theory". The definition os for the notion of center of an associative algebra object, and occurs in Remark 2.1.3. The setting is as follows. We have a symmetric monoidal $\infty$-category $\mathcal{C}$. We write $\mathrm{Alg}(\mathcal{C})$ for the $\infty$-category of associative algebra objects in $\mathcal{C}$ and $\mathrm{LMod}(\mathcal{C})$ for the $\infty$-category of left modules in $\mathcal{C}$. Informally, we think of objects in $\mathrm{LMod}(\mathcal{C})$ as pairs $(A, M)$ where $A$ is an associative algebra object and $M$ is a left $A$-module. Noting that $\mathrm{Alg}(\mathcal{C})$ and $\mathrm{LMod}(\mathcal{C})$ inherit symmetric monoidal structures, we make the following definitions. We define $\mathrm{Alg}^{(2)}(\mathcal{C}) =\mathrm{Alg}(\mathrm{Alg}(\mathcal{C}))$ and $\mathrm{LMod}^{(2)}(\mathcal{C}) = \mathrm{Alg}(\mathrm{LMod}(\mathcal{C}))$. It is known that $\mathrm{Alg}^{(2)}(\mathcal{C})$ is equivalent to the category of $\mathbb{E}_2$-algebra objects. Informally, we think of objects in $\mathrm{LMod}^{(2)}(\mathcal{C})$ as pairs $(A,M)$ where $A$ is an $\mathbb{E}_2$-algebra object and $M$ is an $A$-algebra. We call $\mathrm{LMod}^{(2)}(\mathcal{C})$ the category of central actions in $\mathcal{C}$. Finally, we arrive at the definition I am stuck on. Fix an associative algebra object $M \in \mathrm{Alg}(\mathcal{C})$. We say that a central action $(A,M) \in \mathrm{LMod}^{(2)}(\mathcal{C})$ exhibits $A$ as a center of $M$ if, for every $\mathbb{E}_2$-algebra $B \in \mathrm{Alg}^{(2)}(\mathcal{C})$, the canonical map $$ \mathrm{Map}_{\mathrm{Alg}^{(2)}(\mathcal{C})}(B, A) \to \mathrm{LMod}^{(2)}(\mathcal{C}) \times_{\mathrm{Alg}(\mathcal{C})} \{M\} $$ is a homotopy equivalence. It is not stated what exactly the canonical map is, but it is probably thought of as follows. The factor $\mathrm{Map}_{\mathrm{Alg}^{(2)}(\mathcal{C})}(B, A) \to \mathrm{LMod}^{(2)}(\mathcal{C})$ sends a map $\phi : B \to A$ of $\mathbb{E}_2$-algebras to the central action $(B,M)$ where $B$ acts via $\phi$. The other factor is crystal clear. Perhaps I'm not seeing clearly, but I'm not sure how to parse this definition. It seems to be saying the following. Fix an $\mathbb{E}_2$-algebra $B$. The the data of a central action $(R,M)$ of any $\mathbb{E}_2$-algebra $R$ on $M$ is the same as the data of a map $B \to A$ of $\mathbb{E}_2$-algebras. This doesn't seem quite right to me. Naively translating this into ordinary algebra, it feels quite bizarre. Moreover, Lurie gives another definition for center in Higher Algebra, Definition 5.3.1.6. It is not clear to me that these two definitions are equivalent. Am I just confused? Does this indeed produce a sensible notion of center of an associative algebra? A very vague informal argument will suffice. REPLY [4 votes]: Let us try to figure out what's happening on discrete rings, where $E_2=E_\infty$. The category $\mathrm{LMod}^{(2)}$ is, as you surmised, the category of pairs $(A,B)$ where $A$ is a commutative algebra and $B$ is an associative $A$-algebra (i.e. an algebra object in the monoidal category of $A$-modules). That is we require the multiplication in $B$ to be $A$-bilinear, hence for every elements $b_1,b_2\in B$ and $a_1,a_2\in A$ we have $$ (a_1b_1)(a_2b_2)=(a_1a_2)(b_1b_2)$$ You can rewrite this data as a triple $(A,B,\phi)$ where $A$ is a commutative ring, $B$ is an associative ring and $\phi:A\to B$ is a central ring homomorphism, that is a ring homomorphism whose image is contained in the center. Then for every associative ring $B$ the category $$\mathrm{LMod}^{(2)}\times_{\mathrm{Alg}}\{B\}$$ is simply the category of commutative rings $A$ with a central ring homomorphism to $B$. Lurie is then defining the center of $B$ as the terminal object of this category. It is natural now that this corresponds to the classical notion of the center. Of course, as soon as you pass to $\infty$-categories more complications arise, in particular the center is not going to be a commutative ring anymore but simply an $E_2$-ring. You can however iterate this procedure and obtain the $E_{n+1}$-center of an $E_n$-ring and so obtain as much commutativity as you want. EDIT Actually, I believe that the rotation invariance paper has a typo (as you can see the right hand side of the so-called equivalence is independent of $B$ and in fact it's not even an ∞-groupoid). I think that the correct definition is that the canonical map $$\mathrm{Alg}^{(2)}(\mathcal{C})_{/A}\to \mathrm{LMod}^{(2)}(\mathcal{C})\times_{\mathrm{Alg}(\mathcal{C})}\{B\}$$ sending $[A'\to A]$ to $(A',B)$ where $A'$ acts centrally on $B$ via $A'\to A$ is an equivalence. This is equivalent to the description I gave above.<|endoftext|> TITLE: asymptotic for li(x)-Ri(x) QUESTION [6 upvotes]: Is it true that $$\operatorname{li}(x)-\operatorname{Ri}(x) \sim \frac{1}{2}\operatorname{li}(x^{1/2}) \ (x \to \infty),$$ where $$\operatorname{Ri}(x) = \sum_{n = 1}^\infty \frac{\mu(n)}{n} \operatorname{li}(x^{1/n}) = 1 + \sum_{k = 1}^\infty \frac{(\log x)^k}{k \cdot k!\ \zeta(k+1)}$$ for all $x > 0$? If so, how can one prove the given asymptotic? Note that \begin{align}\label{lirieq} \lim_{x \to \infty} \frac{\operatorname{li}(x) - \operatorname{Ri}(x)}{\frac{1}{2}\operatorname{li}(x^{1/2})} = 1- 2\lim_{x \to \infty} \sum_{n = 3}^\infty \frac{\mu(n)}{n}\frac{\operatorname{li}(x^{1/n})}{\operatorname{li}(x^{1/2})} = 1-2\sum_{n = 3}^\infty \lim_{x \to \infty} \frac{\mu(n)}{n}\frac{\operatorname{li}(x^{1/n})}{\operatorname{li}(x^{1/2})} = 1, \end{align} provided that the given limit can be interchanged with the given sum. However, I am unable to justify interchanging the limit with the sum. REPLY [6 votes]: Yes, the stated asymptotics (and much more) is true. The idea is to truncate $\operatorname{Ri}(x)$ appropriately. Let us use the series representation (see here) $$\operatorname{li}(t)=\gamma+\log\log t+\sum_{k=1}^\infty\frac{(\log t)^k}{k\cdot k!},\qquad t>1.$$ This implies $$\operatorname{li}(t)=\gamma+\log\log t+O(\log t),\qquad 1\log x.$$ As a result, for $x>3$ we get \begin{align*}\sum_{n>\log x} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n})&=O(1)+\sum_{n>\log x}\frac{\mu(n)}{n}\left(\gamma+\log\log x-\log n\right)\\[6pt] &=O\left((\log\log x)^2\right)+\sum_{n=1}^\infty\frac{\mu(n)}{n}\left(\gamma+\log\log x-\log n\right)\\[6pt] &=O\left((\log\log x)^2\right).\end{align*} Here we used that $$\sum_{n=1}^\infty\frac{\mu(n)}{n}=0\qquad\text{and}\qquad\sum_{n=1}^\infty\frac{\mu(n)\log n}{n}=-1$$ by the prime number theorem. To summarize so far, $$\operatorname{Ri}(x)=\sum_{n\leq\log x} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n})+O\left((\log\log x)^2\right),\qquad x>3.$$ The error term on the right hand side can probably be improved, but this is not important. In the sum, we detach the first three terms (corresponding to $n=1,2,3$) and estimate the rest trivially. We get $$\operatorname{Ri}(x)=\operatorname{li}(x)-\frac{1}{2}\operatorname{li}(x^{1/2})+O\left(\frac{x^{1/3}}{\log x}\right),\qquad x>3.$$<|endoftext|> TITLE: Geometric description of a certain sphere bundle QUESTION [10 upvotes]: It appears to be a standard fact in topology that $\mathbb{C}\mathbb{P}^2\#-\mathbb{C}\mathbb{P}^2$ has a structure of a $\mathbb{S}^2$ bundle over $\mathbb{S}^2$. Is there a nice geometric description of the projection to the sphere? This manifold is actually a complex algebraic variety (namely a plane with one point blown up), so the question should make some sense. The best answer would probably be a meromorphic function, but I could not find one. REPLY [8 votes]: I would like to add another geometric description of $\mathbb C\mathbb P^n \#\overline{\mathbb C\mathbb P^n}$ as sphere bundle (same works for $\mathbb C\mathbb P^n \# \mathbb C\mathbb P^n$): Let $S^1 \to S^{2n+1} \to \mathbb C\mathbb P^n$ be the Hopf fibration. Then $S^1$ acts on $\mathbb R^2$ by rotations and the associated vector bundle $E$ is the normal bundle of $\mathbb C\mathbb P^{n-1}$ in $\mathbb C\mathbb P^n$. Thus the total space of $E$ is diffeomorphic to $\mathbb C\mathbb P^n$ with a disc removed. Hence if we glue the disc bundles of $E$ with $\overline E$ (which denotes here the the bundle with the reversed orientation) along their common boundary one obtains $\mathbb C\mathbb P^n\# \overline{\mathbb C\mathbb P^n}$. Moreover one deduces from this description that the connected sum is given as the quotient $(S^{2n-1}\times S^2)/S^1$, where $S^1$ acts on $S^{2n-1}$ such that it induces the Hopf fibrations and on $S^2$ by rotations. This quotient has a projection map to $\mathbb C\mathbb P^{n-1}$ with fibre $S^2$. (This is just the associated $S^2$-bundle to the Hopf fibration)<|endoftext|> TITLE: Eigenvalues and eigenvectors of the matrix with entries $\dbinom{n+1}{2j-i}$ for $i, j = 1, 2, \ldots, n$ QUESTION [19 upvotes]: Let $n$ be a nonnegative integer, and let $B$ be the $n \times n$-matrix (over the rational numbers) whose $\left(i, j\right)$-th entry is $\dbinom{n+1}{2j-i}$ for all $i, j \in \left\{ 1, 2, \ldots, n \right\}$. For example, if $n = 5$, then \begin{equation} B = \left(\begin{array}{rrrrr} 6 & 20 & 6 & 0 & 0 \\ 1 & 15 & 15 & 1 & 0 \\ 0 & 6 & 20 & 6 & 0 \\ 0 & 1 & 15 & 15 & 1 \\ 0 & 0 & 6 & 20 & 6 \end{array}\right) . \end{equation} Question 1. Prove that the eigenvalues of $B$ are $2^1, 2^2, \ldots, 2^n$. (I know how to do this -- I'll write up the answer soon -- but there might be other approaches too.) Question 2. Find a left eigenvector for each of these eigenvalues. What I know is that the row vector $v$ whose $i$-th entry is $\left(-1\right)^{i-1} \dbinom{n-1}{i-1}$ (for $i \in \left\{1,2,\ldots,n\right\}$) is a left eigenvector for eigenvalue $2^1$ (that is, $v B = 2 v$). But the other left eigenvectors are a mystery to me. Question 3. Find a right eigenvector for each of these eigenvalues. For example, it appears to me that the column vector $w$ whose $i$-th entry is $\left(-1\right)^{i-1} / \dbinom{n-1}{i-1}$ (for $i \in \left\{1,2,\ldots,n\right\}$) is a right eigenvector for eigenvalue $2^1$ (that is, $B w = 2 w$). This (if correct) boils down to the identity \begin{equation} \sum_{k=1}^n \left(-1\right)^{k-1} \left(k-1\right)! \left(n-k\right)! \dbinom{n+1}{2k-i} = 2 \left(-1\right)^{i-1} \left(i-1\right)! \left(n-i\right)! \end{equation} for all $i \in \left\{1,2,\ldots,n\right\}$. Note that the entries of $w$ are the reciprocals to the corresponding entries of $v$ ! Needless to say, this pattern doesn't persist, but maybe there are subtler patterns. I am going to put up an answer to Question 1 soon, as a stepping stone for the proof of https://math.stackexchange.com/questions/2886392 , but this shouldn't keep you from adding your ideas or answers. REPLY [4 votes]: The left eigenvectors seem to be related to the Euler polynomials (note that these are referred to in Wikipedia as Eulerian polynomials). For fixed $1\le k\le n$, if the left eigenvector for the eigenvalue $2^k$ is denoted $(v_1,\dots,v_n)$ and normalized to $v_1=1$, then it appears that $$ \frac{\sum_{i=1}^n v_ix^i}{(1-x)^{n+1}}=x+2^kx^2+3^kx^3+\cdots$$ which allows to find the $v_i$ recursively, keeping $k$ and increasing $n$. For $k=n$ (i.e. for the biggest eigenvalue), $\sum_{i=1}^n v_ix^{i-1}$ is the $n$th Euler polynomial.<|endoftext|> TITLE: Proposition in HTT on cofibrations of categories QUESTION [7 upvotes]: Proposition A.3.3.9. in Higher Topos Theory is as follows: Let $S$ be an excellent model category and let $f:C\rightarrow C'$ be a cofibration of small $S$-enriched categories. Then (1) for every combinatorial $S$-enriched model category $A$, the pullback $f^*:A^{C'}\rightarrow A^C$ preserves projective cofibrations and (2) for every projectively cofibrant object $F\in S^C,$ the unit map $F\rightarrow f^*f_!F$ is a projective cofibration. I am OK with all of the proof after the first sentence, which claims that these two properties are clearly invariant under retracts of $f$. After that, the argument proceeds as usual, by proving the claim for transfinite compositions of pushouts of generating cofibrations. But it is not at all clear to me why these two properties persist after a retract. For instance, take a retract $D$ of $C'$, given by functors $p:C'\rightarrow D$ and $q:D\rightarrow C'$ such that $pq$ is the identity functor. Say that $f$ factors through a map $g:C\rightarrow D$, so $g$ is a retract of $f$. Then to show $g^*\cong f^*\circ p^*$ preserves projective cofibrations, it would suffice to show that $p^*$ has the same property. Unfortunately this is usually false. There are a few other ideas that don't seem to work either - is there some nice argument that I'm missing? REPLY [5 votes]: You can argue as follows. Suppose that $g: D \to D'$ is a retract of $f: C \to C'$ (in the category of $S$-enriched categories) via maps $D \stackrel{i}{\to} C \stackrel{r}{\to} D$ and $D' \stackrel{i'}{\to} C' \stackrel{r'}{\to} D'$. Assume that $g^*: A^{C'} \to A^C$ preserves projective cofibrations and that the unit map $F \to f^*f_!F$ is a projective cofibration for every projectively cofibrant $F \in A^C$. We wish to show that $g^*:A^{D'} \to A^D$ preserves cofibrations and that $G \to g^*g_!G$ is a projective cofibration for every projectively cofibrant $G \in A^D$. 1) Consider the Beck-Chevalley transformations $\tau: i_!g^* \Rightarrow f^*i'_!$ and $\sigma: r_!f^* \Rightarrow g^*r'_!$. Then the composed transformation $$ g^* \cong r_!i_!g^* \stackrel{r_!\tau}{\Rightarrow} r_!f^*i'_! \stackrel{\sigma}{\Rightarrow} g^*r'_!i'_! \cong g^* $$ is the identity (indeed, it is the Beck-Chevalley transformation of a square in which two parallel legs are isomorphisms), and hence $g^*$ is a retract of $r_!f^*i'_!$ (in the category of functors $A^{D'} \to A^{D}$). Since $r_!f^*i'_!$ preserves projective cofibrations, so does $g^*$. 2) Consider again the retract diagram $g^* \Rightarrow r_!f^*i'_! \Rightarrow g^*$ above. Pre-composing with $g_!$ we obtain a retract diagram $$g^*g_! \Rightarrow r_!f^*i'_!g_! \cong r_!f^*f_!i_! \Rightarrow g^*g_! ,$$ and one can check that it is compatible with the unit maps $u_g: Id \Rightarrow g^*g_!$ and $u_f: Id \Rightarrow f^*f_!$, i.e., that it renders $u_g:Id \Rightarrow g^*g_!$ a retract of $r_!u_fi_!:Id \cong r_!i_! \Rightarrow r_!f^*f_!i_!$ (in the category of functors $A^{D} \to A^{D}$). It then follow that if $u_f$ is a projective cofibration on every projectively cofibrant functor then $u_g$ a projective cofibration on every projectively cofibrant functor, as desired.<|endoftext|> TITLE: How to show that a hypersurface is a diagonal intersected with hyperplanes? QUESTION [5 upvotes]: Suppose I have a hypersurface $V(F) = \{ \mathbf{x} \in k^n: F(\mathbf{x}) = 0 \}$, where $F$ is a homogeneous form of degree $d > 1$. I would like to show that there exists some diagonal form $D(y_1, \ldots, y_N) = A_1 y_1^d + \ldots + A_N y^d \in k[y_1, \ldots, y_N]$ and some linear forms $L_i(\mathbf{y}) \in k[y_1, \ldots, y_N]$, say $1 \leq i \leq T$, such that $$ V(F) = V(D) \cap V(L_1) \cap \ldots \cap V(L_T). $$ Here $k = \mathbb{Q}, \mathbb{R}$ or $\mathbb{C}$. I seem to recall someone telling me this follows fairly easily from some algebraic geometry (possibly using Veronese embedding) but I couldn't figure it out. Any comments would be appreciated. REPLY [5 votes]: The map from $V(L_1) \cap \dots \cap V(L_T)$ to $\mathbb P^{N-1}$ is a linear map. So an equivalent way of stating this is that there are $N$ linear forms $y_1,\dots,y_N$ in $x_1,\dots,x_n$ such that $\sum_{i=1}^N A_i y_i^d = F$. One way to think about this is to consider, for each $N$, the locus in the space of degree $d$ homogeneous forms that can be written as $\sum_{i=1}^N A_i y_i^d$. If the dimension of this locus for some $N$ is equal to the dimension for $N+1$, it follows that the sum of a generic element of this locus with a generic linear form is a generic element of the locus, so the sum of two generic elements of the locus is a generic element of the locus, so the locus is a dense subset of a linear subspace. However, the space of linear forms raised to the $d$th power is not contained in any linear subspace (this can be expressed as a statement of the Veronese embedding, but can also be proved directly algebraically), so for this $N$ the locus is in fact dense in the whole space. Hence for $2N$ the locus is the whole space. (And their must be such an $N$ because the dimension is bounded.)<|endoftext|> TITLE: Dominated convergence 2.1? QUESTION [5 upvotes]: After this question : Dominated convergence 2.0? I want to know, what about the case when $h\in L^1([0,1])$. The completed question : Let $(f_n)_n$ be a sequence in $C^2([0,1])$ converging pointwise to $g \in L^1([0,1])$ and $\forall x \in [0,1], g(x)\in \mathbb R$. Assume that: $\forall n\in\mathbb N, f_n'' TITLE: Motivated account of the prime number theorem and related topics QUESTION [50 upvotes]: Though my own research interests (described below) are pretty far from analytic number theory, I have always wanted to understand the prime number theorem and related topics. In particular, I often see assertions of things like "the prime number theorem is equivalent to the fact that the Riemann zeta function has no zeros whose real part is at least $1$" and "the Riemann hypothesis is equivalent to the best possible error term in the prime number theorem". However, whenever I have attempted to learn justifications for these slogans, I am immediately confronted with huge masses of complicated and unmotivated formulas. I can verify these calculations line-by-line, but I don't seem to learn anything from them. I always feel that there must be a bigger picture in the back of the mind of the writers, but I don't see it. Question: Can anyone recommend a motivated account of these topics that is accessible to people whose backgrounds are not in analysis? Since they are fundamentally analytic facts, I expect that this will involve learning some analysis. But at a fundamental level I prefer to think either algebraically or geometrically, so I have difficulty attaining enlightenment from complicated formulas/estimates without having the meaning of the various terms explicitly spelled out. My background: I am about 10 years post PhD, and my research is in topology and algebraic geometry. I have a good working knowledge of algebraic number theory (up to class field theory, though I have to admit that I have never carefully studied the proofs of the results in class field theory). I love complex analysis, though I fear that the way I think about that subject is more soft and geometric than analytic (and thus is not really the point of view of people working in analytic number theory). REPLY [13 votes]: Let me record a pedestrian answer here. It all starts with Euler's formula $$ \prod_p\left(1-\frac{1}{p^s}\right)^{-1}=\sum_{n\geq 1}\frac{1}{n^s}=:\zeta(s),\quad s>1, $$ and the observation that since the RHS diverges for $s=1$, so does the LHS, and thus $\sum_p p^{-1}=+\infty$. This is already interesting, since it proves the infinitude of primes, and what is more, shows that the primes are rather dense. Indeed, let us assume for a moment that $p_n\sim cn^\alpha(\log n)^\beta$ for some $\alpha\geq 1$ and $\beta\in \mathbb{R}$, where $p_n$ denotes the $n$-th prime. The prime number theorem is equivalent to the statement that $c=\alpha=\beta=1$. The above observation, combined with $p_n\geq n$, restricts the exponents to $\alpha=1$, $0\leq\beta\leq 1$. Can we do better? Well, a natural idea would be to plug various candidate expansions into the LHS of Euler's identity and check what's compatible with the behaviour of the RHS as $s\to 1$. The behaviour of $\sum_n n^{-s}$ is easy to understand: we can approximate the sum with the integral $\int_1^\infty x^{-s}dx=\frac{1}{s-1}$; the error will give a convergent series for $s\geq \frac{1}{2}$. Thus, $\zeta(s)=(s-1)^{-1}+O(1)$ as $s\to 1$. To understand the asymptotics of the LHS, it is convenient to pass from products to sums: $$ -\sum_n{\log\left(1-\frac{1}{p_n^s}\right)}=\sum_n\frac{1}{p^s_n}+O(1)=\log(s-1)+O(1),\quad s\to 1. $$ Plugging in the candidate expansion for $p_n$, we arrive at $$ \sum_n \frac{1}{n^s(\log n)^{s\beta}}\sim c\log(s-1),\quad s\to 1. $$ It is a nice elementary exercise to check that the left-hand side is asymptotic to $(s-1)^{\beta-1}\Gamma(1-\beta)$ for $\beta<1$, and to $\log (s-1)$ for $\beta=1$. We conclude that we must have $\beta=1$ and $c=1$. (The result we have just proven is due to Chebyshev. Note that no complex analysis was needed thus far.) Of course, in order to prove the PNT we need to remove the a priori assumption $p_n\sim cn^\alpha(\log n)^\beta$, and for that the information about the $s\to 1$ regime is not sufficient. Consider the following toy problem: Compute the asymptotics of the Taylor coefficients of $\exp(x)(1-x)^{-2}$ at the origin. An argument similar to the one above would say that of all "natural" asymptotics, only $a_n\sim e n$ is compatible with the behaviour as $x\to 1$. However, the function $(1+x^2)^{-2}+e(1-x)^{-2}$ has the same behaviour as $x\to 1$, but a completely different asymptotics of the coefficients. Of course, the reason here is that there are other singularities on the unit circle. In this problem, the role of complex variable and the domain of analyticity is very transparent. The PNT is not about Taylor coefficients, but it can be reduced to a problem about the growth of a Fourier transform of an analytic function. To this end, we differentiate the log of Euler's identity (conveniently, it will get rid of logs and make our functions single-valued) to get $$ -\frac{\zeta'(z)}{\zeta(z)}=\sum_p \frac{\log p}{p^z-1}. $$ A natural simplification is to get rid of the $-1$ in the denominator; so, instead of the RHS, we consider $\Phi(z):=\sum_p(\log p)p^{-z}.$ (The difference is analytic for $\Re z>\frac12$, and so it will be unimportant.) Now, $\Phi$ is almost manifestly a Laplace transform: consider the distribution $$\rho(x):=(\log(x))\sum_p\delta(x-p),$$ integrate by parts and change variable: $$ \Phi(z)=\int_1^\infty\rho(x)x^{-z}dx=z\int_1^\infty\theta(x)x^{-z-1}dx=z\int_0^\infty\frac{\theta(e^y)}{e^y}e^{-(z-1)y}dy, $$ where $\theta(x)=\int_1^x \rho(x)=\sum_{p\leq x}\log p$ and $\Re (z-1)>0$. Note that it is elementary to check that PNT is equivalent to $\theta(x)\sim x$, so, we can work with $\theta$, but there are other ways to write $\Phi$ as a Laplace transform, including ones directly involving $\pi(x)$. In Fourier analysis, there's a general principle that the rate of decay of a function at infinity corresponds to smoothness (or analyticity) of its Fourier transform. In particular, the Fourier transform being analytic in the $\epsilon$-neighbourhood of the real line corresponds to the function decaying as $O(\exp(-\epsilon |x|))$, give or take a bit of room. (In our case, the role of the real line is played by $\Re z=1$; translate and rotate to get into the more familiar set-up.) Now, $\Phi(z)$ has a pole at $z=1$, which just corresponds to the purported limit $\lim \theta(x)/x=1$: we have $$ \frac{\Phi(z)}{z}-\frac{1}{z-1}=\int_0^\infty\left(\frac{\theta(e^y)}{e^y}-1\right)e^{-y(z-1)}dy. $$ Also, $\Phi$ is automatically analytic to the right of $\Re z=1$ (due to $\theta$ vanishing identically on the negative axis). If we knew that the left-hand side of the above identity extended analytically to $1-\epsilon<\Re z\leq 1$, then this would mean that $$ \frac{\theta(e^{y})}{e^y}-1=O(e^{-y\epsilon'}) $$ for any $\epsilon'<\epsilon$, that is, $\theta(x)=x+O(x^{1-\epsilon'})$, which is the PNT with an error bound (you can get a bit sharper estimate here). Also, you see that if $\Phi$ had other singularities on the line $\Re z=0$, then $\theta$ would have a different asymptotics. Unfortunately, no-one knows how to prove the above analytic extension, so the rest of the story boils down to showing that there are no singularities on the line $\Re z=1$ itself, and a subtle analytical argument that this is just enough to conclude the PNT without error bounds, or proving that some explicit region around that line is free of zeros, which yields a (sub-power) error bound. The fact that there are no zeros on $\Re z$ is heuristically explained as follows: for $\prod(1-p^{-1+it})^{-1}$ to diverge to zero, the factors $p^{it}$ must conspire in such a way that the "majority" of them points in the negative direction. But then the "majority" of $p^{2it}$ points in the positive direction, meaning that $\prod(1-p^{-1+2it})^{-1}$ diverges to infinity and $\zeta$ has a pole at $1+2it$, which we know it does not. This heuristic is usually packed into an ingenious one-liner which is hard to motivate further. A slick version of the analytic lemma is in Newman's proof, which is the source of most of the material of this answer.<|endoftext|> TITLE: (Euclidean) open orbit in an irreducible real algebraic set QUESTION [5 upvotes]: Let $\tau:GL(n,\mathbb{R}) \rightarrow GL(V)$ be a rational representation of the general linear group of degree $n$ on a finite-dimensional real vector space $V$. Let $C$ be an irreducible real algebraic set in $V$ such that the action of $GL(n,\mathbb{R})$ on $V$ induced by $\tau$ leaves the set $C$ invariant. We endow $C$ with the subspace topology inherited from (Hausdorf) Euclidean topology of $V$. Supose there is a point $v\in C$ such that the orbit $GL(n,\mathbb{R})⋅v$ is an open subset of $C$. My questions is: is it true that the orbit $GL(n,\mathbb{R})⋅v$ must be a Zariski-open subset of $C$? REPLY [7 votes]: No, not at all. Take for $V$ the space of quadratic forms on $\mathbb R^n$, let $v=\sum_ix_i^2$ be the standard form, and $C=V$. Then $GL(n,\mathbb R)v$ is the set of positive definite forms. So, it is Hausdorff open but not Zariski open. NB: This example even works for $n=1$.<|endoftext|> TITLE: Behavior of genus function on a 4-manifold for sums QUESTION [18 upvotes]: Let $X$ be a smooth compact 4-manifold. Then every element of $H_2(X;\mathbb{Z})$ can be represented by a smooth embedded orientable surface and we have the so called genus function $G: H_2(X; \mathbb{Z}) \to \mathbb{Z}_{\geq 0}$ which assigns to a homology class the smallest genus of such a smooth surface needed to represent it. Suppose that $x$ is a nontorision element of $H_2(X; \mathbb{Z})$. Does the sequence $G(x), G(2x), G(3x),...$ limit to infinity? Can there be arbitrarily large zeroes? Is there always a limit? REPLY [11 votes]: In the case that $x\cdot x \neq 0$, topological methods based on the G-signature show that the genus goes to infinity more or less quadratically in $n$. (I'll be more specific below.) This goes back to Rochlin (Two-dimensional submanifolds of four-dimensional manifolds) and Hsiang-Szczarba (On embedding surfaces in 4-manifolds) in the 1970s. Following Rochlin's version (since I don't have the other at hand): if a homology class $\xi$ is divisible by $h$, an odd prime power, then $$ g \geq \left|\frac{(h^2-1)(\xi \cdot \xi)- \sigma(X)}{4 h^2}\right| - \frac{b_2(X)}{2}. $$ Writing $\xi = h \alpha$ we see that the right side grows quadratically in such $h$. (Generally this grows as the square of the largest prime power dividing $n$ where $\xi = n \alpha$; presumably the growth rate of that quantity in $n$ is known.) By looking in a neighborhood (and sticking to prime powers), you can see that you'd expect quadratic growth, but the estimate above looks off by a factor of two. For instance, when it holds, the adjunction formula (as quoted by Marco above) gives a bound that is roughly twice the G-signature bound. Work of Strle (Bounds on genus and geometric intersections from cylindrical end moduli spaces) gives stronger results for surfaces of positive self-intersection in the case that $b_2^+(X) =1$, without the assumption of non-vanishing Seiberg-Witten invariants. See also recent work of Konno (Bounds on genus and configurations of embedded surfaces in 4-manifolds). Finally, in the case of self-intersection $0$, the growth is at most linear (and possibly $0$, as Marco notes). This follows by tubing together parallel copies of a given surface.<|endoftext|> TITLE: About the definition of E8, and Rosenfeld's "Geometry of Lie groups" QUESTION [22 upvotes]: I've been searching the literature for a direct definition of the group $E_8$ (over a general field, but even a definition of just one incarnation would be great). I knew (from talking to people) that there's probably nothing available, but I'm confused about one point. By "direct definition", I mean something other than a definition of $E_8$ as a group of automorphisms of its own Lie algebra. Something promising is the "octo-octonionic projective space" $(\mathbb{O} \otimes \mathbb{O} )\mathbb{P}^2$ -- the group of isometries of the latter is meant to be a form of $E_8$. In his paper on the octonions, John Baez mentions this, but warns that $(\mathbb{O} \otimes \mathbb{O} )\mathbb{P}^2$ can only be defined in terms of $E_8$, so this is circular, and he adds "alas, nobody seems to know how to define [it] without first defining $E_8$. Thus this group remains a bit enigmatic." The existence of the book Geometry of Lie groups by Boris Rosenfeld confuses me. In it, he claims to construct the plane, calling it $(\mathbb{O} \otimes \mathbb{O} ) {\simeq \atop S^2}$ (I cannot even reproduce it well in Latex). See Theorem 7.16 in particular. The problem is that each object, in this book, is claimed to be definable "by direct analogy" with some other object, itself usually not quite defined in full, and so on. I'm having an awful lot of trouble reading Rosenfeld's book. In the end (see 7.7.3) he claims that everything can be carried out over a finite field, yielding a definition of $E_8(q)$. But I cannot find the details anywhere -- the thing is, I don't even know if I'm reading a survey, or a complete treatment with proofs that escape me. Is anybody on MO familiar with Rosenfeld's book? Or is there an alternative reference for this mysterious "octo-octonionic plane"? REPLY [5 votes]: The most direct definition is probably as (the identity component of) the stabilizer of a tensor. More precisely, in The octic $E_8$ invariant by Cederwall and Palmkvist an explicit symmetric 8-form $f$ on $\mathbb{R}^{248}$ is constructed, having the compact $E_8$ (times $\{\pm1\}$) as the stabilizer in $GL(248, \mathbb{R})$. It is shown, moreover, in Simple groups stabilizing polynomials by Garibaldi and Guralnick, that any $E_8$-invariant 8-form that is not a scalar multiple of the fourth power of the Killing form $\kappa$ will do, and that the stabilizer in $GL(248, \mathbb{C})$ is the complex form (times $\mu_8$). The invariant 8-forms can be produced from arbitrary 8-forms by averaging with respect to the Haar measure (and almost all of them are not proportional to $\kappa^4$).<|endoftext|> TITLE: When can the metric be reconstructed (up to scaling) from knowing the conjugate points? QUESTION [5 upvotes]: Let $M$ be a smooth manifold of dimension $\geq 2$. Let $g$ be a complete Riemannian metric on $M$. Let $C \subseteq M \times M$ be the set of pairs of $g$-conjugate points. The set $C$ doesn't generally determine $g$, even up to scaling. For example hyperbolic space and Euclidean space both have no conjugate points. Question: When can the metric $g$ be recovered (up to scaling) from knowledge of the pair $(M,C)$? Does this happen when $M$ is A compact manifold? A compact homogeneous space? A compact symmetric space? Do curvature conditions help? And in the noncompact case, can $g$ sometimes be recovered (up to scaling) from $C$? Note: I've been a bit vague with the special cases I'm suggesting -- e.g. it's not clear if I'm asking whether for $M$ a compact homogeneous space, the conjugate points uniquely determine the metric among all metrics or just among homogeneous metrics. Ultimately I'd be interested in both questions, but I think the latter, more restricted question, is already interesting. EDIT: As an example, I think I've almost convinced myself that for $n \geq 2$, the round metric on $\mathbb R \mathbb P^n$ can be reconstructed (up to scaling) from knowledge of the conjugate points. REPLY [6 votes]: I guess the most famous cases are: (1) Riemannian tori without conjugate points are flat (Hopf, Burago and Ivanov) and (2) Auf-Wiedersehen metrics on the sphere (Green, Berger) are round (i.e. your projective space case). You can then do other things. For example Victor Bangert recently showed that a plane without conjugate point and with the area-growth of a Euclidean metric must be flat. I think you may find something of interest in Besse's book "Manifold all of whose geodesics are closed". Possibly the same conjugate points as the Fubini study metric in CP^n + Kahler implies that the metric is a multiple of the Fubiny-Study metric, etc.<|endoftext|> TITLE: Is each cover of the plane by lines minimizable? QUESTION [12 upvotes]: A cover $\mathcal C$ of a set $X$ by subsets of $X$ is called $\bullet$ minimal if for every $C\in\mathcal C$ the family $\mathcal C\setminus\{C\}$ is not a cover of $X$; $\bullet$ minimizable if $\mathcal C$ contains a minimal subcover of $X$. For example, any cover of the plane by parallel lines is minimal. Problem. Is each cover of the plane $\mathbb R^2$ by lines minimizable? What is the answer for the rational plane $\mathbb Q^2$? Acknowledgement. The problem was motivated by this question of Dominic van der Zypen. Added in Edit. For the rational plane the affirmative answer can be also deduced from the following general Theorem. A countable hypergraph $H=(V,\mathcal E)$ admits a minimal set $\mathcal M\subset \mathcal E$ with $\bigcup \mathcal M=\bigcup \mathcal E$ if each infinite set of vertices $I\subset V$ contains a finite set $F\subset I$ such that the set of edges $\{E\in \mathcal E:F\subset E\}$ is finite. The proof of this theorem is a bit complicated and will be presented in our joint paper with Dominic van der Zypen. We do not know if this theorem holds for arbitrary (not necessarily countable) hypergraphs. On the other hand, @Peter Komjath in his comment to the answer of @bof claims that he can prove that every cover of the real plane by lines is minimizable, using some (difficult) general result on minimal covers of hypergraphs whose edges have the same cardinality and have small intersections. REPLY [9 votes]: Unfortunately, you have two questions in one post. The one about $\mathbb R^2$ is too hard for me. The question about $\mathbb Q^2$ seems to have an easy affirmative answer, unless I'm making some dumb mistake. Let $P=\{p_0,p_1,p_2,\dots\}$ be the set of points, and let $L$ be the set of lines. (In general, the construction seems to work for any countable bipartite graph which is $C_4$-free and has no isolates.) Let $L_0=L.$ Consider the point $p_0.$ Let $L'_0$ be the set of all lines $\ell\in L_0$ such that $p_0\in\ell$ and $L_0\setminus\{\ell\}$ covers $P.$ So $L_0\setminus L'_0$ covers at least $P\setminus\{p_0\}.$ If $L_0\setminus L'_0$ covers $P,$ let $L_1=L_0\setminus L'_0.$ Otherwise, choose some $\ell\in L'_0$ and let $L_1=(L\setminus L'_0)\cup\{\ell\}.$ Continue in this way. I.e., Let $L'_1$ be the set of all lines $\ell\in L_1$ such that $p_1\in\ell$ and $L_1\setminus\{\ell\}$ covers $P,$ etc. Finally, I claim that the set $L_\infty=\bigcap_{n=0}^\infty L_n$ is a minimal cover. Maybe the construction will be clearer in words. At step $n$ we have the set $L_n$ of lines which have not yet been deleted (and this set still covers $P$), and we look at the point $p_n.$ We look at the set of all (undeleted) lines through $p_n.$ If such a line contains some point which is on no other (undeleted) line, we save it; if it contains no such point, we delete it. However, if this results in deleting all lines through $p_n,$ then we save one of them. Note that, after step $n,$ the point $p_n$ is covered by at least one line $\ell_n$ which will never be deleted, because it contains a point which is on no other line. Therefore $L_\infty$ is a cover. It is a minimal cover because, if $\ell$ is in $L_\infty,$ and if $p_n\in\ell,$ then after step $n$ has been done there is some point on $\ell,$ either $p_n$ or some other point, which is in no other surviving line. P.S Now suppose $P=\mathbb R^2$ and $L$ is a cover of $\mathbb R^2$ by lines. I think the same construction will work under some very restrictive assumptions on the cover; namely, that we can enumerate the points as $\mathbb R^2=\{p_\alpha:\alpha\lt\mathfrak c\}$ so that, for each $\alpha,$ either $|\{\ell\in L:p_\alpha\in L\}|\lt\omega$ or else $|\{\ell\in L:p_\alpha\in L\}|\gt|\alpha|.$ This is to prevent $p_\alpha$ from getting uncovered at limit stages.<|endoftext|> TITLE: Double Counting: Motivic Edition QUESTION [19 upvotes]: One of the most important proof techniques in combinatorics is double counting: proving that both sides of an identity count elements of some set in two different ways. This question is an attempt at understanding various forms of double counting for various motivic invariants such as Euler characteristics, counts of $\mathbb F_q$ rational points, cohomology etc. Question: What are your favorite (or instructional) examples of identities that come from computing a motivic invariant for a space in two different ways? I'll start by including some combinatorial examples I could think of. Example 1: The simplest form of this is to do a straightforward $q$-analog of a classical double counting identity. For example Vandermonde's identity $$\binom{m+n}{k}=\sum_{j}\binom{m}{j}\binom{n}{k-j}$$ has a motivic lift to $$[Gr(k,m+n)]=\sum_{j}[Gr(j,m)][Gr(k-j,n)][\mathbb A^{j(m-k-j)}]$$ which gives us the q-analog of Vandermonde's identity. Example 2: The Lagrangian Grassmannian $L(n,2n)$ of $n$-dimensional isotropic subspaces in a symplectic vector space is an example where we can write $[L(n,2n)]$ as a quotient $$[Sp_{2n}(\mathbb F_q)]/[\text{Stab}(\mathbb F_q)]=(1+q)(1+q^2)\cdots(1+q^n)$$ where $\text{Stab}$ is the stabilizer of the transitive action of the Symplectic group on $L(n,2n)$. On the other hand we can also express $[L(n,2n)]$ as a sum over cells which are bundles over grassmannians with fibers being affine spaces. This results in the q-binomial theorem $$[L(n,2n)]=\sum_{j}[Gr(j,n)][\mathbb A^{j(j+1)/2}].$$ (I personally like this example because it shows that the power set $P(\{1,2,\dots, n\})$ can be written as the n-element subsets of $\{1,2,\dots,n,\bar{1},\bar{2},\dots,\bar{n}\}$ which intersect every $\{i,\bar{i}\}$ in precisely one element, and this can be thought of as a kind of $\mathbb F_1$-Lagrangian Grassmannian somehow.) Example 3: Loehr and Warrington defined a statistic on partitions $h_x$ for every $x\in [0,\infty)$ and gave bijective proofs of $$\sum_{\lambda} q^{|\lambda|}t^{h_x(\lambda)}=\prod_{i=1}^{\infty}\frac{1}{1-tq^i}$$ so in particular all $h_x$ have the same distribution on partitions independent of $x$. This has a geometric meaning for irrational $x$ (due to Haiman) since the RHS is the generating function of Poincare polynomials of Hilbert schemes $\operatorname{Hilb}_n(\mathbb A^2)$. All these Hilbert schemes come with an action of a 2 dimensional torus. If we pick a 1 dimensional subtorus with slope parameter $x$ and take Bialynicki-Birula decompositions from its action we get affine cells for every partition $\lambda$ (these index the torus fixed points) whose dimensions are distributed according to $h_x(\lambda)$. So this example is computing the Poincare polynomial of $\operatorname{Hilb}_n(\mathbb A^2)$ in infinitely many ways. REPLY [9 votes]: I will upgrade my comment to an answer. Let $\mathcal{A}$ be an $n$-dimensional hyperplane arrangement defined over the rational numbers $\mathbb{Q}$ (or equivalently by clearing denominators, over $\mathbb{Z}$). Associated to $\mathcal{A}$ is its characteristic polynomial $\chi(\mathcal{A};q)$. There are various ways one can think about $\chi(\mathcal{A};q)$ and these lead to the kind of motivic double counting described in the question (although I don't know if it can be neatly expressed as a single equality of $q$-numbers). On the one hand there is a purely combinatorial definition: $\chi(\mathcal{A};q) = \sum_{F \in L(\mathcal{A})}\mu(F)\,q^{\mathrm{dim}(F)}$, where $L(\mathcal{A})$ is the intersection poset of $\mathcal{A}$, and $\mu$ is the Möbius function of $L(\mathcal{A})$. Technically the intersection poset $L(\mathcal{A})$ depends on a field, but since we assumed $\mathcal{A}$ was defined over $\mathbb{Q}$ we can always take $\mathbb{Q}$ to be our field. We can also think of the characteristic polynomial as (essentially) a Poincaré polynomial of a complex algebraic variety: $\chi(\mathcal{A};q)=q^n\sum_{i}b_i\,(-1/q)^{i}$, where $b_i$ is the $i$th Betti number of the complement $\mathbb{C}^n\setminus \mathcal{A}$. Third, $\chi(\mathcal{A};q)$ also counts points of that "same" variety over finite fields $\chi(\mathcal{A};q)=\#(\mathbb{F}^n_{q}\setminus\mathcal{A})$, where $\mathbb{F}_q$ is a finite field with $q$ elements. Actually we require that $\mathbb{F}_q$ be of sufficiently large characteristic, depending on $\mathcal{A}$, for this to point-counting to hold. There is something vaguely similar to the Weil conjectures here where the topology of a complex algebraic variety determines the point-counting function for that same variety over finite fields. It is also worth mentioning the well known fact that $\chi(\mathcal{A};q)$ contains information about $\mathcal{A}$ viewed as a real arrangement as well: $(-1)^n\chi(\mathcal{A};-1)$ is the number of regions of $\mathcal{A}$ viewed as a real arrangement; and $(-1)^{\mathrm{rank}(\mathcal{A})}\chi(\mathcal{A};1)$ is the number of relatively bounded regions. Finally, as Gjergji mentioned, in some special cases it may be that $\chi(\mathcal{A};q)$ factors nicely into linear terms: for instance, this happens when $\mathcal{A}$ is free. Freeness is a certain deep algebraic property of arrangements introduced by Terao in the 80s; it generalizes the combinatorial property of supersolvability which Stanley introduced in the 70s. We say $\mathcal{A}$ is free if its "module of logarithmic derivations" is a free module over the polynomial ring; in this case the module is generated by homogeneous elements whose degrees $d_1,\ldots,d_n$ are called the exponents of the arrangement. A celebrated theorem of Terao says that for a free arrangement we have $\chi(\mathcal{A};q) = \prod_i (q-d_i)$.<|endoftext|> TITLE: What is the spectral interpretation of the arithmetic zeta function? QUESTION [5 upvotes]: I recently stumbled upon the slides of a talk given by Kedlaya, in which the following appears: For $X$ of finite type over $F_q$, a Weil cohomology theory, mapping $X$ to certain vector spaces $H^i(X)$ over a field of characteristic zero, provides a spectral interpretation of $\zeta(X,s)$ via the formula: $$\zeta(X,s)=\prod_i \det(1-q^{-s}\text{Frob}_q,H^i(X))^{(-1)^{i+1}}$$ Where $\zeta(X,s)$ is the arithmetic zeta function of a scheme $X$ of finite type over $\mathbb{Z}$. This is incredibly intriguing to me, and seems to be intimately related to some objects I'm trying to study - however, this slide really doesn't give me enough information to pull out some juicy math. Where does this formula come from (it seems very close to but not equivalent to the product in the definition of an Artin L-Function), and how does it provide a spectral interpretation of the arithmetic zeta function? If the answer is too big to be self-contained, I'd also appreciate a reference or some buzzwords to google. REPLY [5 votes]: First, it's worth mentioning that the above theorem refers to the (local) Hasse-Weil zeta function of a scheme $X$ of finite type over $\mathbb{F}_q$ rather than the arithmetic zeta function of a scheme $X$ of finite type over $\mathbb{Z}$. Certainly these are related: for $X$ is a scheme of finite type over $\mathbb{Z}$ the arithmetic zeta function $\zeta(X,s)$ is the product of the (local) Hasse-Weil zeta functions $\zeta(X_{\mathbb{F}_p},s)$ for primes p, along with some Archimedean factor. Now what Kedlaya is referring to is Grothendieck's cohomological interpretation of the (local) Hasse-Weil zeta function $\zeta(X,s)$ for $X$ a scheme of finite type over $\mathbb{F}_q$ and the first Weil conjecture, that $\zeta(X,s)$ is a rational function given as the product of characteristic polynomials of the (geometric) Frobenius acting on the $\ell$-adic cohomology of $X$. Put another way, $\zeta(X,s)$ has a spectral interpretation in terms of Frobenius eigenvalues. The idea is that since $\zeta(X,s)$ is a generating function for the number of $\mathbb{F}_{q^n}$-rational points of $X$, and since the number of $\mathbb{F}_{q^n}$-rational points of $X$ is given as the alternating sum of traces of the (geometric) Frobenius acting on the $\ell$-adic cohomology of $X$ by the Grothendieck-Lefschetz trace formula, it takes only a bit of algebra to conclude the cohomological interpretation of $\zeta(X,s)$. A good introductory reference is Zeta Functions in Algebraic Geometry and Milne's Étale cohomology book, or any other book on étale cohomology that covers the Weil conjectures.<|endoftext|> TITLE: Isotrivial Monodromy QUESTION [5 upvotes]: Let $X\to \Delta$ be a projective family, smooth over $\Delta^*$, such that all fibers over $t\in \Delta^*$ are isomorphic. Does the monodromy representation factor through the algebraic automorphism group of the smooth fiber, $Aut(X_t)$? This is certainly false for non-isotrivial families, since Dehn twists on curves are infinite order, whereas smooth curves of genus $>1$ have finite automorphism groups. Example 1: The family $y^2 = x^3+t$ has monodromy of order 6, which is precisely the automorphism group of the Eisenstein elliptic curve. Example 2: The family of smooth quadric surfaces $\mathbb P^1\times \mathbb P^1$ degenerating to a quadric cone has monodromy of order 2, which corresponds to swapping the factors. In the symplectic category, there is a notion of parallel transport, so we have a monodromy map $\pi_1(\Delta^*)\to Symp(X_t)/Ham(X_t)$. Perhaps if the family is algebraically isotrivial, then the monodromy is valued in $Aut(X_t)$ and is homotopy invariant, so there is no quotient? REPLY [3 votes]: I think the answer is yes, depending on how you are defining the monodromy*. I take "isotrivial" to mean that you have a smooth fibre bundle over the punctured plane where the fibres have complex structures and any two fibres are biholomorphic. Take the covering space of $X/X_0$ corresponding to the subgroup of $\pi_1$ given by the kernel of the projection to $\pi_1$ of the base ($\mathbb{Z}$). This is now an isotrivial family over $\mathbb{C}$, which I claim is trivial (see below). The deck group is $\mathbb{Z}$ acting as a translation in $\mathbb{C}$ coupled with a biholomorphism of the fibre, and this automorphism is the monodromy (in particular, you see it's an automorphism). In general, given a fibre bundle where the complex structure varies, you get a map from the base to $J/Diff$, where J is the space of all complex structures. In the isotrivial case, this is the constant map, so lifts to a fibre. The fibre is the orbit of a complex structure under the action of the diffeomorphism group, which is $Diff/Aut$ (as $Aut$ is the stabiliser). In particular, if the base is contractible then this map lifts to a map from the base to $Diff$, which you can use to pullback the complex structure in each fibre to get the trivial family of complex manifolds. If the base is $S^1$ (equivalently $\mathbb{C}^*$) then this map lifts to a map from the interval into $Diff$ whose endpoints map to $Aut$ (starting at the identity). Again, you can think of the automorphism at the endpoint 1 as the monodromy. *For instance, symplectic monodromy will depend (up to Hamiltonian isotopy) on the precise choice of loop in the base and the choice of symplectic form.<|endoftext|> TITLE: What programming language should a professional mathematician know? QUESTION [80 upvotes]: More and more I am becoming convinced that one should know at least one programming language very well as a mathematician of this century. Is my conviction justified, or not applicable? If I am right, then please what languages should someone aspiring to be a mathematician learn? The number out there is so bewildering for a complete novice to judge, and no one else can judge suitable ones than a body of working mathematicians, hence my posting this question here specifically. In particular, this language should be very useful for mathematics applications, should be close in syntax and structure to mathematics, and be mathematically related in other relevant ways. Indeed you may suggest a language that you have found useful/important in other ways not mentioned, but please explain why you make these suggestions clearly. Thank you. PS. I couldn't find sufficiently relevant tags. Please improve as appropriate. REPLY [4 votes]: One of your questions is: What languages should someone aspiring to be a mathematician learn? For an aspiring mathematician -- learn a language that is useful to you outside of academia, since most people who get PhDs do not end up working in academia. Since you could well end up in data science, Python and R are good choices for this, and as it happens Python also seems to be the consensus answer for mathematical research.<|endoftext|> TITLE: Are these two constructions of $K_0(A)$ isomorphic? QUESTION [5 upvotes]: The following question is extracted from this question on MSE, which got no answer so far, probably because it was a bit hidden by another question which a posteriori was totally obvious. Let $A$ be a ring with $1$, that I am happy to suppose commutative if necessary. Let $\mathrm{Proj}(A)$ be the commutative monoid of isomorphism classes $\langle M\rangle$ of finitely generated projective modules, the internal law being given by: $\langle M\rangle+\langle N\rangle =\langle M\times N\rangle$. I know two possible definitions of the group $K_0(A)$. Def 1. $K_0(A)$ is the quotient of the free abelian group on $Proj(A)$ by the relations $ \langle M_2\rangle =\langle M_1\rangle +\langle M_3\rangle $ whenever we have a short exact sequence $0\to M_1\to M_2\to M_3\to 0$. If we denote by $[M]$ the image of $\langle M \rangle $ under the canonical projection, then any element may be written as $[M]-[N]$, and we have $[M]-[N]=0$ if and only if there exists $r,s\geq 0$ such that $M\times A^r\simeq N\times A^s.$ Def 2. $K_0(A)$ is the Grothendieck group (i.e.symmetrization) of the monoid $\mathrm{Proj}(A)$, that is the quotient set of $\mathrm{Proj}(A)\times \mathrm{Proj}(A)$ wrt to the equivalence relation $$(\langle M_1\rangle,\langle N_1\rangle)\sim (\langle M_2\rangle ,\langle N_2\rangle)\\ \iff \exists \ \langle P\rangle\in \mathrm{Proj}(A), \langle M_1\rangle + \langle N_2\rangle +\langle P\rangle=\langle M_2\rangle+\langle N_1\rangle +\langle P\rangle.$$ If $[M]$ denotes this time the class of $(\langle M\rangle,0)$, then any element may be written as $[M]-[N]$, and we have $[M]-[N]=0$ if and only if there exists $n\geq 0$ such that $M\times A^n\simeq N\times A^n$. Question . Do these two constructions yield isomorphic groups ? I do not know any counterexample though, at least amongst all the very few examples of $K_0$ I am aware of. Thanks in advance of any enlightning thoughts. Greg Edit 1 I switched the equivalences between the definitions, since they were misplaced. Edit 2. the equivalence in Def.1 is false. I read this in some reference i found on the web, and believed it without doublechecking at the time being. Thanks to the answers, I realize know that the right equivalence is the same as in Def.2, which solves immediately the question. REPLY [7 votes]: The two constructions give the same result, namely the universal group with a monoid homomorphism from $Proj(A)$ aka the Grothendieck group. The first definition gives this because a short exact sequence with $M_3$ projective always splits so that the relations are simply "$[M_2] = [M_1]+[M_3]$ in the group whenever $[M_2]=[M_1]+[M_3]$ in the monoid". The second definition is simply the explicit construction of the Grothendieck group. Because objects defined by universal properties are unique up to unique isomorphism, the two constructions are isomorphic.<|endoftext|> TITLE: Proof of a combinatorial equation QUESTION [8 upvotes]: How can we use elementary methods to prove that $$\sum_{i = 2}^{n}{{n \choose i} i! n^{n - i}} = \sum_{i = 1}^{n - 1}{{n \choose i}i^i (n - i)^{n - i}}$$ for any integer $n \geq 0$? The values of each side for fixed $n$ are 0, 0, 2, 24, 312, 4720, ... (A001864 - OEIS). REPLY [2 votes]: Here is an alternative. Define the functions $A_n(t)=\sum_k\binom{n}kt(t+k)^{k-1}(n-k)^{n-k}, B_n(t)=\sum_k\frac{n!}{(n-k)!}(t+n)^{n-k}$ and $C_n(t)=\sum_k\binom{n}k(t+k)^k(n-k)^{n-k}$. From $(t+k)^k=t(t+k)^{k-1}+k(t+k)^{k-1}$ and on the basis of Abel's identity, one gets $C_n(t)=A_n(t)+nC_{n-1}(t+1)=(t+n)^n+nC_{n-1}(t+1)$. It is easy to check that $B_n(t)=(t+n)^n+nB_{n-1}(t+1)$. Since both $B_n$ and $C_n$ satisfy the same initial conditions, it follows $B_n(t)=C_n(t)$. So, $B_n(0)=C_n(0)$ gives the desired result.<|endoftext|> TITLE: Bishop-Gromov for Kähler metrics QUESTION [5 upvotes]: Let $(M, g)$ be a (complete) Kähler manifold with Ricci curvature $\geq c$. Is it true that the volume ratio of geodesic balls in $M$ with respect to balls in the corresponding (simply connected) complex space form with Ricci $\equiv c$ is a decreasing function? I suspect the answer is no. Examples would be nice, also showing why the Riemannian proof breaks would help. EDIT: A friend made me notice Gang Liu's paper (https://arxiv.org/pdf/1108.4231v1.pdf), where it is shown that, for real analytic metrics, the volume ratio is decreasing for small values of the radius. REPLY [3 votes]: It looks like the answer is indeed no with the quadric $\mathbb CP^1\times \mathbb CP^1$ a counterexample. Here the corresponding complex model space is $\mathbb CP^2$. Recall that to get an Einstein metric with coefficient $\lambda=1$ we should choose it is as curvature of the anti-canonical bundle $-K$. Now, for $\mathbb CP^1\times \mathbb CP^1$ we have $-K\cong O(2)\times O(2)$. The diagonal $\mathbb CP^1\subset \mathbb CP^1\times \mathbb CP^1$ is geodesic and $-K$ restricts as $O(4)$ to it. It follows that the diameter of $\mathbb CP^1\times \mathbb CP^1$ is larger than that of $\mathbb CP^2$ (where $-K\cong O(3)$). Inddeed the diameter of the former space is equal to the diameter of its diagonal and the diameter of the latter space is equal to the diameter of any line in it. As for where the usual proof breaks down, I don't know.<|endoftext|> TITLE: Difficult trigonometric integral QUESTION [10 upvotes]: This question was also asked here and here. I have faced some difficulties to do the following integral: $$ I=\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta~\sin\theta\int_{0}^{\infty}dr~r^2\frac{3x^2y^2\cos(u r \sin\theta \cos\phi)\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}\mathrm e^{-\frac{r^2}{2}} \tag{1}, $$ where $x$, $y$, and $u$ are real positive constants. I tried at least two ways to solve this integral: First attempt: I began to solve the $r$ integral first. By using Mathematica, then $$ I=\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta~\sin\theta\frac{3x^2y^2(1-u^2\sin^2\theta\cos^2\phi)\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}\mathrm e^{-\frac{u^2}{2}\sin^2\theta\cos^2\phi} \tag{2}. $$ After that, I looked for a solution for $\phi$ integral. My best attempt was: $$ I_\phi(x,y,u,\theta)=\frac{2}{B}\left[B\left(\frac{1}{2}\right)F_1\left(\frac{1}{2},1,-;1;\nu,-\frac{a}{2}\right)-aB\left(\frac{1}{2}\right)F_1\left(\frac{3}{2},1,-;2;\nu,-\frac{a}{2}\right)\right], $$ where $B=x^2\sin^2\theta+x^2y^2\cos^2\theta$, $a=u^2\sin^2\theta$, and $\nu=\frac{x^2-y^2}{x^2+x^2y^2\cot^2\theta}$. In this way, the final results it's something like that: $$ I= \int_{0}^{\pi} \mathrm d \theta~3x^2y^2\sin\theta \cos^2\theta~ I_\phi(x,y,u,\theta). \tag{3}. $$ Eq. $(3)$ cannot be further simplied in general and is the nal result. Second attempt: To avoid the hypergeometric function $F_1$, I tried to start with the $\phi$ integral. In this case, my initial problem is an integral something like that: $$ \int_{0}^{2\pi} \mathrm d \phi \frac{\cos(A \cos\phi)}{a^2\cos^2\phi+b^2\sin^2\phi}. \tag{4} $$ This integral $(4)$ can be solved by series (see Vincent's answer and Jack's answer). However those solutions, at least for me, has not a closed form. This is my final step on this second attempt :( What is the point? It turns out that someone has managed to solve the integral $(1)$, at least the integral in $r$ and $\phi$. The final resuls found by this person was: $$ I_G=\frac{12 \pi x~y}{(1-x^2)^{3/2}}\int_{0}^{\sqrt{1-x^2}} \mathrm dk \frac{k^2 \exp\left(-\frac{u^2}{2}\frac{x^2k^2}{(1-x^2)(1-k^2)}\right)}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}}, $$ where, I belive, $k=\sqrt{1-x^2}\cos\theta$. As you can see in this following code performed in Mathematica IG[x_, y_, u_] := Sqrt[Pi/2] NIntegrate[(12 Pi x y)/(1 - x^2)^(3/2) (v^2 Exp[-(u^2 x^2 v^2)/(2 (1 - x^2) (1 - v^2))])/(Sqrt[1 - v^2] Sqrt[1 - v^2 (1 - y^2)/(1 - x^2)]), {v, 0, Sqrt[1 - x^2]}] IG[.3, .4, 1] ** 4.53251 ** I[x_, y_, u_] := NIntegrate[(r^2 Sin[a] Cos[ u r Sin[a] Cos[b]] 3 x^2 y^2 Cos[a]^2 Exp[-r^2/ 2])/((y^2 Cos[b]^2 + x^2 Sin[b]^2) Sin[a]^2 + x^2 y^2 Cos[a]^2), {r, 0, Infinity}, {a, 0, Pi}, {b, 0, 2 Pi}] I[.3, .4, 1] ** 4.53251 ** the integrals $I$ and $I_G$ are equals. Indeed, since that they emerge from the same physical problem. So, my question is: what are the steps applied for that integral $I$ gives the integral $I_G$? Edit Since my question was not solved yet, I think it is because it is a tough question, I will show a particular case of the integral $I$, letting $u=0$. I hope with this help you help me. In this case, the $r$ integral in $(1)$ is trivial and the integral takes the form: $$ I_P=\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta~\sin\theta\frac{3x^2y^2\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}. \tag{5} $$ The $\phi$ integral can be integrated with the help of Eq. 3.642.1 in Gradstein and Ryzhik's tables of integrals. Thereby, the $I_P$ takes the for: $$ I_P=3xy\int_{0}^{\pi}d\theta\frac{\sin\theta\cos^2\theta}{\sqrt{1+(x^2-1)\cos^2\theta}\sqrt{1+(y^2-1)\cos^2\theta}}. \tag{6}$$ Now the change of variable $k=\sqrt{1-x^2}\cos\theta$ bring expression $(6)$ to the form $$ I_P= \frac{(const) x~y}{(1-x^2)^{3/2}}\int_{0}^{\sqrt{1-x^2}} \mathrm dk \frac{k^2}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}}. $$ Did you notice how $I_G$ and $I_P$ are similar? Do you think a similar approach can be applied to my original problem? Please, let me know. I've solved this problem applying the Schwinger proper-time substitution: $$\frac{1}{q^2}=\int_{0}^{\infty}\mathrm{d\xi}~\mathrm{e^{-q^2\xi}}. $$ REPLY [7 votes]: Here is an outline of the approach I have taken to solve this integral. First rewrite the integral $(1)$ in Cartesian variables: $$I=\int_{-\infty}^{\infty} \mathrm{d}^3v~ \frac{3x^2y^2v_z^2}{y^2v_x^2+x^2v_y^2+x^2y^2v_z^2}\cos(uv_x)\exp\left(-\frac{v_x^2}{2}-\frac{v_y^2}{2}-\frac{v_z^2}{2}\right). $$ Now use the following substitution $$ \frac{1}{y^2v_x^2+x^2v_y^2+x^2y^2v_z^2}=\int_{0}^{\infty}d\tau~\mathrm{ e^{-(y^2v_x^2+x^2v_y^2+x^2y^2v_z^2)\tau}},$$ such that $$ I=\int_{0}^{\infty}d\tau\int_{-\infty}^{\infty} \mathrm{d}^3v~3x^2y^2v_z^2\cos(uv_x) \mathrm{e^{-v_x^2(\tau y^2+1/2)-v_y^2(\tau x^2+1/2)-v_z^2(\tau x^2y^2+1/2)}}. $$ The $(v_x,v_y,v_z)$ can be evaluated with the help of Mathematica. The results gives $$ \int_{-\infty}^{\infty} \mathrm{d}^3v~3x^2y^2v_z^2\cos(uv_x) \mathrm{e^{-\alpha v_x^2-\beta v_y^2-\gamma v_z^2}}=\frac{3\pi^{3/2}}{2x^2y^2}\frac{\exp\left(-\frac{u^2}{4y^2}\frac{1}{\tau+1/2y^2}\right)}{\left(\tau+1/2x^2y^2\right)^{3/2}\left(\tau+1/2x^2\right)^{1/2}\left(\tau+1/2y^2\right)^{1/2}}.$$ Thereby, $$ I= -\frac{3~\mathrm{const}}{x^2y^2~\mathrm{const}}\int_{0}^{\infty}\mathrm{d\tau}\frac{\exp\left(-\frac{u^2}{4y^2}\frac{1}{\tau+1/2y^2}\right)}{\left(\tau+1/2x^2y^2\right)^{3/2}\left(\tau+1/2x^2\right)^{1/2}\left(\tau+1/2y^2\right)^{1/2}}. $$ Now, performing the substitution $\tau=\frac{1-x^2}{2x^2y^2k^2}-\frac{1}{2x^2y^2}$ gives us $$ I=(\mathrm{const})~\frac{3x~y}{(1-x^2)^{3/2}}\int_{0}^{\sqrt{1-x^2}}\mathrm{dk}\frac{k^2\exp\left(-\frac{u^2}{2}\frac{x^2k^2}{\left(1-x^2\right)\left(1-k^2\right)}\right)}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}}, $$ which is the desired integral unless of a constant. ;)<|endoftext|> TITLE: Partition of 4-tuples QUESTION [8 upvotes]: Some $4$-tuples of positive real numbers $(a_1,b_1,c_1,d_1),\dots,(a_n,b_n,c_n,d_n)$ are given, with all $a_i,b_i,c_i,d_i\leq 1$. Can we always partition $\{1,2,\dots,n\}$ into two subsets $X,Y$ so that $$1+\sum_Xa_i\geq \sum_Ya_i\text{ and } 1+\sum_Xb_i\geq \sum_Yb_i$$ and $$\sum_Xc_i\leq 1+\sum_Yc_i\text{ and } \sum_Xd_i\leq 1+\sum_Yd_i?$$ It is shown here that the partition is possible if we replace the $1$'s by $3$'s. What is the best possible value between $1$ and $3$? REPLY [2 votes]: Mike Earnest's answer at MSE may be improved even more, in order to get the estimate of $2$ --- but this is also non-sharp. Indeed, in his last part, we may assume that $x_1\geq x_2\geq x_3\geq x_4$. Next, we may assume that $x_2\geq 0$. Now, if $x_3+x_4\leq 0$, then we may set $x_3'=x_4'=-1$, $x_1'=x_2'=1$, changing each coordinate by at most 2. Asusume now that $x_3+x_4\geq 0$ --- this is more delicate; then $x_3\geq 0$. Choose an index $i\in\{1,2,3\}$ such that $a_i$ is not the strict minimum among $a_1,a_2,a_3$, and $b_i$ is not the strict minimum among $b_1,b_2,b_3$. Then we set $x_i'=-1$ and $x_j'=1$ for all other $j$. Surely, the $c$- and $d$-coorditnates increased by at most 2 (this could happen due to $x_i'$ only!). Now let us show that $a$-coordinate did so as well. Let $a_k=\min(a_1,a_2,a_3)$ with $k\in\{1,2,3\}\setminus\{i\}$, and set $\ell=\{1,2,3\}\setminus\{i,k\}$. Then $$ \sum(x_j'a_j-x_ja_j)=(1-x_4)a_4+(1-x_\ell)a_\ell+(1-x_k)a_k+(-1-x_i)a_i \leq (1-x_4)+(1-x_\ell)+(1-x_k-1-x_i)a_k\leq (2-x_4-x_\ell)+0\leq 2, $$ as required. The $b$-coordinate argument is similar. NB. It seems that such considerations (by making an argument more case distinctive) may lead to a constant of $3/2$. But this still would not give an optimal bound --- at least it seems so! Notice also that this problem has much in common with the famous problem on sign sequences (see, e.g., Chapter 4 in this survey; the method there is more or less the same as in the MSE answer). The difference is that in your problem the coordinates have fixed signs.<|endoftext|> TITLE: Is the function Point -> Julia set "injective"? QUESTION [5 upvotes]: Consider the functions $f_c(z) := z^2 + c$ for $c \in \mathbb C$. For each such function, we may form the associated Julia set. My question: If $c, c' \in \mathbb C$ produce in this way the same Julia set, does this imply $c = c'$? Trivially this is the case if we "consider one more dimension" by taking orbits into account. But if we consider the Julia set only, I can't find the solution. REPLY [6 votes]: You can deduce a positive answer to your question from Theorem 1 in the following paper: P. Atela, J. Hu, Commuting polynomials and polynomials with same Julia set, Internat. J. Bifur. Chaos Appl. Sci. Engrg. 6 (1996), no. 12A, 24 27–2432 (preprint available here: https://arxiv.org/pdf/math/9504210.pdf) The theorem says in particular that, for two centered polynomials of degrees $n,m \geq 2$ respectively, if these polynomials have the same Julia set which is not a circle or an interval, then (up to a symmetry) they are both iterates of the same polynomial. `Centered' means that there is no term of degree $n-1$, resp. $m-1$ (the corresponding coefficients are zero).<|endoftext|> TITLE: Estimate for radius of convergence of solutions given by Cauchy-Kovalevskaya Theorem QUESTION [7 upvotes]: I'm sure you can extract it from the proof, but does anyone know of a reference where the radius of convergence (in terms of radius of convergence of the initial data and PDE) of the solution given by the Cauchy-Kovalevskaya Theorem is written down? On a related, but more speculative note, I'm curious if there are any results along the following lines: Suppose one is given a problem solvable by the Cauchy-Kovalevskaya Theorem along with a real analytic solution with uniform lower bound on its radius of convergence (alternatively which exists in some uniform strip about the hypersurface on which initial data is prescribed). If one approximates the data of the given solution (the right notion is part of the question) by some sequence of real analytic data, can one say anything about the radius of convergence of the corresponding solutions (or size of strip the solutions exist on). REPLY [3 votes]: Theorem 9.4.5 in the first volume of Hörmander's four-volume treatise "The Analysis of Linear Partial Differential Operators" is a quantitative version of the Cauchy-Kovalevsky theorem. Condition (9.4.7) of that theorem gives bounds on the radii of convergence. The theorem is stated with zero initial data. To apply it with non-zero initial data, first move these into the right-hand side of the equation by applying the differential operator to the Taylor polynomial (in $z_n$) of the initial data.<|endoftext|> TITLE: Why is Lebesgue measure theory asymmetric? QUESTION [20 upvotes]: A set $E\subseteq \mathbb{R}^d$ is said to be Jordan measurable if its inner measure $m_{*}(E)$ and outer measure $m^{*}(E)$ are equal.However, Lebesgue mesure theory is developed with only outer measure. A function is Riemann integrable iff its upper integral and lower integral are equal.However, in Lebesgue integration theory, we rarely use upper Lebesgue integral. Why are outer measure and lower integral more important than inner measure and upper integral? REPLY [2 votes]: The asymmetry has only historical reasons. It is possible to develop Lebesgue theory (moreover, all extension theorems and thus the theory of product measures) from the “inner approach”. This was done by Heinz König in a couple of papers and monographs. Although this “inner approach” is for the Lebesgue measure equivalent (more general, in the $\sigma$-finite case, IIRC), it has huge advantages for the non-$\sigma$-finite case, since the "outer approach" loses a lot of information - intuitively, there are “bubbles of measure $\infty$” which lost all information. For example, the product of Haar measures with the “inner” approach is compatible with the Fubini-Tonelli theorem one obtains from the approach by Haar integrals for functions with compact support. (For the “outer” approach this holds only for the $\sigma$-finite case, in general.)<|endoftext|> TITLE: Is canonical model always with canonical singularity QUESTION [8 upvotes]: Let X be a smooth variety, take the proj of canonical ring of X and denote it by Y. Is Y always a canonical variety? I know it's true for general type variety. Thank in advance. REPLY [13 votes]: I believe that $(Y,B)$ is always klt for some boundary $B$. In fact by Theorem 5.2 of https://projecteuclid.org/download/pdf_1/euclid.jdg/1090347529, after passing to a truncation of the pluricanonical rings, there is a klt pair of log general type $(Y',B')$ with isomorphic pluricanonical ring (more precisely $R(K_X)^{(a)}\cong R(K_{X'}+B')^{(b)}$ for appropriate integers $a,b>0$). But then $Y={\rm Proj} R(K_X)\cong {\rm Proj} R(K_{Y'}+B')$ and $Y'\to Y$ is the log canonical model of $(Y',B')$ which has klt singularities as $(Y',B')$ has klt singularities. I also believe that $Y$ may not be canonical. I think one can construct examples where $Y$ is not canonical by considering a surface $S$ of general type with a cyclic group $G$ of order $n$ acting on it such that $S/G$ has non-canonical singularities (eg. $n=3$ and locally $g(x,y)=(\xi x, \xi ^2 y)$ where $\xi$ is a primitive third root of 1) and $E$ an elliptic curve where $G$ acts via a translation of order $n$. Then $X=(S\times E)/G$ should be smooth but $Y=S/G$ is klt but not canonical (I did not check the details!).<|endoftext|> TITLE: What can we say about the Cartesian product of a manifold with its exotic copy? QUESTION [23 upvotes]: Let $M$ be a smooth oriented manifold, and let $M^E$ be an exotic copy, i.e homeomorphic but not diffeomorphic to $M$. Is it true that $M\times M$ is diffeomorphic to $M\times M^E$? I am interested in knowing the answer for closed manifolds. One example which I can think of is an exotic $\mathbb R^4$. But we know that $\mathbb R^8$ has unique smooth structure. So in that case this is true. One of the motivation for asking this question is following: I want to see an example of symplectic manifold $\times$ non-symplectic manifold is symplectic manifold . There are some manifolds whose some exotic structure doesn't admit symplectic structure. Also we know that Cartesian product of two symplectic manifold always admits symplectic structure. If the answer of my original question is positive, then that will imply the example I am looking for. EDIT: A potential example of a symplectic manifold with non-symplectic exotic copy could be: Consider $E$ as elliptic surface. $F$ be fiber. Let $N_1= F\times D^2$. Consider $K$non-fibered knot is $S^3$ with non-monic $\Delta_k$. $T= K\times S^1$. And let $N_2=T\times D^2$ $\subset S^3\times S^1$. Them $E^E = (E-N_1)\cup (S3\times S^1-N_2)$ (gluing along meridian, sometimes it is called knot surgery). Then observe that in homology level we are not changing anything. So Freedman implies that it is homeomorphic with $E$. But Siberg-Witten invariant depends on Alexander polynomial of the knot. So thus we can conclude it has no symplectic structure. Similar ideas are written here https://arxiv.org/pdf/dg-ga/9612014.pdf REPLY [21 votes]: Your question seems to be about simply connected exotic 4-manifolds, in which the answer is yes. That's because $M$ and $M^E$ are h-cobordant (by Wall), say via an h-cobordism W. Then $M \times W$ is an h-cobordism between $M \times M$ and $M\times M^E$, which is trivial by the high-dimensional h-cobordism theorem. For higher dimensional manifolds, I don't think this is true. An example could come from $M = $ an exotic $\mathbb{C}P^4$. But I'd have to think about this for a bit.<|endoftext|> TITLE: The different gradings of a graded ring, and their schemes QUESTION [8 upvotes]: Let $(A,g)$ be a graded commutative ring, where $A$ denotes the commutative ring, and $g$ its grading. What can be said about the set $\mathcal{G}_A := \{ \mathrm{Proj}\Big((A,g)\Big)\ \vert\ g \}$? How do elements of $\mathcal{G}_A$ relate to each other ? Can we endow $\mathcal{G}_A$ with more structure ? REPLY [2 votes]: EDIT. I misread the question - below I'm trying to describe the set of all gradings and not of their ${\rm Proj}$. Note that giving a $\mathbf{Z}$-grading on a $k$-algebra $A$ is equivalent to giving a $\mathbf{G}_m$-action on $X={\rm Spec}(A)$. For example, if $A=k[x]$ for $k$ algebraically closed, then every such action is linearizable, i.e. either is trivial (everything in degree zero), or there exists an $a\in k$ such that $x-a$ is homogeneous with some degree $n$. One can think of the set $\mathcal{G}_A$ in this case as the disjoint union of $\mathbf{Z}\setminus \{0\}$ copies of $\mathbf{A}^1$ and a point corresponding to the zero grading. Suppose that $A = k[x_1,\ldots, x_n]/I$. Then giving a $\mathbf{Z}$-grading/$\mathbf{G}_m$-action on $A$ is equivalent to providing homogeneous decompositions $x_i = \sum_n x_i^n$ of the generators satisfying some relations. In other words, we want to specify a ring homomorphism $\eta: A\to A\otimes k[t^{\pm 1}]$ (whose $\rm Spec$ is the action of $\mathbf{G}_m$) satisfying the usual conditions, given by $\eta(x_i) = \sum_n x_i^n \otimes t^n$. I think you can use this down-to-earth description to endow $\mathcal{G}_A$ with the structure of an ind-scheme, but I suspect that already for $A= k[x,y]$ this will not be a scheme locally of finite type.<|endoftext|> TITLE: Gromov-Hausdorff distance between a disk and a circle QUESTION [32 upvotes]: The Hausdorff distance between the closed unit disk $D^2$ of $\mathbb R^2$ (equipped with the standard Euclidean distance) and its boundary circle $S^1$ is obviously one. Interestingly, the Gromov-Hausdorff distance between $D^2$ and $S^1$ is smaller. I claim that it is $\sqrt 3/2$. Below are partial results to support this. First some notation. A correspondence between two metric spaces $X$ and $Y$ is a subset $R \subset X \times Y$ such that the natural projections of $R$ onto $X$ and $Y$ are both surjective. The distortion of $R$ then is $$ \operatorname{dis}(R) := \sup\{|d_X(x,x')-d_Y(y,y')|\, : \, (x,y), (x',y') \in R\} \ . $$ It can be shown that the Gromov-Hausdorff distance between $X$ and $Y$ is equal to $\frac{1}{2}\inf\operatorname{dis}(R)$, where the infimum is taken over all correspondences between $X$ and $Y$. Here is a correspondence $R$ between $D^2$ and $S^1$ with distortion $\sqrt 3$. In $R$ are all pairs $(x,x)$ if $x \in S^1$ and also all pairs $(x,c)$ if $c$ is the "center" of the sector to which $x$ belongs as indicated in the following figure ($\sqrt 3$ is the distance between two different black center points): Below is a sketch of a proof that $\operatorname{dis}(R) \geq \sqrt 3$ in case $R$ is a correspondence between $D^2$ and $S^1$ for which $(x,x) \in R$ for all $x \in S^1$: Color $D^2$ with three colors. First color three arcs of length $2\pi/3$ on $S^1$ with different colors as in the figure above. Now for any $x \in D^2\setminus S^1$ assign to it the color of $y \in S^1$ if $(x,y) \in R$ (pick one in case there are multiple choices). As an application of Sperner's lemma there are three arbitrarily close points of different color. So these three points correspond to points on different arcs of the boundary. This forces $\operatorname{dis}(R) \geq \sqrt 3$. (The assumption $(x,x) \in R$ for $x \in S^1$ is used for the boundary condition in Sperner's lemma.) Does someone have an idea how to get rid of the condition $(x,x) \in R$ for $x \in S^1$ in the statement above? This would then prove that the Gromov-Hausdorff distance between $D^2$ and $S^1$ is indeed $\sqrt 3/2$. REPLY [3 votes]: Let $P_n=\{V_1,...,V_n\} \subset \partial D^2$ be the set of the vertices of a regular $n-$gon: since the Gromov-Hausdorff distance between $P_n$ and $\partial D^2$ tends to $0$, it is enough to compute $\lim_{n \to \infty} g_n$, where $g_n$ is the Gromov-Hausdorff distance between $P_n$ and $D^2$, so I tried to compute $g_n$ (or at least a lower bound) for some small values of $n$, hoping to see a pattern. First of all observe that, since any correspondence $R$ has the same distortion as its closure (with respect to the product topology), in the case $R\subseteq P_n \times D^2$ we can assume that the sets $A_i=\{ x \in D^2 | (V_i,x) \in R\}$ are closed; moreover, they must be all nonempty and their union must be the whole $D^2$. If we fix a constant $c \le 2$ and try to find a correspondence $R=\bigcup_{i=i}^n \{V_i\} \times A_i$ of distortion less than $c$, we need any two $A_i$ and $A_j$ to be disjoint if $d(V_i,V_j) \ge c$ and, roughly speaking, not too far from each other if $d(V_i,V_j)$ is small (in particular, $i=j$ gives the condition $diam(A_i) TITLE: Counting real zeros of a polynomial QUESTION [19 upvotes]: I recently came across a criteria to count the number of real zeros of a polynomial $P(x)$ with real coefficients. Unfortunately I cannot find the reference! The criteria is the following: Form the matrix M whose entry $M_{i,j}$ (with $0\leq i,j\leq deg(p)-1$) is the coefficient of $X^iY^j$ in the polynomial $$\frac{P(X)P'(Y)-P(Y)P'(X)}{X-Y}.$$ Then all of the roots of $P$ are real if and only if $M$ is positive semi-definite. Moreover the roots are all distinct if $M$ is positive definite. This is clearly a variation on Hermite's criteria which says the same thing, but instead of $M$ as above, uses a Hankel matrix $H$ whose entry $H_{i,j}$ (with $i$ and $j$ in the same range) is the symmetric power polynomial $p_{i+j}$ in the roots of the polynomial which can be generated from the coefficients of $P$ using the Newton identities. The first criteria is much better suited for my specific application and I'd like to cite it, but as I said I can't find the reference. I expect the two matrices are similar, and though it's not immediately obvious to me how, I suspect I can reproduce it with a little effort. The more pressing issue is that I'd like to be able to cite whoever originally discovered this simpler formulation. Does anyone know a reference for this? REPLY [13 votes]: By Remark 9.21 page 340 of the book by Basu, Pollack and Roy on real algebraic geometry the matrix $H$ is the expansion of the Bezoutiant of $P$ and $P'$ in the Horner basis of $P$ instead of the basis of usual monomials which is your $M$. The Horner basis is the sequence of polynomials one actually computes when evaluating $P$ by Horner's scheme.<|endoftext|> TITLE: Are isotopic and conjugate homeomorphisms, conjugate by an element in $\mathrm{Homeo}_0(M)$? QUESTION [7 upvotes]: An answer to this question would also answer Isotopy of periodic homeomorphisms of a surface along periodic homeomorphisms Let $M$ be a topological manifold and let $f,g$ be two orientation preserving homeomorphisms of $M$. Suppose that $f$ and $g$ are isotopic and that they are conjugate in $Homeo^+(M)$. Are they conjugate by a homeomorphism isotopic to the identity? REPLY [4 votes]: No: Let $M$ be the 2-dim. surface consisting of tori welded to each other so that they form a string (sorry, I do not know how to draw here). Let $f$ be the the time 1 flow of a vector field supported in a small part of the $i$-torus, and let $g$ be the time 1 flow of the same vector field, but now supported in the $i+1$-torus. They are isotopic, because each one isotopic to the identity. They are conjugated by the shift of the tori by 1. But this shift can never be isotoped to the identity. If you want $M$ to be compact, you can arrange the tori in a (large) circle. Edit: More details. Suppose that the vector field rotates a circle (which is not involved in the welding) by a maximal angle which characterizes this circle. Then any conjugation has to map this circle to the corresponding one in the next torus. But these two circles are different generators of $\pi_1(M)$. So the conjugating homeomorphism cannot be isotopic to the identity.<|endoftext|> TITLE: Proof of Giroux's correspondence QUESTION [21 upvotes]: It is extensively used and cited the following statement due to Giroux: Given a closed $3$-manifold $M$, there is a $1:1$ correspondence between oriented contact structures on $M$ up to isotopy and open book decompositions of $M$ up to positive stabilization. Given such a contact structure, the existence of an open book supporting the contact structure is proven, for example, in Giroux's Géométrie de contact: de la dimension trois vers les dimensions supérieures. I am asking for a complete proof of the uniqueness part of the result. At this point I would be surprised if somebody provided me with a link to a peer-reviewed paper containing a proof of the result (which is funny because this is widely acknowledged as a "theorem" inside and outside the field of contact geometry). Usual citations include the above paper (which does not contain a proof of the statement), some book that has been "in preparation" for years or even "transparencies from a seminar"! So, links to detailed lecture notes or a proof itself will be appreciated. I know for example the existence of these Lectures on open book decompositions and contact structures by J. Etnyre, but they are somehow sketchy to my taste. I am not an expert in the field and I can't complete all the exercises left to the reader or fill in all the gaps in the "sketches of a proof". REPLY [10 votes]: As far as I know, there is no publicly available written proof of uniqueness. Goodman's thesis pointed out by Chris proves neither uniqueness nor existence. What he did was to provide some of the first steps towards understanding the link between open books and tightness. Before that, he does sketch a proof of the open book theorem but, if I remember correctly, this sketch contains less information than what Giroux wrote in the ICM proceedings. In particular it entirely fails to cite Siebenmann's paper that Giroux cites twice in his uniqueness sketch and is the crucial starting point. This paper has been very hard to find for 30 years, but eventually got published as Les bissections expliquent le théorème de Reidemeister-Singer: Un retour aux sources Annales de la Faculté des sciences de Toulouse: Mathématiques, Série 6: Volume 24 (2015) no. 5 I'm almost certainly the mysterious person that Anubhav Mukherjee mentions in his comment, but writing a proof of this theorem is way beyond the scope of a mathoverflow answer, I'm sorry. I could probably answer more specific questions though.<|endoftext|> TITLE: Why did _Research in the Mathematical Sciences_ change from open access to subscription-based? QUESTION [19 upvotes]: The journal Research in the Mathematical Sciences was founded in 2014 and originally published by SpringerOpen, a division of Springer supporting Open Access journals. In the first article of the introductory issue Ken Ono emphasizes the journal's commitment to Open Access publishing (albeit with a funding structure based on authors paying article-processing fees). But if you go to the old website of the journal, you are now greeted with this message: As of January 1, 2018, Research in the Mathematical Sciences (RMS) has transitioned from an Open Access journal to a subscription-based journal with a hybrid Open Access option. RMS continues to be published by Springer Nature and an archive of all articles previously published in the journal is hosted here. All submissions going forward will be considered for the subscription-based journal. Please see the RMS website for more information. I find it very surprising that within 4 years of existence this math journal had to change its fundamental publishing philosophy. Is there any more information, beyond the above quote, about what happened to Research in the Mathematical Sciences and why it abandoned its original Open Access vision? EDIT: As mentioned in the comments, essentially the same thing happened with Research in Number Theory, and Ken Ono's response also addresses this journal as well. REPLY [26 votes]: this thread was forwarded to me. Together with Springer, the RMS and RNT Editors decided to abandon open access as very few authors have federal funding that pays publication charges. The decision was not based on a low number of submissions (note. the oa goal was to publish just 25-30 papers annually). The decision was made in response to the realities of grant support in mathematics. To this end, Springer agreed to maintain the idea of low cost publication (at the Editorial Board's request), and so I am pleased that both journals have an annual subscription fee of only $99USD. Moreover, RMS and RNT are automatically included in existing bundle subscriptions. There has been an upsurge in submissions, and the editors are pleased with the papers it is receiving, and as a result we anticipate increasing the number of published articles per year to 50-60. For example, in RMS the last few years pure mathematicians who have published include Francis Brown, Bill Duke, Ben Green, Dick Gross, Michael Harris, Kannan Soundararajan, Terry Tao, Richard Taylor, Yuri Manin,...., and in applied mathematics papers by Tom Hou, Andrew Majda, P. Souganidis, Richard Tsai, E. Weinan,... Please consider submitting strong articles to both journals. Best wishes, Ken Ono<|endoftext|> TITLE: Modern references on hyperbolic groups QUESTION [20 upvotes]: Several good references dedicated to hyperbolic groups have been written until 1990, including: Hyperbolic groups, written by M. Gromov. Géométrie et théorie des groupes : les groupes hyperboliques de Gromov, written by M. Coornaert, A. Papadopoulos and T. Delzant. Sur les groupes hyperboliques de M. Gromov, edited by E. Ghys and P. de la Harpe. Since then, fundamental tools have been introduced to study hyperbolic groups. For instance, I have in mind JSJ decompositions. Are there textbooks or surveys on the subject? What are good references dedicated to modern developments on hyperbolic groups? By "modern", I mean subjects not contained in the references mentioned above. I am particularly interested in progresses made in the study of outer automorphism groups of hyperbolic groups. Edit: I am aware that the ideas involved in the study of hyperbolic groups have lots of applications, so that covering them in just a few references is hopeless. Consequently, it would be better to focus on hyperbolic themselves, excluding generalisations (such as relatively / acylindrically hyperbolic groups) or specific examples (such as right-angled Coxeter groups, (automorphisms of) free groups or (automorphisms of) surface groups). REPLY [22 votes]: I think this is a great question, as there is still a need for an authoritative reference about (word-)hyperbolic groups. Since the textbook doesn't exist, I'd like to take the question in a slightly different direction by listing some of the material I think it should cover. (This is inevitably a personal and biased account.) I'll try to include the best references I can think of. The seminal article of Gromov's mentioned in the question contained many assertions about hyperbolic groups, often extensions of Thurston's famous theorems about hyperbolic manifolds. Probably the simplest such statement (which is fundamental in the study of outer automorphism groups of free groups) is now known as Paulin's theorem (cf. 5.4.A of Hyperbolic groups). Paulin's theorem: If $\Gamma$ is a torsion-free hyperbolic group that doesn't split over a cyclic subgroup then $\mathrm{Out}(\Gamma)$ is finite. Another such statement is 5.3.C' from Gromov's article. Subgroup Rigidity Theorem: Let $\Gamma$ be a hyperbolic group and $H$ a one-ended finitely presented group. Then there are only finitely many conjugacy classes of subgroups of $\Gamma$ isomorphic to $H$. In Hyperbolic groups, Gromov suggests that statements like these can be proved using a generalisation of Thurston's arguments using the geodesic flow. Even the problem of constructing a candidate geodesic flow over a hyperbolic group is notoriously difficult. The problem was eventually solved by Mineyev but, to the best of my knowledge, no one so far been able to use Mineyev's geodesic flow to give the Thurstonian proofs of these result that Gromov suggested. Instead, the key tool in proving these results turned out to be the Rips machine: Rips' classification of certain actions of groups on real trees. For actions of finitely presented groups, the Rips machine was developed by Bestvina—Feighn. An account was also given by Misha Kapovich in his book, so the interested reader could look at either of the following. Bestvina & Feighn, Stable actions of groups on real trees. Invent. Math. 121 (1995), no. 2, 287–321 M. Kapovich, Hyperbolic manifolds and discrete groups. Progress in Mathematics, 183. Birkhäuser Boston, Inc., Boston, MA, 2001. xxvi+467 pp For some applications, one needs the Rips machine for finitely generated groups. This was developed initially by Sela and corrected and refined by Guirardel, so the relevant references here are: Sela, Acylindrical accessibility for groups. Invent. Math. 129 (1997), no. 3, 527–565. Guirardel, Actions of finitely generated groups on $\mathbb{R}$-trees. Ann. Inst. Fourier (Grenoble) 58 (2008), no. 1, 159–211. The link with hyperbolic groups is made via what Sela called the Bestvina—Paulin method. Given infinitely many actions of a group $G$ on a suitably nice $\delta$-hyperbolic space $X$, one can pass to a limiting action on a space which is either $X$ itself or a real tree $T$, and in the latter case Rips’ machine applies. As well as Paulin’s original paper, another account of the proof, also modulo the Rips machine, was given by Bridson—Swarup (who corrected a small mistake in Paulin’s proof), and Bestvina gave a very useful account in his survey article on $\mathbb{R}$-trees. So one could look at: Paulin, Outer automorphisms of hyperbolic groups and small actions on $\mathbb{R}$-trees., Arboreal group theory (Berkeley, CA, 1988), 331–343, Math. Sci. Res. Inst. Publ., 19, Springer, 1991. Bridson & Swarup, On Hausdorff-Gromov convergence and a theorem of Paulin, Enseign. Math. (2) 40 (1994), no. 3-4, 267–289. Bestvina, $\mathbb{R}$-trees in topology, geometry, and group theory. Handbook of geometric topology, 55–91, North-Holland, Amsterdam, 2002. For Paulin’s theorem, the only thing one needs from the Rips machine is that it promotes a (nice) action on a real tree to a (nice) action on a simplicial tree. Deeper applications, such as the Subgroup Rigidity Theorem, tend to require a notorious trick called the shortening argument. The idea is that if the actions of $G$ on $X$ were all chosen to be `shortest’ in their conjugacy classes then either $G$ is a free product or the limiting action of $G$ on $T$ isn’t faithful. This trick is notoriously, er, tricky. The first reference is Rips—Sela’s original paper in which they prove the Subgroup Rigidity Theorem in the torsion-free case. (The case with torsion was later handled by Delzant.) Rips & Sela, Structure and rigidity in hyperbolic groups. I. GAFA, 1994. The shortening argument is Theorem 4.3, and the Subgroup Rigidity Theorem is Theorem 7.1. I gave an account of the shortening argument in Theorem 5.1 of Wilton, Solutions to Bestvina and Feighn's exercises on limit groups, Geometric and cohomological methods in group theory, pp. 30–62, LMS Lect. Note Ser. 358, CUP, 2009 Another account of the shortening argument (and many of the theorems mentioned here) adapted to the setting of toral relatively hyperbolic groups was given by Groves in: Groves, Limit groups for relatively hyperbolic groups. II. Makanin-Razborov diagrams. Geom. Topol. 9 (2005), 2319–2358. This toolkit has some further spectacular consequences for the structure of hyperbolic groups. The two biggest are probably the Hopf property and the isomorphism problem. Recall that a group $G$ is said to be non-Hopfian if there is a non-injective epimorphism $G\to G$. In Sela, Endomorphisms of hyperbolic groups. I. The Hopf property. Topology 38 (1999), no. 2, 301–321. Sela proved the Hopf property for all torsion-free hyperbolic groups. The case with torsion is treated in a preprint of Reinfeldt—Weidmann. Sela solved the isomorphism problem for torsion-free rigid hyperbolic groups (such as the hyperbolic 3-manifold groups) in Sela, The isomorphism problem for hyperbolic groups. I. Ann. of Math. (2) 141 (1995), no. 2, 217–283. To extend this to the non-rigid case, one needs JSJ theory for groups, which was introduced by Rips—Sela in Rips & Sela, Cyclic splittings of finitely presented groups and the canonical JSJ decomposition. Ann. of Math. (2), 146 (1997), no. 1, 53–109. As is already apparent from Ian’s answer, this theory has been developed a great deal, and the more recent approaches are frankly a lot simpler than the original Rips—Sela version. In the end, the torsion-free non-rigid case (as well as the toral relatively hyperbolic case) was dealt with in Dahmani & Groves, The isomorphism problem for toral relatively hyperbolic groups. Publ. Math. Inst. Hautes Études Sci. No. 107 (2008), 211–290. while the case with torsion was dealt with in Dahmani & Guirardel, The isomorphism problem for all hyperbolic groups. Geom. Funct. Anal. 21 (2011), no. 2, 223–300. Both of these papers followed Sela’s outline, although major technicalities needed to be overcome. NWMT has alreay said in comments that all this is too difficult for a textbook. In total this is true, but I can imagine an advanced reference book that describes the Rips machine, the Bestvina—Paulin method and the shortening argument, and gives Paulin’s theorem and the Subgroup Rigidity theorem as applications. These are still some of the most amazing and beautiful arguments in the theory of hyperbolic groups, and I’m a little concerned that a lot of current research seems to be moving away from them, rather than attempting to simplify or extend them.<|endoftext|> TITLE: Does every real function have this weak continuity property? QUESTION [51 upvotes]: In my research I came across the following question : Is it true that for every real function $f:\mathbb{R}\to\mathbb{R}$, there exists a real sequence $(x_n)_n$, taking infinitely many values, converging to some real number $c$, such that the sequence $(f(x_n))_n$ converges to $f(c)$ ? REPLY [30 votes]: It has "been known" since 1908 that for any such function, this holds for all but countably many real numbers $c$, even when we additionally require all the sequences to approach $c$ from the same side. In May 1908 William Henry Young presented several results for general functions from $\mathbb R$ to ${\mathbb R},$ including a result implying that, given any such function, all but countably many real numbers $c$ have the property you're asking about. These results (for more about them, see my answer here) may have been joint work with his wife, Grace Chisholm Young, and the results were published in the 1908 paper cited below. Young showed that for co-countably many real numbers $c$ we have $$f(c) \in C^{-}(f,c) \;\; \text{and} \;\; f(c) \in C^{+}(f,c)$$ Definition: Given a function $f: {\mathbb R} \rightarrow {\mathbb R}$ and $c \in {\mathbb R}$, we let $C^{-}(f,c)$ be the set of all extended real numbers $y$ (i.e. $y$ can be $-\infty$ or $+\infty$) for which there exists a sequence $\left\{x_{k}\right\}$ such that for each $k$ we have $x_k < c,$ and we have $x_{k} \rightarrow c$ and $f(x_k) \rightarrow y.$ In other words, $C^{-}(f,c)$ is the set of all numbers (including $-\infty$ and $+\infty$) that can be obtained as a limit of $f$-values when using some sequence converging to $c$ from the left. The right version, $C^{+}(f,c),$ is defined analogously. Incidentally, the requirement in this definition that each $x_k < c$ (and also each $x_k > c)$ allows you to find such sequences converging to $c$ that have infinitely many values. William Henry Young, Sulle due funzioni a più valori costituite dai limiti d'una variabile reale a destra e a sinistra di ciascun punto [On the two functions of multiple values that are determined by the left and right limits of a real variable at each point], Atti della Accademia Reale dei Lincei. Rendiconti. Classe di Scienze fisiche, Matematiche e Naturali (5) 17 #9 (1st semestre) (1908), 582-587. [Paper given at session dated 3 May 1908.]<|endoftext|> TITLE: $(\infty,1)$ 2d TFTs QUESTION [13 upvotes]: 2d topological field theories $Z : \mathrm{Cob}(2) \to \mathrm{Vect}$ are classified by commutative Frobenius algebras. What can be said about $(\infty,1)$ 2d TFTs $Z: \mathrm{Cob}(2) \to \mathcal{S}$ with values in a symmetric monoidal $(\infty,1)$-category $\mathcal{S}$? I am interested in different targets $\mathcal{S}$. I know that $Z(\mathbb{S}^1)$ is naturally an $E_2$-algebra in $\mathcal{S}$, but certainly it has more structure. Please note that I do not refer to fully extended (i.e. 2-1-0-dimensional) TFTs. REPLY [10 votes]: The answer to your question is known when $\mathcal{S}$ is a symmetric monoidal $\infty$-groupoid, by work of Galatius-Madsen-Tillmann-Weiss. In other words: we understand invertible $2$-dimensional TFTs. This might seem like a somewhat silly case, but it's actually a very useful way of testing conjectures. Below I'll sketch how this implies that the answer to Noah's question is no, but let me begin by explaining the setup. Let's a assume that every morphism in the symmetric monoidal $(\infty,1)$-category $\mathcal{S}$ is invertible and that every object of $\mathcal{S}$ is invertible under $\otimes$. (The latter could be replaced by the assumption that every object is isomorphic to the unit $1_{\mathcal{S}}$ - it doesn't really matter since $B(\mathrm{Cob}(2))$ is connected.) A symmetric monoidal $\infty$-groupoid is an $E_\infty$-algebra in spaces, and with the above invertiblity assumption it is an infinite loop space $\mathcal{S} = \Omega^\infty X$ of some connective spectrum $X$. Now every functor $\mathrm{Cob}(2) \to \mathcal{S}$ factors through the groupoidification $\mathrm{Cob}(2) \to (\mathrm{Cob}(2))^{\rm gp}$, which can be modeled as the classifying space of the $\infty$-category $\mathrm{Cob}(2)$. This classifying space was computed by Galatius-Madsen-Tillmann-Weiss as: $$B(\mathrm{Cob}(2)) \simeq \Omega^\infty( \Sigma^{-1} \mathrm{MTSO}_2)$$ Here $\mathrm{MTSO}_2$ is a certain Thom spectrum, which can be studied by standard methods in stable homotopy theory. (I'm happy to elaborate, if you like.) Anyways, putting this together we have that: $$ \mathrm{Fun}_\infty^\otimes(\mathrm{Cob}(2), \mathcal{S}) \simeq \mathrm{Map}_{\mathrm{Sp}}(\tau_{\ge 0} \Sigma^{-1} \mathrm{MTSO}_2, X). $$ Here $\tau_{\ge0}$ is the functor that sends a spectrum to it's connective cover. This could be dropped since $X$ is connective, but is seems more accurate to keep it around as $\mathrm{Cob}(2)$ doesn't know about $\pi_{-1} \Sigma^{-1} \mathrm{MTSO}_2 = \mathbb{Z}$. So why is this useful at all? I'll try to illustrate that be giving a negative answer to the question Noah posed in his post. I'll say traced $E_2^{\rm fr}$-algebra to mean a framed $E_2$-algebra $A$ equipped with an $SO_2$-equivariant trace $\tau:A \to 1$. Let $\mathcal{C}$ be the free symmetric monoidal $(\infty,1)$-category on a traced $E_2^{\rm fr}$-algebra. Since $S^1 \in \mathrm{Cob}(2)$ has this structure, there is a symmetric monoidal functor $F:\mathcal{C} \to \mathrm{Cob}(2)$ sending $A$ to $S^1$. Noah's question can now be reformulated as asking whether for all symmetric monoidal $(\infty,1)$-category $\mathcal{S}$ precomposition with $F$ induces a fully faithful functor $$ \mathrm{Fun}_\infty^\otimes(\mathrm{Cob}(2), \mathcal{S}) \longrightarrow \mathrm{Fun}_\infty^\otimes(\mathcal{C}, \mathcal{S}), \qquad \mathcal{Z} \mapsto \mathcal{Z} \circ F $$ whose image consists of those $\mathcal{Z}': \mathcal{C} \to \mathcal{S}$ such that $\mathcal{Z}'(A \otimes A \xrightarrow{ \mu } A \xrightarrow{ \tau } 1_{\mathcal{C}})$ is a non-degenerate pairing. If $\mathcal{S}$ is an $\infty$-groupoid, then this condition is trivially satisfied, so we should have an equivalence between the functor categories. This implies that $F$ induces an equivalence $B\mathcal{C} \to B\mathrm{Cob}(2)$. However, one can show that this is not true. I'll be very brief on this. Any $E_2^{\rm fr}$-algebra in an $\infty$-groupoid can canonically be trivialised along the unit morphism $1_{\mathcal{S}} \to A$, so we basically just have to give an $\mathrm{SO}_2$-equivariant morphism $1_{\mathcal{S}} \to 1_{\mathcal{S}}$ where both sides are equipped with the trivial action. This implies that: $$ B(\mathcal{C}) \simeq \Omega^\infty \Sigma^{\infty+1} (B\mathrm{SO}_2)_+. $$ So the answer to the question is no since the connective spectra $\Sigma^{\infty+1} (B\mathrm{SO}_2)_+$ and $\tau_{\ge 0} \mathrm{MTSO}_2$ are not equivalent. The above argument is very roundabout and actually quite subtle: there is actually a map of spectra $$ \tau_{\ge 0} \mathrm{MTSO}_2 \to \Sigma^{\infty+1} (B\mathrm{SO}_2)_+ $$ the fiber of which is $\tau_{\ge0} \Sigma^{-1}\mathbb{S}$. (This is known as the Genauer fiber sequence and admits an interpretation on bordism categories.) So in particular this map is a rational equivalence. This might make us hopeful as it looks like we weren't that far off. However, this map goes the wrong way and I actually believe that there is no rational equivalence that goes the correct way around. So let me try to give a more down-to-earth proof that the functor $F:\mathcal{C} \to \mathrm{Cob}(2)$ is not an equivalence on groupoidification. For a traced $E_2^{\rm fr}$-algebra $A$ in $\mathcal{S}$ one can define $\tau(1):1_{\mathcal{S}} \to A \to 1_{\mathcal{S}}$ as the composite of unit and trace. If the traced $E_2^{\rm fr}$-algebra comes from a $2$-dimesional TFT, then this is the value on the $2$-sphere. The $\mathrm{SO}_2$-action on $A$ gives us a $2$-morphism $\alpha:\tau(1) \Rightarrow \tau(1)$. (The Dehn-twist pre- and post-composed with cap and cup.) Since both $1:1_{\mathcal{S}} \to A$ and $\tau:A \to 1_{\mathcal{S}}$ are $\mathrm{SO}_2$-equivariant we have two trivialisations of $\alpha$. Together they yield a $3$-morphism $\gamma: \mathrm{id}_{\tau(1)} \Rrightarrow \alpha \Rrightarrow \mathrm{id}_{\tau(1)}$. In the above I didn't actually use that $\mathcal{S}$ is an $\infty$-groupoid; one can construct this $\gamma$-invariant for any traced $E_2^{\rm fr}$-algebra in any $(\infty,1)$-category. If the traced $E_2^{\rm fr}$-algebra comes from a $2$-dimensional TFT $\mathcal{Z}:\mathrm{Cob}(2) \to \mathcal{S}$, then $\tau(1) = \mathcal{Z}(S^2)$ and $\alpha$ is the Dehn twist along the equator. The $3$-morphism $\gamma: \mathrm{id}_{\mathcal{Z}(S^2)} \Rrightarrow \mathrm{id}_{\mathcal{Z}(S^2)}$ probably corresponds to the $S^1$-family of rotations of $S^2$ around some axis. In particular this $3$-morphism is always of order $2$ since $\pi_1 \mathrm{Diff}^+(S^2) = \pi_1 \mathrm{SO}_3 = \mathbb{Z}/2$. Ok, so now all that's left to show is that there are traced $E_2^{\rm fr}$-algebras where $\gamma$ is not of order $2$. For this purpose let $\mathcal{S}$ be $K(\mathbb{Z},3)$ thought of as a symmetric monoidal $3$-groupoid that has a single object, morphism, $2$-morphism, and $\mathbb{Z}$-many $3$-morphisms. Then we can take the trivial $E_2^{\rm fr}$-algebra $1_{\mathcal{S}}$ in there, equip it with the trivial trace $\tau:1_{\mathcal{S}} \to 1_{\mathcal{S}}$, and give this trace non-trivial coherence data. This ended up much longer than intended, but I hope it's helpful. Let me know if I should clarify anything. I'd be very curious to see a conjectural generators and relations description of $\mathrm{Cob}(2)$ that yields both the right homotopy category and the right classifying space. (Though this seems to be very difficult as the $\Omega^{\infty} \Sigma^{-1} \mathbb{S}$ that turned up above is in some sense not finitely generated...)<|endoftext|> TITLE: Is there a known Turing machine which halts if and only if the Collatz conjecture has a counterexample? QUESTION [22 upvotes]: Some of the simplest and most interesting unproved conjectures in mathematics are Goldbach's conjecture, the Riemann hypothesis, and the Collatz conjecture. Goldbach's conjecture asserts that every even number greater than or equal to 4 can be written as the sum of two prime numbers. It's pretty straightforward how to create a computer program which halts if and only if there exists a counterexample to Goldbach's conjecture: simply loop over all integers, test if each one is a counterexample, and halt if a counterexample is found. For the Riemann hypothesis, there's also a "known" computer program which halts if and only if there exists a counterexample. (Given the usual statement of the Riemann hypothesis, this is not so clear, but Jeffrey C. Lagarias' paper "An Elementary Problem Equivalent to the Riemann Hypothesis" shows that the Riemann hypothesis is equivalent to the statement that a certain sequence of integers $L$ is a lower bound for a certain sequence of real numbers $R$. The sequence $L$ is computable, and $R$ is computable to arbitrary precision, so our computer program only needs to compute all elements of $R$ in parallel, and halt if any element is ever discovered to be smaller than its corresponding element in $L$.) But how about the Collatz conjecture? The Collatz conjecture states that for all positive integers $n$, the "hailstone sequence" $H_n$ eventually reaches $1$. We could try to do the same thing we did with Goldbach's conjecture: loop over all positive integers $n$ and halt if a counterexample is ever found. But there's a problem here: with the Collatz conjecture, given a positive integer $n$, it's not obvious that it's even decidable whether or not $n$ is a counterexample. We can't simply "check whether or not $n$ is a counterexample" like we can with Goldbach's conjecture. So is there a known Turing machine which halts if and only if the Collatz conjecture is false? Of course, a "known Turing machine" doesn't have to be a Turing machine that someone has actually explicitly constructed; if it's straightforward how to write a computer program that would do this, then that counts as a "known Turing machine". On the other hand, saying "it's either the machine which trivially halts, or it's the machine which trivially does not halt" doesn't count as a "known Turing machine"; I'm asking for an answer which mentions one single Turing machine $M$ (with no input or output), such that we know that $M$ halts if and only if the Collatz conjecture is false. REPLY [13 votes]: Let's note that this is not a question of whether Collatz is undecidable. The statement $\neg\mathrm{Con}(PA)$ is undecidable (by $PA$, assuming $PA$ is consistent) but nevertheless $\neg\mathrm{Con}(PA)$ is provably equivalent to a certain Turing machine halting (the one that searches for a proof of a contradiction in PA). Rather, the question is whether there is a $\Pi^0_1$ statement $\varphi$ such that the Collatz problem, which on its face is $\Pi^0_2$, is already known to be equivalent to $\varphi$. Here already known means in particular that we are not allowed to assume that Collatz is or is not provable or disprovable in any particular system, unless we already know that. The best evidence that there is no such $\varphi$ seems to be in the paper mentioned by @Burak: Kurtz, Stuart A.; Simon, Janos, The undecidability of the generalized Collatz problem, Cai, Jin-Yi (ed.) et al., Theory and applications of models of computation. 4th international conference, TAMC 2007, Shanghai, China, May 22--25, 2007. Proceedings. Berlin: Springer (ISBN 978-3-540-72503-9/pbk). Lecture Notes in Computer Science 4484, 542-553 (2007). ZBL1198.03043. Namely, they give a parametrized family of similar problems such that the collection of parameters for which Collatz-for-those-parameters is true, is $\Pi^0_2$-complete and hence not $\Pi^0_1$. They can do this without thereby solving the Collatz problem, just like Matiyasevich et al. could show that solvability of diophantine equation was $\Sigma^0_1$-complete, without thereby solving any particular equation themselves. If Collatz could somehow be simplified to a $\Pi^0_1$ form then quite plausibly the generalized version could too by the same argument (whatever that hypothetical argument would be) but that Kurtz and Simon show will not happen.<|endoftext|> TITLE: Finite-dimensional faithful unitary representations of SL(2,Z) QUESTION [10 upvotes]: Does $SL(2,\mathbb{Z})$ have a finite-dimensional faithful unitary representation? No such representation exists for $SL(2,\mathbb{R})$, but I don't see a reason why one shouldn't exist for $SL(2,\mathbb{Z})$. REPLY [18 votes]: Here a non-explicit proof of the existence of a faithful representation of $\mathrm{SL}_2(\mathbf{Z})$ in $\mathrm{SU}(2)$, using basic algebraic geometry and topology, and relying on the amalgam decomposition of $\mathrm{SL}_2(\mathbf{Z})$. [The basic idea is that if all representations in $\mathrm{SU}(2)$ were non-faithful, by Zariski-density this would also be the case for representations into $\mathrm{SL}_2$. We need to use the particular form of the presentation of $\mathrm{SL}_2(\mathbf{Z})$, since the argument will not carry over representations of $\mathrm{SL}_3(\mathbf{Z})$ in $\mathrm{SU}(3)$.] Let $P_t$ be the set of $2\times 2$ matrices with determinant 1 and trace $t$. Both $P_0$ and $P_1$ are irreducible as algebraic varieties (being $\mathrm{SL}_2$ conjugacy classes). Then for $K$ a field of characteristic zero, $P_0(K)$ is the set of elements of order 4 in $\mathrm{SL}_2(K)$, and $P_1(K)$ is the set of elements of order 6 in $\mathrm{SL}_2(K)$. For every $(g,h)\in P_0\times P_1$, $g^2=h^3$ equals $-I_2$. Hence the set of representations of $$\mathrm{SL}_2(\mathbf{Z})=\langle u,v\mid u^4=v^6=[u^2,v]=[u,v^3]=1\rangle$$ (restricting to those for which the image of $u$ has order 4 and the order of $v$ has order 6) into $\mathrm{SL}_2(K)$ can be naturally identified to $(P_0\times P_1)(K)$. Note that $P_0\times P_1$ is irreducible. Write $P_t^\sharp=P_t(\mathbf{C})\cap\mathrm{SU}(2)$. Then using that $\mathrm{SU}(2)$ is Zariski-dense in $\mathrm{SL}_2(\mathbf{C})$ and describing $P_t(\mathbf{C})$ as a conjugacy class, one deduces that $P_t^\sharp$ is Zariski-dense in $P_t(\mathbf{C})$. So $P_0^\sharp\times P_1^\sharp$ (which is a 4-dimensional real manifold) is Zariski-dense in $(P_0\times P_1)(\mathbf{C})$. For every given nontrivial element $w$ in $\mathrm{SL}_2(\mathbf{Z})$ the set of representations vanishing on $w$ is a proper subvariety of $P_0\times P_1$, hence has dimension $\le 3$. By Zariski density of $P_0^\sharp\times P_1^\sharp$, we deduce that its intersection with $P_0^\sharp\times P_1^\sharp$ is a proper Zariski closed subset (because $\mathrm{SL}_2(\mathbf{Z})$ admits one faithful representation into $\mathrm{SL}_2(\mathbf{C})$, the standard one); in particular it has empty interior (in the ordinary topology). By the Baire theorem, the union over all $w$ is still a proper subset. Hence we deduce the existence of an element of $P_0^\sharp\times P_1^\sharp$ defining a faithful representation.<|endoftext|> TITLE: Grinberg's uniquely hamiltonian 3-connected graphs (Russian paper) QUESTION [7 upvotes]: Many years ago, Grinberg found some uniquely-hamiltonian $3$-connected graphs, and published his results in a paper that has been cited several times as follows. E. Grinberg, Three-connected graphs with exactly one Hamiltonian cycle, Republican Foundation of Algorithms and Programmes, Computing centre. P. Stutschka University, Riga, U.S.S.R. (1986) [in Russian]. However the reference cannot be found (by Google or MathScinet) under this name, so I am guessing that some initial Russian-speaking author translated the details, and subsequent authors have either concurred or taken it on trust. Of course what I really want is the description of the graphs, but I see no way of doing this without the assistance of someone able to navigate the Russian literature. Any assistance would be gratefully received. REPLY [5 votes]: I have now resolved most of the mysteries, and as MO prompts me to answer my own question, I am now doing so even though it feels a bit odd. After some false starts with expired email addresses, I managed to contact Dainis Zeps in Latvia, who kindly filled in the missing details. Basically Zeps and Grinberg were working on unique hamiltonicity, but with Grinberg as the senior of the two. After Grinberg died in 1982, Zeps took some of Grinberg's old notes, rescued a construction for uniquely hamiltonian three-connected graphs, wrote it up, and published it under Grinberg's name. The construction is relatively straightforward (at least after you have seen it). If a graph has a triple of vertices $(x,y,z)$ that satisfy some simple hamiltonicity properties, then two nice things happen You can take two copies of this graph, connect the special triples in a particular way, and get a 3-connected uniquely hamiltonian graph, and The graph you have just built has lots more special triples, and so you can take two of those graphs and so on. They then supplied an initial graph with the right properties. To absolutely no-one's surprise, this initial graph is based on the Petersen graph. Zeps then did some more work on this, including some computations, and now that he felt that he had made enough of an additional contribution, wrote the viXra article that KConrad pointed out to me, now adding himself to the list of authors. Obviously this construction produces graphs with a (cyclic) 3-edge-cut and I think the "special triple" property is essentially exactly what you would need to make a graph with a cyclic 3-edge-cut uniquely hamiltonian. What is interesting is that the smallest of the graphs found by Grinberg/Zeps is just ONE EDGE bigger than the actual computer-determined minimum number of vertices and edges. Doing this by hand, with no computer assistance, and getting so close to the absolute smallest is pretty impressive. Of course this means that the actual smallest graph has no cyclic 3-edge-cut, and indeed it is cyclically 4-edge-connected. It does have a cyclic 4-edge cut and so by looking at the 4 end-points $\{a,b,c,d\}$ we can come up some conditions to make this a "special quadruple". But now we just seem to be heading down a rabbit-hole constructing more and more examples that are not genuinely different.<|endoftext|> TITLE: BSD conjecture for rank 1 elliptic curves QUESTION [14 upvotes]: Let $E/\mathbb{Q}$ be an elliptic curve. The weak Birch and Swinnerton-Dyer conjecture predicts that $$\text{ord}_{s=1}L(E, s)=\text{rank} E(\mathbb{Q}).$$ Thanks to the work of Gross-Zagier and Kolyvagin, we know that this conjecture is true if $\text{ord}_{s=1}L(E, s)\le 1$. What is known in the case $\text{rank} E(\mathbb{Q})\le 1$? Is it known that if $\text{rank} E(\mathbb{Q})=1$, then $L'(E, 1)\neq 0$? Thank you! REPLY [14 votes]: The following theorem is due to Chris Skinner, in this 2014 paper. Let E/Q be an elliptic curve such that rank E(Q) = 1 and the Tate-Shafarevich group Sha(E / Q) is finite, and some other technical assumptions hold. Then $ord_{s = 1} L(E, s) = 1$, and in particular $L'(E, 1) \ne 0$. This is, as far as I know, the best one can do at the moment; if you don't know that Sha (or at least its p-primary part for some p) is finite, then you're stuck.<|endoftext|> TITLE: Physical consequences of cobordism hypothesis? QUESTION [8 upvotes]: Let $C$ be a symmetric monoidal $n$-category. An extended framed $C$-valued TQFT is a symmetric monoidal functor from the framed bordism category $\mathrm{Cob}^{fr}_n(n)$ to $C$. The cobordism hypothesis, first formally written down by Baez--Dolan, states that an extended framed $C$-valued TQFT is determined up to isomorphism by its value at the point. It has been proved by Lurie. I understand that this is an interesting statement from the topological and $n$-categorical points of view. My question is: are there any corollaries of this hypothesis that are of physical interest? Or is there at least some sort of physical motivation/plausibility argument for cobordism hypothesis? REPLY [11 votes]: Yes. The physical motivation is that topological field theories, as examples of quantum field theories, should be fully local, meaning that one should be able to calculate any information about a (fully extended) TQFT $Z$ on a manifold $M$ by cutting $M$ into pieces, formulating $Z$ on these pieces, and gluing. The takeaway in physics is that in any class of systems thought to be described by topological field theories, one should be able to determine the TQFT for a particular system from how the system behaves on a neighborhood of a point. Even though we don't know how to define quantum field theories precisely, physicists know they must satisfy a few important axioms, including locality. This means that all of the data in a quantum field theory formulated on a manifold $M$ satisfies a sheaflike axiom: one must be able to glue the fields together from information on an arbitrary small open cover of $M$. So a principal bundle with connection on $M$ is OK, as is a differential form, but a CW structure is not. Similarly, things such as the Lagrangian or Hamiltonian must be expressible in terms of data satisfying this condition. This implies that information one can calculate about the QFT, such as state spaces or partition functions, also satisfies a gluing axiom: they are determined somehow from the information entering the theory, which we can formulate on small balls around any point, satisfying some kind of gluing condition. In functorial TQFT, we're not in general provided with fields or Hamiltonians, but only things like partition functions and state spaces, and their lower-dimensional analogues. But TQFTs are QFTs, so they must satisfy locality. This motivated Atiyah's original definition: the partition function $Z(M)$ must be local, so we can cut $M$ into a sequence of cobordisms and compose the maps between state spaces to compute partition functions. This is only expressing locality in one direction out of $n$, so we should be able to repeat this process in the other $n-1$ directions to compute the partition function in terms of fully local data, i.e. describing $M$ as glued together from many copies of $\mathbb R^n$, then using whatever $Z(\mathbb R^n)$ means to compute the partition function of $M$. From this perspective, we see $\mathbb R^n$ instead of points, which may seem strange. The reason is that in a cobordism category of Riemannian or Lorentz manifolds, which you'd use for functorial non-topological QFT, one has to add a collar to manifolds in codimension one, so when fully extending, one has a two-dimensional collar in codimension 2, and so on. In this way full locality is seen by $\mathbb R^n$, an $n$-dimensional collar around a point. But for cobordisms between smooth manifolds, a collar is redundant data, so the axioms of fully extended TQFT drop the collars and just use the higher-codimension manifolds, whence $Z(\mathrm{pt})$: the cobordism hypothesis says a TQFT should be determined by fully local data, which would mean cutting any manifold into copies of $\mathbb R^n$, and the axioms of TQFT replace $Z(\mathbb R^n)$ by $Z(\mathrm{pt})$. That's part of the cobordism hypothesis; it also calculates what kinds of objects $Z(\mathrm{pt})$ can be. Unfortunately I'm not sure of the physical motivation for that part. Moving on to consequences in physics. I don't know of a general statement, but I'll discuss an interesting example in the theory of topological phases of matter. These are condensed-matter systems which display surprising topological properties, and condensed-matter theorists are interested in classifying them (in a fixed dimension $n$, say). It's not yet known what the mathematical definition of a topological phase of matter is, but physicists often formulate them in terms of "lattice field theories": discretized QFTs on manifolds equipped with something like a triangulation, and such that the fields, Hamiltonian, etc. are also suitably discretized in terms of purely combinatorial information. (The locality axiom is different in this setting, expressed in terms of graph distance.) It is believed that the low-energy physics of such systems is described by fully extended TQFTs, but even formulating a precise conjecture, never mind a proof, is currently open at both physical and mathematical levels of rigor, as far as I know. Moreover, the low-energy TQFT should classify the lattice system in some strong sense. This provides a (conjectural) approach to classifying topological phases of matter: given an $n$-dimensional TQFT $Z$, build a topological phase with $Z$ as its low-energy TQFT, and then classify $n$-dimensional TQFTs. Of course, explicating the cobordism hypothesis in dimensions greater than 2 is tricky, but this approach appears to be working. For example, the $\mathrm{SO}_3$-homotopy fixed points in the 3-category of $\mathbb C$-linear monoidal categories are conjectured to be spherical fusion categories (one direction is known, due to Douglas-Schommer-Pries-Snyder), and the TQFT associated to a spherical fusion category $\mathsf C$ is the Turaev-Viro-Barratt-Westbury (TVBW) TQFT for $\mathsf C$. In physics, topological phases of matter in (spacetime) dimension 3 on oriented manifolds are believed to be classified by the Levin-Wen model associated to a spherical fusion category, and Kirrilov Jr proves a result which provides strong evidence that the low-energy theory of the Levin-Wen model for $\mathsf C$ is the TVBW TQFT for $\mathsf C$. Another example is the classification of symmetry-protected topological (SPT) phases. These are (conjecturally) the topological phases of matter whose low-energy theories are invertible TQFTs. The cobordism hypothesis implies that these are classified by homotopy classes of maps out of Madsen-Tillmann spectra (though there's now a proof of this fact independent of the cobordism hypothesis, due to Schommer-Pries), and one can compare this classification with other classifications of SPTs. This was taken up by Freed-Hopkins, who found agreement between this and other kinds of classifications for a variety of dimensions and symmertry groups. For some more detail on the motivation for the cobordism hypothesis, as well as some other applications in physics, I recommend Dan Freed's expository article.<|endoftext|> TITLE: Does there exists a group structure on $\circ$ on $(\mathbb{R},\circ)$ such that $(\mathbb{R},\circ)$ is non-isomorphic to $(\mathbb{R},+)$? QUESTION [15 upvotes]: Consider the additive abelian group $(\mathbb{R},+)$. Does there exists a binary operation $\circ:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ such that the following holds $(\Bbb{R},\circ)$ is a group. For all $S\subseteq \mathbb{R}$ the subgroup generated by $S$ in $(\mathbb{R},+)$ is equal (as a set) to the subgroup generated by $S$ in $(\mathbb{R},\circ)$. $(\mathbb{R},+)$ is not isomorphic to $(\mathbb{R},\circ)$. This question is a follow up of this MSE post of mine and due to it I guess that the answer to both the questions are negative. But I can't construct such a group (or prove the existence of such one). Thanks for any help. REPLY [32 votes]: The answer is no: every such group law is isomorphic to the standard law. Let me prove something stronger: the poset structure of the lattice of subgroups $\mathrm{Sub}(\mathbf{R})$ characterizes $\mathbf{R}$ up to group isomorphism. Let us first check it among abelian groups. Let $G$ be a group with $\mathrm{Sub}(G)\simeq\mathrm{Sub}(\mathbf{R})$ as posets. The absence of nontrivial minimal elements characterizes torsion-freeness, so $G$ is torsion-free. Since $G$ is abelian, the absence of nontrivial maximal element characterizes the absence of proper subgroups of finite index, and hence $G$ is divisible. So $G$ is isomorphic to $\mathbf{Q}^{(I)}$ for some cardinal $I$. Cyclic subgroups can be characterized, among abelian groups, as those groups with infinitely many subgroups, and in which every nontrivial subgroup is contained in only finitely many subgroups. The number of cyclic subgroups of $\mathbf{Q}^{(I)}$ for nonempty $I$ is equal to $\max(\aleph_0,|I|)$. So here we have $|I|=c$. This concludes the abelian case. In the nonabelian case, we need two lemmas (of independent interest, and probably proved somewhere in the literature). (1) the poset structure of the lattice of subgroups $\mathrm{Sub}(\mathbf{Z}^2)$ characterizes $\mathbf{Z}^2$ up to group isomorphism. In a lattice $X$ with maximal element $q$, let us say that $x$ satisfies Property $A_p$ ($p$ a given prime) if the interval $[x,q]$ consists of $x$, $q$, and exactly $p+1$ pairwise non-comparable elements. Then in $\mathrm{Sub}(\mathbf{Z}^2)$, it is easy to check that the only subgroup satisfying $A_p$ is $p\mathbf{Z}^2$. Consider $G$ with $\mathrm{Sub}(G)\simeq\mathrm{Sub}(\mathbf{Z}^2)$. Then $G$ has a unique subgroup $G_p$ satisfying $A_p$ for every $p$, and $\bigcap_p G_p=\{1\}$ (since the corresponding property holds in $\mathbf{Z}^2$). In particular, $G_p$ is a normal subgroup. The quotient $G/G_p$ has finitely many subgroups and hence is a finite group. So $G$ is residually finite. If by contradiction $G$ were non-abelian, there would exist $p$ such that $G/G_p$ is non-abelian; take $p$ minimal for this. The finite group $G/G_p$ is nonabelian and has all its proper subgroups cyclic of prime order. The only possible groups having this property have the form $C_q\ltimes C_{q'}$ with $q,q'$ prime and $q$ divides $q'-1$. It has $q'+3$ subgroups ($q'+1$ is we exclude the trivial and whole subgroups), so $q'=p$. Now $G/(G_p\cap G_q)$ inherits from the corresponding group in $\mathbf{Z}^2$, the property of having exactly $(p+3)(q+3)$ subgroups, including $p+q+2$ maximal ones. On the other hand, in $G$, the uniqueness of a subgroup with Property $A_q$ implies that $C_q^3$ is not a quotient of $G$; it follows that $G/(G_p\cap G_q)$ has the form $C_q^2\ltimes C_p$. This group has $q+1$ maximal subgroups that are normal, and $p$ maximal subgroups that are non-normal (the $q$-Sylow subgroups), so the total is $p+q+1$, which does not match: contradiction. Now $G$ is abelian, torsion-free. The (at most) uniqueness of subgroups with Property $A_p$ therefore applies to subgroups of $G$ and hence $G$ has $\mathbf{Q}$-rank $\le 2$. Since $G/G_p\simeq C_p^2$ for all $p$, we deduce $G\simeq\mathbf{Z}^2$. $\Box$ (2) the poset structure of the lattice of subgroups $\mathrm{Sub}(\mathbf{Z})$ characterizes $\mathbf{Z}$ up to group isomorphism. Let $G$ be such a group. Then $G$ is torsion-free. Say that a subgroup has Property B if it is contained in finitely many subgroups. From the property in $\mathbf{Z}$, we deduce that every nontrivial subgroup has Property B, and that the intersection of any two nontrivial subgroups has Property B. We deduce that $G$ has a commensurated subgroup $N$ isomorphic to $\mathbf{Z}$. The germ at infinity of the action on $N$ yields a homomorphism $G\to\mathbf{Q}^\times$. The image has to be contained in $\{\pm 1\}$: otherwise some element of $G$ acts by some element $\neq\pm 1$, and hence generates a cyclic subgroup with trivial intersection with $N$. Also, since $N$ has Property B, we see that $G$ is finitely generated. We deduce that some finite index subgroup of $N$ is normal, say $N$ itself (up to rename). Since $G/N$ has finitely many subgroups, it is finite, and hence $G$ is virtually cyclic and torsion-free (and infinite). This is known to imply that $G$ is infinite cyclic. $\Box$ Now let us conclude. Consider $G$ with $\mathrm{Sub}(G)\simeq\mathrm{Sub}(\mathbf{R})$. So $G$ is torsion-free. By the lemma, we can recognize subgroups of $G$ isomorphic to $\mathbf{Z}$ and $\mathbf{Z}^2$; in particular, we inherit from $\mathbf{R}$ the property that any two infinite cyclic subgroups is contained in a subgroup isomorphic to $\mathbf{Z}^2$. Hence $G$ is abelian. Since the abelian case is done, this ends the proof. Edit: Here's a reference: the book by M. Suzuki, Structure of a Group and the Structure of its Lattice of Subgroups, Ergebnisse der Mathematik und Ihrer Grenzgebiete 10, Springer-Verlag Berlin Heidelberg, 1956 Theorem 3, page 35 (attributed to R. Baer, 1939), says that for any abelian group $G$ of $\mathbf{Q}$-rank $\ge 2$ (i.e., containing a subgroup isomorphic to $\mathbf{Z}^2$), any "projectivity" of $G$ (that is, an isomorphism $\mathrm{Sub}(G)\to\mathrm{Sub}(H)$ for another group $H$), is induced by a unique (up to pre-composition by $x\mapsto -x$) isomorphism $G\to H$. In particular, $H$ is isomorphic to $G$. (While the "in particular" assertion holds for $G=\mathbf{Z}$ by Lemma (2) above, the stronger assertion is false for $G=H=\mathbf{Z}$, as the lattice of subgroups has an uncountable automorphism group then.)<|endoftext|> TITLE: Collapsed partitions and generating functions QUESTION [5 upvotes]: Given $n\in\Bbb{N}$, the number of (unrestricted) integer partitions of $n$ are given by $$\sum_{n\geq0}p(n)x^n=\prod_{j\geq1}\frac1{1-x^j}.$$ Define the collapsed partitions of $n$ to be the partitions of $n$ with multiplicities removed. For example, if $n=4$ then its partitions are $4, 31, 22, 211, 1111$. The collapsed partitions become $4, 31, 2, 21, 1$. Denote the sum of the $k$-th powers of the collapsed partitions of $n$ by $cp_k(n)$. For example, $cp_1(4)=4+3+1+2+2+1+1=14$. The first few values of $cp_1(n)$ are: $$cp_1(1)=1, cp_1(2)=3, cp_1(3)=7, cp_1(4)=14, cp_1(5)=26.$$ Recall the Eulerian polynomials of type $A$ defined by $$\sum_{n\geq0}(n+1)^kx^n=\frac{A_k(x)}{(1-x)^{k+1}}.$$ Experiments prompt me to ask: is this true? $$\sum_{n\geq0}cp_k(n+1)x^n=\frac{A_k(x)}{(1-x)^{k+1}}\prod_{j\geq1}\frac1{1-x^j}.$$ REPLY [4 votes]: It is easy to see that $$cp_k(n) = \sum_{i=1}^n p(n-i)\cdot i^k,$$ where $p(n-i)$ stands for the number of (collapsed) partitions of $n$ that contain $i$ as a part. Since $i^k$ is the coefficient of $x^{i-1}$ in $\frac{A_k(x)}{(1-x)^{k+1}}$, we conclude that $cp_k(n)$ equals the coefficient of $x^{n-1}$ in $\frac{A_k(x)}{(1-x)^{k+1}}\cdot \prod_{j\geq 1}\frac{1}{1-x^j}$. QED<|endoftext|> TITLE: Axioms of length QUESTION [8 upvotes]: Assume I want to define length of plane curves axiomatically. It seems to be reasonable to assume that The length of a unit segment is 1; Congruent curves have equal lengths; Length is additive with respect to concatenation. However this is not enuf to define length completely: many different length-like functionals satisfy these properties. What would be a complete set of axioms? Motivation. I noticed that many (if not all) proofs of the Crofton formula cheat by assuming implicitly that there is a unique length functional that satisfies the above property, which is wrong. The problem is easy to fix, but the proof I see relies on the constructive definition of length; therefore this extra argument has to be repeated in each variation of the Crofton formula, which is not nice. P.S. It seems that the following set of axioms solves the problem (thanks to Taras Banakh): The length of any curve is non-negative and invariant with respect to reparametrizations. The length of a unit segment is 1; Congruent curves have equal lengths; Length is additive with respect to concatenation; Length is lower semi-continuous with respect to pointwise convergence. REPLY [6 votes]: I would suggest the following axioms. The length in a metric space $X$ is a function $\ell:c(X)\to[0,+\infty]$ defined on the family $c(X)$ of all connected compact subsets of $X$ that satisfies the following axioms: 1) $\ell$ is non-degenerated, which means that a continuum $C\in c(X)$ is a singleton if and only if $\ell(C)=0$; 2) $\ell$ is monotone, which means that $\ell(A)\le \ell(B)$ for any continua $A\subset B$ in $X$; 3) $\ell$ is additive, which means that $\ell(A\cup B)=\ell(A)+\ell(B)$ for any continua $A,B\subset X$ with finite non-empty intersection $A\cap B$; 4) $\ell$ is affine, which means that $\ell(f(C))=\lambda\cdot\ell(C)$ for any continuum $C\subset X$, any $\lambda>0$ and any bijective function $f:X\to X$ such that $d(f(x),f(y))=\lambda \cdot d(x,y)$ for all $x,y\in X$; 5) $\ell$ is semicontinuous in the sense that for any $A\in c(X)$ and any $\varepsilon>0$ there a neighborhood $O_A\subset c(X)$ of $A$ in the Vietoris topology such that $\ell(A')\ge \ell(A)-\varepsilon$ for every $A'\in O_A$. I hope that the following theorem of existence and uniqueness holds: Theorem. In each Euclidean space $E$ there exists a length $\ell$. Moreover, two lengths $\ell,\lambda:c(E)\to[0,+\infty]$ are equal if $\ell([a,b])=\lambda([a,b])$ for some distinct points $a,b\in E$. In his survey paper Murat Tuncali writes that the length of continua was studied by Eilenberg, Harrold (1943) and later Buskirk, Nikiel, and Tymchatyn (1992).<|endoftext|> TITLE: Complete atomless Boolean algebras with abelian automorphism group QUESTION [5 upvotes]: Is there any example of a complete atomless Boolean algebra with a non-trivial abelian automorphism group? This is equivalent, by Stone duality, to asking for an extremally disconnected compact Hausdorff space with no isolated points, having abelian homeomorphism group. Note that no such example can be metrizable. I only know of examples of rigid complete Boolean algebras, i.e., those having trivial automorphism groups. For examples, see Jónsson, B., A Boolean Algebra Without Proper Automorphisms, Proc. AMS, 1951. Another rigid example coming from forcing appears in McAloon, K., Consistency Results About Ordinal Definability, Ann. Math. Log, 1971. REPLY [4 votes]: Don Monk's answer confirms my expectation; let me now also provide a more precise statement and a proof: If $A$ is a Boolean algebra with a trivial automorphism group, then the automorphism group of $A\times A$ is canonically isomorphic to $(A,+)$, acting by $a\cdot(b,c)=((1-a)b+ac,(1-a)c+ab)$. By Stone duality, this amounts to proving: if $X$ is a Stone space (totally disconnected, Hausdorff compact topological space) with trivial self-homeomorphism group, then any self-homeomorphism of $X\times\{-1,1\}$ is given by $(x,t)\mapsto (x,u(x)t)$ where $u$ is a continuous function $X\to\{-1,1\}$. To prove the latter, let us first check that for any two clopen subsets $Y,Z$ of $X$ and homeomorphism $h:Y\to Z$, we have $Y=Z$ and $h$ is the identity. Indeed, otherwise there exists $y\in Y$ such that $h(y)\neq y$. By passing to smaller clopen subsets, we can then assume that $Y$ and $Z$ are disjoint. Then we can extend $h$ to a self-homeomorphism of $X$, as equal to $h^{-1}$ on $Z$ and identity elsewhere. (Note that in this part, we use that $X$ is Hausdorff with a basis of clopen subsets, but compactness does not play any role). Then it's easy to conclude. Let $h$ be a self-homeomorphism of $X\times\{-1,1\}$. Call "component" the two subsets $X\times\{t\}$. We find a finite clopen partition of $X$ such that for each part $Y$ of the partition, each of $Y\times\{t\}$ is mapped into a single component. Then by the previous fact on partial homeomorphism, we have $h(y,t)=(y,s(y,t))$ for some $s(y,t)\in\{-1,1\}$ (which is constant when $y$ varies in $Y$). By injectivity, we have $s(y,1)=-s(y,-1)$. So we can write $s(y,t)=u(y)t$. $\Box$ Similarly, for a finite set $F$, the automorphism group of $X\times F$ consists of the $(x,t)\mapsto (x,\sigma(x)(t))$, where $\sigma$ ranges over the continuous functions $X\to\mathfrak{S}(F)$. The corresponding Boolean algebra action can be described accordingly.<|endoftext|> TITLE: What does reduction of structure group of principal bundle say? QUESTION [7 upvotes]: Let $G$ be a Lie group and $\pi:P\rightarrow M$ be a principal $G$ bundle. The notion of reduction of structure group is standard but I will recall here in case some one needs it. Let $f:P(M,G)\rightarrow P'(M',G')$ be a morphism of principal bundles such that $f:P\rightarrow P'$ is an imbedding and $f:G\rightarrow G'$ is a monomorphism. If $M=M'$ and the induced map $f:M\rightarrow M'$ is identity map, we call $P(M,G)$ to be reduced bundle for $P'(M,G').$ Given a principal bundle $P’(M’,G’)$ and a Lie subgroup $G$ of $G’,$ we say the structure group $G’$ is reduced to $G$ if there is a reduced bundle $P(M,G).$ Reduction of structure group says some thing interesting about manifolds involving it. For example, A manifold admits an almost-complex structure if the frame bundle on the manifold, whose fibers are $GL(2n,\mathbb{R})$, can be reduced to the group $GL(n,\mathbb{C})\subset GL(2n,\mathbb{R})$. A manifold is orientable if and only if its frame bundle can be reduced to the special orthogonal group, $SO(n,\mathbb{R})\subset GL(n,\mathbb{R})$. I am interested in knowing similar results about reduction of structure group. Please add references (if possible, a sketch of the proof) for results you quote here. One result in one answer please. REPLY [4 votes]: Let $G$ be a topological group and $M$ be a smooth manifold. Then, a reduction of the structure group of the frame bundle from $\mathrm{GL}_n(\mathbb R)$ to $\mathrm{GL}_n(\mathbb R)\times G$ is equivalent data to a principal $G$-bundle $Q\to M$. Here's a proof sketch: if $H$ and $H'$ are groups, an $(H\times H')$-torsor is the same thing as a product of an $H$-torsor and an $H'$-torsor. So the fiber of the principal $(\mathrm{GL}_n(\mathbb R)\times G)$-bundle we obtained from the frame bundle at some $x\in M$ is a product of a $\mathrm{GL}_n(\mathbb R)$-torsor $P_x$ and a $G$-torsor $Q_x$, and $P_x$ is the $\mathrm{GL}_n(\mathbb R)$ of bases of $T_xM$. Since $Q_x$ varies smoothly, it defines a principal $G$-bundle $Q\to M$.<|endoftext|> TITLE: Realisation of Kac-Moody Lie algebras QUESTION [10 upvotes]: I am reading Infinite dimensional lie algebras by Kac. He starts with a $n \times n$ GCM (Generalized Cartan Matrix) $A$ of rank $l$, then he defines the realization associated with the matrix $A$ which is of dimension $d=2n-l$. I know that in the simple Lie algebra case this dimension is $n$ as $A$ is invertible I don't understand why we are taking the space of dimension $2n-l$? Please explain, When the GCM is not of finite type, what we are getting extra by this definition of Lie algebra? Thanks for your time. REPLY [9 votes]: An equivalent definition of a realisation of a GCM $A=(a_{ij})_{1\leq i,j\leq n}$ of rank $\ell$ is as follows: it is a triple $(\mathfrak h, \Pi, \Pi^{\vee})$ where $\mathfrak h$ is a complex vector space, $\Pi=\{\alpha_i \ | \ 1\leq i\leq n\}\subseteq\mathfrak h^*$ and $\Pi^{\vee}=\{\alpha_i^{\vee} \ | \ 1\leq i\leq n\}\subseteq\mathfrak h$ are indexed subsets of $\mathfrak h^*$ and $\mathfrak h$, respectively, such that (R1) $\Pi$ and $\Pi^{\vee}$ are linearly independent subsets; (R2) $\langle \alpha_j,\alpha_i^{\vee}\rangle= a_{ij}$ for all $i,j\in\{1,\dots,n\}$; (R3) $\mathfrak h$ has minimal dimension for these properties. In other words, the conditions (R1) and (R2) force the dimension of $\mathfrak h$ to be at least $2n-\ell$ (this can be infered from Kac's proof of the existence of a realisation for $A$, see [1, Proposition 1.1] (see also [2, Section 3.5])). The condition (R3) is thus just there to avoid extra unnecessary dimensions, but does not play any role in the theory: one can in fact define a Kac-Moody algebra $\mathfrak g_{\mathcal D}$ from a weaker notion of "realisation of $A$" (called a "Kac-Moody root datum" $\mathcal D$), which only keeps the above condition (R2) (see [3, Chapitre 7], or [2, Section 7.3]). If $\mathcal D$ satisfies (R1) and (R2) but $\mathrm{dim} (\mathfrak h)>2n-\ell$, then $\mathfrak g_{\mathcal D}$ is just a trivial central extension of $\mathfrak g(A)$ that adds the missing dimensions in $\mathfrak h$. On the other hand, there are very good reasons to keep the condition (R1); here are two of them: 1) One can define a gradation of $\mathfrak g(A)$ (or $\mathfrak g_{\mathcal D}$) by the free abelian group $Q:=\bigoplus_{i=1}^n\mathbb Z\alpha_i$: denoting by $e_i,f_i$ the Chevalley generators, one sets $\mathrm{deg}(e_i)=\alpha_i$, $\mathrm{deg}(f_i)=-\alpha_i$, and $\mathrm{deg}(\mathfrak h)=0$, and one then extends as a Lie algebra gradation. However, if $\Pi$ is not linearly independent, then this abstract gradation need not correspond to an eigenspace decomposition for the adjoint action of $\mathfrak h$, as in Kac's book (see [3, Chapitre 7], or [2, Section 3.5]). 2) For every $\lambda\in\mathfrak h^*$, the Kac_Moody algebra $\mathfrak g(A)$ (or $\mathfrak g_{\mathcal D}$) acts on an irreducible highest-weight module $L(\lambda)$ with highest weight $\lambda$ (see [1, Chapter 9]). Moreover, if $\lambda$ is dominant integral (i.e. $\lambda(\alpha_i^{\vee})\in\mathbb N$ for all $i$), then $L(\lambda)$ is integrable, that is, it can be integrated to a representation of the Kac-Moody group associated to $\mathfrak g(A)$. However, if $\Pi^{\vee}$ is not linearly independent, then such a dominant integral weight might not exist. [1] V. Kac, Infinite-dimensional Lie algebras, 3rd edn., Cambridge University Press, Cambridge, 1990. [2] T. Marquis, An introduction to Kac-Moody groups over fields, EMS Textbooks in Mathematics, European Mathematical Society (EMS), Zürich, 2018 [3] B. Rémy, Groupes de Kac–Moody déployés et presque déployés, Astérisque (2002), no. 277, viii+348.<|endoftext|> TITLE: How to check that whether or not a surface is a K3 surface? QUESTION [5 upvotes]: How to check that whether or not a surface in $\mathbb{C}^3$ is a K3 surface? Are there some method or math software to transform the equation of a surface to a normal form? For example, the following is a surface in $\mathbb{C}^3$: \begin{align} 135\, x^2\, y\, z^2 + 360\, x^2\, y\, z + 540\, x^2\, z^2 + 180\, x^2\, z + 90\, x\, y^2\, z^2 + 654\, x\, y^2\, z + 360\, x\, y^2 + 360\, x\, y\, z^2 - 720\, x\, y\, z + 180\, x\, y + 240\, y\, z + 720\, y=0. \end{align} How to transform it to a normal form? Is it a K3 surface? Thank you very much. REPLY [8 votes]: I assume that the equation you've given is of interest to you, so here is my interpretation of Sasha's comment. The equation under consideration has degree $2$ in all three variables, so a compactification of this surface in $\mathbb CP^1\times \mathbb CP^1\times \mathbb CP^1$ is a divisor of degree $(2,2,2)$. So if by any chance it is smooth, then indeed it is a K3 surface. It might happen, however that the surface has singularities in $\mathbb C^3$ or in one of $7=2^3-1$ other charts. So one needs to find the solution of $F_x=F_y=F_z=F=0$ in all these charts. Probably this can be done with a help of some program. Once these singularities are found one needs to check if these are Du Val or not (for this they need to be isolated of course). This should not be super hard since the degree in each variable is $\le 2$. PS. As Jianrong is pointing out, the surface is singular at the point $(0,0,-3)$. In order to see whether it is Du Val or not at this point, we calculate its Taylor series at the point. It turns out the the second term is the following: $$60(-27x^2+42xy+4yz).$$ It is easy to see that this quadratic form has rank three, so we are lucky and the point $(0,0,-3)$ is the simplest possible singularity - an ordinary double point (of course Du Val). Note that, in order to conclude whether this surface is $K3$ or not, we still need to check what kind of singularities it has at infinity (at three planes $x=\infty$, $y=\infty$, $z=\infty$).<|endoftext|> TITLE: Lang's Jacobian identity: slicker, elementary proof? QUESTION [19 upvotes]: In Jeffrey Lang, A Jacobian identity in positive characteristic, J. Commut. Algebra, Volume 7, Number 3 (2015), pp. 393--409, the following result is proven: Theorem 1. Let $p$ be a prime. Let $\mathbf{k}$ be a commutative $\mathbb{F}_p$-algebra. Let $n$ be a nonnegative integer. Let $R$ be the polynomial ring $\mathbf{k}\left[x_1, x_2, \ldots, x_n\right]$. Let $f_1, f_2, \ldots, f_n$ be $n$ polynomials in $R$. Let $\nabla$ be the differential operator $\prod\limits_{i=1}^n \left(\dfrac{\partial}{\partial x_i}\right)^{p-1}$ on $R$. Let $M \in R^{n\times n}$ be the Jacobian matrix of $f_1, f_2, \ldots, f_n$; this is the $n\times n$-matrix over $R$ whose $\left(i,j\right)$-th entry is $\dfrac{\partial f_i}{\partial x_j}$. Then, \begin{align} & \sum\limits_{\left(i_1, i_2, \ldots, i_n\right) \in \left\{0,1,\ldots,p-1\right\}^n} f_1^{i_1} f_2^{i_2} \cdots f_n^{i_n} \nabla\left(f_1^{p-1-i_1} f_2^{p-1-i_2} \cdots f_n^{p-1-i_n} \right) \\ & = \left(-1\right)^n \left(\det M\right)^{p-1} . \end{align} (I have taken the liberty to correct the typo in the paper where the sum ranged over $\left\{1,2,\ldots,p-1\right\}^n$ instead of $\left\{0,1,\ldots,p-1\right\}^n$. Note that later in the paper, in Proposition 1.5, there is a "$0 \leq j \leq n-1$" that should be a "$1 \leq j \leq n-1$"; this suggests blaming a miscommunication between author and editors about which $0$ to replace by a $1$.) Theorem 1 generalizes Glynn's determinant formula (see Section 6 of Hendrik Lenstra, The unit theorem for finite-dimensional algebras, arXiv:1703.07273v1); indeed, it is easy to see that we can obtain the latter formula from Theorem 1 by setting $f_j = \sum\limits_{i=1}^n a_{ij} x_i$. Question. Is there an elementary (e.g., combinatorial, inductive, or Hopf-algebraic) proof of Theorem 1? The proof in Lang's paper relies on a different paper, which in turn relies on some algebraic geometry. I am not sure what is actually used in the proof, but the whole construct seems rather indirect. One approach that looks promising is to consider the $\mathbf{k}$-linear map $\widetilde{\nabla} : R \otimes R \to R$ (all tensors are over $\mathbf{k}$) that sends each pure tensor $a \otimes b$ to $a \nabla\left(b\right)$. Then, the left hand side of Theorem 1 is \begin{equation} \widetilde{\nabla}\left( \prod_{i=1}^n \left(1 \otimes f_i - f_i \otimes 1\right)^{p-1} \right) \end{equation} (where the product is in $R \otimes R$). This is because of the classical fact that $\left(X-Y\right)^{p-1} = \sum\limits_{i=0}^{p-1} X^i Y^{p-1-i}$ in $\mathbb{F}_p\left[X,Y\right]$. But the question is whether $\widetilde{\nabla}$ has any good properties with respect to the Hopf algebra $R$. Also, the operator $\nabla$ can be described rather explicitly on monomials: For any $n$ nonnegative integers $a_1, a_2, \ldots, a_n$, we have \begin{align} & \nabla \left(x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n}\right) \\ & = \begin{cases} \left(-1\right)^n x_1^{a_1-\left(p-1\right)} x_2^{a_2-\left(p-1\right)} \cdots x_n^{a_n-\left(p-1\right)}, & \text{ if } a_i \equiv p-1 \mod p \text{ for all } i ; \\ 0, & \text{ otherwise }. \end{cases} \end{align} REPLY [14 votes]: Awesome question! I haven't looked at Lang's paper yet, so I can't comment on whether this will be a different approach, but it is elementary. I will make use of Glynn's determinant formula at some point later on, so in order to keep this self contained I will start by giving a combinatorial proof of it. Lemma 1: (Glynn's determinant formula) Suppose $A$ is a matrix with entries in a commutative $\mathbb F_p$-algebra $T$. The coefficient of $x_{1}^{p-1}x_{2}^{p-1}\cdots x_n^{p-1}$ in the expansion of $\prod_{i=1}^{n}\left(\sum_{j=1}^n a_{ij}x_j\right)^{p-1}$ is equal to $(\det A)^{p-1}$. Proof: By the Macmahon Master Theorem this coefficient is the same as the coefficient of $x_{1}^{p-1}x_{2}^{p-1}\cdots x_n^{p-1}$ in the series $\det (I-AX)^{-1}\in T[[x_1,x_2,\cdots,x_n]]$ where $X$ is the diagonal matrix with $x_i$ on the diagonal. Notice that this coefficient is unchanged if we multiply by $\det(I-AX)^p$ since this doesn't affect monomials where all exponents are $ (p-1)n$ and will therefore vanish, since this implies that some $r_i\geq p$. We are now ready to complete the proof of Theorem 1 since $$\nabla_{\epsilon}\prod_{i=1}^n \left(\sum_{j=1}^n (\epsilon_j-1)x_j\frac{\partial f_i}{\partial x_j}\right)^{p-1}$$ is $(-1)^n$ times the coefficient of $\epsilon_1^{p-1}\epsilon_2^{p-1}\cdots \epsilon_n^{p-1}$ in $\prod_{i=1}^n \left(\sum_{j=1}^n (\epsilon_j-1)x_j\frac{\partial f_i}{\partial x_j}\right)^{p-1}$. This last coefficient is of course the same as taking the coefficient of $(\epsilon_1-1)^{p-1}\cdots (\epsilon_n-1)^{p-1}$ in the same expression but expanded in monomials $(\epsilon_1-1)^{r_1}(\epsilon_2-1)^{r_2}\cdots (\epsilon_n-1)^{r_n}$. By Glynn's determinant this last coefficient is $x_1^{p-1}x_2^{p-1}\cdots x_n^{p-1}\left(\det M\right)^{p-1}$ and this completes our proof. Remark: This proof, and all the lemmas, adapt easily (essentially just change every occurrence of $(p-1)$ to $(p^m-1)$) to give the following generalization of Lang's identity: For any $m\geq 1$, let's denote by $\nabla^{(m)}$ the linear operator that sends monomials of the form $x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$ to $x_1^{a_1-p^m+1}x_2^{a_2-p^m+1}\cdots x_n^{a_n-p^m+1}$ when $a_i\cong -1\pmod{p^m}$ for all $1\le i\le n$, and to zero otherwise. The following identity holds: $$\begin{align} & \sum\limits_{\left(i_1, i_2, \ldots, i_n\right) \in \left\{0,1,\ldots,p^m-1\right\}^n} f_1^{i_1} f_2^{i_2} \cdots f_n^{i_n} \nabla^{(m)}\left(f_1^{p^m-1-i_1} f_2^{p^m-1-i_2} \cdots f_n^{p^m-1-i_n} \right)= \left(\det M\right)^{p^m-1} . \end{align}$$<|endoftext|> TITLE: Rational perfect power values of $y(y+1)$ QUESTION [7 upvotes]: This is hard, so I am looking for partial results and how hard it is. Let $n>4$. Is it true that the hyperelliptic curve $x^n=y(y+1)$ doesn't have rational point with $x \ne 0$? If necessarily assume $n$ is prime. Integral points on the curve are heavily studied. It is one of the simplest exponential diophantine equations over the rationals. What tools are used for solving it? FLT implies there are no solutions with $y$ being $n$-th power. Added No solution for given $n$ imply Fermat Last Theorem for $n$ as discussed here see (3) with a=1 Assume $u^n-v^n=1$ is counterexample to FLT. Then $(uv)^n=v^n(v^n+1)$ and $(uv,v^n)$ is non trivial rational point on the curve. REPLY [18 votes]: There are no such solutions. Let $x=a/b$ and $y=c/d$ be reduced fractions. Then $a^n/b^n=(c(c+d))/d^2$ and since both sides are reduced fractions we get that $a^n=c(c+d)$ and $b^n=d^2$. From the first equation we deduce that $c=e^n$ and $c+d=f^n$ since $c$ and $c+d$ are co-prime and their product is an $n$-th power. Then $d=f^n-e^n$ so $(f^n-e^n)^2=b^n$. If $n$ is even write $n=2k$ and take square root to get $f^{2k}-e^{2k}=\pm b^k$. If the sign is $+$ then we get $f^{2k}=e^{2k}+b^k$ and since $k>2$ this implies by FLT that one of $e,b,f$ is 0. If $f=0$ then $y=-1$ so $x=0$. $b=0$ is certainly impossible. Finally, $e=0$ means $y=0$. So this option is excluded. If the sign is $-$ we get $e^{2k}=f^{2k}+b^k$ and the same reasoning works. If $n$ is odd, from $(f^n-e^n)^2=b^n$ we deduce that $b$ is a square so $b=g^2$ and then we can extract the root to get $f^n-e^n= \pm g^n$ and proceed in a way similar to the case where $n$ is even.<|endoftext|> TITLE: Number of integer solutions of a linear equation under constraints QUESTION [7 upvotes]: How many positive integer solutions of $$\sum_{i=1}^{k}x_i = N$$ for some positive integer $N$ given the constraints $n_i\leq x_i\leq m_i$ for $i=1,\ldots,k$, where $n_i$ and $m_i$ are positive integers. I know that it can be calculated by finding the coefficient of $y^N$ in the polynomial $\prod_{i=1}^{K}(y^{n_i}+\cdots,y^{m_{i}})$. Is there any compact formula or any upper bound to the number of the positive integer solutions? REPLY [7 votes]: Proposition 1: The number of integer solutions of the equation $$ \sum_{i=1}^{k}x_i = N $$ where $x_i\geq n_i$ for $i=1,\ldots,k$, is given by $$ {\small \binom{N+k-1-n_1-n_2-...-n_k}{k-1} } $$ if the upper index is non-negative and zero otherwise. In the formula above, $\binom{.}{.}$ stands for the generalized binomial coefficients. Now, to tackle the problem as stated, you need to apply Proposition 1 and invoke the inclusion-exclusion principle, in the following sense: For $i=1,...,k$, set as $q_i$: the property of a solution of Proposition 1, to satisfy the condition $$ x_i> m_i \Leftrightarrow x_i\geq m_i+1 $$ If we denote: $N(q_i)$, the number of solutions (provided by Prop. 1) satisfying property $q_i$, $N(q_i q_j)$, the number of solutions (provided by Prop. 1) satisfying both properties $q_i$, $q_j$, ... and generally: $N(q_{i_1}q_{i_2}... q_{i_s})$, the number of solutions (provided by Prop. 1) satisfying all properties $q_{i_1}$, $q_{i_2}$, ..., $q_{i_s}$, then we get -applying Prop. 1- that: $$ {\small N(q_1)=\binom{N+(k-1)-1-m_1-n_2-...-n_k}{k-1} \ \ \ or \ \ \ N(q_1)=0 } $$ $$ {\small N(q_2 q_3)=\binom{N+(k-2)-1-n_1-m_2-m_3-n_4-...-n_k}{k-1} \ \ \ or \ \ \ N(q_2 q_3)=0} $$ ... and generally: $$ {\small N(q_{i_1}q_{i_2}... q_{i_s})=\binom{N+(k-s)-1-\sum_{i\notin I} n_i-\sum_{i\in I} m_i}{k-1} \ \ \ or \ \ \ N(q_{i_1}q_{i_2}... q_{i_s})=0 } $$ where $s$ is the number of properties, $I=\{i_1, i_2, ..., i_s\}\subseteq \{1,2,...,k\}$ and the $N(..)$ function takes zero values whenever the upper index becomes negative. Now all you need to do to obtain a compact formula for the number of solutions satisfying your constraints, is to apply the inclusion-exclusion principle to determine the number of solutions produced by Proposition 1, which have none of the properties $q_i$ for $i=1,2,...,k$. This is given by $$ \binom{N+k-1-\sum n_i}{k-1}-\sum_{i=1}^{k} N(q_i)+\sum_{k\ \geq j > i\geq 1} N(q_i q_j)-...+ \\ +(-1)^s\sum_{k\ \geq i_s>...>i_1\geq 1} N(q_{i_1}q_{i_2}... q_{i_s})+.... +(-1)^k N(q_{1}q_{2}... q_{k}) $$ where in the above formula $s$ is the number of properties and $$ \sum_{k\ \geq j > i \geq 1}=\sum_{i=1}^{k-1}\sum_{j=i+1}^{k} $$ ... etc. Example: As an example of application of the previous method, consider the following special case of the OP: Find the number of (positive) integer solutions of the equation $$\sum_{i=1}^{k}x_i = N$$ for some positive integer $N$, given the constraints $1\leq x_i\leq \alpha$ for $i=1,\ldots,k$ The method described above gives: $$ {\small \binom{N-1}{k-1}-\binom{k}{1}\binom{N-\alpha-1}{k-1}+\binom{k}{2}\binom{N-2\alpha-1}{k-1}-\binom{k}{3}\binom{N-3\alpha-1}{k-1}+\binom{k}{4}\binom{N-4\alpha-1}{k-1}-\cdots } $$ where $\binom{..}{..}$ stands for the generalized binomial coefficients and the summation halts when zero terms appear.<|endoftext|> TITLE: How to be rigorous about combinatorial algorithms? QUESTION [62 upvotes]: 1. The question This may be the worst question I've ever posed on MathOverflow: broad, open-ended and likely to produce heat. Yet, I think any progress that will be made here will be extremely useful to the whole subject. In a nutshell, the question is: Short version. What tools (notations, formalisms, theories) are we missing to bring algorithmic combinatorics to the standard of rigor of modern algebra? To make this more precise, let me explain what I mean by algorithmic combinatorics (and give a few examples). Lots of algorithms appear in mathematics, but we mathematicians tend to avoid them when we are proving things. For example, the Euclidean algorithm computes the gcd of integers; but when it comes to proving things about the gcd (say, Bezout's identity), it is much easier to use induction. In linear algebra, we prefer not to use the Gaussian elimination algorithm either, but instead to extract its main ideas into simple statements and rely on them exclusively, essentially splitting the algorithm back into the little steps it's made of and recombining these steps back into induction arguments. This reluctance to work with algorithms has a reason: Very few mathematicians know how to formally define an algorithm and its execution; even fewer know how to rigorously reason about those. Meanwhile, everyone knows induction. So, as long as algorithmic reasoning can be replaced by induction easily enough, there's little reason to struggle through the algorithmic swamps. This question is about the cases when algorithms cannot be easily avoided. This tends to happen a lot in combinatorics (particularly bijective combinatorics, but also algebraic combinatorics). The algorithms occurring there often have the following properties: The algorithm involves a lot of clobbering. That is, the algorithm doesn't only set each variable once, but rather modifies it several times. This makes it difficult to rewrite the algorithm as a recursive definition (because we have to keep track of "time", i.e., count steps), and also complicates reasoning about the algorithm's execution (since we have to specify what time we are talking about). The well-definedness of the algorithm is not immediately obvious; that is, it may involve non-deterministic steps, or it may "crash" (more formally: throw exceptions). Often the first thing one wants to prove about the algorithm is that it neither is actually non-deterministic (i.e., the choices don't matter to the final result) nor can it "crash". But the mental overhead involved in arguing about the algorithm before this is proven is substantial, and it further complicates any formal models. I will call algorithms with these properties complex. To make things worse, in combinatorics, algorithms are often used as definitions of objects, not just as tools to their understanding, so we cannot just throw away everything we don't care about for a given proof. Let me give a few examples of complex algorithms: The Robinson-Schensted-Knuth (RSK) algorithm. (Various expositions: Knuth, Casselman, van Leeuwen.) At least a dozen variants on this algorithm appear in the literature now, usually more complicated than the original. The standard parenthesis-matching procedure: Start with a word made of parentheses, such as $(()()))((()$. Keep matching parentheses by the standard rule: If you ever find a $($ parenthesis before a $)$ parenthesis such that all parentheses between these two are already matched (this includes the case when there are no parentheses between these two), you can match these two parentheses. Keep doing this until no more matchable parentheses are left. This is a simple example of an algorithm that "looks" non-deterministic, because you can often choose which two parentheses to match, but in truth it can be shown (using the diamond lemma) that the final result does not depend on the choices. This procedure is typically used as a subroutine in more complicated algorithms; see, e.g., Shimozono's Crystals for dummies. The evolution of a box-ball system; see, e.g., §3.1 in Fukuda. Note the wording using boxes, balls and colors -- a staple in this subject, and a pain for formalization. The definition of $q\left( u \right)$ in §2.1 of Erik Aas, Darij Grinberg, Travis Scrimshaw, Multiline queues with spectral parameters. This one has been a particularly pointy thorn in my side -- the proof of Lemma 2.2 is completely ineffable; the only hint we can give is that the reader should draw a picture and stare at it a bit. This question is not about these four examples in particular; I am just using them to illustrate the kind of algorithms I'm talking about and the issues with them. These four examples are relatively simple when compared to more recent constructions, such as type-D analogues of RSK, complicated bijections between Littlewood-Richardson tableaux, and the mess that is the theory of Kohnert diagrams. So here is the precise version of my question: Precise version. What are viable methods of reasoning about complex algorithms that allow for a level of rigor similar to that in the rest of modern mathematics (e.g., algebra, algebraic geometry, analysis, even non-algorithmic combinatorics)? In particular, I'm looking for practical ways to reason about the algorithm itself (syntax); the results of the algorithm (semantics); the way the results change if inputs are modified (uhm, does that have a name?...). The syntax question is the easy one: You can always implement an algorithm in an actual imperative programming language. It is usually easy to translate pseudocode into actual code, so there isn't much of a rigor deficit at this step. But the combinatorics literature badly lacks rigor when dealing with the other two questions; almost every author reasons about their algorithms through wishy-washy "behaviorial" arguments, waving their hands and tracing examples in the hope that they are in some sense representative of the general case. Programming doesn't help here: Rarely does an imperative programming language come with tools for proving the properties of programs. (Some functional languages do, but translating a combinatorial algorithm into functional code in an intuition-preserving way is far from easy.) The result is that a significant part of the contemporary combinatorics literature is unreadable to all but the most mentally agile; and that the community is slow to trust new results with algorithmic proofs, often waiting for independent confirmations through algebraic methods. Ideally, I prefer a way to formalize (at least to the level of rigor accepted in modern algebra) the handwaving that combinatorialists typically do (at least when it isn't genuinely wrong), rather than a completely new toolkit that would require me to reason completely differently. But honestly, the latter wouldn't be bad either, and may even be better in the long run. When I say "viable" and "practical", I mean that formalizing the proofs should not be forbiddingly tedious, hard or requiring creativity. Think of (most) abstract algebra, where finding the proof is the hard part and formalizing it is easy; I want this for combinatorics. 2. Approaches Next I am going to list a few approaches I've seen to the problem, and explain what I believe to be their shortcomings. Unfortunately, my spotty knowledge of computer science makes me the wrong person to comment on these approaches, so I'm making stone soup here -- but this is a question, so I am not feeling guilty. 2.1. Hoare logic The classical way of proving properties of algorithms is called Hoare logic. The rough idea is that you insert an assertion between any two consecutive instructions of the algorithm, and prove that for each instruction, if the state before the (execution of the) instruction satisfies the assertion before the instruction, then the state after the (execution of the) instruction satisfies the assertion after the instruction. (Fine print: Keywords like "if", "while" and "for" come with their special rules about what you need to prove, and in the presence of "while"s, you also need to prove that the algorithm doesn't get stuck in an endless loop.) For example, let us pick apart the Euclidean algorithm (written here in a somewhat defensive style of Python, with the tacit understanding that all variables are integers): def euclid(n, m): # Compute the gcd of two nonnegative integers n and m. u = n v = m while v > 0: if u < v: u, v = v, u # This swaps u and v in Python. u = u - v return u To prove that the returned value is really $\gcd\left(n,m\right)$, we can insert Hoare assertions (traditionally written in braces) as follows (we use == for equality): def euclid(n, m): # Compute the gcd of two nonnegative integers n and m. u = n {u == n.} v = m {u == n and v == m.} while v > 0: {gcd(u, v) == gcd(n, m) and u >= 0 and v >= 0.} if u < v: {u < v and gcd(u, v) == gcd(n, m) and u >= 0 and v >= 0.} u, v = v, u # This swaps u and v in Python. {u >= v and gcd(u, v) == gcd(n, m) and u >= 0 and v >= 0.} {u >= v and gcd(u, v) == gcd(n, m) and u >= 0 and v >= 0.} u = u - v {gcd(u, v) == gcd(n, m) and u >= 0 and v >= 0.} {u == gcd(n, m).} return u (Note that the last Hoare assertion, u == gcd(n, m), follows from the previous Hoare assertion gcd(u, v) == gcd(n, m) and u >= 0 and v >= 0 combined with the fact that v <= 0, which is automatically inferred from the fact that we just escaped from the while-loop. This is part of the Hoare logic for the "while" keyword.) This is all fun when algorithms are as simple as euclid. What if we do this to the RSK algorithm? It turns out that this has already been done -- in Donald Knuth, Permutations, matrices, and generalized Young tableaux, Pacific J. Math., Volume 34, Number 3 (1970), pp. 709--727. More precisely, Knuth considered a subroutine of the RSK algorithm: the insertion of a single number $x$ into what he called a generalized Young tableau $Y$. He calls this subroutine $\operatorname{INSERT}(x)$ (with $Y$ being implicit). His Hoare assertions are rather nice (although he needs to add "$i \geq 1$ and $j \geq 1$" to all of them in order to ensure that $Y$ remains a generalized Young tableau throughout the algorithm). Unfortunately, in order to prove the subsequent claim that the functions $\operatorname{INSERT}$ and $\operatorname{DELETE}$ are inverses of each other (in an appropriate sense), I had to extend the Hoare assertions for each of the two functions to... 2 pages each. Basically, Knuth's Hoare assertions are precisely what is needed to show that $Y$ remains a generalized Young tableau at each step of the algorithm; but they don't let you trace the changes to its entries or argue such things as "the insertion path trends left" (the inequalities (2.5) in Knuth's paper, who just states them as something obvious) and "only the entries $y_{i,r_i}$ are ever modified". For each of these simple observations, you had to add an extra Hoare assertion at the end and percolate it through all the instructions. (See the proof of Proposition A.4 on pp. 4--6 of http://www.cip.ifi.lmu.de/~grinberg/algebra/knuth-comments1.pdf , but beware that I wrote this when I was just starting to learn combinatorics, so I was probably being wasteful.) Perhaps there are tricks to using Hoare logic "economically" and I just don't know them. There are certainly useful modifications that can be made, such as introducing new auxiliary variables inside Hoare assertions, and even new instructions that don't modify the variables of the original algorithm. But it's far from clear how to use this in combinatorics. I would love to see a collection of examples of Hoare logic in mathematics. 2.2. Transforming algorithms into recursive definitions We can try painstakingly transforming a combinatorial algorithm into a recursive definition. For example, we can turn each variable $v$ into a sequence of objects $v_1, v_2, v_3, \ldots$, with $v_i$ being the value of $v$ after step $i$ of the algorithm. And we can introduce another sequence of integers $l_1, l_2, l_3, \ldots$, with $l_i$ being the number of the code-line at which we are after step $i$ of the algorithm. Then, the algorithm transforms into a jointly recursive definition of all these objects. This is a cheap way of formalizing the execution of an algorithm. The advantage of this approach over Hoare logic is that you have all the history of your execution at your disposal rather than just the current state. Unfortunately, actually reasoning about this recursive definition is forbiddingly cumbersome, and one is often forced to prove all the Hoare assertions with the extra complexity of the time parameters. I have never seen an algorithm being rigorously studied using this method in practice. 2.3. Monads? Modern functional programming languages come with a collection of monads that are useful for translating imperative code into functional code. In particular, the state monad lets one write "imperative blocks" inside functional code, which behave as if they had actual mutable state. The maybe monad emulates algorithms that may "crash", and the list monad emulates non-determinism. Unfortunately, this is again just an answer to the question of formalizing the algorithm, but not to the question of formalizing reasoning about it. I would really love to be proven wrong on this point: Are there some known sources on the use of monads in proofs? (I know that the distribution monad was used in Coq-Combi to formalize the "hook walk" in the proof of the hook-length formula, but I have never learnt enough ssreflect/Coq to read that proof.) 2.4. Algebraic circumvention The final trick to cut down on the complexity of algorithms is to find an alternative non-algorithmic (usually algebraic) definition of what they compute, and then forget about the algorithm and just work with that new definition. For example: There are several ways to redefine the RSK algorithm in a more proof-friendly manner. One nice way is shown in Sam Hopkins, RSK via local transformations: Here, the RSK algorithm is re-encoded as a transformation between integer arrays and reverse plane partitions, and this transformation is redefined through a much more convenient recursive definition. Unfortunately, while the new definition is a lot easier to analyze in some aspects, it is not obvious at all that it is equivalent to the old one (and to my knowledge, the proof of this equivalence is scattered across 2 or 3 papers). It is also not suited to proving the connections of RSK with the plactic monoid. The parenthesis-matching procedure can be redefined recursively, too, at the cost of losing the intuition and having to re-prove properties that were previously obvious. I don't know how easy it would be to build the theory of crystals on the basis of such a recursive definition. I think Florent Hivert's Coq-Combi project does something like this for the Littlewood-Richardson rule, but this project was far from straightforward and it is unclear how well the methods scale to deeper results. For box-ball systems, there is the "carrier algorithm" which describes the dynamics of the system in some more algebraic way, although it has an annoying 2-dimensionality that prevents it from being too easy to work with. Combinatorial properties of tableaux can occasionally be deduced from representation theory of quantum groups or Lie algebras. Symmetric groups can be interpreted as Coxeter groups of type A, which provides a whole new viewpoint from which some of their properties become much clearer (but others become a lot more complicated). In general, however, there is no way to take a combinatorial algorithm and "find the algebraic structure behind it"; the cases where this has been done are few. There is no "clarifying" algebraic structure behind most combinatorial objects. Moreover, this is a somewhat escapist approach: We just give up combinatorics and do algebra instead. REPLY [14 votes]: Broadly speaking, there are three approaches to reasoning about software semantics: Denotational semantics provides a mapping from a computer program to a mathematical object representing its meaning. Operational semantics makes use of logical statements about the execution of code, typically using inference rules similar in style to natural deduction for propositional logic. Axiomatic semantics, which includes Hoare logic, is based on assertions about relationships that remain the same each time a program executes. Here's a good book on different semantic formalisms. One approach I'd recommend, perhaps somewhat more practical than others, is something like Dijkstra's predicate transformer semantics, a reformulation of Hoare logic, which is expounded in David Gries' classic book The Science of Programming. I'd have thought anyone who is willing to expend sufficient effort to master this should be able to use it to reason effectively about algorithms (combinatorial or otherwise). The definition of $q(u)$ mentioned in the question looks like a relatively simple candidate for this approach. Perhaps the major challenge for most mathematicians is the fact that becoming competent in a suitable formalism takes some considerable time, and also probably requires as a precursor some basic computer science knowledge in order to learn how to think like a computer scientist (and be able to construct well-designed modular code).<|endoftext|> TITLE: Bijection $f: \mathbb R^n \to \mathbb R^n$ that maps connected onto connected sets must map closed connected onto closed connected sets? QUESTION [15 upvotes]: Willie Wong asked here (MO) and here (MSE) very interesting question. As he phrased it: Let $(X,\tau), (Y,\sigma)$ be two topological spaces. We say that a map $f: \mathcal{P}(X)\to \mathcal{P}(Y)$ between their power sets is connected if for every $S\subset X$ connected, $f(S)\subset Y$ is connected. Question: Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$? When I did some research on that question another question became important to me, and here it is: Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology and for every $S \subset \mathbb R^n$ connected we have that $f(S)$ is connected. Does that imply that if $T \subset \mathbb R^n$ is closed and connected that then $f(T)$ is closed and connected? So, basically, I believe that this is really true, that is, that in requirements that $f$ maps connected sets onto connected sets and that $f$ is a bijection there is hidden a theorem that $f$ maps closed connected sets onto closed connected sets. If this were settled we would be closer to a solution of Willie´s problem, but even if this problem stands on its own it could be of interest to someone. REPLY [14 votes]: This question is equivalent to the question of Willie Wong because of the following theorem of Jones. Theorem (Jones, 1967). Each bijective semicontinuous map from a topological space to a semilocally locally connected Hausdorff space is continuous. A topological space $X$ is semilocally connected if if has a base of the topology consisting of open sets whose complements have finitely many connected components. A function $f:X\to Y$ is called $\bullet$ Darboux if for any connected subspace $C\subset X$ the image $f(C)$ is connected in $Y$; $\bullet$ connected if for any connected subspace $B\subset Y$ the preimage $f^{-1}(B)$ is connected in $Y$; $\bullet$ semiconnected if for any connected closed subset $B\subset Y$ the preimage $f^{-1}(B)$ is connected and closed in $X$. In this terms the problems of @Right and Wong read as follows: Problem (@Right). Is each connected bijection of $\mathbb R^n$ semiconnected? Problem (Wong). Is each connected bijection of $\mathbb R^n$ a homeomorphism? Now I explain why these two problems are equivalent: If the answer to the problem of @Right is affirmative, then any connected bijection $f$ of $\mathbb R^n$ is semiconnected and by Jones Theorem is continuous. By the Invariance of Domain Principle, $f$ is open and hence a homeomorphism. So, we get an affirmative answer to the problem of Wong. If the problem of Wong has an affirmative answer, then any connected bijection $f$ of $\mathbb R^n$ is a homeomorphism and hence it is both semiconnected and Darboux. Concerning (partial) answers to the equivalent problems of Wong and @Right let us mention the following two results: Theorem (Tanaka, Pervine, Levine). A connected Darboux bijection $f:X\to Y$ from a Hausdorff topological space $X$ to a semilocally connected Hausdorff space $Y$ is continuous. and Theorem (Banakhs). A Darboux injective map $f:X\to Y$ between connected metrizable spaces is continuous if one of the following conditions is satisfied: 1) $Y$ is a 1-manifold and $X$ is compact; 2) $Y$ is a 2-manifold and $X$ is a closed $n$-manifold of dimension $n\ge 2$; 3) $Y$ is a 3-manifold and $X$ is a closed $n$-manifold of dimension $n\ge 3$ with finite homology group $H_1(X)$.<|endoftext|> TITLE: Etymology of 'spectrum' in algebraic geometry and algebraic topology QUESTION [9 upvotes]: In algebraic geometry, one has the notion of the spectrum of a commutative ring. These spectra serve as local charts for schemes. In algebraic topology, a spectrum is a sequence of pointed spaces $X_n$ $(n \in \mathbb{N})$ together with the structure maps $S^1 \wedge X_n\rightarrow X_{n+1}$. One can associate a generalized cohomology theory to such a spectrum. My question is whether these two notions are etymologically related. It would be amusing since, in some sense, the field of derived algebraic geometry seeks to merge them into one. REPLY [13 votes]: No, they are not etymologically related. The early development of stable homotopy theory happened simultaneously with the early developments of scheme theory, so certainly neither terminology was influenced by the other. Grothendieck's choice of terminology of the "spectrum" of a ring comes from functional analysis. One can speak of the eigenvalue spectrum of an operator, which can be generalized to the spectrum of a whole family of mutually commuting operators, which can be abstracted to the (Gelfand) spectrum of a commutative $C^\ast$-algebra. The Gelfand-Naimark theorem says that the $C^\ast$-algebra is canonically the algebra of functions on the topological space given by the spectrum, just as in scheme theory. Spectra in topology were introduced by Lima and his advisor Spanier. It is a minor mystery why they chose the name "spectrum", but most likely it is used in the second sense of this definition: Physics. (a) An array of entities, as light waves or particles, ordered in accordance with the magnitudes of a common physical property, as wavelength or mass: often the band of colors produced when sunlight is passed through a prism, comprising red, orange, yellow, green, blue, indigo, and violet. (b) This band or series of colors together with extensions at the ends that are not visible to the eye, but that can be studied by means of photography, heat effects, etc., and that are produced by the dispersion of radiant energy other than ordinary light rays. Compare band spectrum, electromagnetic spectrum, mass spectrum. A broad range of varied but related ideas or objects, the individual features of which tend to overlap so as to form a continuous series or sequence: the spectrum of political beliefs.<|endoftext|> TITLE: existence of riemannian metric on $\text{SL}_3(\mathbb{R})$ with special geodesics QUESTION [7 upvotes]: Is there a left-invariant Riemannian metric on $\text{SL}_3(\mathbb{R})$ for which the geodesics (with respect to the corresponding Levi-Civita connection) through the identity are exactly the integral curves for the invariant, smooth vector fields? REPLY [8 votes]: I believe that a standard computation for the inverse problem in the calculus of variations will show that, up to constant multiples, the only pseudo-Riemannian metric on $\mathrm{SL}(3,\mathbb{R})$ whose geodesics are the $1$-parameter subgroups (which is what you are asking) is the usual bi-invariant one, and that one is not positive definite. In particular, this would answer your question in the negative. For references on the inverse problem in the calculus of variations and a guide to the calculations necessary, I would suggest: Anderson, Ian(1-UTS); Thompson, Gerard(1-TLD), The inverse problem of the calculus of variations for ordinary differential equations. Mem. Amer. Math. Soc. 98 (1992), no. 473, vi+110 pp.<|endoftext|> TITLE: Explicit description of SU(2,2)/U QUESTION [7 upvotes]: Consider the real diagonal $4\times 4$ - matrix $$I_{2,2}={\rm diag}(1,1,-1,-1)$$ and the corresponding special unitary group $$ G={\rm SU}(2,2)=\{g\in {\rm SL}(4,{\mathbb{C}})\ |\ g\cdot I_{2,2}\cdot \bar g ^{\rm tr}=I_{2,2}\}.$$ We regard $G$ as an algebraic group over ${\mathbb{R}}$. It is known that $G$ is quasi-split, that is, $G$ contains a Borel subgroup $B$ defined over ${\mathbb{R}}$. Let $U$ denote the unipotent radical of $B$, it is defined over ${\mathbb{R}}$. Set $X=G/U$. Question. How can one describe the homogeneous space $X$ explicitly (by equations and "inequalities") as a quasi-projective variety over ${\mathbb{R}}$ on which ${\rm SU}(2,2)$ naturally acts? EDIT: I explain why I need explicit equations for $G/U$. I want to twist the desired variety by the 1-cocycle $c=I_{2,2}\in G(\mathbb{R})_2$ and to obtain explicit equations for the twisted variety $Y={}_c X$. This new variety $Y$ over $\mathbb{R}$ is a homogeneous space of the twisted group $_cG={\rm SU}(4)$ for which $U$ is the stabilizer of a $\mathbb C$-point. Of course, $Y$ has no $\mathbb R$-points, because the stabilizer of an $\mathbb R$-point would be a compact form of $U$, but we know that the unipotent group $U$ has no compact forms. REPLY [4 votes]: Although the question has been answered in comments (by Victor Petrov), I prefer to post an answer. I assume that $G={\rm U}(2,2)$ rather than $G={\rm SU}(2,2)$. My variety $G/U$ is the variety $X$ whose real points are the triples $$(W,w,b),$$ where $W\subset \mathbb C^4$ is an isotropic 2-dimensional subspace, $w\in W$ a nonzero vector (which is automatically isotropic), and $b$ is a nonzero element of $\Lambda^2W$. This variety is a $R_{\mathbb C/\mathbb R}\mathbb G_{m,\mathbb C}^2$-torsor over the variety $\mathcal F$ of isotropic flags: the map is $$X\to\mathcal F\colon\quad (W,w,b)\mapsto (W,\langle w\rangle),$$ and the action of $(\mathbb C^\times)^2$ on $X$ is $$ (\lambda,\mu)*(W,w,b)=(W,\lambda w,\lambda\mu b)\quad \text{for } \lambda,\mu\in\mathbb C^\times.$$ By Witt's theorem for Hermitian forms, $G(\mathbb R)$ transitively acts on $X(\mathbb R)$, and my calculations show that the stabilizer of the point $$(\langle e_1,e_2\rangle, e_1, e_1\wedge e_2)\in X(\mathbb R)$$ is a maximal unipotent subgroup of $G$. Thus $X\simeq G/U$. The twisted form of $X$ is the same variety, but for another Hermitian form (and if the form is not hyperbolic, there is no such triples $(W,w,b)$ over $\mathbb R$, as Victor has mentioned). The real points of the variety $\mathcal V$ of Victor's answer are pairs of non-proportional isotropic vectors $(w_1,w_2)$ in $\mathbb C^4$. This variety is a $R_{\mathbb C/\mathbb R}\mathbb G_{a,\mathbb C}$-torsor over $X$: the map is $$\mathcal V\to X\colon\quad (w_1,w_2)\mapsto (\,\langle w_1,w_2\rangle,\, w_1,\, w_1\wedge w_2)$$ and the action of $\mathbb C$ on $\mathcal V$ is $$a*(w_1,w_2)=(w_1, w_2+aw_1)\quad\text{for } a\in\mathbb C.$$ The real points of the variety $\mathcal V'$ of Victor's comment is the set of pairs $(v,u)$, where $v$ is a nonzero isotropic vector in $\mathbb C^4$, and $u$ is a nonzero vector in $v^\perp/\langle v \rangle$ that is isotropic with respect to the induced Hermitian form on $v^\perp$. We construct a $G$-equivariant isomorphism $$\varphi\colon\mathcal V'\to X.$$ Let $(v,u)\in \mathcal V'(\mathbb R)$. We lift $u$ to an isotropic vector $\tilde u\in \mathbb C^4$ and set $$\varphi(v,u)=(\langle v, \tilde u\rangle, v, v\wedge\tilde u)\in X(\mathbb R).$$ In the opposite direction, if we have $(W,w,b)\in X(\mathbb R)$, we choose $y\in W$ such that $b=w\wedge y$, and we set $$\psi(W,w,b)=(w, y+\langle w\rangle)\in\mathcal V'(\mathbb R).$$ Since $\varphi$ and $\psi$ are mutually inverse, we see that $\varphi$ is an isomorphism.<|endoftext|> TITLE: Witt vectors and flat liftings of (non)perfect fields QUESTION [7 upvotes]: My motivating question is as follows: Why does Theorem 2.1. in Deligne-Illusie's classical work on the Hodge degeneration (EUDML link), i.e. the decomposition theorem, does $k$ need to be a perfect field? As far as I see is what they actually need that $W_2(k) \rightarrow \mathbb{Z}/p^2 $ is flat, so my more precise question is: How does flatness of $W_2(k) \rightarrow W_2(\mathbb{F}_p)$ relate to perfectness of $k$? Is there any good reference for (flat) liftings over $\mathbb{Z}/p^2$? REPLY [6 votes]: This answer is a response to your first question about relating Witt vector flatness with perfectness. The second question (about flat liftings to $\mathbf{Z}/p^2$) has a rich literature, see papers that cite Deligne-Illusie (e.g., on mathscinet). Theorem: For any $\mathbf{F}_p$-algebra $R$, the flatness of $\mathbf{Z}/p^2 \to W_2(R)$ is equivalent to the perfectness of $R$. So Deligne-Illusie is in optimal generality. Let me explain the non-trivial implication. Assume $W_2(R)$ is flat over $\mathbf{Z}/p^2$. This implies that the image of multiplication by $p$ on $W_2(R)$ coincides with the kernel of multiplication by $p$. In Witt vector notation, we are saying the following: $(*)$ If $x \in W(R)$ satisfies $px \in V^2W(R)$, then $x \in pW(R) + V^2W(R)$. As $R$ is an $\mathbf{F}_p$-algebra, we have $p = V(1)$, so $px = V(1)x = VF(x)$. So the condition that $px \in V^2W(R)$ appearing above is equivalent to $VF(x) \in V^2 W(R)$. As $V$ is always injective, this just means $F(x) \in VW(R)$. We have also seen here that $pW(R) = VFW(R)$, so $(*)$ simplifies to $(**)$ If $x \in W(R)$ satisfies $F(x) \in VW(R)$, then $x \in VFW(R) + V^2 W(R) \subset VW(R)$. Now let's use $(**)$ to see that $R$ must be perfect, i.e., its Frobenius is bijective. Injectivitity of Frobenius: If $x \in R$ satisfies $x^p = 0$, then $[x] \in W(R)$ satisfies $F([x]) = 0$. Using $(**)$, we get $[x] \in VW(R)$. But $VW(R)$ is the kernel of the restriction map $W(R) \to R$, so the image $x \in R$ of $[x] \in W(R)$ must be $0$. Surjectivity of Frobenius: Given $x \in R$, we have $pV([x]) = V(1)V([x]) = VFV([x]) = V^2F([x]) \in V^2 W(R)$ since $VF = FV$ as $R$ has characteristic $p$. Applying $(**)$ gives $y_1,y_2 \in W(R)$ such that $V([x]) = VF(y_1) + V^2(y_2)$. As $V$ is injective, this gives $[x] = F(y_1) + V(y_2)$. Applying the restriction map down to $R$ (which has kernel $VW(R)$) shows that $x$ is a $p$-th power.<|endoftext|> TITLE: Did Peter May's "The homotopical foundations of algebraic topology" ever appear? QUESTION [25 upvotes]: In the monograph Equivariant Stable Homotopy Theory, Lewis, May, and Steinberger cite a monograph "The homotopical foundations of algebraic topology" by Peter May, as "in preparation." It's their [107]. In his paper "When is the Natural Map $X \rightarrow \Omega\Sigma X$ a Cofibration?" Lewis also cited this monograph, and additionally wrote "Monograph London Math. Soc., Academic Press, New York (in preparation)." Did this monograph ever appear, perhaps under a different name? I could not find anything with the given title. The reason I ask is that LMS wrote in many places things like "details may be found in [107]" and I was looking into something where nitpicky detail might matter. REPLY [32 votes]: An anonymous source told me this question is here. Dylan gave the quick answer and Tyler referred to it. I'll use the question as an excuse to give a pontificating longer answer. When I first planned on writing that, maybe 45 or 50 years ago, I had not yet been converted to model category theory, let alone anything more modern, and what I had in mind would have been very plodding. I've thought hard about the pedagogy, perhaps not to good effect, and Concise and More (or less) Concise give what I came up with. The latter is not as concise as I would like in large part because so much relevant detail seemed missing from both the algebraic and topological foundations of localization and completion, especially in the proper generality of nilpotent rather than simple spaces. Both are available on my web page. http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf http://www.math.uchicago.edu/~may/TEAK/KateBookFinal.pdf As an historical note, in retrospect I came to the conclusion that the standard foundations for classical algebraic topology, as I understood it a half century ago, were given by the mixed model structure on spaces discovered by Mike Cole. The weak equivalences are the same as in the usual Quillen model category, but the fibrations are the Hurewicz fibrations rather than the Serre fibrations. Then the cofibrant spaces are the spaces of the homotopy types of CW complexes, which for sure was everybody's favorite category of spaces in which to work back then.<|endoftext|> TITLE: Good introductory references on moduli (stacks), for arithmetic objects QUESTION [16 upvotes]: I've studied some fundation of algebraic geometry, such as Hartshorne's "Algebraic Geometry", Liu's "Algebraic Geometry and Arithmetic Curves", Silverman's "The Arithmetic of Elliptic Curves", and some chapters of Mumford's "Abelian Varieties". I would like to learn more advanced arithmetic, and I began reading Faltings's chapter, which shows the Faltings's theorem (Mordell conjecture), in Cornell, Silverman's "Arithmetic Geometry". But this chapter needs moduli stacks for curves and abelian varieties. I've heard that Harris, Morrison's "Moduli of Curves" is a good reference. But glancing through some sections, I think this book seems to concern only curves over $\mathbb{C}$. So the first question is: is this book enough for my interesting? That is, do arithmetic geometers use moduli of curves only over $\mathbb{C}$? If not, please suggest to me good references. And I've heard that Mumford's GIT is good, for moduli of abelian varieties. Although these books discusses only moduli schemes, I would like to learn moduli stacks. So the second question is: please suggest to me some references for moduli stacks. And is it a good way to learn moduli stacks, not learning moduli schemes? Finally, I'm completely beginner of this field, but I intuitively think: for a functor $\mathscr{F}$ that takes a scheme $S$ to the set of isomorphism classes of some objects (e.g., smooth projective curves of given genus, or principally polarized abelian varieties of given dimension) over $S$, the fine moduli is the representable scheme of $\mathscr{F}$ (in general, there does not exist), and the coarse moduli is the "almost" representable scheme of $\mathscr{F}$, and the moduli stack is the representable object of $\mathscr{F}$, in some extended category. Is this idea wrong? Thank you very much! REPLY [4 votes]: A very nice introduction to moduli (stack) of elliptic curves is R. Hain - Lectures on Moduli Space of Elliptic Curves (you find it on arXiv): it is completed by 100 and more exercises; a brief appendix is devoted to stacks (I suggest to integrate it with Vistoli's lecture notes on stacks); it doesn't introduce the elliptic curves, therefore you must know it (but this is not the case of the OP).<|endoftext|> TITLE: Maximality of normalizer of $p$-Sylow groups in the symmetric group $S_{p}$ QUESTION [10 upvotes]: Consider the symmetric group $S_{p}$ where $p$ is a prime, then its $p$-Sylow subgroups are isomorphic to the cyclic group $C_{p}$. And it is clear that the normalizer of this cyclic group in the symmetric group is $N_{S_{p}}(C_{p})\cong C_{p}\rtimes F_{p}^{\times}$. My question is about how to show that this normalizer is a maximal subgroup of this symmetric group. Is there a simple group theoretic argument? REPLY [9 votes]: It was proved by Burnside that a permutation group of prime degree must either be solvable, in which case it has a normal Sylow $p$-subgroup and is contained in the affine group, or it is $2$-transitive. See here for a recent straightforward proof of this. The classification of the finite $2$-transitive groups was completed as an application of the classification of finite simple groups (see the wikipedia page or the discussion here for references) and they are listed for example in Section 7.7 of Dixon and Mortimer's book on permutation groups. A look through the list shows that the only examples of prime degree are affine groups, $A_p$, $S_p$, groups $G$ with ${\rm PSL}(d,q) \le G \le {\rm P \Gamma L}(d,q)$ when $(q^d-1)/(q-1)$ is prime, and a few small examples (${\rm PSL}(2,7)$ and ${\rm PSL}(2,11)$ of degrees $7$ and $11$, $M_{11}$, and $M_{23}$). When $(q^d-1)/(q-1)$ is prime $p$, an element of order $p$ in ${\rm PSL}(d,q)$ is a Singer cycle, and its normalizer in ${\rm P \Gamma L}(d,q)$ has order at most $pde$, where $q=r^e$ with $r$ prime, which is smaller than $p(p-1)$. The small examples are all contained in the alternating groups. So the affine group of order $p(p-1)$ is maximal in $S_p$ whenever $p>3$.<|endoftext|> TITLE: Does every manifold admit a Lagrangian Riemannian metric? QUESTION [9 upvotes]: Let $(M,g)$ be a Riemannian manifold. The $LC$ connection associated to the metric gives an $n$ dimensional distribution $D$ for $TM$. Let $\omega$ be the symplectic structure of $TM$ which is obtained by pulling back of the standard structure of the cotangent bundle via the isomorphism between the tangent and cotangent bundle. We say a Riemannian metric is a Lagrangian metric if the distribution $D$ of $TM$ is a Lagrangian distribution. Does every manifold admit a Lagrangian metric? REPLY [12 votes]: The distribution of horizontal subspaces is always Lagrangian not only for Riemannian metrics, but also for the Ehresmann connection associated to Finsler metrics and for the Levi-Civita connection of pseudo-Riemannian metrics. It is not always Lagrangian for general sprays though. In the Riemannian case this is classical (see Klingenberg's book on Riemannian Geometry or Paternain's book on geodesic flows), but you can also see my answer to this question: Intuition for Levi-Civita connection via Hamiltonian flows There is a much more elementary and geometric version of the construction of the connection given in my answer to When is a flow geodesic and how to construct the connection from it . You can complete it by proving (it's really easy!) that the harmonic conjugate---as defined in the answer---of three Lagrangian subspaces is again a Lagrangian subspace.<|endoftext|> TITLE: Existence of certain endomorphism of supersingular elliptic curve QUESTION [9 upvotes]: Let $E$ be a supersingular elliptic curve over an algebraically closed field $K$ of characteristic $p$. Let $R = \operatorname{End}(E)$ be its ring of endomorphisms. Then, it is known $R \otimes_{\mathbb Z} \mathbb Q$ is an order in a quaternion algebra $D$. In particular, $D$ has a multiplicative norm function $N\colon D^* \rightarrow \mathbb Q$, which can be defined as $N(\phi) = \phi \hat \phi$, where $\hat\phi$ denotes the dual of the endomorphism $\phi$, or equivalently, by saying that $N(\phi)$ is the degree of an endomorphism $\phi \in R$. My question is: Does there exist an element $\phi \in R$ such that $N(\phi)$ is singly divisible by $p$, i.e. such that the $p$-adic valuation of $N(\phi)$ is 1? If $E$ is defined over $\mathbb F_p$, then Frobenius is an endomorphism, and has degree $p$, so that gives an affirmative answer, but some supersingular elliptic curves are only defined over $\mathbb F_{p^2}$. In that case, Frobenius is not an endomorphism of $E$, but an isogeny $E \rightarrow E^{(p)}$ where $E^{(p)}$ is the Frobenius twist of $E$. The question is equivalent to: Does there exists a separable isogeny $E^{(p)} \rightarrow E$? Also, I don't know much about quaternion algebras over $\mathbb Q$, but I have read that $D$ is ramified at $p$, and if I understand the consequences of that correctly, that implies that there exists an element $\phi$ of $D$ such that $N(\phi)$ has $p$-adic valuation equal to 1. If that's true, then, the question is whether such an element can be chosen to be in $R$, and not just in $D$. I have also read that $R$ is a maximal order in $D$, which seems like it should help, but I'm not sure how to make use of that fact in a non-commutative setting. REPLY [3 votes]: $\newcommand{\Z}{\mathbb{Z}}$ First, if you want to learn about quaternion algebras, their orders and their relation to supersingular elliptic curves, I suggest John Voight's book. The answer to your question is yes. To see this, you can use the fact that (because $R$ is maximal) $R/p^2R$ contains an element $\pi$ such that $\pi^2 = p$, and therefore satisfying $N(\pi)\equiv \pm p\bmod{p^2}$. Let $\phi\in R$ be such that $\phi\equiv \pi \bmod{p^2R}$, then the valuation at $p$ of $N(\phi)$ is $1$.<|endoftext|> TITLE: Complex Analytic Structure on Moduli Space of Stable Maps QUESTION [5 upvotes]: Suppose $(X,\omega,J)$ is a compact Kähler manifold, and $\beta\in H_2(X,\mathbb Z)$ is given. Then, we can form the space $\overline{\mathcal M}:=\overline{\mathcal M}_{0,0}(X,\beta)$ of stable maps $u:C\to X$ with $C$ a nodal curve of genus 0 and $u_*[C]=\beta$. Let $\mathcal M\subseteq\overline{\mathcal M}$ be the locus where the domain is nonsingular. I know how to show that $\mathcal M$ is a space with a smooth ($C^\infty$) Kuranishi atlas (i.e., roughly, it locally has the structure of the zero set of a smooth vector valued function on a smooth finite dimensional manifold modulo a finite group). Near the points of $\overline{\mathcal M}\setminus\mathcal M$, I don't know how to put a smooth structure (instead, what I know amounts to only a topological implicit atlas, in the sense of https://arxiv.org/abs/1309.2370). My question is the following: since I have assumed $X$ actually Kähler (and not simply symplectic with a compatible almost complex structure), can we show that $\mathcal M$ (or even better, $\overline{\mathcal M}$) has a complex analytic Kuranishi atlas? I would prefer approaches which stay within the realm of differential geometry (though for the case of projective $X$, I would be interested in hearing about algebro-geometric approaches as well). Any references would also be appreciated. REPLY [2 votes]: There's also a really beautiful approach by Salamon-Robbin-Ruan for integrable Js in their paper "The moduli space of regular stable maps" https://people.math.ethz.ch/~salamon/PREPRINTS/smUW.pdf building on the earlier paper by Salamon-Robbin "Construction of the Deligne-Mumford orbifold": https://people.math.ethz.ch/~salamon/PREPRINTS/dmETH.pdf They gives a smooth structure that extends across the boundary in an extremely natural way (I don't remember if they explicitly construct an analytic structure, but because they are working in the case of integrable J, I think it should be possible to extract it from their proof). The idea is the following. Consider the space of holomorphic maps $S^2\to X$. Since a holomorphic map on the sphere is determined by its values on the equator, you can think of this as a submanifold of the space of maps from the circle to $X$. In fact, it's precisely the intersection of the (infinite-dimensional) subspace of maps $S^1\to X$ which extend over the northern hemisphere with the (infinite-dimensional) subspace of maps which extend over the southern hemisphere (this is the kind of thing people mean when they say that Floer theory is the theory of intersections of semi-infinite cycles). Now when your holomorphic map develops a bubble, this bubble will (generically) avoid the equator. So if you think of your moduli space as a space of maps $S^1\to X$ then you don't even notice bubbling. I think their original hope was to use this for nonintegrable Js as well, and overcome all issues with smooth structures at the boundary of moduli spaces. The problem, I believe, is that they need to use Fourier theory and Hardy spaces to set things up in a way they can handle analytically. Why does Fourier theory show up? A simple heuristic: a map from the circle to $\mathbb{C}$ can be thought of as a power series in $z$ and $z^{-1}$ infinite in both directions; those which extend over the unit disc are those where the negative powers of $z$ are absent; those which extend over the north hemisphere of the Riemann sphere are those where the positive powers of $z$ are absent; the only holomorphic maps from $S^2$ to $\mathbb{C}$ are therefore the constants. The coefficients of $z^n$ can be thought of as Fourier coefficients. This kind of description of holomorphic maps as power series only makes sense in the integrable setting.<|endoftext|> TITLE: Roots and relation between polynomials and their derivatives QUESTION [10 upvotes]: This is probably easy but it might be interesting. Here goes $\dots$ Let $P\in\mathbb{R}[x]$ be a polynomial of degree $n>2$ and $P'=\frac{dP}{dx}$. If $x_1, x_2, \dots, x_n$ are the roots of $P(x)$, including multiplicities, consider the multi-variable expression $$V_n(P)=\sum_{1\leq i TITLE: What is the consistency strength of weak Vopenka's principle? QUESTION [13 upvotes]: Weak Vopěnka's principle says that the opposite of the category of ordinals cannot be fully embedded in any locally presentable category. Recall that one form of Vopěnka's principle says that the category of ordinals cannot be fully embedded in any locally presentable category. Adámek and Rosický show that weak Vopěnka's principle follows from Vopěnka's principle and implies the existence of a proper class of measurables, giving a rough indication of its consistency strength. Its interest lies in its equivalence to the principle that any full subcategory of a locally presentable category closed under limits is reflective, or its equivalence to the principle that every orthogonal subcategory of a locally presentable category is reflective. But these results date from Adámek and Rosický's 1994 book Locally Presentable and Accessible Categories. Question: Is a more precise understanding of weak Vopěnka's principle available today? In particular, is its consistency strength well-calibrated? REPLY [14 votes]: The weak Vopěnka principle (WVP) and semi-weak Vopěnka principle (SWVP) are both equivalent to the large cardinal principle "Ord is Woodin". These results now appear in a paper on the arXiv. I kept the proof that Ord is Woodin implies SWVP as part of this MathOverflow answer (see below), but removed the outline of the proof that WVP implies Ord is Woodin because the paper contains a better proof of that now. I will use the form of WVP which says that the opposite of the category of ordinals cannot be fully embedded in the category of graphs. In other words, there is no sequence of graphs $(G_\alpha : \alpha < \mathrm{Ord})$ such that for all $\alpha < \beta$ there is no homomorphism from $G_\alpha$ to $G_\beta$, and for all $\alpha \le \beta$ there is a unique homomorphism from $G_\beta$ to $G_\alpha$. SWVP (see Adámek and Rosický, On Injectivity in Locally Presentable Categories) says that there is no sequence of graphs $(G_\alpha : \alpha < \mathrm{Ord})$ such that for all $\alpha < \beta$ there is no homomorphism from $G_\alpha$ to $G_\beta$, and for all $\alpha \le \beta$ there is at least one homomorphism from $G_\beta$ to $G_\alpha$. For every similarity type consisting of finitary relation symbols, the category of all structures of this type can be fully embedded into the category of graphs (see Hedrlín and Pultr, On full embeddings of categories of algebras,) so replacing graphs by more general relational structures yields equivalent statements of WVP and SWVP. "Ord is Woodin" means that for every class $A$ there is an $A$-strong cardinal. See Kanamori, The Higher Infinite for the definition of $A$-strong cardinal. For maximum generality, we work in GB + AC. (As a special case, the results hold in ZFC for definable classes $A$.) Proof that Ord is Woodin implies SWVP. Assume Ord is Woodin and let $(G_\alpha : \alpha \in \mathrm{Ord})$ be a sequence of graphs such that for all $\alpha \le \beta$ there is at least one homomorphism from $G_\beta$ to $G_\alpha$. We will show that for some ordinal $\kappa$ there is a homomorphism from $G_\kappa$ to $G_{\kappa+1}$. We can write $G_\alpha = G(\alpha)$ where $G$ is a class function with domain Ord. Because Ord is Woodin, some cardinal $\kappa$ is $G$-strong. Take a cardinal $\beta$ greater than $\kappa$ and greater than the von Neumann rank of the graph $G(\kappa+1)$, meaning $G(\kappa+1) \in V_\beta$. Because $\kappa$ is $\beta$-$G$-strong, there is a transitive class $M$ containing $V_\beta$ and an elementary embedding $j : (V;\in) \to (M;\in)$ such that the critical point of $j$ is $\kappa$ and $V_\beta \subset M$ and $j(\kappa) > \beta$ and $j(G) \restriction \beta = G \restriction \beta$. We describe the latter condition by saying that $j$ coheres with $G$. Because $j$ coheres with $G$, the graph $G(\kappa+1)$ is equal to $j(G)(\kappa+1)$, so it is on the sequence $j(G)$, which is an Ord-sequence of graphs in $M$. Note that $j(G(\kappa)) = j(G)(j(\kappa))$ by the elementarity of $j$, so the graph $j(G(\kappa))$ is also on the sequence $j(G)$. We have $j(\kappa) > \kappa+1$, so by our assumption on the sequence $G$ and the elementarity of $j$, we have a homomorphism $j(G(\kappa)) \to G(\kappa+1)$ in $M$ and therefore in $V$. We also have a homomorphism $G(\kappa) \to j(G(\kappa))$ in $V$ given by $j \restriction G(\kappa)$. Composing these, we obtain a homomorphism $G(\kappa) \to G(\kappa+1)$, as desired.<|endoftext|> TITLE: Derived Morita equivalence of associative algebras QUESTION [5 upvotes]: An associative algebra $A$ is said to be Morita equivalent to another one $B$ if there is an equivalence $$\mathsf{Mod}_A\simeq \mathsf{Mod}_B$$ between its corresponding abelian categories of modules. Moreover, whenever $A$ and $B$ are commutative, they are Morita equivalent iff they are isomorphic. On the other hand, we say that $A$ is derived Morita equivalent to $B$ if there is an equivalence $$D(A)\simeq D(B)$$ between its triangulated derived categories. My question is: Q. If $A$ is commutative and derived Morita equivelent to $B$, is $A$ Morita equivalent to $B$? REPLY [5 votes]: This seems to be true by https://arxiv.org/pdf/math/9810134.pdf , theorem 2.7.<|endoftext|> TITLE: Tools for constructing homeomorphisms between 4-manifolds QUESTION [6 upvotes]: (I am a complete amateur in topology, so this is a question out of curiosity.) The question was inspired by this post Fake versus Exotic . What methods can, realistically, be used to construct a homeomorphism between, for example, $\mathbb{C}\mathbb{P}^2\#9\overline{\mathbb{C}\mathbb{P}^2}$ and the Dolgachev surface $E(1)_{2,3}$? In the $Diff$ category there is the Kirby calculus which seems to be efficient enough. (Some very nontrivial diffeomorphisms were found using it, e.g. by Gompf.) My question is, essentially, how things are with $Top$ in comparison. Also, a more specific question: is it possible to use for this purpose the old result of Wall that for simply connected h-cobordant 4-manifolds $M$ and $N$ there is $k$ such that $M\#k(\mathbb{S}^2\times \mathbb{S}^2)\cong N\#k(\mathbb{S}^2\times \mathbb{S}^2)$ (a stable diffeomorphism)? Simply adding $\mathbb{S}^2\times \mathbb{S}^2$ is too bold a move in this case, but maybe it can be useful if done carefully. To make it more clear, I mean $M\#k(\mathbb{S}^2\times \mathbb{S}^2)$ with some additional structure. REPLY [5 votes]: Sadly, it seems that you need pretty much the full strength of Freedman's disc embedding theorem to construct such homeomorphisms. Wall's theorem builds an h-cobordism, essentially starting with the stabilization you mention, and then regluing by a diffeomorphism to get handles to algebraically cancel. But to get them to geometrically cancel, you need to do isotopies guided by Whitney disks. In some circumstances, you can see explicit Casson handles where those disks should go. But you still have to hit those with Freedman's theorem to get the homeomorphism. At some level, Freedman's theorem is based on several complicated limiting arguments (`Bing topology') that produce homeomorphisms rather than diffeomorphisms. It is interesting to compare this with the situation in higher dimensions. Siebenmann's article, Topological manifolds, in the Proceedings of the 1970 ICM (you'll find a pdf readily online) comes as close as one might hope to explaining how this works for a particular homeomorphism that is not isotopic to a PL homeomorphism. Again, a limit is taken that produces homeomorphisms that fail to be PL (and hence smooth). It would be great to see something so explicit in the 4-dimensional case.<|endoftext|> TITLE: Conjectured combinatorial non-equality QUESTION [7 upvotes]: Let $n,k,\ell$ be integers for which $0\leq k<\ell \leq n-6$. For a fixed $n$, think of $k,\ell$ as being allowed to vary. I believe the values $$(n-k-5)(k+1)(k+2)\binom n{k+3}~~~\text{and}~~~(n-\ell-5)(\ell+1)(\ell+2)\binom n{\ell+3}$$ are not equal. A proof they are not equal is the goal, but insight as to why they might not be equal would still be appreciated. Motivation: The integer $(n-k-5)(k+1)(k+2)\binom n{k+3}$, when divided by $2(n-2)$, is the degree of the complex irreducible character of the symmetric group $S_n$ corresponding to the partition $(n-k-3,3,1^k)$. I am a group theorist working on a general conjecture on character degrees; these character degrees if they are distinct will prove my conjecture to be true in the case of both the symmetric and the alternating groups. My work effort: $\bullet$ I ran computer tests for $6\leq n\leq 7000$. For those $n$, the integers are distinct (modulo my ability to write code, anyway). $\bullet$ I thought maybe the set of prime divisors could be used to differentiate the values for distinct $k,\ell$, but there are a large number of values for $n$ where this thought failed. REPLY [9 votes]: It looks like we may simply say which of them is greater. Denoting $k+3=t$ and $f(t)=(n-k-5)(k+1)(k+2)\binom n{k+3}=(n-t-2)(t-1)(t-2)\binom n{t}$ we get $$\frac{f(t+1)}{f(t)}= \frac{t (n - t - 3) (n - t)}{(t - 2) (t + 1) (n - t - 2)}=\frac{n-t-1-\frac{2}{n-t-2}}{t-1-\frac2{t}}. $$ If $t1 $$ $$ \frac{f(n-t+1)}{f(t)}=1- \frac{2 (n - 2) (n - 2 t + 1)}{(t - 2) (t - 1) (n - t - 2) (n - t + 1)}<1. $$ Thus by monotonicity we should have $n-t TITLE: General distributions with the "transportation-cost inequality" property to piece log-concave distributions QUESTION [5 upvotes]: It is now known [Otto et Villani 2000; Cordero et al 2006; etc.] that on an $n$-dimensional smooth Riemannian manifold $X$ and a probability measure $\mu$ on $X$ with density $d\mu \propto e^{-V}dvol$ satisfying the curvature condition $$ \operatorname{Hess}_x(V) + \operatorname{Ric}_x \succeq (1/c) I_n,\forall x \in X, $$ the transportation-cost inequality $$W_2(\nu,\mu) \le \sqrt{2cH(\nu\|\mu)} $$ holds for all other distributions $\nu$ on $X$ absolutely continuous w.r.t $\mu$. Here $W_2$ is the Wasserstein-2 distance induced by the geodesic metric on $X$. Question (A) What are the most general conditions under which the above transportation-cost inequality holds. (B) Can the curvature condition be relaxed to "piecewise" version. That is, what if we instead assume a convex mixture $d\mu = \sum_{i=1}^k \pi_i d\mu_i$ where each $d\mu_i$ is log-concave and satisfies the curvature condition on some piece $X_i$ of $X$ ? (B') Are there concentration inequalities for mixtures of Gaussians ? Partial answer Log-concave distributions satisfying the Bakry-Eméry $\operatorname{CD}(n,\infty)$ curvature condition on manifolds Distributions with densities on compact homogeneous Riemannian manifolds. See this paper of Rothaus. Distributions which can be realized as pushforwards of distributions with some $\text{T}_2(c)$, under Lipschitz maps. If $\mu$ has $\text{T}_2(c)$ property and $\varphi: X \rightarrow Y$ is $L$-Lipschitz, then $\varphi_\#\mu$ has $\text{T}_2(L^2c)$. Finite tensor product $\mu_1 \otimes \mu_2 \otimes \ldots \otimes \mu_k$ of distributions having $\text{T}_2(c)$ also has $\text{T}_2(c)$. REPLY [2 votes]: (A) Nathael Gozlan proved that a distribution $\mu$ satisfies a $T_2$ inequality if and only if all finite tensor products of $\mu$ satisfy a subgaussian concentration property. (B) and (B') For finite mixtures you could certainly get something by viewing them as pushforwards of product measures. However, you would necessarily get something that depends badly on $k$, since any distribution on $\mathbb{R}^n$ --- including those with no good concentration properties --- can be approximated by a mixture of Gaussians.<|endoftext|> TITLE: What would $\mathcal{P} \neq \mathcal{NP}$ tell us about some non-constructive proofs? QUESTION [9 upvotes]: Let me sum up my - hopefully correct - understanding of the travelling salesman problem and complexity classes. It's about decision problems: "[...] a decision problem is a problem that can be posed as a yes-no question of the input values. Decision problems typically appear in mathematical questions of decidability, that is, the question of the existence of an effective method to determine the existence of some object." The travelling salesman problem (TSP) - as a decision problem - is to find an answer to the question: Given an $n \times n$ matrix $W = (w_{ij})$ with $w_{ij} \in \mathbb{Q}$ and a number $L\in \mathbb{Q}$. Is there a permutation $\pi$ of $\{1,\dots, n\}$ such that $$L(\pi) = \sum_{i=1}^{n} w_{\pi(i)\pi(i+1)} < L?$$ with modular addition, i.e. $n+1 = 1$ The answer can be given as a specific example (the output of a constructive "problem solver") which then can be checked for correctness. For TSP we know that a specific example given by a constructive problem solver (e.g. a specific permutation $\pi$) can be checked in polynomial time for $L(\pi) < L$, that means TSP $\in\mathcal{NP}$. But the answer may also be given by just a boolean value YES or NO , which cannot be checked at all. (What would we try to check?) The first kind of answer is given by algorithms that are programmed to read arbitary matrices $W$ and numbers $L$ and give an example $\pi$. These are equivalent to constructive proofs which somehow construct a $\pi$ from given $W$ and $L$, and which may be correct or not. The second kind of answer is given by non-construtive proofs - which nevertheless give an answer. Such a proof also "reads" some general $W$ and $L$ and makes some general considerations about them, e.g. like this: If numbers $x_1, \dots x_n$ can be calculated from $W$ and they relate to $L$ such that $f(x_1,\dots, x_n, L) = 0$ then the answer is YES otherwise NO. My question is: If some day it is proved that TSP $\not\in \mathcal{P}$ (because $\mathcal{P} \neq \mathcal{NP}$ and TSP is $\mathcal{NP}$-hard), what do we learn about hypothetical non-construtive proofs that for given $W$ and $L$ there exist solutions $\pi$ with $L(\pi) < L$ (YES or NO)? Or is the talk about such proofs only a chimera - because they are ill-defined or cannot exist for obvious reasons? Remark 1: Since proofs have no run-time, the things we can learn about them may concern only their length and/or complexity (in general: structure). Remark 2: Very short and simple algorithms may have exponential run-times. To think more specifically about this: Assume there is a proof that proves: If you calculate numbers $x_1(W),\dots, x_m(W)$ of a quadratic matrix $W$ and you find that if $f(x_1,\dots,x_m,L) = 0$ then there is a permutation $\pi$ with $L(\pi) < L$. What could be said about this (hypothetical!) proof, assuming that $\mathcal{P} \neq \mathcal{NP}$? REPLY [2 votes]: I think you may need to formulate your question more precisely. Consider the following "nonconstructive proof": If you do blah-blah-blah and the result is 1 then YES, but if the result is 0 then NO, where "blah-blah-blah," upon closer inspection, amounts to running a Turing machine that exhaustively tries all possible permutations, and outputs 1 if it finds a suitable permutation, but outputs 0 if it does not. This "proof" is pretty short since the Turing machine is rather simple. Secretly, of course, it "constructs" the permutation, but it keeps that information private and does not output it, so from the outside it looks nonconstructive. If this sort of thing counts as a "nonconstructive proof" then we don't need to wait for someone to resolve the P ≠ NP question. We have the proof today. If you don't want this sort of thing to count then you need to specify carefully what you do and do not allow. I conjecture that you may be interested in proof complexity, which is related to but different from computational complexity. But this is just a conjecture.<|endoftext|> TITLE: Transgression image and Serre spectral sequence for tori QUESTION [5 upvotes]: Let $\mathbb{K} \subset \mathbb{T}$ be two tori acting on a topological space $X$ (with all the properties you want). We use the notations $$X_{\mathbb{T}} := (X \times E \mathbb{T}) / \mathbb{T}, \quad X_{\mathbb{K}} := (X \times E \mathbb{K}) / \mathbb{K}$$ for the homotopy quotients. The space $E \mathbb{T}$ can be considered as $E \mathbb{K}$ if provided with the $\mathbb{K}$-action, and therefore there exists a fibration $$X_{\mathbb{K}} \overset{\mathbb{T} / \mathbb{K}}{\longrightarrow} X_{\mathbb{T}}.$$ The Serre spectral sequence associated with this fibration has $E_2$-term $$E_2^{pq} = H^p(X_{\mathbb{T}}) \otimes H^q(\mathbb{T} / \mathbb{K}),$$ and converges to $H^{p+q}(X_{\mathbb{K}})$. I am trying to understand what the transgression map is here, and more precisely what are the images of the generators of $H^*(\mathbb{T} / \mathbb{K})$ in $H^2(B \mathbb{T})$. Moreover, I read somewhere that this $E_2$-term is the Koszul complex of these images (of degree $2$) in the algebra $H^p(X_{\mathbb{T}})$. What does this mean ? Thanks a lot REPLY [2 votes]: The images of the generators for $H^1(\mathbb{T}/\mathbb{K})$ are in some sense Chern classes. The easiest case to make sense of is the codimension one case, i.e., when $\mathbb{T}/\mathbb{K}\cong U(1)$. In this case you can think of $X_{\mathbb{K}}$ as a principal $U(1)$-bundle over $X_\mathbb{T}$. Principal $U(1)$ bundles are classified by their first Chern class, which is an element of $H^2$ of the base, or in this case $H^2(X_{\mathbb{T}})$. The differential $d_2$ sends the generator for $E_2^{01}=H^0(X_{\mathbb{T}})\otimes H^1(U(1))$ to $c_1\otimes 1\in E_2^{20}=H^2(X_{\mathbb{T}})\otimes H^0(U(1))$. In the general case, fix an isomorphism from $\mathbb{T}/\mathbb{K}$ to $U(1)^m$ for some $m$. The picture is similar because a principal $U(1)^m$ bundle is naturally built from $m$ principal $U(1)$-bundles, and you have $m$ first Chern classes in $H^2(X_{\mathbb{T}})$ which are the images of the natural generators for $H^1(U(1)^m)$. I am on shakier ground here, but Koszul resolutions are something to do with homological algebra over commutative rings. View a field $k$ as a module for the polynomial ring $k[x_1,\ldots,x_m]$, with each $x_i$ acting as multiplication by zero, and grade everything so that each $x_i$ has degree two. If you want to make a free resolution for the module $k$, the most economical choice will involve taking a free module of rank $m\choose{j}$ in degree $j$. The differential in this resolution can be described by viewing the whole resolution as a bigraded algebra $\Lambda[y_1,\ldots,y_m]\otimes k[x_1,\ldots,x_m]$, where each $y_i$ has bidegree $(0,1)$ and $x_i$ has bidegree $(2,0)$, and $d(y_i)=x_i$. This bicomplex is identical to the one that you have in the case when $X$ is a single point (i.e., the case of the path-loop fibration over $U(1)^m$). I recommend Matsumura's book `Commutative Ring Theory' if you want to learn more about homological algebra over commutative rings.<|endoftext|> TITLE: Denominators of certain Laurent polynomials QUESTION [10 upvotes]: Consider the following somos-like sequence $$x_n=\frac{x_{n-1}^2+x_{n-2}^2}{x_{n-3}}.$$ It's known that $x_n$ is a Laurent polynomial in $x_0, x_1$ and $x_2$. I got interested in the denominators of the sequence $x_n$. Some initial observations indicate particular structures regarding the exponents of the denominators, so I am tempted to ask: Question: Is this true? $$x_n=P_n(x_0,x_1,x_2)x_0^{1-F_{n-3}}x_1^{1-F_{n-4}}x_2^{1-F_{n-5}},$$ where $P_n$ is some polynomial and $F_n$ is the Fibonacci sequence. REPLY [4 votes]: Suppose that $$\, x_n = c\,x_{n-1}x_{n-2} - x_{n-3} \tag{1}$$ for all $\, n \in \mathbb{Z}\,$ where $\, x_0, x_1, x_2 \,$ are variables and $\, c := (x_0^2 + x_1^2 + x_2^2) / (x_0 x_1 x_2). \,$ Then $$ x_n x_{n-3} = x_{n-1}^2 + x_{n-2}^2 \tag{2}$$ for all $\, n \in \mathbb{Z}.\,$ Define $\, P_n := x_n\, x_0^{e_n} x_1^{e_{n-1}} x_2^{e_{n-2}} \,$ for all $\, n \in \mathbb{Z} \,$ where $\, e_n := -1 + |F_n|. \,$ Notice that $\, x_{2-n}, P_{2-n} \,$ are the same as $\, x_n, P_n \,$ except that $\, x_0 \,$ and $\, x_2 \,$ are swapped. Now we get $$ P_n = (x_0^2 + x_1^2 + x_2^2)\, P_{n-1}P_{n-2} - \big(x_0^{2F_{n-2}} x_1^{2F_{n-3}} x_2^{2F_{n-4}}\big)\, P_{n-3} \tag{3}$$ for all $\, n>3 \,$ from equation $(1)$. Now $\, P_0 = P_1 = P_2 = 1, P_3 = x_1^2 + x_2^2 \,$ and the exponents in equation $(3)$ are non-negative, thus $\, P_n \,$ is a polynomial in $\, x_0,x_1,x_2 \,$ for all $\, n \in \mathbb{Z} \,$ by induction.<|endoftext|> TITLE: 2-Torsion in Jacobians of Curves Over Finite Fields QUESTION [6 upvotes]: Let $C$ be a (smooth, projective) curve over a finite field $\mathbb{F}_q$, and let $J_C(\mathbb{F}_q)$ denote its Jacobian. Suppose the genus $g$ of $C$ is at least $1$. Question 1: Are there curves $C$ for which $J_C(\mathbb{F}_q)$ is isomorphic, as a group, to $(\mathbb{Z}/2\mathbb{Z})^k$ for some $k$? Can one characterize all such curves? Question 2: What is known about upper bounds of the size of the $2$-torsion of $J_C(\mathbb{F}_q)$ compared to the size of $J_C(\mathbb{F}_q)$ itself? Is it necessarily true, say, that the 2-torsion is negligible if some parameter ($q$ or $g$) is large enough? REPLY [9 votes]: I think $y^2=x^9-x$ over $\mathbb{F}_3$ has $J_C(\mathbb{F}_3)$ isomorphic to $(\mathbb{Z}/2)^6$ but please check. The $2$-torsion in $J_C$ over the algebraic closure is $(\mathbb{Z}/2)^{2g}$ (or smaller in characteristic two). On the other hand, $\#J_C(\mathbb{F}_q) \ge (\sqrt{q} -1)^{2g}$, so for $q > 9$, the latter is bigger than the former (and usually much bigger). So the things you want can only exist for small $q$. I don't know a classification.<|endoftext|> TITLE: Del Pezzo surfaces and Picard-Lefschetz theory QUESTION [9 upvotes]: Let $X$ be a smooth compact del Pezzo surface. For instance, one can consider the most classical case of a cubic surface. It is well known that the Picard lattice of $X$ is related to a root system (in the case of a cubic surface the corresponding root system is $E_6$). In particular this relation manifests itself in the action of the Weyl group on $Pic(X)$ permuting the classes of exceptional curves. I suspect that one could see this action via Picard-Lefschetz theory. For that one could consider a moduli space of del Pezzo surfaces and the universal family of del Pezzo surfaces, fibered over it. Then the monodromy action on the middle cohomology of the fiber should be generated by orthogonal reflections, according to Picard-Lefschetz theory. I wonder how to make this picture precise. In particular, what moduli space one should consider here? REPLY [9 votes]: Indeed you can see it this way. This is my symplectic geometer's perspective on it (I blame Paul Seidel's Lecture notes on four-dimensional Dehn twists). Consider the $n$-point blow-up of $\mathbf{CP}^2$ at $n$ general points; you get a moduli space by varying the points and you get isomorphic varieties if the point configurations are related by the action of $PGL(3,\mathbf{C})$, so take as your moduli space the quotient of the space of general point configurations by this $PGL(3)$-action. This is not a fine moduli space because a surface can have automorphisms: if $n\geq 4$ then this automorphism group is finite (the $PGL(3)$-stabiliser of four general points is $S_4$), and if $n\geq 5$ then the automorphism group is generically trivial, so for simplicity let's focus on $n\geq 5$ and just throw away all the surfaces with automorphisms. You get a universal family over this moduli space; using a relatively ample bundle (suitable power of the anticanonical bundle) you get a map from the total space of the family to some fixed projective space. Pull back a Fubini-Study form to get a closed 2-form $\Omega$ on the universal family whose restriction to fibres is a symplectic form. Now you can do symplectic parallel transport along paths in the moduli space: you get a symplectic connection on the total space by taking the $\Omega$-orthogonal complement to the fibres; parallel transport maps preserve the symplectic form on fibres; contractible loops give Hamiltonian monodromies. You therefore get a map from $\pi_1$ of your moduli space to the symplectic mapping class group ($Symp(X)/Ham(X)$). You can further compose this with a map $Symp(X)/Ham(X)\to Aut(H^2(X))$ to the group of automorphisms of cohomology; the image of this latter map will be your Weyl group (because that is the automorphism group of the cohomology lattice preserving its intersection form). How to see the link with Picard-Lefschetz theory? As your $n$ points vary, there is a complex codimension one thing they can fail to be in general position: namely, one of them can pass through the complex line connecting two others. Other things can also happen: a sixth point can pass through the complex conic connecting five others, for example. When one of these things happen, you get a $-2$-sphere by taking the proper transform of the line/conic/whatever. Your relatively ample bundle fails to be ample for this blowup because the anticanonical class annihilates the class of a $-2$-sphere (by adjunction: $-K.C=C^2+2=0$) so the family of Del Pezzos develops a nodal singularity as this degeneration occurs (the minimal resolution of a nodal singularity has exceptional locus a single $-2$-sphere). In terms of symplectic geometry, there is a Lagrangian 2-sphere in the smooth Del Pezzo which is collapsed to the node if you follow the symplectic parallel transport; this is called the vanishing cycle. As this phenomenon is complex codimension one, there is a loop in the moduli space where your $n$ points skirt around this degenerate configuration. The monodromy around this loop is a symplectic Dehn twist in the Lagrangian sphere (as was observed by Arnold). The Dehn twist acts as a reflection in cohomology in the class of the $-2$-sphere: if you have a homology class disjoint from the Lagrangian sphere then it is unaffected by the twist, which is supported near the sphere; the homology class of the sphere gets reversed because the Dehn twist is the antipodal map on the sphere. This is precisely the Picard-Lefschetz formula. Note that the homology class of the Lagrangian sphere and the homology class of the holomorphic $-2$-sphere in the minimal resolution of the singular guy can be identified; I have an old blog-post explaining how this works using small resolutions: http://www.homepages.ucl.ac.uk/~ucahjde/blog/kronheimer-argument.html For a symplectic geometer, the more interesting fact is that you can go beyond the homological monodromy action: if you take the moduli space of $n$ ordered points then there is a universal family (no automorphisms when $n\geq 4$) and the homology action is trivial because the monodromy is generated by squared Dehn twists (when you go around one of the loops I discussed before, you switch two points, so to get back to the identity in homology you need to go around the loop twice) and Picard-Lefschetz tells you that a squared Dehn twist acts as the identity. Nonetheless, the monodromy gives a map from $\pi_1$ of the moduli space to the symplectic mapping class group, and Paul Seidel's early work showed that this is often injective (not only for Del Pezzos). You don't see anything at the level of ordinary smooth mapping class groups (the squared Dehn twist is smoothly isotopic to the identity) so symplectic geometry is remembering more about the algebraic geometry here. Like I said, I learned all of this from Seidel's Lectures on Dehn twists: well worth the read.<|endoftext|> TITLE: Any convergence rule for ${\mathbf X}_k={\mathbf A}{\mathbf X}_{k-1}{\mathbf B}$? QUESTION [6 upvotes]: We know iteration ${\mathbf X}_k=\mathbf{A}{\mathbf X}_{k-1}$ converges if the spectral radius of $\mathbf A$ is smaller than 1 (see here). Is there any known rule for iteration ${\mathbf X}_k={\mathbf A}{\mathbf X}_{k-1}{\mathbf B}$ to converge? Any reference is helpful. Thanks! Here ${\mathbf X}_k$ is a matrix, for example ${\mathbf X}_k$ is $N\times n$, $\bf A$ is $N\times N$ and $\bf B$ is $n \times n$. REPLY [4 votes]: For every square matrix $C$, let $r(C)$ denote its spectral value. We say that a complex number $\lambda$ is a dominant eigenvalue of $C$ if $\lambda$ is the only eigenvalue of $C$ with modulus $r(C)$. a semisimple eigenvalue of $C$ if it is an eigenvalue of $C$ and its algebraic multiplicity coincides with its geometric multiplicity. Theorem. The following assertions are equivalent: (i) The sequence $(X_K)$ converges for every $X_0$. (ii) We have either $r(A)r(B) < 1$, or we have $r(A)r(B) = 1$ and the matrices $A$ and $B$ both have dominant eigenvalues which are semisimple and whose product equals $1$. Proof. "(ii) $\Rightarrow$ (i)" First, let $r(A)r(B) < 1$. By multiplying $A$ and $B$ with appropriate positive numbers that are inverse to each other we may assume that $r(A) < 1$ and $r(B) < 1$. Hence, $X_k = A^kX_0 B^k \to 0$ as $k \to \infty$. If instead $r(A)r(B) = 1$ we may assume without loss of generality that $r(A) = r(B) = 1$. Let $\lambda$ denote the dominant eigenvalue of $A$; then $\lambda$ has modulus $1$ and $\overline{\lambda}$ is the dominant eigenvalue of $B$. By replacing $A$ with $\overline{\lambda}A$ and $B$ with $\lambda B$ we may assume that $A$ and $B$ have $1$ as a dominant eigenvalue and that this eigenvalue is semisimple. Hence, $A^k$ and $B^k$ are both convergent as $k \to \infty$, which implies that $X_k$ is convergent, too. "(i) $\Rightarrow$ (ii)" Assume that (i) holds and that $r(A)r(B) \ge 1$. We have to prove that the second alternative in (ii) holds. Let $\lambda$ and $\mu$ denote eigenvalues of $A$ and $B$ of maximal modulus; let $v$ denote an eigenvector for $\lambda$ of $A$ and let $w$ denote an eigenvector for $\mu$ of the transposed matrix $B^T$ of $B$. If we choose $X_0 = v w^T$, then $X_k = A^k v w^T B^k = (\lambda\mu)^k v w^T$. Since this sequence is convergent and $\lvert\lambda \mu\rvert = r(A) r(B) \ge 1$, it follows that $\lambda \mu = 1$. Since $\lambda$ and $\mu$ were arbitrary eigenvalues of maximal modulus (of matrix $A$ and $B$, respectively), it follows that each of the matrices $A$ and $B$ can have only one eigenvalue of maximal modulus, which is thus a dominant eigenvalue of this matrix. Now, let $\lambda$ and $\mu$ denote the dominant eigenvalues of $A$ and $B$; we have already seen that $\lambda \mu = 1$. If one of those eigenvalues was not semisimple, than we could find a generalised eigenvector of rank $\ge 2$ for it (for $A$ or $B^T$, respectively), and a similar argument as above would show that $(X_k)$ cannot be bounded if $X_0$ is chosen appropriately. Remark. The above argument also shows that $X_k \to 0$ for every $X_0$ if and only if $r(A)r(B) < 1$. This has already been observed by Suvrit in the comments.<|endoftext|> TITLE: Is every polytope combinatorially equivalent to the intersection of a simplex and a linear subspace? QUESTION [7 upvotes]: I wonder whether such a result is known, and if so, whether the proof is trivial. By polytope I mean the convex hull of finitely many points in $\Bbb R^n$. Assume the simplex to be symmetric and centered at the origin, so that the subspace goes through its center. The subspace can have any dimension $k\in\{0,...,n\}$. Question: Is every polytope combinatorially equivalent to the intersection of a simplex and a linear subspace through its center? If this is true, are there estimations for the dimension of the simplex needed? REPLY [10 votes]: The answer is yes. The fact that any polytope is affinely equivalent to a section of a simplex is well-known (see the answer by Tobias Fritz). Now in any simplex with vertices $(v_i)$ we may consider projective transformations via reweighting barycentric coordinates: given positive numbers $(a_i)$, such a transformation is given by $$ \sum \lambda_i v_i \mapsto \frac{\sum a_i \lambda_i v_i}{\sum a_i \lambda_i}.$$ Projective transformations act transitively on the interior of the simplex, so it follows that any polytope is combinatorially equivalent to a section of some regular simplex through its center. What I don't know is whether every polytope is affinely equivalent to a section of some regular simplex through its center.<|endoftext|> TITLE: Criterion for a Banach algebra to be finite dimensional QUESTION [14 upvotes]: Let $A$ be a Banach algebra (say, complex and unital) and suppose that every (closed) commutative subalgebra of $A$ is finite dimensional. Question. Does it follow that $A$ is finite dimensional? Remark. Clearly, every element of $A$ is algebraic (i.e. annihilated by a polynomial) and thus has finite spectrum. If $A$ is semisimple, it therefore follows from a result of Kaplansky that $A$ is finite dimensional (Lemma 7 in "I. Kaplansky: Ring Isomorphisms of Banach Algebras (1954)". So the question is concerned with the non-semi-simple case. REPLY [13 votes]: I think it's true by Dixon's theorem (JLMS 1974) which says (on the third page) that any Banach algebra consisting only of algebraic elements is nilpotent-by-finite. Thanks to this, we may assume $A$ is nilpotent. We moreover assume $A$ is infinite-dimensional and will construct an infinite-dimensional commutative subalgebra. Take the largest $n$ such that the ideal $B := \overline{\rm span}\, A^n$ is infinite-dimensional. For $x\in B$, we denote its left and right multiplication operator on $B$ by $L_x$ and $R_x$. These have finite rank. Now, start with a nonzero $b_1 \in B$ and recursively choose nonzero $b_k \in \bigcap_{i=1}^{k-1}(\ker L_{b_i} \cap \ker R_{b_i})$ that is linearly independent of $\{ b_i : i < k\}$. Then, the algebra generated by $\{ b_1,b_2,\ldots\}$ is commutative and infinite dimensional.<|endoftext|> TITLE: Is each compactification of $\mathbb N$ soft? QUESTION [5 upvotes]: Definition. A compactification $c\mathbb N$ of the countable discrete space $\mathbb N$ is defined to be soft if for any disjoint sets $A,B\subset\mathbb N\subset c\mathbb N$ with $\bar A\cap\bar B\ne\emptyset$ there exists a homeomorphism $h$ of $c\mathbb N$ such that $h(A)\cap B$ is infinite and $h(x)=x$ for all $x\in c\mathbb N\setminus\mathbb N$. Problem. Is each compactification of a countable discrete space soft? Remark. The Stone-Cech compactification $\beta\mathbb N$ of $\mathbb N$ is soft as no disjoint sets $A,B\subset\mathbb N$ have $\bar A\cap\bar B\ne\emptyset$. On the other hand, a compactification $c\mathbb N$ is soft if the space $c\mathbb N$ is Frechet-Urysohn or has sequential square $c\mathbb N\times c\mathbb N$. So, a counterexample if exists should be rather exotic. REPLY [10 votes]: Let $A=\{0,2,4,\dots\}$ be the even numbers and let $B=\{1,3,5,\dots\}$ be the odd numbers. Topologize $A\cup \beta B$ so that $A$ is a sequence limiting to a unique point in $\beta B \setminus B $. This is a compactification of $\mathbb{N}$ that fails to be soft, since any homomorphism of the required form would give a non-trivial sequence in $\beta B$ limiting to a unique point, but no such sequence exists.<|endoftext|> TITLE: Is $\beta\mathbb N$ a unique compactification with the smallest possible permutation group? QUESTION [8 upvotes]: For a compactification $c\mathbb N$ of $\mathbb N$ let $\mathcal H(c\mathbb N,\mathbb N)$ be the group of homeomorphisms $h:c\mathbb N\to c\mathbb N$ such that $h(x)=x$ for all $x\in c\mathbb N\setminus\mathbb N$. The group $\mathcal H(c\mathbb N,\mathbb N)$ determines the subgroup $$S_{\mathbb N,c\mathbb N}:=\{h{\restriction}\mathbb N:h\in\mathcal H(c\mathbb N,\mathbb N)\}$$ in the permutation group $S_{\mathbb N}$ of $\mathbb N$. Observe that for the one-point compactification $\alpha\mathbb N$ of $\mathbb N$ the group $S_{\mathbb N,\alpha\mathbb N}$ coincides with the whole group $S_{\mathbb N}$ whereas for the Stone-Cech compactification $\beta\mathbb N$ of $\mathbb N$ the group $S_{\mathbb N,\beta\mathbb N}$ coincides with the group $S_{<\mathbb N}$ of finitely supported permutations of $\mathbb N$ (i.e., permutations that move only finitely many points of $\mathbb N$). For any compactification $c\mathbb N$ of $\mathbb N$ we have $$S_{<\mathbb N}=S_{\mathbb N,\beta\mathbb N}\subset S_{\mathbb N,c\mathbb N}\subset S_{\mathbb N,\alpha\mathbb N}=S_{\mathbb N}$$so $S_{\mathbb N,c\mathbb N}$ is intermediate between $S_{\mathbb N,\beta\mathbb N}$ and $S_{\mathbb N,\alpha\mathbb N}$. It is easy to prove that the one-point compactification of $\mathbb N$ is the unique compactification $c\mathbb N$ of $\mathbb N$ with $S_{\mathbb N,c\mathbb N}=S_{\mathbb N}$. Question. Is the Stone-Cech compactification a unique compactification of $\mathbb N$ with $S_{\mathbb N,c\mathbb N}=S_{<\mathbb N}$? Remark. Negative answer to this question will imply negative answer to this MO-problem. REPLY [4 votes]: Analyzing the answer of @James Hanson to my preceding question, I realized that this question also has a simple negative answer: the quotient space $c\mathbb N:=\beta\mathbb N/\{p,q\}$ of $\beta\mathbb N$ by any doubleton $\{p,q\}\subset\beta\mathbb N\setminus\mathbb N$ is not homeomorphic to $\beta\mathbb N$ but has the smallest possible permutation group $S_{\mathbb N,c\mathbb N}=S_{<\mathbb N}$. This compactification $c\mathbb N$ also is not soft (according to this definition).<|endoftext|> TITLE: A new cardinal characteristic of the continuum? QUESTION [7 upvotes]: Let $\kappa$ be the smallest cardinality of a family $\mathcal F$ of subsets of $\omega$ such that for any bijective function $f:A\to B$ between disjoint infinite subsets of $\omega$ there exists a set $F\in\mathcal F$ such that the set $\{x\in A\cap F:f(x)\notin F\}$ is infinite. It can be shown that $\mathfrak s\le \kappa\le\mathfrak c$, where $\mathfrak s$ is the splitting number. Question. Is it consistent that $\kappa<\mathfrak c$? Or $\kappa=\mathfrak c$ in ZFC? Remark. The affirmative answer to the first part of this question will imply a consistent negative answer to this MO-problem. REPLY [8 votes]: Recall that $\mathbf{non}(\mathcal{B})$ is the least cardinality of a non-meager subset of $\mathbb{R}$; the choice of presentation of $\mathbb{R}$ does not matter for this, so we take $\mathbb{R} = \mathcal{P}(\omega)$. It is well-known that consistently, $\mathbf{non}(\mathcal{B}) < \mathfrak{c}$; indeed, start from a model of CH and add $\aleph_2$-many Cohen reals. Theorem. $\kappa \leq \mathbf{non}(\mathcal{B})$. Proof. Suppose $\mathcal{F} \subseteq \mathcal{P}(\omega)$ is nonmeager. It suffices to show that $\mathcal{F}$ works for the definition of $\kappa$. Let $A, B$ be disjoint infinite subsets of $\omega$. Let $X$ be the set of all $F \subseteq \omega$ such that there are infinitely many $n \in A \cap F$ with $f(n) \not \in F$. I claim that $X$ is comeager. Indeed, for each finite $J \subseteq A$, let $X_J$ be the set of all $F \subseteq \omega$ such that there is some $n \in A \backslash J$ with $n \in A \cap F$ and $f(n) \not \in F$. Each $X_J$ is open dense, and $X = \bigcap_J X_J$. Thus, consistently $\kappa < \mathfrak{c}$. Incidentally, the following is perhaps a cleaner formulation of $\kappa$: Claim. Let $\lambda$ be the least cardinality of a family $\mathcal{G}$ of subsets of $\omega$, such that for each increasing sequence $(x_n: n < \omega)$, there is some $F \in \mathcal{G}$ such that for infinitely many even $n < \omega$, $x_{n} \in F$ but $x_{n+1} \not \in F$. Then $\kappa = \lambda$. Proof. It is straightforward to check that $\lambda \leq \kappa$: given $\mathcal{F}$ witnessing the definition of $\kappa$ and $(x_n: n < \omega)$, write $A = \{x_{2n}: n < \omega\}$, $B = \{x_{2n+1}: n < \omega\}$ and let $f: A \to B$ be $x_{2n} \mapsto x_{2n+1}$. So we show $\kappa \leq \lambda$. Let $\mathcal{G}$ witness the definition of $\lambda$; let $\mathcal{F} = \mathcal{G} \cup \{\omega \backslash S: S \in \mathcal{G}\}$. We show that $\mathcal{F}$ is as in the definition of $\kappa$. Suppose $A, B$ are infinite disjoint sets and $f: A \to B$ is a bijection. Note that since $\mathcal{F}$ is closed under complements, we are free to replace $(A, B, f)$ by $(B, A, f^{-1})$, if desired. Choose $A' \subseteq A$ infinite such that either for all $x \in A'$, $f(x) > x$, or else for all $x \in A'$, $f(x) < x$. After possibly interchanging $A$ and $B$, we can suppose that for all $x \in A'$, $f(x) > x$. Write $B' = f[A']$. Define an increasing sequence $(x_n: n < \omega)$ inductively so that for all $n$, $x_{2n} \in A'$ and $x_{2n+1} = f(x_{2n}) \in B'$. Namely, let $x_0 = \min(A')$, let $x_1 = f(x_0)$, and having defined $x_{2n-1}$, let $x_{2n}$ be the least element of $A'$ bigger than $x_{2n-1}$, and let $x_{2n+1} = f(x_{2n})$. Choose $F \in \mathcal{F}$ such that there are infinitely many even $n < \omega$ with $x_n \in F$ but $x_{n+1} \not \in F$. Then for each such $n$, $x_n \in A \cap F$ but $f(x_n) = x_{n+1} \not \in F$, so $F$ is as desired. Remark. For each infinite $I \subseteq \omega \times \omega$, we get a variant notion $\kappa_I$ of $\kappa$, namely $\kappa_I$ is the least cardinality of a family $\mathcal{G}$ of subsets of $\omega$, such that for each increasing sequence $(x_n: n < \omega)$, there is some $F \in \mathcal{G}$ such that for infinitely many $(n, m) \in I$, $x_n \in F$ but $x_m \not \in F$ (possibly $\kappa_I = \infty$). It is easy to check that $\kappa = \kappa_{\{(2n, 2n+1): n < \omega\}}$ and $\mathfrak{s} = \kappa_{\{(n, n+1): n < \omega\}}$. The above proof shows that each $\kappa_I \leq \mathbf{non}(\mathcal{B})$, provided that for each finite $J \subset \omega$, there are infinitely many $(n, m) \in I$ with $n, m \not \in J$. It is interesting to ask which of these $\kappa_I$'s can be separated...<|endoftext|> TITLE: Is the smallest $L_\alpha$ with undefinable ordinals always countable? QUESTION [7 upvotes]: Let $\mathfrak{t}$ be the least ordinal such that $L_{\mathfrak{t}}$ has undefinable ordinals; i.e. there is an $\alpha<\mathfrak{t}$ such that $L_{\mathfrak{t}}$ cannot define $\alpha$. This ordinal is quite large, but may have countable bounds under certain conditions. Because this ordinal is at most $\omega_1$ ($L_{\omega_1}$ is an uncountable structure and therefore must have undefinable sets in every uncountable subset, like $\omega_1$), it is definitely definable because its definition only quantifies over formulae bounded in $L_{\alpha}$, which are overall bounded in $L_{\omega_1}$. Here's some data on it: It must be a limit ordinal; $L_{\alpha+1}$ defines $\alpha$ (as the largest ordinal) and thus $L_\alpha$ allowing for statements like "$\varphi^{L_\alpha}(\beta)$". Therefore, if $\beta\in\alpha$ is definable in $L_\alpha$ by the formula $\varphi$, then $\beta$ is definable by $\varphi^{L_\alpha}$. It must be larger than $\omega$ because all finite numbers are clearly definable in $L_\omega=V_\omega$ It must be larger than $\zeta$ (the supremum of eventually writable ordinals) because: Every turing machine is definably encoded as a natural number in $L_\omega$ If $\alpha>\omega$, then $L_\alpha$ can define $L_\omega$ Klev's $\mathcal{O}^{++}$ is therefore definable in $L_\alpha$ for $\alpha>\omega$ on any $n$ such that $\mathcal{O}^{++}(n)\in L_\alpha$ If every ordinal below $\mathfrak{t}$ is eventually writable, then for every $\alpha<\mathfrak{t}$, $\alpha$ can be defined in $L_{\mathfrak{t}}$ as $\mathcal{O}^{++}(n)$ for some $n$ That's a contradiction, so there is some ordinal below $\mathfrak{t}$ which is not eventually writable, one of which must therefore be $\zeta$ If there is some countable $\alpha$ such that $\mathcal{P}(L_\alpha)\cap L\models\text{MK}$, for example an $L$-inaccessible $\alpha$, then $\alpha$ is an upper bound to $\mathfrak{t}$ (that is, $L_\alpha$ has some undefinable ordinals). This is because: If $\mathcal{P}(L_\alpha)\cap L\models\text{MK}$, then $\mathcal{P}^{L}(L_\alpha)\models\text{MK}$. Therefore $L\models\mathcal{P}(L_\alpha)\models\text{MK}$. (Working in $L$ for this bullet) Because $\text{MK}$ proves that $V$ has undefinables in every uncountable subclass, $\mathcal{P}(L_\alpha)$ also satisfies that $V$ has undefinables in every uncountable subclass. (Still working in $L$) $V^{\mathcal{P}(L_\alpha)}=L_\alpha$, and therefore in every $L_\alpha$-uncountable subset of $L_\alpha$, for example $\omega_1^{L_\alpha}$, there are $L_\alpha$-undefinables. (No longer working in $L$) $L_\alpha^L=L_\alpha$, so $\omega_1^{L_\alpha}$ has some $L_\alpha$-undefinables, and therefore $\mathfrak{t}$ is smaller than $\alpha$. EDIT: Thanks to user Miha Habič (you should check out some of his brilliant work if you haven't already), we know that this ordinal is for sure countable. Any countable ordinal $\alpha$ which is larger than an $L_{\omega_1}$-undefinable ordinal with $L_\alpha\prec L_{\omega_1}$ is clearly an upper bound. So here's the question, just how big is it? Is this ordinal admissible? Computably inaccessible? Computably mahlo? Just how big is it? Is it larger than $\Sigma$, the least ordinal which is not accidentally writable? REPLY [10 votes]: ${\mathfrak t}$ is the least $\beta$ such that there is a $\gamma<\beta$ with $L_\gamma \prec L_\beta$. That ${\mathfrak t} \leq$ the least such $\beta$ is obvious. On the other hand, if $X \subset L_{\mathfrak t}$ is $\subseteq$-least with $X \prec L_{\mathfrak t}$, then $X \not= L_{\mathfrak t}$; hence if $\sigma \colon L_\gamma \cong X$, then either $\sigma$ is the identity and thus $\gamma<{\mathfrak t}$ (as desired) or $\sigma$ is not the identity, in which case $L_{{\rm crit}(\sigma)} \prec L_{{\sigma}({\rm crit}(\sigma))}$, so that ${\mathfrak t} \leq \sigma({\rm crit}(\sigma)) < {\mathfrak t}$ by the other direction (giving a contradiction). In particular, $L_{\mathfrak t}$ is a model of ZFC${}^-$ (ZFC w/o the power set axiom), in fact of ZFC${}^-$ plus ``every set is at most countable,'' but ${\mathfrak t}$ is not the least $\beta$ such that $L_\beta$ is a model of ZFC${}^-$ and it is much bigger than $\Sigma$, the supremum of the accidentally writable ordinals.<|endoftext|> TITLE: Quantum functional analysis QUESTION [10 upvotes]: Can one explain some philosophy behind "quantum functional analysis" (or "quantized functional analysis") which was initiated and developed by such researchers as: Ruan Z.-J., Pisier J., Effros E.G., Haagerup U., et al.... The main notion of this subject is a quantum space: $(E, \{\|\cdot\|_n\})$ -- some normed space, equipped with "quantum norms" $\|\cdot\|_n, \, n\in \mathbb{N}$ (in turn, these quantum norms $\{\|\cdot\|_n\}$ are not the norms on $E$ but on matrices $M_n(E)$). It appears that classical bounded operators in $(E, \|\cdot\|_1)$ may have good ("quantum brothers") properties in whole quantum space, and as far as I understood they are called "completely bounded" operators. So quantum functional analysis is a theory about "such" operators, about matrices of operators (which are themselves operators) and from the book A.J. Helemsky I've read that it is a beautiful theory, "much simpler" than the classical functional analysis. Also there should be a relationship with quantum mechanics, since there is a notion of "quantization" of normed spaces (but I don't know anything more than that). I was wondering, can one here post some beautiful results of this theory? (for example, problem of Halmos about polynomially contracting operators was negatively resolved by Pisier using techniques from this theory). Among these beautiful results those ones which speak about classical analysis are extremely beautiful. Are there many extremely beautiful results? REPLY [16 votes]: Okay, I'll take this one. First let me say that the English term is "completely bounded" (or "complete isometry", etc.). About the term "quantum". The general principle is that analyzing some aspect of a physical system typically involves very different kinds of mathematical structures, depending on whether the system is classical or quantum. Thus if you study a classical system using topological techniques, or measure theory, or Lie groups, or graphs, the analogous analysis of a quantum system would probably involve C*-algebras, or von Neumann algebras, or quantum groups, or operator systems. (For instance, here's a paper of mine that talks about how operator systems arise in quantum error correction in the same way that graphs arise in classical error correction.) There is no simple rule for translating classical structures into quantum structures, but we have a pretty large dictionary and one can identify broad themes. My take is that operator spaces fit into this scheme as the quantum analog of relations on a set. See Section 2 of the paper linked above. I'm not aware of any really compelling direct connections with real-world quantum mechanics in this instance, but the word "quantum" may be appropriate just because the subject fits into the general scheme of mathematical quantization which does include many direct links. Now, as to the "beautiful results" you ask for. Well, there's a lot of good stuff so the selection would be idiosyncratic. One general principle is that when you're studying things related to Hilbert space (single operators, C*-algebras, etc.) it is often the case that you get much nicer results if you assume the "complete" (at all matrix levels) version of your hypothesis. For example, if two C*-algebras are linearly isometric, are they isomorphic as C*-algebras? Not in general, but yes if they are completely linearly isometric, i.e., linearly isometric at all matrix levels. (I think this is folklore.) The result of Pisier that you refer to was actually a counterexample, showing that a polynomially bounded operator need not be similar to a contraction. The positive result is due to Paulsen and says that any completely polynomially bounded operator is similar to a contraction. Here's another nice result, this one due to Zhong-Jin Ruan, one of the giants of the subject. B. Johnson proved that a locally compact group $G$ is amenable if and only if the Banach algebra $L^1(G)$ is amenable. Ruan showed that this happens if and only if the Fourier algebra $A(G)$ is completely amenable. A good place to learn more is the book Operator Spaces by Effros and Ruan. References: G. Pisier, A polynomially bounded operator on Hilbert space which is not similar to a contraction, J. Amer. Math. Soc. 10 (1997), 351–369. V. Paulsen, Every completely polynomially bounded operator is similar to a contraction, J. Funct. Anal. 55 (1984), 1–17. Z-J. Ruan, The operator amenability of A(G), Amer. J. Math. 117 (1995), 1449–1474.<|endoftext|> TITLE: Localization of a model category QUESTION [5 upvotes]: Let $M$ be a very nice model category (cofibrantly generated, combinatorial or cellular and left proper simplicial model category). Let $f: X\rightarrow Y$ and $g: X\rightarrow Z $ be two morphisms in $M$ such that $g$ is weak equivalence. Suppose that the map $r: Z\rightarrow Y\cup_{X} Z $ is a weak equivalence in the localized model category $\mathrm{L}_{\{ f\}}M$. Is it true that $\mathrm{L}_{\{ f\}}M=\mathrm{L}_{\{ r\}}M$ ? REPLY [6 votes]: No. Let $M$ be the category of simplicial sets with the Kan model structure. Let $S^1$ be $\Delta^1$ with its endpoints identified and let $f : \Delta^1 \to S^1$ be the obvious map. Let $g : \Delta^1 \to \Delta^0$ be the unique map. Then $r : \Delta^0 \to \Delta^0$ is the identity map, so it is a weak equivalence even in $M$, so $L_{\{r\}}M = M$, but $L_{\{f\}}M \neq M$ since $S^1$ is not equivalent to $\Delta^1$ in $M$.<|endoftext|> TITLE: Why did Voevodsky consider categories "posets in the next dimension", and groupoids the correct generalisation of sets? QUESTION [63 upvotes]: Earlier today, I stumbled upon this article written by V. Voevodsky about the "philosophy" behind the Univalent Foundations program. I had read it before around the time of his passing, and one passage that I remember vividly is this, for which I have little in the way of rigorous justification: The greatest roadblock for me was the idea that categories are “sets in the next dimension.” I clearly recall the feeling of a breakthrough that I experienced when I understood that this idea is wrong. Categories are not “sets in the next dimension.” They are “partially ordered sets in the next dimension” and “sets in the next dimension” are groupoids. — Voevodsky, The Origins and Motivations of Univalent Foundations If I had to guess, I'd say that the noninvertibility of morphisms in categories corresponds to some form of partial ordering, whereas a groupoid carries no such information since all morphisms are in fact isomorphisms, but surely there's something deeper at play that caused someone like Voevodsky to consider this realisation a "breakthrough" that accompanied a significant "roadblock". REPLY [40 votes]: The other answers are quite good and nothing is wrong with them, but lest a wrong impression be given (e.g. by the implicit suggestion that the idea of "categories as sets in the next dimension" in early work on higher categories was "wrong"), I want to add that while Voevodsky is of course correct from a certain point of view, there is another valid point of view according to which categories are, indeed, "sets in the next dimension". The point is that you have to have two orthogonal axes of "dimension". David Corfield mentioned this briefly in a comment: instead of just $n$-categories, consider $(n,r)$-categories: categories with (potentially) nontrivial $k$-morphisms for $0\le k\le n$, but where all $k$-morphisms are invertible for $k>r$. Here "0-morphisms" are objects, and it doesn't make sense to ask for them to be invertible, so $r\ge 0$. Thus for instance: A (1,1)-category is a 1-category in the usual sense: objects (0-morphisms) and arrows (1-morphisms), not required to be invertible. A (1,0)-category is a groupoid: objects and arrows, but the arrows are required to be invertible. A (2,2)-category is a 2-category in the usual sense: objects, arrows, and 2-cells, none required to be invertible. A (2,1)-category is a 2-category all of whose 2-cells are invertible, but whose 1-morphisms may not be, i.e. a category (perhaps weakly) enriched over groupoids. A (2,0)-category is a 2-groupoid: a 2-category all of whose 1-morphisms and 2-morphisms are (perhaps weakly) invertible. An $(\infty,0)$-category is an $\infty$-groupoid: it has cells of all dimension, all invertible. An $(\infty,1)$-category is what Lurie's school calls an "$\infty$-category": it has morphisms of all dimension, all invertible except the 1-morphisms. A (0,0)-category is a set: only objects, no invertibility requirements. Now you might think from the definition of $(n,r)$-category that you would have to have $r\le n$. But in fact there is a natural way to extend it to the case $r=n+1$. For when $r\le n$, we can identify $(n,r)$-categories (up to equivalence) as $(n+1,r)$-categories in which any two parallel $(n+1)$-morphisms are equal. Then we can take the latter when $r=n+1$ as a definition of $(n,n+1)$-category. This gives: A (0,1)-category is a 1-category in which any two parallel 1-morphisms are equal, i.e. (up to equivalence) a poset. A (1,2)-category is a category enriched over posets, or equivalently a 2-category in which all hom-categories are posets. Now we can see how the "raising dimension" step that Voevodsky was thinking of takes sets to groupoids and posets to categories: just add one to $n$ in $(n,r)$. $(0,0) \mapsto (1,0)$ $(0,1) \mapsto (1,1)$ But there's another natural "raising dimension" step that does indeed take sets to categories: add one to both $n$ and $r$. $(0,0) \mapsto (1,1)$ Since in common usage "$n$-category" for $n\ge 1$ refers to an $(n,n)$-category, it is therefore very natural to say that it is sets, i.e. (0,0)-categories, deserve the name "0-categories", whereas posets should be called (0,1)-categories. So I think Voevodsky's phrasing of his breakthrough may have been a bit too dogmatic. The point isn't that it's "wrong" to regard categories as sets in the next dimension (or at least a next dimension). The point is that it's also valid to regard groupoids as sets in the next dimension, and that this latter point of view is extremely fruitful, leading in particular to univalent foundations but also (somewhat independently) to the recent boom in $(\infty,1)$-category and $(\infty,n)$-category theory.<|endoftext|> TITLE: Analogue of Urysohn metrization for Lawvere metric spaces? QUESTION [7 upvotes]: Urysohn proved that any regular, Hausdorff, second-countable space $X$ is metrizable, i.e. there exists a metric space whose underlying topological space is $X$. But what if we ask the same question for Lawvere metric spaces? Definition: Let $(X,d)$ be a Lawvere metric space. For any $\epsilon>0$ and point $x\in X$, define the $\epsilon$-ball emanating from $x$, denoted $B(x,\epsilon)$, to be the set $$B(x,\epsilon):=\{x'\in X\mid d(x,x')<\epsilon\}.$$ Define the induced topology on $X$ to be the set of those subsets $U\subseteq X$ with the property that for all $x\in U$ there exists $\epsilon>0$ with $B(x,\epsilon)\subseteq X$. Example: The Sierpinski space is not Hausdorff, so it is not metrizable. But it is Lawvere metrizable. Indeed, let $S=\{o,c\}$, let $d(o,c)=2$ and $d(c,o)=0$. Then taking $\epsilon=1$ we have $B(o,1)=\{o\}$, so the singleton set $\{o\}$ is open. But for all $\epsilon>0$, we have $B(c,\epsilon)=\{o,c\}$, so $\{c\}$ is not open. Question: Do you know of a characterization of those topological spaces that are Lawvere metrizable? REPLY [7 votes]: According to this SE-post, a Lawvere metric on a set $X$ is a function $d:X\times X\to[0,+\infty)$ satisfying two axioms: 1) $d(x,x)=0$ and 2) $d(x,z)\le d(x,y)+d(y,z)$ for all $x,y,z\in X$. Then the following theorem can be considered as a counterpart of the Urysohn metrization theorem (I strongly suspect that this theorem was known to Lawvere). Theorem. Each topological space $X$ with countable base is metrizable by a Lawvere metric. Proof. Fix a countable base $\{U_n\}_{n\in\omega}$ of the topology of $X$. For every $n\in\omega$ consider the Lawvere metric $f_n:X\times X\to \{0,1\}$ defined by $$f(x,y)=\begin{cases}1&\mbox{if $x\in U_n$ and $y\notin U_n$};\\ 0&\mbox{otherwise}. \end{cases} $$ The Lawvere metrics $f_n$, $n\in\omega$ compose another Lawvere metric $$f=\max_{n\in\omega}\frac1{2^n}f_n$$ which generates the topology of $X$. Indeed, for any open set $U\subset X$ and any $x\in U$ we can find $n\in\omega$ with $x\in U_n\subset U$ and conclude that $B(x,\frac1{2^n})\subset U_n\subset U$. On the other hand, let us show that a set $U\subset X$ is open if for any $x\in U$ there exists $\varepsilon>0$ with $B(x,\varepsilon)\subset U$. Let $\Omega_x:=\{n\in\omega:x\in U_n\}$. Choose $m\in\omega$ with $\frac1{2^m}<\varepsilon$ and consider the open neighborhood $V:=\bigcup\{U_n:n\in\Omega_x,\;n\le m\}$ of $x$. Observe that for every $y\in V$ and every $n\le m$ we have $f_n(x,y)=0$. So, $d(x,y)<\frac1{2^m}<\varepsilon$ and $x\in V\subset B(x,\varepsilon)\subset U$, which means that $x$ is an interior point of $U$ and $U$ is open. Remark. The proof essentially uses the Lawvere metrizability of the Sierpinski two-point space and the fact that each $T_0$-space with countable base embeds into the countable product of the Sierpinski two-point spaces.<|endoftext|> TITLE: $p$-groups with trivial $H^3$ QUESTION [15 upvotes]: Let $Q_8$ be the group of quaternions of order $8$. It is a non-abelian $2$-group such that $H^3(Q_8,\mathbb{Z})=0$, where $\mathbb{Z}$ has the trivial action. For a proof, see the book "Homological Algebra" of Cartan and Eilenberg, Chapter XII, Section 7 (Examples), where the case of cyclic groups and generalized quaternions is also considered. I am curious if more examples of this kind exist (for other primes). More precisely, let $p>2$ be an odd prime. Does there exists a finite (non-abelian) $p$-group $G$ such that $H^3(G,\mathbb{Z})=0$? I could not find anything even for groups of order $p^3$. REPLY [10 votes]: For $G$ a finite group, $H^3(G,\mathbb{Z})$ is isomorphic to the Schur multiplier, and you’ll find lots of examples using that as a search term (also, “Schur-trivial” is sometimes used to mean “having trivial Schur multiplier”). For an example of order $p^3$, see The integral cohomology rings of groups of order $p^3$ by Gene Lewis (Trans AMS, 132(2), p. 501-529, (1968)). The semidirect product $C_{p^2}\rtimes C_p$, with a generator of $C_p$ acting on $C_{p^2}$ by $x\mapsto x^{p+1}$ is an example for every $p$.<|endoftext|> TITLE: CW complex of iterated loop spaces QUESTION [14 upvotes]: In Milnor's book Morse Theory, it is proved that the loop space $\Omega S^n$ of the n sphere has the homotopy type of a CW complex with one cell each in the dimensions 0, n-1, 2n-2, 3n-3, ... Or more generally, given non conjugate points p, q on a complete Riemannian Manifold M, the path space $\Omega(M,p,q)$ (of all continuous path joining p to q) has the homotopy type of a countable CW complex which contains one cell of dimension $d$ for each geodesic from p to q of Morse index $d$. For iterated loop spaces $\Omega^k M$, do we have a similar theorem? Say is there any known result concerning the CW structure of $\Omega^k S^n$? REPLY [19 votes]: By a result of Milnor, the space of maps from a finite CW complex to any CW complex is homotopy equivalent to a CW complex. This gives a general reason why spaces like $\Omega^k M$ have a CW structure. There is a well-known construction from which you can, at least in principle, get an explicit cell structure for spaces of the form $\Omega^k \Sigma^k X$, where $X$ is a based path-connected CW-complex. This includes $\Omega^k S^n$ as a special case (for $n>k$). The general construction goes as follows. For a finite set $i$, let $F(\mathbb R^k; i)$ be the space of injective maps from $i$ to $\mathbb R^k$. The assignment $i\mapsto F(\mathbb R^k; i)$ gives a contravariant functor from the category of finite sets and injections to the category of topological spaces. Similarly we have a covariant functor between same categories, $i\mapsto X^i$, where the functor structure is given by basepoint-inclusions. Given a pair of functors like this, one may form the coend $$F(\mathbb R^k; i)\otimes_i X^i.$$ A classic theorem of (I think) Milgram, May and Segal asserts that this coend is homotopy-equivalent to $\Omega^k\Sigma^k X$. This endows the homotopy type of $\Omega^k\Sigma^k X$ with a natural filtration, and it also can be used to equip it with a CW structure, since the spaces $F(\mathbb R^k; i)$ can be endowed with CW structures compatible with maps between them. One can get different CW models for the homotopy type of $\Omega^k\Sigma^k X$ by using cellular approximations of the functor $i\mapsto F(\mathbb R^k; i)$. Specific cellular approximations were constructed by Milgram, Barratt-Eccles, Jeff Smith, and possibly some others. This paper of Clement Berger gives a nice historical survey of constructions of this type. When $k=1$ there is a particularly small CW model, as you indicated. In other cases it is not going to be so simple to enumerate the cells. Nevertheless, this construction is useful for many purposes. For example, it was used to describe the homology of $\Omega^k\Sigma^k X$ (with field coefficients) as a functor of the homology of $X$.<|endoftext|> TITLE: Iteration of a morphism and flatness QUESTION [10 upvotes]: Let $A$ be a Noetherian local ring, $f:A \rightarrow A$ be a local ring morphism. Assume some power of $f$ is a flat morphism, must $f$ be flat as well? Motivation: Kunz's theorem shows the result is true for a positive characterestic ring $A$ and its Frobenius morphism. REPLY [12 votes]: Yes. Assume $f^n$ is flat for some $n>1$. Then since $f^n$ is local, it is faithfully flat. For any $A$-module $M$, put $M_1:=A\otimes_{f,A} M$ and recursively $M_i:=(M_{i-1})_1$. Let $u:E\to F$ be an injective $A$-module homomorphism. We need to show that $u_1:E_1\to F_1$ is injective. Let $K$ be its kernel. We have a sequence $K_{n-1}\to E_n\to F_n$ with zero composite, but by assumption $E_n\to F_n$ is injective, hence $K_{n-1}\to E_n$ is zero. Thus, $K_{n}\to E_{n+1}$ is also zero, but since $f^n$ is faithfully flat this implies that $K\to E_{1}$ is zero. QED<|endoftext|> TITLE: Irreducible representation of the product of two groups and tensor product QUESTION [6 upvotes]: Let $G_1, G_2$ be two lie groups, $V$ be a finite dimensional (continuous) irreducible complex representation of $G_1 \times G_2$, must $V \cong V_1 \otimes V_2$ for some irreducible representation $V_i$ of $G_i$? If $G_i$ are compact, this is true by Peter-Weyl theorem. REPLY [9 votes]: If the field is $\mathbb C$, There are many ways of seeing this. For $i=1,2$ we may replace $G_i$ by its Zariski closure in $GL(V_i)$ without changing the hypotheses or the conclusion. But if an algebraic subgroup $G\subset GL(V)$ is irreducible, then it is reductive (the unipotent radical will have a fixed space which is G invariant and hence zero). We now use the fact that the reductive group $G_i$ has a Zariski dense compact group $K_i$; we may thus replace $G_i$ by the compact $K_i$ where you have accepted the result.<|endoftext|> TITLE: Can the supremum of continuous functions be discontinuous at every point of an interval? QUESTION [10 upvotes]: Pether Luthy gave an example of a sequence of continuous real valued functions whose supremum was discontinuous on a set of positive measure. But does it exist a sequence of continuous real valued functions $f_n:\mathbb{R}\to\mathbb{R}$ such that $f(x) = \sup_{n \in \mathbb{N}} f_n(x)$ is a discontinuous function at every point of a subinterval of $\mathbb{R}$ ? If such a sequence does not exist, how is it possible to prove it? REPLY [29 votes]: Since the function $f$ is supremum of a set of continuous functions, it is lower-semicontinuous.1 Every lower semicontinuous function belongs to the first Baire class.2 If $f\colon \mathbb R\to\mathbb R$ is of the first Baire class, then the set $D_f$ of the points of discontinuity is a meager set.3 In particular, $D_f$ cannot be an interval. 1Theorem 10.3 in van Rooij-Schikhof: A Second Course on Real Functions. Mathematics Stack Exchange: To show that the supremum of any collection of lower semicontinuous functions is lower semicontinuous or Show that the supremum of a collection of lower semicontinuous function is lower semicontinuous. 2Theorem 10.6 and Exercise 11.E in van Rooij-Schikhof; Show that lower semicontinuous function is the supremum of an increasing sequence of continuous functions on Mathematics Stack Exchange 3Theorem 11.4 in van Rooij-Schikhof; MathOverflow: Points of continuity of Baire class one functions<|endoftext|> TITLE: Should I publish a paper if its results overlap significantly with an earlier paper? QUESTION [29 upvotes]: I have a preprint X that is sitting in the ArXiv for which I am not sure if it is still worth publishing. It turns out the paper I wrote has considerable overlap with another preprint Y after one of its authors informed me about it through email. Consider the following: Paper Y was posted in the ArXiv just a month before I posted mine, and I was not aware of its existence previously. Prior to posting paper X, I looked for as many papers with related results to include in my discussion, but due to the differences in the terminologies used, Google did not show paper Y in the results. In order to find paper Y on Google or ArXiv, one would have to use a different set of keywords. It appears that we have been working on the same problem, they just finished first and had a month of lead. I am aware that alternate proofs are sometimes of interest and are therefore worth publishing, and I am trying to determine if this is the case with my paper. Since I wrote paper X independently and without knowledge of the existence of paper Y, my proof was essentially different and indeed the two papers have very different motivations for the constructions used. However, while the methods used as different, they have the same "flavor" and the key results in both papers use different versions of the same pre-existing theorem. If there is anything drastically different from the results of both papers, it would be the length. My proofs are shorter than theirs, use less lemmas and mathematical machinery. Paper Y derives a few more corrolaries which are not in paper X, but even if one compares the length of both papers using only the parallel or similar portions, my paper is still significantly shorter than theirs. I can also make the argument, understandably a subjective one, that my methods are simpler. However, results of Paper Y are, to a certain degree, more general than mine and therefore, technically my results follow from theirs. My questions are as follows: Should I still submit my paper to a journal? If yes, how should I deal with the existence of paper Y in my own paper? I feel obliged to cite it, but I'm not really sure how to discuss the similarity and differences between the results. Should I tell the editor about the situation? Should I make other assumptions on the reason behind the author of Paper Y for sending me an email about their paper, other than to inform me that our papers have the same results? If you were in the shoes of the author of paper Y, what would you prefer that I do? REPLY [10 votes]: In the few times in my career I have been in a similar situation, I have withdrawn my paper before publication, but gone on to modify it to be different enough from the other's (and my) prior work that it justifies publication. The lesson is: don't always view a paper as a complete, fixed, immutable work of scholarship but instead a working space, amenable to revision and improvement. Viewing your work this way may prod you to discover a new result (or a new approach to a known result) so your prior work wasn't "in vain."<|endoftext|> TITLE: Is each Parovichenko compact space homeomorphic to the remainder of a soft compactification of $\mathbb N$? QUESTION [6 upvotes]: Definition 1. A compactification $c\mathbb N$ of the discrete space $\mathbb N$ is called soft if for any disjoint sets $A,B\subset\mathbb N$ with $\bar A\cap\bar B\ne\emptyset$ there exists a homeomorphism $h:c\mathbb N\to c\mathbb N$ such that $h(x)=x$ for all $x\in c\mathbb N\setminus\mathbb N$ and the set $\{x\in A:h(x)\in B\}$ is infinite. Definition 2. A compact Hausdorff space $X$ is called Parovichenko (resp. soft Parovichenko) if $X$ is homeomorphic to the remainder $c\mathbb N\setminus\mathbb N$ of some (soft) compactification $c\mathbb N$ of $\mathbb N$? Remark 1. By a classical Parovichenko Theorem, each compact Hausdorff space of weight $\le\aleph_1$ is Parovichenko. Hence, under CH a compact Hausdorff space is Parovichenko if and only if it has weight $\le\mathfrak c$. By a result of Przymusinski, each perfectly normal compact space is Parovichenko. On the other hand, Bell constructed an consistent example of a first-countable compact Hausdorff space, which is not Parovichenko. More information and references on Parovichenko spaces can be found in this survey of Hart and van Mill (see $\S$3.10), Problem 1. Is each Parovichenko compact space soft Parovichenko? Remark 2. The Stone-Cech compactification $\beta\mathbb N$ of $\mathbb N$ is soft, but there are simple examples of compactifications which are not soft. A compactification $c\mathbb N$ of $\mathbb N$ is soft if for any disjoint sets $A,B\subset\mathbb N$ with $\bar A\cap\bar B\ne\emptyset$ there are sequences $\{a_n\}_{n\in\omega}\subset A$ and $\{b_n\}_{n\in\omega}\subset B$ that converge to the same point $x\in\bar A\cap\bar B$. This implies that a compactification $c\mathbb N$ is soft if the space $c\mathbb N$ is Frechet-Urysohn or has sequential square. This also implies that each first-countable Parovichenko space is soft Parovichenko (more generally, a Parovichenko space $X$ is soft Parovichenko if each point $x\in X$ has a neighborhood base of cardinality $<\mathfrak p$). Problem 2. Is each (Frechet-Urysohn) sequential Parovichenko space soft Parovichenko? The following concrete version of Problem 1 describes an example of a Parovichenko space for which we do not know if it is soft Parovichenko. Problem 3. Let $X$ be a compact space that can be written as the union $X=A\cup B$ where $A$ is homeomorphic to $\beta\mathbb N\setminus\mathbb N$, $B$ is homeomorphic to the Cantor cube $\{0,1\}^\omega$ and $A\cap B\ne\emptyset$. Is the space $X$ soft Parovichenko? REPLY [6 votes]: Here is a partial answer: the Continuum Hypothesis implies that all Parovichenko spaces are soft-Parovichenko; the proof is a bit long, so I put it in a PDF-file on my website. Also, I retract my claim in the comments that all compactifications with $\omega_1+1$ as a remainder are soft. It is true, in ZFC, that $\omega_1+1$ is soft-Parovichenko but "all compactifications with remainder $\omega_1+1$ are soft" is equivalent to $\mathfrak{t}>\omega_1$. Added 2018-11-12: The note linked to above now contains a, consistent, example of a Parovichenko space that is not soft-Parovichenko. The example is the ordered space $\omega_1+1+\omega_1^\ast$.<|endoftext|> TITLE: Tuples of 2x2-matrices simultaneously conjugate to matrices with integer entries QUESTION [14 upvotes]: I am interested in the following question: Given an $n$-tuple of matrices $(A_1, \dots, A_n)\in SL(2,\mathbb R)^n$, does there exist a matrix $B\in SL(2,\mathbb R)$ such that $BA_jB^{-1}\in SL(2,\mathbb Z)$ for any $j$? If $n=1$, i.e., in the case of a single matrix $A\in SL(2,\mathbb R)$, it is quite easy to see that $A$ is conjugate to a matrix with integer entries if and only if $tr A\in \mathbb Z$. So, the interesting case is really $n\ge 2$ and I would like to have a characterization of $n$-tuple of matrices simultaneously conjugate to ones in $SL(2,\mathbb Z)$. Preferably, the criterion should be easy to check (of course, if there is one.) REPLY [5 votes]: We assume $n=2$ and we consider $A,B\in SL_2(\mathbb{R})$ s.t. $A,B$ are not simultaneously triangularizable over $\mathbb{C}$ (that is $\det(AB-BA)\not= 0$). Let $C,D\in SL_2(\mathbb{Z})$ s.t. $(A,B)$ and $(C,D)$ are in the same class of simultaneous similarity. By a result due to Friedland, such a class modulo $GL_n(\mathbb{C})$ depends only on the values of $tr(A),tr(A^2),tr(B),tr(B^2),tr(AB)$. Here $\det(A)=\det(B)=1$; then, as wrote Andreas, it suffices to know $tr(A),tr(B),tr(AB)$. Thus a necessary condition for the existence of $(C,D)$ is $(*)$ $tr(A),tr(B),tr(AB)\in\mathbb{Z}$. Conversely, under the above conditions $(*)$, assume that there are $C,D\in SL_2(\mathbb{Z})$ satisfying $tr(A)=tr(C),tr(B)=tr(D),tr(AB)=tr(CD)$. There is $P\in GL_2(\mathbb{C})$ s.t. $P^{-1}AP,P^{-1}BP\in SL_2(\mathbb{Z})$; clearly, we can choose $P\in GL_2(\mathbb{R})$ and even $P$ s.t. $\det(P)=\pm 1$. If $P=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ then let $Q=\begin{pmatrix}-a&b\\-c&d\end{pmatrix}$; then $\det(Q)=-\det(P)$ and $Q^{-1}AQ,Q^{-1}BQ\in SL_2(\mathbb{Z})$. Finally we can choose $P\in SL_2(\mathbb{R})$. The question is whether $C, D$ always exist. Let $a,b,c\in\mathbb{Z}$. We can always find $A,B\in SL_2(\mathbb{C})$ satisfying $(1)$ $tr(A)=a,tr(B)=b,tr(AB)=c$, but it's false in $SL_2(\mathbb{Z})$. EDIT 1. $\textbf{Remark}$. Let $C,D\in SL_2(\mathbb{Z})$ satisfying $(1)$; if we want to simplify the form of $C,D$ (at least that of $C$) to simplify the required sufficient condition concerning a, b, c, then we must go through its conjugacy class wrt. $GL_2(\mathbb{Z})$. Unfortunately, to a fixed trace, can correspond several such conjugacy classes. For example, to $tr(C)=-1$ corresponds a unique class, that of $\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$; yet, when $tr(C)=12$, there are at least $2$ conjugacy classes, that of $\begin{pmatrix}0&-1\\1&12\end{pmatrix}$ and $\begin{pmatrix}7&2\\17&5\end{pmatrix}$. Note that, in the previous examples, the characteristic polynomial is irreducible. Let $f\in\mathbb{Z}[x]$ be monic irreducible of degree $2$. A result due to Latimer, Mac Duffee says $\textbf{Theorem}$. The conjugacy classes of matrices $C\in M_2(\mathbb{Z})$ with $\chi_C=f$ are in bijection with the $\mathbb{Z}[u]$-ideal classes in $\mathbb{Q}(u)$ where $f(u)=0$. $\textbf{Conclusion}$. When $f$ is irreducible, we can find the classes with the help of the software Magma. When $f$ is reducible, $C$ is similar, over $\mathbb{Z}$, to a matrix in the form $\begin{pmatrix}\epsilon&b\\0&\epsilon\end{pmatrix}$ where $\epsilon=\pm 1,b\not= 0.$ Let $U=\{(a,b,c)\in\mathbb{Z}^3;$ there are $C,D\in SL_2(\mathbb{Z})$ not simultaneously triangularizable and s.t. $tr(C)=a,tr(D)=b,tr(CD)=c\}$. Since the number of representatives for $C$ depends on $a$, I think that there are no algebraic representation of the fact that $(a,b,c)\in U$; we have to work on explicit values of a, b, c. $\square$ For example, if we consider the standard class associated to $a$, then the required condition is as follows $(C,D)$ are simult. similar to $C=\begin{pmatrix}0&-1\\1&a\end{pmatrix},D=\begin{pmatrix}p&q\\r&b-p\end{pmatrix}$ and the condition can be easily written $p(b-p)-qr=1,-r+q+a(b-p)=c$. It is a system of $2$ equations in the unknowns $p,q$ (considering $r$ as a parameter). It is not difficult to prove (discuss according to parity of $ar-b$) that a necessary and sufficient condition for the existence of integer solutions is There is $r\in\mathbb{Z}$ s.t. $(ar-b)^2-4(-abr+cr+r^2+1)$ is a square, that is $(2)$ there is $r\in\mathbb{Z}$ s.t. $(a^2-4)r^2+(2ab-4c)r+b^2-4$ is a square. Note that the coefficients of $r^2,r,1$ are independent. Of course, if we consider the standard class of $D$ (instead of that of C), then the condition becomes $(2')$ there is $r\in\mathbb{Z}$ s.t. $(b^2-4)r^2+(2ab-4c)r+a^2-4$ is a square. There are instances s.t. the solutions in $a,b,c,$ of $(2)$ and $(2')$ are not the same. EDIT 2. $\textbf{The case when $A,B$ are simultaneously triangularizable}$ and where $tr(A)=a,tr(B)=b,tr(AB)=c$ are integers. Then, necessarily, $a,b,c$ are linked by a unique relation $a^2+b^2+c^2-abc-4=0$ (according to the @user44191 's comment below). The above relation can be realized by integers iff $(b^2-4)(c^2-4)$ is a square. In the sequel, we suppose that $a,b,c$ satisfy the above two conditions. CASE 1. $A$ or $B$ (for example $A$) has distinct eigenvalues $p,q$. To a fixed $a,b,c$ are associated $3$ possible similarity classes, over $\mathbb{C}$, for the couple $(A,B)$ $[diag(p,q),\begin{pmatrix}r&1\\0&s\end{pmatrix}],[diag(p,q),\begin{pmatrix}r&0\\1&s\end{pmatrix}],[diag(p,q),\begin{pmatrix}r&0\\0&s\end{pmatrix}]$. In a second step, one calculates the similarity classes associated to $tr(A)$ over $\mathbb{Z}$ and we seek the conditions about $a,b,c$ (for the $3$ classes) as in the detailed example (the case of the standard class) in the first part of the post. CASE 2. $A$ and $B$ have double eigenvalues $\pm 1$. 2.1. $A$ or $B$ (for example $A$) is equal to $\pm I_2$; then the condition $b\in\mathbb{Z}$ suffices. 2.2. $A,B$ is similar over $\mathbb{C}$ to $\pm\begin{pmatrix}1&1\\0&1\end{pmatrix},\pm\begin{pmatrix}1&u\\0&1\end{pmatrix}$ where $u\not= 0$. Proceed over $\mathbb{Z}$ as in the first part of the post.<|endoftext|> TITLE: Inequivalent compact closed symmetric monoidal structures on the same category QUESTION [11 upvotes]: I am looking for interesting examples of categories admitting multiple monoidally inequivalent closed (or compact closed) symmetric monoidal structures. We know how to construct disconnected toy models [1], but none of direct practical interest. I also seem to recall from [2] that the category of $G$-sets admits two inequivalent such structures, but I am curious to know whether this is a sporadic occurrence or whether there is a large family of interesting examples lurking somewhere out there. [1]: see e.g. https://arxiv.org/abs/1803.00708 [2]: Borceux, Francis. Handbook of Categorical Algebra REPLY [15 votes]: A pretty interesting class of examples comes about by classifying compact monoidal groupoids. Given a group $G$, a $G$-module $M$, and a (normalized) 3-cocycle $a: G \times G \times G \to M$, one can manufacture a compact monoidal groupoid whose category of objects is $G$, whose morphisms are ordered pairs $(g, m) \in G \times M$ where we define $\text{dom}(g, m) = \text{cod}(g, m) = g$ and where endomorphism composition is defined by addition in $M$, and where we define the tensor product by $(g, m) \otimes (h, n) = (g h, m + g n)$. It's the 3-cocycle that furnishes the associativity data, and we get monoidally inequivalent groupoids whenever the 3-cocycles are not cohomologous. This was observed by Joyal and Street in their paper Braided Monoidal Categories. Now you were asking about the symmetric monoidal case (where we now assume $G$ is abelian). These are also known as Picard groupoids. In the simplified scenario where we demand strict associativies and consider only the case where $G$ acts trivially on $M$, in which case the underlying category becomes the product $K G \times B M$ of the evident discrete monoidal category $K G$ with the evident one-object category $B M$, any symmetric bilinear pairing $G \otimes G \to M$ can be used to manufacture a symmetry isomorphism for a symmetric monoidal structure on $K G \times B M$, and these examples are generally symmetric-monoidally inequivalent. I can't tell how far away such examples are from the "toy"examples" you have in mind, but Picard groupoids are surely of interest -- see for example applications to 2-stage Postnikov systems of spectra here. REPLY [7 votes]: The category $Cat$ of small categories admits two inequivalent tensor products: the Cartesian product and the funny product. This generalizes up to 2-cat, 3-cat, etc. By the way, a word of advise to those like me with a murky memory: googling "cat funny product" is going to just give you pictures of cats. REPLY [4 votes]: The large class of examples I have in mind, though I am not sure if it meets your compactness requirement (definition?) are the Lax tensor products on $n$-Cat. It has been constructed in the cases $n=2$ and $n=\omega$ by Gray in the case $n=2$ and Crans, Steiner, and Verity independently in the case $n=\omega$. Edit: To clarify, these are all biclosed, and their right adjoints are the $n$-categories whose objects are strict $n$-functors and whose higher cells are (op)lax natural transformations and (op)lax modifications between them.<|endoftext|> TITLE: Does $2^x-3p^y=5$ (with $p$ an odd prime) have only finitely many positive integer solutions? QUESTION [10 upvotes]: Let $p$ be an odd prime. Does the equation $$2^x-3p^y=5$$ only have finitely many solutions in positive integers $x$ and $y$? REPLY [25 votes]: The solutions of your equation can be injected into the solutions of the $S$-unit equation over $\mathbb{Q}$, where $S=\{\infty,2,3,5,p\}$. As the latter is known to have finitely many solutions by the results of Siegel, Mahler, Lang (see Chapter 5 in Bombieri-Gubler: Heights in Diophantine geometry), your equation also has finitely many solutions. By a result of Beukers and Schlickewei (see Theorem 5.2.1 in the above mentioned book), the number of solutions is at most $2^{72}$ for any given prime $p$. In fact the earlier work of Evertse (Inventiones, 1984) yields the better bound $3\cdot7^{7}$ for the number of solutions (apply Theorem 1 with $\lambda=1/5$, $\mu=3/5$, $S=\{\infty,2,p\}$).<|endoftext|> TITLE: The Image of a Derivation is Contained in the Jacobson Radical QUESTION [7 upvotes]: Let $A$ be a finite-dimensional unital commutative associative algebra over a field $K$ of characteristic $0$. Is it true that for any derivation $D$ of $A$ we have $D(A) \subseteq J(A)$ where $J(A)$ is the Jacobson radical of $A$? Please help me with this question or give some references. REPLY [8 votes]: I suppose you want $D$ to be a $K$-derivation, otherwise there are obvious counter-examples with $A=K$. Then the answer is yes, but for somewhat stupid reasons. First of all, $A$ is a product of a finite number of local rings, and $D$ respects this decomposition: if $e\in A$ is idempotent, from $e^2=e$ one gets $(2e-1)De=0$, hence $De=0$, and therefore $D(Ae)\subset Ae$. Thus one can assume that $A$ is local, with maximal ideal $\mathfrak{m}$. Let $x\in\mathfrak{m}$; there exists an integer $n\geq 1$ such that $x^n=0$ but $x^{n-1}\neq 0$. Then $nx^{n-1}D(x)=0$, which means that $D(x)$ cannot be invertible, hence $D(x)\in\mathfrak{m}$. Edit : As observed by @Keith Kearnes, the argument is completed as follows: we have shown $D(\mathfrak{m})\subset \mathfrak{m}$, so that $D$ induces a $K$-derivation of $A/\mathfrak{m}$. But $A/\mathfrak{m}$ is a finite extension of $K$, so any $K$-derivation of $A/\mathfrak{m}$ is zero. But this means $D(A)\subset \mathfrak{m}$.<|endoftext|> TITLE: The characteristic varieties of the complement of the braid arrangement QUESTION [6 upvotes]: The characteristic varieties $V_d^i(X)$ of a (sufficiently nice) space $X$ are the cohomology jumping loci for 1-dimensional (complex) local systems on $X$. Assume that $H_1(X;\mathbb{Z}) \cong \mathbb{Z}^n$ for some $n > 0$. Then $$V^i_d(X) = \{\ \rho\in \text{Hom}(\pi_1(X), \mathbb{C}^*)\ \ |\ \ \text{dim}\ H^i(X; \mathbb{C}_{\rho}) \geq d \ \}$$ where $\mathbb{C}_{\rho}$ is the 1-dimensional complex local system on $X$ associated to the character $\rho$. These loci are Zariski closed in the algebraic torus $\text{Hom}(\pi_1(X), \mathbb{C}^*) = (\mathbb{C}^*)^n$. This question is concerned with the characteristic varieties $V^1_d$ of the complement of the braid arrangement $X_k\subset \mathbb{C}^k$. General results of Arapura imply that $V_1^1(X_k)$ is a union of subtori of $(\mathbb{C}^*)^n$, some of which may be translated away from the identity $\bf{1}\in$ $(\mathbb{C}^*)^n$. In the late 90s, Cohen--Suciu showed that the components of $V_1^1(X_k)$ that contain $\bf{1}$ are two-dimensional and gave an explicit description of them. In about 2009, Settepanella found the remaining components of $V_1^1(X_k)$. $\bf{Question}$: Is anything known about the structure of the characteristic varieties $V_d^1(X_k)$ with $d\geq 2$? In other words, for fixed $k$ and $d$, is anything known about the set of 1-dimensional local systems on $X_k$ whose cohomology is at least d-dimensional? REPLY [4 votes]: Let $T$ be an irreducible component of $V^1_d(X_k)$ with $d\ge 2$. If $T$ contains $\mathbb{1}$, then $T=\{\mathbb{1}\}$. It is probably the case that all components of $V^1_d(X_k)$ pass through the identity, but I don't think that has been established, except for small values of $k$.<|endoftext|> TITLE: Descartes' rule of signs for infinite series QUESTION [7 upvotes]: Consider the function given by $$f(x)=1-a_1x-a_2x^2-a_3x^3-\cdots$$ where each $a_k\geq0$ and some $a_j>0$. If $f(x)$ is a polynomial then Descartes' Rule of signs tells us there is exactly one positive zero, i.e. root of $f(x)=0$. Assume $f(x)$ is a (real) power series with radius of convergence $0 TITLE: What is $\mathbb{Q}_1$, the "field of $1$-adic numbers"? QUESTION [21 upvotes]: (Disclaimer: I'm totally ignorant about $\mathbb{F}_1$ theory) There are now (several) working definitions of the "field with one element" $\mathbb{F}_1$ (not literally a field, of course), and certain mathematical objects "over $\mathbb{F}_1$" are recognized. For example, $\mathbb{F}_1$-vector spaces are pointed sets, $\mathrm{GL}_n(\mathbb{F}_1)$ is the symmetric group $\mathfrak{S}_n$ on $n$ elements (or was it $n-1$? $n+1$?). As far as I remember there's even a notion of "algebraic variety over $\mathbb{F}_1$": they are related to toric varieties. So, my question: Say $\mathbb{F}_p$ is to $\mathbb{Q}_p$ as $\mathbb{F}_1$ is to some object that we call $\mathbb{Q}_1$. How can $\mathbb{Q}_1$ be described explicitely in terms of more conventional mathematical objects? The same question could be asked for $\mathbb{Z}_1$, $\mathbb{C}_1$, $\Omega_1$. REPLY [7 votes]: Completions of a field are associated to places of that field. So for instance, $\mathbb Q(i)$ has two different completions of residue characteristic $5$. But $\mathbb F_1$ is not supposed to be a place of $\mathbb Q$. It is supposed to be the base field of $\mathbb Q$, which is different.<|endoftext|> TITLE: Topological dimension of $p$-adic manifolds QUESTION [10 upvotes]: What is the topological dimension of a (locally analytic) $p$-adic manifold over a non Archimedean field $K$? Is the topological dimension of $K^n$, $n$? REPLY [10 votes]: $p$-adic numbers are locally compact, Hausdorff and totally disconnected (see this nLab page), hence they are zero-dimensional. This means that---at least naively---topological dimension of $p$-adic manifolds doesn't work as you'd expect from real or complex manifolds. However, there are ways to do analytic geometry over the $p$-adic numbers, see e.g. this stackexchange question and the references on this nLab page.<|endoftext|> TITLE: Cell structures of Dold manifold and Wu manifold QUESTION [10 upvotes]: In Dold's 1956 paper Erzeugende der Thomschen Algebra N, Dold studied the Dold manifold $P(m,n)=(S^m\times\mathbb{CP}^n)/\tau$ where $\tau$ acts as $-1$ on $S^m$ and a complex conjugation on $\mathbb{CP}^n$. In section B of that paper, Dold gave the cell structure of $P(m,n)$, the cells were denoted by $(C_i,D_j)$ where $C_i$ has dimension $i$ and $D_j$ has dimension $2j$, $(C_i,D_j)$ has dimension $i+2j$. In equation (4) of that paper, Dold gave the boundary relations between the cells: $$\partial(C_i,D_j)=(1+(-1)^{i+j})(C_{i-1},D_j),$$ $$\partial(C_0,D_j)=0$$ for $i=1,2,\dots,m$, $j=0,1,\dots,n$. I determine the cell structure of the 5d Dold manifold $P(1,2)$, following Dold's results, $$\partial(C_1,D_0)=\partial(C_1,D_2)=0,$$ $$\partial(C_1,D_1)=2(C_0,D_1),$$ $$\partial(C_0,D_j)=0$$ for $j=0,1,2$. So there is exactly one cell in each dimension from 0 to 5 and there is only one nonzero boundary map from the 3-cell to the 2-cell. Both Dold manifold $P(1,2)$ and Wu manifold $X=SU(3)/SO(3)$ generate $\Omega_5^{SO}$, actually they are bordant. But $P(1,2)$ and $X$ have different cohomology groups $$H^*(P(1,2),\mathbb{Z}/2)=\mathbb{Z}/2[c,d]/(c^2,d^3)$$ where $|c|=1$, $|d|=2$. $$H^*(X,\mathbb{Z}/2)=\Lambda(w_2(X),w_3(X)).$$ My question: What is the cell structure of Wu manifold? Is there a nonzero boundary map from the 2-cell to the 1-cell? Thank you! REPLY [5 votes]: The 1965 paper Simply connected five-manifolds of Barden contains a simple topological description of the Wu manifold. The Wu manifold decomposes into two copies of the (unique) orientable non-trivial $D^3$-bundle over $S^2$. Since a $D^3$-bundle over $S^2$ decomposes into one 0-handle and one 2-handle, we deduce that the Wu manifold decomposes into one 0-handle, one 2-handle, one 3-handle, and one 5-handle. So there is a cell decomposition with one 0-cell, one 2-cell, one 3-cell, and one 5-cell.<|endoftext|> TITLE: Limit of homeomorphisms from square to square QUESTION [13 upvotes]: Let $\square=[0,1]\times[0,1]$ be the unit square and $f\colon\square\to \square$ is a continuous map that fixes the points on the boundary. Assume $f$ is a limit of homeomorphisms $\square\to \square$. (By Moore's theorem it is equivalent to the condition that for any point $p\in \square$ the inverse image $f^{-1}\{p\}$ is connected and its complement is connected.) Is it true that for most segments in $\square$, their inverse images are Jordan arcs? Say, given $s\in [0,1]$ consider the vertical unit segment $I_s$ defined by $x=s$. Is it true that for a dense G-delta set of values $s\in [0,1]$ the inverse image $$J_s=f^{-1}I_s$$ is a Jordan arc? Comments: If $f$ is injective on $J_s$ then $J_s$ is a Jordan arc (evident). One may expect that $f$ is injective for most of values $s$, but this is not the case --- even if all $J_s$ are Jordan arcs, the map $f$ might map an arc in $J_s$ to one point for all $0 TITLE: Transportation-cost inequality for pushforward measure QUESTION [7 upvotes]: Let $X=(X,d_X)$ and $Y=(X,d_Y)$ be metric spaces and $\varphi: X\rightarrow Y$ be an $L$-Lipschitz map, with $0 \le L < \infty$. Suppose $\mu$ is a probability measure on $X$ which satisfies Talagrand transportation-cost inequality, namely There exists a constant $c_\mu > 0$ such that $$ W(\nu,\mu) \le \sqrt{2c_\mu H(\nu\|\mu)}, $$ for every other probability measure $\nu$ on $X$. Here, $H$ denotes relative entropy (KL divergence) and $W$ denotes the Wasserstein $2$-distance on the space of probability measures on $X$ (with finite 2nd moment). Question Does the pushforward $\varphi_\#\mu$ of $\mu$ under $\varphi$ satisfy such an inequality ? My guess is that it does, with constant $c_{\varphi_\#\mu} \le L c_\mu$. REPLY [3 votes]: The result you want (but of course with $c_{\varphi_\#\mu} \le L^2 c_\mu$ rather than $c_{\varphi_\#\mu} \le L c_\mu$) is Lemma 2.1 in the paper at https://arxiv.org/pdf/math/0410172<|endoftext|> TITLE: Thrice intersecting closed geodesic on genus 2 orientable closed surface QUESTION [6 upvotes]: Does there exist a closed geodesic on a closed genus 2 orientable surface (with hyperbolic metric) that self-intersects at only one point thrice? REPLY [5 votes]: Yes, such curves exist on closed hyperbolic surfaces. As mentioned by Sam Nead, one can think of such a curve as lying on a subsurface which is a 4-holed sphere or 2-holed torus (genus one with two boundary components). I'll first point out that one of the possible configurations cannot occur inside of a 3-holed sphere. The argument uses Gauss-Bonnet, and is the same as given in Example 7 of this paper. Labeling the three angles $\alpha, \beta, \gamma$, then $\alpha+\beta+\gamma=\pi$. But we get a geodesic triangle on the back side which implies $\alpha+\beta+\gamma < \pi$, a contradiction. There are two possible configurations of immersed curve on an orientable surface with a single triple point. We may immerse a regular neighborhood of such curves in the plane in these two ways: The first lives inside a 4-holed sphere, the second in a 2-holed torus. We'll show that the second configuration is realizable. It sits on a genus 2 surface like this: The purple and blue curves intersect this immersed curve each in one point. Suppose that we can realize both resolutions of this immersed curve as geodesics in a hyperbolic metric. Then there must be some surface with the triple point. We can pinch the blue and purple curve to get a noded Riemann surface / hyperbolic surface with 4 cusps, in the boundary of moduli space (say in the Deligne-Mumford compactification). The curve will look like this: The picture should be thought of a lying on the Riemann sphere with four punctures. A noded surface is created by identifying the blue and purple points in pairs. One can see that this is realized as a single triple point intersection in the hyperbolic metric by symmetry. On the other hand, one may also realize both resolutions, by "squeezing" the two blue punctures together, limiting to a configuration without a triple point. Hence, both configurations are realized by noded surfaces in the boundary of moduli space. Perturbing to the interior of moduli space, we see that both configurations are realized, and hence a surface with a triple point exists.<|endoftext|> TITLE: The mirror of the Landau--Ginzburg model given by elliptically fibered K3 QUESTION [7 upvotes]: Let $f:X\rightarrow \mathbb{P}^1$ be an elliptically fibered K3 surface. Choose a coordinate on $\mathbb{P}^1$ and consider $X\backslash f^{-1}(\infty)\rightarrow \mathbb{C}$ as a Landau--Ginzburg model. What is the mirror dual of this LG model? REPLY [4 votes]: In general, if $X$ is a compact smooth $n$-dimensional Calabi-Yau manifold, and $D\subset X$ is an ample (or numerically effective) divisor, then the mirror of $X$ is usually a degeneration of the mirror $X^\vee$ of $X$, namely it's a singular Calabi-Yau variety $Y$. It is expected that we should have the following equivalences $D^b\mathit{Coh}(Y)\cong D^\mathit{perf}\mathcal{W}(X\setminus D);$ $\mathit{Perf}(Y)\cong D^\mathit{perf}\mathcal{F}(X\setminus D),$ where $\mathit{Perf}(Y)\subset D^b\mathit{Coh}(Y)$ is the triangulated subcategory of perfect complexes (it differs from $D^b\mathit{Coh}(Y)$ since $Y$ is singular), $\mathcal{F}(X\setminus D)$ is the Fukaya category of compact Lagrangian submanifolds in $X\setminus D$, and $\mathcal{W}(X\setminus D)$ is the wrapped Fukaya category, where certain non-compact Lagrangians are allowed as its objects. A special case of the above expectation is stated as Conjecture 6.20 in the paper of Harder-Katzarkov: https://arxiv.org/abs/1708.01181, where $X\rightarrow\mathbb{P}^1$ is a Calabi-Yau fibered Calabi-Yau manifold, and $D$ is a union of (smooth) fibers, which is the case that you are interested in. In this case, by removing the fiber at infinity, one gets a Landau-Ginzburg model $X\setminus D\rightarrow\mathbb{C}$. Sometimes one can remove even more stuff from $X$, which may or may not affect your LG model on the categorical level. As a special case, we have the work of Seidel on fiberwise compactifications of Lefschetz fibrations: https://arxiv.org/abs/1504.06317. If you treat this LG model as your A-side, then there is another Fukaya category, namely the partially wrapped Fukaya category $\mathcal{W}_\Lambda(X\setminus D)$ with the stop $\Lambda\subset\partial_\infty(X\setminus D)$ determined by the fibration $X\setminus D\rightarrow\mathbb{C}$. For its definition, see the work of Sylvan (https://arxiv.org/abs/1604.02540) and Ganatra-Pardon-Shende (https://arxiv.org/abs/1706.03152). This category should be mirror to the so-called categorical resolution of $\mathit{Perf}(Y)$, in the sense that $\mathcal{F}(X\setminus D)$ is proper but not smooth as an $A_\infty$-category, while $\mathcal{W}_\Lambda(X\setminus D)$ is smooth with appropriate choice of $\Lambda$, so it provides a "resolution" of $\mathcal{F}(X\setminus D)$ in the sense of noncommutative geometry. From the point of view of algebraic geometry, if $Y$ contains only rational singularities, and $\widetilde{Y}\rightarrow Y$ is a usual resolution of singularities, then $D^b\mathit{Coh}(\widetilde{Y})$ provides a categorical resolution of $\mathit{Perf}(Y)$. Thus $\widetilde{Y}$ should be considered as the expected mirror of the LG model $X\setminus D\rightarrow\mathbb{C}$. For the more general case of irrational singularities, the construction of a categorical resolution is more involved, see the work of Kuznetsov-Lunts: https://arxiv.org/abs/1212.6170. You might also find it interesting to have a look at the work of Polishcuhk-Lekili on the homological mirror symmetry of nodal stacky curves: https://arxiv.org/abs/1705.06023, where the Auslander order plays the role of a categorical resolution.<|endoftext|> TITLE: Countable support product of Sacks forcings and selective ultrafilters QUESTION [11 upvotes]: If $U$ is a selective ultrafilter on $\omega$, then $U$ generates an ultrafilter in $V^{\mathbb S}$, where ${\mathbb S}$ is Sacks forcing. The same is true with ${\mathbb S}$ being replaced by ${\mathbb S}_n$, the product of $n$ copies of Sacks forcing, $n<\omega$ (Halpern and Pincus, 1981), and I can see a proof of that. How about ${\mathbb S}_\omega$, the full (= ctble.) support product of $\omega$ copies of Sacks forcing? People refer to R. Laver, "Products of infinitely many perfect trees," 1984, for a positive answer, but I don't find this answer in that paper (cf. Theorem 6 of that paper which states a weaker result). Is it true that if $U$ is a selective ultrafilter on $\omega$, then $U$ generates an ultrafilter in $V^{{\mathbb S}_\omega}$? REPLY [8 votes]: The collection of possible large sets is analytic, namely: $\{A\subset \omega: \forall i<\omega\ \exists U_i\subset T_i \text{ $U_i$ is perfect and } f\restriction \bigcup_{n\in A}\Pi_{i<\omega} U_i(n) \text{ is constant}\}$ (here we can assume the length of the roots of $T_i$ goes to infinity so the coloring $f$ is coded by a real). The finish by a theorem of Mathias: if $P\subset [\omega]^\omega$ is analytic, and $U$ is Ramsey, then there exists $A\in U$ such that $[A]^\omega\subset P$ or $[A]^\omega\cap P =\emptyset$. The first option must appear by density of the set under $\subseteq$. I saw this argument first from Olga Yiparaki's thesis: On some tree partitions. To trigger another problem, it may be interesting to think about what happens if the ultrafilter is merely a P-point. Then I think we need to get to the nitty-gritty of Laver's proof.<|endoftext|> TITLE: Set of homeomorphic fixed points that is dense, but not equal to whole space QUESTION [8 upvotes]: If $(X,\tau)$ is a topological space, let $FH(X)$ denote the collection of $x\in X$ such that there is a non-identity homeomorphism $\varphi:X\to X$ with $\varphi(x) = x$. What is an example of a $T_2$-space $(X,\tau)$ such that $FH(X)$ is dense in $X$, but $FH(X)\neq X$? REPLY [4 votes]: Such an example can be constructed unifying two pathological examples of Cook and van Mill. Example (Cook, 1967): There exists a non-degenerated metric continuum $K$ such that any continuous map $f:K\to K$ is either constant or the identity. Example (van Mill, 1983): There exists a metrizable separable Boolean group $G$ that admits no homeomorphism of $G$ has no fixed points. The van Mill group $G$, being separable, contains a dense countable subgroup $G_\omega$. The countable subgroup $G_\omega$, being Boolean, can be written as the countable union $G_\omega=\bigcup_{n\in\omega}G_n$ of an increasing sequence $(G_n)_{n\in\omega}$ of finite Boolean subgroups. Since each subgroup $G_n$ is complemented in $G_{n+1}$, we can fix a homomorphism $r_n:G_{n+1}\to G_n$ such that $r_n(x)=x$ for all $x\in G_n$. Now take the Cook continuum $K$ and fix two distinct points $a,b\in K$. Consider the space $$U:=(K\times G\times\{\omega\})\cup\bigcup_{n\in\omega}K\times G_n\times\{n\}\subset K\times G\times(\omega+1)$$where the ordinal $\omega+1=\omega\cup\{\omega\}$ is endowed with the order topology (which is compact and metrizable). On the space $U$ consider the equivalence relation $$ \begin{aligned} \sim:&=\{((x,g,\omega),(y,g,\omega)):x,y\in K,\;g\in G\}\cup\\ &\cup\bigcup_{n\in\omega}\bigcup_{g\in G_{n+1}}\{((a,g,n+1),(b,r_n(g),n)),((b,r_n(g),n),(a,g,n+1))\}. \end{aligned} $$ Consider the quitient space $X:=U/_\sim$ and the corresponding quotient map $q:U\to X$. The space $X$ contains a closed copy $G':=q(K\times G\times \{\omega\})$ of the van Mill's group $G$ and the complement $X\setminus G'$ is a connected dense set, which is the union of countably many copies $K_{n,g}:=q(K\times \{(g,n)\})$, $g\in G_n$, of the Cook continuum $K$. Each $K_{n,g}$ intersects the continua $K_{n+1,p}$ where $p\in r_n^{-1}(g)$. It can be shown that the space $X$ has the required property: the set $FH(X)$ is dense not not coincide with $X$. More precisely, $FH(X)$ is equal to the complement $X\setminus G'$ of the copy $G'$ of the group $G$ in $X$. Indeed, for any $s\in G_{n+1}$ with $r_n(s)=0$ the translation $x\mapsto x+s$ of the group $G$ extends to a homeomorphism of $X$ that does no move points of the union $\bigcup_{i\le n}\bigcup_{g\in G_i}K_{i,h}$. So, $FH(X)\supset X\setminus G'$. On the other hand, any homeomorphism of $X$ that fixes some point of the group $G'$ is identity on $G'$ and extends unquely to the identity homeomorphism of $X$ (because of the rigidity properties of the Cook continuum $K$).<|endoftext|> TITLE: Vector space objects in schemes - confusion QUESTION [7 upvotes]: Let $R$ be the ring $\mathbf{C}\times\mathbf{C}$, and consider the affine line $\mathbf{A}^1_R$. $\mathbf{A}^1_R$ can be given the structure of additive group scheme over $R$, denoted $(\mathbf{G}_a)_R$. $(\mathbf{G}_a)_R$ carries a functorial multiplicative action of $R$ making it into an $R$-module object in schemes: the free $R$-module scheme of rank $1$. On the other hand, $R$ is a $\mathbf{C}$-vector space of dimension $2$. Is $(\mathbf{G}_a)_R$ a $\mathbf{C}$-vector space object in schemes, of dimension $2$? I am having difficulties to gain intuition about this, because $(\mathbf{G}_a)_R$ is, as a scheme, a disjoint union of two copies of $\mathbf{A}^1_{\mathbf{C}}$. How does one endow it with a $\mathbf{C}$-vector space object structure? (likely a very simple question. I’m just a little too confused about something trivial) REPLY [10 votes]: Briefly, $\mathbb{A}^1_R$ is not a vector space over $\mathbb{A}^1_\mathbb{C}$ in a natural way. Strictly speaking, saying that $\mathbb{A}^1_R$ is a ring object in schemes in not precise. What is correct is that the morhpism $\mathbb{A}^1_R\to \mathrm{Spec} R$ (adjoint to $R\to R[T]$) is a ring object in the category of schemes over $R$ (i.e. schemes equipped with a morphism into $\mathrm{Spec} R$). [This means that there are addition and product morphisms $+,\cdot:\mathbb{A}^1_R\times_R\mathbb{A}^1_R\to \mathbb{A}^1_R$ and $R$-sections $0,1:\mathrm{Spec} R\to \mathbb{A}^1_R$ such that the ring axioms (phrased as diagrams) hold.] In order to speak about $\mathbb{A}^1_R$ as a "module over $\mathbb{A}^1_{\mathbb{C}}$" in some sense, you need to work in the category of schemes over $\mathbb{C}$, where $\mathbb{A}^1_{\mathbb{C}}\to\mathrm{Spec } \mathbb{C}$ is a ring object. With this notation, the $2$-dimensional ($\mathbb{A}^1_{\mathbb{C}}\to \mathrm{Spec}\mathbb{C}$)-vector space object in schemes over $\mathbb{C}$ would be $$ \mathbb{A}^1_{\mathbb{C}}\times_{\mathrm{Spec}\mathbb{C}}\mathbb{A}^1_{\mathbb{C}}=\mathbb{A}^2_{\mathbb{C}} $$ equipped with the evident morphism into $\mathrm{Spec}\mathbb{C}$. In particular, it is not isomorphic to $\mathbb{A}^1_R\to \mathrm{Spec}\mathbb{C}$ (adjoint to $x\mapsto (x,x):\mathbb{C}\to R[T]$) in the category of schemes over $\mathbb{C}$. Therefore, $\mathbb{A}^1_R\to \mathrm{Spec} \mathbb{C}$ is not a 2-dimensional vector space over $\mathbb{A}^1_{\mathbb{C}}\to \mathrm{Spec}\mathbb{C}$ (or a vector space of any dimension in the said category). A more advanced way to see what is happening is by looking on: the sheaf on (the large Zariski site of) $\mathrm{Spec} R$ represented by $\mathbb{A}^1_R\to \mathrm{Spec} R$, compared to: the sheaf on $\mathrm{Spec} \mathbb{C}$ represented by $\mathbb{A}^1_R\to \mathrm{Spec} \mathbb{C}$. The first is easily seen to be naturally isomorphic to the sheaf of rings $\mathcal{O}_{\mathrm{Spec} R}$ (it assigns $Y\to \mathrm{Spec R}$ to $\Gamma(Y,\mathcal{O}_Y)$), which is essentially the same as saying that $\mathbb{A}^1_R\to \mathrm{Spec} R$ is a ring object in the category of $R$-schemes. The second one is just a sheaf of sets and does not posses any natural structure of an $\mathcal{O}_{\mathbb{A}^1_{\mathbb{C}}}$-module (which amounts to an ($\mathbb{A}^1_{\mathbb{C}}\to \mathrm{Spec}\mathbb{C}$)-vector space structure on $\mathbb{A}^1_R\to \mathrm{Spec}\mathbb{C}$). Rather, it is the set sheaf $\mathcal{O}_{\mathbb{A}^1_{\mathbb{C}}}\coprod \mathcal{O}_{\mathbb{A}^1_{\mathbb{C}}}$. [Edit: You can obtain from $\mathbb{A}^1_R$ a two-dimensional "$\mathrm{A}^1_\mathbb{C}$-vector space in the category of $\mathbb{C}$-schemes", i.e. $\mathbb{A}^2_{\mathbb{C}}$, by applying Weil restriction relative to $\mathrm{Spec} R\to \mathrm{Spec }\mathbb{C}$.]<|endoftext|> TITLE: Bounding probability densities on a Wasserstein-2 geodesic QUESTION [5 upvotes]: Consider two probability measures which are supported on a bounded domain $\Omega$ with density functions $p_0$ and $p_1$. It is well-known that for the Wasserstein-2 distance, there exists uniquely a geodesic $\mathcal{L}_t, 0\le t\le 1,$ connecting $p_0$ and $p_1$. That is to say, $Law(\mathcal{L}_0)=p_0,\ Law(\mathcal{L}_1)=p_1$,and for any $0 \le s \le t\le1$, $$ W_2(\mathcal{L}_s,\mathcal{L}_t)=(t-s)W_2(\mathcal{L}_0,\mathcal{L}_1). $$ We use $p_t$ to denote the density function of $\mathcal{L}_t $. My question is that, under what conditions on $p_0$ and $p_1$, there exists a constant $\lambda>0$ such that for all $0\le t\le1$, inequality $ 1/\lambda< p_t < \lambda $ holds universally on $\Omega$? REPLY [5 votes]: It turns out that for most convex domains $\Omega$, one can find smooth probability measures $\rho_0$ and $\rho_1$ which are supported on $\Omega$, have strictly positive density everywhere on $\Omega$, but whose barycenter (in the $2$-Wasserstein sense) has non-convex support (so has support strictly smaller than $\Omega$). This phenomena was recently discovered by Santambrogio and Wang [1]. As such, I'm not sure of sufficient conditions to ensure that a lower bound on the density holds throughout the displacement interpolation. To the best of my knowledge, this question is still open. Given this, the natural question is whether it is possible to establish lower bounds for $d \rho_t$ on its support. The densities of $\rho_0$ and $\rho_1$ do not go to zero at the boundary $\partial \Omega$, so one might imagine that this continues to be the case for $\rho_t$ (although the support is evolving). By contrast, it might also be the case that $d \rho_t$ vanishes continuously at some points within the domain. Presumably both are possible.  Later edit:  After thinking about it further, I realized that Caffarelli's $C^2$ estimate can be used to obtain a lower bound for $d \rho_t$ on its support. As such, the following can be considered as a lower bound of the densities, with the caveat that the support along displacement interpolation can be smaller than the initial or target measures.  Proposition: Suppose that $\Omega$ is a smooth and strongly convex domain and that $\rho_0$ and $\rho_1$ are smooth probability measures supported in $\Omega$. Suppose further that both measures are absolutely continuous with respect to the Lebesgue measure and satisfy $1/C \leq d \rho_i \leq C$ for some $C \geq 1.$ Let $\rho_t$ be 2-Wasserstein geodesic between $\rho_0$ and $\rho_1$. Then for $0 \leq > t \leq 1$, there is a constant $C^\prime>0$ so that $$ d\rho_t > > C^\prime$$ on the support of $\rho_t$ (which might be strictly smaller than $\Omega$). Proof: By Brenier's theorem [2], optimal transport from $\rho_0$ to $\rho_1$ is induced by the sub-differential of some convex function $u$. Furthermore, the 2-Wasserstein geodesic between $d \rho_0$ and $d \rho_1$ is induced by displacement interpolation, which transports mass along straight lines. As such, we have that$$ \rho_t = (\nabla u_t )_\sharp \rho_0.$$ In this formula, $u_t =(1-t) \frac{x^2}{2}+ t \cdot u$ and the sharp notation indicates pushforward of measures. It may seem odd, but $ \nabla \frac{x^2}{2}$ serves as a convenient way to write the identity map. The goal is to derive a lower bound for $d \rho_t $ on its support. Using the change of variables formula, we have the following:  $$ d \rho_t = \frac{d \rho_0}{\det(\nabla^2 u_t)} $$  From this, we see that we need to bound the Jacobian of $\nabla u_t$, which is simply the determinant of the Hessian matrix $$\det \left(\frac{\partial^2 u_t}{\partial x^i \partial x^j} \right). $$ For convenience, we will use the notation $ H(u)$ for the Hessian matrix  $\frac{\partial^2 u}{\partial x^i \partial x^j} $. In order to get the desired estimate, we need to use the structure of the optimal transport. In particular, Brenier's theorem shows that the potential $u$ is a weak solution to the Monge-Ampere equation $$ \det \left( H(u) \right) = \frac{d \rho_0(x)}{d \rho_1( \nabla u (x))} \hspace{2in} (\ast)$$ with appropriate boundary conditions (which are not important here). From our assumptions, we can estimate the right hand side as $$     \frac{1}{ C^2 }     \leq  \frac{d \rho_0(x)}{d \rho_1( \nabla u (x))} \leq C^2, $$ so we have uniform bounds on $\det H(u)$. However, this is not enough. What we actually want is upper bounds on $\det H(u_t)$. Using the fact that the Hessian of $\frac{x^2}{2}$ is the identity matrix, we can expand $\det(H(u_t))$ to obtain the following: $$\det( H(u_t))= \det((1-t) I + t H(u) ) = \Pi_{i=1}^n \left( (1-t)+t \lambda_i \right ).$$ Here, $ \lambda_i$s are the eigenvalues of $H(u)$. From this, we can see that in order to bound $\det(H(u_t))$, we need to bound the eigenvalues of $H(u)$, not simply the determinant. To do this, we use a deep result of Caffarelli. Theorem (Caffarelli [3]): Suppose that $\Omega$ is strongly convex and $\rho_0$ and $\rho_1$ are smooth probability measures supported in $\Omega$ satisfying $1/C \leq > d \rho_i \leq C$ for some $C \geq 1$. Then there exists a $C^\prime$ so that the convex function $u$ solving $(\ast)$ satisfies$$\| u > \|_{C^2} \leq C^\prime.$$ This establishes an a priori $C^2$ estimate for solutions to the Monge-Ampere equation $(\ast)$. Even more strikingly, it implies that $u$ is smooth, uniformly strongly convex, and in fact a classical solution to $(\ast)$. For our purposes, with a uniform $C^2$ estimate on $u$, we can bound all the eigenvalues of $H(u)$, which gives a bound on the Jacobian of $\nabla u_t$. Together with the assumption that $d \rho_0 > 1/C$, this implies a lower bound on the density of $d  \rho_t$, wherever it is nonzero. $ \hspace{2in} \square $ To give an intuitive explanation for this result, we can imagine optimal transport moving a large body of water. Santambroggio and Wang's result shows that dry spots might emerge during this transport, even if there are no dry spots for the initial and final configurations. However, the argument above shows that under additional assumptions, dry spots cannot emerge from the water smoothly receding. Instead, it's more akin to "Moses parting the Red Sea," where there are tall walls of water around the dry areas. For those not familiar with the imagery, I've included a painting by Cornelis de Wael.   [1] Santambrogio, Filippo; Wang, Xu-Jia, Convexity of the support of the displacement interpolation: counterexamples, Appl. Math. Lett. 58, 152-158 (2016). ZBL1345.49056. [2] Brenier, Yann, Décomposition polaire et réarrangement monotone des champs de vecteurs. (Polar decomposition and increasing rearrangement of vector fields), C. R. Acad. Sci., Paris, Sér. I 305, 805-808 (1987). ZBL0652.26017. [3]Caffarelli, Luis A., The regularity of mappings with a convex potential, J. Am. Math. Soc. 5, No. 1, 99-104 (1992). ZBL0753.35031.<|endoftext|> TITLE: Stationarity and Fodor's lemma for a (nice) poset? QUESTION [8 upvotes]: The notion of a stationary set is peculiar in that it applies to subsets of certain very particular posets -- ordinals or powersets. At least to a non-set-theorist, the situation seems to beg for the relevant properties of these posets to be abstracted. I'm wondering if this has been done before. As evidence that it's at least possible, here's one possible approach. The Setting: Definition: Let $\kappa$ be an uncountable regular cardinal. Say that a poset $P$ is $\kappa$-generated by $A \subseteq P$ if the following conditions hold: $P$ is directed and has $\kappa$-small directed joins. Each $a \in A$ is compact in $P$, i.e. $a \leq \vee_i p_i \Rightarrow \exists i\, a \leq p_i$ if the join is directed. For each $p \in P$, the set $A_{\leq p} := \{a \in A \mid a \leq p\}$ is $\kappa$-small. For each $p \in P$, the join $\vee A_{\leq p}$ exists in $P$ and is equal to $p$. Examples: Let $\kappa$ be an uncountable regular cardinal. The wellorder $\kappa$ is $\kappa$-generated by itself. For any set $A$, the restricted powerset $P_\kappa(A)$ is $\kappa$-generated by $A$. For any join-semilattice $A$, the $\kappa$-small Ind-completion of $A$ is $\kappa$-generated by $A$. Some familiar concepts: Definition: Let $\kappa$ be an uncountable regular cardinal and let $P$ be a poset $\kappa$-generated by $A$. A subset $S \subseteq P$ is club if it is closed under $\kappa$-small directed colimits and every $p \in P$ has an upper bound in $S$, and stationary if it meets every club set nontrivially. A filter $\mathcal F \subseteq P(P)$ is fine if for every $a \in A$ it contains the set $P_{\geq a} := \{p \in P \mid p \geq a\}$. A filter $\mathcal F \subseteq P(P)$ is normal if for every $A$-indexed family $(X_a)_{a \in A}$ with $X_a \in \mathcal F$, the diagonal intersection $\Delta_{a \in A} X_a := \{p \in P \mid p \in \cap_{a \leq p} X_a\}$ is in $\mathcal F$. Observation: If $\mathcal F$ is fine and $\kappa$-complete, then $\mathcal F$ contains each $P_{\geq p}$ for $p \in P$. Some familiar theorems: In this setting, we can recover some basic results about stationarity. Applied to the examples (1) and (2) above, one gets some familiar facts. Proposition: Let $\kappa$ be an uncountable regular cardinal and $P$ a poset $\kappa$-generated by $A$. The club sets generate the minimal $\kappa$-complete fine normal filter $\mathcal F_{club} \subseteq P(P)$. Fodor's lemma holds: if $f: P \to A$ is regressive in that $f(s) \leq s$ for each $s \in S$ where $S \subseteq P$ is stationary, then there is a stationary $S_0 \subseteq S$ such that $f|_{S_0}$ is constant. The club filter is generated by the sets $Cl_f = \{p \in P \mid \forall a \in b,\, f(b) \leq p\}$ for $f: P_\omega(A) \to P$. Questions: Are the notions of clubness / stationarity / fineness / normality / diagonal intersection with respect to a poset in the literature somewhere? If so, what are they used for? Is Fodor's lemma known in some form for some general class of posets? REPLY [13 votes]: You may look at the paper Regressive functions and stationary sets by Karsten Steffens (In: Müller G.H., Scott D.S. (eds) Higher Set Theory (Proc. Conf., Math. Forschungsinst., Oberwolfach, 1977), pp. 423–435. Lecture Notes in Mathematics 669 (1978)): Review from Mathscinet: The notions of closed unbounded sets, regressive functions, stationary sets and diagonal intersections are generalized to a wide range of partially ordered sets. The main theme is a generalization of Fodor's theorem and conditions are given under which the generalized theorem holds.<|endoftext|> TITLE: Analogues of the Riemann zeta function that are more computationally tractable? QUESTION [15 upvotes]: Many years ago, I was surprised to learn that Andrew Odlyzko does not consider the existing computational evidence for the Riemann hypothesis to be overwhelming. As I understand it, one reason is as follows. Define the Riemann–Siegel theta function by $$\vartheta(t) := -\frac{t}{2}\log\pi + \arg \Gamma\left(\frac{2it+1}{4}\right)$$ and define the Hardy function $Z(t) := e^{i\vartheta(t)} \zeta(1/2 + it)$. Then $Z(t)$ is real when $t$ is real, and the Riemann hypothesis implies that for $t$ sufficiently large, $Z(t)$ has no positive local minimum or negative local maximum, so its zeros are interlaced with its minima and maxima. On the other hand, as I believe Lehmer ("On the roots of the Riemann zeta-function," Acta Mathematica 95 (1956), 291–298) was the first to point out, there exist "Lehmer pairs" of zeros of $Z(t)$ that are unusually close together, which may be regarded as "near counterexamples" to the Riemann hypothesis. Harold Edwards has suggested that Lehmer pairs "must give pause to even the most convinced believer in the Riemann hypothesis." There is a relationship between Lehmer pairs and large values of $Z(t)$. It is known that $Z(t)$ is unbounded, but it approaches its asymptotic growth rate very slowly. As Odlyzko has explained (see Section 2.9 in particular), there is reason to believe that current computations are not yet exhibiting the true asymptotic behavior of $Z(t)$. So one could argue that the existing computational data about Lehmer pairs is still in the realm of the "law of small numbers." A related observation concerns $S(t) := \pi^{-1}\arg\zeta(1/2 + it)$. Let me quote from Chapter 22 of John Derbyshire's book Prime Obsession, where among other things he reports on a conversation he had with Odlyzko. For the entire range for which zeta has so far been studied—which is to say, for arguments on the critical line up to a height of around $10^{23}$—$S$ mainly hovers between $-1$ and $+1$. The largest known value is around 3.2. There are strong reasons to think that if $S$ were ever to get up to around $100$, then the RH might be in trouble. The operative word there is "might"; $S$ attaining a value near $100$ is a necessary condition for the RH to be in trouble, but not a sufficient one. Could values of the $S$ function ever get that big? Why, yes. As a matter of fact, Atle Selberg proved in 1946 that $S$ is unbounded; that is to say, it will eventually, if you go high enough up the critical line, exceed any number you name! The rate of growth of $S$ is so creepingly slow that the heights involved are beyond imagining; but certainly $S$ will eventually get up to $100$. Just how far would we have to explore up the critical line for $S$ to be that big? Andrew: "Probably around $T$ equals $10^{10^{10,000}}$." Way beyond the range of our current computational abilities, then? "Oh, yes. Way beyond." In light of what I learned from another MO question of mine, about fake integers for which the Riemann hypothesis fails, I got to wondering—If exploring $Z(t)$ and $S(t)$ for the actual Riemann zeta function is hitting our computational limits, could we perhaps gain some insight by computationally studying other zeta functions? More specifically: Are there $L$-functions in the Selberg class for which there are analogues of $Z(t)$ and $S(t)$ which are computationally more tractable than the Riemann zeta function, for which we could computationally explore the analogue of the "$S\approx100$" regime? (Incidentally, I don't understand what is significant about the $S\approx100$ regime. Anybody know?) Are there Beurling generalized number systems for which the analogue of RH fails but which can be shown computationally to mimic the empirically observed behavior of $Z(t)$ and $S(t)$ (including, I guess, the GUE phenomenon)? REPLY [5 votes]: Let $p_k$ be the $k$-th prime number, and pick a sequence of primes $q_k$, such that $q_k\sim p_k^{3/2}$. Let $G$ be the arithmetic semigroup consisting of all integers not divisible by one of the $q_k$. Then $\zeta_G(s)=\zeta(s)\prod_{k=1}^\infty(1-q_k^{-s})$. Now $$ \sum_{k=1}^\infty\log(1-q_k^{-s})-\log(1-p_k^{-3s/2}) =\sum_{k=1}^\infty \frac{p_k^{3s/2}-q_k^s}{(p_k^{3/2}q_k)^s}+H(s), $$ where $H$ is uniformly bounded in $\Re s>1/3+\delta$. From the prime number theorem for short intervals we see that we can pick the sequence $q_k$ in the such a way that $|p_k-q_k^{3/2}|\frac{1}{5}$. We conclude that $\zeta_G(s)=\zeta(s)\zeta(3s/2)^{-1}H(s)$, where $H$ is holomorphic and zero free in $\Re s>\frac{1}{3}$. Assuming RH we find that in the half plane $\Re\;s>\frac{1}{3}$ we have that $\zeta_G$ behaves exactly as $\zeta$, but it has an additional zero at $\frac{2}{3}$. So if you believe in GUE for $\zeta$, then this function satisfies GUE, but not RH. If you define $S(t)$ as the error term in the approximate formula for $N(t)$, then $\zeta_G$ and $\zeta$ share $S(t)$, but $\zeta_G$ does not satisfy RH. However, this example does not say anything about $Z(t)$, because although the function $H(s)$ is holomorphic and zero free, at one point in the computations we took derivatives, thus $H(s)$ might grow exponentially with $|\Im s|$. It is not clear whether you can pick the sequence $q_k$ in such a way that $\zeta_G$ has only polynomial growth. A more natural example to study would be Selberg $\zeta$-functions. These functions behave pretty much like the Riemann $\zeta$-function. Selberg $\zeta$-functions satisfy RH, with possibly some real exceptional zeroes (and there are cases where these exceptions actually occur). However, in this case $S(t)$ is much larger than in Riemann's case, so this might not be such a good analogue.<|endoftext|> TITLE: Monoidal Equivalence for Drinfeld--Jimbo Quantum Groups QUESTION [9 upvotes]: For $U_q(\frak{g})$ the Drinfeld--Jimbo quantum group, its category of representations is equivalent to the category of representations of $U(\frak{g})$, or equivalently the category of Lie algebra representations of $\frak{g}$. Both categories have an obvious monoidal structure, what is not obvious is if this is an equivalence of monoidal categories. Edit: As Phil's comment below, this is not a monodial equivalence. As the linked answer says, the problem is that the associators in both cases are different. What is the easiest way to see that this is so? The discussion about 6j symbols and coordinates on stacks is unfortunately lost on me. In fact, for modules over a Hopf algebra the assoicators look easy to the naive untrained eye, why is it about the assoiators on either category that is non-trivial? REPLY [5 votes]: For simplicity let’s just do the $\mathfrak{sl}(2)$ case. Let X be the 2-dimensional natural representation. From the fusion rules, $\mathrm{Hom}(X \otimes X,1)$ and $\mathrm{Hom}(1, X \otimes X)$ are one-dimensional. Choose a map in each normalized such that the zig-zag is the identity: $$X = X \otimes 1 \rightarrow X \otimes (X \otimes X) \rightarrow (X \otimes X) \otimes X \rightarrow 1 \otimes X = X.$$ Note that this fixes the maps up to rescaling them in opposite ways. Now compute the quantum dimension (aka the circle): $$1 \rightarrow X \otimes X \rightarrow 1.$$ This is a numerical invariant of such a monoidal category and a calculation shows that it is $q+q^{-1}$. Thus the monoidal categories are not equivalent except for replacing q by its inverse. In the classical case these maps are the determinant map and the copairing associated to the determinant map. I.e. since $\det(e_1, e_1) = 0$, $\det(e_1, e_2) = 1$, $\det(e_2, e_1) = -1$, $\det(e_2, e_2) = 0$, the copairing sends 1 to $e_1 \otimes e_2 - e_2 \otimes e_1$. Thus the value of the circle is $$\det(e_1 \otimes e_2 - e_2 \otimes e_1) = 1+1 = 2.$$ (Some side notes for experts. If $X$ were not self-dual you'd have to replace some of the $X$'s above with $X^{*}$ which results in some trickiness, the upshot of which is you either need a pivotal structure, or (following ENO) you compute $\mathrm{qdim}(X)\mathrm{qdim}(X^{*})$ to get a monoidal invariant. This wouldn't quite let you distinguish everything because of switching q and -q. When $X$ is self-dual something very tricksy is going on with the signs here... I will update this bullet when I understand it better.) REPLY [5 votes]: First of all, the fact that those are equivalent as mere categories is true only for (locally-) finite dimensional representations, and when $q$ is generic enough. This part is somewhat "easy". Then, this is a deep theorem by Drinfeld and Kazhdan--Lusztig that they become equivalent as (braided) monoidal categories, if on the $U(\mathfrak g)$ side one keeps the same tensor product, but introduces a highly non trivial associator coming from the so-called KZ equation. Hence you're right that the associator for modules over an Hopf algebra is trivial, but then on the quantum side the coproduct, hence the formula defining the tensor product of modules, is complicated. On the non-quantum side, one keeps the same simple coproduct/tensor product, but at the cost of having to put a highly non-trivial associator, which of course is not equivalent to the identity. A nice exposition can be found there: https://arxiv.org/abs/0711.4302 Edit: to clarify a point raised in OP's edit: your first use of the word "associator" here means what people also call fusion rules which are non-trivial even if the associator in the sense of associativity constraints of monoidal categories (which is what your second use of the word "associator" means) are identities. Hence the bottom line is that the fusion rules of $U_q$ and $U$ are different, because their coproducts are different, and that the former is obtained from the latter by plugging this KZ associator.<|endoftext|> TITLE: Largest Eigenvalue of a Matrix with Special Form in terms of n QUESTION [5 upvotes]: In one step of solving a difficult problem, I would like to know the largest eigenvalue of a matrix with this pattern: $$A_n = \begin{bmatrix} 0 & 0 & 0 & 0 &\dots & 0 \\ 0 & 1 & 1 & 1&\dots & 1 \\ 0 & 1 &2 &2 &\dots &2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & 3 & \dots& n-1 \end{bmatrix}$$ In other words, we can "peel" the "L" shaped layers from upper left corner and all layer consist of the same entry. This matrix is symmetric, so all eigenvalues are real. It is of rank $n-1$, and numerical experiment demonstrates that it is positive semidefinite. I want to know the largest eigenvalue in terms of $n$, or have a lower bound on the largest eigenvalue in terms of $n$. I performed numerical experiment on various $n$ and it seems like we can quickly get $\lambda_{max}\approx 0.4n^2$ for large $n$. However, I don't know how to derive this result. I tried to look at the characteristic polynomial, but I could not find a pattern. So anyone has an idea about deriving a (lower) bound on the largest eigenvalue of this matrix? Thanks! REPLY [6 votes]: Your matrix has entries given by $a_{ij}=\min(i,j)$, where $0\le i,j\le n-1$. Have a look at Section 3 of this paper of mine for a derivation of explicit bounds.<|endoftext|> TITLE: A trace-constrained maximization problem in the cone of positive definite matrices QUESTION [9 upvotes]: Let $A\in\mathbb{R}^{n\times n}$ be a matrix having eigenvalues with strictly negative real part (in other words, $A$ is supposed to be Hurwitz stable). Let $\mathrm{tr}(\cdot)$ denote the trace operator and $I$ the identity matrix. Let $>$ be the standard partial order in the cone of positive definite matrices. Consider the following optimization problem $$\tag{$\star$}\label{prob} m(A) :=\sup_{\substack{X\in\mathbb{R}^{n\times n}\\ X>0,\ \mathrm{tr}(X)=1}} \mathrm{tr}\left(X(I+P)^{-1}\right), $$ where $P$ is the (unique) positive definite solution of the following Lyapunov equation $AP+PA^\top=-X$. My question. Suppose that $A$ is upper triangular, does the following inequality $$ m(A)\ge \frac{2\,\mathrm{tr}\,A}{2\,\mathrm{tr}\,A-1} $$ always hold true? A special case. Notice that if $A+A^\top<0$, then the answer is in the affirmative. To see this, just pick $P^\star= -\frac{1}{2\mathrm{tr}(A)}I$ and observe that $X^\star=-(AP^\star+P^\star A^\top)$ is positive definite and satisfies $\mathrm{tr}(X^\star)=1$. Numerical evidences. Quite surprisingly, after runnning an extensive number of numerical simulations, it seems that the answer is in the affirmative for any (upper triangular Hurwitz stable) $A$. More precisely it seems that (modulo numerical errors of magnitude $\sim 10^{-6}$) the conjectured inequality actually holds with equality, that is $$ m(A) = \frac{2\,\mathrm{tr}\,A}{2\,\mathrm{tr}\,A-1}. $$ However, this fact does not seem trivial to prove (I spent quite some time thinking about this, but I didn't manage to prove it); so any help in clarifying this conjecture is greatly appreciated. Thanks! An (perhaps useful?) equivalent formulation. By plugging $X=-AP-PA^\top$ into the trace functional in \eqref{prob}, the latter can be rewritten as \begin{align}\tag{$\star\star$} m(A) &=\sup_{\substack{X\in\mathbb{R}^{n\times n}\\ X>0,\ \mathrm{tr}(X)=1}} -\mathrm{tr}\left((AP+PA^\top)(I+P)^{-1}\right)\notag \\ &=\sup_{\substack{X\in\mathbb{R}^{n\times n}\\ X>0,\ \mathrm{tr}(X)=1}} -2\mathrm{tr}\left(A(I+P^{-1})^{-1}\right)\\ &=-2\inf_{\substack{X\in\mathbb{R}^{n\times n}\\ X>0,\ \mathrm{tr}(X)=1}} \mathrm{tr}\left(A(I+P^{-1})^{-1}\right)\notag\\ &=-2\inf_{\substack{P\in\mathbb{R}^{n\times n}\\ P>0,\ \mathrm{tr}(AP)=-\frac{1}{2}\\ AP+PA^\top<0}} \mathrm{tr}\left(A(I+P^{-1})^{-1}\right)\\ &\overset{(\#)}{=}-2\inf_{\substack{P\in\mathbb{R}^{n\times n}\\ P>0,\ \mathrm{tr}(AP)=-\frac{1}{2}\\ AP+PA^\top<0}} \mathrm{tr}\left(A-A(I+P)^{-1}\right) \\ &=-2\,\mathrm{tr}\,A + 2\sup_{\substack{P\in\mathbb{R}^{n\times n}\\ P>0,\ \mathrm{tr}(AP)=-\frac{1}{2}\\ AP+PA^\top<0}} \mathrm{tr}\left(A(I+P)^{-1}\right), \\\label{prob-eq} \end{align} where in (#) I used the Woodbury matrix identity. REPLY [2 votes]: Not a full answer, but some ideas. From the last line of your equivalent formulation, we need to solve $$ J^{*} = \sup_{\substack{P\in\mathbb{R}^{n\times n}\\ P>0,\ \mathrm{tr}(AP)=-\frac{1}{2}}} \mathrm{tr}\left(A(I+P)^{-1}\right), $$ which, after letting $Q:=I + P$, and $a:=\text{tr}(A)$, becomes $$J^{*} = \sup_{\substack{Q\in\mathbb{R}^{n\times n}\\ Q>I,\ \mathrm{tr}(AQ)=a-\frac{1}{2}}} \mathrm{tr}\left(AQ^{-1}\right).$$ If $A$ were negative diagonal then surely $\mathrm{tr}\left(AQ^{-1}\right)$ would have been concave, making the above a convex optimization problem. (Question: does concavity still hold for Hurwitz $A$?) In any case, the first order optimality conditions give: $\lambda Q A Q = A$, where $\lambda >0$ is the Lagrange multiplier, and $\mathrm{tr}(AQ)=a-\frac{1}{2}$. Taking trace of $\lambda Q A Q = A$ gives $J^{*}=\lambda(a-1/2)$, where $\lambda>0$ remains to be determined (not sure yet how).<|endoftext|> TITLE: A question about the vanishing of motivic cohomology in negative Tate twist QUESTION [9 upvotes]: Let $DM_{\text{gm}}$ be the category of Voevodsky´s geometric motives. Let $p,q\in \mathbb{Z}$ be integers with $p<0$. Is it true that $$\text{Hom}_{DM_{\text{gm}}}(M_{\text{gm}}(X),\mathbb{Z}(p)[q])=0,$$ where $M(X)$ is the motive of a smooth scheme $X$ over a field $k$ and $\mathbb{Z}(p)$ is the Tate motive? REPLY [5 votes]: Here's an easy way to see this (which is more or less an elaboration of Mikhail's answer): Suppose $i > 0$ and $n\in \mathbb Z$. We have the Thom isomorphism $$ H^n(X, \mathbb Z(-i)) \cong H^{n+2i}_X(\mathbb A^i_X, \mathbb Z(0)). $$ Weight $0$ motivic cohomology is just Zariski cohomology, so $$ H^n(X,\mathbb Z(0)) = \begin{cases} \mathbb Z^{\pi_0(X)} & \text{if }n=0,\\ 0 & \text{otherwise.}\end{cases} $$ Now the long exact sequence for cohomology with support shows that $H^*_X(\mathbb A^i_X,\mathbb Z(0))=0$.<|endoftext|> TITLE: Analogue of the original Birch–Swinnerton-Dyer conjecture for abelian varieties QUESTION [9 upvotes]: $\newcommand{\Q}{\Bbb Q} \newcommand{\N}{\Bbb N} \newcommand{\R}{\Bbb R} \newcommand{\Z}{\Bbb Z} \newcommand{\C}{\Bbb C} \newcommand{\F}{\Bbb F} \newcommand{\p}{\mathfrak{p}} $ Let $A$ be an abelian variety over a number field $F$. It is expected that the $L$-function of $A$ has analytic continuation to $\Bbb C$ and satisfies a functional equation relating $s$ to $2-s$. In that setting, the (generalized) Birch–Swinnerton-Dyer conjecture states that $$\mathrm{ord}_{s=1}(L(A_{/F},s)) = \mathrm{rk}_{\Z}(A(F)) =: r.$$ Originally, the conjecture for an elliptic curve $E$ over $\Q$ was $$\exists C>0,\quad \prod_{p \leq x} \dfrac{|E(\F_p)|}{p} \sim C \;\mathrm{log}(x)^r \qquad (x \to \infty).$$ My question is to know what is the analogue of the original conjecture, in the framework of abelian varieties over number fields. My first guess would to replace to LHS by $$\prod_{N(\p) \leq x} L_{\p}(A_{/F}, N(\p)^{-1}),$$ where $L_{\p}(A_{/F},s)$ is the local factor of the L-function of $A$ at $\p$. But I'm not sure what the RHS should be. Typically, how does it depend on the dimension of $A$ or on the degree of the number field? REPLY [4 votes]: $\newcommand{\p}{\mathfrak{p}}$By Theorem 6.3 of this paper by Keith Conrad, strong conjectures about $L(A,s)$ (stronger than GRH for this $L$-function, but still "believable"), imply that $$ \prod_{N\p\le x}L_{\p}(A,N\p^{-1}) \sim C (\log x)^r $$ where $r$ is the order of vanishing at $s=1$, which by the usual BSD should be the rank of $A(K)$. See also this question about the relation between the original and the modern formulation of BSD.<|endoftext|> TITLE: Can two non-equivalent polytopes of same dimension have the same graph? QUESTION [7 upvotes]: By a polytope I mean the convex hull of finitely many points. The graph of a polytope is the graph isomorphic to its 1-skeleton. By equivalence of polytopes I mean combinatorial equivalence, i.e. their face lattices are isomorphic. I know that two polytopes can have isomorphic graphs while being non-equivalent, e.g. neighborly polytopes. However, all examples I know of are polytopes of different dimension. So I wonder: Question: Can there be two non-equivalent polytopes of the same dimension with the same graph? Especially, are all $k$-neighborly polytopes of the same dimension equivalent? REPLY [4 votes]: There are many non-equivalent neighborly polytopes, already in dimension $d=4$. See for example Arnau Padrol. "Many Neighborly Polytopes and Oriented Matroids"<|endoftext|> TITLE: Kernels of homomorphisms of group schemes QUESTION [5 upvotes]: Let $S$ be some base scheme, $H$ a finite flat group scheme over $S$, and $\alpha: \mu_p \to H$ a homomorphism of group schemes ($p$ a prime). Is the kernel of $\alpha$ necessarily flat over $S$? (I know that kernels of general homomorphisms of FFGS $G \to H$ need not be flat, but I don't know of a counterexample when $G = \mu_p$.) REPLY [8 votes]: This holds for any homomorphism $f: G\to H$ with $G$ of multiplicative type and of finite type, and $H$ separated and finitely presented. Here I assume that by "finite flat" you mean "finite locally free". Reference: SGA3, IX, Thm 6.8.<|endoftext|> TITLE: Are all real-closed subfields of $\overline{\mathbb{Q}}$ conjugate? QUESTION [8 upvotes]: Let $\overline{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$. The absolute galois group $G_\mathbb{Q}$ of $\mathbb{Q}$ acts on the set of real-closed subfields of $\overline{\mathbb{Q}}$. Does it act transitively? The real-closed subfields are in bijection with the involutions of $G_\mathbb{Q}$ under the Galois correspondence, so another way to ask the question would be, Do the involutions of $G_{\mathbb{Q}}$ form a single conjugacy class? I am asking out of curiosity. I have been unable to locate the answer in any of my texts on real fields, or via internet search, but please forgive me if it is well known. Because the order on a real-closed field is unique, and $\mathbb{Q}$ is order-dense in its real closure, the real-closed subfields $K$ of $\overline{\mathbb{Q}}$ have trivial automorphism group, and it follows that the stabilizer of each $K$ for the action of $G_\mathbb{Q}$ is just the involution fixing $K$ pointwise; thus the action is almost free, and this made me curious if it is transitive. REPLY [12 votes]: Any real-closed subfield $R\subseteq\overline{\mathbb Q}$ is a real closure of $\mathbb Q$ (being real closed and algebraic over $\mathbb Q$). Thus, by uniqueness of real closures, any two such fields are isomorphic, and an isomorphism of $R$ to $R'$ extends to an isomorphism of $R(i)=\overline{\mathbb Q}$ to $R'(i)=\overline{\mathbb Q}$, i.e., an automorphism of $\overline{\mathbb Q}$.<|endoftext|> TITLE: DGLA controlling deformation of holomorphic curves QUESTION [9 upvotes]: Suppose $C$ is a compact Riemann surface and $X$ is a compact Kähler manifold. Suppose $f:C\to X$ is a stable holomorphic map. Then, the deformations of $f$ are controlled by the complex $L^\bullet = R\Gamma(C,df:T_C\to f^*T_X)$. Explicitly, this complex may be realized using the Dolbeault resolution of $T_C$ and $f^*T_X$. In this realization, there are three terms in this complex: $L^0 = \Omega^0(C,T_C)$, $L^1 = \Omega^{0,1}(C,T_C)\oplus\Omega^0(C,f^*T_X)$ and $L^2 = \Omega^{0,1}(C,f^*T_X)$ with the differentials $L^0\to L^1$ and $L^1\to L^2$ given by a sum of pushforward by $df$ and the canonical $\bar\partial$ operator on a holomorphic vector bundle. By some general philosophy (for example in the deformation theory book by Kontsevich-Soibelman), $L^\bullet$ should carry the structure of a differential graded Lie algebra (DGLA) such that the deformations of $f$ over a local Artin ring $(A,\mathfrak m)$ with residue field $\mathbb C$ can be seen as solutions $\omega\in L^1\otimes\mathfrak m$ to the Maurer-Cartan equation $d\omega + \frac12[\omega,\omega] = 0$ modulo the gauge action of $\exp(L^0\otimes\mathfrak m)$. Can we realize the DGLA structure in this case explicitly? In particular, what is the explicit expression for the bracket $[\cdot,\cdot]:L^1\otimes L^1\to L^2$? I am able to see that the degree zero bracket $L^0\otimes L^0\to L^0$ should be simply the usual commutator Lie bracket of vector fields. REPLY [10 votes]: Firstly, I assume you mean deformations of $C$ over $X$ ("deformations of $f$" is ambiguous, as it could mean fixing neither or both of $C$ and $X$). The DGLA philosophy is then that there should exist some DGLA quasi-isomorphic to the explicit realisation of the complex $L$ you wrote down. It doesn't guarantee a DGLA structure on $L$ itself, though it will transfer a non-canonical $L_{\infty}$ structure. In this case, you can reinterpret the problem as trying to deform $\mathcal{O}_C$ as a sheaf of $f^{-1}\mathcal{O}_X$-algebra. The DGLA you want should then be an explicit model for $\mathbf{R}\Gamma(C,\mathbf{R}\mathrm{Der}_{f^{-1}\mathcal{O}_X}(\mathcal{O}_C))$. At this point, you encounter the problem that free algebra resolutions and flabby sheaf resolutions don't interact well. One explicit model is given by first forming the Harrison complex (or a natural analogue for holomorphic functions) $\mathrm{Harr}_{f^{-1}\mathcal{O}_X}(\mathcal{O}_C)$ (a sheaf of DGLAs), then take a nice open cover $\mathfrak{U}$ of $C$ and form a Cech complex $\check{C}(\mathfrak{U},\mathrm{Harr}_{f^{-1}\mathcal{O}_X}(\mathcal{O}_C))$, giving a cosimplicial DGLA. Then apply Thom-Whitney cochains to give a DGLA. You'll find various related constructions in several works by Iacono, Manetti and Fiorenza, as well as Ciocan-Fontanine's derived Hilbert schemes and some of my early papers.<|endoftext|> TITLE: Short arc of small curvature whose every planar shadow crosses itself QUESTION [14 upvotes]: Let $\cal{A}$ denote the family of all twice-differentiable simple open arcs $A$ in $\mathbb{R}^3$ satisfying the following properties: (1) at each point of every arc $A\in\cal{A}$ the curvature of $A$ is at most $1$; (2) no perpendicular projection of $A$ to any plane is an injection. Question 1. What is the infimum of the length $|A|$ of $A\in\cal{A}$ ? Question 2. Is there an arc $A_{min}$ of minimum length among all arcs in $\cal{A}$? If so, is it unique (up to isometry)? Remark. This question is related to the notion of rope knots, see: Hans Stricker, Some questions about ideal knots REPLY [2 votes]: This doesn't seem close to tight, but there's an easy lower bound of $\pi$. If you have a curve shorter than $\pi$, choose any projection direction perpendicular to the curve at its midpoint. The resulting projected curve will be monotonic and therefore non-self-intersecting.<|endoftext|> TITLE: Countable version of Erdös-Lovasz-Faber conjecture QUESTION [5 upvotes]: Let $X$ be an infinite set, and let $(A_n)_{n\in\omega}$ be a collection of subsets of $X$ with the following properties: $|A_m\cap A_n| \leq 1$ for $m\neq n\in \omega$, and $|A_n|=\aleph_0$ for all $n\in \omega$. We consider the following statement: (EFL$_\omega$:) There is $f:X\to \omega$ such that for all $n\in\omega$ the restriction $f|_{A_n}:A_n\to\omega$ is a bijection. Questions. Is (EFL$_\omega$) true? Or does (EFL$_\omega$) imply the original Erdös-Faber-Lovasz conjecture? REPLY [7 votes]: If I understand it correctly, it's false. Let $x \notin A_0 = \{1,2,\dots \}$. Then let $A_i$ all meet at $x$, and also each meet $A$ at $i$ (add extra elements as necessary; they should be irrelevant). Then $f(x) \neq f(i)$ for any $i$, so $f(x) \notin f(A)$. This didn't work in the finite case because the sets meeting $A$ cannot exhaust $A$, as the number of such sets is strictly less than the number of elements of $A$. When working over infinite sets, this is no longer true.<|endoftext|> TITLE: Connectivity of suspension-loop adjunction QUESTION [7 upvotes]: Let $X$ be a $k$-connected spectrum for $k \in \Bbb{Z}$. I want to deduce how connected the counit of $(\Sigma^\infty, \Omega^\infty)$- adjunction is, that is, how connected is the map $$ \Sigma^\infty\Omega^\infty X \to X. $$ Any help would be appreciated. REPLY [9 votes]: Indeed, as John Klein shows, the map is $(2k+2)$-connected. Let me offer an alternative proof of the fact that, for $X$ a $k$-connective spectrum, $k\geq 0$, the homomorphism $\pi_i\Sigma^\infty\Omega^\infty X\rightarrow\pi_i X$ is an isomorphism for $i\leq 2k+1$. As indicated by John, this homomorphism is surjective for all $i$ and it suffices to show that the canonical homomorphism $\pi_i\Omega^\infty X\rightarrow \pi_i\Sigma^\infty\Omega^\infty X$ is an iso for $i\leq 2k+1$. We will actually prove that this is more generally true replacing $\Omega^\infty X$ with a $k$-connected space $Y$ with abelian fundamental group and vanishing Whitehead products, e.g. a single loop space. I though that someone might eventually find this useful, even if it is not what you're asking about. The Freudenthal suspension theorem shows that, for an $k$-connected space $Y$, the suspension operator \[ \Sigma_*\colon \pi_i Y \longrightarrow \pi_{i+1}\Sigma Y \] is an isomorphism for $i\leq 2k$ and an epimorphism for $i=2k+1$. This and the fact that $\Sigma^j Y$ is $(j+k)$-connected clearly shows that the natural map from the homotopy groups of $Y$ to those of its suspension spectrum $\Sigma^\infty Y$ \[ \pi_i Y \longrightarrow \pi_{i}\Sigma^\infty Y, \] which is a transfinite composition of suspension operators, is an iso for $i\leq 2k$ and epi for $i=2k+1$. Moreover, by the theorems of Blakers and Massey on the homotopy groups of triads, we have an exact sequence in the critical dimension \[ \pi_{k+1}Y\otimes \pi_{k+1}Y\stackrel{[-,-]}\longrightarrow \pi_{2k+1}Y\stackrel{\Sigma_{*}}\twoheadrightarrow \pi_{2k+2}\Sigma Y \] where the first arrow is the Whitehead product. This still makes sense for $k=0$ using the non-abelian tensor product of groups and the Brown-Loday non-abelian Van Kampen theorems. In this case $[-,-]$ is the commutator product. If Whitehad products in $\pi_*Y$ vanish and the fundamental group is abelian, then the suspension operator is also an isomorphism for $i=2k+1$, and we get an isomorphism between the homotopy groups of $Y$ and $\Sigma^\infty Y$ in this dimension too.<|endoftext|> TITLE: analogue of Theorem of Mattuck for Abelian varieties over $\mathbf{F}_q(\!(t)\!)$ QUESTION [7 upvotes]: By a theorem of Mattuck [Abelian Varieties over $p$-Adic Ground Fields, Annals of Mathematics, Second Series, Vol. 62, No. 1 (Jul., 1955), pp. 92-119], for an Abelian variety $A$ of dimension $g$ over a $p$-adic local field $K$ with ring of integers $O_K$, there is an exact sequence $0 \to O_K^g \to A(K) \to \{\mathrm{finite}\} \to 0$. Is there an analogue of this theorem for Abelian varieties over $\mathbf{F}_q(\!(t)\!)$ (e.g. with $O_K^g$ replaced by a pro-$p$-group)? The only thing I found is a remark in [Milne, Arithmetic Duality Theorems, Remark I.3.6, bottom of p. 45]. For elliptic curves, there is Proposition 10.2.26 (Filtration of $E(K)$) in [Liu, Algebraic Geometry and Arithmetic Curves, 2nd ed.]. REPLY [4 votes]: By [Serre, Lie Algebras and Lie Groups], p. 116, Theorem and p. 118, Corollary 2, there is an open subgroup of $A(K)$ which is a pro-$p$-group. Therefore, it suffices to show that the torsion subgroup of $A(K)$ is finite, because then there is an exact sequence (algebraically and topologically) $$0 \to P \to A(K) \to F \to 0$$ with $P$ a torsion-free pro-$p$-group and $F$ finite (shrink the pro-$p$-group such that it contains no torsion). By potential semistable reduction, one can assume $A/K$ semi-stable, and since it is clear for tori, $A/K$ of good reduction. Then use [Clark, Pete L. and Xarles, Xavier: Local bounds for torsion points on abelian varieties. In: Can. J. Math., 60(3) (2008), 532–555], Proposition 10 to show that the torsion subgroup of $A(K)$ is finite. Note that in contrast to $p$-adic local fields, $A(K)/p$ is not finite, but uncountably infinite. (Use flat cohomology and the Kummer sequence for $A/K$ giving $$0 \to A(K)/p \to \mathrm{H}^1_\mathrm{fppf}(K,A[p]) \to \mathrm{H}^1(K,A)[p] \to 0$$ and [Milne, Arithmetic Duality Theorems], Theorem III.7.8.) By [Serre, Galois cohomology], §1.1 exercise 1), $P$ is isomorphic to an infinite product of copies of $\mathbf{Z}_p$.<|endoftext|> TITLE: Submanifolds whose second fundamental form has constant rank in every direction QUESTION [5 upvotes]: Let $M$ be a submanifold of a Riemannian manifold $\widetilde{M}$. Let $A$ be the second fundamental form of $M$. Suppose that, for all $p \in M$, the linear map $A(v, \cdot)\: \colon T_{p}M \to N_{p}M$ has rank $k$ for every nonzero $v \in T_{p}M$. Does this condition define some well-known class of submanifolds (if any)? REPLY [5 votes]: Sometimes, this is an open condition, so that it does not impose any differential equations on the submanifold. For example, consider the case of a surface $\Sigma^2$ in a $4$-manifold $M^4$. The condition that the rank of $A(v,\cdot):T_p\Sigma\to N_p\Sigma$ be equal to $2$ for every nonzero $v\in T_p\Sigma$ is an open condition on the submanifold $\Sigma$, so once one has such a submanifold, any $C^2$-nearby submanifold will also have this property. Probably, someone has a name for this class of surfaces, but I don't think that it is well-known. There are many examples of such surfaces in $\mathbb{R}^4$. For example, let $\Sigma\subset\mathbb{C}^2$ be a holomorphic curve without flexes regarded as a smooth surface in $\mathbb{R^4}=\mathbb{C}^2$.<|endoftext|> TITLE: The homology groups of the smooth locus of a singular variety QUESTION [14 upvotes]: Let $X$ be a complex irreducible variety and denote its smooth locus by $X^{smooth}$. I would like to know what can be said about the induced maps $H_i(X^{smooth};\mathbb{Q})\rightarrow H_i(X;\mathbb{Q})$ for $i$ small when the codimension of the singular locus is large. Of course, since the ambient space $X$ may not be a manifold it is not possible to perturb cycles away from the singular locus by the usual transversality argument. Moreover, $X$ may be contractible without $X^{smooth}$ having the same property; think about the affine cone over a projective variety. So what I am really looking forward to is for the inclusion to induce surjections $H_i(X^{smooth};\mathbb{Q})\rightarrow H_i(X;\mathbb{Q})$ for small $i$'s within a range determined by the codimension of the singular locus. You can assume that $X$ is normal if it helps. I am aware of similar results when it comes to low-dimensional homotopy groups: https://arxiv.org/pdf/1412.1483.pdf https://arxiv.org/pdf/1412.0272.pdf but I couldn't find anything about comparing the rational homology of the smooth locus with that of the ambient variety. Can anyone help in giving direction in this regard? REPLY [14 votes]: I am adding some additional details to the comment above, since somebody else asked me about this recently. Results about extensions of cohomology classes to all of $X$ from an open subset $U=X\setminus Z$ (or dually, proving that homology classes are obtained by pushforward from an open subset) are usually called Purity Theorems in algebraic geometry. In the setting that both $X$ and $Z$ are smooth and pure-dimensional, there is a very strong Purity Theorem in étale cohomology. One reference is Theorem 16.1 and Corollary 16.2, p. 108 of Milne's book. J. S. Milne Lectures on étale cohomology https://www.jmilne.org/math/CourseNotes/LEC.pdf Taking inverse limits of $\mathbb{Z}/\ell^r\mathbb{Z}$ coefficients and then inverting $\ell$, this gives a theorem in $\ell$-adic cohomology. If $X$ is defined over $\mathbb{C}$, this gives a theorem in singular cohomology with coefficients in a characteristic $0$ field by comparison theorems. Then using the Universal Coefficients Theorem, this gives the corresponding theorem in homology with $\mathbb{Q}$-coefficients, as you ask. However, that is all in the smooth case. For low degree cohomology and its algebraic avatars (étale fundamental groups, Picard groups, Brauer groups, ...), there are other Purity Theorems that usually require much less than smoothness. One typical hypothesis is that $X$ is everywhere locally a complete intersection (LCI). Two great references are SGA 2 and Grothendieck's three exposes, "Le groupe de Brauer" in "Dix exposes sur la cohomologie des schemas". First, regarding purity for connectedness, i.e., $\pi_0$, there is $S2$ extension which is greatly generalized in Hartshorne's Connectedness Theorem. This says that if you remove a Zariski closed subset $Z$ from a pure-dimensional variety $X$ that is LCI, or even just $S2$, this does not change $\pi_0$ provided that $Z$ everywhere has codimension $\geq 2$ (regardless of singularities of $X$ and $Z$ beyond the $S2$ hypothesis). The hypothesis on the codimension is necessary -- just consider a quadric hypersurface of rank $2$ and $Z$ is its codimension $1$ singular locus. The next result is purity for the étale fundamental group. The Purity Theorem is Théorème X.3.4, p. 118, of SGA 2. MR2171939 (2006f:14004) Grothendieck, Alexander Cohomologie locale des faisceaux cohérents et théorèmes de Lefschetz locaux et globaux Séminaire de Géométrie Algébrique du Bois Marie, 1962. Augmenté d'un exposé de Michèle Raynaud. With a preface and edited by Yves Laszlo. Revised reprint of the 1968 French original. Documents Mathématiques (Paris) 4. Société Mathématique de France, Paris, 2005. https://arxiv.org/abs/math/0511279 If $X$ is LCI and if the codimension of $Z$ is everywhere at least $3$, then the pushforward map of étale fundamental groups from $U$ to $X$ is an isomorphism. The hypothesis on the codimension is necessary -- just consider a quadric hypersurface of rank $3$ and $Z$ is its codimension $2$ singular locus, cf. Exercise II.6.5 in Hartshorne's Algebraic geometry. The next Purity Theorem is for Picard groups (roughly an $H^2$ result rather than the $H^1$ result coming from fundamental groups). The first results were proved by Auslander-Buchsbaum for $X$ smooth (their famous theorem, later considered also by Serre, about factoriality of regular local rings). The LCI case was conjectured by Samuel and proved by Grothendieck: Théorème XI.3.13 and Corollaire XI.3.14 of loc. cit. Here the hypothesis is that $Z$ has codimension at least $4$. The hypothesis on the codimension is necessary -- just consider a quadric hypersurface of rank $4$ and $Z$ is its codimension $3$ singular locus, cf. Exercise II.6.5 in Hartshorne's Algebraic geometry. For Brauer groups (e.g., related to torsion in $H^3$), the first Purity Theorem is due to Grothendieck, Théorème 6.1, p. 135 of "Le groupe de Brauer, III". MR0244271 (39 #5586c) Grothendieck, Alexander Le groupe de Brauer. III. Exemples et compléments. Dix exposés sur la cohomologie des schémas, 88–188 Adv. Stud. Pure Math., 3, North-Holland, Amsterdam, 1968. For $X$ in characteristic $0$, this theorem has a smoothness hypothesis on $X$, but the only hypothesis on $Z$ is that it everywhere has codimension $\geq 2$. In case of a complex variety $X$ that is LCI but not necessarily smooth, Grothendieck made a number of conjectures at the end of SGA 2. Those were eventually proved by Hamm and Lê, and then reproved by Goresky-MacPherson using stratified Morse theory. One reference is the theorem on p. 199 of their book. MR0932724 (90d:57039) Goresky, Mark; MacPherson, Robert Stratified Morse theory. Ergebnisse der Mathematik und ihrer Grenzgebiete, 14. Springer-Verlag, Berlin, 1988. Because Goresky and MacPherson are proving such a general and precise version of the theorem, it is a bit difficult to parse. However, in the main case that $X$ is LCI, the point is that if $Z$ everywhere has codimension $\geq c$, then we can choose the general linear section $H$ of that theorem to have codimension $b=\text{dim}(X)-c+1$, so that $H$ is disjoint from $Z$. Then the integer $\widehat{n}$ of that theorem simply equals $\text{dim}(X)-b=c-1$ (Edit this is not $\text{dim}(X)-c$ as previously written). Thus, applying the theorem first to $X$ and then to $U=X\setminus Z$ (since $H$ is contained in $U$), the result is that the pushforward map on homotopy groups, $$\pi_i(U)\to \pi_i(X),$$ is an isomorphism for $i< \widehat{n}=c-1$, and the map is a surjection if $i$ equals $\widehat{n}=c-1$. By Hurewicz, the same thing holds if we replace $\pi_i(-)$ by $H_i(-)$. If you use $\mathbb{Q}$-coefficients and then use the Universal Coefficients Theorem, you get the analogous result in cohomology. You can also get cohomology result with torsion coefficients, e.g., for a complex variety $X$ that is LCI, for a Zariski closed subset $Z$ of $X$ whose codimension is everywhere $\geq 4$, the map of Brauer groups from $X$ to $X\setminus Z$ is an isomorphism. (Does anybody know a good example proving this codimension hypothesis is best possible? Quadratic hypersurfaces have trivial Brauer group, so that does not seem to work.) Thus, for every complex variety $X$ that is LCI and whose singular locus $Z$ of $X$ everywhere has codimension $\geq c$, the pushforward map, $$H_i(X^{\text{smooth}},\mathbb{Q}) \to H_i(X,\mathbb{Q}),$$ is an isomorphism for $i TITLE: HOMFLYPT vs. Jones vs. Alexander polynomial? QUESTION [6 upvotes]: I'm searching for examples (perhaps the simplest one?) to show that the HOMFLYPT polynomial is stronger than the Jones and Alexander polynomial, respectively. Any ideas what is the 1st knot in the knot table (or any other bigger knot) that the HOMFLYPT polynomial distinguishes it, but the Alexander (resp. Jones) polynomial does not? REPLY [7 votes]: I will give an example that is likely not the simplest one. The example comes from the paper Behavior of knot invariants under genus 2 mutation by Dunfield, Garoufalidis, Shumakovitch, and Thistlethwaite (link here). The construction of their example uses cabled mutation. A mutation in a knot diagram removes a round disk intersecting the diagram transversely in four points symmetrically spaced about the boundary, rotates the disk by $180^\circ$, then replaces it back into the diagram. The tangle inside the round disk consists of two arcs, and we say the mutation preserves these arcs if it switches the endpoints of arc one and switches the endpoints of arc two. A cabled mutation of the diagram takes a blackboard framed $n$-cable of the knot diagram and the same disk as before (now intersecting the knot diagram transversely in $4n$ points), and then removes the disk and performs the same arc-preserving rotation as before. See definition 2.10 in the linked paper for a more precise definition of cabled mutation. The authors prove that cabled mutation preserves the Alexander and Jones polynomials, but does not preserve the HOMFLY-PT polynomial. Let $K_{75}$ be the knot with the $75$ crossing diagram below. Think of the bottom half of the picture as a round disk in the cabled mutation. Let $\tau$ be the $180^\circ$ rotation that preserves the arcs of the bottom tangle (when it has one strand for each arc instead of three). Finally, let $K_{75}^\tau$ be the cabled mutant of $K_{75}$. The HOMFLY-PT polynomials of $K_{75}$ and $K_{75}^\tau$ are different; the coefficients of these two polynomials are given on pages 119 and 120 of the linked paper. Since $K_{75}^\tau$ is a cabled mutant of $K_{75}$, they have the same Alexander and Jones polynomials. It seems unlikely that this is the smallest such example. In this StackExchange post, Kyle Miller followed Danny Ruberman's suggestion from the comments above and found no such examples among knots with twelve or fewer crossings.<|endoftext|> TITLE: A remarkable almost-identity QUESTION [34 upvotes]: OEIS sequence A210247 gives the signs of $\text{li}(-n,-1/3) = \sum_{k=1}^\infty (-1)^k k^n/3^k$, also the signs of the Maclaurin coefficients of $4/(3 + \exp(4x))$. Mikhail Kurkov noticed that it appeared that $a(n+28) = -a(n)$ for this sequence. It's not quite true: the first $n$'s for which this is not true are $578, 1143,$ and $1736$. But still it's remarkably close to true. A nice illustration of the Strong Law of Small Numbers? Is there any explanation for the almost-identity, or is it just coincidence? Is it still true for large $n$ that $a(n+28)$ is usually $-a(n)$? REPLY [53 votes]: Consider $F(z) = 4/(3+\exp(4z))$ as a function of the complex variable $z$. It is meromorphic and has simple poles where the denominator vanishes. Namely when $4z = \log 3 + (2k +1)\pi i$ for integers $k$. The poles with the smallest magnitude of $z$ occur when $4z = \log 3 \pm \pi i$. We can compute the Taylor series coefficients of $F$ by looking at $$ \frac{1}{2\pi i} \int_{|z|= r} F(z) z^{-n} \frac{dz}{z}, $$ starting with $r$ suitably small. Now we can estimate the integral asymptotically by taking larger values of $r$, and accounting for poles that are encountered. As noted above, the smallest poles are at $(\log 3 \pm \pi i)/4$ and these will account for the leading asymptotics of these coefficients. Now, the argument of $(\log 3 \pm \pi i)/4$ is $\pm 1.23438\ldots $ which is very nearly $11 \pi/28=1.23419\ldots$. This accounts for the observed phenomenon.<|endoftext|> TITLE: Is there a non-trivial group $C$ such that $A*C \cong B*C$ implies $A \cong B$? QUESTION [38 upvotes]: This is a crosspost from this MSE question from a year ago. Finite groups are cancellable from direct products, i.e. if $F$ is a finite group and $A\times F \cong B\times F$, then $A \cong B$. A proof can be found in this note by Hirshon. In the same note, it is shown that $\mathbb{Z}$ is not cancellable, but if we only allow $A$ and $B$ to be abelian, it is (see here). I would like to know if there are any groups that can be cancelled from free products rather than direct products. That is: Is there a non-trivial group $C$ such that $A*C \cong B*C$ implies $A \cong B$? It is certainly not true that every group is cancellable in free products. For example, if $A$, $B$, $C$ are the free groups on one, two, and infinitely many generators respectively, then $A*C \cong C \cong B*C$ but $A\not\cong B$. Many non-examples can be constructed this way, but they are all infinitely generated. As is discussed in the original MSE question, it follows from Grushko's decomposition theorem that if $A$, $B$, and $C$ are finitely generated, then $A*C \cong B*C$ implies $A \cong B$. REPLY [18 votes]: Ian Agol's answer only uses two properties of $\mathbb{Z}/2\mathbb{Z}$: it is freely indecomposable, and it is not isomorphic to a proper subgroup of itself. There are many other groups which have these two properties, e.g. finite groups. A group which is not isomorphic to a proper subgroup of itself is called co-Hopfian. As I recently learnt on MSE, all co-Hopfian groups are freely indecomposable: if $H$ and $K$ are non-trivial, their free product $H\ast K$ has the proper isomorphic subgroup $H\ast (kh)K(kh)^{-1}$ where $h \in H$ and $k \in K$ are non-identity elements. So we can conclude the following: For any co-Hopfian group $C$, if $A\ast C \cong B\ast C$, then $A\cong B$. Examples of co-Hopfian groups include finite groups, $\mathbb{Q}$, $\mathbb{Q}/\mathbb{Z}$, and fundamental groups of closed hyperbolic manifolds. This last class of examples leads to the following observation: Let $M$, $M'$, and $N$ be closed hyperbolic manifolds. If $M\# N$ is diffeomorphic to $M'\# N$, then $M$ is diffeomorphic to $M'$. Proof: In dimension two, the assumptions imply $M$ and $M'$ have the same genus and are therefore diffeomorphic. In dimensions greater than two we have $$\pi_1(M)\ast\pi_1(N) \cong \pi_1(M\# N) \cong \pi_1(M'\# N) \cong \pi_1(M')\ast\pi_1(N).$$ As $\pi_1(N)$ is co-Hopfian, we see that $\pi_1(M) \cong \pi_1(M')$, so $M$ and $M'$ are diffeomorphic by Mostow rigidity. $\,\square$ Note, there are many examples of manifolds $M$, $M'$, and $N$ for which $M\# N$ is diffeomorphic to $M'\# N$, but $M$ is not diffeomorphic to $M'$, e.g. $M = T^2$, $M' = \mathbb{RP}^2\#\mathbb{RP}^2$, and $N = \mathbb{RP}^2$. For an orientable example, take $M = S^2\times S^2$, $M' = \mathbb{CP}^2\#\overline{\mathbb{CP}^2}$, and $N = \overline{\mathbb{CP}^2}$.<|endoftext|> TITLE: Are relative curves $X \to S$ determined by their fibers? QUESTION [7 upvotes]: Consider relative curves $X \to S$, defined to be flat, integral, projective schemes of relative dimension 1 over $S$. When are these objects determined by their fibers? So if $X,Y$ are $S$-schemes with isomorphic fibers $X_s \cong Y_s, \forall s\in S$, when can we conclude that $X \cong Y$ (as $S$-schemes)? I am most interested in the case where $S$ is a Dedekind scheme, but more general cases are interesting as well. I do not want to stipulate that the $S$-schemes are smooth, but I could tolerate the stipulation to be normal or regular. The quasiprojective case would be interesting too. In the cases when I can conclude $X \cong Y$, how might I go about constructing a relatively explicit isomorphism, if I have relatively explicit isomorphisms of the fibers? [EDIT:] Thanks to all those who added comments. In light of the useful information they provided, I would like to refine the question to the following: Can a relative curve $X \to \mathrm{Spec}(\mathcal{O}_K)$ over a number ring (sometimes called an arithmetic surface) be determined by its fibers, when the fibers have genus 0? What about higher genus? [EDIT again:] Thanks to all for their comments and patience. In light of the additional useful information provided in the comments, I would like to rerefine the question to the following: Can a regular relative curve $X \to \mathrm{Spec}(\mathcal{O}_K)$ over a number ring with class number 1 be determined by its fibers, when the fibers have genus 0? REPLY [10 votes]: This isn't a complete answer but it should give you some ideas or at least things to google. This is really a question about moduli problems. First of all perhaps we should define what we mean by "determined by the fibers" (Here for simplicity I will interpret "fibers" as being "geometric fibers"). Let $X\rightarrow S$ be a flat proper morphism of relative dimension 1. By flatness the arithmetic genus will be constant, lets call the genus $g$. For simplicity let us also assume that the geometric fibers of $X\rightarrow S$ are stable, the key point being that the automorphism groups are finite. In particular we assume $g\ge 2$. In general I will call a morphism of schemes $X\rightarrow S$ a stable curve if it is flat proper with stable curves as fibers. In this case there is a nice geometric object associated to such families of curves, namely, the "moduli stack of stable curves of genus $g$", which we will denote by $\mathcal{M}_g$ (it's more commonly referred to as $\overline{\mathcal{M}_g}$, with $\mathcal{M}_g$ being reserved for the open substack of smooth curves). The key property of such a moduli stack is that there is a bijection between: $$\{\text{$S$-isom classes of stable curves $X/S$}\}\stackrel{\sim}{\longrightarrow} \text{Hom}(S,\mathcal{M}_g)$$ (This is essentially the defining property of the moduli stack $\mathcal{M}_g$) Note that a moduli stack is not a scheme, but it sits inside the category (really a 2-category) of algebraic stacks, which contains the category of schemes as a full subcategory, and so the Hom on the right side is taken in this sense. Even though $\mathcal{M}_g$ isn't a scheme, there is often (and in this case) a scheme which can be thought of as a best approximation for $\mathcal{M}_g$. This scheme we will denote by $M_g$, and is called the "coarse moduli scheme of stable curves of genus $g$". This scheme has the key property that for any algebraically closed field $k$, there is a bijection: $$\{\text{$k$-isom classes of stable curves $X/k$}\}\stackrel{\sim}{\longrightarrow}\text{Hom}(\text{Spec }k,M_g)$$ Moreover, there is a natural map $p : \mathcal{M}_g\rightarrow M_g$ compatible with the key properties mentioned above. In particular, a given stable curve $X/S$ determines a map $S\rightarrow\mathcal{M}_g$, and hence a map $S\rightarrow M_g$ by composing with $p$. Now if $S$ is reduced (and such that the closed points are dense in $S$, thanks Ariyan) and $M_g$ is separated (I think $M_g$ is always separated?), then any map $S\rightarrow M_g$ is determined by where it sends closed points, and so I would say that the data of "the (geometric) fibers of $X/S$" should perhaps be interpreted as the data of the map $S\rightarrow M_g$. On the other hand the isomorphism class of $X/S$ is equivalent to the map $S\rightarrow\mathcal{M}_g$, so we are interested in asking whether a particular map $S\rightarrow M_g$ has a unique lift to a map $S\rightarrow\mathcal{M}_g$. This is certainly not always true. One remark is that the map $p : \mathcal{M}_g\rightarrow M_g$ "forgets information" (namely, the automorphism groups of the objects it parametrizes), and so above a point in $M_g$ some choices are required in choosing a "point" of $\mathcal{M}_g$ lying above it. A fundamental intuitive example for this failure is that if you fix some curve $C$ with nontrivial automorphisms and $S$ is a "line segment", you can consider the trivial family $C\times S$ with fiber $C$ over $S$. From this trivial family, you can glue the fiber over one endpoint of $S$ to the fiber over the other endpoint of $S$ to produce a family with fibers $C$ over $S$-with-glued-endpoints (ie, a circle). The possible gluings correspond to automorphisms of $C$, and different gluings will generally give you nonisomorphic families over the circle (at least if the automorphism groups are discrete?). This can be carried out explicitly and quite easily by considering twists of elliptic curves. For example, the isotrivial elliptic curve given by $y^2 = x^3-1$ over $\text{Spec }\mathbb{C}[t,t^{-1}]$ has the same fibers (equivalently, same $j$-invariant) as the elliptic curve given by $ty^2 = x^3-1$ (over the same base), and yet it can be checked that they are not isomorphic over $\mathbb{C}[t,t^{-1}]$ (one needs to pass to $\mathbb{C}[\sqrt{t},\sqrt{t^{-1}}]$ for them to become isomorphic). In both pictures the issue comes down to the existence of nontrivial automorphisms, and indeed at least for the moduli of stable curves, the map $p : \mathcal{M}_g\rightarrow M_g$ is an isomorphism at points (ie, stable curves) which have no nontrivial automorphisms. Thus for some $X/S$ such that the fibers have no automorphisms, then the associated map $S\rightarrow M_g$ will have a unique lift to a map $S\rightarrow\mathcal{M}_g$, simply because the image of $S\rightarrow M_g$ will be contained in a subscheme above which $p$ is an isomorphism. The above point works for arbitrary bases $S$, though we require the fibers to have no nontrivial automorphisms. If one is happy to restrict the base $S$ to schemes with trivial etale fundamental group (e.g. $\mathbb{A}^1_\mathbb{C}$), or at least etale fundamental groups of order coprime to the order of the automorphism groups of the fibers, then I expect it may be possible to remove the condition on the nonexistence of nontrivial automorphisms (Note that the fundamental (counter)example described above was only possible because the base was not simply connected, and this holds in the case with $dim(S) = 0$ due to the theory of twists). However, I'm just returning from a rather long vacation-from-math and I'm not sure if this is true for $dim(S) > 0$.<|endoftext|> TITLE: What are some mathematical consequences of the study of 6D $\mathcal N = (2,0)$ SCFT? QUESTION [18 upvotes]: Arguments made in physics apparently predict the existence of a family of six-dimensional $\mathcal N = (2,0)$ superconformal field theories (Wikipedia, nLab, PhysicsOverflow) sometimes called Theory $\mathcal X$. I don't really know what that would entail; apparently we have partial information about this theory, suggesting that it's difficult to describe even in a physics sense (e.g. non-Lagrangian), let alone to mathematically formalize. But I have also heard from mathematicians who are interested in Theory $\mathcal X$ for what I assume are entirely mathematical reasons. I came away with the impression that even though we can't construct it, there are ways to study it to yield interesting results in pure mathematics; but I don't know any examples of such results. So my question is: what are some purely mathematical takeaways from the story of Theory $\mathcal X$? And, if it's known, what would be some expected mathematical consequences of a construction of Theory $\mathcal X$ in a physics sense? I get the impression that various dimensional reductions of Theory $\mathcal X$ should include several commonly-studied TQFTs and QFTs, so in a sense studying Theory $\mathcal X$ generalizes the study of those TQFTs. So one possible answer is that there could be theorems about those TQFTs whose proofs were inspired by some conjectured aspect of Theory $\mathcal X$ — but is this accurate? REPLY [11 votes]: In general, mathematical outputs of SUSY field theories often become more accessible after performing some twist, and the same is true of the 6d (2,0) SCFT. Considering the theory on $\Sigma\times M_4$, it admits a twist (first studied by Beem-Rastelli I believe) that's holomorphic along $\Sigma$ and topological along $M_4$. Some discussion of the difficulties in mathematically describing this twist in perturbation theory can be found in these notes from a talk of Kevin Costello: https://math.berkeley.edu/~qchu/Notes/6d/Day%205,%20Talk%202,%20Costello.pdf One interesting application of this twist to representation theory is the AGT correspondence. Specializing $M_4=\mathbb{R}^4$, the rough idea is that after putting suitable $\Omega$-backgrounds in the $\mathbb{R}^4$ direction, the degrees of freedom of the 6d theory localize to $\Sigma$ where we find a Toda field theory. We then get several relationships between this Toda field theory living on $\Sigma$ and the 4d $\mathcal{N}=2$ gauge theory gotten by compactifying on $\Sigma$. In particular, the conformal blocks of the Toda field theory should match the partition function of the 4d $\mathcal{N}=2$ theory, and vertex operator insertions on $\Sigma$ should correspond to including certain defects in the 4d $\mathcal{N}=2$ theory. In the math literature, this connection (at least for pure gauge theory) was established through work of Maulik-Okounkov and Schiffman-Vasserot who showed that the affine W-algebra (i.e. the local observables of the above Toda field theory) acts on the equivariant cohomology of the moduli of instantons on $\mathbb{A}^2$. Let me also very briefly comment that there's a 6d string theory (dubbed the little string theory) that becomes the (2,0) SCFT in the limit of infinite string mass. This string theory appears to be just as mathematically rich as its SCFT limit. These slides: https://member.ipmu.jp/yuji.tachikawa/stringsmirrors/2016/main/Aganagic%20v2.pdf from Mina Aganagic's 2016 String-Math talk contain some statements in this direction.<|endoftext|> TITLE: Examples of notably long or difficult proofs that only improve upon existing results by a small amount QUESTION [106 upvotes]: I was recently reading Bui, Conrey and Young's 2011 paper "More than 41% of the zeros of the zeta function are on the critical line", in which they improve the lower bound on the proportion of zeros on the critical line of the Riemann $\zeta$-function. In order to achieve this, they define the (monstrous) mollifier: $$ \psi=\sum_{n\leq y_1}\frac{\mu(n)P_1[n]n^{\sigma_0-\frac{1}{2}}}{n^s}+\chi(s+\frac{1}{2}-\sigma_0)\sum_{hk\leq y_2}\frac{\mu_2(h)h^{\sigma_0-\frac{1}{2}}k^{\frac{1}{2}-\sigma_0}}{h^sk^{1-s}}P_2(hk). $$ This is followed by a nearly 20-page tour-de-force of grueling analytic number theory, coming to a grinding halt by improving the already known bounds from $40.88\%$ to... $41.05\%$. Of course any improvement of this lower bound is a highly important achievement, and the methods and techniques used in the paper are deeply inspirational. However, given the highly complex nature of the proof and the small margin of improvement obtained over already existing results, I was inspired to ask the following question: What are other examples of notably long or difficult proofs that only improve upon existing results by a small amount? REPLY [7 votes]: The following example is described in The Man Who Loved Only Numbers by Paul Hoffman. There is a reasonably short proof, found by Esther Klein (later Szekeres) in 1932, that given 5 points in the plane, there exist 4 of them that form a convex quadrilateral. The minimum number of points needed such that there always exists 5-element subset forming a convex pentagon is 9. There is a conjecture that the minimum number of points needed such that there exists an $n$-element subset forming a convex $n$-gon is $2^{n-2} + 1$. It is known it has to be at least that. This is called the "Happy Ending Problem" (apparently because George Szekeres had impressed Esther so much with his proof that there is a minimum finite number for each $n$ that he won her hand in marriage). For the case $n = 6$, the best that Erdős achieved was 71 points, a somewhat distant cry from the conjectured 17. This was sometime in the 1930's. After the memorial service for Erdős in 1996, Ronald Graham and his wife Fan Chung thought it was high time that someone have another crack at it, after 60 years -- and were very excited that during a long flight to New Zealand, they managed to lower it down by just a single point, to 70. This required introducing new ideas. (As Graham explains, Kleitman and Pachner soon lowered it further to 65. There were subsequent further improvements until finally George Szekeres and his collaborator Peters reached 17, with the help of computers, in a paper published in 2006.)<|endoftext|> TITLE: Does Zorn's Lemma imply a physical prediction? QUESTION [22 upvotes]: A friend of mine joked that Zorn's lemma must be true because it's used in functional analysis, which gives results about PDEs that are then used to make planes, and the planes fly. I'm not super convinced. Is there a direct line of reasoning from Zorn's Lemma to a physical prediction? I'm thinking something like "Zorn's Lemma implies a theorem that says a certain differential equation has a certain property, and that equation models a phenomena that indeed has that property". REPLY [4 votes]: Pretty much all mathematics of relevance to physics can apparently be formalised in a strongly finitist foundation. Working in ... a fragment of quantifier-free primitive recursive arithmetic (PRA) with the accepted functions limited to elementary recursive functions. Elementary recursive functions are the functions constructed from some base arithmetic functions by composition and bounded primitive recursion. Feng Ye's Strict Finitism and the Logic of Mathematical Applications (free draft copy, see also this review (paywall)) builds enough mathematics to treat pretty much all of applied mathematics/theoretical physics (say, that part of theoretical physics that is experimentally confirmed :-) So as others have pointed out, logical axioms of the strength of full Zorn's lemma are far from necessary.<|endoftext|> TITLE: Categorifications of the real numbers QUESTION [27 upvotes]: For the purposes of this question, a categorification of the real numbers is a pair $(\mathcal{C},r)$ consisting of: a symmetric monoidal category $\mathcal{C}$ a function $r\colon \mathrm{ob}(\mathcal{C})\to\mathbb{R}$ such that: $r(X\otimes Y) = r(X) r(Y)$ for all objects $X$ and $Y$ of $\mathcal{C}$ $r(\mathbb{1}) = 1$, where $\mathbb{1}$ is the monoidal unit $X\cong X'\implies r(X)=r(X')$ Some examples of categorifications of $\mathbb{R}$ are: finite sets and cardinality; finite-dimensional vector spaces and dimension; topological spaces (with the homotopy type of a CW-complex, say) and Euler characteristic. However, in these examples the map $r$ factors through $\mathbb{N}$ or $\mathbb{Z}$. I am interested in examples where the values of $r$ are not so restricted. Question: What categorifications of $\mathbb{R}$ are there where $r$ can take all values in $\mathbb{R}$, or perhaps all values in $(0,\infty)$ or $(1,\infty)$? I am especially interested in examples that already appear somewhere in the mathematical literature. I am also especially interested in examples where $\mathcal{C}$ is symmetric monoidal abelian and r(A)+r(C) = r(B) for every short exact sequence $0\to A\to B\to C\to 0$. REPLY [5 votes]: Via the notion of groupoid cardinality (generalizing ordinary set-theoretic cardinality), finite non-empty groupoids can be seen as a categorification of the positive rationals, and more general groupoids as a categorification of the non-negative reals: John Baez and James Dolan, From Finite Sets to Feynman Diagrams, in Mathematics Unlimited—2001 and Beyond, Springer, 2001. (arXiv link) An example that Baez and Dolan discuss is the groupoid of finite sets and bijections, which has cardinality $e$. (The notion of Euler characteristic of a category, mentioned in Gregory Arone's answer, can be seen as a further generalization of groupoid cardinality, see Example 2.7 of Leinster's article.) REPLY [5 votes]: Can we do an example along these lines: An appropriate collection of metric spaces, with $\otimes$ the Cartesian product, and $r$ a fractal dimension? Or more precisely, $r(X) = \exp(\dim(X))$. Perhaps the arrows are weakly contracting $d(f(x),f(y) \le d(x,y)$. And perhaps we want the finite-dimenaional (so that $\dim(X) = \infty$ is disallowed) fractals in the sense of Taylor (so that $\dim(X \times Y) = \dim(X)+\dim(Y)$).<|endoftext|> TITLE: Vopenka's principle is equivalent to the existence of a strong compactness cardinal for any "logic"? QUESTION [7 upvotes]: According to Cantor's attic, Vopenka's principle is equivalent to the existence of a strong compactness cardinal for any "logic". But I can't find a definition of what a "logic" is either there or in any of the cited references. Question: What is a "logic" in the sense of this statement? Some examples are given: apparently infinitary logic $L_{\kappa\kappa}$ is a "logic", and infinitary higher(but finite)-order logic $L_{\kappa\kappa}^n$ is a "logic". Thus Vopenka's principle should imply the existence of a proper class of strongly compact cardinals, and even a proper class of extendible cardinals. But I don't think it's supposed to be as simple as "Vopenka's principle is equivalent to a proper class of extendibles", so there must be more "logics" than these. The next thing I can think of is some sort of infinitary-order infinitary logic $L^\alpha_{\kappa\kappa}$. It would also make sense to consider structures with infinitary operations. I don't know if there's a large cardinal principle associated to strong compactness for either of these sorts of logic. And then of course there are "logics" such as various flavors of type theory (which higher order logic starts to resemble anyway!) but just for cultural reasons, I doubt that "logic" is meant to encompass anything along these lines. REPLY [4 votes]: https://www.jstor.org/stable/2273786?seq=1#page_scan_tab_contents Is the article where it is from. It seems to have never been added to the library, which would be my fault. https://projecteuclid.org/download/pdf_1/euclid.pl/1235417266 is the first chapter of the textbook where I personally found it, although I would use the JSTOR article in the library of Cantor's Attic after this post. The massive textbook I used is called "Model-Theoretic Logics" by the "Ω-Group" (the coolest pen-name for a group of mathematicians). In Part F, this equivalence is proven. Chapter XVII is "Set-theoretic definability of logics" (written by Väänänen, one of my heroes) which is where this comes from. The definition is quite nuanced, but it is a great read. I recommend this textbook. What Is a Logic? (loose definition) We start with a "vocabulary." $\tau$ is often used for this type of object; it is defined as a (nonempty) class of constant symbols, finitary predicate symbols, and finitary function symbols (simple enough). However, you also have to have "sort symbols" for every function symbol and every constant symbol, and furthermore you have to have sort symbols for every argument of every relation symbol and every input of every function symbol. The sort symbols are just for well-defining "terms" and their syntactics. The "$\tau$-terms" are built up of just function symbols applied to constant symbols, each of which are then given a "sort" to help define things later. Using the sort, you can well-define the number of inputs in a given "formula" made out of $\tau$. An "abstract logic" $\mathcal{L}$ gives every vocabulary $\tau$ a class $\mathcal{L}(\tau)$ of formulae and "atomic $\tau$-formulae of $\mathcal{L}$." More important is that it gives every $\tau$ a relation $\models^\tau$ which defines semantics: how a structure interprets a formula of $\mathcal{L}$. The kind of logic we are talking about has been formalized by Väänänen, but it suggests that the definitions of $\models^\tau$ and deciding what things are atomic formulae are recursive somewhat (synctatic) and furthermore there is a set $A$ such that every formula of $\mathcal{L}$ is in $A$ and that $\mathcal{L}$ can easily describe this set $A$ (the syntax of $\mathcal{L}$ is representable). There is a lot more to this definition that I haven't talked about, so this is a very rough summary. I urge you to take a look at the book instead of just taking my word (I am in no ways an expert).<|endoftext|> TITLE: Summing Bernoulli numbers QUESTION [6 upvotes]: Consider the Bernoulli numbers denoted by $B_n$, which are rational numbers. It is known that the harmonic numbers $H_n=\sum_{k=1}^n\frac1k$ are not integers once $n>1$. I am curious about the following: Question: If $n>0$, will $\sum_{k=0}^nB_k$ ever be an integer? REPLY [12 votes]: It can never be an integer for $n>0$. There is a result by K.G.C. von Staudt and independently by T. Clausen that $$B_n+\sum_{p\in \mathbb{P}\, ,\, p-1|n}\frac{1}{p}\in \mathbb Z$$ [1] T. Clausen. Lehrsatz aus einer Abhandlung uber die Bernoullischen Zahlen. Astr. Nachr., 17:351–352, 1840 [2] K. G. C. von Staudt. Beweis eines Lehrsatzes die Bernoulli'schen Zahlen betreffend. J. Reine Angew. Math., 21:372–374, 1840 Using this we see that the integrality of $\sum_{k=0}^n B_k$ is equivalent to the integrality of $$\sum_{k=2}^n\left(\sum_{p\in \mathbb P \, , \, p-1|k}\frac{1}{p}\right)=\sum_{p\in \mathbb P}\frac{\lfloor\frac{n}{p-1}\rfloor}{p}$$ If $q$ is the largest prime $\le n$ we have $\lfloor \frac{n}{q-1}\rfloor =1$ therefore our expression has $q$-valuation $-1$, so is not an integer. REPLY [8 votes]: It is never an integer. By Bertrand postulate there exists a prime $p=2s+1$ between $n/2$ and $n$, it divides the denominator of $B_{p-1}$ and not of other summands in your sum, by von Staudt - Clausen Theorem.<|endoftext|> TITLE: A stronger form of the Dirichlet Theorem on prime numbers in arithmetic sequences QUESTION [7 upvotes]: Question 1. Let $a,b>1$ be two natural numbers. Is there a prime number $p\in 1+b\mathbb N$ such that $a+p\mathbb Z$ is a generator of the multiplicative group of the field $\mathbb Z/p\mathbb Z$? In need this fact for establishing some properties of the Golomb topology on positive integers. Added in Edit: Francesco Polizzi presented a simple counterexample to Question 1: $a=4$ and $b\ge 3$. In this case $a$ is a square. So, we can ask a more refined version of Question 1 (which is still sufficient for my purposes). Question 1'. Let $a,b\in\mathbb N$ and $a\notin\{n^k:n,k\in\mathbb N\setminus\{1\}\}$. Is there a prime number $p\in 1+b\mathbb N$ such that $a+p\mathbb Z$ is a generator of the multiplicative group of the field $\mathbb Z/p\mathbb Z$? REPLY [16 votes]: This is a hybrid of Dirichlet's theorem with Artin's conjecture on primitive roots. Artin's primitive root conjecture says that if $a \in \mathbf{Z}$ is not a perfect square or $-1$, then there are infinitely many p such that a is a generator mod p (and, further, that the set of such p has some specific non-zero relative density inside the set of primes, depending on a). The primitive root conjecture is presently open, I'm afraid, so there is no hope of proving this stronger statement. However, if you're content with a conditional proof, then Artin's primitive root conjecture is known to follow from the Generalised Riemann Hypothesis by work of Hooley (whose 90th birthday it is today, incidentally); and Hooley's work has been extended by subsequent authors to consider the case of primes in arithmetic progressions. Here is a paper by Moree which gives an explicit formula (Theorem 4 of the paper) for the density of primes $p$ such that $p = 1 \bmod b$ and $a$ is a generator mod $p$, for any non-zero integers $a, b$. In particular, Moree shows (mid-way down page 14 of the paper) that there are infinitely many primes $p$ with this property if and only if the following are satisfied: $a \ne -1$; $a$ is not an $\ell$-th power for any prime $\ell$ dividing $b$; $b$ is not divisible by the integer $\Delta$, where $\Delta$ is defined as follows: write $a = u v^2$, where $u$ is square-free; then $\Delta = u$ if $u = 1 \bmod 4$ and $\Delta = 4u$ otherwise. (If the last condition is not satisfied, then $a$ will be a square modulo $p$ for every $p = 1 \bmod b$, so it cannot be a generator. The examples by Francesco and Don show that this condition is really needed.)<|endoftext|> TITLE: How nearly abelian are nilpotent groups? QUESTION [13 upvotes]: It is not uncommon to read that "nilpotent groups are 'close to abelian'."1,2 Can this sentiment be made precise in the sense of the Turán and Erdős definition of "the probability that two elements of $G$ commute," discussed in the MO question, "Measures of non-abelian-ness"? More precisely, Q. What the probability that two randomly selected elements of a nilpotent group commute? What is the max and min over all nilpotent groups? For example, the nilpotent Quaternion group $Q_8$ is $62.5\%$ abelian, if I calculated correctly:                     $Q_8$: $24$ out of $64$ entries are non-abelian. So it is $40/64=62.5\%$ abelian. Update. @BenjaminSteinberg immediately observed that the $Q_8$ example establishes the upper bound of $5/8$, and shortly thereafter cited literature that shows that there is no positive lowerbound. 1Princeton Companion to Mathematics, "Gromov's Polynomial-Growth Theorem," p.702.: 'nilpotent groups are "close to abelian."' 2Wikipedia: Nilpotent group: 'a nilpotent group is a group that is "almost abelian."' REPLY [13 votes]: Here are a few general remarks (concerning finite groups). The relevant results can mostly be found in the 2006 Journal of Algebra paper by Bob Guralnick and myself (a link to a preprint version is given in one of Benjamin Steinberg's comments). An extra-special $p$-group of order $p^{3}$ has commuting probability $\frac{p^{2}+p-1}{p^{3}},$ which is only slightly more than $\frac{1}{p},$ so according to that measure, for a large prime $p,$ such a group is quite far from Abelian. The asymptotic behaviour of the number of $p$-groups of order $p^{n}$ was shown by C. Sims (and maybe G. Higman) to be much the same as that of the number of $p$-groups of nilpotence class $2$ and order $p^{n},$ so perhaps it is not surprising that $p$-groups of class $2$ already exhibit a high measure of non-commutativity (this is also borne out by a theorem of P. Neumann, of which Guralnick and I were unaware when we wrote our paper, which was pointed out by a comment by Sean Eberhard to an earlier MO question on this topic-in fact, the question mentioned in Joseph O'Rourke's first comment). Note that the largest Abelian subgroup of an extraspecial $p$-group of order $p^{2n+1}$ has order $p^{n+1},$ which is not much more than the square root of the group order. But perhaps such examples illustrate that the commuting probability is too crude a measure of non-Abelianness, since a group of nilpotence class $2$ is in some ways not too far from Abelian.<|endoftext|> TITLE: Galois categories for topological spaces? QUESTION [7 upvotes]: Can the theory of Galois categories (as developed in SGA1) be modified to produce the usual fundamental group of a topological space (maybe assumed to be path connected and locally path connected)? Recall from SGA 1 that the theory of Galois categories is developed to construct the étale fundamental group of a connected locally noetherian scheme. Given a Galois category $\mathcal{C}$, the main result is that a choice of fiber functor $F$ determines an equivalence between $\mathcal{C}$ and the category of finite $\pi := \operatorname{Aut}(F)$-sets. Can one prove an analogous result that applies to a path connected+locally path connected topological space? Except in special cases, one can't literally use Galois categories since a topological space can certainly admit connected covers of infinite degree. But maybe we can modify the construct by removing some finiteness conditions? Ideally, if the answer is "yes", I'd be great to have a written reference developing this idea. REPLY [3 votes]: To expand on Niels' answer: given a connected semilocally simply connected topological space $X$ and a point $x\in X$, the fiber functor $F_x:\mathrm{Cov}_X\rightarrow\mathrm{Set}$ from the category of covering spaces of X yields an equivalence of categories $F_x:\mathrm{Cov}_X\xrightarrow{\sim}\pi_1(X,x)\mathrm{Set}$ with permutation representations of the fundamental group $\pi_1(X,x)$ by sending a covering space $\pi:Y\rightarrow X$ to the permutation representation of $\pi_1(X,x)$ on the fiber $\pi^{-1}(x)$ induced by monodromy. However, $\mathrm{Cov}_X$ is very much not a Galois category: according to SGA1 or Stacks, a Galois category is equivalent to the category of finite (continuous) permutation representations of a profinite group, whereas $\mathrm{Cov}_X$ is equivalent to the category of possibly infinite permutation representations of a (discrete) group. I suppose $\mathrm{Cov}_X$ could be called an infinite Galois category, although this conflicts somewhat with the notion of infinite Galois category used to define the pro-étale fundamental group. On the other hand if you want to work with an honest Galois category: given a connected topological space $X$ and a point $x\in X$, the fiber functor the fiber functor $F_x:\mathrm{FCov}_X\rightarrow\mathrm{Fin}$ from the category of finite covering spaces of X yields an equivalence of categories $F_x:\mathrm{FCov}_X\xrightarrow{\sim}\widehat{\pi}_1(X,x)\mathrm{Fin}$ with finite (continuous) permutation representations of the profinite completion $\widehat{\pi}_1(X,x)$ of the fundamental group. Note that we do not need to assume that $X$ is semilocally simply connected for this to work: in the presence of a universal cover of $X$ the fundamental group is equivalently the automorphism group of the universal cover, but in general the (profinite) fundamental group can be recovered as the automorphism group of the fiber functor. Why would one want to consider only finite covers of topological spaces? One reason is to formulate comparison theorems: for instance given a scheme $X$ (locally) of finite type over $\mathbb{C}$ and a point $x\in X$ we have an equivalence of Galois categories $\mathrm{FEt}_X\xrightarrow{\sim}\mathrm{FCov}_{X(\mathbb{C})}$, but this is obviously false at the level of infinite covering spaces.<|endoftext|> TITLE: Reference request: birational automorphism group is finite QUESTION [9 upvotes]: I am interested in having a look at the proof of the following fact: If $X$ is a smooth variety of general type, then $\mathrm{Aut(X)}$ is finite. I know that this is proved in "On algebraic groups of birational transformations" by Matsumura. Unfortunately, this paper is not available at the library of my institution, nor I could find it online. Could anyone point at a way to find such paper, or an alternative reference? REPLY [7 votes]: See $\S$11.7 or more specifically Theorem 11.12 in Algebraic Geometry An Introduction to Birational Geometry of Algebraic Varieties by Shigeru Iitaka.<|endoftext|> TITLE: A simultaneous generalization of the Grunwald-Wang and Dirichlet Theorems on primes QUESTION [6 upvotes]: By Grunwald-Wang Theorem, if for some odd number $n$ the equation $x^n=a$ has no solutions in $\mathbb Z$, then the equation $x^n=a\mod p$ has no solutions for some prime number $p$. I am interested if we can always choose $p$ is the arithmetic sequence $1+n\mathbb N$ (at least for prime $n$). Question. Let $b$ be a prime number and $a\in\mathbb N\setminus\{n^{k+1}:n,k\in\mathbb N\}$. Is it true that the equality $x^b=a\mod p$ has no solutions for some prime number $p\in 1+b\mathbb N$? Remark. This question has affirmative answer under GRH (a Generalized Riemann Hypothesis), see this MO-problem and the answer of David Loeffler who cites the paper of Moree. So, the problem is to find an affirmative answer to my question without GRH. For my purposes it suffices to assume that the number $a$ is taken from the set $d\mathbb N$ for some $d\in\mathbb N$. REPLY [5 votes]: $\newcommand{\Z}{\mathbf{Z}}$ $\newcommand{\Q}{\mathbf{Q}}$ $\newcommand{\F}{\mathbf{F}}$ $\newcommand{\OK}{\mathcal{O}_K}$ EDIT. To prove the existence of at least one prime (or infinitely many primes) meeting the OP's requirement, there is a much simpler argument, see the answer by a so-called friend Don. My answer is more about computing the density of such primes. The answer is yes, this is a consequence of the Cebotarev theorem for Galois extensions of number fields. Let $b \geq 2$ be an integer, and let $a \in \Z \backslash \{0\}$. Let $L$ be the splitting field of the polynomial $P=X^b-a$ over $\Q$. Then $L$ contains the $b$-th roots of unity and thus the cyclotomic field $K=\Q(\zeta_b)$. Let $G$ be the Galois group of $L/K$. The action of $G$ on the roots of $P$ is easy to understand. Namely, if $\alpha$ is a root of $P$ (in other words $\alpha$ is a $b$-th root of $a$), then every other root is of the form $\alpha'=\zeta_b^k \alpha$ for some $k \in \Z/b\Z$. It follows that for any $\sigma \in G$ we have $\sigma(\alpha)=\zeta_b^{\lambda(\sigma)} \alpha$ for some $\lambda(\sigma) \in \Z/b\Z$. This provides a group morphism $\lambda : G \to \Z/b\Z$ which does not depend on the choice of $\alpha$, and is injective since $G$ fixes the $b$-th roots of unity. Now if $b$ is odd and for any prime divisor $p$ of $b$, the integer $a$ is not a $p$-th power in $\Z$, then $\lambda$ is surjective (see Lang, Algebra, Chapter 6, Thm 9.4), so in this case the action of $G$ on the roots of $P$ is simply transitive. (The situation is more complicated in the case $b$ is even, since $\sqrt{-1} \in \Q(\zeta_4)$ and $\sqrt{2} \in \Q(\zeta_8)$ for example. You could try to look at Jacobson--Vélez, The Galois group of a radical extension of the rationals. At least, this is also true when $b=2$, in which case the argument below applies.) Let us recall some algebraic number theory. Let $p$ be a prime not dividing $b$. Let $\Phi_b$ be the cyclotomic polynomial and $\overline{\Phi_b} \in \F_p[X]$ its reduction mod $p$. It is known that $\overline{\Phi_b}$ is a product of $r$ distinct irreducible polynomials of degree $e$, where $e$ is the order of $p$ in $(\mathbf{Z}/b\mathbf{Z})^\times$ and $re=\varphi(b)$, see Demazure, Cours d'algèbre, Prop 9.17. In particular $\overline{\Phi_b}$ has a root in $\F_p$ if and only if $p \equiv 1$ mod $b$. Accordingly $p$ is the product of distinct prime ideals $\pi_1,\ldots,\pi_r$ in the ring of integers $\OK$ of $K$, and each such ideal has norm $p^e$. In particular, the prime ideals with $e \geq 2$ will have density zero (with respect to the norm), and we only need to care about prime ideals above the primes $p \equiv 1$ mod $b$. For such a prime $p$, note that $\OK/\pi \cong \F_p$ for any prime $\pi$ above $p$, so the equation $x^b = a$ has a root mod $p$ if and only if it has a root in $\OK$ modulo some (or any) prime $\pi$ above $p$. Now we want to apply the Cebotarev theorem to the Galois extension $L/K$. By the above, we have to count the number of $\sigma$ in $G$ which fix at least one root of $P$, and this happens if and only if $\sigma = 1$. We get that the density of prime ideals $\pi$ of $\OK$ such that $x^b=a$ has a root modulo $\pi$ is equal to $1/b$. It follows that the density of primes $p \equiv 1$ mod $b$ such that $a$ is not a $b$-th power mod $p$ is equal to $1-1/b$. More generally, one can prove as an exercise that given a divisor $d$ of $b$, the density of primes $p \equiv 1$ mod $b$ such that $a^{(p-1)/b}$ has order $d$ in $(\Z/p\Z)^\times$ is equal to $\varphi(d)/b$. This says in particular that $a^{(p-1)/b}$ is a primitive $b$-th root of unity in $\Z/p\Z$ for at least $\varphi(b)/b$ of the primes $p \equiv 1$ mod $b$. Alternatively, we could have used the Cebotarev theorem for the Galois extension $L/\Q$. It is an instructive exercise since the Galois group is more complicated: it is isomorphic to the semi-direct product $(\Z/b\Z) \rtimes (\Z/b\Z)^\times$. Its action on the roots of $P$ is given (up to isomorphism) by affine transformations $x \mapsto \alpha x+\beta$ with $\alpha \in (\Z/b\Z)^\times$ and $\beta \in \Z/b\Z$. If $b$ is an odd prime we get that the density of primes $p$ such that $x^b=a$ has no root mod $p$ is equal to $1/b$. Using the fact that each such prime must be $\equiv 1$ mod $b$ (if $b$ is prime to $p-1$ then $x \mapsto x^b$ is a bijection of $\Z/p\Z$) together with Dirichlet's theorem for primes congruent to $1$ mod $b$, we recover the above result. But I find the first argument easier, and it works for arbitrary $b \geq 2$. REPLY [5 votes]: Perhaps I miss the point, but: It seems the OP already knows that, under his hypothesis, there is some prime $p$ for which $x^b \equiv a\pmod{p}$ has no solution. He is only asking that $p$ be allowed to be taken from the progression $1\bmod{b}$. But this extra restriction on $p$ is not an extra restriction at all! As François points out, if $p-1$ is coprime to $b$ then it is easy to see that $x\mapsto x^{b}$ is a bijection on the $(p-1)$-element group $(\mathbb{Z}/p\mathbb{Z})^{\times}$, and so also on the field $\mathbb{Z}/p\mathbb{Z}$. So the only $p$ that can arise above have $p-1$ not coprime to $b$. Since $b$ is assumed prime, $p\equiv 1\pmod{b}$. EDITED TO ADD: Here's a fairly simple way to see that there are some primes for which $a$ is not a $b$th power. If $b$ is prime and $a \in \mathbb{Z}$ is not a $b$th power (as follows from the OP's assumptions), then $x^b-a$ is irreducible over $\mathbb{Q}$. This is an elementary exercise in Galois theory. (By more intricate arguments, one can completely characterize irreducible binomials over an arbitrary field --- this is a theorem attached to the name of Capelli.) But it's also well-known that an irreducible polynomial of degree $>1$ cannot have a root modulo every prime (or even "almost" every prime); see: Irreducible polynomials with a root modulo almost all primes<|endoftext|> TITLE: Hausdorff distance is a lower (or upper bound) for what probability metric? QUESTION [7 upvotes]: In a metric space $X=(X, d)$, given a probability measure $\mu$ and two subsets $A$ and $B$ of positive measure, it's not hard to prove that $$ d(A, B) \le W(\mu|_A, \mu|_B), $$ where $d(A, B):= \inf_{a \in A,\;b \in B}d(x,y)$ is the distance between $A$ and $B$. $\mu|_A$ defined by $\mu|_A(C):= \mu(A\cap C)/\mu(A)$ is the conditional measure induced by $A$ $W(\mu|_A,\mu|_B)$ is the Wasserstein distance between $\mu|_A$ and $\mu|_B$. Now, consider the Hausdorff distance between $A$ and $B$, defined by $$ d_H(A,B) := \max\left(\sup_{b \in B}\inf_{a \in A}d(a,b),\sup_{a \in A}\inf_{b \in B}d(a,b)\right) = \inf\{\epsilon \ge 0 | A \subseteq B^\epsilon \text{ and } B\subseteq A^\epsilon\}, $$ where $A^\epsilon := \{x \in X | \exists a \in A\text{ with }d(a,x)\le \epsilon\} = \cup_{a \in A}\operatorname{Ball}_\epsilon(a)$ is the $\epsilon$-blowup of $A$. Question Is $d_H(A,B)$ a lower bound (or upper bound) for some probability metric ? Which one ? Between what and what ? Edit A user mentioned "$L^\infty$ Wasserstein distance" in the comments. This comment seems to have gone unnoticed until now. Indeed, Exercise 36 of OTAM asks to prove that $d_H(supp(\mu),supp(\nu)) \le W_\infty(\mu,\nu)$ for every pair of measures $\mu$ and $\nu$ on a Polish space. This solves one half of my question. The other half has been solved (in the negative) by another user. REPLY [6 votes]: A general note is that the answer depends heavily on the properties of $\mu$. First a note that in general $d_H(A,B) \not \le C \cdot W_p(\mu|_A,\mu|_B)$ for $p\in[1,\infty)$ and some $C>0$. Though it's true for the case $p = \infty$. Here the example: Let $\mu_\lambda = (1-\lambda) \delta_x + \lambda \delta_y$ for distinct $x,y \in X$. Choose $A=\{x\}$ and $B=\{x,y\}$. Then $\mu_\lambda|_A$ is $\delta_x$ and $\mu_\lambda|_B$ is $\mu_\lambda$. Thus $d_H(A,B) = d(x,y)$ and $W_p(\mu|_A,\mu|_B) = \lambda^{\frac{1}{p}}d(x,y)$. For the case $p=\infty$, i.e. $W_\infty(\mu,\nu) = \inf \|d\|_{L^\infty(\pi)} $, it is true. But there is no lower bound for the Hausdorff distance $d_H(\cdot,\cdot)$ w.r.t. to any Wasserstein distance $W_p$ if there is at least one non-isolated point (as $W_p \le W_\infty$ it suffices to get counterexamples for $p=\infty$). Choose $x_n \to x$ and $\mu = \frac{1}{2}\delta_x + \sum_{n\in \mathbb{B}} \lambda_n \delta_{x_n}$ for some $\lambda_n \ge 0$ with $\sum_{n\in \mathbb{B}} \lambda_n = \frac{1}{2}$. Assume, in addition, $\lambda_1 > 2 \lambda_n$ for $n>1$. Now Choose $A = \{x, x_1\}$ and $B_n = \{x_n,x_1\}$. Then $d_H(A,B_n) = d(x,x_n) \to 0$. However, $W_\infty(\mu|_A, \mu|_{B_n}) \ge \inf \{d(x,x_1),d(x_n,x_1)\}$ because $\mu|_A(\{x\}) > \frac{1}{2} > \mu|_{B_n}(\{x_n\})$.<|endoftext|> TITLE: Is the identity function a unique multiplicative homeomorphism of $\mathbb N$? QUESTION [12 upvotes]: Endow the set $\mathbb N$ of positive integers with the topology $\tau$ generated by the base consisting of arithmetic progressions $a+b\mathbb N_0$ where $\mathbb N_0=\{0\}\cup\mathbb N$, where $a,b\in\mathbb N$. This topology is often referred to as the Furstenberg topology or the profinite topology. The space $\mathbb N_\tau:=(\mathbb N,\tau)$ is homeomorphic to the space $\mathbb Q$ of rational numbers (being a second-countable regular countable space without isolated points). So $\mathbb N_\tau$ has many non-trivial homeomorphisms. But I know no non-trivial homeomorphism of $\mathbb N_\tau$ which would be multiplicative is the sense that $f(x\cdot y)=f(x)\cdot f(y)$ for any $x,y\in\mathbb N$. Problem 1. Is the identity function a unique multiplicative homeomorphism of $\mathbb N_\tau$? An affirmative answer to this problem follows from an affirmative answer to Problem 2. Are there prime numbers $a,b,c$ such that for any $d\in\mathbb N$ there exists $n\in\mathbb N$ such that $a^n\equiv 1\!\!\!\mod\! d\;\;$ but $\;\;b^n\not\equiv 1\!\!\!\mod\! c$? REPLY [19 votes]: No. First observe that the automorphisms of the semigroup $\mathbf{N}^*$ (which you denote $\mathbb{N}$) are induced by permutations of primes. Consider the automorphism $f$ induced by the transposition $(2,3)$ (thus, mapping $2^a.3^b.c$ to $2^b.3^a.c$, $c$ coprime to 6). I claim that $f$ is continuous. Indeed, consider any convergent sequence $m_i\to m$. Thus, $m_i=m+r_i$ with $r_i$ tending to 0 in the profinite completion of $\mathbf{Z}$ (that is, for every $n\ge 1$ there exists $i_0$ such that $n$ divides $r_i$ for all large $i$). Write $m=2^a.3^b.c$ with $c$ coprime to 6. There exists $i_0$ such that $2^{a+1}3^{b+1}$ divides $r_i$ for all $i\ge i_0$. So, for $i\ge i_0$, $m_i=2^a.3^b.c+2^{a+1}3^{b+1}.t_i$ for some $t_i$. So $m_i=2^a3^b(c+6t_i)$ with $c$ coprime to 6. Thus $f(m_i)=2^b3^a(c+6t_i)=f(m)+r'_i$, with $r'_i=2^{b-a}3^{a-b}r_i$. Thus $r'_i$ also tends to 0 in the profinite completion of $\mathbf{Z}$. Thus $f(m_i)$ tends to $f(m)$. Hence, $f$ is continuous. The argument adapts immediately to any finitely supported permutation of the set of primes.<|endoftext|> TITLE: Simple Subalgebras of Simple Lie Algebras QUESTION [8 upvotes]: Given a complex simple Lie algebra $\mathfrak{g}$ of rank $n\in\mathbb{N}$ with $n$ sufficiently large (say $n\ge10$), is there a way to determine whether $\mathfrak{g}$ contains a simple subalgebra of a prescribed type with rank "close" to $n$? For example, if $\mathfrak{g}$ is of type $B_n$ with $n\ge10$, does $\mathfrak{g}$ contain a subalgebra of type $C_{n-2}$? REPLY [4 votes]: Simple algebras of rank >8 are classical, so you are asking is there a representation (linear, orthogonal or symplectic) of a given dimension of a prescribed Lie algebra. This amounts to the question what is the minimal dimension of a nontrivial linear, orthogonal or symplectic representation of the Lie algebra. A table in Bourbaki answers this.<|endoftext|> TITLE: Research in applied algebra QUESTION [7 upvotes]: I am in my final year of my doctoral study in Mathematics, where my research topic is $p$-groups, specifically classification of $p$-groups by coclass. My work involves a great deal of computation in GAP. I really like programming and have knowledge in C, MatLab and Mathematica. So far all my research is in pure math but for my post-doctoral research I would like to research some applications of algebra/group theory. I don't have sufficient knowledge in this regard though I have heard that genomics and crystallography both rely on applied algebra. I would appreciate learning of some areas/fields where I can apply computational applied algebra as well as institutions/centers and/or scholars whom I could contact. Regarding location, I am open to any place but in specific I am looking for some positions in USA & Canada, Europe & UK or in Australia & New Zealand. REPLY [9 votes]: In the UK, there is the Applied Algebra and Geometry Research Network. You could browse the list of former speakers and abstracts for ideas. The University of St Andrews has a strong group in Combinatorics and Algebra, with some members (such as Rosemary Bailey) working on computations and applications. In Ireland, Graham Ellis's group at NUI Galway is very active in the field of computational algebra. In Leipzig, Germany there is the Max Planck Institute for Mathematics in the Sciences, where in particular Bernd Sturmfels' group works on applications of algebra to non-linear models<|endoftext|> TITLE: Is $-\det\big[\big(\frac{i^2+j^2}p\big)\big]_{1\le i,j\le (p-1)/2}$ always a square for each prime $p\equiv 3\pmod 4$? QUESTION [13 upvotes]: Let $p$ be an odd prime and let $S_p$ denote the determinant $$\det\left[\left(\frac{i^2+j^2}p\right)\right]_{1\le i,j\le (p-1)/2}$$ with $(\frac{\cdot}p)$ the Legendre symbol. By Theorem 1.2 of my paper arXiv:1308.2900 available from http://arxiv.org/abs/1308.2900, $-S_p$ is a quadratic residue modulo $p$. Here I ask a further question. QUESTION. Is it true that for each prime $p\equiv3\pmod4$ the number $-S_p$ is always a positive square divisible by $2^{(p-3)/2}$? Define $a_p=\sqrt{-S_p}/2^{(p-3)/4}$ for any prime $p\equiv3\pmod4$. Then \begin{gather*}a_3=a_7=a_{11}=1,\ a_{19}=2,\ a_{23}=1,\ a_{31}=29,\ a_{43}=254, \\a_{47}=367,\ a_{59}=9743,\ a_{67}=305092,\ a_{71}=29,\ a_{79}=1916927. \end{gather*} I have computed the values of $a_p$ for all primes $p\equiv3\pmod4$ with $p<2000$. Based on the numerical data, I conjecture that the above question has an affirmative answer but I'm unable to prove this. Any ideas towards the solution? Your comments are welcome! REPLY [13 votes]: It can be seen that $S_p$ is not divisible by $p$. The argument about the decomposition of the matrix as $\frac{2}{i\sqrt{p}}A^2$ suggested above implies that $-S_p$ is a square in $\mathbb{Q}[\zeta_p][\sqrt{\lambda_p}]$, where $\lambda_p=-2i\sqrt{p}$. Writing $-S_p=(a\sqrt{\lambda_p}+b)^2$ with $a,b\in\mathbb{Q}[\zeta_p]$, we have $ab=0$ (since $S_p^2\in\mathbb{Z}\subset\mathbb{Q}[\zeta_p]$ and $\sqrt{\lambda_p}\not\in\mathbb{Q}[\zeta_p]$). The case $b=0$ is impossible as then $S_p$ is divisible by $p$, which follows from the fact that the norm of $\lambda_p$ is divisible by an odd power of $p$. Thus $-S_p=b^2$ with $b\in\mathbb{Q}[\zeta_p]$, which implies $-S_p$ is a square since $S_p$ is not divisible by $p$.<|endoftext|> TITLE: Embedding Riemannian manifolds into some infinite dimensional manifolds? QUESTION [12 upvotes]: First of all I am new to the field of embedding one manifold into another other. I have recently come across with the paper "Embedding Riemannian manifolds by their heat kernel" by P. BERARD, G. BESSON, S. GALLOT (published in Geometric and Functional Analysis in 1984), who prove that one can embed a closed Riemannian manifold with certain assumptions on its Ricci curvature and its diameter into $\ell^2$. My question is if there is other results on isometric embedding of a closed Riemannian manifold without any assumption on its Ricci curvature and its diameter into some infinite dimensional manifold (Banach, Hilbert, or Frechet manifolds)? Or into some $L^2$ space? The point is the ambient space must be an infinite dimensional manifold, not finite dimension. Or if not, what other "weaker" assumptions have to be imposed on the manifold? Thank you. REPLY [13 votes]: There is a more general result. Fix an even Schwartz function $\newcommand{\bR}{\mathbb{R}}$ $w:\bR\to[0,\infty)$. Let $\Delta$ be the Laplacian of the compact connected Riemann manifold $(M,g)$, $\dim M=m$. Its eigenvalues are $$0=\lambda_0< \lambda_1\leq \lambda_2\leq \cdots$$ where each eigenvalue appears as many times as its multiplicity. Fix an orthonormal eigen-basis $(\Psi_k)_{k\geq 0}$ of $L^2(M,g)$, $$\Delta\Psi_k=\lambda_k\Psi_k. $$ For each $\newcommand{\ve}{\varepsilon}$ $\ve >0$ define $\Xi_\ve: M\to L^2(M,g)$ by setting $$\Xi_\ve(p)= \left(\frac{\ve^{m+2}}{d_m}\right)^{\frac{1}{2}}\sum_{k\geq 0}w\bigl(\,\ve \sqrt{\lambda_k}\,\bigr)^{\frac{1}{2}}\Psi_k(p)\Psi_k, $$ where $$d_m:=\frac{2\pi^{\frac{m}{2}}}{m \Gamma(\frac{m}{2})}\int_0^\infty w(r) r^{m+1} dr.$$ Then for $\ve>0$ sufficiently small the map $\Xi_\ve$ is an embedding. Moreover, as $\ve\to 0$ the induced metric converges to the original metric. No assumption on the metric $g$ is required. Note that when $w$ is compactly supported the above sum consists of finitely many terms so $\Xi_\ve$ is actually an embedding into a finite dimensional space. The result of Berard-Besson-Gallot corresponds to $w(r)=e^{-r^2}$. For details see this paper.<|endoftext|> TITLE: $p$-adic sums of $p$ terms QUESTION [11 upvotes]: My question is inspired by this riddle: Let $p \geq 5$ be prime, and let $$ 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 {p-1} = \frac a b $$ where $a/b$ is the fraction expressed in lowest terms. Show that $p^2$ divides $a$. Another way to express this claim is that we are giving a $p$-adic approximation to the sum $$ \sum_{k=0}^{p-1} \frac {1} {k+1} = \sum_{k=1}^p \frac {1} {k} = \frac {1} {p} + E $$ where $|E|_p \leq p^{-2}$. This suggests a more general question: Let $f$ be a rational function with coefficients in $\mathbb {Q}_p$. Is there an algorithm faster than direct summation for estimating (being close in the $p$-adic metric) the sum $\sum _{k=0} ^{p-1} f (k)$? Say, something that's polynomial in $\log p$? In particular, is there a way of estimating the harmonic sum as in this riddle with error $\sim p^3$? REPLY [3 votes]: This is not a complete answer, but the following result might useful. Basically it says that if the poles of $f(x)$ are negative integers, then the time to compute $\sum_{k=0}^{p-1}f(k)$ is bounded by a constant times the time to compute binomial coefficients ${kp\choose p}$ modulo powers of $p$. Claim: Suppose $f(x)\in\mathbb{Q}(x)$ has poles contained in the negative integers, and let $n\in\mathbb{Z}$ be given. Then there is a polynomial in $p$, $p^{-1}$, and binomial coefficients ${kp\choose p}$, whose coefficients are rational numbers independent of $p$, which is congruent to $\sum_{m=0}^{p-1}f(m)$ modulo $p^n$ for all sufficiently large $p$. Proof sketch: Using a partial fraction decomposition, it suffices to prove the claim for $f(x)=x^b$, $b\geq 0$, and for $f(x)=(x+a)^{-b}$, $a,b\geq 1$. For $f(x)=x^b$, we can use the formula for power sums, and the sum is a polynomial in $p$. For $f(x)=(x+a)^{-b}$, the sum agrees, up to a rational function of $p$ (which can be expanded in a Laurent series), with $$ \sum_{j=1}^{p-1}\frac{1}{j^b}. $$ Write $e_{i,p}$ for the $i$-th elementary symmetric function evaluated at $1^{-1},2^{-1},\ldots,(p-1)^{-1}$. For every integer $j$, we have $$ \begin{align*} {jp\choose p}&=j\sum_{i=0}^p (j-1)^i p^i e_{i,p}\\ &\equiv j\sum_{i=0}^{n+b-1}(j-1)^i p^i e_{i,p} \mod p^{n+b}. \end{align*} $$ This is a linear system whose coefficient matrix is Vandermonde, hence we can solve for the terms $p^i e_{i,p}$ modulo $p^{n+b}$ as a linear combination of binomial coefficients ${jp\choose p}$, and the coefficients are independent of $p$. For $i\leq b$, we get an expression for $e_{i,p}$ modulo $p^n$ as a linear combination of ${jp\choose p}$. Finally, we can use Newton's formula to expression $\sum_{j=1}^{p-1}\frac{1}{j^b}$ as a polynomial (with coefficients independent of $p$) in the $e_{i,p}$, $1\leq i\leq b$. I wrote a Sage script to compute these expressions. A random example: for all $p$ sufficiently large, $$ \sum_{n=0}^{p-1}\frac{n}{(n+1)^2(n+2)}= -2 \, p^{2} - \frac{1}{36} \, {\left(\frac{{3 p \choose p}}{p} - \frac{6 \, {2 p \choose p}}{p} + \frac{9}{p}\right)}^{2} + 2 \, p + \frac{{3 p \choose p}}{3 \, p^{2}} - \frac{{2 p \choose p}}{p^{2}}+O(p^3). $$ I'll mention that such sums can also be written in terms of $p$-adic zeta values. My software does more optimization here, so the expressions tend to be simpler. For example: $$ \sum_{n=0}^{p-1}\frac{n}{(n+1)^2(n+2)}=-p^{-2}+2p-2p\zeta_p(3)-2p^{2}+2p^{3}-4p^{3}\zeta_p(5)-2p^4 + O(p^{5}). $$<|endoftext|> TITLE: Parabolic Kazhdan-Lusztig polynomial coincide? QUESTION [5 upvotes]: Let $(W,S)$ be a Coxeter system. For any subset $I\subseteq S$, we can have the parabolic Kazhdan-Lusztig polynomial $P_{x,w}^I(q)$ with respect to $I$. Now consider $I\subseteq J\subseteq S$. Both $(W,S)$, $(W_J,J)$ are Coxeter systems. Since $I\subseteq J$, we get the parabolic Kazhdan-Lusztig polynomial $\overline{P}_{x,w}^I(q)$ with respect to $I$ when considering the Coxeter system $(W_J,J)$. Does $P_{x,w}^I(q)=\overline{P}_{x,w}^I(q)$ for all $x,w\in W_J$? REPLY [5 votes]: Yes, that's true. The standard recursive constructions will give you this fact easily, because the only group elements involved in $P_{x,w}^I$ are those which are $\leq w$ w.r.t. the Bruhat order. If $w\in W_J$, then all those elements are themselves contained in $W_J$.<|endoftext|> TITLE: Reference for Kakutani result on power sum bases of symmetric functions QUESTION [9 upvotes]: Numerical semigroups are additive submonoids $A$ of the natural numbers such that the greatest common divisor of all elements of $A$ is 1. The complement of a numerical semigroup in $\mathbb{N}$ is finite and is called the genus of the numerical semigroup. The sequence A007323 in the OEIS gives the number of numerical semigroups with a given genus $g$. Zagier writes in the comments that this also gives the number of sets of $g$ power sum symmetric functions which form a $\mathbb{Q}$-algebra basis for the field of symmetric functions in $g$ variables. He attributes this result to Kakutani but says he forgot the reference. Does anyone know the source of this result? REPLY [8 votes]: I think Zagier must have been thinking of the following papers of Kakeya, instead of Kakutani Kakeya, S.: On fundamental systems of symmetric functions. I, II. Jap. J. Math.2, 69–80 (1925) ; 4, 77–85 (1927) where the following theorem is established: Theorem: Suppose that the complement of the sequence $k_1,k_2,\dots,k_n$ is a numerical semigroup, then the power sums $p_{k_1},p_{k_2},\dots,p_{k_n}$ generate the field $\mathbb Q(p_1,p_2,\dots,p_n)$ of symmetric functions in $n$ variables. He made the conjecture that the converse also holds: if $Q(p_{k_1},p_{k_2},\cdots,p_{k_n})\cong Q(p_1,p_2,\cdots,p_n)$ then $k_1,k_2,\cdots,k_n$ must be the complement of a numerical semigroup. However this is still open in general (for example see section 6 of this recent paper of Dvornicich and Zannier) therefore the two descriptions of sequence A007323 are only conjecturally the same.<|endoftext|> TITLE: Buying more absoluteness for countable transitive models? QUESTION [5 upvotes]: Let $M$ be a countable transitive model of (enough of) ZFC. Mostowski's Absoluteness Theorem says that $\Pi^1_1$ statements are absolute between $M$ and larger models, in particular, between $M$ and the universe $V$. For general $M$, this cannot be extended to $\Sigma^1_2$ statements, see Andrés Caicedo's answer here: Failure of Shoenfield's Absoluteness. My question is: Can we have absoluteness of $\Sigma^1_2$ (and beyond) statements between $M$ and $V$ for some countable transitive models $M$? (Possibly at the expense of some local or global large cardinal assumptions.) Or, is it the case that for any countable transitive model $M$, there is a $\Sigma^1_2$ (lightface) statement true in $V$ which fails in $M$? REPLY [8 votes]: First, it is consistent that there is a ctm but no $\Sigma^1_2$-correct ctm, for example if $V$ is the minimal model with a ctm. If there is a model of ZFC $M$ containing all the reals (e.g., if there is an inaccessible), then there is a projectively correct ctm: take the transitive collapse $H$ of a countable elementary substructure of $M$. Since reals don't move in the collapse, $H$ is projectively correct with real parameters. (By this argument there is a projectively correct model of ZFC if and only if there is a projectively correct ctm.) As Andrés points out, one can also obtain correct models from determinacy hypotheses, though these hypotheses are much stronger than the existence of such models. E.g., if $\Pi^1_1$-determinacy holds then there is a $\Sigma^1_2$-correct ctm (e.g., $L_\kappa$ where $\kappa$ is the least Silver indiscernible).<|endoftext|> TITLE: Existence of newforms which are non-ordinary at a given prime QUESTION [8 upvotes]: Let $f$ be a newform of weight $k \geq 2$ and level $N \geq 1$ without complex multiplication. A prime $p$ is said to be ordinary for $f$ if the $p$-th Fourier coefficient $a_p(f)$ is a $p$-adic unit (to make sense of this in general, one needs to choose a prime ideal above $p$ in the field $K_f$ of Fourier coefficients of $f$, equivalently an embedding of $K_f$ into $\overline{\mathbf{Q}_p}$). Non-ordinary primes seem to be rare. For example, if $f$ has coefficients in $\mathbf{Z}$, then a very rough guess is that $a_p(f)$ modulo $p$ is randomly distributed, so we may expect that \begin{equation*} \# \{p \leq x : a_p \equiv 0 \textrm{ mod } p \} \stackrel{?}{\approx} \sum_{p \leq x} \frac{1}{p} \sim \log \log x. \end{equation*} On the other hand, for the modular form $\Delta = \sum_{n \geq 1} \tau(n) q^n$, it seems not to be known that there are infinitely many primes $p$ such that $\tau(p) \not\equiv 0$ mod $p$, so estimating the number of ordinary or non-ordinary primes is difficult in general. If we consider elliptic curves, we may ask whether every prime $p$ is non-ordinary for some elliptic curve $E$ over $ \mathbf{ Q } $, which means that the reduction of $E$ mod $p$ is supersingular. In fact, Deuring has shown that given any integer $a$ such that $|a|<2\sqrt{p}$, there exists an elliptic curve $E_p$ over the finite field $\mathbf{F}_p$ with exactly $p+1-a$ points over $\mathbf{F}_p$. Now take any elliptic curve $E$ over $\mathbf{Q}$ whose reduction mod $p$ is $E_p$ and use the modularity theorem. We get a modular form $f$ of weight $2$ with integral coefficients satisfying $a_p(f)=a$, and we may choose it to be non-CM (in fact, we get infinitely many such modular forms). My question is whether this kind of result is known or even expected in higher weight. Certainly, if you have a finite set of newforms without CM, the above heuristics suggest that there exist infinitely many primes which are ordinary for all these newforms. But I don't know what happens for an infinite set of newforms, even if this set is "thin" in some sense. Here is another example of question which arises. Fix a weight $k \geq 3$. Is every prime $p$ non-ordinary for some non-CM newform of weight $k$ and level not dividing $p$? At least, are there infinitely many such primes? Hida's theory implies that if $p \geq 5$ and $k$ is equal to $\{3,4,5,\ldots,10,14\}$ modulo $p-1$, then all newforms of weight $k$ and level $1$ are non-ordinary at $p$. This gives a positive answer to the question for some primes $p$ which are small with respect to $k$ (in particular, these primes satisfy $p \leq k-2$). One may try to use the theory of congruences between modular forms, like the theory of Hida families, but usually one starts with a modular form which is ordinary at $p$, and I don't know to which extent the theory has been generalized to the non-ordinary case. REPLY [6 votes]: Given p and k, it's clear we can find a CM-type newform of weight k and some level which is supersingular at p (just choose an imaginary quadratic field in which p is inert, and some sufficiently large conductor away from p). So it suffices to find a second newform that is congruent to the first one mod p and is not of CM type. This can always be done (after adding some auxiliary primes to the level) by the answer to this old question of mine: Congruences between CM and non-CM modular forms.<|endoftext|> TITLE: Character formula for Lie superalgebras QUESTION [7 upvotes]: The Weyl character formula and the denominator identity play important roles in the representation theory of classical simple Lie algebras and Kac-Moody Lie algebras over $\mathbb{C}$ Can you suggest any reference for similar formulas for Lie super-algebras? I suppose there are such formulas, at least for classes of (say classical simple) Lie super-algebras. REPLY [2 votes]: Since you ask for formulas for the character, I will first assume that you are interested in finite dimensional representations. If $\mathfrak{g}$ is a basic classical Lie superalgebra, such as $\mathfrak{sl}(m|n)$, one distinguishes between typical weights and atypical weights. A weight is typical if the irreducible representation $L(\lambda)$ is projective, i.e. does not have any nontrivial extensions. For atypical weights one can further distinguish between the degree of atypicality $1,2, \ldots$. For typical weights Kac already gave a character formula in Kac, V. G.: Characters of typical representations of classical Lie superalgebras. For atypical weights no closed formula is known in general. For weights which have atypicality 1 there is a character formula obtained by Kac and Wakimoto in Kac, Victor G.; Wakimoto, Minoru: Integrable highest weight modules over affine superalgebras and number theory. This covers in particular the case when $\mathfrak{g}$ is a simple exceptional Lie superalgebra or $\mathfrak{osp}(2|2n)$ or $\mathfrak{sl}(n|1)$ since in these cases dominant integral weights are either typical or have atypicality 1. This leaves basically the cases where $L(\lambda)$ is an irreducible representation of atypicality $\geq 2$ of $\mathfrak{gl}(m|n)$, $\mathfrak{osp}(m|2n)$, $\mathfrak{p}(n)$ and $\mathfrak{q}(n)$ for general $m,n$. In the $\mathfrak{gl}(m|n)$ case the character problem was succesfully first solved by Serganova in Serganova, Vera: Kazhdan-Lusztig polynomials and character formula for the Lie superalgebra $\mathfrak{gl}(m|n)$ However this does not give you a closed formula for the character. It is basically an algorithmic solution. Her approach is very similar to the usual one in category $\mathcal{O}$: Write down an (infinite) resolution $$ 0 \leftarrow M^0 \leftarrow M^1 \leftarrow \ldots $$ of $L(\lambda)$ which has a filtration with quotients isomorphic to Kac modules (the universal highest weight modules in these representation categories). The character of $L(\lambda)$ is then $$ ch L(\lambda) = \sum_{i=0}^{\infty} (-1)^i ch M^i $$ where the characters of the $M^i$ can be easily calculated since the characters of Kac modules are known (by Kac himself). Then finding the character of $L(\lambda)$ amounts to determine the coefficients $b_{\lambda,\mu}$ in $$ ch L(\lambda) = \sum_{\mu} b_{\lambda, \mu} ch V(\mu) $$ where $V(\mu)$ is the Kac module. The coefficient $b_{\lambda,\mu} = K_{\lambda,\mu}(-1)$ is the value at $-1$ of a certain Kazhdan-Lusztig polynomial. A nice overview article of Serganova's work is Gruson's Bourbaki article Gruson, Caroline: Sur les representations de dimension finie de la super algebre de Lie $\mathfrak{gl}(m,n)$. Later Brundan gave a different approach using categorification techniques in Kazhdan-Lusztig polynomials and character formulae for the Lie superalgebra $\mathfrak{gl}(m|n)$ A closed formula for the character was then obtained in Su, Yucai; Zhang, R. B.: Character and dimension formulae for general linear superalgebra. They actually obtained a closed formula for the character by reworking some calculations of Brundan. The formula however is so complicated (with an immense amount of cancellations) that this is not comparable to the nice situation of the Weyl character formula for a semisimple Lie algebra. In some cases (when the weight is a so-called Kostant weight) the combinatorics collapses and one can get simple formulas. This can be understood in terms of KL theory since Kostant weights are precisely those, in which the occurring KL polynomials are monomials. Algorithmic or Kazhdan-Lusztig type solutions for the character problem are known also for other Lie superalgebras such as $\mathfrak{osp}(m|2n)$ (due to Gruson-Serganova and later by Cheng-Lam-Wang) and $\mathfrak{q}(n)$ (due to Penkov-Serganova and Brundan), but in general no closed formulas are known for general weights. As far as I am aware the problem of finding the character in the periplectic $\mathfrak{p}(n)$-case is open so far. Of course the character problem has been studied for irreducible modules in category $\mathcal{O}$ as well. As others have mentioned, relevant names here are Brundan, Cheng, Lam, Wang and many others. For the infinite case I would recommend the overview article by Brundan Brundan, John: Representations of the general linear Lie superalgebra in the BGG category $\mathcal O$.<|endoftext|> TITLE: What is to Stone space of the free sigma-algebra on countably many generators? QUESTION [5 upvotes]: I asked the question on MSE. https://math.stackexchange.com/questions/2898377/what-is-the-stone-space-of-the-free-sigma-algebra-on-countably-many-generators The answer I got, however, seems disputed. I just thought that someone here could answer the question for sure. Many thanks. REPLY [5 votes]: You got a wrong answer on Math Stackexchange from Daron. The free Boolean algebra on countably many generators is the Boolean algebra of clopens of $2^\omega$ (topologized with the product topology), and the free $\sigma$-algebra on countably many generators is the Baire $\sigma$-algebra of $2^\omega$, which, as $2^\omega$ is metrizable, is the same as the Borel $\sigma$-algebra. A good source for this is Halmos's book Lectures on Boolean Algebras. To see that Daron's claim is wrong, observe that $\mathcal{P}(\mathbb{N})$ is a complete Boolean algebra, but the Borel $\sigma$-algebra of $2^\omega$ is not, because the least upper bound of the family of all singletons contained in a non-Borel subset of $2^\omega$ does not exist. The algebras $\mathcal{P}(\mathbb{N})$ and $\mathrm{Borel}(2^\omega)$ cannot be isomorphic because any Boolean algebra isomorphic to a complete Boolean algebra is complete. The "$\sigma$-Stone space" of $\mathrm{Borel}(2^\omega)$, i.e. the set of ultrafilters closed under countable intersection, or equivalently the $\{0,1\}$-valued countably additive measures, does have a nice description, as it is isomorphic to $2^\omega$ itself. Unfortunately, the ordinary Stone space of $\mathrm{Borel}(2^\omega)$ does not have a nice description, except tautologous rephrasings of the usual one, such as saying it is the space of $\{0,1\}$-valued finitely-additive Borel measures on $2^\omega$. I think this follows from the fact that there exist models of ZF where the axiom of choice fails and there are no nonprincipal ultrafilters on $\omega$. Any ultrafilter on $\mathrm{Borel}(2^\omega)$ that is not closed under countable intersections can be used to define a non-principal ultrafilter on $\omega$.<|endoftext|> TITLE: Is the top Stiefel-Whitney number of a topological manifold the Euler characteristic mod two? QUESTION [14 upvotes]: Recall that the Stiefel-Whitney classes of a smooth manifold are defined to be those of its tangent bundle - this definition doesn't extend to topological manifolds as they don't have a tangent bundle. Wu's theorem states that for a closed smooth manifold, $w = \operatorname{Sq}(\nu)$. The expression $\operatorname{Sq}(\nu)$ makes sense for a closed topological manifold and therefore serves as a definition for the Stiefel-Whitney classes on such a manifold. Recall that if $M$ is a closed smooth $n$-dimensional manifold, then $w_n(M)$ is equal to the mod $2$ reduction of $e(M)$, see Corollary 11.12 of Milnor and Stasheff's Characteristic Classes. In particular, the Stiefel-Whitney number $\langle w_n(M), [M]\rangle$ is the mod $2$ reduction of the Euler characteristic. Is this still true for closed topological manifolds? Let $M$ be a closed topological $n$-dimensional manifold. If $w_n(M)$ is the top Stiefel-Whitney class of $M$, as defined above, is the Stiefel-Whitney number $\langle w_n(M), [M]\rangle$ the mod $2$ reduction of $\chi(M)$? REPLY [3 votes]: Elaborating on Mark's comment, associated to a manifold of type CAT=PL,DIFF,TOP or a menagerie of others we have a $\mathbb{R}^n$ bundle with structure group $CAT$. In all these cases, we can form the fiberwise one point compactification, and then identify these to get the Thom space. The normal proof of the Thom isomorphism goes through with the obvious notion of orientability of these bundles. Recall that to define Stiefel-Whitney classes for a vector bundle, we only use Steenrod operations and the mod 2 Thom isomorphism. Since every disk bundle is oriented mod 2, we can use the exact same definition. As well, if we have an integrally oriented disk bundle we can use the exact same definition of Euler class, as a pullback of the Thom class by the zero section. An oriented CAT manifold is easily seen to have an integrally oriented $\mathbb{R}^n$ bundle, so we have an Euler class. By picking a section of the disk bundle with isolated singularities (where it hits the zero section), we may mimic the proof in the DIFF case since the total index of the section of the disk bundle is the Euler characteristic of the manifold. This means the Euler class evaluates on the fundamental class to the Euler characteristic. Then the same proof as the DIFF case shows that the CAT Stiefel-Whitney class is the mod 2 reduction of the CAT Euler class. And, again, the same proof as in the DIFF case shows that Stiefel-Whitney classes can be defined via your formula with the Wu classes, so this definition agrees with yours.<|endoftext|> TITLE: Categorifying skein algebras? QUESTION [5 upvotes]: We can obtain the Jones polynomial by the Temperly-Lieb algebra and the HOMFLYPT polynomial from the Hecke algebra. Were there attempts to categorify the algebras itself and obtain the Khovanov homology or HOMFYLPT homology from there? When googling, one can find a lot of papers containing certain categorifications of algebras but I find it hard to pinpoint which of these arise most naturally regarding my question. REPLY [2 votes]: Much of the research in knot homology has been about categorifying these algebras! Khovanov's paper math/0103190 is devoted to defining and studying the Temperley-Lieb 2-category which is a categorification of the Temperley-Lieb category. (The TL algebras arise and endomorphisms of objects in the TL category.) Later, many authors have studied the category of Soergel bimodules, which is a categorification of the Hecke algebra. This approach was used by Khovanov-Rozansky math/0510265 to define HOMFLYPT homology.<|endoftext|> TITLE: Generalization of Drinfeld double to comodule algebras QUESTION [10 upvotes]: Let $ \mathcal C $ be a monoidal category. Then $ \mathcal C $ is both a left and right module category over itself. Moreover, the Drinfeld centre of $ \mathcal C $ can be defined as the category of functors from $ \mathcal C $ to itself which commute with these module category structures. $$ Z(\mathcal C) = Fun_{\mathcal C | \mathcal C}(\mathcal C, \mathcal C) $$ Suppose that $\mathcal C = H\text{-mod}$ where $H$ is a Hopf algebra. Then a well-known result identifies $ Z(\mathcal C) = D(H)\text{-mod}$ where $D(H) $ is the Drinfeld double of $ H$. I am interested in the generalization of this result to the setting of more general module categories. Let $ \mathcal C $ be a monoidal category and let $ \mathcal M $ be a $ \mathcal C$-module category. Then we can consider the category of functors from $ \mathcal M $ to itself, compatible with the module structure: $$ \mathcal D = Fun_{\mathcal C}(\mathcal M, \mathcal M) $$ Assume that $ \mathcal C = H\text{-mod}$ and $ \mathcal M = A\text{-mod} $ where $H$ is a Hopf algebra, $ A$ is an algebra and $ A $ is also an $ H$-comodule (this comodule structure gives rise to the action of $ \mathcal C $ on $ \mathcal M $). Question: Under this setup, can we realize $ \mathcal D $ as the module category of some algebra constructed from $ H $ and $ A $? My question is motivated from the theory of lattice models in condensed matter physics. In the paper Models for gapped boundaries and domain walls by Kitaev and Kong, the authors consider a Levin-Wen model with input $ \mathcal C, \mathcal M $ as above. In this model, the category of functors $ \mathcal D $ is the category of boundary excitations. Moreover, in section 4 of this paper, the author construct an algebra whose module category is claimed to be equivalent to $ \mathcal D $. (Actually they just claim that simple objects correspond.) So I was wondering if these ideas had been developed in the mathematics literature. REPLY [2 votes]: Section 1 (in particular Prop 1.23) of On module categories over finite-dimensional Hopf algebras by Andruskiewitsch-Mombelli come close to an answer to your question: namely, they show that if $H$ is a finite dimensional Hopf algebra, and $K,L$ f.d. comodules algebra algebra over $H$, then $Rep\ H$-linear right exact functor from $K-mod$ to $L-mod$ are $K-L$-bmodules in $H$-comodules. They show this directly, but there is a general explanation. Note that your question has to be special to Hopf algebra (as opposed to general tensor categories) because the relation between $Rep\ H$ and $K-mod$ is somehow "external". One point of view I find illuminating, and hopefully might be useful to you if you want variant of this result, is the following: a comodule algebra $K$ is an $H^*$-module algebra hence becomes an algebra internal to the tensor category $C=Rep\ H^*$, so that you can talk about the category $K-mod_C$ of equivariant/internal $K$-modules. But it turns out $C$ and $D=Rep\ H$ are related by some sort of categorical Koszul duality, and this is where Hopf-ness is used crucially, because $C$ and $D$ are augmented by their fiber functor, ie Vect is a module over those. Long story short, the assignment $$M \longmapsto Fun_C(Vect,M)$$ gives a 2-functorial equivalence between (appropriate adjectives) $C$-module categories and $D$-module categories. Now the above result follows from: this equivalence maps $K-mod_C$, to just $K-mod$ as a $D$-module category, hence to answer your question we might as well compute $C$-module endofunctors of $K-mod_C$ but now we are in an "internal" situation, hence $C$-module functor from $K-mod_C$ to itself are given by internal $K$-bimodule (this is Eilenberg-Watts theorem ), i.e. $K$-bimodule in $H$-comodules.<|endoftext|> TITLE: Proving Positivity for Schubert Calculus QUESTION [10 upvotes]: In study of the cohomology ring of the Grassmannians, which is usually known as Schubert calculus, one usually deals with a distinguished basis known as the Schubert basis $\{\sigma_\lambda\}$. One of the most properties of this basis is positivity, the fact that for any two basis elements $\sigma_\lambda$ and $\sigma_\nu$, the multiplication constants $$ \sigma_\lambda \bullet \sigma_\mu = \sum_{\nu} c_{\lambda,\mu}^{\nu} \sigma_{\nu} $$ satisfy the positivity condition $$ c_{\lambda,\mu}^{\nu} \geq 0, ~~~~~~~~ \text{ for all } \nu. $$ Searching the literature, there seem to be a number of different proofs of this property, the relation between which is not always clear. Can people out there offer an opinion on which is the most insightful approach to proving positivity and what are the advantages/disadvantages or intuitons offered by the other approaches. REPLY [18 votes]: I would say there are three basic reasons for / proofs of positivity. Geometry. [Kleiman 1973] proves that the number one's trying to compute is the number of points in a transverse intersection of cycles. Ergo, a nonnegative number. Combinatorics. Present the cohomology ring of $Gr(k,n)$ as a quotient of that of $Gr(k,\infty)$, taking Schubert classes to Schubert classes or to zero. The latter ring is a polynomial ring containing the Schur polynomials in $x_1,\ldots,x_k$, which give the Schubert classes. There are various combinatorial proofs of the Littlewood-Richardson rule for multiplication of Schur polynomials. Once you know this rule is correct, then you know the coefficients are nonnegative. Representation theory. The Schur polynomials also give the characters of "polynomial" representations of $GL(k)$, multiplication corresponds to tensor product, and decomposing in the Schur polynomial basis corresponds to decomposing the representation. Here, the coefficients are dimensions of intertwining spaces, thus nonnegative. Perhaps the principal way to judge advantage/disadvantages of these approaches is to ask how or whether they generalize beyond the original question you ask, concerning cohomology of Grassmannians. #1 generalizes in many ways, in particular to other homogeneous spaces and other cohomology theories (equivariant, $K$, quantum, etc.) #3 generalizes to representations of other groups. My favorite connection between #1 and #2 is Ravi Vakil's "geometric Littlewood-Richardson rule". Probably the best connection of #2 and #3 is via the theory of crystals. In both cases one can retrodict some of the combinatorial theory. The best connection I know of #1 and #3 is Belkale's construction of a tensor invariant, given an intersection point of three Schubert cycles, such that the tensor invariants constructed form a basis. Which approach is the most insightful... I guess I have to admit a predilection for the geometry. Certainly there are many generalizations of the original problem for which we have geometric proofs of positivity but no combinatorial proofs. Maybe the simplest one involves cohomology of $d$-step flag manifolds, where $d\geq 4$ (the $d=3$ case only solved last year, in a way that doesn't extend to $d\geq 4$).<|endoftext|> TITLE: What does the $p$-adic closure of an arithmetic lattice look like? QUESTION [10 upvotes]: Let $\Gamma$ be an arithmetic lattice in a linear algebraic $\mathbb{Q}$-group $\mathbf{G}$, that is, $\Gamma$ is a subgroup of $\mathbf{G}(\mathbb{Q})$ that is commensurable with $\mathbf{G}(\mathbb{Z})$. For a prime $p$, we can consider $\Gamma$ as a subspace of $\mathbf{G}(\mathbb{Q}_p)$. My question is: What does the closure of $\Gamma$ in $\mathbf{G}(\mathbb{Q}_p)$ with respect to the $p$-adic topology look like? The closure of $\mathbb{Z}$ in $\mathbb{Q}_p$ is the ring of $p$-adic integers $\mathbb{Z}_p$. So it seems plausible to me that, for example, the closure of the lattice $\Gamma = SL(n,\mathbb{Z})$ in the $p$-adic topology of $SL(n,\mathbb{Q}_p)$ would be $SL(n,\mathbb{Z}_p)$. Is this correct? Also, what about other lattice, for example, what is the closure of a congruence subgroup $$\Gamma(c) := \{ g \in SL(n,\mathbb{Z}) : g - I_n \equiv 0 \;\text{ mod } c\, \} \subset SL(n,\mathbb{Z})$$ in the $p$-adic topology of $SL(n,\mathbb{Q}_p)$? REPLY [8 votes]: Suppose $G$ is $\mathbb Q$ simple (i.e. has no connected normal algebraic subgroups which are defined over $\mathbb Q$) and is simply connected (i.e. $G(\mathbb C)$ is simply connected). Assume also that $G(\mathbb R)$ is not compact. With these assumptions, the closure of an arithmetic lattice in $G({\mathbb Z}_p)$ is an open subgroup. This statement is known as strong approximation. More generally, if $G(\mathbb Z)$ is Zariski dense in $G$, and $G(\mathbb C)$ is connected and simply connected, then the closure of a finite index subgroup of $G(\mathbb Z)$ is open in $G(\mathbb Z _p)$. Examples are $G=SL_n$ and $Sp_{2n}$. But not $PGL_n$ (this is not simply connected). In your example, the closure of $SL(n,\mathbb Z)$ is indeed $SL(n,\mathbb Z _p)$; this can be proved by using the fact that $SL(n,\mathbb Z), SL(n,\mathbb Z _p)$ are generated by unipotent elements.<|endoftext|> TITLE: Serre's remark on group algebras and related questions QUESTION [27 upvotes]: I've recently heard about an idea of Serre that for each finite group $G$ there exists a group scheme $X$ such that for each field $K$ the group $X(K)$ is naturally isomorphic to the unit group of $K[G]$. Unfortunately, the article where this fact was mentioned gave no reference, so I ask you if you know how to construct such a scheme. Of course, an interesting question would be: what about a set of $R$-points of $X$, where $R$ is a ring, how is it related to $R[G]$? And can this be generalized somehow to arbitrary groups? In the form given above it sounds not really possible as for $K[\mathbb Z]$ the group of units is isomorphic to $K^*\times \mathbb Z$ and one can hardly imagine a group scheme whose group of $K$-points is isomorphic to $\mathbb Z$. By the way, why there is no group scheme whose group of points is isomorphic to $\mathbb Z$? Or it exists? REPLY [24 votes]: It's fairly easy to do this for finite groups. In fact, the functor $R \mapsto R[G]$ is naturally representable by a ring scheme: the underlying set functor is represented by $\mathbb A^n$ where $n = |G|$, and the ring structure comes from the functor of points $R \mapsto R[G]$. Write $Y$ for this ring scheme (say over $\operatorname{Spec} \mathbb Z$). Now the unit group can be constructed as the closed subset $V \subseteq Y \times Y$ of pairs $(x,y)$ such that $xy = 1$. It is closed because it is the pullback of the diagram $$\begin{array}{ccc}V & \to & Y \times Y\\\downarrow & & \downarrow \\ 1 & \hookrightarrow & Y\end{array},$$ where the right vertical map is the multiplication morphism on $Y$. This shows that $R \mapsto R[G]^\times$ is representable. It naturally becomes a group scheme, again by the functor of points point of view. $\square$ In the infinite case, this construction doesn't work, because the functor $R \mapsto R[G]$ is not represented by $\mathbb A^G$ (the latter represents the infinite direct product $R \mapsto R^G$, not the direct sum $R \mapsto R^{(G)}$). I have no idea whether the functor $R \mapsto R^{(G)}$ (equivalently, the sheaf $\mathcal O^{(G)}$) is representable, but I think it might not be. On the other hand, in the example you give of $G = \mathbb Z$, the functor on fields $$K \mapsto K[x,x^{-1}]^\times = K^\times \times \mathbb Z$$ is representable by $\coprod_{i \in \mathbb Z} \mathbb G_m$, but this does not represent the functor $R \mapsto R[x,x^{-1}]^\times$ on rings for multiple reasons. Indeed, it is no longer true that $R[x,x^{-1}]^\times = R^\times \times \mathbb Z$ if $R$ is non-reduced, nor does $\coprod \mathbb G_m$ represent $R \mapsto R^\times \times \mathbb Z$ if $\operatorname{Spec} R$ is disconnected. These problems do not cancel out, as can already be seen by taking $R = k[\varepsilon]/(\varepsilon^2)$.<|endoftext|> TITLE: Outline of the proof that Cech cohomology and singular cohomology coincide on any locally contractible space QUESTION [8 upvotes]: If $X$ is paracompact and locally contractible, then singular cohomology and Cech cohomology of $X$ coincide, with coefficients in any abelian group. I hear that this is a classical result but I fail to see a clear proof in the literature. Dan Petersen says here that it is in Spanier's book. (I am assuming it is the Algebraic topology book by Spanier). I checked it twice roughly but somehow I could not find it. It would be a good idea to give some outline even if it is there. Can someone give an outline of how this proof goes? There is some proof by using a double cochain complex here. Is there a proof that does not use this double cochain complex set up? Please do not assume any familiarity with spectral sequences. REPLY [9 votes]: It is in Spanier's book, but you have to do two steps. On p. 334 he proves that Cech and Alexander-Spanier cohomology coincide, on p. 340 he proves that Alexander-Spanier and singular cohomology coincide.<|endoftext|> TITLE: Lie algebra of a compact Lie group and derivations of the Hopf algebra of representative functions QUESTION [5 upvotes]: Let $\mathcal{G}$ be a compact (real) Lie group. We know that the Lie algebra $\mathfrak{g}$ of $\mathcal{G}$ is, by definition, the space of all left-invariant (smooth) vector fields over $\mathcal{G}$ with bracket given by the commutator. We also know that it is isomorphic, as a Lie algebra, with $T_e\mathcal{G}$ (the tangent space to $\mathcal{G}$ at the neutral element $e$). Consider the Hopf algebra of representative functions $H:=\mathcal{R}_{\mathbb{R}}(\mathcal{G})$ associated to $\mathcal{G}$ and recall that $H\subseteq \mathcal{C}^\infty(\mathcal{G})$ is a dense subspace with respect to the supremum norm (Proposition I.3.12 of Brocker, Dieck, Representations of Compact Lie Groups together with Peter-Weyl Throem). The Lie algebra $\mathcal{P}(H^\circ)$ of primitive elements of its finite dual is isomorphic the Lie algebra of left-invariant derivations ${^{H}\mathsf{Der}_{\mathbb{R}}(H,H)}$ (i.e., derivations $\delta:H\to H$ such that $\Delta\delta=(H\otimes \delta)\Delta$). I would expect to have an isomorphism $\mathfrak{g}\cong {^{H}\mathsf{Der}_{\mathbb{R}}(H,H)}$. Q1. Is this true? I didn't find it anywhere in the literature, whence I am trying to provide one by myself. Let $X$ be a left-invariant vector field and let $\varphi$ be a smooth function on $\mathcal{G}$. Then $\boldsymbol{X}(\varphi):\mathcal{G}\to \mathbb{R}$ given by $\boldsymbol{X}(\varphi)(g) = X_g(\varphi)$ ($X_g\in T_g\mathcal{G}$) is a smooth function and hence we have an assignment $\boldsymbol{X}:\mathcal{C}^\infty(\mathcal{G})\to \mathcal{C}^\infty(\mathcal{G})$. One can verify that this induces a Lie algebra map $$\mathfrak{g}\to {^{H}\mathsf{Der}_{\mathbb{R}}(H,H)}: X\mapsto \boldsymbol{X}.$$ To provide a candidate inverse to this morphism, I consider a left-invariant derivation $\delta$ and for every $g$ in $\mathcal{G}$ the function $\delta_g:\mathcal{R}_{\mathbb{R}}(\mathcal{G})\to \mathbb{R}, \varphi\mapsto\delta(\varphi)(g)$. Now, I would expect to be able to extend such a function to the whole $\mathcal{C}^\infty(\mathcal{G})$, maybe resorting to the Continuous Linear Extension Theorem (see Theorem 5.7.6 in Foundations of Applied Mathematics, Volume I: Mathematical Analysis by Humpherys, Jarvis, Evans), but I didn't manage to. Q2. Could somebody suggest a way to do this? Q3. Is there some reference in which this is treated in some detail? OT: I already asked it on MSE, but maybe it could be that this question is more suitable for MO. REPLY [2 votes]: Some years ago we did something that -I think- answers the algebraic version of your question [FS]. If you have a Hopf algebra $H$, then the counit gives you a map $\varepsilon_* :\mathrm{Hom}(H,H)\to \mathrm{Hom}(H,k)$. This map has a nice and natural splitting $$H^*\to \mathrm{End}(H)$$ $$f\mapsto F$$ where $F(h)=h_1f(h_2)$. This splitting actually lift to a splitting of the maps $$\varepsilon_*:\mathrm{Hom}(H^{\otimes n},H)\to \mathrm{Hom}(H^{\otimes n},k)$$ for all $n$, in a way that it is compatible with Hochschild coboundary, and stable under Gerstenhaber bracket (see the proof of Theorem 1.5 in the reference above). In particular, $\mathrm{Der}(H,k)$ injects into $\mathrm{Der}(H,H)$. For $H=\mathcal O(G)$, $\mathrm{Der}(H,k)$ (with $k$ viewed as $H$-module via $\varepsilon$) is -almost by definition- ${\frak g}=T_e(G)$, and $\mathrm{Der}(H,H)$ is the algebraic version of ${\frak X}(G)$. The compatibility with Gerstenhaber structure says that these maps are Lie algebra maps. So you get a "Hopf"-way of defining "the unique left invariant vector field" associated to an element in the tangent space of the identity, and for $H=\mathcal O(G)$ you get a precise map $$\mathrm{Der}(\mathcal O(G),\mathcal O(G)) \cong \mathcal O(G)\otimes \mathrm{Der}(\mathcal O(G),k) =\mathcal O(G)\otimes\frak g.$$ Regarding the $C^\infty$-problems, I think that if you write the formula " $F(h)=h_1f(h_2)$" in a diagrammatic way, you should be able to do the $C^\infty$-case. [FS] M. Farinati - A. Solotar, G-Structure on the cohomology of Hopf algebras, Proc. Amer. Math. Soc. 132 (2004), 2859-2865.] (also in https://arxiv.org/abs/math/0207243)<|endoftext|> TITLE: Inclusion of multiplicative group of one local field into the idele class group is a closed embedding, but inclusion of more than one isn't? QUESTION [12 upvotes]: Let $k$ be a global field, $J$ its idele group, and $C = J/k^\times$ the idele class group. For any place $v$ of $k$, we have the familiar closed embedding $k_v^\times \hookrightarrow J$. More generally, we can let $S$ be a finite set of places of $k$, consider $P = \prod_{v \in S} k_v^\times$ with the product topology, and still obtain a closed embedding $P \hookrightarrow J$. This much is clear from the definition of the topology on $J$. I've seen it asserted various places that the continuous injection $k_v^\times \rightarrow C$ is still a closed embedding, and always treated this statement as 'obvious' without much thought. Recently, while reading the Artin-Tate notes (Ch. X, section 2, p.76), I learned the corresponding continuous injection $P \rightarrow C$ is not a topological embedding when $|S| > 1$, and that the image is not even closed! I found this surprising and upon reflection realized that I do not know how to prove either statement. Worse yet, I can't find any discussion of the matter in the standard references (e.g. Cassels-Fröhlich, Neukirch, Weil's Basic Number Theory, ...) Can someone please explain to me why we get a closed embedding when $|S| = 1$, and what goes wrong in the more general case? (bonus points for explicit examples). EDIT Thanks to GH from MO for addressing the "embedding" part of this question in the special case that $S$ contains the archimedean places. I expect that this is the basic explanation for what's going on here. However, the question is not fully answered: I'm still very interested in why the map $P \rightarrow C$ is closed when $|S| = 1$ but not otherwise. It seems to me that a slightly different argument is needed here - I'm not sure what it should look like. It would be good to remove the restriction about the archimedean places - after all, one is frequently interested in the case where $S$ consists of a single non-archimedean place, or in the case where there is more than one archimedean place. Here's my attempt at extending the argument to this case (copied from my comment on the answer): My guess is the answer to the second question goes something like this. Let $|S| = s_\infty + s_f$ with $s_\infty$ the number of archimedean places in $S$. We can force the log of the absolute value of $t$ to be close to $0$ at the $r-s_\infty$ archimedean places outside of $S$. The logs of absolute values of $S$-units span a lattice of dimension $r + s_f - 1 = (r - s_\infty) + |S| - 1$, and we win if we force the image of $t$ to be $0$ in this lattice. So we have "enough dimensions to work with" outside of $S$ iff $|S| = 1$. But this isn't quite rigorous and I'm confused how to make it so (The statement I'm looking for is something like "the projection of a $d$-dimensional lattice in $\mathbb{R}^n$ to an $m$-dimensional subspace has discrete image iff $m \geq d$ - neither direction is obvious to me.) REPLY [5 votes]: Let $J^1$ be the ideles of idelic norm one; the global points $k^*$ is contained in $J^1$. In the number field case, we have a surjection $J/k^* \rightarrow \mathbb R _{>0}$ with kernel $J^1/k^*$ (the latter is compact). Now let $v$ be a place and $O_v^*$ the unit group of $k_v^*$ (if $v$ is complex, $O_v^*=S^1$ ; if $v$ is real, $O_v^*=\{\pm 1\}$; if $v$ is non-archimedean, $O_v^*$ is the group of units). Then under the above quotient map $J/k^*$ the inclusion of $k_v^*$ in $J$ induces a map $k_v^*\rightarrow \mathbb R_{>0}$ whose kernel is the compact group $O_v^* \subset J^1/k^*$. The map $O_v^* \rightarrow J^1/k^*$ is a continuous injection of compact Hausdorff spaces and is hence a homeomorphism onto the image.Since $k_v^*$ is the product of $O_v^*$ with a closed subgroup of $R_{>0}$ (in the archimedean case, it is all of $\mathbb R_{>0}$ and in the non-archimedean case it is a discrete subgroup of $\mathbb R_{>0}$ generated by the uniformizing parameter of $k_v^*$) it follows that we need only check that the map $k_v^* \rightarrow \mathbb R_{>0}$ has closed image. The image is always a closed subgroup and hence if $Card S=1$, $k_v^* \rightarrow J/k^*$ is a closed embedding. In the non-archimedean case, the image is of the form $\mid \pi \mid ^{\mathbb Z}$ with $\pi $ a uniformizing parameter. If $S$ has two non-archimedean places (which lie above different primes of $\mathbb Q$), the image is of the form $\mid \pi _v \mid ^{\mathbb Z} \mid \pi _w\mid ^{\mathbb Z} \subset \mathbb R_{>0}$ and is not a closed subgroup in $\mathbb R_{>0}$ and hence the embedding of $k_v^*\times k_w^*$ is not closed.<|endoftext|> TITLE: Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices? QUESTION [15 upvotes]: (Disclaimer : I know very well that $SO(N)$ has a Lie algebra of dimension $N(N-1)/2$ etc. This absolutely not the point of my question.) To make my problem more understandable, I start with the example of $SO(2)$. All $SO(2)$ matrices $M$ can be written as ($\theta\in [0,2\pi[$) $$ M=\begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta&\cos\theta\end{pmatrix}. $$ Using the basis of $2\times2$ real matrices $\sigma_0=\begin{pmatrix}1 & 0\\0&1\end{pmatrix}$, $\sigma_1=\begin{pmatrix}0 & 1\\1&0\end{pmatrix}$, $\sigma_2=\begin{pmatrix}0 & 1\\-1&0\end{pmatrix}$, $\sigma_3=\begin{pmatrix}1 & 0\\0&-1\end{pmatrix}$, one find that $$M=\cos\theta\;\sigma_0+\sin\theta\;\sigma_2.$$ Clearly, $M$ does not have components along $\sigma_1$ and $\sigma_3$, so the dimension of the smallest linear subspace of $\mathrm{M}_2(\mathbb{R})$ that contains $SO(2)$ is $2$. How to articulate the reasoning (for the cases $N>2$ in particular) is not completely clear. I guess that we can say that the component along $\sigma_0$ and $\sigma_2$ are independent because $\cos\theta$ and $\sin\theta$ are independent functions (in a functional analysis sense). Assuming that made sense, we can try to increase $N$. For example, an $SO(3)$ matrix can be written as $$ M=\left(\begin{matrix} \cos\varphi\cos\psi - \cos\theta\sin\varphi\sin\psi & -\cos\varphi\sin\psi - \cos\theta\sin\varphi\cos\psi & \sin\varphi\sin\theta\\ \sin\varphi\cos\psi + \cos\theta\cos\varphi\sin\psi & -\sin\varphi\sin\psi + \cos\theta\cos\varphi\cos\psi & -\cos\varphi\sin\theta\\ \sin\psi\sin\theta & \cos\psi\sin\theta & \cos\theta \end{matrix}\right)\, $$ with $(\phi,\psi)\in [0,2\pi[^2$ and $\theta\in [0,\pi[$. Now, if I look at each matrix element one by one, they are all independent in a functional sense.$^*$ Does that mean that the ''dimension of the matrix space'' that $SO(3)$ matrices live in is $9$ ? Is there a way to generalize that to arbitrary $N$ ? In the end, does any of what I wrote above make any sense ? $^*$ It is slightly more subtle than that for the $\theta$ dependence, because in the end I am interested in doing integral over the Haar measure, which means that one should look at $x=\cos(\theta)\in[-1,1]$. But $x$ and $\sqrt{1-x^2}$ are orthogonal, so all should be fine. REPLY [17 votes]: Let $S$ be the span of $SO(N)$ . Then it's obvious that if $A \in S$ and $D_1, D_2 \in SO(N)$ then $D_1^{-1} A D_2 \in S$ . Therefore it's enough to show that $B := diag(1,0,...0) \in S$ . If $N$ is odd and $> 1$ then we can write $$B = \frac{1}{2} (\left[\begin{matrix}1&0\cr0&I_{N-1}\end{matrix}\right] + \left[\begin{matrix}1&0\cr0&-I_{N-1}\end{matrix}\right])$$ . If $N$ is even and $> 2$ then we can write $$B = \frac{1}{4} (\left[\begin{matrix}1&0&0\cr0&1&0\cr0&0&I_{N-2}\end{matrix}\right] + \left[\begin{matrix}1&0&0\cr0&1&0\cr0&0&-I_{N-2}\end{matrix}\right] + \left[\begin{matrix}1&0&0\cr0&-1&0\cr0&0&E\end{matrix}\right] + \left[\begin{matrix}1&0&0\cr0&-1&0\cr0&0&-E\end{matrix}\right])$$ , where we choose $E \in O(N-2)$ with $det E = -1$ .<|endoftext|> TITLE: Stable Cohomotopy as $K \mathbb{F}_1$ QUESTION [13 upvotes]: Various classical results suggest that stable cohomotopy may usefully be regarded as being the algebraic K-theory over the "field with one element" $\mathbb{F}_1$: $$ K \mathbb{F}_1 \;\simeq\; \mathbb{S} $$ A summary table with pointers is on the nLab here. Guillot has made this observation the starting point in his study of "Adams operations in cohomotopy" (arXiv:math/0612327), on which Jack Morava then based his note "Power operations and Absolute Geometry" (2012, pdf). Both of these seem to remain unpublished, though. Has anyone picked this up? Are there any further developments along these lines? REPLY [4 votes]: Every symmetric monoidal category has a $K$-theory spectrum, and the $K$-theory spectrum of the category of finite sets and bijections is just the sphere spectrum. I believe that everything that you have mentioned can be done in a well-defined way form this point of view, without needing to refer to $\mathbb{F}_1$. I think that Stefan Schwede has written about $\beta$-rings in these terms, although I do not remember where. You pointed to a page on the nLab involving the Atiyah-Segal completion theorem and related things. As these can be formulated uniformly for a large class of groups $G$, it is best to discuss them in the context of global stable homotopy theory, which is another body of work of Schwede. I am not sure whether he has tied these things together.<|endoftext|> TITLE: Closed orientable surfaces have even Euler characteristic QUESTION [9 upvotes]: It is of course completely standard that closed orientable surfaces have even Euler characteristic. What is the most elementary proof of this? More specifically, suppose I have a finite simplicial complex $K$ with vertices $V$, edges $E$ and faces $F$. I suppose that each edge is contained in precisely two faces, and that the link of each vertex is a cycle of size at least three. This implies that $|K|$ is a closed surface, and it is not hard to see that $3|F|=2|E|$, so $|F|$ is even. We want to show that $|V|-|E|+|F|$ is even, or equivalently that $|V|=|E|\pmod{2}$. This is not true in general if $K$ is not orientable, so we need to say something about orientations. Let $D$ be the set of directed edges, and let $\chi\colon D\to D$ reverse direction. Let $S$ be the set of permutations $\sigma\colon D\to D$ such that For any $(u,v)\in D$ we have $\sigma(u,v)=(v,w)$ and $\sigma(v,w)=(w,u)$ and $\sigma(w,u)=(u,v)$ for some $w$ such that $\{u,v,w\}\in F$ If $u,v,w$ are as above, and $x$ is the other vertex such that $\{u,v,x\}\in F$, then $\sigma(v,u)=(u,x)$ and $\sigma(u,x)=(x,v)$ and $\sigma(x,v)=(v,u)$. We find that $S$ bijects with the set of orientations of $|K|$. Also, we have $\sigma^3=1$ and we can describe the composite $\rho=\sigma\chi$ as follows: the edge $\rho(u,v)$ starts at $u$, and is the next edge round $u$ after $(u,v)$ in clockwise order. Given $K$ and $\sigma$ as above, it seems that there should be some very direct combinatorial argument to show that $|V|=|E|\pmod{2}$, but I am not seeing one. REPLY [15 votes]: The set of vertices can be split as the union $V=V_{\text{odd}}\cup V_{\text{even}}$ of vertices with odd and even degrees, respectively. Since the sum of degrees over all vertices is the same as twice the number of edges, we know that $|V_{\text{odd}}|= 0\pmod 2$. Therefore we want to establish that $|V_{\text{even}}|=|E|\pmod 2$. The permutation $\sigma$ is the product of $|F|$ cycles of length 3, so $\operatorname{sign}(\sigma)=1$. The permutation $\chi$ is the product of $|E|$ cycles of length 2 so $\operatorname{sign}(\chi)=(-1)^{|E|}$. And finally the permutation $\rho$ is the product of $|V_{\text{odd}}|$ cycles of odd length and $|V_{\text{even}}|$ cycles of even length, therefore $\operatorname{sign}(\rho)=(-1)^{|V_{even}|}$. Since $\rho=\sigma \chi$ we must have $$\operatorname{sign}(\rho)=\operatorname{sign}(\sigma)\operatorname{sign}(\chi)\implies |E|=|V_{\text{even}}|\pmod 2$$<|endoftext|> TITLE: On spaces with finite homological dimension QUESTION [5 upvotes]: Let $X$ be a connected $CW$-complex, such $\pi_1(X)$ is torsion-free and $H_k(X,\mathbb Z) = 0$ for all $k \geq N$ and some $N \in \mathbb N$. Then $(1)$ Does it follow that $X$ is homotopy-equivalent to its $N$-skeleton, i.e $X \simeq X^{(N)}$ ? $(2)$ If $(1)$ is false, does it follow that $X \simeq X^{(k)}$ for some $k \in \mathbb N$ ? $(3)$ If $(2)$ is also false, at the very least, does it follow that $X \simeq Y$ for some finite-dimensional $CW$-complex $Y$ ? It is clear to me that the question has a negative answer when $H_k$ is replaced by $\pi_k$ (take a $CW$-model of $BG$ for a group $G$ containing elements of finite order). I am also very certain that this question has been asked before and I just couldn't find the corresponding thread, so I have no problem with this question being marked as duplicate. Edit: One could then possibly find a finitely dominated $CW$-complex $X$, with $\pi_1(X)$ not torsion-free, such that its Wall finiteness obstruction element $w(X) \in \widetilde{K_0}(\mathbb Z[\pi_1(X)])$ is non-trivial. Any finitely dominated $CW$-complex has finite homological-dimension (so it satisfies the assumptions), but if its finiteness obstruction doesn't vanish, it doesn't have the homotopy type of a finite complex. Conversely, if $X$ is finitely-dominated and $w(X) = 0$, then it has the homotopy type of a finite-dimensional complex. In particular, this must be the case if $\widetilde{K_0}(\mathbb Z[\pi_1(X)]) =0$. It is conjectured that $\widetilde{K_0}(\mathbb Z[G]) = 0$ if $G$ is torsion-free, hence the assumption in the statement of the question. Edit 2: Mark has given an example of a $CW$-complex $X$ with finite homological dimension, but infinite cohomological dimension (and thus, has given a negative answer to my original questions) . It would be still interesting to know if there exist counter-examples $X$, such that for any $\pi_1(X)$-module $M$, both the homology $H_*(X,M)$ and the cohomology $H^*(X,M)$ have groups only in finitely many dimensions. REPLY [8 votes]: You can take $X=BG$ where $G$ is a torsion-free, acyclic group of infinite cohomological dimension. Acyclic means $H_k(BG;\mathbb{Z})=0$ for $k>0$, and infinite cohomological dimension implies infinite geometric dimension, so $BG$ is not homotopy equivalent to any finite dimensional CW complex $Y$. Such a group therefore gives a (rather extreme) counter-example to (1), (2) and (3). I believe an infinite direct product of copies of Higman's group $H$ gives such a $G$. It is obviously torsion-free (since $H$ is), and is acyclic since homology commutes with direct limits and a finite product of acyclic groups is acyclic. Since $H$ is of type $FP_\infty$ (iirc, the presentation complex is a model for $BH$), we can apply Benjamin Steinberg's answer here to conclude that $G$ has infinite cohomological dimension. REPLY [3 votes]: For an elementary example, give $B^{n+1}$ the CW structure with a $0$-cell, a boundary $n$-cell, and an $n+1$-cell, then take $X=\bigvee_nB^{n+1}$. Then $X$ is contractible but $X^{(n)}$ is homotopy equivalent to $S^n$ (from the $B^{n+1}$ term) so $X^{(n)}$ is never homotopy equivalent to $X$. It would be better to look for $Y$ with $X^{(n)}\subseteq Y\subseteq X^{(n+1)}$ such that the inclusion $Y\to X$ is a homotopy equivalence. I suspect that that is not possible either, although it would be a bit more delicate to provide an example. Better still, we can ask whether there is a CW complex $Y$ obtained from $X^{(n)}$ by attaching some $n+1$-cells, such that the inclusion $X^{(n)}\to X$ extends over $Y$, and the resulting map $Y\to X$ is a homotopy equivalence. The attaching maps for $Y$ will lie in the relative homotopy group $\pi_{n+1}(X,X^{(n)})$, which maps to $H_{n+1}(X,X^{(n)})$. If $X$ is simply connected then you can use the relative Hurewicz theorem and I think it works out that you can always find such a $Y$ provided that $H_{n+1}(X)$ is free abelian (perhaps zero) and $H_k(X)=0$ for $k\leq n$. If $X$ is not simply connected then you need to worry about the Wall finiteness obstruction.<|endoftext|> TITLE: Naive question on local cohomology QUESTION [6 upvotes]: Let $X$ be a smooth, projective variety and $Z_1, Z_2$ two smooth, projective subvarieties in $X$ of the same dimension. Let $E$ be a locally free sheaf on $X$. Recall, there are natural morphims: $$g_i:H^k_{Z_i}(E) \xrightarrow{f_i} H^k_{Z_1 \cup Z_2}(E) \to H^k(E)$$ for $i=1,2$. Let $\alpha_i \in H^k_{Z_i}(E)$ for $i=1,2$. Is it true that $g_1(\alpha_1)=g_2(\alpha_2)$ if and only if $f_1(\alpha_1)=f_2(\alpha_2)$? (If necessary assume that $Z_1 \cap Z_2$ is of codimension at least $2$ in both $Z_1$ and $Z_2$). Any idea/reference will be most welcome. REPLY [3 votes]: I don't think this has much of a chance. The group $H^{k}(E)$ could easily be zero while the local cohomology is non-zero, so $g_1(\alpha_1)=g_2(\alpha_2)$ for any two classes, while if you have a single class $\alpha$ for which $f_1(\alpha)\neq 0$ then what you are hoping for doesn't work. More particularly, let $X=\mathbb P^n$, $E=\mathscr O_X$, and $Z_i\in\mathbb P$ different single points. Then for $k=n$ both $f_1$ and $f_2$ are injective, but their images only intersect in $0$, so $f_1(\alpha_1)=f_2(\alpha_2)$ only if $\alpha_1=\alpha_2=0$. My guess is that this fails more times than it is true. Notice that in this example $Z_1\cap Z_2=\emptyset$. The more natural thing to look at is the Mayer-Vietoris sequence which tells you that $$\dots \to H^{k}_{Z_1\cap Z_2}(E) \to H^k_{Z_1}(E) \oplus H^k_{Z_2}(E) \xrightarrow{\ f_1-f_2 \ } H^k_{Z_1 \cup Z_2}(E) \to H^{k+1}_{Z_1\cap Z_2}(E) \to \dots$$ is exact. Perhaps you can use that for what you needed this.<|endoftext|> TITLE: A possible characterization of sphere or projective space QUESTION [5 upvotes]: Is there a compact Riemanian manifold $M$ not diffeomorphic to sphere or real or complex or quaternion projective space which admit a diffeomorphism $f$ with the property that $$\forall x \in M, \quad d(f(x), x)=diam(M)$$ where $d$ is the metric arising from the Riemannian metric and diameter of $M$ with repect to this metric is denoted by $diam(M)$ REPLY [10 votes]: The article by X. Liu and Sh. Deng "The antipodal sets of compact symmetric spaces" gives many examples, e.g. $\mathrm{SU}(2n)$, $\mathrm{Spin}(5)$, $\mathrm{Spin}(7)$,.... All those spaces have unique antipodal points which implies that the antipodal map is a diffeomorphism, see below. Here the antipodal points at $p$ are defined as $A(p)=\{ x\in M ~|~d(x,p)=\mathrm{diam}M\}$. If a homogeneous space, equipped with a left-invariant metric has unique antipodal points then by compactness one gets a homeomorphism $f$ with the desired properties. By left-invariant of the metric the map $t\mapsto d(\exp(tX)x_0,\exp(tX)y_0)$ is constant so that $f(\exp(tX)x) = \exp(tX)f(x)$ which shows $f$ is differentiable. Here $\exp(X)$ denotes the isometry generated by the Killing field $X$.<|endoftext|> TITLE: Every unorientable 4-manifold has a $Pin^c$, $Pin^{\tilde c+}$ or $Pin^{\tilde c-}$ Structure QUESTION [5 upvotes]: The precise statement on J. W. Morgan's "The Seiberg-Witten Equations and Applications to the Topology of Smooth Four-Manifolds (MN-44)" that 4-manifold $X$ admits a Spinc structure (Lemma 3.1.2) seems to be that every 4-orientable manifold X admits a $Spin^c$ structure. This means that we impose the $w_1(X)=0$ for the 4-orientable manifold X. However, for 4-unorientable manifold $M$, we may modify the statement to different structures. In addition to, $Spin^c=\frac{(Spin \times U(1))}{\mathbb{Z}/2\mathbb{Z}},$ we can have: $$Pin^c=\frac{(Pin^+ \times U(1))}{\mathbb{Z}/2\mathbb{Z}}=\frac{(Pin^- \times U(1))}{\mathbb{Z}/2\mathbb{Z}},$$ $$Pin^{\tilde c+}=\frac{(Pin^+ \ltimes U(1))}{\mathbb{Z}/2\mathbb{Z}},$$ $$Pin^{\tilde c-}=\frac{(Pin^- \ltimes U(1))}{\mathbb{Z}/2\mathbb{Z}}.$$ See for example this Ref, Annals of Physics 394, 244-293 (2018) and References therein. It seems that I can improve John Morgan's statement to show that Every unorientable 4-manifold has either a $Pin^c$, $Pin^{\tilde c+}$ or $Pin^{\tilde c-}$ Structure. (?) e.g. My approach is based on improving the map \begin{equation*} H^1(X;Pin^c) \to H^1(X; O(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{} H^2(X;\mathbb{Z}_2), \end{equation*} \begin{equation*} H^1(X;Pin^{\tilde c+}) \to H^1(X; O(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{} H^2(X;\mathbb{Z}_2), \end{equation*} \begin{equation*} H^1(X;Pin^{\tilde c-}) \to H^1(X; O(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{} H^2(X;\mathbb{Z}_2). \end{equation*} The last maps of all three need to have appropriate constraints between $c_1$ and $w_2(M)$, $w_1(M)$ and $w_1^2(M)$. Question: I wonder whether there exists any math literature show the similar results like mine above? Or are my statements obviously true? (Or obviously wrong?) REPLY [6 votes]: $\newcommand{\RP}{\mathbb{RP}}\newcommand{\Z}{\mathbb Z}$ The conjecture is false: $\RP^4\amalg(\RP^2\times\RP^2)$ is an unorientable 4-manifold that has no pin$c$, pin$\tilde c-$, or pin$\tilde c+$ structure. It suffices to find three unorientable 4-manifolds $A$, $B$, and $C$, such that $A$ isn't pin$c$, $B$ isn't pin$\tilde c-$, and $C$ isn't pin$\tilde c+$. Then $A\amalg B\amalg C$ doesn't admit any of the three structures: a $G$-structure is a reduction of the principal bundle of frames, so a $G$-structure on a manifold induces a $G$-structure on each connected component. First, $A = \RP^2\times\RP^2$ doesn't have a pin$c$ structure; this is discussed here. Then, $B = \RP^4$ has no pin$\tilde c-$-structure. Let $H^*(-; \Z_{w_1})$ denote integer cohomology twisted by the orientation bundle. From Shiozaki-Shapourian-Gomi-Ryu, Lemma D.7, we learn that a manifold $M$ has a pin$\tilde c-$-structure iff $w_2 + w_1^2\in H^2(M;\Z/2)$ admits a lift across the mod 2 reduction map $$\tag{$*$} H^2(M;\Z_{w_1})\longrightarrow H^2(M; (\Z/2)_{w_1}) = H^2(M;\Z/2).$$ Twisted Poincaré duality tells us that if $M$ is a closed $n$-manifold, there are isomorphisms $H_k(M;\Z)\cong H^{n-k}(M;\Z_{w_1})$, so $H^2(\RP^4;\Z_{w_1})\cong H_2(\RP^4;\Z) = 0$. A quick computation shows that if $x\in H^1(\RP^4;\Z/2)$ is the generator, then $w_2(\RP^4) + w_1(\RP^4)^2 = x^2$; in particular, it's nonzero, so it can't lift to $H^2(\RP^4;\Z_{w_1})$. Therefore $\RP^4$ admits no pin$\tilde c-$-structure. Finally, $C = \RP^2\times\RP^2$ doesn't have a pin$\tilde c+$-structure. Lemma D.8 of the Shiozaki-Shapourian-Gomi-Ryu paper tells us that a manifold $M$ has a pin$\tilde c+$-structure iff $w_2\in H^2(M;\Z/2)$ admits a lift across ($*$). Let $x$ denote the generator of $H^1(-;\Z/2)$ of the first $\RP^2$ and $y$ be that for the second $\RP^2$. Using the usual CW structure on $\RP^2$ and the product CW strucure on $\RP^2\times\RP^2$, you can check that $H_2(\RP^2\times\RP^2;\Z)\cong\Z/2$ and the reduction mod 2 map $H_2(\RP^2\times\RP^2;\Z)\to H_2(\RP^2\times\RP^2;\Z/2)$ sends the nonzero element of $H_2(\RP^2\times\RP^2;\Z)$ to the Poincaré dual of $x+y$. The Poincaré duality isomorphisms $H_2(\RP^2\times\RP^2;\Z)\cong H^2(\RP^2\times\RP^2;\Z_{w_1})$ and $H_2(\RP^2\times\RP^2;\Z/2)\cong H^2(\RP^2\times\RP^2;\Z/2)$ are natural with respect to change-of-coefficients, which means that the reduction mod 2 map ($*$) for $M = \RP^2$ sends the nonzero element of $H^2(\RP^2\times\RP^2;\Z_{w_1})\cong\Z/2$ to $x+y$. However, $w_2(\RP^2\times\RP^2) = x^2+xy+y^2$, so it's not in the image of ($*$), and therefore $\RP^2\times\RP^2$ has no pin$\tilde c+$-structure.<|endoftext|> TITLE: A new determinant question for primes $p\equiv3\pmod4$ QUESTION [7 upvotes]: Let $p$ be an odd prime, and let $(\frac{\cdot}p)$ denote the Legendre symbol. Motivated by my question http://mathoverflow.net/questions/310301, here I introduce the matrices $A^+_p$ and $A^-_p$ whose definitions are as follows: $$A^+_p=[a_{ij}^+]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^+=\left(\frac jp\right) \ \text{and}\ a_{ij}^+=\left(\frac{i+j}p\right)\ \text{for}\ i>1,$$ $$A^-_p=[a_{ij}^-]_{1\le i,j\le (p-1)/2}\ \text{with}\ a_{1j}^-=\left(\frac jp\right) \ \text{and}\ a_{ij}^-=\left(\frac{i-j}p\right)\ \text{for}\ i>1.$$ QUESTION: Let $p\equiv3\pmod4$ be a prime. Is it true that $\det A_p^-=(-1)^{(p-3)/4}?$ When $p>3$, is it true that $\det A_p^+=-2^{(p-3)/2}$? Based on my computation, I conjecture that the question has a positive answer. REPLY [7 votes]: The conjectures are true. If you negate the first row of $A_p^{-}$ you get a cofactor of the matrix in "Chapman's evil determinant". In particular you can get the answer from the same matrix decomposition used by Vsemirnov in his paper: On the evaluation of R. Chapman’s “evil determinant”. I couldn't deduce the result for $A_p^{+}$ from Vsemirnov's paper so I'm sketching a different approach here. The proof essentially consist of (1) using quadratic Gauss sums to write the matrix as a (rank one perturbation of) a product of shifted Vandermonde matrices, and (2) repeated use of the Lagrange interpolation formula to evaluate various sums of rational functions evaluated at roots of unity. Here are the details: Throughout this post $p$ is a prime $3\pmod{4}$ and $\zeta=e^{\frac{2\pi i}{p}}$. Lemma 1: Given $a_1,a_2,\dots,a_{\frac{p-1}{2}},b_1,b_2,\dots,b_{\frac{p-1}{2}}\in \mathbb Z/p\mathbb Z$ we have $$\left[\left(\frac{a_i +b_j}p\right)\right]_{1\le i,j\le \frac{p-1}{2}}=\frac{1}{i\sqrt{p}}\left(J+2M_aM_b^{\top}\right)$$ where $M_a=[\zeta^{a_i k^2}]_{1\le i,k\le \frac{p-1}{2}}$ and $M_b=[\zeta^{b_jk^2}]_{1\le j,k\le \frac{p-1}{2}}$ and $J$ is the matrix with all entries $1$ which can also be written as $\textbf{1}^{\top}\textbf{1}$, with $\textbf{1}$ being the row vector of all $1$'s. Proof: This follows entry-wise from the quadratic Gauss sum $$\left(\frac{a_i +b_j}p\right)=\frac{1}{i\sqrt p}\left(1+2\sum_{k=1}^{\frac{p-1}{2}}\zeta^{(a_i+b_j)k^2}\right)=\frac{1}{i\sqrt p}\left(1+2\sum_{k=1}^{\frac{p-1}{2}}\zeta^{a_ik^2}\cdot \zeta^{k^2 b_j}\right).$$ Corollary 1: By the matrix determinant lemma, we have $$\det \left[\left(\frac{a_i +b_j}p\right)\right]_{1\le i,j\le \frac{p-1}{2}}=\frac{1}{\left(i\sqrt p\right)^{\frac{p-1}{2}}}\left(2^{\frac{p-1}{2}}\det M_a \det M_b+\textbf{1}^{\top}\textbf{Adj} (2M_aM_b^{\top}) \textbf{1}\right)$$ $$=\frac{1}{\left(i\sqrt p\right)^{\frac{p-1}{2}}}\left(2^{\frac{p-1}{2}}\det M_a \det M_b+2^{\frac{p-3}{2}}\sum_{r=1}^{\frac{p-1}{2}}\det M_a^{(r)}\det M_b^{(r)}\right)$$ Where we use the notation $M^{(r)}$ to denote the matrix obtained from $M$ by replacing the $r$-th row by $\textbf{1}$. The matrix $A_p^{+}$ corresponds to the choice $(a_1,\dots,a_{\frac{p-1}{2}})=(0,2,3,\dots,\frac{p-1}{2})$ and $(b_1,\dots,b_{\frac{p-1}{2}})=(1,2,\dots,\frac{p-1}{2})$. Computing $\det M_a,\det M_b$ or $\det M_a^{(r)},\det M_b^{(r)}$ for these vectors is an evaluation of Vandermonde determinants or simple instances of Schur polynomials, so I am leaving it as an exercise, for the sake of brevity. What you will end up getting from corollary 1 on these specific vectors, is the following: $$\det A_p^{+}=$$ $$\frac{\left(\prod_{k_1 TITLE: Spec Z is simply connected space? QUESTION [6 upvotes]: Does it follows from conductor's property in Dedekind domains? Where I can find a reference? Konstantin REPLY [10 votes]: Here's a nice way to see that $\mathrm{Spec}(\mathbb{Z})$ is étale simply connected. Since $\mathrm{Spec}(\mathbb{Z})$ is a normal scheme every connected finite étale cover of $\mathrm{Spec}(\mathbb{Z})$ is of the form $\mathrm{Spec}(\mathcal{O}_K)$ for some number field $K$, as $\mathcal{O}_K$ is the normalization of $\mathbb{Z}$ in $K$ over $\mathbb{Q}$. Now suppose that $K$ has degree $n=r_1+2r_2$ where $r_1$ and $r_2$ are the numbers of real and complex embeddings respectively of $K$, and let $\Delta_K$ be the discriminant of $K$. Then by the Minkowski bound (see Neukirch) every class in the ideal class group of $K$ contains an integral ideal of norm at most the Minkowski constant $M_K=\sqrt{\vert\Delta_K\vert}(\frac{4}{\pi})^{r_2}\frac{n!}{n^n}$. But by definition an integral ideal of $\mathcal{O}_K$ has norm at least $1$ so that $1\leq M_K$ and hence $\sqrt{\vert\Delta_K\vert}\geq(\frac{\pi}{4})^{r_2}\frac{n^n}{n!}\geq(\frac{\pi}{4})^{n/2}\frac{n^n}{n!}$; in particular if $K$ is a nontrivial extension of $\mathbb{Q}$ we have $\vert\Delta_K\vert>1$, and since a prime $p\in\mathbb{Z}$ ramifies in $K$ precisely if $p$ divides $\Delta_K$ it follows that the finite étale cover $\mathrm{Spec}(\mathcal{O}_K)\rightarrow\mathrm{Spec}(\mathbb{Z})$ is somewhere unramified unless it is trivial, which implies $\widehat{\pi}_1(\mathrm{Spec}(\mathbb{Z}))=0$. In general for $\mathcal{O}_K$ the ring of integers of a number field $K$ the maximal Abelian quotient $\widehat{\pi}^\mathrm{ab}_1(\mathrm{Spec}(\mathcal{O}_K))$ is isomorphic to the narrow class group $\mathrm{Cl}^+_K=I_K/P^+_K\simeq\mathrm{Gal}(H^+_K/K)$ where where $H^+_K$ is the narrow class field of $K$ (the maximal Abelian extension of $K$ which is unramified outside the finite places), and $I_K/P^+_K$ is the group of fractional ideals of $\mathcal{O}_K$ modulo the group of principal fractional ideals $(\alpha)=\alpha\mathcal{O}_K$ whose generator $\alpha\in K$ is totally positive for each real embedding.<|endoftext|> TITLE: Reference on Persistent Homology QUESTION [40 upvotes]: I will be teaching a course on algebraic topology for MSc students and this semester, unlike previous ones where I used to begin with the fundamental group, I would like to start with ideas of singular homology as in Vick's book. I am quite new to the ideas of persistent homology and have not done a single computations in this field. But, I like to learn on the subject. More is that I like to lead the course that I will be teaching so that towards the end, I can give some taste of persistent homology to students. But, I am not sure if there is any well written set of lecture notes on the material, or should we dive into the literature and start with some papers!?! The course involves of $3/2\times 30$ hours of lectures. Do you think this is possible or should I use some simplicial approaches instead? or you think it is more suitable to give this as a task to students to start as a project and discover the ideas for themsevles?!?! I would be very grateful for any advise in terms of addressing to main references on the subject. I also would be grateful if you can give me some advise on the history of the subject; for instance when people decided to use homology to study biological problems and whether or not the main stream researchers in biology or data analysis really consider these kind of tools?! REPLY [36 votes]: Since this area is developing rather quickly, there is a dearth of canonical references that would satisfy basic pedagogical requirements. If I were teaching a course on this material right now, I would probably use Oudot's nice book if the students had sufficient background, and the foundational paper of Zomorodian-Carlsson if they did not. I haven't read Jose's recent article mentioned in Joe's answer, but here is what I remember of the good old days (with apologies to all the important stuff that got missed). 1992: Frosini introduces "size functions", which we would today consider equivalent to 0-dimensional persistent homology. 1995: Mischaikow + Mrozek publish a computer-assisted proof of chaos in the Lorenz equations; a key step involves computing Conley indices, which are relative homology classes. This produces considerable interest in machine computation of homology groups of spaces from finite approximations (eg large cell complexes). 1999: Robbins publishes this paper emphasizing that functoriality helps approximate the homology of an underlying space from Cech complexes of finite samples; meanwhile Kaczynski, Mischaikow and Mrozek publish their book on efficient homology computation via simple homotopy type reductions of cell complexes. 2002: Edelsbrunner, Letscher and Zomorodian introduce persistence from a computational geometry viewpoint; as written, their algorithm works only for subcomplexes of spheres and only with mod-2 coefficients. 2005: Zomorodian and Carlsson reinterpret persistence of a filtration via the representation theory of graded modules over graded pid's, thus giving an algorithm for all finite cell complexes over arbitrary field coefficients; they also introduce the barcode, which is a perfect combinatorial invariant of certain tame persistence modules. 2007: de Silva and Ghrist use persistence to give a slick solution to the coverage problem for sensor networks. Edelsbrunner, Cohen-Steiner and Harer show that the map $$\text{[functions X to R]} \to \text{[barcodes]}$$ obtained by looking at sublevel set homology of nice functions on triangulable spaces is 1-Lipschitz when the codomain is endowed with a certain metric called the bottleneck distance. This is the first avatar of the celebrated stability theorem. 2008: Niyogi, Smale and Weinberger publish a paper solving the homology inference problem for compact Riemannian submanifolds of Euclidean space from finite uniform samples. Carlsson, with Singh and Sexton, starts Ayasdi, putting his money where his math is. 2009: Carlsson and Zomorodian use quiver representation theory to point out that getting finite barcodes for multiparameter persistence modules is impossible, highlighting dimension 2 as the new frontier for theoretical work in the field. 2010: Carlsson and de Silva, by now fully immersed in the quiver-rep zone, introduce zigzag persistence. The first software package for computing persistence (Plex, by Adams, de Silva, Vejdemo-Johansson,...) materializes. 2011: Nicolau, Levine and Carlsson discover a new type of breast cancer using 0-dimensional persistence on an old, and purportedly well-mined, tumor dataset. 2012: Chazal, de Silva, Glisse and Oudot unleash this beastly reworking of the stablity theorem. Gone are various assumptions about tameness and sub-levelsets. They show that bottleneck distance between barcodes arises from a certain "interleaving distance" on the persistence modules. This opens the door for more algebraic and categorical interpretations of persistence, eg Bubenik-Scott. 2013: Mischaikow and I retool the simple homotopy-based reductions to work for filtered cell complexes, thus producing the first efficient preprocessor for the Zomorodian-Carlsson algorithm along with a fast (at the time!) software package Perseus. 2015: Lesnick publishes a comprehensive study of the interleaving distance in the context of multiparameter persistence modules. 2018: MacPherson and Patel concoct bisheaves to attack multi-parameter persistence geometrically for fibers of maps to triangulable manifolds. Good luck with your course!<|endoftext|> TITLE: Vertex cover number vs matching number QUESTION [5 upvotes]: Let $G=(V,E)$ be a finite, simple, undirected graph. A matching is a set $M\subseteq E$ of pairwise disjoint edges. A vertex cover is a set $C\subseteq V$ of vertices such that $C\cap e \neq \emptyset$ for all $e\in E$. The matching number $\mu(G)$ of $G$ is the maximum size that a matching can have, and the vertex cover number $\tau(G)$ is the minimum size that a vertex cover can have. If $M$ is a maximal matching with respect to $\subseteq$, then $C=\bigcup M$ is a vertex cover by maximality of $M$, and $|C|=2|M|$, so we have $\tau(G)\leq 2\cdot \mu(G)$ for all graphs $G$. Question. What is the value of $$\inf\{c\in \mathbb{R}: \tau(G)\leq c\cdot \mu(G) \text{ for all finite graphs } G\}?$$ REPLY [3 votes]: Minimum $c$ equals 2 as is seen from the complete graph on a large number $n$ of vertices: $\tau$ equals $n-1$, $\mu$ equals $[n/2]$.<|endoftext|> TITLE: Has it been proved that weak solutions to the Navier-Stokes equations are non-unique, and does this prove that the Navier-Stokes are not valid? QUESTION [13 upvotes]: This preprint claims that, for finite kinetic energy initial solutions, uniqueness of weak solutions to the Navier-Stokes equations doesn't hold: https://arxiv.org/abs/1709.10033 What's the current consensus of the community? Is the proof considered to be correct? Does this imply that the Navier-Stokes equations are not a valid model of fluid flow, or do we need a similar result for the smooth solutions of NS before we have to abandon/modify them? REPLY [10 votes]: To avoid people finding this question via google and wondering about the correctness of this paper, I want to point out that the paper has now been accepted by the Annals; see here.<|endoftext|> TITLE: $2$-determined Hausdorff spaces QUESTION [7 upvotes]: Is there an infinite Hausdorff space $(X,\tau)$ with the following property? If $x\neq y \in X$ and $f:\{x,y\}\to X$ is a map, then there is exactly one continuous function $f': X\to X$ such that $f'|_{\{x,y\}} = f$. REPLY [5 votes]: Using the technique of van Mill it is possible to prove the following Theorem. There exists a subset $Z$ in the complex plane $\mathbb C$ such that 1) for any complex numbers $x,a,b,c,d\in Z$ with $a\ne b$ and $c\ne d$ we have $h_{a,b,c,d}(Z)\subset Z$ where $h_{a,b,c,d}(z):=c+\frac{d-c}{b-a}(z-a)$; 2) for any homeomorphism $h:Z\to Z$ there exist unique complex numbers $a,b,c,d\in Z$ with $a\ne b$ and $c\ne d$ such that $h=h_{a,b,c,d}$; 3) Under CH (more generally, under $\acute{\mathfrak n}=\mathfrak c$) for any non-constant continuous map $f:Z\to Z$ there exist unique points $a,b,c,d\in Z$ with $a\ne b$ and $c\ne d$ such that $f=h_{a,b,c,d}$. Here by $\acute{\mathfrak n}$ denotes the smallest cardinality of an infinite cover of $[0,1]$ by pairwise disjoint closed sets. By the classical Sierpi$\acute{\mathfrak n}$ski Theorem, $\aleph_1\le \acute{\mathfrak n}\le\mathfrak c$. More information on the small uncountable cardinal $\acute{\mathfrak n}$ can be found in this MO-post. Remark. Under $\acute{\mathfrak n}=\mathfrak c$, the space $Z$ from Theorem has the property required in the question of Dominic van der Zypen: for any distinct points $a,b\in Z$ and point $c,d\in Z$ the map $h_{a,b,c,d}$ is a unique continuous self-map $f:Z\to Z$ such that $f(a)=c$ and $f(b)=d$.<|endoftext|> TITLE: Presentation for a Finite Etale Cover of an (Affine) Elliptic Curve QUESTION [8 upvotes]: I posted this question on MSE a few days ago, but I did not get much interest in it. So I thought I would try my luck here. If you are interested in answering the question, there is a bounty over on MSE. https://math.stackexchange.com/questions/2911318/presentation-for-a-finite-etale-cover-of-an-affine-elliptic-curve I will write the question first, then try and explain myself more clearly after: Question: How can one find a presentation for a finite etale cover of an affine piece of an elliptic curve? If $E/\mathbb{C}$ is an elliptic curve, then it is homeomorphic to a torus. For this reason we know $\pi_{1}(E) \cong \mathbb{Z}\times \mathbb{Z}$. If we puncture $E/\mathbb{C}$, then we obtain an affine curve, let us denote it $X$. Moreover, $X$ can be thought of as the elliptic curve minus the point at infinity. Since this is affine, we can give a presentation for it, say, $$X:= \text{Spec}(\mathbb{C}[x,y]/\langle y^{2} - f(x) \rangle)$$ for $f(x) = x(x-1)(x-2)$. We know $\pi_{1}^{et}(X) \cong \hat{\pi}_{1}(X(\mathbb{C})^{an})$ and $X(\mathbb{C})^{an}$ is homeomorphic to a punctured torus. The fundamental group of a punctured torus is the free group on two generators $F_{2}$. The profinite completion of which is non-trivial. So, there are algebraic covers of such an affine curve. How can we get our hands on them? Is there a presentation which in some sense is in terms of $f(x)$? Does anyone have a reference for a discussion on such a construction? Thanks in advance :) REPLY [4 votes]: If you are just looking for examples you can consider the following construction, which is geometric although I don't care to write down the equations explicitly. Recall, that if $C$ is a compact Riemann surface and $F$ is a reduced divisor of degree $f$ then there is a $f$-fold cyclic cover of $C$, branched at $F$, which we denote by $D$. That is, there is a degree $f$ map: $$ \pi\colon D \rightarrow C $$ which has branching of order $f$ at each of the points of $F$. (Remark: this cover depends on a choice of $f$-th tensor root of the line bundle $\mathcal{O}_C(E)$. There are always $f$-th roots as the map: $$ (-)^{\otimes f}\colon\mathrm{Pic}_1(C) \rightarrow \mathrm{Pic}_f(C) $$ is surjective.) Using this we can construct the following étale covers of the punctured torus. Start by considering an étale $d$-sheeted cover of your elliptic curve $E$: $$ \phi\colon C \rightarrow E. $$ (Thus, $C$ is a genus 1 curve as well.) Let $x\in E$ be a point and consider a reduced divisor of degree $f$, $F\subset\phi^{-1}(x)$, which is supported on the preimage of $x$ in $C$. Taking the cyclic cover branched at $F$ we get: $$ D\xrightarrow{\pi}C\xrightarrow{\phi}E. $$ Then the composition $\phi\circ \pi$ is a degree $d\cdot f$ map which is étale away from the point $x\in E$.<|endoftext|> TITLE: Pathology in Complex Analysis QUESTION [35 upvotes]: Complex analysis is the good twin and real analysis the evil one: beautiful formulas and elegant theorems seem to blossom spontaneously in the complex domain, while toil and pathology rule the reals. ~ Charles Pugh People often like to talk about elegant "miracles" in Complex Analysis. However, what's are "pathological" objects/properties in Complex Analysis? EDIT (09/13/18): Also posted as https://math.stackexchange.com/questions/2912320/most-pathological-object-in-complex-analysis EDIT: Changed the wording of the question. REPLY [5 votes]: Another complex dynamics example: Suppose $0 < \lambda < \frac{1}{e}$. The Julia set of $\lambda e^z$ can be divided into a set $E$ of "endpoints" and a collection of "hairs" connecting these endpoints to $\infty$. Mayer proved in 1990 that $E$ is totally separated, but $E \cup \{\infty\}$ is connected.<|endoftext|> TITLE: Is there an explicit Dold-Thom theorem? QUESTION [15 upvotes]: The Dold-Thom theorem tells us that we can recover the reduced homology of a pointed space $(X,x)$ via taking homotopy groups of the symmetric product: $$\pi_i(\mathrm{Sym}^{\infty}(X,x)) \cong H_i(X,x;\mathbb{Z}).$$ The usual proof of this theorem generally involves verifying that the left-hand side satisfies the axioms of a (reduced) homology theory, and as such is pretty abstract. The only application I know is constructing Eilenberg-MacLane spaces as infinite symmetric products of Moore spaces. (There's also a factorization-homology proof due to Bandklayder, which is perhaps a bit more useful for my question but is definitely outside my wheelhouse.) I'm wondering if there are any explicit convergence results: we have inclusions $$\mathrm{Sym}^0X \hookrightarrow \mathrm{Sym}^1X \hookrightarrow \mathrm{Sym}^2X \hookrightarrow \cdots,$$ so how far do we have to go in this sequence before the homotopy groups stabilize? To make this question explicit, suppose that $X$ is a finite simplicial complex, such that $X$ has at most $n$ nondegenerate simplices in each dimension up to $d$. Is there a function $f(i,n,d)$ such that for all such $X$ we have $$\pi_i(\mathrm{Sym}^{f(i,n,d)}(X,x)) \cong H_i(X,x;\mathbb{Z})?$$ Part of my difficulties in thinking about this problem are that I don't know how to do many computations of $\mathrm{Sym}^n X.$ I know that $\mathrm{Sym}^n(S^1) \simeq S^1$ and that we know a little bit about the algebraic topology of $\mathrm{Sym}^g(\Sigma_g)$, where $\Sigma_g$ is the Riemann surface of genus $g$ (this perhaps betrays my exposure to Heegaard-Floer homology). In general, I believe one can compute a simplicial decomposition of $\mathrm{Sym}^nX$, though it's pretty complicated to write down and I'd appreciate a reference to automated computations. But I don't know how to compute the homotopy groups of an arbitrary abstract simplicial complex in an algorithmic fashion, so this has been pretty inaccessible to me. REPLY [10 votes]: The homology of symmetric products has been studied in lots of ways by many people for 70 years. If one enters `homology of symmetric products' into MathSciNet one gets 480 Math Reviews. A good starting point might be to look at papers from the 1960's and 70's by Nakaoka, Dold, and Milgram. See, e.g. Dold's 1959 paper in the Annals called "The homology of symmetric products and other functors of complexes". (I am sure some good approximation to your specific function f(i,n,d) can be read off from known results.) By the way, another important application of the Dold Thom Theorem is that it gives a rather natural infinite factorization of the Hurewicz map from homotopy to homology. [I am rather fond of this filtration "in the stable range": see my papers on Whitehead's Conjecture to see why.]<|endoftext|> TITLE: Does the Gauss-Bonnet theorem apply to non-orientable surfaces? QUESTION [28 upvotes]: I hesitated for a long time to ask such an elementary-seeming question on Math Overflow, but when I asked and bountied it on Math SE, I found that a few experts seem to disagree on the answer, and I didn't get enough responses to indicate a statistically significant consensus. A cursory online search gives that about half the sources include the requirement of orientability in the Gauss-Bonnet theorem, and half don't. (I listed eight on the "no" side in the SE question; a Google search yields many more on the "yes" side.) Ted Shifrin claims that It is absolutely a necessity, as to define the [global] integral $\iint_M K dA$ requires an orientation.... (The far abstracted version of Gauss-Bonnet refers to the Euler class of the tangent bundle of an oriented 2n-dimensional Riemannian manifold. Orientability is needed there, too, to define the Pfaffian of the curvature matrix.) Sunghyuk Park, on the other hand, gives an answer claiming that with any non-orientable surface you can consider its orientable double cover, and for that double cover all three terms (the Gaussian curvature surface integral, the boundary geodesic curvature line integral, and the Euler characteristic term) all double, so that the theorem remains true. Ted Shifrin concedes that the theorem might hold for closed non-orientable surfaces, but claims that the boundary line integrals actually cancel out instead of doubling. So what's the deal? Does the Gauss-Bonnet theorem hold for (a) any compact non-orientable surface, (b) only closed non-orientable surfaces, or (c) no (nontrivial) generic class of non-orientable compact surfaces? REPLY [16 votes]: The answer is already given in the comments (by Ryan Budney and Mizar). But I think it makes sense to clear this confusing point. The classical Gauss-Bonnet formula is [e.g. https://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem ] $$\int_M K dA+\int_{\partial M} k_g ds=2\pi \chi(M).$$ In this formula nothing requires orientation of $M$! $dA$ is the area element, $ds$ is the line element on the boundary, $K$ is the Gauss curvature and $k_g$ is the geodesic curvature of the boundary. Note that while the sign of the geodesic curvature generally depends on a choice of a normal to the curve, in this particular situation this choice is predetermined (it is the inner normal). There is a way to prove this equality without introducing any orientation, but even if you only have a proof for the oriented case the doubling arguments (mentioned in the comments) trivially extends it to non-orientable surfaces. The orientability mess comes from not too faithful to the original generalizations of this formula to higher dimensions. If I am not mistaken, the first generalization [The Gauss-Bonnet Theorem for Riemannian Polyhedra Carl B. Allendoerfer and Andre Weil, Transactions of the American Mathematical Society Vol. 53, No. 1 (Jan., 1943), pp. 101-129] does not really presuppose orientability, but it is difficult to spot this as it is almost perfectly obscured by the notation. But the subsequent development follows the lines of Chern [A Simple Intrinsic Proof of the Gauss-Bonnet Formula for Closed Riemannian Manifolds Shiing-Shen Chern Annals of Mathematics Second Series, Vol. 45, No. 4 (Oct., 1944), pp. 747-752]; in this approach what is integrated is a differential form (rather then a density) which, of course, requires orientation. In the Chern method, what is actually computed is not the Euler characteristic of the manifold but the Euler class of its tangent bundle. Which is the same thing except the Euler class only makes sense for orientable vector bundles, hence the restriction. This restriction is convenient but not necessary for the Gauss-Bonnet formula (regardless of dimension).<|endoftext|> TITLE: Show that there exist $k\in\{1,2,\cdots,n\}$ such that $\frac{1}{n}\sum_{i=1}^{n}\left(\{kx_{i}\}-\frac{1}{2}\right)^2>\frac{1}{12}-\frac{1}{6n}$ QUESTION [12 upvotes]: The following question has post mathsatck :Nice problem, perhaps from a question of expectation.The problem seemed so elementary, but I had tried to solve it for a long time and eventually failed.I 'd like to get your help or help out here.Thanks! Show that: for any real numbers $x_{1},x_{2},\cdots,x_{n}$, there exist $k\in\{1,2,\cdots,n\}$ such that $$\dfrac{1}{n}\sum_{i=1}^{n}\left(\{kx_{i}\}-\dfrac{1}{2}\right)^2>\dfrac{1}{12}-\dfrac{1}{6n},$$ where $\{x\}=x-\lfloor x\rfloor$. REPLY [14 votes]: Here's an elementary proof of the inequality $$ (1) \qquad\qquad \sum_{k=1}^{N-1} \left(1-\frac{k}{N}\right)B_2(\{kx\}) \ge \frac1{12N} - \frac1{12}. \qquad\qquad\phantom{(1)} $$ This is nearly the same as the inequality cited by Joe Silverman (with $N=K+1$), but with the lower bound improved from $-\frac1{12}$ to $\frac1{12N} - \frac1{12}$. This is best possible; equality holds iff $x = m/N$ for some integer $m$ with $\gcd(m,N)=1$. The slight improvement by $\frac1{12N}$ seems to be exactly what's needed to prove the inequality posed by function sug. Proof: We prove for any $x_1,\ldots,x_N \in \bf R$ the inequality $$ (2) \qquad\qquad\qquad \sum_{i=1}^N \sum_{j=1}^N B_2(\{x_i-x_j\}) \geq \frac1{6N}, \qquad\qquad\qquad\phantom{(2)} $$ with equality iff the $x_i \bmod 1$ are a permutation of $\{x_0 + (m/N): 0 \leq m < N\}$ for some $x_0$. To recover (1) from (2), set $x_k = kx$, remove the terms with $i=j$ (which contribute $N/6$ to the total), and divide by 2N. Permuting the $x_i$ does not change the double sum in (2). If the $x_i \bmod 1$ are a permutation of $\{x_0 + (m/N): 0 \leq m < N\}$ then the sum in (2) is $N \sum_{i=0}^{N-1} B_2(i/N)$, which indeed equals $1/6N$. We next show that this is minimal. Permute the $x_i$ so that $x_1 \le x_2 \le x_3 \le \cdots \le x_N \le x_1 + 1$, and write the sum in (2) as $$ \sum_{i=1}^N \sum_{j=1}^N B_2(\{x_{i+j}-x_j\}), $$ with the index $i+j$ taken $\!\!\mod N$. Then for each $i$ the fractional parts $\{x_{i+j}-x_j\}$ in the inner sum total $i$. Since $B_2$ is convex upwards, this inner sum is minimized when $\{x_{i+j}-x_j\} = i/N$ for each $j$. This is the case when each $x_m \equiv x_0 + m/N$, and that sufficient condition is also necessary at least when $i=1$. This completes the proof. $\Box$<|endoftext|> TITLE: Vector bundle structure of conformal block bundle QUESTION [7 upvotes]: My question is about the conformal block bundle, which (following Kohno's "Conformal Field Theory and Topology") is constructed as follows: Consider the projection map onto the first $n$ coordinates \begin{equation} \pi: (\mathbb{C}P^1)^{n+1} \to (\mathbb{C}P^1)^n. \end{equation} Let $D_i$ denote the hyperplane of $(\mathbb{C}P^1)^{n+1}$ given by setting $z_i=z_{n+1}$ where $(z_1,\dots, z_{n+1})$ are coordinates for $(\mathbb{C}P^1)^{n+1}$. Let $\mathcal{M}_{D_1, \dots, D_n}(U)$ denote the space of meromorphic functions on $\pi^{-1}(U)$ with poles only on the hyperplanes $D_1,\dots, D_n$. Consider $\mathfrak{g} \otimes \mathcal{M}_{D_1, \dots, D_n}(U)$ as a Lie algebra with bracket defined by \begin{equation} [X \otimes f, Y \otimes g]=[X,Y] \otimes fg. \end{equation} Given an element $f \in \mathcal{M}_{D_1, \dots, D_n}(U)$, we can expand $f$ as a Laurent series on the hyperplane $D_i$ \begin{equation} f_{D_i}(z)=\sum_{k=-N}^\infty a_n(z_1, \dots, z_n) (z-z_i)^k \end{equation} where $a_n(z_1, \dots, z_n)$ is holomorphic in $z_1, \dots, z_n$. Fixing $(z_1, \dots, z_n) = (p_1, \dots, p_n)$, we can view $f_{D_i}$ as an element of $\mathbb{C}((t_i))$ where we define $t_i=z - z_i$. This gives us a canonical inclusion map \begin{equation} \tau_i : \mathfrak{g} \otimes \mathcal{M}_{D_1, \dots, D_n}(U) \hookrightarrow \hat{\mathfrak{g}}_i \cong \mathfrak{g} \otimes \mathbb{C}((t_i)) \oplus \mathbb{C}c. \end{equation} To each $t_i$, associate an integrable highest weight $\hat{\mathfrak{g}}_i$-module $V_{\lambda_i}$ of highest weight $\lambda_i$. We can define an action of $\hat{\mathfrak{g}}_i$ on $\bigotimes_{i=1}^n V_{\lambda_i}$ by \begin{equation} X \cdot (v_1 \otimes \cdots \otimes v_n) = \sum_{i=1}^n v_1 \otimes \cdots \otimes \tau_i (X) v_i \otimes \cdots \otimes v_n. \end{equation} Now, consider the trivial bundle \begin{equation} E:= \text{Conf}_n(\mathbb{C}P^1) \times \text{Hom}_{\mathbb{C}}\left(\bigotimes_{i=1}^n V_{\lambda_i},\mathbb{C}\right) \to \text{Conf}_n(\mathbb{C}P^1). \end{equation} For each open set $U \subseteq \text{Conf}_n(\mathbb{C}P^1)$, denote by $\mathcal{E}_{\lambda_1, \dots, \lambda_n}(U)$ the space of smooth sections $\sigma: U \to E$ satisfying \begin{equation} \sum_{i=1}^n (\sigma(p_1, \dots, p_n))(v_1 \otimes \cdots, \otimes \tau_i(f)v_i \otimes \cdots \otimes v_n)=0 \end{equation} for all $f \in \mathfrak{g} \otimes \mathcal{M}_{D_1, \dots, D_n}(U)$ and $v_1 \otimes \cdots \otimes v_n \in \bigotimes_{i=1}^n V_{\lambda_i}$. Question: Why is the sheaf of smooth sections $\mathcal{E}_{\lambda_1, \dots, \lambda_n}(U)$ locally-free? Ultimately, I'd like to understand why we can realise the disjoint union of the space of conformal blocks as a vector bundle over $\text{Conf}_n(\mathbb{C}P^1)$. Any help is greatly appreciated. REPLY [9 votes]: I also had the same question when I was reading Kohno's book. My opinion is that that book is only an introduction to the topic of conformal blocks and is not rigorous. For rigorous treatment you need to read the classical paper Conformal Field Theory on Universal Family of Stable Curves with Gauge Symmetries by Tsuchiya, Ueno, Yamada. There are also many articles and monographs explaining their work. You may try Mukhopadhyay, or Ueno. As a survey of TUY's work I also like the last chapter of Bakalov-Kirillov. The existence of a vector bundle structure is to me a very deep result. The argument of TUY could be summarized as follows. Basically you are considering a family of Riemann surfaces, i.e. a proper submersion $\pi:\mathcal C\rightarrow\mathcal B$ such that each fiber is a compact Riemann surface. In your case $\mathcal B=\text{Conf}_n(\mathbb CP^1)$ and $\mathcal C=\mathbb CP^1\times\mathcal B$. You want to introduce a vector bundle structure on the collection of conformal blocks (also called spaces of vacua). It is even not trivial to show that the dimensions of conformal blocks are independent of the fibers. To prove all these things one considers the sheaf of vacua (sheaf of conformal blocks) and its dual sheaf: sheaf of covacua. These are sheaves of $\mathcal O_{\mathcal B}$-modules where $\mathcal O_{\mathcal B}$ is the structure sheaf of $\mathcal B$. One then shows that the sheaf of covacua $\mathcal V$ is a coherent sheaf, and that there exists a connection on $\mathcal V$. There is a standard and elementary result in the literature of D-modules saying that any coherent sheaf with a connection is locally free, i.e. it behaves like a holomorphic vector bundle. (This is also proved in the above mentioned references.) This proves the existence of vector bundle structure and in particular, the constance of the dimensions of conformal blocks.<|endoftext|> TITLE: 2-transitive subgroups of a symmetric group QUESTION [6 upvotes]: I wonder does the following statement holds: 1. For any sufficiently large $n$ the symmetric group $S_n$ contains at least one 2-transitive subgroup other than $S_n$ itself and $A_n$? Actually I'm interested in whether every symmetric group $S_n$, for sufficiently large $n$, contains a transitive subgroup whose order is divisible by $n(n-1)$. If the first statement holds then it obviously implies the second. I would appreciate any comments on the above two questions. REPLY [14 votes]: It has been pointed out in comments that the answer to Question 1 is no. The answer to Question 2 is also no. Let $p$ be a prime with $p \equiv 1 \bmod 72$, and let $n = 2p+1$. Then $n \equiv 3 \bmod 9$, so $n$ is not prime and is not a proper power. Also, $n \equiv 3 \bmod 8$, so $n+1$ is not a power of $2$. I claim that no transitive subgroup $G$ of $S_n$ apart from $A_n$ and $S_n$ has order divisible by $n(n-1)$. Such a $G$ contains an element $g$ of order $p$, and it would have to be primitive because the maximal imprimitive subgroups of $S_n$ are wreath products of the form $S_a \wr S_b$ with $ab=n$, and their orders are not divisible by $p$ (note that $n$ odd implies $a,b \ge 3$). Now by an old theorem of Jordan, if $g$ was a single $p$-cycle then we would have $G=A_n$ or $S_n$, so we can assume that $g$ is a product of two disjoint $p$-cycles. So $G$ is either 2-transitive, or its point stabilizer has two orbits of the same length $p$, and then $G$ is $3/2$-transitive. $3/2$-transitive groups are classified in Corollary 3 of a paper by Liebeck, Praeger and Saxl. The list consists of $2$-transitive groups, Frobenius groups, affine groups, and almost simple groups of even degree or degree $21$. A Frobenius group with an element of order $p$ would have to be affine, and is ruled out by our choice of $n$, which is odd, not prime, and not a proper power. So we are just left with the $2$-transitive groups. For sufficiently large $n$, the only such groups of odd degree (apart from $A_n$ and $S_n$) are the Suzuki groups ${\rm Sz}(2^k)$, which are of degree $2^{2k}+1$, and so cannot occur in this context, the unitary groups $U_3(2^k)$ of degree $2^{3k}+1$, ditto, and the groups containing $L_k(q)$ of degree $n=(q^k-1)/(q-1)$. Groups in the last class have order dividing $$q^{k(k-1)/2}(q^{k}-1)(q^{k-1}-1)\cdots(q-1)e,$$ where $q=r^e$ with $r$ prime. Our choice of $n$ ensures that $n+1$ is not a power of $2$, which rules out $q=2$, and for $q>2$ , we see that all prime divisors of $|G|$ are less than $p = (n-1)/2$.<|endoftext|> TITLE: Points of the big Zariski site QUESTION [12 upvotes]: It's relatively simple to show that the geometric morphisms $ \mathbf{Set} \to \mathrm{Sh}(\mathbf{CRing}^\mathrm{op}_{\mathrm{fp}}, \mathrm{Zar})$ correspond to local rings. More precisely, since the site has finite limits, the inverse image functors are induced by functors $\mathbf{CRing}^\mathrm{op}_{\mathrm{fp}} \to \mathbf{Set}$ that preserve finite limits and map covers to epimorphic families, and these are precisely the functors (isomorphic to) $\mathbf{CRing}(-, \mathcal{O})$ for local rings $\mathcal{O}$. But we often consider Zariski sheaves on larger sites. For example, one might assume a small inaccessible cardinal $\kappa$ and consider the site $(\mathbf{CRing}^\mathrm{op}_{\kappa}, \mathrm{Zar})$ of all $\kappa$-small affine schemes. The functors $\mathbf{CRing}(-, \mathcal{O})$ still define (inverse image parts of) geometric points, and every point whose inverse image part preserves all limits of affine schemes must be of this form. But in general, the inverse image part of a geometric point is only required to preserve finite limits, so a priori there might be additional points. Is there an argument that every geometric point of such a site is given as above? Or can there be points not of that form? REPLY [7 votes]: Let's simplify and consider the presheaf topos. I asked the same question over at the nForum a while back. There Marc Hoyois reminded me of the following quite general fact: The category of topos-theoretic points of $\mathrm{PSh}(\mathcal{C})$ is the ind-completion of $\mathcal{C}^{\mathrm{op}}$. With this fact in mind, we can … … recover what you said in your first paragraph, since the ind-completion of the category of finitely presented rings is the category of all rings. … see that there's no reason to expect that every point of the (pre-)Zariski topos defined using one of the larger sites is given by a ring. This situation is even more pronounced from the point of view of classifying toposes. The usual (pre-)Zariski topos, defined using finitely presented rings, classifies the theory of rings. In contrast, I don't believe that a nontautologous answer to the question "which theory does the topos defined using one of the larger sites classify?" has ever been written down. For the purposes of algebraic geometry, the restriction to the rather small site of finitely presentables can be a bit cumbersome. A bit of discussion is included in Section 15 of these notes of mine.<|endoftext|> TITLE: How do I calculate the modular fusion category from a given Lie algebra and level in Chern-Simons theory? QUESTION [7 upvotes]: In Chern-Simons theory, one has modular fusion categories that are labelled by a Lie algebra and a "level", e.g. $SU(2)_2$ ("$SU(2)$ level $2$"). Physically this modular fusion category describes the TQFT/anyon statistics of the Chern-Simons gauge theory. Now I think of a modular fusion category as an $F$ and an $R$ tensor, where the $F$ tensor has $6$ indices labelled by simple objects and $4$ fusion space indices, and $R$ has $3$ indices labelled by simple objects and $2$ fusion space indices. Similarly, one can think of a Lie algebra in terms of the "structure coefficients" of the Lie bracket in some chosen basis, yielding a tensor with three indices. The level I'm not sure how to formulate in such a constructive language (this is somehow part of the question). My question: Is there a constructive way to calculate the $F$ and $R$ tensor from the Lie algebra structure coefficients? Constructive in the sense that e.g. there is a computer program that takes tensors as inputs and tensors as outputs. And what would that construction look like? REPLY [5 votes]: I stumbled over this older question. I actually wrote a program, that takes the type of algebra (A,B,...,G), the rank, level, and appropriate root of unity as an input. It uses the associated quantum group to calculate the F- and R-symbols, as well as the modular data. It's currently slow. You can download the package from GitHub. The package is based on arxiv.org/abs/1004.5456. More info on the quantum group construction can be found in f.i. Bakalov & Kirillov, Lectures on Tensor Categories and Modular Functors<|endoftext|> TITLE: Early examples of mathematicians publishing (from home) in a foreign language? QUESTION [11 upvotes]: Today this is common, but how exactly did it start? I am looking for examples in various languages, and suggest: Exclude Latin (as more “ancient” or “international” than “foreign”) Exclude French after, say, Huygens (as well-known: Leibniz, Euler, Jacobi, Dirichlet, etc.) Exclude anything resulting from a visit or permanent move to the foreign country (thus Cauchy or Riemann in Italian, World War refugees in English, etc.) Exclude translations not originated by the author. Do not otherwise exclude (early) English and German! Bend these rules if necessary. Mathematical physics or celestial mechanics qualify. This ties in with various questions at hsm, but I am hoping for a wider pool of knowledge here. So far I am only aware of isolated(?) examples by Abel, Plücker, Lie, Lorentz, without clear lineage to the present. REPLY [7 votes]: The obvious examples would include pretty much anyone who was not born speaking any of the "Congress" languages. They (or should I say "we", being Polish) would publish their mathematics in a more widely spoken/read language in order to increase the chance of its being read and understood. Let me however give some less obvious answers involving two Polish students of Sophus Lie (still having to do with dissemination of the ideas, perhaps tailoring to the intended audience). Kazimierz Zorawski (1866-1953; PhD 1891) published over 70 works in his lifetime, roughly half of them in German and half in Polish (understandable). However, there are also 3 works in Czech (1914-1915). He was a member of the Czech Academy (since 1910), but otherwise had no Czech connection. Another student of Lie, Lucjan Emil Boettcher (1872-1937; PhD 1898), published mainly in Polish (with some early and late exceptions in German and French), but his most cited paper is "Glavnyshiye zakony skhodimosti iteratsiy i ikh prilozheniya k' analizu" ("Main laws of convergence of iterations and their applications to analysis"). He published it in 1903-1904 (in parts) in Russian, in "Bulletin de la Societe Physico-Mathematique de Kasan". REPLY [3 votes]: A surprisingly recent example: Don Zagier: Een ongelijkheid tegengesteld aan die van Cauchy Proc. Koninkl. Ned. Akad. v. Wetensch. (Indag. Math.) 80 (1977) 349-351 There are also several papers in French in Zagier's list.<|endoftext|> TITLE: Does Koszul duality between $Comm$ and $Lie$ imply the power series identity $\exp(\ln(1-z))-1 = -z$? QUESTION [19 upvotes]: To a symmetric sequence $V_\bullet$ of vector spaces, associate the generating function $F_V(z) = \sum_n \frac{\dim(V_n)}{n!} z^n$. Then $$F_{Comm_\ast}(z) = \exp(z)-1 \qquad F_{Lie}(z) = \ln(1-z)$$ where $Comm_\ast$ is the reduced commutative operad and $Lie$ is the Lie operad. Notice that On the one hand, these power series are inverse up to a sign. On the other hand, these operads are Koszul dual (and perhaps the sign corresponds to the shift that appears in Koszul duality?). Similarly, $$F_{Ass_\ast}(z) = \frac{z}{1-z}$$ where $Ass_\ast$ is the reduced associative operad. On the one hand, this power series is its own inverse up to a sign. On the other hand, this operad is Koszul self-dual. Question: Is this a coincidence? Or is there some deeper connection between (1) and (2)? More concretely, is it the case (under certain conditions, perhaps) that Koszul dual operads have inverse generating functions, up to some sign? REPLY [18 votes]: Let me flesh out the answer a little. The general statement is given by Theorem 7.5.1 in the book Algebraic Operads by Loday and Vallette. First a definition. Let $P = P(E,R)$ be a quadratic operad, with generators $E$ (f.gen. s.t. $E(0) = 0$) and $R \subset E \circ E$ quadratic relations. Let $P^{(r)}(n)$ be the subspace of operations of weight $r$, where $E$ is of weight $1$. There is a generating series, aka Hilbert-Poincaré series: $$f^P(x,y) = \sum_{r \ge 0, n \ge 1} \frac{\dim P^{(r)}(n)}{n!} y^r x^n.$$ The theorem states that if $P$ is Koszul, with dual $P^!$, then there is a functional equation: $$f^{P^!}(f^P(x,y),-y) = x.$$ Remark: as Nicholas Kuhn explained, this equality follows from the acyclicity of the Koszul complex, the product $P^¡ \circ P$ with the Koszul differential. $\newcommand{\Com}{\mathsf{Com}}\newcommand{\Lie}{\mathsf{Lie}}$ Apply this to $P = \Lie$, $P^! = \Com$. It's well-known that $\Com(n) = \Com^{(n-1)}(n)$ is of dimension $1$ for $n \ge 1$, while $\Lie(n) = \Lie^{(n-1)}(n)$ is of dimension $(n-1)!$ for $n \ge 1$. So in particular you get \begin{align} f^\Com(x,1) & = \sum_{n \ge 1} \frac{x^n}{n!} = \exp(x) - 1,\\ f^\Lie(x,-1) & = \sum_{n \ge 1} \frac{(-1)^{n-1} x^n}{n} = \ln(1+x) \end{align} Apply the functional equation to $y = -1$ and you get $\exp(\ln(1-x))-1=x$.<|endoftext|> TITLE: Is there a kind of Poincare duality for Borel equivariant cohomology? QUESTION [11 upvotes]: Let $G$ be a finite (or discrete) group, $M$ a $d$-dimensional manifold with smooth $G$-action (I am interested in the case where the action is not free, so $M/G$ is not a manifold). For an Abelian group $A$, let $\mathcal{C}^n(M,A)$ be the group of $n$-cochains on $X$ with $A$ coefficients. We can treat this as a $G$-module (Abelian group with compatible $G$-action) with the $G$-action inherited from the $G$-action on $M$. Now for any $G$-module $B$, we can introduce the Abelian group of "group cochains" $\mathcal{C}^m(G, B)$ that are used to define group cohomology. For example, we can define $\mathcal{C}^m(G, B) = \mathrm{Hom}_G(F_n, B)$, where $$\cdots F_n \to F_{n-1} \to \cdots \to F_0 \to \mathbb{Z} \to 0$$ is a projective $\mathbb{Z}[G]$-resolution of the integers. In particular, $\mathcal{Q}^{m,n} = \mathcal{C}^m(G, \mathcal{C}^n(X,A))$ defines a double complex. The total cohomology of this double complex computes the equivariant cohomology $H^{\bullet}( (M \times EG)/G, A)$. Now suppose I instead define $\widetilde{\mathcal{Q}}^{m,n} = \mathcal{C}^m(G, \mathcal{C}_{d-n}(M,A))$, where $\mathcal{C}_{\bullet}(M,A)$ denotes the group of $(d-n)$-chains on $X$ with $A$ coefficients. What is the interpretation of the total cohomology of this complex? It's not equivariant homology, because we are still doing "cohomology in G". Under what circumstances is there a Poincare duality relating this "equivariant (?)-ology" to the equivariant cohomology? At the level of cellular chains/cochains of $M$ it seems to hold. But there are subtleties about what kind of cellulations should be allowed, given the $G$ action. REPLY [7 votes]: This kind of thing shows up quite naturally in parameterised stable homotopy theory. Let me translate an idea I know from there into the language in this question. Cap product gives a map $$C^{p}(M ; A) \otimes C_q(M, \mathbb{Z}) \to C_{q-p}(M;A),$$ which is $G$-equivariant and is a chain map when we sum over $(p,q)$ and give the two sides their natural differential. With the cup product in group cohomology this yields maps $$C^n(G ; C^p(M ; A)) \otimes C^m(G ;C_q(M, \mathbb{Z})) \to C^{n+m}(G ; C^p(M ; A) \otimes C_q(M, \mathbb{Z})) \to C^{n+m}(G;C_{q-p}(M;A))$$ and again summing up over all indices, and flattening (n,m) and (p,q), it gives a map of double complexes. Thus if I give you a cycle $ \xi \in C^0(G ;C_d(M, \mathbb{Z}))$ then there is an induced map of double complexes $$\xi_\# : \mathcal{Q}^{n,p} \to \widetilde{\mathcal{Q}}^{n,p}.$$ Furthermore, if $\xi$ is a fundamental cycle for $M$ under the evaluation map $C^0(G ;C_d(M, \mathbb{Z})) \to C_d(M, \mathbb{Z})$, then the map induced by $\xi_\#$ on spectral sequences for these double complexes is, on $E_2$-pages, $$H^n(G ; H^p(M;A)) \to H^n(G ; H_{d-p}(M;A))$$ induced by the map on coeffciients which caps with the fundamental class: this is an isomorphism by ordinary Poincare duality for $M$, and so $\xi_\#$ is a quasiisomorphism. It remains for me to give you such a cycle $\xi$. The spectral sequence for the double complex $C^*(G ;C_\bullet(M, \mathbb{Z}))$ has $$E_2^{p,q} = H^p(G ; H_q(M;\mathbb{Z}))$$ so as long as $M$ is oriented and the $G$-action preserves this orientation then we can form the class $\xi' :=[M] \in H^0(G ; H_d(M;\mathbb{Z}))$. Some contemplation shows that the differentials in this spectral sequence go $$d_r : E_r^{p,q} \to E_r^{p+r, q+r-1}$$ from which it follows that this $\xi'$ is a permanent cycle. If $\xi$ is any cycle in the double complex representing $\xi'$, then it has the properties I used in the third paragraph.<|endoftext|> TITLE: Relationship between Smith's special homology groups and equivariant homology theory QUESTION [16 upvotes]: EDIT: Tyler Lawson's answer was so nice that I was inspired to rewrite the notes discussed below to use Bredon homology in the definition of the Smith special homology groups. The original version is available for readers who prefer being low-tech; just click on the abstract button and a download button for this will become available. Let $p$ be a prime, let $G$ be a finite $p$-group, and let $X$ be a "reasonable" finite-dimensional $G$-space (let me be vague about what reasonable means, but certainly a finite-dimensional simplicial complex upon which $G$ acts simplicially counts). Around 1940, P. Smith proved a number of theorems that relate the algebraic topology of $X$ to the algebraic topology of the fixed point set $X^G$. For instance, one of his theorems says that if $X$ is mod-$p$ acyclic, then so is $X^G$ (which implies in particular that $X^G$ is nonempty). I have notes on Smith theory on my page of notes here that give more details. The original proofs of the main results of Smith theory use the "special homology groups" of Smith, which are defined algebraically and whose geometric meaning seems to me to be quite mysterious. I describe these homology groups in the notes above. When Borel introduced equivariant homology via the Borel construction, one of his original applications was to give a new proof of Smith's main results. These days, equivariant homology (both in the original Borel flavor and also using Bredon's more powerful definitions) is one of the fundamental tools in the theory of transformation groups. Question: Is there a way to relate Smith's special homology groups to these modern developments? How are they related to equivariant homology? From my reading of the literature, they seem to have been mostly discarded from the toolbox of people working in the subject. Is it now understood that whatever information they contain is also contained in other places? REPLY [15 votes]: Smith's special homology groups are special instances of Bredon homology. Here's roughly how it goes, taken from Peter May's "A generalization of Smith theory". A Bredon coefficient system $M$ is some kind of functorial assignment of abelian groups to $G$-sets of the form $G/H$, or some kind of functor from $G$-sets to abelian groups that takes disjoint unions to direct sums. In particular, when $G = \Bbb Z/p$ we unravel this and find that it an abelian group $A$ with $G$-action, a second abelian group $B$ acted on trivially, and a $G$-map $A \to B$. Associated to a simplicial complex with "good" $G$-action (the stabilizer of a simplex fixes the simplex--this is equivalent to your description in terms of $X/G$), you get a complex computing the Bredon homology groups $H_*(X;M)$. Here are some Bredon coefficient systems: The trivial map $0 \to \Bbb F_p$. This computes the homology $H_*(X^G;\Bbb F_p)$. The identity map $\Bbb F_p \to \Bbb F_p$. This computes the homology $H_*(X/G;\Bbb F_p)$. The augmentation $\Bbb F_p[G] \to \Bbb F_p$. This computes the homology $H_*(X;\Bbb F_p)$. Let $\tau = 1 - t$ where $t$ is the generator of $\Bbb Z/p$. Then there's a sub-coefficient-system $\tau^k \Bbb F_p[G] \to 0$ of the previous one (it's the $k$'th power of the augmentation ideal, because $\tau$ generates the augmentation ideal). These Bredon homology groups compute Smith's special homology groups. Most of the manipulations that one does to obtain the results of Smith theory are specializations of standard relationships and exact sequences that one gets from functoriality and long exact sequences for Bredon homology. It simply expresses the proof that you already wrote in slightly more standardized language. You also asked about relationships to modern developments. Bredon (co)homology is certainly part of the standard toolbox in equivariant homotopy theory. In principle, these homology and cohomology theories are naturally part of equivariant stable homotopy theory. One reason why you don't see Bredon theory mentioned as often anymore is that much of the research focus in equivariant stable theory is on studying the "genuine" equivariant stable category rather than the "naive" stable category (don't blame me for the naming), because things like transfer maps and duality theorems live in the former rather than the latter. Bredon cohomology theories are residents of the naive stable category. They only come from the genuine stable category if they have transfer maps making them into something called Mackey functors. I get a little bit confused about the difference between Bredon homology and cohomology, but unless I've made a mistake the cohomology version of the above argument does lift to Mackey functors and so it should be something that can be phrased in terms of the genuine stable category. (I'm being cagey because I've been burned about subtle issues on similar questions before and I haven't verified things carefully.) I don't have a better explanation why there aren't as many new developments in Smith theory--in particular, whether it's a problem of interest or generalizability.<|endoftext|> TITLE: Raynaud's universal Tate elliptic curves QUESTION [8 upvotes]: In the end of Section 9.2 of Bosch's book Lectures on Formal and Rigid Geometry, a rigid $S$-space $E_Q$ is constructed, for a variable $Q$ replacing the classical parameter $q\in k$. (Here $k$ is a non-archimedean field and $S=\mathrm{Spf}\mathbb Z[[Q]]$). Then $E_Q$ may be viewed as the family of all Tate elliptic curves by looking at the morphism $Z[[Q]] \to k^\circ, Q\mapsto q$. (1) It looks interesting but the reference is missing in Bosch's book. Which one of Raynaud's paper is about this construction? (2) Could we in some sense further generalize it by considering more than one variables, i.e. $S'=\mathbb Z[[Q_1,\dots, Q_k]]$? Notice that $S'$ seems still satisfy the condition $(N)$ required by Bosch (see p 162) REPLY [3 votes]: Q1: Michel Raynaud, Géométrie analytique rigide d’après Tate, Kiehl..., Bull. Soc. Math. France, Mémoire 39-40, p. 319-327 (1974).<|endoftext|> TITLE: Does ampleness descend along finite maps? QUESTION [6 upvotes]: First, let me emphasize that for $X$ a not-necessarily proper variety, we say that a line bundle $L$ on $X$ is ample, if for some positive integer $n$, $L^{\otimes n}$ arises as $j^*O(1)$ for some (not-necessarily closed) immersion $j:X\rightarrow \mathbb{P}^n$. Now, let $X$ be a variety and $L$ a line bundle on $X$, and let $f:X^{nor}\rightarrow X$ be the normalization map. Suppose that $f^*L$ is ample. Is it true that $L$ is ample? REPLY [5 votes]: There is a non-quasi-affine variety $X$ with quasi-affine normalization. See Tag 0271. Then $\mathcal{O}_X$ is a counter example.<|endoftext|> TITLE: Residues of $\frac{1}{\prod_{i=1}^n (x-P_i)^{e_i}}$ QUESTION [12 upvotes]: This is a problem occurring in my research about deformations of $\mathbb{Z}/p^n$-covers over a ring of power series. Given an algebraically closed field $k$ of characteristic $p>0$, suppose $1< e_i 0$. Thus, $\omega$ is a derivative of some rational functions if and only if $x^{p-e_1}(x-Q)^{p-e_2}$ is. The later is a derivative if and only if all the $kp-1$ coefficients are equal to $0$. Suppose $e_1+e_2 \ge p+2$. Then $2p-(e_1+e_2) \le p-2$. Thus it is clearly a derivative since all the $kp-1$ coefficients are equal to $0$. Suppose $e_1+e_2 < p+2$. Then $2p-(e_1+e_2) > p-2$ and the $p-1$th coefficient is not zero when $Q$ is different from zero. Update: Gjergji Zaimi gave a counter example where $p=7, n=7$ and all $e_i=2$. Another counter-example is $p=7, n=4$ with $e_1=e_2=e_3=2, e_4=6$. So my conjecture is false! My question right now is whether there is a sufficient condition on $e_i$'s for $\frac{1}{\prod_{i=1}^n (x-P_i)^{e_i}}$ to be a derivative of some rational functions. REPLY [8 votes]: If we restrict to the case when $e_1=e_2=\cdots=e_n=2$ the right condition is $\sum_{i=1}^n e_i=n+kp$ or $n+kp+1$ for some $k\geq 1$. This provides counterexamples to the stated conjecture (see below) and it shows that the right condition is more complicated than just one inequality. Here is a proof of my claim: The residue in this case has a simple formula $$\operatorname{Res}_{P_i}\left(\frac{1}{\prod_{i=1}^n(x-P_i)^2}\right)=\frac{2}{\prod_{j\neq i}(P_i-P_j)^2}\left(\sum_{j\neq i}\frac{1}{P_j-P_i}\right)$$ from which we conclude that $$\operatorname{Res}_{P_i}=0 \iff \sum_{j\neq i}\frac{1}{P_j-P_i}=0.$$ Using the fact that $$\frac{d^2}{dx^2}\left(\prod_{i=1}^n (x-P_i)\right)=\frac{1}{2}\sum_{i=1}^n\left(\prod_{j\neq i}(x-P_j)\sum_{j\neq i}\frac{1}{x-P_j}\right)$$ we notice that $\sum_{j\neq k}\frac{1}{P_j-P_k}=0$ is equivalent to the second derivative of $\prod_{i=1}^n(x-P_i)$ vanishing at $P_k$. Since the degree of the second derivative is $n-2$, the only way it can vanish at $n$ distinct points is if it is equal to $0$. So we have proven that $$\operatorname{Res}_{P_i}=0 \quad \text{for all i}\iff \frac{d^2}{dx^2}\left(\prod_{i=1}^n (x-P_i)\right)=0$$ By looking at the leading coefficient we see that the degree can only be $0$ or $1\pmod{p}$. Or, in other words, $\sum_{i=1}^n (e_i-1)\in \{p,p+1,2p,2p+1,\dots\}$. To show that each of these degrees work you can take $P_i$'s to be the roots of polynomials $x^{kp}-x-1$ and $x^{kp+1}-1$, respectively (both polynomials have distinct roots, and their second derivatives vanish). For an explicit counter example to your conjecture look at $p=5, n=7$ with all $e_i=2$. We have $\sum e_i=14\neq 1\pmod 5$. Yet there exists no choice of distinct $P_i$ for which $\operatorname{Res}_{P_i}=0$ for all $i$.<|endoftext|> TITLE: Number of (distinct) knots with a bounded number of crossings QUESTION [7 upvotes]: The title pretty much covers it: are there good (asymptotic) estimates on the number of knot types whose projection has at most $N$ crossings? Similar question with "projection" replaced by "alternating projection". REPLY [14 votes]: Let $k(n)$ denote the number of prime knots with $n$ crossings, $l(n)$ the number of prime links with $n$ crossings, $a(n)$ the number of alternating prime links with $n$ crossings, and $ak(n)$ the number of prime alternating knots (all unoriented and unordered). Clearly from inclusion of sets $ak(n)\leq a(n)\leq l(n)$ and $ak(n)\leq k(n)\leq l(n)$. Then we have $$2.68 \leq \liminf_{n\to \infty} k(n)^{\frac1n} \leq \liminf_{n\to \infty} l(n)^{\frac1n} \leq 10.398...,$$ where the left is due to Welsh based on Ernst-Sumners (and only counts the growth of 2-bridge knots), and the right estimate is due to Stoimenow. For alternating knots, one has the same lower bound since 2-bridge knots are alternating. So one has $$2.68 \leq \liminf_{n\to \infty} ak(n)^{\frac1n} \leq \lim_{n\to \infty} a(n)^{\frac1n} = 6.14793...,$$ the right upper bound coming from Sundberg-Thistlethwaite. From estimates on the growth of prime knots with $n$ crossings, one should be able to obtain upper bounds on the growth of all knots with $n$ crossings via the prime decomposition. Improved lower bounds over the prime growth are trickier, since we don’t know that crossing number is additive under connect sum.<|endoftext|> TITLE: Handlebody decomposition of 4-spheres without 3-handles QUESTION [16 upvotes]: There used to be many candidates for an exotic 4-sphere, but a lot of them are now known to be the standard smooth $S^4$. The ones of Cappell-Shaneson (maybe not all of them?) were described in terms of handlebody decompositions where there are no 3-handles, but I think the proofs that these were diffeomorphic to $S^4$ involved altering the decompositions with a cancellation trick by introducing some 3-handles. I am not sure what is left over, but I seek the following: 1) Is there a potentially exotic 4-sphere which is prescribed by an explicit 4-handlebody decomposition with only 0,1,2-handles and a single 4-handle? 2) Of the standard Cappell-Shanelson spheres with a given handlebody decomposition involving no 3-handles, is it known that we can prove [edit: some of them] they're standard without the trick involving 3-handles? The reason I ask is because I'd like to study certain differential 2-forms on these spheres which "play well" with the handles, and to understand the deformation of these 2-forms as we change the handlebody decompositions. REPLY [9 votes]: Not really an answer, but may be helpful... First, I would not say that the problem gets easy by introducing a canceling 2/3-handle pair. Akbulut, Kirby and Gompf worked several years on this. S. Akbulut, and R. Kirby, A potential smooth counterexample in dimension 4 to the Poincaré conjecture, the Schoenflies conjecture, and the Andrews-Curtis conjecture. Topology 24 (1985), 375–390. R. Gompf, Killing the Akbulut-Kirby 4-sphere, with relevance to the Andrews-Curtis and Schoenflies problems. Topology 30 (1991), 97–115. Even to just understand the results will take some time. See for example the introduction to R. Gompf, More Cappell-Shaneson spheres are standard. Algebr. Geom. Topol. 10 (2010), 1665–1681. for some explanation on the history. In the same paper, it is also explained exactly which Cappell-Shaneson spheres are standard. The handle decompositions of the others should be an answer to your first question. However, it is conjectured by Gompf that in fact, all Cappell-Shaneson spheres are standard. The difference between allowing to introduce canceling 2/3-handle pairs and not allowing is given by the Andrew-Curtis conjecture. If you have a handle decomposition without 3-handles of a homotopy sphere you get a balanced presentation of the trivial group. If you can prove that these handle decomposition is a decomposition of the standard sphere without introducing 3-handles then the corresponding presentation of the trivial group given by the handle decomposition is Andrews-Curtis trivial. The Andrews-Curtis conjecture (which is open) says that every balanced presentation of the trivial group is Andrews-Curtis trivial. This conjecture is believed to be false, many potential counterexamples are known. Probably the presentations given by the handle decompositions of the Cappell-Shaneson spheres are counterexamples. The original work of Gompf contains more information. R. Gompf, Killing the Akbulut-Kirby 4-sphere, with relevance to the Andrews-Curtis and Schoenflies problems. Topology 30 (1991), 97–115.<|endoftext|> TITLE: Profinite completion of finitely presented groups QUESTION [11 upvotes]: Let $G$ be a finitely presented group, $\widehat{G}$ be the profinite completion of $G$, and $f: G\rightarrow \widehat{G}$ be the natural map. My question is: Is there an example of $G$ for which $\text{Im} f$ is not finitely presented? REPLY [17 votes]: Yes. Take the Baumslag-Solitar group $$G=\mathrm{BS}(2,3)=\langle t,x\mid tx^2t^{-1}=x^3\rangle$$ Then $G$ is finitely presented; the image of $G$ in its profinite completion (i.e., the largest residually finite quotient of $G$) is $\mathbf{Z}[1/6]\rtimes_{2/3}\mathbf{Z}$, which is not finitely presented. (Here $(2,3)$ can be replaced with any coprime pair $(n,m)$ with $n,m\ge 2$.) Here's a similar example where I can provide details. Fix $n\ge 2$. Define $$H_n:\langle t,x,y|txt^{-1}=x^n,t^{-1}yt=y^n,[x,y]=1\rangle;$$ let $u$ be the group endomorphism of $H_n$ mapping $(t,x,y)\mapsto (t,x^n,y)$. It well-defined, since the triple of images satisfies the relators, and is clearly surjective (since the image of $t^{-1}xt$ is $t^{-1}x^nt=x$). It is not injective, because $[t^{-1}xt,y]$ belongs to the kernel; to show that this element is not trivial can be obtained by observing that the presentation describes $H_n$ as an HNN-extension of $\mathbf{Z}^2=\langle x,y\mid [x,y]=1\rangle$ with an isomorphism $\langle x,y^n\rangle\to \langle x^n,y\rangle$ mapping $(x,y^n)$ to $(x^n,y)$. As in every surjective endomorphism $u$ of a finitely generated group, all elements in $L_u=\bigcup_m\mathrm{Ker}(u^m)$ belong to the kernel of the profinite completion homomorphism. In the present case, all elements $[t^{-m}xt^m,y]$ belong to $L_u$. But in the quotient by these additional relators, the elements $t^{-m}xt^m$ generate a copy of $\mathbf{Z}[1/n]$, the elements $t^{m}yt^{-m}$ generate another one, and they commute with each other. This allows to show that the quotient (by these additional relators) is isomorphic to $$\Gamma_n=\mathbf{Z}[1/n]^2\rtimes_A\mathbf{Z},$$ where the action is by the diagonal matrix $A=\mathrm{diag}(n,n^{-1})$. The group $\Gamma_n$ is residually finite, and is easily deduced to be the largest residually finite quotient of $H_n$. Since the kernel of the quotient homomorphism $H_n\to\Gamma_n$ is the strictly increasing union of normal subgroups $\mathrm{Ker}(u^m)$, it is not finitely presented (the latter observation also follows from classical results of Bieri-Strebel).<|endoftext|> TITLE: Block matrices and their determinants QUESTION [7 upvotes]: For $n\in\Bbb{N}$, define three matrices $A_n(x,y), B_n$ and $M_n$ as follows: (a) the $n\times n$ tridiagonal matrix $A_n(x,y)$ with main diagonal all $y$'s, superdiagonal all $x$'s and subdiagonal all $-x$'s. For example, $$ A_4(x,y)=\begin{pmatrix} y&x&0&0\\-x&y&x&0\\0&-x&y&x \\0&0&-x&y\end{pmatrix}. $$ (b) the $n\times n$ antidigonal matrix $B_n$ consisting of all $1$'s. For example, $$B_4=\begin{pmatrix} 0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}.$$ (c) the $n^2\times n^2$ block-matrix $M_n=A_n(B_n,A_n(1,1))$ or using the Kronecker product $M_n=A_n(1,0)\otimes B_n+I_n\otimes A_n(1,1)$. Question. What is the determinant of $M_n$? UPDATE. For even indices, I conjecture that $$\det(M_{2n})=\prod_{j,k=1}^n\left[1+4\cos^2\left(\frac{j\pi}{2n+1}\right)+4\cos^2\left(\frac{k\pi}{2n+1}\right)\right]^2.$$ This would confirm what Philipp Lampe's "perfect square" claim. REPLY [6 votes]: Flip the order of the Kronecker products to get $M'=A_n(I_n,I_n)+B_n\otimes T_n$, where $T_n=A_n(1,0)$. Note that $\det M=\det M'$. Since all blocks are polynomial in $A$, they commute, and therefore the determinant of $M'$ is $\det(f(T_n))$, where $f(x)=\det(A_n(1,1)+xB_n)$. That is, $f(x)=\det(B_n)\det(x+A_n(1,1)B_n)$ is $(-1)^{n\choose 2}$ times the characteristic polynomial of $H=-A_n(1,1)B_n$. Let $t_n(x)$ be the characteristic polynomial of $T_n$. By repeated cofactor expansion on the first row, $t_n(x)=xt_{n-1}(x)+t_{n-2}(x)$. The initial conditions then imply that the $t_n$ is the $(n+1)^{th}$ Fibonacci polynomial. The roots of $t_n$ are $2i\cos(k\pi/(n+1))$ for $k=1,\dots,n$. The eigenvalues of $H$ are worked out in "The eigenvalues of some anti-tridiagonal Hankel matrices". When $n$ is odd they are $1$ and $\pm\sqrt{3+2\cos(\frac{2k\pi}{n+1})}$ for $k=1,\dots,\frac{n-1}{2}$. When $n$ is even they are $\pm\sqrt{1+4\cos^2(\frac{(2k+1)\pi}{n+1})}$ for $k=0,\dots,\frac{n}{2}-1$. By a quick diagonalization argument $\det(f(T_n))$ is the resultant of $f$ and $t_n$. This plus some trig gives $$ \det(M_{2n})=\prod_{j,k=1}^n\left[1+4\cos^2\left(\frac{j\pi}{2n+1}\right)+4\cos^2\left(\frac{k\pi}{2n+1}\right)\right]^2 $$ and $$ \det(M_{2n-1})=\prod_{j=1}^{n-1}\left[1+4\cos^2\left(\frac{j\pi}{2n}\right)\right]^2\prod_{k=1}^{n-1}\left[1+4\cos^2\left(\frac{j\pi}{2n}\right)+4\cos^2\left(\frac{k\pi}{2n}\right)\right]^2. $$<|endoftext|> TITLE: Image of the norm map for degree $3$ galois extension over $\mathbb{Q}$ QUESTION [8 upvotes]: I want to construct a cyclic division algebra of degree $3$ over some degree $3$ Galois extension $E$ of $\mathbb{Q}$. So the construction is as follows: As a set $D=E\oplus uE \oplus u^2 E$ where $u$ is an indeterminate. Addition is defined component wise while multiplication is defined as $e.u=u\sigma(e)$ where $\sigma\in Gal(E/\mathbb{Q})$ is fixed and $u^3=a$ for some fixed $a\in \mathbb{Q}\setminus N_{E/\mathbb{Q}}(E)$ where $N_{E/\mathbb{Q}}$ is the norm map. I want to find particular $a$ for some extension for some calculation purpose in $D^*$. I started with $E=\frac{\mathbb{Q}[x]}{x^3+x^2-2x-1}$ and calculated that $N_{E/\mathbb{Q}}(a+bx+cx^2)= a^3-a^2b-2ab^2+b^3+5ac^2-abc-b^2c+6ac^2-2bc^2+c^3$. But it is difficult to find one element which is not norm of any element of the extension in this manner. So here is my question: Can you find explicitly one $a\in\mathbb{Q}$ such that $N_{E/\mathbb{Q}}(\alpha)\neq a$ for all $\alpha\in E$. REPLY [7 votes]: There is a general strategy to tackle such question. Since the norm is multiplicative, it make sense to look for a prime $p$ not in the image. Then, we can ask about the number of times $p$ divides a general norm. The norm map extends to prime ideals of the integer ring, and if the norm $N_{E/F}(q)$ is $p^3$ for every such ideal over $p$, then $p$ can not possibly be a norm, since its ideal is even not a norm of a fractional ideal of $\mathcal{O}_E$. This happens exactly for those primes $p$ which are inert in $E$, namely those primes modulo which the defining polynomial of $E$ remains irreducible. Indeed, the norm $N_{E/F}(q)$ coincide the the size of $\mathcal{O}_E/q$ which is $p^3$ for inert $q$, the quotient being $\mathbb{F}_{p^3}$. In our case, for example, the polynomial $x^3+x^2-2x-1$ reduces mod $2$ to the polynomial $x^3+x^2+1$ which is irreducible, so $2$ is an inert prime and the number $2$ can not be realized as a norm. So you can choose $a=2$. edit: Apparently a similar answer was put in the comments while I written this, sorry for the double answer.<|endoftext|> TITLE: On the upper bound of $\sum_{i=1}^{n}x^m_{i}$ subject to the conditions $\sum_{i=1}^{n}x_{i}=0$ and $\sum_{i=1}^{n}x^2_{i}=n$ QUESTION [5 upvotes]: The following question has been posted on mathematics stackexchange: inequalities problem, perhaps arising from a question on expectations. Let $x_{1},x_{2},\cdots,x_{n}$ are real numbers, and such $$\begin{cases} x_{1}+x_{2}+\cdots+x_{n}=0\\ x^2_{1}+x^2_{2}+\cdots+x^2_{n}=n \end{cases}$$ Let $\alpha_{m}=\displaystyle\dfrac{1}{n}\sum_{i=1}^{n}x^m_{i}$ See Mitrinovic D.S Analytic inequalities (Springer 1970) Page 347. M.LAKSHMANAMURTI proved that $$\alpha_{m}\le\dfrac{(n-1)^{m-1}+(-1)^m}{n(n-1)^{(m/2)-1}}.$$ I am interested in the details of the proof or a published reference. REPLY [3 votes]: This is addressed in the following paper: Rivin, Igor, Counting cycles and finite dimensional $L^{p}$ norms, Adv. Appl. Math. 29, No. 4, 647-662 (2002). ZBL1013.05042.<|endoftext|> TITLE: Separability of compact quantum groups QUESTION [9 upvotes]: In the theory of compact quantum groups due Woronowicz, we assume usually that the C*-algebra of the compact quantum group is separable. Is the assumption essential in the theory? Will it eventually make sense to develop the theory of nonseparable compact quantum groups? What has gone wrong? REPLY [7 votes]: A Haar measure on a compact quantum group without requiring separability was constructed in The Haar measure on a compact quantum group (1995).<|endoftext|> TITLE: Proof of Minkowski theorem using harmonic analysis QUESTION [19 upvotes]: I am trying to properly write a proof of Minkowski's theorem in a self-contained way and understandable by (good) undergraduates. Theorem (Minkowski) Let $L$ be a lattice of $\mathbb{R}^n$ and $C$ a convex body, symmetric relatively to the origin and with $\mathrm{vol}{C} > 2^n \det(L)$. Then there is a nontrivial point of $L$ lying in $C$. I would like to illustrate the use of Poisson formula and of analytic tools to approximate counting functions, for instance as a first contact with analytic number theory or trace formulas. Here are the steps : 1. Poisson formula For $f$ continuous, with integrable Fourier transform, and both of moderate growth, $$\sum_{x \in L} f(x) = \mathrm{covol}(L)^{-1} \sum_{\xi \in \widehat{L}} \widehat{f}(\xi)$$ This follows from elementary Fourier analysis and is fine. 2. Procedure We want to approximate the counting function of $C \cap L$ as a spectral side in the Poisson formula above. For this, we require a function $f$ such that $f \leqslant \mathbf{1}_C$ $\widehat{f} \geqslant 0$ $f$ admissible for Poisson formula Provided such a function, the result then essentially follows from the computations $$|C \cap L| = \sum_{x \in L} \mathbf{1}_C(x) \geqslant \sum_{x \in L} f(x) = \mathrm{covol}(L)^{-1} \sum_{\xi \in \widehat{L}} \widehat{f}(\xi) \geqslant \mathrm{covol}(L)^{-1} \widehat{f}(0)$$ 3. Construction of $\mathbf{f}$ Now it remains to be able to provide such an $f$. Naively the characteristic function of $C$ would work formally, but does not satisfy the continuity property. In order to get positivity and regularity, $\mathbf{1}_{C/2}\star \mathbf{1}_{C/2}$ is a good try but does not respect the growth conditions in Poisson (at least it seems hard to prove it and it is not true in general, however I am ready to suppose any simplifying assumption on $C$ if it could help). So here is the question: is there any simple construction of such a function $f$, or modification (e.g. mollifying) of the above ones? REPLY [12 votes]: $$ \newcommand{\Vol}{\mathrm{Vol}\,} \newcommand{\supp}{\mathrm{supp}\,} \newcommand{\eps}{\varepsilon} \renewcommand{\phi}{\varphi} \newcommand{\R}{{\mathbb R}} \newcommand{\Z}{{\mathbb Z}} \newcommand{\<}{\langle} \newcommand{\>}{\rangle} $$ This is in fact fairly standard, but filling in the technical details can be a little tricky for a non-expert. To slightly simplify the notation, I write $K:=\frac12\,C$, and assume that $L=\Z^n$, so that the Poisson formula gets the shape $$ \sum_{z\in\Z^n} f(z) = \sum_{u\in\Z^n}\hat f(u) \tag{1}; $$ the general-lattice case follows by applying a suitable linear transformation, or by a very slight modification of the argument below. Technically, for (1) to hold, it suffices to assume that both $f$ and $\hat f$ are continuous and satisfy $|f(z)|\ll\min\{1,|z|^{-n-1}\}$ and $|\hat f(u)|\ll\min\{1,|u|^{-n-1}\}$. We show that if $\Vol(K)>1$, then for any given $\eps>0$, the set $(1+\eps)C$ contains a non-zero point of the integer lattice; clearly, this will imply the assertion. Fix a smooth function $h\colon\R^n\to\R$ satisfying $$ h(-z) = h(z),\quad z\in\R^n, \tag{2} $$ $$ 1_{(1-\eps)K} \le h \le 1_{(1+\eps)K}, \tag{3} $$ and $$ \int_{\R^n} h(z)\,dz = \Vol(K). \tag{4} $$ Ideologically, $h$ is a smoothed indicator function of $K$; I explain below how such a function can be constructed. By (2), $$ \overline{\hat h(u)} = \int_{\R^n} h(z)\, e^{2\pi i\}\,dz = \int_{\R^n} h(-z) e^{-2\pi i\}\,dz = \int_{\R^n} h(z) e^{-2\pi i\}\,dz = \hat h(u); $$ that is, $\hat h$ is real-valued. Let $f:=h\ast h$. Since $\hat h$ is real, $\hat f=(\hat h)^2$ is also real and, indeed, non-negative. Furthermore, $\supp f\subseteq\supp h+\supp h\subseteq (1+\eps)C$ by (3), and $$ f(z) = \int_{\R^n} h(x)h(z-x)\,dx \le \int_{\R^n} h(x)\,dx = \Vol(K),\quad z\in\R^n $$ by (4), showing that $$ f \le 1_{(1+\eps)C}\,\Vol(K). \tag{5} $$ Finally, since $h$ is smooth and finitely supported, we have $$ |\hat h(u)| \ll \min\{1,|u|^{-n-1}\} $$ whence also $$ |\hat f(u)| = (\hat h(u))^2 \ll \min\{1,|u|^{-n-1}\}. $$ We thus can apply (1). In view of (5), the left-hand side is $$ \sum_{z\in\Z^n} f(z) \le \Vol(K) \sum_{z\in\Z^n} 1_{(1+\eps)C}(z) = |(1+\eps)C\cap\Z^n|\,\Vol(K), $$ while, in view of (4), the right-hand side is $$ \sum_{u\in\Z^n} \hat f(u) = \sum_{u\in\Z^n} (\hat h(u))^2 \ge (\hat h(0))^2=(\Vol(K))^2. $$ Therefore $$ |(1+\eps)C\cap\Z^n| \ge \Vol(K) > 1, $$ as wanted. Why does a smooth function $h$ satisfying (2)$-$(4) exist? Start with a smooth function $\phi\colon\R^n\to\R_{\ge 0}$ such that $\phi(-z)=\phi(z)$, $\supp\phi\subseteq K$, and $\int_{\R^n}\phi(z)\,dz=1$, let $\phi_\eps(z):=\eps^{-n}\phi(\eps^{-1}z)$, and define $h:=1_K\ast\phi_\eps$. Then (2) is immediate to verify, (4) follows from $$ \int_{\R^n} (1_K\ast \phi_\eps)(z)\,dz = \int_{\R^n} 1_K(z)\,dz \cdot \int_{\R^n} \phi_\eps(z)\,dz = \Vol(K) \cdot \int_{\R^n} \phi(x)\,dx, $$ and smoothness follows from $|\hat h(u)|=|\widehat{1_K}(u)||\widehat{\varphi_\eps}(u)|$ and $\widehat{\phi_\eps}(u)=\hat\phi(\eps u)$. Finally, to get (3) observe that \begin{multline*} h(z) = \int_K \phi_\eps(z-x)\,dx = \eps^{-n} \int_K\phi(\eps^{-1}(z-x))\,dx \\ = \int_{\eps^{-1}(z-K)} \phi(y)\,dy \le \int_{\R^n} \phi(y)\,dy = 1, \end{multline*} and that if $z\notin(1+\eps)K$, then $\eps^{-1}(z-K)\cap K=\varnothing$ whence, indeed, $h(z)=0$; if $z\in(1-\eps)K$, then $\eps^{-1}(z-K)\supseteq K$ (hint: this uses convexity of $K$), whence $h(z)=1$.<|endoftext|> TITLE: Entropy and total variation distance QUESTION [17 upvotes]: Let $X$, $Y$ be discrete random variables taking values within the same set of $N$ elements. Let the total variation distance $|P-Q|$ (which is half the $L_1$ distance between the distributions of $P$ and $Q$) be at most $\epsilon$. What is a standard, easily provable bound on the difference between the entropies of $X$ and $Y$? (It is easy to see that such a bound must depend not only on $\epsilon$ but also on $N$.) REPLY [6 votes]: $\newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\R}{\mathbb{R}}$ Here is yet another answer providing the exact bound. This answer is perhaps a bit more elementary than the excellent answer given by user Algernon. Another advantage of this approach is that it produces the exact bound for any convex function $f$ in place of the function $p\mapsto p\ln p$. To preserve the history of the question, I have also retained my previous answer, which used different (if somewhat similar) ideas and provided a suboptimal bound. Take indeed any convex function $f\colon[0,1]\to\R$ and consider the difference \begin{equation} \De:=\sum_1^N f(p_i)-\sum_1^N f(q_i) \end{equation} between the "generalized" entropies of $Q=(q_i)_{i=1}^N$ and $P=(p_i)_{i=1}^N$. We want to find the exact upper bound on $\De$ subject to the given conditions on $(P,Q)$. In what follows, $(P,Q)$ is a point satisfying these conditions. Without loss of generality (wlog), for some $k\in\{1,\dots,N\}$ we have $p_i\ge q_i$ for $i\le k$, $p_i\le q_i$ for $i\ge k+1$, and $q_1\ge\cdots\ge q_k$, so that \begin{equation} \ep=\sum_1^k(p_i-q_i)=\sum_{k+1}^N(q_i-p_i)>0. \end{equation} Let $p_i^*:=q_i$ for $i=2,\dots,k$ and $p_1^*:=q_1+\ep[=\sum_1^k p_i-\sum_2^k q_i\le1]$. Then the vector $(p_1^*,\dots,p_k^*)$ majorizes (in the Schur sense) the vector $(p_1,\dots,p_k)$ and still satisfies the condition $p_i^*\ge q_i$ for $i\le k$. Also, since $f$ is convex, $\sum_1^k f(p_i)$ is Schur convex in $(p_1,\dots,p_k)$. So, wlog $(p_1,\dots,p_k)=(p_1^*,\dots,p_k^*)$. In particular, $p_1>q_1$ and $q_m\ge p_m$ for all $m=2,\dots,N$. Moreover, wlog $p_m=0$ for any $m=2,\dots,N$. Indeed, take any $m=2,\dots,N$ with $p_m>0$ and replace $p_1,q_1,p_m,q_m$ respectively by $p_1+t,q_1+t,p_m-t,q_m-t$, where $t:=p_m\in(0,1-p_1]$; then all the conditions on $P,Q$ will still hold. After this replacement, $\De$ will change by the sum of the nonnegative expressions $[f(p_1+t)-f(q_1+t)]-[f(p_1)-f(q_1)]$ and $[f(q_m)-f(p_m)]-[f(q_m-t)-f(p_m-t)]$; this nonnegativity follows by the convexity of $f$. Making such replacements for each $m=2,\dots,N$ with $p_m>0$, we will change $\De$ by a nonnegative amount, and will also get $p_m=0$ for all $m=2,\dots,N$ indeed. Thus, wlog \begin{gather} p_1=1=q_1+\ep,\quad p_i=0\ \forall i=2,\dots,N,\quad \sum_2^N q_i=\ep. \end{gather} Since $\sum_2^N f(q_i)$ is Schur convex in the $q_i$'s, wlog \begin{align} \De&=f(1)-f(q_1)+\sum_2^N f(0)-\sum_2^N f(q_i) \\ &\le f(1)-f(1-\ep)+(N-1)f(0)-(N-1)f(\tfrac\ep{N-1})=:\De_{f;\ep,N}. \end{align} The bound $\De_{f;\ep,N}$ on $\De$ is obviously exact, since it is attained when $p_1=1=q_1+\ep$ and $q_2=\cdots=q_N=\tfrac\ep{N-1}$. In the particular case when $f(p)=p\ln p$ (with $f(0)=0$), the exact bound $\De_{f;\ep,N}$ becomes $H(\ep)+\ep\ln(N-1)$, where $H(\ep):=\ep\ln\frac1\ep+(1-\ep)\ln\frac1{1-\ep}$.<|endoftext|> TITLE: Map from the Multiset Monad to the Giry Monad: From Data to Probabilities QUESTION [6 upvotes]: The Mulitiset monad, aka the free commutative monoid monad or "Bag" monad, takes a set to the set of all Multisets for that set. A Multiset is like a set, but can have duplicates. It is used in computer science as the Bag monad, where the Bag is a data structure that can take in symbols or readings from a set and pop them out in random order (ie there is no ordering for the data). There is a monad that captures this, $(M, \mu, \eta)$ where the product $\mu: M \cdot M \rightarrow M$ works by taking a Multiset of Multisets, dissolving the inner container walls, and making one big Multiset. The Giry Monad, $(G, \mu_G, \eta_G)$ is meant to capture probability measures. The functor, $G$, takes a set to the set of probability measures on that set. The product axiom is described in a comment here and goes as follows: A probability measure on an affine space has an average value (also called expectation or integral), which is a point in that affine space. Apply this to the affine space of probability measures on X. In other words, a measure on the space of measures determines a (weighted) average of those measures. Every Multiset has an associated probability measure that gives the probability of finding one of the set elements in that Multiset. This must mean that there is a transformation from the Multiset monad to the Giry monad. Could someone write down, in detail, what this transformation is and give some interpretations of it too. Some restrictions have been suggested. We restrict to finite multisets Because $M$ is on Set and $G$ is usually on a category of nice topological spaces, we will restrict $G$ to Set by viewing a set as a discrete topological space. I will accept any answer with these restrictions. If you have a general answer, please also post that. REPLY [7 votes]: The intuition described in the question can be made precise as a natural map from the finite-multiset monad to a certain monad of (non-normalised) measures. However, this map doesn’t factor through the discrete Giry monad; in fact, there is no monad morphism from either the finite-multiset monad or the non-empty-finite-multiset monad to the discrete Giry monad. Specifically, define monads on $\newcommand{\Set}{\mathbf{Set}}\Set$ by: $MX$ is the set of finite multisets on $X$, with monad structure as defined in the question; $HX$ is the set of finitely supported measures on $X$, with monad structure defined as for the Giry monad; $GX \subseteq HX$ is the set of finitely supported probability measures on $X$, with the induced monad structure (i.e. $G$ is a version of the Giry monad). (Note these are not the original Giry monad, since these are on sets not spaces, and hence require the restriction to finitely or countably supported measures in order to define the monad multiplication.) Then the natural transformation $\alpha : M \to H$ sending a finite multiset $m$ from $X$ to the measure $\mu(x) = |x \in m|$ is a monad morphism. Indeed, it is a monomorphism, and can be seen as including $M$ as the submonad of integer-valued finitely supported measures within $H$. Similarly the inclusion natural transformation $\iota : G \to H$ is a monad morphism by definition. However, there is no natural transformation $M \to G$ (so, a fortiori, no transformation $H \to G$): for any set $X$ with two distinct elements $x, y$, the empty multiset in $X$ is in the image of both $Mi_x$ and $Mi_y : M1 \to MX$ (where $i_x : 1 \to X$ picks out $x$, and $i_y$ similarly), so its image in $GX$ should be in the image of both $Gi_x$, $Gi_y$; but these maps have disjoint images (they pick out the point measures concentrated on $x$ and $y$ respectively). Restricting to the monads $M'$, $H'$ of non-empty multisets / nonzero measures, there is at least a natural transformation $\nu : H' \to G$, sending a measure to its normalised probability measure (and hence a composite transformation $M' \to G$). However, this is \emph{not} a monad map, nor is the composite $\nu \alpha : M \to G$. Concretely, $\nu \alpha$ doesn’t preserve monad multiplication (and so $\nu$ can’t): given a set $X$ with distinct elements $x,y$, $\{\{x,x\},\{y\}\} \in M'M'X$ goes to the half-and-half measure if you normalise the components then multiply in $G$, but to the two-thirds/one-third measure if you multiply in $M$ and then normalise it. In fact, there is no monad morphism $M' \to G$. Suppose given any natural transformation $\beta : M' \to G$. A symmetry/naturality argument shows that any multiset from $X$ with all nonzero multiplicities equal must be sent by $\beta$ to the uniform distribution on the corresponding subset of $X$. But any multiset (like $\{x,x,y\}$ above) arises by multiplication from a multiset with all nonzero multiplicities 1, whose elements each involve only one element. So if $\beta$ preserved multiplication, it would have to send any multiset $m$ to the uniform distribution on the set of elements occurring in $m$. This determines $\beta$ fully; but then this can’t preserve multiplication, since the uniform distributions don’t form a submonad of $G$.<|endoftext|> TITLE: On non-representability of certain hom schemes QUESTION [13 upvotes]: Let $k$ be an algebraically closed field. It is well-known that the isom-sheaf Isom$(\mathbb{A}^1_k,\mathbb{A}^1_k)$ is not representable by an algebraic space. (To be clear, the functor Isom$(\mathbb{A}^1,\mathbb{A}^1)$ associates to a $k$-scheme $S$, the set of isomorphisms $\mathbb{A}^1_S\to \mathbb{A}^1_S$ of $S$-schemes.) Now, I expect that the non-representability of this isom-functor implies the non-representability of many hom-functors. My question is about how to make this precise. Let X be a (positive-dimensional) variety and let $f:\mathbb{A}^1_k\to X$ be a finite morphism. How does one show that the hom-functor $\mathrm{Hom}_k(\mathbb{A}^1_k,X)$ is not representable by an algebraic space? What have I tried? Well, there is a natural morphism Isom$(\mathbb{A}^1,\mathbb{A}^1) \to \mathrm{Hom}(\mathbb{A}^1_k,X)$ which sends $g$ to $f\circ g$. My expectation is that this is an open or closed immersion of functors. But how to make this precise? REPLY [10 votes]: Welcome, new contributor. Let $k$ be a field. Let $Y$ be a finite type, separated $k$-scheme such that the $k$-algebra $\mathcal{O}_Y(Y)$ is a $k$-vector space of infinite dimension. For instance, this holds for every finite type, affine $k$-scheme of positive dimension. Let $X$ be a finite type, separated $k$-scheme that is not everywhere finite, étale over $\text{Spec}(k)$. The Hom functor $\text{Hom}_{k-\text{Sch}}(Y,X)$ is limit preserving because the Yoneda functor of $X$ is limit preserving. Thus, if the Hom functor is representable, it is representable by a $k$-scheme that is locally finite type. The claim is that the Hom functor is not representable by a $k$-scheme that is locally finite type. If the Hom functor is representable by a locally finite type $k$-scheme, then after base change to the algebraic closure of $k$, also the Hom scheme of the base change is representable by a scheme that is (ed. locally) finite type over the algebraic closure. Thus, without loss of generality, assume that $k$ is algebraically closed. Since $X$ is not everywhere étale, there exists a $k$-point $x$ of $X$ at which the $k$-vector space $\Omega_{X/k,x}\otimes_{\mathcal{O}_{X,x}} k = \mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2$ has positive dimension as a $k$-vector space. Denote by $$c_x:Y\to X$$ the constant $k$-morphism with image $x$. Then the Zariski tangent space to the Hom functor at $c_x$ equals $$\text{Hom}_{\mathcal{O}_Y}(c_x^*\Omega_{X/k},\mathcal{O}_Y) = \mathcal{O}_Y \otimes_k \left(\mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2\right).$$ This is a $k$-vector space of infinite dimension. If the Hom scheme were representable by a locally finite type $k$-scheme, then the Zariski tangent space at $c_x$ would be a finite-dimension $k$-vector space. Thus, the Hom scheme is not representable.<|endoftext|> TITLE: Seifert fiberings of zero euler number which are semi-bundles QUESTION [5 upvotes]: Let M be a closed oriented manifold which has the structure of a "semi-bundle" (See Section 1.2. of Hatcher's notes on three-manifolds) over an interval I. Assume that M is Seifert fibered over a base B and that the Euler number of the Seifert fibering is zero. Suppose finally that M has a unique Seifert fibering, so B is uniquely determined. Is it always the case that B is non-orientable? REPLY [4 votes]: Let $\Sigma$ be the orientable generic fiber of the semi-bundle structure on $M$. I only consider the case $\chi(\Sigma) < 0$ for simplicity: the cases $\chi(\Sigma) \geq 0$ should be worked out by hand; in many cases the fibration is not unique there. In that case the answer is yes, because of a stronger fact: If an orientable Seifert manifold $M \to B$ contains an orientable separating incompressible surface $\Sigma$ with $\chi(\Sigma)<0$, then the base orbifold $B$ is non-orientable. The proof goes as follows. Since $\Sigma$ is incompressibile, it is isotopic to a horizontal or vertical surface. Since $\chi(\Sigma)<0$, it cannot be vertical, so it is horizontal. The fibration now induces an orbifold covering map $\pi\colon\Sigma \to B$. Since $\Sigma$ is separating in $M$, one sees easily that some deck transformation of $\pi$ must be orientation-reversing. The fact that I stated is actually equivalent your assertion, since every horizontal surface is a fiber in a (semi-)bundle structure for $M$.<|endoftext|> TITLE: Factorization of colimits through slices? QUESTION [5 upvotes]: I could swear I remember a result of the following form: Suppose we have a pair of functors $$C\xrightarrow{F}D\xrightarrow{G} X,$$ with $X$ cocomplete. then we obtain a functor $$D\to X$$ sending $$d\mapsto \operatorname{colim}_{(F\downarrow d)}(G\circ \pi_d)$$ where $\pi_d:(F\downarrow d)\to D$ is the projection functor sending $(F(c)\to d) \mapsto F(c)$. Then there is a canonical isomorphism $$\operatorname{colim}_D\left (\operatorname{colim}_{(F\downarrow d)}(G\circ \pi_d)\right) \cong \operatorname{colim}_C (G\circ F).$$ Is this true? Do you know of a source? Is there a name for this kind of result? REPLY [6 votes]: The functor $\text{colim}_{(F\downarrow -)}(G\circ\pi_-)\colon D\to X$ is a left Kan extension of $G\circ F$ along $F$. The corresponding natural transformation $\ell_F^{G\circ F}\colon G\circ F\to \text{colim}_{(F\downarrow -)}(G\circ\pi_-)\circ F$ is defined by $$ \ell_F^{G\circ F}(c)=\varphi^{F(c)}(id_{F(c)}), $$ where $\varphi^{F(c)}$ is a colimiting cocone of $G\circ\pi_{F(c)}$. Verification of the universality is long but straightforward. Then it remains to note that for any functors $T$ and $S$ there exists an isomorphism $$ \varinjlim\text{Lan}_TS\cong\varinjlim S. $$ Seems that your statement is just one of the great amount of interesting properties of Kan extensions, so I doubt that it has a name. Some basic properties of pointwise Kan extensions are described in the first volume of Borceux's handbook.<|endoftext|> TITLE: Localization of $\infty$-categories QUESTION [8 upvotes]: In ordinary category theory, the localization $C[S^{-1}]$ at a class of morphisms $S$ (with possibly some assumptions on $S$) is a category $C[S^{-1}]$ together with a map $L:C \to C[S^{-1}]$ such that any functor $C \to D$ that sends $S$ to isomorphisms factors uniquely through $L$. Does a generalization of this notion exist in $\infty$-category theory? Can one simply carry out the construction as one does for ordinary categories? It's unclear to me how one would define the mapping space in the localized category. A reference would be sufficient! REPLY [3 votes]: There are a number of good references for this. Section 5 of Danny Stevenson's paper on simplicial localization and covariant model structures contains an extremely simple explicit construction of the localization which models, in simplicial sets, the construction suggested by Hurkyl. One just pushes out $C$ with a free-living isomorphism for each arrow in $S$, along the canonical map from the disjoint union of the elements of $S$ to $C$. Of course, one then has to take a fibrant replacement to get an $\infty$-category on the nose. I would argue that this gives the most elementary understanding of this fundamental construction, relative to DK localization or the use of marked simplicial sets. Why "fundamental?" Stevenson also gives a short proof of a theorem known in some form to Dwyer and Kan, and due in this context to Joyal: every $\infty$-category arises as a localization of its ordinary category of simplices. Chapter 7 of Cisinski's recent book is also all about localization, based on the same construction that appears in Stevenson, and in particular its interaction with limits and continuous functors, which imports much of classical homotopical algebra into the $\infty$-context. This was also in part the goal of Mazel-Gee's thesis, which focused on $\infty$-model categories.<|endoftext|> TITLE: Searching for a proof for a series identity QUESTION [5 upvotes]: The below identity I have found experimentally. Question. Is this true? If so, may you provide a "slick" (or any) proof. $$6\sum_{k=1}^{\infty}\frac{k^2q^k}{(1-q^k)^2}+12\left(\sum_{k=1}^{\infty}\frac{kq^k}{1-q^k}\right)^2=\sum_{k=1}^{\infty}\frac{(5k^3+k)q^k}{1-q^k}.$$ REPLY [11 votes]: Follow the comments of Lucia and note that $$\sum_{n\ge 1}\frac{n^2q^n}{(1-q^n)^2}=q\frac{\,d}{\,dq}\sum_{n\ge 1}\frac{nq^n}{1-q^n}.$$ I believe the identity actually is the well known $$q\frac{\,d}{\,dq}L=\frac{L^2-M}{12},$$ where $$L=1-24\sum_{n\ge 1}\frac{nq^n}{1-q^n}\;\mbox{ and}\; M=1+240\sum_{n\ge 1}\frac{n^3q^n}{1-q^n}.$$ You can find it in here https://en.wikipedia.org/wiki/Eisenstein_series. REPLY [7 votes]: $$ \sum_{k=1}^\infty \frac{k^2 q^k}{(1-q^k)^2} = \sum_{n=1}^\infty \sigma(n) n q^n$$ $$ \sum_{k=1}^\infty \frac{k q^k}{1-q^k} = \sum_{n=1}^\infty \sigma(n) q^n$$ $$ \left(\sum_{k=1}^\infty \frac{k q^k}{1-q^k}\right)^2 = \sum_{n=1}^\infty \sum_{m=1}^{n-1} \sigma(m) \sigma(n-m) q^n $$ $$ \sum_{k=1}^\infty \frac{k^3 q^k}{1-q^k} = \sum_{n=1}^\infty \sigma_3(n) q^n $$ so your identity is saying $$ 6 n \sigma(n) + 12 \sum_{m=1}^{n-1} \sigma(m)\sigma(n-m) = 5 \sigma_3(n) + \sigma(n)$$ Hmm, surely that's got to be known. Adding number-theory to the tags.<|endoftext|> TITLE: On the Fourier-Laplace transform of compactly supported distributions QUESTION [7 upvotes]: Let $\mathcal{E}'(\mathbb{R})$ be the space of all compactly supported distributions on $\mathbb{R}$. For $f\in \mathcal{E}'(\mathbb{R})$, let $\widehat{f}$ denote the entire extension of the Fourier transform of $f$. Question: If $f_n\stackrel{n\rightarrow\infty}{\longrightarrow}f$ in $\mathcal{E}'(\mathbb{R})$, then does $\widehat{f}_n$ converge to $\widehat{f}$ uniformly on compact subsets of $\mathbb{C}$ as $n\rightarrow\infty$? REPLY [2 votes]: Let me expand Nate Eldredge's comments. A more elementary proof than the one provided by Jochen Wengenroth, still requiring no computations, is as follows: $\mathcal{E}'$ is the dual space of $C^\infty$, which is endowed with the Fréchet space topology given e.g. by the seminorms $\|f\|_k:=\|f\|_{C^k(B_k(0))}$. The fact that $f_n\to f$ in $\mathcal{E}'$ means that $\langle f_n,\phi\rangle\to\langle f,\phi\rangle$ for any $\phi\in C^\infty$. Now the uniform boundedness principle holds also for linear maps from a Fréchet space to (say) a normed vector space. The proof is the same as that for the Banach space version; for details, see Theorem 2.6 in Rudin's Functional Analysis (2nd edition). Hence the $f_n$'s are equicontinuous, i.e. there exists $k\in\mathbb{N}$ and $C>0$ such that $$ |\langle f_n,\phi\rangle|\le C\|\phi\|_{C^k(B_k(0))}. $$ This gives $|\widehat{f_n}(\xi)|\le C\|e^{-2\pi i\langle\xi,\cdot\rangle}\|_{C^k(B_k(0))}\le C'(1+|\xi|^k)e^{2\pi k|\xi|}$ for $\xi\in\mathbb{C}$. Also, $\widehat{f_n}(\xi)\to\widehat{f}(\xi)$ for every $\xi\in\mathbb{C}$ (since it corresponds to evaluation of $f_n$ on the test function $e^{-2\pi i\langle\xi,\cdot\rangle}$). So you have a locally equibounded sequence of pointwise converging holomorphic functions and this implies your claim. This clearly works in any dimension.<|endoftext|> TITLE: Cyclic cubic extensions and Kummer theory QUESTION [10 upvotes]: The Galois cohomology group $H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z})$ classifies cyclic cubic extensions $K/\mathbb{Q}$ (specifically: the non-trivial elements correspond to Galois cubic field extensions $K/\mathbb{Q}$ together with a choice of isomorphism $\mathrm{Gal}(K/\mathbb{Q}) \cong \mathbb{Z}/3\mathbb{Z}$). Let $k = \mathbb{Q}(\mu_3)$. There are restriction and corestriction maps $$\mathrm{Res}: H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}) \to H^1(k, \mathbb{Z}/3\mathbb{Z}), \quad \mathrm{Cores}: H^1(k, \mathbb{Z}/3\mathbb{Z}) \to H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}).$$ Restriction followed by corestriction is multiplication by $2$ on $H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z})$. As each element is $3$-torsion, it follows that $\mathrm{Res}$ is injective and that $\mathrm{Cores}$ is surjective. But as $\mu_3 \subset k$, it follows from Kummer theory that $$H^1(k, \mathbb{Z}/3\mathbb{Z}) \cong H^1(k, \mu_3) \cong k^*/k^{*3}.$$ Composing with corestriction, we therefore obtain a surjective map $$f: k^{*}/k^{*3} \to H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}).$$ Can the map $f$ be made explicit? Namely, given a non-cube $a \in k^*$, what is the cyclic cubic extension of $\mathbb{Q}$ induced by $f$? I know that the corestriction $H^1(k, \mu_3) \cong k^*/k^{3*} \to \mathbb{Q}^*/\mathbb{Q}^{*3} \cong H^1(\mathbb{Q}, \mu_3)$ is just usual norm map. But this doesn't seem to help here. REPLY [16 votes]: It's just the map $$x \mapsto y = \frac{x}{x^{\sigma}},$$ where the corresponding degree three extension of $\mathbb{Q}$ is the degree three subfield of $k(y^{1/3})$. The point is that it is obvious from the restriction map that $$H^1(\mathbb{Q},\mathbb{Z}/3 \mathbb{Z}) = (k^{\times}/k^{\times 3})^{G = {\chi}},$$ where $\chi$ is the non-trivial character of $G = \mathrm{Gal}(\mathbb{Q}(\zeta_3)/\mathbb{Q}) = \mathbb{Z}/2 \mathbb{Z}$. (Added Here $M^{\chi}$ means what is says on the tin. If $\sigma \in G$ and $m \in M^{\chi}$, then $\sigma m = \chi(\sigma) m$.) And basic Kummer theory also says that degree 3 cyclic extensions $K$ of $\mathbb{Q}$ have the form $K(\zeta_3) = \mathbb{Q}(\zeta_3)(\alpha^{1/3})$ where $$\sigma \alpha = \alpha^{-1} \mod k^{\times}/k^{\times 3}.$$ The same basic structure holds mutatis mutandis with $\mathbb{Q}$ replaced by any number field $F$, and $3$ replaced by $p$, and $G = \chi$ where now $\chi$ is the mod-p cyclotomic character of $G = \mathrm{Gal}(F(\zeta_p)/F)$, which is the canonical (possibly trivial) map $G \rightarrow (\mathbb{Z}/p \mathbb{Z})^{\times}$. And now the map from $k^{\times}/k^{\times p}$ is just the projection to the $\chi$-eigenspace. Added: If you want an explicit polynomial, you can, of course, use Galois theory to do so. In fact, everything in this question one can (and I do) teach in the introductory undergraduate Galois theory course. To spell out the elementary details, you want an element of $k(y^{1/3})$ which is fixed by the order two element $\sigma \in \mathrm{Gal}(k(y^{1/3})/\mathbb{Q})$ (there is an obvious splitting from $\mathrm{Gal}(k/\mathbb{Q}) \rightarrow \mathrm{Gal}(k(y^{1/3}/\mathbb{Q})$). The obvious element to take is thus $$z = y^{1/3} + \sigma y^{1/3} = y^{1/3} + y^{-1/3},$$ which is a root of $$T^3 - 3 T - (y + y^{-1}) = T^3 - 3 T - \left(\frac{x}{x^{\sigma}} + \frac{x^{\sigma}}{x}\right) = T^3 - 3 T - \frac{Tr(x^2)}{N(x)} \in \mathbb{Q}[T].$$<|endoftext|> TITLE: Are mapping class groups of orientable surfaces good in the sense of Serre? QUESTION [16 upvotes]: A group G is called ‘good’ if the canonical map $G\to\hat{G}$ to the profinite completion induces isomorphisms $H^i(\hat{G},M)\to H^i(G,M)$ for any finite $G$-module $M$. I’ve had multiple academics in my department claim to me that the orientation-preserving mapping class group of an orientable surface of genus $g$ with $n$ boundary components satisfies this property, though none have been able to give me a reference. I think the easiest way to prove this would be to exhibit a finite index solvable subgroup, which for mapping class groups is equivalent to a finite index abelian subgroup (A theorem in Sullivan’s ‘genetics of homotopy theory’ says this implies that G is good). I was thinking of the subgroup of homeomorphisms which act trivially on cohomology with $\mathbb{Z}/n\mathbb{Z}$ coefficients, but I have very little experience with solvable groups and couldn’t even figure out if this was a reasonable guess. So, does anyone know if there exists a finite index solvable subgroup of the mapping class groups? Or is there a different proof that the mapping class groups are good? Thanks for any help. REPLY [20 votes]: The braid groups are good (which are mapping class groups of punctured disks) by Proposition 3.5 of Grunewald, F.; Jaikin-Zapirain, A.; Zalesskii, P. A., Cohomological goodness and the profinite completion of Bianchi groups., Duke Math. J. 144, No. 1, 53-72 (2008). ZBL1194.20029. (You can find a version here but missing the statement of this Proposition.) The summary of the proof is that if $G$ is good, and $H$ is commensurable with $G$, then $H$ is also good (Lemma 3.2). Hence we may pass to the pure braid group $P_n$. This is an extension of $P_{n-1}$ by $F_{n-1}$ (essentially the Birman exact sequence). Hence by induction, $P_n$ is good using Lemma 3.3 (residually finite extensions of good groups by certain good groups such as finitely generated free groups are also good). By a similar inductive argument, the mapping class groups of the $n$-punctured sphere and torus are good. Moreover, the mapping class group of the genus $2$ surface is a central extension of the mapping class group of the $6$-punctured sphere by $\mathbb{Z}/2$ (generated by the hyperelliptic involution). Hence by Lemma 3.3 the mapping class group of the genus $2$ surface (and also with $n$ punctures by induction) is good. Mapping class groups of surfaces with boundary are central extensions of the (pure) punctured surface mapping class groups, so by similar reasoning they are good if the punctured ones are. Thus one sees that the general case would follow from the closed case. REPLY [18 votes]: This is an open and probably very difficult question. There have been purported proofs (for instance, this one), but they have all had fatal flaws. The mapping class group is definitely not virtually abelian (or virtually solvable; for other readers, the relevant fact [alluded to by the OP] is that all solvable subgroups of the mapping class group are virtually abelian). For instance, it contains huge numbers of nonabelian free subgroups. One easy-to-state example due to Ishida is that the subgroup generated by two Dehn twists about curves that intersect at least twice is free. See Ishida, Atsushi, The structure of subgroup of mapping class groups generated by two Dehn twists. Proc. Japan Acad. Ser. A Math. Sci. 72 (1996), no. 10, 240-241.<|endoftext|> TITLE: What is the largest known Dehn function of f.p. subgroup of a f.p. group with quadratic Dehn function? QUESTION [16 upvotes]: Is it true that the Baumslag-Solitar groups, say, $BS(1,n)$, $|n|\ge 2$, are finitely presented groups with largest Dehn functions (namely, exponential growth) known to be inside finitely presented groups with quadratic Dehn functions? REPLY [5 votes]: As mentioned in the comments, Sapir and Olshanskii recently proved in Algorithmic problems in groups with quadratic Dehn functions that: Theorem. For every recursive function $f$, there exist finitely presented groups $H \leq G$ such that $G$ has quadratic Dehn function and such that $H$ has Dehn function at least $f$. However, I would like to mention a construction that predates Cornulier and Tessera's article cited in the previous answer: In their article Finitely presented subgroups of automatic groups and their isoperimetric functions, Baumslag, Bridson, Miller III, and Short proved that: Theorem. There exist a biautomatic group $B$ and a finitely presented subgroup $G \leq B$ such that $G$ is not of type $FP_3$ and its isoperimetric function is strictly exponential. Moreover one can arrange for $B$ to be the fundamental group of a closed manifold of non-positive curvature. In this case, the overgroup has quadratic Dehn function but it can also be chosen to be CAT(0). In this direction, I probably should mention the very recent preprint Superexponential Dehn functions inside CAT(0) groups in which finitely presented subgroups with superexponential Dehn functions are constructed in CAT(0) groups.<|endoftext|> TITLE: Are infinite groups in which most elements have order $\leq 2$ commutative? QUESTION [23 upvotes]: The starting point of this question is the following: If $G$ is a group such that all elements have order at most $2$, then $G$ is commutative. If $G$ is any group, let $G_{>2}$ denote the set of elements $g\in G$ such that $g^2 \neq 1_G$ where $1_G$ denotes the neutral element of the group. Question. If $G$ is infinite, directly indecomposable, and $|G_{>2}|<|G|$, is $G$ necessarily commutative? REPLY [5 votes]: Here is some thougths on infinite finitely generated groups. Let $sq\colon G\to G$ be the square function mapping $g$ to $g^2$. Hence the set of elements of order at most $2$ is equal to $sq^{-1}(1)$. Then, as proved in other answers, if $G-sq^{-1}(1)$ is small, then $G$ is abelian. In the specific case of finitely generated groups, more can be said. Observe that such groups are countable, and hence "small" equal finite in this context. If $G$ is finitely generated, then $sq(G)$ is infinite as soon as $G$ is infinite. In particular, if there exists $F$ a finite subset of $G$ such that $G-sq^{-1}(F)$ is finite, then $G$ is finite. If $G$ is finitely generated and there exists $F$ a finite subset of $G$ and $g\in G$ such that $G-\bigl(sq^{-1}(F)\cup gsq^{-1}(F)\bigr)$ is finite, then $G$ is virtually abelian. In this setting, it is not possible to conclude that $G$ is abelian. Counter-examples include the infinite dihedral group (with $F=\{1\}$) and generalized dicyclic group $Dic(A,x)$ (with $F=\{x^2\}$ not containing $1$). The proof of 1. consists of applying carefully Dicman’s Lemma, which says that if $N$ is a finite normal subset of $G$ consisting of torsion elements, then $\langle N\rangle$ is a finite normal subgroup. The proof of 2. is more convoluted and use arguments on random walks due to Tointon. See https://arxiv.org/abs/2010.06020 for the details and some related statements.<|endoftext|> TITLE: What non-standard model of arithmetic does Hofstadter reference in GEB? QUESTION [17 upvotes]: Following some of the coolest bits of Hofstadter's Gödel, Escher, Bach, extensions of the standard model of arithmetic are described. A ways in, the paragraph "Supernatural Addition and Multiplication" mentions: ... It turns out that you can "index" the supernaturals in a simple and natural way by assocating with each supernatural number a trio of ordinary integers... [e.g. (9,-8,3)] ... Under some indexing schemes, it is very easy to calculate the index triplet for the sum of two supernaturals, given the indices of the two numbers to be added. Under other indexing schemes, it is very easy to calculate the index triplet for the product of two supernaturals, given the indices of the two numbers to be multiplied. But under no indexing scheme is it possible to calculate both. -- Douglas Hofstadter; Gödel, Escher, Bach; Chapter XIV The description is mystifying me and also seems to have mystified at least one Wikipedia talk page, where you can also find a fuller quote. Having done some basic digging on this, I'm confused. It looks like what's being discussed is a countable non-standard model of arithmetic. However, the "index triplet" isn't something I can find reference to anywhere. Is there a plausible candidate for such a "simple and natural" indexing? The theorem referenced sounds a lot like Tennenbaum's theorem, but that theorem says more strongly that neither addition nor multiplication is computable (cf discussion here). Is there some other candidate for what theorem is being referred to here? An old Math Forum post has some hints, but I don't have access to the linked articles at the moment. REPLY [10 votes]: My first guess is that the triples come from the fact that nonstandard countable models of PA look like $$\mathbb N + \mathbb Z\times\mathbb Q$$ and elements of $\mathbb Q$ can be represented by pairs. As for the separate computability of $+$ (addition) and $\cdot$ (multiplication), I am speculating but perhaps the idea is something like this: Let the theories $T_1$ and $T_2$ be obtained from $\mathrm{Th}(\mathbb N,+)$ and $\mathrm{Th}(\mathbb N,\cdot)$, respectively, by adding axioms saying that a new constant symbol $c$ is "infinite". Then $T_1$ and $T_2$ each have computable models. [Computable model just means that the operations $+$ and $\cdot$ are computable functions.] For $T_1$ you could add axioms $\varphi_n$ expressing $c\ge n$ as $$\exists x(\underbrace{1+\dots+1}_{n}+x=c)$$ For $T_2$, you could say that there are at least $n$ primes that divide $x$. But in $T_2$ you do not get a linear ordering, so I don't know whether using triples is sensible for models of $T_2$.<|endoftext|> TITLE: Two graphs with the same number of walks but without a common equitable partition QUESTION [5 upvotes]: Consider two undirected graphs $G$ and $H$ of the same order (same number of vertices). If $G$ and $H$ have a common equitable partition, then it is known (see e.g., Chapter 6 in 1) that these graphs must have the same number of walks of any length. (after all, they have the same iterated degree sequence.) I am interested in finding two graphs of the same order which have the same number of walks of any length (i.e., they have the same main eigenvalues and main angles); yet do not have a common equitable partition; and have a different number closed walks of a certain length (i.e. they are not cospectral). Any such an example would be appreciated. 1. Edward R. Scheinerman and Daniel H. Ullman. Fractional Graph Theory: a Rational Approach to the Theory of Graphs. John Wiley & Sons, 1997. https://www.ams.jhu.edu/ers/wp-content/uploads/sites/2/2015/12/fgt.pdf. REPLY [2 votes]: It appears that one way to obtain a desired example pair $G$ and $H$ is to let $G$ be the disjoint union $G_1\cup G_2$ and $H$ be the disjoint union of $H_1\cup H_2$, such that $G_1$ and $H_1$ have a common equitable partition, yet are not co-spectral; and $G_2$ and $H_2$ have the same number of walks of any length, yet they have no common equitable partition. Clearly, the resulting graphs $G$ and $H$ will not be co-spectral and similarly won't have a common equitable partition. Nevertheless, the property of that $G_1$ and $H_1$, and $G_2$ and $H_2$, have the same number of walks of any length is preserved by the union. To be more explicit: For $G_1$ and $H_1$ one can take, e.g., $C_6$ (cycle of length $6$) and $2C_3$ (two disjoint triangles). These graphs are known not to be co-spectral and have a common equitable partition (see e.g., Fig. 3 in [1]). For $G_2$ and $H_2$ one can just take the pair of co-spectral graphs with co-spectral complements given in Fig. 4 in [2]. That is, $G_2$ is $C_6\cup K_1$ (cycle of length $6$ with an additional isolated vertex) and $H_2$ is a star with one central vertex from which $3$ paths of length 2 start. Co-spectral graphs with co-spectral complements are known to have the same number of walks of any length (see Theorem 3 in [3]). Furthermore, since $G_2$ has an isolated vertex whereas $H_2$ does not, they cannot have a common equitable partition (as their degree sequences must coincide.) This results in graphs $G$ and $H$ consisting of 13 vertices. Perhaps smaller examples can be found. [1] Fractional isomorphism of graphs, M. V.Ramanaa, E. R.Scheinerman and D. Ullman, Vol. 132, Issues 1–3, pp. 247-26, 1994. https://doi.org/10.1016/0012-365X(94)90241-0 [2] Enumeration of cospectral graphs, W. H. Haemers and E. Spence, European Journal of Combinatorics, Vol. 5, Issue 2, pp. 199-211, 2004. https://doi.org/10.1016/S0195-6698(03)00100-8 [3] Cospectral graphs and the generalized adjacency matrix, E.R. van Dam, W.H. Haemers and J.H. Koolen, Linear Algebra and its Applications, Vol. 423, pp. 33–41. 2007.https://doi.org/10.1016/j.laa.2006.07.017<|endoftext|> TITLE: Is an open subscheme of a rationally connected variety, rationally connected? QUESTION [8 upvotes]: Let $X$ be a projective, irreducible variety over an algebraically closed field (of characteristic zero) which is rationally connected. Is it true that any open dense subvariety of $X$ is rationally connected? If so, can we say that same for rationally chain connectedness? By rationally (chain) connected we do not assume properness (similar to the definition in Kollar's "Rational curves on algebraic varieties"). We simply ask that two general points in the variety are connected by a (chain of) rational curve. Any idea/reference is most welcome. REPLY [10 votes]: If I have understood the OP's definitions correctly, the answer is no for rational chain connectedness. Let $E$ be a smooth genus $1$ curve in $\mathbb{P}^2$ and let $X \subset \mathbb{P}^3$ be the cone on $E$, with vertex $x_0$. For any $x_1$ and $x_2$ in $X$, the lines joining $x_1$ to $x_0$ and $x_0$ to $x_2$ are in $X$, so $x_1$ and $x_2$ are connected by a chain of rational curves. However, let $Y = X \setminus \{ x_0 \}$. Let $\pi : Y \to E$ be the projection. For any rational curve $C \subset Y$, the map $\pi : C \to E$ maps a genus $0$ curve to a genus $1$ curve, so it is constant. Thus, all rational curves of $Y$ lie in fibers of $\pi$, and points in different fibers of $\pi$ cannot be connected by a chain of rational curves. Disclaimer: I'm not very familiar with the literature on rational connectedness, so I might have misguessed a definition.<|endoftext|> TITLE: higher Casimirs for $\mathfrak{sl}$ QUESTION [7 upvotes]: The Wikipedia universal enveloping algebra suggest a way to obtain higher Casimir operators (e.g. generators of the center of $\mathfrak{U(g)}$ for $\mathfrak{g}$ semisimple) by evaluating certain determinant: $$ \det\, (t\mathrm{I} - \mathrm{ad}_X) = \sum_{i=0}^{\dim \mathfrak{g}} p_i(X) t^i $$ However if one does that for $\mathfrak{sl}_3$ then the resulting polynomial has only degrees (in $t$) 0,1,2,4,6,8 and it's coefficients seem to be just powers of the quadratic Casimir operator. If one tries to do the same for the defining representation $\mathbb{C}^3$ (replacing $\mathrm{ad}_X$ by $\rho(X)$ in the above formula) then one obtains quadratic as well as cubic invariant polynomials. In the proof of Harish-Chandra isomorphism as presented e.g. in (1) there is construction of elements of $Z(\mathfrak{U(g)}$ using traces of matrices from representations of $\mathfrak{g}$. Something like $$ \sum\mathrm{tr}(\rho(X_{i_1})\rho(X_{i_2})\ldots \rho(X_{i_n}))X_{i_1}^*X_{i_2}^*\ldots X_{i_n}^* $$ where $X_i$ form basis for $\mathfrak{g}$ and $X_i^*$ form dual basis with respect to Killing form. Q1: What is going on here? Q2: Is it true that for a semi-simple complex Lie algebra and it's smallest nontrivial representation one obtains in this way all generators of the $Z(\mathfrak{U(g)}$? Q3: Does the approach through determinant give the same operators that appear in the proof of the H-Ch isomorphism? (1) Cohomological Induction and Unitary Representations by Knapp, Vogan REPLY [3 votes]: For $\mathfrak{sl}(3)$ the coefficients are not powers of the Casimir, though the polynomial is one in the square of $t$.<|endoftext|> TITLE: Unbounded version of continuous functional calculus QUESTION [5 upvotes]: For a normal operator $T$ on a Hilbert space ${\cal H}$, it is well known that for any continuous complex valued function $f$ on the spectrum of $T$, we have a well-defined operator $f(T) \in B({\cal H})$. I have heard that this extends to the unbounded case, but I can't find a precise statement. So if $D$ is a densely-defined unbounded normal operator on ${\cal H}$, does the unbounded version of functional calculus work for any continuous function on the spectrum of $D$? REPLY [6 votes]: You find the precise abstract statment in the Internet Seminar lecture notes of Markus Haase, especially in Chapter 4. See https://www.math.uni-kiel.de/isem21/en/course/phase1 By the way, the answer is yes.<|endoftext|> TITLE: How big can the index inside the root lattice of the lattice generated by a subset of roots be? QUESTION [6 upvotes]: Let $\Phi$ be an irreducible crystallographic root system in a Euclidean vector space $V$. Let $S\subseteq \Phi$ be some subset of roots for which $\mathrm{Span}_{\mathbb{R}}(S)=V$. Question: How big can $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]$ be? If $\Phi$ is of Type A, then I believe we always have $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]=1$ because of total unimodularity. But for instance, if $\phi=D_4$ and $S=\{\alpha_1,\alpha_1+2\alpha_2+\alpha_3+\alpha_4,\alpha_3,\alpha_4\}$, where $\alpha_2$ corresponds to the trivalent node, then I get $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]=2$. (With the standard realization of $D_4$ this is $S=\{(1,-1,0,0),(1,1,0,0),(0,0,1,-1),(0,0,1,1)\}$.) In particular I'd like to know if $[\mathrm{Span}_{\mathbb{Z}}(\Phi):\mathrm{Span}_{\mathbb{Z}}(S)]$ is absolutely bounded or not. REPLY [4 votes]: As suggested in my second comment, one approach to the answer is via Borel–de Siebenthal theory, which classifies the subgroups $H$ that can arise as in my first comment as those whose Dynkin diagrams arise by deleting a root, other than the lowest root, from the extended Dynkin diagram. (Deleting the lowest root always gives back the original group.) The desired index, which is the same as the order of the centre of the resulting group $H$, is the coefficient of the deleted vertex in the highest root, hence is absolutely bounded (by 6, which occurs only in case $\Phi$ is of type $\mathsf E_8$).<|endoftext|> TITLE: monochromatic subset QUESTION [8 upvotes]: Suppose we have $n^2$ red points and $n(n-1)$ blue points in the plane in general position. Is it possible to find a subset $S$ of red points such that the convex hull of $S$ does not contain any blue points, where $|S|>n^{\epsilon}$? REPLY [7 votes]: By coloring the Horton set with two colors, periodically mod 3 according to the $x$-coordinate, Devillers et al. obtained arbitrarily large bicolored point sets with no monochromatic empty convex $5$-gon (that is, monochromatic $5$-hole). Using the fact that every set of 10 points in general position in the plane contains a $5$-hole, the bicolored Horton set has no monochromatic $10$-island (a set of 10 points of the same color such that their convex hull contains no other points of the Horton set). So for sufficiently large $n$ the answer is no. See http://www.eurogiga-compose.eu/posezo/horton_set for some additional references about the Horton set. Edit: The Horton sets have size $2^m$, and the coloring by Devillers et al. has one third of the points red and two thirds blue. Take $m$ such that $2^m>3n^2$. To get $n^2$ red and $n(n-1)$ blue points, we take a second copy of the same Horton set of size $2^m$, with one third of the points blue and two thirds red, and place the two copies side by side, so that their convex hulls are disjoint. Then we cut off an appropriate number of points from the left copy ($n^2-2n$ if $n$ is divisible by $3$) and the right copy ($n^2+n$ if $n$ is divisible by $3$), by two vertical lines (or, in addition, remove a constant number of other points from the right side of the left copy or from the left side of the right copy to solve some divisibility issues). This will keep the maximum size of a monochromatic island constant (at most 19).<|endoftext|> TITLE: When is the category of models of a limit theory a topos? QUESTION [15 upvotes]: If $\mathcal{E}$ is a Grothendieck topos on a small base, then it is locally presentable, and hence is equivalent to the category of models of some limit theory. Is there a characterization of limit theories $\mathcal{T}$ such that the category of models of $\mathcal{T}$ is a topos? The category of models of a $\mathcal{T}$ is locally presentable and hence a reflective subcategory of presheaves on $\mathcal{T}$. If there is a characterization of these theories, is it automatic that the reflector is finitely continuous, and hence gives the theories that underly Grothendieck toposes? I know that Johnstone characterizes the product theories whose models are toposes. Here $\mathcal{T}$ must have no pseudoconstants and be sufficiently unary. I guess I'm curious if anyone has generalized this result. REPLY [14 votes]: My collaborator Julia Ramos González and I are working on this question precisely in these days. A part of the answer is already cointained in a paper by Carboni, Pedicchio and Rosický: Syntactic characterizations of various classes of locally presentable categories, Journal of Pure and Applied Algebra, 161 (2001) pp 65-90. Putting together Thm. 5 and 19 one gets that: A finitely presentable category is a Grothendieck topos if and only if the full subcategory of finitely presentable objects is extensive and pro-exact. Please, read also the paragraph that comments Thm. 19. Recall that the full subcategory of finitely presentable objects is essentially the limit theory that presents the locally (finitely) presentable category, i.e. $$\mathcal{K} \cong \text{Lex}(\text{Pres}(\mathcal{K})^{\circ}, \text{Set}). $$ 3 March 2019. As I was mentioning in the previous version of this answer, together with Julia, we worked on a generalization of this statement to the infinitary case and related the site-theoretic presentation with the limit-theory presentation. The result of this investigation, Gabriel-Ulmer duality for topoi and its relation with site presentations, is now on the arXiv:1902.09391.<|endoftext|> TITLE: Codimension-1 subgroups of 3-manifold groups QUESTION [5 upvotes]: Let $G$ be a finitely generated group and let $H$ be a subgroup of $G$. $H$ is a codimension-1 subgroup of $G$ if $C_{G}/H$ has more than one end, where $C_{G}$ is the Cayley graph of $G$. Do all codimension-1 subgroups of a $3$-manifold group correspond to the fundamental group of an immersed surface in the $3$-manifold? REPLY [5 votes]: Let me make some remarks on this. As far as I know, the terminology "codimension-1 subgroup" originated from a paper of Micah Sageev Sageev, Michah, Ends of group pairs and non-positively curved cube complexes, Proc. Lond. Math. Soc., III. Ser. 71, No. 3, 585-617 (1995). ZBL0861.20041. MR1347406. A subgroup $H< G$ of a finitely generated group $G$ is codimension-1 if the number of ends $e(G,H) > 1$. If $\Gamma$ is a Cayley graph for $G$, then $e(G,H)$ is the number of ends of the graph $\Gamma/H$. In this case, Sageev calls the group "semi-splittable". Then the main theorem of the paper is Theorem Suppose $G$ is a finitely generated group. Then $G$ acts essentially on a cubing if and only if $G$ is semi-splittable. A cubing is a complete non-positively curved (or CAT(0)) cube complex (where "cube" refers to what is more commonly known as a "hypercube"). If it is finite-dimensional, then "essential" is equivalent to the existence of an unbounded orbit. In a CAT(0) cube complex, one has "hyperplanes" which intersect each cube in a codimension-1 cube (setting one coordinate to = 0, if a cube is parameterized as $[-1,1]^n$) or the empty set. Now the stabilizer of a hyperplane in the theorem gives rise to a codimension-1 subgroup of $G$, and conversely, a codimension-1 subgroup will stabilize a hyperplane in Sageev's construction. Note that the construction of the cube complex from the subgroup is not canonical, but relies on some choices (these choices have been codified in the notion of "wall spaces"). Hopefully this helps explain the choice of terminology. Now, assume $G$ is the fundamental group of a connected closed irreducible 3-manifold, and assume that $G$ has a codimension-1 subgroup (in particular, $G$ is infinite). Then $G=\pi_1(M)$, where $M$ is an aspherical 3-manifold by the sphere theorem. Corresponding to a codimension-1 subgroup $H1$, one has that $N$ has more than one end. Hence $H_2(N;\mathbb{Z}) \neq 0$. Choose a non-zero homology class $z\in H_2(N)$, then we may find an embedded closed surface $\Sigma \subset N$ such that $[\Sigma]=z$ (meaning that the map induced by the inclusion of the fundamental class of $\Sigma$ into $H_2(N)$ is $z$). If $\Sigma$ is not $\pi_1-$injective, then we may use the loop theorem to compress $\Sigma$ to get a surface of larger Euler characteristic. So by induction, we may assume that (each component of) $\Sigma$ is incompressible and hence is $\pi_1-$injective. If $\Sigma$ is not connected, we may take a component which is still homologically non-trivial. This surface cannot be a sphere because $N$ is also irreducible (again by the sphere theorem). Thus, $\Sigma$ is a $\pi_1-$injective surface. Corollary 5.3 in Sageev's paper states a slightly weaker version of this result. So we may associate to $H$ a $\pi_1$-injective surface, but in a highly non-canonical way. This argument is due to Swarup. Conversely, any surface subgroup of an aspherical 3-manifold group is codimension-1 (and actually may be realized by an embedding in the corresponding cover by a theorem of Freedman-Hass-Scott).<|endoftext|> TITLE: The probability that two elements of a finite nonabelian simple group commute QUESTION [7 upvotes]: It is mentioned in here (last paragraph of the first page) that Dixon proved the following result: the probability that two elements of a finite nonabelian simple group commute is at most $\frac{1}{12}$. I cannot seem to find this proof anywhere online. Do you know a proof of this fact? Or do you have a reference for it? (I am actually more interested in the original proof) REPLY [8 votes]: You can find Dixon's argument here (Google Books). (J.D..Dixon, Solution to Problem 176, Canadian Mathematical Bulletin 16 (1973), p302.)<|endoftext|> TITLE: Frobenius coordinate expansion of character QUESTION [5 upvotes]: Let $\lambda$ be the partition of integer $d$. The Frobenius coordinate of $\lambda$ is given $$ (a_1 ,\ldots,a_{d(\lambda)}|b_1,\ldots,b_{d(\lambda)}),$$ where $d(\lambda)$ denote the diagonal of $\lambda$. Let $a_i'= a_i +\frac12$ and $b_i'= b_i +\frac12$ are modified Frobenius coordinates. By a classical theorem of Frobenius, it states that $f(\lambda)=\frac{|C_{2,1,\ldots,1}|}{dim \lambda}\chi_{2,1,\ldots,1}^{\lambda}=\frac12\sum_{i=1}^{d(\lambda)}(a_i')^2-(b_i')^2$ Notice that the above example we treat the L.H.S as a function of $\lambda$ as we fix the partition of the form $2,1^{d-2}$ If there is any generalisation of the above theorem? That is if we replace $(2,1,\ldots,1)$ with any other partition of $d$ then say $\mu$ then what will be $$\frac{|C_{2^{d/2}}|}{dim \lambda}\chi_{2^{d/2}}^{\lambda}$$ For example say $\mu$ is of the form $2^{d/2}$ in Frobenius coordinate? If it is already studied can anyone please site the reference. REPLY [3 votes]: There is a generalisation of the character formula, although it is usually stated in terms of contents rather than Frobenius coordinates. Also, it applies to conjugacy classes labelled by partitions $(1^{m_1} 2^{m_2} \cdots)$, where $m_2, m_3, \ldots$ are fixed and $m_1$ varies with $n$ (as in the example you gave, where there is one part of size $2$, and $n-2$ parts of size 1). So in particular, it does not apply directly to the partitions $(2^{n/2})$. For the benefit of the uninitiated, let me state right away that $\frac{|C_\mu| \chi_{\mu}^\lambda}{dim(\lambda)}$ is a central character. The sum of all elements of $S_n$ of cycle type $\mu$ is a central element of the group algebra of $S_n$, and the central character is the scalar by which the central element acts on the irreducible representation labelled by $\lambda$. Central characters are important in modular representation theory, which might provide a little motivation for why the quantity in question is a natural thing to consider. Suppose that we have a partition $\lambda = (\lambda_1, \ldots, \lambda_l)$. Suppose that $d$ is the largest integer such that $\lambda_d \geq d$ (the Durfee size of $\lambda$). Then the Frobenius coordinates of $\lambda$ are $(\alpha_1, \ldots, \alpha_d | \beta_1, \ldots, \beta_d)$, where $\alpha_i = \lambda_i - i$ and $\beta_i = \lambda_i^\prime - i$ (here $\lambda^\prime$ is the transpose partition of $\lambda$). So the $\alpha_i$ and $\beta_i$ count the number of boxes in past the diagonal in the $i$-th row and $i$-th column respectively (and the number of each is $d$). On the other hand, the contents of a box in row $j$ and column $i$ is $i-j$. It is not difficult to see that the multiset of contents of $\lambda$ consists of $0$ with multiplicity $d$, and $\{1,2,\ldots, \alpha_i\}$ for each $i$ together with $\{-1, -2, \ldots, -\beta_i\}$ for each $i$. So in particular, the sum of $r$-th powers of the contents of $\lambda$ can be expressed as polynomials in the $\alpha_i$ and $\beta_i$ by Faulhaber's Formula. For example, the sum of the contents ($r=1$) is $$\sum_i (1 + 2 + \cdots + \alpha_i) - (1 + 2 + \cdots + \beta_i) = \sum_i {\alpha_i + 1 \choose 2} - {\beta_i + 1 \choose 2},$$ which is precisely the quantity appearing in the example you gave. We write $cont(\lambda)$ for the multiset of contents of $\lambda$. Now, for a partition $\mu = (\mu_1, \mu_2, \cdots, \mu_k)$, let $\bar{\mu} = (\mu_1 - 1, \mu_2 - 1, \ldots, \mu_k - 1)$ be the partition obtained from $\mu$ by subtracting 1 from each nonzero part of $\mu$ (and ignoring trailing zeros). For example, if $\mu = (2,1, \ldots, 1)$ as in your example, then $\bar{\mu} = (1)$. Theorem There exists a family $f_{\rho}$ of elements of $\mathbb{Q}[n] \otimes \Lambda$ (where $n$ is a polynomial variable) and $\Lambda$ is the ring of symmetric functions with the property that $$\frac{|C_\mu| \chi_{\mu}^\lambda}{dim(\lambda)} = f_{\bar{\mu}}(|\lambda|, cont(\lambda))$$ By this notation I mean that we evaluate $f_{\bar{\mu}}$ at $n=|\lambda|$ and its symmetric function variables at $cont(\lambda)$. It turns out that $f_{(1)} = p_1$ (the first power-sum symmetric function) which doesn't actually depend on $n$. Then putting this together, we see $$\frac{|C_{(2,1,\ldots,1)}| \chi_{(2,1,\ldots,1)}^\lambda}{dim(\lambda)} = p_1(cont(\lambda)) = \sum_i {\alpha_i + 1 \choose 2} - {\beta_i + 1 \choose 2},$$ which recovers the result you stated. In general, you can take $f_{\bar{\mu}}$, express it as a polynomial in $n$ and the power-sums $p_1, p_2, \ldots \in \Lambda$, and upon evaluating at the contents of $\lambda$ you will obtain a polynomial in $n$ and the Frobenius coordinates of $\lambda$. For example, $$f_{(2)} = p_2 - {n \choose 2},$$ which tells us that the central character corresponding to the conjugacy class $(3,1,\ldots, 1)$ may be computed using the formula $$\left( \sum_{i} \frac{\alpha_i(\alpha_i + 1)(2\alpha_i + 1)}{6} + \frac{\beta_i (\beta_i + 1)(2\beta_i + 1)}{6} \right) - {|\lambda| \choose 2}.$$ There is a lot more to be said about the $f_\rho$. For example, they form a $\mathcal{R}$-basis of $\mathcal{R} \otimes \Lambda$, wher $\mathcal{R}$ is the subring of $\mathbb{Q}[n]$ consisting of integer-valued polynomials. But in this post I worked over $\mathbb{Q}$ because I wanted to use power-sum symmetric functions to turn contents-evaluaion into polynomials in Frobenius coordinates. References you might find helpful (each of which contains an explanation of the above theorem): Corteel, Sylvie; Goupil, Alain; Schaeffer, Gilles, Content evaluation and class symmetric functions, Adv. Math. 188, No. 2, 315-336 (2004). ZBL1059.05104. Section 5.4 of Ceccherini-Silberstein, Tullio; Scarabotti, Fabio; Tolli, Filippo, Representation theory of the symmetric groups. The Okounkov-Vershik approach, character formulas, and partition algebras., Cambridge Studies in Advanced Mathematics 121. Cambridge: Cambridge University Press (ISBN 978-0-521-11817-0/hbk). xv, 412 p. (2010). ZBL1230.20002. (although this is based on the material of the above paper) Section 3 of my own paper (which doesn't appear in the citation engine, so I'm providing an arXiv link instead) Stable Centres I: Wreath Products<|endoftext|> TITLE: Manifolds with negative dimension – Definition, References QUESTION [6 upvotes]: Does the concept of differential manifold with negative dimension make sense, in differential geometry? If yes, how is it defined? Do you have any reference to recommend? My problem was born in Einstein warped-product manifolds, i.e. to admit a type of metric, the fiber-manifold must have a negative dimension, but I do not know if that manifold makes sense in differential geometry, General Relativity or String Theory… Thank you for any help REPLY [5 votes]: Smooth manifolds of negative dimension are defined in derived geometry. Recall that if A→M and B→M are two transversal submanifolds of codimension a and b respectively, then their intersection C is again a submanifold, of codimension a+b. Derived geometry explains how to remove the transversality condition and make sense out of a nontransversal intersection C as a derived smooth manifold of codimension a+b. In particular, dim C = dim M - a - b, and the latter number can be negative. See Spivak, "Derived Smooth Manifolds". "Simplicial approach to derived differential manifolds" by Borisov and Noel simplifies Spivak's foundations considerably.<|endoftext|> TITLE: Partitions, $q$-polynomials and generating functions QUESTION [6 upvotes]: Recall the integer partition function $P(n)$ with generating function $$\sum_{n\geq0}P(n)x^n=\prod_{k=1}^{\infty}\frac1{1-x^k}.$$ Let $[n]_q=\frac{1-q^n}{1-q}$ denote the $q$-analogue of the integer $n$ and let $\lambda=(\lambda_1,\lambda_2,\dots)\vdash n$ be a partition of $n$. Now, define the function $$\Psi_q(n)=\sum_{\lambda\vdash n}\,\,\sum_{j\geq1}\,\, [\lambda_j]_q.$$ For example, $\Psi_q(3)=[3]_q+([2]_q+[1]_q)+([1]_q+[1]_q+[1]_q)=q^2+2q+6$. In particular, when $q=1$, we obtain $\Psi_q(n)=nP(n)$ with generating function $$\sum_{n\geq1}nP(n)x^n=x\frac{d}{dx}\prod_{k=1}^{\infty}\frac1{1-x^k}.$$ I would like to ask Question 1. Is there a generating function for the sequence of $q$-polynomials $\Psi_q(n)$? Question 2. What is the combinatorial interpretation of the coefficients of $\Psi_q(n)$? For example, the constant term of $\Psi_q(n)$ enumerates the number of $1$'s in all partitions of $n$. REPLY [3 votes]: A generating function of sorts is given by $$ \sum_{n\geq 1}\Psi_q(n)x^n = P(x)\sum_{m\geq 0}q^m\sum_{k\geq m+1} \frac{x^k}{1-x^k}, $$ where $P(x)=\prod_{i\geq 1}(1-x^i)^{-1}$.<|endoftext|> TITLE: Convolution with semigroup: does this belong to the Sobolev space $W^{1,1}$? QUESTION [6 upvotes]: Let $X$ be a Banach space, $T(t)$ be a strongly continuous semigroup on $X$, and $f\in L^1(0,\tau;X)$. It has been implied that the integral $$v(t)=\int_0^t T(t-s)f(s)ds,\quad t\in [0,\tau]$$ is not always an element of $W^{1,1}(0,\tau;X)$. That seems odd to me. Can anyone think of an example? REPLY [8 votes]: In other words, if $A$ is the infinitesimal generator of $T$, the mild solution of the abstract inhomogeneous Cauchy problem $$\begin{cases}\dot v =A v +f\\v(0)=0\end{cases}$$ needs not to be $W^{1,1}_{loc}(\mathbb{R}_+, X)$. For instance an $f\in C^0(\mathbb{R}_+,X)$ of the form $f(t):=T(t)x$ for some $x=x(\theta)\in X$, gives $v(t)=tT(t)x=tf(t)$ which has no reason to be in $W^{1,1}_{loc}(\mathbb{R}_+, X)$. To justify the preceding claim it is sufficient to show that $f(t)$ itself, the mild solution to the homogeneous Cauchy problem $\dot f=Af$ with $f(0)=x$, may fail to be $W^{1,1}$ (even at any point). Consider e.g. $X:=L^1(\mathbb{S}^1)$, $1$-periodic one variable $L^1_{loc}$ functions; $T :\mathbb{R}_+\times X\to X$ the left translation semigroup $T(t)x:=x(\cdot+t)$, whose infinitesimal generator is $A:=\partial_\theta$, with domain $D(A)=W^{1,1}(\mathbb{S}^1)$; Then, for $x\in X$, $f(t):=T(t)x=x(\cdot+t)$ is, of course, the mild solution to $\dot f(t)=\partial_\theta f(t)$ with initial data $f(0)=x$, and defines a continuous path $f:\mathbb{R}\ni t \mapsto x(\cdot+t)\in X$. Saying, for some open interval $I$, that $f\in W^{1,1}(I;X)$ means there is $h\in L^1(I;X)\sim L^1(I\times \mathbb{S}^1)$ such that $f(t')-f(t)=\int_t^{t'} h(s,\cdot)ds$ in $X$, that is $x(\theta+t')-x(\theta+t)=\int_t^{t'} h(s,\theta)ds$ a.e., whence $x\in W^{1,1}(\mathbb{S}^1)$.<|endoftext|> TITLE: Radial symmetry of the first eigenfunction QUESTION [7 upvotes]: Let $M$ be a simply connected space form (i.e. $\mathbb R^n$, sphere, or hyperbolic space) and $B$ be a ball in $M$. Let $\phi$ be the first Laplacian eigenfunction on $B$, with respect to the Dirichlet boundary condition $\phi=0$ on $\partial B$. In Cheng's remarkable paper "Eigenvalue comparison theorems and its geometric applications", it is remarked that (p. 290) "since all simply connected space forms are two-point homogeneous, $\phi$ is a radial function". I think two-point homogeneity means that there always exists an isometry mapping one pair of points to any other pair of points which have the same distance apart (is it?). I don't understand how this condition is used. I think the reason why $\phi$ is radial is that we can apply the spherical mean of it to obtain $\widetilde \phi$, which is also an eigenfunction as rotation preserves the metric. If $\widetilde \phi\ne \phi$, then $h=\phi-\widetilde \phi$ is a first eigenfunction that changes sign, which is a contradiction. But I don't see where the "two-point homogeneous condition" is used. (Is there an alternate proof?) If my argument is correct it should also apply to balls centered at 0 of the warped product space $dr^2+f(r)^2 d\theta^2$. Is this well-known? If so, is there a quick reference? REPLY [3 votes]: For me it is a consequence of the Alexandrov moving plane method. In the euclidean case you have the following result due to Gidas, Ni and Nirenberg, see Theorem 2.34 of the book of Han and Lin, Elliptic Partial Differential equations, Theorem: Suppose $u\in C^2(B)\cap C(\overline{B})$ is a positive solution of $$\Delta u =f(u) $$ with $$u=0 \hbox{ on } \partial B .$$ If $f$ is Lip then $u$ is radially symettic. The proof rely on the moving plane method which consists in applying the maximum principle to some "reflected" solution. Hence, Cheng say that as soon as you can build such a re flexion between any pair of points the result is still true. Of course it works only for the first eigenvalue since, you need the positivity. REPLY [2 votes]: The fact that the first eigenfunction is radial follows from the uniqueness (up to scale) of the first eigenfunction, that the eigenvalue equation is invariant under isometries fixing $B$ and (1) the fact that the isotropy group at the orgin of the ball is transitive. Fact (1) is shown below. Now if $\Delta_g u = \lambda_1 u$ with $\|u\|_{L^2(B)}=1$ and $\psi:B\to B$ is an isometry then $\tilde u (y) := u(\psi(y))$ satisfies the same equation and its $L^2$-norm is the same. Since the eigenspace is one-dimensional this means $u=\tilde u$. Because the isotropy group at the center $x$ of $B$ acts transitive on each $\partial B_s(x)\subset B=B_r(x)$ it holds $u(y)=u(z)$ for all $y,z\in B$ with $d_g(y,x)=d_g(z,x)$. This shows $u$ must be radial. The whole argument does not require linearity of the equation only a form of invariance under scaling is needed (i.e. if $u$ is a solution so is $\alpha u$). In such a case it suffices to show that the Rayleigh quotient is invariant under isometries and that the space of solution is one-dimensional. An example where this argument applies is the first eigenfunction of the p-Laplace equation $\operatorname{div}(|\nabla|^{p-2}\nabla u) = \lambda_1 u $. However, if the equation is linear then via averaging one can show that for each eigenvalue there is a radial solution. (1) The fact that $B=B_r(x)$ is a ball in a two-point homogeneous space $(M,g)$ means that for all $y,z \in B$ with $d_g(y,x)=d_g(z,x)$ there is an isometry $\psi$ of $M$ with $\psi(x)=x$ and $\psi(y)=\psi(z)$. Since $x$ is fixed and $\psi$ is distance preserving it holds $\psi(B)=B$, i.e. $\psi$ fixes the ball.<|endoftext|> TITLE: Isomorphic equivariant sheaves are equivariantly isomorphic on a toric variety QUESTION [6 upvotes]: Let $X$ be a toric variety containing the $n$-torus $T\overset{i}{\hookrightarrow} X$. The action of $T$ extends naturally to an action on the sheaf $i_*\mathcal{O}_T$ by $$(\alpha\cdot f)(x):=f(\alpha^{-1}x) \; \; \forall \, f\in (i_*\mathcal{O}_T)(U),\forall \alpha\in T$$ on each invariant open set $U\subseteq X$. Now, given a coherent $\mathcal{O}_X$-module $\mathcal{F}\leq i_*\mathcal{O}_T$ we say that it is equivariant if we can restrict the above action of $T$ to $\mathcal{F}$. I whant to know if the following is true Two equivariant sheaves $\mathcal{F_1}$ and $\mathcal{F_2}$ are isomorphic as $\mathcal{O}_X$-modules if and only if they are isomorphic as $T$-equivariant $\mathcal{O}_X$-modules, i.e, there is an isomorphism commuting with the action of $T$. This appear to be a consequence of Theorem 9.II in Chapter 1 of the book Toroidal Embeddings I of Mumford et al. but it can as well be just a typo (there is no proof for that part). REPLY [7 votes]: Edit. The short answer is that this fails for every toric variety except when $T$ equals all of $X$. I edited the original answer to prove this, and also to prove the positive result below by @S. carmeli. Denote by $X^{(1)}$ the finite set of irreducible components of the closed subset $X\setminus T$ of $X$. Definition 1. A principal $T$-invariant Weil divisor is an element in the kernel $\text{PDiv}^T(X)$ of the following map to the Weil divisor class group of $X$, $$ 0 \to \text{PDiv}^T(X) \to \bigoplus_{D\in X^{(1)}}\mathbb{Z}\cdot [D] \to \text{Cl}(X). $$ Every $T$-invariant Weil divisor of $\text{PDiv}^T(X)$ is a principal Cartier divisor whose associated invertible sheaf is isomorphic to $\mathcal{O}_X$ as an $\mathcal{O}_X$-module. Moreover, since every ideal sheaf $\mathcal{O}_X(-\underline{D})\subset \mathcal{O}_X\subset i_*\mathcal{O}_T$ has a natural $T$-linearization, also the invertible sheaf of every element of $\text{PDiv}^T(X)$ has a natural $T$-linearization. Definition 2. For every principal $T$-invariant Weil divisor, the character of each $T$-equivariant generator of the associated invertible sheaf is independent of the choice of generator up to $T$-equivariant units on $X$. The character of the principal $T$-invariant Weil divisor is the image under the associated group homomorphism, $$ \text{char}_X: \text{PDiv}^T(X) \to \text{Hom}_{k-\text{GpSch}}((T,1),(\mathbb{G}_{m,k},1))/\text{Hom}_{k-\text{Sch}}((X,1),(\mathbb{G}_{m,k},1)), $$ where we make all of the schemes pointed by distinguishing the group identity in $T\subset X$. Lemma 3. The group homomorphism $\text{char}_X$ is an isomorphism. Proof. Every character $$\chi:(T,1)\to (\mathbb{G}_{m,k},1),$$ is a regular morphism from $T$ to $\mathbb{A}^1_k$, and thus it determines (and is uniquely determined by) a rational function on $X$. The associated principal divisor of $\chi$ is an element of $\text{PDiv}^T(X)$ whose image under $\text{char}_X$ equals the image of $\chi$. Thus $\text{char}_X$ is surjective. For a principal $T$-invariant Weil divisor whose image under $\text{char}_X$ is trivial, there exists a $T$-equivariant generator of the invertible sheaf whose character equals the character of a unit on $X$. Scaling the generator by the inverse of the unit, there exists a $T$-equivariant generator whose character is trivial. Thus, the principal $T$-equivariant divisor is the divisor of a $T$-equivariant rational function whose restriction to $T$ is the constant rational function $1$. Since $T$ is schematically dense in $X$, the rational function is everywhere equal to $1$. The principal divisor of the rational function $1$ is the zero divisor. QED Proposition 4. For every pair $(\mathcal{F}_1,\mathcal{F}_2)$ of nonzero, $T$-invariant, coherent subsheaves of $i_*\mathcal{O}_T$ that are isomorphic as $\mathcal{O}_X$-modules, there exists a unique divisor $D\in \text{PDiv}^T(X)$ such that $\mathcal{F}_1$ equals $\mathcal{F}_2(\underline{D})$ as subsheaves of $i_*\mathcal{O}_T$. For every character $\chi$ of $T$ that maps to $\text{char}(D)$, the $T$-linearized $\mathcal{O}_X$-modules $\mathcal{F}_1$ and $\mathcal{F}_2\otimes \chi$ are isomorphic. Proof. Via adjointness of $i_*$ and $i^*$, there is a natural equivalence between the set of nonzero, equivariant, coherent subsheaves of $i_*\mathcal{O}_T$ and the set of equivalence classes of pairs $(\mathcal{F},\phi)$ of a torsion-free, coherent $\mathcal{O}_X$-module $\mathcal{F}$ together with an isomorphism $$\phi:i^*\mathcal{F}\xrightarrow{\cong} \mathcal{O}_T$$ that admits a $T$-linearization. Since $X$ is normal, every such pair is of the form $(\mathcal{I}\cdot \mathcal{G},\psi|_{\mathcal{I}\cdot \mathcal{G}})$ where $(\mathcal{G},\psi)$ is such a pair with $\mathcal{G}$ reflexive, and where $\mathcal{I}$ is the ideal sheaf of a $T$-invariant closed subscheme $Z$ of $X$ that everywhere has codimension $\geq 2$ (the subscheme $Z$ is empty if $\mathcal{F}$ is already reflexive). Every $\mathcal{O}_X$-module isomorphism from $\mathcal{F}_1$ to $\mathcal{F}_2$ induces an isomorphism between the reflexive hulls $\mathcal{G}_1$ and $\mathcal{G}_2$ as well as between the ideal sheaves $\mathcal{I}_1$ and $\mathcal{I}_2$, so that also $Z_1$ equals $Z_2$ as closed subschemes of $X$. Therefore, $\mathcal{F}_1$ is isomorphic to $\mathcal{F}_2$ if and only if $\mathcal{G}_1$ is isomorphic to $\mathcal{G}_2$ and $Z_1$ equals $Z_2$. Thus, without loss of generality, assume now that $\mathcal{F}_1$ and $\mathcal{F}_2$ are reflexive. For reflexive pairs $(\mathcal{F}_1,\phi_1)$ and $(\mathcal{F}_2,\phi_2)$, the composition $\phi_{1,2}:=\phi_1^{-1}\circ \phi_2$ is an equivariant isomorphism from $i^*\mathcal{F}_2$ to $i^*\mathcal{F}_1$. For every prime Weil divisor $D$ of $X$ contained in $X\setminus T$, both $\mathcal{F}_1$ and $\mathcal{F}_2$ are locally free of rank $1$ on a neighborhood of the generic point of $D$. Thus, the valuation of $\phi_{1,2}$ at $D$ is well-defined. Summing over all such prime Weil divisors gives a Weil divisor associated to $\phi_{1,2}$, $$\text{Div}(\phi_{1,2}) = \sum_{D\subset X\setminus T} \text{val}_D(\phi_{1,2}) \underline{D}.$$ The pushforward from the smooth locus of $X$ of the invertible sheaf of this Cartier divisor is a reflexive, coherent $\mathcal{O}_X$-module of rank $1$, say $\mathcal{O}_X(\text{Div}(\phi_{1,2}))$. Moreover, this $\mathcal{O}_X$-module has a natural $T$-linearization as a $T$-equivariant subsheaf of $i_*\mathcal{O}_T$. Since they are reflexive, the $\mathcal{O}_X$-modules $\mathcal{F}_1$ and $\mathcal{F}_2$ are isomorphic if and only if they are isomorphic away from codimension $2$, and this holds if and only if $\mathcal{O}_X(\text{Div}(\phi_{1,2}))$ is isomorphic to $\mathcal{O}_X$, i.e., if and only if $\text{Div}(\phi_{1,2})$ is an element of $K$. In this case, $\mathcal{F}_1$ is isomorphic to $\mathcal{F}_2(\text{Div}(\phi_{1,2}))$ as subsheaves of $i_*\mathcal{O}_T$, and thus also as $T$-linearized $\mathcal{O}_X$-modules. By the definition of $\text{char}(\text{Div}(\phi_{1,2}))$, for every character $\chi$ of $T$ mapping to $\text{char}(\text{Div}(\phi_{1,2})$, the $T$-linearized invertible sheaf $\mathcal{O}_X(\text{Div}(\phi_{1,2}))$ is isomorphic to the $T$-linearized invertible sheaf $\mathcal{O}_X\otimes_k \chi$. Thus, $\mathcal{F}_1$ is isomorphic to $\mathcal{F}_2\otimes \chi$ as a $T$-linearized, $\mathcal{O}_X$-module. QED Corollary 5. For a toric variety $(X,T)$, the subgroup $\text{Hom}_{k-\text{Sch}}((X,1),(\mathbb{G}_{m,k},1))$ equals all of $\text{Hom}_{k-\text{GpSch}}((T,1),(\mathbb{G}_{m,k},1))$ if and only if $T$ equals all of $X$. Thus, for every toric variety such that $T$ is a proper open subset of $X$, there exists a pair $(\mathcal{F}_1,\mathcal{F}_2)$ of nonzero, $T$-invariant, coherent subsheaves of $i_*\mathcal{O}_T$ that are isomorphic as $\mathcal{O}_X$-modules yet are not isomorphic as $T$-linearized, $\mathcal{O}_X$-modules. Proof. Since $X$ is normal and $T$ is an open affine, the complement of $T$ in $X$ has pure codimension $1$. Thus, if $T$ is a proper open subset of $X$, then there exists a prime Weil divisor $D$ that is an irreducible component of $X\setminus T$. Since $X$ is normal, the local ring of $X$ at the generic point of $D$ is a discrete valuation ring; denote the associated valuation by $$\text{val}_D:k(T)^\times/k^\times\twoheadrightarrow \mathbb{Z}.$$ Since $D$ is $T$-invariant, the valuation is uniquely determined by its values on the monomials of $k[T]$. If every monomial has zero valuation, then the valuation is zero. Thus, some monomial has nonzero valuation. Up to replacing the monomial by its inverse, assume that the monomial has negative valuation. Then the monomial is an element of $\text{Hom}_{k-\text{GpSch}}((T,1),(\mathbb{G}_{m,k},1))$ that is not in $\text{Hom}_{k-\text{Sch}}((X,1),(\mathbb{G}_{m,k},1))$. QED For any particular toric variety, it is straightforward to compute $\text{Hom}_{k-\text{Sch}}((X,1),(\mathbb{G}_{m,k},1))$, $\text{PDiv}^T(X)$, and $\text{char}_X$, thus classifying all counterexamples. For instance, for $X=\mathbb{A}^n_k$, resp. $\mathbb{P}^n_k$, with its natural structure of toric variety, the group $\text{Hom}_{k-\text{Sch}}((X,1),(\mathbb{G}_{m,k},1))$ is trivial, so that the group homomorphism $\text{char}$ is an isomorphism to the character group of $T$, $$\text{char}_{\mathbb{A}^n_k}:\text{PDiv}^T(\mathbb{A}^n_k)\xrightarrow{\cong} \text{Hom}_{k-\text{GpSch}}((T,1),(\mathbb{G}_{m,k},1)),$$ $$\text{char}_{\mathbb{P}^n_k}:\text{PDiv}^T(\mathbb{P}^n_k)\xrightarrow{\cong} \text{Hom}_{k-\text{GpSch}}((T,1),(\mathbb{G}_{m,k},1)).$$ Here are the simplest counterexamples. Let $X$ equal $\mathbb{A}^1_k = \text{Spec}\ k[s]$, and let $T$ equal $D(s)$ with its natural inclusion $i$, and let the group operation on $T$ be the standard one, $$s_1\bullet s_2 = s_1s_2.$$ The sheaf homomorphism $i^\#$ makes $\mathcal{O}_X$ into a $T$-equivariant sheaf. On the level of rings, the injective $k[s]$-module homomorphism is $$k[s]\hookrightarrow k[s,s^{-1}].$$ For every integer $d\geq 0$, consider the ideal $$I_d = \langle f_d \rangle \subset k[s], \ \ f_d = t^d.$$ As a $k[s]$-module, this is principal, and the set of principal generators is just the set of multiples of $f_d$ by elements of $k^\times$. In particular, this module is isomorphic to the ring as a module, so that all $\mathcal{O}_X$-modules $\widetilde{I}_d$ are isomorphic. Since $f_d$ is a monomial, the ideal sheaf $\widetilde{I}_d \subset \mathcal{O}_X \subset i_*\mathcal{O}_T$ is equivariant. Since $\widetilde{I}_d$ is principal, it is isomorphic to $\mathcal{O}_X$ as an $\mathcal{O}_X$-module. Under every $k[s]$-module isomorphism, the generator $f_d$ of $I_d$ maps to a generator of $k[s]$, i.e., to a scalar multiple of the generator $f_0=1$. Scalar multiples of $1$ are (nonzero) $T$-invariant elements of $k[s]$. Thus, $I_d$ is equivariantly isomorphic to $k[s]$ if and only if the generator $f_d$ is a $T$-invariant element. The coordinate $s$ on $\mathbb{A}^1_k$ identifies the $k$-algebra $\mathbb{A}^1_k(\text{Spec}\ k)$ with $k$, and that identification extends to an identification of the group $T(\text{Spec}\ k)$ with the multiplicative group $k^\times$. For every $a\in k^\times$, $$(a\cdot f_d)(s) = f_d(as) = a^d s^d = a^d f(s).$$ Thus, the character of $f_d$ is $d$ times the tautological character of $T$. Therefore, if $d$ is nonzero, the generator $f_d$ is not $T$-invariant. Even though the equivariant subsheaf $\widetilde{I}_d\subset \mathcal{O}_X \subset i_*\mathcal{O}_T$ is isomorphic to $\mathcal{O}_X$ as a $\mathcal{O}_X$-module, there is no equivariant $\mathcal{O}_X$-module isomorphism of $\widetilde{I}_d$ with $\mathcal{O}_X$.<|endoftext|> TITLE: The Disco Ball Problem QUESTION [14 upvotes]: Let me first give some of a background as to where I got this problem. I had a math teacher ask me a few months ago: "How many 1 unit by 1 unit squares could one fit on a sphere with a radius of 32 units?" The only problem was that it was a high school math course, and that I, as a high school student, had no way of answering the question in its true sense, as the teacher simply wanted a division of surface area. But the problem remains unanswered, and it has been bothering me for some time now. How many mirrors can you fit on a disco ball? This isn't a question for the construction of a disco ball simply the construction of a theoretical "perfect" disco ball given the parts. My teacher didn't realize that the problem answer she was expecting was for an entirely different problem relating to how many areas of 1 unit squared you could create from the surface. Given my severely limited math knowledge, I am simply unable to answer a problem as complex as this. Here is the problem in its entirety: If you have a sphere of radius $r$ and squares of length $l$, then how many squares can you fit on it in tangent with the sphere with only the centers of each square touching the sphere, and with no square overlap? This is a disco ball so the parameters get a bit tight: The squares can only be attached by their centers The squares can't overlap The squares can't go into the sphere The squares can't be broken or bent (don't want bad luck) The squares go on the outside of the sphere I don't know what math this would take to solve this, but I know that it has a solution lurking in the voids of some imagination if so gifted to me. What this question might mean to a mathematician: In this given situation the square mirrors could be interchanged with spheres without any adverse effects to the problem. So where this problem might find a solution could be in the already well-defined field of sphere packing, the only caveat being the fact that it's packing with set radii ratios and where non-touching spheres are discarded. REPLY [2 votes]: There are two interpretations of "no overlap". One is suggested by the original problem, in that two mirror tiles do not physically intersect, while still having their centers attached to the sphere. For a sphere of radius 1/2 times 1/$\sqrt{3}$, there are at least two ways to glue three unit tiles for this definition of overlap: one where the tiles form a triangular prism, and one where one tile is given a 45 degree twist. Both of these do not overlap in the first sense, but the second one fails to not overlap in a second sense, where spherical projections of each tile are supposed to be disjoint. While both problems are of interest, it would be good to resolve this ambiguity and focus on just one version. This is one reason why I think the original version was not good for this forum: the problem is not clear enough. I encourage the poster to think more and revise the problem further toward greater clarity. I took the trouble to make some computations for the first sense of overlap and small numbers of tiles. I suspect that the radii computed are the smallest that can accommodate the given number of unit square tiles, but the amount of effort to prove it for any number greater than two tiles is substantial. I invite readers to contribute a brief proof even for $t=3$, and remember to use overlap in the first sense. $$t = (1,2,3, 4, 5, 6, 12, 18)$$ $$r= 1/2 (\epsilon,\epsilon,1/\sqrt{3}, 1/\sqrt{\phi},1,1,\sqrt{3},1+ \sqrt{2})$$ For larger values of t the two senses of overlap seem to converge, and that the smallest radius having a disco ball of $t$ tiles in either sense approach each other fairly rapidly. However, as there is still work to do on packing unit squares into large squares of nonintegral side length, I hesitate to declare any results on this problem as other than temporary. Gerhard "Do Not Overlap With Ambiguity" Paseman, 2018.09.19.<|endoftext|> TITLE: What's the point of "created limits"? QUESTION [7 upvotes]: Seeing as there's an nLab page about creation of limits, I take it that at least some people think this is an important notion. There's also a discussion here about what the "correct" definition of this notion is. However, and maybe this is just my ignorance speaking, but I haven't actually found a reason to worry about this notion, so I basically just ignore it. I mean, okay, a diagram $D$ that lacks a limit in one category $\mathbf{C}$ might gain a limit in a category with more objects, or arrows, or both. More generally, given a functor $F : \mathbf{C} \rightarrow \mathbf{D}$, the composite $F \circ D$ might or might not have a limit, quite independently of whether or not $D$ does. But so what? I don't quite get what the notion "creation of limits" really teaches us. Question. Can someone who finds this to be a helpful notion explain what they find helpful about it? Is there, for example, a theorem about creation of limits that's ubiquitous but impossible to see without this notion? Is there an important definition that can't be elegantly stated unless creation of limits is invoked? What's the point of this idea? REPLY [9 votes]: The main point of “creating limits” is things like: Theorem. If $C$ has all limits of shape $I$ and $F : D \to C$ creates such limits, then $D$ has all such limits and $F$ preserves and reflects them. In other words, the definition is pretty much designed to be a tractable and concrete solution to the implication “(limits in $C$) + ???? $\Rightarrow$ (limits in $D$, and preservation/reflection)” Pragmatically, it isolates the structure common to the construction/analysis of limits in groups, topological spaces, algebras for monads, and many other categories of “structured objects over some category where limits are already understood”. It does require constructing the limits in $D$ anyway, but it guides you to a particular tractable way of constructing them, and then shows that by constructing them this way, you get preservation + reflection for free. Regarding the two different definitions of creation of limits, see Displayed Categories, Ahrens, Lumsdaine 2017, arXiv:1705.04296 for my personal (maybe slightly idiosyncratic) take on why the standard definition using equality of objects really is very reasonable and natural.<|endoftext|> TITLE: When a localization of a category is (non-)reflective? QUESTION [7 upvotes]: Let $S$ be a class of maps of a category $\mathcal{C}$. The localization of $\mathcal{C}$ at $S$ is reflective when the localization functor $\mathcal{C} \to \mathcal{C}[S^{-1}]$ has a fully faithful right adjoint. Question 1: Are there any general conditions on $S$ which imply that $\mathcal{C}[S^{-1}]$ is reflective? Suppose that $\mathcal{C}$ is locally presentable and that $S$ is a small set. Then the full subcategory $\mathcal{C}$ consisting of $S$-local objects is reflective. Let $\widetilde{S}$ be the class of $S$-local maps. Then $S \subseteq \widetilde{S}$ and $\mathcal{C}[S^{-1}]$ is reflective if and only if $\mathcal{C}[S^{-1}] = \mathcal{C}[\widetilde{S}^{-1}]$. Question 2: What are natural/useful examples of non-reflective localizations $\mathcal{C}[S^{-1}]$ (where $\mathcal{C}$ is locally presentable and $S$ is a set)? The localization functor $\mathcal{C} \to \mathcal{C}[\widetilde{S}^{-1}]$ satisfies a modified version of the universal property of $\mathcal{C}[S^{-1}]$. Namely, we require all functors in the definition of a localization to be left adjoints. This definition makes sense for an arbitrary category $\mathcal{C}$ and an arbitrary class of maps $S$. Question 3: Is there a name for functors satisfying this universal property? Does it appear in the literature? I want to call such a functor a reflective localization of $\mathcal{C}$ at $S$, but this terminology clashes with the ordinary notion of a reflective localization. Let us call it a quasireflective localization at $S$. Every reflective localization at $S$ is a quasireflective localization at $S$ by this proposition, but a quasireflective localization at $S$ (if it exists) is a reflective localization at $\widetilde{S}$ rather than at $S$. Thus, we can ask a weaker version of Question 2: Question 2': What are examples of quasireflective localizations at $S$ which are not reflective localizations at $S$. REPLY [2 votes]: My answer comments questions Q1 and Q2 and is connected to the fact that Fosco gave an answer to the case in which $\mathcal{K}[S^{-1}]$ is not concrete. I will try to cover the other case, when $\mathcal{K}[S^{-1}]$ is concrete. I claim that this is a very strong requirement on $S$. When a cocomplete category is concrete, quite often it is also locally presentable (Top is the most famous counterexample to this statement, which is still quite true). Thus, imagine to have a locally presentable $\mathcal{K}[S^{-1}]$ for which such a faithful right adjoint exist. $$L: \mathcal{K} \leftrightarrows \mathcal{K}[S^{-1}]: i $$ Since both the categories are locally presentable, $i$ must be accessible for a big enough cardinal (Thm 2.23 LPAC). Thm. An accessibly embedded reflective subcategory of a locally presentable category $\mathcal{A} \hookrightarrow \mathcal{K}$ is always of the form $\mathcal{K}[S^{-1}]$ for a small set $S$. This result appears as Thm 1.39 in Locally Presentable and Accessible categories. This means that quite often, the requirement for the existence of the faithful right adjoint is a strong requirement on the category $\mathcal{K}[S^{-1}]$, which in fact has to be equivalent to $\mathcal{K}[\bar{S}^{-1}]$ where $\bar{S}$ is (essentially) small. May I conclude even that $\bar{S} \subset S$? Probably yes, in concrete examples $\bar{S}$ can even be chosen to lie inside $S$, this means that most of the information contained in $S$ is superfluous. This is due to the structure of the proof of 1.39. All in all, putting together this answer with the other, the motto (which has no ambition of being a theorem) is: either $S$ is small or $\mathcal{K}[S^{-1}]$ is too big. Of course, this super-general motto, has many counterexample. The easiest one is when $\mathcal{K}$ is connected and $S$ is everything, in this case $\mathcal{K}[S^{-1}] \cong 1$, which is a quite small category. Thus, one should be careful when applying this motto, that still in real life examples will turn out to be quite true. This argument can even be strengthened under Vopenka principle. In fact, under Vopenka every reflective subcategory of a locally presentable category must be locally presentable (Cor 6.24 LPAC), thus the illegitimate upgrade from concrete to locally presentable comes for free.<|endoftext|> TITLE: About a canonical model structure on topologically enriched categories QUESTION [6 upvotes]: Consider the discrete combinatorial model structure on the category of $\Delta$-generated spaces: all maps are cofibrations and fibrations and the weak equivalences are the homeomorphisms. Applying Proposition A.3.2.4 of (Higher Topos Theory, Lurie) By mimicking the construction of the canonical model structure on the category of small categories, one obtains a model structure on the category of topologically enriched small categories and enriched functors such that the weak equivalences $F:C\to D$ are the local homeomorphisms such that the underlying functor $F_0:C_0\to D_0$ is essentially surjective. Like in the non-enriched case, is this model structure unique (I mean for this class of weak equivalences) ? Is it "canonical" in this sense ? EDIT: The discrete model structure on the category of $\Delta$-generated spaces is only accessible, not combinatorial because the underlying category is not locally finitely presentable. REPLY [6 votes]: Yes, the canonical model structure is unique. The uniqueness of the canonical model structure on $Cat$ was nicely exposited by Chris Schommer-Pries on the Secret Blogging Seminar back in the day. Let's go through and mimic the proof there. Explicit description of the canonical model structure: Let $(C,W,F)$ denote the cofibrations, weak equivalences, and fibrations of the canonical model structure. The cofibrations are generated by $\emptyset \to pt$, along with $\Sigma X \to \Sigma Y$ for every map $X \to Y$ in $Top_\Delta$, where $\Sigma X$ is the $Top_\Delta$-enriched category with two objects $0,1$ and $Hom(0,0) = Hom(1,1) = pt$, $Hom(0,1) = X$, and $Hom(1,0) = \emptyset$. Therefore $C$ consists of the injective-on-objects enriched functors. $C \cap W$ consists of the injective-on-objects enriched equivalences. So it is generated by an inclusion $pt \to E$ where $E$ is the walking isomorphism. $W$ consists of the enriched equivalences. $W \cap F$ consists of the surjective-on-objects enriched equivalences. $F$ consists of the isofibrations. Uniqueness of this model structure: Let $(C',W,F')$ be the cofibrations, weak equivalences, and fibrations of a model structure on $Cat_{Top_\Delta}$ with the same weak equivalences as the canonical model structure. We'll use "cofibration", "$\hookrightarrow$", "fibration", and $\twoheadrightarrow$ in the sense of this model structure, and "canonical cofibration" and "canonical fibration" to refer to the notions from the canonical model structure. $\emptyset \hookrightarrow pt \in C'$. For there must be some nonempty cofibrant object $X$, and then $0 \to pt$ is a retract of $\emptyset \hookrightarrow X$. $C' \supseteq C$, $C' \cap W \supseteq C \cap W$, $F' \cap W \subseteq F \cap W$, $F' \subseteq F$. For since $\emptyset \to pt \in C'$, we have that every $ f \in F' \cap W$ is surjective on objects. Since $f$ is also an enriched equivalence. we have $f \in F \cap W$, i.e. $F' \cap W \subseteq F \cap W$. The other inclusions follow formally. If $C' \neq C$, then $E \to pt \in C' \cap W$ (where again $E = (0 \cong 1)$ is the walking isomorphism). If $C' \neq C$, then by (2), $C' \not \subseteq C$, so there is a cofibration $A \hookrightarrow B \in C' \setminus C$ which is not injective on objects. Pick $x,y \in A$ which map to the same object in $B$. There is a unique enriched functor $A \to E$ sending $x$ to 0 and $y$ to 1. Then by pushout we obtain a cofibration $E \hookrightarrow E \cup_A B$ sending 0 and 1 to the same point. The codiscretification map $E \cup_A B \hookrightarrow (E\cup_A B)^\flat$ is injective on objects, i.e. a canonical cofibration, and hence a cofibration. Composing, we obtain a cofibration $E \hookrightarrow E \cup_A B \hookrightarrow (E\cup_A B)^\flat$ sending 0 and 1 to the same point, and sending the isomorphism between them to the identity. So $E \hookrightarrow pt$ is a retract of this map and hence also a cofibration. Since $E \to pt \in W$ is an enriched equivalence of categories, the claim follows. If $C \neq C'$, then every fibrant object $X$ has no non-identity isomorphisms. $X \to pt$ must lift against $E \to pt$ by (3). $C = C'$, i.e. the model structures agree. Every object must be equivalent to a fibrant object. But enriched equivalences preserve the property of having objects with non-identity automorphisms. So because there exist $Top_\Delta$-enriched categories with objects with non-identity automorphisms, this contradicts (4) if $C \neq C'$. This argument generalizes to any cartesian enriching category $V$ such that there exist objects in $V$-categories with nontrivial automorphisms. I'd like to say this includes any cartesian enriching category which is not a poset, but I'm not quite sure.<|endoftext|> TITLE: Is the evaluation map from harmonic forms on the torus surjective on flat neighbourhoods? QUESTION [5 upvotes]: In a nutshell: Given a metric on the torus $\mathbb{T}^n$, can we extend any element $\sigma \in \bigwedge^k T_p^*\mathbb{T}^n$ to a global harmonic form? Let $\mathbb{T}^n$ be the $n$-Torus. Fix $1 TITLE: Connections between martingales and Fourier analysis QUESTION [12 upvotes]: I have had this strange feeling recently that somehow, the theory of martingales we study in probability, and the theory of Fourier analysis are very alike. But I am not able to formalize my thoughts. To illustrate, let us focus on $f\in L_1(\mathbb T,\mathcal B_{\mathbb T},Leb)$. Define $S_N(f)(x)=\sum_{k=-N}^N \hat{f}(k)e^{ikx}$, the $N$-th partial sum of the Fourier series of $f$. It seems to me that $\{S_N(f)\}_N$ is essentially like a random walk, as the increment $\hat{f}(N)e^{iNx}+\hat{f}(-N)e^{-iNx}$ at the $N$-th stage has mean $0$, and it is orthogonal to $\{1,e^{\pm x},...,e^{\pm (N-1)x}\}$, which would perhaps translate to independence and thus it seems plausible that $\{S_N(f)\}_N$ is a martingale. Of course, independence, conditional expectations, etc. are probably not meaningful terms in classical analysis, and this is one thing I am unable to formalize. Now let me come to the convergence results. It is well known that in martingale theory, for $p>1$, an $L_p$-bounded martingale converges a.s. and in $L_p$. In Fourier analysis, we have the result that $L_p$ norm convergence of $S_N(f)$ to $f$ (if $f\in L_p$) is equivalent to $\sup_N ||S_N||<\infty$ where $S_N:L_p\to L_p$ is being treated as a linear operator. So here we have the analogy with $L_p$ bounded martingales. Further, for $p>1$, we have $S_N(f)\to f$ a.e., again in analogy. The results for both Fourier series and martingales fail when $p=1$ and you need more conditions like uniform integrability of martingales, which, I think, translates to $\sum_n |\hat{f}(n)|<\infty$. I also do not know what happens to this little jump from 1 to $p$ that makes both these two results work/fail. I can feel these two theories go parallelly but it seems quite mysterious to me. Maybe there is a connection which I cannot see? REPLY [5 votes]: To add to what Yemon wrote in his comment, and to give a concrete reference: there is some known connection giving Martingales based proofs of Littlewood-Paley type inequalities. This is discussed, for example, in PAUL-ANDRÉ MEYER Démonstration probabiliste de certaines inégalités de Littlewood-Paley Séminaire de probabilités (Strasbourg), tome 10 (1976) (There are a total of five exposes, all available on Numdam. The first is http://www.numdam.org/item?id=SPS_1976__10__125_0 .)<|endoftext|> TITLE: Has dynamics on $G/\Gamma$ ever been used to prove interesting things about $\Gamma$? QUESTION [13 upvotes]: Fix a Lie group $G$ and a discrete subgroup $\Gamma \subset G$. Homogeneous dynamics is about studying the actions of subgroups $H \subset G$ on the quotient $G/\Gamma$. Does anyone know of an example of a question about $\Gamma$ that was answered by considering these dynamics? REPLY [5 votes]: Kahn and Markovic solved the surface subgroup problem and Ehrenpreis conjecture making use of exponential mixing of the geodesic flow on compact hyperbolic manifolds. The geodesic flow may be thought of as homogeneous dynamics on a diagonal subgroup $H< G$ acting on $G/\Gamma$, although the proof by Cal Moore only refers to the unit tangent bundle (which is where the geodesic flow lives). At least in the case of $PSL_2(\mathbb{R})$ (for hyperbolic surfaces), this is the same thing. In turn, the surface subgroup problem tells us many more interesting things about $\Gamma$ when $\Gamma$ is a closed hyperbolic 3-manifold group.<|endoftext|> TITLE: Combinatorial constructions found by computer QUESTION [16 upvotes]: In preparation for a talk I am giving to our undergraduate mathematics society, I am trying to collect examples of combinatorial constructions that were found by computer. Thus my question is the following. What are the most notable examples of combinatorial constructions found by computer search? For example, one fertile area for such results is lower bounds on Ramsey numbers, for example this recent paper by Exoo and Tatarevic. (But since I am looking for constructions, computer-assisted upper bounds on Ramsey numbers don't count.) Another non-example would be Aubrey de Grey's construction of a unit-distance graph on 1581 vertices with chromatic number 5 (thereby showing that the chromatic number of the plane is at least 5), since the graph was constructed by hand. REPLY [2 votes]: Emmanuel Jeandel, Michael Rao, An aperiodic set of 11 Wang tiles, https://arxiv.org/abs/1506.06492<|endoftext|> TITLE: Ostrowski's Theorem for topological rings? QUESTION [16 upvotes]: Ostrowski's theorem classifies all absolute values on a number field $K$. Questions: More generally, can one classify all Hausdorff topologies on $K$ making $K$ into a topological field? In particular, is every Hausdorff topology on a number field $K$ making $K$ into a topological field induced by an absolute value? It would already be interesting to understand this when $K= \mathbb Q$. On the other hand, I'd be interested to understand this question for more general fields and rings as well. For "large" fields / rings, I imagine one might need to consider valuations in more general value groups as well. But I don't know a generally-accepted definition of "archimedean valuation" not over $\mathbb R$, so I'm not quite sure how to formulate a potentially-correct statement saying that "every topology comes from a generalized absolute value" in this context. REPLY [4 votes]: The following relevant classification result, due to Kowalsky and Dürbaum [2], appears in Appendix B of Engler and Prestel [1]. Let $(K,\tau)$ be a topological field. Then $\tau$ is called a V-topology if for every neighbourhood $W\ni0$, there exists a neighbourhood $U\ni0$ such that $(K\smallsetminus W)(K\smallsetminus W)\subseteq K\smallsetminus U$ (that is, for any $x,y\in K$, if $xy\in U$, then $x\in W$ or $y\in W$). Theorem: The V-topologies on a given field are exactly the topologies induced by valuations (with arbitrary value groups) or by archimedean absolute values. Note that nonarchimedean absolute values are also covered, being special cases of valuations (with, confusingly enough, value groups that are archimedean, i.e., rank 1). In the special case of global fields (including number fields), all valuations have rank 1, i.e., they are equivalent to nonarchimedean absolute values (e.g., see Thm. 2.1.4 and Cor. 3.2.5 in [1]). Thus: Corollary: If $K$ is a global field, the topologies on $K$ induced by absolute values are exactly the V-topologies. References: [1] Antonio J. Engler and Alexander Prestel, Valued fields, Springer, 2005. [2] Hans-Joachim Kowalsky and Hansjürgen Dürbaum, Arithmetische Kennzeichnung von Körpertopologien, Journal für die reine und angewandte Mathematik 191 (1953), 135­–152.<|endoftext|> TITLE: Distribution of sum of two permutation matrices QUESTION [7 upvotes]: Determinant and permanent of sum of two $n\times n$ permutation matrices can be arbitrarily different. What is the distribution of determinant of sum and difference of two $n\times n$ permutation matrices? What is the distribution of permanent of sum and difference of two $n\times n$ permutation matrices? How much do the distributions differ? REPLY [5 votes]: I will abuse notation by identifying a permutation and the matrix it represents. We can denote by $E(\sigma), O(\sigma)$ the number of even and odd cycles that $\sigma$ decomposes into. Given two permutations $\sigma_1,\sigma_2$ we can compute the following: $$\det(\sigma_1+\sigma_2)=\left\{ \begin{array}{ll} (-1)^{E(\sigma_1)}2^{O(\sigma_1\sigma_2^{-1})} & \mbox{if } E(\sigma_1\sigma_2^{-1})=0 \\ 0 & \mbox{otherwise } \end{array} \right. $$ $$\operatorname{per}(\sigma_1+\sigma_2)=2^{E(\sigma_1\sigma_2^{-1})+O(\sigma_1\sigma_2^{-1})}$$ $$\operatorname{per}(\sigma_1-\sigma_2)=\left\{ \begin{array}{ll} 2^{E(\sigma_1\sigma_2^{-1})} & \mbox{if } O(\sigma_1\sigma_2^{-1})=0 \\ 0 & \mbox{otherwise } \end{array} \right. $$ and trivially $\det(\sigma_1-\sigma_2)=0$ since the vector of all 1's is always in the kernel of $\sigma_1-\sigma_2$. These calculations follow from noticing that the matrices decompose as direct sums of smaller matrices corresponding to each cycle of $\sigma_1\sigma_2^{-1}$. Distributions of cycle statistics like these are easy to obtain with the exponential formula. From here we can count the number of occurrences of each value. Let's start with $\det(\sigma_1+\sigma_2)$. The exponential generating function for odd cycles (or cyclic permutatons of odd size) on $\{1,2,\dots,n\}$ is $x+\frac{x^3}{3}+\cdots=\frac{1}{2}\left(\log(1+x)-\log(1-x)\right)$. This is because there are $(n-1)!$ odd cycles when $n$ is odd, and $0$ otherwise. By the exponential formula, the generating function of permutations that consist of only odd cycles, together with a statistic $t$ that keeps track of the number of cycles, is $$e^{\frac{t}{2}\left(\log(1+x)-\log(1-x)\right)}=\left(\frac{1+x}{1-x}\right)^{\frac{t}{2}}$$ By substituting $t=2s$ we get $\left(\frac{1+x}{1-x}\right)^{s}$. The coefficient $a_{k,n}$ of the monomial $s^kx^n$ is given exactly by $\frac{1}{n!}$ times the number of permutations on $n$ letters that decompose into $k$ odd cycles and no even cycles, times a factor of $2^k$. Therefore the number of permutation pairs $(\sigma_1,\sigma_2)$ for which $\det(\sigma_1+\sigma_2)=-2^k$ is the same as the number of permutation pairs for which $\det(\sigma_1+\sigma_2)=2^k$ and is given by $\frac{(n!)^2a_{k,n}}{2}$. Here we used the fact that $(-1)^{E(\sigma)}$ is the sign of $\sigma$, and the number of permutations with sign $-1$ is the same as those with sign $+1$. For $\operatorname{per}(\sigma_1+\sigma_2)$ we are looking at $2^{\text{number of cycles}}$ over all permutations. So we start with the generating function of cycles which is $x+\frac{x^2}{2}+\cdots=-\log(1-x)$. So the exponential generating function $$e^{t(-\log(1-x))}=\frac{1}{(1-x)^t}$$ has as coefficient of $t^kx^n$ the number of permutations on $n$ letters with precisely $k$ cycles, divided by $n!$. Substituting $t=2s$ we get $\frac{1}{(1-x)^{2s}}$, and we denote by $b_{k,n}$ the coefficient of $s^kx^n$. This coefficient is equal to $\frac{1}{n!}$ times the number of permutations on $n$ letters with precisely $k$ cycles, times $2^k$. Therefore the number of permutation pairs $(\sigma_1,\sigma_2)$ with $\operatorname{per}(\sigma_1+\sigma_2)=2^k$ is exactly $(n!)^2b_{k,n}$. Finally for $\operatorname{per}(\sigma_1-\sigma_2)$ we want to look at permutations with only even cycles. The exponential generating function of even cycles is given by $\frac{x^2}{2}+\frac{x^4}{4}+\cdots=-\frac{1}{2}\log(1-x^2)$. Similarly to above the generating function $$e^{2s\left(-\frac{1}{2}\log(1-x^2)\right)}=\frac{1}{(1-x^2)^s}$$ has coefficients $c_{k,n}$ for monomials $s^kx^n$ which are equal to $\frac{1}{n!}$ times the number of permutations on $n$ letters which decompose into exactly $k$ even cycles, times $2^k$. So the number of permutation pairs $(\sigma_1,\sigma_2)$ with $\operatorname{per}(\sigma_1-\sigma_2)=2^k$ is exactly $(n!)^2c_{k,n}$. REPLY [3 votes]: If $A$ is the matrix for a permutation that is a single cycle of size $m$, then the eigenvalues of $A$ are the $m$'th roots of unity, and $\det(I+A)$ is the product of $1+\omega$ over the $m$'th roots of unity, which is $0$ if $m$ is even and $2$ if $m$ is odd. Thus for a permutation that is a product of $r$ disjoint cycles, $\det(I+A) = 0$ if any of the cycles is odd, $2^r$ if they are all even. For two permutation matrices $A$ and $B$ corresponding to permutations $\sigma$ and $\pi$, we have $\det(A+B) = \det(A) \det(I+A^{-1} B) = 0$ if $\sigma^{-1} \pi$ has any odd cycles, $2^r$ if $\sigma^{-1}\pi$ has only $r$ even cycles and $\sigma$ is even, $-2^r$ if $\sigma^{-1}\pi$ has only $r$ even cycles and $\sigma$ is odd.<|endoftext|> TITLE: is signed distance function real analytic for real analytic domains QUESTION [7 upvotes]: If $\Omega$ is a real analytic domain in $\mathbb R^n$, is the signed distance function, $f$, defined by \begin{equation} f(x)=\begin{cases}d(x,\partial \Omega )&{\mbox{ if }}x\in \Omega \\-d(x,\partial \Omega )&{\mbox{ if }}x\in \Omega ^{c}\end{cases} \end{equation} is real analytic? where \begin{equation} d(x,\partial \Omega ):=\inf _{y\in \partial \Omega }d(x,y)\end{equation} where inf denotes the infimum. Thanks in advance! REPLY [7 votes]: The answer is yes, that is $f(x)$ is real analytic in a neighborhood of any point on $\partial\Omega$. First recall that if $f$ and $g$ are real analytic functions (of several variables), then $f+g$, $f\cdot g$, $f/g$ (when $g\neq 0$), $f\circ g$, and the inverse map $f^{-1}$, if $f$ is a diffeomorphism, are all real analytic. You can find proofs in the book [1]. If $\partial\Omega$ is locally the image of a real analytic embedding $\Phi:\mathbb{R}^{n-1}\supset U\to\mathbb{R}^n$, then $N(x)$, the unit normal vector orthogonal to the image of $D\Phi(x)$ in the interior direction of $\Omega$ is also real analytic. Indeed, $D\Phi$ is real analytic and we find a normal vector by solving linear equations involving $D\Phi(x)$ so there is a real analytic normal vector $M(x)$. Possibly $M$ is not unit, but $N(x)=M(x)/|M(x)|$ is real analytic, because it is obtained from $M$ by applying to $N$ operations (listed above) that preserve analyticity. The mapping $\Psi:U\times(-\varepsilon,\varepsilon)\to\mathbb{R}^n$, $\Psi(x,t)=\Phi(x)+tN(x)$ is a real analytic diffeomorphism (if $U$ and $\varepsilon$ are small enough) so the inverse mapping $\Psi^{-1}:W\to U\times(-\varepsilon,\varepsilon)$, defined in a neighborhood of a point on $\partial\Omega$ is also real analytic. If $\pi:U\times(-\varepsilon,\varepsilon)\to(-\varepsilon,\varepsilon)$ is the projection on the $t$ component, then $\pi\circ\Psi^{-1}$ is real analytic and it remains to observe that the signed distance satisfies $$ f(x)=\pi\circ\Psi^{-1} \quad \text{in} \quad W $$ if $W$ is small. That follows immediately from the fact that the distance to the boundary is measured (near the boundary) along the normal line. [1] Krantz, S. G.; Parks, H. R. A primer of real analytic functions. Basler Lehrbücher [Basel Textbooks], 4. Birkhäuser Verlag, Basel, 1992.<|endoftext|> TITLE: Classification of closed 3-manifolds with finite first homology group? QUESTION [6 upvotes]: I am interested in a topological classification of connected closed 3-manifold $M$ that have finite homology group $H_1(M)$. Since $H_1(M)$ is the abelization of the fundamental group $\pi_1(M)$, each closed 3-manifold with finite homotopy group has finite homology group. It is known that each closed 3-manifold with finite homotopy group $\Gamma$ is a spherical 3-manifold (i.e., is the orbit space $S^3/_\sim$ of the 3-sphere, endowed with a free action of the group $\Gamma$). Question. Is each closed 3-manifold with trivial homology group a spherical 3-manifold? Equivlalently, is the fundamental group $\pi_1(M)$ of a closed 3-manifold finite if its first homology group $H_1(M)$ is finite? REPLY [6 votes]: Pick any knot in the three-sphere, and perform any Dehn surgery on it with some coefficient $p/q \neq 0$. This means that you remove the tubular neighborhood of the knot and you glue it back in a different way, parametrized by $p/q$. The manifold you get has $H_1(M,\mathbb Z) = \mathbb Z/_{p\mathbb Z}$. You get plenty of distinct 3-manifolds in this way. For instance, if the knot is hyperbolic, you get plenty of closed hyperbolic manifolds if $p$ or $q$ is sufficiently large. You can also require that $p=1$ and find plenty of homology spheres.<|endoftext|> TITLE: Is the lattice generated by finitely many subspaces in a finite-dimensional vector space finite? QUESTION [7 upvotes]: Let $V$ be a finite-dimensional vector space, let $U_1,\dots,U_n$ be subspaces, and let $L$ be the lattice they generate; namely, the smallest collection of subspaces containing the $U_i$ and closed under intersections and sums. Is $L$ finite? This is well-known to hold if $n\le3$: if $n=3$ then there are at most $28$ elements in $L$, indpendently of $V$'s dimension. Note that I suspect the answer to be "no" if $V$ is allowed to be infinite-dimensional: there exist infinite modular lattices generated by $4$ elements; here $L$ is a bit more than modular ("arguesian", see Is the free modular lattice linear?) and finiteness of free arguesian lattices doesn't seem to be known. It would be nice to have an example. REPLY [8 votes]: The answer is in fact "no", and appears in another MO post, How many subspaces are generated by three or more subspaces in a Hilbert space? : starting from four points in $P^2(\mathbb R)$, infinitely many points may be generated by intersecting lines and joining points.<|endoftext|> TITLE: Which mathematical definitions should be formalised in Lean? QUESTION [131 upvotes]: The question. Which mathematical objects would you like to see formally defined in the Lean Theorem Prover? Examples. In the current stable version of the Lean Theorem Prover, topological groups have been done, schemes have been done, Noetherian rings got done last month, Noetherian schemes have not yet been done (but are probably not going to be too difficult, if anyone is interested in trying), but complex manifolds have not yet been done. In fact I think we are nearer to perfectoid spaces than complex manifolds -- maybe because algebra is closer to the axioms than analysis. But actually we also have Lebesgue measure (it's differentiability we're not too strong at), and today we got modular forms. There is a sort of an indication of where we are. What else should we be doing? What should we work on next? Some background. The Lean theorem prover is a computer program which can check mathematical proofs which are written in a sufficiently formal mathematical language. You can read my personal thoughts on why I believe this sort of thing is timely and important for the pure mathematics community. Other formal proof verification software exists (Coq, Isabelle, Mizar...). I am very ignorant when it comes to other theorem provers and feel like I would like to see a comparison of where they all are. Over the last year I have become increasingly interested in Lean's mathematics library, because it contains a bunch of what I as a number theorist regard as "normal mathematics". No issues with constructivism, the axiom of choice, quotients by equivalence relations, the law of the excluded middle or anything. My impression that most mathematicians are not particularly knowledgeable about what can actually be done now with computer proof checkers, and perhaps many have no interest. These paragraphs are an attempt to give an update to the community. Let's start by getting one thing straight -- formalising deep mathematical proofs is extremely hard. For example, it would cost tens of millions of dollars at least, i.e. many many person-years, to formalize and maintain (a proof is a computer program, and computer programs needs maintaining!) a complete proof of Fermat's Last Theorem in a theorem prover. It would certainly be theoretically possible, but it is not currently clear to me whether any funding bodies are interested in that sort of project. But formalising deep mathematical objects is really possible nowadays. I formalised the definition of a scheme earlier this year. But here's the funny thing. 15 months ago I had never heard of the Lean Theorem Prover, and I had never used anything like a theorem prover in my life. Then in July 2017 I watched a live stream (thank you Newton Institute!) of Tom Hales' talk in Cambridge, and in particular I saw his answer to Tobias Nipkow's question 48 minutes in. And here we are now, just over a year later, with me half way through perfectoid spaces, integrating Lean into my first year undergraduate teaching, and two of my starting second year Imperial College undergraduate students, Chris Hughes and Kenny Lau, both much better than me at it. The links are to their first year undergraduate projects, one a complete formal proof of Sylow's theorems and the other an almost finished formalization of the local Langlands conjectures for abelian algebraic groups over a p-adic field. It's all open source, we are writing the new Bourbaki in our spare time and I cannot see it stopping. I know many people don't care about Bourbaki, and I know it's not a perfect analogy, but I do care about Bourbaki. I want to know which chapters should get written next, because writing them is something I find really good fun. But why write Bourbaki in a computer language? Well whether you care or not, I think it's going to happen. Because it's there. Whether it happens in Lean or one of the other systems -- time will tell. Tom Hales' formal abstracts project plans to formalise the statements of new theorems (in Lean) as they come out -- look at his blog to read more about his project. But to formalise the statements of hard theorems you have to formalise the definitions first. Mathematics is built on rigorous definitions. Computers are now capable of understanding many more mathematical definitions than they have ever been told, and I believe that this is mostly because the mathematical community, myself included, just didn't ever realise or care that it was happening. If you're a mathematician, I challenge you to formalise your best theorem in a theorem prover and send it to Tom Hales! If you need hints about how to do that in Lean, come and ask us at the Lean Zulip chat. And if if it turns out that you can't do it because you are missing some definitions, you can put them down here as answers to this big list question. We are a small but growing community at the Lean prover Zulip chat and I am asking for direction. REPLY [3 votes]: One of the outstanding problems in finite element analysis is how to come up with good sets of basis functions to discretize mixed problems. By mixed problems I mean partial differential equations that can be derived through minimization of some convex functional, but with a constraint. A classic example is the Stokes problem, which describes the flow of very viscous fluids: find a velocity field $u$ in a nice domain $\Omega$ that minimizes the free energy dissipation $$J(u) = \int_\Omega\left(\mu\nabla u : \nabla u - f\cdot u\right)dx,$$ where $\mu$ is the viscosity and $f$ the body forces on the fluid, with the added constraint of incompressibility: $$\nabla\cdot u = 0.$$ This can be expressed instead through finding a critical point of the Lagrangian $$L(u, p) = J(u) - \int_\Omega p\nabla\cdot u\, dx$$ where $p$ is the pressure field. For unconstrained problems, we can triangulate the domain and take the solution $u$ to be a polynomial of pretty much any degree in each triangle and everything converges nicely as you refine the mesh. Constrained problems like Stokes are much harder! The discretization for the velocity and pressure fields have to be carefully chosen in order to get a non-singular finite-dimensional linear system. The most common choice is to take the velocity to be piecewise quadratic and the pressure to be piecewise linear in each triangle; this is called the Taylor-Hood element. But there are many other elements with various tradeoffs. The proof that a given element pair works -- that it satisfies a discrete Ladyzhenskaya-Babuska-Brezzi condition -- is usually confusing and arduous. It would be cool if these proofs could be formalized. It would be publication-worthy if the techniques used could enable the discovery of new element pairs. More generally, one could try to formalize finite element exterior calculus.<|endoftext|> TITLE: Theorems similar to Tischler fibering theorem QUESTION [5 upvotes]: Tischler theorem states that the existence of a nowhere vanishing closed $1$-form in a compact manifold $M$ implies that the manifold fibers over $S^1$. Do you know any other diffential topology results of this kind? By this kind I mean $$ \text{Existence of some diffential form/s} \implies \text{Topological consequences on $M$}$$ REPLY [4 votes]: On a $2k+1$-dimensional manifold, a 1-form $\alpha$ such that $\alpha\wedge (d\alpha)^k \neq 0$ at each point gives a contact structure. I believe that it is still open which $2k+1$-manifolds for $k>1$ admit a contact structure ( all 3-manifolds have a contact structure). On a 3-manifold, a nowhere vanishing $1-$form $\alpha$ such that $\alpha\wedge d\alpha=0$ is equivalent to having a foliation ($ker\alpha$ gives an integrable plane field by the Frobenius theorem). This does not give much topological information, since every 3-manifold admits a (2-dimensional smooth orientable) foliation. However, if there is also a closed 2-form $\omega$ such that $\omega \wedge \alpha >0$ (i.e. nowhere vanishing), then the foliation is taut. This is a non-trivial topological condition, as many 3-manifolds do not admit a taut foliation, and holds for fibered 3-manifolds, so strictly generalizes Tischler's theorem in the 3-dimensional case in some sense.<|endoftext|> TITLE: Riemann sum formula for definite integral using prime numbers QUESTION [11 upvotes]: I had asked this question in MSE. It got lot of upvotes but no answer (except one which was too long to be posted as a comment) hence I am posting it in MO. While answering another question in MSE I had used the following result which I thought was a trivial consequence of the prime number theorem and equidistribution. However, I realized from the comments that many people thought that this was not either true or counter intuitive. Hence I am posting this as a question looking for a proof or disproof. Let $p_k$ be the $k$-th prime and $f$ be a continuous function Riemann integrable in $(0,1)$ such that $$\lim_{n \to \infty}\frac{1}{n}\sum_{r = 1}^{n}f\Big(\frac{r}{n}\Big) = \int_{0}^{1}f(x)dx. $$ Then, $$ \lim_{n \to \infty}\frac{1}{n}\sum_{r = 1}^{n}f\Big(\frac{p_r}{p_n}\Big) = \int_{0}^{1}f(x)dx. $$ My approach: It was was based on showing that as $n \to \infty$, the ratios $p_r/p_n$ approached equidistribution in $(0,1)$ hence the integral follows as a property of equidistributed sequence. Motivation: There are several identities, limits etc on prime numbers which can be easily proven using this simple formula. REPLY [19 votes]: The statement follows from the prime number theorem. By an approximation argument, we can assume that $f$ is continuously differentiable on $[0,1]$. Then, $$\sum_{r=1}^{n}f\left(\frac{p_r}{p_n}\right)=\int_0^1 f(x)\,d\pi(p_n x)=nf(1)-\int_0^1 f'(x)\pi(p_n x)\,dx.$$ We estimate the last integral for fixed $f$: $$\int_0^1 f'(x)\pi(p_n x)\,dx=\int_{\frac{1}{\log n}}^1 f'(x)\pi(p_n x)\,dx+o(n).$$ In the last integral, we have by the prime number theorem, $$\pi(p_n x)\sim\frac{p_n x}{\log(p_n x)}\sim\frac{p_n x}{\log n}\sim nx,$$ so that \begin{align*}\int_0^1 f'(x)\pi(p_n x)\,dx &=\int_{\frac{1}{\log n}}^1 f'(x)\bigl(nx+o(nx)\bigr)\,dx+o(n)\\ &=\int_{0}^1 f'(x)\bigl(nx+o(nx)\bigr)\,dx+o(n)\\ &=n\int_0^1 f'(x)x\,dx+o(n)\\ &=nf(1)-n\int_0^1 f(x)\,dx + o(n). \end{align*} Putting everything together, $$\sum_{r=1}^{n}f\left(\frac{p_r}{p_n}\right)=n\int_0^1 f(x)\,dx + o(n),$$ that is, $$\frac{1}{n}\sum_{r=1}^{n}f\left(\frac{p_r}{p_n}\right)=\int_0^1 f(x)\,dx + o(1).$$<|endoftext|> TITLE: Trace on $\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k)$ QUESTION [11 upvotes]: I asked this question on Math StackExchange, but it did not receive an answer, despite my offering a bounty to attract attention. I am unsure whether it is appropriate for this venue, but I thought that I would try my luck. Below I have reproduced the question with some modifications. Let $\mathcal{S}(\mathbb{R}^k)$ denote the $k$-dimensional Schwartz space with the usual topology, and let $\mathcal{S}'(\mathbb{R}^k)$ denote its strong dual (i.e. the space of tempered distributions equipped with the topology of uniform convergence on bounded sets). Let $\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}(\mathbb{R}^k)$ denote the completed projective tensor product of $\mathcal{S}(\mathbb{R}^k)$ and $\mathcal{S}'(\mathbb{R}^k)$. Note that since both the Schwartz space and the space of tempered distributions are nuclear, the projective tensor product coincides with the injective tensor product. If $f\in\mathcal{S}(\mathbb{R}^k)$ and $g\in\mathcal{S}'(\mathbb{R}^k)$, then we can define $$\operatorname{Tr}(f\otimes \bar{g}) := \overline{\langle{g, \bar{f}}\rangle}_{\mathcal{S}'-\mathcal{S}},$$ where $\langle{\cdot,\cdot}\rangle_{\mathcal{S}'-\mathcal{S}}$ denotes the duality pairing. Now if the duality pairing were a continuous map $$\mathcal{S}(\mathbb{R}^k) \times \mathcal{S}'(\mathbb{R}^k) \rightarrow \mathbb{C},$$ then by the universal property of the $\pi$-tensor product, we would obtain a unique continuous map $$\operatorname{Tr}: \mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k) \rightarrow \mathbb{C}$$ with the property that $\operatorname{Tr}(f\otimes \bar{g})$ is as above. Unfortunately, the duality pairing is not continuous, it is only separately continuous--this is a general feature of non-normable locally convex spaces. Therefore, the preceding approach fails, which leads me to my question. Question 1. Is there a "canonical" way to define a trace $\operatorname{Tr}$ on $\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k)$ (i.e. a linear map such that $\operatorname{Tr}(f\otimes\bar{g}) = \overline{\langle{g,\bar{f}}\rangle}$)? It seems that such a map $\operatorname{Tr}$ cannot be continuous $\mathcal{S}(\mathbb{R}^k)\hat{\otimes}\mathcal{S}'(\mathbb{R}^k) \rightarrow \mathbb{C}$, otherwise, since the canonical bilinear map $$\mathcal{S}(\mathbb{R}^k)\times\mathcal{S}'(\mathbb{R}^k) \rightarrow \mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}' (\mathbb{R}^k), \qquad (f,g) \mapsto f\otimes g$$ is continuous, we would have the continuity of the evaluation map. Question 2. If the answer to Question 1 is no, is there a non-canonical way of defining a trace $\operatorname{Tr}$ on $\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi}\mathcal{S}'(\mathbb{R}^k)$ in such a way that if $\gamma\in\mathcal{S}(\mathbb{R}^k) \mathbin{\hat{\otimes}_\pi} \mathcal{S}'(\mathbb{R}^k)$ and can be identified with an element of trace-class operators on $L^2(\mathbb{R}^k)$, then $\operatorname{Tr}$ coincides with the usual definition of trace? REPLY [4 votes]: In the "classical theory of topological vector spaces" the questions like this are intricated (in my opinion, this is an artifical complexity, the Nature can't be so complicated). But in the theory of stereotype spaces they become simple: for each stereotype space $X$ (including ${\mathcal S}({\mathbb R}^k)$, since it is also stereotype as a Fréchet space) the pairing $$ (x,f)\in X\times X^\star\mapsto f(x)\in{\mathbb C} $$ is a continuous bilinear map in the stereotype sense, and, as a corollary, it can be extended to a continuous linear functional on the "projective stereotype tensor product" $\circledast$ (this is an analog of $\hat{\otimes}_\pi$ in the stereotype theory) $$ \operatorname{cont}: X\circledast X^\star\to {\mathbb C}. $$ This functional is called a "contraction", you can look at the details in my paper of 2003 (page 265). If you want to define a trace for all operators $\varphi:X\to X$ that are images of the tensors $\alpha\in X\circledast X^\star$ under the Grothendieck transformation $X\circledast X^\star\to{\mathcal L}(X)$, then your space $X$ must have the stereotype approximation property. As far as I know, nobody was interested up to now, whether the space ${\mathcal S}({\mathbb R}^k)$ has the stereotype approximation, but at first glance this is true: one can try to use the same trick as I did in my paper of 2018 for the space ${\mathcal C}(G)$ of continuous functions on a locally compact group $G$. Also Albrecht Pietsch writes in his Nuclear locally convex spaces, 10.3.2, that the Hermite polynomials form a basis in the space ${\mathcal S}'({\mathbb R})$ (which coincides with the stereotype dual space ${\mathcal S}({\mathbb R})^\star$). I think something similar must be true for each ${\mathcal S}'({\mathbb R}^k)$ (with arbitrary $k\in{\mathbb N}$), and if so, then ${\mathcal S}'({\mathbb R}^k)={\mathcal S}({\mathbb R}^k)^\star$ and ${\mathcal S}({\mathbb R}^k)$ have the stereotype approximation property.<|endoftext|> TITLE: What are the "smallest" topoi? QUESTION [9 upvotes]: Yesterday I was talking to somebody from the Haskell community. Late in the night we found ourselves discussing possible topoi. Lets order topoi (up to equivalence, ...) by number of objects/morphisms The smallest topos is a point (this is the only finite topos) Then there is FinSet (each topos has a functor from FinSet, and this is either "injective" on objects or the topoi in question is trivial) Next we (might) have Set Questions: Are there topoi "in between" FinSet & Set? (There are more Set-like topos, limiting by cardinality of the set, see nlab.) Are there more toposes "in between" FinSet & Set? Is each topos with countable many objects automatically equivalent to FinSet? (take the minimum over all equivalent topoi ...) Edits: added "definitions" of Set & FinSet (via linking nlab) I noted that the nlab entry already answers the first questions, so I made the question more precise added a rough definition of smallnes REPLY [13 votes]: Here are some examples, that should show you that there is a lot of countable toposes and lot of things in between finite sets and sets, too much to actually hope to list or classifies. First as I said a lots of usual construction of Grothendieck toposes have a finistic version, which produces elementary toposes (in fact, bounded toposes over FinSet most of the time). For example given a finite category $C$ the category of finite presheaves on $C$ is a topos, in fact you can even take a topology and look at finite sheaves. You can have a look to chapter 5.2 in the third volume of Borceux's handbook of categorical algebra for some details on these. If you have an inaccesible cardinal $\lambda$, then the category of sets of cardinality less than $\lambda$ is an essentially small elementary topos with a natural number objects. As mentioned by Todd Trimble, elementary toposes are a bit weaker than ZF because they don't allow for unbounded replacement. concretely it means that you can construct essentially small full subcategory of the category of sets that are elementary topos without invoking an inaccessible cardinal. If you take any set $X$, the full subcategory of sets that appears as element of $\mathcal{P}^{n}(X)= \mathcal{P} (\mathcal{P}( \dots \mathcal{P}(X)) \dots )$ for some $n$ ($\mathcal{P}(X)$ being the set of subsets of $X$) is an elementary topos. If $X$ is finite, this is justs the category of finite sets, but if $X$ is infinite it will have a natural number object, but it will be uncountable. This produces a lot of concrete example of full subcategory of the category of sets that are small toposes. Every time you have a model of ZF, its "category of sets" is a topos. Because of Löwenheim–Skolem there exists countable model of ZF, so this produces example of countable toposes admiting natural number objects. There exists a "free topos" which admit a unique logical morphisms to any topos, and a "free topos with a natural number object" which admit a unique logical morphisms preserving the natural number object to any topos with a natural number object, see there as well as the reference given there for more details. It admit a syntactical description which shows that it is countable (informally: its objects can be describe as certain formula involving the operation that you have in a topos, like taking power objects and finite limits, and there is only a countable number of these, same for the maps).<|endoftext|> TITLE: Explicit family of number rings $\mathcal{O}_{K_n}$ requiring $n$ generators? QUESTION [9 upvotes]: Could someone provide or point me to a family of number rings $\mathcal{O}_{K_n}$ that require $n$ generators (as $\mathbb{Z}$-algebra)? Second best would be a family requiring $f(n)$ generators for a strictly increasing and positive function $f:\mathbb{N}\to\mathbb{N}$. I would also be interested in seeing several explicit examples of number rings requiring 3 or 4 generators REPLY [20 votes]: An earlier question on this type of topic was asked by Zev Chonoles in 2010 at Which number fields are monogenic? and related questions and I want to draw your attention to the comment there by BCnrd for a nice geometric analogy. When answering that question I did not address the part Zev asked about finding rings of integers needing lots of generators as a $\mathbf Z$-algebra, so I'll do that here. Pick an integer $r \geq 1$. Here is a sufficient condition on a number field $K$ that forces $\mathcal O_K$ to need more than $r$ generators as a $\mathbf Z$-algebra: there is a prime $p$ such that (i) $[K:\mathbf Q] > p^r$ and (ii) $p$ splits completely in $\mathcal O_K$. Example 1. Taking $r = 1$ and $p=2$, to find a number field $K$ such that $\mathcal O_K$ needs at least two generators as a $\mathbf Z$-algebra ($K$ is not "monogenic") it is sufficient to find a $K$ such that $[K:\mathbf Q] > 2$ and $2$ splits completely in $\mathcal O_K$, such as a cubic field in which $2$ splits completely. Dedekind found the first example of such a field: $\mathbf Q(\alpha)$ where $\alpha^3 - \alpha^2 - 2\alpha - 8 = 0$. A method of constructing infinitely many such $K$ is to use the cubic subfield of the cyclotomic field $\mathbf Q(\zeta_p)$ for primes $p$ such that $p \equiv 1 \bmod 3$ and $2^{(p-1)/3}\equiv 1 \bmod p$; that means $p$ splits completely in $\mathbf Q(\sqrt[3]{2},\zeta_3)$, and there are infinitely many such $p$ (and hence infinitely many such cubic fields) since the density of such $p$ is $1/6$ by the Chebotarev density theorem. The first few such $p$ are $31$, $43$, $109$, and $127$. For example, using PARI and Galois theory, the cubic subfield of $\mathbf Q(\zeta_{31})$ is $\mathbf Q(\alpha)$ where $\alpha$ is a root of $$x^3 + x^2 - 10x - 8$$ and $2$ splits completely in this cubic field (e.g., PARI says this cubic polynomial splits completely over $\mathbf Q_2$) so the ring of integers of this cubic field needs at least $2$ generators as a $\mathbf Z$-algebra. (This cubic field is different from Dedekind's, e.g., Dedekind's cubic field has $1$ real embedding while this cubic field has $3$ real embeddings, of Dedekind's cubic field has discriminant $-503$ while this cubic field has discriminant $961 = 31^2$.) Example 2. Taking $r = 2$, to find a number field $K$ such that $\mathcal O_K$ needs at least three generators as a $\mathbf Z$-algebra it is sufficient to find a $K$ such that $[K:\mathbf Q] > 4$ and $2$ splits completely in $\mathcal O_K$, such as a quintic field in which $2$ splits completely. A way to construct such fields is to use the quintic subfield of $\mathbf Q(\zeta_p)$ for primes $p$ such that $p \equiv 1 \bmod 5$ and $2^{(p-1)/5}\equiv 1 \bmod p$; that means $p$ splits completely in $\mathbf Q(\sqrt[5]{2},\zeta_5)$, and there are infinitely many such $p$ since their density is $1/20$ by Chebotarev. The first few such $p$ are $151$, $241$, $251$, and $431$. Using PARI and Galois theory as in the previous example, the quintic subfield of $\mathbf Q(\zeta_{151})$ is $\mathbf Q(\alpha)$ where $\alpha$ is a root of $$x^5 + x^4 - 60x^3 - 12x^2 + 784x + 128$$ and $2$ splits completely in this quintic field, so the integers of the quintic field need at least $3$ generators as a $\mathbf Z$-algebra. I hope from these examples you see the pattern by which, for each $r$, you can use a subfield of degree $d$ in $\mathbf Q(\zeta_p)$ for infinitely many primes $p \equiv 1 \bmod d$ to get infinitely many number fields whose ring of integers requires more than $r$ generators as a $\mathbf Z$-algebra. It is time to prove (i) and (ii) above are sufficient conditions for $\mathcal O_K$ to require more than $r$ generators as a $\mathbf Z$-algebra. Let $K$ be a number field such that $\mathcal O_K$ has at most $r$ generators as a $\mathbf Z$-algebra. We will show either condition (i) or (ii) has to break down, or more simply if (ii) holds then (i) does not: if $\mathcal O_K$ has at most $r$ generators as a $\mathbf Z$-algebra and a prime $p$ splits completely in $\mathcal O_K$ then we will show $[K:\mathbf Q] \leq p^r$. That $\mathcal O_K$ has at most $r$ generators as a $\mathbf Z$-algebra is the same as saying there is a surjective ring homomorphism $\mathbf Z[x_1,\ldots,x_r] \twoheadrightarrow \mathcal O_K$. Reducing both sides modulo $p$ for an arbitrary prime $p$, there is a surjective ring homomorphism $\mathbf {\mathbf F}_p[x_1,\ldots,x_r] \twoheadrightarrow \mathcal O_K/(p)$. Let $I$ be the kernel, so $\mathbf {\mathbf F}_p[x_1,\ldots,x_r]/I \cong \mathcal O_K/(p)$ as rings. Assume $p$ splits completely in $K$ (infinitely many primes split completely in each number field, so this assumption is not crazy). Then $\mathcal O_K/(p) \cong {\mathbf F}_p^n$ as rings, where $n = [K:\mathbf Q]$, so there is a ring isomorphism $\mathbf {\mathbf F}_p[x_1,\ldots,x_r]/I \cong {\mathbf F}_p^n$. The product ring $\mathbf F_p^n$ has $n$ maximal ideals, each with residue field $\mathbf F_p$, so there are $n$ maximal ideals in ${\mathbf F}_p[x_1,\ldots,x_r]$ that contain $I$ and have residue field $\mathbf F_p$. A maximal ideal in ${\mathbf F}_p[x_1,\ldots,x_r]$ with residue field $\mathbf F_p$ has the form $(x_1 - a_1,\ldots,x_r-a_r)$ where $a_i \in \mathbf F_p$. There are $p^r$ such ideals in total, so necessarily $n \leq p^r$. That is exactly the negation of condition (i) at the start. Update: This proof extends to the relative case. If $E/F$ is an extension of number fields then a sufficient condition for $\mathcal O_E$ to need more than $r$ generators as an $\mathcal O_F$-algebra is that there is a (nonzero) prime ideal $\mathfrak p$ in $\mathcal O_F$ such that (i) $[E:F] > ({\rm N}\mathfrak p)^r$ and (ii) $\mathfrak p$ splits completely in $\mathcal O_E$. In the above proof, "ring homomorphism" has to be replaced by "$\mathcal O_F$-algebra homomorphism".<|endoftext|> TITLE: Why did Dedekind claim that $\sqrt{2}\cdot\sqrt{3}=\sqrt{6}$ hadn't been proved before? QUESTION [29 upvotes]: In a letter to Lipschitz (1876) Dedekind doubts that $\sqrt{2}\cdot\sqrt{3}=\sqrt{6}$ had been proved before: quoted from Leo Corry, Modern algebra, German original: Why did Dedekind doubt that $(\sqrt{a}\cdot\sqrt{b})^2=(\sqrt{a})^2\cdot(\sqrt{b})^2$ had been proved (which lies at the heart of his vicious circle)? What's wrong with this identity which easily follows from the associativity and commutativity of multiplication: $$(m\cdot n)\cdot(m\cdot n) = (m\cdot m)\cdot(n\cdot n)$$ He argues that rational and irrational numbers are of a different kind, and thus $(m\cdot n)^2 = m^2\cdot n^2$ (which had been proved for the rationals) may not be "applied without scruples" to the irrationals. But why – from his point of view – had $(m\cdot n)^2 = m^2\cdot n^2$ been proved only for the rationals? Couldn't it – for example – have been proved already in the system of Euclid's Elements by elementary geometrical considerations, regardless of rational or irrational, i.e. for arbitrary lengths: The proof that the fat green circle intersects the horizontal line at the same point as the fat red line (which is constructed as the parallel to the thin red line going through 1) may have been complicated for Euclid, but it seems possible. If not so: Why couldn't it have been proved by Euclid? And finally: What was Dedekind's alternative proof, eventually? REPLY [7 votes]: I don't think Dedekind's complaint can be that a proof in an axiomatic system must be invalid if the objects in the system have not been constructed from simpler objects. This obviously leads to an infinite regress. Today we typically take "set" to be undefined, but Dedekind would surely have been familiar with treatments of Euclidean geometry in which "point" is an undefined term. I suspect that Dedekind was complaining that people were not working in a full axiomatic system for the reals, but rather were using only the elementary axioms and did not realize that the completeness axiom was needed. Note that in the quote, the argument he criticizes uses only elementary properties such as commutativity and associativity. Without completeness, we don't have any way of proving the existence of square roots, but the textbook treatments he complains about presumably don't admit that the theorem $\sqrt{2}\sqrt{3}=\sqrt{6}$ is conditional on the unproved existence of these roots. Why couldn't it have been proved by Euclid? Euclid couldn't have proved it to modern standards of rigor because Euclid couldn't have proved that the circles in your diagram intersected. This is equivalent to lacking the completeness property of the reals.<|endoftext|> TITLE: Clustering distance QUESTION [5 upvotes]: Is there a good notion of distance between partitions of a (fixed, finite) set? The context is this: suppose I have a clustering algorithm, which clusters points using some method or other. Now, I perturb the positions of the points, the clustering changes, and I want some quantitative estimate of how much it has changed. When there are two clusters $K_1, K_2$ which morph into $L_1, L_2,$ then it seems reasonable to look at the minimum of the sizes of $K_1 \Delta L_1$ and $K_1 \Delta L_2,$ but for more clusters it seems less clear. REPLY [4 votes]: The variation of information seems to be the sort of thing you're looking for.<|endoftext|> TITLE: Hilbert representation of a bilinear form QUESTION [5 upvotes]: Let $\sigma:\mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}$ be a bilinear symmetric form which is non-degenerate in the sense that for every $0\neq u\in \mathbb{R}^n$ there is $v\in \mathbb{R}^n$ with $\sigma\left(u,v\right)\ne 0$. It is well-known and easy to see that there is a basis $e_1,...,e_n$ and $m \in \overline{0,n+1}$ such that $\sigma\left(u,v\right)= \sum_{k=1}^{m} u_kv_k-\sum_{k=m}^{n} u_kv_k$, where $v=(v_1,...,v_n)$ and $u=(u_1,...,u_n)$ with respect to $e_1,...,e_n$. Endow $\mathbb{R}^n$ with an inner product $\langle u,v\rangle=\sum_{k=1}^{m} u_kv_k$, where $v=(v_1,...,v_n)$ and $u=(u_1,...,u_n)$ with respect to $e_1,...,e_n$. Then $T:\mathbb{R}^n\to \mathbb{R}^n$ defined by $Te_k=e_k$, for $1\le k\le m$ and $Te_k=-e_k$, for $m\le k\le n$ is an isometry, such that $\sigma\left(u,v\right)=\langle Tu,v\rangle$. I am wondering if the following infinite-dimensional version holds: Let $V$ be a real vector space, and let $\sigma$ be a symmetric and non-degenerate bilinear form on $V$. Can we find a Hilbert space $H$, a linear injection $S:V\to H$ and a surjective isometry $T:H\to H$ such that $\sigma\left(u,v\right)=\langle TSu,Sv\rangle_{H}$, for every $u\in V$? REPLY [6 votes]: The answer is no. Let be construct such a $(V,\sigma)$. Let $X$ be the set of functions $\mathbf{N}\to\mathbf{R}$. Let $Y\subset X$ be a subset such that $Y$ is linearly independent $Y$ contains all Dirac functions $n\mapsto\delta_{m,n}$ for every $g\in X$, there exists $f\in Y$ such that $\limsup(f/(|g|+1))=\infty$. Note that the last condition forces $Y$ to be uncountable. The existence of $Y$ is checked by considering a maximal linearly independent subset containing all Dirac functions. Now consider the space $V$ with basis indexed by $\mathbf{N}\sqcup Y$, namely $(e_n)_{n\in\mathbf{N}}$ and $(z_f)_{f\in Y}$, and symmetric bilinear form $$\sigma(e_n,e_m)=\delta_{m,n},\quad \sigma(e_n,z_f)=f(n),\quad\sigma(z_f,z_g)=0.$$ I claim that $\sigma$ is non-degenerate; I'll use the first two axioms. Indeed, let $v=\sum_na_ne_n+\sum_fb_fz_f$ belong to the kernel of $\sigma$. Then $0=\sigma(v,z_f)=\sum_na_nf(n)=0$ for all $f\in Y$. Choosing $f=\delta_n$, we deduce $a_n=0$, for all $n$. So $v=\sum_fb_fz_f$. Then $0=\sigma(v,e_n)=\sum_{f\in Y}b_ff(n)$ for all $n$, which means that $\sum_{f\in Y}b_ff=0$. Since $Y$ is linearly independent, we deduce $b_f=0$ for all $f$, and hence $v=0$. Now suppose the existence of $H$, $S:V\to H$ and $T$ as required. Write $E_n=S(e_n)$ and $Z_f=S(z_f)$, $C_f=\|TZ_f\|$, $g(n)=\|E_n\|$. Then for all $f\in Y$ and $n$, we have $f(n)=\sigma(z_f,e_n)=\langle TZ_f,E_n\rangle$. So $|f(n)|\le C_fg(n)$ for all $n$. Hence $\limsup (f/(g+1))<\infty$. Since this holds for each $f\in Y$, we contradict the third axiom, which ends the proof. However, if $V$ has countable dimension, the answer is yes. Indeed, one can construct, by induction, an orthogonal basis and conclude in the same way as in finite dimension. For the induction, one first observes that every finite-dimensional subspace $W$ is contained in a non-degenerate one. This allows to write $V$ as an increasing union of non-degenerate subspaces, and thereby construct the desired basis.<|endoftext|> TITLE: What parts of the theory of quasicategories have been simplified since the publication of HTT? QUESTION [37 upvotes]: It has been almost ten years since Lurie published Higher Topos Theory, where (following Joyal and probably others) he set up foundations for higher category theory via quasicategories. My impression is that this field is not yet completely settled, and so one might expect there to have been significant simplifications of this material since. I know of one possible example (though I am not sure if it is best described as a simplification, rather than an alternative route), namely the proof of the equivalence between quasicategories and simplicial categories given in https://arxiv.org/abs/0911.0469. What other foundational advances have happened after the publication of HTT? REPLY [7 votes]: A significant technical improvement has been found by J. Shah in the theory of Kan extensions. Unfortunately I do not know of an exposition that does only the classical case, but reading the proof of theorem 10.3 in Jay Shah Parametrized higher category theory and higher algebra: Exposé II - Indexed homotopy limits and colimits in the case $S=\Delta^0$ we can obtain a much simpler proof of the fact that left Kan extensions in ∞-categories can be computed pointwise if there exist enough colimits.<|endoftext|> TITLE: A name for a group with finite abelization? QUESTION [5 upvotes]: Let us recall that a group $G$ is called perfect if it coincides with its commutator subgroup $G'$, or equivalently, if its abelianization $G/G'$ is trivial. Question. Is there any name for a group $G$ whose abelianization $G/G'$ is finite? REPLY [6 votes]: Prompted by the OP, I'm writing my comment as an answer. I suggest “almost perfect” group. A google search indicates this terminology has been used; for instance, here: core.ac.uk/download/pdf/61487184.pdf .<|endoftext|> TITLE: Ideals on $\mathbb N$ and large sets that have small intersection QUESTION [8 upvotes]: Let $\mathcal I$ be a (non-principal) ideal of subsets of $\mathbb N$. Suppose that every family $\mathcal{A} \subset \wp(\mathbb N)\setminus \mathcal I$ with the following property is countable: $$A,B\in \mathcal A, A\neq B \Rightarrow A\cap B \in \mathcal I. $$ Let us observe that maximal ideals have this property as only the empty family or the singletons meet this requirement. Are there further examples of ideals with this property? (As observed by Leonetti every such ideal must be non-meagre when regarded as a subset of the Cantor set.) REPLY [3 votes]: In topological language: for any closed subset, $F$, of $\beta\mathbb{N}\setminus\mathbb{N}$ the family $I_F=\{A:A^*\cap F=\emptyset\}$ is an ideal. Your property translates into: the closed set $F$ is a ccc space. The example by Monk can be made in this way too: take a map $s$ from $\beta\mathbb{N}\setminus\mathbb{N}$ onto the Cantor set $C$ and take $F$ such that $s:F\to C$ is irreducible and onto, then $F$ is sparable and hence ccc. The example by Blass corresponds to the closure of a countable set in $\beta\mathbb{N}\setminus\mathbb{N}$, again a separable subspace. In general: take any compact ccc space, $X$, of weight at most $\mathfrak{c}$ and embed it into the Tychonoff cube $[0,1]^\mathfrak{c}$. There is a continuous map $\sigma$ from $\beta\mathbb{N}$ onto that cube. Now take $F$ such that $\sigma:F\to X$ is irreducible and onto; then $F$ is ccc. For a non-separable example let $X$ be the Stone space of the measure algebra.<|endoftext|> TITLE: Diameter of a quotient of the infinite dimensional sphere QUESTION [20 upvotes]: Suppose a group $\Gamma$ acts by isometries on the Hilbert space $\mathbb{H}^\infty$ and it fixes the origin. So $\Gamma$ acts on the unit sphere $\mathbb{S}^\infty$ as well. Assume that the action $\Gamma$ on $\mathbb{S}^\infty$ has no dense orbits. Is there a universal constant $\varepsilon >0$ such that there are two orbits of $\Gamma$ on distance at least $\varepsilon$ from each other? In other words, is there a constant $\varepsilon>0$ such that $$\mathrm{diam}\, (\mathbb{S}^\infty/\Gamma) >0 \quad\Longrightarrow\quad \mathrm{diam}\, (\mathbb{S}^\infty/\Gamma) > \varepsilon\ ?$$ Comments. The statement is related to the following results in finite dimension. It was proved in [Gre] that, on the $n$-sphere for $n\ge 2$, there is such a lower bound $\varepsilon_n>0$ (choose it as optimal). It is expected (see the introduction of this arXiv paper by Gorodski and Lytchak) that $\inf_{n\ge 2}\varepsilon_n>0$. This is announced to hold by Claudio Gorodski, Christian Lange, Alexander Lytchak and Ricardo Mendes (it is not published yet). Namely, for some universal constant $\varepsilon>0$, for any $n\ge 2$ and for any isometric group action of any compact group on the unit $n$-sphere, the orbit space of the action is either a point or has diameter $\ge\varepsilon$. [Gre] S. J. Greenwald, Diameters of spherical Alexandrov spaces and curvature one orbifolds, Indiana Univ. Math. J. 49 (2000), no. 4, 1449–1479. REPLY [16 votes]: There is no such universal constant $\epsilon > 0$. Work with the complex Hilbert space $L^2[0,1]$ (which of course is also a real Hilbert space). Fix $n \in \mathbb{N}$. Let $\Gamma_0$ be the set of continuous piecewise linear increasing bijections from $[0,1]$ to itself. [1] It is a group with composition as product. [2] It acts by isometries of $L^2[0,1]$ by the map $f \mapsto \sqrt{\phi'}\cdot (f\circ\phi)$ for $f \in L^2[0,1]$ and $\phi \in \Gamma_0$. Also let $\Gamma_1 \subset L^\infty[0,1]$ consist of the measurable functions from $[0,1]$ to $\mathbb{T}_n = \{e^{2\pi i k/n}: 0 \leq k < n\}$, identifying functions which differ on a null set. This is a group under pointwise product and [3] it acts isometrically by multiplication on $L^2[0,1]$. Let $\Gamma$ be the group of isometries of $L^2[0,1]$ generated by $\Gamma_0$ and $\Gamma_1$ under these actions. ([4] This is a semidirect product of $\Gamma_0$ and $\Gamma_1$.) [5] The $\Gamma_0$ action takes the unit vector $1_{[0,1]}$ to any piecewise constant strictly positive unit vector $f \in L^2[0,1]$. (If $f$ takes the value $c$ on an interval $I$, let $\phi$ have slope $c^2$ on this interval.) [6] These functions are dense in the positive part of the unit sphere. [7] Applying the action of $\Gamma_1$ then gets us to arbitrarily close to any unit vector in $L^2[0,1]$ whose argument lies almost everywhere in $\mathbb{T}_n$. [8] It follows that the distance from $1_{[0,1]}$ to any other orbit is at most $\alpha = |1 - e^{\pi i/n}|$ ($\approx \frac{\pi}{n}$ for large $n$). [9] It follows straightforwardly that the same is true for any positive unit vector in place of $1_{[0,1]}$, and then [10] that the distance between any two orbits is at most $2\alpha$. [11] On the other hand, the distance from the orbit of $1_{[0,1]}$ to the vector $e^{\pi i/n}\cdot 1_{[0,1]}$ is at least the distance from $(1,0) \in \mathbb{R}^2$ to the line through the origin of slope $\frac{\pi}{n}$ (again approximately $\frac{\pi}{n}$ for large $n$), so this orbit is not dense and since the action is isometric no orbit is dense. Edit: maybe people want more details. [1] The composition of two continuous functions is continuous, of two PL functions is PL, of two increasing functions is increasing, of two bijections is a bijection. The inverse of a continuous PL increasing bijection is a continuous PL increasing bijection. [2] $\|\sqrt{\phi'}\cdot (f\circ \phi)\|_2^2 = \int_0^1 \phi'|f\circ\phi|^2\, dt = \int_0^1 |f|^2\, dt = \|f\|_2^2$. [3] If $h \in \Gamma_1$ then $\|hf\|_2^2 = \int_0^1 |hf|^2\, dt = \int_0^1 |f|^2\, dt = \|f\|_2^2$ since $|h| = 1$ a.e. [4] This isn't needed, but anyway if $\phi \in \Gamma_0$ and $h \in \Gamma_1$ then $\sqrt{\phi'}\cdot (hf\circ \phi) = (h\circ \phi)\cdot \sqrt{\phi'}\cdot(f\circ\phi)$. [5] Let $f$ be a piecewise constant strictly positive unit vector in $L^2[0,1]$. Then $f = a_0\cdot 1_{[0,t_1)} + \cdots + a_k\cdot 1_{[t_k,1)}$ a.e. for some $0 < t_1 < \cdots < t_k < 1$ and some $a_0, \ldots, a_k > 0$. The unit norm condition means that $\sum_{i=1}^k a_i^2(t_{i+1} - t_i) = 1$. Now define $\phi: [0,1] \to \mathbb{R}$ so that $\phi(0) = 0$, $\phi$ is continuous, and $\phi$ is linear with slope $a_i^2$ on $[t_{i-1},t_i]$. The unit norm condition just detailed shows that $\phi(1) = 1$, i.e., $\phi$ is a continuous PL increasing bijection. We have $\sqrt{\phi'}\cdot (1_{[0,1]}\circ \phi) = \sqrt{\phi'}$, which takes the value $a_i$ constantly on $(t_{i-1},t_i)$. So $1_{[0,1]}$ is taken to $f$. [6] First, positive piecewise constant functions can uniformly approximate any continuous function on $[0,1]$, and since the positive continuous functions are dense for the $L^2$ norm in the positive part of $L^2[0,1]$, this shows that positive piecewise constant functions are dense in the positive part of $L^2[0,1]$. Given a positive $f \in L^2[0,1]$ with unit norm, find a sequence $(f_k)$ of positive piecewise constant functions with $f_k \to f$ in $L^2[0,1]$. Then $\|f_k\|_2 \to 1$ so $\frac{1}{\|f_k\|_2}f_k \to f$. Thus, any positive unit vector is approximated by positive piecewise constant unit vectors. [7] Since multiplying by $h \in \Gamma_1$ is an isometry, if $f = h|f| \in L^2[0,1]$ is a unit vector whose argument $h$ lies in $\mathbb{T}_n$ a.e. and $g$ is a positive piecewise constant unit vector which is close to $|f|$, then $hg$ will be equally close to $f$. [8] Given any unit vector $f = h|f| \in L^2[0,1]$, we can find $\tilde{h} \in \Gamma_1$ such that $|h(t) - \tilde{h}(t)| \leq \alpha$ a.e. As we just saw that the orbit of $1_{[0,1]}$ comes arbitrarily close to $\tilde{h}|f|$, it follows that the distance from $f$ to this orbit is at most $\|f - \tilde{h}|f|\|_2 = \|(h - \tilde{h})|f|\|_2 \leq \alpha \|f\|_2 = \alpha$. [9] We saw already that any positive unit vector $f$ is approximated by elements in the orbit of $1_{[0,1]}$. Since the action is isometric, this means that $1_{[0,1]}$ is approximated by elements in the orbit of $f$. Again by isometric action, since we can take $1_{[0,1]}$ to within $\alpha'$ of any unit vector, for any $\alpha' > \alpha$, the same is then true of $f$. [10] Any two unit vectors lie within $\alpha'$ of the orbit of $1_{[0,1]}$, for any $\alpha' > \alpha$. So (by isometric action again) one lies within $2\alpha'$ of the orbit of the other. [11] The argument of any vector $f$ in the orbit of $1_{[0,1]}$ lies pointwise a.e. in $\mathbb{T}_n$. So $|f(t) - e^{\pi i/n}| \geq \beta$ pointwise, where $\beta$ is the distance from $(1,0) \in \mathbb{R}^2$ to the line through the origin of slope $\frac{\pi}{n}$ (= the distance from $e^{\pi i/n} \in \mathbb{C}$ to the union of the lines through the origin of slopes $\frac{2k\pi}{n}$). Thus $\|f - e^{\pi i/n}\cdot 1_{[0,1]}\|_2 \geq \beta$.<|endoftext|> TITLE: “Algebraization" of $p$-adic fields QUESTION [11 upvotes]: Part 1: a single finite place. Let $K$ be a finite extension of $\mathbf{Q}_p$. Does there exist a number field $F/\mathbf{Q}$ and a finite place $v$ lying over $p$, such that for the completion of $F$ at $v$ we have a topological isomorphism of topological field extensions of $\mathbf{Q}_p$: $$F_v\simeq K ?$$ In imprecise words, does any $p$-adic field come as a completion of a number field? Part 2: a family of $p$-adic fields. (ASKED AS A SEPARATE QUESTION, here) Let $(K_p)_p$ be a collection of $p$-adic fields. Assume the first part of the question has positive answer. Does there exist a condition on $(K_p)_p$ such that there is one number field $K$ recovering each $K_p$ as a completion? REPLY [10 votes]: The answer to part 1 is yes. Given $K/\mathbb{Q}_p$, let $\alpha\in K$ be a primitive element, with minimal monic polynomial $f(x)=x^n+\sum_{i=1}^n a_i x^{n-i}$, $a_i\in\mathbb{Q}_p$. So we have $K\cong\mathbb{Q}_p[x]/(f(x))$. Krasner's lemma implies there exists a positive integer $N$ such that if $b_i\in\mathbb{Q}_p$ satisfy $v_p(b_i-a_i)\geq N$, then the polynomial $g(x)=x^n+\sum_i b_i x^{n-i}$ satisfies $\mathbb{Q}_p[x]/(f(x))\cong\mathbb{Q}_p[x]/(g(x))$. In particular, since $\mathbb{Q}$ is dense in $\mathbb{Q}_p$, we can choose the $b_i$ to be rational. Then $F:=\mathbb{Q}[x]/(g(x))$ is a finite extension of $\mathbb{Q}$, and $F\otimes\mathbb{Q}_p\cong\mathbb{Q}_p[x]/(g(x))\cong K$. In general, tensoring a number field with $\mathbb{Q}_p$ gives the product of the completions of that number field at the primes lying over $p$, so in this case there is only one prime $v$ of $F$ over $p$, and $F_v\cong K$. I do not know an answer to part 2, but I will mention some necessary conditions on the family $(K_p)$: only finitely many can be ramified, only finitely many degrees occur, and $\{p:K_p=\mathbb{Q}_p\}$ has positive lower density.<|endoftext|> TITLE: What's the maximum probability of associativity for triples in a nonassociative loop? QUESTION [24 upvotes]: In a finite nonabelian group, the probability that two randomly chosen elements commute cannot exceed 5/8. One easy proof also makes it easy to find the smallest groups that attain this bound, namely the two nonabelian groups of order 8: The 5/8 theorem. This makes me wonder: how large can the probability be that three randomly chosen elements $a,b,c$ of a finite nonassociative loop associate, i.e. obey $(ab)c = a(bc)$? You can prove the 5/8 theorem for groups by separately settling two questions: What is the largest possible fraction of elements of a noncommutative finite group that lie in the center? (Answer: 1/4) Given a noncentral element of a finite group, what's the largest possible fraction of elements that commute with it? (Answer: 1/2) The nonabelian groups of order 8 achieve both these upper bounds. We could try a similar strategy for my question, attempting to settle these: What is the largest possible fraction of elements $a$ in a finite loop such that $(ab)c = a(bc)$ for all elements $b,c$? If an element $a$ of a finite loop does not have $(ab)c = a(bc)$ for all elements $b,c$, what is the largest possible fraction of elements $b$ such that $(ab)c = a(bc)$ for all $c$? If a pair $a,b$ does not have $(ab)c = a(bc)$ for all elements $c$, what is the largest possible fraction of elements $c$ such that $(ab)c = a(bc)$? Unfortunately I don't know how to settle these. Since the quaternion 8-group $$Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$$ attains the 5/8 bound for commutativity of pairs in a nonabelian group, one might hope that the octonion 16-loop $$O_{16} = \{\pm 1, \pm e_1, \dots, \pm e_7\}$$ attains the maximum probability of associativity for triples in a nonassociative loop. Does it? I'm afraid I haven't even worked out the probability that a triple in $O_{16}$ associates, though it would be easy to do. REPLY [5 votes]: This is a bit too long for a comment, so it's an answer. The Loops package for Gap, by Gabor Nagy and Petr Vojtechovsky contains implementations of all the nonassociative Moufang loops of order $\leq 64$ and order equal to $81$ or $243$. So I wrote a gap script to calculate the association probabilities of triples of elements. I made it as simple-minded as possible to reduce the possibility of bugs, and because I'd never written anything in Gap before. None of the association probabilities exceeded $\frac{43}{64}$, the association probability for the octonion loop, so the conjecture is correct for these particular Moufang loops (the script took about half an hour on my laptop). Since the package also has some Bol loops, I checked them, and the left Bol loop of order 8 with the following Cayley table has association probability $\frac{13}{16} = \frac{52}{64} > \frac{43}{64}$: 1 2 3 4 5 6 7 8 --------------- 1|1 2 3 4 5 6 7 8 2|2 1 4 3 7 8 5 6 3|3 4 1 2 6 5 8 7 4|4 3 2 1 8 7 6 5 5|5 6 7 8 1 2 3 4 6|6 8 5 7 3 1 4 2 7|7 5 8 6 2 4 1 3 8|8 7 6 5 4 3 2 1 and many of the left Bol loops of order 16 also have association probabilities exceeding $\frac{43}{64}$. Therefore, if the conjecture is correct for Moufang loops, the proof must use an argument that fails for left Bol loops.<|endoftext|> TITLE: Measures and differential forms on manifolds QUESTION [13 upvotes]: Let $M$ be a differentiable manifold. Let $\mu$ be a (probability) measure on $M$. What are the conditions under which $\mu$ is given by a differential form on $M$? I imagine some sort of compatibility of the topology or the differentiable structure of $M$ with the $\sigma$-algebra of $\mu$ would be required. (Apologies if the question is too elementary for this forum. A pointer to the relevant result in the literature would suffice.) REPLY [5 votes]: Any smooth manifold has a canonical σ-ideal of negligible subsets, and μ must vanish on these. Apart from that, the Lie derivative of μ with respect to any smooth vector field must exist. This is how smooth measures are defined by Ramanan in Definition 1.9 of Chapter 3 of Global Calculus, for example. Remark 2.8 in Chapter 8 there explains how this definition is equivalent to the traditional definition of a smooth measure as a smooth section of the line bundle of densities.<|endoftext|> TITLE: Has the arcsine law been generalised to higher order divisor functions? QUESTION [11 upvotes]: The arcsine law for the distribution of the logarithms of the divisors of an integer $n$ states that $$ \frac{1}{x}\sum_{n\leq x}\frac{1}{d(n)}\sum_{\substack{q|n\\q\leq n^{A}}}1\sim \frac{2}{\pi}\arcsin \sqrt A $$ for $01$ has also appeared in the literature, or if this is known but perhaps unpublished? REPLY [3 votes]: There's an extension of the Deshouillers-Dress-Tenenbaum theorem by Bareikis and Manstavičius, http://doai.io/10.4064/aa126-2-5, which almost treats the question you ask. I'm not sure why the condition $f(p^\ell) \ll 1$ is imposed there; as far as the method goes the condition $f(p^\ell) \ll \ell^C$ ($C$ absolute) should be fine.<|endoftext|> TITLE: Nash isometric embedding theorem with keeping the symplectic structures of our ambient spaces QUESTION [5 upvotes]: I apologize in advance if this question has an obvious answer. Let $(M,g)$ be a Riemannian manifold. Then the tangent bundle $TM$ carries a natural symplectic structure $\omega_g$. In fact $\omega_g$ is the pull back of the canonical symplectic structure of the cotangent bundle via the obvious diffeomorphism between $TM$ and $T^* M$ which is defined by the inner product $g$. The standard structure of $T\mathbb{R}^n=\mathbb{R}^n \times \mathbb{R}^n$ is denoted by $\omega$. For every Riemannian manifold $(M,g)$, is there an isometric embedding $j$ of $M$ in some $\mathbb{R}^n$ such that $j^*(\omega)=\omega_g?$ REPLY [8 votes]: This is always the case (using "naturality" as Paul Bryan suggested in the comments). Let $f: M \to N$ be a smooth map between Riemannian manifolds $(M,g)$ and $(N, h)$. Let $g^\flat: TM \to T^*M$ denote the musical isomorphism induced by $g$ (and similarly $h^\flat$). Now, $f$ is an isometry if and only if $h_{f(m)} (T_m f (X), T_m f (\cdot)) = g_m (X, \cdot)$ holds for all $X \in T_m M$. In other words, $g^\flat (X) = f^* (h^\flat (T_m f(X)))$ or shorter $$g^\flat = T^* f \circ h^\flat \circ Tf,$$ where $T^*f: T^* N \to T^* M$ is the cotangent lift of $f$ (often called a point transformation). Using the fact that cotangent lifts are always symplectic maps, we have $$\omega_g = (g^\flat)^* \omega_M = (Tf)^* (h^\flat)^* \omega_N = (Tf)^* \omega_h,$$ where $\omega_M$ and $\omega_N$ denote the canonical symplectic forms on $T^* M$ and $T^* N$, respectively.<|endoftext|> TITLE: What is the definition of the function T used in Atiyah's attempted proof of the Riemann Hypothesis? QUESTION [107 upvotes]: In Michael Atiyah's paper purportedly proving the Riemann hypothesis, he relies heavily on the properties of a certain function $T(s)$, known as the Todd function. My question is, what is the definition of $T(s)$? Atiyah states that this function is defined in his paper "The Fine Structure Constant", but I can't seem to find a copy of the paper. So can anyone tell me how Atiyah defined $T$ in that paper? REPLY [25 votes]: Here's a public paper of the "the fine structure constant" by Atiyah. It doesn't seem to be the original, but a copy: https://drive.google.com/file/d/1WPsVhtBQmdgQl25_evlGQ1mmTQE0Ww4a/view See the section 3.4, the Todd function is defined there. [Edit] As pointed by @T_M in the comments, here you would find the main properties of the T function.<|endoftext|> TITLE: Spectral and derived deformations of schemes QUESTION [8 upvotes]: I'd like to understand how ordinary schemes deform or lift to spectral and derived schemes in two basic examples as well as what the structure of the space of deformations in general is. Let $S = (X, \mathcal{O}_X)$ be a scheme, where $X$ is the underlying topological space, and $\mathcal{O}_X$, its structure sheaf of commutative rings. Call a spectral scheme $S^\infty = (X, \mathcal{O}^\infty_X)$ a spectral deformation (or lift) of $S$ if $\pi_0(\mathcal{O}^\infty_X) = \mathcal{O}_X$, i.e., if $S$ is the scheme underlying the spectral scheme $S^\infty$. Let $\Sigma^\infty(S)$ be the collection of all spectral deformations of $S$. Call it the spectral deformation space of $S$. (1) What is $\Sigma^\infty(Spec\ R)$ for a commutative ring $R$? This is equivalent to asking for a description of the collection of all $\mathbb{E}_\infty$ rings whose $\pi_0$ is $R$. Examples for some elementary rings, e.g., $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{F}_p$, $\mathbb{Q}_p$, $\mathbb{\bar{Q}}$ would be specially welcome. (2) What is $\Sigma^\infty(\mathbb{P}^n_k)$? Here $\mathbb{P}^n_k$ is the $n$-dimensional projective space over a field $k$. (3) What is the structure of $\Sigma^\infty(S)$? Is $\Sigma^\infty$ functorial for morphisms of schemes? I'd also like to ask the same questions for the derived deformation space $\Sigma^{\triangledown}(S)$ of $S$ obtained by replacing spectral schemes by derived schemes in the discussion above, with a simple example in which it differs from $\Sigma^\infty(S)$. REPLY [6 votes]: In general, these are incredibly hard questions. It seems to me that one natural question to ask (if you are interested in $\pi_0$ of ring spectra) would be about understanding even periodic $\mathbf{E}_\infty$-rings $A$ with $\pi_0 A = R$. Alternatively, you could attempt to understand those $\mathbf{E}_\infty$-rings $A$ with $\pi_\ast A = R_\ast$, where $R_\ast$ is some graded ring. In what follows, I'll attempt to address some of these questions (in particular, this is not a complete answer to your question). Let me begin with some generalities. If you restrict to asking for $K(1)$-local $\mathbf{E}_\infty$-rings $A$ with $\pi_0 A = R$, then there are a few things you can say. For instance, if $p=0$ in $R$, then the collection of such $A$ is necessarily empty if $R$ is nonzero. If $R$ does not have a lift of Frobenius, then the collection of such $A$ is again empty. (Examples of such non-liftability results, in the $K(d)$-local $\mathbf{E}_n$-case for varying $d$ and $n$, are in the paper referred to above; there's an updated version on my webpage, which hasn't yet found its way onto the arXiv. Also see Schwanzl-Vogt-Waldhausen's "Adjoining roots of unity to E∞ ring spectra in good cases -- a remark".) There are also some positive results: suppose $R_\ast$ is an even-periodic graded $p$-torsion free ring such that $R_0$ is the $p$-adic completion of a smooth $\mathbf{Z}_p$-algebra, which is equipped with: a formal group classified by a map $MUP_0 \to R_0$ for which the induced formal group over $R_0/p$ is of height $1$, and an action of $\mathbf{Z}_p^\times$ and a compatible $p$-derivation (a "$\theta$-algebra structure"). Then there is a(n even periodic, complex oriented) $K(1)$-local $\mathbf{E}_\infty$-ring $A$ such that $\pi_\ast A = R_\ast$. This can be deduced from the results in the following paper of Lawson's and Naumann's: https://arxiv.org/abs/1101.3897. Admittedly, the conditions specified above seem like a lot; however, I know of no way to get past this. I also do not know how one might generalize this to constructing $K(n)$-local $\mathbf{E}_\infty$-rings for $n\geq 2$. The obstruction stems from the fact that we just do not know as much about power operations at heights $\geq 2$ as we do at height $1$. (This is not to downplay the work of Rezk, Zhu, and others on height $2$ power operations.) Let me now give some examples. Fix an $\mathbf{E}_\infty$-ring $A$. Suppose $\pi_0 A = \mathbf{Q}$. Then $A$ is an $\mathbf{E}_\infty$-$\mathbf{Q}$-algebra, and these are very well understood. Indeed, $\mathbf{E}_\infty$-$\mathbf{Q}$-algebras are the same as commutative dg-$\mathbf{Q}$-algebras. The same is true of any $\mathbf{E}_\infty$-$R$-algebras over any (ordinary) $\mathbf{Q}$-algebra $R$ (hence, in particular, $\mathbf{Q}_p$ and $\overline{\mathbf{Q}}$). Suppose $\pi_0 A = \mathbf{F}_p$. If $A$ is $p$-local, then a result of Hopkins and Mahowald implies that $A$ is an $\mathbf{E}_\infty$-$\mathbf{F}_p$-algebra. There can be many such $A$, because Steenrod operations exist. For example, take the $\mathbf{F}_2$-algebras $A_1 = \mathbf{F}_p \otimes^\mathbf{L}_{\mathbf{Z}} \mathbf{F}_p$ (the fiber product in "derived schemes") and $A_2 = \mathbf{F}_p \otimes_{\mathbb{S}} \mathbf{F}_p$ (the fiber product in "spectral schemes"). Then $\pi_\ast A_1 = \mathbf{F}_p[t]/t^2$ with $|t|=1$ (so $\pi_0 A_1 = \mathbf{F}_p$), but $\pi_\ast A_2$ is the mod $p$ dual Steenrod algebra (so $\pi_0 A_2 = \mathbf{F}_p$ as well), but one is clearly much larger than the other! This gives an example showing that (in your notation) $\Sigma^\infty(\mathbf{F}_p) \supsetneq \Sigma^\triangledown(\mathbf{F}_p)$. Suppose $\pi_0 A = \mathbf{Z}$. Then, there are a ton of possibilities for $A$. For instance, $A$ could be $\mathbb{S}, MU, KU, KO, TMF, Tmf, ...$ (and their connective covers). Again, even periodicity would be a reasonable condition to impose. Then, the number of examples is cut down significantly (in the list provided above, for instance, you're only left with $KU$). Hopefully someone else more knowledgeable in these topics can provide a more comprehensive and satisfactory answer to your question.<|endoftext|> TITLE: Interesting properties in $...\to K(\mathbb{Z}_4,1) \overset{f}{\to} K(\mathbb{Z}_2,1)\overset{g}{\to}K(\mathbb{Z}_2,2) \to ...$ QUESTION [5 upvotes]: Let $K(G,n)$ be the Eilenberg Maclane space. Consider the map from $$ K(\mathbb{Z}_2,1) \to K(\mathbb{Z}_4,1) \overset{f}{\to} K(\mathbb{Z}_2,1)\overset{g}{\to}K(\mathbb{Z}_2,2) \to \dots, $$ It looks that we can represent the map from $K(\mathbb{Z}_2,1)\to K(\mathbb{Z}_2,2)$ relating to the generator of cohomology group $$ q \in H^2(K(\mathbb{Z}_2,1),\mathbb{Z}_2)=H^2(B\mathbb{Z}_2,\mathbb{Z}_2)=\mathbb{Z}_2. $$ Say $$ p \in H^1(K(\mathbb{Z}_2,1),\mathbb{Z}_2)=H^1(B\mathbb{Z}_2,\mathbb{Z}_2)=\mathbb{Z}_2, $$ then $q$ and $p$ are group cocycles related by $$ q = p \cup p. $$ Roughly, say due to the exact sequence above, we may say the kernel (in $K(\mathbb{Z}_2,1)$) of the map $g$, such that $$ q = p \cup p =0, $$ matches the image (in $K(\mathbb{Z}_2,1)$) of $$ f. $$ Can we see explicitly that $q = p \cup p$ may be non-zero in $H^2(K(\mathbb{Z}_2,1),\mathbb{Z}_2)=H^2(B\mathbb{Z}_2,\mathbb{Z}_2)=\mathbb{Z}_2$, but $q =0$ when we pull back via $f$ from the $K(\mathbb{Z}_2,1)$ to $K(\mathbb{Z}_4,1)$? Can this be demonstrated explicitly at the level of group cocycle? REPLY [9 votes]: Represent $p$ by the identity map $id: \mathbb{Z}_2 \to \mathbb{Z}_2$. Then $(p\cup p)(a,b) = p(a)p(b)$ is non-zero only on the 2-chain $(1,1)$. Namely, as a polynomial mod 2, $(p\cup p)(a,b) = ab$. When lifted to $\mathbb{Z}_4$, we get $p\cup p(a,b)=ab \mod 2$. Consider the function $\gamma: \mathbb{Z}_4\to \mathbb{Z}_2$ defined by $\gamma(x) = \frac{x^2-x}{2} \mod{2}$, namely $\gamma(0)=0,\gamma(1)=0,\gamma(2)=1,\gamma(3)=1$. Then $$d\gamma(x,y)=\gamma(x+y)-\gamma(x)-\gamma(y)=xy,$$ by $(x+y)^2-x^2-y^2 + (x +y)-x-y=2xy$. Note that $\gamma$ is really well defined because always $x^2 \equiv x \mod 2$ and you can see that if you add a multiple of $4$ to $x$ it don't changes the residue mod $2$ of the answer. So $d\gamma = p\cup p$ in $C^2(\mathbb{Z}_4,\mathbb{Z}_2)$.<|endoftext|> TITLE: Factoring polynomials over the abelian closure of the rationals QUESTION [12 upvotes]: What algorithms are known to perform the following task? Input: a univariate polynomial over the rationals $f \in \mathbb{Q}[t]$. Output: the factorization of $f$ into irreducible factors over the abelian closure $\mathbb{Q}^{\mathrm{ab}} = \varinjlim_{n\in\mathbb{N}} \mathbb{Q}(\zeta_n)$ (where $\zeta_n := \exp(2i\pi/n)$ is an $n$-th root of unity), where elements of $\mathbb{Q}^{\mathrm{ab}}$ are expressed as combinations of roots of unity. I can show that this is, indeed, Church-Turing computable by way of the following argument (which yields an exceedingly slow algorithm): One can perform exact computations with elements of the algebraic closure $\mathbb{Q}^{\mathrm{alg}}$ (i.e., perform rational operations, comparisons, factorizations of polynomials, etc.). See, e.g., here and here. One can also compute the Galois group (over $\mathbb{Q}$) of an element of $\mathbb{Q}^{\mathrm{alg}}$ (represented as in the first point through, among other things, its minimal polynomial over $\mathbb{Q}$): see here. This makes it possible to check whether an algebraic number belongs to $\mathbb{Q}^{\mathrm{ab}}$. In particular, one can factor $f$ over $\mathbb{Q}^{\mathrm{alg}}$ into distinct roots and, for every subset of these roots, check whether all of their symmetric functions belong to $\mathbb{Q}^{\mathrm{ab}}$: the smallest possible such subsets provide the desired factorization. Given an element $\alpha\in\mathbb{Q}^{\mathrm{ab}}$ (represented as an algebraic number as in the first point), one can test whether $\alpha \in \mathbb{Q}(\zeta_n)$, so one can find such an $n$ by enumerating all natural numbers until one is found, at which point expressing $\alpha$ as a sum of powers of $\zeta_n$ amounts merely to solving a system of linear equations over $\mathbb{Q}$. This is… not exactly efficient. I'm hoping one can do better. Is there such an algorithm in the literature? REPLY [4 votes]: I will try to sketch an algorithm which avoids the computation of the Galois group, since such a computation is very expensive. The idea is to look at the factorisations of the reductions of $f$ modulo sufficiently many primes $p$. This is the same idea as in the computation of the Galois group, but the point is that we don't need the whole Galois group, only its maximal abelian quotient, so things should be simpler. We may assume that $f \in \mathbb{Z}[x]$ is monic and irreducible. Let $K = \mathbb{Q}[x]/(f)$ be the number field defined by $f$ and let $L$ be the Galois closure of $K$. We embed both $L$ and $\mathbb{Q}^{\mathrm{ab}}$ into some algebraic closure $\mathbb{Q}^{\mathrm{alg}}$. Let $K^{\mathrm{ab}} = K \cap \mathbb{Q}^{\mathrm{ab}}$ and $L^{\mathrm{ab}} = L \cap \mathbb{Q}^{\mathrm{ab}}$ be the maximal abelian subfields of $K$ and $L$ (sorry for the unusual notation...). Let $f = g_1 g_2 \cdots g_s$ be the factorisation of $f$ over $\mathbb{Q}^{\mathrm{ab}}$. By Galois theory, we know that each $g_i$ has coefficients in $L^{\mathrm{ab}} \cap K = K^{\mathrm{ab}}$. Definition. A bound for a finite abelian extension $F$ of $\mathbb{Q}$ is a pair $(N,G)$ where $N \geq 1$ is an integer and $G$ is a subgroup of $(\mathbb{Z}/N\mathbb{Z})^\times$ such that $F \subset \mathbb{Q}(\zeta_N)^G$. Note that for any abelian $F/\mathbb{Q}$ we always have the trivial bound $(|\Delta_F|,\{1\})$ where $\Delta_F$ is the (absolute) discriminant of $F$. Also, it is not hard to see that among all the bounds for $F$ there is a unique minimal one $(N,G)$. The integer $N$ is the conductor of $F$, and $G$ is the norm group, namely the kernel of $(\mathbb{Z}/N\mathbb{Z})^\times \to \mathrm{Gal}(F/\mathbb{Q})$. So starting with $f$, we know that $K^{\mathrm{ab}}$ is bounded by $(|\Delta_K|,\{1\})$ (it is possible to simply compute the discriminant of $f$, but this may lead to a worse bound). So we start with $(N,G)=(|\Delta_K|,\{1\})$ and we want to improve the bound. In fact, since $K^{\mathrm{ab}}$ is contained in both $K$ and $\mathbb{Q}(\zeta_N)$, we may take $G$ to be the image of the power map $x \mapsto x^\delta$ in $(\mathbb{Z}/N\mathbb{Z})^\times$, where $\delta= \operatorname{gcd}(\deg(f),\varphi(N))$. Now we look at the reduction of $f$ modulo some primes, similarly as in Algorithm 4.4.3 in Cohen, Advanced topics in computational number theory. Let $p$ be a prime not dividing $N$. Let $\bar{f}$ be the reduction of $f$ mod $p$, and let $\bar{f}=f_1 f_2 \cdots f_r$ be its factorization in $\mathbb{F}_p[x]$. The $f_i$ are pairwise distinct monic irreducible polynomials, say of degree $d_i$. If $K$ were abelian then we would know that $d_1=\ldots=d_r=d$ where $d$ is the order of $p$ in $\mathrm{Gal}(K/\mathbb{Q})$ (seen as a quotient of $(\mathbb{Z}/N\mathbb{Z})^\times$). We are not in this case, but still know by multiplicativity of the inertia degree that the order $d$ of $p$ in $\mathrm{Gal}(K^{\mathrm{ab}}/\mathbb{Q})$ divides each $d_i$. So we adjoin to the group $G$ the element $p^h$ where $h$ is the gcd of $d_1,\ldots,d_r$. Doing this for enough primes $p$, we should get a reasonable bound $(N,G)$. Now we can replace $(N,G)$ by a minimal equivalent bound $(N',G')$ where $N'$ divides $N$. (Precisely, $N'$ is the minimal divisor of $N$ such that $G$ contains the kernel of reduction mod $N'$, and $G'$ is the image of $G$ mod $N'$.) At this point it remains to factor $f$ over $K'=\mathbb{Q}(\zeta_N)^G$. By the theory of Gauss sums, we know a primitive element $\alpha$ in $K'$ (written in terms of $\zeta_N$) and its minimal polynomial, so we simply factor $f$ over the number field $\mathbb{Q}(\alpha)$. A natural question is how many primes to use. In fact, if we are only interested in having an algorithm, this doesn't matter: we could simply have used the trivial bound. To get a more efficient algorithm, we may use some primes but it's not clear to me, for example, how many primes are needed to ensure that we get $K'=K^{\mathrm{ab}}$. In general, it is known that the norm group of an abelian extension is generated by the prime powers $p^d$ as above when $p$ runs through the primes less than an explicit bound $B$, see Algorithm 4.4.5 in Cohen. But the bound $B$ is a priori exponential in the size of the discriminant, while there is a much better polynomial bound assuming GRH, so some choices need to be made. Another difficulty is that the gcd $h$ used above need not equal $d$. Maybe the Cebotarev theorem (an effective version!) could be used to prove this doesn't matter, but this may be a non-trivial task and I haven't thought about it. So I would make the following compromise: simply use a certain amount of primes without caring whether $K'=K^{\mathrm{ab}}$ or not. At the end, if really needed, one could determine over which minimal extension the factorisation of $f$ actually takes place. The "right" number of primes will depend on the parameters but could be determined experimentally or by finding some heuristics.<|endoftext|> TITLE: Escaping from a centralizer QUESTION [15 upvotes]: Let $G = Sym(n)$, $n$ even. Let $H TITLE: What's special about elliptic cohomology? QUESTION [7 upvotes]: Apologies for any basic mistakes in this question; I'm a beginner to this theory and don't have anyone at my institution to consult for advice. What I mean is, if you take an elliptic curve $E$ over $\mathbb{Q}$ say, you should get associated an cohomology theory by taking some deformation of it/putting it in a family and completing. (At least as far as I know, the deformation is necessary, since otherwise LEFT fails at supersingular primes. Is this correct?) It seems to me that you could pretty much do this with any formal group over $\mathbb{Z}$; just find some deformation so that the heights are all generically $1$, then apply LEFT. (Or is this maybe harder to do than I think? Seems it wouldn't be, though, just from stacky considerations.) My question is, is there some intrinsic interest in the cohomology theories actually arising from elliptic curves, as opposed to these ones? Or is the reason for the interest entirely because of the geometry of the universal tmf, and its speculative connections to loop groups, the Witten genus, etc.? Or maybe is the principal interest in the direction of searching for a good equivariant theory which encodes all the elliptic curve's geometry, instead of just its formal Neron model? REPLY [3 votes]: Here's a partial answer which I find somewhat satisfying: elliptic genera have a natural geometric definition as genera which kill the projective total spaces of even-dimensional complex bundles over compact oriented manifolds, and are closely related to spin geometry and elliptic operators on free loop spaces. This gives a principled way to distinguish elliptic genera from others, coming from geometry. I would still be interested to see if this distinction is possible in purely homotopical terms, or if this is a case where the content really is purely geometrical (and "just happens" to link to the algebraizability of the associated formal group!).<|endoftext|> TITLE: Derived equivalences of Dyck paths QUESTION [10 upvotes]: Call two Dyck paths $D_1$ and $D_2$ derived equivalent in case their corresponding Nakayama algebras are derived equivalent (The Dyck path of a Nakayama algebra with a linear quiver is just the top boundary of its Auslander-Reiten quiver, so we can identify Nakayama algebras with a linear quiver with Dyck paths). Derived categories of Nakayama algebras appeared in the literature (see for example https://www.sciencedirect.com/science/article/pii/S0001870813000182 ) but it seems to be wide open to classify Dyck paths with respect to derived equivalences (or?). Call a Nakayama algebra (or the corresponding Dyck path) bouncing in case the Kupisch series is of the form $[a_1+1,a_1,...,3,2,a_2+1,a_2,...,2,...,a_r+1,a_r,...,3,2,1]$. The truth of the conjecture in What are the periodic Dyck paths? would imply the following: In case a Dyck path $D_1$ is derived equivalent to a bouncing Dyck path $D_2$, then also $D_1$ is bouncing. Thus, it would give that being bouncing is a derived invariant. Question 1: Do we know whether it is true that being bouncing is a derived invariant? That is: In case a Nakayama algebra $D_1$ is derived equivalent to a bouncing Nakayama algebra $D_2$, then also $D_1$ is bouncing? Question 2: Is it known which bouncing Nakayama algebras with a fixed number of simple modules are derived equivalent to eachother? How many equivalence classes (up to derived equivalence) exist of bouncing Nakayama algebras? REPLY [6 votes]: Question 1 has a positive answer by the answer of Gjergji Zaimi in this thread: What are the periodic Dyck paths? . Here a positive answer to question 1 and 2 using a complicated classification result that uses several other deep results from the literature (so more elementary proofs of Question 2 are welcome): Question 1 and 2 have a positive answer by the classification of iterated tilted algebras of Dynkin type $\mathcal{A}$. See the main theorem of https://www.tandfonline.com/doi/abs/10.1080/00927878108822697 , which implies here that the bouncing Dyck paths correspond to Nakayama algebras with a linear quiver having only relations of length two. (edit: I think I found an elementary answer to Question 2 using tilting modules. I might post it here soon)<|endoftext|> TITLE: Normal Fuchsian subgroups QUESTION [5 upvotes]: I've been working with Fuchsian groups and from geometrical motivations finding a cocompact normal Fuchsian subgroups of $PSL(2,\mathbb{R})$ would have intresting properties for my research. It is known that $SL(2,\mathbb{R})$ has no connected normal subgroups other than the its centre { Id,-Id }, however it might have discrete normal subgroups. I been working with quaternion generated cocompact Fuchsian subgroups which are in addition purely hyperbolic, all its elements have Trace bigger 2, and I couln't find any normal subgroups of $SL(2,\mathbb{R})$ there, I suspect that I have to include some elliptic elements to improve my chances. To add up: it possible for $PSL(2,\mathbb{R})$ to admit discrete normal subgroups? how about cocompact normal subgroups? Note: Notice that I'm not talking about normal subgroups of Fuchsian subgroups. REPLY [8 votes]: Let $\Gamma$ be a discrete subgroup of a connected Lie group $G$. Suppose that $\Gamma$ is normal. For given $\gamma\in\Gamma$ the conjugacy class is the image of $G$ under the continuous map $x\mapsto x\gamma x^{-1}$, therefore it is connected. Since it lies in $\Gamma$, it consists of one point only, and as $x=1$ occurs, this point is $\gamma$. This means that $\Gamma$ lies in the center of $G$. In the case $G=SL_2({\mathbb R})$ the center is $\{ \pm 1\}$. So $\Gamma $ is either trivial or $\{\pm 1\}$.<|endoftext|> TITLE: Does geometrization of Alexandrov 3-spaces follow from that of 3-orbifolds? QUESTION [8 upvotes]: Galaz-Garcia and Guijarro proved the geometrization of closed (compact, boundaryless) Alexandrov 3-spaces. Part of the strategy was to use the so-called ramified double cover $\tilde{X}$ of the space $X$. This ramified cover is a smooth $3$-manifold. Being this the case, the space $X$ would be isometric to a Riemannian $3$-orbifold. I don't quite follow why then, it's not immediate that the geometrization of $X$ follows from the geometrization of $3$-orbifolds? EDIT: Please note that My question is not whether geometrization can be deduced from "the geometrization of the ramified cover" but, once it has been deduced that the space itself is isometric to a Riemannian 3-orbifold, whether this information together with the geometrization of 3-orbifolds can be used to conclude. REPLY [5 votes]: In principle, yes. Note, however, that topologically singular Alexandrov 3-spaces are homeomorphic to non-orientable orbifolds. We could not find an appropriate reference for the geometrization in the non-orientable case (where topological singularities are present). To the best of my knowledge, the geometrization statements in the literature are for orientable orbifolds.<|endoftext|> TITLE: Is there a size 2 generating set of the signed symmetric group $B_n$? QUESTION [5 upvotes]: The signed symmetric group $B_n$ is a permutation group where the underlying set is $B_n=\{\sigma \in S_{A_n}| \forall x \in A_n, \sigma(-x)= -\sigma(x)\}$ with $A_n=\{-n,-(n-1),-(n-2),\cdots,-1,1,\cdots, n-1,n\}$. $B_n$ can be generated by the 3 permutations $(1,2)(-1,-2) ; (1,2,\cdots, n)(-1,-2,\cdots, -n)$ and $(-i,i)$ for some $i$. Is there a generating set of $B_n$ with only 2 elements? REPLY [10 votes]: Yes. Take two $a,b$ generators of $S_n$, where $b$ has odd order and fixes point $1$. For example, if $n$ is even, let $a=(1,2)$, $b=(2,3,\ldots,n)$. If $n$ is odd let $a=(1,2,3,4)$, $b=(3,4,\ldots,n)$. Let $\bar{a},\bar{b}$ be their natural images$^\dagger$ in $B_n$. Then $B_n = \langle \bar{a}, \bar{b}(1,-1) \rangle$. This is because $\bar{b}$ and $(-1,1)$ commiute and have coprime order, so both $\bar{b}$ and $(-1,1)$ are powers of $\bar{b}(-1,1)$ and hence $\langle \bar{a}, \bar{b}(1,-1) \rangle=\langle \bar{a}, \bar{b},(1,-1) \rangle =B_n$. $\dagger$ : For a permutation $\alpha$ of $\{1,2,\ldots,n\}$ let $\alpha'$ be the permutation of $\{-1,-2,\ldots,-n\}$ defined by $\alpha(-i) := -\alpha(i)$ for $1 \le i \le n$. By the natural embeddding $S_n \to B_n$ I mean the map defined by $\alpha \mapsto \alpha\alpha'$.<|endoftext|> TITLE: $K_X+B \equiv 0$ implies $K_X + B \sim_\mathbb{Q} 0$? QUESTION [5 upvotes]: Let $(X,B)$ be a projective log canonical pair (here I mean $B \geq 0$). Assume that the coefficients of $B$ are rational, and that $K_X+B \equiv 0$. Is it true that $K_X + B \sim_\mathbb{Q} 0$? I know it is true if $(X,B)$ is klt, and "The moduli b-divisor of an lc-trivial fibration" by Ambro has a proof of it. I am curious whether the result extends to singularities that are lc and not klt, and, if so, where I could find a reference. REPLY [10 votes]: For lc pair or slc pair, it is true. This is Gongyo’s result. See [J. ALGEBRAIC GEOMETRY 22 (2013) 549–564]. BTW, the relative version is also true, which is not a trivial generalization of the absolute case. It is proved by Hacon and Xu [On Finiteness of B-representation and Semi-log Canonical Abundance].<|endoftext|> TITLE: Extreme points of convex compact sets QUESTION [17 upvotes]: Preparing to a lecture on Krein--Milman theorem I read in W. Rudin's Functional analysis textbook (1973) that it is unknown whether any convex compact set in any topological vector space has an extreme point. Is it still unknown? REPLY [22 votes]: A counterexample is given in the following paper: Roberts, James W. "A compact convex set with no extreme points." Studia Mathematica 60.3 (1977): 255-266. I did not see the original paper, but an exposition can be found in Section 5.6 of the book "Metric linear spaces" by Rolewicz.<|endoftext|> TITLE: Has Apéry's proof of the irrationality of $\zeta(3)$ ever been used to prove the irrationality of other constants? QUESTION [19 upvotes]: Apéry's proof of the irrationality of $\zeta(3)$ astounded contemporary mathematicians for its wealth of new ideas and techniques in proving the irrationality of a known constant. It is often the case that such new Theorems, whose substance contains completely novel approaches to an old problem, inspire solutions to other, similar problems by providing the tools necessary to tackle them. However, in the case of Apéry's Theorem, it seems like the new techniques presented in the proof are not easily used within other contexts. This has led me to ask the question: Has Apéry's proof of the irrationality of $\zeta(3)$ ever been used to prove the irrationality of other constants? Since the proof contains numerous new ideas, I also put forward the following corollary question: Which techniques employed in Apéry's proof of the irrationality of $\zeta(3)$ have been used within other proofs, whether in the field of irrationality/transcendence theory or other fields? REPLY [9 votes]: As Frits Beukers write in http://www.staff.science.uu.nl/~beuke106/caen.pdf "Ironically all generalisations tried so far did not give any new interesting results. Only through a combination of miracles such generalisations seem to work, which in practice means that we fall back to $\zeta(2)$ or $\zeta(3)$ again". Nevertheless Apéry-like numbers do appear in somewhat unexpected places: In the study of the moments $$W_n(s)=\int\limits_{[0,1]^n}\left |\sum_{k=1}^ne^{2\pi i x_k}\right|d\vec{x}$$ of the distance traveled by a walk in the plane with unit steps in random directions. Namely for 3- and 4-step short random walks we have (see https://link.springer.com/article/10.1007/s11139-011-9325-y ) $$W_3(2k)=\sum_{j=0}^k\binom{k}{k}^2\binom{2j}{j},$$ and $$W_4(2k)=\sum_{j=0}^k\binom{k}{k}^2\binom{2j}{j}\binom{2(k-j)}{k-j}.$$ A striking feature that connects the 3- and 4-step random walk densities to irrationality proofs is their modularity: https://arxiv.org/abs/1103.2995 (Densities of short uniform random walks, by J.M. Borwein, A> Straub, J. Wan and W. Zudilin). See Beuker's "Irrationality Proofs Using Modular Forms": http://carmasite.newcastle.edu.au/wadim/zw/Beukers-Asterisque1987.pdf Positivity of rational functions and their diagonals (an article by Armin Straub and Wadim Zudilin): https://arxiv.org/abs/1312.3732 In some series for $1/\pi$:https://icerm.brown.edu/materials/Slides/tw-14-5/Apery_numbers_and_their_experimental_siblings_]_Armin_Straub,_University_of_Illinois_at_Urbana-Champaign.pdf Calabi-Yau differential equations: https://www.cambridge.org/core/journals/proceedings-of-the-edinburgh-mathematical-society/article/generalizations-of-clausens-formula-and-algebraic-transformations-of-calabiyau-differential-equations/D5A27594AC2F57F9B933A7716506C027 (Generalizations of Clausen's Formula and algebraic transformations of Calabi–Yau differential equations, by G. Almkvist, D. van Straten and W. Zudilin).<|endoftext|> TITLE: Random reals preserving Cohen reals QUESTION [5 upvotes]: Suppose we have a model (of $\mathsf{ZFC}$) $M$, and that $x\in 2^\omega$ is random over $M$, and that $y\in 2^{\omega}$ is Cohen over $M$. My question is whether $y$ is also Cohen over $M[x]$. In other words, if I have a real that's Cohen over a model, is it still Cohen over the model that results from performing Random forcing? REPLY [9 votes]: That depends on the particular random real $x$ and Cohen real $y$. On the one hand, I could first choose $x$ random over $M$ and then choose $y$ Cohen over $M[x]$. Then $y$ is also Cohen over the submodel $M$, so it's an example where the answer to your question is yes. On the other hand, I could first choose $y$ Cohen over $M$ and then choose $x$ random over $M[y]$. Then $x$ is also random over the submodel $M$. I'll show that $y$ is not Cohen over $M[x]$. Partition $\omega$ into intervals $I_n$ of rapidly increasing length. (In fact, it suffices to take $I_n$ of length $n$, but "rapidly" avoids the need for arithmetic.) Define a symmetric binary relation $R$ on $2^\omega$ by putting $aRb$ iff $a$ and $b$ agree on infinitely many of these intervals. Note that, for any $a$, the set of $b$'s $R$-related to $a$ is comeager but has Lebesgue measure 0. Note also that $R$ is a low-level Borel set with code in $M$ (provided you chose the sequence of $I_n$'s in $M$). Since $x$ is random over $M[y]$, we have that $x$ and $y$ are not $R$-related. But then $y$ cannot be Cohen over $M[x]$.<|endoftext|> TITLE: When is the etale cohomology of $\mathrm{Sym}^n(X)$ isomorphic to the $\Sigma_n$-invariants in the étale cohomology of $X^n$? QUESTION [15 upvotes]: Suppose $X$ is a smooth projective variety defined over an arbitrary algebraically closed field $k$, and consider the action of $\Sigma_n$ on the $n$-fold product $X^n$. Is it true that $H_{\acute{e}t}^i(\mathrm{Sym}^n(X),\mathbb{Q}_\ell)\cong H_{\acute{e}t}^i(X^n,\mathbb{Q}_\ell)^{\Sigma_n}$? In particular, what happens in the case where $\operatorname{char} k=p>0$? In Grothendieck's Toh\^oku paper Sec. 5.2, he determines sufficient conditions to ensure that, for a topological space $X$ with a finite group $G$ acting on it (not necessarily faithfully), $H^i(X/G,\mathcal{A})\cong H^i(X,\mathcal{A})^G$ for a sheaf $\mathcal{A}$ (Cor. to Prop. 5.2.3). In characteristic zero, comparison theorems allow me to appeal to this result. In positive characteristic, if the variety lifts to characteristic zero, then I can make the same argument, but it seems like there ought to be a direct proof of this fact. I am particularly interested in when $X$ is a surface, but would be happy to know of any general results (with references) similar to Grothendieck's result above. REPLY [3 votes]: The following is a particular case of (SGA 4.3, XVII Th. 5.5.21) : Let $X$ be a quasi-projective scheme over an algebraically closed field $k$. Then for any $n \geq 0$ and any $r \geq 1$ we have $$ R \Gamma_c(\mathrm{Sym}_{k}^n(X), \mathbb{Z}/r\mathbb{Z}) = L \Gamma^n ( R\Gamma_c(X, \mathbb{Z}/r\mathbb{Z})). $$ The functor $L \Gamma^n$ is the left derived functor of the non-additive functor $\Gamma^n$, which coincides on flat modules with the ``symmetric tensor'' functor. The $R \Gamma_c$ denotes higher direct image with compact supports. (SGA 4.3, XVII Th. 5.5.21) gives a more general statement, in a relative situation, with more general coefficients.<|endoftext|> TITLE: Non-Hamiltonian actions in physics QUESTION [5 upvotes]: I was reading the following article when I came across the interesting sentence "non-Hamiltonian [symplectic group] actions also occur in physics" I took a cursory look at the article cited but nothing jumped out at me to back up the claim. Does anyone have a good example of such a thing? Ideally a classical system, if possible, as I don't have much contact with the quantum world :) REPLY [4 votes]: The master thesis Nonholonomic Dynamical Systems by Brett Ryland contains several examples of non-Hamiltonian systems from classical physics: the dynamics of a laser and the evolution of a gas flame (page 14), the rattleback (page 29).<|endoftext|> TITLE: Pull-back divisor being Cartier QUESTION [7 upvotes]: Let $\pi \colon X \rightarrow Y$ be a projective morphism with connected fibers between normal quasi-projective varieties. Let $N$ be a $\mathbb{Q}$-Cartier divisor on $Y$ so that $\pi^*(N)$ is Cartier. Does it follows that $N$ is itself Cartier? REPLY [5 votes]: Following the clarification in the comments, I am interpreting the question as follows. Question. For an effective Weil divisor $N$ on $Y$, for an effective Cartier divisor $A$ on $X$, for a positive integer $\ell$ such that the effective Weil divisor $\ell N$ is Cartier and such that the pullback effective Cartier divisor $\pi^*(\ell N)$ equals $\ell A$ as effective Cartier divisors, is $N$ Cartier? The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano. The minimal resolution of a cone over a smooth plane cubic is a ruled surface. Let $C\subset \mathbb{P}^2$ be a smooth, plane cubic (a genus $1$ curve). Let $Y$ be the projective cone in $\mathbb{P}^3$ over $C$. The minimal desingularization of $Y$, $$\pi:X\to Y,$$ is the closure in $Y\times C$ of the graph of the linear projection from $Y$ to $C$. Denote by $\rho$ the projection, $$ \rho:X\to C.$$ This morphism is a $\mathbb{P}^1$-bundle. The morphism $\rho$ maps the exceptional locus $E$ of $\pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $\rho$. The normal sheaf $\mathcal{O}_X(\underline{E})|_E$ is isomorphic to $\mathcal{O}_{\mathbb{P}^2}(-1)|_C$. Effective Weil divisors "torsion-equivalent" to a conical hyperplane class. Let $H\subset C$ be the restriction to $C$ of a general hyperplane in $\mathbb{P}^2$. Let $D\subset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $\ell>1$ in the Picard group of $C$, i.e., $\ell H - \ell D$ is the divisor of a rational function $f$ on $C$. Let $M\subset Y$, resp. $N \subset Y$, be the cone over $H$, resp. $D$. Note that $\ell N$ and $\ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $\mathcal{O}_{\mathbb{P}^3}(\ell)|_Y$. In fact, $\ell$ is the smallest positive integer such that $\ell N$ is a Cartier divisor on $Y$. Denote by $\widetilde{M}\subset X$, resp. by $\widetilde{N}\subset X$, the strict transform under $\pi$ of $M$, resp. of $N$. Note that $\ell \widetilde{M}-\ell\widetilde{N}$ equals the Cartier divisor of $f\circ \rho$, as does the total pullback under $\pi$ of the Cartier divisor $\ell N-\ell M$. Thus, the coefficient of $E$ in the pullback of $\ell M$ equals the coefficient of $E$ in the pullback of $\ell N$. Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $\widetilde{M}$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $\ell M$ equals $\ell \widetilde{M} + \ell E$. Therefore also $\ell N$ equals t$\ell \widetilde{N} +\ell E$. Although $N$ is not Cartier on $Y$, the pullback of $\ell N$ equals $\ell A$ on $X$ for the effective Cartier divisor $A=\widetilde{N}+E$.<|endoftext|> TITLE: Is there a closed non-smoothable 4-manifold with zero Euler characteristic? QUESTION [10 upvotes]: I will just repeat the title: Is there a closed non-smoothable 4-manifold with zero Euler characteristic? I am guessing yes simply based on other existence theorems I have seen for 4-manifolds. REPLY [12 votes]: Alternatively, one can start with the $E_8$-manifold and connect sum with (five copies of) $S^1\times S^3$, and appeal to Donaldson's diagonalisation theorem instead. More precisely, the (negative) $E_8$-plumbing $P$ bounds the Poincaré homology sphere $Y$; by Freedman's theorem, $Y$ is also the boundary of a contractible topological 4-manifold $W$, and gluing $P\cup_Y -W$ yields a closed, simply connected topological 4-manifold $X$ with intersection form $-E_8$; $\chi(X) = 10$. The connected sum $X\#5(S^1\times S^3)$ has Euler characteristic 0, by Mike's computation, and the intersection form is still $-E_8$. Hence, by Donaldson's theorem, $X\#5(S^1\times S^3)$ does not have a smooth structure, since its intersection form is negative definite but not diagonal. REPLY [11 votes]: You can also get this from Donaldson's theorem by a similar device. Take a non-diagonalizable definite form with even rank $2n$, and realize it (Freedman again) by a simply connected manifold. Then $W\ \#\ (n+1) (S^1 \times S^3)$ is not smoothable and has Euler characteristic $0$. The argument is that if it were smoothable, then you could surger away the fundamental group and realize that intersection form by a simply connected manifold.<|endoftext|> TITLE: Surjective order-preserving map $f:{\cal P}(X)\to \text{Part}(X)$ QUESTION [7 upvotes]: Let $X$ be a set, and let $\text{Part}(X)$ denote the collection of all partitions of $X$. For $A, B\in \text{Part}(X)$ we set $A\leq B$ if $A$ refines $B$, that is for all $a\in A$ there is $b\in B$ such that $a\subseteq b$. This relation defines a partial order on $\text{Part}(X)$. If $X$ is an infinite set, is there a surjective order-preserving map $f:{\cal P}(X)\to \text{Part}(X)$ (where ${\cal P}(X)$ denotes the power-set of $X$, ordered by $\subseteq$)? REPLY [6 votes]: The positive solution uses an equivalent of the Axiom of Choice: for every infinite set $A$ there is a bijection $f:A\to A\times A$. In the basic Fraenkel Model (section 4.3 in Jech's Axiom of Choice) there is an infinite set where there is no surjection as in the question. Let $A$ be the infinite set of atoms in the model. Step 1. Every subset of $A$ in the model is finite or cofinite. Let $X\subseteq A$ and let $E\subseteq A$ be finite such that $\operatorname{fix} E\subseteq\operatorname{sym} X$. If $X\subseteq E$ then $X$~is finite. If $X\not\subseteq E$ then $A\setminus E\subseteq X$: fix $x\in X\setminus E$ and let $y\in A\setminus E$ be arbitrary; the permutation~$\pi$ that interchanges $x$ and $y$ is in $\operatorname{fix} E$, so $\pi X=X$, but this means that $y=\pi(x)\in\pi X=X$. Step 2. Let $P$ be partition of $A$ in the model and let $E\subseteq A$ be finite with $\operatorname{fix} E\subseteq \operatorname{sym} P$. Assume there are $a$ and $b$ in $A\setminus E$ that are distinct and $P$-equivalent. Let $c\in A\setminus E$ be arbitrary and not equal to~$a$. Let $\pi$ be the permutation that interchanges $b$ and $c$; then $\pi\in\operatorname{fix} E$, hence $\pi P=P$. Then $a,b\in X$ for some $X\in P$, and hence $a,c\in\pi X$. As $\pi X\in\pi P=P$ this shows that $a$ and $c$ are $P$-equivalent as well. It follows that either there is an element of $P$ that contains $A\setminus E$, or all elements of $A\setminus E$ determine one-element members of $P$. Step 3. Assume $f:\mathcal{P}(A)\to\operatorname{Part}(A)$ is an order-preserving surjection. And let $E\subseteq A$ be finite such that $\operatorname{fix} E\subseteq\operatorname{sym} F$. Assume $X$ and $Y$ are finite such that $X\cap E=Y\cap E$ and $|X\setminus E|=|Y\setminus E|$, and such that $f(X)$ and $f(Y)$ are partitions that contain $\{\{a\}:a\in A\setminus E\}$. Then $f(X)=f(Y)$. To see this let $\pi$ be a permutation that maps $X\setminus E$ to $Y\setminus E$ and vice versa and leaves all other points in place. Then $\pi\in\operatorname{fix} E$, so that $\pi(f(X))=f(X)$ and $\pi(f(Y)=f(Y)$. The ordered pair $\langle X,f(X)\rangle$ is in $f$, hence $\langle \pi(X),\pi(f(X))\rangle$ is in~$\pi f$, but $\pi f=f$ and $\pi X=Y$ so that $\langle Y,f(X)\rangle\in f$, that is, $f(Y)=f(X)$. A similar statement holds when $X$ and $Y$ are infinite, hence cofinite, and $X\cap E=Y\cap E$, and $|A\setminus(E\cup X)|=|A\setminus(E\cup Y)|$. Step 4. Now consider the set $\Pi(E)$ of partitions of~$A$ that contain $\{\{a\}:a\in A\setminus E\}$; this is essentially the set of partitions of~$E$ where each is augmented with $\{\{a\}:a\in A\setminus E\}$. The argument above also shows the following: if $X$ and $Y$ are finite such that $X\cap E=Y\cap E$, $|X\setminus E|\le|Y\setminus E|$ and $f(X),f(Y)\in\Pi(E)$ then $f(X)\le f(Y)$. This is so because we can find a permutation $\pi$ in $\operatorname{fix} E$ such that $\pi(X)\subseteq Y$. Likewise: if $X$ and $Y$ are infinite such that $X\cap E=Y\cap E$, $|A\setminus(X\cup E)|\le|A\setminus(Y\cup E)|$ and $f(X),f(Y)\in\Pi(E)$ then $f(X)\ge f(Y)$. Finally: if $X$ is finite and $Y$ is infinite such that $X\cap E=Y\cap E$ and $f(X),f(Y)\in\Pi(E)$ then $f(X)\ge f(Y)$. In conclusion: each subset of $E$ determines a chain in the set~$\Pi(E)$. Step 5. If we enlarge $E$ then $\operatorname{fix} E$ becomes smaller, so we can choose $E$ as large as we please. Let $n$ be a natural number. The number of partitions of the set $\{1,2,\ldots,4n\}$ into four sets of size $n$ is equal to $$ \binom{4n}{n}\binom{3n}{n}\binom{2n}{n} $$ This number is larger than $$ 3^n\cdot 2^n\cdot\frac{4^n}{2n+1} = \frac{24^n}{2n+1} $$ For $n\ge9$ we have $24^n/(2n+1)>16^n$. This gives us our contradiction: if necessary enlarge $E$ so that $|E|=4n$ for some $n\ge9$. Then, by 4 above, the map $f$ divides $\Pi(E)$ into, at most, $2^{4n}=16^n$ chains. On the other hand the partitions of $E$ into four pieces of size $n$ form an antichain of cardinality more than $24^n/(2n+1)$, which in turn is larger than $16^n$.<|endoftext|> TITLE: Smooth functions on subsets of $\mathbb{R}^n$ QUESTION [7 upvotes]: I am teaching a course in basic differential topology, and, following e.g. Milnor, I defined functions of class $C^k$ on subsets of the Euclidean space $\mathbb{R}^n$ as follows. Let $f\colon X\to \mathbb{R}$ be a function, where $X\subseteq \mathbb{R}^n$. Then $f$ is of class $C^k$ if for every point $x_0\in X$ there exist an open neighbourhood $U$ of $x_0$ in $\mathbb{R}^n$ and a function $F\colon U\to\mathbb{R}$ of class $C^k$ such that $f|_{X\cap U}=F|_{X\cap U}$. A student in the audience asked the following question: Is it true that a function $f\colon X\to\mathbb{R}$ is $C^\infty$ if and only if it is $C^k$ for every $k\in\mathbb{N}$? (The ``only if'' statement of course is trivial). Well, I was not able to answer to the question... REPLY [6 votes]: The answer is yes for functions defined on closed sets $X\subset\mathbb{R}^n$. In Section 1.5.5 in [1] we have a necessary and a sufficient condition of the existence of an extension to a $C^m$ function for a finite $m$ and in Section 1.5.6 in [1] we have a necessary and sufficient condition for the existence of an extension to $C^\infty$. It turns out that if the condition for the existence of an extension to $C^m$ is satisfied for all finite $m$, then it is precisely the condition for the existence of an extension to $C^\infty$. [1] Narasimhan, R. Analysis on real and complex manifolds, North-Holland Mathematical Library, 35. North-Holland Publishing Co., Amsterdam, 1985.<|endoftext|> TITLE: Obsessive editing/revising of math papers QUESTION [35 upvotes]: I've been wanting to ask this question, because I have no insights into the way other mathematicians prepare papers (for eventual publication). How much are editing, revising, updating, adding to, etc., part of the "normal" process of process of drafting math papers? Specifically, papers of moderate (10-15 pages) length. I've noticed I tend to do this for several months, and the thought has occurred to me that perhaps I'm being too "fussy" and that I'm wasting time. REPLY [9 votes]: Unless you have a need to get the paper published asap, I recommend you put it aside for a month or so instead of looking through it repeatedly. When you next read the paper, many of your preconceptions and assumptions will have been forgotten and you will be much more likely to notice things like missing definitions, logical gaps, and sections of proofs that are going to confuse readers.<|endoftext|> TITLE: What is the relationship between the $\ell$-adic cohomology of a DM stack and that of its coarse moduli space? QUESTION [5 upvotes]: Let $\mathscr{X}$ be a smooth proper DM stack over a field $k$ (perhaps assumed to be separably closed and/or of char. $0$) and let $\pi \colon \mathscr{X} \rightarrow X$ be its coarse moduli space. What are some general results on the relationship between $H^i_{\mathrm{et}}(\mathscr{X}, \underline{\mathbf{Z}_\ell})$ and $H^i_{\mathrm{et}}(X, \underline{\mathbf{Z}_\ell})$? By the usual spectral sequence argument, I guess I'm asking for what some general results are about the pushforwards $R^i \pi_* \underline{\mathbf{Z}_\ell}$. If I'm not mistaken, proper base change should tell us that these are (constructible? lcc?) $\ell$-adic sheaves with stalks $H^i_{\mathrm{et}}(\mathscr{X}_x, \underline{\mathbf{Z}_\ell})$. I'm happy for results that work in significantly less generality, or to know what some interesting conditions on $\mathscr{X}$ are which make this question easier. Conversely, I'd love to hear something that works when $\underline{\mathbf{Z}_\ell}$ is replaced with some other (lcc etc.) $\ell$-adic sheaf. My motivation for this question came from the case where $\mathscr{X} = [Y/G]$ for $Y$ a smooth projective variety (even a hypersurface) over $\mathbf{C}$ and $G$ a finite cyclic group acting with non-discrete fixed points, and I wanted to compute torsion in the singular cohomology with $\mathbf{Z}$-coefficients. REPLY [3 votes]: The result you want is this. Let $f : \mathscr{X} \to S$ be a proper tame DM stack with $S$ a scheme, and $g : S' \to S$ any morphism of schemes. Let $\mathscr{F}$ be a torsion sheaf on $\mathscr{X}$. Then the natural base change morphism $$g^\ast R^i f_\ast \mathscr{F} \to R^if'_\ast g'^\ast \mathscr{F}$$ is an isomorphism. For a proof, see Theorem A.0.8 of Abramovich-Corti-Vistoli. They also provide a description of the stalks of the higher pushforwards, which is something we discussed yesterday.<|endoftext|> TITLE: mod p etale cohomology of the special fiber and the generic fiber QUESTION [11 upvotes]: Let $O$ be a valuation ring with fraction field $K$ of characteristic zero and residue field $O/m=k$ of characteristic $p>0$, and $X$ be a proper smooth scheme over $O$. Then can we control the mod $p$ etale cohomology of the special fiber by that of the generic fiber? Namely, do we have $\dim_{\Bbb F_p}H^i_{et}(X_k,\Bbb F_p) \leq\dim_{\Bbb F_p}H^i_{et}(X_K,\Bbb F_p)$ in general? I am interested in the case $O=O_{\Bbb C_p}$, and it turns out mod p etale cohomology vanish for $i>\text{dim}(X_k)$ by Artin-schreier sequence while the $2\text{dim}(X_K)$'s etale cohomology can be nonzero for the generic fiber, see this answer. So they are not of the same dimension in general and it's natural to ask for a bound. Also, smooth base change theorem is good for $\Bbb Z/\ell$ ($\ell \not=p$) but don't hold if the coefficient sheaf has $p$-torion. From BMS we know $\text{dim}_{\Bbb F_p}H^i_{et}(X_K,\Bbb F_p) \leq \text{dim}_{k}H^i_{dR}(X_k)$, so maybe a lower bound is also possible in our setting. Edit: The case $i=0,1$ is true. What about the case $i=2$? Moreover, $\text{dim} _{\Bbb　F_p}H^i_{et}(X,\Bbb F_p)$ is finite and less than $\text{dim}_k H^i(X_k, O_k)$, see Lemma 0A3L for a related semi-linear algebra result (so maybe we need to understand the Frobenius action). Edit: Note that $LHS \leq \text{dim}_{k}H^i_{k}(X_k,O_{X_k})\geq \text{dim}_{C}H^i(X_C,O_{X_C}) \leq \text{dim}_{C}H^i_{sing}(X_K^{an},K) \leq RHS. $ The only $\geq$ is due to upper semi-continuities, and we use an abstract isomorphism between $\Bbb C$ and $K$ to apply singular cohomologies, Hodge-de Rham decomposition and the universal coefficient theorem. So if $\text{dim}_{k}H^i_{k}(X_k,O_{X_k})=\text{dim}_{C}H^i(X_C,O_{X_C})$ then what we want holds, however this is not true in general. For example, one considers lifting of a singular Enriques surface in char $2$, but even in this case the inequality still holds (because the second Betti number is $10$, which is much larger than $h^{2,0}=1$). REPLY [2 votes]: There was a nonsense answer here earlier. It is now removed.<|endoftext|> TITLE: Are finite presentations of arithmetic groups computable? QUESTION [10 upvotes]: In this famous paper by Borel and Harish-Chandra, Arithmetic Subgroups of Algebraic Groups, it is proved that, in characterisitic zero, arithmetic groups are finitely presented. I have an extremely vague idea of how the argument goes, but I was wondering if anything was known about how this would be done algorithmically. To be specific, (following work by Grunewald and Segal) an arithmetic group $\Gamma \leq \mathrm{GL}(n,\mathbb C)$ is said to be expicitly given if we have the following data: a system $S$ of polynomial equations defining the algebraic group $V(S)= \mathcal G$ (which contains $\Gamma$), a known upper bound $k$ for the index $[\mathrm{GL}(n,\mathbb Z)\cap \mathcal G:\Gamma]$, and a procedure which given $g \in \mathrm{GL}(n,\mathbb Z)\cap \mathcal G$ will decide if $g \in \Gamma$. Then my question is: If $\Gamma$ is an explicitly given arithmetic group, is it possible to compute the presentation of $\Gamma$ or, even better, to get the orbihedron whose fundamental group is $\Gamma$, which is used in the construction of Borel and Harish-Chandra? If there is a reference for this, or if this is known to be impossible, that would be awesome! REPLY [3 votes]: You might want to study the work of Detinko, Flannery, and co-authors. For example: Detinko, A.; Flannery, D. L.; Hulpke, A., Zariski density and computing in arithmetic groups, ZBL06825254. Detinko, A. S.; Flannery, D. L.; Hulpke, A., Algorithms for arithmetic groups with the congruence subgroup property., J. Algebra 421, 234-259 (2015). ZBL1319.20040. (moving comment up to the main body): They find things like the congruence depth (the index of the principal congruence subgroup), so the co-volume, and then (at least in the special linear and symplectic case you can use the Minkowski model (the PSD cone) of the symmetric space to construct the fundamental polyhedron (the bound tells you when to stop), so pretty much what you are asking for. In the real hyperbolic case, there is the very interesting paper of Mark and Paupert: Presentations for cusped arithmetic hyperbolic lattices, by Alice Mark and Julien Paupert.<|endoftext|> TITLE: Intersection of nested open ball in complete metric spaces is nonempty? QUESTION [8 upvotes]: My question is that whether the following statement is true or not. In a complete metric space $(X, d)$, if a sequence of open balls $\{B(x_i, r_i)\}_{i=1}^\infty$ satisfies $$ \exists \epsilon > 0 ~~s.t.~ B(x_{i+1}, (1+\epsilon)r_{i+1}) \subset B(x_i, r_i), \forall i \ge 1 \tag{1} $$ then $\bigcap_{i=1}^\infty B(x_i, r_i) \neq \emptyset$. Here are several related facts: If we further require that $r_i \rightarrow 0$ as $i\rightarrow \infty$, then $\bigcap_{i=1}^\infty B(x_i, r_i) \neq \emptyset$ holds. There is a example shows that $\bigcap_{i=1}^\infty B(x_i, r_i) = \emptyset$ if we replace $(1)$ by $\overline{B(x_{i+1}, r_{i+1})} \subset B(x_i, r_i)$ for all $i\ge 1$ where $\overline{B}$ is the closure of $B$ In a complete metric space, $B(x, r) \subset B(x', r')$ may hold for $r > r'$. As Choi pointed it out, the above statement is similar to Cantor's intersection lemma. Recall that Cantor's intersection lemma in a complete metric space is given as follows: In a complete metric space $(X, d)$, if a sequence of closed set $\{C_i\}_{i=1}^\infty$ satisfies $$ C_{i+1} \subset C_i, \forall i\ge 1 ~~\&~~ diameter(C_i) \rightarrow 0~ as~ i\rightarrow \infty $$ then $\bigcap_{i=1}^\infty C_i \neq \emptyset$ However, we don't require $diameter(B(x_i, r_i)) \rightarrow 0$ as $i\rightarrow \infty$. And there is indeed a example shows Cantor's inetersection lemma fails if we drop the diameter restriction. Please refer to Nested closed balls with empty intersection REPLY [8 votes]: I think this statement is true. Suppose we had a counterexample $\{B(x_i,r_i)\}_{i=1}^\infty$ satisfying condition (1) for some $\epsilon > 0$ but whose intersection was empty. Observe that any subsequence will still be a counterexample. For each $i$, the point $x_i$ does not belong to some $B(x_j,r_j)$, as otherwise the intersection of all the balls would be nonempty. So by passing to a subsequence we can ensure that $x_i \not\in B(x_{i+1},r_{i+1})$ for all $i$. Note that this forces $r_{i+1} < r_i$, as $x_{i+1}$ does belong to $B(x_i,r_i)$. By Cantor's intersection lemma, the $r_i$ cannot decrease to zero. So they must decrease to some $r > 0$. Without loss of generality we can now assume that $(1 + \epsilon)r > r_1$. But for any $i$ we have $x_i \in B(x_1,r_1)$, i.e., $d(x_i,x_1) < r_1 < (1+\epsilon)r$, which means that $x_1$ belongs to every $B(x_i,(1+\epsilon)r_i)$. But then by condition (1) $x_1$ belongs to every $B(x_i,r_i)$, a contradiction.<|endoftext|> TITLE: What is the asymptotic growth of $\sum_{k=1}^n 2^{\omega_k}$? QUESTION [5 upvotes]: Question: Let $\omega_k$ be the number of distinct prime divisors of k. What is the asymptotic growth of $C_n := \sum_{k=1}^n 2^{\omega_k}$? Thank you for considering this elementary question. Below I give some motivation for this problem and some of my progress. Motivation: There are a couple. The first is purely number theoretic. Note that because $2^{\omega_k} = \sum_{d | k} |\mu(d)|$, we have $C_n = \sum_{k=1}^n \sum_{d|k} |\mu(d)|,$ and thus $C_n$ is the asymptotic growth of absolute values of the Möbius function (in other contexts this relates to the Prime Number Theorem, see here). The second motivation comes from geometric group theory. The commensurability index of $A, B \leq G$ (all groups) is $[A : A \cap B][B: A \cap B]$. The full commensurability growth function assigned to a pair $A \leq B$ is defined to be $$ C_n(A,B) = \# \{ \Delta \leq B : c(\Delta, A) \leq n \}. $$ This is a generalization of the subgroup growth function to pairs and it can be infinite in natural settings. The question above is the case $G(\mathbb{Z}) = \mathbb{Z} \leq \mathbb{R} = G(\mathbb{R})$ of the following Problem: Compute the full commensurability growth function for the pairs $G(\mathbb{Z}) \leq G(\mathbb{R})$ where $G$ is a unipotent linear algebraic group. Some progress: Write $f \preceq g$ if there exists $C > 0$ such that $f(n) \leq C g(C n)$. In Proposition .3, it is shown that $n (\log(n))^{\log(2)} \leq C_n \preceq n(\log(n))$. REPLY [10 votes]: As you observe, $$C_n=\sum_{k=1}^n\sum_{d\mid k}|\mu(d)|=\sum_{k=1}^n\sum_{d\mid k\text{ squarefree}}1.$$ Exchanging the order of summation, $$C_n=\sum_{d\leq n\text{ squarefree}}\sum_{d\mid k\leq n}1=\sum_{d\leq n\text{ squarefree}}\left\lfloor\frac{n}{d}\right\rfloor=n\sum_{d\leq n\text{ squarefree}}\frac{1}{d}+O(n).$$ Now, the squarefree numbers have density $\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$, so by summation by parts it's easy to find $\sum_{d\leq n\text{ squarefree}}\frac{1}{d}\sim\frac{1}{\zeta(2)}\log n$, hence giving $C_n\sim\frac{1}{\zeta(2)}n\log n$. You should be able to get an $O(n)$ bound on the error this way, and using the hyperbola method you should get some more precise estimates.<|endoftext|> TITLE: Nash isometric embedding for noncompact manifolds QUESTION [11 upvotes]: It seems that the smooth isometric embedding theorem by Nash is true also for noncompact manifolds. Is it true that any (complete, connected) Riemannian manifold $(M^n,g)$ admits a proper smooth isometric embedding $\iota:M^n\hookrightarrow\mathbb R^N$ into some Euclidean space? This is equivalent to ask that $\iota(M)$ is a closed subset of $\mathbb R^N$. Remark. It is pretty trivial to modify the proof of Whitney's embedding theorem to obtain that any manifold admits a proper embedding into some Euclidean space (if we don't care about obtaining the best dimension for which this is possible...). But of course completeness is a necessary condition to find a proper isometric embedding $\iota$. Is completeness also sufficient? Are weaker results known? E.g. is it true if $\operatorname{inj}(M)>0$ and/or the sectional curvature $\operatorname{sec}(M)$ is bounded? REPLY [4 votes]: A slight modification of Anton Petrunin's answer, namely Müller's idea brought up in Igor Belegradek's comment, allows one to deduce the desired result starting from the classical statement of the isometric embedding theorem, i.e. Any Riemannian manifold $(M^n,g)$ admits a smooth isometric embedding into a Euclidean space. Fix $p\in M$ and take again a smoothing $\phi$ of $\frac{1}{4}\operatorname{dist}_p$ with $|\phi-\frac{1}{4}\operatorname{dist}_p|\le 1$ and $|\nabla\phi|_g\le\frac{1}{2}$ (see below for the details). Now apply the classical Nash embedding theorem to the Riemannian manifold $(M^n,g-d\phi\otimes d\phi)$, obtaining an isometric embedding $F$. Then $(F,\phi)$ is the desired proper isometric embedding of $(M^n,g)$, since $\operatorname{dist}_p$ (and thus $\phi$) is a proper map by completeness. Construction of $\boldsymbol{\phi}$. Take a locally finite open cover $\{U_j\}$ of $M$ such that $\overline{U}_j$ is compact and lies in a coordinate chart, together with a subordinated partition of unity $\{\rho_j\}$. Since $|\nabla\operatorname{dist}_p|_g\le 1$ a.e., we can approximate $\frac{1}{4}\operatorname{dist}_p$ on $U_j$ by a smooth function $\phi_j$ with $$|\nabla\phi_j|_g\le\frac{3}{8},\qquad|\phi_j-\frac{1}{4}\operatorname{dist}_p|\le\min\bigg\{1,\frac{1}{8N_j'\|\nabla\rho_j\|_{L^\infty}}\bigg\},$$ where $N_k:=\#\{\ell:U_k\cap U_\ell\neq\emptyset\}$ and $N_j':=\max\{N_k\mid U_j\cap U_k\neq\emptyset\}$. Finally, $\phi:=\sum\rho_j\phi_j$ has $|\phi-\frac{1}{4}\operatorname{dist}_p|\le 1$ and, being $$\nabla\phi=\sum_k\rho_k\nabla\phi_k+\sum_k\nabla\rho_k(\phi_k-\frac{1}{4}\operatorname{dist}_p),$$ on each $U_j$ (and thus on $M$) we have $|\nabla\phi|_g\le\frac{3}{8}+\sum_{k:U_j\cap U_k\neq\emptyset}\frac{1}{8N_k'}\le\frac{1}{2}$ (as $N_k'\ge N_j$).<|endoftext|> TITLE: Breaking a morphism with generic fiber $\mathbb{F}_n$ QUESTION [6 upvotes]: Assume we are working over $\mathbb{C}$, and we have a projective morphism with connected fibers $f: X \rightarrow Z$ whose geometric generic fiber $X_\overline{\eta}$ is isomorphic to a Hirzebruch surface $\mathbb{F}_n$. Thus, $X_\overline{\eta}$ admits a morphism to $\mathbb{P}^1$, and this is defined over some finite extension of $K(Z)$. So, we know that, up to a generically finite base change and birational modification of the main component of the fiber product, we get a morphism $f': X' \rightarrow Z'$ that factors as $g': X' \rightarrow Y'$ and $h': Y' \rightarrow Z'$, where $g'$ and $h'$ are both generically $\mathbb{P}^1$-bundles. My question is the following. Given the setup above, are there cases when we know that the finite base extension is not needed? My naive hopes rely on two facts. First, the morphism $\mathbb{F}_n \rightarrow \mathbb{P}^1$ is defined over $\mathrm{Spec}(\mathbb{Z})$, and so we can base change it to $K(Z)$. Second, if the base $Z$ is a curve, by the theorem of Graber-Harris-Starr we know that $X_\eta$ has a $K(Z)$-point. For instance, is it reasonable to get something in the direction I want if the base is a curve? REPLY [4 votes]: If $n>0$, and if you have a rational section (for instance when $Z$ is a curve), then you do not need the finite extension. The reason is that the field $K(Z)$ is perfect (as you work in characteristic zero), and that the Galois group acts on $\mathbb{F}_n$ preserving the exceptional curve (unique curve of negative self-intersction) and the fibration (the fibres are the only curve of self-intersection $0$). The morphism to the curve is then defined over $K(Z)$, and the base is rational as it has a point, so the generic fibre is already isomorphic to $\mathbb{F}_n$ over $K(Z)$. If $n=0$, then you really need the finite extension in general. Take for instance the variety $X=\{([w:x:y:z],t)\in \mathbb{P}^3 \times \mathbb{A}^1\mid xy-z^2-w^2p(t)=0\}$ for a polynomial $p$ of large degree with no multiple root and consider the morphism $f\colon X\to Z=\mathbb{A}^1$ given by the second projection (you can obtain a projective example by putting this in the suitable $\mathbb{P}^2$-bundle over $\mathbb{P}^1$). The generic fibre is isomorphic to a smooth quadric, not isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$, but after a finite extension that consists of adding a square root of $f$, you get a fibre which is isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$.<|endoftext|> TITLE: Enlarging a subdegree-finite "almost transitive" permutation group to a transitive one? (follow-up) QUESTION [6 upvotes]: Consider a permutation group $G$ acting on an infinite set $X$. Assume $G$ has finitely many orbits, and every point stabiliser $G_x$ has finite orbits. Can we always find a permutation $\tau\in\operatorname{Sym}(X)$ (not necessarily of finite order or finite support) such that $H=\langle G,\tau\rangle$ is transitive, while every point stabiliser $H_x$ still has finite orbits? As noted in this previous question some obvious choice of $\tau$ will not work in general. REPLY [4 votes]: It seems that a counterexample looks as follows (it is somewhat siimilar with the counterexample to your previous question). Let $k$ be a large integer. Take an infinite tree $T=(V,E)$, where all degrees equal $k+1$. Let $G$ be its group of automorphisms. $G$ acts on $V\cup E$ with two obvious orbits, and all orbits of the stabilizers are finite. I could find only quite technical proof that this example works; prehaps, there are easier ones. Define the metric $d(\cdot,\cdot)$ on $V\cup E$ identifying each edge with its midpoint (the length of every edge is $1$). E.g., the distance between a vertex and an incident edge is $1/2$. Assume that we took a transitive group $H\geqslant G$ with finite stabilizers' orbits. Let $G_v$ and $G_e$ be stabilizers of a vertex $v$ and an edge $e$ in $G$, and $H_v$ and $H_e$ be those in $H$. Then $H_e$ and $H_v$ are conjugates, so there is a cardinality-preserving bijection of their sets of orbits. Say that the radius of $H_v$-orbit is the maximal distance from $v$ to its element; the same for $H_e$-orbits. $H_v$-orbits are unions of $G_v$-orbits which have cardinalities $(k+1),(k+1),k(k+1),k(k+1),\dots,k^n(k+1),k^n(k+1),\dots$. Similarly, $H_e$-orbits are unions of $G_e$-orbits whose cardinalities are $2,2k,2k,2k^2,\dots,2k^n,2k^n,\dots$ (notice here that an $H_v$-, and hence $H_e$-, orbit cannot have just $2$ elements). So the cardinality of any $H_v$-orbit $\Omega_v$ has the form either $k^n+O(k^{n-1})$ or $2k^n+O(k^{n-1})$, and that of the corresponding $H_e$-orbit $\Omega_e$ may have the form either $2k^n+O(k^{n-1})$ or $4k^n+O(k^{n-1})$. Hence this (common) cardinality is $2k^n+O(k^{n-1})$. This means that the radius of $\Omega_v$ is an integer $r$< and that of $\Omega_e$ is either $r$ or $r+1/2$. Case 1. $\Omega_v$ and $\Omega_e$ have the same radius $r$. Consider now some $\tau\in H$ mapping $v$ to $e$; every $H_v$-orbit $\Omega_v$ is mapped to some $H_e$-orbit $\Omega_e$ of the same radius $r$. Notice that a dominating part $\partial\Omega_v$ of $\Omega_v$ consists of far vertices $v'$ with $d(v,v')=r$ and far edges $e'$ with $d(v,e')=r-1/2$; similarly, a dominating part $\partial \Omega_e$ of $\Omega_e$ consists of far edges $e'$ with $d(e,e')=r$. So most of such permutations $\tau$ map a good proportion of far vertces and edges (for $v$) to far edges (for $e$). See footnote for the explanation of the term `most of'. Now consider some orbit' radius $r$. Take a vertex $v$ ans some $\tau$ mapping $v\mapsto e$. Let $\Omega_v$ and $\Omega_e$ be $H_v$- and $H_e$-orbits of radius $r$. Under $\tau$, most of $k^{r-1}(k+1)$ far edges for $v$ map to far edges for $e$. Let $e'$ be an edge incident to $v$, and let $\Omega_{e'}$ he $H_{e'}$-orbit of radius $r$. Notice that $k^r$ of far edges for $e'$ are also far edges for $v$. So, most of them also map to far edges/vertices for $\tau(e')$, so thare are almost $k^r$ common far edges for $e$ and $\tau(e')$. This may happen only if $\tau(e')$ is a vertex incident to $e$. But $e'$ can be chosen in $k+1$ ways, while $e$ has only two endpoints. A contradiction. Case 2. Assume that the radius of $\Omega_e$ is $r+1/2$. By symmetry, we may assume that $d(v,e)=r+1/2$, so that $v\in\Omega_e$. Due to the cardinalities, $\Omega_e$ should contain also either all vertices at distance $r-1/2$ from $e$, or all edges $e'$ at distance $r-1$ from $e$. In the former case, we get that an edge at distance $r-1/2$ and an edge at distance $r+1/2$ are equivalent modulo $H_v$, so $e\in\Omega_v$, which is impossible. In the latter case, we have $\sigma\in H$ with $\sigma(e')=v$, $\sigma(e)=e$. But, since $e$ lies in $\Omega_{e'}$ corresponding to $\Omega_e$, we get $e\in\Omega_v$ again. Footnote. "Most of permutations $\tau$" is regarded in the following sense. If we consider all permutations $\tau$ mapping some $x\mapsto y$, there are only finitely many $t$ into which may a fixed $z$ map. If we are interested in images of a finite set of such $z$'s, there are finitely many tuples of images, and they are `equally distributed' (since all $\tau$ realizing one tuple form a coset of their joint stabilizer). Now we may speak on the probability in this sense.<|endoftext|> TITLE: Motive of a conic without points QUESTION [8 upvotes]: Let $C\subset \mathbb{P}^2$ be a smooth conic without $k$-points. Call the Chow $k$-motive in zero-dimensional if it is a sum of $M\mathbb{L}^n$ where $M$ is an Artin motive, i.e. a part of a motive of zero-dimensional scheme. Q. How to see that a motive of $C$ is not (or is) zero-dimensional? P.S. By Chow $k$-motive I understand the category of motives $Chow_{\mathbb{Q}}$ defined by rational equivalence. REPLY [7 votes]: This is true with $\mathbb Q$-coefficients; see the proposition below. The reason it's difficult is that it's hard to compute Chow groups over non-algebraically closed fields. However, we have the following: Lemma 1. Let $k \subseteq \ell$ be a separable algebraic field extension, and let $X$ be a proper geometrically integral $k$-scheme. Then the pullback map $\operatorname{Pic}(X) \to \operatorname{Pic}(X_\ell)$ is injective. Proof. See e.g. this post. $\square$ Lemma 2. Let $k \subseteq \ell$ be a finite field extension, and let $X$ be any $k$-scheme. Then the pullback map $\operatorname{CH}_*(X) \otimes_\mathbb Z \mathbb Q \to \operatorname{CH}_*(X_\ell) \otimes_\mathbb Z \mathbb Q$ is injective. Proof. If $\pi \colon X_\ell \to X$ is the map induced by $k \to \ell$, then $\pi_* \pi^*$ is multiplication by $[\ell:k]$; see e.g. [Ful, Ex. 1.7.4]. (This is in fact true for any finite flat morphisms of schemes.) $\square$ However, there exist Severi-Brauer varieties $X$ for which the pullback $\operatorname{CH}_*(X) \to \operatorname{CH}_*(X_\ell)$ (with integral coefficients) is not injective, as is explained in this post. Let $C$ be a smooth conic without rational points, and let $k \to \ell$ be the minimal splitting field (which is a separable quadratic extension of $k$). Lemma 3. We have $\operatorname{CH}^*(C) = \mathbb Z[2h] \subseteq \mathbb Z[h]/(h^2) = \operatorname{CH}^*(C_\ell)$. Proof. The pullback $\operatorname{CH}^*(C) \to \operatorname{CH}^*(C_\ell)$ is injective by Lemma 1 above. The line bundle $\omega_C^{-1}$ has class $2h$. If $\mathscr L$ is a line bundle with class $h$, then Riemann-Roch gives an isomorphism $C \stackrel\sim\to \mathbb P^1$, contradicting the choice of $C$. $\square$ Lemma 4. The image of the pullback $\operatorname{CH}^*(C \times C) \to \operatorname{CH}^*((C\times C)_\ell)$ equals $$\mathbb Z[h_1+h_2,2h_1] \subseteq \mathbb Z[h_1,h_2]/(h_1^2,h_2^2).$$ Proof. Clearly $h_1 + h_2 = [\Gamma_{\operatorname{id}_C}]$ and $2h_1 = \omega_C^{-1} \boxtimes \mathcal O_C$ are contained in $\operatorname{CH}^*(C \times C)$. Moreover, they span $\operatorname{CH}^1(C \times C) = \operatorname{Pic}(C \times C)$, since $h_1$ is not defined on $C \times C$. Indeed, if $\mathscr L$ is a line bundle with class $h_1$, then it defines a map $C \times C \to \mathbb P^1$, whose image under the $2$-uple embedding is the first projection $C \times C \to C \subseteq \mathbb P^2$. This again contradicts the choice of $C$. Finally, the intersection $(h_1 + h_2) \cdot 2h_1 = 2h_1h_2$ is a degree $2$ zero-cycle on $C \times C$. There are no zero-cycles of odd degree, for their pushforward under either projection would be a divisor of odd degree on $C$. $\square$ Remark. I don't know if the pullback $\operatorname{CH}_*(C \times C) \to \operatorname{CH}^*((C \times C)_\ell)$ is injective. This is true for divisors (and for dimension $2$ cycles), and it is true rationally. However, there could be torsion zero-cycles that disappear on $(C \times C)_\ell$. Remark. We will use the ring isomorphism $\operatorname{CH}^*(X \times \mathbb P^1) \cong \operatorname{CH}^*(X) \otimes \mathbb Z[h]/h^2$ of [Ful, Ex. 8.3.4] for any smooth $k$-variety $X$ (it is true as groups in much greater generality). Then the pushfoward $\operatorname{CH}^*(X \times \mathbb P^1) \to \operatorname{CH}^*(X)$ is given by $\alpha \otimes 1 \mapsto 0$ and $\alpha \otimes h \mapsto \alpha$ for $\alpha \in \operatorname{CH}^*(X)$. Proposition. The motives $C$ and $\mathbb P^1$ are isomorphic rationally. Proof. With rational coefficients, we have $\operatorname{CH}_\mathbb Q^*(C) \cong \operatorname{CH}_\mathbb Q^*(\mathbb P^1) \cong \mathbb Q[h]/(h^2)$, and similarly for all finite products involving $C$ and $\mathbb P^1$. The class $\phi = \tfrac{1}{2}[\omega_C^{-1} \boxtimes \mathcal O_{\mathbb P^1}(2)] \in \operatorname{Corr}(C, \mathbb P^1)$ corresponds to $h_1+h_2$. We now easily compute \begin{align*} \phi \circ \phi^\top &= \pi_{13,*}(\pi_{12}^* \phi^\top \cdot \pi_{23}^* \phi) = \pi_{13,*}((h_1+h_2)(h_2+h_3)) \\ &= \pi_{13,*}(h_1h_3 + h_2(h_1 + h_3) ) = h_1 + h_3 \\ &= [\Gamma_{\operatorname{id}_{\mathbb P^1}}] \in \operatorname{CH}^*_\mathbb Q(\mathbb P^1 \times \mathbb P^1), \end{align*} and similarly for the other composition $\phi^\top \circ \phi$. $\square$ Remark. I think that the motive $C$ does not occur integrally as a direct summand of $\mathbb P^n \times Z$ for any $n \in \mathbb Z_{\geq 0}$ and any zero-dimensional smooth $k$-scheme $Z$. For example, if $Z = \operatorname{Spec} m$ for a Galois extension $k \subseteq m$ with group $G$, then $H^0(C_\bar k, \mathbb F_p)$ should occur as a direct summand of $$H^0((\operatorname{Spec}_\ell)_\bar k, \mathbb F_p) = \mathbb F_p[G]$$ for any prime $p \not\mid \operatorname{char} k$. But if $p \mid \#G$, then the unique trivial subrepresentation $\mathbb F_p \cdot (\sum_{\sigma \in G} \sigma) \subseteq \mathbb F_p[G]$ does not split as a direct summand. If $\operatorname{char} k = 0$, this forces $G = 1$, i.e. $Z = \operatorname{Spec} k$. Then it remains to compute $\operatorname{CH}^*(C \times \mathbb P^n)$ as above and see that no $\phi$ in there satisfies $\phi \circ \phi^\top = 1$, even modulo projectors. This is not a very good argument, so I would be interested to see if someone has a cleaner obstruction for the integral version of the question. References. [Ful] W. Fulton, Intersection theory, second edition. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. 2. Springer-Verlag (1998). ZBL0885.14002.<|endoftext|> TITLE: How Much Flesh to the Bones does an Initial Online Publication need? QUESTION [7 upvotes]: Background of my question is the following: I have found a solution for my question Smoothness Conditions for Planar “Mock-parametric” Spline Interpolation and while developing the solution, I encountered aspects of spline interpolation that I haven't seen documented: calculation of interpolating splines via an iterative formula, that would also allow the calculation of splines from point-sequences that are infinite in both directions a general solution for optimizing linear functionals on interpolating splines now, as I am no professional mathematician and haven't published anything yet, I would like to communicate my results in a lightweight process, which would mean giving enough detail to make my findings checkable and applicable. Question: is acceptable to make an online publications sketchy, i.e. providing arguments instead of proofs, in first version and, to deliver what is missing in later editing e.g. upon feedback? I'm thinking about putting the results on arXiv, but I would also be happy to put them on MO in a CW contribution; direct communication to interested people would also be an option to me. Any suggestions in that respect are welcome REPLY [7 votes]: I personally regard an arXiv preprint as a usual article, just not refereed. One should be able to refer to it, hence it should contain full proofs. After you put your results with full proofs on arXiv, you will be able to write a short answer to your question (not necessarily as a CW) with a reference to your preprint. This acceptable; see this answer. REPLY [6 votes]: I presume that with "online" publication you have arXiv in mind. (For a journal it does not really make a difference whether it is online or not.) With regards to your question, the key difference between arXiv and a journal is not so much the absence of a refereeing process, but the fact that on arXiv you can post multiple revised versions. With a journal publication you get a chance to revise it after the referee reports come in, but once it's accepted no further revisions are possible. So yes, you can submit incomplete papers to arXiv, and complete them later. The arXiv moderators will hold you to a certain standard, but if you are not submitting a proof of the Riemann hypothesis you should be OK. There are risks and benefits with an early submission to arXiv. One benefit is that you can protect your priority. Another benefit is that you might receive helpful feedback on your findings. One risk is that a sloppy first version may damage your reputation a professional researcher. Another risk is that someone may build on your results (with proper attribution, of course), perhaps generalizing them or applying them in a different context, while you may have wanted to do that yourself.<|endoftext|> TITLE: Stronger versions of Wilson's Theorem QUESTION [17 upvotes]: Problem Let $c \in \mathbb{N}$ $;$ $\exists$ a prime $p$ for which: $$p^c \mid (p-1)!+1$$ Does $\exists$ $M$ $\in$ $\mathbb{N}$ $;$ $\forall$ $c \geqslant M$ $;$ $\nexists$ $p$ satisfying the above? When $c$ = $1$ The statement is equivalent to Wilson's Theorem. For every prime $p$: $$p \mid (p-1)!+1$$ Proof: $\forall$ $x \in {1,2,...,p-1}$ $\exists!$ $ x' \in {1,2,...,p-1}$ ; $x \cdot x'\equiv 1 \pmod{p}$ $x=x' \iff p \mid x^2-1 \iff x = 1, x=p-1$ $(p-1)! = 1 \cdot (p-1) \cdot \prod{(x \cdot x')} \equiv 1^n \cdot (p-1) \equiv p-1 \pmod{p}$ $\implies p \mid (p-1)!+1$ QED When $c$ = $2$ We have the statement: $$p^2 \mid (p-1)!+1$$ The only known primes that satisfy this are $5$, $13$ and $563$. $(5-1)!+1 = 25 = 5^2$ $(13-1)!+1 = 479001601 = 13^2 \cdot 2834329$ $563^2 \mid (563-1)!+1 \approx 1.128 \cdot 10^{1303}$ Such primes $p$ are known as Wilson Primes. It is conjectured that there are infinitely many Wilson Primes. However, if there exists a fourth Wilson prime $p$, then $p>2 \cdot 10^{13}$. When $c \geqslant 3$ There are no known primes for which $p^3 \mid (p-1)!+1$ as if there is, then $p$ also has to be a Wilson Prime. $(5-1)!+1 = 25 \equiv 25 \pmod{5^3}$ $(13-1)!+1 = 479001601 \equiv 676 \pmod{13^3}$ $(563-1)!+1 \equiv 91921010 \pmod{563^3}$ It is most likely due to following evidence that there exists an upper bound $M$ for which: $$c \geqslant M \implies p^c \nmid (p-1)!+1$$ where $M \geqslant 3$. We consider $(p-1)!+1 \pmod{p^c}$ We assume that every remainder divisible by $p$ (Wilson's Theorem) is equally probable. Thus, the probability of required remainder $0$ is $\frac{1}{p^{c-1}}$ Thus, probable number of primes for given constant $c$ is: $$\sum{\frac{1}{p^{c-1}}} = P(c-1)$$ where $P(x)$ is the Prime Zeta Function When $c=2$, the expected number of Wilson primes is $P(1)$. $$P(1)=\sum{\frac{1}{p}}$$ This sum is divergent. Thus, it is probable that there exist infinitely many Wilson primes. Proof: Define $N(x)$ to be the number of positive integers $n \leqslant x$ for which $p_i \nmid n$, where $i > j$ for constant $j$ and $p_i$ is the $i$th smallest prime. Then, we write: $$n=k^2m$$ where $m$ is square-free. As $m$ is square-free, and the only primes that divide it are $p_i$ for $1 \leqslant i \leqslant j$, it has $2^j$ possibilities. $n^2 \leqslant x \implies n \leqslant \sqrt{x}$, thus giving $n$ a maximum of $\sqrt{x}$ possibilities. $$\implies N(x) \leqslant 2^j\sqrt{x}$$ Assume the contrary, then for some $j$: $$\sum_{i=j+1}^\infty \frac{1}{p_i} < \frac{1}{2}$$ We also have $x-N(x)$ is the number of numbers less than or equal to $x$ divisible by one or more of $p_i$ for $i>j$. $$x-N(x) \leqslant \sum_{i=j+1}^\infty \frac{x}{p_i} < \frac{x}{2}$$ $$\implies 2^j\sqrt{x} > \frac{x}{2}$$ which is untrue for $x \geqslant 2^{2j+2}$ Thus the sum diverges. The divergence is similar to $\log{\log{x}}$ (Which is very slow). Probable Answer To Problem When $c \geqslant 3$, the sum converges and is less than $1$. When $c=3$, $P(c-1) \approx 0.45$ When $c=4$, $P(c-1) \approx 0.17$ When $c=5$, $P(c-1) \approx 0.07$ When $c=6$, $P(c-1) \approx 0.03$ When $c=7$, $P(c-1) \approx 0.002$ We now go on to show why there most probably exists a constant $M$ such as the one in the problem. Consider: $$\sum_{i=3}^\infty P(i-1)$$ We have: $$\sum_{i=3}^\infty P(i-1) < \sum{\frac{1}{n(n+1)}} = \sum{\biggl(\frac{1}{n}-\frac{1}{n+1}\biggl)} = 1$$ Thus, the probable sum of the number of primes that satisfy the statement for $c \geqslant 3$, including a prime $p$, $n-2$ times, if the maximum $c$ satisfied is $n$, is less than $1$. However, if the answer to the problem is false, then, the sum would be infinite. Thus, it is highly unlikely for there to be a solution for $c \geqslant 3$ as the probable answer is less than $1$ but the actual answer would be a positive integer. However, it is almost impossible for the answer to the problem to be false, as the probable answer is less than $1$ but the actual answer would be infinite! Any of the following: Any progress or insight Answers conditional on conjectures Polynomial or logarithmic non-trivial bounds on $M$ will be accepted and appreciated. P.S. This question is also in Mathematics Stack Exchange. Link: https://math.stackexchange.com/questions/2651733/stronger-versions-of-wilsons-theorem REPLY [10 votes]: Short expansion of my comment on what's immediately achievable when assuming the $abc$-conjecture. Proporistion. Assuming the $abc$-conjecture, we have $$v_p((p-1)!+1)=o(p).$$ Proof. One formulation of the $abc$-conjecture goes as follows (straight from Wikipedia): ABC conjecture. For every positive real number $\varepsilon$, there is a constant $K_\varepsilon$ such that for all triples $(a, b, c)$ of coprime positive integers, with $a+b=c$: $$c < K_\varepsilon\cdot \text{rad}(abc)^{1+\varepsilon}.$$ Set $a=(p-1)!$, $b=1$, and $c=a+b=p^{v_p((p-1)!+1)}c'$, where $p\not|c'$. Fix a positive real number $\varepsilon$. Then, assuming the $abc$-conjecture, there exists a constant $K_\varepsilon$ such that \begin{align} p^{v_p((p-1)!+1)}c'<&\ K_\varepsilon\cdot\text{rad}((p-1)!p^{v_p((p-1)!+1)}c')^{1+\varepsilon}\\ =&\ K_\varepsilon\cdot(p\cdot\text{rad}(c')\prod_{q\lt p}q)^{1+\varepsilon}\\ <&\ K_\varepsilon\cdot(c'\prod_{q\le p}q)^{1+\varepsilon}\\ \Rightarrow p^{v_p((p-1)!+1)}<&\ K_\varepsilon\cdot c'^\varepsilon(\prod_{q\le p} q)^{1+\varepsilon}\\ =&\ K_\varepsilon \cdot \big(\frac{(p-1)!+1}{p^{v_p((p-1)!+1)}}\big)^\varepsilon(\prod_{q\le p}q)^{1+\varepsilon}\\ \Rightarrow p^{(1+\varepsilon)v_p((p-1)!+1)}<&\ K_\varepsilon \cdot ((p-1)! + 1)^\varepsilon(\prod_{q\le p}q)^{1+\varepsilon}. \end{align} We now take logs on both sides, and apply Stirling's Formula and the Prime Number Theorem: \begin{align} (1+\varepsilon)v_p((p-1)!+1)\log{p}<&\ \varepsilon p\log{p} + O(p) + (1+\varepsilon)(p+o(p))\\ \Rightarrow v_p((p-1)!+1) <&\ \varepsilon p + o(p). \end{align} Since this is true for all positive real numbers $\varepsilon$, we get the proposition.<|endoftext|> TITLE: Hodge decomposition and degeneration of the spectral sequence QUESTION [22 upvotes]: I am teaching a course on Hodge theory and I realised that I don't understand something basic. Let first $X$ be a compact Kahler manifold. Let $H^{p,q}(X)=H^q(X,\Omega^p_X)$ where $\Omega^p_X$ is the sheaf of holomorphic $p$-forms. Let also $\Omega^*_X$ denote the holomorphic de Rham complex. Then there two statements: 1) The spectral sequence coming from the stupid filtration on $\Omega^*_X$ degenerates at $E_1$. This is equivalent to saying that $H^*(X,\mathbb{C})$ has a filtration such that the associated graded space is canonically identified with the direct sum of $H^{p,q}(X)$. 2) $H^k(X,\mathbb{C})$ is canonically isomorphic to the direct sum of $H^{p,q}(X)$ with $p+q=k$. The 2nd statement implies the 1st (since for degeneration of the spectral sequence it is enough to check that the two spaces have the same dimension). I want to understand to what extent the 1st statement implies the 2nd. In fact, the theory of harmonic forms implies that $H^{p,q}(X,\mathbb{C})$ is equal to $F^p(H^{p+q}( X,\mathbb{C}))\cap \overline{F^q(H^{p+q}( X,\mathbb{C}))}$ where $F^*$ stands for the Hodge filtration (induced by the stupid filtration on the de Rham complex). My question is whether it is possible to prove this fact without using harmonic forms if we already know the degeneration of the spectral sequence? For example, assume that $X$ is a complete smooth algebraic variety over $\mathbb{C}$. Then Deligne and Illusie give an algebraic proof of the degeneration of the spectral sequence using reduction to characteristic $p$. Is it true that the Hodge decomposition holds in this case as well? How to prove this? (I know that if $X$ is not necessarily Kahler then it might happen that the spectral sequence degenerates but the Hodge decomposition fails, but these examples are not algebraic, so the question still makes sense). Basically, I want to understand whether it is possible to prove the existence of Hodge decomposition without ever using the Dolbeaut complex (or any explicit resolution of the holomorphic or algebraic de Rham complex)? REPLY [8 votes]: I am far from being an expert in this area, but I will try to present my understanding of this subject. First of all, this is true that Hodge decomposition holds for smooth proper varieties over $\mathbf C$. However, the standard proof (that is explained below) uses the projective case as a black box. Secondly, degeneration of Hodge-to-de Rham spectral sequence does imply that there is a filtration on $H^{n}(X,\mathbf C)$ s.t. the associated graded pieces are isomorphic to $H^{p,q}(X)$. In particular, it means that there is an abstract isomorphism of $\mathbf C$ vector spaces $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^q(X,\Omega^p_{X/\mathbf C})$. But I don't know any way to pick a canonical one just assuming degeneration of the spectral sequence. Moreover, there is an example of a smooth compact non-Kahler complex manifold X, s.t. the associated Hodge-to-de Rham spectral sequence degenerates but the the Hodge filtration on the de Rham cohomology $H^n_{dR}(X)$ doesn't define a pure Hodge structure. An explicit situation of such a manifold is a Hopf surface $X=(\mathbf C^2 - \{0,0\})/q^{\mathbf Z}$, where $q\in \mathbf C^*$ and $|q|<1$ (action is diagonal). Then one can prove that Hodge-to-de Rham spectral sequence degenerates for $X$ but $H^1(X,\mathcal O_X)=0\neq \mathbf C=H^0(X,\Omega^1_{X/\mathbf C})$. Hence the Hodge filtration doesn't define a pure Hodge structure on $H^1_{dR}(X)$ because of the failure of hodge symmetry. Although, it doesn't imply that there is no canonical isomorphism $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^q(X,\Omega^p_{X/\mathbf C})$, it shows that it's unlikely that the 2nd statement can be a consequence of the first. All of that being said, I will explain how to construct pure Hodge structures on cohomology of any proper smooth variety. Let us start with $X$ being a smooth algebraic variety over $\mathbf C$. Then one can define a descending filtration on the algebraic de Rham complex $\Omega^{\bullet}_{X/\mathbf C}$ as $F^i\Omega^{\bullet}_{X/\mathbf C}=\Omega_{X/\mathbf C}^{\geq i}$. This induces a descending filtration $F^{\bullet}H^n_{dR}(X)=F^{\bullet}H^n(X,\mathbf C)$. The main result is that this construction defines a pure Hodge structure under the properness assumption on X. Theorem 1: Let $X$ be a smooth and proper variety over $\mathbf C$, then a pair $(H^n(X,\mathbf Z), F^{\bullet}H^n(X,\mathbf C))$ is a pure Hodge structure of weight n. Moreover we have isomorphisms $F^{p}H^n(X,\mathbf C) \cap \overline{F^{n-p}H^n(X,\mathbf C)}\cong H^{q}(X,\Omega^p_{X/\mathbf C})$. Remark 1: General theory of pure Hodge structures implies that there is a canonical isomorphism $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^{q}(X,\Omega^p_{X/\mathbf C})$. Remark 2: Although we have a canonical decomposition $H^n(X,\mathbf C)\cong \oplus_{p+q=n}H^{q}(X,\Omega^p_{X/\mathbf C})$, a priori there is no morphism $H^q(X,\Omega^p_{X/\mathbf C}) \to H^n(X,\mathbf C)$. I don't know any purely algebraic construction of these maps, which doesn't use the fact that Hodge filtration defines a pure Hodge structure as an input. I believe this is the crucial place where theory of harmonic forms comes into play. Once you have such morphisms, Theorem $1$ is basically equivalent to the degeneration of the Hodge-to-de Rham spectral sequence. Remark 3: I think (I may be wrong) that there is no purely algebraic proof of Theorem $1$ as well. However, what I will show below is that if one takes Theorem 1 in the projective case as an input (this is the situation where $X^{an}$ is a Kahler manifold, so one can use theory of harmonic forms), then he/she can deduce the general form of Theorem $1$ in a purely algebraic way. Before going to explain this proof, I need to mention a lemma which eventually be the key step to reduce a case of a proper variety to the case of a projective one. Lemma 1: Let $f:X \to Y$ be a proper birational morphism between smooth connected varieties over a field $k$, then the induced morphism $H^p(f^*):H^p(Y,\Omega^q_{Y/k}) \to H^p(X,\Omega^q_{X/k})$ is injective for any $p$ and $q$. Proof: Recall that $H^p(f^*)$ is induced by a morphism $$ \Omega^q_{Y/k} \to \mathbf Rf_*f^*\Omega^q_{Y/k} \to \mathbf Rf_*\Omega^q_{X/k}. $$ Compose it with the Trace map $$ \operatorname{Tr_f}:\mathbf Rf_*\Omega^q_{X/k} \to \Omega^q_{Y/k}. $$ (Here we use that $\operatorname{dim}X =\operatorname{dim}Y$ because in general the Trace map is map from $:\mathbf Rf_*\Omega^q_{X/k}$ to $\Omega^{q-d}_{Y/k}[-d]$ where $d=\operatorname{dim}X -\operatorname{dim}Y$). Now note that the formation of Trace map is compatible with restrictions to open subsets. Hence, the composition $Tr_f\circ f^*:\Omega^q_{Y/k} \to \Omega^q_{Y/k}$ is the identity morphism on an open dense subset $U\subset Y$ (because $f$ is birational and Y is normal!). Since $\Omega^q_{Y/k}$ is a locally free sheaf, we conclude that $Tr_f\circ f^*$ is also the identity morphism. Therefore, $H^p(Tr_f\circ f^*)=H^p(Tr_f)\circ H^p(f^*)$ is the identity morphism as well. So $H^p(f^*)$ is injective as desired. Now let's come back to the proof of Theorem 1 in the proper case. Step 1: Chow lemma and Hironaka's resolution of singularities allow us to find a smooth and projective variety $X'$ together with a map $f:X' \to X$ s.t. $f$ is birational and proper. Then there is a canonical morphisms of Hodge-to-de Rham spectral sequences for $X$ and $X'$. Namely, let $$ E_1^{p,q}=H^q(X,\Omega^p_{X/\mathbf C}) \Rightarrow H^{p+q}_{dR}(X). $$ $$ \downarrow{f^*} $$ $$ E_1'^{p,q}=H^q(X',\Omega^p_{X'/\mathbf C}) \Rightarrow H^{p+q}_{dR}(X'). $$ Note that Lemma $1$ guarantees that $f^*$ is injective on the first page. Since we know that $E'$ degenerates on the first page ($d_1'^{p,q}=0$ for any $p,q$) we conclude that $d_1^{p,q}=0$ (because $f^*$ is compatible with differentials and injective). So $E_2^{p,q}=E_1^{p,q}$ and $E_2'^{p,q}=E_1'^{p,q}$, therefore $f^*$ is also injective on the second page. Then the same argument shows that $d_2^{p,q}=0$ for any $p,q$. Keep going to show that $E$ degenerates on the first page ($E_1^{p,q}=E_{\infty}^{p,q}$). Step 2: Two consequences from the Step 1. I) $F^pH^n_{dR}(X)\cap \overline{F^{n-p+1}H^n_{dR}(X)}=0$. We know that that his equality holds for $X'$ (b/c $X'$ is projective) and we know that Hodge filtration $F^{\bullet}H^n_{dR}(-)$ is functorial in $X$. Hence, argument in the Step $1$ says that $$ F^pH^n_{dR}(X)\cap \overline{F^{n-p+1}H^n_{dR}(X)} \subset F^pH^n_{dR}(X')\cap \overline{F^{n-p+1}H^n_{dR}(X')}=0. $$ II) We have a canonical isomorphism $F^pH^n_{dR}(X)/F^{p+1}H^n_{dR}(X)\cong H^{q}(X,\Omega^p_{X/\mathbf C})$. This is a direct consequence of the degeneration of the Hodge-to-de Rham spectral sequence. Step 3: The last thing we need to check is that $F^pH^n_{dR}(X)+\overline{F^{n-p+1}H^n_{dR}(X)}=H^n_{dR}(X)$. Since $F^pH^n_{dR}(X)$ and $\overline{F^{n-p+1}H^n_{dR}(X)}$ are disjoint inside $H^n_{dR}(X)$, it suffices to prove the equality $$ \operatorname{dim}(F^pH^n_{dR}(X))+\operatorname{dim}(\overline{F^{n-p+1}H^n_{dR}(X)})=\operatorname{dim}(H^n_{dR}(X)). $$ Ok, Consequence II from Step $2$ implies that $$ \operatorname{dim}(F^pH^n_{dR}(X))=\sum_{i\geq p}h^{i,n-i}, $$ $$\operatorname{dim}(\overline{F^{n-p+1}H^n_{dR}(X)} = \sum_{i\geq n-p+1}h^{i,n-i}, $$ $$ \operatorname{dim}(H^n_{dR}(X))=\sum_{i\geq 0}h^{i,n-i} \text{, where } h^{p,q}= \operatorname{dim}H^q(X,\Omega^p_{X/\mathbf C}). $$ Since we know that $F^pH^n_{dR}(X)$ and $\overline{F^{n-p+1}}H^n_{dR}(X)$ are disjoint we conclude that $$ \sum_{i\geq p}h^{i,n-i} +\sum_{i\geq n-p+1}h^{i,n-i} \leq \sum_{i\geq 0}h^{i,n-i}. $$ Subtract $\sum_{i\geq n-p+1}h^{i,n-i}$ to obtain an inequality $$ \sum_{i\geq p}h^{i,n-i} \leq \sum_{i\leq n-p}h^{i,n-i} (*) $$ Our goal is to prove that this is actually an equality. Here comes the miracle! We use an additional symmetry between hodge numbers of interchange left and right hand sides of this inequality to actually show that it is an equality! Namely, we apply Serre duality to see that $h^{p,q}=h^{d-p,d-q}$ where $d=\operatorname{dim} X$. Then we can rewrite both hand sides of $(*)$ in a different way. $$ \sum_{i\geq p}h^{i,n-i}=\sum_{i\geq p}h^{d-i,d-n+i}=\sum_{j\leq d-p=(2d-n)-(d-n+p)} h^{j,(2d-n)-j}, $$ $$ \sum_{i\leq n-p}h^{i,n-i}=\sum_{i\leq n-p}h^{d-i,d-n+i}=\sum_{j\geq d-n+p}h^{j,2d-n-j}. $$ But now observe that inequality (*) (applied to $n'=2d-n,p'=N-n+p$) guarantees that $$ \sum_{i\leq n-p}h^{i,n-i}=\sum_{j\geq d-n+p}h^{j,2d-n-j} \leq \sum_{j\geq d-n+p}h^{j,2d-n-j} =\sum_{i\leq n-p}h^{i,n-i}. $$ Therefore $\sum_{i\geq p}h^{i,n-i}=\sum_{i\leq n-p}h^{i,n-i}$! And we showed that this is equivalent to the equality $$ F^pH^n_{dR}(X)+\overline{F^{n-p+1}H^n_{dR}(X)}=H^n_{dR}(X). $$ In other words we showed that $(H^n(X,\mathbf Z), F^{\bullet}H^n(X,\mathbf C))$ is a pure Hodge structure of weight $d$.<|endoftext|> TITLE: Question on PDEs which are related to certain geometric problems (e.g. Calabi conjecture, Gauduchon conjecture) QUESTION [5 upvotes]: There are interesting symmetric functions $P_k$ arising from differential geometry and PDEs, where $P_k$ is given by \begin{equation} \begin{aligned} P_k(\lambda) = \prod_{1\leq i_{1}<\cdots < i_{k}\leq n} (\lambda_{i_{1}}+\cdots+ \lambda_{i_{k}}). \end{aligned} \end{equation} In addition, $\lambda_{i_{1}}+\cdots+ \lambda_{i_{k}}>0$ for any $1\leq i_{1}<\cdots < i_{k}\leq n$. The operator $P_1$ is related to Calabi conjecture in K\"ahler geometry, while $P_{n-1}$ is related to Gauduchon conjecture in Hermitian geometry. These two geometric problems can be reduced to solving fully nonlinear elliptic equations on the given compact complex manifold. My question: Are there books or literatures could present the detailed discussion about these functions $P_k$ or $\log P_k$ ($1\leq k TITLE: About the Hausdorff dimension of removable singularities of PDE QUESTION [6 upvotes]: There are some interesting phenomenons about removable singularities (or extension problems). In the theory of functions of several complex variables, we know the classical Hartogs theorem: Let $f$ be a holomorphic function on a set $G \setminus K$, where $G$ is an open subset of $\mathbb{C}^n$ ($n \ge 2$) and $K$ is a compact subset of $G$. If the complement $G \setminus K$ is connected, then $f$ can be extended to a unique holomorphic function on $G$. As a corollary, roughly speaking, holomorphic function can be extended in the codimension 2 case. Naturally, I wonder what's the largest Hausdorff dimension of removable singularities to elliptic equations perhaps under some assumptions, like boundedness. Are there some results? For example, we know a bounded harmonic function in a ball without the center can be extended harmonically to the whole ball. But what if the region is a ball without a closed set of Hausdorff dimension 1, 2 and so on? Is there an optimal dimension? In fact, I am more concerned about the problem for minimal surface equations and Monge–Ampère equations. REPLY [2 votes]: I am not familiar with the current studies on the Hausdorff dimension of the singular set of solutions to PDE, but I know that in [1] a necessary and sufficient condition for the holding of Hartogs phenomenon for a linear system of partial differential operators was stated and proved. The author does not use the methods of geometric measure theory: however, he deals with general compact singularities for the solutions, including the ones with non-zero Lebesgue measure. [1] Gaetano Fichera (1983), "Sul fenomeno di Hartogs per gli operatori lineari alle derivate parziali [Hartogs phenomenon for certain linear partial differential operators]" (Italian), Rendiconti dell' Istituto Lombardo di Scienze e Lettere. Scienze Matematiche e Applicazioni, Series A., 117: 199–211, MR0848259, Zbl 0603.35013.<|endoftext|> TITLE: Do two new special points in any triangle exist? QUESTION [10 upvotes]: There are some special points in any triangle, as Fermat point, symmedian point, incenter, Morley center, et cetera. Let $P$ be a point on the plane, $PA$, $PB$, $PC$ meet $BC$, $CA$, $AB$ at $A'$, $B'$, $C'$ respectively. From my construction by geogebra sofware. I proposed a conjecture: In any triangle exist two points $P$ so that: $AA'=BB'=CC'$. My question: Is the conjecture above correct? My geogebra: The Red locus: If $P$ lie on red locus then $AA'=CC'$. The Blue locus: If $P$ lie on red locus then $AA'=BB'$. The Pink locus: If $P$ lie on pink locus then $CC'=BB'$ See also: Triangle centers REPLY [10 votes]: Your conjecture is true: these two points are sometimes called the equicevian points of $ABC.$ REPLY [7 votes]: These two points are the focuses of Steiner circumconic https://en.m.wikipedia.org/wiki/Steiner_ellipse.<|endoftext|> TITLE: Limiting representation theory of quantum groups at roots of unity and $SL(2,\mathbb{C})$ QUESTION [32 upvotes]: Let $V_N$ denote the $N$-dimensional representation of the quantum group $U_q(\mathfrak s\mathfrak l_2)$. I am told that in the limit $N\to\infty$ with $q=e^{2\pi i/n}$ and $N/n\to\alpha\in(0,1)$, the representation $V_N$ of $U_q(\mathfrak s\mathfrak l_2)$ "converges" in some precise sense to an infinite-dimensional representation of the complex Lie group $\operatorname{SL}_2(\mathbb C)$. What is the precise formulation of this result, and where is it discussed in the literature? [My searches so far have not found anything substantial.] REPLY [11 votes]: This is a very interesting question. I have also made some search but i have not found this result explicitly mentioned somewhere in the literature. However, i remember i have heard such a claim in the past. So i will try to record my thoughts on the problem. Here is the way i understand it: Since you are considering the limits $N\to \infty$ and $N/n\to\alpha\in(0,1)$, this implies that $n\to\infty$ thus $q\to 1$. In this limit: $U_q(\mathfrak{sl}_2)\stackrel{q\to 1}{\longrightarrow} U(\mathfrak{sl}_2)$ and so you get a representation of the UEA and thus of the lie algebra $\mathfrak{sl}_2$. Since the corresponding lie group $SL(2,\mathbb{C})$, is simply connected, its category of representations is equivalent to the category of representations of the lie algebra $\mathfrak{sl}_2$ (in the sense that there is a bijection -up to isomorphism- between the lie group and the lie algebra representations). Thus, we finally arrive at a representation (up to isomorphism) of the lie group $SL(2,\mathbb{C})$. And since $N\to\infty$, the limit -if it exists- will be an infinite dim representation. Let us try to examine the situation a little closer and attempt to establish whether -and when- the limit is well defined and exists : The "convergence" of the quantum group to the lie algebra: A non-trivial point in the preceding argument is how to obtain a rigorous formulation of the limit: $U_q(\mathfrak{sl}_2)\stackrel{q\to 1}{\longrightarrow} U(\mathfrak{sl}_2)$ This is a subtle point, which -imo- does not have a unique answer; it depends on the description one uses for the quantum group $U_q(\mathfrak{sl}_2)$. Generally, we can either view $U_q(\mathfrak{sl}_2)$ as a $\mathbb{C}[[h]]$-algebra and write $q=e^{h/2}$ and $K=q^H$ (in which case $q\to 1\Leftrightarrow h\to 0$ and we use Del' Hospital for the $E$-action limit and power series expansion for the $K$-action limit), or we can view $U_q(\mathfrak{sl}_2)$ as a suitable quotient of a larger $\mathbb{C}$-algebra, which is however well defined at $q=1$. Both methods are well-known. See for example: - How $U_{q}(\mathfrak{sl}_{2})$ becomes the universal enveloping algebra $U(\mathfrak{sl}_{2})$ of $\mathfrak{sl}_{2}$, - Quantized Enveloping Algebras at $q=1$, - Exponentiations over the quantum algebra $U_{q}(\mathfrak{sl}_{2})$ (see the discussion and the computations of section 10, p. 65) The "convergence" of the representation: Let us study the limit on the irreducible, f.d. representations of $U_{q}(\mathfrak{sl}_{2})$ with $q=e^{2\pi i/n}$: (First recall that if $q$ is not a root of unity, then the f.d., irred, are highest weight representations. They are parameterized by $\varepsilon=\pm 1$ and the positive integers, i.e. we will denote such $N$-dim, irred, modules as $V_{\varepsilon, N-1}$. The notion of the $N\to\infty$ limit here is well defined in the following sense: The $V_{1,N-1}$ modules here have matrix elements which are not generally continuous functions of $q$ but they can be handled with methods similar to those mentioned above to show that $V_{1,N-1}\stackrel{q\to 1}{\longrightarrow}V_{N-1}$ where $V_{N-1}$ are the usual $N$-dim, irred, highest weight reps of $\mathfrak{sl}_2$. The $V_{-1,N-1}$ modules "vanish" when $q\to 1$. One can be more precise at that point but it would be irrelevant to the rest. We can equally well consider that we are taking the $N\to\infty$ limit considering the $V_{1,N-1}$ family only). Now, let us come back to the root of unity case: $q=e^{2\pi i/n}$. It is well known that, the irreducible modules now have an upper bound in their dimension: ${\small e=\begin{cases} n, & \text{$n$: odd} \\ n/2, & \text{$n$: even.} \end{cases}}$. It is also well known that the $N$-dim irred, representations, now fall into two classes: First, representations for which $N TITLE: What is the geometric significance of fibered category theory in topos theory? QUESTION [25 upvotes]: Often in topos theory, one starts with a geometric morphism $f: \mathcal Y \to \mathcal X$, but quickly passes to the Grothendieck fibration $U_f: \mathcal Y \downarrow f^\ast \to \mathcal X$, which is "$\mathcal Y$ regarded as an $\mathcal X$-topos". This maneuver takes one outside of the category of toposes and geometric morphisms, and thus tends to remove me from thinking of toposes geometrically, as generalized spaces. It seems there should be some geometric interpretation lurking here -- after all, Grothendieck fibrations are "categories varying over a base", and a topos is "a base that can be varied over". But although $\mathcal Y \downarrow f^\ast$ is a topos, the fibration $U_f: \mathcal Y \downarrow f^\ast \to \mathcal X$ is not (the direct image of) a geometric morphism! In fact it has a right adjoint $F^\ast$ which has a further right adjoint $F_\ast: \mathcal Y \downarrow f^\ast \to \mathcal X$. We in fact have a totally connected geometric morphism $F: \mathcal Y \downarrow f^\ast \to \mathcal X$. Moreover, the fibers of $U_f$ are toposes, but the reindexing functors are most naturally viewed as the inverse images of étale geometric morphisms. So we don't naturally have a "topos fibered in toposes" it seems. For one more perspective, $\mathcal Y \downarrow f^\ast$ is a cocomma object in the 2-category of toposes. As such, $\mathcal Y \downarrow f^\ast \leftarrow \mathcal Y$ is the free co-fibration on $f$ in the 2-category of toposes. I think this is the point of departure for Rosebrugh and Wood. But I have no geometric intuition for what a co-Grothendieck fibration is. And anyway, the functor $U_f$ doesn't even live in the category of toposes. I'm not really sure what to make of this. So here are some Questions: Let $f: \mathcal Y \to \mathcal X$ be a geometric morphism. Is there a geometric interpretation of the fibration $U_f: \mathcal Y \downarrow f^\ast \to \mathcal X$? Does the totally connected geometric morphism $F: \mathcal Y \downarrow f^\ast \to \mathcal X$ enjoy some sort of universal property with respect to $f$? (for example, does the inclusion of the category of toposes and totally connected geometric morphisms into toposes and all geometric morphisms have a left adjoint which is being applied here?) Is there a fibration in the 2-category of toposes lurking around here somewhere (see the Elephant B4.4 for some discussion of these)? Is there a general geometric interpretation of an opfibration over a topos whose fibers are toposes, with étale geometric morphisms for reindexing? Addendum: Here are a few places where "indexed" notions come up where the viewpoint that "everything is indexed over the codomain" seems kind of inflexible to me: A proper geometric morphism $f: \mathcal Y \to \mathcal X$ is one such that $f_\ast$ preserves $\mathcal X$-indexed filtered colimits of subterminal objects. It's a bit awkward, for example, to even discuss the composition of proper morphisms if you're tied to the view that $\mathcal Y$ is internal to $\mathcal X$. A separated geometric morphism is one such that the diagonal is proper; in particular, a Hausdorff topos is one such that $\mathcal X \to \mathcal X \times \mathcal X$ is proper. It seems very unnatural to me in this context to think of $\mathcal X$ as being primarily an object "internal to $\mathcal X \times \mathcal X$". Locally connected geometric morphisms also use indexing in one form of their definition. So do local geometric morphisms. Let me also point out there there is at least one place in Sketches of an Elephant where $\mathcal Y \downarrow f^\ast$ plays a role qua topos -- in Ch C3.6 on local geometric morphisms, where it's referred to as a "scone". See in particular Example 3.6.3(f), Lemma 3.6.4, and Corollary 3.6.13 The scone is the dual construction $f^\ast \downarrow \mathcal Y$. But in the end of Ch C3.6, Johnstone does consider $\mathcal Y \downarrow f^\ast$ qua topos, and shows that it is related to totally connected geometric morphisms in the same way that the scone is related to local geometric morphisms. REPLY [3 votes]: See my notes on fibered categories (arXiv:1801.02927) where in the second part I expose the fibered view of gemetric morphisms explaining that bounded geometric morphisms to a base topos SS are equivalent to Grothendieck toposes fibered over SS as proved in Jean-Luc Moens's Thesis from 1982 in Louvain-la-Neuve. In my opinion this is the real justification of geometric morphisms since the usual one as generalized continuous maps is a mere analogy. There is no systematic translation of notions from topology from spaces to locales and/or toposes! Moreover, developing analysis in this setting would be impossible or at least most inconvenient: just imagine to define exp or sin as locale morphisms!<|endoftext|> TITLE: "Cyclic" continuum QUESTION [6 upvotes]: On p. 221 of http://topology.auburn.edu/tp/reprints/v08/tp08113.pdf, I found the following definition: "A curve is said to be cyclic if its first Čech cohomology group with integer coefficients does not vanish". Here, a curve means a homogeneous metric continuum of dimension 1. Can someone explain this definition in different, more elementary, terms, and give some examples to illustrate the meaning? REPLY [5 votes]: Perhaps the first Čech cohomology group with integer coefficients may seem not elementary. Then use the characterization of it as the first Borsuk's cohomotopy group (or it appears as the Brushlinsky's group in the well-known text on Homotopy Theory by Hu). This group's elements are simply the homotopy classes of mappings into $\ \mathbb S^1.$ Space $\ S^1\ $ is a topological group, it induces the group structure on the homotopy classes.<|endoftext|> TITLE: Research semester in math QUESTION [14 upvotes]: Admittedly, this is a soft question. My own experience in mathematical research has been of long periods of research, mostly characterized in long "blocks" and sporadic breakthrough. How does this image fit with research semesters, in which many researchers (veterans, early careers, postdocs and students) gather at a single institution, e.g., ICERM, IPAM etc., for a few months? What kind of research do one usually do in these sorts of semesters? Should one get there with a well defined project a-priori? Do you find this format effective for mathematical research? REPLY [5 votes]: Generally one applies for a short-term position to perform the (expected) final research (and writing) on a topic one has been working on for a while, for instance to explore ramifications, extensions, and so on. It is a rare venue (like the Institute for Advanced Studies) that hires you for an extended period based more on "promise" than well-identified needed execution. My recommendation is to do what lots of great mathematicians do: think about several projects intermittently. When one project seems particularly promising, or has succumbed to part of your analyses, highlight that one and apply for a short internship to explore it more fully. Since you asked: The benefits of such a short-term research position include: The opportunity to work with others, especially experts in a field. Get professional exposure, meeting a wider range of mathematicians than you would otherwise. Opportunity to travel, especially to foreign countries (where short-term positions are easier to get than permanent positions). Opportunity to focus intensely on a technical problem, without teaching or administration duties. Temporarily solving the "two-body" problem, i.e., to follow a spouse to his/her place of employment.<|endoftext|> TITLE: A proof required for this identity QUESTION [21 upvotes]: Experiments support the below identity. Question. Is this true? Combinatorial proof preferred if possible. $$\sum_{m=0}^n\binom{n-\frac13}m\binom{n+\frac13}{n-m}(1+6m-3n)^{2n+1} =\left(\frac43\right)^n\frac{(3n+1)!}{n!}.$$ In View of MTyson's suggestion (see below), a generalized question can be asked: Question. Is this true? Combinatorial proof preferred if possible. $$\sum_{m=0}^n\binom{n-y}m\binom{n+y}{n-m}(y+2m-n)^{2n+1} =y\prod_{k=1}^n4(k^2-y^2).$$ REPLY [5 votes]: "show that the a priori degree $\le n$ polynomial $\sum_{i+j=n}{n−x\choose i}{n+x\choose j}(i−j)^k$ is only degree $k$". That is fairly obvious when $k<2n+1$ because it suffices to check it for integer $x\in\{-n,\dots,n\}$. We will just show that for every $a,b\ge 0$ with $a+b\le k$, the sum $\sum_{i+j=n}{n−x\choose i}{n+x\choose j}{i\choose a}{j\choose b}$ is a polynomial of degree $\le a+b$ in $x$. When $|x|\le n$, this sum has a simple combinatorial meaning: choose some $n$ balls out of $2n$ ($i$ out of first $n-x$, $j$ out of last $n+x$) and color $a$ in the first group red and $b$ in the last group blue. But we can then compute it as ${n-x\choose a}{n+x\choose b}{2n-a-b\choose n-a-b}$ (choose colored balls first and add the remaining ones afterwards). That the leading coefficient is right is not immediately obvious to me but, perhaps, someone can see it too.<|endoftext|> TITLE: Connecting torsors by a rational curve QUESTION [10 upvotes]: Assume that $k$ is an infinite field. Let $G$ be a finite (constant) group, let $V$ be a faithful $G$-representation over $k$, $U$ a non-empty open subset of $V$ where the action is free. The map $\pi:U\to U/G$ is a $G$-torsor, and moreover, every $G$-torsor over $k$ arises as a fiber of $\pi$, and the set of $p\in (U/G)(k)$ such that $\pi^{-1}(p)$ is isomorphic to a given torsor is dense in $U/G$. This is actually true over every field extension $K/k$. Let $\alpha,\beta$ be two $G$-torsors over $k$, and assume that there exists a $G$-torsor $P\to Z$ where $Z$ is an open subset of $\mathbb{P}^1$, and assume that $\alpha$ and $\beta$ both arise as fibers of $P\to Z$. Can I find a map from some open subset $Z'$ of $\mathbb{P^1}$ to $U/G$ such that the induced torsor above $Z'$ admits $\alpha$ and $\beta$ as fibers? The converse is of course obvious. Note that if $U/G$ is retract rational (condition which does not depend on $V$) but only on $G$), then any two points on $U/G$ may be connected by an open subset of a rational curve, so everything is trivial in this case. Unfortunately the only examples that I have so far belong to this trivial class. One might of course ask a similar question for an arbitrary linear algebraic group $G$ (in this case $V$ must be chosen generically free and not just faithful, and the quotient will be a rational quotient). Added: the same question, but with a chain of rational curves in place of a single one. REPLY [6 votes]: Welcome new contributor. That is true, even for $G$ a geometrically reductive group scheme, not merely for finite groups. The proof is elementary, but it is a bit long. It is related to the problems of weak approximation and strong approximation, just for affine space in your original question. Strong approximation is elementary for affine space. However, there are many other schemes that satisfy weak approximation and strong approximation (sometimes by deep arguments), and the analogous result also holds for those schemes. Let $S$ be a scheme. Let $\pi_S:V_S\to S$ be a smooth morphism, resp. a smooth group $S$-scheme, a smooth simplicial $S$-scheme, etc. Definition 1. For every $S$-scheme $Z$, a form of $V_S$ over $Z$ is a smooth morphism $\pi:V_Z\to Z$, resp. a smooth $Z$-group scheme, a smooth simplicial $Z$-scheme, etc., such that there exists a smooth, surjective morphism $Z'\to Z$ and an isomorphism of $V_Z\times_Z Z'$ with $V_S\times_S Z'$ as smooth $Z'$-schemes, resp. as smooth group $Z'$-schemes, smooth simplicial $Z'$-schemes, etc. Definition 2. Let $\mathcal{C}$ be a class of affine $S$-schemes. All forms of $V_S/S$ are said to satisfy strong approximation, resp. weak approximation, with respect to $\mathcal{C}$ if for every affine $S$-scheme $Z$ in the class $\mathcal{C}$, for every form $\pi:V_Z\to Z$ of $V_S/S$, for every closed subscheme $i:\Pi\hookrightarrow Z$ that is Artinian, and for every $Z$-morphism $\sigma_\Pi:\Pi\to V_Z$, there exists a $Z$-morphism $\sigma:Z\to V_Z$ that restricts to $\sigma_\Pi$, resp. after replacing $Z$ by an open subscheme containing $\Pi$, there exists a $Z$-morphism $\sigma$ that restricts to $\sigma_\Pi$. Forms of $V_S/S$ satisfy very weak approximation with respect to $\mathcal{C}$ if every form of $V_S/S$ over an infinite field has a Zariski dense set of rational points and for every $Z/S$ in $\mathcal{C}$, for every form $V_Z$ of $V_S$ over $Z$, for every Artinian closed subscheme $\Pi$ of $Z$ whose residue fields are infinite, and for every closed subset $W\subset V_Z$ whose fiber $W_p\subset V_{Z,p}$ is nowhere dense for every point $p\in \Pi$, after replacing $Z$ by an open subscheme containing $\Pi$, there exists a $Z$-morphism $\sigma$ such that $\sigma(\Pi)$ is disjoint from $W$. Lemma 3. If forms of $V_S/S$ satisfy strong approximation, then they satisfy weak approximation. If forms of $V_S/S$ over every infinite field have a Zariski dense set of rational points and if they satisfy weak approximation, then they satisfy very weak approximation. Proof. The first claim is obvious. For the second claim, use Zariski density of points to find a rational point on each fiber $V_{Z,p}$ that is in the complement of $W_p$. Next, use $Z$-smoothness of $V_Z$ to extend that rational point to a section $\sigma_\Pi$. QED Let $R$ be an irreducible scheme, and let $Q\subset R$ be a proper closed subset. Corollary 4. Assume that $V_S/S$ satisfies very weak approximation for class $\mathcal{C}$. For every $Z/S$ in class $\mathcal{C}$, for every form $V_Z/Z$ of $V_S/S$, for every Artinian, closed subscheme $\Pi$ of $Z$ with infinite residue fields, and for every morphism $f:V_Z\to R$ such that $f_p:V_{Z,p}\to R$ is dominant for every $p\in \Pi$, there exists a section $\sigma$ of $\pi$ such that $f\circ \sigma(\Pi)$ is disjoint from $Q$. Proof. Define $W$ to be the inverse image closed subset $f^{-1}(Q)$. By the hypothesis, for every $p\in \Pi$, the fiber $W_p$ is a proper closed subset of the fiber $V_{Z,p}$. QED There are many examples of morphisms that satisfy strong approximation, resp. weak approximation for a specified class of affine schemes. One example that satisfies strong approximation for all affine schemes is affine space itself. Every form of the additive group scheme $\mathbb{A}^r_S$ over $Z$ is a geometric vector bundle $$\pi:V_Z\to Z$$, i.e., $\text{Spec}_Z(\text{Sym}^\bullet_{\mathcal{O}_Z}(\mathcal{E}))$ for a locally free $\mathcal{O}_Z$-module $\mathcal{E}$ of finite rank $r$. Let $$i:\Pi \hookrightarrow Z$$ be a closed subscheme that is Artinian. Let $$\sigma_{\Pi}:\Pi \to V_Z$$ be a $Z$-morphism, i.e., $\pi\circ \sigma_{\Pi}$ equals $i$. Strong Approximation for Affine Space. There exists a section $\sigma:Z\to V_Z$ of $\pi$ whose restriction to $\Pi$ equals $\sigma_{\Pi}$. Also, rational points are dense on every fiber over an infinite field. Proof. The fact that $k$-points of $\mathbb{A}^r_k$ are Zariski dense for every infinite field is elementary; the main claim is strong approximation. The $\mathcal{O}_Z(Z)$-module $\mathcal{E}(Z)$ is locally free of finite rank $r$. Thus, we can check locally that for every $\mathcal{O}_Z(Z)$-module $N$, the adjointness homomorphism, $$\text{Hom}_{\mathcal{O}_Z(Z)}(\mathcal{E}(Z),\mathcal{O}_Z(Z))\otimes_{\mathcal{O}_Z(Z)} N \to text{Hom}_{\mathcal{O}_Z(Z)}(\mathcal{E}(Z),N)$$ is an isomorphism. In particular, setting $N$ equals to $\mathcal{O}_{\Pi}(\Pi)$, it follows that every $\mathcal{O}_Z(Z)$-module homomorphism from $\mathcal{E}(Z)$ to $\mathcal{O}_{\Pi}(\Pi)$ is the restriction of a homomorphism to $\mathcal{O}_Z(Z)$. Now use the universal property of the relative Spec and of the symmetric algebra. QED Forms Associated to a Torsor. Let $S$ be a scheme. Let $G$ be an affine, flat, finitely presented $S$-group scheme. Let $V_S \to S$ be an affine, smooth $S$-scheme together with an $S$-action of $G$, $$\mu_V:G\times_S V_S \to V_S.$$ For every $S$-scheme $E$, the base change $V_S\times_S E \to E$ is a smooth, affine $E$-scheme. For every action of $G$ on $E$, $$\mu_E:G\times_S E \to E,$$ there is an associated diagonal action, $$(\mu_V,\mu_E):G\times_S (V_S\times_S E) \to V_S\times_S E$$ that is equivariant for both projections. For every $S$-scheme $Z$ and for every $G$-torsor over $Z$, $$(E\to Z,\mu_E:G\times_S E \to E),$$ by fppf descent for affine morphisms, there is a unique smooth, affine $Z$-scheme, $$V_Z\to Z,$$ whose base change $V_Z\times_Z E$ is $G$-equivariantly isomorphic to $V_S\times_S E$ as an $E$-scheme. Thus, $V_Z/S$ is a form of $V_S/S$. Since $V\times_S E$ is smooth over $E$, also $V_Z$ is smooth over $B$. The projection morphism, $$q_{V,E}:V_S\times_S E \to V_Z,$$ is a categorical quotient of $V_S\times_S Z$ by the $G$-action $(\mu_V,\mu_E)$. The natural projection $$\text{pr}_V:V_S\times_S E \to V_S$$ is $G$-equivariant. Quotient Hypothesis. The $S$-scheme $V_S$ has geometrically irreducible fibers. There exists a categorical quotient $q_V:V_S\to V_S/G$ of the $G$-action. There exists a dense open subscheme $U/G$ of $V_S/G$ whose inverse image $U$ is a $G$-torsor over $U/G$ that is dense in every $S$-fiber of $V_S$. Nota bene. The hypotheses hold for a generically free linear representation $V_S$ of a finite or geometrically reductive group scheme $G$ (it can also hold for certain representations of non-reductive group schemes). Now assume that $S$ equals $\text{Spec}(k)$ for an infinite field $k$, assume that $Z$ is affine in a specified class $\mathcal{C}$, let $\Pi\subset Z$ be a closed subscheme that is Artinian, and let $Q\subset V/G$ denote the closed complement of the image of $U$. Corollary 6. If $V_S/S$ satisfies very weak approximation for class $\mathcal{C}$, then there exists a section $\sigma$ of $V_Z\to Z$ such that the composition $f\circ \sigma$ maps $\Pi$ into the image of $U$. Proof. This is just Corollary 4 with $R$ set equal to $V/G$. QED Proposition 7. Let $k$ be an infinite field. Let $G$ be a finite type, affine group $k$-scheme. Let $V_k$ be a smooth, affine $k$-scheme that satisfies very weak approximation for class $\mathcal{C}$ and that satisfies the Quotient Hypothesis. For every affine $k$-scheme $Z$ in class $\mathcal{C}$, for every $G$-torsor $E$ over $Z$, and for every Artinian, closed subscheme $\Pi\subset Z$, there exists an open subscheme $Z'\subset Z$ containing $\Pi$ and a $k$-morphism $s:Z'\to U/G$ such that the $s$-pullback to $Z'$ of the $G$-torsor $U\to U/G$ is isomorphic as a $G$-torsor to the restriction over $Z'$ of the $G$-torsor $E$. Proof. With respect to the previous corollary, define $Z'$ to be the open complement of $(f\circ \sigma)^{-1}(Q)$, and define $s$ to be the restriction to $Z'$ of $f\circ \sigma$. QED Corollary 8. For $k$ an infinite field, for every geometrically reductive group $k$-scheme $G$, for every generically free linear representation $V$ of $G$ with free locus $U$, for every affine $k$-scheme $Z$, for every Artinian, closed subscheme $\Pi$ of $Z$, for every $G$-torsor $E$ over $Z$, there exists an open subscheme $Z'$ of $Z$ containing $\Pi$ and there exists a $k$-morphism $s:Z'\to U/G$ such that the $s$-pullback of $U$ is isomorphic as a $G$-torsor to the restriction over $\Pi$ of $E$. Proof Every generically free linear representation of a geometrically reductive satisfies the Quotient Hypothesis. Thus, the corollary follows from Proposition 8, from Strong Approximation for Affine Space, and from Lemma 3. QED<|endoftext|> TITLE: How does the parity of $n$ affect the properties of $\mathbb{R}^n$? QUESTION [8 upvotes]: Does the parity of the dimension of $\mathbb{R}^n$ affect its structure/properties? As in, does it make a difference if $n$ is even or odd? REPLY [4 votes]: If the dimension of a vector space is odd, then all (orientation-preserving) rotations in odd dimensions fix some axis. Many of the differences between even-dimensional and odd-dimensional geometry relate to this fact. For example, The lack of symplectic structure in odd dimensions follows from the Lie-algebra version of the above statement: all odd-dimensional antisymmetric maps are degenerate. The $-1$ map doesn't fix any axis, so it cannot be orientation-preserving in odd dimensions. Synge's theorem states that if $M$ is compact, Riemannian, and has positive sectional curvature, then there is a conclusion which depends on the pairity of its dimension. The proof makes essential use of the above fact. (See Lemma 3.8 in "Riemannian Geometry" by do Carmo.)<|endoftext|> TITLE: Homotopy type of a specific discrete monoid QUESTION [6 upvotes]: Consider the discrete monoid $M$ of nondecreasing continuous maps from $[0,1]$ to itself preserving the extremities. Note that the monoid is right-cancellative ($x.z=y.z$ implies $x=y$, since $z$ is always onto). What do we know about the homotopy type of this monoid (viewed as a one-object category) ? In particular, about its homotopy groups ? My background on this subject is very small. By a paper from Dusa McDuff (On the classifying space of discrete monoids), every path-connected space has the same homotopy type as the classifying space of some monoid, and the fundamental group of $BM$ is the groupification of $M$. EDIT: I am a bit confused between the English meaning and the French meaning of nondecreasing. The monoid I am talking about is not a group because a nondecreasing map preserving extremities is not necessarily one-to-one. I hope that this clarification will be helpful. REPLY [11 votes]: This space is contractible, and so all of its homotopy groups are trivial. Define two elements in $M$ by: $$ \begin{align*} A(x) &= \begin{cases} 2x &\text{if }x \leq 1/2\\1 &\text{if }x \geq 1/2\end{cases}\\ B(x) &= \begin{cases} 0 &\text{if }x \leq 1/2\\2x-1 &\text{if }x \geq 1/2\end{cases} \end{align*} $$ Define three monoid homomorphisms $Id, U, V: M \to M$ by: $$ \begin{align*} (Id(f))(x) &= f(x)\\ (Uf)(x) &= \begin{cases} \tfrac{1}{2}f(2x) &\text{if }x \leq 1/2\\x &\text{if }x \geq 1/2\end{cases}\\ (Vf)(x) &= x \end{align*} $$ For any $f \in M$, we have the following identities: $$ \begin{align*} A \circ (Uf) &= f \circ A\\ B \circ (Uf) &= (Vf) \circ B \end{align*} $$ As a result, we can reinterpret this in terms of the one-object category $M$: we get three functors $Id,U,V: M \to M$ and natural transformations $A: U \to I$ and $B: U \to V$. Upon taking geometric realization, we get a space $BM$, these functors turn into continuous maps $Id, U, V:BM \to BM$ and homotopies from $U$ to $Id$ and from $U$ to $V$. However, $V$ is a constant map, and so this says that the homotopy type of $BM$ is contractible. (I believe that I learned this from somewhere in a paper of Lurie's, but I can't find it currently.)<|endoftext|> TITLE: On the existence of integer square root of a $3 \times 3$ positive definite matrix QUESTION [5 upvotes]: As far as I know, a real square matrix $M$ has a real square root if $M$ is positive semidefinite, i.e., if all eigenvalues are nonnegative. And, in fact, its square root is unique. I have read some research papers on how to solve the square root of a $3 \times 3$ positive definite matrix using Cayley-Hamilton, the minimal polynomial, and diagonalization. However, when does a $3 \times 3$ integer matrix $M$ have an integer square root? Trivially, $M$ must be positive definite to make sure its square root exists and is real. Also, $\det(M)$ must be a a perfect square. Other than that, I am stuck. Please help me with this. Or just give me a hint or a lead. Thank you. REPLY [3 votes]: Besides the fact that $\det M$ needs to be a perfect square, there is an other set of necessary conditions: among the diagonal entries $m_{ii}$ and the principal $2\times2$ minors $m_{ii}m_{jj}-m_{ij}^2$, none of them may be $\equiv7$ mod $8$. Writing $A=\sqrt M$, this is because on the one hand $$m_{ii}=a_{i1}^2+a_{i2}^2+a_{i3}^2.$$ And on the other hand $$m_{ii}m_{jj}-m_{ij}^2=\sum_{k,\ell=1}^3(a_{ij}a_{k\ell}-a_{i\ell}a_{kj})^2.$$<|endoftext|> TITLE: Why the Thom spectrum of $-\xi$ (or more generally of a virtual bundle) is defined as it is? QUESTION [8 upvotes]: As the title suggests, I'm trying to find motivation on the definition of the Thom spectrum of $-\xi$, or more generally on the definition of the Thom spectrum of a virtual bundle. In this paper by S. Bauer (middle of page 7) he defines the Thom spectrum for a virtual bundle of the form $\xi - \Bbb R^m\times X$ as the desuspension: $$ Th(\xi-\Bbb R^m\times X) = \sum^{-m}\Bbb S\wedge X^{\xi}$$ with this definition, let $\eta$ be a bundle over $X$ such that $\xi\oplus \eta\cong X\times \Bbb R^N$, then we have we have that $$Th(-\xi) = \sum^{-N}\Bbb S\wedge X^{\eta}$$ I'm wondering what is the reason of this definition and if it was somehow related with the concept of duality morphism (Rudyak, page 47, def 2.3) My intuition is that with this definition we should have that the two spectra are dual in the sense above. This would explain for example why Bauer in the paper mentioned above, or Crabb & Knapp in this paper (page 90, Lemma 1.1) claim that there is a pairing given by cap product: $$\widetilde{h}^0(X^{-\xi})\times \widetilde{h}^r(X^{\xi})\to \widetilde{h}^r(X)$$ since we could interpret the first one as an homology group and then cap product is well defined. So I tried proving that we have this aforementioned duality. In fact we have something resembling a duality between $Th(-\xi)$ and $Th(\xi)$: $$Th(-\xi)\wedge Th(\xi)=\sum^{-N}\Bbb S\wedge X^{\eta}\wedge \Bbb S\wedge X^{\xi}$$ $$ = \sum^{-N}\Bbb S\wedge X^{\eta}\wedge X^{\xi}$$ $$= \sum^{-N}\Bbb S\wedge (X\times X)^{\eta \times \xi}$$ but my problem is that $(X\times X)^{\eta \times \xi}$ is not quite $X_+\wedge \sum^N\Bbb S$, instead we have $$X_+\wedge \sum^N\Bbb S = \Delta^*(X\times X)^{\eta \times \xi}\wedge \Bbb S$$ where $\Delta \colon X \to X\times X$ is the diagonal map, and this prevents me for proving that we have an actual duality. I'm slightly confused here so I apologise in advance if something is not super precise. REPLY [13 votes]: Just a quick answer to explain the original reason behind the definition and why our modern understanding of Thom spectra vindicates it. Let $X$ be a space and $\xi$ a virtual vector bundle over $X$. The definition you give is the only possible definition of $X^\xi$ such that $X^\xi$ coincide with the classical notion of Thom spectrum (the suspension spectrum of the one-point compactification of $\xi$) when $\xi$ is a vector bundle The formula $X^{\xi\oplus \mathbb{R}}\cong \Sigma X^\xi$ (which is known to be true for the case when $\xi$ is a vector bundle) holds. This definition is in fact very useful, because it allows us to state Atiyah duality in a very simple and elegant way: Theorem (Atiyah duality): Let $M$ be a $d$-dimensional closed smooth manifold. Then the Spanier-Whitehead dual of $\Sigma^\infty_+M$ is the Thom spectrum $M^{-TM}$ where $TM$ is the tangent bundle over $M$. This theorem can be seen as a generalization of Poincaré duality for nonorientable manifolds (and homology theories): it's saying that for every spectrum $E$ there are natural isomorphisms $$\tilde{E}^{-*}(M^{-TM})\cong E_*M$$ When $TM$ is $E$-orientable, the Thom isomorphism gives us the classical statement of Poincaré duality. There is also a version for manifolds with boundary recovering Lefschetz duality. I think that the map that's puzzling you to comes from the map of spectra $$\Sigma^\infty_+X\to X^\xi\wedge X^{-\xi}\cong (X\times X)^{\xi \boxplus -\xi}$$ induced by the map of spaces equipped with a virtual vector bundle $$(X,0)\to (X\times X,\xi\boxplus -\xi)$$ given by the diagonal $X\to X\times X$ (since the pullback of $\xi\boxplus -\xi$ along the diagonal is $\xi\oplus -\xi\cong 0$). Note that this map does not come from a duality map. Finally let me mention that it turns out that the Thom spectrum can also be defined as $$X^\xi=\mathrm{hocolim}_{x\in X} \mathbb{S}^{\xi_x}$$ where $\mathbb{S}^{\xi_x}$ is the suspension spectrum of the 1-point compactification of the fiber $\xi_x$ over $x\in X$. This definition is very convenient for proving a lot of properties. For example see this paper using it to study the multiplicative structures on Thom spectra.<|endoftext|> TITLE: Are there Prüfer sequences for rooted forests? QUESTION [7 upvotes]: One well-known, extremely slick proof of Cayley's tree enumeration theorem is the use of Prüfer sequences. Cayley also proved a version for forests, namely that the number of forests with $n$ labelled vertices that consist of $s$ distinct trees such that $s$ specified vertices belong to distinct trees is $sn^{n-s-1}$; see https://core.ac.uk/download/pdf/82105567.pdf . Is there a generalization of Prüfer sequences that corresponds to this quantity? Or more generally, is there a nice way to enumerate all $sn^{n-s-1}$ such forests, given $s$ and $n$? REPLY [5 votes]: We can prove this using ordinary Prüfer sequences; no generalization is needed. First I'll describe Prüfer's correspondence, as a bijection from trees on $\{0,1,2,\ldots,n\}$, where $n\ge1$, to sequences $a_1 \ldots a_{n-1}$ of elements of $\{0,1,2,\ldots,n\}$: If $n=1$ then $a_1\ldots a_{n-1}$ is empty. Otherwise we let $b_1$ be the greatest leaf of the tree and let $a_1$ be the unique vertex adjacent to $b_1$. (Prüfer's correspondence is usually described with the least leaf instead of the greatest leaf, but the greatest leaf works better in this application.) We then remove $b_1$ and its incident edge from the tree and let $b_2$ be the greatest leaf of the remaining tree, and let $a_2$ be the vertex adjacent to $b_2$. We continue in this way until only two vertices remain and we have constructed $a_1,a_2,\dots, a_{n-1}$. To prove Cayley's formula it is enough to count forests of $s$ trees with vertex set $\{1,2,\dots,n\}$ in which vertices $1,2,\dots, s$ are all in different trees. By adding a new vertex 0 adjacent to vertices $1, 2, \dots, s $ but no others, we see that the problem is equivalent to counting trees with vertex set $\{0,1,\dots,n\}$ in which vertex 0 is adjacent to vertices $1, 2, \dots, s $ but no others. I claim that the Prüfer codes for these trees are the sequences $a_1 a_2 \ldots a_{n-1}$ such that (1) $a_i\in \{1,2,\dots, n\}$ for $i=1,2,\dots, n-s-1$; (2) $a_{n-s}\in\{1,2,\ldots,s\}$; and (3) $a_i = 0$ for $i=n-s+1, n-s+2, \ldots, n-1$. The Prüfer code for any such tree has these properties because in the process of Prüfer's correspondence, no element of $\{0,1,2,\ldots,s\}$ will be the greatest leaf until all other vertices have been removed. Thus the last $s-1$ values of $ b_i$ will be (in order) $s, s-1, s-2, \ldots, 2$. (The two remaining vertices, 1 and 0, are never $b_i$.) The corresponding values of $a_i$ are 0 since these vertices are adjacent only to 0 (after the earlier $b_i$'s have been removed), and no other vertices are adjacent to 0 so no other $a_i$ is 0. Moreover, $a_{n-s}$ must be an element of $\{1,2,\ldots,s\}$ since after $b_{n-s}$ is removed, the only remaining vertices are $0,1,\ldots, s$, (and $b_{n-s}$ can't be adjacent to 0). Conversely, any sequence satisfying (1), (2), and (3) is the Prüfer code for a tree of the kind that we want to count. This requires some additional justification, which I'll omit. The number of such sequences is clearly $sn^{n-s-1}$.<|endoftext|> TITLE: Critical points and high homotopy groups QUESTION [6 upvotes]: Is there any known or interesting relation between critical points (possibly degenerate, or maybe only nondegenerate) of a function on a manifold and generators/relations of high homotopy groups? I only know about the relation to representations of a fundamental group.... REPLY [3 votes]: Bott periodicity (which computes all the homotopy group of a Lie group was proved by Morse Theory (so, the study of critical points, admittedly on the various path spaces, not quite the manifold itself). See the nice write-up by Aaron Mazel-Gee REPLY [3 votes]: I am not sure how it is related to your question, but it is possible to construct a mapping $f\in C^1(\mathbb{S}^{n+1},\mathbb{S}^n)$ for all $n\geq 4$, that generates a nontrivial element in $\pi_{n+1}(\mathbb{S}^n)$, but such that all points of the mapping are critical, that is $\operatorname{rank}Df(x) TITLE: Is every 3-dim self-dual Galois representation a symmetic square of 2-dim representation? QUESTION [7 upvotes]: Basically, my question as in the title. Here the Galois representation I consider is an $\ell$-adic Galois representation (comes from geometry). And by the word "self-dual" I mean that representation is isomorphic to its dual representation (transpose inverse in language of matrix) up to a Tate twist. If the answer is positive, please let me know the idea or reference. If the answer is negative, I am wondering if there is any extra condition(s) to make my statement to be true? Thanks Edit: Thanks for the comments from @David Loeffler and @David E Speyer, I think the "correct" statement should be: Is any essentially self-dual 3 dimensional $\ell$-adic Galois representation isomorphic (up to a quadratic character) to a symmetric square of a 2-dimensional Galois representation? REPLY [17 votes]: The correct statement is: any 3-dimensional selfdual Galois representation is isomorphic to a quadratic twist of the adjoint of some 2-dimensional representation. (The quadratic twist is really necessary, as one can see from a variant of David Speyer's example.) If $\rho: G_F \to GL_3(\overline{\mathbf{Q}}_\ell)$ is a Galois representation, where $F$ is a number field, and $\rho \cong \rho^*$, then the image of $\rho$ has to be contained in the subgroup $O_3(\overline{\mathbf{Q}}_\ell)$ of matrices preserving an orthogonal form. The quotient $O_3 / SO_3$ has order 2, so there is some quadratic character $\nu$ such that $\rho \otimes \nu$ takes values in $SO_3$. There is an exceptional isomorphism $Ad: PGL_2 \cong SO_3$, so $\rho$ must be $Ad(\tau)$ for some 2-dimensional projective representation $\bar\tau$. You want to know if $\bar\tau$ lifts to an actual representation; the obstruction to this lies in some $H^2$, and by a theorem of Tate this vanishes, so $\bar\tau$ is the image in $PGL_2$ of some $\tau: G_F \to GL_2(\overline{\mathbf{Q}}_\ell)$. (Note that $\tau$ is non-unique, and it is far from obvious a priori that $\tau$ can be chosen to be geometric if $\rho$ is. This kind of lifting-with-local-conditions question has been studied in detail in Patrikis' thesis.) This statement has an analogue on the automorphic side: Theorem A of Ramakrishnan's paper "An exercise concerning the selfdual cusp forms on GL(3)" states: Let F be a number field, and $\Pi$ a cuspidal, selfdual automorphic representation of $GL_3(\mathbb{A}_F)$. Then there exists a non-dihedral cuspidal automorphic representation $\pi$ of $GL_2/F$, and an idele class character $\nu$ of F with $\nu^2 = 1$, such that $\Pi \cong Ad(\pi) \otimes \nu$. EDIT. I just realised I had missed something: you are taking a (slightly unusual) definition of "selfdual". You aren't requiring that $\rho^* = \rho$ but just that $\rho^* = \rho \otimes \kappa^n$ for some $n$ (where $\kappa$ is the cyclotomic character), so $\rho$ is "selfdual up to Tate twists". However if you make the even more general assumption that $\rho^* = \rho \otimes \chi$ for any character $\chi$ (this is what people call "essentially selfdual"), then comparing determinants we have $\chi^3 \det(\rho)^2 = 1$, and hence $\chi$ is a square in the group of characters of $G_F$. So we can twist $\rho$ to make it self-dual on the nose, and apply the previous argument to that; and the conclusion is still that $\rho$ is a twist of the adjoint of something (or, if you prefer, a twist of the symmetric square of something). EDIT 2. (Let's not keep doing this, if you want to change the question then open a new question.) You now ask if any essentially selfdual 3-d representation is a quadratic twist of a symmetric square. This is not clear to me. What is clear from the above arguments is the following: if $A_\ell$ is the abelian group of characters $G_F \to \overline{\mathbf{Q}}_\ell^\times$, and $\Sigma$ is any set of representatives for the quotient $A_\ell / 2 \cdot A_\ell$, then any ess. selfdual $\rho$ can be written as $\operatorname{Sym}^2(\tau) \otimes \chi$ for some 2-dimensional $\tau$ and some (unique) $\chi \in \Sigma$. So your new question reduces to a purely one-dimensional one: is every class in $A_\ell / 2A_\ell$ represented by a quadratic character? But I don't see any reason why this should be true in general: I don't see why $A_\ell[2]$ should surject onto $A_\ell / 2A_\ell$.<|endoftext|> TITLE: Is it possible that both a graph and its complement have small connectivity? QUESTION [19 upvotes]: Let $G=(V,E)$ be a simple graph with $n$ vertices. The isoperimetric constant of $G$ is defined as $$ i(G) := \min_{A \subset V,|A| \leq \frac n2} \frac{|\partial A|}{|A|} $$ where $\partial A$ is the set edges $uv \in E$, with $v \in A$ and $u \in A^c$. Is there a graph $G$ such that $$i(G) + i(G^c) < \frac 12$$ where $G^c$ denotes the complement of $G$? REPLY [15 votes]: It is almost obvious. Let $i(A)=\frac{|\partial_G A|}{|A|}$ and $j(B)$ be the same ratio for a vertex set $B$ and the complement of $G$. Let $x=|A\cap B|, y=|A\cap B^c|, z=|A^c\cap B|, t=|A^c\cap B^c|$. The edges going between $A\cap B$ and $A^c\cap B^c$ are boundary edges for both $A$ and $B$ and so are the edges between $A^c\cap B$ and $A\cap B^c$. Thus, we have $$ i(A)(x+y)+j(B)(x+z)\ge xt+yz $$ If $i(A)+j(B)<1$, the left hand side is strictly less than $x+\max(y,z)$, which is at most $xt+yz$ if $t,y,z>0$. Note that if $t=0$, then $x=0$ because of the condition $|A|,|B|\le \frac n2$, so if $y,z>0$ in this case, we still have a contradiction. Finally, if, say, $y=0$, then $x>0$ (otherwise $A=\varnothing)$ and $t\ge \frac n2$ (because $|B|\le\frac n2$), so the RHS is at least $\frac n2$ while the LHS is less than $\max(|A|,|B|)\le \frac n2$ and we get a contradiction again.<|endoftext|> TITLE: Representations of the automorphism group of graphs via spectral graphs theory QUESTION [11 upvotes]: Given a (simple) graph $G=(V,E)$ with $V=\{1,...,n\}$ and let $A$ be its adjacency matrix. I am interested in the representation theory (over $\Bbb R$) of the automorphism group $\def\Aut{\mathrm{Aut}}\Aut(G)$. One way to generate such representations is via spectral graph theory. If $\lambda$ is an eigenvalue of $A$ and $\{e_1,...,e_m\}$ is an orthonormal basis of the associated eigenspace $\def\Eig{\mathrm{Eig}}\Eig_\lambda(G)$, then the rows of the matrix $$U:=\begin{pmatrix} | & & |\\ e_1 & \cdots & e_m \\ | & & | \end{pmatrix}$$ can be interpreted as the positions $v_i\in\Bbb R^m,i=1,...,n$ of the vertices of $G$ in a graph embedding. What is special about this embedding is, that it realizes all the symmetries of $G$. This means, for each automorphism $\phi\in\Aut(G)$, there is a linear map $M_\phi\in\mathrm{GL}(m,\Bbb R)$ with $v_{\phi(i)}=M_\phi v_i$. This gives a real representation $\Aut(G)\to\mathrm{GL}(m,\Bbb R),\phi\mapsto M_\phi$. My questions are: Was this construction of real representations of $\Aut(G)$ already studied somewhere in the literature? and especially: When are these respresentations (real) irreducible? This answer mentions graphs with trivial symmetry group but large eigenspaces, which therefore cannot provide irreducible representations. However, I am interested in graphs with a lot of symmetries, especially arc-transitive graphs. In all the cases I studied, all the representations turned out to be irreducible. Update A similar question was asked on Math.StackExchange and received an interesting answer from C. Godsil. Especially the last parenthesized sentence leaves space for an interesting counter-example. The searchable terminology seems to be "reducible/irreducible eigenspaces of graphs". At least this lead me to the following paper G. Berkolaiko, W. Liu: Eigenspaces of Symmetric Graphs are not Typically Irreducible (2018) However, I am not aware of a direct connection to the problem stated here, partially because the paper's terminology is not very familiar to me, yet. REPLY [2 votes]: Recently I found an arc-transitive graph (and then many more) for which some eigenspaces are reducible, something that I believed might not occure. The example is the Shrikhande graph, a strongly regular graph with parameters $(16,6,2,2)$. The spectrum consists of eigenvalues $6^1, -2^6, 2^9$ (multiplicities in the exponent), where only the eigenspace of $2$ is reducible. I cannot tell you how exactly the eigenspace decomposes since my knowledge about this stems from computing the characters and the Frobenius-Schur indicator as explained here. There are other examples: e.g. $C_{10}\times C_{10}$ and some circulant graphs that have one suspiciously large eigenspace that decomposes. I have not investigated for what parameters these reducible eigenspaces occure, I just know that they are not always present. What is interesting though, is, that the eigenspace to the second largest eigenvalue seems to be always irreducible. This is interesting since this eigenspace is related to the algebraic connectivity and is a central object of my research. I will have to investigate whether this is always true. Update Even the eigenspace of the second-largest eigenvalue does not have to be irreducible. I found some examples by computing Frobenius-Schur indicators for various arc-transitive graphs. However, counter-examples seem to be rare.<|endoftext|> TITLE: Can the graph of a symmetric polytope have more symmetries than the polytope itself? QUESTION [16 upvotes]: I consider convex polytopes $P\subseteq\Bbb R^d$ (convex hull of finitely many points) which are arc-transitive, i.e. where the automorphism group acts transitively on the 1-flags (incident vertex-edge pairs). Especially, this includes that the polytope is vertex- and edge-transitive. The graph of a polytope is the graph isomorphic to its 1-skeleton. Question: Are there arc-transitive polytopes, where the graph has more symmetries than the polytope? When weakening the requirement of arc-transitivity, there are examples: A rhombus is edge- but not vertex-transitive. However, its graph is vertex-transitive. (Thanks to Henrik for the comment) There are vertex-transitive neighborly polytopes other than a simplex, but none of these can be edge-transitive. Their graphs are complete and are therefore edge-transitive. REPLY [5 votes]: Partial progress: Let $V$ be the vertex set of $P$, let $E$ be the set of directed edges and let $X$ be the set of ordered pairs of distinct elements of $V$. Let $G$ be the group of combinatorial symmetries of the edge graph and let $\Gamma \subset G$ be the group of geometric symmetries of the polytope. So it is assumed that $E$ is a single orbit for both $G$ and $\Gamma$ acting on $X$. I claim there must be some other $G$-orbit on $X$ which splits into more than one $\Gamma$ orbit. In particular, we must have more than one $G$-orbit on $X$, which means that neighborly polytopes won't work. Without loss of generality, we may assume that $P$ spans $\mathbb{R}^d$ and the centroid of $P$ is at $\vec{0}$, so action of $\Gamma$ extends uniquely to a linear action on $\mathbb{R}^d$. Proof: Suppose to the contrary that $G$ and $\Gamma$ have the same orbits on $X$. Let $\mathbb{R} V$ be the permutation representation on $V$. It is well known that the dimension of $\mathrm{Hom}_G(\mathbb{R} V, \mathbb{R} V)$ is $|V^2/G| = |X/G| + 1$, and likewise for $\mathrm{Hom}_{\Gamma}$. So the hypothesis on orbits implies that $\mathrm{Hom}_G(\mathbb{R} V, \mathbb{R} V) = \mathrm{Hom}_{\Gamma}(\mathbb{R} V, \mathbb{R} V)$. As a corollary, any $\Gamma$-subrepresentation $W$ of $\mathbb{R}V$ is also a $G$-subrepresentation, because we can choose a $\Gamma$ equivarient projection $\mathbb{R}V \to W$, and then this projection will also be $G$-equivariant. The map taking the basis vector $e_v$ of $\mathbb{C} V$ to the vertex $v$ of the polytope $P$ gives a $\Gamma$-equivariant linear surjection from $\mathbb{R} V$ to $\mathbb{R}^d$. So $\mathbb{R}^d$ can be identified with a $\Gamma$ summand of $\mathbb{R} V$. But every $\Gamma$ summand is also a $G$-summand, so the $\Gamma$ action extends to a $G$ action, contradiction. $\square$. So we want a graph $(V,E)$ with arc-transitive symmetry group $G$, and a subgroup $\Gamma$ of $G$ which is still arc-transitive but has more orbits on $X$. Such graphs definitely exist. As one example, let $(V,E)$ be the Hamming $n$-cube, whose symmmetry group is $S_n \ltimes C_2^n$ (here $C_2$ is the cyclic group of order $n$.) If $H$ is a transitive but not $k$-transitive subgroup of $S_n$ for some $k$, then $H \ltimes C_2^n$ has more orbits on $X$, but all edges of $(V,E)$ remain a single orbit. But I haven't succeeded yet in embedding an example like this as the edge graph of a polytope.<|endoftext|> TITLE: Complex Multiplication and algebraic integers QUESTION [13 upvotes]: Let $q=e^{2\pi i\tau}$ and $$E_2(\tau) = 1 - 24 \sum_{n=1}^\infty\frac{nq^n}{1-q^n}$$ be the Eisenstein Series of weight $2$ and let $E_2^*(\tau) = E_2(\tau) - \frac{3}{\pi\cdot Im(\tau)}$ be the corresponding almost holomorphic modular form. Then let $$\eta(\tau)=e^{\pi i\tau /12}\cdot\prod_{n=1}^\infty(1-q^n)$$ be the Dedekind $\eta$-Function. Question: How can I prove the following statement from here: Let $\tau$ be any Complex Multiplication point. By basic theorems of complex multiplication, if you choose a suitable period $\omega(\tau)$, $E_4(\tau)/\omega(\tau)^4$, $E_6(\tau)/\omega(\tau)^6$, and $\sqrt{D}E_2^*(\tau)/\omega(\tau)^2$ (with $E_2^*(\tau)=E_2(\tau)-3/(\pi \cdot Im(\tau))$ and $D$ the discriminant of $\tau$) will be algebraic numbers of known degree, and if you choose $\omega(\tau)=\eta(\tau)^2$, they will even be algebraic integers. Partial Solution for $E_2^*$: Francois Brunault pointed out that this statement can be found in Prop. 5.10.6 on p. 202 of Cohen/Strömberg's book Modular Forms: A Classical Approach. Unfortunately, the proof of this proposition starts with "we only prove algebraicity, not the integrality". Who can help with proving the integrality? Since I am no expert in complex multiplication, I am looking for a rather detailed answer, or for a reference. Complete Solution for $E_4$ and $E_6$: The statement that $\frac{E_4}{\eta^8}$ is an algebraic integer follows from $$\left(\frac{E_4}{\eta^8}\right)^3 = j(\tau)$$ and the statement that $\frac{E_6}{\eta^{12}}$ is an algebraic integer follows from $$\left(\frac{E_6}{\eta^{12}}\right)^2 = j(\tau)-1728$$ with the absolute invariant $j(\tau)=\frac{1728E_4^3}{E_4^3-E_6^2}=\frac{E_4^3}{\eta^{24}}$ which is an algebraic integer (see Silverman, Advanced Topics in the Arithmetic of Elliptic Curves, Theorem 6.1, p. 140). EDIT: Complete Solution for $E_2^*$: The answer of Michael Griffin (see below) can now be found in more details in the appendix of this arXiv-preprint. REPLY [6 votes]: I have seen this statement about $E_2^*$ tossed around off-handedly by experts a number of times, but never seen a complete proof referenced. The tools to prove it are (mostly) in Masser's "Elliptic functions and transcendence", Appendix 1. There, Masser gives a formula for certain non-holomorphic modular functions. One of these fomrulas (Lemma A3) can be re-written as $$E_2^*(\tau)\left(\frac{\pi}{\omega_1}\right)^2=-\frac{3S}{\sqrt{D} ~\tau},$$ where $(\omega_1,\omega_2)$ are choices of periods of a CM elliptic curve with rational equation, $D$ is the discriminant of the CM point $\tau=\frac{\omega_1}{\omega_2}$, and $S$ is a sum of division points on the curve. If $\tau$ satisfies the reduced, integral quadratic, $C\tau^2+B\tau+A$, then Masser points out that by a theorem of Baker, $(AC)^2\wp$ is an algebraic integer. Moreover, the norm of $\tau$ is $A/C$, and so it's clear the only additional primes that could divide the denominator are divisors of $AC$. On page 118 of Masser, he offers formulas for the function $$\gamma(\tau)=\frac{E_2^*(\tau)E_4(\tau)}{6E_6(\tau) j(\tau)}-\frac{7j(\tau)-6912}{6j(\tau)(j(\tau)-1728)}$$ at CM points, in terms of singular moduli of $j(\tau)$. The Gross-Zagier formula (which gives a factorization for the norm of differences of singular moduli) can then be used to show that no primes that split in the CM field can divide the denominator. Any prime dividing $A$ or $C$ is either split or ramified in the CM field. None of the split primes can divide the denominators. A more careful use of the Gross-Zagier formula shows that the ramified primes appear no more than expected, and so $\sqrt{D}E_2^*(\tau)\left(\frac{\pi}{\omega_1}\right)^2$ must be an algebraic integer.<|endoftext|> TITLE: A Modern Proof of Erdos and Renyi's 1959 Random Graph Paper? QUESTION [12 upvotes]: In their paper, Erdos and Renyi consider a random graph with a fixed number of edges, as opposed to the more modern approach of adding each edge independently with probability $p$. From what I understand (both intuitively and from searching around) their results on connectivity still hold if one use the more standard $G(n,p)$ model. I have attempted to generalize their proof ideas to this setting, but was unsuccessful. I was hoping that someone could either sketch or refer me to a proof of their connectivity results in the $G(n,p)$ setting. Ideally this proof would be in the same spirit as the original proof, but I'd accept any paper that works. REPLY [5 votes]: The most satisfying solution I've seen so far are these notes by Sabastian Roch (see section 2.2.3, and in particular claim 2.25). He first argues that the only components besides the giant one are isolated vertices, and he then shows that there is a positive probability of isolated vertices existing (with a refined estimation being performed as part of the exercises of the notes).<|endoftext|> TITLE: Multiple of identity plus compact QUESTION [15 upvotes]: Is there an example of a bounded operator $T\in\mathcal{B}(H)$, where $H$ is a separable complex Hilbert space, such that no restriction to an infinite dimensional closed subspace is multiple of identity plus compact? Edit: What I mean by "restriction" is $T_{|M}:M\to H$, where $M$ is an infinite dimensional subspace of $H$, and by "multiple of identity" I mean a complex multiple of the inclusion map $i:M\to H$. REPLY [6 votes]: I think this is essentially Theorem 3.1 in: A. Brown, C. Pearcy, Compact restrictions of operators, Acta. Sci. Math (32) 1971, 271-282 (link) In fact, something a bit stronger and in more generality is true (see Proposition 3.2 in this paper): Proposition: Let $X$ is a Banach space and $T\in\mathcal{B}(X)$ such that there exists $\lambda\in\partial\sigma(T)\setminus\sigma_p(T)$. Then there exists $e\in X$ such that for any $\epsilon>0$ we can find $Y$ an infinite dimensional and infinite co-dimensional subspace of $X$ with the property that $(\lambda I-T)_{|Y}:Y\to Y+[e]$, and $(\lambda I-T)_{|Y}$ is compact and has norm less than $\epsilon$.<|endoftext|> TITLE: How to visualize finiteness of class number? QUESTION [11 upvotes]: As the question title asks for, how do others "visualize" the finiteness of class number with algebro-geometric insight? I just think of it as a result in algebraic number theory and not one in algebraic geometry. Bonus points for pictures. REPLY [8 votes]: The first question to ask is how you visualize a number field, and the second is how you visualize a class group. There's more than one way to do each of these. Schemes What I think your question intended to refer to is the viewpoint of schemes. Let $K$ be a degree $d$ number field with ring of integers $\mathcal{O}_K$. The scheme $\operatorname{Spec} \mathcal{O}_K$ is one-dimensional and regular, which suggests that it ought to be a non-singular curve. The closed points of this scheme, which ought to correspond to points in the usual geometric sense (i.e., points on a Riemann surface), are the non-zero prime ideals of $\mathcal{O}_K$. There's also a generic point, the zero prime ideal, which corresponds to generic behavior on the curve (where generic means "at all but finitely many primes"). The points of this scheme are of a really different nature than the points we're used to looking at. The closed points, for example, can't be directly interpreted in terms of the real or complex numbers, because every morphism from $\operatorname{Spec} \mathbf{R}$ or $\operatorname{Spec} \mathbf{C}$ to $\operatorname{Spec} \mathcal{O}_K$ factors through the generic point. In fact, limiting ourselves to any one field, regardless of its characteristic, leads to similar problems. What makes $\operatorname{Spec} \mathcal{O}_K$ so useful is that it is a single object that unifies different aspects of the behavior of $K$. Geometrically, a line bundle over a scheme $X$ is a scheme $L$ together with a morphism $\pi \colon L \to X$ such that, for some open cover of $X$, say $\{U_i\}_{i \in I}$, we have isomorphisms $\pi^{-1}(U_i) \cong \mathbf{A}^1_{U_i}$ such that the induced automorphisms of $\mathbf{A}^1_{U_i \cap U_j}$ are fiberwise linear transformations. Algebraically, a line bundle is a rank one locally free sheaf (of modules over the structure sheaf of $X$). For an affine scheme $A$, this condition translates to a line bundle being a projective module $M$ over $A$. The Picard group of a scheme is the group of line bundles under tensor product. When $X$ is $\operatorname{Spec} \mathcal{O}_K$, then $\operatorname{Pic} X$ is the class group of $K$. The Picard group, and therefore the class group, is an obstruction to global triviality. What this means in the arithmetic situation is perhaps not immediately clear, but the geometry is a good guide here. So let's consider an invertible (i.e., rank one locally free) sheaf $\mathcal{L}$ on a scheme $X$. There is a notion of a meromorphic section of $\mathcal{L}$. If $X$ is irreducible with generic point $\eta$ and satisfies mild technical hypotheses, then a meromorphic section of $\mathcal{L}$ is just a section of $\mathcal{L}_\eta$, the restriction to the generic point. Such a section determines an isomorphism $\mathcal{L}_\eta \cong \mathcal{O}_{X,\eta}$ and may be used to identify $\mathcal{L}$ with a subsheaf of the sheaf $\mathcal{K}_X$ of rational functions on $X$. Any one such identification leads to many others by post-multiplication by rational functions on $X$. Under some more mild technical hypotheses, the image of $\mathcal{L}$ is of the form $\mathcal{O}_X(D)$ for some Cartier divisor $D$, and the linear equivalence class of $D$ is independent of the chosen meromorphic section. If we specialize all of the above to $\operatorname{Spec} \mathcal{O}_K$, then we get a geometric interpretation of non-principal ideal classes. A line bundle $\mathcal{L}$ now corresponds to a rank one projective module $M$ over $\mathcal{O}_K$. A meromorphic section is an element of $M \otimes_{\mathcal{O}_K} K$; such a section is the same as a homomorphism $K \to M \otimes_{\mathcal{O}_K} K$, and such a homomorphism is an isomorphism because $M$ is rank one. Using the isomorphism in the reverse direction allows us to embed $$M \hookrightarrow M \otimes_{\mathcal{O}_K} K \cong K$$ and thereby identify $M$ with a fractional ideal of $K$. As above, we may also identify $M$ with a Cartier divisor. For $\mathcal{O}_K$, such a divisor has the form $$\prod_i \mathfrak{p}_i^{e_i}$$ for some set of prime ideals $\{\mathfrak{p}_i\}_{i \in I}$. The linear equivalence class of this Cartier divisor is independent of the section, which simply means that $M$ gives a well-defined class group element. In even more geometric terms: If $\mathcal{L}$ were a trivial bundle, it would have a nowhere vanishing section; when it is not trivial, the linear equivalence class of $D$ tells us $\mathcal{L}$ is wrapped around $X$ by telling us how sections can cross zero. (Think of a section of a Möbius strip.) A non-trivial ideal class in $\mathcal{O}_K$ tells us the same thing, but now "crossing zero" means "vanishing modulo a power of a prime ideal." The class group is a measurement of how many different ways this can happen, so it is a kind of measurement of the internal complexity of $\mathcal{O}_K$, just like, say, the Picard group of a curve. The finiteness of the class group is not obvious from this perspective. After all, the Picard group of a curve over a field may be infinite! The scheme-theoretic perspective is a good one if you are looking to generalize to, say, arithmetic surfaces. But in order to have any hope of finiteness, you need something more attuned to the situation at hand than just the language of schemes. Geometry of numbers The most classical way to study a number field geometrically is by the geometry of numbers. The field $K$ embeds in the real and complex numbers; if $K \cong \mathbf{Q}[x]/(f)$, then the real embeddings $K \to \mathbf{R}$ are defined by sending $x$ to a real root of $f$, and the complex embeddings $K \to \mathbf{C}$ by sending $x$ to a complex root of $f$. ("Complex root" will mean "non-real root in the complex numbers".) Taken together, these define an interesting embedding of $K$ in $\mathbf{R}^d$ called the Minkowski embedding. A quick way to state this embedding is that it is the canonical map $$\mathcal{M} \colon K \to (K \otimes_{\mathbf{Q}} \mathbf{C})^\sigma,$$ where $\sigma$ is the complex conjugation operator on $\mathbf{C}$. The codomain is called Minkowski space. Explicitly, suppose the real roots of $f$ are $x_1, \dots, x_{r_1}$ and the complex roots are $x_{r_1 + 1}, \dots, x_{r_1 + 2r_2}$ with $\bar x_{r_1 + i} = x_{r_1 + r_2 + i}$ for all $1 \le i \le r_1$. The tensor breaks up as a product over the roots of $f$: $$K \otimes_{\mathbf{Q}} \mathbf{C} \cong \mathbf{C}[x]/(f) \cong \prod_{i=1}^d \mathbf{C}[x]/(x - x_i) \cong \mathbf{C}^d,$$ where the isomorphisms are of $\mathbf{C}$-algebras. Complex conjugation on $\mathbf{C}[x]/(f)$ conjugates the scalars, hence the coefficients of $f$, hence the roots; so on the first $r_1$ coordinates of $\mathbf{C}^d$, it is complex conjugation, while on the last $2r_2$, it swaps coordinates $r_1 + i$ and $r_1 + r_2 + i$ and conjugates both. It is easy to check that $\sigma$ is a real linear transformation and an involution, and that its eigenvalues are $\pm 1$, whence the isomorphism of Minkowski space with $\mathbf{R}^d$. The significance of the Minkowski embedding is that, for any non-zero proper fractional ideal $\mathfrak{a}$ of $K$, $\mathcal{M}(\mathfrak{a})$ is a rank $d$ lattice. That is, suppose we give $K \otimes_{\mathbf{Q}} \mathbf{C}$ the inner product $\langle x, y\rangle = \operatorname{tr}(x\bar y)$, where $\operatorname{tr}(z)$ is the trace of the linear operator "multiply by $z$." Then the restriction of this inner product to $\mathcal{M}(\mathfrak{a})$ is a positive definite bilinear form. Equivalently, the inner product induces a topology, and $\mathcal{M}(\mathfrak{a})$ is a discrete subgroup in this topology. It's obvious that different ideals define different lattices: $\mathcal{M}(\mathcal{O}_K) \subsetneq \mathcal{M}((2))$. In fact, the latter is precisely the image of $\mathcal{M}(\mathcal{O}_K)$ under the map that doubles each coordinate. More generally, if $\mathfrak{a}$ is principal with generator $a \in K$, then $\mathcal{M}(\mathfrak{a})$ is the image of $\mathcal{M}(\mathcal{O}_K)$ under the map defined by multiplication by $a$. But if $\mathfrak{a}$ is not principal, then this is false: There is no $a \in K$ for which $\mathcal{M}(\mathfrak{a})$ is the image of $\mathcal{M}(\mathcal{O}_K)$ under multiplication by $a$. This failure is precisely quantified by the class group: The set of lattices in Minkowski space which are defined by fractional ideals, modulo the relation given by multiplication by elements of $K$, is the set of class group elements. One way to visualize the class group, therefore, is as a set of lattices modulo an equivalence relation. We can do this quite explicitly for quadratic fields. For instance, suppose $f(x) = x^2 - D$ with $D$ positive, squarefree, and $3 \bmod 4$. The ring of integers of $K = \mathbf{Q}[x]/(f)$ is generated by $1$ and the image of $x$, and the roots of $f$ are $\pm\sqrt{D}$. The image of $\mathcal{O}_K$ is therefore generated by the vectors $(1, 1)$ and $(\sqrt{D}, -\sqrt{D})$, and there is an obvious fundamental domain which is a rectangle with sides at $45^\circ$ to the axes. On the other hand, if $D = 15$, then $(2, 1 + \sqrt{15})$ is a non-trivial ideal class. We have $\mathcal{M}(2) = (2, 2)$ and $\mathcal{M}(1 + \sqrt{15}) = (1 + \sqrt{15}, 1 - \sqrt{15}) \approx (4.87, -2.87)$. The latter vector makes an angle of about $-30^\circ$ with the first axis. The fact that this ideal has non-trivial ideal class means that there is no element of $K$ that transforms this lattice into $\mathcal{M}(\mathcal{O}_K)$. The finiteness of the class group now has the following geometric interpretation: There are only finitely many different equivalence classes of lattices. The proof is essentially a compactness argument. Suppose we find a way to parametrize lattices; suppose further that we can attach an invariant to each lattice and that this invariant lies in a compact group; and suppose lastly that different ideal classes lie in different components of this group. Then the group of components, being compact, must be finite. This is made precise by working with adeles; the invariant is the unit norm idele class group, as mentioned in the comments to François Brunault's answer. Historically, the finiteness of the class group was known decades before adeles were introduced, so there is surely a way to understand finiteness without understanding the adeles. But the adeles are an extremely powerful language, and they are so useful that they seem to have obliterated the older approach. They are well worth taking the time to learn.<|endoftext|> TITLE: Prime plus square equals prime QUESTION [8 upvotes]: Furstenberg–Sárközy's theorem states that if a set of positive integers has positive upper density, then there exists (infinitely many) pair of elements of the set, whose difference is a perfect square. Since prime numbers have zero density, this theorem does not apply to them. And I was wondering if there is a result in this direction regarding prime numbers. Explicitily - are there infinitely many pairs of prime numbers $(p,q)$ such that $p-q$ is a perfect square REPLY [4 votes]: As Joel mentions, this follows from the work of Tao and Ziegler. Alternatively, this can be directly deduced from the density of the primes and the known bounds on the Furstenberg–Sárközy theorem. Indeed, the bounds of Pintz, Steiger, and Szemeredi from the 1980's are suffient. See: J. Pintz, W. L. Steiger, and E. Szemeredi, On sets of natural numbers whose difference set contains no squares, J. London Math. Soc. (2) 37 (1988), 219–231 See also the even stronger bounds in the more recent work of Bloom and Maynard: https://arxiv.org/pdf/2011.13266.pdf.<|endoftext|> TITLE: An infinite dimensional local domain whose chains of primes are finite QUESTION [7 upvotes]: Does there exist a local domain of infinite dimension in which every chain of prime ideals is finite? Of course, such a ring must be neither noetherian nor catenary. (This question arose while trying to compare different definitions of catenarity, and more precisely while trying to understand what may cause codimensions in topological spaces to be infinite.) REPLY [11 votes]: Choose a field $k$ and a Noetherian $k$-algebra $R$ of infinite dimension. (I know you know such a thing exists.) Let $R' \subset R[x]$ be the set of polynomials $f = \sum a_i x^i$ whose constant term is constant, i.e., $a_0 \in k \subset R$. Then we have $$ \text{Spec}(R') = \text{Spec}(R[x]) \amalg_{\text{Spec}(R)} \text{Spec}(k) $$ by Tag 0B7J. Observe that $R'[1/x] = R[x, 1/x]$ is Noetherian. Observe that for any prime $\mathfrak p \subset R$ the prime $\mathfrak p' = R' \cap \mathfrak p R[x]$ is contained in the maximal ideal $\mathfrak m = \text{Ker}(R' \to k) = \sqrt{xR'}$ (small detail omitted). Thus the local ring of $R'$ and $\mathfrak m$ is an example. Namely, its punctured spectrum is Noetherian of infinite dimension (as the localization of $R'$ at $\mathfrak p'$ is the same as the localization of $R[x]$ at $\mathfrak p R[x]$ which is flat over $R_\mathfrak p$ and hence has dimension $\geq \dim(R_\mathfrak p)$. Enjoy!<|endoftext|> TITLE: Does harmonic map heat flow of a curve always fully converge to a geodesic? QUESTION [7 upvotes]: Consider a smooth closed curve $u_0$ in a compact Riemannian manifold $(M,g)$. Let $u_0$ evolve by harmonic map heat flow, $\partial_tu=\nabla_{\partial_su}\partial_su$, and call the result $u(t)$. Since the circle is 1-dimensional, a miracle happens and we get a gradient estimate for free, so the flow exists for all time. By standard arguments, there is a sequence $t_k\to\infty$ such that $u(t_k)$ converges to a geodesic $u_\infty$ in the $C^\infty$ topology. Furthermore, one can see that $u_\infty$ is homotopic to $u_0$ because $u(t_k)$ is eventually within a small enough neighborhood of $u_\infty$. However, what's not clear to me is if the flow provides this homotopy. That is, does $\lim_{t\to\infty}u(t)=u_\infty$ in the $C^\infty$ topology, not just up to a subsequence? Hartman has proved this when $M$ has nonpositive sectional curvature for general harmonic maps with bounded image. I'm wondering if this is true for curves without a curvature assumption on the target. REPLY [9 votes]: The situation is actually quite complicated, it seems. In the case where the target manifold is real analytic, Leon Simon's results in Asymptotics for a Class of Non-Linear Evolution Equations, with Applications to Geometric Problems implies the desired convergence. However, using Topping's construction in Section 5 of Rigidity in the Harmonic Map Heat Flow (note Remark 6 which states that the construction is essentially independent of the dimension of the domain), one concludes that, in general, the convergence really is only up to subsequence: the flow can fail to converge even in $C^0$. In a paper by Choi and Parker, it was however shown that convergence holds for a "generic" set of target manifolds, which they call manifolds with "bumpy metrics".<|endoftext|> TITLE: Should I inform the editor about a generalized result of a result in a paper under review? QUESTION [21 upvotes]: Hoping that my question is appropriate for MO, I would like to ask the following question: I have sent one of the editors of a very good math journal a paper of mine which contains a main result, call it theorem A. The editor wrote me that he sent my paper to referee and will contact me when he will get the referee's report (this was 7 weeks ago). A week ago I have noticed that my Theorem A can be generalized to Theorem B, and Theorem B has a nice application, which Theorem A does not have. What should I do: (1) Should I be patient and wait for the referee's report, and only after receiving it decide what to do with my Theorem B? (2) Should I inform the editor about my Theorem B and let him decide whether to send it to the referee or not? On the one hand, I do not want to disturb the editor (and referee), but on the other hand, perhaps the referee will be glad to see my Theorem B? (if he is not sure if my Theorem A is good enough to be published in their very good math journal, but Theorem B is). This question is slightly similar, but it asks about fixing an error, not about generalizing a result. Thank you very much! Edit: I have just found this question, which is quite similar to mine, though there are differences between the two. REPLY [15 votes]: Theorem B is the result of additional work of yours after submission of your paper of A. I would consider writing a follow-up paper later about that work. The additional advantage is that it adds a publication in your name. If my paper A as submitted would be refused, then I would write a paper A+B and submit it, probably to another journal. But anyway, before trying to add additional work B in paper A I would thoroughly check the journal's guidelines for submission to see whether that would fit or not.<|endoftext|> TITLE: What is a geodesic in Outer space? QUESTION [6 upvotes]: The Culler-Vogtmann Outer space $\text{CV}_n$ is an analogue of Teichmuller space for the group $\text{Out}(F_n)$. Is there any notion of a geodesic path in $\text{CV}_n$? Are there different competing definitions of geodesic? If so, what would be a simple example of a geodesic path vs. a non-geodesic one, say on $\text{CV}_2$? REPLY [8 votes]: Besides the geodesic paths of the asymmetric metric $d(\cdot,\cdot)$ that are mentioned in other answers (namely paths such that $d(\gamma(s),\gamma(t)) = t-s$ if $s \le t$), there is another class of paths with many uses known as Stallings fold paths. You can see some discusions of them in the outer space context, with applications, in these lecture notes of Bestvina, these notes of Kapovich and Myasnikov, and this issue of the AMS Memoirs by Handel and myself.<|endoftext|> TITLE: A nowhere-zero point in a linear mapping conjecture QUESTION [5 upvotes]: I found a very interesting problem in the Open Problem Garden, which I am surprised is not as well-known as I would think it would be: Prove that If $p>3$ is prime and $A$ is an invertible $n \times n$ matrix with entries in ${\mathbb Z_p}$, then there are column vectors $x,y \in {\mathbb Z_p}^n$ which have no coordinates equal to zero such that $Ax=y$. It is easy to show that this conjecture is false when $p=2,3$. Also, when $p$ is a little larger than $n$, this conjecture has been proven true. However, the case when $n$ is large relative to $p$ seems not only difficult to prove, but also it seems to require nontrivial computational resources to even get a sense of what is going on. Has any work been done to try to computationally understand this problem? In particular, is this conjecture known to be true for say $p=5$, $n=15$? REPLY [2 votes]: This is the Alon-Jaeger-Tarsi conjecture first stated in 1981 and resolved very recently (for $p\ge 83$) by János Nagy and Péter Pál Pach.<|endoftext|> TITLE: Which bounded sequence can be realized as the Fourier Series of a probability measure on the circle? QUESTION [15 upvotes]: Given a finite Borel measure $\mu$ on $\mathbb{S}^1 = \mathbb{R}/\mathbb{Z}$, define its Fourier coefficients by $$ \hat\mu(n) = \int e^{2i\pi nx} d\mu(x) \qquad\forall n\in \mathbb{Z}.$$ Clearly, $(\hat\mu(n))_n$ is bounded. What sufficient conditions on a bounded sequence $(a_n)_n$ are known that ensure that there is a finite measure with $\hat\mu(n)=a_n$ ? same with probability measures instead of finite ones; same with necessary conditions. I would guess that characterizations are out of reach, but maybe I am wrong? Added in Edit: Yemon Choi rightfully asks what kind of sufficient or necessary condition I am after. Any is good for my culture, but I am especially interested in sufficient condition that enable one to construct measure satisfying constraints on Fourier coefficient. To be honest, one of my goals was to understand why it is not easy to disprove Furstenberg's $\times 2$, $\times 3$ conjecture by simply picking Fourier coefficients $(c_n)_n$ such that $c_{2^p3^qm}=c_m$ (and $c_0=1$) inside the set of Fourier series of probability measures. I think I am starting to get the point. REPLY [10 votes]: The most elegant solution exists for problem 2: the necessary and sufficient condition for $a_n=\hat{\mu}(n)$ for a positive measure is that he sequence $(a_n)$ is non-negative semi-definite, which means that all Toplitz forms $$\sum_{i,j=0}^na_{i-j}z_i\overline{z}_j\geq 0$$ for all integers $n$ and complex $z_i$. This is a theorem of Caratheodory and Toplitz (see, for example N. Akhiezer, Classical moment problem, Ch V, section 1). For probability measures one has to add to this $a_0=1$. For 1, the best necessary and sufficient condition is that $a_n$ is a difference of two on-negative semi-definite sequences, which is not very effective, of course. Of course these conditions are not always easy to check, but one can derive many simple necessary conditions. Bochner's theorem is a continuous analog of this (for Fourier transforms of non-periodic measures).<|endoftext|> TITLE: Why must the essential image break the principle of equivalence? QUESTION [5 upvotes]: I'm having trouble understanding why the "essential image" is defined the way it is. The nlab article gives the following definition: (A concrete realization of) the essential image of a functor $F: A\to B$ between categories or $n$-categories is the smallest replete subcategory of the target $n$-category $B$ containing the image of $F$ This obviously breaks the principle of equivalence as stated later in the article. When I think about how to define the essential image, the first thing that comes to mind is if we remember the identification of the functor as a new functor from the source to the essential image - then we don't have to break equivalence: Say: The "essential image" of a functor $F:A\to B$ is the initial object in the category of triples $(D, i:D\hookrightarrow B, F':A \to D)$ such that $F= i\circ F'$ (with morphisms the same as in $Cat/B$ that also require post composing them with $F'$ of the domain gives the $F'$ of the codomain) It exists as the classical image of $F$ satisfies the universal property. All the different choices of an initial object form a contractible groupoid whose skeleton yields a subcategory of the classical essential image, as the classical image satisfies the universal property and is contained in the classical essential image (but this is not obvious these are equivalent, I suspect in general they are not, but came short of examples). Questions: How large is the difference between this definition and the classical one? To break this down a little: a. For what categories we have different subcategories of the two constructions (I believe this highly depends on AC to build pathological cases) b. (Aside) Is this difference between the definitions equivalent to AC? Which definition behaves better (this is vague in purpose as there are many interpretations to the question that give different points of view on "which" definition is "the right one"). An argument for this new one is that this definition respects the principle of equivalence. An argument for the classical definition is simplicity - the classical definition is a subcategory and not an equivalence class of subcategories equipped with extra information)? Other arguments I can think about can come from higher categorial properties: For example, gluing essential images (in both definitions) of different functors into a pseudofunctor - when can you do this? REPLY [10 votes]: There are really two different issues here. In this answer I'm going to deal with the one which I think is mainly a distraction, namely the difference between giving a concrete construction and characterizing something only up to isomorphism. As the nLab page says, one thing that violates the principle of equivalence is the property of being equal to the essential image. But this is not really anything special about the essential image. Consider, for instance, the product category, which is generally defined to have $\mathrm{ob}(A\times B) := \mathrm{ob}(A) \times \mathrm{ob}(B)$ and so on; then the property of "being equal to the product category" also violates the principle of equivalence. The violation here is not in the constructions; the violation is in the words "being equal to" that we choose to apply to them. If you like, you can always turn a particular construction with a universal property into a definition that appears to be more equivalence-invariant. For instance, instead of defining the product of two sets to be the set $A\times B$ of ordered pairs $(a,b)$ with $a\in A$ and $b\in B$, you can define a product to be a set equipped with projections $\pi_1 :P\to A$ and $\pi_2:P\to B$ satisfying the universal property of the product. However, you then have to prove that such an object exists, which you do by giving the particular construction anyway. (From a homotopy-type-theoretic perspective, this is kind of like the difference between $\mathrm{isContr}(A) := \sum_{a:A} \prod_{x:A} (a=x)$ and $\mathrm{isContr}(A) := A \times \prod_{x:A} \prod_{y\in A} (x=y)$.) The second approach does prevent us from saying non-invariant things like "is equal to the product", but since (as you point out) it's rather more cumbersome, I'm inclined to instead point the finger at the words "is equal to" rather than at the choice to give a particular construction rather than a definition by universal property. (In particular, there are other ways to prevent ourselves from saying non-invariant things, such as by simply removing the predicate "is equal to" from the language except in cases when it's invariant, or by redefining "equal" to mean "equivalent" as in the univalence axiom.) In any case, when comparing two notions of image, we should compare apples to apples. The notion of essential image on the nLab page — the smallest replete subcategory containing all the objects and arrows that are the $F$-image of some object or arrow in $A$ — should be compared to a similar notion of image, e.g. the smallest (arbitrary) subcategory containing all those objects and arrows. Both notions of image have a corresponding universal property: they are initial objects in the category of factorizations you describe, where $\hookrightarrow$ means respectively an isofibration that is injective on objects and arrows (in the 'essential' case) or a mere functor that is injective on objects and arrows (in the 'inessential' case). So if we wanted to compare oranges to oranges instead, we could compare these two universal properties. But comparing apples to oranges is a distraction. A more interesting question, I think, is the extent to which both notions of image do, or do not, respect equivalences of their input data. This question is not addressed on the current nLab page at all.<|endoftext|> TITLE: Fundamental Theorem of Category Theory appropriate for undergraduates? QUESTION [14 upvotes]: I have been self-studying category theory (mostly from Lawvere and Schanuel’s Conceptual Mathematics). In my experience in undergraduate mathematics, many one or two quarter classes in mathematics build up to a “Fundamental Theorem” that is a somewhat deep result in the subject, is accessible to undergraduate level students in the time available, and serves as a kind of “capstone” which brings together the techniques and theorems studied in the course. For example, an introduction to abstract algebra and group theory may have the fundamental theorem of finite abelian groups, and an introduction to analysis may have the fundamental theorem of calculus. Is there an analogous result in Category theory that meets these criteria? Does category theory have too many prerequisites for reaching such a result? What would be a good result to conclude a short introduction to category theory with? REPLY [14 votes]: The Special Adjoint Functor Theorem has already been recommended, and I agree with that suggestion. I also nominate the idea that "All concepts are Kan Extensions" as a capstone. Classically, you might finish the course with MacLane's coherence theorem, but I prefer to end with something that has lots of applications students would appreciate. Another example (that I might select, as a homotopy theorist) would be Giraud's theorem. A good resource is Emily Riehl's book Category Theory in Context, which is aimed at undergraduates and finishes with an Epilogue titled "Theorems in Category Theory". REPLY [7 votes]: As mentioned above I believe the Yoneda lemma is a good candidate. It requires an understanding of categories, functors, exponentials, natural transformations and their interplay -- how to parametrize natural transformations between the contravariant representable functors of a category $\mathcal{C}$ in terms of arrows in $\mathcal{C}$ and vice-verse. As mentioned by Todd Trimble above the proof is short and sweet, probably easier than most standard capstone proofs, but I agree that it is 'deep' -- it elucidates something fundamental about categories themselves rather than how category theory can be applied elsewhere. It does miss out on machinery like limits/colimits which are mentioned in many adjoint functor theorems; perhaps that 'right adjoints preserve limits' could be an additional 'capstone' theorem in the opposite direction, slightly easier for an undergraduate class than the standard adjoint functor theorems.<|endoftext|> TITLE: Moduli space of linear partial differential equations QUESTION [13 upvotes]: Is there a way to view "the space of all possible linear PDE's" as an algebraic variety with singularities? This is in connection with a quote from someone on the web that I saw a long time ago. At that time I had contacted the author, but they chose not to answer. The quote: In some sense, the space of all possible linear PDE's can be viewed as a singular algebraic variety, where Hormander's theory applies only to generic (smooth) points and the most interesting and heavily studied PDE's all lie in a lower-dimensional subvariety and mostly in the singular set of the variety. Any pointers/refs on any of the points made in the quote would be gratefully received... REPLY [14 votes]: Hormander showed that there is a generic set of scalar linear PDE's that can be studied using general techniques, known as microlocal analysis. This can be linked to algebraic geometry as follows: Any scalar linear partial differential operator of order $k$ on an open set in $\mathbb{R}n$ can be written as $$ Pu = \sum_{|\alpha|\le k} a^\alpha\partial_\alpha u, $$ where each coefficient $a^\alpha$ is a smooth function, $\alpha = (\alpha_1, \dots, \alpha_n)$ and $\partial_\alpha = (\partial_1)^{\alpha_1}\cdots(\partial_n)^{\alpha_n}$. If this is studied using the Fourier transform, then a natural object to study turns out to the principal symbol $$ \sigma(x,\xi) = \sum_{|\alpha| = k} a^\alpha(x)\xi_\alpha, $$ where $\xi = (\xi_1, \dots, \xi_n) \in \mathbb{R}^n$ and $\xi_\alpha = (\xi_1)^{\alpha_1}\cdots(\xi_n)^{\alpha_n}$. For each $x$, this is a homogeneous polynomial of degree $k$ and therefore its zero set is a real algebraic variety on $\mathbb{R}P^{n-1}$. This is known as the characteristic variety. Hormander proved, if the characteristic variety is generic in a suitable sense, regularity estimates, local existence of solutions, and many other things about solutions to equations defined using such operators. However, PDEs most studied have symbols lying in a subvariety of very high codimension, and the techniques used by Hormander are used outside the field of microlocal analysis in only a few specialized areas (e.g., scattering theory, inverse problems). The PDEs with most impact are elliptic, hyperbolic, and parabolic PDEs. Elliptic and most hyperbolic PDEs are generic in Hormander's sense, but parabolic PDEs are not.<|endoftext|> TITLE: smooth equidimensional fibers over a smooth base QUESTION [5 upvotes]: This is probably a simple-minded question, but I haven't been able to prove it or find a counterexample. This old question seems to dance around my question, but I don't think any of the answers address exactly my situation. (Please tell me if I am wrong!) Suppose $\varphi: X\to Y$ is a surjective morphism of algebraic varieties (reduced, irreducible, separated schemes, finite type over an algebraically closed field), and furthermore assume that $Y$ is smooth. Also, I can assume that $\varphi$ is finitely presented. If the fiber $X_y$ is smooth and equidimensional of dim $n$ for any $y\in Y$, is the morphism flat? I know that equidimensionality alone does not mean flat, but I wonder if the smoothness assumptions are enough. Obviously, I want to conclude that $X$ is smooth and this is enough. The equidimensionality assumption rules out the blow-up examples in the above cite problem, and the normalization of a node on a curve is ruled out by smoothness of $Y$. I would be happy with a counterexample though. REPLY [2 votes]: One can also show directly that $X$ is smooth (assuming it is irreducible or even just equidimensional). The problem is local on $X$ and $Y$ so we may assume that $X\subset \mathbb{C}^N$ is affine of codimension $k=N-n$ and $Y=\mathbb{C}^d$. Choose polynomials $f_1,...,f_m, t_1,...,t_d\in \mathbb{C}[X_1,...,X_N]$ such that $$ X\ = \ \{x\in\mathbb{C}^N\,:\, f_1(x)=\cdots = f_m(x)=0\} $$ and $\varphi(x)=(t_1(x),...,t_d(x))$ for all $x\in X$. Fix $p\in X$ and assume $0=f(p)\in\mathbb{C}^d$. Then $$ X_0\ = \ \{x\in\mathbb{C}^N\,:\, f_1(x)=\cdots =f_m(x)=t_1(x)=\cdots =t_d(x)=0\ \} $$ Since $p$ is a smooth point of $X_0$, after reordering the indices if necessary, there exist $r\leq m$ and $s\leq d$ such that $r+s=k+d$ and the following set is linearly independent. $$ \{df_1(p),...,df_r(p),dt_1(p),...,dt_s(p)\} $$ Since $s\leq d$ we see $r\geq k$. But $X$ has pure codimension $k$ so $r\leq k$. Thus $r=k$ and $\{df_1(p),...,df_k(p)\}$ is linearly independent so $p\in X$ is a smooth point.<|endoftext|> TITLE: Spin cobordism v.s. KO theory in low or in any dimensions QUESTION [8 upvotes]: It seems that from this webpage, the spin cobordism is equivalent to KO theory in low dimension. If we denote the $p$-torsion part (mean $\mathbb{Z}_{p^n}$ for some $n$) $$\Omega_d(BG)_p.$$ Question 1: Then do we have $$\Omega_d^{spin}(BG)_p = ko_d(BG)_p?$$ for $p=2$ and free part, for $d\le 7$? (how about higher $d>7$?) And $$ \Omega_d^{spin}(BG)_p = \Omega_d^{SO}(BG)? $$ for $p \neq 2$ and $p$ is an odd prime? Namely, the 2-torsion and free part of $Mspin$ and $KO$ is the same. If there is an odd $p$ torsion, we need to consider localization at odd prime by $MSO$ cohomology. Is this correct? Question 2: If this is a statement about the spectra, not just about stable homotopy groups, and thus within these spin cobordism and ko theory, do they completely coincide for any dimensions $d$, instead of just $d \leq 7$? REPLY [4 votes]: First of all, let me say that the page you are quoting is a little bit misleading if not inaccurate on the Anderson-Brown-Peterson splitting. $ko\langle 4n(J)\rangle $ should read $\Sigma ^{4n(J)} ko$ and $ko\langle 4n(J)-2\rangle $ should read $\Sigma ^{4n(J)-4} ko\langle 2\rangle $ See, e.g. https://pdfs.semanticscholar.org/c377/7dd83a6ba0d959c5c62d633ed4109cddb660.pdf With this correction, at 2, $Mspin \wedge BG$ splits whose bottom piece is $ko\wedge BG$, other pieces are at least 7-connected since the "next bottom" piese is $\Sigma ^8ko \wedge BG$ as is pointed out by ArunDebray, corresponding to the partition $J=(2)$. Thus the answer to your questions, at the prime 2 is that We have an isomorphism up to $d\leq 7$ The map is always surjective, but the kernel is in general non-trivial for $d\geq 8$.<|endoftext|> TITLE: Does T-T Moh's paper really contain a gap? QUESTION [21 upvotes]: It is well known now that Yitang Zhang's work on Jacobian conjecture collapsed because his advisor's work earlier contains unjustified claims. I am wondering what specifically is unclear about his paper. From fellow researchers I heard his paper is unreadable, and the last claim in his paper on Jacobi conjecture may even be wrong (among other things, he claim that there are no counter examples of degree less than 100). Later, a paper by his student appeared which claimed to have repaired the gap. Unfortunately, it has been suggested to me both papers are unreadable. While Yitang Zhang suggested in an interview that Moh's paper is "neither correct nor incorrect", Moh claim that the paper is indeed correct (page 8) without giving any explanation on the main idea of his paper. Yitang Zhang's opinion may be quoted at here: 张益唐：他认为他是对的，而且谁都相信他是对的，但是，他没有证出来。他告诉我他的研究是对的，我照着他说的路子就都做出来了，但回过头来才发现，没有证据证明他是对的。我也不认为他是错的，但他还没有拿出证据证明他是对的。 The passage may be translated as: Yitang Zhang: He believed he was correct, and everyone believed he was correct. But he did not prove the result he claimed. He told me his research is correct, and I proved (Jacobi conjecture?) by what he suggested. However, when I reckon the proof, I discovered there is no evidence suggesting he is correct. I also do not believe he is wrong; but he has not provided evidence to show it is correct. What is the current consensus? I ask this because a colleague in my department got seriously stumped on the paper. If the paper is correct and there is no gap in the paper, I imagine Prof. Zhang or anyone else in his situation would be eager to publish the result. So I am genuinely confused. Q1: Does Moh's paper really contain a gap? Q2: If Moh's paper is complete and correct as he claimed, is there any follow up work explaining the numerical reason behind the $(99,66)$ degree case? REPLY [6 votes]: Yes, there appears to be a follow-up work by Yansong Xu on the (99,66)-case: Intersection Numbers and the Jacobian Conjecture, in turn followed by The Jacobian Conjecture: Approximate roots and intersection numbers by Guccione-Guccione-Horruitiner-Valqui. As their abstract says, they have obtained "nearly the same formulas for the intersection number of Jacobian pairs (as Xu), but with an inequality instead of an equality".<|endoftext|> TITLE: Physical intuition behind prequantization spaces QUESTION [8 upvotes]: Given a symplectic manifold $(M,\omega)$ with integral symplectic form, that is $$\omega \in \text{Im}(H_2(M,\mathbb{Z}) \to H_2(M,\mathbb{R})),$$ one can form a so-called prequantization space, that is a $S^1$ principal bundle $$ \pi : (V, \alpha) \to (M,\omega),$$ where $\alpha$ is an $S^1$-invariant $1$-form on $V$ satisfying $$\pi^* \omega = d \alpha.$$ This makes $(V,\alpha)$ into a contact manifold. What is the physical intuition behind this construction ? I know that it corresponds to the notion of geometric quantisation, but I have trouble seeing why $(V,\alpha)$ could represent a "quantum" space associated with $(M,\omega)$. For instance, what is the meaning of the fibres of $\pi$ (indentified with the Reeb flow) ? REPLY [10 votes]: If you think instead of the prequantum line bundle (i.e. the complex line bundle associated to your prequantum circle bundle using the standard representation of the circle on $\mathbb{C}$) then the sections of this prequantum line bundle are the wavefunctions in quantum mechanics (so the circle bundle is capturing something about the phase). Of course, most of the time in quantum mechanics, your symplectic form is exact (e.g. a cotangent bundle) and you don't need to worry about these being sections of a bundle (they're just complex-valued functions). However, when your symplectic manifold is not exact then you need to figure out whether you want wavefunctions to be functions or sections of some bundle. I believe the following reasoning explains why you should pick the prequantum line bundle. If we take the wavefunctions to be sections of some complex line bundle (and pick a unitary connection on the bundle) then we can try to quantise the observable $F$ by associating the operator $\nabla_{V_F}+2\pi iF$ on the space of wavefunctions (where $V_F$ is the Hamiltonian vector field associated to $F$). Now the commutator of two such operators involves a curvature term (from the commutator of the covariant derivatives) and because Dirac tells you that commutators should agree with Poisson brackets, this tells us that the curvature of the bundle should be the symplectic form. This tells you which bundle to pick. In the end, you want your space of wavefunctions to be something more like the space of functions of position (not of both position and momentum); you therefore pick a "polarisation" of your symplectic manifold (something like a Lagrangian foliation) and restrict attention to sections of the prequantum bundle which are covariantly constant along the polarisation. For example, in a cotangent bundle, you could polarise using the Lagrangian foliation by cotangent fibres, and your wavefunctions are precisely the functions on the base manifold; or if you're on Euclidean space, you could also polarise using a horizontal Lagrangian foliation and you'd get wavefunctions of momentum (related to the wavefunctions of position by Fourier transform). I learned this stuff (and much more) from Tyurin's beautiful book "Quantization, classical and quantum field theory and theta functions"; it's on pages 1-2.<|endoftext|> TITLE: Multiple of a flat family of subschemes is flat QUESTION [5 upvotes]: Let $X$ be a fixed curve (e.g. a Noetherian, projective scheme of dimension 1, of finite type over an algebraically closed field $k$) and let $S$ be an arbitrary parameter scheme over $k$. Let $D \subset X \times S$ be a flat family over $S$ of subschemes of $X$, of relative dimension 0 and degree $d$, with ideal sheaf $\mathcal I$. Consider now the clolsed subscheme $D' \subset X \times S$ defined by the ideal $\mathcal I^n$. In my mind, $D' = nD$ in some geometric sense. Is $D'$ again flat over $S$? REPLY [5 votes]: I am just posting my comment as an answer. No, that is not true. Let $X$ be $\text{Spec}\ k[t,u]/\langle tu\rangle $. Let $S$ be $\text{Spec}\ k[ϵ]/\langle ϵ^2 \rangle.$ Let $I$ be $\langle t,u−ϵ\rangle.$ Then $\mathcal{O}_{X×S}/I^2$ is a direct sum of two copies of $\mathcal{O}_S$ (generated by $1$ and $u$) plus one summand $\mathcal{O}_S/ϵ\mathcal{O}_S$ (generated by $t$). It is not $\mathcal{O}_S$-flat.<|endoftext|> TITLE: Defining abstract varieties and their morphisms over a finitely generated subfield of the base field QUESTION [12 upvotes]: Let $k$ be an algebraically closed field. By a finitely generated subfield of $k$ I mean a subfield $k_0\subset k$ that is finitely generated over the prime subfield of $k$ (that is, over $\mathbb Q$ or $\mathbb F_p$). I need a reference or a proof for the following (well-known? evident?) proposition: Proposition. Let $$f\colon X\to V$$ be a morphism of $k$-varieties. Then the triple $(X,Y,f)$ can be defined over a finitely generated subfield of $k$. In other words, there exists a finitely generated subfield $k_0\subset k$ and a morphism of $k_0$-varieties $$f_0\colon X_0\to Y_0$$ such that $(X_0,Y_0,f_0)\times_{k_0} k$ is isomorphic to $(X,Y,f)$. A referee asked me to prove this assertion. I know how to prove this for affine and projective varieties, but not for abstract varieties. REPLY [10 votes]: A variety $V$ is a finite union of open affine varieties $V_i$, and because $V$ is separated (usually part of the definition of variety) the intersections $V_i\cap V_j$ are also affine. Now $V$ can be reconstructed from the affine varieties $V_i,V_i\cap V_j$ and the maps of affine varieties $V_i\cap V_j\to V_i$. Obviously, this system is defined over a finitely generated field. Hence $X$ and $V$ are defined over a finitely generated field. The graph of $f$ is a closed subvariety of $X\times V$, and so is defined by a coherent sheaf of ideals in the structure sheaf of $X\times V$, which is obviously defined over a finitely generated field.<|endoftext|> TITLE: What is the correct definition of localisation of a category? QUESTION [16 upvotes]: Disclaimer: I wasn't sure if this was an appropriate question for MathOverflow, and so I've also asked this on StackExchange. There appears to be a discrepancy in the literature regarding the definition of a localisation of a category. Let $\mathcal{C}$ be a category and let $S$ be a class of morphisms. The classical literature, for example Gabriel & Zisman, define a localisation of $\mathcal{C}$ by $S$ as follows: GZ1) A category $\mathcal{C}[S^{-1}]$ along with a functor $Q: \mathcal{C} \longrightarrow \mathcal{C}[S^{-1}]$ which makes the elements of $S$ isomorphisms. GZ2) If a functor $F: \mathcal{C} \longrightarrow \mathcal{D}$ makes the elements of $S$ isomorphisms, then there is a functor $G: \mathcal{C}[S^{-1}] \longrightarrow \mathcal{D}$ such that $F$ factors through $Q$ in the sense that $F = G \circ Q$. Moreover, $G$ is unique up to natural isomorphism. Gabriel & Zisman then state the following lemma: For each category $\mathcal{D}$, the functor $$- \circ Q: \text{Fun}(\mathcal{C}[S^{-1}], \mathcal{D}) \longrightarrow \text{Fun}(\mathcal{C}, \mathcal{D})$$ is an isomorphism of categories from $\text{Fun}(\mathcal{C}[S^{-1}], \mathcal{D})$ to the full subcategory of $\text{Fun}(\mathcal{C}, \mathcal{D})$ consisting of functors which make elements of $S$ invertible. Gabriel & Zisman then claims that this lemma is just a restatement of the conditions GS1 and GS2 above in more precise terms. On the other hand, Kashiwara & Shapira define a localisation of the category $\mathcal{C}$ by $S$ as follows: KS1) A category $\mathcal{C}[S^{-1}]$ along with a functor $Q: \mathcal{C} \longrightarrow \mathcal{C}[S^{-1}]$ which makes the elements of $S$ isomorphisms; KS2) If a functor $F: \mathcal{C} \longrightarrow \mathcal{D}$ makes the elements of $S$ isomorphisms, then there is a functor $G: \mathcal{C}[S^{-1}] \longrightarrow \mathcal{D}$ such that $F$ factors through $Q$ in the sense that $F = G \circ Q$. KS3) If $G_{1}$ and $G_{2}$ are two objects of $\text{Fun}(\mathcal{C}[S^{-1}], \mathcal{D})$, then the natural map $$ - \circ Q: \text{Hom}_{\text{Fun}(\mathcal{C}[S^{-1}], \mathcal{D})}(G_{1}, G_{2}) \longrightarrow \text{Hom}_{\text{Fun}(\mathcal{C}, \mathcal{D})}(G_{1} \circ Q, G_{2} \circ Q) $$ is a bijection. However, Kashiwara & Shapira then make the claim that condition KS3 implies that the $G$ in KS2 is unique up to unique isomorphism. These seem to be contradictory. Gabriel & Zisman claims that their definition makes $G$ unique up to isomorphism. Kashiwara & Shaipira claims that their definition makes $G$ unique up to unique isomorphism. Ordinarily this wouldn't be a problem - obviously one is free to define your terms however you please. But on the face of it, it would appear that GZ1+GZ2 is equivalent to KS1+KS2+KS3, yet each text makes a different claim about the uniqueness of G. When I attempted to prove the uniqueness of G, I was able to show that it was unique up to isomorphism, but not that the isomorphism was unique as Kashiwara & Shapira claim. In fact, if $F$ has an automorphism besides the identity, then it would seem that there would necessarily be multiple distinct isomorphisms between $G_{1}$ and $G_{2}$. This is something that I have seen in a number of other texts besides these two. And even worse, some texts seem to use one definition or the other in their proofs. Am I just missing something and these are equivalent? Any clarification here is appreciated. Thanks REPLY [4 votes]: See Chapter 6 of the book Derived Categories for various notions of localization, including the Ore conditions. (Also available at the arXiv at https://arxiv.org/abs/1610.09640.) Here's a sample of the type of stuff in loc. cit. It captures the 2-categorical nature of localization<|endoftext|> TITLE: Open book decompositions of $T^3$ QUESTION [5 upvotes]: Please pardon my ignorance on the subject of open books, I'm a noob. I would like to know some explicit descriptions of open book decompositions of the three torus $T^3$. Are there examples with connected binding? Is there a direct topological proof that shows that every 3-manifold admits an open book decomposition? Ideally, such a proof might start with a Heegaard decomposition or a surgery description of a given 3-manifold, and then proceed to give an algorithm for converting this to an open book decomposition. In such a case, I suppose I could then follow such a proof to answer my first question regarding $T^3$. REPLY [4 votes]: As mentioned in a comment, Ken Baker has a nice visualization of an open book decomposition of $T^3$ due to Jeremy Van-Horn Morris. One way to get open book structures is to start with a fibered universal knot or link, and then pull back the open book structure from $S^3$ to a branched cover which is branched over that link. The Whitehead link, the Borromean rings, and the figure 8 knot are all fibered links that are known to be universal. In fact, every open book decomposition of a 3-manifold pulls back as a branched cover of $S^3$ of the standard open book decomposition of the unknot.<|endoftext|> TITLE: Does there always exist a non-rational algebraic integer in a number field whose discriminant divides its norm? QUESTION [5 upvotes]: Let $K$ be a number field of degree $n$ over the rationals. Under what conditions does there exist a non-rational algebraic integer $\alpha $ in $K$ such that the discriminant of $\alpha $ divides the norm of $\alpha$? This question was first asked on Math StackExchange, Question 2923849, two weeks ago. REPLY [6 votes]: Consider the cyclotomic field $K=\mathbb{Q}(\zeta_5)$. Proposition. There is no $\alpha \in \mathcal{O}_K$ of degree 4 such that $D(\alpha)$ divides $N(\alpha)$. Proof. Assume such an $\alpha$ exists. Since the discriminant of $K$ is $\Delta_K=5^3$, we must have $5^3|D(\alpha)$ and thus $5^3|N(\alpha)$. Let $\pi$ be the unique prime ideal above $5$ in $\mathcal{O}_K$, so that $5\mathcal{O}_K=\pi^4$. Then $\alpha \in \pi^3$. But we have surjective maps \begin{equation*} \frac{\mathcal{O}_K}{\mathbb{Z}[\alpha]} \to \frac{\mathcal{O}_K}{\mathbb{Z}+\alpha\mathcal{O}_K} \to \frac{\mathcal{O}_K}{\mathbb{Z}+\pi^3}. \end{equation*} The last group has cardinality $5^2$ since the image of $\mathbb{Z}$ in $\mathcal{O}_K/\pi^3$ is isomorphic to $\mathbb{Z}/5\mathbb{Z}$. We deduce that the index of $\mathbb{Z}[\alpha]$ in $\mathcal{O}_K$ is divisible by $5^2$. It follows that $5^4 \Delta_K | D(\alpha)$ and thus $5^7 | N(\alpha)$. Repeating the process, we get $N(\alpha)=0$, a contradiction. I guess that if you take a quartic field $K$ with no intermediate subfield and in which some prime $p$ is sufficiently ramified, then the same reasoning will give you an example of number field for which the answer to your question is negative.<|endoftext|> TITLE: Is being close to a Halting set computable? QUESTION [11 upvotes]: Let $\Phi$ be a universal Turing machine and let $S$ be the set on which it halts. I’m curious about if its decidable to check if a number is close to $S$. There are two notions of distance that come to mind: the additive distance and the Hamming distance. The additive distance, $d_+(x,S)$ is the smallest number $n$ such that at least one of $x+n$ and $x-n$ is in $S$. The Hamming distance, $d_h(x,S)$, is the minimum number of bit flips requires to transform $x$ into an element of $S$. For the purposes of this question, consider numbers as beginning with an infinite string of $0$’s and bits before the first non-zero bit can also be flipped. These functions can’t be computable because the inverse image of $0$ gives a Halting set. Is it computable to check if $d(x,S)=k$ or if $d(x,S) TITLE: Blocking $a\to b\to c$ in a DAG with bounded degrees QUESTION [7 upvotes]: (This is an (easy-looking) toy question for this one.) Question. Find the smallest $\alpha$ satisfying the following: Let $G=(V,E)$ be a finite directed acyclic graph, where each in- and out-degree is at most $2$. Then it is possible to remove at most $\alpha|V|$ vertices so that the remaining graph contains no (directed) path with $3$ vertices. I seem to have troubles even with this setup; if it can be answered, then, surely, a more general question would be about the graphs with all (in- and out-) degrees bounded by $k$. What I know. ($1$) $\alpha\geq 1/2$. This is achieved on every graph with $4n$ vertices $v_1,\dots,v_{4n}$ and edges $v_i\to v_{i+1}$ and $v_i\to v_{i+2}$. ($2$) $\alpha\leq 3/5$. This can be shown by induction on $|V|$. Say that the rank of a vertex is the maximal length (=number of vertices) of a path ending at that vertex. If all vertices are of rank $1$, then the graph has no edges. Otherwise, let $s$ be a rank $2$ vertex, with $v\to s$ an incoming edge. Now one can remove the other neighbor of $v$ (if it exists), all the out-neighbors of $s$, thus making $v$ and $s$ ``safe''. So one may forget about $v$ and $s$, and proceed by induction. UPD: ($2'$) $\alpha\leq 4/7$: see an answer by Mikhail Tikhomirov. Any better (upper or lower) bound is welcome! REPLY [3 votes]: We can obtain $\alpha \leq 4/7$ as follows. Process vertices in a topological order and divide them into three sets $V_0, V_1, V_2$ as follows: if all edges leading into $v$ start in $V_2$, then $v \in V_0$ (in particular, all vertices with in-degree 0 go to $V_0$); if there is an edge $u \to v$ with $u \in V_1$, then $v \in V_2$; otherwise, $v \in V_1$ (in this case, there is an edge from $V_0$ and no edge from $V_1$). We claim that removing $V_2$ breaks all paths of length 3. Indeed, suppose that there is a path $v \to u \to w$ confined to $V_0 \cup V_1$. All edges leaving $V_1$ go to $V_2$, hence $v, u \in V_0$. But simultaneously $u \not \in V_0$ by construction, a contradiction. Further, we claim $|V_1| \leq 2|V_0|$ and $|V_2| \leq 2|V_1|$. Indeed, for any $v \in V_1$ there is an edge $u \to v$ with $u \in V_0$, and all such edges are distinct, but the number of these edges is at most $2|V_0|$. The second bound is completely analogous. The bounds above imply $|V| = |V_0| + |V_1| + |V_2| \geq |V_2|/4 + |V_2|/2 + |V_2| = 7|V_2|/4$, hence $|V_2| \leq 4|V|/7$. This upper bound still doesn't seem tight (for instance, we didn't at all use the fact that all in-degree are $\leq 2$). I believe $\alpha = 1/2$ is attainable. Will update later hopefully.<|endoftext|> TITLE: Does the hypergraph of subgroups determine a group? QUESTION [20 upvotes]: A hypergraph is a pair $H=(V,E)$ where $V\neq \emptyset$ is a set and $E\subseteq{\cal P}(V)$ is a collection of subsets of $V$. We say two hypergraphs $H_i=(V_i, E_i)$ for $i=1,2$ are isomorphic if there is a bijection $f:V_1\to V_2$ such that $f(e_1) \in E_2$ for all $e_1\in E_1$, and $f^{-1}(e_2) \in E_1$ for all $e_2\in E_2$. If $G$ is a group, denote by $\text{Sub}(G)$ the collection of the subgroups of $G$. Are there non-isomorphic groups $G,H$ such that the hypergraphs $(G, \text{Sub}(G))$ and $(H, \text{Sub}(H))$ are isomorphic? REPLY [11 votes]: For large prime $p$, there are uncountably many non-isomorphic Tarski monsters of exponent $p$. For these groups $G$, the subgroups lattice consists of basically a partition of $G\smallsetminus\{1\}$ into countably many subsets of cardinal $p-1$ (so the subgroups are the whole group, $\{1\}$ and the union of $\{1\}$ with any component of the partition. These hypergraphs are obviouly isomorphic.<|endoftext|> TITLE: The determinant of a $4\times4$ matrix associated to some specific polynomial as follow QUESTION [7 upvotes]: Let $f\in \mathbb{R}[x_1,x_2,x_3,x_4]$ defined by $$f_a(x_1,x_2,x_3,x_4)=\prod_{1\leqslant i TITLE: Geodesics on hyperbolic surfaces whose closures have arbitrary Hausdorff dimension QUESTION [11 upvotes]: Consider the geodesic flow on $X = \Gamma \backslash \text{PSL}(2,\mathbf{R})$, the unit tangent bundle of a hyperbolic surface, where $\Gamma$ is a lattice. I have heard that, for any real number $\alpha \in [1,3]$, there exists a orbit of the geodesic flow whose closure in $X$ has Hausdorff dimension $\alpha$. Where can I find a proof of this? REPLY [2 votes]: Show that for a Bernoulli system, there exists ergodic (Bernoulli) measures of any given entropy (between 0 and full entropy). Pick such a measure with appropriate entropy as you would like. Recall that especially in such systems, the entropy relates to the Minkowski dimension (and also the Hausdorff dimension, by a result of Furstenberg). Use the ergodic theorem a-la Furstenberg to generate a generic point whose orbit is dense in the support. Use a Bernouliocity theorem (a-la Adler-Weiss) to transfer everything to the modular surface (such theorem is achieved in practice by constructing actual Markov partitions, so you save the metric structure by preserving entropy). There is a minor technicality in the fact that you might need to use countable rather than finite encoding (due to the fact your lattice might be non-uniform), so you would need to massage a bit of the arguments, but everything is essentially well-known (and actually, this encoding method show you you are able to choose geodesics with bounded height so by Dani's correspondence, you may only deal with endpoints which are BA). Some good sources for the encoding (for the modular surface) are C. Series' articles, or S. Katok's book. There is a chance that Series' articles are actually dealing with the general case.<|endoftext|> TITLE: The valuation of j-functions vs number of isomorphisms for an elliptic curve QUESTION [21 upvotes]: Gross and Zagier prove the following fantastic result in their paper "Singular Moduli": Let $R$ be a discrete valuation ring over $\mathbb Z_p$ with uniformizer $\pi$ such that $k = R/\pi$ is algebraically closed and normalize the valuation so that $v(\pi) = 1$. Now, let $E_1,E_2$ be two distinct (ie, non isomorphic) elliptic curves over $R$ with complex multiplication with $j$-invariants $j_i = j(E_i)$. These are algebraic integers in $R$ and we can talk about their reduction mod $\pi$. Define $$i(n) = \frac{|\operatorname{Isom}_{R/\pi^n}(E_1,E_2)|}{2}.$$ Then Gross-Zagier show that: $$v(j_1-j_2) = \sum_{n\geq 1}i(n).$$ Unfortunately, their proof is a very explicit case by case analysis (depending on both $n,p$) of both sides of the equation. This proof is very unsatisfying to me because of how uniform the statement of the result is (it doesn't depend on p or the Elliptic curves, it doesn't even distinguish between the curves with extra automorphisms and those without). However, I suspect that there should be a very nice uniform proof using maybe a moduli space (stack..?) of Elliptic curves. For instance, we easily see that $v(j_1-j_2) \geq 1 \iff i(1) \geq 1$ because over an algebraically closed field the $j$-invariant determines the isomorphism class. Does anyone know such a uniform proof? (The paper I am referring to is here, the theorem is 2.3 on page 196.) REPLY [17 votes]: Yes, let's use the fact that the moduli stack of elliptic curves is etale-locally a scheme. We could also use the formal deformation space of the elliptic curve mod $\pi$ (of course we may assume $E_1 \cong E_2 \mod \pi$). We pick an etale-local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an $R$-point. (This many involve henselizing or completing $R$, which obviously won't affect things.) Over the moduli stack of pairs of elliptic curves, and thus over any etale-local model, the scheme of isomorphisms is a disjoint union of smooth closed curves (Proof is deformation theory - the deformation space of an isomorphism is one-dimensional and always maps injectively to the deformation space of a pair of curves, because it is just the deformation space of one curves). The sum $i(n)$ is the sum over isomorphisms mod $\pi$ of half the power of $\pi$ they lift too, which is a sum over these curves of half the intersection number with $\operatorname{Spec} R$. In other words, each of these curves is locally defined by an equation in the moduli stack, and we restrict that equation to $R$ and look at its degree, then sum these degrees and divide by two. On the other hand, the valuation of $j(E_1)-j(E_2)$ is the intersection number with the divisor of $j(E_1)- j(E_2)$. This divisor is supported at the union of all the curves in the Isom scheme. So it suffices to show that the divisor of $j(E_1)-j(E_2)$ is half the sum of all the curves in the Isom scheme. Because these are divisors supported at the same union of curves, it suffices to check at their mutual generic points. But these generic points are generic elliptic curves with an automorphism of order two, so the Isom divisor appears with multiplicity two while the $j$ divisor has multiplicity one.<|endoftext|> TITLE: What is the minimum of this quantity on $S^{n-2}\times S^{n-2}$? QUESTION [17 upvotes]: My question is to find the minimum of the following expression: $$A(x,y) = \sum_{1\leq i TITLE: consecutive prime gaps and explicit bound QUESTION [5 upvotes]: I am aware of the theorem that $p_{n+1}- p_n \leq n^{0.525}$ which is true for all sufficiently large numbers due to Baker, but if i want to make the implicit "for all sufficiently large numbers" explicit, is it known that $p_{n+1}-p_n \leq c n^{\alpha}$ for all $n \geq 1$ and for small $c$, lets say $c \leq 2$ and $\alpha \leq 0.55$ ? Any ref that can give me the explicit numbers or a way to construct them would be great. Thank you, also i posted the question yesterday on MSE REPLY [8 votes]: The result you quote is due to Baker-Harman-Pintz (2000). I am not aware of any concrete effective version of this result, but if you increase the exponent $0.525$ to $2/3$, then such a variant is available by the work of Dudek. See also my response to this MO question.<|endoftext|> TITLE: Euler factors of L-function at bad primes QUESTION [12 upvotes]: This is of course a very-well known problem, but still let me ask the questions my way. Let $L(s)$ be a "motivic" $L$-function, whatever that means: in particular, it has an Euler product (including at "bad" primes), and a (possibly conjectural) functional equation of standard type with an arithmetic conductor $N$ occurring as $N^{s/2}$ in the factor at infinity. Let $p$ be a prime, denote by $L_p(s)$ the denominator of the Euler factor at $p$, by $d_p$ the degree of $L_p(s)$ in $p^{-s}$, and by $v_p(N)$ the valuation of $N$ at $p$. Finally let $d$ be the degree of the $L$-function, equal both to the number of $\Gamma_{\mathbb R}(s)$ factors and to the generic degree in $p^{-s}$ of $L_p(s)$. I define a prime to be good if it does not divide $N$ (note that if $L$ comes from a variety with bad reduction at $p$, it may still be a good prime in this sense). For the following statements, please tell me if they are true, or only sometimes true, sometimes false (and in that case possibly some extra info): 1) If $p$ is good then $d_p=d$. 2) If $d_p=d$ then $p$ is good. 3) $v_p\ge d-d_p$ 4) I know that one can have $v_p>d-d_p$, but can one give a reasonably simple criterion for equality ? REPLY [15 votes]: 1) True. 2) True. 3) True. 4) They are equal in the case of "tame ramification", e.g. if the degree of $K(\mu_p)$ over $K$, $K$ the coefficient field of the motive, is greater than $d$. One just has to recall that for $V$ the $\ell$-adic Galois representation associated to the motive, $\dim V$ is the degree of the $L$-function, the local $L$-factor is the characteristic polynomial of $\operatorname{Frob}_p$ on the invariants of the inertia group at $p$, and a prime is good if and only if the inertia group acts nontrivially. Furthermore $v_p$ is the Artin conductor which consists of the codimension of the inertia invariant subspace (i.e. $d-d_p$) plus the Swan conductor. The Swan conductor vanishes if and only if the wild inertia group acts trivially. The last claim depends on something that I think is still conjectural, depending on the theory of motives, which is that the characteristic polynomials of elements of the Weil group are defined over $K$. From this, it follows immediately, as these characteristic polynomials have degree $d$, and for elements of teh wild inertia group, roots $p$-power roots of unity.<|endoftext|> TITLE: A conjecture of Littlewood QUESTION [17 upvotes]: The following is a conjecture due to Littlewood. For any set of distinct non-zero integers $n_1,\ldots,n_k$ the inequality $$\int_0^{2\pi}|1+e^{in_1x}+\cdots+e^{in_kx}| \, dx\geq C\log k$$ holds. Has this proven to be true or false? Update 1. An extension to finite fields can be found here REPLY [25 votes]: This was proved by S. Konyagin [7] and independently by McGehee, Pigno, and Smith [13] in 1981. A short proof is available in [5]. [7] S.V. Konjagin, On a problem of Littlewood, Mathematics of the USSR, Izvestia, 18 (1981), 205–225. http://mi.mathnet.ru/eng/izv1556 [5] R.A. DeVore and G.G. Lorentz, Constructive Approximation, Springer-Verlag, Berlin, 1993. [13] O.C. McGehee, L. Pigno, and B. Smith, Hardy’s inequality and the L1 norm of exponential sums, Ann. Math. 113 (1981), 613–618. https://www.jstor.org/stable/2007000<|endoftext|> TITLE: Big finitistic dimension of finite dimensional algebra QUESTION [5 upvotes]: Let $A$ be a finite dimensional algebra. The big finitistic dimension of $A$ is $$\operatorname{FinDim}(A)=\sup\{\operatorname{pd}(M)\mid M\in \text{Mod-}A \text{ and } \operatorname{pd}(M)<\infty\},$$ where $\operatorname{pd}(M)$ is the projective dimension of $M$, $\text{Mod-}A$ is the category of right $A$-modules. Problem: Does there exist a finite dimensional algebra $A$ such that $\operatorname{FinDim}(A)=\infty?$ http://cn.arxiv.org/abs/1804.09801 (see Definition 4.1.) REPLY [11 votes]: You quoted a very recent arxiv article, where it is stated that this is an open problem. No progress has been made towards this open problem since then. The only big progress in the questions sourounding the finitistic dimension conjecture was made 1992 in https://link.springer.com/article/10.1007/BF02100610 where it was proven that the first finitistic dimension conjecture is wrong.<|endoftext|> TITLE: About normal minimal subgroups not in the Frattini QUESTION [9 upvotes]: In Neukirch--Schmidt--Wingberg, "Cohomology of Number Fields", Second edition, page 624, Exercise 2, it is stated the following fact. $\textbf{Claim}$: If $N$ is a normal subgroup, minimal among normal subgroups, of a group $G$ not contained in the Frattini's subgroup of $G$, then $G$ is a semi-direct product over $N$, i.e. there exists $H$ subgroup of $G$ with $H \cap N=\{id\}$ and $HN=G$. Now, if $N$ is a simple non-abelian normal subgroup of $G$, the above Claim implies that $G$ is a semi-direct product over $N$. Indeed the Frattini of $G$ is nilpotent, and since $N$ is non-abelian, it cannot be contained in a nilpotent group, otherwise $N$ itself would be nilpotent, and nilpotent and simple implies having prime order and hence abelian. Moreover $N$ is clearly minimal, since $N$ itself does not have non-trivial proper normal subgroup. But then I see a clear contradiction with Lemma 4.3 of this paper of Lucchini, Menegazzo and Morigi https://projecteuclid.org/download/pdf_1/euclid.ijm/1258488162. Could somebody explain (in case I did not perform a mistake in reasoning) which of the two publications has a mistake? REPLY [13 votes]: The first claim is false: We will give an elementary proof using GAP. Consider the following code: U := SmallGroup(720, 765); N := Group([ (1,2,9,3)(4,6,10,8), (3,4,9,10)(5,8,6,7) ]); P := SylowSubgroup(U, 2); Running isSimple(N) verifies that N is a simple group, and running IsNormal(U,N) verifies that N is normal in U. We claim that U does not split over N: The subgroup N has index 2 and therefore if it had a complement, that would mean there is a cyclic subgroup of size 2 that is disjoint from N. This is impossible. Indeed, Zuppos(P) gives a list of all cyclic subgroups of P and every cyclic subgroup of size 2 is in N. Here is some quick code to check this (prints all true): list := Zuppos(P); for i in list do if Order(i) = 2 then Print(IsSubgroup(N,Group([i]))); fi; od; Hence U does not split over N, as N is normal and all 2-Sylow subgroups are conjugate. (In fact, one can run the above code over all of U and use almost no theory whatsoever.) Finally, N is not contained in the Frattini since it is not inside the maximal subgroup of U containing P, since P and N collectively generated U by index considerations. PS: U never splits, as N is the only non-trivial normal subgroup of U (just run NormalSubgroups(U)). The group U is the Mathieu group of degree 10 and N is the alternating group on 6 letters. Moreover, P is the Semidihedral group SD16.<|endoftext|> TITLE: Discriminant locus of elliptic K3 surfaces QUESTION [7 upvotes]: Given a complex elliptic K3 surface $\pi\colon X\rightarrow \mathbb P^1$, its discriminant locus is the divisor $$D = \sum_{i = 1}^s n_i P_i$$ on $\mathbb P^1$ such that $n_i$ is equal to the Euler-Poincaré characteristic of the fiber $\pi^{-1}(P_i)$, where the sum runs over the points $P_i \in \mathbb P^1$ such that $\pi^{-1}(P_i)$ is singular. It is well known that $\deg D = 24$. Conversely, given an effective divisor $D$ of degree $24$ on $\mathbb P^1$, when is it the discriminant locus of a complex elliptic K3 surface? I am particularly curious about the minimal possible $s$. The maximal Euler-Poincaré characteristic of a singular fiber is $20$, so $s \geq 2$. But in case, say, $n_1 = 20$, then the fibration is of type $I_{14}^*,I_1,I_1,I_1,I_1$ (see Schütt-Schweizer), so indeed $s = 5$. Are smaller $s$ possible? REPLY [9 votes]: The minimal $s$ is $3$. It is attained by several elliptic K3's, including $y^2 = x^3 + (t^2-t)^4$ which has IV* fibers at $t = 0, 1, \infty$ and no other singular fibers. The comment by Ariyan Javanpeykar gives one argument that $s$ can be no smaller. (See postscript. This uses characteristic zero; in small positive characteristic $s$ can be as small as $1$, e.g. in characteristic 2 the elliptic K3 surface $y^2 + y = x^3 + t^9$ has only one reducible fiber, at $t = \infty$.) P.S. There are other ways to prove $s>2$; for example it follows from Szpiro's inequality, which has an elementary proof via the Mason-Stothers theorem (polynomial ABC). See MO 190530, Are there nonisotrivial elliptic curves over $\mathbb{G}_m$?, for this and some related ideas. (That question is related because ${\bf G}_m({\bf C}) = {\bf CP}^1 - \{0, \infty\}$ and if an elliptic surface $\pi: X \to {\bf P}^1$ has $s \leq 2$ then one can choose the coordinate on ${\bf P}^1$ so that each bad fiber maps to $0$ or $\infty$.)<|endoftext|> TITLE: Adventure with infinite series, a curiosity QUESTION [10 upvotes]: It is easily verifiable that $$\sum_{k\geq0}\binom{2k}k\frac1{2^{3k}}=\sqrt{2}.$$ It is not that difficult to get $$\sum_{k\geq0}\binom{4k}{2k}\frac1{2^{5k}}=\frac{\sqrt{2-\sqrt2}+\sqrt{2+\sqrt2}}2.$$ Question. Is there something similarly "nice" in computing $$\sum_{k\geq0}\binom{8k}{4k}\frac1{2^{10k}}=?$$ Perhaps the same question about $$\sum_{k\geq0}\binom{16k}{8k}\frac1{2^{20k}}=?$$ NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern. REPLY [3 votes]: Let $$f(x):=\sum_{k\ge 0}\binom{2k}{k}x^k=\frac{1}{\sqrt{1-4x}}$$ with $|x|<1/4$. We have for $N,j\in\mathbb{Z}_+$, \begin{align} \frac{1}{N}\sum_{r=1}^Ne(-{jr}/{N})f\left(xe({r}/{N})\right)&=\frac{1}{N}\sum_{r=1}^Ne(-{jr}/{N})\sum_{k\ge 0}\binom{2k}{k}x^ke({kr}/{N})\\ &=\sum_{k\ge 0}\binom{2k}{k}x^k\frac{1}{N}\sum_{r=1}^Ne({(k-j)r}/{N}), \end{align} where $e(x):=e^{2\pi{\rm i}x}$. Hence clearly, $$\sum_{\substack{k\ge 0\\ k\equiv j\pmod N}}\binom{2k}{k}x^k=\frac{1}{N}\sum_{r=1}^N\frac{e(-{jr}/{N})}{\sqrt{1-4xe(r/N)}},$$ holds for all $N,j\in\mathbb{Z}_+$, which is a more general result.<|endoftext|> TITLE: Fourier expansion at inequivalent cusps QUESTION [8 upvotes]: Let $\Gamma\subset SL(2,\mathbb{R})$ be a Fuchsian group of the first kind. Let $c_1, c_2$ be inequivalent cusps of $\Gamma.$ Consider $f\in M_k(\Gamma)$ a weight $k$ holomorphic automorphic form, and suppose the Fourier expansion of $f$ at the cusp $c_1$ is known. Given the above expansion, is there an algorithm to compute (even numerically) the Fourier expansion of $f$ at the cusp $c_2$? Thanks! REPLY [3 votes]: Since this has not yet been mentioned explicitly, I would like to add that the adelic language is a powerful tool for computing Fourier expansions at arbitrary cusps. Specifically, if $f$ is a cuspidal holomorphic newform then the Fourier coefficients of $f$ at a given cusp are given by values of the Whittaker newform of $f$ at certain adelic matrices. The Whittaker newform associated to $f$ is a global object which factors as a product of local newforms. Concretely, the local newforms are functions on $\mathrm{GL}_2(\mathbb{Q}_p)$ depending only on the local automorphic representation associated to $f$. Since Loeffler and Weinstein have given an algorithm (implemented e.g. in Sage) to compute these local representations, it should be possible to compute the local newforms explicitly. For more details, you can look at the recent article of Corbett and Saha, On the order of vanishing of newforms at the cusps, especially Section 3. EDIT. Hao Chen has also developed some algorithms to solve this problem in his PhD thesis, Chapter 4.<|endoftext|> TITLE: Brauer group of a curve over non-algebraically closed field QUESTION [12 upvotes]: It is a famous consequence of Tsen's theorem that a smooth curve over an algebraically closed field has trivial Brauer group. But what about curves over non algebraically closed fields? Let us fix a smooth, projective curve $X$ over some field $k$. If $X$ has a rational point $x\in X(k)$, then the natural map $\text{Br}(k)\to\text{Br}(X)$ has a retraction $\text{Br}(X)\to\text{Br}(k)$, thus it is injective. Moreover, it is widely known that $\text{Br}(X)$ injects in $\text{Br}(k(X))$. But what about closed points? A Brauer class that is trivial on each closed point is globally trivial? Precise question: Is the map $$\text{Br}(X)\to\prod_{x\in X^1}\text{Br}(k(x))$$ injective? If not, can we describe its kernel? Observe that, for $k$ algebraically closed, the injectivity above is precisely $\text{Br}(X)=0$. REPLY [3 votes]: Here's an explicit example (joint with A. Landesman). Let $k$ be an algebraically closed field of characteristic not $2$ or $3$, and let $X/k$ be a non-supersingular K3 with Neron-Severi rank $\geq 5$. Then $X$ admits an elliptic fibration $f: X \to \mathbf{P}^1$. Observation: Let $\ell$ be a prime not equal to $\operatorname{char} k$. Then there is an exact sequence $$ 0 \to \operatorname{Pic}(X) \otimes \mathbf{Z}_\ell \to H^2(X,\mathbf{Z}_{\ell}(1)) \to T_{\ell}\operatorname{Br}(X) \to 0.$$ By Hodge theory, the middle term has rank $22$ and therefore by the assumption that $X$ is not supersingular, $\operatorname{Br}(X) \neq 0$. Let $E$ be the generic fiber of $f : X \to \mathbf{P}^1$, it is a smooth elliptic curve over $K := k(\mathbf{P}^1)$. We claim that $\operatorname{Br}(E) \neq 0$. Indeed, it is enough to show that $\operatorname{Br}(X) \subseteq \operatorname{Br}(E)$. Let $\alpha$ be a Brauer class on $X$. If $\alpha|_E = 0$, then $\alpha$ must die on some open $U \subseteq X$. But $X$ is a regular integral scheme and so $\operatorname{Br}(X) \hookrightarrow \operatorname{Br}(U)$. Hence $\alpha = 0$ and we are done.<|endoftext|> TITLE: When are all the fibers of a morphism reduced? QUESTION [6 upvotes]: This is a sort-of follow up to this question, which I asked before I became confused about if things were reduced. More specifically, suppose $\varphi:X\to Y$ is a surjective morphism of finite presentation between algebraic varieties (reduced, irreducible, separated schemes, finite type over an algebraically closed field), and furthermore assume that $Y\cong \mathbb{A}^r$ is an affine space. I am able to show that the underlying reduced schemes of all the fibers $X_y$ for $y\in Y$ are smooth and all equidimensional of dimension $k>0$ (In my case $k=r$, so $\dim(X)=2r$), and also that each fiber is generically reduced. I even know that the generic fiber is reduced, so I know $X_y$ is reduced for $y\in U\subset Y$ in some nonempty open. Can I conclude that the fibers are reduced? More generally, under relatively simple circumstances, one can get that fibers over a Zariski-open are reduced. Is there some simple criteria as to when this may be strengthened to all fibers? (short of assuming $\varphi$ is smooth, for example.) EDIT: As pointed out by Snowball, $X$ being Cohen-Macaulay would suffice. Perhaps a better formulation of the question is Is $X$ necessarily Cohen-Macaulay, and if not are their known counterexamples? REPLY [4 votes]: Let me repeat the assumptions to make sure we agree. Say $f : X \to Y$ is a morphism of varieties over an algebraically closed field $k$ such that (a) $Y$ is affine and smooth of dimension $m$, (b) $(X_y)_{red}$ is smooth of fixed dimension $n$ for all $y \in Y(k)$, (c) the maximal open $V_y \subset X_y$ which is a reduced scheme is dense in $X_y$ for all $y \in Y(k)$. Claim: $f$ is smooth. Step 1. Let $V \subset X$ be the open locus where the morphism $f$ is smooth. By assumptions (a), (b), (c) we see that $V_y$ is the fibre of $V \to Y$ over $y$. Namely, $f$ is smooth in points of $V_y$ by 2.8 of the paper by de Jong on alterations. The converse inclusion is obvious. Step 2. Let $\nu : X' \to X$ be the normalization morphism. Then $V$ is an open subscheme of $X'$. For $y \in Y(k)$ we consider $(X'_y)_{red} \to (X_y)_{red}$. This is a finite morphism which is an isomorphism over the dense open $V_y$. Also every irreducible component of $(X'_y)_{red}$ has dimension $n$ by Krull's height theorem. Hence $(X'_y)_{red} \to (X_y)_{red}$ is birational and hence an isomorphism as the target is normal. Step 3. In particular the assumptions are true for $f' : X' \to Y$. If $Z \subset Y$ is a smooth effective Cartier divisor, then we can consider the morphism $(f')^{-1}(Z) \to Z$. Using that $X'$ is normal, it is straightforward to show that $(f')^{-1}(Z)$ is reduced, using the criterion $(R_0) + (S_1)$ for reducedness. If $m > 1$, then for any $y \in Y(k)$ we can pick $Z$ such that $(f')^{-1}(Z)$ is irreducible by a Bertini theorem (a la Jouanolou). Step 4. By induction on $m$ we see that $(f')^{-1}(Z) \to Z$ is smooth. Hence all the fibres of $f'$ are smooth. Hence $X'$ is smooth. Since we have seen above that $X' \to X$ is a bijection on closed points, it suffices to show that no tangent vectors get collapsed. Such a tangent vector would have to be vertical. But this would mean that $(X_y)_{red}$ cannnot be smooth. Answer to first comment: the fibres are nonempty by assumption (b) or they are all empty if $n < 0$ and then the result is true also. Answer to second comment: forgot to say $y \in Z$. The induction works because we've checked (f')^{-1}(Z) is a variety (except in the case $n = 1$ you get that it might be a disjoint union of varieties). Anyway, others can add more details to this answer if they so desire.